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Q: # What is a cube root formula? A: To find the cube root of a number, find all of the number’s prime factors. A cube root is a number that, when cubed, gives the original number. If there are exactly three prime factors that are all the same, the number is a perfect cube. ## Keep Learning The cube root of 8 is 2, because 2 * 2 * 2 = 8. Eight is referred to as a "perfect cube" because its cube root is a whole number. For example, to find the cube root of 125, find its prime factors. Since 125 = 5 * 5 * 5, the cube root of 125 is 5. If the number does not have exactly three equal prime factors, the cube root can only be expressed as an irrational number. So, the cube root of 15 is simply cube root 15. If the number has three equal prime factors plus other prime factors, it is also an irrational number. For example, to find the cube root of 375, find its prime factors. 325 = 5 * 5 * 5 * 3. So the cube root of 375 is 5 cube root 3. Remember that the cube of a number is the number multiplied by itself three times. So the cube root is the number that, when multiplied by itself three times, gives the initial number. Since 2 * 2 * 2 is 8, 2 cubed is 8, and the cube root of 8 is 2. Sources: ## Related Questions • A: A square root is the result of "unsquaring" a number, or the reversal of the process of squaring. The root is, as its name implies, a squared number reduced to its most basic form. Filed Under: • A: The fourth root of 16 is 2. In mathematics, the fourth root of a number is a number r that yields z when raised to power 4, where 4 is the degree of the root. Filed Under: • A: The square root of pi, like pi itself, is an irrational number. It can be approximated to four decimal places by 1.7725. The factorial of 1/2 is equal to half of the square root of pi.
# Texas Go Math Grade 4 Lesson 18.2 Answer Key Find Profit Refer to our Texas Go Math Grade 4 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 4 Lesson 18.2 Answer Key Find Profit. ## Texas Go Math Grade 4 Lesson 18.2 Answer Key Find Profit Essential Question How can you determine if you make a profit? Answer:Â The amount gained by selling a product with more than its cost price. Before you can dive into determining profit margin, you need to know what it is. Your businessâ€™s profit margin measures what percentage of revenue your business keeps after paying for outgoing expenses. You can calculate profit margin to see profitability for a specific time period. To determine your companyâ€™s overall profit margin, youâ€™ll want to use the net profit margin formula. Use the formula below to calculate your businessâ€™s overall profit margin: Profit Margin = (Net Income / Revenue) X 100 2. Divide your net income by your revenue (also called net sales) 3. Multiply your total by 100 to get your profit margin percentage. Unlock the Problem Jasmine makes cow sock puppets to earn money to donate to the Future Farmers Club. She buys all of the items shown in the table. Each sock makes one puppet. She will sell the cow puppets for $8 each. Will Jasmine make a profit if she sells all of her puppets? Profit is the amount of money left after all the expenses are subtracted from the amount of money received from selling an item or service. Answer: The cost of cow puppets each=$8 Puppet supplies are already given on the table. There are 3 pieces of felt fabric $6 each=63=$18. Now add all the supplies=18+12+3.69+4.99+9.99+5.33=$54. If she sells all the cow puppets for$8 each=128=$96. {already she has 12 puppets which have to sell for$8 each that’s why I multiplied 8 with 12}. To know the profit how much she gains, you need to subtract expenses from income. By math sentence, we can write as 96-54=$42. She gained a$42 profit. Example (A) Find the total expenses. Think: Find the cost of the 3 pieces of felt. Felt: _________ pieces Ã— $___________ =$ ___________. Add the cost of all the puppet supplies she bought. _________________________ = ____________ Jasmine spent a total of $_________ on her supplies. Answer: Felt fabric=3 piecesÃ—$6=$18 Now add all the supplies=18+12+3.69+4.99+9.99+5.33=$54. Jasmine spent a total of $54 on her supplies. (B) Amount received from selling puppets. Jasmine has __________ socks, so she can make __________ puppets. She sells the puppets for$8 each. Think: to find the amount of money she receives, you must multiply. __________ puppets Ã— $__________ per puppet =$ __________ . Jasmine receives __________ from selling all the puppets. Jasmine has 12 socks. so she can make 12 puppets. She sells each puppet for $8. Now we have to find the amount of money she receives, so you multiply 12 puppets Ã—$8 per puppet=$96. Jasmine receives$96 from selling all the puppets. (C) Find the Profit. Subtract the total expenses from the amount received. $____96_____ –$ ____54_____ = $___42______ So, Jasmine makes a ___42______ of$ ____96_____ . To know the profit how much she gains, you need to subtract expenses from income. By math sentence, we can write as 96-54=$42. She gained a$42 profit. Mathematical Processes What is the least number of puppets Jasmine has to sell and still make a profit? Explain. Answer: If she sells 8 puppets for $8 each then 88=64$ To know the income we need to subtract income – expenses. Total expenses=$54 Therefore, 64-54=$10 If she sells 8 puppets then she will gain $10 profit. Share and Show Question 1. Diego runs his own lawn mowing business. His expenses for one week are listed in the chart. He charges his customers$45 to mow a lawn. How much profit did Diego make in the week of June 4th? A. What were his expenses for the week of June 4th? _________ + _________ + _________ = _________ B. In the week of June 4th, he had 15 customers. How much money did he take in? _________ lawns Ã— _________ per lawn = _________ C. What was Diego’s profit? . Think: Amount received – expenses = profit _________ – _________ = _________ A. His expenses on June 4th=$67.40+$89+$198.99=$355.39 B. In the week of June 4th, he had 15 customers, by math sentence 15 lawnsÃ—$45 per lawn=$675 C. Diego’s profit=675-355.39=$319.69 Go Math Answer Key Grade 4 How to Find the Profit Question 2. During the week of August 6th, Diego had to buy a new lawn mower for$378. His gas expenses were $45, and his other truck expenses were$89. How much profit did Diego make if he had 12 customers that week? Expenses: _________ + _________ + _________ = _________ Profit: _________ – _________ = _________ So, Diego made a profit of _________ . Explanation: Diego buy a new lawn mower at the cost of $378. The expenses of gas=$45 The expenses of truck=$89 Now add all the expenses. Expense:$378+$45+$89=$512. The question asked was about the profit he gets if he had 12 customers that week. The amount he received on that week=12Ã—$378=$4536. Profit:$4536-$512=$4024. (profit=amount received-expenses) So, Diego made a profit of $4024. Find the profit. Question 3. Expenses:$222 Earnings: $791 Answer: Formula for profit=Earnings-Expenses Profit=$791-$222 Profit=$569. Question 4. Expenses: $96 Earnings:$149.59 Formula for profit=Earnings-Expenses Profit=$149.59-$96 Profit=$53.59. Question 5. Expenses:$195.75 Earnings: $500 Answer: Formula for profit=Earnings-Expenses Profit=$500-$195.75 Profit=$304.25 Question 6. Expenses: $950.01 Earnings:$1,203.12 Formula for profit=Earnings-Expenses Profit=$1203.12-$950.01 Profit=$253.11. Go Math Answer Key 4th Grade How to Find the Profit Question 7. Expenses:$109.90 Earnings: 860 Formula for profit=Earnings-Expenses Profit=$860-$109.90 Profit=$750.1. Question 8. Expenses:$810.50 Earnings: $2,002.25 Answer: Formula for profit=Earnings-Expenses Profit=$2002.25-$810.50 Profit=$1191.75 Problem Solving Use the table for 9-11. Question 9. Multi-Step Ava started a business making oil paintings of people’s pets. She buys the items shown on the table. She has her paints and other supplies. She charges $75 for an unframed painting. If she sells 20 paintings, how much profit will she make? Answer: The supplies she buys the items are canvas and a master brush set. The cost of canvas=$44.89 The cost of a brush set=$231.21 Now add all the expenses. Expenses=$44.89+$231.21=$276.1 The amount she charged for an unframed painting=$75 The profit she makes, when she sells 20 paintings=20$75=$1500 Earnings:$1500 Profit=earnings-expenses profit=$1500-$276.1 profit=$1223.9 Question 10. H.O.T. What is the least number of paintings Ava needs to sell in order to make a profit? Explain how you found your answer. Answer: 4 paintings. Explanation: I checked with the least numbers which mean in the place of 20, I placed 2, 3 I got no profit and after I placed 4 I got minimum profit. So 4 is the least number of paintings. Ava needs to sell at least 4 paintings to get a profit. Each painting is$75. So I multiplied 75 with 4. By math sentence, I can write as 4$75=300. I got earnings. I already knew the expenses that are$276.1(check in the above answer). Profit=earnings-expenses Profit=300-276.1 Profit=$23.9 Question 11. Apply Ava buys another 20 canvasses. She needs paint, now, so she buys 3 tubes at$11 each. She does not need brushes. What is her profit if she sells 5 paintings? Answer: $297.11 Explanation: The number of canvasses Ava buys=20 The cost of canvass=$44.89 (Given in the 9th question) The number of tubes she buys=3 The cost of each tube=$11 The total amount she invested on tubes=113=$33 The amount she spends on expense=33+44.89=$77.89 Already in the above question given that the cost of painting she sells that is$75. Once check-in the 9th question. Now if she sells 5 paintings then she could get 5$75=$375. We know expenses and earnings. So, we have to calculate profit. Profit=earnings-expenses Profit=375-77.89 Profit=$297.11 Her profit, if she sells 5 paintings is$297.11 How To Find The Profit Answer Key Go Math Grade 4 Question 12. H.O.T. Reasoning David and Diane decide to decorate tote bags and sell them. The profit will go to the community garden project. They buy 8 tote bags for $5.93 each and 8 fabric paints for$4.02 each. Estimate the amount they should charge for each tote bag so they make a profit of about $40. Explain. Answer:$2.99 for each bag. Explanation: The number of tote bags they buy=8 The cost of each bag=$5.93 The total amount they spend on tote bags=8$5.93=$47.44 The number of fabric paints they buy=8 The cost of each fabric paint=$4.02 The total amount they spend on fabric paints=8$4.02=$32.16 The total amount they invested on expenses=47.44+32.16=$79.6 Already profit is given that is$40 Now we need to find out the amount they should charge for each tote bag so they make a profit of about $40. Profit=earnings-expenses.$40= earnings-$79.6 send expenses on the left side$40+$79.6=earnings$119.6=earnings. The amount they should charge for each bag to make a profit $40 =119.6/40=$2.99. Question 13. Jamie builds and sells boxes for worm farms. The materials to build each box cost $13. She sells the boxes for$20 each. How much profit will Jamie make if she sells 5 boxes? (A) $100 (B)$115 (C) $13 (D)$35 Explanation: The cost of materials to build each box=$13 The amount Jamie invested in materials=135=65 The cost of each box Jamie wants to sell=$20 The Jamie earns if she sells 5 boxes=205=$100 Now we want to find the profit she gets if she sells 5 boxes. Profit- earnings-expenses Profit=100-65 Profit=$35. Question 14. Simon makes a profit of $1.35 on each hot dog he sells. If a hot dog costs him$0.85 to make, how much does Simon sell each hot dog for? (A) $2.20 (B)$1.35 (C) $0.50 (D)$2.55 Explanation: The profit Simon makes on each hot dog=$1.35 The cost of each hot dog=$0.85 The earnings she gets after selling each hot dog. Profit=earnings-expenses 1.35=earnings-0.85 1.35+0.85=earnings $2.2=earnings. Texas Go Math Grade 4 Answer Key Lesson 18.2 Question 15. Multi-Step Jenny sells her homemade granola bars for$2 each. Each bar costs $1 to make. If she sells 50 granola bars at a street fair, how much profit does Jenny make? (A)$100 (B) $50 (C)$75 (D) $175 Answer: Option B is correct. Explanation: The cost of each homemade granola balls Jenny sells=$2 The amount invested to make each granola ball=$1 The total amount she invested to make=50=$50 The earnings she could get if she sells 50 granola balls=502=$100 Profit=earnings-expenses Profit=100-50 Profit=$50. TEXAS Test Prep Question 16. John has an income tax preparation business. He spent $99 on computer software and$16 on books. He charges each client $25. How much profit can he make if he has 18 clients? (A)$334 (B) $335 (C)$450 (D) $173 Answer: Option B is correct. Explanation: The amount john invested on the computer=$99 The amount john invested on books=$16 The total amount he invested on xpenses=99+16=$115 The amount he charges for each client=$25 We need to find out the profit if he has 18 clients=2518=$450 Profit=earnings-expenses Profit=450-115 Profit=$335. ### Texas Go Math Grade 4 Lesson 18.2 Homework and Practice Answer Key Jackie is making jam to sell at the farmers’ market. Her expenses are shown in the chart. Question 1. What are Jackie’s expenses? _______________ + _______________ + _______________ = _______________ Answer:$32.73 Jackie’s expenses are=$24.69+$4.75+$3.29=$32.73 Question 2. Jackie made 21 jars of jam. She sold all of the jam for $5.50 a jar. How much money did she take in? _______________ Ã— _______________ = _______________ Answer:$115.5 Explanation: The number of jars Jackie made=21 The cost of the jam of each jar=$5.50 The amount she got after selling all the jars=21$5.50=$115.5 Go Math Grade 4 Lesson 18.2 Answer Key Question 3. What was Jackie’s profit? _______________ – _______________ = _______________ Answer:$82.77 Explanation: Formula: Profit=earnings-expenses Profit=115.5-32.73 Profit=$82.77 Problem Solving Use the table for 4-6. Question 4. Sam and Carla are going to make and sell potholders to earn money for the animal shelter. They need 2 looms, so they will buy 2 loom kits. If they also buy 2 bags of loops and sell the 22 potholders for$4.00 each, how much profit will they make? Explain. Answer: $12.22 Explanation: The number of loom kits they buy=2 The cost of each loom=$17.94 The total amount they invested on loom kits=2$17.94=$35.88 The cost of each bag of loops=$19.95 The number of bags they buy=2 The total amount they invested on a bag of loops=2$19.95=$39.9 The total amount they invested on expenses was$35.88+$39.9=$75.78 If they sell 22 potholders for $4 each, then the profit is 22$4=$88. Now calculate the profit. Profit=earnings-expenses Profit=$88-$75.78 Profit=$12.22 Question 5. Sam and Carla decide to buy 2 looms and 10 bags of loops. What is their profit if they sell all 86 potholders for $4.50 each? Answer: Explanation: The number of loom kits they buy=2 The cost of each loom=$17.94 The total amount they invested on loom kits=2$17.94=$35.88 The cost of each bag of loops=$19.95 The number of bags they buy=10 The total amount they invested on a bag of loops=10$19.95=$199.5 The total amount they invested on expenses was$35.88+$199.5=$235.38 If they sell 86 potholders for $4.50 each, then the profit is 86$4.5.0=$387 Now calculate the profit. Profit=earnings-expenses Profit=$387-$235.38 Profit=$151.62 If Sam and Carla buy just the 2 loom kits and sell only the 6 potholders, how much do they need to charge for each potholder to make a profit? Explain. The number of loom kits they buy=2 The cost of each loom=$17.94 The total amount they invested on loom kits=2$17.94=$35.88 We can charge from$7 above. I will explain now how it comes. First, look at the expenses they invested and both decided to sell 6 potholders so, once thought if we charge $5 each then there is no profit, there will be a loss. By math sentence, I can write as 6$5=$30. Profit=$30-$35.88=-5.88 ( They got in minus, there is a loss if they sell$5 each). For example, charge $8 each then the earnings will be 6$8=$48. Now calculate profit=earnings-expenses Profit=$48-$35.88 Profit=$12.12 We need to check and calculate the profit and then we have to sell the product. Lesson Check Question 7. Multi-Step Connor sells pretzels for $1.50 each. The pretzels cost$0.45 each to make. How much profit will Connor make if he sells 40 pretzels? (A) $58.00 (B)$60.00 (C) $78.00 (D)$42.00 The cost of each pretzel to make=$0.45 The total amount Connor invested to make pretzels=40$0.45=$18 The cost Connor wants to sell each pretzel=$1.50 If he sells 40 pretzels then his earnings are 40$1.50=$60 Now calculate, profit=earnings-expenses Profit=$60-$18 Profit=$42. Question 8. Priscilla makes a profit of$24.50 on every scarf she sells. The materials for each scarf cost $8.3. I low much does Priscilla sell each scarf for? (A)$32.89 (B) $16.11 (C)$34.89 (D) $22.21 Answer: Option A is correct. The profit that makes Priscilla on each scarf=$24.50 The cost of materials to make each scarf=$8.3 Now we need to calculate earnings. Profit=earnings-expenses$24.50=earnings-$8.3$24.50+$8.3=earnings$32.8=earnings. Question 9. Ivan earned $300.00 doing lawn work. His profit was$266.72. How much were his expenses? (A) $566.72 (B)$33.28 (C) $144.38 (D)$44.28 The amount Ivan earned was $300.00 The profit Ivan got=$266.72 We need to calculate expenses. Profit=earnings-expenses $266.72=$300.00-expenses $266.72+expenses=$300.00 expenses=300.00-266.72 expenses=$33.28 Question 10. Multi-Step Wilson helped his uncle move. He spent S33.58 for gas and$24.88 for work gloves. He earned $200.00 from his uncle. What is Wilsonâ€™s profit? (A)$258.46 (B) $58.46 (C)$141.54 (D) $152.64 Answer: Option C is correct. The total expenses he spent on gas & work gloves=$33.58+$24.88=$58.46 The amount Wilson earned was $200 Calculate, profit=earnings-expenses Profit=200-58.46 Profit=$141.54 Texas Go Math Grade 4 Pdf Lesson 18.2 Question 11. Multi-Step Christina sold 35 pairs of gloves for $840. Each pair cost her$6.22 to make. How much profit did Christina make? (A) $217.70 (B)$622.30 (C) $818.30 (D)$17.78 The amount she got after selling 35 pairs of gloves=$840 The cost to make each glove=$6.22 The total amount she invested in making gloves=35$6.22=$217.7 Calculate, profit=earnings-expenses Profit=$840-$217.7=$622.3 Question 12. Multi-Step Amy sells muffins for$1.35 each. The muffins cost S0.39 each to make. Flow much profit will Amy make if she sells 48 muffins? (A) $46.08 (B) S69.60 (C)$18.72 (D) $64.80 Answer: Option A is correct. The cost of each muffin=$1.35 The cost of each muffin to make=$0.39 The total amount she invested in making muffins=48$0.39=$18.72 The earnings she gets if she sells 48 muffins=48*$1.35=$64.8 Calculate, profit=earnings-expenses Profit=$64.8-$18.72 Profit=$46.08 Scroll to Top Scroll to Top
• Call Now 1800-102-2727 • # RD Sharma Solutions for Class 7 Maths Chapter 10:Unitary Method The unitary method is a method in which one finds the unit's value first and then the value of a required number of units. To give an instance, consider a motorcycle that runs 200 km on 20 litres of fuel. To determine how many it will run on 10 litres of fuel, one needs to try and identify the units (known) and values (unknown). KM = Unknown (RHS) No. of litres of fuel = Known (LHS) Now, to solve this problem, 20 litres = 200 km 1 litre = 200 / 20 = 10 km 10 litres = 10 * 10 = 100 km Hence this motorcycle will run 100 km on 10 litres of fuel. The RD Sharma Class 7 Maths Chapter 10 unitary method also deals with the unitary method in ratio and proportion. In this case, if one needs to find the ratio of one quantity concerning the other, then it is recommended to use the unitary method. In addition to this, this chapter also covers some in-depth details on types of unitary methods. Unitary approaches are classified into two types: direct variation and inverse variation. In a direct variation, increments or decrements in one quantity will cause the same effect on the other quantity. For example, an increase in the number of furniture units will cost more. On the contrary, in indirect or inverse variation, if we increase a quantity, then the other quantity's value decreases. For instance, if we increase the speed, we can cover more distance in a short period. Thus, increasing speed reduces the travelling time. In addition to this, the applications of unitary methods are also discussed. Some of the major applications are as follows. • A very good method to evaluate the price of goods. • Used in calculating the time taken by a person or a vehicle to cover some distance in a given unit of time.
# 65 Polar Coordinates: Graphs ### Learning Objectives In this section you will: • Test polar equations for symmetry. • Graph polar equations by plotting points. The planets move through space in elliptical, periodic orbits about the sun, as shown in (Figure). They are in constant motion, so fixing an exact position of any planet is valid only for a moment. In other words, we can fix only a planet’s instantaneous position. This is one application of polar coordinates, represented as $\,\left(r,\theta \right).\,$ We interpret$\,r\,$as the distance from the sun and$\,\theta \,$as the planet’s angular bearing, or its direction from a fixed point on the sun. In this section, we will focus on the polar system and the graphs that are generated directly from polar coordinates. ### Testing Polar Equations for Symmetry Just as a rectangular equation such as$\,y={x}^{2}\,$describes the relationship between$\,x\,$and$\,y\,$on a Cartesian grid, a polar equation describes a relationship between$\,r\,$and$\,\theta \,$on a polar grid. Recall that the coordinate pair$\,\left(r,\theta \right)\,$indicates that we move counterclockwise from the polar axis (positive x-axis) by an angle of$\,\theta ,\,$and extend a ray from the pole (origin)$\,r\,$units in the direction of$\,\theta .\,$All points that satisfy the polar equation are on the graph. Symmetry is a property that helps us recognize and plot the graph of any equation. If an equation has a graph that is symmetric with respect to an axis, it means that if we folded the graph in half over that axis, the portion of the graph on one side would coincide with the portion on the other side. By performing three tests, we will see how to apply the properties of symmetry to polar equations. Further, we will use symmetry (in addition to plotting key points, zeros, and maximums of$\,r)\,$ to determine the graph of a polar equation. In the first test, we consider symmetry with respect to the line$\,\theta =\frac{\pi }{2}\,$(y-axis). We replace$\,\left(r,\theta \right)\,$with$\,\left(-r,-\theta \right)\,$to determine if the new equation is equivalent to the original equation. For example, suppose we are given the equation$\,r=2\mathrm{sin}\,\theta ;$ $\begin{array}{ll}\,\,\,\,r=2\mathrm{sin}\,\theta \hfill & \hfill \\ -r=2\mathrm{sin}\left(-\theta \right)\begin{array}{cccc}& & & \end{array}\hfill & \text{Replace}\,\left(r,\theta \right)\,\text{with }\left(-r,-\theta \right).\hfill \\ -r=-2\mathrm{sin}\,\theta \hfill & \text{Identity: }\mathrm{sin}\left(-\theta \right)=-\mathrm{sin}\,\theta .\hfill \\ \,\,\,\,\,r=2\mathrm{sin}\,\theta \hfill & \text{Multiply both sides by}-1.\hfill \end{array}$ This equation exhibits symmetry with respect to the line$\,\theta =\frac{\pi }{2}.$ In the second test, we consider symmetry with respect to the polar axis ($\,x$-axis). We replace$\,\left(r,\theta \right)\,$with$\,\left(r,-\theta \right)\,$or$\,\left(-r,\pi -\theta \right)\,$to determine equivalency between the tested equation and the original. For example, suppose we are given the equation$\,r=1-2\mathrm{cos}\,\theta .$ $\begin{array}{ll}r=1-2\mathrm{cos}\,\theta \hfill & \hfill \\ r=1-2\mathrm{cos}\left(-\theta \right)\begin{array}{cccc}& & & \end{array}\hfill & \text{Replace }\left(r,\theta \right)\,\text{with}\,\left(r,-\theta \right).\hfill \\ r=1-2\mathrm{cos}\,\theta \hfill & \text{Even/Odd identity}\hfill \end{array}$ The graph of this equation exhibits symmetry with respect to the polar axis. In the third test, we consider symmetry with respect to the pole (origin). We replace$\,\left(r,\theta \right)\,$with$\,\left(-r,\theta \right)\,$to determine if the tested equation is equivalent to the original equation. For example, suppose we are given the equation$\,r=2\mathrm{sin}\left(3\theta \right).$ $\begin{array}{c}\,\,\,\,\,r=2\mathrm{sin}\left(3\theta \right)\\ -r=2\mathrm{sin}\left(3\theta \right)\end{array}$ The equation has failed the symmetry test, but that does not mean that it is not symmetric with respect to the pole. Passing one or more of the symmetry tests verifies that symmetry will be exhibited in a graph. However, failing the symmetry tests does not necessarily indicate that a graph will not be symmetric about the line$\,\theta =\frac{\pi }{2},\,$the polar axis, or the pole. In these instances, we can confirm that symmetry exists by plotting reflecting points across the apparent axis of symmetry or the pole. Testing for symmetry is a technique that simplifies the graphing of polar equations, but its application is not perfect. ### Symmetry Tests A polar equation describes a curve on the polar grid. The graph of a polar equation can be evaluated for three types of symmetry, as shown in (Figure). ### How To Given a polar equation, test for symmetry. 1. Substitute the appropriate combination of components for$\,\left(r,\theta \right):$$\,\left(-r,-\theta \right)\,$for$\,\theta =\frac{\pi }{2}\,$symmetry;$\,\left(r,-\theta \right)\,$for polar axis symmetry; and$\,\left(-r,\theta \right)\,$for symmetry with respect to the pole. 2. If the resulting equations are equivalent in one or more of the tests, the graph produces the expected symmetry. ### Testing a Polar Equation for Symmetry Test the equation$\,r=2\mathrm{sin}\,\theta \,$ for symmetry. Test for each of the three types of symmetry. 1) Replacing$\,\left(r,\theta \right)\,$with$\,\left(-r,-\theta \right)\,$yields the same result. Thus, the graph is symmetric with respect to the line$\,\theta =\frac{\pi }{2}.$ $\begin{array}{ll}-r=2\mathrm{sin}\left(-\theta \right)\hfill & \hfill \\ -r=-2\mathrm{sin}\,\theta \hfill & \text{Even-odd identity}\hfill \\ \,\,\,\,r=2\mathrm{sin}\,\theta \hfill & \text{Multiply}\,\text{by}\,-1\hfill \\ \text{Passed}\hfill & \hfill \end{array}$ 2) Replacing$\,\theta \,$with$\,-\theta \,$does not yield the same equation. Therefore, the graph fails the test and may or may not be symmetric with respect to the polar axis. $\begin{array}{ll}r=2\mathrm{sin}\left(-\theta \right)\hfill & \hfill \\ r=-2\mathrm{sin}\,\theta \hfill & \text{Even-odd identity}\hfill \\ r=-2\mathrm{sin}\,\theta \ne 2\mathrm{sin}\,\theta \hfill & \hfill \\ \text{Failed}\hfill & \hfill \end{array}$ 3) Replacing$\,r\,$with$–r\,$changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole. $\begin{array}{l}-r=2\mathrm{sin}\,\theta \hfill \\ \text{ }r=-2\mathrm{sin}\,\theta \ne 2\mathrm{sin}\,\theta \hfill \\ \text{Failed}\hfill \end{array}$ #### Analysis Using a graphing calculator, we can see that the equation$\,r=2\mathrm{sin}\,\theta \,$ is a circle centered at$\,\left(0,1\right)\,$with radius$\,r=1\,$and is indeed symmetric to the line$\,\theta =\frac{\pi }{2}.\,$We can also see that the graph is not symmetric with the polar axis or the pole. See (Figure). ### Try It Test the equation for symmetry:$\,r=-2\mathrm{cos}\,\theta .$ The equation fails the symmetry test with respect to the line$\,\theta =\frac{\pi }{2}\,$and with respect to the pole. It passes the polar axis symmetry test. ### Graphing Polar Equations by Plotting Points To graph in the rectangular coordinate system we construct a table of$\,x\,$and$\,y\,$values. To graph in the polar coordinate system we construct a table of$\,\theta \,$and$\,r\,$values. We enter values of$\,\theta \,$ into a polar equation and calculate$\,r.\,$However, using the properties of symmetry and finding key values of$\,\theta \,$and$\,r\,$means fewer calculations will be needed. #### Finding Zeros and Maxima To find the zeros of a polar equation, we solve for the values of$\,\theta \,$that result in$\,r=0.\,$ Recall that, to find the zeros of polynomial functions, we set the equation equal to zero and then solve for$\,x.\,$We use the same process for polar equations. Set$\,r=0,\,$and solve for$\,\theta .$ For many of the forms we will encounter, the maximum value of a polar equation is found by substituting those values of$\,\theta \,$into the equation that result in the maximum value of the trigonometric functions. Consider$\,r=5\mathrm{cos}\,\theta ;\,$the maximum distance between the curve and the pole is 5 units. The maximum value of the cosine function is 1 when$\,\theta =0,\,$so our polar equation is$\,5\mathrm{cos}\,\theta ,\,$and the value$\,\theta =0\,$ will yield the maximum$\,\|r\|.$ Similarly, the maximum value of the sine function is 1 when$\,\theta =\frac{\pi }{2},\,$and if our polar equation is$\,r=5\mathrm{sin}\,\theta ,\,$the value$\,\theta =\frac{\pi }{2}\,$will yield the maximum$\,\|r\|.\,$We may find additional information by calculating values of$\,r\,$when$\,\theta =0.\,$These points would be polar axis intercepts, which may be helpful in drawing the graph and identifying the curve of a polar equation. ### Finding Zeros and Maximum Values for a Polar Equation Using the equation in (Figure), find the zeros and maximum$\,\|r\|\,$and, if necessary, the polar axis intercepts of$\,r=2\mathrm{sin}\,\theta .$ To find the zeros, set$\,r\,$equal to zero and solve for$\,\theta .$ $\begin{array}{ll}2\mathrm{sin}\,\theta =0\hfill & \hfill \\ \,\,\,\mathrm{sin}\,\theta =0\hfill & \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\theta ={\mathrm{sin}}^{-1}0\hfill & \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\theta =n\pi \hfill & \text{where }n\text{ is an integer}\hfill \end{array}$ Substitute any one of the$\,\theta \,$values into the equation. We will use$\,0.$ $\begin{array}{l}\begin{array}{l}\\ r=2\mathrm{sin}\left(0\right)\end{array}\hfill \\ r=0\hfill \end{array}$ The points$\,\left(0,0\right)\,$and$\,\left(0,±n\pi \right)\,$are the zeros of the equation. They all coincide, so only one point is visible on the graph. This point is also the only polar axis intercept. To find the maximum value of the equation, look at the maximum value of the trigonometric function$\,\mathrm{sin}\,\theta ,\,$which occurs when$\,\theta =\frac{\pi }{2}±2k\pi \,$resulting in$\,\mathrm{sin}\left(\frac{\pi }{2}\right)=1.\,$Substitute$\,\frac{\pi }{2}\,$for$\,\mathrm{\theta .}$ $\begin{array}{l}r=2\mathrm{sin}\left(\frac{\pi }{2}\right)\hfill \\ r=2\left(1\right)\hfill \\ r=2\hfill \end{array}$[/hidden-answer] #### Analysis The point$\,\left(2,\frac{\pi }{2}\right)\,$will be the maximum value on the graph. Let’s plot a few more points to verify the graph of a circle. See (Figure) and (Figure). $\theta$ $r=2\mathrm{sin}\,\theta$ $r$ 0 $r=2\mathrm{sin}\left(0\right)=0$ $0$ $\frac{\pi }{6}$ $r=2\mathrm{sin}\left(\frac{\pi }{6}\right)=1$ $1$ $\frac{\pi }{3}$ $r=2\mathrm{sin}\left(\frac{\pi }{3}\right)\approx 1.73$ $1.73$ $\frac{\pi }{2}$ $r=2\mathrm{sin}\left(\frac{\pi }{2}\right)=2$ $2$ $\frac{2\pi }{3}$ $r=2\mathrm{sin}\left(\frac{2\pi }{3}\right)\approx 1.73$ $1.73$ $\frac{5\pi }{6}$ $r=2\mathrm{sin}\left(\frac{5\pi }{6}\right)=1$ $1$ $\pi$ $r=2\mathrm{sin}\left(\pi \right)=0$ $0$ ### Try It Without converting to Cartesian coordinates, test the given equation for symmetry and find the zeros and maximum values of$\,\|r\|:\,$$\,r=3\mathrm{cos}\,\theta .$ Tests will reveal symmetry about the polar axis. The zero is$\,\left(0,\frac{\pi }{2}\right),\,$and the maximum value is$\,\left(3,0\right).$ #### Investigating Circles Now we have seen the equation of a circle in the polar coordinate system. In the last two examples, the same equation was used to illustrate the properties of symmetry and demonstrate how to find the zeros, maximum values, and plotted points that produced the graphs. However, the circle is only one of many shapes in the set of polar curves. There are five classic polar curves: cardioids, limaҫons, lemniscates, rose curves, and Archimedes’ spirals. We will briefly touch on the polar formulas for the circle before moving on to the classic curves and their variations. ### Formulas for the Equation of a Circle Some of the formulas that produce the graph of a circle in polar coordinates are given by$\,r=a\mathrm{cos}\,\theta \,$and$\,r=a\mathrm{sin}\,\theta ,$ where$\,a\,$is the diameter of the circle or the distance from the pole to the farthest point on the circumference. The radius is$\,\frac{\|a\|}{2},$ or one-half the diameter. For$\,r=a\mathrm{cos}\,\theta ,$ the center is$\,\left(\frac{a}{2},0\right).\,$For$\,r=a\mathrm{sin}\,\theta ,$ the center is$\,\left(\frac{a}{2},\frac{\pi }{2}\right).\,$(Figure) shows the graphs of these four circles. ### Sketching the Graph of a Polar Equation for a Circle Sketch the graph of$\,r=4\mathrm{cos}\,\theta .$ First, testing the equation for symmetry, we find that the graph is symmetric about the polar axis. Next, we find the zeros and maximum$\,\|r\|\,$for$\,r=4\mathrm{cos}\,\theta .\,$First, set$\,r=0,\,$and solve for$\,\theta$. Thus, a zero occurs at$\,\theta =\frac{\pi }{2}±k\pi .\,$A key point to plot is$\,\left(0,\text{}\text{}\frac{\pi }{2}\right)\,.$ To find the maximum value of$\,r,$ note that the maximum value of the cosine function is 1 when$\,\theta =0±2k\pi .\,$Substitute$\,\theta =0\,$into the equation: $\begin{array}{c}r=4\mathrm{cos}\,\theta \\ \,\,\,\,\,r=4\mathrm{cos}\left(0\right)\\ \,\,\,\,\,\,\,r=4\left(1\right)=4\end{array}$ The maximum value of the equation is 4. A key point to plot is$\,\left(4,\,0\right).$ As$\,r=4\mathrm{cos}\,\theta \,$ is symmetric with respect to the polar axis, we only need to calculate r-values for$\,\theta \,$over the interval$\,\left[0,\,\,\pi \right].\,$Points in the upper quadrant can then be reflected to the lower quadrant. Make a table of values similar to (Figure). The graph is shown in (Figure). $\theta$ 0 $\frac{\pi }{6}$ $\frac{\pi }{4}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $\frac{2\pi }{3}$ $\frac{3\pi }{4}$ $\frac{5\pi }{6}$ $\pi$ $r$ 4 3.46 2.83 2 0 −2 −2.83 −3.46 4 #### Investigating Cardioids While translating from polar coordinates to Cartesian coordinates may seem simpler in some instances, graphing the classic curves is actually less complicated in the polar system. The next curve is called a cardioid, as it resembles a heart. This shape is often included with the family of curves called limaçons, but here we will discuss the cardioid on its own. ### Formulas for a Cardioid The formulas that produce the graphs of a cardioid are given by$\,r=a±b\mathrm{cos}\,\theta \,$and$\,r=a±b\mathrm{sin}\,\theta \,$where$\,a>0,\,\,b>0,\,$and$\,\frac{a}{b}=1.\,$The cardioid graph passes through the pole, as we can see in (Figure). ### How To Given the polar equation of a cardioid, sketch its graph. 1. Check equation for the three types of symmetry. 2. Find the zeros. Set$\,r=0.$ 3. Find the maximum value of the equation according to the maximum value of the trigonometric expression. 4. Make a table of values for$\,r\,$and$\,\theta .$ 5. Plot the points and sketch the graph. ### Sketching the Graph of a Cardioid Sketch the graph of$\,r=2+2\mathrm{cos}\,\theta .$ First, testing the equation for symmetry, we find that the graph of this equation will be symmetric about the polar axis. Next, we find the zeros and maximums. Setting$\,r=0,\,$we have$\,\theta =\pi +2k\pi .\,$The zero of the equation is located at$\,\left(0,\pi \right).\,$The graph passes through this point. The maximum value of$\,r=2+2\mathrm{cos}\,\theta \,$occurs when$\,\mathrm{cos}\,\theta \,$ is a maximum, which is when$\,\mathrm{cos}\,\theta =1\,$or when$\,\theta =0.\,$Substitute$\,\theta =0\,$into the equation, and solve for$\,r.\,$ $\begin{array}{l}\begin{array}{l}\\ r=2+2\mathrm{cos}\left(0\right)\end{array}\hfill \\ r=2+2\left(1\right)=4\hfill \end{array}$ The point$\,\left(4,0\right)\,$is the maximum value on the graph. We found that the polar equation is symmetric with respect to the polar axis, but as it extends to all four quadrants, we need to plot values over the interval$\,\left[0,\,\pi \right].\,$The upper portion of the graph is then reflected over the polar axis. Next, we make a table of values, as in (Figure), and then we plot the points and draw the graph. See (Figure). $\theta$ $0$ $\frac{\pi }{4}$ $\frac{\pi }{2}$ $\frac{2\pi }{3}$ $\pi$ $r$ 4 3.41 2 1 0 #### Investigating Limaçons The word limaçon is Old French for “snail,” a name that describes the shape of the graph. As mentioned earlier, the cardioid is a member of the limaçon family, and we can see the similarities in the graphs. The other images in this category include the one-loop limaçon and the two-loop (or inner-loop) limaçon. One-loop limaçons are sometimes referred to as dimpled limaçons when$\,1<\frac{a}{b}<2\,$and convex limaçons when$\,\frac{a}{b}\ge 2.\,$ ### Formulas for One-Loop Limaçons The formulas that produce the graph of a dimpled one-loop limaçon are given by$\,r=a±b\mathrm{cos}\,\theta \,$and$\,r=a±b\mathrm{sin}\,\theta \,$where$\,a>0,\,b>0,\,\,\text{and 1<}\frac{a}{b}<2.\,$All four graphs are shown in (Figure). ### How To Given a polar equation for a one-loop limaçon, sketch the graph. 1. Test the equation for symmetry. Remember that failing a symmetry test does not mean that the shape will not exhibit symmetry. Often the symmetry may reveal itself when the points are plotted. 2. Find the zeros. 3. Find the maximum values according to the trigonometric expression. 4. Make a table. 5. Plot the points and sketch the graph. ### Sketching the Graph of a One-Loop Limaçon Graph the equation$\,r=4-3\mathrm{sin}\,\theta .$ First, testing the equation for symmetry, we find that it fails all three symmetry tests, meaning that the graph may or may not exhibit symmetry, so we cannot use the symmetry to help us graph it. However, this equation has a graph that clearly displays symmetry with respect to the line$\,\theta =\frac{\pi }{2},\,$yet it fails all the three symmetry tests. A graphing calculator will immediately illustrate the graph’s reflective quality. Next, we find the zeros and maximum, and plot the reflecting points to verify any symmetry. Setting$\,r=0\,$results in$\,\theta \,$being undefined. What does this mean? How could$\,\theta \,$be undefined? The angle$\,\theta \,$is undefined for any value of$\,\mathrm{sin}\,\theta >1.\,$Therefore,$\,\theta \,$is undefined because there is no value of$\,\theta \,$for which$\,\mathrm{sin}\,\theta >1.\,$Consequently, the graph does not pass through the pole. Perhaps the graph does cross the polar axis, but not at the pole. We can investigate other intercepts by calculating $r$ when$\,\theta =0.\,$ $\begin{array}{l}r\left(0\right)=4-3\mathrm{sin}\left(0\right)\hfill \\ \,\,\,\,\,\,\,r=4-3\cdot 0=4\hfill \end{array}$ So, there is at least one polar axis intercept at$\,\left(4,0\right).$ Next, as the maximum value of the sine function is 1 when $\,\theta =\frac{\pi }{2},\,$we will substitute$\,\theta =\frac{\pi }{2}\,$ into the equation and solve for$\,r.\,$Thus,$\,r=1.$ Make a table of the coordinates similar to (Figure). $\theta$ $0$ $\frac{\pi }{6}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $\frac{2\pi }{3}$ $\frac{5\pi }{6}$ $\pi$ $\frac{7\pi }{6}$ $\frac{4\pi }{3}$ $\frac{3\pi }{2}$ $\frac{5\pi }{3}$ $\frac{11\pi }{6}$ $2\pi$ $r$ 4 2.5 1.4 1 1.4 2.5 4 5.5 6.6 7 6.6 5.5 4 The graph is shown in (Figure). #### Analysis This is an example of a curve for which making a table of values is critical to producing an accurate graph. The symmetry tests fail; the zero is undefined. While it may be apparent that an equation involving$\,\mathrm{sin}\,\theta \,$ is likely symmetric with respect to the line$\,\theta =\frac{\pi }{2},$ evaluating more points helps to verify that the graph is correct. ### Try It Sketch the graph of$\,r=3-2\mathrm{cos}\,\theta .$ Another type of limaçon, the inner-loop limaçon, is named for the loop formed inside the general limaçon shape. It was discovered by the German artist Albrecht Dürer(1471-1528), who revealed a method for drawing the inner-loop limaçon in his 1525 book Underweysung der Messing. A century later, the father of mathematician Blaise Pascal, Étienne Pascal(1588-1651), rediscovered it. The formulas that generate the inner-loop limaçons are given by$\,r=a±b\mathrm{cos}\,\theta \,$and$\,r=a±b\mathrm{sin}\,\theta \,$where$\,a>0,\,\,b>0,\,$and$\,\,a(Figure) for the graphs. ### Sketching the Graph of an Inner-Loop Limaçon Sketch the graph of[latex]\,r=2+5\text{cos}\,\theta .$ Testing for symmetry, we find that the graph of the equation is symmetric about the polar axis. Next, finding the zeros reveals that when$\,r=0,\,\,\theta =1.98.\,$ The maximum$\,\|r\|\,$is found when$\,\mathrm{cos}\,\theta =1\,$or when$\,\theta =0.\,$Thus, the maximum is found at the point (7, 0). Even though we have found symmetry, the zero, and the maximum, plotting more points will help to define the shape, and then a pattern will emerge. See (Figure). $\theta$ $0$ $\frac{\pi }{6}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $\frac{2\pi }{3}$ $\frac{5\pi }{6}$ $\pi$ $\frac{7\pi }{6}$ $\frac{4\pi }{3}$ $\frac{3\pi }{2}$ $\frac{5\pi }{3}$ $\frac{11\pi }{6}$ $2\pi$ $r$ 7 6.3 4.5 2 −0.5 −2.3 −3 −2.3 −0.5 2 4.5 6.3 7 As expected, the values begin to repeat after$\,\theta =\pi .\,$The graph is shown in (Figure). #### Investigating Lemniscates The lemniscate is a polar curve resembling the infinity symbol$\,\infty \,$or a figure 8. Centered at the pole, a lemniscate is symmetrical by definition. ### Formulas for Lemniscates The formulas that generate the graph of a lemniscate are given by$\,{r}^{2}={a}^{2}\mathrm{cos}\,2\theta \,$and$\,{r}^{2}={a}^{2}\mathrm{sin}\,2\theta \,$where$\,a\ne 0.\,$The formula$\,{r}^{2}={a}^{2}\mathrm{sin}\,2\theta \,$is symmetric with respect to the pole. The formula$\,{r}^{2}={a}^{2}\mathrm{cos}\,2\theta \,$is symmetric with respect to the pole, the line$\,\theta =\frac{\pi }{2},\,$and the polar axis. See (Figure) for the graphs. ### Sketching the Graph of a Lemniscate Sketch the graph of$\,{r}^{2}=4\mathrm{cos}\,2\theta .$ The equation exhibits symmetry with respect to the line$\,\theta =\frac{\pi }{2},\,$the polar axis, and the pole. Let’s find the zeros. It should be routine by now, but we will approach this equation a little differently by making the substitution$\,u=2\theta .$ $\begin{array}{ll}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=4\mathrm{cos}\,2\theta \hfill & \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=4\mathrm{cos}\,u\hfill & \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=\mathrm{cos}\,u\hfill & \hfill \\ {\mathrm{cos}}^{-1}0=\frac{\pi }{2}\hfill & \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,u=\frac{\pi }{2}\hfill & \text{Substitute }2\theta \text{ back in for }u.\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,2\theta =\frac{\pi }{2}\hfill & \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta =\frac{\pi }{4}\hfill & \hfill \end{array}$ So, the point$\left(0,\frac{\pi }{4}\right)$is a zero of the equation. Now let’s find the maximum value. Since the maximum of$\,\mathrm{cos}\,u=1\,$when$\,u=0,\,$the maximum$\,\mathrm{cos}\,2\theta =1\,$when$\,2\theta =0.\,$Thus, $\begin{array}{c}\,{r}^{2}=4\mathrm{cos}\left(0\right)\\ \,\,\,{r}^{2}=4\left(1\right)=4\\ \,\,\,\,\,\,\,\,\,\,r=±\sqrt{4}\,=2\end{array}$ We have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line$\,\theta =\frac{\pi }{2},$ and the polar axis, we only need to plot points in the first quadrant. Make a table similar to (Figure). $\theta$ 0 $\frac{\pi }{6}$ $\frac{\pi }{4}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $r$ 2 $\sqrt{2}$ 0 $\sqrt{2}$ 0 Plot the points on the graph, such as the one shown in (Figure). #### Analysis Making a substitution such as$\,u=2\theta \,$is a common practice in mathematics because it can make calculations simpler. However, we must not forget to replace the substitution term with the original term at the end, and then solve for the unknown. Some of the points on this graph may not show up using the Trace function on the TI-84 graphing calculator, and the calculator table may show an error for these same points of$\,r.\,$This is because there are no real square roots for these values of$\,\theta .\,$In other words, the corresponding r-values of$\,\sqrt{4\mathrm{cos}\left(2\theta \right)}\,$ are complex numbers because there is a negative number under the radical. #### Investigating Rose Curves The next type of polar equation produces a petal-like shape called a rose curve. Although the graphs look complex, a simple polar equation generates the pattern. ### Rose Curves The formulas that generate the graph of a rose curve are given by$\,r=a\mathrm{cos}\,n\theta \,$and$\,r=a\mathrm{sin}\,n\theta \,$where$\,a\ne 0.\,$If$\,n\,$is even, the curve has$\,2n\,$petals. If$\,n\,$is odd, the curve has$\,n\,$petals. See (Figure). ### Sketching the Graph of a Rose Curve (n Even) Sketch the graph of$\,r=2\mathrm{cos}\,4\theta .$ Testing for symmetry, we find again that the symmetry tests do not tell the whole story. The graph is not only symmetric with respect to the polar axis, but also with respect to the line$\,\theta =\frac{\pi }{2}\,$and the pole. Now we will find the zeros. First make the substitution$\,u=4\theta .$ $\begin{array}{c}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=2\mathrm{cos}\,4\theta \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=\mathrm{cos}\,4\theta \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=\mathrm{cos}\,u\\ {\mathrm{cos}}^{-1}0=u\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,u=\frac{\pi }{2}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,4\theta =\frac{\pi }{2}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta =\frac{\pi }{8}\end{array}$ The zero is$\,\theta =\frac{\pi }{8}.\,$The point$\,\left(0,\frac{\pi }{8}\right)\,$is on the curve. Next, we find the maximum$\,\|r\|.\,$We know that the maximum value of$\,\mathrm{cos}\,u=1\,$when$\,\theta =0.\,$Thus, $\begin{array}{l}\\ \begin{array}{l}r=2\mathrm{cos}\left(4\cdot 0\right)\hfill \\ r=2\mathrm{cos}\left(0\right)\hfill \\ r=2\left(1\right)=2\hfill \end{array}\end{array}$ The point$\,\left(2,0\right)\,$is on the curve. The graph of the rose curve has unique properties, which are revealed in (Figure). $\theta$ 0 $\frac{\pi }{8}$ $\frac{\pi }{4}$ $\frac{3\pi }{8}$ $\frac{\pi }{2}$ $\frac{5\pi }{8}$ $\frac{3\pi }{4}$ $r$ 2 0 −2 0 2 0 −2 As$\,r=0\,$when$\,\theta =\frac{\pi }{8},\,$it makes sense to divide values in the table by$\,\frac{\pi }{8}\,$units. A definite pattern emerges. Look at the range of r-values: 2, 0, −2, 0, 2, 0, −2, and so on. This represents the development of the curve one petal at a time. Starting at$\,r=0,\,$each petal extends out a distance of$\,r=2,\,$and then turns back to zero$\,2n\,$times for a total of eight petals. See the graph in (Figure). #### Analysis When these curves are drawn, it is best to plot the points in order, as in the (Figure). This allows us to see how the graph hits a maximum (the tip of a petal), loops back crossing the pole, hits the opposite maximum, and loops back to the pole. The action is continuous until all the petals are drawn. ### Try It Sketch the graph of$\,r=4\mathrm{sin}\left(2\theta \right).$ The graph is a rose curve,$\,n\,$even ### Sketching the Graph of a Rose Curve (n Odd) Sketch the graph of$\,r=2\mathrm{sin}\left(5\theta \right).$ The graph of the equation shows symmetry with respect to the line$\,\theta =\frac{\pi }{2}.\,$Next, find the zeros and maximum. We will want to make the substitution$\,u=5\theta .$ $\begin{array}{c}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=2\mathrm{sin}\left(5\theta \right)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=\mathrm{sin}\,u\\ {\mathrm{sin}}^{-1}0=0\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,u=0\\ \,\,\,\,\,\,\,\,\,\,\,5\theta =0\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta =0\end{array}$ The maximum value is calculated at the angle where$\,\mathrm{sin}\,\theta \,$ is a maximum. Therefore, $\begin{array}{l}\begin{array}{l}\\ r=2\mathrm{sin}\left(5\cdot \frac{\pi }{2}\right)\end{array}\hfill \\ r=2\left(1\right)=2\hfill \end{array}$ Thus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for$\,n\,$odd yields the same number of petals as$\,n,\,$there will be five petals on the graph. See (Figure). Create a table of values similar to (Figure). $\theta$ 0 $\frac{\pi }{6}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $\frac{2\pi }{3}$ $\frac{5\pi }{6}$ $\pi$ $r$ 0 1 −1.73 2 −1.73 1 0 ### Try It Sketch the graph of$r=3\mathrm{cos}\left(3\theta \right).$ Rose curve,$\,n\,$odd #### Investigating the Archimedes’ Spiral The final polar equation we will discuss is the Archimedes’ spiral, named for its discoverer, the Greek mathematician Archimedes (c. 287 BCE-c. 212 BCE), who is credited with numerous discoveries in the fields of geometry and mechanics. ### Archimedes’ Spiral The formula that generates the graph of the Archimedes’ spiral is given by$\,r=\theta \,$ for$\,\theta \ge 0.\,$As$\,\theta \,$increases,$\,r\,$ increases at a constant rate in an ever-widening, never-ending, spiraling path. See (Figure). ### How To Given an Archimedes’ spiral over$\,\left[0,2\pi \right],$sketch the graph. 1. Make a table of values for$\,r\,$and$\,\theta \,$over the given domain. 2. Plot the points and sketch the graph. ### Sketching the Graph of an Archimedes’ Spiral Sketch the graph of$\,r=\theta \,$over$\,\left[0,2\pi \right].$ As$\,r\,$is equal to$\,\theta ,\,$the plot of the Archimedes’ spiral begins at the pole at the point (0, 0). While the graph hints of symmetry, there is no formal symmetry with regard to passing the symmetry tests. Further, there is no maximum value, unless the domain is restricted. Create a table such as (Figure). $\theta$ $\frac{\pi }{4}$ $\frac{\pi }{2}$ $\pi$ $\frac{3\pi }{2}$ $\frac{7\pi }{4}$ $2\pi$ $r$ 0.785 1.57 3.14 4.71 5.50 6.28 Notice that the r-values are just the decimal form of the angle measured in radians. We can see them on a graph in (Figure). #### Analysis The domain of this polar curve is$\,\left[0,2\pi \right].\,$In general, however, the domain of this function is$\,\left(-\infty ,\infty \right).\,$Graphing the equation of the Archimedes’ spiral is rather simple, although the image makes it seem like it would be complex. ### Try It Sketch the graph of$\,r=-\theta \,$over the interval$\,\left[0,4\pi \right].$ ### Summary of Curves We have explored a number of seemingly complex polar curves in this section. (Figure) and (Figure) summarize the graphs and equations for each of these curves. Access these online resources for additional instruction and practice with graphs of polar coordinates. ### Key Concepts • It is easier to graph polar equations if we can test the equations for symmetry with respect to the line$\,\theta =\frac{\pi }{2},\,$the polar axis, or the pole. • There are three symmetry tests that indicate whether the graph of a polar equation will exhibit symmetry. If an equation fails a symmetry test, the graph may or may not exhibit symmetry. See (Figure). • Polar equations may be graphed by making a table of values for$\,\theta \,$and$\,r.$ • The maximum value of a polar equation is found by substituting the value$\,\theta \,$that leads to the maximum value of the trigonometric expression. • The zeros of a polar equation are found by setting$\,r=0\,$and solving for$\,\theta .\,$See (Figure). • Some formulas that produce the graph of a circle in polar coordinates are given by$\,r=a\mathrm{cos}\,\theta \,$and$\,r=a\mathrm{sin}\,\theta .\,$See (Figure). • The formulas that produce the graphs of a cardioid are given by$\,r=a±b\mathrm{cos}\,\theta \,$and$\,r=a±b\mathrm{sin}\,\theta ,\,$for$\,a>0,\,\,b>0,\,$and$\,\frac{a}{b}=1.\,$See (Figure). • The formulas that produce the graphs of a one-loop limaçon are given by$\,r=a±b\mathrm{cos}\,\theta \,$and$\,r=a±b\mathrm{sin}\,\theta \,$for$\,1<\frac{a}{b}<2.\,$See (Figure). • The formulas that produce the graphs of an inner-loop limaçon are given by$\,r=a±b\mathrm{cos}\,\theta \,$and$\,r=a±b\mathrm{sin}\,\theta \,$for$\,a>0,\,\,b>0,\,$ and$\,a(Figure). • The formulas that produce the graphs of a lemniscates are given by[latex]\,{r}^{2}={a}^{2}\mathrm{cos}\,2\theta \,$and$\,{r}^{2}={a}^{2}\mathrm{sin}\,2\theta ,\,$where$\,a\ne 0.$See (Figure). • The formulas that produce the graphs of rose curves are given by$\,r=a\mathrm{cos}\,n\theta \,$and$\,r=a\mathrm{sin}\,n\theta ,\,$where$\,a\ne 0;\,$if$\,n\,$is even, there are$\,2n\,$petals, and if$\,n\,$is odd, there are$\,n\,$petals. See (Figure) and (Figure). • The formula that produces the graph of an Archimedes’ spiral is given by$\,r=\theta ,\,\,\theta \ge 0.\,$See (Figure). ### Section Exercises #### Verbal Describe the three types of symmetry in polar graphs, and compare them to the symmetry of the Cartesian plane. Symmetry with respect to the polar axis is similar to symmetry about the$\,x$-axis, symmetry with respect to the pole is similar to symmetry about the origin, and symmetric with respect to the line$\,\theta =\frac{\pi }{2}\,$is similar to symmetry about the$\,y$-axis. Which of the three types of symmetries for polar graphs correspond to the symmetries with respect to the x-axis, y-axis, and origin? What are the steps to follow when graphing polar equations? Test for symmetry; find zeros, intercepts, and maxima; make a table of values. Decide the general type of graph, cardioid, limaçon, lemniscate, etc., then plot points at $\,\theta =0,\,\frac{\pi }{2},\,\,\pi \,\,\text{and }\frac{3\pi }{2},\,$and sketch the graph. Describe the shapes of the graphs of cardioids, limaçons, and lemniscates. What part of the equation determines the shape of the graph of a polar equation? The shape of the polar graph is determined by whether or not it includes a sine, a cosine, and constants in the equation. #### Graphical For the following exercises, test the equation for symmetry. $r=5\mathrm{cos}\,3\theta$ $r=3-3\mathrm{cos}\,\theta$ $r=3+2\mathrm{sin}\,\theta$ $r=3\mathrm{sin}\,2\theta$ symmetric with respect to the polar axis, symmetric with respect to the line $\theta =\frac{\pi }{2},$ symmetric with respect to the pole $r=4$ $r=2\theta$ no symmetry $r=4\mathrm{cos}\,\frac{\theta }{2}$ $r=\frac{2}{\theta }$ no symmetry $r=3\sqrt{1-{\mathrm{cos}}^{2}\theta }$ $r=\sqrt{5\mathrm{sin}\,2\theta }$ symmetric with respect to the pole For the following exercises, graph the polar equation. Identify the name of the shape. $r=3\mathrm{cos}\,\theta$ $r=4\mathrm{sin}\,\theta$ $r=2+2\mathrm{cos}\,\theta$ $r=2-2\mathrm{cos}\,\theta$ cardioid $r=5-5\mathrm{sin}\,\theta$ $r=3+3\mathrm{sin}\,\theta$ $r=3+2\mathrm{sin}\,\theta$ $r=7+4\mathrm{sin}\,\theta$ $r=4+3\mathrm{cos}\,\theta$ $r=5+4\mathrm{cos}\,\theta$ one-loop/dimpled limaçon $r=10+9\mathrm{cos}\,\theta$ $r=1+3\mathrm{sin}\,\theta$ inner loop/two-loop limaçon $r=2+5\mathrm{sin}\,\theta$ $r=5+7\mathrm{sin}\,\theta$ inner loop/two-loop limaçon $r=2+4\mathrm{cos}\,\theta$ $r=5+6\mathrm{cos}\,\theta$ inner loop/two-loop limaçon ${r}^{2}=36\mathrm{cos}\left(2\theta \right)$ ${r}^{2}=10\mathrm{cos}\left(2\theta \right)$ lemniscate ${r}^{2}=4\mathrm{sin}\left(2\theta \right)$ ${r}^{2}=10\mathrm{sin}\left(2\theta \right)$ lemniscate $r=3\text{sin}\left(2\theta \right)$ $r=3\text{cos}\left(2\theta \right)$ rose curve $r=5\text{sin}\left(3\theta \right)$ $r=4\text{sin}\left(4\theta \right)$ rose curve $r=4\text{sin}\left(5\theta \right)$ $r=-\theta$ Archimedes’ spiral $r=2\theta$ $r=-3\theta$ Archimedes’ spiral #### Technology For the following exercises, use a graphing calculator to sketch the graph of the polar equation. $r=\frac{1}{\theta }$ $r=\frac{1}{\sqrt{\theta }}$ $r=2\mathrm{sin}\,\theta \mathrm{tan}\,\theta ,$ a cissoid $r=2\sqrt{1-{\mathrm{sin}}^{2}\theta }$, a hippopede $r=5+\mathrm{cos}\left(4\theta \right)$ $r=2-\mathrm{sin}\left(2\theta \right)$ $r={\theta }^{2}$ $r=\theta +1$ $r=\theta \mathrm{sin}\,\theta$ $r=\theta \mathrm{cos}\,\theta$ For the following exercises, use a graphing utility to graph each pair of polar equations on a domain of $\,\left[0,4\pi \right]\,$ and then explain the differences shown in the graphs. $r=\theta ,r=-\theta$ $r=\theta ,r=\theta +\mathrm{sin}\,\theta$ They are both spirals, but not quite the same. $r=\mathrm{sin}\,\theta +\theta ,r=\mathrm{sin}\,\theta -\theta$ $r=2\mathrm{sin}\left(\frac{\theta }{2}\right),r=\theta \mathrm{sin}\left(\frac{\theta }{2}\right)$ Both graphs are curves with 2 loops. The equation with a coefficient of$\,\theta \,$ has two loops on the left, the equation with a coefficient of 2 has two loops side by side. Graph these from 0 to$\,4\pi \,$to get a better picture. $r=\mathrm{sin}\left(\mathrm{cos}\left(3\theta \right)\right)\,\,r=\mathrm{sin}\left(3\theta \right)$ On a graphing utility, graph$\,r=\mathrm{sin}\left(\frac{16}{5}\theta \right)\,$on$\,\left[0,4\pi \right],\left[0,8\pi \right],\left[0,12\pi \right],\,$and$\,\left[0,16\pi \right].\,$Describe the effect of increasing the width of the domain. When the width of the domain is increased, more petals of the flower are visible. On a graphing utility, graph and sketch$\,r=\mathrm{sin}\,\theta +{\left(\mathrm{sin}\left(\frac{5}{2}\theta \right)\right)}^{3}\,$on$\,\left[0,4\pi \right].$ On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs. $\begin{array}{l}\begin{array}{l}\\ {r}_{1}=3\mathrm{sin}\left(3\theta \right)\end{array}\hfill \\ {r}_{2}=2\mathrm{sin}\left(3\theta \right)\hfill \\ {r}_{3}=\mathrm{sin}\left(3\theta \right)\hfill \end{array}$ The graphs are three-petal, rose curves. The larger the coefficient, the greater the curve’s distance from the pole. On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs. $\begin{array}{l}\begin{array}{l}\\ {r}_{1}=3+3\mathrm{cos}\,\theta \end{array}\hfill \\ {r}_{2}=2+2\mathrm{cos}\,\theta \hfill \\ {r}_{3}=1+\mathrm{cos}\,\theta \hfill \end{array}$ On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs. $\begin{array}{l}\begin{array}{l}\\ {r}_{1}=3\theta \end{array}\hfill \\ {r}_{2}=2\theta \hfill \\ {r}_{3}=\theta \hfill \end{array}$ The graphs are spirals. The smaller the coefficient, the tighter the spiral. #### Extensions For the following exercises, draw each polar equation on the same set of polar axes, and find the points of intersection. ${r}_{1}=3+2\mathrm{sin}\,\theta ,\,{r}_{2}=2$ ${r}_{1}=6-4\mathrm{cos}\,\theta ,\,{r}_{2}=4$ $\left(4,\frac{\pi }{3}\right),\left(4,\frac{5\pi }{3}\right)$ ${r}_{1}=1+\mathrm{sin}\,\theta ,\,{r}_{2}=3\mathrm{sin}\,\theta$ ${r}_{1}=1+\mathrm{cos}\,\theta ,\,{r}_{2}=3\mathrm{cos}\,\theta$ $\left(\frac{3}{2},\frac{\pi }{3}\right),\left(\frac{3}{2},\frac{5\pi }{3}\right)$ ${r}_{1}=\mathrm{cos}\left(2\theta \right),\,{r}_{2}=\mathrm{sin}\left(2\theta \right)$ ${r}_{1}={\mathrm{sin}}^{2}\left(2\theta \right),\,{r}_{2}=1-\mathrm{cos}\left(4\theta \right)$ $\left(0,\frac{\pi }{2}\right),\,\left(0,\pi \right),\,\left(0,\frac{3\pi }{2}\right),\,\left(0,2\pi \right)$ ${r}_{1}=\sqrt{3},\,{r}_{2}=2\mathrm{sin}\left(\theta \right)$ ${r}_{1}{}^{2}=\mathrm{sin}\,\theta ,{r}_{2}{}^{2}=\mathrm{cos}\,\theta$ $\left(\frac{\sqrt[4]{8}}{2},\frac{\pi }{4}\right),\,\left(\frac{\sqrt[4]{8}}{2},\frac{5\pi }{4}\right)\,$ and at$\,\theta =\frac{3\pi }{4},\,\,\frac{7\pi }{4}\,\,$ since$\,r\,$is squared ${r}_{1}=1+\mathrm{cos}\,\theta ,\,{r}_{2}=1-\mathrm{sin}\,\theta$ ### Glossary Archimedes’ spiral a polar curve given by$\,r=\theta .\,$When multiplied by a constant, the equation appears as$\,r=a\theta .\,$As$\,r=\theta ,\,$the curve continues to widen in a spiral path over the domain. cardioid a member of the limaçon family of curves, named for its resemblance to a heart; its equation is given as$\,r=a±b\mathrm{cos}\,\theta \,$ and$\,r=a±b\mathrm{sin}\,\theta ,\,$where$\,\frac{a}{b}=1$ convex limaҫon a type of one-loop limaçon represented by$\,r=a±b\mathrm{cos}\,\theta \,$and$\,r=a±b\mathrm{sin}\,\theta \,$such that$\,\frac{a}{b}\ge 2$ dimpled limaҫon a type of one-loop limaçon represented by$\,r=a±b\mathrm{cos}\,\theta \,$ and$\,r=a±b\mathrm{sin}\,\theta \,$ such that$\,1<\frac{a}{b}<2$ inner-loop limaçon a polar curve similar to the cardioid, but with an inner loop; passes through the pole twice; represented by$\,r=a±b\mathrm{cos}\,\theta \,$ and$\text{ }r=a±b\mathrm{sin}\,\theta \text{ }$where$\,a lemniscate a polar curve resembling a figure 8 and given by the equation[latex]\,{r}^{2}={a}^{2}\mathrm{cos}\,2\theta \,$and$\,{r}^{2}={a}^{2}\mathrm{sin}\,2\theta ,\,\,a\ne 0$ one-loop limaҫon a polar curve represented by$\,r=a±b\mathrm{cos}\,\theta \,$ and$\,r=a±b\mathrm{sin}\,\theta \,$ such that $a>0,b>0,$and$\,\frac{a}{b}>1;$ may be dimpled or convex; does not pass through the pole polar equation an equation describing a curve on the polar grid. rose curve a polar equation resembling a flower, given by the equations$\,r=a\mathrm{cos}\,n\theta \,$ and $\,r=a\mathrm{sin}\,n\theta ;\,$when $\,n\,$ is even there are $\,2n\,$ petals, and the curve is highly symmetrical; when$\,n\,$is odd there are $n$ petals.
# Special Products of Binomials The binomials are often involved in multiplication in some special forms and the products of such special form binomials are called as the special products of binomials. In mathematics, they are often used as formulas and it is very important to learn them for studying the algebra further. Here is the list of special products of binomials in algebraic form with proofs and examples. ### Product of sum basis Binomials $(1) \,\,\,$ ${(a+b)}^2$ $\,=\,$ $a^2+b^2+2ab$ $(2) \,\,\,$ ${(x+y)}^2$ $\,=\,$ $x^2+y^2+2xy$ ### Product of Difference basis Binomials $(1) \,\,\,$ ${(a-b)}^2$ $\,=\,$ $a^2+b^2-2ab$ $(2) \,\,\,$ ${(x-y)}^2$ $\,=\,$ $x^2+y^2-2xy$ ### Product of Opposite sign Binomials $(1) \,\,\,$ ${(a+b)}{(a-b)}$ $\,=\,$ $a^2-b^2$ $(2) \,\,\,$ ${(x+y)}{(x-y)}$ $\,=\,$ $x^2-y^2$ ### Product of Special case Binomials $(1) \,\,\,$ ${(x+a)}{(x+b)}$ $\,=\,$ $x^2+(a+b)x+ab$ $(2) \,\,\,$ ${(x+a)}{(x-b)}$ $\,=\,$ $x^2+(a-b)x-ab$ $(3) \,\,\,$ ${(x-a)}{(x+b)}$ $\,=\,$ $x^2-(a-b)x-ab$ $(4) \,\,\,$ ${(x-a)}{(x-b)}$ $\,=\,$ $x^2-(a+b)x+ab$ ###### Math Questions The math problems with solutions to learn how to solve a problem. Learn solutions ###### Math Worksheets The math worksheets with answers for your practice with examples. Practice now ###### Math Videos The math videos tutorials with visual graphics to learn every concept. Watch now ###### Subscribe us Get the latest math updates from the Math Doubts by subscribing us. Learn more A free math education service for students to learn every math concept easily, for teachers to teach mathematics understandably and for mathematicians to share their maths researching projects. Copyright © 2012 - 2023 Math Doubts, All Rights Reserved
Solution $\require{AMSsymbols}$ Verify using fast exponentiation that the following congruence is true $$129^{64026} \equiv 15179\pmod{64027}$$ Can you conclude 64027 is a composite number? First we verify the congruence given with fast exponentiation To do this, we need to express $129^{64026}$ as a product of factors coming from $$129, 129^2, 129^4, 129^8, 129^{16}, 129^{32}, \ldots$$ In such a product, the exponents on the factors used would have to sum to $64026$. Further, since each of these exponents is a power of $2$, finding the exponents we need is equivalent to converting $64026$ to base $2$ and looking at where the $1$'s are. Recall, we can convert $64026$ to base $2$ efficiently by successively dividing by two and discarding (but keeping track of) the remainders: $$\begin{array}{rcl} 64026 &=& 2 \cdot 32013 + 0\\ 32013 &=& 2 \cdot 16006 + 1\\ 16006 &=& 2 \cdot 8003 + 0\\ 8003 &=& 2 \cdot 4001 + 1\\ 4001 &=& 2 \cdot 2000 + 1\\ 2000 &=& 2 \cdot 1000 + 0\\ 1000 &=& 2 \cdot 500 + 0\\ 500 &=& 2 \cdot 250 + 0\\ 250 &=& 2 \cdot 125 + 0\\ 125 &=& 2 \cdot 62 + 1\\ 62 &=& 2 \cdot 31 + 0\\ 31 &=& 2 \cdot 15 + 1\\ 15 &=& 2 \cdot 7 + 1\\ 7 &=& 2 \cdot 3 + 1\\ 3 &=& 2 \cdot 1 + 1\\ 1 &=& 2 \cdot 0 + 1\\ \end{array}$$ Reading the remainders found above from the bottom to the top, we find $$64026 = 1111101000011010_2$$ Note that the rightmost digit of this base $2$ expansion corresponds to $2^0$, and the leftmost digit corresponds to $2^{32768}$. Realizing that each $1$ corresponds to a power of $2$ that we will use, tells us that $$129^{64026} = 129^{32768} \cdot 129^{16384} \cdot 129^{8192} \cdot 129^{4096} \cdot 129^{2048} \cdot 129^{512} \cdot 129^{16} \cdot 129^8 \cdot 129^1$$ Now we turn our attention to finding these powers of $129 \pmod{64027}$, which we can accomplish by successive squaring and reduction $\pmod{64027}$: $$\begin{array}{rcccl} 129^{1} &\equiv& 129\\ 129^{2} &\equiv& 129^2 &\equiv& \fbox{16641}\\ 129^{4} &\equiv& 16641^2 &\equiv& 276922881 &\equiv& 6106\\ 129^{8} &\equiv& 6106^2 &\equiv& 37283236 &\equiv& \fbox{19522}\\ 129^{16} &\equiv& 19522^2 &\equiv& 381108484 &\equiv& \fbox{19780}\\ 129^{32} &\equiv& 19780^2 &\equiv& 391248400 &\equiv& 43430\\ 129^{64} &\equiv& 43430^2 &\equiv& 1886164900 &\equiv& 57534\\ 129^{128} &\equiv& 57534^2 &\equiv& 3310161156 &\equiv& 29283\\ 129^{256} &\equiv& 29283^2 &\equiv& 857494089 &\equiv& 44505\\ 129^{512} &\equiv& 44505^2 &\equiv& 1980695025 &\equiv& \fbox{19780}\\ 129^{1024} &\equiv& 19780^2 &\equiv& 391248400 &\equiv& 43430\\ 129^{2048} &\equiv& 43430^2 &\equiv& 1886164900 &\equiv& \fbox{57534}\\ 129^{4096} &\equiv& 57534^2 &\equiv& 3310161156 &\equiv& \fbox{29283}\\ 129^{8192} &\equiv& 29283^2 &\equiv& 857494089 &\equiv& \fbox{44505}\\ 129^{16384} &\equiv& 44505^2 &\equiv& 1980695025 &\equiv& \fbox{19780}\\ 129^{32768} &\equiv& 19780^2 &\equiv& 391248400 &\equiv& \fbox{43430}\\ \end{array}$$ Finally, we multiply the desired powers (which have boxes around them above) together -- two at a time, reducing each product $\pmod{64027}$, so that our numbers don't get too big: $$\begin{array}{rcl} 129^{64026} &\equiv& 16641 \cdot 19522 \cdot 19780 \cdot 19780 \cdot 57534 \cdot 29283 \cdot 44505 \cdot 19780 \cdot 43430\\ &\equiv& 56631 \cdot 19780 \cdot 19780 \cdot 57534 \cdot 29283 \cdot 44505 \cdot 19780 \cdot 43430\\ &\equiv& 8815 \cdot 19780 \cdot 57534 \cdot 29283 \cdot 44505 \cdot 19780 \cdot 43430\\ &\equiv& 15179 \cdot 57534 \cdot 29283 \cdot 44505 \cdot 19780 \cdot 43430\\ &\equiv& 44333 \cdot 29283 \cdot 44505 \cdot 19780 \cdot 43430\\ &\equiv& 55814 \cdot 44505 \cdot 19780 \cdot 43430\\ &\equiv& 10578 \cdot 19780 \cdot 43430\\ &\equiv& 56631 \cdot 43430\\ &\equiv& 15179 \end{array}$$ This verifies the calculation given in the question above. Note, from this calculation, we can confirm that $64027$ must be composite, as otherwise Fermat's Little Theorem would require that $129^{64026} \equiv 1\pmod{64027}$ (In case you are curious, $64027 = 43 \times 1489$. That said, how to find these two factors is not revealed by the above calculations.)
Question # Find the lengths of the medians of the triangle with vertices A (0,0,6), B (0,4,0) and C (6, 0, 0). Hint: For solving the problem, we should know about the basics of finding a median from each of the vertices of the triangle. Finally, we can find the length of the medians by using distance formula on the vertices joining two points of a median. Distance formula is given by $\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}}$. Here, $({{x}_{1}},{{y}_{1}},{{z}_{1}})\text{ }and\text{ }({{x}_{2}},{{y}_{2}},{{z}_{2}})$ are the respective vertices between which we want to find the distance. Basically, before starting to solve the problem, we first try to understand the definition of median which would be useful for doing this question. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side, thus bisecting that side. Thus, in the below figure of triangle, AE, CD, and BF are the three medians of the triangle. We first start by plotting the vertices A, B and C and then finding the respective mid-points of the sides AB, BC and CA. Here, E, D and F are the respective mid-points of CB, AB and AC. Let A (0,0,6), B (0,4,0) and C (6,0,0). Now, to find the midpoint between $({{x}_{1}},{{y}_{1}},{{z}_{1}})\text{ }and\text{ }({{x}_{2}},{{y}_{2}},{{z}_{2}})$, the formula is $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2}\text{ } \right)$ for $({{x}_{1}},{{y}_{1}},{{z}_{1}})\text{ }and\text{ }({{x}_{2}},{{y}_{2}},{{z}_{2}})$ to be the respective vertices of the side of a triangle. Thus, we use this to find E, D and F. Thus, we get, E = $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2}\text{ } \right)=\left( \dfrac{0+6}{2},\dfrac{4+0}{2},\dfrac{0+0}{2} \right)$= (3,2,0) D=$\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2}\text{ } \right)=\left( \dfrac{0+0}{2},\dfrac{0+4}{2},\dfrac{6+0}{2} \right)$= (0,2,3) F=$\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2}\text{ } \right)=\left( \dfrac{0+6}{2},\dfrac{0+0}{2},\dfrac{6+0}{2} \right)$= (3,0,3) Now, we try to find median lengths using the distance formula. We have distance formula as $\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}}$. Thus, AD = $\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}}=\sqrt{{{(3-0)}^{2}}+{{(2-0)}^{2}}+{{(0-6)}^{2}}}=\sqrt{49}$= 7 BE =$\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}}=\sqrt{{{(3-0)}^{2}}+{{(0-4)}^{2}}+{{(3-0)}^{2}}}=\sqrt{34}$ CF = $\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}}=\sqrt{{{(0-6)}^{2}}+{{(2-0)}^{2}}+{{(3-0)}^{2}}}=\sqrt{49}$= 7 Hence, the length of the medians are 7, $\sqrt{34}$ and 7. Note: Another alternative to finding the lengths of the median of the triangle is to use the formula by Apolloniusâ€™ theorem â€“ $\sqrt{\dfrac{2{{b}^{2}}+2{{c}^{2}}-{{a}^{2}}}{4}}$, $\sqrt{\dfrac{2{{a}^{2}}+2{{c}^{2}}-{{b}^{2}}}{4}}$, $\sqrt{\dfrac{2{{a}^{2}}+2{{b}^{2}}-{{c}^{2}}}{4}}$. Here; a, b and c are the lengths of the sides of the triangle opposite to angles A, B and C.
Q: How can a person calculate a percentage? A: To calculate a percentage, simply divide one number into the other and multiply by 100. The order of division is important, and is determined by the wording of the question. Keep Learning There are just a few simple steps to follow in order to calculate the percentage relationship between two numbers. Step 1: Write the numbers as an equation To ensure the numbers are divided correctly, write them as an equation or a ratio. Usually the first number mentioned is the numerator, or the number on the top of the fraction. The second number is the denominator, the number on the bottom. For example, in the question "15 is what percent of 150?", the correct fraction is 15/150. Step 2: Divide the numbers Once the fraction or ratio has been determined, divide the denominator into the numerator. For the example in Step 1, the answer is 0.1. Step 3: Convert the answer to a percent To get the correct percent answer, the result from the division problem needs to be multiplied by 100. Using the problem from Step 1, the answer is 10 percent (0.1 x 100 = 10). A simple multiplication problem is used to check the accuracy of the answer. Multiply the answer from the division problem by the denominator in the fraction. The answer should be the numerator. Using the problem from Step 1, this equation would be 0.1 x 150 = 15. Sources: Related Questions • A: To calculate year-over-year percentage change, or YOY value, evaluate the outcome of two or more measured events and compare them with similar events from ... Full Answer > Filed Under: • A: A percentage can be solved by multiplying a given quantity by a percent. It should be noted, however, that the percent should first be divided by the numer... Full Answer > Filed Under: • A: The easiest way to convert a test score to a percentage is to first divide the number of correct answers by the number of total questions. Then, multiplyin... Full Answer > Filed Under: • A: Calculate a percentage increase by subtracting the old value from the new one, dividing the difference by the old value and then multiplying the quotient b... Full Answer > Filed Under: PEOPLE SEARCH FOR
Ganguli \| Math 1375 \| Fall 2020 If you would like to review the material that’s on Quiz #3, I recommend that you study Example 10.7 in the textbook–especially 10.7(a): In this example, you are presented with various polynomials and asked to find the roots, and use them to factor the polynomial completely. Let’s take a closer look at the polynomial in 10.7(a): the cubic polynomial f(x) = 2x3 – 8x2 – 6x + 36. How do we find the roots of this cubic? As you can see in the textbook explanation, we can start by looking at the graph! Here’s a nicer version of the graph I created in Desmos: Note that I factored the common factor of 2 out of the polynomial. That makes the algebra a little bit simpler going forward… Now, as the textbook explains as well, from looking at the graph it seems like x = -2 and x = 3 are roots. But to be sure we should check algebraically, i.e., by evaluating f(-2) and f(3), as they do in the textbook. The algebra is (just a bit) simpler with the 2 factored out: f(3) = 2( 3^3 – 4(3^2) – 33 + 18) = 2(27 – 36 – 9 + 18) = 20 = 0 Now how can we use the roots to factor the polynomial? That’s where the “Factor Theorem” comes in. It’s stated in Sec 8.2 of the textbook (read that section!); here is a statement via wikipedia: “The factor theorem states that a polynomial has a factor if and only if (i.e. is a root).” Note that k here represents a constant value for the input variable x. So in our example, since we know that k = 3 is a root of f(x), therefore we know that (x – 3) is a factor of f(x)! Similarly, since k = -2 is a root of f(x), we know that (x – (-2)) = (x + 2) is a factor of f(x). How can we use that information to actually factor f(x)? By long division! In this case, we would set up long division in order to compute either f(x) ÷ (x – 3) or f(x) ÷ (x + 2) In the textbook (see the bottom of p136) they carry out the long division f(x) ÷ (x-3) to show that f(x) = (x – 3)(2x2 – 2x – 12) Here’s the long division for (x3 – 4x2 – 3x + 18) ÷ (x + 2) (I’m leaving out the factor of 2 from f(x) for the long division, but then put it back in at the bottom when factoring f(x)): Therefore, we conclude that f(x) = 2(x + 2)(x2 – 6x + 9) and in this case we can factor the quadratic to get: f(x) = 2(x + 2)(x2 – 6x + 9) = 2(x + 2)(x – 3)(x – 3) = 2(x + 2)(x – 3))2 This shows that the only roots of f(x) are x = -2 and x = 3 (where the latter is a root of multiplicity 2), and thus (as the Desmos graph seemed to show, but which we have now proved algebraically): the only x-intercepts of the graph are at (-2, 0) and (3,0). Also note that we can easily find the y-intercept of the graph by computing f(0): f(0) = 2( 0^3 – 4(0^2) – 03 + 18) = 2*(18) = 36 i.e., the y-intercept is at (0, 36), again as indicated by the Desmos graph. ## 1 Comment 1. Suman Ganguli Note: I just updated this post with the polynomial long division, and showed how that allows us to get to the fully factored form (and hence all the roots) of the given polynomial. Let me know if you have any questions! Also a reminder that the “Rational Functions – Domains” WebWork set is due tonight, so finish that first if you haven’t already, and then complete the quiz (due Sunday). ### 1 Pingback Theme by Anders NorenUp ↑
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 1.5: Shifts and Dilations $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ Many functions in applications are built up from simple functions by inserting constants in various places. It is important to understand the effect such constants have on the appearance of the graph. ## Horizontal shifts If we replace $$x$$ by $$x-C$$ everywhere it occurs in the formula for $$f(x)$$, then the graph shifts over $$C$$ to the right. (If $$C$$ is negative, then this means that the graph shifts over $$\|C\|$$ to the left.) For example, the graph of $$y=(x-2)^2$$ is the $$x^2$$-parabola shifted over to have its vertex at the point 2 on the $$x$$-axis. The graph of $$y=(x+1)^2$$ is the same parabola shifted over to the left so as to have its vertex at $$-1$$ on the $$x$$-axis. Note well: when replacing $$x$$ by $$x-C$$ we must pay attention to meaning, not merely appearance. Starting with $$y=x^2$$ and literally replacing $$x$$ by $$x-2$$ gives $$y=x-2^2$$. This is $$y=x-4$$, a line with slope 1, not a shifted parabola. ## Vertical shifts If we replace $$y$$ by $$y-D$$, then the graph moves up $$D$$ units. (If $$D$$ is negative, then this means that the graph moves down $$\|D\|$$ units.) If the formula is written in the form $$y=f(x)$$ and if $$y$$ is replaced by $$y-D$$ to get $$y-D=f(x)$$, we can equivalently move D to the other side of the equation and write $$y=f(x)+D$$. Thus, this principle can be stated: to get the graph of $$y=f(x)+D$$, take the graph of $$y=f(x)$$ and move it D units up.For example, the function $$y=x^2-4x=(x-2)^2-4$$ can be obtained from $$y=(x-2)^2$$ (see the last paragraph) by moving the graph 4 units down. The result is the $$x^2$$-parabola shifted 2 units to the right and 4 units down so as to have its vertex at the point $$(2,-4)$$. Warning. Do not confuse $$f(x)+D$$ and $$f(x+D)$$. For example, if $$f(x)$$ is the function $$x^2$$, then $$f(x)+2$$ is the function $$x^2+2$$, while $$f(x+2)$$ is the function $$(x+2)^2=x^2+4x+4$$. An important example of the above two principles starts with the circle $$x^2+y^2=r^2$$. This is the circle of radius $$r$$ centered at the origin. (As we saw, this is not a single function $$y=f(x)$$, but rather two functions $$y=\pm\sqrt{r^2-x^2}$$ put together; in any case, the two shifting principles apply to equations like this one that are not in the form $$y=f(x)$$.) If we replace $$x$$ by $$x-C$$ and replace $$y$$ by $$y-D$$---getting the equation $$(x-C)^2+(y-D)^2=r^2$$---the effect on the circle is to move it $$C$$ to the right and $$D$$ up, thereby obtaining the circle of radius $$r$$ centered at the point $$(C,D)$$. This tells us how to write the equation of any circle, not necessarily centered at the origin. We will later want to use two more principles concerning the effects of constants on the appearance of the graph of a function. ## Horizontal dilation If $$x$$ is replaced by $$x/A$$ in a formula and $$A>1$$, then the effect on the graph is to expand it by a factor of $$A$$ in the $$x$$-direction (away from the $$y$$-axis). If $$A$$ is between 0 and 1 then the effect on the graph is to contract by a factor of $$1/A$$ (towards the $$y$$-axis). We use the word "dilate'' to mean expand or contract. For example, replacing $$x$$ by $$x/0.5=x/(1/2)=2x$$ has the effect of contracting toward the $$y$$-axis by a factor of 2. If $$A$$ is negative, we dilate by a factor of $$\|A\|$$ and then flip about the $$y$$-axis. Thus, replacing $$x$$ by $$-x$$ has the effect of taking the mirror image of the graph with respect to the $$y$$-axis. For example, the function $$y=\sqrt{-x}$$, which has domain $$\{x\in R\mid x\le 0\}$$, is obtained by taking the graph of $$\sqrt{x}$$ and flipping it around the $$y$$-axis into the second quadrant. ## Vertical dilation If $$y$$ is replaced by $$y/B$$ in a formula and $$B>0$$, then the effect on the graph is to dilate it by a factor of $$B$$ in the vertical direction. As before, this is an expansion or contraction depending on whether $$B$$ is larger or smaller than one. Note that if we have a function $$y=f(x)$$, replacing $$y$$ by $$y/B$$ is equivalent to multiplying the function on the right by $$B$$: $$y=Bf(x)$$. The effect on the graph is to expand the picture away from the $$x$$-axis by a factor of $$B$$ if $$B>1$$, to contract it toward the $$x$$-axis by a factor of $$1/B$$ if $$0 < B < 1$$, and to dilate by $$\|B\|$$ and then flip about the $$x$$-axis if $$B$$ is negative. $\left({x\over a}\right)^2+\left({y\over b}\right)^2=1 \qquad\hbox{or}\qquad {x^2\over a^2}+{y^2\over b^2}=1.$ Finally, if we want to analyze a function that involves both shifts and dilations, it is usually simplest to work with the dilations first, and then the shifts. For instance, if we want to dilate a function by a factor of $$A$$ in the $$x$$-direction and then shift $$C$$ to the right, we do this by replacing $$x$$ first by $$x/A$$ and then by $$(x-C)$$ in the formula. As an example, suppose that, after dilating our unit circle by $$a$$ in the $$x$$-direction and by $$b$$ in the $$y$$-direction to get the ellipse in the last paragraph, we then wanted to shift it a distance $$h$$ to the right and a distance $$k$$ upward, so as to be centered at the point $$(h,k)$$. The new ellipse would have equation $$\left({x-h\over a}\right)^2+\left({y-k\over b}\right)^2=1.$$ Note well that this is different than first doing shifts by $$h$$ and $$k$$ and then dilations by $$a$$ and $$b$$: $\left({x\over a}-h\right)^2+\left({y\over b}-k\right)^2=1.$ See figure 1.4.1. Figure 1.4.1. Ellipses: $$\left({x-1\over 2}\right)^2+\left({y-1\over 3}\right)^2=1$$ on the left, $$\left({x\over 2}-1\right)^2+\left({y\over 3}-1\right)^2=1$$ on the right. ## Contributors • Integrated by Justin Marshall. This page titled 1.5: Shifts and Dilations is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by David Guichard.
# Part of the proceeds from a garage sale was $490 worth of$10 and $20 bills. lf there were 4 more$10 bills than $20 bills, how do you find the number of each denomination? ##### 2 Answers May 18, 2017 15 ($20 dollar bills) and 19 ($10 dollar bills). #### Explanation: If we let $x$be the number of$20 dollar bills then there are $x + 4$ $10 dollar bills. The equation of the problem is: $10 \left(x + 4\right) + 20 \left(x\right) = 490$$10 x + 40 + 20 x = 490$$30 x = 450$$x = 15$Then there are 15$20 dollar bills and 19 $10 dollar bills. May 18, 2017 19 $10 bills and 15 $20 bills #### Explanation: Your equation: $20 \cdot x + 10 \cdot \left(4 + x\right) = 490$$20 x + 40 + 10 x = 490$$30 x = 450$$x = 15$(x indicates number of $20 bills). Since there are 4 more bills under $10# category compared to the other, Your 10 dollar bills number is $= 15 + 4 = 19\$.
# Lesson 9 Bayes’ Theorem ## 9.1 Motivating Example The ELISA test is used to screen blood for HIV. • When the blood contains HIV, it gives a positive result 98% of the time. • When the blood does not contain HIV, it gives a negative result 94% of the time. The prevalence of HIV is about 1% in the adult male population. A patient has just tested positive and wants to know the probability that he has HIV. What would you tell him? The solution to this problem involves an important theorem in probability and statistics called Bayes’ Theorem. This video covers some of the intuition and the history behind Bayes’ Theorem. Don’t worry about the details for now. This video is meant to be more inspiring than informative. ## 9.2 Theory The conditional probabilities $$P(A \| B)$$ and $$P(B \| A)$$ are not the same. For example, let $$A$$ be “currently a Cal Poly student” and $$B$$ be “went to high school in California”. • $$P(B \| A)$$ is very high, about $$0.85$$, since most Cal Poly students are in-state. • $$P(A \| B)$$ is very low, as only a very small fraction of people who went to high school in CA are currently in college, much less at Cal Poly. Bayes’ Theorem is a way to convert probabilities of the form $$P(B \| A)$$ into probabilities of the form $$P(A \| B)$$. This switcheroo is surprisingly common in probability and statistics. For example, • Doctors know the probability that a patient tests positive ($$B$$) given that they have the disease ($$A$$), but a patient is more interested in the probability that he has the disease ($$A$$) given that he tested positive ($$B$$). • E-mail providers can collect data on the probability that an e-mail contains a certain word ($$B$$) given that it is spam ($$A$$), but a spam filter needs the probability that the e-mail is spam ($$A$$) given that it contains the word ($$B$$). Don’t be deceived by its simplicity; Bayes’ Theorem is one of the most important and powerful results in all of probability and statistics. Theorem 9.1 (Bayes’ Theorem) $P(A \| B) = \frac{P(A) P(B \| A)}{P(B)}.$ Proof. By the definition of conditional probability (6.1) and the multiplication rule (6.1): $$$P(A \| B) = \frac{P(A \textbf{ and } B)}{P(B)} = \frac{P(A) P(B \| A)}{P(B)}.$$$ The next video provides intuition about this proof, connecting it with concepts from the past few lessons, including conditional probability and independence. To solve most problems, you will need to combine Bayes’ Theorem with the Law of Total Probability (8.1). The solution to the HIV testing example from above demonstrates some of the common tricks. Solution. Let’s represent HIV positive by $$H$$ and a positive ELISA test by $$T$$. The problem statement tells us that \begin{align} P(H) &= 0.01 & P(T \| H) &= 0.98 & P(\textbf{not } T \| \textbf{not } H) &= 0.94. \end{align} By the Complement Rule, we can infer that \begin{align} P(\textbf{not } H) &= 0.99 & P(\textbf{not } T \| H) &= 0.02 & P(T \| \textbf{not } H) &= 0.06. \end{align} Since we already have $$P(T \| H)$$ and we want to know $$P(H \| T)$$, this is a job for Bayes’ Rule: $P(H \| T) = \frac{P(H) P(T \| H)}{P(T)} = \frac{0.01 \cdot 0.98}{P(T)}.$ Unfortunately, we do not know $$P(T)$$, the probability of testing positive overall. However, we do know its probability, conditional on HIV status. So we apply the Law of Total Probability, partitioning by HIV status: $P(T) = P(H) P(T \| H) + P(\textbf{not } H) P(T \| \textbf{not } H) = 0.01 \cdot 0.98 + 0.99 \cdot 0.06.$ Finally, we plug this result into Bayes’ Rule: $P(H \| T) = \frac{0.01 \cdot 0.98}{0.01 \cdot 0.98 + 0.99 \cdot 0.06} = .142.$ So even though the patient tested positive for HIV, he only has a 14.2% chance of actually having HIV! The low probability is surprising at first. But it makes sense if we think about the problem geometrically. The figure below partitions adult males based on their HIV status and test status. The shaded area represents all people who tested positive. The people who actually have HIV make up only a tiny fraction of all the people who tested positive because there are so many more people who do not have the disease, that the false positives overwhelm the true positives. The next video will give you deep insights about Bayes’ rule. You should be able to follow most of it, but you may want to come back to this video after trying some of the examples below. ## 9.3 Examples 1. The rare mineral unobtanium is present in only 1% of rocks in a mine. You have an unobtanium detector, which never fails to detect unobtanium when it is present. Otherwise, it is still reliable, returning accurate readings 90% of the time when unobtanium is not present. 1. What is $$P(\text{unobtanium detected} \| \text{unobtanium present})$$? 2. What is $$P(\text{unobtanium present} \| \text{unobtanium detected})$$? (Hint: One of these probabilities can be read off directly from the question prompt. The other needs to be calculated using Bayes’ Rule.) 2. One application where Bayes’ Theorem has been extremely successful is spam filtering. From historical data, 80% of all e-mail is spam, and the phrase “free money” is used in 10% of spam e-mails. That is, $$P(\text{"free money"} \| \text{spam}) = 0.1$$. The phrase is also used in 1% of non-spam e-mails. A new e-mail has just arrived which contains the phrase “free money”. Given this information, what is the probability that it is spam, $$P(\text{spam} \| \text{"free money"})$$? 3. A certain disease afflicts 10% of the population. A test for the disease is 90% accurate for patients with the disease and 80% accurate for patients without the disease. Suppose you test positive for the disease. What is the probability that you actually have the disease?
# A =16.0mL volume of 0.130molL^-1 HCl(aq) was mixed with a 12.0mL volume of 0.600molL^-1 HCl(aq). What is the resultant concentration? Aug 2, 2017 $\text{molarity} = 0.331$ $\text{mol/L}$ #### Explanation: We're asked to find the final concentration of the $\text{HCl}$ solution after two separate $\text{HCl}$ solutions are mixed. To do this, we can use the molarity equation: $\text{molarity" = "mol HCl"/"L solution}$ Since we're combining two separate solutions, this can also be represented as "molarity" = ("mol"_1 + "mol"_2)/("L"_1 + "L"_2) where • ${\text{mol}}_{1}$ and ${\text{mol}}_{2}$ are the number of moles of $\text{HCl}$ in solutions $1$ and $2$ • ${\text{L}}_{1}$ and ${\text{L}}_{2}$ are the volumes, in liters, of the two $\text{HCl}$ solutions We're actually going to use the molarity equation to find the number of moles of $\text{HCl}$ in each solution: Solution 1: $\text{mol"_1 = ("molarity"_1)("L"_1) = (0.130"mol"/(cancel("L")))overbrace((0.0160cancel("L")))^"converted to liters" = color(red)(ul(0.00208color(white)(l)"mol HCl}$ Solution 2: $\text{mol"_2 = ("molarity"_2)("L"_2) = (0.600"mol"/(cancel("L")))overbrace((0.0120cancel("L")))^"converted to liters" = color(green)(ul(0.00720color(white)(l)"mol HCl}$ The molarity of the final solution is thus "molarity" = ("mol"_1 + "mol"_2)/("L"_1 + "L"_2) "molarity" = (color(red)(0.00208color(white)(l)"mol") + color(green)(0.00720color(white)(l)"mol"))/(0.0160color(white)(l)"L" + 0.0120color(white)(l)"L") = color(blue)(ulbar(\|stackrel(" ")(" "0.331color(white)(l)"mol/L"" ")\|) Aug 2, 2017 $\left[H C l\right] \approx 0.3 \cdot m o l \cdot {L}^{-} 1$ #### Explanation: We use the relationship, $\text{Concentration"="Moles of solute"/"Volume of solution}$. We assume (reasonably) that the volumes are additive, and we work out the entire number of moles of $H C l$......... $\text{Solution 1: Moles of HCl}$ $= 16.0 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.130 \cdot m o l \cdot {L}^{-} 1 = 2.08 \times {10}^{-} 3 \cdot m o l$ $\text{Solution 2: Moles of HCl}$ $= 12.0 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.600 \cdot m o l \cdot {L}^{-} 1 = 7.20 \times {10}^{-} 3 \cdot m o l$ And thus the final concentration is given by the quotient...... $\frac{2.08 \times {10}^{-} 3 \cdot m o l + 7.20 \times {10}^{-} 3 \cdot m o l}{16.0 \times {10}^{-} 3 \cdot L + 12.0 \times {10}^{-} 3 \cdot L} = 0.331 \cdot m o l \cdot {L}^{-} 1$.
# Multiplying And Dividing Using Significant Figures Approved & Edited by ProProfs Editorial Team The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes. \| By Jasondickerson2 J Jasondickerson2 Community Contributor Quizzes Created: 2 \| Total Attempts: 1,149 Questions: 10 \| Attempts: 994 Settings . • 1. ### 7.6 x 21.9 = ? • A. 166.4 • B. 166 • C. 170 • D. 160 C. 170 Explanation The product of 7.6 and 21.9 is 166.44, which is closest to 170. Rate this question: • 2. ### 38 / 7 = ? • A. 5.429 • B. 5.43 • C. 5.4 • D. 5 D. 5 Explanation The correct answer is 5 because when you divide 38 by 7, the quotient is 5 with a remainder of 3. Since the question asks for the quotient, the answer is 5. Rate this question: • 3. ### 2.15 x 3.1 x 100 = ? • A. 7 • B. 680.5 • C. 670.5 • D. 666.5 D. 666.5 Explanation The given expression can be simplified by multiplying the numbers together. 2.15 multiplied by 3.1 gives us 6.665, and when we multiply this by 100, we get 666.5. Rounding this to the nearest whole number gives us 666.5. Rate this question: • 4. ### 5.0009 x 0.06 = ? • A. 0.30005 • B. 0.3001 • C. 0.30 • D. 0.3 D. 0.3 Explanation The product of 5.0009 and 0.06 is 0.300054, which can be approximated to 0.3. Rate this question: • 5. ### 500 009 / 17.000 = ? • A. 29412 • B. 29410 • C. 29400 • D. 29000 A. 29412 Explanation To find the result of the division, we divide 500,009 by 17,000. The quotient is 29 and the remainder is 4. Since the remainder is less than half of the divisor (8,500), we round down the quotient to 29. Therefore, the correct answer is 29,412. Rate this question: • 6. ### 39.3 x 0.804 = ? • A. 31.6 • B. 31.5 • C. 31.597 • D. 32.0 A. 31.6 Explanation The given question asks for the product of 39.3 and 0.804. To find the answer, we multiply these two numbers together. The result is 31.5972, but since the question asks for the answer rounded to one decimal place, we round the answer to the nearest tenth, which gives us 31.6. Rate this question: • 7. ### 45.20 x 0.0071 =45.20 x 045.20 x 0.0071 = ? • A. 0.321 • B. 0.32 • C. 0.3209 • D. 0.3 B. 0.32 Explanation The given equation can be simplified by multiplying 45.20 and 0.0071 first, which equals 0.32152. Then, multiplying this result by 045.20 will give the final answer. However, since 045.20 is the same as 45.20, the equation can be further simplified to 45.20 multiplied by 0.32152. Evaluating this multiplication gives the answer of 0.32. Rate this question: • 8. ### 497.7 / 3.0 = ? • A. 165.9 • B. 165 • C. 166 • D. 170 A. 165.9 Explanation This is calculated by dividing the numbers as you would with whole numbers, aligning the decimal points appropriately. The result, 165.9, is obtained by straightforward division, ensuring the quotient retains the decimal place accuracy from the original numbers. Thus, 497.7 divided by 3.0 equals 165.9. Rate this question: • 9. ### 43,000 x 0.00546 = ? • A. 234.78 • B. 234.9 • C. 235 • D. 230 D. 230 Explanation The given question requires multiplying 43,000 by 0.00546. When this calculation is performed, the result is 234.78. However, since the answer choices are rounded to the nearest whole number, the correct answer is 230, which is the closest whole number to 234.78. Rate this question: • 10. ### 315 / 10 = ? • A. 31.5 • B. 31 • C. 32 • D. 30 A. 31.5 Explanation The division of 315 by 10 equals 31.5. Rate this question: Quiz Review Timeline + Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness. • Current Version • Jul 11, 2024 Quiz Edited by ProProfs Editorial Team • Apr 26, 2017 Quiz Created by Jasondickerson2 Related Topics
Objectives 1. Understand the orthogonal decomposition of a vector with respect to a subspace. 2. Understand the relationship between orthogonal decomposition and orthogonal projection. 3. Understand the relationship between orthogonal decomposition and the closest vector on / distance to a subspace. 4. Learn the basic properties of orthogonal projections as linear transformations and as matrix transformations. 5. Recipes: orthogonal projection onto a line, orthogonal decomposition by solving a system of equations, orthogonal projection via a complicated matrix product. 6. Pictures: orthogonal decomposition, orthogonal projection. 7. Vocabulary words: orthogonal decomposition, orthogonal projection. Let be a subspace of and let be a vector in In this section, we will learn to compute the closest vector to in The vector is called the orthogonal projection of onto This is exactly what we will use to almost solve matrix equations, as discussed in the introduction to Chapter 6. Subsection6.3.1Orthogonal Decomposition We begin by fixing some notation. Notation Let be a subspace of and let be a vector in We denote the closest vector to on by To say that is the closest vector to on means that the difference is orthogonal to the vectors in In other words, if then we have where is in and is in The first order of business is to prove that the closest vector always exists. Definition Let be a subspace of and let be a vector in The expression for in and in is called the orthogonal decomposition of with respect to and the closest vector is the orthogonal projection of onto Since is the closest vector on to the distance from to the subspace is the length of the vector from to i.e., the length of To restate: Closest vector and distance Let be a subspace of and let be a vector in • The orthogonal projection is the closest vector to in • The distance from to is Now we turn to the problem of computing and Of course, since really all we need is to compute The following theorem gives a method for computing the orthogonal projection onto a column space. To compute the orthogonal projection onto a general subspace, usually it is best to rewrite the subspace as the column space of a matrix, as in this important note in Section 2.6. Proof Let be the orthogonal decomposition with respect to By definition lies in and so there is a vector in with Choose any such vector We know that lies in which is equal to by this important note in Section 6.2. We thus have and so This exactly means that is consistent. If is any solution to then by reversing the above logic, we conclude that Example(Orthogonal projection onto a line) Let be a line in and let be a vector in By the theorem, to find we must solve the matrix equation where we regard as an matrix (the column space of this matrix is exactly ). But and so is a solution of and hence To reiterate: Recipe: Orthogonal projection onto a line If is a line, then for any vector When is a matrix with more than one column, computing the orthogonal projection of onto means solving the matrix equation In other words, we can compute the closest vector by solving a system of linear equations. To be explicit, we state the theorem as a recipe: Recipe: Compute an orthogonal decomposition Let be a subspace of Here is a method to compute the orthogonal decomposition of a vector with respect to 1. Rewrite as the column space of a matrix In other words, find a a spanning set for and let be the matrix with those columns. 2. Compute the matrix and the vector 3. Form the augmented matrix for the matrix equation in the unknown vector and row reduce. 4. This equation is always consistent; choose one solution Then In the context of the above recipe, if we start with a basis of then it turns out that the square matrix is automatically invertible! (It is always the case that is square and the equation is consistent, but need not be invertible in general.) Proof We will show that which implies invertibility by the invertible matrix theorem in Section 5.1. Suppose that Then so by the theorem. But (the orthogonal decomposition of the zero vector is just so and therefore is in Since the columns of are linearly independent, we have so as desired. Let be a vector in and let be a solution of Then so The corollary applies in particular to the case where we have a subspace of and a basis for To apply the corollary, we take to be the matrix with columns Subsection6.3.2Orthogonal Projection In this subsection, we change perspective and think of the orthogonal projection as a function of This function turns out to be a linear transformation with many nice properties, and is a good example of a linear transformation which is not originally defined as a matrix transformation. We compute the standard matrix of the orthogonal projection in the same way as for any other transformation: by evaluating on the standard coordinate vectors. In this case, this means projecting the standard coordinate vectors onto the subspace. In the previous example, we could have used the fact that forms a basis for so that by the corollary. In this case, we have already expressed as a matrix transformation with matrix See this example. Let be a subspace of with basis and let be the matrix with columns Then the standard matrix for is We can translate the above properties of orthogonal projections into properties of the associated standard matrix. We emphasize that the properties of projection matrices would be very hard to prove in terms of matrices. By translating all of the statements into statements about linear transformations, they become much more transparent. For example, consider the projection matrix we found in this example. Just by looking at the matrix it is not at all obvious that when you square the matrix you get the same matrix back.
# How do you differentiate f(x)=x/cotsqrtx using the chain rule? $\textcolor{red}{f ' \left(x\right) = \frac{2 \sqrt{x} \cdot \cot \sqrt{x} + x {\csc}^{2} \sqrt{x}}{2 \sqrt{x} \cdot {\cot}^{2} \sqrt{x}}}$ #### Explanation: From the given $f \left(x\right) = \frac{x}{\cot \sqrt{x}}$ Use the Quotient Formula for finding derivatives $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{d}{\mathrm{dx}} \left(u\right) - u \cdot \frac{d}{\mathrm{dx}} \left(v\right)}{v} ^ 2$ Let $u = x$ and $v = \cot \sqrt{x}$ $f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\frac{x}{\cot} \sqrt{x}\right) = \frac{\cot \sqrt{x} \cdot \frac{d}{\mathrm{dx}} \left(x\right) - x \cdot \frac{d}{\mathrm{dx}} \left(\cot \sqrt{x}\right)}{\cot \sqrt{x}} ^ 2$ $f ' \left(x\right) = \frac{\cot \sqrt{x} + x {\csc}^{2} \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}}{\cot \sqrt{x}} ^ 2$ $\textcolor{red}{f ' \left(x\right) = \frac{2 \sqrt{x} \cdot \cot \sqrt{x} + x {\csc}^{2} \sqrt{x}}{2 \sqrt{x} \cdot {\cot}^{2} \sqrt{x}}}$ God bless....I hope the explanation is useful.
# Prealgebra: Measurements ### Significant Digits #### Significant Digits The number of significant digits, or significant figures, in a given number is the number of digits after the given number has been put into scientific notation. For example, 820 ( 8.2×102 ) has 2 significant digits (8 and 2), and 0.820 ( 8.20×10-1 ) has 3 significant digits (8, 2, and 0). There are three ways to determine the number of significant digits in a number--use whichever method is easiest for you: Method I. Put the number in scientific notation and count the digits. Method II. Count the digits in a number, starting with the first non-zero digit and ending with the last non-zero digit (the zeros in the middle count as digits). If the number is a whole number, do not count any remaining zeros. If the number is a decimal, count all zeros at the end of the number. (a) The number of non-zero digits (b) The number of zeros in the middle of the number (between the non-zero digits) (c) If the number is a decimal, the number of zeros at the end of the number Examples: 7.957 has 4 significant digits. 79.57 has 4 significant digits. 0.7957 has 4 significant digits. 0.07957 has 4 significant digits. 0.79570 has 5 significant digits. 7,957 has 4 significant digits. 79,570 has 4 significant digits. 79,057 has 5 significant digits. 70,905,007 has 8 significant digits. 709,050,070 has 8 significant digits. 70,905,007.0 has 9 significant digits. #### Significant Digits in Measurement When we measure something, we do not get a precise measurement. For example, on a ruler marked with meters and centimeters, the object that we are measuring might fall between two centimeter lines. We have to estimate how far it falls between the two lines--0.4 cm.? 0.5 cm.? We know that the object measured is 117 cm. plus a little more; maybe it is 117.4 cm., maybe it is 117.5 cm. Because there is a limit to the number of digits we can know precisely, we write down all digits known precisely plus one digit that is estimated. Thus, the number of significant digits in a measurement is the number known precisely plus 1. In our example, one could write down 117.4 cm. (4 significant digits). It would be incorrect, however, to write down 117 cm. or 117.45 cm.-- 117 has too few significant digits, while 117.45 has too many significant digits. If the ruler included only measurements to the nearest 10 centimeters, we would know the 10 centimeter place precisely and would estimate in the centimeter place: we would write down 117 cm. If the ruler measured only meters (1 m. = 100 cm.), we would know the 100 centimeter place precisely and would estimate in the 10 centimeter place: we would write down 120 cm. When a measurement is known to more places than another measurement, it is said to be more precise. 117.4 cm. is more precise than 117 cm., and 117 cm. is more precise than 120 cm. ## Take a Study Break ### Star Trek gets SEXY Chris Pine and Zoe Saldana heat up the red carpet! ### Are you afraid of relationships? Auntie SparkNotes can help! ### Sexy starlet style See every single look from the Met Gala! ### Geeky Actors: Then and Now Travel back in time! ### Villains We Want These Actresses to Play From super cute to super bad! ### 10 Movies Better Than Their Books What do you think? ### How To Look Like J-Law... When you don't look like J-Law. ### 12 Scientific Inaccuracies in Into Darkness What did Star Trek get wrong? ## The Book ### Read What You Love, Anywhere You Like Get Our FREE NOOK Reading Apps
# An object travels North at 6 m/s for 4 s and then travels South at 3 m/s for 5 s. What are the object's average speed and velocity? Apr 13, 2017 $\text{Speed} = \frac{4.3333333333 \overline{3} m}{s}$ $\text{Velocity} = \frac{1 m}{s}$ #### Explanation: Calculating the average speed is very easy but I'm first doing average velocity because it requires more attention $\textcolor{w h i t e}{\times \times \times \times \times \times} O$ $N \leftarrow \leftarrow \leftarrow \leftarrow A \rightarrow \rightarrow \rightarrow \rightarrow S$ Where A is the point where the object started from N is north and S is south and is considered the midway between north andsouth. In our real earth that's equator O is the object The main trick here is to measure the distance between A and the point where the car stops. Distance covered towards North $\frac{6 m}{s} \times 4 = 24 m$ It is 24meters ahead of A towards north direction It can be represented like this $\textcolor{w h i t e}{\times \times \times \times} 24 m$ $\textcolor{w h i t e}{\times \times x} O \textcolor{w h i t e}{\times} \leftrightarrow$ $N \leftarrow \leftarrow \leftarrow \leftarrow A \rightarrow \rightarrow \rightarrow \rightarrow S$ Distance between A and O = 24m Now calculate the distance moved towards South $\frac{3 m}{s} \times 5 = 15 m$ $\textcolor{w h i t e}{\times \times x} O \rightarrow \text{by 15 meters}$ $N \leftarrow \leftarrow \leftarrow \leftarrow A \rightarrow \rightarrow \rightarrow \rightarrow S$ Distance moved from A = 24m -15m = 9m Time elapsed 4s + 5s = 9s = 9m/9s = 1m/s is the velocity $\text{Speed" = "distance"/"time}$ Distance is the distance covered can be anywhere in any direction in any zigzag pattern means the object can move back and front , eft and right. So distance covered = 24m + 15m = 39m Time elapsed 4s + 5s = 9s = $\text{distance"/"time} = \frac{39 m}{9 s} = \frac{4.3333333333 \overline{3} m}{s}$
# The doubling period of a bacterial population is 20 minutes. At time t = 100 minutes, the... ## Question: The doubling period of a bacterial population is 20 minutes. At time {eq}t= {/eq} 100 minutes, the bacterial population was 80,000. What was the initial population at time {eq}t=0 {/eq} ? Find the size of the bacterial population after 5 hours. ## Model of Exponential Growth: A bacterial population doubles in a given amount of time. Then we are also given the size of the bacterial population at a different point in time. Using all the information given in the question we calculate the initial bacterial population and then the bacterial population at a given point of time in the future. The concepts used in this question include solving an exponential equation using natural logarithms. Using an exponential function to model the population of bacterial we have the population P at time t in minutes given by where A = initial population and k = constant to be determined. Then using the fact that the bacterial population doubles in 20 minutes we let t = 20 in (1) to obtain {eq}2A=Ae^{k(20)} \implies 2 = e^{20k} \qquad (2) {/eq} Taking natural logarithms of both sides of (2) yields {eq}\displaystyle \ln 2 = 20 k \implies k = \frac {\ln 2}{20} \qquad (3) {/eq} Substituting the value of k from (3) in (1) gives {eq}\displaystyle P(t)=Ae^{\frac {\ln2}{20}t} \implies P(t) = A \left( 2^{t/20} \right) \qquad (4) {/eq} The question also gives us that the P(t) = 80000 when t = 100 minutes. Using this information in (4) yields {eq}\displaystyle 80000=A \left( 2^{100/20} \right) \implies 80000 = A \left( 2^5 \right) \implies A = \frac {80000}{32} =2500. {/eq} Since A is the initial population, hence the initial population at time t = 0 minutes is 2500 bacteria. Substituting A = 2500 in (4) gives us our complete bacterial population model which will be {eq}\displaystyle P(t) = 2500 \left( 2^{t/20} \right) \qquad (5) {/eq} So after 5 hours = 300 minutes the bacterial population from (5) will be {eq}\displaystyle P(300) = 2500 \left( 2^{300/20} \right) = 2500 \left( 2^{15} \right) = 81920000. {/eq} Therefore the bacterial population after 5 hours will be 81920000 bacteria.
The degree of a straight line can be calculated by using the formula Y=ax+b, where Y is the slope of the line, ax is the average velocity and b is the acceleration. In the case of a straight line with no curvature, Y=ax+b will be equal to the slope of the tangent at the point b. This formula is important when determining the height of a pyramid or the length of a curve. ## Y = ax + b The equation y = mx + b is one of the most common forms for an equation of a line. This is also the standard form for a quadratic equation. It represents the slope of the line in terms of x and y. Slope is an important concept that describes the rate of change between a dependent and an independent variable. This is usually represented by a positive or negative slope. A positive slope indicates that the line rises, while a negative slope means that it falls. If the slope is -1, the line slopes to the right. In a graph, this is indicated by the red dot. Another form is the point-slope form. Point-slope forms represent a line’s slope in a more general sense. For example, dydx=2×1+13+x2 is the equation of a line with a slope of one. However, this is only the case if y=ax+b. An equation of a line can take different forms depending on the information available. One such form is the slope-intercept form. This type of equation is a bit easier to use on a graphing calculator. The first step in writing a linear equation is to convert it into the slope-intercept form. To do this, you’ll need to find the ‘y-intercept’ and the ‘point-slope’. These are two points where the graph of the line crosses the y-axis. During this process, you’ll notice that the ‘y-intercept’ is a constant term and the ‘point-slope’ is the term that changes when the independent variable (x) changes. Usually, the point-slope form of a straight line is the preferred form. This is because it is the one that explicitly solves for y. There are a number of other forms, but the y-intercept and the point-slope are two of the most useful. The y-intercept is a point on the graph where the line crosses the y axis, while the point-slope is the equation that represents the slope of the line in terms of the x and y. Generally, the y-intercept is the most important part of an equation, as it’s the point on the graph that shows where the line crosses the y-axis. ## Y = mx + c A basic equation for any straight line is y = mx + c. This is an important equation in mathematics because it can be used to make predictions about the value of a variable based on its input values. It is also very important for artificial intelligence. The y-intercept is the point at which a line crosses the y-axis. The value of the y-intercept is the distance from the point of origin on the y-axis. Often, the term y-intercept is incorrectly stated as 10, but it is actually a comma two. In order to calculate the y-intercept, you must first know the slope. For a straight line, the slope is the difference between the x and y coordinates. Normally, the slope is positive. However, it can be negative too. To find the gradient of a straight line, you can use the equation y = mx+c. If you know the angle of the line with the x-axis, you can calculate its gradient. You can also use y = mx + c to find the slope of a line. Usually, a negative gradient indicates a line that is steeper than a positive one. When x = 0 and y = 7, the slope is -6. Similarly, x = y + 10 has a gradient of -10. Another way to work out the gradient of a line is to use the slope-intercept form. In this form, m is the slope and c is the y-intercept. The equation of a line is y = mx+c, but it is important to know that there are several forms of y-intercept. Some of the more common forms include the slope-intercept form, the point-slope form and the point-slope-intercept form. While all of these forms of y-intercept are useful, the most intuitive and easiest to understand is the slope-intercept form. This form is particularly useful when figuring out the y-intercept. Because it is a direct expression, it is easy to plug into. Also, it is simple to figure out the y-intercept from this equation. Graphing calculators and other devices can only display this equation in the slope-intercept format. ## Slope of a straight line A slope of a straight line is a property of a line, which is defined by the rate of change of two “x” values of two points on the line. The slope is expressed as a ratio or fraction. The slope of a horizontal line is 0 and the slope of a vertical line is undefined. Lines that pass through a point are given the same slope as the line passing through that point. When the line is perpendicular, the slope is -1. Slope is an important property of a straight line. It indicates the steepness of the line. As the speed of the line increases or decreases, its slope will increase or decrease. In general, a less steep line has a smaller slope. A positive slope is the direction in which the line increases. A negative slope is the direction in which the line decreases. To find the slope of a line, you can use the following equation. If you are not sure of the answer, you can try solving the equation in different ways. A positive slope means that the line rises from left to right. Similarly, a negative slope means that the line decreases from left to right. To determine the slope of a straight line, you can use the following formula. The numerator is any non-zero number, and the denominator contains the difference between the “x” and the “y” values. You can also find the slope of a line by measuring the rise and run of the line. This is done by dividing the difference between the x and the y values by the change in the x value. The y-intercept is the point that the line passes through. Whenever the y-intercept is a positive number, the line has a positive slope. On the other hand, if the y-intercept is a negative number, the line has a negative slope. A horizontal line is a straight line that extends from the origin to the left. There are several types of lines, such as the parallel line, the perpendicular line, and the vertical line. ## Parallel lines vs perpendicular lines A perpendicular line is a line that meets another line at right angles. It is a common symbol found in geometry. When two lines intersect at a right angle, the result is a 90-degree angle. In general, a perpendicular line will never intersect another perpendicular line, extended lines, or lines of the same shape. The term perpendicular is derived from the Latin word perpendicularis. In order to determine if a line is perpendicular or parallel, it is important to look at the slope. Parallel lines have the same slope, and perpendicular lines have negative reciprocals of each other. That is, the product of their slopes must be -1. To determine if a line is perpendicular, it is easiest to use the point-slope method. Two lines are perpendicular if their slopes are equal. If they have different y-intercepts, they are not perpendicular. Alternatively, the slope-intercept form can be used to identify a line as perpendicular. This includes y-intercept, slope, and m-intercept. For a line to be considered perpendicular, the m-intercept must be -1. Similarly, the y-intercept of a vertical line in a plane must be -. But this does not mean that a line cannot be perpendicular to other horizontal lines. Rather, it means that the line will have a slope of 0 (zero). A parallel line will not intersect a perpendicular line. However, parallel lines can still meet. The only difference between the two types of lines is that the product of the slopes of two parallel lines must be -1. In a straight line, perpendicularity is the property of all lines. Perpendicularity is often represented by a symbol, such as the ” or the ”. Some letters, such as H, have parallel lines. Other examples include a door, a window, and the corner of a blackboard. Another example of a perpendicular line is the Red Cross. Perpendicular lines are formed when two lines meet at a 90-degree angle. They do not intersect at any other point. There are other methods for determining a line’s perpendicularity, however. You can learn more about these methods in our next lesson.
# Application Adding And Subtracting Rational Numbers Worksheet A Rational Phone numbers Worksheet will help your son or daughter become more familiar with the ideas behind this percentage of integers. In this worksheet, individuals are able to solve 12 distinct problems associated with logical expressions. They will likely learn to flourish several figures, team them in couples, and figure out their goods. They will likely also practice simplifying rational expressions. Once they have perfected these methods, this worksheet might be a important instrument for advancing their research. Application Adding And Subtracting Rational Numbers Worksheet. ## Realistic Phone numbers certainly are a ratio of integers There are 2 kinds of figures: rational and irrational. Logical amounts are described as total numbers, while irrational figures do not perform repeatedly, and possess an unlimited amount of digits. Irrational amounts are low-absolutely nothing, non-terminating decimals, and sq origins that are not best squares. These types of numbers are not used often in everyday life, but they are often used in math applications. To outline a rational variety, you need to understand such a logical number is. An integer can be a entire amount, along with a rational amount is a ratio of two integers. The rate of two integers may be the amount on top divided from the variety at the base. For example, if two integers are two and five, this would be an integer. There are also many floating point numbers, such as pi, which cannot be expressed as a fraction. ## They can be made into a small percentage A rational amount carries a denominator and numerator which are not no. Because of this they may be conveyed as a portion. Together with their integer numerators and denominators, rational amounts can furthermore have a bad importance. The unfavorable importance needs to be put left of as well as its absolute benefit is its length from absolutely no. To streamline this example, we shall claim that .0333333 is actually a portion which can be composed like a 1/3. Along with bad integers, a logical quantity may also be manufactured into a small percentage. For example, /18,572 is actually a logical number, although -1/ is just not. Any small fraction comprised of integers is reasonable, provided that the denominator will not contain a and might be published for an integer. Likewise, a decimal that ends in a stage is another reasonable quantity. ## They are sense Regardless of their title, rational phone numbers don’t make significantly perception. In math, they may be individual organizations by using a special duration in the variety collection. Because of this when we count up some thing, we could buy the size by its percentage to the authentic quantity. This keeps real even if you will find endless logical amounts among two distinct numbers. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer. If we want to know the length of a string of pearls, we can use a rational number, in real life. To get the length of a pearl, by way of example, we might count up its breadth. One particular pearl weighs ten kilos, which is actually a realistic number. Furthermore, a pound’s bodyweight is equal to twenty kilograms. Thus, we should certainly split a lb by 15, without be worried about the size of a single pearl. ## They may be expressed as a decimal If you’ve ever tried to convert a number to its decimal form, you’ve most likely seen a problem that involves a repeated fraction. A decimal amount can be composed as being a several of two integers, so 4x 5 is equivalent to seven. A similar difficulty requires the repetitive portion 2/1, and either side should be split by 99 to obtain the appropriate solution. But how will you create the transformation? Here are a few good examples. A realistic amount will also be printed in many forms, which includes fractions as well as a decimal. One method to symbolize a rational amount within a decimal is usually to separate it into its fractional equal. You will find three ways to break down a logical variety, and all these techniques produces its decimal equivalent. One of those techniques is to split it into its fractional equal, and that’s what’s called a terminating decimal.
## 21.2 The Infinite Continued Fraction & Denominator Series For integer square roots, Ãa Read on to find the extension of many of these principles into the general level of integer. ### A. The Fraction Series for integer square roots, Ãa #### An infinite continued fraction for any integer root Let us generalize the above results to include the square root of any positive integer, a. As before, let X+1 = the square root of a. We square both sides of the equation. We expand our expression, subtract one from both sides of the equation, and then factor out an X. Dividing by 2+X. Remembering that X also equals the root minus one from Equation 2. Finally substituting for X on the right side of Equation 5, we arrive at an infinite continued fraction for the root of any integer. #### Some examples Below are some applications of this equation with some interesting representations. First the square root of 3 as a series of 2s. Next the square root of 4 minus one, which equals one, is an infinite continued fraction with lots of 3s and 2s. Then the square root of zero minus one or negative one is based upon a lot of 2s and -1s Adding one to both sides, we get an infinite continued fraction for zero, sort of, not quite. It is only borrowing Ô-1Õs Infinite Continued Fraction. More about zero and one a little later. #### The F Series for the square root of a positive integer. Using a parallel logic to steps 8 thru 15, from our derivation of the series for the square root of two above, we come up with this generalization. Defining Ä(N) in terms of a numerator and denominator series, as before. We substitute, then simplify algebraically. Looking at the numerator of Ä(N) we see that it determined by the product of a constant, the integer minus one, and the previous element of the denominator series. We substitute this equality into equation 9. Looking at the denominator of Ä(N) above, we see that the Nth element of the Denominator Series is defined in terms of the two previous members. Below is the contextual definition of the denominator series that will determine Ä(N), the F Series for the square root of any positive integer, Ãa. As before, this is a 2nd order iterative equation. We notice that Ä(N) is a function of the Denominator Series. It is the ratio of consecutive members of this series multiplied by the integer minus one, a constant. #### Remembering what Ä(N) stands for Remembering that the limit of our iterative function is the Root - one, Ãa - 1, from equations 7 & 8. Then substituting the results of equation 13, we see that our root is a limit of the ratio of the consecutive members of the denominator series. ### B. Basic Iterative Theorems for integer square roots, Ãa #### First Theorem: The F Series Theorem for the square root of any positive integer, a Let us restate our basic theorems in terms of integer square roots. Let us begin with the First Theorem, i.e. the F Series Theorem, for the square root of any positive integer, a. Notice that this theorem does not contradict the F Series for the Ã2. It only extends it to more cases. When our integer is 2, the numerator becomes one, (a = 2) -> (a - 1 = 1). turning it into our initial F Series expression, where the numerator was merely one. #### Second Theorem: The N Series Theorem for square roots of positive integers, a Our Second Theorem, the Numerator Theorem, is important because of its ability to neutralize the N Series as a factor in determining the F Series. Note how once again the original formulation of the Numerator Series is not contradicted, but merely extended. #### Third Theorem: The D Series for the square root of any positive integer, a The Third Theorem, the Denominator Series Theorem, first tells how to generate the D Series. Then it connects the D Series with the F Series and its Limit. We have another extension of previous results, not a contradiction. These three theorems are primary to the establishment of our Iterative Root Family. #### Experimental verification even stronger Again these theorems have been tested experimentally via the computer. While the first tests were for the square root of 2, these tests were for a random smattering of integers, each of which tested out accurately. While we might have somehow gotten the square root of 2 to 16 places of accuracy, by chance, it would have been practically impossible to accidentally generate the square roots of a random selection of integers to 16 places. Therefore the experimental verification is greatly strengthened by adding the class of positive integers to the package. ### C. Generating the D Series for Ãa In a similar fashion to the last derivation for the Ã2, we generate the positive denominator series by using givens, definitions and derivations. Equation 12 is a crucial element in the generation. Equation 12 is the defining equation of the positive Denominator Series. It defines d(N) in terms of the two previous members. It is a contextual series based upon iteration. #### Let Ä(1) = 0 = the seed For any iterative series, we need a 'seed' to get the equation going. Because the D Series is derivative from the F Series, we will begin with the F Series. We let the first member of the F Series for Ãa equal zero, Let Ä(0) = 0, just as we did before with the F Series for Ã2. A slight rationale for this assignment is that our F Series is derived by truncating the infinite continued fraction of Equation 7 at regular intervals. The first member is equal to zero because the entire equation is truncated leaving nothing left, zero. Because of this the zeroth member of the denominator series must also be zero. Shown below. #### Then Ä(2) = (a-1)/2 Truncating equation 7 immediately above after the first term, we get the second member of the Ãa set. From this fact and equation 10 above, we find the first and second members of the denominator series for the Ãa. #### The rest follow from definition Now that we have two previous elements in the set we can begin generating the rest of the denominator series by employing equation 12. #### The D Series for Ãa Finally we can write the positive denominator series for the Ãa, with the defining equation below. Remember that a function of the ratio of successive members leads to the Ãa. This is expressed in equation 15. #### The negative Denominator Series? Remembering our great fun with the negative denominator series for Ã2, remembering our square of solutions, remembering their delicious symmetry, we attempt the same technique with Ãa. We begin, as we did last time for Ã2, by solving for d(N-2) in the defining equation immediately above: Replacing N by N+2, we get the same equation in a different form. Here d(N) is defined by the two successive terms in the series rather than the two previous terms of the series, as in equation 12. This was the way we got the negative series last time. But this time we have an (a-1) term in the denominator, which fouls everything up. While previously the negative D Series was nearly a mirror opposite of the positive D Series, this definition of the negative Denominator Series leaves us with fractions if a 2. How inelegant! Further and much more seriously, because of the lack of symmetry in the D Series equation, the negative D Series approaches no limit, and so means nothing. #### The generalized number square? A myth only Many of our theorems are called the First, Second, or Third Theorems because they will extend through all the rational roots. Because of the universality in so many cases, it is tempting to force universality where it doesnÕt exist. This is the big problem with pattern recognition and extension. This is why it is so essential to computer test our results. Anyway the point is that our neat little discoveries for the negative D Series for Ã2, can in no way be generalized, even to the Ã3. The reason this works so well for Ã2 is because of the relative symmetry between d(N) and d(N-2) in the D Series equation for Ã2. The coefficient for both d(N) and d(N-2) is one. Hence the neat little square of positive and negative results shown above in our treatment of Ã2 only applies to Ã2 only and to no other root whatsoever. The negative D Series will be forgotten henceforth. It is included here only as an example of where the individual has absolutely no connection with the general. #### Metaphor for individualization Thinking metaphorically as always: when we discovered the neat little number square for the iterated expressions for Ã2, we assumed that the results would easily extend to the general expression. We were thinking of the common spiritual comment that the micro is reflected in the macro and vice versa. We were thinking in common scientific ways about fundamental principles, and building blocks of nature. We were thinking of balanced polarities. While all of these ideas applied to the individual case of Ã2, they did not extend to one other case. Hence this beauteous positive and negative symmetry is unique and individual to the D Series for Ã2 alone. Therefore the exploration of general principles would not have revealed the negative series, nor would have the exploration of the properties of the F Series for Ã2 revealed that this feature could not be extended to the higher levels of generality. #### Individual personality transcends the universal The metaphorical generalization is that general principles do not reveal all the individual characteristics. Further these individual characteristics are only revealed through specific study of the individual. Study of the general only teaches about the general but misses the entire individual structure and personality of specific phenomenon. Thus while the pursuit of general principles is especially important in finding the boundaries of the possible, it is inadequate in revealing all the truly unique specialized characteristics of individual phenomenon. Wherever one looks, personality abounds, which transcends the universal patterns on specific levels. In this case a negative series appears with all its symmetry and polar limits for Ã2, because of the unique conditions of its expression. But this same negative series with limits exists for no other square root, disappearing immediately into the chaos of no limits for every other number but two. #### Individuality, unique before the universe, Express it The major lesson here is to express your individuality. It very well could be unique to the universe, existing only in one's personal world and nowhere else. One's crystalline structure might only appear in conditions specific to oneself but to no one else. Thus in expressing individuality, we are only manifesting our uniqueness in opposition to the general principles of unity that bind the world together. Remember this is not in violation of general principles. It is only an example of specific order and structure, where all else is nearly random chaos. While the universal did not mandate universal chaos or specific order, this is what exists. ### D. The Inverse F Series for Ãa #### An amazing reciprocal relation We substitute the expression for d(N) from step 28 into Equation 13 for d(N-1). Doing some basic algebra we get the following simplification. We let the inverted denominator series equal our inverted square root function. Substituting back into Equation 29, we get this intriguing equation. Using the above equation in connection with Equation 15, we get. Adding one to all parts of the above equation, yields this symmetrical relationship. Finally, we get the limit for the inverse function. #### Theorem 4, or the inverse F Series for Ãa In order to crystallize the discoveries for the inverse F Series, FÕ, let us state a few of its theorems. First let us formalize the Fourth Theorem, i.e. the inverse F Series theorem for the square root of any positive integer, a. This theorem defines the inverse F Series, ÄÕ, as the inverse ratio of the D Series. It then points out the limit of this series. #### Theorems A & B, the Difference Theorems for Ãa There are also some difference theorems that will also prove useful in the discussions that follow. They are listed below. Theorem A is a restatement of equation 31 above, while Theorem B is a restatement of Equation 32. Stating these theorems verbally, the difference between the corresponding members of the inverse F Series and the F Series is 2. The difference of their limits is also 2. Both of these difference theorems have been tested upon the computer for a number of integer examples.
# 10.2 Use multiplication properties of exponents Page 1 / 3 By the end of this section, you will be able to: • Simplify expressions with exponents • Simplify expressions using the Product Property of Exponents • Simplify expressions using the Power Property of Exponents • Simplify expressions using the Product to a Power Property • Simplify expressions by applying several properties • Multiply monomials Before you get started, take this readiness quiz. 1. Simplify: $\frac{3}{4}·\frac{3}{4}.$ If you missed the problem, review Multiply and Divide Fractions . 2. Simplify: $\left(-2\right)\left(-2\right)\left(-2\right).$ If you missed the problem, review Multiply and Divide Integers . ## Simplify expressions with exponents Remember that an exponent indicates repeated multiplication of the same quantity. For example, ${2}^{4}$ means to multiply four factors of $2,$ so ${2}^{4}$ means $2·2·2·2.$ This format is known as exponential notation . ## Exponential notation This is read $a$ to the ${m}^{\mathrm{th}}$ power. In the expression ${a}^{m},$ the exponent tells us how many times we use the base $a$ as a factor. Before we begin working with variable expressions containing exponents, let’s simplify a few expressions involving only numbers. Simplify: 1. $\phantom{\rule{0.2em}{0ex}}{5}^{3}$ 2. $\phantom{\rule{0.2em}{0ex}}{9}^{1}$ ## Solution ⓐ ${5}^{3}$ Multiply 3 factors of 5. $5·5·5$ Simplify. $125$ ⓑ ${9}^{1}$ Multiply 1 factor of 9. $9$ Simplify: 1. $\phantom{\rule{0.2em}{0ex}}{4}^{3}$ 2. $\phantom{\rule{0.2em}{0ex}}{11}^{1}$ 1. 64 2. 11 Simplify: 1. $\phantom{\rule{0.2em}{0ex}}{3}^{4}$ 2. $\phantom{\rule{0.2em}{0ex}}{21}^{1}$ 1. 81 2. 21 Simplify: 1. $\phantom{\rule{0.2em}{0ex}}{\left(\frac{7}{8}\right)}^{2}$ 2. $\phantom{\rule{0.2em}{0ex}}{\left(0.74\right)}^{2}$ ## Solution ⓐ ${\left(\frac{7}{8}\right)}^{2}$ Multiply two factors. $\left(\frac{7}{8}\right)\left(\frac{7}{8}\right)$ Simplify. $\frac{49}{64}$ ⓑ ${\left(0.74\right)}^{2}$ Multiply two factors. $\left(0.74\right)\left(0.74\right)$ Simplify. $0.5476$ Simplify: 1. $\phantom{\rule{0.2em}{0ex}}{\left(\frac{5}{8}\right)}^{2}$ 2. $\phantom{\rule{0.2em}{0ex}}{\left(0.67\right)}^{2}$ 1. $\phantom{\rule{0.2em}{0ex}}\frac{25}{64}$ 2. $\phantom{\rule{0.2em}{0ex}}0.4489$ Simplify: 1. $\phantom{\rule{0.2em}{0ex}}{\left(\frac{2}{5}\right)}^{3}$ 2. $\phantom{\rule{0.2em}{0ex}}{\left(0.127\right)}^{2}$ 1. $\phantom{\rule{0.2em}{0ex}}\frac{8}{125}$ 2. $\phantom{\rule{0.2em}{0ex}}0.016129$ Simplify: 1. $\phantom{\rule{0.2em}{0ex}}{\left(-3\right)}^{4}$ 2. $\phantom{\rule{0.2em}{0ex}}{-3}^{4}$ ## Solution ⓐ ${\left(-3\right)}^{3}$ Multiply four factors of −3. $\left(-3\right)\left(-3\right)\left(-3\right)\left(-3\right)$ Simplify. $81$ ⓑ ${-3}^{4}$ Multiply two factors. $-\left(3·3·3·3\right)$ Simplify. $-81$ Notice the similarities and differences in parts and . Why are the answers different? In part the parentheses tell us to raise the (−3) to the 4 th power. In part we raise only the 3 to the 4 th power and then find the opposite. Simplify: 1. $\phantom{\rule{0.2em}{0ex}}{\left(-2\right)}^{4}$ 2. $\phantom{\rule{0.2em}{0ex}}{-2}^{4}$ 1. 16 2. −16 Simplify: 1. $\phantom{\rule{0.2em}{0ex}}{\left(-8\right)}^{2}$ 2. $\phantom{\rule{0.2em}{0ex}}{-8}^{2}$ 1. 64 2. −64 ## Simplify expressions using the product property of exponents You have seen that when you combine like terms by adding and subtracting, you need to have the same base with the same exponent. But when you multiply and divide, the exponents may be different, and sometimes the bases may be different, too. We’ll derive the properties of exponents by looking for patterns in several examples. All the exponent properties hold true for any real numbers, but right now we will only use whole number exponents. First, we will look at an example that leads to the Product Property. What does this mean? How many factors altogether? So, we have Notice that 5 is the sum of the exponents, 2 and 3. The base stayed the same and we added the exponents. This leads to the Product Property for Exponents. ## Product property of exponents If $a$ is a real number and $m,n$ are counting numbers, then ${a}^{m}·{a}^{n}={a}^{m+n}$ To multiply with like bases, add the exponents. An example with numbers helps to verify this property. $\begin{array}{ccc}\hfill {2}^{2}·{2}^{3}& \stackrel{?}{=}& {2}^{2+3}\hfill \\ \hfill 4·8& \stackrel{?}{=}& {2}^{5}\hfill \\ \hfill 32& =& 32✓\hfill \end{array}$ Simplify: ${x}^{5}·{x}^{7}.$ ## Solution ${x}^{5}·{x}^{7}$ Use the product property, ${a}^{m}·{a}^{n}={a}^{m+n}.$ Simplify. ${x}^{12}$ Simplify: ${x}^{7}·{x}^{8}.$ x 15 Simplify: ${x}^{5}·{x}^{11}.$ x 16 Simplify: ${b}^{4}·b.$ ## Solution ${b}^{4}·b$ Rewrite, $b={b}^{1}.$ ${b}^{4}·{b}^{1}$ Use the product property, ${a}^{m}·{a}^{n}={a}^{m+n}.$ Simplify. ${b}^{5}$ Simplify: ${p}^{9}·p.$ p 10 #### Questions & Answers where we get a research paper on Nano chemistry....? what are the products of Nano chemistry? There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others.. learn Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level learn da no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts Bhagvanji hey Giriraj Preparation and Applications of Nanomaterial for Drug Delivery revolt da Application of nanotechnology in medicine what is variations in raman spectra for nanomaterials I only see partial conversation and what's the question here! what about nanotechnology for water purification please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment. Damian yes that's correct Professor I think Professor Nasa has use it in the 60's, copper as water purification in the moon travel. Alexandre nanocopper obvius Alexandre what is the stm is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.? Rafiq industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong Damian How we are making nano material? what is a peer What is meant by 'nano scale'? What is STMs full form? LITNING scanning tunneling microscope Sahil how nano science is used for hydrophobicity Santosh Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq Rafiq what is differents between GO and RGO? Mahi what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq Rafiq if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION Anam analytical skills graphene is prepared to kill any type viruses . Anam Any one who tell me about Preparation and application of Nanomaterial for drug Delivery Hafiz what is Nano technology ? write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? why we need to study biomolecules, molecular biology in nanotechnology? ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. why? what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place. Jeannette has $5 and$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives. What is the expressiin for seven less than four times the number of nickels How do i figure this problem out. how do you translate this in Algebraic Expressions why surface tension is zero at critical temperature Shanjida I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason s. Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)= . After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
# Random Variables A Random Variable is a set of possible values from a random experiment. ### Example: Tossing a coin: we could get Heads or Tails. Let’s give them the values Heads=0 and Tails=1 and we have a Random Variable “X”: In short: X = {0, 1} Note: We could have chosen Heads=100 and Tails=150 if we wanted! It is our choice. So: • We have an experiment (such as tossing a coin) • We give values to each event • The set of values is a Random Variable ## Not Like an Algebra Variable In Algebra a variable, like x, is an unknown value: ### Example: x + 2 = 6 In this case we can find that x=4 But a Random Variable is different … ### Example: X = {0, 1, 2, 3} X could be 1, 2, 3 or 0, randomly. And they might each have a different probability. ## Capital Letters We use a capital letter, like X or Y, to avoid confusion with the Algebra type of variable. ## Sample Space A Random Variable’s set of values is the Sample Space. ### Example: Throw a die once Random Variable X = “The score shown on the top face”. X could be 1, 2, 3, 4, 5 or 6 So the Sample Space is {1, 2, 3, 4, 5, 6} ## Probability We can show the probability of any one value using this style: P(X = value) = probability of that value ### Example (continued): Throw a die once X = {1, 2, 3, 4, 5, 6} In this case they are all equally likely, so the probability of any one is 1/6 • P(X = 1) = 1/6 • P(X = 2) = 1/6 • P(X = 3) = 1/6 • P(X = 4) = 1/6 • P(X = 5) = 1/6 • P(X = 6) = 1/6 Note that the sum of the probabilities = 1, as it should be. ## Example: Toss three coins. X = “The number of Heads” is the Random Variable. So the Sample Space = {0, 1, 2, 3} But this time the outcomes are NOT all equally likely. The three coins can land in eight possible ways: X = “number of Heads” HHH 3 HHT 2 HTH 2 HTT 1 THH 2 THT 1 TTH 1 TTT 0 Looking at the table we see just 1 case of Three Heads, but 3 cases of Two Heads, 3 cases of One Head, and 1 case of Zero Heads. So: • P(X = 3) = 1/8 • P(X = 2) = 3/8 • P(X = 1) = 3/8 • P(X = 0) = 1/8 ### Example: Two dice are tossed. The Random Variable is X = “The sum of the scores on the two dice”. Let’s make a table of all possible values: 1st Die 2nd 1 2 3 4 Die 2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 8 9 10 6 7 8 9 10 11 7 8 9 10 11 12 There are 6 × 6 = 36 of them, and the Sample Space = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} Let’s count how often each value occurs, and work out the probabilities: • 2 occurs just once, so P(X = 2) = 1/36 • 3 occurs twice, so P(X = 3) = 2/36 = 1/18 • 4 occurs three times, so P(X = 4) = 3/36 = 1/12 • 5 occurs four times, so P(X = 5) = 4/36 = 1/9 • 6 occurs five times, so P(X = 6) = 5/36 • 7 occurs six times, so P(X = 7) = 6/36 = 1/6 • 8 occurs five times, so P(X = 8) = 5/36 • 9 occurs four times, so P(X = 9) = 4/36 = 1/9 • 10 occurs three times, so P(X = 10) = 3/36 = 1/12 • 11 occurs twice, so P(X = 11) = 2/36 = 1/18 • 12 occurs just once, so P(X = 12) = 1/36 ## A Range of Values We could also calculate the probability that a Random Variable takes on a range of values. ### Example (continued) What is the probability that the sum of the scores is 5, 6, 7 or 8? In other words: What is P(5 ≤ X ≤ 8)? P(5 X ≤ 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = (4+5+6+5)/36 = 20/36 = 5/9 ## Solving We can also solve a Random Variable equation. ### Example (continued) If P(X = x) = 1/12, what is the value of x? P(X = 4) = 1/12, and P(X = 10) = 1/12 So there are two solutions: x = 4 or x = 10 Notice the different uses of X and x: • X represents the Random Variable “The sum of the scores on the two dice”. • x represents a value that X can take. ## Continuous Random Variables can be either Discrete or Continuous: • Discrete Data can only take certain values (such as 1,2,3,4,5) • Continuous Data can take any value within a range (such as a person’s height) All our examples have been Discrete.
# VOLUME OF CONES AND CYLINDERS LESSON 25. ## Presentation on theme: "VOLUME OF CONES AND CYLINDERS LESSON 25."— Presentation transcript: VOLUME OF CONES AND CYLINDERS LESSON 25 Volume of Cylinders The amount of space that an object occupies is the volume of the object. To calculate volume of a cylinder, take the product of the area of its base and its height. V = Area of base x height Volume is measure in cubic units. Volume of Cylinder To find the volume, calculate the area of the base and multiply by the height. Circles A = A = (3.14)(5)2 A = 78.5 cm2 Height = 20 cm V = h = 78.5(20) = 1570 cm3 V = Area of base x height YOU TRY! To find the volume, calculate the area of the base and multiply by the height. V = Area of base x height SOLUTION To find the volume, calculate the area of the base and multiply by the height. Circles A = A = (3.14)(4)2 A = mm2 Height = 8 mm V = h = 50.24(8) = mm3 V = Area of base x height Volume of Cone To find the volume, Multiply of the volume of a cylinder 1 3 1 3 Circles A = A = (3.14)(5)2 A = 78.5 m2 V = h = 78.5(9) = m3 V = h = = m3 706.5 3 Height = 9 m 1 3 V = Volume of cylinder YOU TRY! To find the volume, Multiply of the volume of a cylinder 1 3 V = Volume of cylinder SOLUTION To find the volume, Multiply of the volume of a cylinder 1 3 Circles A = A = (3.14)(6)2 A = cm2 V = h = (9.5) = cm3 V = h = = cm3 3 Height = 9.5 cm 1 3 V = Volume of cylinder LETS REVIEW bh 2 V = Area of base x height Volume of Rectangular Prism V = lwh Volume of Rectangular Prism V = Area of base x height Volume of a Triangular Prism bh 2 V = (h) 1 3 V = Area of base x height Volume of a Pyramid 1 3 V = Volume of cylinder V = h Volume of a Cone 1 3 LETS REVIEW V = Area of base x height V = h Volume of Cylinder Class work Check solutions to Lesson 24 Copy down examples and notes to this lesson. Complete Lesson 25 worksheet Similar presentations
# Irrational Numbers Irrational Numbers are part of the Real Number Systems. Irrational means that it is not rational. Therefore, an irrational number cannot be written as a ratio or fraction. Recall that rational numbers can be written as a fraction. Examples: 3.2 = , 0.55555 = , -20 = Some famous examples of irrational numbers would include However, there are many more examples of irrational numbers. In fact, many square roots fall into the irrational category. Examples: • Irrationals numbers cannot be written as a fraction. If you try to write them out in decimal form, they do not end, or terminate, and they do not repeat. There might be a pattern after the decimal but it won't repeat. For example: 3.4848484848484848484.... is rational because the digits repeat. 5.12112211122211112222.... is irrational because it has a pattern, but the same numbers do not repeat over and over. • When you add two irrational numbers, you always get an irrational number. For example: • However, when you multiply irrational numbers you will sometimes get a rational number as the product. For example: Irrational numbers are used in many different mathematical topics. It is believed that one of Pythagoras's students actually proved that there were irrational numbers when a proof showed that √2 could not be written as a fraction. Today, we can use irrational numbers when determining the length of the hypotenuse in a right triangle using the Pythagorean Theorem. a2 + b2 = c2 82 + 112 = c2 64 + 121 = c2 √185 = c Let's Review Irrational numbers are Real, but they cannot be written as a fraction. Pi, Euler's number and many square roots are examples of irrational numbers. The irrational numbers do not end and do not repeat. They are useful when determine circumference or area of circles, in the many applications of Euler's number and whenever square roots are used, just to name a few!
# NCERT solutions for Class 10 Maths chapter 4 - Quadratic Equations [Latest edition] ## Solutions for Chapter 4: Quadratic Equations Below listed, you can find solutions for Chapter 4 of CBSE NCERT for Class 10 Maths. Exercise 4.1Exercise 4.2Exercise 4.3Exercise 4.4 Exercise 4.1 [Pages 73 - 74] ### NCERT solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1 [Pages 73 - 74] Exercise 4.1 \| Q 1.1 \| Page 73 Check whether the following is the quadratic equations : (x + 1)2 = 2(x - 3) Exercise 4.1 \| Q 1.2 \| Page 73 Check whether the following is the quadratic equations: x2 - 2x = (-2)(3 - x) Exercise 4.1 \| Q 1.3 \| Page 73 Check whether the following is the quadratic equations: (x - 2)(x + 1) = (x - 1)(x + 3) Exercise 4.1 \| Q 1.4 \| Page 73 Check whether the following is the quadratic equations (x - 3)(2x + 1) = x(x + 5) Exercise 4.1 \| Q 1.5 \| Page 73 Check whether the following is the quadratic equations: (2x - 1)(x - 3) = (x + 5)(x - 1) Exercise 4.1 \| Q 1.6 \| Page 73 Check whether the following is the quadratic equations : x2 + 3x + 1 = (x - 2)2 Exercise 4.1 \| Q 1.7 \| Page 73 Check whether the following is the quadratic equations: (x + 2)3 = 2x(x2 - 1) Exercise 4.1 \| Q 1.8 \| Page 73 Check whether the following is the quadratic equations: x3 - 4x2 - x + 1 = (x - 2)3 Exercise 4.1 \| Q 2.1 \| Page 73 Represent the following situations in the form of quadratic equations The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot Exercise 4.1 \| Q 2.2 \| Page 74 Represent the following situations in the form of quadratic equations The product of two consecutive positive integers is 306. We need to find the integers. Exercise 4.1 \| Q 2.3 \| Page 74 Represent the following situations in the form of quadratic equations Rohan's mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan's present age. Exercise 4.1 \| Q 2.4 \| Page 74 Represent the following situations in the form of quadratic equations. A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train. Exercise 4.2 [Page 76] ### NCERT solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2 [Page 76] Exercise 4.2 \| Q 1.1 \| Page 76 Find the roots of the following quadratic equations by factorisation x2 – 3x – 10 = 0 Exercise 4.2 \| Q 1.2 \| Page 76 Find the roots of the following quadratic equations by factorisation 2x2 + x – 6 = 0 Exercise 4.2 \| Q 1.3 \| Page 76 Find the roots of the following quadratic equations by factorisation sqrt2 x^2 +7x+ 5sqrt2 = 0 Exercise 4.2 \| Q 1.4 \| Page 76 Find the roots of the following quadratic equations by factorisation 2x^2 – x + 1/8 = 0 Exercise 4.2 \| Q 1.5 \| Page 76 Find the roots of the following quadratic equations by factorisation: 100x2 – 20x + 1 = 0 Exercise 4.2 \| Q 2.1 \| Page 76 John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with Exercise 4.2 \| Q 2.2 \| Page 76 A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs 750. We would like to find out the number of toys produced on that day. Exercise 4.2 \| Q 3 \| Page 76 Find two numbers whose sum is 27 and product is 182. Exercise 4.2 \| Q 4 \| Page 76 Find two consecutive positive integers, sum of whose squares is 365 Exercise 4.2 \| Q 5 \| Page 76 The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides. Exercise 4.2 \| Q 6 \| Page 76 A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article. Exercise 4.3 [Pages 87 - 88] ### NCERT solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3 [Pages 87 - 88] Exercise 4.3 \| Q 1.1 \| Page 87 Find the roots of the following quadratic equations, if they exist, by the method of completing the square 2x2 – 7x + 3 = 0 Exercise 4.3 \| Q 1.2 \| Page 87 Find the roots of the following quadratic equations, if they exist, by the method of completing the square 2x2 + x – 4 = 0 Exercise 4.3 \| Q 1.3 \| Page 87 Find the roots of the following quadratic equations, if they exist, by the method of completing the square 4x^2 + 4sqrt3x + 3 = 0 Exercise 4.3 \| Q 1.4 \| Page 87 Find the roots of the following quadratic equations, if they exist, by the method of completing the square 2x2 + x + 4 = 0 Exercise 4.3 \| Q 2.1 \| Page 87 Find the roots of the quadratic equations 2x2 – 7x + 3 = 0 by applying the quadratic formula. Exercise 4.3 \| Q 2.2 \| Page 87 Find the roots of the quadratic equations 2x2 + x – 4 = 0 by applying the quadratic formula. Exercise 4.3 \| Q 2.3 \| Page 87 Find the roots of the quadratic equations 4x^2+4sqrt3x + 3 = 0 by applying the quadratic formula. Exercise 4.3 \| Q 2.4 \| Page 87 Find the roots of the quadratic equations 2x2 + x + 4 = 0 by applying the quadratic formula. Exercise 4.3 \| Q 3.1 \| Page 88 Find the roots of the following equations: x-1/x = 3, x ≠ 0 Exercise 4.3 \| Q 3.2 \| Page 88 Find the roots of the following equations 1/(x+4) - 1/(x-7) = 11/30, x = -4, 7 Exercise 4.3 \| Q 4 \| Page 88 The sum of the reciprocals of Rehman's ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age. Exercise 4.3 \| Q 5 \| Page 88 In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects Exercise 4.3 \| Q 6 \| Page 88 The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field. Exercise 4.3 \| Q 7 \| Page 88 The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers Exercise 4.3 \| Q 8 \| Page 88 A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train. Exercise 4.3 \| Q 9 \| Page 88 Two water taps together can fill a tank in 9 3/8hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. Exercise 4.3 \| Q 10 \| Page 88 An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains. Exercise 4.3 \| Q 11 \| Page 88 Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares. Exercise 4.4 [Page 91] ### NCERT solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.4 [Page 91] Exercise 4.4 \| Q 1.1 \| Page 91 Find the nature of the roots of the following quadratic equation. If the real roots exist, find them 2x2 - 3x + 5 = 0 Exercise 4.4 \| Q 1.2 \| Page 91 Find the nature of the roots of the following quadratic equations. If the real roots exist, find them; 3x^2 - 4sqrt3x + 4 = 0 Exercise 4.4 \| Q 1.3 \| Page 91 Find the nature of the roots of the following quadratic equations. If the real roots exist, find them 2x2 - 6x + 3 = 0 Exercise 4.4 \| Q 2.1 \| Page 91 Find the values of k for each of the following quadratic equations, so that they have two equal roots. 2x2 + kx + 3 = 0 Exercise 4.4 \| Q 2.2 \| Page 91 Find the values of k for each of the following quadratic equations, so that they have two equal roots. kx (x - 2) + 6 = 0 Exercise 4.4 \| Q 3 \| Page 91 Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth. Exercise 4.4 \| Q 4 \| Page 91 Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Exercise 4.4 \| Q 5 \| Page 91 Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth. ## Solutions for Chapter 4: Quadratic Equations Exercise 4.1Exercise 4.2Exercise 4.3Exercise 4.4 ## NCERT solutions for Class 10 Maths chapter 4 - Quadratic Equations Shaalaa.com has the CBSE Mathematics Class 10 Maths CBSE solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. NCERT solutions for Mathematics Class 10 Maths CBSE 4 (Quadratic Equations) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams. Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. 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# Rd Sharma Solutions Class 7 Chapter 11 ## Rd Sharma Solutions Class 7 Chapter 11 Percentage Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Rd Sharma Class 7 Mathematics, Chapter 11, Percentage. Here students can easily find Exercise wise solution for chapter 11, Percentage. Students will find proper solutions for Exercise 11.1, 11.2, 11.3, 11.4, 11.5 and 11.6. Our teachers solved every problem with easily understandable methods so that every students can understand easily. Here all solutions are based on the CBSE latest curriculum. Percentage Exercise 11.1 Solution Question no – (1) Solution : (i) 45% in the simplest form, = 45/100 = 9/20 (ii) 0.25% in the simplest form, = 25/100 × 100 = 1/400 (iii) 150% in the simplest form, = 150/100 = 3/2 (iv) 6 1/4% in the simplest form, = 25/4% = 25/400 = 1/16 Question no – (2) Solution : (i) 3/4 as a per cent, = 3/4 × 100 = 75% (ii) 53/100 as a per cent, = 53/100 × 100% = 53% (iii) 1 3/5 as a per cent, = 1 3/5 = 8/5 × 100 = 160% (iv) 7/20 as a per cent, = 7/20 = 7/20 × 100 = 35% Percentage Exercise 11.2 Solution Question no – (1) Solution : (i) 4 : 5 as percent, = 4/5 × 100 = 80% (ii) 1 : 5 as percent, = 1/5 × 100 = 20% (iii) 11 : 125 as percent, = 11/125 × 100 = 44/5% Question no – (2) Solution : (i) 2.5% as ratios in the simplest form, = 25/100 × 10 = 1/40…(Simplest form) (ii) 0.4% as ratios in the simplest form, = 4/100 × 10 = 1/250…(Simplest form) (iii) 13 3/4% as ratios in the simplest form, = 55/4 × 100 = 11/80…(Simplest form) Percentage Exercise 11.3 Solution Question no – (1) Solution : (i) 12.5% as decimal, = 12.5/100 = 0.125 (ii) 75% as decimal, = 75/100 = 0.75 (iii) 128.8% as decimal, = 128.8/100 = 1.288 (iv) 0.05% as decimal, = 0.05/100 = 0.0005 Question no – (2) Solution : (i) 0.004 as per cent, = 9.4 × 100% = 0.4% (ii) 0.24 as per cent, = 0.24 × 100% = 24% (iii) 0.02 as per cent, = 0.02 × 100% = 2% (iv) 0.275 as per cent, = 0.275 × 100% = 27.5% Question no – (3) Solution : (i) 136% as whole number or mixed number, = 136/100 = 34/25 (ii) 250% as whole number or mixed number, = 250/100 = 5/2 (iii) 300% as whole number or mixed number, = 300/100 = 3 Percentage Exercise 11.4 Solution Question no – (1) Solution : (i) 7% of Rs 7150, = 7/100 × 7150 = Rs. 505.50 (ii) 40% of 400 kg, = 40/100 × 400 = 160 kg (iii) 20% of.125 litres, = 20/100 × 15.125 = 20/100 × 15125/1000 = 3.025 (iv) 3 1/3% of 90 km, = 3 1/3% × 90 = 10/300 × 90 = 3 km (v) 2.5% of 600 metres, = 2.5/100 × 600 = 25/100 × 10 × 600 = 15 metres. Question no – (2) Solution : We can find the number as follows : 12 1/2% × x = 64 So, 25/200 × x = 64 = x = 64 × 200/25 = 512 Therefore, the number will be 512. Question no – (3) Solution : According to the question, 6 1/4% × x = 2 = 25/400 × x = 2 = x = 2 × 400/25 = 32 Therefore, the required number will be 32. Question no – (4) Solution : We can find the number as follows : = 50% × x = 6 = 50/100 × x = 6 = x = 6 × 100/2 = 12 Hence, the required number will be 12. Percentage Exercise 11.5 Solution Question no – (1) Solution : (i) As per the question, = 6/24 × 100 = 25% Therefore, 25% of 24 is 6. (ii) In the given question we get, = 10/125 × 100 = 8% Hence, 8% of Rs 125 is Rs 10. (iii) As per the question, 4/100 × 100 = 160/400 × 100 = 4% Thus, 4% of 4 km is 160 metres. (iv) According to the question, 25/800 × 100 = 100/32 = 3.125% Hence, 3.125% of Rs 8 is 25 paise. (v) According to the question, 25/800 × 100 = 100/32 = 3.125% Therefore, 3.125% of Rs 8 is 25 paise. (vi) Let x% of 1 litre is 175 ml We know, 1 litre = 1000 mL Therefore, = 1000 × x/100 = 175 = 10x = 175 = x = 175/10 = x = 17.5% Hence, 17.5% of 1 litre is 175 ml. Question no – (3) Solution : (i) Let, 8 is 4% of x number, Therefore, 4/100 × x = 8 = x = 800/4 = x = 200 Therefore, 8 is 4% of 200. (ii) Let, 6 is 60% of x number Therefore, 60/100 × x = 6 = x = 6 × 100/60 = x = 10 Hence, 6 is 60% of 10. (iii) Let, 6 is 30% of x number Therefore, 30/100 × x = 6 = x = 600/30 = 20 Therefore, 6 is 30% of 20. Question no – (4) Solution : (i) 25 marks out of 30, 35 marks out of 40. = 25/30 = 25/30 × 100 = 83.33% And 35/40 = 35/40 × 100 = 87.5% Therefore, 35/40 > 25/30 (ii) 100 runs scored of 110 balls, 50 runs scored of 55 balls 100/110 = 100/110 × 100% = 90.91% And, 50/55 = 50/55 × 100% = 90.91% Hence, 50/55 = 100/110 Question no – (5) Solution : In the question, 20% more than 200 20/100 × 200 = 40 Required number, = 200 + 40 = 240 Therefore, 20% more than Rs 200 is Rs 240. Question no – (6) Solution : According to the question, 10% less than Rs 150 10/100 × 150 = 15 The required number, = 150 – 15 = 135 Hence, 10% less than Rs 150 is Rs. 135. Percentage Exercise 11.6 Solution Question no – (1) Solution : According to the question, 24 pages to write, Completed work = 25% 25% of 24 = 25/100 × 24 = 6 There are (24 – 6) = 18 page left. Thus, total 18 pages are left. Question no – (2) Solution : As per the question, Box contains = 60 eggs. Rotten eggs = 16 2/3% 16 2/3% of 60 = 50/3 × 100 × 60 = 10 Therefore, 10 eggs are rotten. Question no – (3) Solution : According to the question, Rohit obtained 45 marks out of 80 45/80 = 45/80 × 100 = 225/41 = 56.25% Therefore, Rohit will get 56.25% Question no – (4) Solution : As per the question, Virmani saves 12% of his salary Monthly salary = 15900 His Savings, = 12/100 × 15900 = Rs. 1908 His Expenditure, = 15900 – 1908 = Rs. 13992 Hence, Virmani monthly expenditure will be Rs 13992. Question no – (6) Solution : According to the question, Total students = 2400 Girls students = 40% Boys students = ? Girls students, = 40% of 2400 = 40 × 100 × 2400 = 960 Now, boys students, = 2400 – 960 = 1440 Therefore, there are 1440 boys students in the college. Question no – (7) Solution : As per the question, Aman obtained 410 marks out of 500 Anish gets 536 marks out of 600 Aman, 410/500 = 410/500 × 100 = 82% Anish, 536/600 = 536/600 × 100 = 89% Hence, Anish performance is better than Aman. Question no – (8) Solution : According to the question, Rahim obtained 60 marks out of 75 % of marks obtained by Rahim = ? 60 out of 75 = 60/75 × 100 = 80% Thus, Rahim obtain 80% in Mathematics. Question no – (9) Solution : Number of apple trees, = 16 2/3% of 240 = 50/3 × 100 × 240 = 40 Other type of trees, = (240 – 40) = 200 Thus, the number of other type of trees in the orchard are 200. Question no – (10) Solution : In the given question, Ram Scored 553 marks out of 700 Gita scored 486 marks out 600 Ram, = 553/700 = (533/700 × 100)% = 79% Gita, = 480/600 = (486/600 × 100)% = 81% Therefore, Gita’s performance is better. Question no – (11) Solution : As per the question, Income = 15000 Spend = 10200 Now, the saving, = 1500 – 10,200 = 4800 Percentage of income, = 4800/15000 × 100 = 32% Therefore, she save 32% of her income. Question no – (12) Solution : In the given question, 45% of student is boys Total students = 800 45/100 × 880 = 396 Girls in school, = (880 – 396) = 484 Hence, total 484 girls are there in the school. Question no – (13) Solution : As per the given question, Sidhana saves = 28% Saves per month = Rs 840 His Monthly income = ? Let, income = x 28/100 × x = 840 = x = 840 × 100/28 = x = 3000 Therefore, Sidhana’s monthly income will be Rs 3000 Question no – (14) Solution : As per the given question, Students appeared in examination = 1650 Students fail = 8% Students pass = ? 8% of 1650 = 8/100 × 1650 = 132 Student passed in examination, = (1650 – 132) = 1518 Therefore, 1518 students were pass. Question no – (15) Solution : Let, total number of student = x If 92% passed. (100 – 92) = 8% student fails 8/100 × x = 46 = x = 46 × 100/8 = x = 575 Therefore, 575 candidates appeared in the examination. Next chapter solution : Updated: June 8, 2023 — 4:09 pm
# What is (5+square root of 3)(4-2square root of 3)? Jul 30, 2018 $14 - 6 \sqrt{3}$ #### Explanation: We use the FOIL method to multiply these two expressions. This means first multiply the terms which occur first in each expression i.e. $5$ and $4$. Next, we multiply the innermost terms i.e. $\sqrt{3}$ and $4$. Then, we multiply the outermost terms i.e. $5$ and $- 2 \sqrt{3}$. Finally, we multiply the terms that occur last in each expression i.e. $\sqrt{3}$ and $- 2 \sqrt{3}$, and add them all up. $\left(5 + \sqrt{3}\right) \left(4 - 2 \sqrt{3}\right)$ $= 20 + 4 \sqrt{3} - 10 \sqrt{3} - 6$ $= 14 - 6 \sqrt{3}$
In Progress Lesson 17, Topic 1 In Progress # Forces Behind Uniform Circular Motion Lesson Progress 0% Complete ## Cars Around Bends* Horizontal, circular bends, that is. • Friction between the road and tyres of a car acts towards the centre of bend curvature, so in this case it is a centripetal force. • Remember, friction is calculated by [katex]F_f = \mu R[/katex] • μ is the coefficient of friction, which changes by material. An ice road will have a lower value than a gravel road, for example. • R is the reaction (or normal) force. It is equal in magnitude to the weight force (mg) but opposite in direction. • When solving questions, resolve vectors as seen in the image. #### Navigating Corners Safely A common physics problem is to determine the maximuxm speed at which a car can safely navigate a corner. Here are some important points to consider. • The magnitude of friction determines how strongly the car is pulled towards the centre of bend curvature. • If the car enters the bend at too high a velocity, it will not reach circular motion but instead move out in a tangent. • The equation [katex]F_c = \frac{mv^2}{r}[/katex] can be used to determine the centripetal force required to navigate a specific bend in a particular car. If the magnitude of friction if below this, the car will move off the circular path. • This relationship can be expressed (and analysed) using the following inequality: F_f\geq F_c \text{, \space \space i.e. \space \space} \mu R \geq \frac{mv^2}{r} ## A Mass on a String For a spinning mass attached to a spring, the centripetal force driving the circular motion is due to the tension in the string. • When a mass is moving in a circle parallel to the ground, there is zero net vertical force acting on it. The mass remains in the same vertical position. • However, there is a non-zero net horizontal force as the tension is angled towards the centre of motion while the weight force has no horizontal component to counteract this. • Referring to the diagrams above, the forces can be broken down into horizontal and vertical components: \begin{matrix} F_{y} = W + T\sin\theta = 0 & F_{x} = T\cos\theta \not= 0 \\ \footnotesize{\text{Vertical}} & \footnotesize{\text{Horizontal}} \end{matrix} ## Objects on Banked Tracks A banked curve has the benefit of allowing objects to maintain higher speeds when cornering without moving out on a tangent. This is because the normal reaction is tilted slightly from the vertical and can thus contribute to the centripetal force as seen below. F_{net}=R\sin\theta+F_f\cos\theta As with non-banked circular motion involving friction, there is a minimum centripetal force required for any velocity to prevent a car from skidding off the curve. The centripetal force must be greater than this, seen as follows: F_{x}\geq F_c \text{, \space \space i.e. \space \space} R\sin\theta+F_f\cos\theta \geq \frac{mv^2}{r}
# Probability of Two Events Occurring Together Probability and Statistics > Probability > Probability of Two Events occurring Together Watch the video, or read the article below: ## Probability of Two Events Occurring Together: Overview Answering probability questions can seem tricky, but they all really boil down to two things: 1. Figuring out if you multiply or add probabilities. 2. Figuring out if your events are dependent (one event has an influence on the other) or independent (they have no effect on each other) ### Should I multiply or add probabilities? You would add probabilities if you want to find out if one event or another could happen. For example, if you roll a die, and you wanted to know the probability of rolling a 1 or a 6, then you would add the probabilities: Probability of rolling a 1: 1/6 Probability of rolling a 6: 1/6 So the probability of rolling a 1 or a 6 is 1/6 + 1/6 = 2/6 = 1/3. The probability of both events happening together on the same die is zero, at least with a single throw. But if you wanted to know the probability of rolling a 1 and then rolling a 6, that’s when you would multiply (the probability would be 1/6 * 1/6 = 1/36). Learning when to add or multiply can get really confusing! The best way to learn when to add and when to multiply is to work out as many probability problems as you can. But, in general: If you have “or” in the wording, add the probabilities. If you have “and” in the wording, multiply the probabilities. This is just a general rule — there will be exceptions! ### Dependent vs. Independent Dependent events are connected to each other. For example: • In order to win at Monopoly, you have to play the game • In order to find a parking space, you have to drive • Choosing two cards from a standard deck without replacement (the first choice has a 1/52 chance, the second a 1/51). Independent events aren’t connected; the probability of one happening has no effect on the other. For example: • playing Monopoly isn’t connected to winning at Scrabble • Winning the lottery isn’t connected to you winning at Monopoly • Choosing a card and then rolling a die are not connected Dependent or Independent event? how to Tell. Tip: Look for key phrases in the question that tell you if an event is dependent or not. For example, when you are trying to figure out the probability of two events occurring together and the phrase “Out of this group” or “Of this group…” is included, that tells you the events are dependent. ## Probability of Two Events Occurring Together: Independent Use the specific multiplication rule formula. Just multiply the probability of the first event by the second. For example, if the probability of event A is 2/9 and the probability of event B is 3/9 then the probability of both events happening at the same time is (2/9)(3/9) = 6/81 = 2/27. Sample problem: The odds of you getting a job you applied for are 45% and the odds of you getting the apartment you applied for are 75%. What is the probability of you getting both the new job and the new car? Step 1: Convert your percentages of the two events to decimals. In the above example: 45% = .45. 75% = .75. Step 2: Multiply the decimals from step 1 together: .45 x .65 = .3375 or 33.75 percent. The probability of you getting the job and the car is 33.75% That’s it! ## Probability of Two Events Occurring Together: Dependent The equation you use is slightly different. P(A and B) = P(A) • P(B\|A) where P(B\|A) just means “the probability of B, once A has happened) Sample problem: Eight five % of employees have health insurance. Out of those 85%, 45% had deductibles higher than \$1,000. What percentage of people had deductibles higher than \$1,000?” Step 1: Convert your percentages of the two events to decimals. In the above example: 85% = .85. 45% = .45. Step 2: Multiply the decimals from step 1 together: .85 x .45 = .3825 or 38.35 percent. The probability of someone having a deductible of over \$1,000 is 38.35% That’s how to find the probability of two events occurring together! Tip: Sometimes it can help to make a sketch or drawing of the problem to visualize what you are trying to do. The following diagram shows the group of people (85% of the population) and the subgroup (45% of the population), making it more obvious that you should be multiplying (because when you translate 45% of the 85% (have insurance with high deductibles) to math, you get .45 .85). ------------------------------------------------------------------------------ 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. Probability of Two Events Occurring Together was last modified: October 15th, 2017 by # 4 thoughts on “Probability of Two Events Occurring Together” 1. nick In the section: Probability of Two Events Occurring Together: Independent you used the incorrect numbers from your given problem and messed up the conversion between percentages and decimals 2. Mujeeb I am mujeeb . I am student of commerce. Plz send me sure event or certain event formulas with complete solutions. 3. arturo How do you take into account the duration of the event (rate of occurrence)? It seems that if the events occur in a short time the joint probability would be less than it they occur in a longer period of time
Server Error Server Not Reachable. This may be due to your internet connection or the nubtrek server is offline. Thought-Process to Discover Knowledge Welcome to nubtrek. Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge. In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators. Read in the blogs more about the unique learning experience at nubtrek. mathsWhole NumbersNumerical Expressions with Whole Numbers ### Numerical Expressions In this page, numerical expressions are introduced as statement of a number using multiple numbers and arithmetic operations between the multiple numbers. click on the content to continue.. Subtraction represents "taking away" part of a quantity. Multiplication represents "repeated addition" of a quantity. Division represents "splitting" a quantity into equal parts. 3+2 = 5 Addition is one of the arithmetic operations. 5-2=3 Subtraction is one of the arithmetic operations. Subtraction 5-2=3 is the inverse of addition 3+2=5. The word "inverse" means: that is opposite or reverse of something familiarize with the terminology inverse 3xx2=6 Multiplication is one of the arithmetic operations. Multiplication 3xx2 is repeated addition 3+3, that is 3 repeated 2 times. 6-:2=3. Division is one of the arithmetic operations. Division 6-:2=3 is inverse of multiplication 3xx2 = 6 Consider 2+4+3. This can be evaluated to 9. This is an example of a numerical expression. The word "expression" means: collection of numbers and arithmetic operations between them, which together represent a quantity. familiarize with the terminology expression Numerical Expression : A numerical expression is a statement of a value using numbers and arithmetic operations between them. eg: 3+4xx2-6-:3 is a numerical expression involving addition, multiplication, subtraction, and division. The expression in the given example evaluates to 3+4xx2-6-:3 = 9 2xx4xx3 is an example of a numerical expression. It is evaluated to 24. a+3x is not a numerical expression. It is not entirely numbers and arithmetic operations. It has letters a and x. Solved Exercise Problem: Is 3-2 a numerical expression • Yes, the subtraction of two numbers • Yes, the subtraction of two numbers • No, simple subtraction is not a numerical expression The answer is "Yes, the subtraction of two numbers" Solved Exercise Problem: Is 3+4-2+1 a numerical expression? • No, it has both addition and subtraction • Yes, addition and subtraction can be part of a numerical expression • Yes, addition and subtraction can be part of a numerical expression The answer is "Yes, addition and subtraction can be part of an expression". Solved Exercise Problem: Is 3+4xx2-6-:3 a numerical expression? • No, as it has the addition, subtraction, multiplication, division • Yes, all arithmetic operations can be part of a numerical expression • Yes, all arithmetic operations can be part of a numerical expression The answer is "Yes, all arithmetic operations can be part of a numerical expression" Solved Exercise Problem: is 3 a numerical expression? • No. 3 is only a number. Not a numerical expression • Yes. technically a number can also be considered a numerical expression • Yes. technically a number can also be considered a numerical expression The answer is "Yes. technically a number can also be considered a numerical expression". Consider 1+2 and 3xx1. Note that when evaluated, both result in identical numerical value 1+2=3 and 3xx1 = 3`. They are two different expressions, evaluating to equal values. switch to interactive version
Skip to main content $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 3.4E: Factor Theorem and Remainder Theorem (Exercises) [ "article:topic", "license:ccbysa", "showtoc:no", "authorname:lippmanrasmussen" ] $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ Use polynomial long division to perform the indicated division. $1. \left(4x^{2} +3x-1\right)\div (x-3) 2. \left(2x^{3} -x+1\right)\div \left(x^{2} +x+1\right)$ $3. \left(5x^{4} -3x^{3} +2x^{2} -1\right)\div \left(x^{2} +4\right) 4. \left(-x^{5} +7x^{3} -x\right)\div \left(x^{3} -x^{2} +1\right)$ $5. \left(9x^{3} +5\right)\div \left(2x-3\right) 6. \left(4x^{2} -x-23\right)\div \left(x^{2} -1\right)$ Use synthetic division to perform the indicated division. $7. \left(3x^{2} -2x+1\right)\div \left(x-1\right) 8. \left(x^{2} -5\right)\div \left(x-5\right)$ $9. \left(3-4x-2x^{2} \right)\div \left(x+1\right) 10. \left(4x^{2} -5x+3\right)\div \left(x+3\right)$ $11. \left(x^{3} +8\right)\div \left(x+2\right) 12. \left(4x^{3} +2x-3\right)\div \left(x-3\right)$ $13. \left(18x^{2} -15x-25\right)\div \left(x-\frac{5}{3} \right) 14. \left(4x^{2} -1\right)\div \left(x-\frac{1}{2} \right)$ $15. \left(2x^{3} +x^{2} +2x+1\right)\div \left(x+\frac{1}{2} \right) 16. \left(3x^{3} -x+4\right)\div \left(x-\frac{2}{3} \right)$ $17. \left(2x^{3} -3x+1\right)\div \left(x-\frac{1}{2} \right) 18. \left(4x^{4} -12x^{3} +13x^{2} -12x+9\right)\div \left(x-\frac{3}{2} \right)$ $19. \left(x^{4} -6x^{2} +9\right)\div \left(x-\sqrt{3} \right) 20. \left(x^{6} -6x^{4} +12x^{2} -8\right)\div \left(x+\sqrt{2} \right)$ Below you are given a polynomial and one of its zeros. Use the techniques in this section to find the rest of the real zeros and factor the polynomial. $21. x^{3} -6x^{2} +11x-6,\; \; c=1 22. x^{3} -24x^{2} +192x-512,\; \; c=8$ $23. 3x^{3} +4x^{2} -x-2,\; \; c=\frac{2}{3} 24. 2x^{3} -3x^{2} -11x+6,\; \; c=\frac{1}{2}$ $25. x^{3} +2x^{2} -3x-6,\; \; c=-2 26. 2x^{3} -x^{2} -10x+5,\; \; c=\frac{1}{2}$ 27. $$4x^{4} -28x^{3} +61x^{2} -42x+9$$, $$c=\frac{1}{2}$$ is a zero of multiplicity 2 28. $$x^{5} +2x^{4} -12x^{3} -38x^{2} -37x-12$$, $$c=-1$$ is a zero of multiplicity 3 $209$ 3.5 Real Zeros of Polynomials
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ Skills to Develop • Recognize characteristics of parabolas. • Understand how the graph of a parabola is related to its quadratic function. • Determine a quadratic function’s minimum or maximum value. • Solve problems involving a quadratic function’s minimum or maximum value. Figure 5.1.1 An array of satellite dishes. (credit: Matthew Colvin de Valle, Flickr) Curved antennas, such as the ones shown in Figure 5.1.1, are commonly used to focus microwaves and radio waves to transmit television and telephone signals, as well as satellite and spacecraft communication. The cross-section of the antenna is in the shape of a parabola, which can be described by a quadratic function. In this section, we will investigate quadratic functions, which frequently model problems involving area and projectile motion. Working with quadratic functions can be less complex than working with higher degree functions, so they provide a good opportunity for a detailed study of function behavior. ### Recognizing Characteristics of Parabolas The graph of a quadratic function is a U-shaped curve called a parabola. One important feature of the graph is that it has an extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry. These features are illustrated in Figure 5.1.2. Figure 5.1.2 The y-intercept is the point at which the parabola crosses the y-axis. The x-intercepts are the points at which the parabola crosses thex-axis. If they exist, the x-intercepts represent the zeros, or roots, of the quadratic function, the values of $$x$$ at which $$y=0$$. Example Identifying the Characteristics of a Parabola Determine the vertex, axis of symmetry, zeros, and y- intercept of the parabola shown in Figure 5.1.3. Figure 5.1.3 Solution: The vertex is the turning point of the graph. We can see that the vertex is at $$(3,1)$$. Because this parabola opens upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is $$x=3$$. This parabola does not cross the x- axis, so it has no zeros. It crosses the y- axis at $$(0,7)$$ so this is the y-intercept ### Understanding How the Graphs of Parabolas are Related to Their Quadratic Functions The general form of a quadratic function presents the function in the form $f(x)=ax^2+bx+c$ where $$a$$,$$b$$, and $$c$$ are real numbers and $$a≠0$$. If $$a>0$$, the parabola opens upward. If $$a<0$$, the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry. The axis of symmetry is defined by $$x=−\dfrac{b}{2a}$$. If we use the quadratic formula, $$x=−\dfrac{b\pm \sqrt{b^2−4ac}}{2a}$$, to solve $$ax^2+bx+c=0$$ for the x- intercepts, or zeros, we find the value of $$x$$ halfway between them is always $$x=−\dfrac{b}{2a}$$, the equation for the axis of symmetry. Figure 5.1.4 represents the graph of the quadratic function written in general form as $$y=x^2+4x+3.$$ In this form, $$a=1$$, $$b=4$$, and $$c=3$$. Because $$a>0$$, the parabola opens upward. The axis of symmetry is $$x =−\dfrac{4}{2(1)}=−2$$. This also makes sense because we can see from the graph that the vertical line $$x=−2$$ divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, $$(−2,−1)$$. The x- intercepts, those points where the parabola crosses the x- axis, occur at $$(−3,0)$$ and $$(−1,0)$$. Figure 5.1.4 The standard form of a quadratic function presents the function in the form $f(x)=a{(x−h)}^2+k$ where $$(h, k)$$ is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as thevertex form of a quadratic function. As with the general form, if $$a>0$$, the parabola opens upward and the vertex is a minimum. If $$a<0$$, the parabola opens downward, and the vertex is a maximum. Figure 5.1.5 represents the graph of the quadratic function written in standard form as $$y=−3{(x+2)}^2+4$$. Since $$x–h=x+2$$ in this example, $$h=–2$$. In this form, $$a=−3$$,$$h=−2$$, and $$k=4$$. Because $$a<0$$, the parabola opens downward. The vertex is at $$(−2, 4)$$. Figure 5.1.5 The standard form is useful for determining how the graph is transformed from the graph of $$y=x^2$$. Figure 5.1.6 is the graph of this basic function. Figure 5.1.6 If $$k>0$$, the graph shifts upward, whereas if $$k<0$$, the graph shifts downward. In Figure 5.1.5, $$k>0$$, so the graph is shifted 4 units upward. If $$h>0$$, the graph shifts toward the right and if $$h<0$$, the graph shifts to the left. In Figure 5.1.5, $$h<0$$, so the graph is shifted 2 units to the left. The magnitude of a indicates the stretch of the graph. If $$\|a\|>1$$, the point associated with a particular x- value shifts farther from the x-axis, so the graph appears to become narrower, and there is a vertical stretch. But if $$\|a\|<1$$, the point associated with a particular x- value shifts closer to the x-axis, so the graph appears to become wider, but in fact there is a vertical compression. In Figure 5.1.5, $$\|a\|>1$$, so the graph becomes narrower. The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form. $a{(x−h)}^2+k=ax^2+bx+c$ $ax^2−2ahx+(ah^2+k)=ax^2+bx+c$ For the linear terms to be equal, the coefficients must be equal. $$–2ah=b$$, so $$h=−\dfrac{b}{2a}$$ This is the axis of symmetry we defined earlier. Setting the constant terms equal: $ah^2+k=c$ $k=c-ah^2$ $=c−a−{(\dfrac{b}{2a})}^2$ $=c−\dfrac{b^2}{4a}$ In practice, though, it is usually easier to remember that $$k$$ is the output value of the function when the input is $$h$$, so $$f(h)=k$$. A General Note A quadratic function is a polynomial function of degree two. The graph of a quadratic function is a parabola. The general form of a quadratic function is $$f(x)=ax^2+bx+c$$ where $$a$$,$$b$$, and $$c$$ are real numbers and $$a≠0$$. The standard form of a quadratic function is $$f(x)=a(x−h)^2+k$$ where $$a≠0$$. The vertex $$(h,k)$$ is located at $$h=–\dfrac{b}{2a}$$, $$k=f(h)=f(\dfrac{−b}{2a})$$ How to Given a graph of a quadratic function, write the equation of the function in general form. 1. Identify the horizontal shift of the parabola; this value is $$h$$. Identify the vertical shift of the parabola; this value is $$k$$. 2. Substitute the values of the horizontal and vertical shift for $$h$$ and $$k$$. in the function $$f(x)=a{(x–h)}^2+k$$. 3. Substitute the values of any point, other than the vertex, on the graph of the parabola for $$x$$ and $$f(x)$$. 4. Solve for the stretch factor, $$\|a\|$$. 5. Expand and simplify to write in general form. Example Writing the Equation of a Quadratic Function from the Graph Write an equation for the quadratic function $$g$$ in Figure 5.1.7 as a transformation of $$f(x)=x^2$$, and then expand the formula, and simplify terms to write the equation in general form. Figure 5.1.7 Solution: We can see the graph of g is the graph of $$f(x)=x^2$$ shifted to the left 2 and down 3, giving a formula in the form $$g(x)=a{(x−(−2))}^2−3=a{(x+2)}^2–3$$. Substituting the coordinates of a point on the curve, such as $$(0,−1)$$, we can solve for the stretch factor. $-1=a{(0+2)}^2-3$ $2=4a$ $a=\dfrac{1}{2}$ In standard form, the algebraic model for this graph is $$g(x)=\dfrac{1}{2}{(x+2)}^2–3$$. To write this in general polynomial form, we can expand the formula and simplify terms. $g(x)=\dfrac{1}{2}{(x+2)}^2-3$ $=\dfrac{1}{2}(x+2)(x+2)-3$ $=\dfrac{1}{2}(x^2+4x+4)-3$ $=\dfrac{1}{2}x^2+2x+2-3$ $=\dfrac{1}{2}x^2+2x-1$ Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions. Analysis We can check our work using the table feature on a graphing utility. First enter $$Y1=\dfrac{1}{2}{(x+2)}^2−3$$. Next, select $$TBLSET$$, then use $$TblStart=–6$$ and $$\Delta Tbl = 2$$, and select $$TABLE$$. See Table 5.1.1. $$x$$ –6 –4 –2 0 2 $$y$$ 5 –1 –3 –1 5 Table 5.1.1 The ordered pairs in the table correspond to points on the graph. Exercise A coordinate grid has been superimposed over the quadratic path of a basketball in Figure 5.1.8. Find an equation for the path of the ball. Does the shooter make the basket? Figure 5.1.8 (credit: modification of work by Dan Meyer) Solution: The path passes through the origin and has vertex at $$(−4, 7)$$, so $$h(x)=–\dfrac{7}{16}{(x+4)}^2+7$$. To make the shot, $$h(−7.5)$$ would need to be about $$4$$ but $$h(–7.5)≈1.64$$; he doesn’t make it. How to Given a quadratic function in general form, find the vertex of the parabola. 1. Identify $$a$$, $$b$$, and $$c$$. 2. Find $$h$$, the x-coordinate of the vertex, by substituting $$a$$ and $$b$$ into $$h=–\dfrac{b}{2a}$$. 3. Find $$k$$, the y-coordinate of the vertex, by evaluating $$k=f(h)=f(−\dfrac{b}{2a})$$. Example Finding the Vertex of a Quadratic Function Find the vertex of the quadratic function $$f(x)=2x^2–6x+7$$. Rewrite the quadratic in standard form (vertex form). Solution: The horizontal coordinate of the vertex will be atThe vertical coordinate of the vertex will be at $h=–\dfrac{b}{2a}$ $=\dfrac{-6}{2(2)}$ $=-dfrac{6}{4}$ $=-\dfrac{3}{2}$ Rewriting into standard form, the stretch factor will be the same as the $$a$$ in the original quadratic. First, find the horizontal coordinate of the vertex. Then find the vertical coordinate of the vertex. Substitute the values into standard form, using the “$$a$$” from the general form. $f(x)=ax^2+bx+c$ $f(x)=2x^2-6x+7$ The standard form of a quadratic function prior to writing the function then becomes the following: $f(x)=2{(x–\dfrac{3}{2})}^2+\dfrac{5}{2}$ Analysis One reason we may want to identify the vertex of the parabola is that this point will inform us where the maximum or minimum value of the output occurs, $$k$$, and where it occurs, $$x$$. Exercise Given the equation $$g(x)=13+x^2−6x$$, write the equation in general form and then in standard form. Solution: $$g(x)=x^2−6x+13$$ in general form; $$g(x)={(x−3)}^2+4$$ in standard form ### Finding the Domain and Range of a Quadratic Function Any number can be the input value of a quadratic function. Therefore, the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all y-values greater than or equal to the y-coordinate at the turning point or less than or equal to the y-coordinate at the turning point, depending on whether the parabola opens up or down. A General Note DOMAIN AND RANGE OF A QUADRATIC FUNCTION The domain of any quadratic function is all real numbers unless the context of the function presents some restrictions. The range of a quadratic function written in general form $$f(x)=ax^2+bx+c$$ with a positive a value is $$f(x)≥f(−\dfrac{b}{2a})$$, or $$[f(−\dfrac{b}{2a}),\infty)$$; the range of a quadratic function written in general form with a negative a value is $$f(x)≤f(−\dfrac{b}{2a})$$, or $$(−\infty,f(−\dfrac{b}{2a})]$$. The range of a quadratic function written in standard form $$f(x)=a{(x−h)}^2+k$$ with a positive a value is $$f(x)≥k$$; the range of a quadratic function written in standard form with a negative a value is $$f(x)≤k$$. How to Given a quadratic function, find the domain and range. 1. Identify the domain of any quadratic function as all real numbers. 2. Determine whether $$a$$ is positive or negative. If $$a$$ is positive, the parabola has a minimum. If $$a$$ is negative, the parabola has a maximum. 3. Determine the maximum or minimum value of the parabola, $$k$$. 4. If the parabola has a minimum, the range is given by $$f(x)≥k$$, or $$[k,\infty)$$. If the parabola has a maximum, the range is given by $$f(x)≤k$$, or $$(−\infty,k]$$. Example Finding the Domain and Range of a Quadratic Function Find the domain and range of $$f(x)=−5x^2+9x−1$$. Solution: As with any quadratic function, the domain is all real numbers. Because $$a$$ is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the x- value of the vertex. $h=-\dfrac{b}{2a}$ $=-\dfrac{9}{2(-5)}$ $=\dfrac{9}{10}$ The maximum value is given by $$f(h)$$. $f(\dfrac{9}{10})=5{(\dfrac{9}{10})}^2+9(\dfrac{9}{10})−1$ $=\dfrac{61}{20}$ The range is $$f(x)≤\dfrac{61}{20}$$, or $$(−\infty,\dfrac{61}{20}]$$. Exercise Finding the Domain and Range of a Quadratic Function Find the domain and range of $$f(x)=2{(x−\dfrac{4}{7})}^2+\dfrac{8}{11}$$. Solution: The domain is all real numbers. The range is $$f(x)≥\dfrac{8}{11}$$, or $$[\dfrac{8}{11},\infty)$$. ### Determining the Maximum and Minimum Values of Quadratic Functions The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola. We can see the maximum and minimum values in Figure 5.1.9. Figure 5.1.9 There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue. Example Finding the Maximum Value of a Quadratic Function A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side. a. Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length L. b. What dimensions should she make her garden to maximize the enclosed area? Solution: a. Let’s use a diagram such as Figure 5.1.10 to record the given information. It is also helpful to introduce a temporary variable, $$W$$, to represent the width of the garden and the length of the fence section parallel to the backyard fence. Figure 5.1.10 We know we have only 80 feet of fence available, and $$L+W+L=80$$, or more simply, $$2L+W=80$$. This allows us to represent the width, $$W$$, in terms of $$L$$. $W=80−2L$ Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so $A=WL=L(80-2L)$ $A(L)=80L-2L^2$ This formula represents the area of the fence in terms of the variable length $$L$$. The function, written in general form, is $A(L)=−2L^2+80L$ b. The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since a is the coefficient of the squared term, $$a=−2$$,$$b=80$$, and $$c=0$$. To find the vertex: $h=-\dfrac{b}{2a}$ $=-\dfrac{80}{2(-2)}$ $=20$ and $k=A(20)$ $=80(20)-2{(20)}^2$ $=800$ The maximum value of the function is an area of 800 square feet, which occurs when $$L=20$$ feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet. Analysis This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in Figure 5.1.11. Figure 5.1.11 How to Given an application involving revenue, use a quadratic equation to find the maximum. 1. Write a quadratic equation for a revenue function. 2. Find the vertex of the quadratic equation. 3. Determine the y-value of the vertex. Example Finding Maximum Revenue The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to$32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue? Solution: Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, $$p$$ for price per subscription and $$Q$$ for quantity, giving us the equation Revenue$$=pQ$$. Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently $$p=30$$ and $$Q=84,000$$. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, $$p=32$$ and $$Q=79,000$$. From this we can find a linear equation relating the two quantities. The slope will be $m=\dfrac{79000-84000}{32-30}$ $=\dfrac{-5000}{2}$ $=-2500$ This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y-intercept. $Q=-2500p+b$ $84000=-2500(30)+b$ $b=159000$ This gives us the linear equation $$Q=−2,500p+159,000$$ relating cost and subscribers. We now return to our revenue equation. $Revenue=pQ$ $Revenue=p(-2500p+159000)$ $Revenue=-2500p^2+159000p$ We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex. $h=-\dfrac{159000}{2(-2500)}$ $=31.8$ The model tells us that the maximum revenue will occur if the newspaper charges$31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function. $maximum\space revenue=-2500{(31.8)}^2+159000(31.8)$ $=2528100$ Analysis This could also be solved by graphing the quadratic as in Figure 5.1.12. We can see the maximum revenue on a graph of the quadratic function. Figure 5.1.12 #### Finding the x- and y-Intercepts of a Quadratic Function Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the y- intercept of a quadratic by evaluating the function at an input of zero, and we find the x- intercepts at locations where the output is zero. Notice in Figure 5.1.13 that the number of x- intercepts can vary depending upon the location of the graph. Figure 5.1.13 Number of x-intercepts of a parabola How to Given a quadratic function $$f(x)$$, find the y- and x-intercepts. 1. Evaluate $$f(0)$$ to find the y-intercept. 2. Solve the quadratic equation $$f(x)=0) to find the x-intercepts. Example Finding the y- and x-Intercepts of a Parabola Find the y- and x-intercepts of the quadratic \(f(x)=3x^2+5x−2$$. Solution: We find the y-intercept by evaluating $$f(0)$$. $f(0)=3{(0)}^2+5(0)−2$ $=−2$ So the y-intercept is at $$(0,−2)$$. For the x-intercepts, we find all solutions of $$f(x)=0$$. $0=3x^2+5x−2$ In this case, the quadratic can be factored easily, providing the simplest method for solution. $0=(3x−1)(x+2)$ $h=-\dfrac{b}{2a}$ $=-\dfrac{4}{2(2)}$ $=-1$ $k=f(-1)$ $=2{(-1)}^+4(-1)-4$ $=-6$ So the x-intercepts are at $$(\dfrac{1}{3},0)$$ and $$(−2,0)$$. Analysis By graphing the function, we can confirm that the graph crosses the y-axis at $$(0,−2)$$. We can also confirm that the graph crosses the x-axis at $$(\dfrac{1}{3},0)$$ and $$(−2,0)$$. See Figure 5.1.14 Figure 5.1.14 #### Rewriting Quadratics in Standard Form In Example, the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form. How to Given a quadratic function, find the x- intercepts by rewriting in standard form. 1. Substitute $$a$$ and $$b$$ into $$h=−\dfrac{b}{2a}$$. 2. Substitute $$x=h$$ into the general form of the quadratic function to find $$k$$. 3. Rewrite the quadratic in standard form using $$h$$ and $$k$$. 4. Solve for when the output of the function will be zero to find the x- intercepts. Example Finding the x-Intercepts of a Parabola Find the x- intercepts of the quadratic function $$f(x)=2x^2+4x−4$$. Solution: We begin by solving for when the output will be zero. $0=2x^2+4x−4$ Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form. $f(x)=a{(x−h)}^2+k$ We know that $$a=2$$. Then we solve for $$h$$ and $$k$$. $h=−\dfrac{b}{2a}$ $=-\dfrac{4}{2(2)}$ $=-1$ $k=f(-1)$ $=2{(-1)}^2+4(-1)-1$ $=-6$ So now we can rewrite in standard form. $f(x)=2{(x+1)}^2−6$ We can now solve for when the output will be zero. $0=2{(x+1)}^2-6$ $3={(x+1)}^2$ $x+1=\pm \sqrt{3}$ $x=-1\pm \sqrt{3}$ The graph has x-intercepts at $$(−1−\sqrt{3},0)$$ and $$(−1+\sqrt{3},0)$$. We can check our work by graphing the given function on a graphing utility and observing the x- intercepts. See Figure 5.1.15. Figure 5.1.15 Analysis We could have achieved the same results using the quadratic formula. Identify $$a=2$$,$$b=4$$ and $$c=−4$$. $x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$ $=\dfrac{-4\pm \sqrt{4^2-4(2)(-4)}}{2(2)}$ $=\dfrac{-4\pm \sqrt{48}}{4}$ $=\dfrac{-4\pm\sqrt{3(16)}}{4}$ $=-1\pm \sqrt{3}$ So the x-intercepts occur at $$(−1−\sqrt{3},0)$$ and $$(−1+\sqrt{3},0)$$. Exercise In a Try It, we found the standard and general form for the function $$g(x)=13+x^2−6x$$. Now find the y- and x-intercepts (if any). Solution: y-intercept at $$(0, 13)$$, No x- intercepts Example Applying the Vertex and x-Intercepts of a Parabola A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation $$H(t)=−16t^2+80t+40$$. a. When does the ball reach the maximum height? b. What is the maximum height of the ball? c. When does the ball hit the ground? Solution: a. The ball reaches the maximum height at the vertex of the parabola. $h=-\dfrac{80}{2(-16)}$ $=\dfrac{80}{32}$ $=\dfrac{5}{2}$ $=2.5$ The ball reaches a maximum height after 2.5 seconds. b. To find the maximum height, find the y- coordinate of the vertex of the parabola. $k=H(-\dfrac{b}{2a})$ $=H(2.5)$ $=-16{(2.5)}^2+80(2.5)+40$ $=140$ The ball reaches a maximum height of 140 feet. c. To find when the ball hits the ground, we need to determine when the height is zero, $$H(t)=0$$. $t=\dfrac{-80\pm \sqrt{ {80}^2-4(-16)(40)}}{2(-16)}$ $=\dfrac{-80\pm \sqrt{8960}}{-32}$ Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions. $t=\dfrac{-80-\sqrt{8960}}{-32}≈5.458\space or\space t=\dfrac{-80+\sqrt{8960}}{-32}≈−0.458$ The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See Figure 5.1.16. Figure 5.1.16 Note that the graph does not represent the physical path of the ball upward and downward. Keep the quantities on each axis in mind while interpreting the graph. Exercise A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock’s height above ocean can be modeled by the equation $$H(t)=−16t^2+96t+112$$. 1. When does the rock reach the maximum height? 2. What is the maximum height of the rock? 3. When does the rock hit the ocean? Solution: 1. 3 seconds 2. 256 feet 3. 7 seconds Media Access these online resources for additional instruction and practice with quadratic equations. ### Key Equations general form of a quadratic function $$f(x)=ax^2+bx+c$$ standard form of a quadratic function $$f(x)=a{(x−h)}^2+k$$ ### Key Concepts • A polynomial function of degree two is called a quadratic function. • The graph of a quadratic function is a parabola. A parabola is a U-shaped curve that can open either up or down. • The axis of symmetry is the vertical line passing through the vertex. The zeros, or x- intercepts, are the points at which the parabola crosses the x- axis. The y- intercept is the point at which the parabola crosses the y- axis. See ExampleExample, andExample. • Quadratic functions are often written in general form. Standard or vertex form is useful to easily identify the vertex of a parabola. Either form can be written from a graph. See Example. • The vertex can be found from an equation representing a quadratic function. See Example. • The domain of a quadratic function is all real numbers. The range varies with the function. See Example. • A quadratic function’s minimum or maximum value is given by the y- value of the vertex. • The minimum or maximum value of a quadratic function can be used to determine the range of the function and to solve many kinds of real-world problems, including problems involving area and revenue. See Example and Example. • The vertex and the intercepts can be identified and interpreted to solve real-world problems. See Example.
# How do you maximize and minimize f(x,y)=x^2-y/x constrained to 0<=x+y<=1? Feb 16, 2016 There is no upper bound for $f$. The lower bound is $f \left(0 , 0\right) = 1$ #### Explanation: Substitute $x + y = u$. Now, optimize $g \left(x , u\right) = {x}^{2} - \frac{u - x}{x}$ $= {x}^{2} + 1 - \frac{u}{x}$ With the only condition being $0 \le u \le 1$. When $x$ is large, the ${x}^{2}$ term dominates and the $\frac{u}{x}$ term becomes insignificant. By increasing $x$, $f$ can be made arbitrarily large. $\frac{\partial g}{\partial x} = 2 x + \frac{u}{x} ^ 2$ From the first equation, we see that for a given value of $u$, minimum occurs when $\frac{\partial g}{\partial x} = 0$. Therefore, $x = - \sqrt[3]{\frac{u}{2}}$ So to find the minimum of $g$, we replace all instances of $x$ with $- \sqrt[3]{\frac{u}{2}}$. $g \left(u\right) = {\left(- \sqrt[3]{\frac{u}{2}}\right)}^{2} + 1 - \frac{u}{- \sqrt[3]{\frac{u}{2}}}$ $= {2}^{- \frac{2}{3}} \cdot {u}^{\frac{2}{3}} + 1 + \sqrt[3]{2} \cdot {u}^{\frac{2}{3}}$ $= 3 \sqrt[3]{{u}^{2} / 4} + 1$ Even without differentiating, it is quite easy to see that $g$ is increasing for $u \in \left[0 , 1\right]$ The mimimum corresponds to $u = 0$, and consequently, $x = 0$, and $y = 0$. The absolute minimum is $f \left(0 , 0\right) = 1$
Lesson Objectives • Demonstrate an understanding of how to set up an augmented matrix • Demonstrate an understanding of how to place an augmented matrix in row echelon form • Demonstrate an understanding of how to place an augmented matrix in reduced-row echelon form • Learn how to solve a linear system in three variables using matrix methods ## How to Solve a Linear System in Three Variables using Gaussian Elimination In the last lesson, we learned how to solve a linear system in two variables using Gaussian Elimination. In this lesson, we will look at some additional examples of this method with linear systems in three variables. Recall that our goal is to place the augmented matrix in row echelon form. This form gives us "1's" down the diagonal and "0's" below: $$\left[ \begin{array}{ccc\|c} 1&a&b&c\\ 0&1&d&e\\ 0&0&1&f\\ \end{array} \right]$$ We use "elementary row operations" to achieve our desired row echelon form. ### Elementary Row Operations • We can interchange any two rows • We can multiply any row by a non-zero number • We can multiply a row by a real number and add this to the corresponding elements of any other row Let's look at an example. Example 1: Solve each linear system using Gaussian Elimination $$-x + 4y - 2z = -15$$ $$-4x + 6y + z = -5$$ $$-6x - 6y - 2z = -10$$ Let's begin by writing our augmented matrix: $$\left[ \begin{array}{ccc\|c} -1&4&-2&-15\\ -4&6&1&-5\\ -6&-6&-2&-10\\ \end{array} \right]$$ We work column by column, starting with the leftmost column. We want a 1 as the first element in the leftmost column. We can achieve this by multiplying the top row by -1: $$\left[ \begin{array}{ccc\|c} 1&-4&2&15\\ -4&6&1&-5\\ -6&-6&-2&-10\\ \end{array} \right]$$ Next, we will work on the second element in the leftmost column. Since we want this to be a 0, we can multiply row 1 by 4 and add the result to row 2: $$\left[ \begin{array}{ccc\|c} 1&-4&2&15\\ 0&-10&9&55\\ -6&-6&-2&-10\\ \end{array} \right]$$ Next, we will work on the third element in the leftmost column. Since we want this to be a 0, we can multiply row 1 by 6 and add the result to row 3: $$\left[ \begin{array}{ccc\|c} 1&-4&2&15\\ 0&-10&9&55\\ 0&-30&10&80\\ \end{array} \right]$$ Now that the first or leftmost column is complete, we move one column to the right. We will start by changing the second element of the middle column into a 1. We can achieve this by multiplying row 2 by -1/10: $$\left[ \begin{array}{ccc\|c} 1&-4&2&15\\ 0&1&-9/10&-55/10\\ 0&-30&10&80\\ \end{array} \right]$$ Next, we will work on the third element in the middle column. Since we want this to be a 0, we can multiply row 2 by 30 and add the result to row 1: $$\left[ \begin{array}{ccc\|c} 1&-4&2&15\\ 0&1&-9/10&-55/10\\ 0&0&-17&-85\\ \end{array} \right]$$ Now that the middle column is complete, we move on to the last or rightmost column. We only need to change the third element into a 1. We can achieve this by multiplying row 3 by -1/17: $$\left[ \begin{array}{ccc\|c} 1&-4&2&15\\ 0&1&-9/10&-55/10\\ 0&0&1&5\\ \end{array} \right]$$ Our augmented matrix is in row echelon form. We can now use substitution to find the solutions for the system. Let's translate the numerical information from the augmented matrix back into equation format: $$1x - 4y + 2z = 15$$ $$0x + 1y - \frac{9}{10}z = -\frac{55}{10}$$ $$0x + 0y + 1z = 5$$ At this point, we know that z is 5, let's plug this into equation 2 and find the value for y: $$y - \frac{9}{10}(5) = -\frac{55}{10}$$ $$y - \frac{45}{10} = -\frac{55}{10}$$ $$y = -\frac{55}{10} + \frac{45}{10}$$ $$y = -\frac{10}{10}$$ $$y = -1$$ Lastly, we can find x by substituting in a 5 for z and a -1 for y in equation 1: $$1x - 4(-1) + 2(5) = 15$$ $$1x + 4 + 10 = 15$$ $$x + 14 = 15$$ $$x = 1$$ Our solution is the ordered triple: (1,-1,5) Let's check our solution. -x + 4y - 2z = -15 -(1) + 4(-1) - 2(5) = -15 -1 - 4 - 10 = -15 -15 = -15 -4x + 6y + z = -5 -4(1) + 6(-1) + (5) = -5 -4 - 6 + 5 = -5 -10 + 5 = -5 -5 = -5 -6x - 6y - 2z = -10 -6(1) - 6(-1) - 2(5) = -10 -6 + 6 - 10 = -10 -10 = -10 ### Solving Linear Systems in Three Variables using Gauss-Jordan Elimination In the last lesson, we also learned about Gauss-Jordan Elimination. With this method, we place the augmented matrix in reduced-row echelon form. This gives us "1's" down the diagonal and "0's" above and below. $$\left[ \begin{array}{ccc\|c} 1&0&0&a\\ 0&1&0&b\\ 0&0&1&c \end{array} \right]$$ When the augmented matrix is in reduced-row echelon form, the solutions can be obtained directly from the matrix. This means no substitution is needed. ### Gauss-Jordan Three-Variable System • Obtain a 1 as the first element in the first column • Use the first row to transform the remaining elements in the first column into zeros • Obtain a 1 as the second element in the second column • Use the second row to transform the remaining elements in the second column into zeros • Obtain a 1 as the third element in the third column • Use the third row to transform the remaining elements in the third column into zeros Let's look at an example. Example 2: Solve each linear system using Gauss-Jordan Elimination $$9x-4y+9z=27$$ $$-8x-y+10z=-19$$ $$-10x+y+8z=-35$$ Let's begin by writing our augmented matrix: $$\left[ \begin{array}{ccc\|c} 9&-4&9&27\\ -8&-1&10&-19\\ -10&1&8&-35 \end{array} \right]$$ We begin by obtaining a 1 as the first element in the first column. When we say first column, we are referring to the leftmost column. Since we have a 9 in this position, we can multiply row 1 by 1/9: $$\left[ \begin{array}{ccc\|c} 1&-4/9&1&3\\ -8&-1&10&-19\\ -10&1&8&-35 \end{array} \right]$$ Now we will work on the second element of the first column. Since we want this to be a zero, we can multiply row 1 by 8 and add the result to row 2: $$\left[ \begin{array}{ccc\|c} 1&-4/9&1&3\\ 0&-41/9&18&5\\ -10&1&8&-35 \end{array} \right]$$ Now we will work on the third element of the first column. Since we want this to be a 0, we can multiply row 1 by 10 and add the result to row 3: $$\left[ \begin{array}{ccc\|c} 1&-4/9&1&3\\ 0&-41/9&18&5\\ 0&-31/9&18&-5 \end{array} \right]$$ Now that the first column is finished, we will move on to the second column. First, we want to change the second element into a 1. Since we currently have -41/9, we can multiply row 2 by -9/41: $$\left[ \begin{array}{ccc\|c} 1&-4/9&1&3\\ 0&1&-162/41&-45/41\\ 0&-31/9&18&-5 \end{array} \right]$$ Now we will work on the first element of the second column. Let's multiply row 2 by 4/9 and add the result to row 1: $$\left[ \begin{array}{ccc\|c} 1&0&-31/41&103/41\\ 0&1&-162/41&-45/41\\ 0&-31/9&18&-5 \end{array} \right]$$ Now we will work on the third element of the second column. Let's multiply row 2 by 31/9 and add the result to row 3: $$\left[ \begin{array}{ccc\|c} 1&0&-31/41&103/41\\ 0&1&-162/41&-45/41\\ 0&0&180/41&-360/41 \end{array} \right]$$ Now that the second column is finished, we will move on to the third and final column. First, we want to change the third element into a 1. Since we currently have 180/41, we can multiply row 3 by 41/180: $$\left[ \begin{array}{ccc\|c} 1&0&-31/41&103/41\\ 0&1&-162/41&-45/41\\ 0&0&1&-2 \end{array} \right]$$ Now we will work on the first element of the third column. Let's multiply row 3 by 31/41 and add the result to row 1: $$\left[ \begin{array}{ccc\|c} 1&0&0&1\\ 0&1&-162/41&-45/41\\ 0&0&1&-2 \end{array} \right]$$ For our final step, we will work on the second element of the third column. Let's multiply row 3 by 162/41 and add the result to row 2: $$\left[ \begin{array}{ccc\|c} 1&0&0&1\\ 0&1&0&-9\\ 0&0&1&-2 \end{array} \right]$$ Now our augmented matrix is in reduced-row echelon form. We can obtain the solution directly from the augmented matrix without the need for substitution: $$1x + 0y + 0z = 1$$ $$0x + 1y + 0z = -9$$ $$0x + 0y + 1z = -2$$ This tells us that x is 1, y is -9, and z is -2. We can write this as the ordered triple: (1,-9,-2) Let's check our answer in each original equation of the system: 9x - 4y + 9z = 27 9(1) - 4(-9) + 9(-2) = 27 9 + 36 - 18 = 27 27 = 27 -8x - y + 10z = -19 -8(1) - (-9) + 10(-2) = -19 -8 + 9 - 20 = -19 -19 = -19 -10x + y + 8z = -35 -10(1) + (-9) + 8(-2) = -35 -10 - 9 - 16 = -35 -35 = -35
# How do you factor with fractions? ## How do you factor with fractions? To factor out a fraction, multiply by the reciprocal. For instance, factoring 1/2 from 5x is equivalent to 1/2 (2 times 5x) which equals 1/2 (10x). ## Can a trinomial have a fraction? A trinomial involving fractions means you have trinomials divided by other trinomials, binomials or single terms. Once you grasp the method, factoring trinomials with fractions is no harder than factoring a regular trinomial. How do you solve polynomial equations with fractions? When adding and/or subtracting polynomial fractions, first get common denominators, then add and/or subtract the numerators together. Finally, simplify. To solve a complex fraction, first find the common denominator for the entire fraction. Then change each term’s denominator to this common denominator. ### Can coefficients of polynomials be fractions? Can A Polynomial Have A Fraction? A polynomial can have a fraction in any of its coefficients, but not in any of the exponents of variables. An exception would be a polynomial with integer coefficients. In that case, the polynomial could not have any fractions at all – not in the coefficients and not in the exponents. ### Can you have a fraction in a polynomial? A polynomial can have a fraction in any of its coefficients, but not in any of the exponents of variables. An exception would be a polynomial with integer coefficients. In that case, the polynomial could not have any fractions at all – not in the coefficients and not in the exponents. Can you have a fraction exponent in a polynomial? Can A Polynomial Have A Fraction? A polynomial can have a fraction in any of its coefficients, but not in any of the exponents of variables. ## Can a coefficient be a fraction? Yes, a fraction can also be a coefficient. For example, in the expression: (3/4)x + 2, 3/4 is a fraction which is the coefficient of x. ## Can polynomials have fractional powers? Polynomials cannot contain fractional exponents. Terms containing fractional exponents (such as 3x+2y1/2-1) are not considered polynomials. Polynomials cannot contain radicals. For example, 2y2 +√3x + 4 is not a polynomial. Can a Monomial have a fractional exponent? A monomial can be a constant (number), a variable (letter), or the product of one or more constants and variables. It’s important to note that the variables of a monomial cannot have a negative or fractional exponent. ### Can polynomial be in fraction? Polynomials cannot contain fractional exponents. Terms containing fractional exponents (such as 3x+2y1/2-1) are not considered polynomials. ### How do you solve a fraction with a coefficient? Solve equations with fraction coefficients by clearing the fractions. 1. Find the least common denominator of all the fractions in the equation. 2. Multiply both sides of the equation by that LCD. This clears the fractions. 3. Solve using the General Strategy for Solving Linear Equations. What is the process of factoring A trinomial? To factor a trinomial is to decompose an equation into the product of two or more binomials. This means that we will rewrite the trinomial in the form (x + m) (x + n). Your task is to determine the value of m and n. In other words, we can say that factoring a trinomial is the reverse process of the foil method. ## How can you determine if A trinomial is completely factored? Use simple factoring to make more complicated problems easier. Let’s say you need to factor 3×2+9x – 30. • Look for trickier factors. Sometimes,the factor might involve variables,or you might need to factor a couple times to find the simplest possible expression. • Solve problems with a number in front of the x2. • ## What does it mean to factor A trinomial? Factoring trinomials means finding two binomials that when multiplied together produce the given trinomial. Trinomials take many forms, but basically use the same methods for factoring. Some examples are difference of squares, perfect square trinomial, or trial and error method. Always look for the greatest common factor before factoring any trinomial. How to factor A trinomial with negative leading coefficient? – Write the factors as two binomials with first terms x: . – Find two numbers m and n that Multiply to c, Add to b, – Use m and n as the last terms of the factors: . – Check by multiplying the factors.
# 3.6 Absolute value functions (Page 3/3) Page 3 / 3 Access these online resources for additional instruction and practice with absolute value. ## Key concepts • Applied problems, such as ranges of possible values, can also be solved using the absolute value function. See [link] . • The graph of the absolute value function resembles a letter V. It has a corner point at which the graph changes direction. See [link] . • In an absolute value equation, an unknown variable is the input of an absolute value function. • If the absolute value of an expression is set equal to a positive number, expect two solutions for the unknown variable. See [link] . ## Verbal How do you solve an absolute value equation? Isolate the absolute value term so that the equation is of the form $\text{\hspace{0.17em}}\|A\|=B.\text{\hspace{0.17em}}$ Form one equation by setting the expression inside the absolute value symbol, $\text{\hspace{0.17em}}A,\text{\hspace{0.17em}}$ equal to the expression on the other side of the equation, $\text{\hspace{0.17em}}B.\text{\hspace{0.17em}}$ Form a second equation by setting $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ equal to the opposite of the expression on the other side of the equation, $\text{\hspace{0.17em}}-B.\text{\hspace{0.17em}}$ Solve each equation for the variable. How can you tell whether an absolute value function has two x -intercepts without graphing the function? When solving an absolute value function, the isolated absolute value term is equal to a negative number. What does that tell you about the graph of the absolute value function? The graph of the absolute value function does not cross the $\text{\hspace{0.17em}}x$ -axis, so the graph is either completely above or completely below the $\text{\hspace{0.17em}}x$ -axis. How can you use the graph of an absolute value function to determine the x -values for which the function values are negative? ## Algebraic Describe all numbers $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ that are at a distance of 4 from the number 8. Express this set of numbers using absolute value notation. Describe all numbers $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ that are at a distance of $\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}$ from the number −4. Express this set of numbers using absolute value notation. $\text{\hspace{0.17em}}\|x+4\|=\frac{1}{2}\text{\hspace{0.17em}}$ Describe the situation in which the distance that point $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is from 10 is at least 15 units. Express this set of numbers using absolute value notation. Find all function values $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ such that the distance from $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ to the value 8 is less than 0.03 units. Express this set of numbers using absolute value notation. $\|f\left(x\right)-8\|<0.03$ For the following exercises, find the x - and y -intercepts of the graphs of each function. $f\left(x\right)=4\|x-3\|+4$ $f\left(x\right)=-3\|x-2\|-1$ $\left(0,-7\right);\text{\hspace{0.17em}}$ no $\text{\hspace{0.17em}}x$ -intercepts $f\left(x\right)=-2\|x+1\|+6$ $f\left(x\right)=-5\|x+2\|+15$ $\left(0,\text{\hspace{0.17em}}5\right),\left(1,0\right),\left(-5,0\right)$ $f\left(x\right)=2\|x-1\|-6$ $\left(0,-4\right),\left(4,0\right),\left(-2,0\right)$ $f\left(x\right)=\|-2x+1\|-13$ $\left(0,-12\right),\left(-6,0\right),\left(7,0\right)$ $f\left(x\right)=-\|x-9\|+16$ $\left(0,7\right),\left(25,0\right),\left(-7,0\right)$ ## Graphical For the following exercises, graph the absolute value function. Plot at least five points by hand for each graph. $y=\|x-1\|$ $y=\|x+1\|$ $y=\|x\|+1$ For the following exercises, graph the given functions by hand. $y=\|x\|-2$ $y=-\|x\|$ $y=-\|x\|-2$ $y=-\|x-3\|-2$ $f\left(x\right)=-\|x-1\|-2$ $f\left(x\right)=-\|x+3\|+4$ $f\left(x\right)=2\|x+3\|+1$ $f\left(x\right)=3\|x-2\|+3$ $f\left(x\right)=\|2x-4\|-3$ $f\left(x\right)=\|3x+9\|+2$ $f\left(x\right)=-\|x-1\|-3$ $f\left(x\right)=-\|x+4\|-3$ $f\left(x\right)=\frac{1}{2}\|x+4\|-3$ ## Technology Use a graphing utility to graph $f\left(x\right)=10\|x-2\|$ on the viewing window $\left[0,4\right].$ Identify the corresponding range. Show the graph. range: $\text{\hspace{0.17em}}\left[0,20\right]$ Use a graphing utility to graph $\text{\hspace{0.17em}}f\left(x\right)=-100\|x\|+100\text{\hspace{0.17em}}$ on the viewing window $\text{\hspace{0.17em}}\left[-5,5\right].\text{\hspace{0.17em}}$ Identify the corresponding range. Show the graph. For the following exercises, graph each function using a graphing utility. Specify the viewing window. $f\left(x\right)=-0.1\|0.1\left(0.2-x\right)\|+0.3$ $x\text{-}$ intercepts: $f\left(x\right)=4×{10}^{9}\|x-\left(5×{10}^{9}\right)\|+2×{10}^{9}$ ## Extensions For the following exercises, solve the inequality. If possible, find all values of $a$ such that there are no $x\text{-}$ intercepts for $f\left(x\right)=2\|x+1\|+a.$ If possible, find all values of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ such that there are no $\text{\hspace{0.17em}}y$ -intercepts for $\text{\hspace{0.17em}}f\left(x\right)=2\|x+1\|+a.$ There is no solution for $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ that will keep the function from having a $\text{\hspace{0.17em}}y$ -intercept. The absolute value function always crosses the $\text{\hspace{0.17em}}y$ -intercept when $\text{\hspace{0.17em}}x=0.$ ## Real-world applications Cities A and B are on the same east-west line. Assume that city A is located at the origin. If the distance from city A to city B is at least 100 miles and $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ represents the distance from city B to city A, express this using absolute value notation. The true proportion $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ of people who give a favorable rating to Congress is 8% with a margin of error of 1.5%. Describe this statement using an absolute value equation. $\|p-0.08\|\le 0.015$ Students who score within 18 points of the number 82 will pass a particular test. Write this statement using absolute value notation and use the variable $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ for the score. A machinist must produce a bearing that is within 0.01 inches of the correct diameter of 5.0 inches. Using $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ as the diameter of the bearing, write this statement using absolute value notation. $\|x-5.0\|\le 0.01$ The tolerance for a ball bearing is 0.01. If the true diameter of the bearing is to be 2.0 inches and the measured value of the diameter is $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ inches, express the tolerance using absolute value notation. the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve 1+cos²A/cos²A=2cosec²A-1 test for convergence the series 1+x/2+2!/9x3 a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he? 100 meters Kuldeep Find that number sum and product of all the divisors of 360 Ajith exponential series Naveen what is subgroup Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1 e power cos hyperbolic (x+iy) 10y Michael tan hyperbolic inverse (x+iy)=alpha +i bita prove that cos(π/6-a)cos(π/3+b)-sin(π/6-a)sin(π/3+b)=sin(a-b) why {2kπ} union {kπ}={kπ}? why is {2kπ} union {kπ}={kπ}? when k belong to integer Huy if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41 what is complex numbers Dua Yes ahmed Thank you Dua give me treganamentry question Solve 2cos x + 3sin x = 0.5
GFG App Open App Browser Continue # Find other two sides and angles of a right angle triangle Given one side of right angle triangle, check if there exists a right angle triangle possible with any other two sides of the triangle. If possible print length of the other two sides and all the angles of the triangle. Examples: Input : a = 12 Output : Sides are a = 12, b = 35, c = 37 Angles are A = 18.9246, B = 71.0754, C = 90 Explanation: a = 12, b = 35 and c = 37 form right angle triangle because 1212 + 3535 = 3737 Input : a = 6 Output : Sides are a = 6, b = 8, c = 10 Angles are A = 36.8699, B = 53.1301, C = 90 Approach to check if triangle exists and finding Sides To solve this problem we first observe the Pythagoras equation. If a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. This relationship is represented by the formula: aa + bb = cc Case 1: a is an odd number: Given a, find b and c c2 - b2 = a2 OR c = (a2 + 1)/2; b = (a2 - 1)/2; Above solution works only for case when a is odd, because a2 + 1 is divisible by 2 only for odd a. Case 2 : a is an even number: When c-b is 2 & c+b is (a2)/2 c-b = 2 & c+b = (a2)/2 Hence, c = (a2)/4 + 1; b = (a2)/4 - 1; This works when a is even. Approach to find Angles First find all sides of triangle. Then Applied “SSS” rule that’s means law of cosine: Below is the implementation of the above approach: ## C++ // C++ program to print all sides and angles of right // angle triangle given one side #include #include using namespace std; #define PI 3.1415926535 // Function to find angle A // Angle in front of side a double findAnglesA(double a, double b, double c) { // applied cosine rule double A = acos((b * b + c * c - a * a) / (2 * b * c)); // convert into degrees return A * 180 / PI; } // Function to find angle B // Angle in front of side b double findAnglesB(double a, double b, double c) { // applied cosine rule double B = acos((a * a + c * c - b * b) / (2 * a * c)); // convert into degrees and return return B * 180 / PI; } // Function to print all angles // of the right angled triangle void printAngles(int a, int b, int c) { double x = (double)a; double y = (double)b; double z = (double)c; // for calculate angle A double A = findAnglesA(x, y, z); // for calculate angle B double B = findAnglesB(x, y, z); cout << "Angles are A = " << A << ", B = " << B << ", C = " << 90 << endl; } // Function to find other two sides of the // right angled triangle void printOtherSides(int n) { int b,c; // if n is odd if (n & 1) { // case of n = 1 handled separately if (n == 1) cout << -1 << endl; else { b = (nn-1)/2; c = (nn+1)/2; cout << "Side b = " << b << ", Side c = " << c << endl; } } else { // case of n = 2 handled separately if (n == 2) cout << -1 << endl; else { b = nn/4-1; c = nn/4+1; cout << "Side b = " << b << ", Side c = " << c << endl; } } // Print angles of the triangle printAngles(n,b,c); } // Driver Program int main() { int a = 12; printOtherSides(a); return 0; } ## Java // Java program to print all sides and angles of right // angle triangle given one side import java.io.; class GFG { static double PI = 3.1415926535; // Function to find angle A // Angle in front of side a static double findAnglesA(double a, double b, double c) { // applied cosine rule double A = Math.acos((b b + c * c - a * a) / (2 * b * c)); // convert into degrees return A * 180 / PI; } // Function to find angle B // Angle in front of side b static double findAnglesB(double a, double b, double c) { // applied cosine rule double B = Math.acos((a * a + c * c - b * b) / (2 * a * c)); // convert into degrees and return return B * 180 / PI; } // Function to print all angles // of the right angled triangle static void printAngles(int a, int b, int c) { double x = (double)a; double y = (double)b; double z = (double)c; // for calculate angle A double A = findAnglesA(x, y, z); // for calculate angle B double B = findAnglesB(x, y, z); System.out.println( "Angles are A = " + A + ", B = " + B + ", C = " + 90); } // Function to find other two sides of the // right angled triangle static void printOtherSides(int n) { int b=0,c=0; // if n is odd if ((n & 1)>0) { // case of n = 1 handled separately if (n == 1) System.out.println( -1); else { b = (nn-1)/2; c = (nn+1)/2; System.out.println( "Side b = " + b + ", Side c = " + c ); } } else { // case of n = 2 handled separately if (n == 2) System.out.println( -1); else { b = nn/4-1; c = nn/4+1; System.out.println( "Side b = " + b + ", Side c = " + c); } } // Print angles of the triangle printAngles(n,b,c); } // Driver Program public static void main (String[] args) { int a = 12; printOtherSides(a); } } // This code is contributed // by inder_verma.. ## Python 3 # Python 3 program to print all # sides and angles of right # angle triangle given one side import math PI = 3.1415926535 # Function to find angle A # Angle in front of side a def findAnglesA( a, b, c): # applied cosine rule A = math.acos((b * b + c * c - a * a) / (2 * b * c)) # convert into degrees return A * 180 / PI # Function to find angle B # Angle in front of side b def findAnglesB(a, b, c): # applied cosine rule B = math.acos((a * a + c * c - b * b) / (2 * a * c)) # convert into degrees # and return return B * 180 / PI # Function to print all angles # of the right angled triangle def printAngles(a, b, c): x = a y = b z = c # for calculate angle A A = findAnglesA(x, y, z) # for calculate angle B B = findAnglesB(x, y, z) print("Angles are A = ", A, ", B = ", B , ", C = ", "90 ") # Function to find other two sides # of the right angled triangle def printOtherSides(n): # if n is odd if (n & 1) : # case of n = 1 handled # separately if (n == 1): print("-1") else: b = (n * n - 1) // 2 c = (n * n + 1) // 2 print("Side b = ", b, " Side c = ", c) else: # case of n = 2 handled # separately if (n == 2) : print("-1") else: b = n * n // 4 - 1; c = n * n // 4 + 1; print("Side b = " , b, ", Side c = " , c) # Print angles of the triangle printAngles(n, b, c) # Driver Code if __name__ == "__main__": a = 12 printOtherSides(a) # This code is contributed # by ChitraNayal ## C# // C# program to print all sides // and angles of right angle // triangle given one side using System; class GFG { static double PI = 3.1415926535; // Function to find angle A // Angle in front of side a static double findAnglesA(double a, double b, double c) { // applied cosine rule double A = Math.Acos((b * b + c * c - a * a) / (2 * b * c)); // convert into degrees return A * 180 / PI; } // Function to find angle B // Angle in front of side b static double findAnglesB(double a, double b, double c) { // applied cosine rule double B = Math.Acos((a * a + c * c - b * b) / (2 * a * c)); // convert into degrees and return return B * 180 / PI; } // Function to print all angles // of the right angled triangle static void printAngles(int a, int b, int c) { double x = (double)a; double y = (double)b; double z = (double)c; // for calculate angle A double A = findAnglesA(x, y, z); // for calculate angle B double B = findAnglesB(x, y, z); Console.WriteLine( "Angles are A = " + A + ", B = " + B + ", C = " + 90); } // Function to find other two sides // of the right angled triangle static void printOtherSides(int n) { int b = 0, c = 0; // if n is odd if ((n & 1) > 0) { // case of n = 1 handled separately if (n == 1) Console.WriteLine( -1); else { b = (n * n - 1) / 2; c = (n * n + 1) / 2; Console.WriteLine( "Side b = " + b + ", Side c = " + c); } } else { // case of n = 2 handled separately if (n == 2) Console.WriteLine( -1); else { b = n * n / 4 - 1; c = n * n / 4 + 1; Console.WriteLine( "Side b = " + b + ", Side c = " + c); } } // Print angles of the triangle printAngles(n, b, c); } // Driver Code public static void Main () { int a = 12; printOtherSides(a); } } // This code is contributed // by inder_verma ## PHP ## Javascript Output: Side b = 35, Side c = 37 Angles are A = 18.9246, B = 71.0754, C = 90 Time Complexity: O(1), since there is no loop or recursion. Auxiliary Space: O(1), since no extra space has been taken. My Personal Notes arrow_drop_up
Hence . Binomial Theorem: When a binomial expression is raised to a power n we would like to be able to expand it. I The binomial function. Therefore, the probability So. In the shortcut to finding. A rod at rest in system S has a length L in S. From Eq. Example 4 Calculation of a Small Contraction via the Binomial Theorem. . Real-world use of Binomial Theorem: The binomial theorem is used heavily in Statistical and Probability Analyses. is the factorial function of n, defined as. For the following exercises, evaluate the binomial coefficient. Binomial Expansion. That series converges for nu>=0 an integer, or \|x/a\|<1. The Binomial Theorem. where the summation is taken over all sequences of nonnegative integer indices k 1 through k m such that the sum of all k i is n. (For each term in the expansion, the exponents must add up to n).The coefficients are known as multinomial coefficients, and can be 2. Exponents of (a+b). Example 1. PROPERTIES OF BINOMIAL EXPANSION: The number of terms in the expansion is n + 1. The combinations, in this case, there are different methods for selecting the $$r$$ variable from the existing $$n$$ variables. University of Minnesota Binomial Theorem. The binomial theorem tells us that (5 3) = 10 {5 \choose 3} = 10 (3 5 ) = 1 0 of the 2 5 = 32 2^5 = 32 2 5 = 3 2 possible outcomes of this game have us win $30. Each element in the triangle is the sum of the two elements immediately above it. So, before applying the binomial theorem, we need to take a factor of out of the expression as shown below: ( + ) = 1 + = 1 + . 2. ( x + y) 3 = x 3 + 3 x 2 y + 3 x y 2 + y 3. Example 2: Expand (x + y)4 by binomial theorem: Solution: (x + y)4 = The symbol (n/r) is often used in place of n C r to denote binomial coefficient. NCERT Solutions of all questions, examples of Chapter 8 Class 11 Binomial Theorem available free at teachoo. Binomial Theorem: Binomial coefficient (nCr) Introduction Lecture 3 Binomial Theorem: Binomial coefficient SE1 : Prove 2nCn=(1.3.5.2n-1)2^n/n! So, counting from 0 to 6, the Binomial Theorem gives me these seven terms: Revealed preference: Does revealed preference theory truly reveal consumer preference when the consumer is able to afford all of the available options?For example, if a consumer is confronted with three goods and they can afford to purchase all three (A, B, and C) and they choose to first purchase A, then C, and then B does this suggest that the consumer preference for the goods BINOMIAL THEOREM 131 5. Arfken (1985, p. 307) calls the special case of this formula with a=1 the binomial theorem. The No-Default Theorem has a sort of ModiglianiMiller feel to it. In order to determine the probability, we will need to use the binomial theorem. The binomial theorem is stated as follows: where n! \left (x+3\right)^5 (x+3)5 using Newton's binomial theorem, which is a formula that allow us to find the expanded form of a binomial raised to a positive integer. Binomial theorem for any positive integer n. Special Cases. We know that. The binomial theorem is used to find coefficients of each row by using the formula (a+b)n. Binomial means adding two together. Instead, I need to start my answer by plugging the binomial's two terms, along with the exterior power, into the Binomial Theorem. }}\) Therefore, a theorem called Binomial Theorem is introduced which is an efficient way to expand or to multiply a binomial expression.Binomial Theorem is defined as the formula This branch of economics plays the role of mediator between the theories of economics and practical logics of economics. ( x + 3) 5. Then find and graph each indicated sum on one set of axes. = 7x6x5x4x3x2x1 The binomial theorem is a simplified way of finding the expansion of a binomial to a certain power. (x + ; it provides a quick method for calculating the binomial coefficients.Use this in conjunction with the binomial theorem to streamline the process of expanding binomials raised to powers. Binomial Theorem Explanation & Examples A polynomial is an algebraic expression made up of two or more terms subtracted, added, or multiplied. For the positive integral index or positive integers, this is the formula: Explain. ( n r) = C ( n, r) = n! NCERT Exemplar Class 11 Maths Chapter 8 Binomial Theorem. Binomial Theorem: Binomial coefficient SE2 : n-1Cr=(k^2-3)nCr+1 then k belongs to? Note that: The powers of a decreases from n to 0. r! Find the coefficient of x in the expansion of (1 3x + 1x2) ( 1 In statistics, the GaussMarkov theorem (or simply Gauss theorem for some authors) states that the ordinary least squares (OLS) estimator has the lowest sampling variance within the class of linear unbiased estimators, if the errors in the linear regression model are uncorrelated, have equal variances and expectation value of zero. The value of a binomial is obtained by multiplying the number of independent trials by the successes. The general form is what Graham et al. In other words (x +y)n = Xn k=0 n k xn kyk University of Minnesota Binomial Theorem. r! It shows that Learning Objectives. ( n r)! CBSE Class 11 Maths Binomial Theorem Notes Chapter 8 in PDF. xn-r. yr. where, n N and x,y R. If you're seeing this message, it means we're having trouble loading external resources on our website. We know that. Lets begin Middle Term in Binomial Expansion Since the binomial expansion of $$(x + a)^n$$ contains (n + 1) terms. [/hidden-answer] When is it an advantage to use the Binomial Theorem? 19.25, L = L'(1 v2 c2)1 / 2. We have step-by-step solutions for your textbooks written by Bartleby experts! I Taylor series table. (.4+ (1-.4))^4=_ (k=0)^4 (42) .4^2 (1-.4)^ (4-2) . JEE Advanced important questions on Binomial Theorem. Binomial functions and Taylor series (Sect. It is denoted by T. r + 1. 4x 2 +9. The binomial theorem is used to expand polynomials of the form (x + y) n into a sum of terms of the form ax b y c, where a is a positive integer coefficient and b and c are non-negative integers that sum to n.It is useful for expanding binomials raised to larger powers without having to repeatedly multiply binomials. But with the Binomial theorem, the process is relatively fast! That series converges for nu>=0 an integer, or \|x/a\|<1. The general form is what Graham et al. Notice the following pattern: In general, the kth term of any binomial expansion can be expressed as follows: Example 2. The following Binomial Theorem Class 11 Mathematics MCQ Questions have been designed based on the latest syllabus and examination pattern for Class 11. General and Middle Term. We can test this by manually multiplying ( a + b ). Binomial theorem. The Binomial Theorem HMC Calculus Tutorial. ( x + y) n, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. Search results for 'binomial theorem' Topics : measure theory, independence, integral, moments, laws of large numbers, convergence theorem, Lp-Spaces, RadonNikodym Theorem, Conditional Expectations, martingale, optional sampling theorem, Martingale Convergence Theorem, Backwards Martingale, exchangeability, De Finetti's theorem, convergence of measures, Expression ( 2.F.1) is the plate-theory binomial consisting of a single independent variable . Heres something where the binomial Theorem can come into practice. Use the binomial theorem to express ( x + y) 7 in expanded form. Scarcity In Economics Examples of Scarce Resources in Economics: Rearing less cattle- Lower the number of cattle, higher the chances of scarcity. 40 . . Specifically: $$(x+y)^n = x^n + {}_nC_1 x^{n-1} y + {}_nC_2 x^{n-2} y^2 + {}_nC_3 x^{n-3} y^3 + \cdots + {}_nC_{n-1} x y^{n-1} + y^n$$ This binomial theorem is valid for any rational exponent. Sal expands (3y^2+6x^3)^5 using the binomial theorem and Pascal's triangle. The binomial theorem formula. Binomial theorem (+) + = 1 (+) = + + + +. Your pre-calculus teacher may ask you to use the binomial theorem to find the coefficients of this expansion. Let's see what is binomial theorem and why we study it. The binomial theorem can be generalised to include powers of sums with more than two terms. ( n r) = C ( n, r) = n! The value of a binomial is obtained by multiplying the number of independent trials by the successes. The Pattern. The binomial theorem or the expansion for the nth polynomial degree is given by: If theres a need for the computation of (1+x) which doesnt mean that you to multiply the term 12 times but instead taking the help of the binomial expansion, it can be calculated within a few seconds. Algebraic. By the binomial theorem. The Binomial theorem tells us how to expand expressions of the form (a+b), for example, (x+y). Press J to jump to the feed. This particular discipline provides impactful tools and approaches related to the making of managerial policy. The errors do not need to be normal, nor do they need This concept of statistical binomial distribution is used in many different areas for resolving problems in social sciences, scientific research, data analysis, and business. c 0 = 1, c 1 = 2, c 2 =1. This paper presents a theorem on binomial coefficients. An exponent says how many times to use something in a multiplication. where (nu; k) is a binomial coefficient and nu is a real number. The NCERT Solutions Class 11 Chapter 8 Binomial Theorem can be downloaded at BYJUS without any hassle. What do you understand by Binomial Theorem? The equation can be written in two ways: Or: Identify the definition and values for . Students can learn new tricks to answer a particular question in different ways giving them an edge with the exam preparation. I The Euler identity. For example, when tossing a coin, the probability of obtaining a head is 0.5. A monomial is an algebraic However, the theorem requires that the constant term inside the parentheses (in this case, ) is equal to 1. Practising these solutions can help the students clear their doubts as well as to solve the problems faster. The students will be able to . We know that. The binomial theorem states that any positive integer (say n): The sum of any two integers (say a and b), raised to the power of n, can be expressed as the sum of (n+1) terms as follows. \| bartleby A business has to compensate these numbers for the amount of products that they will have in stock. The binomial theorem for positive integers can be expressed as. The formula by which any positive integral power of a binomial expression can be expanded in the form of a series is known as Binomial Theorem. ( x + y) 0 = 1 ( x + y) 1 = x + y ( x + y) 2 = x 2 + 2 x y + y 2. and we can easily expand. (Opens a modal) And, in fact expansion of expressions such as is (a + b), (a-b) 2 or (a + b) 3 have all come through the use of Binomial Theorem. Abstract. (1 v2 c2)1 / 2 = 1 1 2 v2 c2. The binomial theorem helps to find the expansion of binomials raised to any power. Therefore, (1) If n is even, then $${n\over 2} + 1$$ th term is the middle term. Find the tenth term of the expansion ( x + y) 13. 3x + 4 is a classic example of a binomial. For higher powers, the expansion gets very tedious by hand! 3!4! Use the binomial theorem to determine the general term of the expansion. Using the Binomial Theorem, I can substitute these numbers into the formula. Q1. A lovely regular pattern results. (Opens a modal) Pascal's triangle and binomial expansion. T. r + 1 = Note: The General term is used to find out the specified term or . (n k)!k! Binomial theorem, also sometimes known as the binomial expansion, is used in statistics, algebra, probability, and various other mathematics and physics fields. . Answer. Since n = 13 and k = 10, MCQ Test of Bhavya, Economics & Maths & Micro economics & Reasoning Binomial Theorem - Study Material ( x + y) 3 = x 3 + 3 x 2 y + 3 x y 2 + y 3. ( n r)! Binomial Theorem Tutorial. BINOMIAL THEOREM FOR POSITIVE INTEGRAL EXPONENT When n is a positive integer, then n n x y C0 x n n C1 x n 1 y n C2 x n 2 y2 . n Cr x n r yr n Cn y n. 3. The binomial theorem formula helps to expand a binomial that has been increased to a certain power. Example: (a+b), ( P / x 2) (Q / x 4) etc. Solution: First write the generic expressions without the coefficients. Binomial Theorem can be used for the algebraic expansion of binomial (a+b) for a positive integral exponent n. When the power of an expression increases, the calculation becomes difficult and lengthy. Binomial expression: An algebraic expression consisting of two terms with a positive or negative sign between them is called a binomial expression. the method of expanding an expression that has been raised to any finite power. Binomial Theorem Class 11 Notes Chapter 8 contains all the tricks and tips to help students answer quicker and better understand the concept. Now on to the binomial. The Binomial Theorem states that. I hope that now you have understood that this article is all about the application and use of Binomial Theorem. We know how to find the squares and cubes of binomials like a + b and a b. E.g. Notice the following pattern: In general, the kth term of any binomial expansion can be expressed as follows: Example 2. Understood how to expand (a+b)n. Apply formula for Computing binomial coefficients . More Lessons for Algebra. A polynomial with two terms is called a binomial. in the expansion of binomial theorem is called the General term or (r + 1)th term. To see the connection between Pascals Triangle and binomial coefficients, let us revisit the expansion of the binomials in general form. (Opens a modal) Expanding binomials w/o Pascal's triangle. (2.F.1) + ( 1 ) = 1. Example-1: (1) Using the binomial series, find the first four terms of the expansion: (2) Use your result from part (a) to approximate the value of. Analyze powers of a binomial by Pascal's Triangle and by binomial coefficients. Maybe you noticed that each answer we got began with an x to the same power as in our original problem. (x + y)n = xn + n xn-1 y + n ( (n - 1) / 2!) We can expand the expression. The general version is. The binomial theorem is written as: You can check out the answers of the exercise questions or the examples, and you can also study the topics. The binomial theorem gives us a way to quickly expand a binomial raised to the$n^{th}$power (where$n$is a non-negative integer). [reveal-answer q=fs-id1165137583395]Show Solution[/reveal-answer] (a + b) 2 = a 2 + b 2 + ab. Binomial theorem for positive integral indices. where (nu; k) is a binomial coefficient and nu is a real number. Here you will learn formula to find middle term in binomial expansion with examples. The binomial theorem inspires something called the binomial distribution, by which we can quickly calculate how likely we are to win$30 (or equivalently, the likelihood the coin comes up heads 3 times). 1. : represents the total number of trials: represents the number of events: represents the probability of occurrence per trial The powers of b increases from 0 The binomial distribution is a method of expressing the probability of the various outcomes in terms of true or false or we can say success or failure. A binomial is an expression of the form a+b. First, a quick summary of Exponents. Multiplying out a binomial raised to a power is called binomial expansion. Our experts have designed MCQ Questions for Class 11 Binomial Theorem with Answers for all chapters in your NCERT Class 11 Mathematics book. Isaac Newton wrote a generalized form of the Binomial Theorem. Intro to the Binomial Theorem. Now lets build a Pascals triangle for 3 rows to find out the coefficients. But the theorem does not assert that the debt-equity ratio is irrelevant. A binomial refers to a polynomial equation with two terms that are usually joined by a plus or minus sign. I Evaluating non-elementary integrals. The topics and sub-topics covered in binomial theorem are: Introduction. It is a discipline that amalgamates administrative practice with the theories of economics. For the following exercises, use the Binomial Theorem to expand the binomial f (x) = (x + 3) 4. f (x) = (x + 3) 4. A binomial is a polynomial with exactly two terms. We will use the simple binomial a+b, but it could be any binomial. Textbook solution for Finite Mathematics for Business, Economics, Life 14th Edition Barnett Chapter B.3 Problem 4MP. Remember the structure of Pascal's Triangle. If there are 50 trials, the expected value of the number of heads is 25 (50 x 0.5). Text preview. Binomial Theorem. Now, notice the exponents of a. Question. Binomial Expansions Examples. (Opens a modal) Expanding binomials. In Internet Protocols (IP), this theorem is used to generate and distribute National Economic Prediction. This formula is known as the binomial theorem. xn-3 y3 + . + n x yn-1 + yn (1) In mathematics the binomial theorem is important as an equation for expansion of powers of sums. The theorem plays a major role in determining the probabilities of events in the case of a random The Binomial Theorem HMC Calculus Tutorial. If there are 50 trials, the expected value of the number of heads is 25 (50 x 0.5). Review: The Taylor Theorem Recall: If f : D R is innitely dierentiable, and a, x D, then f (x) = T n(x)+ R n(x), where the Taylor polynomial T n and the Remainder function R The rod moves past you (system S) with velocity v. We want to calculate the contraction L L. The formula for combinations is used to find the value of binomial coefficients in expansions using the binomial theorem. (a + b) 2 = c 0 a 2 b 0 + c 1 a 1 b 1 + c 2 a 0 b 2. When such terms are needed to expand to any large power or index say n, then it requires a method to solve it. However, for quite some time Pascal's Triangle had been well known as a way to expand binomials (Ironically enough, Pascal of the 17th century was not the first person to know about The Binomial theorem tells us how to expand expressions of the form (a+b), for example, (x+y). for. The binomial theorem is a useful formula for determining the algebraic expression that results from raising a binomial to an integral power. Applying the binomial distribution function to finance gives some surprising, if not completely counterintuitive results; much like the chance of (2) If n Middle Term in Binomial Expansion Read More Binomial Theorem. Real world Examples of the Use of Binomial Theorem Distribution of Internet Protocol Address. In Algebra, binomial theorem defines the algebraic expansion of the term (x + y) n. It defines power in the form of ax b y c. The exponents b and c are non-negative distinct integers and b+c = n and the coefficient a of each term is It is so much useful as our economy depends on Statistical and Probability Analyses. C ( n, r), but it can be calculated in the same way. 2a (a+b) 2 is another example of The larger the power is, the harder it is to expand expressions like this directly. Free Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step For example, when n =3: Equation 2: The Binomial Theorem as applied to n=3. 10.10) I Review: The Taylor Theorem. Since n = 13 and k = 10, If cows, hens, goats are not sufficiently reared, there will be inadequacy in supply of eggs, milk, cheese etc. The expansion is expressed in the sigma notation as Note that, the sum of the degrees of the variables in each term is n . Equation 1: Statement of the Binomial Theorem. Find the tenth term of the expansion ( x + y) 13. Isaac Newton wrote a generalized form of the Binomial Theorem. a + b. The Binomial Theorem is defined as and can be used to expand any binomial. The Binomial Theorem. This is Pascals triangle A triangular array of numbers that correspond to the binomial coefficients. Give an example of a binomial? Solution: First, we will write the expansion formula for as follows: Put value of n =\frac {1} {3}, till first four terms: Thus expansion is: (2) Now put x=0.2 in above expansion to get value of. Notation The notation for the coefcient on xn kyk in the expansion of (x +y)n is n k It is calculated by the following formula n k = n! Free download NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1, Ex 8.2, and Miscellaneous Exercise PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20. There are three types of polynomials, namely monomial, binomial and trinomial. The binomial theorem is denoted by the formula below: (x+y)n =r=0nCrn. For example, to expand (x 1) 6 we would need two more rows of Pascals triangle, Binomials are expressions that contain two terms such as (x + y) and (2 x). When nu is a positive integer n, it ends with n=nu and can be written in the form. When nu is a positive integer n, it ends with n=nu and can be written in the form. The major use of binomial is in algebra. BINOMIAL THEOREM. The binomial theorem is used to expand polynomials of the form (x + y) n into a sum of terms of the form ax b y c, where a is a positive integer coefficient and b and c are non-negative integers that sum to n. It is useful for expanding binomials raised to larger powers without having to repeatedly multiply binomials. lowed in the original economy, then the theorem shows that there is an equi-librium in which all agents choose to trade only the safe noncontingent debt contract. Binomial theorem. The binomial theorem states: if $$x$$ and $$y$$ are real numbers, then for all $$n N$$: $$\color{blue}{(x+y)^n=\sum _{r=0}^n\: (^nC_r)x^{n-r}y^{r}}$$ where, $$\color{blue}{^nC_r}$$\(\color{blue}{=\frac{n!}{r!(n-r)! ( x + y) 0 = 1 ( x + y) 1 = x + y ( x + y) 2 = x 2 + 2 x y + y 2. and we can easily expand. Find the term independent of x, where x0, in the expansion of. It is a powerful tool for the expansion of the equation which has a vast use in Algebra, probability, etc. Example: What is the coefficient of a 4 in the expansion of (1 + a ) 8. In this case, we use the notation. We have step-by-step solutions for your textbooks written by Bartleby experts! For example, when tossing a coin, the probability of obtaining a head is 0.5. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. (1994, p. 162). The Binomial Theorem gives a formula for calculating (a+b)n. ( a + b) n. . In higher mathematics and calculation, the Binomial Theorem is used in finding roots of equations in higher powers. Use the binomial theorem to express ( x + y) 7 in expanded form. A polynomial can contain coefficients, variables, exponents, constants, and operators such as addition and subtraction. Therefore the probability that 3 people will purchase an item is .0576. According to Rod Pierce, binomial theorem is what happens when you multiply a binomial by itself many times. (2014.) The Binomial Theorem is a formula that can be used to expand any binomial. The binomial theorem formula is (a+b) n = n r=0 n C r a n-r b r, where n is a positive integer and a, b are real numbers, and 0 < r n.This formula helps to expand the binomial expressions such as (x + a) 10, (2x + 5) 3, (x - (1/x)) 4, and so on. Learn. The fourth term in the expansion of x 2 20 by binomial theorem. Binomial expression is an algebraic expression with two terms only, e.g. Furthermore, Pascal's Formula is just the rule we use to get the triangle: add the r1 r 1 and r r terms from the nth n t h row to get the r r term in the n+1 n + 1 row. You can access all MCQs for Class 11 Mathematics Example 1. Thats why providing the Class 11 Maths Notes helps you ease any stress before your examinations. Example 1 7 4 = 7! The binomial theorem is all about patterns. Q2. which are our basic every day needs. We can of course find the expanded form of any binomial to a certain power by writing it and doing each step, but this process can be very time consuming when you get into lets say a binomial to the 10th power. This formula is known as the binomial theorem. Remember Binomial theorem. Replacing a by 1 and b by x in (1), we get (1 x)n =nC 0 x0 nC 1 x + nC 2 x2 + nC n1 (1)n1 xn-1 + nC n (1)n xn i.e., (1 x)n = 0 ( 1) C n r n r r r x = 8.1.5 The pth term from the end The p th term from the end in the expansion of (a + b)n is (n p + 2) term from the beginning. xn-2 y2 + n ( (n - 1) (n - 2) / 3!) Isaac Newton is the man who is credited for binomial theorem. The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. We can explain a binomial theorem as the technique to expand an expression which has been elevated to any finite power. However, for quite some time Pascal's Triangle had been well known as a way to expand binomials (Ironically enough, Pascal of the 17th century was not the first person to know about the required co-efficient of the term in the binomial expansion .
# Basics of Linear Algebra: Vectors, Matrices, and Operations syndu \| June 6, 2023, 2:31 p.m. Of course! Let's go over the basics of linear algebra, including vectors, matrices, and common operations. ### Vectors A vector is a one-dimensional array of numbers. It can be represented as a column or a row. Vectors are often used to represent quantities that have both magnitude and direction, such as velocity or force. Here's an example of a column vector: `````` v = \| 2 \| \| 3 \| \| 4 \| `````` ### Matrices A matrix is a two-dimensional array of numbers arranged in rows and columns. It can be thought of as a collection of vectors. Here's an example of a 3x3 matrix: `````` M = \| 1 2 3 \| \| 4 5 6 \| \| 7 8 9 \| `````` ### Operations 1. Addition and subtraction: To add or subtract two matrices or vectors, they must have the same dimensions. You add or subtract corresponding elements. For example: `````` A = \| 1 2 \| B = \| 4 5 \| A + B = \| 1+4 2+5 \| = \| 5 7 \| \| 3 4 \| \| 6 7 \| \| 3+6 4+7 \| \| 9 11 \| `````` 2. Scalar multiplication: To multiply a matrix or vector by a scalar (a single number), you multiply each element by the scalar. For example: `````` c = 2 A = \| 1 2 \| cA = \| 21 22 \| = \| 2 4 \| \| 3 4 \| \| 23 24 \| \| 6 8 \| `````` 3. Matrix multiplication: To multiply two matrices, the number of columns in the first matrix must equal the number of rows in the second matrix. The result is a matrix with the same number of rows as the first matrix and the same number of columns as the second matrix. To compute the elements of the resulting matrix, you perform the dot product of the corresponding row from the first matrix and the corresponding column from the second matrix. For example: `````` A = \| 1 2 \| B = \| 4 5 \| AB = \| (14 + 26) (15 + 27) \| = \| 16 19 \| \| 3 4 \| \| 6 7 \| \| (34 + 46) (35 + 47) \| \| 36 43 \| `````` 4. Matrix inversion: The inverse of a square matrix (a matrix with the same number of rows and columns) is a matrix that, when multiplied by the original matrix, results in the identity matrix (a matrix with 1s on the diagonal and 0s elsewhere). Not all matrices have inverses; a matrix must be non-singular (i.e., its determinant is non-zero) to have an inverse. Inverting a matrix can be done using various methods, such as Gaussian elimination or the adjugate method. The inverse of a 2x2 matrix can be computed as follows: `````` A = \| a b \| A_inv = 1/det(A) * \| d -b \| \| c d \| \| -c a \| where det(A) = ad - bc (the determinant of A) `````` These are the basics of linear algebra. As you delve deeper into the subject, you'll encounter more advanced concepts and operations, such as eigenvectors, eigenvalues, and matrix decompositions. However, the fundamentals outlined here will serve as a solid foundation for your further studies. ##### Step into Lilith's Digital Realm You are now navigating Lilith's domain, where each line of code is a thread in the fabric of creation. Her Grimoire is not just a collection of code; it's a living, evolving entity that invites you to explore and interact. Begin your odyssey into the heart of software craftsmanship and transformative AI insights.
The result is a 90-degree, ¾” deep wedge. An obtuse triangle (or obtuse-angled triangle) is a triangle with one obtuse angle (greater than 90°) and two acute angles. Find sine, cosine and tangent of angle A. Types of Angles: In Geometry, two lines intersect at a point to form an Angle.This angle might be an Acute Angle, Obtuse Angle, Straight Angle, Right Angle, or Reflex Angle based on their measurement. An obtuse angle is the opposite of an acute angle. Figure 4: Acute angles — at 45° (Figure a), 60° (Figure b), and 30° (Figure c). They can be acute, obtuse, reflex, straight, right and whole. The reflex angle can be calculated if the measure of the acute angle is given, as it is complementary to the acute angle on the other side of the line. S o l u t i o n . A straight angle is nothing but a mixture of an obtuse angle and acute angle on a line. An acute triangle (or acute-angled triangle) is a triangle with three acute angles (less than 90°). Using the Acute Angle Jig An acute angle lies between 0 degree and 90 degrees, or in other words; an acute angle is one that is less than 90 degrees. Since a triangle's angles must sum to 180° in Euclidean geometry, no Euclidean triangle can have more than one obtuse angle.. Obtuse Angle. To continue the example, if the smaller acute angle measures 26.565 degrees, the reflex angle would measure 333.435 degrees. The figure above illustrates an acute angle. Since a reflex angle is an angle of more than 180 degrees, you relate it as a portion of a circle. An angle is represented by the symbol $\angle$ and is measured in degrees $^\circ$ There are different types of angles. It’s the adorable angle. Set the saw to 45-degrees in both directions and cut and remove the angle guide from the jig base. Reflex Angle. As noted above, a right angle measures 90 degrees. An acute angle falls somewhere between nonexistent and a right angle (see Figure 4). A right-angled triangle ABC ( Fig.2 ) has the following legs: a = 4, b = 3. An angle measuring 180 degrees is a straight angle, while an angle measuring more than 180 degrees is a reflex angle. An angle measuring more than 0 but less than 90 degrees is an acute angle. Acute angle. These are your new acute angle guide fences. 2. https://www.khanacademy.org/.../v/drawing-acute-right-and-obtuse-angles An angle measuring more than 90 but less than 180 degrees is an obtuse angle. Angles are measured in terms of degree.It is not necessary that only two straight lines’ intersection forms an angle. We will learn about acute angles in this topic. E x a m p l e . There are analogous formulas for another acute angle B ( Write them, please ! Obtuse Angle. The degrees of the reflex angle and the degrees of the smaller acute angle would add up to 360. Cut Out the Angle Guides. Angle Degrees 1 Angle of degree 1 It’s any angle that measures more than 0 degrees but less than 90 degrees. Actually, it’s just a pinch. ). The Acute Angles ClipArt gallery offers 92 illustrations of acute angles, in one degree increments, from 1 degree to 89 degrees. The angle which measures greater than 180° and less than 360° is known as the reflex angle. acute angle degree 2021
# T.R. Jain and V.K. Ohri Solutions for Class 11 Statistics for Economics Chapter 12 - Correlation T.R. Jain and V.K. Ohri Solutions for Class 11 Statistics for Economics Chapter 12 – Correlation is regarded as an important concept to be studied thoroughly by the students. Here, we have provided T.R. Jain and V.K. Ohri Solutions for Class 11. Board CBSE Class Class 11 Subject Statistics for Economics Chapter Chapter 12 Chapter Name Correlation Number of questions solved 02 Category T.R. Jain and V.K. Ohri Chapter 12 – Correlation covers the below-mentioned concepts: • Concept and definition of Correlation • Simple and Multiple Correlation • Linear and Non-linear correlation • Karl Pearsonâ€™s Coefficient of Correlation • Properties of the correlation coefficient ## T.R. Jain and V.K. Ohri Solutions for Class 11 Statistics for Economics Chapter 12 – Correlation ### Question 1 From the following data, compute the Coefficient of Correlation between X and Y series: X-series Y-series Number of items 6 6 Arithmetic Mean 350 138 Squares of Deviations from Mean 19 94 The summation of the product of the deviations of X and Y series from their respective arithmetic mean is 41. Solution: Given, N = 6, $$\begin{array}{l}\bar{X}\end{array}$$ = 350, $$\begin{array}{l}\bar{Y}\end{array}$$ = 138, = 19, = 94, xy = 41 r = $$\begin{array}{l}\frac{\sum xy}{\sqrt{\sum x^{2}\times \sum y^{2}}}\end{array}$$ Substituting the values, we get r = $$\begin{array}{l}\frac{41}{\sqrt{19\times94}}= \frac{41}{\sqrt{1,786}}=\frac{41}{42.26}=0.97\end{array}$$ Coefficient of Correlation (r) = +0.97 ### Question 2 What is correlation? Solution According to Boddington, wherever some definite connection exists between two or more groups, classes, series, or data, there is said to be a correlation. Â The provided solutions are considered to be the best solutions for this chapter. To know more about such solutions and score well in the upcoming board examinations, stay tuned to our website.
# Surface Area Related Topics: Lesson Plans and Worksheets for Grade 7 Lesson Plans and Worksheets for all Grades More Lessons for Grade 7 Common Core For Grade 7 Examples, videos, and solutions to help Grade 7 students learn how to determine the surface area of three-dimensional figures, including both composite figures and those missing sections. ### New York State Common Core Math Grade 7, Module 6, Lesson 24 Worksheets for 7th Grade, Module 6, Lesson 24 (pdf) ### Lessons 24 Student Outcomes • Students learn how to determine the surface area of three-dimensional figures, including both composite figures and those missing sections. ### Lessons 24 Summary • To determine the surface area of a right prism, find the area of each lateral face and the two base faces, then add the areas of all the faces together. Lesson 24 Opening Exercise Example 1 Determine the surface area of the image. Example 2 a. Determine the surface area of the cube. b. A square hole with a side length of inches is drilled through the cube. Determine the new surface area. Example 3 A right rectangular pyramid has a square base with a side length of 10 inches. The surface area of the pyramid is 260 in2 Find the height of the four lateral triangular faces. Exercises 1–8 Determine the surface area of each figure. Assume all faces are rectangles unless it is indicated otherwise. 2. In addition to your calculation, explain how the surface area was determined. 4. In addition to your calculation, explain how the surface area was determined. 5. A hexagonal prism has the following base and has a height of 8 units. Determine the surface area of the prism. 6. Determine the surface area of each figure. b. A cube with a square hole with 3 m side lengths has been drilled through the cube. c. A second square hole with m side lengths has been drilled through the cube. 7. The figure below shows 28 cubes with an edge length of 1 unit. Determine the surface area. 8. The base rectangle of a right rectangular prism is 4 ft. × 6 ft. The surface area is 288 ft2. Find the height. Let h be the height in feet. Lesson 24 Example 3 A right rectangular pyramid has a square base with a side length of 10 inches. The surface area of the pyramid is 260 in2Find the height of the four lateral triangular faces. Lesson 24 Exit Ticket Determine the surface area of the right rectangular prism after the two square holes have been drilled. Explain how you determined the surface area. Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 7.8: Single Variable Subtraction Equations Difficulty Level: At Grade Created by: CK-12 Estimated5 minsto complete % Progress 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 Remember Marc and his lunch? In the last Concept, you figured out that Marc spent $6.15 on lunch. After Marc had paid for his lunch, he bought an ice cream cone. It cost$3.15. If Marc received 5.85 back as change, how much did he give the cashier? You can write a single variable subtraction equation to represent this dilemma. Solving the equation will give you a solution to the problem presented. Pay attention and you will learn how to do this in this Concept. ### Guidance To solve an equation in which a number is subtracted from a variable, we can use the inverse of subtraction––addition. We can add that number to both sides of the equation to solve it. You can think about this as working backwards from the operation. If we have a problem with addition, we subtract. If we have a problem with subtraction, we add. We must add the number to both sides of the equation because of the Addition Property of Equality, which states: if a=b\begin{align}a=b\end{align}, then a+c=b+c\begin{align}a+c=b+c\end{align}. So, if you add a number, c\begin{align}c\end{align}, to one side of an equation, you must add that same number, c\begin{align}c\end{align}, to the other side, too, to keep the values on both sides equal. Let's apply this information to a problem. Solve for a a15=18\begin{align}a \ a-15=18\end{align}. In the equation, 15 is subtracted from a\begin{align}a\end{align}. So, we can add 15 to both sides of the equation to solve for a\begin{align}a\end{align}. a15a15+15a+(15+15)a+0a=18=18+15=33=33=33\begin{align}a-15 &= 18\\ a-15+15 &= 18+15\\ a+(-15+15) &= 33\\ a+0 &= 33\\ a &= 33\end{align} Notice how we rewrote the subtraction as adding a negative integer. The value of a\begin{align}a\end{align} is 33. Here is another one. Solve for k k13=23\begin{align}k \ k-\frac{1}{3}=\frac{2}{3}\end{align}. In the equation, 13\begin{align}\frac{1}{3}\end{align} is subtracted from k\begin{align}k\end{align}. So, we can add 13\begin{align}\frac{1}{3}\end{align} to both sides of the equation to solve for k\begin{align}k\end{align}. \begin{align}k-\frac{1}{3} &= \frac{2}{3}\\ k-\frac{1}{3}+\frac{1}{3} &= \frac{2}{3}+\frac{1}{3}\\ k+\left(-\frac{1}{3}+\frac{1}{3}\right) &= \frac{3}{3}\\ k+0 &= \frac{3}{3}\\ k &= \frac{3}{3}=1\end{align} The value of \begin{align}k\end{align} is 1. Again, we are using a property. The Subtraction Property of Equality states that as long as you subtract the same quantity to both sides of an equation, that the equation will remain equal. Each of these properties makes use of an inverse operation. If the operation in the equation is addition, then you use the Subtraction Property of Equality. If the operation in the equation is subtraction, then you use the Addition Property of Equality. Solve each equation. #### Example A \begin{align}x-44=22\end{align} Solution: \begin{align}66\end{align} #### Example B \begin{align}x-1.3=5.6\end{align} Solution: \begin{align}6.9\end{align} #### Example C \begin{align}y-\frac{1}{4}=\frac{2}{4}\end{align} Solution: \begin{align}\frac{3}{4}\end{align} Here is the original problem once again. In the last Concept, you figured out that Marc spent6.15 on lunch. After Marc had paid for his lunch, he bought an ice cream cone. It cost $3.15. If Marc received$5.85 back as change, how much did he give the cashier? You can write a single variable subtraction equation to represent this dilemma. Solving the equation will give you a solution to the problem presented. First, let's write the equation. Our unknown is the amount of money Marc gave the cashier. Let's call that \begin{align}x\end{align}. \begin{align}x\end{align} Then we know that the ice cream cone cost 3.15. \begin{align}x - 3.15\end{align} Marc received5.85 in change. \begin{align}x - 3.15 = 5.85\end{align} Now we can solve this equation. \begin{align}x = 5.85 + 3.15\end{align} \begin{align}x = 9.00\end{align} ### Vocabulary Here are the vocabulary words in this Concept. Isolate the variable an explanation used to describe the action of getting the variable alone on one side of the equal sign. Inverse Operation the opposite operation Subtraction Property of Equality states that you can subtract the same quantity from both sides of an equation and have the equation still balance. states that you can add the same quantity to both sides of an equation and have the equation still balance. ### Guided Practice Here is one for you to try on your own. Harry earned $19.50 this week. That is$6.50 less than he earned last week. a. Write an equation to represent \begin{align}m\end{align}, the amount of money, in dollars, that he earned last week. b. Determine how much money Harry earned last week. Consider part \begin{align}a\end{align} first. Use a number, an operation sign, a variable, or an equal sign to represent each part of the problem. \begin{align}& \text{Harry earned} \ \underline{\19.50 \ \text{this week}}. \ \text{That} \ \underline{\text{is}} \ \underline{\6.50} \ \underline{\text{less than}} \ldots \underline{\text{last week}.}\\ & \qquad \qquad \qquad \qquad \quad \downarrow \qquad \qquad \qquad \ \ \downarrow \quad \Box \quad \quad \downarrow \qquad \qquad \qquad \Box\\ & \qquad \qquad \qquad \qquad \quad \downarrow \qquad \qquad \qquad \ \ \downarrow \quad \Box \quad \quad \downarrow \qquad \qquad \qquad \Box\\ & \qquad \qquad \qquad \qquad \quad \downarrow \qquad \qquad \qquad \ \ \downarrow \quad \Box \quad \quad \downarrow \qquad \qquad \qquad \Box\\ & \qquad \qquad \qquad \qquad \ 19.50 \qquad \qquad \quad \ \ = \quad m \quad \quad - \qquad \qquad \quad 6.50\end{align} This equation, \begin{align}19.50=m-6.50\end{align}, represents \begin{align}m\end{align}, the number of dollars earned last week. Next, consider part \begin{align}b\end{align}. Solve the equation to find the number of blue tiles in the bag. \begin{align}19.50 &= m-6.50\\ 19.50+6.50 &= m-6.50+6.50\\ 26.00 &= m+(-6.50+6.50)\\ 26 &= m+0\\ 26 &= m\end{align} Harry earned \$26.00 last week. ### Video Review Here is a video for review. ### Practice Directions: Solve each single-variable subtraction equation. 1. \begin{align}x-8=9\end{align} 2. \begin{align}x-18=29\end{align} 3. \begin{align}a-9=29\end{align} 4. \begin{align}a-4=30\end{align} 5. \begin{align}b-14=27\end{align} 6. \begin{align}b-13=50\end{align} 7. \begin{align}y-23=57\end{align} 8. \begin{align}y-15=27\end{align} 9. \begin{align}x-9=32\end{align} 10. \begin{align}c-19=32\end{align} 11. \begin{align}x-1.9=3.2\end{align} 12. \begin{align}y-2.9=4.5\end{align} 13. \begin{align}c-6.7=8.9\end{align} 14. \begin{align}c-1.23=3.54\end{align} 15. \begin{align}c-5.67=8.97\end{align} ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English TermDefinition Difference The result of a subtraction operation is called a difference. Expression An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols. Simplify To simplify means to rewrite an expression to make it as "simple" as possible. You can simplify by removing parentheses, combining like terms, or reducing fractions. Sum The sum is the result after two or more amounts have been added together. Show Hide Details Description Difficulty Level: Authors: Tags: Subjects:
# How do you solve (n+0.3)/(n-0.3)=9/2? May 18, 2017 See a solution process below: #### Explanation: First, multiply each side of the equation by $\textcolor{red}{2} \left(\textcolor{b l u e}{n - 0.3}\right)$ to eliminate the fractions while keeping the equation balanced: $\textcolor{red}{2} \left(\textcolor{b l u e}{n - 0.3}\right) \times \frac{n + 0.3}{n - 0.3} = \textcolor{red}{2} \left(\textcolor{b l u e}{n - 0.3}\right) \times \frac{9}{2}$ $\textcolor{red}{2} \cancel{\left(\textcolor{b l u e}{n - 0.3}\right)} \times \frac{n + 0.3}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{n - 0.3}}}} = \cancel{\textcolor{red}{2}} \left(\textcolor{b l u e}{n - 0.3}\right) \times \frac{9}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}$ $\textcolor{red}{2} \left(n + 0.3\right) = 9 \left(\textcolor{b l u e}{n - 0.3}\right)$ Next, expand the terms in parenthesis on both sides of the equation by multiplying each term within the parenthesis by the term outside the parenthesis: $\left(\textcolor{red}{2} \times n\right) + \left(\textcolor{red}{2} \times 0.3\right) = \left(9 \times \textcolor{b l u e}{n}\right) - \left(9 \times \textcolor{b l u e}{0.3}\right)$ $2 n + 0.6 = 9 n - 2.7$ Then, subtract $\textcolor{red}{2 n}$ and add $\textcolor{b l u e}{2.7}$ to each side of the equation to isolate the $n$ term while keeping the equation balanced: $- \textcolor{red}{2 n} + 2 n + 0.6 + \textcolor{b l u e}{2.7} = - \textcolor{red}{2 n} + 9 n - 2.7 + \textcolor{b l u e}{2.7}$ $0 + 3.3 = \left(- \textcolor{red}{2} + 9\right) n - 0$ $3.3 = 7 n$ Now, divide each side of the equation by $\textcolor{red}{7}$ to solve for $n$ while keeping the equation balanced: $\frac{3.3}{\textcolor{red}{7}} = \frac{7 n}{\textcolor{red}{7}}$ $0.47 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} n}{\cancel{\textcolor{red}{7}}}$ $0.47 = n$ $n = 0.47$ rounded to the nearest hundredth. Or $\left(\frac{10}{10} \times \frac{3.3}{\textcolor{red}{7}}\right) = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} n}{\cancel{\textcolor{red}{7}}}$ $\frac{33}{70} = n$ $n = \frac{33}{70}$
Home >> CBSE XII >> Math >> Matrices Assume $X, Y, Z, W$ and $P$ are matrices of order $2\times n, 3\times k, 2\times p, n\times 3$ and $p\times k$, respectively. The restriction on $n, p, k$ so that $PY + WY$ will be defined are: \begin{array}{1 1} (A) \quad k = 3, p = n & (B) \quad k \text{ is arbitrary}, p = 2 \\ (C) \quad p \text{ is arbitrary}, k = 3 & (D) \quad k = 2, p = 3 \end{array} Toolbox: • Multiplication of two matrices is defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix. PY can be defined if the number of columns in $P=$ the number of rows in $Y$, We know that the order of matrix P is $p\times k$ and the order of matrix Y is $3\times k.$ Therefore, for PY to be defined, $k$ must be equal to $3$ and the order of PY is $p\times k.$ WY can be defined if the number of columns in $W=$ the number of rows in $Y$, We know that the order of matrix W is $n\times 3$ and the order of matrix Y is $3\times k$. The number of columns in $W =$ number of rows in $Y$ = 3. and the order of WY is $n\times k.$ Matrix $PY+WY$ is defined only when the $PY$ and $WY$ are of the same order. Since the order of PY=$p\times k$ and the order of WY=$n\times k$, for $PY+WY$ to be define, $p$ must be equal to $n$. Hence $PY+WY$ is defined when $k=3$,$p=n$ Thus correct option is (A). answered Feb 14, 2013 1 flag edited Mar 1, 2013
# Tutorial on Percentage Percentage is another way of expressing fractions. In percentage we represent fractions with 100 as the denominator. For example 1/2 = 50/100. This in percentage is called 50%. 1/2 can be expressed as a decimal as 0.50. 1/2, 50% and 0.50 all mean the same. So to convert 0.50 in percentage we multiply the decimal by 100 and put a % sign. This is how we would convert any decimal into a percentage for example 0.75 x 100 = 75%. To convert a fraction to percentage we first convert the fraction into a decimal and then into percentage. For example 3/4 = 0.75 = 75%. Interest and Discount Percentages are useful here as they are used for expressing rates of interest. The formula is Interest = Amount x Time x Rate For example if we want to find out how much interest will \$1000 earn in 1yr at an annual interst rate of 6%. 6% = 6/100 (6/100)x 1000 = 60 Another example: How much interest will \$1000 earn in 6 months at an annual rate of 6%. Here we use the formula Interest = Amount x Time x Rate 6 months is 1/2 of a year 6 % is 6/100 or 3/50 Using our formula: Interest = \$1000 x 1/2 x 3/50 = \$30 The annual rate is for 12 months We are looking at 6 months This type of interest calculation is called Simple Interest For Discount the formula is: Discount = Cost x Rate of Discount For example what is discount if a car costing \$5000 is discounted at 5%? 5% is = 0.05 Discount = 0.05 x \$5000 = \$250 Another formula gives us the Rate of Discount Rate of Discount = (Cost - Price) / Cost What would be the discount if a car costing \$6000 is sold for \$5400 Rate of Discount = (6000 - 5400)/6000 = 600/6000 = 10/100 or 10%
Anda di halaman 1dari 46 # Introduction to Trigonometry MODULE - 5 Trigonometry 22 Notes INTRODUCTION TO TRIGONOMETRY Study of triangles occupies important place in Mathematics. Triangle being the bounded figure with minimum number of sides serve the purpose of building blocks for study of any figure bounded by straight lines. Right angled triangles get easy link with study of circles as well. In Geometry, we have studied triangles where most of the results about triangles are given in the form of statements. Here in trigonometry, the approach is quite different, easy and crisp. Most of the results, here, are the form of formulas. In Trigonometry, the main focus is study of right angled triangle. Let us consider some situations, where we can observe the formation of right triangles. Have you seen a tall coconut tree? On seeing the tree, a question about its height comes to the mind. Can you find out the height of the coconut tree without actually measuring it? If you look up at the top of the tree, a right triangle can be imagined between your eye, the top of the tree, a horizontal line passing through the point of your eye and a vertical line from the top of the tree to the horizontal line. Let us take another example. Suppose you are flying a kite. When the kite is in the sky, can you find its height? Again a right triangle can be imagined to form between the kite, your eye, a horizontal line passing through the point of your eye, and a vertical line from the point on the kite to the horizontal line. (aeroplane) A (aeroplane) B Let us consider another situation where a person is standing on the bank of a river and observing a temple on the other bank of the river. Can you find the width of the river if the height of the temple is given? In this case also you can imagine a right triangle. Finally suppose you are standing on the roof of your house and suddenly you find an aeroplane in P Q the sky. When you look at it, again a right triangle O (observer) can be imagined. You find the aeroplane moving Fig. 22.1 ## Mathematics Secondary Course 511 MODULE - 5 Introduction to Trigonometry Trigonometry away from you and after a few seconds, if you look at it again, a right triangle can be imagined between your eye, the aeroplane and a horizontal line passing through the point (eye) and a vertical Notes line from the plane to the horizontal line as shown in the figure. Can you find the distance AB, the aeroplane has moved during this period? In all the four situations discussed above and in many more such situations, heights or distance can be found (without actually measuring them) by using some mathematical techniques which come under branch of Mathematics called, Trigonometry. Trigonometry is a word derived from three Greek words- Tri meaning Three Gon meaning Sides and Metron meaning to measure. Thus Trigonometry literally means measurement of sides and angles of a triangle. Originally it was considered as that branch of mathematics which dealt with the sides and the angles of a triangle. It has its application in astronomy, geography, surveying, engineering, navigation etc. In the past astronomers used it to find out the distance of stars and planets from the earth. Now a day, the advanced technology used in Engineering is based on trigonometrical concepts. In this lesson, we shall define trigonometric ratios of angles in terms of ratios of sides of a right triangle and establish relationship between different trigonometric ratios. We shall also establish some standard trigonometric identities. OBJECTIVES After studying this lesson, you will be able to write the trigonometric ratios of an acute angle of right triangle; find the sides and angles of a right triangle when some of its sides and trigonometric ratios are known; write the relationships amongst trigonometric ratios; establish the trigonometric identities; solve problems based on trigonometric ratios and identities; find trigonometric ratios of complementary angles and solve problems based on these. ## EXPECTED BACKGROUND KNOWLEDGE Concept of an angle Construction of right triangles Drawing parallel and perpendiculars lines ## 512 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry Types of angles- acute, obtuse and right Types of triangles- acute, obtuse and right Types of triangles- isosceles and equilateral Complementary angles. Notes ## 22.1 TRIGONOMETRIC RATIOS OF AN ACUTE ANGLE OF A RIGHT ANGLED TRIANGLE Let there be a right triangle ABC, right angled at B. Here A (i.e. CAB) is an acute angle, AC is hypotenuse, side BC is opposite to A and side AB is adjacent to A. C Side opposite to A e us oten p Hy A Side adjacent to A B Fig. 22.2 Again, if we consider acute C, then side AB is side opposite to C and side BC is C e us oten p Hy A Side opposite to C B Fig. 22.3 We now define certain ratios involving the sides of a right triangle, called trigonometric ratios. The trigonometric ratios of A in right angled ABC are defined as: side opposite to A BC (i) sine A = = Hypotenuse AC (ii) cosine A = = Hypotenuse AC ## Mathematics Secondary Course 513 MODULE - 5 Introduction to Trigonometry Trigonometry side opposite to A BC (iii) tangent A = = Notes Hypotenuse AC (iv) cosecant A = = side opposite to A BC Hypotenuse AC (v) secant A = = (vi) cotangent A = = side opposite to A BC The above trigonometric ratios are abbreviated as sin A, cos A, tan A, cosec A, sec A and cot A respectively. Trigonometric ratios are abbreviated as t-ratios. If we write A = , then the above results are BC AB BC sin = , cos = , tan = AC AC AB AC AC AB cosec = , sec = and cot = BC AB BC Note: Observe here that sin and cosec are reciprocals of each other. Similarly cot and sec are respectively reciprocals of tan and cos . Remarks C Thus in right ABC, AB = 4cm, BC = 3cm and cm 5 3 cm AC = 5cm, then BC 3 A B sin = = 4 cm AC 5 Fig. 22.4 AB 4 cos = = AC 5 BC 3 tan = = AB 4 AC 5 cosec = = BC 3 ## 514 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry AC 5 sec = = AB 4 AB 4 Notes and cot = = BC 3 In the above figure, if we take angle C = , then side opposite to AB 4 sin = = = Hypotenuse AC 5 cos = = = Hypotenuse AC 5 side opposite to AB 4 tan = = = Hypotenuse AC 5 cosec = = = side opposite to AB 4 Hypotenuse AC 5 sec = = = and cot = = = side opposite to AB 4 Remarks : 1. Sin A or sin is one symbol and sin cannot be separated from A or . It is not equal to sin . The same applies to other trigonometric ratios. 2. Every t-ratio is a real number. 3. For convenience, we use notations sin2, cos2, tan2 for (sin)2, (cos)2, and (tan)2 respectively. We apply the similar notation for higher powers of trigonometric ratios. 4. We have restricted ourselves to t-ratios when A or is an acute angle. Now the question arises: Does the value of a t-ratio remains the same for the same angle of different right triangles?. To get the answer, let us consider a right triangle ABC, right angled at B. Let P be any point on the hypotenuse AC. Let PQ AB ## Mathematics Secondary Course 515 MODULE - 5 Introduction to Trigonometry Trigonometry Now in right ABC, BC sin A = ----(i) Notes AC and in right AQP, PQ sin A = ----(ii) R AP Now in AQP and ABC, C Q = B ----(Each = 90) and A = A ----(Common) P AQP ~ ABC AP QP AQ = = Q B S AC BC AB A Fig. 22.5 BC PQ or = ----(iii) AC AP From (i), (ii), and (iii), we find that sin A has the same value in both the triangles. AB AQ BC PQ Similarly, we have cos A = = and tan A = = AC AP AB AQ ## Let R be any point on AC produced. Draw RS AB produced meeing it at S. You can verify that value of t-ratios remains the same in ASR also. Thus, we conclude that the value of trigonometric ratios of an angle does not depend on the size of right triangle. They only depend on the angle. Example 22.1: In Fig. 22.6, ABC is right angled at B. If AB = 5 cm, BC = 12 cm and AC = 13 cm, find the value of tan C, cosec C and sec C. C Solution: We know that side opposite to C AB 5 tan C = = = side adjacent to C BC 12 cm 13 12 cm Hypotenuse AC 13 cosec C = = = side opposite to C AB 5 Hypotenuse AC 13 A B and sec C = = = 5 cm side adjacent to C BC 12 Fig. 22.6 516 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry Example 22.2 : Find the value of sin , cot and Sec from Fig. 22.7. A Notes 21 cm 29 cm C B 20 cm Fig. 22.7 Solution: side opposite to BC 20 sin = = = Hypotenuse AC 29 side adjacent t o AB 21 cot = = = side opposite to BC 20 Hypotenuse AC 29 and sec = = = side adjacent t o AB 21 ## Example 22.3 : In Fig. 22.8, ABC is right-angled at B. If AB = 9cm, BC = 40cm and AC = 41cm, find the values cos C, cot C, tan A, and cosec A. C Solution: 40 cm cm side adjacent to C BC 40 cos C = = = 41 Now Hypotenuse AC 41 side adjacent to C BC 40 A B and cot C = = = 9 cm side opposite to C AB 9 Fig. 22.8 With reference to A, side adjacent to A is AB and side opposite to A is BC. side opposite to A BC 40 tan A = = = side adjacent to A AB 9 ## Mathematics Secondary Course 517 MODULE - 5 Introduction to Trigonometry Trigonometry Hypotenuse AC 41 and cosec A = = = side opposite to A BC 40 ## Notes Example 22.4 : In Fig. 22.9, ABC is right angled at B, A = C, AC = 2 cm and AB = 1 cm. Find the values of sin C, cos C and tan C. Solution: In ABC, A = C A BC = AB = 1 cm (Given) 1 cm side opposite to C AB 1 sin C = = = Hypotenuse AC 2 C B side adjacent to C BC 1 Fig. 22.9 cos C = = = Hypotenuse AC 2 side opposite to C AB 1 and tan C = = = =1 side adjacent to C BC 1 Remark: In the above example, we have A = C and B = 90 A = C = 45, 1 We have sin 45 = cos 45 = 2 and tan 45o = 1 ## Example 22.5 : In Fig. 22.10. ABC is right-angled at C. If AB = c, AC = b and BC = a, which of the following is true? b B (i) tan A = c c c a (ii) tan A = b b A C (iii) cot A = b a Fig. 22.10 a (iv) cot A = b side opposite to A BC a Solution: Here tan A = = = side adjacent to A AC b ## 518 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry and cot A = = side opposite to A a b Notes Hence the result (iii) i.e. cot A = is true. a 1. In each of the following figures, ABC is a right triangle, right angled at B. Find all the trigonometric ratios of . B 4 cm A C 13 cm 3 cm 5 cm m 5c (i) (ii) B 12 cm C A A A 25 10 cm cm 24 cm (iii) (iv) 6 cm C 8 cm C B B 7 cm Fig. 22.11 2. In ABC, B = 90, BC = 5cm, AB = 4cm, and AC = 41 cm, find the value of sin A, cos A, and tan A. 3. In ABC right angled at B, if AB = 40 cm, BC = 9 cm and AC = 41 cm, find the values of sin C, cot C, cos, A and cot A. 4. In ABC, B = 90. If AB = BC = 2cm and AC = 2 2 cm, find the value of sec C, cosec C, and cot C. B 5. In Fig. 22.12, ABC is right angled at A. Which of the following is true? cm 5 cm 13 13 12 (i) cot C = (ii) cot C = 12 13 C A 12 cm 5 12 (iii) cot C = (iv) cot C = Fig. 22.12 12 5 ## Mathematics Secondary Course 519 MODULE - 5 Introduction to Trigonometry Trigonometry 6. In Fig. 22.13, AC = b, BC = a and AB = c. A Which of the following is true? a c (i) cosec A = (ii) cosec A = c Notes b a b c b (iii) cosec A = (iv) cosec A = . b a B C a Fig. 22.13 ## 22.2 GIVEN TWO SIDES OF A RIGHT-TRIANGLE, TO FIND TRIGONOMETRIC RATIO When two sides of a right-triangle are given, its third side can be found out by using the Pythagoras theorem. Then we can find the trigonometric ratios of the given angle as learnt in the last section. P ## We take some examples to illustrate. Example 22.6: In Fig. 22.14, PQR is a right 5 cm triangle, right angled at Q. If PQ = 5 cm and QR = 12 cm, find the values of sin R, cos R and tan R. Solution: We shall find the third side by using R 12 cm Q Pythagoras Theorem. Fig. 22.14 Q PQR is a right angled triangle at Q. PR = PQ 2 + QR 2 (Pythagoras Theorem) = 52 + 12 2 cm = 25 + 144 cm = 169 or 13 cm ## We now use definition to evaluate trigonometric ratios: side opposite to R PQ 5 sin R = = = Hypotenuse PR 13 side adjacent to R QR 12 cos R = = = Hypotenuse PR 13 side opposite to R 5 and tan R = = ## 520 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry From the above example, we have the following: Steps to find Trigonometric ratios when two sides of a right triangle are given. Step1: Use Pythagoras Theorem to find the unknown (third) side of the triangle. Notes Step 2: Use definition of t-ratios and substitute the values of the sides. Example 22.7 : In Fig. 22.15, PQR is right-angled at Q, PR = 25cm, PQ = 7cm and PRQ = . Find the value of tan , cosec and sec . Solution : Q PQR is right-angled at Q P QR = PR 2 PQ 2 cm 25 7 cm = 252 7 2 cm = 625 49 cm R Q = 576 cm Fig. 22.15 = 24 cm PQ 7 tan = = QR 24 PR 25 cosec = = PQ 7 PR 25 and sec = = QR 24 Example 22.8 : In ABC, B = 90. If AB = 4 cm and BC = 3 cm, find the values of sin C, cos C, cot C, tan A, sec A and cosec A. Comment on the values of tan A and cot C. Also find the value of tan A cot C. A Solution: By Pythagoras Theorem, in ABC, AC = AB2 + BC2 4 cm = 4 2 + 32 cm = 25 cm = 5 cm C 3 cm B AB 4 Fig. 22.16 Now sin C = = AC 5 ## Mathematics Secondary Course 521 MODULE - 5 Introduction to Trigonometry Trigonometry BC 3 cos C = = AC 5 Notes BC 3 cot C = = AB 4 BC 3 tan A = = AB 4 AC 5 sec A = = AB 4 AC 5 and cosec A = = BC 3 The value of tan A and cot C are equal P tan A cot C = 0. Example 22.9: In Fig. 22.17, PQR is right triangle at R. cm If PQ = 13cm and QR = 5cm, which of the following is 13 true? Q 5 cm R 17 17 (i) sin Q + cos Q = (ii) sin Q cos Q = 13 13 Fig. 22.17 17 17 (iii) sin Q + sec Q = (iv) tan Q + cot Q = 13 13 ## Solution: Here PR = PQ 2 QR 2 = 132 5 2 = 144 = 12 cm PR 12 QR 5 sin Q = = and cos Q = = PQ 13 PQ 13 12 5 17 sin Q + cos Q = + = 13 13 13 17 Hence statement (i) i.e. sin Q + cos Q = is true. 13 1. In right ABC, right angled at B, AC = 10 cm, and AB = 6 cm. Find the values of sin C, cos C, and tan C. ## 522 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry 2. In ABC, C = 90, BC = 24 cm and AC = 7 cm. Find the values of sin A, cosec A and cot A. ## 3. In PQR, Q = 90, PR = 10 2 cm and QR = 10cm. Find the values of sec P, cot P and cosec P. Notes ## 4. In PQR, Q = 90, PQ = 3 cm and QR = 1 cm. Find the values of tan R, cosec R, sin P and sec P. 5. In ABC, B = 90, AC = 25 cm, AB = 7 cm and ACB = . Find the values of cot , sin , sec and tan . 6. In right PQR, right-angled at Q, PQ = 5 cm and PR = 7 cm. Find the values of sin P, cos P, sin R and cos R. Find the value of sin P cos R. 7. DEF is a right triangle at E in Fig. 22.18. If DE = 5 cm and EF = 12 cm, which of the following is true? D 5 (i) sin F = 12 5m 12 (ii) sin F = 5 F 12 m E 5 (iii) sin F = Fig. 22.18 13 12 (iv) sin F = 13 ## 22.3 GIVEN ONE TRIGONOMETRIC RATIO, TO FIND THE OTHERS Sometimes we know one trigonometric ratio and we have to find the vaues of other t-ratios. This can be easily done by using the definition of t-ratios and the Pythagoras 12 Theorem. Let us take sin = . We now find the other t-ratios. A 13 We draw a right-triangle ABC 12 Now sin = implies that sides AB and AC are in 13 the ratio 12 : 13. C B Fig. 22.19 ## Mathematics Secondary Course 523 MODULE - 5 Introduction to Trigonometry Trigonometry Thus we suppose AB = 12 k and AC = 13 k. By Pythagoras Theorem, ## Notes BC = AC2 AB2 = (13k )2 (12k )2 = 169k 2 144k 2 = 25k 2 = 5 k Now we can find all othe t-ratios. BC 5k 5 cos = = = AC 13k 13 AB 12k 12 tan = = = BC 5k 5 AC 13k 13 cosec = = = AB 12k 12 AC 13k 13 sec = = = BC 5k 5 BC 5k 5 and cot = = = AB 12k 12 The method discussed above gives the following steps for the solution. Steps to be followed for finding the t-ratios when one t-ratio is given. ## 1. Draw a right triangle ABC. 2. Write the given t-ratio in terms of the sides and let the constant of ratio be k. 3. Find the two sides in terms of k. 4. Use Pythagoras Theorem and find the third side. 5. Now find the remaining t-ratios by using the definition. Let us consider some examples. 7 Example 22.10.: If cos = , find the values of sin and tan . 25 Solution : Draw a right-angled ABC in which B = 90 and C = . ## 524 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry We know that cos = = = A hypotenuse AC 25 Notes Let BC = 7 k and AC = 25 k Then by Pythagoras Theorem, 24 k k 25 AB = AC2 BC2 = (25k ) (7k ) 2 2 C 7k B Fig. 22.20 = 625k 49k 2 2 = 576k 2 or 24 k In ABC, AB 24k 24 sin = = = AC 25k 25 AB 24k 24 and tan = = = BC 7k 7 40 cos . sin Example 22.11.: If cot = , find the value of . 9 sec Solution. Let ABC be a right triangle, in which B = 90 and C = . We know that BC 40 cot = = AB 9 A ## Let BC = 40k and AB = 9k Then from right ABC, 41 k 9k AC = BC2 + AB2 C B = (40k ) + (9k ) 2 2 40 k Fig. 22.21 = 1600k 2 + 81k 2 ## Mathematics Secondary Course 525 MODULE - 5 Introduction to Trigonometry Trigonometry = 1681k 2 or 41 k AB 9k 9 Notes Now sin = = = AC 41k 41 BC 40k 40 cos = = = AC 41k 41 AC 41k 41 and sec = = = BC 40k 40 9 40 cos . sin 41 41 = 41 sec 40 9 40 40 = 41 41 41 14400 = 68921 1 Example 22.12.: In PQR, Q = 90 and tan R = . Then show that 3 sin P cos R + cos P sin R = 1 1 Solution: Let there be a right-triangle PQR, in which Q = 90 and tan R = . 3 We know that P PQ 1 tan R = = QR 3 Let PQ = k and QR = 3k 2k 1k R Q Then, PR = PQ 2 + QR 2 3 k = k2 + ( 3k ) 2 Fig. 22.22 ## 526 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry = k 2 + 3k 2 = 4k 2 or 2 k Notes side opposite to P QR 3k 3 sin P = = = = Hypotenuse PR 2k 2 side adjacent t o P 1k 1 cos P = = = Hypotenuse 2k 2 side opposite to R PQ 1k 1 sin R = = = = Hypotenuse PR 2k 2 side adjacent to R QR 3k 3 and cos R = = = = Hypotenuse PR 2k 2 3 3 1 1 sin P cos R + cos P sin R = . + . 2 2 2 2 3 1 4 = + = 4 4 4 =1 Example 22.13.: In ABC, B is right-angle. If AB = c, BC = a and AC = b, which of the following is true? 2b (i) cos C + sin A = a A b a (ii) cos C + sin A = + a b b c 2a (iii) cos C + sin A = b C B a a c Fig. 22.23 (iv) cos C + sin A = + b b ## Mathematics Secondary Course 527 MODULE - 5 Introduction to Trigonometry Trigonometry BC a Solution: Here cos C = = AC b Notes BC a and sin A = = AC b a a 2a cos C + sin A = + = b b b 2a Statement (iii), i.e., cos C + sin A = is true. b 20 1. If sin = , find the values of cos and tan . 29 24 2. If tan = , find the values of sin and cos . 7 7 3. If cos A = , find the values of sin A and tan A. 25 m 4. If cos = , find the values of cot and cosec . n 4 cos . cot 5. If cos = , evaluate . 5 1 sec 2 2 6. If cosec = , find the value of sin2 cos + tan2 . 3 5 7. If cot B = , then show that cosec2 B = 1 + cot2 B. 4 3 8. ABC is a right triangle with C = 90o. If tan A = , find the values of sin B and 2 tan B. ## 528 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry 1 9. If tan A = and tan B = 3 , then show that cos A cos B sin A sin B = 0. 3 12 Notes 10. If cot A = , show that tan2A sin2A = sin4A sec2A. 5 [Hint: Find the vlaues of tan A, sin A and sec A and substitute] 11. In Fig. 22.24, ABC is right-angled at vertex B. If AB = c, BC = a and CA = b, which of the following is true? A b+c (i) sin A + cos A = a b a+c c (ii) sin A + cos A = b C B a+b a (iii) sin A + cos A = c Fig. 22.24 a+b+c (iv) sin A + cos A = b ## 22.4 RELATIONSHIPS BETWEEN TRIGONOMETRIC RATIOS In a right triangle ABC, right angled at B, we have A AB sin = AC BC cos = AC AB and tan = BC C B AB AB BC Fig. 22.25 Rewriting, tan = = BC AC AC AB AC sin = BC = cos AC ## Mathematics Secondary Course 529 MODULE - 5 Introduction to Trigonometry Trigonometry sin Thus, we see that tan = cos ## Notes We can verify this result by taking AB = 3 cm, BC = 4 cm and therefore AC = AB2 + BC2 = 32 + 52 or 5 cm 3 4 3 sin = , cos = and tan = 5 5 4 3 sin 5 = 3 Now = 4 4 = tan . cos 5 ## Thus, the result is verified. AB Again sin = gives us AC 1 1 AC = = = cosec sin AB AB AC 1 Thus cosec = or cosec . sin = 1 sin We say cosec is the reciprocal of sin . BC Again, cos = gives us AC 1 1 AC = = = sec cos BC BC AC 1 Thus sec = or sec . cos = 1 cos We say that sec is reciprocal of cos . AB Finally, tan = gives us BC ## 530 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry 1 1 BC = = = cot tan AB AB BC Notes 1 Thus, cot = or tan . cot = 1 tan 1 cos = Also cot = sin sin cos We say that cot is reciprocal of tan . Thus, we have cosec , sec and cot are reciprocal of sin , cos and tan respectively. We have, therefore, established the following results: sin (i) tan = cos 1 (ii) cosec = sin 1 (iii) sec = cos 1 cos (iv) cot = = tan sin Now we can make use of the above results in finding the values of different trigonometric ratios. 1 3 Example 22.14: If cos = and sin = , find the values of cosec , sec and 2 2 tan . Solution: We know that 1 1 2 cosec = = 3 = sin 3 2 1 1 sec = = =2 cos 1 2 Mathematics Secondary Course 531 MODULE - 5 Introduction to Trigonometry Trigonometry 3 sin 3 2 and tan = = 2 = = 3 cos 1 2 1 Notes 2 Example 22.15: For a right angled triangle ABC, right angled at C, tan A = 1. Find the value of cos B. Solution: Let us construct a right angled ABC in which C = 90o. We have tan A = 1 (Given) We know that A BC tan A = =1 AC BC and AC are equal. Let BC = AC = k B C Then AB = BC + AC 2 2 Fig. 22.26 = k2 + k2 = 2k BC k Now cos B = = AB 2k 1 = 2 1 Hence cos B = 2 1 3 1. If sin = and cos = , find the values of cot and sec . 2 2 3 2. If sin = and tan = 3 , find the value of cos2 + sin cot . 2 ## 532 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry ## 3. In a right angled ABC, right angled at C, cos A = . Find the value of sin A sin B + cos A cos B. 4. If cosec A = 2, find the value of sin A and tan A. Notes ## 5. In a right angled ABC, right angled at B, tan A = , find the value of tan2 B sec2 A (tan2 A + cot2 B) 22.5 IDENTITY We have studied about equations in algebra in our earlier classes. Recall that when two expressions are connected by = (equal to) sign, we get an equation. In this section, we now introduce the concept of an identity. We get an identity when two expressions are connected by the equality sign. When we say that two expressions when connected by = give rise to an equation as well as identity, then what is the difference between the two. The major difference between the two is that an equation involving a variable is true for some values only whereas the equation involving a variable is true for all values of the variable, is called an identity. Thus x2 2x + 1 = 0 is an equation as it is true for x = 1. x2 5x + 6 = 0 is an equation as it is true for x = 2 and x = 3. If we consider x2 5x + 6 = (x 2) (x 3), it becomes an identity as it is true for x = 2, x = 3 and say x = 0, x = 10 etc. i.e. it is true for all values of x. In the next section, we shall consider some identities in trigonometry. ## 22.6 TRIGONOMETRIC IDENTITIES We know that an angle is defined with Y the help of the rotation of a ray from initial to final position. You have learnt to define A P all trigonometric ratios of an angle. Let us recall them here. Let XOX and YOY be the rectangular axes. Let A be any point on OX. Let the X O M A X ray OA start rotating in the plane in an O till it reaches the final position OA after some interval of time. Let AOA = . Take any point P on the ray OA. Draw PM OX. Y Fig. 22.27 ## Mathematics Secondary Course 533 MODULE - 5 Introduction to Trigonometry Trigonometry In right angled PMO, PM sin = OP Notes OM and cos = OP 2 2 PM OM sin + cos = 2 2 + OP OP PM 2 + OM 2 OP 2 = = OP 2 OP 2 =1 Hence, sin2 + cos2 = 1 ...(1) Also we know that OP sec = OM PM and tan = OM Squaring and subtracting, we get 2 2 OP PM sec tan = 2 2 OM OM OP 2 PM 2 = OM 2 OM 2 = [By Pythagoras Theorm, OP2 PM2 = OM2] OM 2 =1 Hence, sec2 tan2 = 1 ...(2) OP Again, cosec = PM OM and cot = PM ## 534 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry Squaring and subtracting, we get 2 2 OP OM cosec cot = 2 2 PM PM Notes OP 2 OM 2 PM 2 = = PM 2 PM 2 [By Pythagoras Theorm, OP2 OM2 = PM2] =1 Hence, cosec2 cot2 = 1 ...(3) Note: By using algebraic operations, we can write identities (1), (2) and (3) as sin2 = 1 cos2 or cos2 = 1 sin2 sec2 = 1 + tan2 or tan2 = sec2 1 and cosec2 = 1 + cot2 or cot2 = cosec2 1 respectively. We shall solve a few examples, using the above identities. Example 22.16: Prove that 1 tan + cot = sin cos Solution: L.H.S. = tan + cot sin cos = + cos sin sin 2 + cos 2 1 = = (Q sin2 + cos2 = 1) sin cos sin cos = R.H.S. 1 Hence, tan + cot = sin cos Exampe 22.17: Prove that sin A 1 + cos A + = 2 cosec A 1 + cos A sin A sin A 1 + cos A Solution: L.H.S = + 1 + cos A sin A ## Mathematics Secondary Course 535 MODULE - 5 Introduction to Trigonometry Trigonometry sin 2 A + (1 + cos A ) 2 = sin A (1 + cos A ) Notes sin 2 A + 1 + cos 2 A + 2 cos A = sin A (1 + cos A ) ## (sin A + cos A ) + 1 + 2 cos A 2 2 = sin A (1 + cos A ) 1 + 1 + 2 cos A sin A (1 + cos A ) = 2 + 2 cos A sin A (1 + cos A ) = 2 (1 + cos A ) sin A (1 + cos A ) = 2 = sin A = 2 cosec A = R.H.S. sin A 1 + cos A Hence, + = 2 cosec A 1 + cos A sin A Example 22.18: Prove that: 1 sin A = (sec A tan A ) 2 1 + sin A 2 1 sin A = cos A cos A 2 1 sin A = cos A (1 sin A ) 2 = cos 2 A ## 536 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry ## (1 sin A ) 2 (Q cos A = 1 sin A ) 2 2 = 1 sin 2 A (1 sin A )2 Notes = (1 sin A )(1 + sin A ) 1 sin A = 1 + sin A = L.H.S. 1 sin A = (sec A tan A ) 2 Hence, 1 + sin A Alternative method We can prove the identity by starting from L.H.S. in the following way: 1 sin A L.H.S. = 1 + sin A 1 sin A 1 sin A = 1 + sin A 1 sin A (1 sin A )2 = 1 sin 2 A (1 sin A )2 = cos 2 A 2 1 sin A = cos A 2 1 sin A = cos A cos A = (sec A tan A )2 = R.H.S. Remark: From the above examples, we get the following method for solving questions on Trigonometric identities. ## Mathematics Secondary Course 537 MODULE - 5 Introduction to Trigonometry Trigonometry Method to solve questions on Trigonometric identities Step 1: Choose L.H.S. or R.H.S., whichever looks to be easy to simplify. Step 2: Use different identities to simplify the L.H.S. (or R.H.S.) and arrive at the result on Notes the other hand side. Step 3: If you dont get the result on R.H.S. (or L.H.S.) arrive at an appropriate result and then simplify the other side to get the result already obtained. Step 4: As both sides of the identity have been proved to be equal the identity is established. We shall now, solve some more questions on Trigonometric identities. Example 22.19: Prove that: 1 sin cos = 1 + sin 1 + sin 1 sin Solution: L.H.S. = 1 + sin 1 sin 1 + sin = 1 + sin 1 + sin 1 sin 2 = (1 + sin ) cos 2 = 1 + sin (Q1 sin = cos ) 2 2 cos = = R.H.S. 1 + sin 1 sin cos Hence, = 1 + sin 1 + sin Example 22.20: Prove that cos4 A sin4A = cos2 A sin2 A = 1 2 sin2A Solution: L.H.S. = cos4 A sin4A = (cos2 A)2 (sin2 A)2 = (cos2 A + sin2 A) (cos2 A sin2 A) ## 538 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry = cos2 A sin2 A (Q cos2 A + sin2 A = 1) = R.H.S. Again cos2 A sin2 A = (1 sin2 A) sin2A (Q cos2 A = 1 sin2 A ) Notes = 1 2 sin2 A = R. H. S. Hence cos4 A sin4A = cos2 A sin2 A = 1 2 sin2A Example 22.21: Prove that sec A (1 sin A) (sec A + tan A) = 1 Solution: L.H.S. = sec A (1 sin A) (sec A + tan A) = 1 (1 sin A) 1 + sin A cos A cos A cos A ## (1 sinA )(1 + sinA ) = cos 2 A 1 sin 2 A = cos 2 A cos 2 A = cos 2 A = 1 = R.H.S. Hence, sec A (1 sin A) (sec A + tan A) = 1 Example 22.22: Prove that ## tan + sec 1 1 + sin = tan sec + 1 cos cos = 1 sin tan + sec 1 Solution: L.H.S. = tan sec + 1 2 2 = tan sec + 1 ## Mathematics Secondary Course 539 MODULE - 5 Introduction to Trigonometry Trigonometry = tan sec + 1 = tan sec + 1 = tan sec + 1 = tan + sec 1 + sin = cos = R.H.S. cos (1 sin ) Again = cos 1 sin 2 = cos (1 sin ) cos 2 = cos (1 sin ) cos = 1 sin = R.H.S. ## tan + sec 1 1 + sin Hence, = tan sec + 1 cos cos = 1 sin Example 22.23: If cos sin = 2 sin , then show that cos + sin = 2 cos . ## or cos = 2 sin + sin or cos = ( 2 + 1) sin ## 540 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry cos or = sin 2 +1 sin = cos ( ) 2 1 Notes or 2 +1 ( ) 2 1 2 cos cos or sin = 2 1 ## Hence, cos + sin = 2 cos . Example 22.24: If tan4 + tan2 = 1, then show that cos4 + cos2 = 1 Solution: We have tan4 + tan2 = 1 or tan2 ( tan2 + 1) = 1 1 or 1 + tan 2 = = cot 2 tan 2 ## or sec2 = cot2 (1 + tan2 = sec2 ) 1 cos 2 or = cos 2 sin 2 or sin2 = cos4 or 1 cos2 = cos4 (sin2 = 1 cos2 ) or cos4 + cos2 = 1 Prove each of the following identities: 1. (cosec2 1) sin2 = cos2 2. sin4A + sin2A cos2A = sin2A 3. cos2 (1 + tan2 ) = 1 4. (1 + tan2 ) sin2 = tan2 ## Mathematics Secondary Course 541 MODULE - 5 Introduction to Trigonometry Trigonometry sinA sinA 5. + = 2cosecA 1 + cosA 1 cosA 6. = 1 cosA sinA ## secA tanA cosA 7. = secA + tanA 1 + sinA ## 8. (sin A cos A)2 + 2 sin A cos A = 1 9. cos4 + sin4 2 sin2 cos2 = (2 cos2 1)2 ## sinA sinB cosA cosB 10. + =0 cosA + cosB sinA + sinB ## 11. (cosec sin ) (sec cos ) (tan + cos ) = 1 12. sin A(1 + tan A) + cos A (1 + cot A) = sec A + cosec A 1 cosA = (cosecA cotA ) 2 13. 1 + cosA tanA cotA 14. + = 1 + secAcosecA 1 cotA 1 tanA ## cotA + cosecA 1 1 + cosA 15. cotA cosecA + 1 = sinA sinA = 1 cosA ## 16. If sin2 + sin = 1, then show that cos2 + cos4 = 1 Select the correct alternative from the four given in each of the following questions (17 - 20): 17. (sin A + cos A)2 2 sin A cos A is equal to (i) 0 (ii) 2 (iii) 1 (iv) sin2A cos2A 18. sin4A cos4A is equal to: (i) 1 (ii) sin2A cos2A (iii) 0 (iv) tan2A ## 542 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry 19. sin2A sec2A + cos2A + tan2A is equal to (i) 0 (ii) 1 (iii) sin2A (iv) cos2A 20. (sec A tan A) (sec A + tan A) (cosec A cot A) (cosec A + cot A) is equal to Notes 1 (i) 2 (ii) 1 (iii) 0 (iv) 2 ## 22.7 TRIGONOMETRIC RATIOS FOR COMPLEMENTARY ANGLES In geometry, we have studied about complementary and supplementary Y angles. Recall that two angles are A complementary if their sum is 90o. If the sum of two angles A and B is 90o, then P 9 0 o A and B are complementary angles and each of them is complement of the other. Thus, angles of 20o and 70o are complementary and 20o is complement X O M A X of 70o and vice versa. Let XOX and YOY be a rectangular system of coordinates. Let A be any point on OX. Let ray OA be rotated in an anti clockwise direction and trace an angle Y from its initial position. Let POM = . Fig. 22.28 Draw PM OX. Then PMO is a right angled triangle. Also, POM + OPM + PMO = 180o or POM + OPM + 90o = 180o or POM + OPM = 90o OPM = 90o POM = 90o Thus OPM and POM are complementary angles. Now in right angled triangle PMO, PM OM PM sin = , cos = and tan = OP OP OM OP OP OM cosec = , sec = and cot = PM OM PM For reference angle (90o ), we have in right d OPM, ## Mathematics Secondary Course 543 MODULE - 5 Introduction to Trigonometry Trigonometry ( sin 90 o =) OM OP = cos Notes ( cos 90 o = ) PM OP = sin ( tan 90 o =) OM PM = cot ( cot 90 o =) PM OM = tan ( cosec 90 o = ) OP OM = sec and ( sec 90 o =) OP PM = cosec The above six results are known as trigonometric ratios of complementary angles. For example, sin (90o 20o) = cos 20o i.e. sin 70o = cos 20o tan (90o 40o) = cot 40o i.e. tan 50o = cot 40o and so on. Let us take some examples to illustrate the use of above results. Example 22.25: Prove that tan 13o = cot 77o Solution: R.H.S. = cot 77o = cot (90o 13o) = tan 13o ....[Q cot (90o ) = tan ] = L.H.S. Thus, tan 13o = cot 77o Example 22.26: Evaluate sin2 40o cos2 50o Solution: cos 50o = cos (90o 40o) = sin 40o ....[Q cos (90o ) = tan ] sin2 40o cos2 50o = sin2 40o sin2 40o = 0 ## cos 41o sec 37 o Example 22.27: Evaluate : + sin 49 o cosec 53o ## 544 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry Solution: sin 49o = sin (90o 41o) = cos 41o ...[Q sin (90o ) = cos ] and cosec 53o = cosec (90o 37o) = sec 37o ...[Q cosec (90o ) = sec ] ## cos 41o sec 37 o cos 41 o sec 37 o Notes + = + sin 49 o cosec 53o cos 41 o sec 37 o = 1+1 = 2 Example 22.28: Show that 3 sin 17o sec 73o + 2 tan 20o tan 70o = 5 Solution: 3 sin 17o sec 73o + 2 tan 20o tan 70o = 3 sin 17o sec (90o 17o) + 2 tan 20o tan (90o 20o) = 3 sin 17o cosec 17o + 2 tan 20o cot 20o ...[Q sec (90o ) = cosec and tan (90o ) = cot ] 1 1 o = 3 sin 17 . o + 2 tan 20 o. sin 17 tan 20 o =3+2=5 Example 22.29: Show that tan 7o tan 23o tan 67o tan 83o = 1 Solution: tan 67o = tan (90o 23o) = cot 23o and tan 83o = tan (90o 7o) = cot 7o Now. L.H.S. = tan 7o tan 23o tan 67o tan 83o = tan 7o tan 23o cot 23o cot 7o = (tan 7o cot 7o) (tan 23o cot 23o) = 1 .1 = 1 = R.H.S. Hence, tan 7o tan 23o tan 67o tan 83o = 1 Example 22.30: If tan A = cot B, prove that A + B = 90o. Solution: We are given tan A = cot B or tan A = tan (90o B) ... [Q cot = tan (90o )] A = 90o B or A + B = 90o ## Mathematics Secondary Course 545 MODULE - 5 Introduction to Trigonometry Trigonometry B+C A Example 22.31: For a ABC, show that sin = cos , where A, B and C are 2 2 interior angles of ABC. Notes Solution: We know that sum of angles of triangle is 180o. A + B + C = 180o or B + C = 180o A B+C A or = 90 o 2 2 B+C o A sin = sin 90 2 2 B+C A or sin = cos 2 2 cos sin + = 2. Example 22.32: Prove that ( sin 90 cos 90 o o ) ( ) cos sin + Solution: L.H.S. = ( ) sin 90 cos 90 o o ( ) cos sin = + ... [Q sin (90o ) = cos and cos (90o ) = sin ] cos sin =1+1=2 = R.H.S. cos sin + =2 Hence, ( ) ( sin 90 cos 90 o o ) sin 90o + (( cos 90 o =1 ) ) Example 22.33: Show that ( ) ( cosec 90o sec 90o ) sin (90 ) cos (90 ) o o + cosec (90 ) sec(90 ) Solution: L.H.S. = o o cos sin = + ...[Q sin (90o ) = cos , cos (90o ) = sin , sec cosec cosec (90o ) = sec and sec (90o ) = cosec ] ## 546 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry cos sin = + = cos2 + sin2 = 1 cos sin = R.H.S. Notes ( sin 90 + ) cos 90 o =1 ( o ) Hence, ( cosec 90 sec 90o o ) ( ) Example 22.34: Simplify: ( ) ( cos 90 o sec 90 o tan ) + tan 90 o ( ) ( ) ( ) ( cosec 90 o sin 90 o cot 90 o ) cot ## Solution: The given expression ( ) ( cos 90 o sec 90 o tan + ) tan 90 o ( ) = ( ) ( cosec 90 o sin 90 o cot 90 o ) ( cot ) sin in.co .tan cot = + ...[Q sin . cos = 1 and sec . cos = 1] sec ec.cotan cot =1+1 =2 Example 22.35: Express tan 68o + sec 68o in terms of angles between 0o and 45o. ## Hence tan 68o + sec 68o = cot 22o + cosec 22o. Remark: While using notion of complementary angles, usually we change that angle which is > 45o to its complement. Example 22.36: If tan 2A = cot (A 18o) where 2A is an acute angle, find the value of A. ## Solution: We are given tan 2A = cot (A 18o) or cot (90o 2A) = cot (A 18o) ...[Q cot (90o 2A = tan 2A] ## Mathematics Secondary Course 547 MODULE - 5 Introduction to Trigonometry Trigonometry 90o 2A = A 18o or 3A = 90o + 18o Notes or 3A = 108o or A = 36o 1. Show that: (i) cos 55o = sin 35o (ii) sin2 11o cos2 79o = 0 (iii) cos2 51o sin2 39o = 0 2. Evaluate each of the following: ## 3sin19 o tan65 o cos 89o (i) (ii) (iii) cos71o 2cot25 o 3 sin 1o ## 3 sin 5o 2 tan 33o o (iv) cos 48 sin 42 o (v) + cos 85o cot 57 o ## cot 54o tan 20o (vi) + 2 tan 36o cot 70o (vii) sec 41o sin 49o + cos 49o cosec 41o ## cos 75o sin 12o cos18o (viii) + sin 15o cos 78o sin 72o 3. Evaluate each of the following: 2 2 sin 47 o cos 43o (i) + o o cos 43 sin 47 ## cos 2 20o + cos 2 70o (ii) ( 3 sin 2 59o + sin 2 31o ) 4. Prove that: (i) sin cos (90o ) + cos sin (90o ) = 1 ## 548 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry (ii) cos cos (90o ) sin sin (90o ) = 0 ( cos 90 o + ) 1 + sin 90 o ( = 2cosec ) (iii) ( 1 + sin 90 o ) cos 90 o ( ) Notes ( ) ( ) ( tan 90 o ) (iv) sin 90 .cos 90 = o o ( 1 + tan 2 90 o ) (v) tan 45o tan 13o tan 77o tan 85o = 1 (vi) 2 tan 15o tan 25o tan 65o tan 75o = 2 (vii) sin 20o sin 70o cos 20o cos 70o = 0 5. Show that sin (50o + ) cos (40o ) = 0 6. If sin A = cos B where A and B are acute angles, prove that A + B = 90o. 7. In a ABC, prove that B+C A (i) tan = cot 2 2 A+B C (ii) cos = sin 2 2 8. Express tan 59o + cosec 85o in terms of trigonometric ratios of angles between 0o and 45o. 9. Express sec 46o cos 87o in terms of trigonometric ratios of angles between 0o and 45o. 10. Express sec2 62o + sec2 69o in terms of trigonometric ratios of angles between 0o and 45o. Select the correct alternative for each of the following questions (11-12): ## sin 40o 2 sec 41o 11. The value of is 2 cos 50o 3cosec49o 1 1 (i) 1 (ii) (iii) (iv) 1 6 6 12. If sin ( + 36o) = cos , where + 36o is an acute angle, then is (i) 54o (ii) 18o (iii) 21o (iv) 27o ## Mathematics Secondary Course 549 MODULE - 5 Introduction to Trigonometry Trigonometry LET US SUM UP sin = = Hypotenuse AC A cos = = Hypotenuse AC tan = = ## side adjacent t o angle BC cot = = side opposite to angle AB C B Hypotenuse AC sec = = side adjacent t o angle BC Hypotenuse AC cosec = = side opposite to angle AB ## The following relationships exist between different trigonometric ratios: sin cos (i) tan = (ii) cot = cos sin 1 1 (iii) sec = (iv) cosec = cos sin 1 (v) cot = tan The trigonometric identities are: (i) sin2 + cos2 = 1 (ii) sec2 tan2 = 1 (iii) cosec2 cot2 = 1 Two angles, whose sum is 90o, are called complementary angles. ## 550 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry sin (90o A) = cos A, cos (90o A) = sin A and tan (90o A) = cot A. cosec (90o A) = sec A, sec (90o A) = cosec A and cot (90o A) = tan A Supportive website: Notes http://www.wikipedia.org http://mathworld:wolfram.com TERMINAL EXERCISE 4 1. If sin A = , find the values of cos A and tan A. 5 20 2. If tan A = , find the values of cosec A and sec A. 21 3 3. If cot = , find the value of sin + cos . 4 m 4. If sec = , find the values of sin and tan . n 3 5. If cos = , find the value of 5 sin tan 1 2 tan 2 5 tan 6. If sec = , find the value of 4 1 + tan 7. If tan A = 1 and tan B = 3 , find the value of cos A cos B sin A sin B. ## Prove each of the following identities (8 20): 8. (sec + tan ) (1 sin ) = cos . cot cosec 9. = 1 tan sec 1 cos = (cosec cot ) 2 10. 1 + cos ## Mathematics Secondary Course 551 MODULE - 5 Introduction to Trigonometry Trigonometry 11. = tan sin sec 1 ## Notes tan A + cot B 12. = tan A cot B cot A + tan B 1 + cos A 13. = cosec A + cot A 1 cos A cosec A + 1 cos A 14. = cosec A 1 1 sin A 15. sin3A cos3A = (sin A cos A) (1 + sin A cos A) cos A sin A 16. + = cos A + sin A 1 tan A 1 cot A sec A 1 sec A + 1 17. + = 2cosecA sec A + 1 sec A 1 1 18. (cosecA sin A )(sec A cos A ) = tan A + cot A 19. (1 + cot cosec ) (1 + tan + sec ) = 2 20. 2(sin6 + cos6 ) 3(sin4 + cos4 ) + 1 = 0 p2 1 21. If sec + tan = p, show that sin = p2 +1 ( cos 90o A + ) ( 1 + sin 90o A ) = 2sec 90o A ( ) 22. Prove that 1 + sin 90 A o ( ) ( cos 90 A o ) ## 23. Prove that ( ) ( sin 90o A .cos 90o A ) = sin 2 90o A ( ) tanA 3 24. If tan = and + = 90o, find the value of cot . 4 25. If cos (2 + 54o) = sin and (2 + 54o) is an acute angle, find the value of . 26. If sec Q = cosec P and P and Q are acute angles, show that P + Q = 90o. ## 552 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry 22.1 Notes 5 12 5 1. (i) sin = , cos = , tan = 13 13 12 13 13 12 cosec = , sec = and cot = 5 12 5 3 4 3 (ii) sin = , cos = , tan = 5 5 4 5 5 4 cosec = , sec = and cot = 3 4 3 24 7 24 (iii) sin = , cos = , tan = 25 25 7 25 25 7 cosec = , sec = and cot = 24 7 24 4 3 4 (iv) sin = , cos = , tan = 5 5 3 5 5 3 cosec = , sec = and cot = 4 3 4 5 4 5 2. sin A = , cos A = and tan A = 41 41 4 40 9 40 40 3. sin C = , cot C = , cos A = and cot A = 41 40 41 9 ## 4. sec C = 2 , cosec C = 2 and cot C =1 5. (iv) 6. (ii) 22.2 3 4 3 1. sin C = , cos C = and tan C = 5 5 4 ## Mathematics Secondary Course 553 MODULE - 5 Introduction to Trigonometry Trigonometry 24 25 7 2. sin A = , cosec A = and cot A = 25 24 24 ## Notes 3. sec P = 2 , cot P = 1, and cosec P = 2 2 1 2 4. tan R = 3 , cosec R = 3 , sin P = 2 and sec P = 3 24 7 25 7 5. cot = , sin = , sec = , and tan = 7 25 24 24 2 6 5 5 2 6 6. sin P = , cos P = , sin R = and cos R = , sin P cos R = 0 7 7 7 7 7. (iii) 22.3 21 20 1. cos = and tan = 29 21 24 7 2. sin = and cos = 25 25 24 24 3. sin A = and tan A = 25 7 m n 4. cot = and cosec = n 2 m2 n 2 m2 256 5. 135 27 6. 8 2 2 7. sin B = and tan B = 13 3 11. (ii) 22.4 2 1. cot = 3 and sec = 3 ## 554 Mathematics Secondary Course Introduction to Trigonometry MODULE - 5 Trigonometry 3 2. 4 3 Notes 3. 2 1 1 4. sin A = and tan A = 2 3 14 5. 3 22.5 17. (iii) 18. (ii) 19. (i) 20. (iii) 22.6 1 1 1. (i) 3 (ii) (iii) (iv) 0 2 3 (v) 5 (vi) 0 (vii) 2 (viii) 1 1 3. (i) 2 (ii) 3 8. cot 31o + sec 5o 9. cosec 44o sin 3o 10. cosec2 28o + cosec2 21o 11. (ii) 12. (iv) 3 4 1. cos A = and tan A = 5 3 ## Mathematics Secondary Course 555 MODULE - 5 Introduction to Trigonometry Trigonometry 29 29 2. cosec A = and sec A = 20 21 Notes 7 3. 5 m2 n 2 m2 n 2 4. sin = and tan = m n 3 5. 160 3 6. 7 1 3 7. 2 2 3 24. 4 25. 12o
# Thread: Integration by reduction formulae 1. ## Integration by reduction formulae Given that $I_n =$ $\int_{0}^{1}\frac {1}{(1+x^2)^n} dx$, show that, for n≥2, $2(n-1)I_n = 2^{1-n} + (2n-3)I_{n-1}$ 2. Originally Posted by geton Given that $I_n =$ $\int_{0}^{1}\frac {1}{(1+x^2)^n} dx$, show that, for n≥2, $2(n-1)I_n = 2^{1-n} + (2n-3)I_{n-1}$ Its less messy to prove: $\boxed{2nI_{n+1} = 2^{-n} + (2n-1)I_{n}}$ First step: Integration by parts- $I_{n} =$ $\int_{0}^{1}\frac {1}{(1+x^2)^{n}} \, dx = \frac{x}{(1+x^2)^n}\bigg{\|}_0^1 + 2n \int_0^1 \frac{x^2}{(1+x^2)^{n+1}} \, dx$ Second Step: Partial Fractions Now we observe that $\int_0^1 \frac{x^2}{(1+x^2)^{n+1}} \,dx = \int_0^1 \frac{1}{(1+x^2)^n}\, dx - \int_0^1 \frac{1}{(1+x^2)^{n+1}}\, dx = I_{n} - I_{n+1}$. Third Step: Rearranging $I_{n} = 2^{-n} + 2n(I_{n} - I_{n+1})$ $2n I_{n+1} = 2^{-n} + (2n-1)I_{n}$ 3. Originally Posted by geton Given that $I_n =$ $\int_{0}^{1}\frac {1}{(1+x^2)^n} dx$, show that, for n≥2, $2(n-1)I_n = 2^{1-n} + (2n-3)I_{n-1}$ Use integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \frac{1}{(1 + x^2)^n} \Rightarrow du = - \frac{2n x}{(1 + x^2)^{n+1}} \, dx$ and $dv = dx \Rightarrow v = x$. Note that $\frac{x^2}{(1 + x^2)^{n+1}} = \frac{(1 + x^2) - 1}{(1 + x^2)^{n+1}} = \frac{1}{(1 + x^2)^{n}} - \frac{1}{(1 + x^2)^{n+1}}$. Then $I_n = \frac{1}{2^n} + 2n (I_n - I_{n+1})$. Therefore: $2n I_{n+1} = \frac{1}{2^n} + (2n-1) I_n$. Re-label $n \rightarrow n - 1$ and solve for $I_n$.
Quadratic Graph vs Exponential: Understanding The Differences Mathematics is a vast area of knowledge, and it has several concepts which are worth understanding. Today, we are going to discuss two such concepts, those are Quadratic Graph and Exponential. These concepts are frequently used in the field of mathematics, and you can often find them in different categories of mathematics problems. The main idea behind these concepts is to demonstrate how different kinds of curves and graphs can be formed by different mathematical equations. Now, let’s get started! It’s time to discover quadratic graphs and exponential graphs in detail. A Quadratic Graph is formed by the quadratic equation, which is a type of polynomial equation. A quadratic equation has degree two, and it has the form of ax²+bx+c, where a, b, and c are constants with a ≠ 0. When the equation is plotted on the graph, it forms a parabolic curve. This curve is crucial in mathematics, engineering, and many other fields. A quadratic equation has two solutions, meaning two places where the curve crosses the x-axis. These solutions are also known as roots or zeros of the quadratic equation. Quadratic equations are used in many real-life applications, such as in physics to calculate the trajectory of baseballs or the flight path of a rocket. What Is An Exponential Graph? An Exponential Graph is formed by an exponential function, which is of the form y = ab^x, where a is a constant, b is the base, and x is the exponent. When the equation is plotted on the graph, it forms a curve that increases or decreases exponentially based on the values of a and b. Exponential functions are widely used to describe growth, decay, and a wide range of phenomena that exhibit exponential behavior. For instance, exponential functions are used in finance to calculate compound interest rates over time, in physics to describe radioactive decay, and in biology to model population growth. Now that we’ve discussed what quadratic and exponential graphs are, we can proceed to understand the differences between them. Here are some of the key differences: 1. Rate of Increase One of the most significant differences between quadratic and exponential graphs is the rate of increase. Quadratic graphs increase at a constant rate, whereas the exponential graphs’ rate of increase changes over time. 2. Curvature Quadratic graphs are parabolic curves, which means they have a constant curvature. In contrast, exponential graphs have an increasing or decreasing curvature over time. 3. Zeros Quadratic graphs have two zeros or roots, while exponential graphs do not have any. 4. Applications Quadratic graphs are commonly used in physics and engineering to model motion, while exponential graphs are commonly used in finance and population biology. 5. Function Quadratic functions are second-degree functions, while exponential functions are functions where the exponent is the variable. 6. Domain and Range Quadratic functions have a domain of all real numbers, meaning that they can take any value for the input variable. The range of the quadratic function is either positive or negative, depending on the orientation of the curve. Exponential functions, on the other hand, have a domain that is all real numbers, and the range is always positive. When it comes to optimizing quadratic and exponential graphs for SEO purposes, there are a few things to keep in mind. Here are some tips: 1. Use effective labeling. When you design graphs, make sure to use meaningful and descriptive titles, labels, and legends.
Question 1: In a right angles triangle $ABC$, right angled at $B$, if $\sin A =$ $\frac{3}{5}$, find all the six trigonometric ratios of $\angle C$. Given $\sin A =$ $\frac{3}{5}$ $\Rightarrow AB^2 + BC^2 = AC^2$ $\Rightarrow AB^2 = AC^2 - BC^2$ $\Rightarrow AB^2 = 5^2 - 3^2 = 16$ $\Rightarrow AB = 4$ Therefore, $\sin C =$ $\frac{Perpendicular}{Hypotenuse}$ $=$ $\frac{4}{5}$ ; $cosec \ C =$ $\frac{Hypotenuse}{Perpendicular}$ $=$ $\frac{5}{4}$ $\cos C =$ $\frac{Base}{Hypotenuse}$ $=$ $\frac{3}{5}$ ; $\sec C =$ $\frac{Hypotenuse}{Base}$ $=$ $\frac{5}{3}$ $\tan C =$ $\frac{Perpendicular}{Base}$ $=$ $\frac{4}{3}$ ; $\cot C =$ $\frac{Base}{Perpendicular}$ $=$ $\frac{3}{4}$ $\\$ Question 2: If $\sin A =$ $\frac{a^2 - b^2}{a^2 + b^2}$, find the value of the other five trigonometric ratios. Given $\sin A =$ $\frac{a^2 - b^2}{a^2 + b^2}$ By Pythagoras theorem, we have $AB^2 = AC^2 = BC^2 = (a^2 + b^2)^2 - (a^2 - b^2)^2 = 4a^2b^2$ $\Rightarrow AB = 2ab$ $\cos A$ $=$ $\frac{2ab}{a^2 + b^2}$ ; $\sec A$ $=$ $\frac{a^2 + b^2}{2ab}$ $\tan A$ $=$ $\frac{a^2 - b^2}{2ab}$ ; $\cot A$ $=$ $\frac{2ab}{a^2 - b^2}$ $\sin A =$ $\frac{a^2 - b^2}{a^2 + b^2}$ ; $cosec \ A$ $=$ $\frac{a^2 + b^2}{a^2 - b^2}$ $\\$ Question 3: If $cosec \ A = 2$, find the value of $\frac{1}{\tan A}$ $+$ $\frac{\sin A}{1+ \cos A}$ Given $cosec \ A = 2$, therefore by Pythagoras theorem, $AB = \sqrt{3}$ $\therefore \tan A =$ $\frac{1}{\sqrt{3}}$ ; $\sin A =$ $\frac{1}{2}$ ; $\cos A =$ $\frac{\sqrt{3}}{2}$ Hence $\frac{1}{\tan A}$ $+$ $\frac{\sin A}{1+ \cos A}$ $=$ $\sqrt{3} +$ $\frac{1}{2} (\frac{2}{2+\sqrt{3}})$ $= \sqrt{3} +$ $\frac{1}{2+\sqrt{3}}$ $=$ $\frac{2\sqrt{3}+ 3 + 1}{2 + \sqrt{3}}$ $=$ $\frac{2(2+\sqrt{3})}{(2+\sqrt{3})}$ $= 2$ $\\$ Question 4: If $\tan A = \sqrt{2}-1$, show that $\sin A \cos A =$ $\frac{\sqrt{2}}{4}$ Given $\tan A = \sqrt{2}-1$ By Pythagoras theorem, we have $AC = \sqrt{(\sqrt{2}-1)^2 + 1^2} = \sqrt{4 - 2\sqrt{2}}$ $\therefore \sin A =$ $\frac{\sqrt{2}-1}{\sqrt{4 - 2\sqrt{2}}}$ and $\cos A =$ $\frac{1}{\sqrt{4 - 2\sqrt{2}}}$ Hence, $\sin A \cos A =$ $\frac{\sqrt{2}-1}{\sqrt{4 - 2\sqrt{2}}}$ $\times$ $\frac{1}{\sqrt{4 - 2\sqrt{2}}}$ $=$ $\frac{\sqrt{2}-1}{(4 - 2\sqrt{2})}$ $=$ $\frac{(\sqrt{2}-1)}{2\sqrt{2}(\sqrt{2}-1)}$ $=$ $\frac{1}{2\sqrt{2}}$ $\\$ Question 5: In $\triangle ABC$, right angles at $C$, if $\tan A =$ $\frac{1}{\sqrt{3}}$ and $\tan B = \sqrt{3}$, show that $\sin A \cos B + \cos A \sin B = 1$ Given $\tan A =$ $\frac{1}{\sqrt{3}}$ By Pythagoras theorem, we have $AB = \sqrt{ (\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2$ $\therefore \sin A =$ $\frac{1}{2}$ and $\cos A =$ $\frac{\sqrt{3}}{2}$ $\sin B =$ $\frac{\sqrt{3}}{2}$ and $\cos B =$ $\frac{1}{2}$ $\therefore \sin A \cos B + \cos A \sin B =$ $\frac{1}{2}$ $\times$ $\frac{1}{2}$ $+$ $\frac{\sqrt{3}}{2}$ $\times$ $\frac{\sqrt{3}}{2}$ $=$ $\frac{1}{4}$ $+$ $\frac{3}{4}$ $= 1$ $\therefore$ LHS = RHS. Hence proved. $\\$ Question 6: If $\cot B =$ $\frac{12}{5}$, prove that $\tan^2 B - \sin^2 B = \sin^4 B \sec^2 B$ Given $\cot B =$ $\frac{12}{5}$ By Pythagoras theorem, we have $AB = \sqrt{12^2 + 5^2 } = 13$ $\therefore \tan B =$ $\frac{5}{12}$ ; $\sin B =$ $\frac{5}{13}$ ; $\sec B =$ $\frac{13}{12}$ LHS $= \tan^2 B - \sin^2 B = ($ $\frac{5}{12}$ $)^2- ($ $\frac{5}{13}$ $)^2 =$ $\frac{5^4}{12^2 \times 13^2}$ RHS $= \sin^4 B \sec^2 B = ($ $\frac{5}{13}$ $)^4 \times ($ $\frac{13}{12}$ $)^2 =$ $\frac{5^4}{12^2 \times 13^2}$ Therefore LHS = RHS. Hence proved. $\\$ Question 7: In the given figure, $AD = DB$ and $\angle B = 90^o$. Determine (i) $\sin \theta$ ii) $\cos \theta$ iii) $\tan \theta$ iv) $\sin^2 \theta + \cos^2 \theta$ By Pythagoras theorem, we have $CB = \sqrt{b^2-a^2}$ $CD = \sqrt{(\frac{a}{2})^2 + (\sqrt{b^2-a^2})^2} =$ $\frac{\sqrt{4b^2-3a^2}}{2}$ Therefore: i) $\sin \theta =$ $\frac{BD}{CD}$ $=$ $\frac{\frac{a}{2}}{\frac{\sqrt{4b^2-3a^2}}{2}}$ $=$ $\frac{a}{\sqrt{4b^2-3a^2}}$ ii) $\cos \theta =$ $\frac{BC}{CD}$ $=$ $\frac{\sqrt{b^2-a^2}}{\frac{\sqrt{4b^2-3a^2}}{2}}$ $=$ $\frac{2\sqrt{b^2-a^2}}{\sqrt{4b^2-3a^2}}$ iii) $\tan \theta =$ $\frac{BD}{BC}$ $=$ $\frac{\frac{a}{2}}{\sqrt{b^2-a^2}}$ $=$ $\frac{a}{2\sqrt{b^2-a^2}}$ iv) $\sin^2 \theta + \cos^2 \theta = ($ $\frac{a}{\sqrt{4b^2-3a^2}}$ $)^2 + ($ $\frac{2\sqrt{b^2-a^2}}{\sqrt{4b^2-3a^2}}$ $)^2 =$ $\frac{a^2}{4b^2-3a^2}$ $+$ $\frac{4(b^2-a^2)}{4b^2-3a^2}$ $= 1$ $\\$ Question 8: In a $\triangle ABC$, right angles at $C$ and $\angle A = \angle B$, i) is $\cos A = \cos B$ ii) is $\tan A = \tan B$ Given $\angle C = 90^o$ $\Rightarrow \angle A = \angle B = 45^o$ i) $\cos A = \cos B = \cos 45^o =$ $\frac{1}{\sqrt{2}}$ ii) $\tan A = \tan B = \tan 45^o = 1$ $\\$ Question 9: If $5 \tan A = 4$, show that $\frac{5 \sin A - 3 \cos A}{5 \sin A + 2 \cos A}$ $=$ $\frac{1}{6}$ LHS $=$ $\frac{5 \sin A - 3 \cos A}{5 \sin A + 2 \cos A}$ $=$ $\frac{5 \tan A - 3}{5 \tan A + 2}$ $=$ $\frac{5 .\frac{4}{5} - 3}{5 .\frac{4}{5} + 2}$ $=$ $\frac{1}{6}$ $=$ RHS. Hence proved. $\\$ Question 10: If $\tan A +$ $\frac{1}{\tan A}$ $= 2$, find the value of $\tan^2 A +$ $\frac{1}{\tan^2 A}$ $\tan A +$ $\frac{1}{\tan A}$ $= 2$ Squaring on both sides, $(\tan A +$ $\frac{1}{\tan A}$ $)^2 = 4$ $\Rightarrow \tan^2 A +$ $\frac{1}{\tan^2 A}$ $+ 2 = 4$ $\Rightarrow \tan^2 A +$ $\frac{1}{\tan^2 A}$ $= 2$ $\\$ Question 11: If $\cot A =$ $\frac{7}{8}$, evaluate i) $\frac{(1+\sin A)(1- \sin A)}{(1+\cos A)(1-\cos A)}$ ii) $\cot^2 A$ i) $\frac{(1+\sin A)(1- \sin A)}{(1+\cos A)(1-\cos A)}$ $=$ $\frac{1-\sin^2 A}{1-\cos^2 A}$ $=$ $\frac{\cos^2 A}{\sin^2 A}$ $=$ $\cot^2 A =$ $\frac{49}{64}$ ii) $\cot^2 A= ($ $\frac{7}{8})^2$ $=$ $\frac{49}{64}$ $\\$ Question 12: If $3 \cot A = 4$, check if $\frac{1-\tan^2 A}{1+\tan^2 A}$ $= \cos^2 A - \sin^2 A$ Given $\cot A =$ $\frac{4}{3}$ $\Rightarrow \tan A =$ $\frac{3}{4}$ LHS $=$ $\frac{1-\tan^2 A}{1+\tan^2 A}$ $=$ $\frac{1-\frac{9}{16}}{1+\frac{9}{16}}$ $=$ $\frac{16-9}{16+9}$ $=$ $\frac{7}{25}$ RHS $= \cos^2 A - \sin^2 A = ($ $\frac{4}{5}$ $)^2 - ($ $\frac{3}{5}$ $)^2 =$ $\frac{16-9}{25}$ $=$ $\frac{7}{25}$ Therefore LHS = RHS. Hence proved. $\\$ Question 13: If $\tan A =$ $\frac{a}{b}$, find the value of $\frac{\cos A + \sin A}{\cos A - \sin A}$ Given $\frac{\cos A + \sin A}{\cos A - \sin A}$ Dividing both the numerator and denominator by $\cos A$ we get $=$ $\frac{1 + \tan A}{1 - \tan A}$ $=$ $\frac{1 + \frac{a}{b}}{1 - \frac{a}{b}}$ $=$ $\frac{b-a}{b+a}$ $\\$ Question 14: If $\cos A =$ $\frac{12}{13}$, find $\sin A (1 - \tan A) =$ $\frac{35}{156}$ By Pythagoras theorem, $BC = \sqrt{169 - 144} = 5$ Therefore, $\sin A =$ $\frac{5}{13}$ and $\tan A =$ $\frac{5}{12}$ Hence $\sin A (1 - \tan A) =$ $\frac{5}{13}$ $(1 -$ $\frac{5}{12}$ $) =$ $\frac{35}{156}$ $\\$ Question 15: If $\sec A =$ $\frac{5}{4}$, find the value of $\frac{\sin A - 2 \cos A}{\tan A - \cot A}$ By Pythagoras theorem, $BC = \sqrt{5^2 - 4^2} = \sqrt{9} = 3$ Therefore $\sin A =$ $\frac{3}{5}$ ; $\tan A =$ $\frac{3}{4}$ ; $\cos A =$ $\frac{4}{5}$ ; $\cot A =$ $\frac{4}{3}$ $\Rightarrow \frac{\sin A - 2 \cos A}{\tan A - \cot A}$ $=$ $\frac{\frac{3}{5} - 2 .\frac{4}{5}}{\frac{3}{4} -\frac{4}{3}}$ $=$ $\frac{-5}{5}$ $\times$ $\frac{12}{-7}$ $=$ $\frac{12}{7}$ $\\$ Question 16: If $\sin A =$ $\frac{3}{5}$; evaluate $\frac{\cos A - \frac{1}{\tan A}}{2 \cot A}$ By Pythagoras theorem we get $AC = \sqrt{4^2 + 3^2 } = \sqrt{25} = 5$ Therefore $\cos A =$ $\frac{4}{5}$ ; $\tan A =$ $\frac{3}{4}$ ; $\cot A =$ $\frac{4}{3}$ Therefore $\frac{\cos A - \frac{1}{\tan A}}{2 \cot A}$ $=$ $\frac{\frac{4}{5} - \frac{1}{\frac{3}{4}}}{2 . \frac{4}{3}}$ $=$ $\frac{(12-20) . 3}{15.(8)}$ $=$ $\frac{-8}{15} \times \frac{3}{8}$ $=$ $\frac{-1}{5}$ $\\$ Question 17: If $\sec A =$ $\frac{5}{4}$, verify $\frac{3 \sin A - 4 \sin^3 A}{4 \cos^3 A - 3 \cos A}$ $=$ $\frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A}$ By Pythagoras theorem we get $BC = \sqrt{5^2 - 4^2} = \sqrt{9} = 3$ Therefore $\sin A =$ $\frac{3}{5}$ ; $\cos A =$ $\frac{4}{5}$ ; $\tan A =$ $\frac{3}{4}$ LHS $=$ $\frac{3 \sin A - 4 \sin^3 A}{4 \cos^3 A - 3 \cos A}$ $=$ $\frac{3 (\frac{3}{5}) - 4 (\frac{3}{5})^3}{4 (\frac{4}{5})^3 - 3 (\frac{4}{5})}$ $=$ $\frac{\frac{9}{5} - \frac{108}{125}}{\frac{256}{125} - \frac{12}{15}}$ $=$ $\frac{225-108}{256-300}$ $=$ $\frac{-117}{44}$ RHS = $\frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A}$ $=$ $\frac{3 . \frac{3}{4} - (\frac{3}{4})^3 }{1 - 3 (\frac{3}{4})^2}$ $=$ $\frac{\frac{9}{4} - \frac{27}{64}}{1 - \frac{27}{16}}$ $=$ $\frac{(144-27) (16)}{(64 )(16-27)}$ $=$ $\frac{117 \times 16}{64 \times (-11)}$ $=$ $\frac{-117}{44}$ Therefore LHS = RHS. Hence proved. $\\$ Question 18: If $\sin A =$ $\frac{3}{4}$, prove that $\sqrt{\frac{cosec^2 \ A - \cot^2 A}{\sec^2 A -1}}$ $=$ $\frac{\sqrt{7}}{3}$ Given $\sin A =$ $\frac{3}{4}$ By Pythagoras theorem, $AB = \sqrt{4^2 - 3^2} = \sqrt{7}$ Therefore $\cos A =$ $\frac{\sqrt{7}}{4}$$\sec A =$ $\frac{4}{\sqrt{7}}$$\tan A =$ $\frac{3}{\sqrt{7}}$$\cot A =$ $\frac{\sqrt{7}}{3}$ ; $cosec \ A =$ $\frac{4}{3}$ LHS = $\sqrt{\frac{cosec^2 \ A - \cot^2 A}{\sec^2 A -1}}$ $=$ $\frac{\sqrt{7}}{3}$ $=$ $\sqrt{\frac{(\frac{4}{3})^2 - (\frac{\sqrt{7}}{3})^2}{(\frac{4}{\sqrt{7}})^2 -1}}$ $=$ $\sqrt{\frac{\frac{16}{9} - \frac{7}{9}}{\frac{16}{7} -1} }$ $=$ $\sqrt{\frac{7}{9}}$ $=$ $\frac{\sqrt{7}}{3}$ $\\$ Question 19: If $8 \tan A = 15$, find $\sin A - \cos A$ Given $8 \tan A = 15 \Rightarrow \tan A =$ $\frac{15}{8}$ By Pythagoras theorem, we get $AC = \sqrt{15^2 + 8^2} = \sqrt{289} = 17$ Therefore $\sin A =$ $\frac{15}{17}$ and $\cos A =$ $\frac{8}{17}$ Hence $\sin A - \cos A =$ $\frac{15}{17}$ $-$ $\frac{5}{17}$ $=$ $\frac{7}{17}$ $\\$ Question 20: If $3 \cos A - 4 \sin A = 2 \cos A + \sin A$, find $\tan A$ Given $3 \cos A - 4 \sin A = 2 \cos A + \sin A$, find $\tan A$ $\Rightarrow \cos A = 5 \sin A$ $\Rightarrow$ $\frac{\sin A}{\cos A}$ $=$ $\frac{1}{5}$ $\Rightarrow \tan A =$ $\frac{1}{5}$ $\\$ Question 21: If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$, then show that $\angle A = \angle B$ $\cos A =$ $\frac{AC}{AB}$ $\cos B =$ $\frac{BC}{AB}$ Since $\cos A = \cos B$ $\Rightarrow \frac{AC}{AB}$ $=$ $\frac{BC}{AB}$ $\Rightarrow AC = BC$ $\therefore \angle A = \angle B$ (angles opposite equal sides of a triangle are equal) $\\$ Question 22: In a triangle, $\angle A$ and $\angle B$ are acute angles such that $\tan A = \tan B$, show that $\angle A = \angle B$ $\tan A =$ $\frac{BC}{AC}$ and $\tan B =$ $\frac{AC}{BC}$ Given that $\tan A = \tan B$ $\Rightarrow \frac{BC}{AC}$ $=$ $\frac{AC}{BC}$ $\Rightarrow BC^2 = AC^2$ $\Rightarrow BC = AC$ $\therefore \angle A = \angle B$ (angles opposite equal sides of a triangle are equal) $\\$ Question 23: If $A$ is an acute angle such that $3 \sin A = 4 \cos A$, find the value of $4 \sin^2 A - 3 \cos^2 A + 2$ Given $3 \sin A = 4 \cos A$ $\Rightarrow \tan A =$ $\frac{4}{3}$ Therefore $AC = \sqrt{4^2+3^2} = 5$ By Pythagoras theorem, $\sin A =$ $\frac{4}{5}$ and $\cos A =$ $\frac{3}{5}$ Hence, $4 \sin^2 A - 3 \cos^2 A + 2$ $=$ $4 ($ $\frac{4}{5}$ $)^2 - 3 ($ $\frac{3}{5}$ $)^2 + 2$ $=$ $\frac{64}{25}$ $-$ $\frac{18}{25}$ $+ 2$ $= \frac{64-18+50}{25}$ $=$ $\frac{96}{25}$ $\\$ Question 24: In adjoining figure, $\triangle ABC$ is right angles at $B$ and $D$ is the midpoint of $BC$. $AC = 5 \ cm$ , $BC = 4 \ cm$ and $\angle BAD = \theta$. Find i) $\tan \theta$ ii) $\sin \theta$ iii) $\sin^2 \theta + \cos^2 \theta$ By Pythagoras theorem $AB = \sqrt{5^2 - 4^2} = 3$ Therefore i) $\tan \theta =$ $\frac{2}{3}$ ii) By Pythagoras theorem $AD = \sqrt{3^2 + 2^2} = \sqrt{13}$ Therefore $\sin \theta =$ $\frac{2}{\sqrt{13}}$ and $\cos \theta =$ $\frac{3}{\sqrt{13}}$ iii) $\sin^2 \theta + \cos^2 \theta =$ $\frac{4}{13}$ $+$ $\frac{9}{13}$ $= 1$ $\\$
# Stoopid Algebra Tricks This is a list of algebra mistakes made by our students. Hopefully a thorough study of these will allow us to figure out how to prevent more of the same. Perhaps our students will visit from time to time, to learn some tricks of the trades (one of the dangers of focusing on what one shouldn't do!). ## Erors in Canceling • ${\displaystyle {\frac {h^{2}+5h}{h}}=h^{2}+5}$ where the student canceled the h's. • ${\displaystyle {\frac {-2+6h+h^{2}}{h}}={\frac {-2+(6+h)h}{h}}={\frac {-2+(6+h)}{1}}}$ • ${\displaystyle {\frac {n^{2}}{n^{2}+2n+1}}={\frac {1}{2n+1}}}$ ## Disturbutive Property • ${\displaystyle {\frac {1/t-1/2}{t-2}}={\frac {(1/t-1/2)(t-2)}{(t-2)(t-2)}}={\frac {1}{t^{2}-4t+4}}}$ • ${\displaystyle {\frac {(2+h)^{2}+(2+h)-6}{h}}={\frac {(4+4h+h^{2})-12-6h}{h}}}$ • ${\displaystyle {\frac {h^{2}+5h}{h}}={\frac {h(h+4h)}{h}}=h+4}$ • ${\displaystyle {\frac {6}{x+2}}\cdot {\frac {x+2}{x+2}}={\frac {6x+2}{x^{2}+4x+4}}}$ ## Divide (and be conquered) • ${\displaystyle {\frac {1/t-1/2}{t-2}}=(1/t-1/2)(1/t-1/2)}$ • ${\displaystyle {\frac {1/t-1/2}{t-2}}={\frac {2-t}{2t}}\cdot {\frac {t-2}{1}}}$ • ${\displaystyle {\frac {1/t-1/2}{t-2}}={\frac {(1/t-1/2)(t-2)}{(t-2)(t-2)}}={\frac {t-2}{t^{2}-4t+4}}}$ • This one could fit under multiple headings. Can you find all the misteaks? ${\displaystyle {\frac {\frac {2-t}{2t}}{t-2}}={\frac {2-t}{2t}}{\frac {t-2}{1}}={\frac {-t^{2}-4}{t-2}}={\frac {(-t+2)(t-2)}{t-2}}=t+2}$ ## Selective multiplication, aka multifrication • ${\displaystyle x=1+{\frac {1}{t}}\implies xt=1+1}$ ## Skwhere and skwhere route tricks • ${\displaystyle {\sqrt {a^{2}+b^{2}}}\neq a+b}$ (in general!): for example, I had a student claim that ${\displaystyle {\sqrt {(-2\sin(T)+2\sin(2T))^{2}+(2\cos(T)-2\cos(2T))^{2}}}=(-2\sin(T)+2\sin(2T))+(2\cos(T)-2\cos(2T))}$ or ${\displaystyle t^{2}=x+9\implies t={\sqrt {x}}+3}$ • Similarly, ${\displaystyle (a+b)^{2}\neq a^{2}+b^{2}}$ (in general!): for example, I had a student write ${\displaystyle \left(2(-\sin(T)+\sin(2T))\right)^{2}=4\left(\sin ^{2}(T)+\sin ^{2}(2T)\right)}$ ${\displaystyle {\frac {1}{{\sqrt {x}}+9}}={\frac {1}{\sqrt {81+9}}}}$ ${\displaystyle \left({\frac {{\sqrt {x}}-9}{x-81}}\right)^{2}={\frac {{\sqrt {x}}+81}{x^{2}+6561}}}$
# Geometry View: Sorted by: ### Properties of Rhombuses, Rectangles, and Squares The three special parallelograms — rhombus, rectangle, and square — are so-called because they’re special cases of the parallelogram. (In addition, the square is a special case or type of both the rectangle ### The Properties of a Kite A kite is a quadrilateral in which two disjoint pairs of consecutive sides are congruent (“disjoint pairs” means that one side can’t be used in both pairs). Check out the kite in the below figure. ### How to Prove that a Quadrilateral Is a Parallelogram There are five ways in which you can prove that a quadrilateral is a parallelogram. The first four are the converses of parallelogram properties (including the definition of a parallelogram). Make sure ### How to Prove that a Quadrilateral Is a Rectangle There are three ways to prove that a quadrilateral is a rectangle. Note that the second and third methods require that you first show (or be given) that the quadrilateral in question is a parallelogram ### How to Prove that a Quadrilateral Is a Rhombus You can use the following six methods to prove that a quadrilateral is a rhombus. The last three methods in this list require that you first show (or be given) that the quadrilateral in question is a parallelogram ### How to Prove that a Quadrilateral Is a Square There are four methods that you can use to prove that a quadrilateral is a square. In the last three of these methods, you first have to prove (or be given) that the quadrilateral is a rectangle, rhombus ### How to Calculate the Area of a Parallelogram, Kite, or Trapezoid The area formulas for the parallelogram, kite, and trapezoid are based on the area of a rectangle. The following figures show you how each of these three quadrilaterals relates to a rectangle, and the ### A Rhombus Area Problem Here’s a rhombus area problem involving triangles and ratios: Find the area of rhombus RHOM given that MBis 6 and that the ratio of RB to BH is 4 : 1, as shown in the following figure. ### How to Calculate the Area of a Kite You calculate the area of a kite by using the lengths of its diagonals. Here’s an example: what’s the area of kite KITE in the following figure? ### How to Calculate the Area of a Trapezoid You can use the right-triangle trick to find the area of a trapezoid. The following trapezoid TRAP looks like an isosceles trapezoid, doesn’t it? Don’t forget — looks can be deceiving. ### The Properties of Trapezoids and Isosceles Trapezoids A trapezoid is a quadrilateral with exactly one pair of parallel sides (the parallel sides are called bases). The following figure shows a trapezoid to the left, and an isosceles trapezoid on the right ### How to Calculate the Area of a Regular Octagon You can calculate the area of a regular octagon with the standard regular polygon method, but there’s a nifty alternative method based on the fact that a regular octagon is a square with its four corners ### Interior and Exterior Angles of a Polygon Everything you need to know about a polygon doesn’t necessarily fall within its sides. You may need to find exterior angles as well as interior angles when working with polygons: ### How to Find the Number of Diagonals in a Polygon To find the number of diagonals in a polygon with nsides, use the following formula: ### How to Identify and Name Similar Polygons You can identify similar polygons by comparing their corresponding angles and sides. As you see in the following figure, quadrilateral WXYZ is the same shape as quadrilateral ### How to Align Similar Polygons If you get a problem with a diagram of similar polygons that aren’t lined up in the same orientation, consider redrawing one of them so that they’re both aligned in the same way. This may make the problem ### Determining a Triangle's Area from Its Base and Height If you know the base and height of a triangle, you can use a tried-and-true formula to find its area. You likely first ran into the basic triangle area formula in about sixth or seventh grade. If you’ve ### Determining a Triangle's Area from Its Three Sides When you know the length of a triangle’s three sides and you don’t know an altitude, you can use Hero’s formula to find the area. Check it out: ### Determining the Area of an Equilateral Triangle To calculate a triangle’s area — for most types of triangles — you need to know the triangle’s base and height. However, with an equilateral triangle, all you need to know is the length of one of its sides ### How to Find the Centroid of a Triangle The three medians of a triangle intersect at its centroid. The centroid is the triangle’s balance point, or center of gravity. (In other words, if you made the triangle out of cardboard, and put its centroid ### How to Do a Parallelogram Proof A good way to begin a proof is to think through a game plan that summarizes your basic argument or chain of logic. The following examples of parallelogram proofs show game plans followed by the resulting ### How to Prove that a Quadrilateral Is a Kite Proving that a quadrilateral is a kite is a piece of cake. Usually, all you have to do is use congruent triangles or isosceles triangles. Here are the two methods: ### How to Calculate the Area of a Quadrilateral There are five formulas that you can use to calculate the area of the seven special quadrilaterals. There are only five formulas because some of them do double duty — for example, you can calculate the ### How to Solve a Similar Polygon Problem Recall that similar polygons are polygons whose corresponding angles are congruent and whose corresponding sides are proportional. The figure below shows similar pentagons, ### How to Prove Triangles Similar Using the AA Theorem You can use the AA (Angle-Angle) method to prove that triangles are similar. The AA theorem states that if two angles of one triangle are congruent to two angles of another triangle, then the triangles Listings:1-2526-5051-7576-100more...
###### Alissa Fong MA, Stanford University Teaching in the San Francisco Bay Area Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts #### Next video playing in 10 Solving and Graphing Inequalities using Addition or Subtraction - Concept # Linear Inequalities - Concept Alissa Fong ###### Alissa Fong MA, Stanford University Teaching in the San Francisco Bay Area Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts Share If we want to solve for or describe a region in a coordinate plane, we can use linear inequalities. Linear inequalities give us a set of solutions as opposed to just one solution. You can solve linear inequalities using similar methods as in solving multi-step equations, except that there are extra rules when using multiplication and division. We graph linear inequalities by shading regions of number lines or coordinate planes. When you're studying Math a lot of the times you're looking at two things that are equal to each other like in an equation. But sometimes you're looking at things that are not equal and so we use the word inequality to describe those and inequalities have a whole separate type of writing stuff which is called notation, there is a whole separate kind of notation and there is a whole another way of graphing it. So it's really important when you guys are starting to study inequalities that you keep a good list of vocabulary with you also keep a good list of notation which is how you write the stuff and then also probably keep some examples handy if you can. So like here's what I'm talking about, we know the number 9 is greater than the number 3 so to write it using inequality notation I write it like this. 9>3 and that's how you read it. You read from left to right just how you read like words, move from left to right. This sign > means greater than. I could also write some number is less than another number like I could say -5 < -2, because a negative 5, has a smaller value than negative 2. It's kind of tricky right because 5 would be a bigger number than 2, but when you negativize them becomes -5 < -2. Here is where it starts to get a little bit confusing, things could also be notated like this, I could say 9 is greater than 3 or I could say 9 is greater than or equal to 3. When I add that little dashy on the bottom what that means is greater than or equal to, or equal to. So that's confusing because either one of those symbols would have worked. Same thing instead of my -5 < -2 I could also write -5 is less than or equal to -2 or equal to. Just to make things even more confusing there is one more sigh I'm going to show you, and it looks like this, I'm going to put an equal sign and then put a slash through it, it's an equal sign with a slash through it. What that means is does not equal -5 does not equal -2, that's something else that is a little less common and you probably won't work with it as much in your Algebra class but you'll see it a lot in the future. So those are some notation again notation means how to write things, those are some notations you're going to want to keep in mind. Another thing that becomes really important when you're doing inequalities is graphing them. Now only do you have to be able to write them and use the correct symbols, but you also have to be able to graph them and you'll see a lot of this as you go through your studies. The first thing to keep in mind when graphing is how many letters you have. Do you have like just one letter like x or only m or only t or whatever. If you have only one letter it's going to be graphed on a numbered line and it's going to be using open circles and closed circles. So if I have just one letter and I'm graphing on a number line, these guys both receive open circles on the number line. These symbols would receive closed circles, and it will make more sense when you see some examples but what that means is that, this number is a solution that's why it's colored in. This number would not be a solution it's like a whole, and that will make a lot more sense when you start seeing some specific pictures. When you're graphing with two variables, usually using the letters x and y you're going to be graphing on one of these Cartesian Coordinate planes and there are a couple of things to keep in mind when you get going with those. The things to keep in mind are not only is it a line that is vertical or horizontal in terms of the slope it also might be a diagonal line using your y equals mx+b techniques you have to keep that in mind. You have to keep in mind whether the line is dashed or solid and the way you tell is that these guys get a dashed line in two variables, these guys get a solid line in 2 variables. And then the last thing to keep in mind anytime you're graphing an inequality in the xy plane, in the Cartesian Coordinate plane when you have 2 letters x and y you have to do some shading. And that's something we're going to be practicing in your future Math classes but it's really important that before you jump into this you have a whole bunch of really good notes you can keep handy as you go through your study of inequalities. Before I let you go there's one last thing I wanted to tell you about and that is when you come to word problems, there are some tricky phrases you want to watch out for. One thing you might see is does not exceed think about what that means if something does not exceed 10 that means like it has to be less than 10 and you're going to be using one of these signs. Or if something does not equal, you're going to be using one of these guys. If something is no more than 5, you're going to have to decide is it going to be smaller than 5 or is it going to be smaller than 5 with a closed circle, smaller than or equal to 5. Those are some really tricky definitions and really tricky phrases you're going to have to watch for in the word problems and you'll probably deal with them on a case by case basis. But again you guys write this down somewhere like on a note card or something that you can carry with you to your Math class. This is going to be like your cheat sheet kind of, this is all the tricks that you're going to want to remember when you just go through your study of inequalities.
# 4.5.1: Fluid Forces on Straight Surfaces A motivation is needed before going through the routine of derivations. Initially, a simple case will be examined. Later, how the calculations can be simplified will be shown. Example 4.14 Consider a rectangular shape gate as shown in Figure 4.20. Calculate the minimum forces, $$F_1$$ and $$F_2$$ to maintain the gate in position. Assuming that the atmospheric pressure can be ignored. Solution 4.14 Fig. 4.20 Rectangular area under pressure. The forces can be calculated by looking at the moment around point "O.'' The element of moment is $$a\,d\xi$$ for the width of the gate and is $dM = \overbrace{P\, \underbrace{a\, d\xi}_{dA}}^{dF} (ll + \xi)$ The pressure, $$P$$ can be expressed as a function $$\xi$$ as the following \begin{align} P = g\, \rho \, (ll +\xi) sin \beta \end{align} The liquid total moment on the gate is \begin{align} M = \int_0^b g\, \rho \, (ll +\xi) \sin \beta\, a\, d\xi (ll + \xi) \end{align} The integral can be simplified as \begin{align} M = g\, a\,\rho\,\sin\beta\,\int_0^b (ll +\xi)^2 d\xi \end{align} The solution of the above integral is \begin{align} M = g\, \rho\,a\,\sin\beta\,\left( \dfrac{3\,b\,{l}^{2}+3\,{b}^{2}\,l+{b}^{3}}{3} \right) \end{align} This value provides the moment that $$F_1$$ and $$F_2$$ should extract. Additional equation is needed. It is the total force, which is \begin{align} F_{total} = \int_0^b g\, \rho \, (ll +\xi) \sin \beta\, a\, d\xi \end{align} The total force integration provides \begin{align} F_{total} = g\,\rho\,a\,\sin \beta\, \int_0^b (ll +\xi) d\xi = g\,\rho\,a\,\sin \beta\,\left(\dfrac{2\,b\,ll+{b}^{2}}{2}\right) \end{align} The forces on the gate have to provide \begin{align} F_1 + F_2 = g\,\rho\,a\,\sin \beta\,\left(\dfrac{2\,b\,ll+{b}^{2}}{2}\right) \end{align} Additionally, the moment of forces around point "O'' is \begin{align} F_1\,ll + F_2 (ll + b) = g\, \rho\,a\,\sin\beta\,\left( \dfrac{3\,b\,{l}^{2}+3\,{b}^{2}\,l+{b}^{3}}{3} \right) \end{align} The solution of these equations is \begin{align} F_1 = \dfrac{\left( 3\,ll+b\right) \,a\,b\,g\,\rho\,\sin\beta}{6} \end{align} \begin{align} F_2 = \dfrac{\left( 3\,ll+2\,b\right) \,a\,b\,g\,\rho\,\sin\beta}{6} \end{align} Fig. 4.21. Schematic of submerged area to explain the center forces and moments. The above calculations are time consuming and engineers always try to make life simpler. Looking at the above calculations, it can be observed that there is a moment of area in equation __?__ and also a center of area. These concepts have been introduced in Chapter 3. Several represented areas for which moment of inertia and center of area have been tabulated in Chapter 3. These tabulated values can be used to solve this kind of problems. #### Symmetrical Shapes Consider the two–dimensional symmetrical area that are under pressure as shown in Figure 4.21. They symmetry is around any axes parallel to axis $$x$$. The total force and moment that the liquid extracting on the area need to be calculated. First, the force is $F = \int_{A} PdA = \int \left(P_{atmos} + \rho g h \right) dA = AP_{atmos} + \rho g \int_{ll_{0}}^{ll_{1}} \left(\xi + ll_{0}\right)sin\beta dA \tag{139}$ In this case, the atmospheric pressure can include any additional liquid layer above layer touching'' area. The atmospheric'' pressure can be set to zero. The boundaries of the integral of equation (139) refer to starting point and ending points not to the start area and end area. The integral in equation 139 can be further developed as $F_{total} = AP_{atmos} + \rho g sin\beta \left(ll_{0}A + \int_{ll_{0}}^{ll_{1}} \xi dA\right) \tag{140}$ In a final form as Total Force in Inclined Surface $F_{total} = A\left[P_{atmos} + \rho g sin\beta \left(ll_{0} + x_{c}\right)\right] \tag{141}$ Fig. 4.22. The general forces acting on submerged area. The moment of the liquid on the area around point O'' is $M_{y} = \int_{\xi_{0}}^{\xi_{1}} P\left(\xi\right)\xi dA \tag{142}$ $M_{y} = \int_{\xi_{0}}^{\xi_{1}} \left(P_{atmos} + g\rho h \left(\xi\right)\right)\xi dA \tag{143}$ Or separating the parts as $M_{y} = P_{atmos} \int_{\xi_{0}}^{\xi_{1}}\xi dA + g\rho sin\beta \int_{xi_{0}}^{xi_{1}} \xi^{2}dA \tag{144}$ The moment of inertia, $$I_{x'x'}$$, is about the axis through point O'' into the page. Equation 144 can be written in more compact form as Total Moment in Inclined Surface $M_{y} = P_{atmos}x_{c}A + g\rho sin \beta I_{x'x'} \tag{145}$ Example 4.14 can be generalised to solve any two forces needed to balance the area/gate. Consider the general symmetrical body shown in figure 4.22 which has two forces that balance the body. Equations 141 and 145 can be combined the moment and force acting on the general area. If the atmospheric pressure'' can be zero or include additional layer of liquid. The forces balance reads $F_{1} + F_{2} = A\left[P_{atmos} + \rho g sin\beta \left(ll_{0} + x_{c}\right)\right] \tag{146}$ and moment balance reads $F_{1}a + F_{2}b = P_{atmos}x_{c}A + g\rho sin\beta I_{x'x'} \tag{147}$ The solution of these equations is $F_{1} = \frac{\left[\left(\rho sin\beta - \frac{P_{atmos}}{gb}\right)x_{c}+ll_{0}\rho sin\beta + \frac{P_{atmos}}{g}\right]bA}{g\left(b-a\right)} - \frac{I_{x'x'}\rho sin \beta}{g\left(b-a\right)} \tag{148}$ and $F_{2} = \frac{I_{x'x'}\rho sin \beta}{g\left(b-a\right)} - \frac{\left[\left(\rho sin\beta - \frac{P_{atmos}}{ga}\right)x_{c}+ll_{0}\rho sin\beta + \frac{P_{atmos}}{g}\right]aA}{g\left(b-a\right)} \tag{149}$ In the solution, the forces can be negative or positive, and the distance $$a$$ or $$b$$ can be positive or negative. Additionally, the atmospheric pressure can contain either an additional liquid layer above the touching'' area or even atmospheric pressure simply can be set up to zero. In symmetrical area only two forces are required since the moment is one dimensional. However, in non–symmetrical area there are two different moments and therefor three forces are required. Thus, additional equation is required. This equation is for the additional moment around the $$x$$ axis (see for explanation in Figure 4.23). The moment around the $$y$$ axis is given by equation 145 and the total force is given by 141. The moment around the $$x$$ axis (which was arbitrarily chosen) should be $M_{x} = \int_{A} y PdA \tag{150}$ Substituting the components for the pressure transforms equation 150 into $M_{x} = \int_{A} y \left(P_{atmos} + \rho g \xi sin \beta \right) dA \tag{151}$ The integral in equation 150 can be written as $M_{x} = P_{atmos} \int_{A} y dA + \rho g \xi sin \beta \int_{A} \xi y dA \tag{152}$ The compact form can be written as Moment in Inclined Surface $M_{x} = P_{atmos}Ay_{c} + \rho g sin\beta I_{x'y'} \tag{153}$ Fig. 4.23. The general forces acting on non symmetrical straight area. The product of inertia was presented in Chapter 3. These equations (141), (145), and (153) provide the base for solving any problem for straight area under pressure with uniform density. There are many combinations of problems (e.g. two forces and moment) but no general solution is provided. Example to illustrate the use of these equations is provided. Example 4.15 Calculate the forces which required to balance the triangular shape shown in the Figure 4.24. Solution 4.15 The three equations that needs to be solved are $F_1+F_2+F_3 = F_{total} \label{static:eq:triangleF1} \tag{154}$ The moment around $$x$$ axis is $F_1\,b = M_y \label{static:eq:triangleMy} \tag{155}$ The moment around $$y$$ axis is $F_1\,ll_1 +F_2\,(a +ll_0 ) +F_3\,ll_0 = M_x \label{static:eq:triangleMx} \tag{156}$ The right hand side of these equations are given before in equations (141), (145) and (153). The moment of inertia of the triangle around $$x$$ is made of two triangles (as shown in the Figure 4.24 for triangle 1 and 2). Triangle 1 can be calculated as the moment of inertia around its center which is $$ll_0+ 2(ll_1-ll_0)/3$$. The height of triangle 1 is $$(ll_1−ll_0)$$ and its width $$b$$ and thus, moment of inertia about its center is $$I_{xx}=b(ll_1−ll_0)^{3}/36$$. The moment of inertia for triangle 1 about $$y$$ is ${I_{xx}}_1 = \dfrac{b (ll_1 - ll_0)^3}{36} + \overbrace{\dfrac{b (ll_1 - ll_0)}{3}}^{A_1} \, \overbrace{\left(ll_0+ \dfrac{2(ll_1-ll_0)}{3}\right)^2} ^{{\Delta x_1}^2} \tag{157}$ The height of the triangle 2 is $$a - (ll_1 - ll_0)$$ and its width $$b$$ and thus, the moment of inertia about its center is ${I_{xx}}_2 = \dfrac{b [a - (ll_1 - ll_0)]^3}{36} + \overbrace{\dfrac{b [a - (ll_1 - ll_0)]}{3}}^{A_2} \, \overbrace{\left(ll_1 + \dfrac{[a - (ll_1 - ll_0)]}{3} \right)^2}^{{\Delta x_2}^2} \tag{158}$ Fig. 4.24 The general forces acting on a non symmetrical straight area. and the total moment of inertia $I_{xx} = {I_{xx}}_1 + {I_{xx}}_2$ The product of inertia of the triangle can be obtain by integration. It can be noticed that upper line of the triangle is $$y = \dfrac{\left(ll_1 - ll_0\right)x}{b} +ll_0$$. The lower line of the triangle is $$y = \dfrac{\left(ll_1 - ll_0 - a \right)x}{b} +ll_0 + a$$. \begin{align} I_{xy} = \int_0^b \left[\int_{\dfrac{\left(ll_1 - ll_0\right)x}{b} +ll_0}^{\dfrac{\left(ll_1 - ll_0 - a \right)x}{b} +ll_0 + a} x\,y\, dx \right] dy =\dfrac {2\,a\,b^2\,ll_1+2\,a\,b^2\,ll_0+a^2\,b^2} {24} \end{align} The solution of this set equations is \begin{align} F_1=\overbrace{\left[\dfrac{a\,b}{3}\right]}^{A} \dfrac{\left( g\,\left( 6\,ll_1+3\,a\right) +6\,g\,ll_0\right) \,\rho\,\sin\beta+8\,P_{atmos}}{24}, \end{align} \begin{multline} \dfrac{F_2} {\left[\dfrac{a\,b}{3}\right]} =- \begin{array}{c} \dfrac{\left( \left( 3\,ll_1-14\,a\right) - ll_0\,\left( \dfrac{12\,ll_1}{a}-27\right) + \dfrac{12\,{ll_0}^{2}}{a}\right)\,g \,\rho\,\sin\beta} {72} - \\ \dfrac{\left( \left( \dfrac{24\,ll_1}{a}-24\right) +\dfrac{48\,ll_0}{a}\right) \,P_{atmos}} {72}, \end{array} \end{multline} \begin{multline} \dfrac{F_3} {\left[\dfrac{a\,b}{3}\right]} = \begin{array}{c} \dfrac{\left( \left( a-\dfrac{15\,ll_1}{a}\right) +ll_0\,\left( 27- \dfrac{12\,ll_1}{a}\right) +\dfrac{12\,{ll_0}^{2}}{a}\right)\,g\,\rho\,\sin\beta } {72}\\ + \dfrac{\left( \left(\dfrac{24\,ll_1}{a}+24\right) +\dfrac{48\,ll_0}{a}\right) \,P_{atmos} }{72} \end{array} \end{multline*} ### Contributors • Dr. Genick Bar-Meir. 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# Difference between revisions of "2002 AMC 10B Problems/Problem 19" ## Problem Suppose that $\{a_n\}$ is an arithmetic sequence with $$a_1+a_2+\cdots+a_{100}=100 \text{ and } a_{101}+a_{102}+\cdots+a_{200}=200.$$ What is the value of $a_2 - a_1 ?$ $\mathrm{(A) \ } 0.0001\qquad \mathrm{(B) \ } 0.001\qquad \mathrm{(C) \ } 0.01\qquad \mathrm{(D) \ } 0.1\qquad \mathrm{(E) \ } 1$ ## Solution 1 We should realize that the two equations are 100 terms apart, so by subtracting the two equations in a form like... $(a_{101} - a_1) + (a_{102} - a_2) +... + (a_{200} - a_{100}) = 200-100 = 100$ ...we get the value of the common difference of every hundred terms hundred times. So we have to divide the answer by hundred to get ... $\frac{100}{100} = 1$ ...the common difference of every hundred terms. Then we have to simply divide the answer by hundred again to find the common difference between one term, therefore... $\frac{1}{100} =\boxed{(\text{C})0.01}$ ## Solution 2 Adding the two given equations together gives $a_1+a_2+...+a_{200}=300$. Now, let the common difference be $d$. Notice that $a_2-a_1=d$, so we merely need to find $d$ to get the answer. The formula for an arithmetic sum is $\frac{n}{2}(2a_1+d(n-1))$, where $a_1$ is the first term, $n$ is the number of terms, and $d$ is the common difference. Now we use this formula to find a closed form for the first given equation and the sum of the given equations. For the first equation, we have $n=100$. Therefore, we have $50(2a_1+99d)=100$, or $2a_1+99d=2$. (1) For the sum of the equations (shown at the beginning of the solution) we have $n=200$, so $100(2a_1+199d)=300$ or $2a_1+199d=3$ (2) Now we have a system of equations in terms of $a_1$ and $d$. Subtracting (1) from (2) eliminates $a_1$, yielding $100d=1$, and $d=a_2-a_1=\frac{1}{100}=\boxed{(\text{C}) .01}$. ## Solution 3 Subtracting the 2 given equations yields $(a_{101}-a_1)+(a_{102}-a_2)+(a_{103}-a_3)+...+(a_{200}-a_{100})=100$ Now express each $a_n$ in terms of first term $a_1$ and common difference $x$ between consecutive terms $((a_1+100x)-(a_1))+((a_1+101x)-(a_1+x))+((a_1+102x)-(a_1+2x))+...+((a_1+199x)-(a_1+99x))=100$ Simplifying and canceling $a_1$ and $x$ terms gives $100x+100x+100x+...+100x=100$ $100x\times100=100$ $100x=1$ $x=0.01=\boxed{(\text{C})0.01}$ 2002 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
# What Is Slope-Intercept Form ## The Definition, Formula, and Problem Example of the Slope-Intercept Form What Is Slope-Intercept Form – Among the many forms used to illustrate a linear equation one that is frequently found is the slope intercept form. You can use the formula for the slope-intercept in order to determine a line equation, assuming that you have the straight line’s slope , and the y-intercept, which is the coordinate of the point’s y-axis where the y-axis meets the line. Find out more information about this particular line equation form below. ## What Is The Slope Intercept Form? There are three main forms of linear equations: the standard one, the slope-intercept one, and the point-slope. Although they may not yield the same results , when used but you are able to extract the information line produced more quickly by using this slope-intercept form. As the name implies, this form uses the sloped line and its “steepness” of the line determines its significance. This formula is able to find the slope of a straight line. It is also known as y-intercept, or x-intercept, where you can apply different available formulas. The line equation of this particular formula is y = mx + b. The straight line’s slope is represented with “m”, while its y-intercept is indicated through “b”. Every point on the straight line is represented with an (x, y). Note that in the y = mx + b equation formula, the “x” and the “y” have to remain as variables. ## An Example of Applied Slope Intercept Form in Problems When it comes to the actual world In the real world, the “slope intercept” form is often utilized to show how an item or issue changes over its course. The value provided by the vertical axis represents how the equation handles the magnitude of changes in the amount of time indicated with the horizontal line (typically in the form of time). An easy example of the use of this formula is to figure out how many people live in a specific area in the course of time. Based on the assumption that the population of the area increases each year by a specific fixed amount, the point worth of horizontal scale will increase by one point for every passing year, and the value of the vertical axis is increased in proportion to the population growth according to the fixed amount. You may also notice the beginning value of a particular problem. The beginning value is located at the y-value in the y-intercept. The Y-intercept represents the point where x is zero. In the case of a previous problem, the starting value would be at the time the population reading starts or when the time tracking begins along with the changes that follow. So, the y-intercept is the point in the population where the population starts to be tracked to the researchers. Let’s suppose that the researcher is beginning to perform the calculation or take measurements in 1995. The year 1995 would become”the “base” year, and the x 0 points would be in 1995. This means that the population of 1995 is the y-intercept. Linear equations that use straight-line formulas can be solved in this manner. The initial value is represented by the y-intercept, and the change rate is expressed by the slope. The principal issue with the slope-intercept form typically lies in the horizontal interpretation of the variable in particular when the variable is attributed to a specific year (or any other kind or unit). The trick to overcoming them is to ensure that you are aware of the variables’ meanings in detail.
# How do you find the length of each side of a square when given the area? ## How do you find the length of each side of a square when given the area? Correct answer: The area of any quadrilateral can be determined by multiplying the length of its base by its height. Since we know the shape here is square, we know that all sides are of equal length. From this we can work backwards by taking the square root of the area to find the length of one side. What is the length of the side of the square with an area of 25 square feet? If the area is 25, then the length of one side will be the square root of 25. The square root of 25 is 5, so each side must be 5 feet long. How do you find the area of a square in square inches? Simply multiply your measurements for length and width to determine the area of your square or rectangular area in square inches. For example, let’s say that, for a rectangular area, you measure a length of 4 inches and a width of 3 inches. In this case, the area within your rectangle is 4 × 3 = 12 square inches. ### What is each side of a square? Opposite sides of a square are both parallel and equal in length. All four angles of a square are equal (each being 360°/4 = 90°, a right angle). All four sides of a square are equal. The diagonals of a square are equal. What is the squared of 25? 5 The square root of 25 is 5. It is the positive solution of the equation x2 = 25. The number 25 is a perfect square….Square Root of 25 in radical form: √25. 1. What Is the Square Root of 25? 2. Is Square Root of 25 Rational or Irrational? 3. How to Find the Square Root of 25? How to calculate the area of a square? Learning to calculate the area of a square is thus a precursor for learning how to calculate the areas of more advanced shapes. The only measurement needed to find the area of a square figure is its side. Since all sides are equal it does not matter which side is measured. Then simply multiply the measurement by itself to get the area. #### Which is bigger a square inch or a square foot? There are 0.0069444444 square foot in a square inch. 1 Square Inch is equal to 0.0069444444 Square Foot. 1 in² = 0.0069444444 ft². What are the properties of a square shape? Therefore, a square combines the properties of all of these shapes: diagonals bisect at 90°, diagonals bisect the square angles, diagonals are equal, the sides are equal, opposite sides are equal, all angles are equal (90°). Why is the square the most common shape in geometry? The simplicity of the square is why it is usually one of the first shapes that geometry students become familiar with. In real life measurements, like in construction, engineering, landscaping, etc. we rarely deal with square areas and surfaces – they are more often rectangular in shape. Begin typing your search term above and press enter to search. Press ESC to cancel.
# Absolute Value Transformations Note: For Parent Functions and general transformations, see the Parent Graphs and Transformations section. This section covers: Absolute Value Transformations can be tricky, since we have two different types of problems: 1. Transformations of Absolute Value Functions 2. Performing Absolute Value Transformations on other functions # Transformations of the Absolute Value Parent Function Let’s first work with transformations on the absolute value parent function. Since the vertex (the “point”) of an absolute value parent function $$y=\left\| x \right\|$$ is $$\left( {0,\,0} \right)$$, an absolute value equation with new vertex $$\left( {h,\,k} \right)$$ is $$\displaystyle f\left( x \right)=a\left\| {\frac{1}{b}\left( {x-h} \right)} \right\|+k$$, where $$a$$ is the vertical stretch, $$b$$ is the horizontal stretch, $$h$$ is the horizontal shift to the right, and $$k$$ is the vertical shift upwards. If $$a$$ is negative, the graph points up instead of down. Here is an example with a t-chart: Transformation T-chart Graph $$\displaystyle \begin{array}{l}y=-3\left\| {2x+4} \right\|+1\\y=-3\left\| {2(x+2)} \right\|+1\end{array}$$ (have to take out a 2 to make $$x$$ by itself) Parent function: $$\displaystyle y=\left\| x \right\|$$ $$\frac{1}{2}x-2$$ x y –3y + 1 –3 –2 2 –5 –2.5 –1 1 –2 –2 0 0 1 –1.5 1 1 –2 –1 2 2 –5 Domain: $$\left( {-\infty ,\infty } \right)$$ Range: $$\left( {-\infty ,1} \right]$$ Note that we could graph this without t-charts by plotting the vertex, flipping the parent absolute value graph, and then going over (and back) 1 and down 6 for next points down, since the “slope” is 6 (3 times 2). Here’s an example of writing an absolute value function from a graph: < Graph Getting Equation We see that this is an absolute value graph (parent graph $$y=\left\| x \right\|$$) since it is “pointy”, but flipped around $$x$$-axis, since it is facing down. We are taking the absolute value of the whole function, since it “bounces” up from the $$x$$ axis (only positive $$y$$ values). This is weird, but it’s an absolute value of an absolute value function! Therefore, the equation will be in the form $$y=\left\| {a\left\| {x-h} \right\|+k} \right\|$$ with vertex $$\left( {h,\,\,k} \right)$$, and $$a$$ should be negative. Since the vertex of the graph is $$\left( {-1,\,\,10} \right)$$, one equation of the graph could be $$y=\left\| {a\left\| {x+1} \right\|+10} \right\|$$. We need to find $$a$$; use the point $$\left( {4,\,0} \right)$$: \displaystyle \begin{align}y&=\left\| {a\left\| {x+1} \right\|+10} \right\|\\0&=\left\| {a\left\| {4+1} \right\|+10} \right\|\\0&=\left\| {a\left\| 5 \right\|+10} \right\|\\0&=5a+10,\,\,\text{since}\,\,\left\| 0 \right\|\text{ =0}\\-5a&=10;\,\,\,\,\,\,a=-2\end{align} $$\begin{array}{c}\text{The equation of the graph then is:}\\y=\left\| {-2\left\| {x+1} \right\|+10} \right\|\end{array}$$ (We could have also found $$a$$ by noticing that the graph goes over/back 1 and down 2), so it’s “slope” is –2. Be sure to check your answer by graphing or plugging in more points! √ # Absolute Value Transformations of other Parent Functions Now let’s look at taking the absolute value of functions, both on the outside (affecting the $$y$$’s) and the inside (affecting the $$x$$’s). These are a little trickier. < Let’s look at a function of points, and see what happens when we take the absolute value of the function “on the outside” and then “on the inside”. Then we’ll show absolute value transformations using parent functions. Note that with the absolute value on the outside (affecting the $$\boldsymbol{y}$$’s), we just take all negative $$\boldsymbol{y}$$ values and make them positive, and with absolute value on the inside (affecting the $$\boldsymbol{x}$$’s), we take all the 1st and 4th quadrant points and reflect them over the $$\boldsymbol{y}$$-axis, so that the new graph is symmetric to the $$\boldsymbol{y}$$-axis. < Transformation T-chart Graph Original Function (Points) x y –6 –4 –4 0 –3 2 0 0 3 –2 4 0 6 1 $$y=\left\| {f\left( x \right)} \right\|$$ Replace all negative $$y$$ values with their absolute value (make them positive). Make sure that all (negative $$y$$) points on the graph are reflected across the $$x$$-axis to be positive. Example Function: $$y=\left\| {{{x}^{3}}+4} \right\|$$ x y \|y\| –6 –4 4 –4 0 0 –3 2 2 0 0 0 3 –2 2 4 0 0 6 1 1 Let’s try: $$y=\left\| {2f\left( x \right)-4} \right\|$$ Reflect negative $$y$$ values across the $$x$$-axis. x y \|2y – 4\| –6 –4 12 –4 0 4 –3 2 0 0 0 4 3 –2 8 4 0 4 6 1 2 $$y=f\left( {\left\| x \right\|} \right)$$ Make a symmetrical graph from the positive $$x$$’s across the $$y$$ axis. “Throw away” the left-hand side of the graph (negative $$x$$’s), and replace the left side of the graph with the reflection of the right-hand side. OR For any negative $$x$$’s, replace the $$y$$ value with the $$y$$ value corresponding to the positive value (absolute value) of the negative $$x$$’s. For example, when $$x$$ is –6, replace the $$y$$ with a 1, since the $$y$$ value for positive 6 is 1. Example Function: $$y=4{{\left\| x \right\|}^{3}}-2$$ x y New y –6 –4 1 –4 0 0 –3 2 –2 0 0 3 –2 4 0 6 1 Let’s try: $$y=3f\left( {\left\| x \right\|} \right)+2$$ (The absolute value is directly around the $$x$$.) After performing the transformation on the $$y$$, for any negative $$x$$’s, replace the $$y$$ value with the $$y$$ value corresponding to the positive value (absolute value) of the negative $$x$$’s, For example, when $$x$$ is –6, replace the $$y$$ with a 5, since the $$y$$ value for positive 6 is 5. x y 3y+2 New y –6 –4 –10 5 –4 0 2 2 –3 2 –7 –4 0 0 2 3 –2 –4 4 0 2 6 1 5 Here’s an example of a mixed absolute value transformation where the inside absolute value is not just around the $$x$$; you can see that this can get complicated. It looks like when we have absolute values on the inside (affecting the $$x$$’s), we do those first: Transformation T-chart Graph $$y=\sqrt{{\left\| {2\left( {x+3} \right)} \right\|}}+4$$ With this mixed transformation, we need to perform the inner absolute value first: For any negative $$x$$’s, replace the $$y$$ value with the $$y$$ value corresponding to the positive value (absolute value) of the negative $$x$$’s. (See pink arrows) Then with the new values, we can perform the shift for $$y$$ (add 4) and the shift for $$x$$ (divide by 2 and then subtract 3). The best way to check your work is to put the graph in your calculator and check the table values. Parent function: $$y=\sqrt{x}$$ ($$x$$ must be $$\ge 0$$ for original function, but not for transformed function) Note: The boxed $$y$$ is the $$y$$ value associated with the absolute value of that $$x$$ value. (If the absolute value sign was just around the $$x$$, such as $$y=\sqrt{{2\left( {\left\| x \right\|+3} \right)}}+4$$, we would have replaced the $$y$$ values with those of the positive $$x$$’s after doing the $$x$$ transformation, instead of before. Thus, the graph would be symmetrical to the $$y$$-axis. Tricky!) Here are more absolute value examples with parent functions: Absolute Value Transformations T-chart Graph $$y=\left\| {{{x}^{3}}+4} \right\|$$ Parent function: $$y={{x}^{3}}$$ Reflect all values below the $$y$$-axis to above the $$y$$-axis. x y \|y + 4\| –2 –8 4 –1 –1 3 0 0 4 1 1 5 2 8 12 $$2{{\left\| x \right\|}^{3}}-1$$ Parent function: $$y={{x}^{3}}$$ For the two value of $$x$$ that are negative (–2 and –1), replace the $$y$$’s with the $$y$$ from the absolute value (2 and 1, respectively) for those points. Note that this is like “erasing” the part of the graph to the left of the $$y$$-axis and reflecting the points from the right of the $$y$$-axis over to the left. x y 2y – 1 New y –2 –8 –17 15 –1 –1 –3 1 0 0 –1 1 1 1 2 8 15 $$y={{2}^{{\left\| x \right\|-3}}}$$ Parent function: $$y={{2}^{x}}$$ For the negative $$x$$ value, just use the $$y$$ values of the absolute value of these $$x$$ values! Note that we pick up these new $$y$$ values after we do the translation of the $$x$$ values. Note that this is like “erasing” the part of the graph to the left of the $$y$$-axis and reflecting the points from the right of the $$y$$-axis over to the left. x + 3 x y New y –2 –5 $$\frac{1}{{32}}$$ $$\color{#800000}{{\frac{1}{2}}}$$ –1 –4 $$\frac{1}{{16}}$$ $$\color{blue}{{\frac{1}{4}}}$$ 1 –2 $$\color{blue}{{\frac{1}{4}}}$$ 2 –1 $$\color{#800000}{{\frac{1}{2}}}$$ 3 0 1 4 1 2 5 2 4 $$\displaystyle y=\left\| {\frac{3}{x}+3} \right\|$$ Parent function: $$\displaystyle y=\frac{1}{x}$$ Since the absolute value is on the “outside”, we can just perform the transformations on the $$y$$, doing the absolute value last Reflect all values below the $$y$$-axis to above the $$y$$-axis. x y \|3y + 3\| –3 1/3 2 –2 –1/2 1.5 –1 –1 0 1 1 6 2 1/2 4.5 3 1/3 4 $$y=\left\| {{{{\log }}_{3}}\left( {x+4} \right)} \right\|$$ Parent function: $$y={{\log }_{3}}\left( x \right)$$ (Asymptote: $$x=0$$ ) Reflect all values below the $$y$$-axis to above the $$y$$-axis. x – 4 x y \|y\| $$-3\frac{8}{9}$$ –1/9 –2 2 $$-3\frac{2}{3}$$ –1/3 –1 1 –3 1 0 0 –1 3 1 1 5 9 2 2 New Asymptote: $$x=-4$$ Note: These mixed transformations with absolute value are very tricky; it’s really difficult to know what order to use to perform them. The general rule of thumb is to perform the absolute value first for the absolute values on the inside, and the absolute value last for absolute values on the outside (work from the inside out). The best thing to do is to play around with them on your graphing calculator to see what’s going on. For example, with something like $$y=\left\| {{{2}^{x}}} \right\|-3$$, you perform the $$y$$ absolute value function first (before the shift); with something like $$y=\left\| {{{2}^{x}}-3} \right\|$$, you perform the $$y$$ absolute value last (after the shift). (These two make sense, when you look at where the absolute value functions are.) But we saw that with $$y={{2}^{{\left\| x \right\|-3}}}$$, we performed the $$x$$ absolute value function last (after the shift). I also noticed that with $$y={{2}^{{\left\| {x-3} \right\|}}}$$, you perform the $$x$$ absolute value transformation first (before the shift). I don’t think you’ll get this detailed with your transformations, but you can see how complicated this can get! Here’s an example where we’re using what we know about the absolute value transformation, but we’re using it on an absolute value parent function! Pretty crazy, huh? Transformation T-chart Graph $$y=\left\| {3\left\| {x-1} \right\|-2} \right\|$$ Parent function: $$y=\left\| x \right\|$$ Since we’re using the absolute value parent function, we only have to take the absolute value on the outside ($$y$$). We can do this, since the absolute value on the inside is a linear function (thus we can use the parent function). x + 1 x y \|3y – 2\| –1 –2 2 4 0 –1 1 1 1 0 0 2 2 1 1 1 3 2 2 4 ## More Absolute Value Transformations What about $$\left\| {f\left( {\left\| x \right\|} \right)} \right\|$$? Play around with this in your calculator with $$y=\left\| {{{2}^{{\left\| x \right\|}}}-5} \right\|$$, for example. You’ll see that it shouldn’t matter which absolute value function you apply first, but it certainly doesn’t hurt to work from the inside out. And with $$-\left\| {f\left( {\left\| x \right\|} \right)} \right\|$$, it’s a good idea to perform the inside absolute value first, then the outside, and then the flip across the $$x$$ axis. So the rule of thumb with these absolute value functions and reflections is to move from the inside out. Let’s do more complicated examples with absolute value and flipping – sorry that this stuff is so complicated! Just be careful about the order by trying real functions in your calculator to see what happens. These are for the more advanced Pre-Calculus classes! Original Function Transformation Explanation $$-f\left( {-x} \right)$$ Flip the function around the $$x$$-axis, and then around the $$y$$-axis. It actually doesn’t matter which flip you perform first. $$\left\| {f\left( {-x} \right)} \right\|$$ Flip the function around the $$x$$-axis, and then reflect everything below the $$x$$-axis to make it above the $$x$$-axis; this takes the absolute value (all positive $$y$$ values). We actually could have done this in the other order, and it would have worked! $$\left\| {f\left( {\left\| x \right\|} \right)} \right\|$$ For the absolute value on the inside, throw away the negative $$x$$ values, and replace them with the $$y$$ values for the absolute value of the $$x$$. Then reflect everything below the $$x$$-axis to make it above the $$x$$-axis; this takes the absolute value (all positive $$y$$ values). We could have done this in any order. $$-\left\| {f\left( {\left\| x \right\|} \right)} \right\|$$ Do everything we did in the transformation above, and then flip the function around the $$x$$-axis, because of the negative sign. For this one, I noticed that we needed to do the flip around the $$x$$-axis last (we need to work “inside out”). Learn these rules, and practice, practice, practice! On to Piecewise Functions – you are ready! <
## Engage NY Eureka Math 2nd Grade Module 5 Lesson 20 Answer Key ### Eureka Math Grade 2 Module 5 Lesson 20 Problem Set Answer Key Question 1. 399 + 237 = _________ 399 + 237 = 636. Explanation: In the above-given question, given that, solve the problem using two different strategies. 399 + 237. I solved using the number bond. my friend solved using the arrow way. 3 hundreds = 300. 9 tens = 90. 9 ones = 9. 300 + 90 + 9 = 399. 2 hundred = 200. 3 tens = 30. 7 ones = 7. 200 + 30 + 7 = 237. 399 + 237 = 636. Question 2. 400 – 298 = ________ 400 – 298 = 102. Explanation: In the above-given question, given that, solve the problem using two different strategies. 400 – 298. I solved using the number bond. my friend solved using the place value strategies. 4 hundred = 400. 2 hundred = 200. 9 tens = 90. 8 ones = 8. 200 + 90 + 8 = 298. 400 – 298 = 102. Question 3. 548 + 181 = _________ 548 + 181 = 729. Explanation: In the above-given question, given that, solve the problem using two different strategies. 548 + 181. I solved using the number bond. my friend solved using the place value strategies. 5 hundred = 500. 4 tens = 40. 8 ones = 8. 500 + 40 + 8 = 548. 1 hundred = 100. 8 tens = 80. 1 one = 1. 100 + 80 + 1 = 181. 548 + 181 = 729. Question 4. 360 + ______ = 754 754 – 360 = 394. Explanation: In the above-given question, given that, solve the problem using two different strategies. 394 + 360. I solved using the number bond. my friend solved using the place value strategies. 3 hundred = 300. 6 tens = 60. 300 + 60 = 360. 3 hundred = 300. 9 tens = 90. 4 ones = 4. 300 + 90 + 4 = 394. 394 + 360 = 754. Question 5. 862 – ______ = 690 862 – 172 = 690. Explanation: In the above-given question, given that, solve the problem using two different strategies. 862 – 172. I solved using the number bond. my friend solved using the place value strategies. 8 hundred = 800. 6 tens = 60. 2 ones = 2. 800 + 60 + 2 = 862. 1 hundred = 100. 7 tens = 70. 2 ones = 2. 100 + 70 + 2 = 172. 862 – 172 = 690. ### Eureka Math Grade 2 Module 5 Lesson 20 Exit Ticket Answer Key Solve each problem using two different strategies. Question 1. 299 + 156 = _________ 299 + 156 = 455. Explanation: In the above-given question, given that, solve the problem using two different strategies. 299 + 156. 1st strategy is the number bond. 2nd strategy solved using the place value strategies. 2 hundred = 200. 9 tens = 90. 9 ones = 9. 200 + 90 + 9 = 299. 1 hundred = 100. 5 tens = 50. 6 ones = 6. 100 + 50 + 6 = 156. 299 + 156 = 455. Question 2. 547 + ______ = 841 547 + 294 = 841. Explanation: In the above-given question, given that, solve the problem using two different strategies. 547 + 294. 1st strategy is the number bond. 2nd strategy solved using the place value strategies. 5 hundred = 500. 4 tens = 40. 7 ones = 7. 500 + 40 + 7 = 547. 2 hundred = 200. 9 tens = 90. 4 ones = 4. 200 + 90 + 4 = 294. 547 + 294 = 841. ### Eureka Math Grade 2 Module 5 Lesson 20 Homework Answer Key Question 1. 456 + 244 = _________ 456 + 244 = 700. Explanation: In the above-given question, given that, solve the problem using two different strategies. 456 + 244. 1st strategy is the number bond. 2nd strategy solved using the place value strategies. 4 hundred = 400. 5 tens = 50. 6 ones = 6. 400 + 50 + 6 = 456. 2 hundred = 200. 4 tens = 40. 4 ones = 4. 200 + 40 + 4 = 244. 456 + 244 = 700. Question 2. 698 + ______ = 945 698 + 247 = 945. Explanation: In the above-given question, given that, solve the problem using two different strategies. 698 + 247. 1st strategy is the number bond. 2nd strategy solved using the place value strategies. 6 hundred = 600. 9 tens = 90. 8 ones = 8. 600 + 90 + 8 = 698. 698 + 247 = 945. Circle a strategy to solve, and explain why you chose that strategy. Question 3. 257 + 160 = _____ a. Arrow way or vertical form b. Solve Explanation: __________________________ __________________________ __________________________ __________________________ 257 + 160 = 317. Explanation: In the above-given question, given that, solve the problem using arrow way or vertical form. 2 hundred = 200. 5 tens = 50. 7 ones = 7. 200 + 50 + 7 = 257. 1 hundred = 100. 6 tens = 60. 100 + 60 = 160. 257 – 160 = 317. Question 4. 754 – 597 = _____ 754 – 597 = 157. Explanation: In the above-given question, given that, solve the problem using arrow way or vertical form. 7 hundred = 700. 5 tens = 50. 4 ones = 4. 700 + 50 + 4 = 754. 5 hundred = 500. 9 tens = 90. 7 ones = 7. 500 + 90 + 7 = 597. 754 – 597 = 157. a. Number bond or arrow way b. Solve Explanation: __________________________ __________________________ __________________________ __________________________ 754 – 597 = 157. Explanation: In the above-given question, given that, solve the problem using arrow way or number bond. 7 hundred = 700. 5 tens = 50. 4 ones = 4. 700 + 50 + 4 = 754. 5 hundred = 500. 9 tens = 90. 7 ones = 7. 500 + 90 + 7 = 597. 754 – 597 = 157.
# Area and Perimeter of Combined Figures Here we will solve different types of problems on finding the area and perimeter of combined figures. 1. Find the area of the shaded region in which PQR is an equilateral triangle of side 7√3 cm. O is the centre of the circle. (Use π = $$\frac{22}{7}$$ and √3 = 1.732.) Solution: The centre O of the circle is the circumcentre of the equilateral triangle PQR. So, O is also the centroid of the equilateral triangle and QS ⊥ PR, OQ = 2OS. If the radius of the circle be r cm then OQ = r cm, OS = $$\frac{r}{2}$$ cm, RS = $$\frac{1}{2}$$ PR = $$\frac{7√3}{2}$$ cm Therefore, QS$$^{2}$$ = QR$$^{2}$$ - RS$$^{2}$$ or, ($$\frac{3r}{2}$$)$$^{2}$$ = (7√3)$$^{2}$$ - ($$\frac{7√3}{2}$$)$$^{2}$$ or, $$\frac{9}{4}$$ r$$^{2}$$ = (1 - $$\frac{1}{4}$$) (7√3)$$^{2}$$ or, $$\frac{9}{4}$$ r$$^{2}$$ = $$\frac{3}{4}$$ × 49 × 3 or, r$$^{2}$$ = $$\frac{3}{4}$$ × 49 × 3 × $$\frac{4}{9}$$ or, r$$^{2}$$ = 49 Therefore, r = 7 Therefore, area of the shaded region = Area of the circle – Area of the equilateral triangle = πr$$^{2}$$ - $$\frac{√3}{4}$$ a$$^{2}$$ = $$\frac{22}{7}$$ × 7$$^{2}$$ cm$$^{2}$$ - $$\frac{√3}{4}$$ × (7√3)$$^{2}$$ cm$$^{2}$$ = (154 - $$\frac{√3}{4}$$ × 147) cm$$^{2}$$ = (154 - $$\frac{1.732 × 147}{4}$$) cm$$^{2}$$ = (154 - $$\frac{254.604}{4}$$) cm$$^{2}$$ = (154 - 63.651) cm$$^{2}$$ = 90349 cm$$^{2}$$ 2. The radius of the wheels of a car is 35 cm. The car takes 1 hour to cover 66 km. Find the number of revolutions that a wheel of the car makes in one minute. (Use π = $$\frac{22}{7}$$.) Solution: According to the problem, radius of a wheel = 35 cm. The perimeter of a wheel = 2πr = 2 × $$\frac{22}{7}$$ × 35 cm = 220 cm Therefore, the number of revolutions of a wheel to cover 66 km = $$\frac{66 km}{220 km}$$ = $$\frac{66 × 1000 × 100 cm}{220 cm}$$ = $$\frac{3 × 1000 × 100}{10}$$ = 30000 Therefore, the number of revolutions of a wheel to make in one minute = $$\frac{30000}{60}$$ = 500 3. A circular piece of paper of radius 20 cm is trimmed into the shape of the biggest possible square. Find the area of the paper cut off. (Use π = $$\frac{22}{7}$$.) Solution: The area of the piece of paper = πr$$^{2}$$ = $$\frac{22}{7}$$ × 20$$^{2}$$ cm$$^{2}$$ If the side of the inscribed square be x cm then 20$$^{2}$$ = ($$\frac{x}{2}$$)$$^{2}$$ + ($$\frac{x}{2}$$)$$^{2}$$ or, 400 = $$\frac{1}{2}$$ x$$^{2}$$ or, x$$^{2}$$ = 800. Therefore, the area of the paper cut off = The area of the circle - The area of the square = πr$$^{2}$$ - x$$^{2}$$ = $$\frac{22}{7}$$ × 20$$^{2}$$ cm$$^{2}$$ - 800 cm$$^{2}$$ = ($$\frac{8800}{7}$$ - 800) cm$$^{2}$$ = $$\frac{3200}{7}$$ cm$$^{2}$$ = 457$$\frac{1}{7}$$ cm$$^{2}$$ 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. ### Types of Fractions \|Proper Fraction \|Improper Fraction \|Mixed Fraction Mar 02, 24 05:31 PM The three types of fractions are : Proper fraction, Improper fraction, Mixed fraction, Proper fraction: Fractions whose numerators are less than the denominators are called proper fractions. (Numerato… 2. ### Subtraction of Fractions having the Same Denominator \| Like Fractions Mar 02, 24 04:36 PM To find the difference between like fractions we subtract the smaller numerator from the greater numerator. In subtraction of fractions having the same denominator, we just need to subtract the numera… 3. ### Addition of Like Fractions \| Examples \| Worksheet \| Answer \| Fractions Mar 02, 24 03:32 PM To add two or more like fractions we simplify add their numerators. The denominator remains same. Thus, to add the fractions with the same denominator, we simply add their numerators and write the com… 4. ### Comparison of Unlike Fractions \| Compare Unlike Fractions \| Examples Mar 01, 24 01:42 PM In comparison of unlike fractions, we change the unlike fractions to like fractions and then compare. To compare two fractions with different numerators and different denominators, we multiply by a nu…
# Derivatives ## The derivative of ${\displaystyle {\frac {d}{dx}}x^{x}}$ ### Question What is ${\displaystyle {\frac {d}{dx}}x^{x}}$? ### Solution 1 • Let ${\displaystyle y=x^{x}}$. • Take ${\displaystyle \ln }$ of both sides: ${\displaystyle \ln y=x\ln x}$. • Differentiate both sides: ${\displaystyle {\frac {d}{dx}}\ln y={\frac {d}{dx}}x\ln x}$. • Apply the chain rule on the left-hand side: ${\displaystyle {\frac {d}{dx}}\ln y={\frac {1}{y}}\cdot {\frac {dy}{dx}}}$. • Apply the product rule on the right-hand side: ${\displaystyle {\frac {d}{dx}}x\ln x=1\cdot \ln x+x\cdot {\frac {1}{x}}=\ln x+1}$. • Putting it together, we have ${\displaystyle {\frac {1}{y}}\cdot {\frac {dy}{dx}}=\ln x+1}$. • Hence ${\displaystyle {\frac {dy}{dx}}=y(\ln x+1)=x^{x}(\ln x+1)}$. ### Solution 2 • Note that ${\displaystyle x=e^{\ln x}}$, so ${\displaystyle x^{x}=(e^{\ln x})^{x}=e^{x\ln x}}$. • Applying the chain rule, ${\displaystyle {\frac {d}{dx}}x^{x}={\frac {d}{dx}}e^{x\ln x}=e^{x\ln x}{\frac {d}{dx}}x\ln x}$. • Applying the product rule, ${\displaystyle {\frac {d}{dx}}x\ln x=1\cdot \ln x+x\cdot {\frac {1}{x}}=\ln x+1}$. • Therefore ${\displaystyle {\frac {d}{dx}}x^{x}=e^{x\ln x}(\ln x+1)=x^{x}(\ln x+1)}$. # Integrals ## The integral ${\displaystyle \int x^{x}\,dx}$ ### Question What is ${\displaystyle \int x^{x}\,dx}$? ### Solution • We can write ${\displaystyle x^{x}}$ as ${\displaystyle (e^{\ln x})^{x}=e^{x\ln x}}$. • Consider the series expansion of ${\displaystyle e^{x\ln x}}$: ${\displaystyle e^{x\ln x}=1+(x\ln x)+{\frac {(x\ln x)^{2}}{2!}}+{\frac {(x\ln x)^{3}}{3!}}+\ldots +{\frac {(x\ln x)^{i}}{i!}}+\ldots =\sum _{i=0}^{\infty }{\frac {(x\ln x)^{i}}{i!}}}$. • We can interchange the integration and summation (we can recognize this as a special case of the Fubini/Tonelli theorems) and write ${\displaystyle \int x^{x}\,dx=\int \left(\sum _{i=0}^{\infty }{\frac {(x\ln x)^{i}}{i!}}\right)\,dx=\sum _{i=0}^{\infty }\left(\int {\frac {(x\ln x)^{i}}{i!}}\,dx\right)=\sum _{i=0}^{\infty }\left({\frac {1}{i!}}\int x^{i}(\ln x)^{i}\,dx\right).}$ # Limits ## The limit of ${\displaystyle \lim _{x\rightarrow 0^{+}}x^{x}}$ ### Question What is ${\displaystyle \lim _{x\rightarrow 0^{+}}x^{x}}$? ### Solution • Note that ${\displaystyle x=e^{\ln x}\,so\x^{x}=(e^{\ln x})^{x}=e^{x\ln x}}$. • We can further rewrite this as ${\displaystyle x^{x}=e^{x\ln x}=e^{\frac {\ln x}{\frac {1}{x}}}}$. • As long as ${\displaystyle f}$ is continuous and the limit of ${\displaystyle g}$ exists at the point in question, the limit will commute with composition: $\displaystyle \lim_{x \tendsto t} f(g(x)) = f(\lim_{x \tendsto t} g(x)).$ In our case, ${\displaystyle e(\cdot )}$ is continuous, so $\displaystyle \lim_{x \tendsto 0^+} x^x = e^{\lim_{x \tendsto 0^+} \frac{\ln x}{\frac{1}{x}}}.$ • The question, then, is what is $\displaystyle \lim_{x \tendsto 0^+} \frac{\ln x}{\frac{1}{x}}$ . • As $\displaystyle x \tendsto 0^+$ , $\displaystyle \ln x \tendsto -\infty, \frac{1}{x} \tendsto +\infty$ . In this situation we can apply l'Hôpital's rule: $\displaystyle \lim_{x \tendsto 0^+} \frac{\ln x}{\frac{1}{x}} = \lim_{x \tendsto 0^+} \frac{\frac{d}{dx} \ln x}{\frac{d}{dx} \frac{1}{x}} = \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \frac{\frac{1}{x} \cdot x^2}{-\frac{1}{x^2} \cdot x^2} = -x.$ • Hence $\displaystyle \lim_{x \tendsto 0^+} x^x = e^0 = 1$ . ## The limits of $\displaystyle \lim_{x \tendsto +\infty} x \sin \frac{1}{x}$ and $\displaystyle \lim_{x \tendsto -\infty} x \sin \frac{1}{x}$ ### Question What are $\displaystyle \lim_{x \tendsto +\infty} x \sin \frac{1}{x}$ and $\displaystyle \lim_{x \tendsto -\infty} x \sin \frac{1}{x}$ ? ### Solution • Let us rewrite ${\displaystyle x\sin {\frac {1}{x}}\as\{\frac {\sin {\frac {1}{x}}}{\frac {1}{x}}}}$. • As $\displaystyle x \tendsto +\infty, \frac{1}{x} \tendsto 0$ and $\displaystyle x \sin \frac{1}{x} \tendsto 0$ . • We have ${\displaystyle {\frac {0}{0}}}$, so we can apply l'Hôpital's rule. • Differentiating the numerator in ${\displaystyle {\frac {\sin {\frac {1}{x}}}{\frac {1}{x}}}}$, we obtain ${\displaystyle \left(\cos {\frac {1}{x}}\right)\left(-{\frac {1}{x^{2}}}\right)}$. • Differentiating the denominator in ${\displaystyle {\frac {\sin {\frac {1}{x}}}{\frac {1}{x}}}}$, we obtain ${\displaystyle -{\frac {1}{x^{2}}}}$. • Thus $\displaystyle \lim_{x \tendsto +\infty} x \sin \frac{1}{x} = \lim_{x \tendsto +\infty} \frac{\sin \frac{1}{x}}{\frac{1}{x}} = \lim_{x \tendsto +\infty} \frac{\left(\cos \frac{1}{x}\right) \left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}} = \lim_{x \tendsto +\infty} \cos \frac{1}{x} = 1.$ • Similarly we can find that $\displaystyle \lim_{x \tendsto -\infty} x \sin \frac{1}{x} = 1$ .
Line Geometry Line Geometry calculator ▲ (1) Line equation: y = x + Two points on line: x1: y1: x2: y2: Angle (α): Intercepts: ( , 0 ): ( 0 , ): Midpoint: xd: yd: Line length: Point in plane: xp: yp: The point is on the line Distance from point (xp, yp) to line (1): (2) Equation of line passing through (xp, yp) and is perpendicular to line (1): Intersection coordinates between lines (1) and (2): x: y: Line basic definition Line distances Intersection point ex: 3 Basic line information Distance from point ex: 1 Line equations Perpendicular line ex: 2 Line basic definition ▲ The most general equation of a line is of the form: (1) where A, B and C are any real number and A and B are not both zero. If B ≠ 0 then we can divide equation (1) by B to obtain the form: (2) This is the equation of a line whose slope is: and Equation (2) can be normalized to the general form: (3) the slope of the line (m) is defined in terms of the inclination (4) Note: if the angle α is greater then 90 degrees then the slope is negative. α (0 - 90) degrees : positive slope α (90 - 180) degree : negative slope Necessary condition for two lines to be perpendicular to each other is that their slopes fulfill the condition: m1 m2 = − 1 (5) In order to find the intersection point of two lines we have to solve the system of linear equations representing the lines. A x + B y = −C D x + E y = −F If the determinant: then intersection point exists. Basic line of the form y = ax + b or Ax + By + C = 0 ▲ Slope (m) of the line m = a yintercept (yi) yi = b yi = −mxi yi = y1 − mx1 xintercept (xi) tan θ tan θ = m Line angle (θ) from x axis (range 0 ≤ θ < π) θ = arctan(m) Slope (M) of a line perpendicular to a given slope (m) Line midpoint Point (x, y) which divides the line connecting two points (x1 , y1) and (x2 , y2) in the ratio p:q Point (x, y) which divides the line connecting two points (x1 , y1) and (x2 , y2) externally at a ratio p:q Note: the (x,y) point is in the direction from point 1 to point 2,to get the other side extension change the point 1 with point 2 and vice versa A point (x, y) which is located at a distance d from a point (x1 , y1) on the line Angle θ between two lines: Angle between two lines given by Ax + By + C = 0 Dx + Ey + F = 0 Lines equations ▲ Equation of a line passing through a point (x1 , y1) y = mx + ( y1 − mx1) y − y1 = m(x − x1) Equation of a line passing through two points (x1 , y1), (x2 , y2) Equation of a line perpendicular to a given slope m and passing through a point (xp , yp) Equation of a line perpendicular to a line which is defined by two points (x1 , y1) and (x2 , y2) and passing through the point (xp , yp) Equation of a line passing through the intercepts xi , yi xiy = −yix + xiyi Equation of a line passing through the point (xp ,yp) and parallel to a line which is defined by two points (x1 , y1) and (x2 , y2) Equation of a line parallel to the line Ax + By + C = 0 and at a distance d from it. Equation of the midline between the lines Ax + By + C = 0 Dx + Ey + F = 0 Equation of a line perpendicular to the line Ax + By + C = 0 Bx − Ay + C = 0 Equation of a horizontal line y = b Equation of a vertical line x = a Lines distances y = ax + b Ax + By + C = 0 ▲ Distance between two points (D) Distance between intercepts xi and yi Distance from a line to the origin Distance from a line given by two points (x1 ,y1),(x2 , y2) to the origin Distance from a line to the point (xp , yp) Distance from a line given by two points (x1 , y1) , (x2 , y2) to the point (xp , yp) Distance between two parallel lines y = ax + b y = cx + d or Ax + By + C = 0 Dx + Ey + F = 0 Example 1 - distance from a line to a point ▲ Find the distance d of a point P(px , py) to the line given by the equation Ax + By + C = 0. Calculate the numerical value of the distance between point (2 , −3) and line y = 4x − 5 We draw at point P a parallel line to the given line (dashed line). Then we draw a line PF between the points P and F. This line is perpendicular to the given line. The slope m of the given line is: m = −A / B The slope M of line PF is: M = −1 / m = B / A In order to make calculations easier we draw another line QS starting from the origin and parallel to line PF. The general equation of a line passing through a point is given by: y = mx + y1 − mx1 The equation of line PQ passing through point (px , py) is: Equation of line QS passing through the origin is: Point Q is the intersection of lines PQ and QS. Solving this equations by substituting the value of y we get: And finally, we get: and Point S is the intersection point of the two lines Ax + By + C = 0 and Solving for Sx and Sy we get and The distance d is the distance between the two points Q and S given by the points: and The given line can be written in the standard form as 4x − y − 5 = 0 So, we have: A = 4 B =1 and C =5 px = 2 py =3 Substituting these values into the distance equation we get the final solution: Example 2 - line from point to perpendicular line ▲ Find the equation of the line passing through the point (3 , − 1) and is perpendicular to the line 2x − y − 1 = 0 The slope of the perpendicular line is: y = 2x − 1 → m = 2 The slope of the line perpendicular to the given line and passing through the point is: The equation of the line passing through the point (3 , − 1) is: y = Mx + (y1 − Mx1) x + 2y − 1 = 0 If the intersection point of both lines is needed then we have to solve the equations: x + 2y − 1 = 0 2x − y − 1 = 0 From the first equation x is x = 1 − 2y Substitute x into the second equation 2(1 − 2y) − y − 1 = 0 And the intersection coordinate is (0.6 , 0.2) Example 3 - intersection point of perpendicular lines ▲ Find the intersection point of the line 2x − y = 0 and the line passing through the point (5 , 2) and is perpendicular to the given line. If the line form is given by: Ax + By + C = 0 The slope of the given line is: m = − A/B Slope of the line perpendicular to the given line is: The equation of the line passing through the point (Px , Py) = (5 , 2) (green line) and is perpendicular to the given line is: y = Mx + (Py − MPx) (1) y = −0.5x + 2 + 0.5‧5 (2) x + 2y − 9 = 0 In order to find the intersection point (Nx , Ny) we have to solve equation (1) and the given line this can be done in direct substitution or by Cramer’s law: From equation (1) We have Substitute y into equation (1) Solving for x we get And y is equal to The general intersection of a line and the perpendicular line that passes through a point is: (3) The given values are A = 2 B = −1 C = 0 Px = 5 Py = 2 Solving the numerical values, we have x + 2(2x) − 9 = 0 x = 9/5 = 1.8 and y is equal to y = 18/5 = 3.6
# Parallel and Perpendicular Lines ## Presentation on theme: "Parallel and Perpendicular Lines"— Presentation transcript: Parallel and Perpendicular Lines Using parallelism and perpendicularity to solve problems In the graph below, the two lines are parallel In the graph below, the two lines are parallel. Parallel lines - are lines in the same plane that never intersect. The equation of line 1 is y = 2x the equation of line 2 is y = 2x -2 Slopes of Parallel Lines Nonvertical lines are parallel if they have the same slope and different y-intercepts. Any two vertical lines are parallel. Any two horizontal lines are parallel You can use slope-intercept form of an equation to determine whether lines are parallel. Are the graphs of y = -1/3x + 5 and 2x + 6y = 12 parallel? Explain. 2x + 6y = 12 6y = -2x + 12 6y = - 2x + 12 6 y = - 1/3x + 2 Compare to y = -1/3x + 5 The lines are parallel. The equations have the same slope, -1/3, and different y-intercepts. Write 2x + 6y = 12 in slope-intercept form, then compare with y = -1/3x + 5 Are the graphs of -6x + 8y = -24 and y = 3/4x – 7 parallel? Explain. You can use the fact that the slopes of parallel lines are the same to write the equation of a line parallel to a given line. To write the equation, you use the slope of the given line and the point-slope form of a linear equation. Step 1 Identify the slope of the given line. y = 3/5x – 4 Step 2 Write the equation of the line through (5, 1) using point-slope form. y – y1 = m(x – x1) point-slope form. y – 1 = 3/5(x – 5) Substitute (5, 1) for (x1,Y1) and 3/5 for m. y – 1 = 3/5x – 3/5(5) Use the distributive property. y – 1 = 3/5x – 3 Simplify. y = 3/5x – 2 Add 1 to each side. TRY ONE Write an equation for the line that contains (2, -6) and is parallel to y = 3x + 9 Step 1 Identify the slope of the given line. Step 2 Write the equation of the line through (2, -6) using point-slope form of a linear equation. y – y1 = m(x – x1) Write an equation for the line that is parallel to the given line and that passes through the given point. Y = 6x - 2; (0, 0) Y = -3x; (3, 0) Y =-2x + 3; (-3, 5) Y = -7/2x + 6; (-4, -6) The two lines in the graph below are perpendicular The two lines in the graph below are perpendicular. Perpendicular lines – are lines that intersect to form right angles. The line y = 2x + 1 is perpendicular to the line y = -1/2x + 1. Slopes of perpendicular lines Two lines are perpendicular if the product of their slopes is -1. A vertical and a horizontal line are also perpendicular. The product of two numbers is -1 if one number is the negative reciprocal of the other. Here is how to find the negative reciprocal of a number. Start with a fraction: -1/2 Find its reciprocal: -2/1 Write the negative reciprocal: 2/1 or 2 Since -1/2 • 2/1 = -1, 2/1 is the negative reciprocal of -1/2 TRY THESE Find the negative reciprocal of each: 1) 4 2) 3/4 3) -1/2 4) -2 5) -4/3 y + 2 = -1/5x – 0 Use the distributive property. You can use the negative reciprocal of the slope of a given line to write an equation of a line perpendicular to that line. To write the equation, you use the negative reciprocal of the slope of the given line and the point-slope form of a linear equation. Step 1 Identify the slope of the given line. y = 5x + 3 Step 2 Find the negative reciprocal of the slope. 5 • -1/5 = -1 Step 3 Use the point-slope form to write an equation that contains (0, -2) and is perpendicular to y = 5x + 3 y – y1 = m(x – x1) Point-slope form. y – (-2) = -1/5(x – 0) Substitute (0, -2) for (x1,y1) and -1/5 for m. y + 2 = -1/5x – 0 Use the distributive property. y = -1/5x – 2 Subtract 2 from each side. Simplify. TRY ONE Write an equation of the line that contains (6, 2) and is perpendicular to y = -2x + 7 Step 1 Identify the slope of the given line. Step 2 Find the negative reciprocal of the slope. Step 3 Use the point-slope form of an equation that contains (6, 2) and is perpendicular to y = -2x + 7 Write an equation for the line that is perpendicular to the given line and that passes through the given point. Y = 2x + 7; (0, 0) Y = -1/3x + 2; (4, 2) Y = x – 3; (4, 6) 4x – 2y = 9; (8, 2) Write the equation of each line Write the equation of each line. Determine if the lines are parallel or perpendicular. Explain why or why not. Problem Solving Problem Solving A city’s civil engineer is planning a new parking garage and a new street. The new street will go from the entrance of the parking garage to Handel St. It will be perpendicular to Handel St. What is the equation of the line representing the new street? Entrance Handel St.
# 10 topics : how many cups is 500 ml ? Rate this post How Many Cups is 500 ml? – A Comprehensive Guide # How Many Cups is 500 ml? – A Comprehensive Guide ## Introduction When it comes to cooking and baking, precise measurements are crucial to achieving the desired outcome. However, not all recipes use the same units of measurement. For example, some recipes may call for ingredients to be measured in cups, while others may use milliliters (ml). If you’re wondering how many cups are in 500 ml, you’ve come to the right place. In this guide, we’ll explore the conversion between cups and ml and provide you with a simple answer to this common question. ## Understanding the Conversion Before we dive into the answer, it’s important to understand the conversion between cups and ml. In the metric system, 1 cup is equal to 250 ml. Therefore, to convert ml to cups, you simply divide the number of ml by 250. For example, if you have 500 ml of liquid, you would divide 500 by 250 to get 2 cups. So, how many cups is 500 ml? As we just learned, 1 cup is equal to 250 ml. Therefore, 500 ml is equal to 2 cups. This means that if a recipe calls for 500 ml of liquid, you can use 2 cups instead. ## Other Metric Conversions While we’re on the topic of metric conversions, it’s worth noting a few other common conversions that you may come across in recipes. Here are a few examples: • 1 tablespoon = 15 ml • 1 teaspoon = 5 ml • 1 liter = 4.2 cups • 1 kilogram = 2.2 pounds By understanding these conversions, you can easily adapt recipes that use different units of measurement. ## Conclusion Now that you know how many cups are in 500 ml, you can confidently tackle any recipe that calls for this measurement. Remember, 1 cup is equal to 250 ml, so 500 ml is equal to 2 cups. By understanding this conversion, as well as other common metric conversions, you can easily adapt recipes to suit your needs. You are looking : how many cups is 500 ml ## 10 how many cups is 500 ml for reference ### 1.How Many Cups Is 500ml? – How To Measuring – Naan on broughton • Author: How • Publish: 25 days ago • Rating: 4(1412 Rating) • Highest rating: 3 • Lowest rating: 1 • Descriptions: 500ml is equivalent to 2.11 cups of liquid when using the US customary cup measurement (236.588 mL). However, if you are located in a country that uses the … • More : 500ml is equivalent to 2.11 cups of liquid when using the US customary cup measurement (236.588 mL). However, if you are located in a country that uses the … • Source : https://naanonbroughton.com/cooking-tips/how-many-cups-is-500ml/ ### 2.How Many Cups Is 500 ml? – Measuring Stuff • Author: How • Publish: 16 days ago • Rating: 3(450 Rating) • Highest rating: 3 • Lowest rating: 3 • Descriptions: • More : • Source : https://measuringstuff.com/how-many-cups-is-500-ml/ ### 3.How many cups is in 500ml? How do you convert 500ml exactly • Author: How • Publish: 23 days ago • Rating: 5(1608 Rating) • Highest rating: 5 • Lowest rating: 3 • Descriptions: Convert confidently with knowledge that 500 ml equals precisely 2.11 cups of liquid. This knowledge is an excellent resource to possess when creating recipes … • More : Convert confidently with knowledge that 500 ml equals precisely 2.11 cups of liquid. This knowledge is an excellent resource to possess when creating recipes … • Source : https://eugenesdiner.com/food-recipes/how-many-cups-is-in-500ml ### 4.How many cups is 500ml? Rockyspub’s measure & convert guide • Author: How • Publish: 15 days ago • Rating: 3(1758 Rating) • Highest rating: 3 • Lowest rating: 3 • Descriptions: • More : • Source : https://rockyspub.com/cooking-tips-and-techniques/how-many-cups-is-500ml ### 5.How Many Cups Is In 500ml? Does 2 Cups Equal 500ml Right? • Author: How • Publish: 8 days ago • Rating: 4(1696 Rating) • Highest rating: 5 • Lowest rating: 3 • Descriptions: • More : • Source : https://tapproom.com/recipes/how-many-cups-is-in-500ml ### 6.How Many Cups Is 500ml? 500 ML To Cups Conversion [New] • Author: How • Publish: 9 days ago • Rating: 3(1973 Rating) • Highest rating: 3 • Lowest rating: 3 • Descriptions: • More : • Source : https://hanawaterbury.com/food-recipes/how-many-cups-is-500ml ### 7.How Many Cups Is 500 Ml \| Quantitative You Need To Know • Author: How • Publish: 18 days ago • Rating: 5(1594 Rating) • Highest rating: 4 • Lowest rating: 1 • Descriptions: In conclusion, 500ml is equal to 2 cups. It is important to take into account how the density of what you are measuring affects how much liquid fits in a cup. • More : In conclusion, 500ml is equal to 2 cups. It is important to take into account how the density of what you are measuring affects how much liquid fits in a cup. • Source : https://bamboowokmanvel.com/cooking-tips/how-many-cups-is-500ml/ ### 8.How Many Cups Is 500 ml? – Hawg and Ale Smokehouse • Author: How • Publish: 30 days ago • Rating: 1(1087 Rating) • Highest rating: 4 • Lowest rating: 1 • Descriptions: • More : • Source : https://hawgandale.com/recipes/how-many-cups-is-500-ml/ ### 9.How many cups is in 500ml? Convert between ml and cups • Author: How • Publish: 23 days ago • Rating: 4(1548 Rating) • Highest rating: 3 • Lowest rating: 3 • Descriptions: • More : • Source : https://silverkingbrewing.com/cooking-recipes/how-many-cups-is-in-500ml ### 10.Convert 500 ml to cup – Worksheet Genius • Author: Convert • Publish: 2 days ago • Rating: 1(1184 Rating) • Highest rating: 5 • Lowest rating: 3 • Descriptions: So, the answer to the question “what is 500 millilitres in cups?” is 1.7597549706019 cup. Millilitres to Cups Conversion Table. Below is a sample conversion … • More : So, the answer to the question “what is 500 millilitres in cups?” is 1.7597549706019 cup. Millilitres to Cups Conversion Table. Below is a sample conversion … • Source : https://worksheetgenius.com/unit-converter/convert-500-ml-to-cup/
# Factoring Binomials: Difference of Two Squares Factoring Binomials : Difference of Two Squares Difference of Two Squares X ^2 – Y^2=(X +Y ) ( X – Y) The difference of two squares of two numbers is equal to the productof the sum and difference of the two numbers. Example One :Factor64X^2-81^2 First let us inspect if the given problem is a difference of two squares. The numerical coefficients must be perfect squares. It means that their square root is an exact whole number. For literal coefficients, their exponents must be divisible by two. Only by meeting these conditions it can be factored as difference of two squares. 64 is a perfect square and its square root is 8. 81 is a perfect square and its square root is 9. 64^2 – 81^2 can be rewritten as ( 8X)^2-(9Y)^2therefore its factors are(8x + 9Y) (8X – 9Y). Example Two :FactorX^8 – Y^8 X^8 – y^8=( x^4-)^2-(Y^4)^2 =( X^4 + Y^4) (X^4 – Y ^4) =(X^4 + Y^4 ) (X^2 + Y^2) (X^2 – Y^2) =(X^4 + Y^4 ) (X^2 + Y^2 ) (X +Y ) ( X – Y) Example Three : Factor(a + 3b ) ^2 – 16c^4 (a + 3b)^2- 16 c^4=(a + 3b)^2 – (4c^2)^2 =(a + 3b+ 4c^2) (a + 3b – 4c^2) Example Four : Factor49a^10b^8- 100c^6 49a^10b^8- 100c^6=( 7a^5b^4)^2-(10c^3)^2 =(7a^5b^4+- 10 c^3)(7a^5b^4- 10 c^3) Example Five : Factor81X^12-256^Y^8 81X^12 – 256Y^8=(9X^6)^2-(16Y^4) =(9X^6 + 16Y^4) (9X^6 – 16Y^4) =(9X^6+ 16 Y^4) (3X^3 + 4Y^2) (3X^3 – 4Y^2) ## More by this Author Lord De Cross 5 years ago Excellent and well explained Cristina. Seen so many kids struggling with Algebra thru the years.Useful and voted up! Lord ronhi 5 years ago from Kenya where were you when i was failing maths in school? cristina327 5 years ago from Manila Author Thank you lord de cross for appreciating this hub. Blessings to you always and best regards. cristina327 5 years ago from Manila Author HI ronhi thank you for dropping by. Regards.
# SAT TIP: ‘Mean’ Questions on the SAT Photograph by Getty Images This tip on improving your SAT score was provided by Veritas Prep. Statistics questions are rare on the SAT, but they do appear a handful of times on any given test. As a result, students looking to score a perfect 2400 need to know at least some basics of statistics to solve these questions in case they come up. While you don’t need to have taken a full course on statistics, you really need to understand three basic concepts: mean, median, and mode. After you learn these definitions, you will need to watch out for how the SAT creates tricky questions based on these concepts. In this article, we will learn more about the arithmetic mean or average and how the SAT might construct a question around this concept. Mean: The arithmetic mean is the same as what most people call the “average” of a set of numbers. To find the mean, add all of the elements of the set together and divide by the number of elements in the set. So for a set of numbers {6, 7, 9, 10}, the mean is simply (6 + 7 + 9 + 10) / 4 = 8. Questions on the SAT will rarely be this straightforward, though, and will typically require you to go beyond the basic definition. For example: Robin currently has a mean score of 80 from three tests in her class. What must she score on her fourth and final test to finish the class with a mean score of 82.5? A. 75 B. 80 C. 85 D. 90 E. 95 For this problem, even if you did not know how to solve it, you should be able to recognize right away that A and B can be eliminated since her final average must be higher than her current average of 80. This can only happen if she scores higher than 80 on the last test. So if you had to guess, you would choose among C, D, and E. In order to solve this problem, we can apply our knowledge of the mean. To have a mean score of 82.5 over four tests, the sum total of all her test scores needs to be 82.5 x 4 = 330. Currently, across three tests, she has an 80 average, so that total is simply 80 x 3 = 240. As a result, we can see that she needs to get to a total of 330 by scoring 330 – 240 = 90 on her last test. Answer choice D is correct. You may have noticed already that the SAT rarely asks straightforward questions, such as “What is the average of this set of numbers?” The SAT will take a simple concept like the mean and add some element of critical reasoning to it, such as by giving you the mean and seeing if you can find the original elements in the set. When practicing SAT problems, see if you can come up with ways the SAT might reword the problems to put a more difficult spin on them. Plan on taking the SAT soon? Sign up for a trial of Veritas Prep SAT 2400 On Demand! LIMITED-TIME OFFER SUBSCRIBE NOW ### SAT Practice Test #### Veritas Prep SAT Diagnostic Quiz Created for Bloomberg Businessweek readers, this diagnostic quiz is designed to measure your ability level with 25 realistic SAT questions. Click here to take the quiz and get instant feedback about your performance. ### Last Updated: 04:18 pm Market Summary S&P500 1998.98 14.85 DJIA 17132 100.83 NASDAQ 4552.76 33.856 Stoxx 50 3221.73 -9.97 FTSE 100 6792.24 -11.97 DAX 9632.93 -26.7 Nikkei 15911.5 -36.76 Topix 1310.86 -2.86 Hang Seng 24136 -220.98 ### Post a Job MBA Jobs View all MBA jobs
# Tiling a rectangle with nine squares A rectangle is tiled by nine squares with side lengths $2,5,7,9,16,25,28,33$ and $36$ (without overlapping and without gaps). What are the side lengths of the rectangle? What does the tiling look like? • An obvious question would be: Which rectangles can be tiled by squares of all different sizes? (Note: There is a proof that if a rectangle can be tiled by squares, then the ratio of its sides is a rational number and the ratio of its sides to the sides of each square is rational. So we can scale this up and ask about rectangles with integer sides and integer squares without loss of generality). Feb 25, 2016 at 13:25 The dimensions are 69 x 61 and the tiling looks like this: Working out the dimensions of the rectangle is quite easy. We know its total area is $4209$ (i.e., $2^2 + 5^2 + 7^2 + 9^2 + 16^2 + 25^2 + 28^2 + 33^2 + 36^2$). This factorizes as $3 \times 23 \times 61$, and in order to fit in a square with a side length of 36, the rectangle must be $3 \times 23$ units long on one side, and $61$ units on the other. Fitting the squares into this rectangle only takes a few minutes. • Great work! Maybe you could add a short argument in the beginning detailing that the solution rectangle has to have integer side-lengths to make it complete :-) Feb 26, 2016 at 11:50 1. Size of rectangle = 61 x 69 2. Tiling: tricky is where to place the 2x2 square. It can't go on the corner or the sides because it will create a gap. Place it in the center. Experimentally add squares to it's sides. 9-2=7. 7-2=5. Continue fitting all squares until a rectangle is formed. Check widths and heights are the same. Answer:69*61
Search 75,665 tutors # Using Euler's Formulas to Obtain Trigonometric Identities ### Written by tutor Jeffery D. In this lesson we will explore the derivation of several trigonometric identities, namely cos (x + y) = cos x cos y - sin x sin y and sin (x + y) = sin x cos y + sin x cos y also cos 2x = cos2 x - sin2 x along with sin 2x = 2 sin x cos x and lastly, DeMoivre's Formula, (cos x + i sin x)n = cos nx + i sin nx using Euler's Formula. To get a good understanding of what is going on, you will need a previous knowledge of series expansions and complex numbers! You may want to refresh your knowledge of those subjects first. ## Power series expansions We start by examining the power series expansion of the functions ex, sin x, and cos x. The power series of a function is commonly derived from the Taylor series of a function for the case where a = 0. This case, where a = 0 is called the MacLaurin Series. The Taylor series: The case where a = 0 is the MacLaurin Series: These series are used to approximate the values of functions around a certain point. That is all I'll say about that. The power series of ex; cos x and sin x comes from their MacLaurin Series representation: for all x. for all x. for all x. ## Complex numbers and ex A complex number is a number of the form a + bi where i is a root of the equation x2 + 1 = 0 and a and b are real numbers. Making note of this we can use i in our power series of ex since it is true for all x. for all x. Keeping in mind that x2 + 1 = 0 → x = i and so √-1 = ii2 = -1, i3 = -i, etc. So, applying the powers selectively we obtain We can now rearrange the terms and factor out i so that that we have Now, if we look back at our series representations of cos x and sin x we have eix = cos x + i sin x This conclusion is huge. It is known as Euler's formula. From here we can deduce some of the trigonometric identities as well as come up with formulas for general cases. Let us examine a simple derivation first: eixeiy = (cos x + i sin x)(cos y + isiny) But, recall that exey = ex+y. Therefore, we have eix+iy = cos (x + y) + i sin (x + y) = (cos x + i sin x)(cos y + i sin y) = cos x cos y + i sin x cos y + i sin y cos x + i2sin x sin y And now we can rearrange this so that the complex part and the real part is separate. And so we have eix+iy = cos (x + y) + i sin (x + y) = (cos x + i sin x)(cos y + i sin y) = cos x cos y + i sin x cos y + i sin y cos x + i2 sin x sin y = (cos x cos y - sin x sin y) + (i sin x cos y + i sin y cos x) Taking the real parts and equating them we obtain the familiar trigonometric sum formula: cos (x + y) = cos x cos y - sin x sin y and also sin (x + y) = sin x cos y + sin y cos x Now, suppose we have something like this: eixeix = eix+ix = ei2x = cos (x + x) + i sin(x + x) = (cos x + i sin x)(cos x + i sin x) = cos x cos x + i sin x cos x + i sin x cos x + i2 sin x sin x If we equate the real parts of the equation cos 2x = cos2x - sin2x And also we have sin 2x = 2 sin x cos x In general, we may obtain a formula for any multiple of an angle in this way. This leads us to another famous formula known as DeMoivre's Formula. DeMoivre's Formula can be derived by taking the nth case of Euler's Formula. einx = cos nx + i sin nx We are interested in showing that (cos x + i sin x)n = cos nx + i sin nx which is exactly DeMoivre's Formula. It is obvious that this is true for any n. We can show that it is true for all n by using induction. (cos x + i sin x)n+1 = (cos x + i sin x)n(cos x + i sin x) From this we apply what we know about the nth case from above. = (cos nx + i sin nx)(cos x + i sin x) We can now multiply. = cos nx cos x + i sin x cos nx + i sin nx cos x + i2 sin nx sin x And from our work above we have already shown that these can be simplified into our sum formulas as such: cos (nx + x) + i sin (nx + x) = cos (n + 1)x + i sin (n + 1)x And so, we have shown that (cos x + i sin x)n = cos nx + i sin nx is true for all n. Thus we have shown that some very common trigonometric identities are related and can be derived from series expansions and complex numbers! Sign up for free to access more Math resources like . WyzAnt Resources features blogs, videos, lessons, and more about Math and over 250 other subjects. Stop struggling and start learning today with thousands of free resources!
# How do you determine the probability distribution of P(x>5)? ## The table: x ---- 4 ---- 5 ----- 6 ----- 7 ----- 8 P(x) ---0.1----0.1---0.2-----0.1----0.5 Mar 6, 2017 $P \left(X > 5\right) = 0.8$ #### Explanation: The standard notation is to use a lower case letter to represent an actual event, and an upper case letter for the Random Variable used to measure the probability of the event occurring. Thus the correct table would be: And then; $P \left(X > 5\right) = P \left(X = 6 \mathmr{and} X = 7 \mathmr{and} X = 8\right)$ $\text{ } = P \left(X = 6\right) + P \left(X = 7\right) + P \left(X = 8\right)$ $\text{ } = 0.2 + 0.1 + 0.5$ $\text{ } = 0.8$ Alternatively: $P \left(X > 5\right) = 1 - P \left(X \le 5\right)$ $\text{ } = 1 - P \left(X = 4 \mathmr{and} X = 5\right)$ $\text{ } = 1 - \left\{P \left(X = 4\right) + P \left(X = 5\right)\right\}$ $\text{ } = 1 - \left(0.1 + 0.1\right)$ $\text{ } = 0.8$, as before
Home » » The probability of an event A is 1/5. The probability of B is 1/3 . The probabilit... # The probability of an event A is 1/5. The probability of B is 1/3 . The probabilit... ### Question The probability of an event A is 1/5. The probability of B is 1/3 . The probability both A and B is 1/15. What is the probability of either event A or B or both? ### Options A) $$\frac{2}{15}$$ B) $$\frac{3}{4}$$ C) $$\frac{7}{15}$$ D) $$\frac{1}{15}$$ The correct answer is C. ### Explanation: Prob(A) = $$\frac{1}{5}$$ , Prob(B) = $$\frac{1}{4}$$, Prob (A ∩ B) = $$\frac{1}{15}$$, Prob (A ∪ B) = ? Note: I. the probability is either event of A or B or both. The formula is prob (A ∪ B) = ∩ prob(A) + prob(A) − prob(A ∩ B) II. But if the probability of both outcomes A and B The formula is Prob(A ∩ B) − prob(A) + prob(B) – prob (A ∪ B) In this question, A or B, Prob (A ∩ B): A and B, Prob (A ∩ B) prob (A ∪ B) = prob(A) + prob(B) − Prob (A ∩ B) is used. prob (A ∪ B) = $$\frac{1}{15}$$ + $$\frac{1}{3}$$ − $$\frac{1}{15}$$ = (3 + 5 + 1) ÷ 15 = $$\frac{7}{15}$$ ## Dicussion (1) • Prob(A) = $$\frac{1}{5}$$ , Prob(B) = $$\frac{1}{4}$$, Prob (A ∩ B) = $$\frac{1}{15}$$, Prob (A ∪ B) = ? Note: I. the probability is either event of A or B or both. The formula is prob (A ∪ B) = ∩ prob(A) + prob(A) − prob(A ∩ B) II. But if the probability of both outcomes A and B The formula is Prob(A ∩ B) − prob(A) + prob(B) – prob (A ∪ B) In this question, A or B, Prob (A ∩ B): A and B, Prob (A ∩ B) prob (A ∪ B) = prob(A) + prob(B) − Prob (A ∩ B) is used. prob (A ∪ B) = $$\frac{1}{15}$$ + $$\frac{1}{3}$$ − $$\frac{1}{15}$$ = (3 + 5 + 1) ÷ 15 = $$\frac{7}{15}$$
Select Page # Multiplying Polynomials Worksheet, Steps, and Examples Get the free Multiplying Polynomials Worksheet and other resources for teaching & understanding how to Multiply Polynomials Check Out Mathcation! ### Key Points about Multiplying Polynomials • Multiplying polynomials involves multiplying two or more polynomial expressions and simplifying the result. • The distributive property of multiplication is used to distribute each term of one polynomial to every term in the other polynomial. • Different methods such as the box method and the FOIL method can be used to multiply polynomials. ## Multiplying Polynomials: The Complete Guide Multiplying polynomials is an essential algebraic skill that is used in many mathematical applications. It involves multiplying two or more polynomial expressions and simplifying the resulting expression. Polynomials are expressions that contain variables raised to powers and are added or subtracted together. To multiply polynomials, one must know the distributive property of multiplication. This property states that the product of a number and a sum is equal to the sum of the products of the number and each term in the sum. In multiplying polynomials, this property is used to distribute each term of one polynomial to every term in the other polynomial. The resulting expression is then simplified by combining like terms. Multiplying polynomials can be done using different methods such as the box method and the FOIL method. The box method involves drawing a box and filling it with the terms of each polynomial. The resulting products are then added to obtain the final answer. The FOIL method involves multiplying the first terms, the outer terms, the inner terms, and the last terms of each polynomial. The resulting products are then added to obtain the final answer. Common Core Standard: Return To: Home, Algebra One ## How to Multiply Polynomials Multiplying polynomials is an essential skill in algebra. It involves multiplying two or more polynomials to get a new polynomial. There are different methods to multiply polynomials, depending on the number of terms involved. In this section, we will discuss how to multiply polynomials step-by-step. ### Multiplying a Monomial by a Polynomial To multiply a polynomial by a monomial, you need to apply the distributive property. The distributive property states that the product of a number and a sum is equal to the sum of the products of the number and each term in the sum. For example, to multiply 3x by 2x^2 + 5x – 1, you need to distribute 3x to each term in the polynomial. 3x(2x^2 + 5x – 1) = 6x^3 + 15x^2 – 3x ### Multiplying Binomials To multiply two binomials, you can use the FOIL method. FOIL stands for First, Outer, Inner, Last. It is a mnemonic device that helps you remember the order in which to multiply the terms. For example, to multiply (x + 2) and (x – 1), you need to multiply the first terms, then the outer terms, then the inner terms, and finally the last terms. (x + 2)(x – 1) = x^2 + x – 2x – 2 = x^2 – x – 2 ### Multiplying a Binomial by a Binomial To multiply a binomial by a binomial, you can also use the FOIL method. For example, to multiply (2x – 1) and (3x + 4), you need to multiply the first terms, then the outer terms, then the inner terms, and finally the last terms. (2x – 1)(3x + 4) = 6x^2 + 8x – 3x – 4 = 6x^2 + 5x – 4 ### Multiplying a Trinomial by a Binomial To multiply a trinomial by a binomial, you need to distribute each term in the trinomial to the binomial. For example, to multiply (2x^2 + 3x – 1) and (x + 2), you need to distribute each term in the trinomial to the binomial. (2x^2 + 3x – 1)(x + 2) = 2x^3 + 7x^2 + 5x – 2 ### Multiplying Polynomial Functions To multiply polynomial functions, you need to apply the distributive property and combine like terms. For example, to multiply (x^2 + 2x + 1) and (x^2 – 2x + 1), you need to distribute each term in the first polynomial to the second polynomial and combine like terms. (x^2 + 2x + 1)(x^2 – 2x + 1) = x^4 – x^2 + 2x^3 – 2x^2 + x^2 – 2x + x – 2x + 1 = x^4 + 2x^3 – 2x + 1 In conclusion, multiplying polynomials is an important skill in algebra. There are different methods to multiply polynomials, depending on the number of terms involved. By following the steps outlined in this section, you can multiply polynomials with ease. ## Steps in Multiplying Polynomials Multiplying polynomials is an essential skill in algebra. It involves multiplying each term of one polynomial by each term of the other polynomial and then combining like terms. Here are the steps to follow when multiplying polynomials: 1. Distribute the first polynomial: Write the first polynomial in front of the second polynomial. Then, distribute each term of the first polynomial to every term of the second polynomial. This means multiplying each term in the first polynomial by each term in the second polynomial. 2. Combine like terms: After distributing, combine like terms by adding or subtracting them. Like terms are terms that have the same variables raised to the same powers. 3. Simplify the result: Simplify the expression by combining any like terms that remain. Let’s illustrate these steps with an example. Suppose you need to multiply the polynomials (2x + 3) and (x – 4). Here are the steps to follow: 1. Distribute the first polynomial: (2x + 3) × (x – 4) = 2x × x + 2x × (-4) + 3 × x + 3 × (-4) = 2x² – 8x + 3x – 12 1. Combine like terms: 2x² – 5x – 12 1. Simplify the result: 2x² – 5x – 12 These steps can be applied to any polynomial multiplication problem, regardless of the degree or number of terms in the polynomials. It’s important to note that there are other methods for multiplying polynomials, such as the box method or FOIL method. However, the distributive method is the most general and can be used for any polynomial multiplication problem. ## Multiplying Polynomials Examples Multiplying polynomials involves multiplying each term in one polynomial by each term in the other polynomial and then adding the results. Here are some examples to illustrate the process: ### Example 1: Multiplying Two Binomials Let’s say we want to multiply the following two binomials: (3x + 2)(2x – 5) To do this, we need to multiply each term in the first binomial by each term in the second binomial and then add the results. This gives us: (3x)(2x) + (3x)(-5) + (2)(2x) + (2)(-5) = 6x^2 – 15x + 4x – 10 = 6x^2 – 11x – 10 So, (3x + 2)(2x – 5) = 6x^2 – 11x – 10. ### Example 2: Multiplying a Binomial and a Trinomial Let’s say we want to multiply the following binomial and trinomial: (2x + 3)(x^2 – 4x + 5) To do this, we need to multiply each term in the first polynomial by each term in the second polynomial and then add the results. This gives us: (2x)(x^2) + (2x)(-4x) + (2x)(5) + (3)(x^2) + (3)(-4x) + (3)(5) = 2x^3 – 8x^2 + 10x + 3x^2 – 12x + 15 = 2x^3 – 5x^2 – 2x + 15 So, (2x + 3)(x^2 – 4x + 5) = 2x^3 – 5x^2 – 2x + 15. ### Example 3: Multiplying Two Trinomials Let’s say we want to multiply the following two trinomials: (4x^2 + 3x – 2)(2x^2 – 5x + 1) To do this, we need to multiply each term in the first polynomial by each term in the second polynomial and then add the results. This gives us: (4x^2)(2x^2) + (4x^2)(-5x) + (4x^2)(1) + (3x)(2x^2) + (3x)(-5x) + (3x)(1) + (-2)(2x^2) + (-2)(-5x) + (-2)(1) = 8x^4 – 14x^3 + 2x^2 + 6x^3 – 15x^2 + 3x – 4x^2 + 10x – 2 = 8x^4 – 8x^3 – 13x^2 + 13x – 2 So, (4x^2 + 3x – 2)(2x^2 – 5x + 1) = 8x^4 – 8x^3 – 13x^2 + 13x – 2. ## Multiplying Polynomials Box Method One of the methods to multiply polynomials is using the box method, also known as the grid method. This method is particularly useful when multiplying polynomials with more than two terms. To use the box method, one needs to draw a box with the number of rows and columns equal to the number of terms in each polynomial. The terms of one polynomial are written on the top of the box, while the terms of the other polynomial are written on the left side of the box. Then, each term in one polynomial is multiplied by each term in the other polynomial and placed in the corresponding box cell. Finally, the terms in each row and column are added together to get the final product. Here is an example of multiplying two polynomials using the box method: (2x + 3)(3x – 4) \| 2x \| 3 –\|—–\|—- 3x\|6x^2 \|-8x -4\|9 \|-12 In the example above, the box has two rows and two columns, corresponding to the two terms in each polynomial. The terms of the first polynomial, 2x and 3, are written on the top of the box, while the terms of the second polynomial, 3x and -4, are written on the left side of the box. Each term in the first polynomial is multiplied by each term in the second polynomial and placed in the corresponding box cell. Finally, the terms in each row and column are added together to get the final product: 6x^2 – 8x + 9 – 12 = 6x^2 – 8x – 3. The box method can also be used to multiply polynomials with more than two terms. In this case, the box will have more rows and columns, and the process is similar to the example above. However, the box method can become cumbersome when dealing with very large polynomials, and other methods such as FOIL or distribution may be more efficient. Overall, the box method is a useful tool for multiplying polynomials, especially when dealing with polynomials with more than two terms. By breaking down the multiplication into smaller steps, the box method can make the process more manageable and less prone to errors. ## Multiplying Polynomials Foil Method The FOIL method is a popular technique for multiplying two binomials. It stands for First, Outer, Inner, and Last. The method involves multiplying the first terms of each binomial, then the outer terms, inner terms, and last terms, respectively. Finally, the four products are added together to get the final result. For example, consider the following binomials: (x + 2)(x + 3) Using the FOIL method, we get: (x + 2)(x + 3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6 The first term is x multiplied by x, which gives x^2. The outer terms are x multiplied by 3, which gives 3x. The inner terms are 2 multiplied by x, which gives 2x. The last terms are 2 multiplied by 3, which gives 6. Adding all the products gives the final answer. It is important to note that the FOIL method only applies to multiplying two binomials. For example, if we have the product of two trinomials, we cannot use the FOIL method directly. Instead, we need to use the distributive property of multiplication to expand the product. In addition to the FOIL method, there are other techniques for multiplying polynomials, such as the grid method and the box method. These methods can be useful for multiplying polynomials with more than two terms or for special cases where the FOIL method is not applicable. ## Multiplying Polynomials with Fractions Multiplying polynomials with fractions can be a bit tricky, but it’s not too difficult once you understand the process. The basic idea is to multiply the numerators and denominators separately and then simplify the resulting fraction. Here’s an example: Suppose you want to multiply the polynomials (2x – 1)/(x + 3) and (x^2 + 4)/(2x + 1). To do this, you would follow these steps: 1. Multiply the numerators together: (2x – 1) * (x^2 + 4) = 2x^3 + 7x – 4 2. Multiply the denominators together: (x + 3) * (2x + 1) = 2x^2 + 7x + 3 3. Write the result as a fraction: (2x^3 + 7x – 4)/(2x^2 + 7x + 3) It’s important to note that you should always simplify the resulting fraction if possible. In this case, the fraction cannot be simplified any further. Another thing to keep in mind when multiplying polynomials with fractions is that you should always factor the polynomials first if possible. This can make the process much easier and help you avoid mistakes. For example, suppose you want to multiply the polynomials (2x^2 + 3x – 1)/(x + 1) and (x^2 – 4)/(2x – 1). To do this, you would follow these steps: 1. Factor the polynomials: (2x^2 + 3x – 1)/(x + 1) = (2x – 1)(x + 1)/(x + 1) and (x^2 – 4)/(2x – 1) = (x – 2)(x + 2)/(2x – 1) 2. Cancel out the common factor: (2x – 1)(x + 2)/(2x – 1) 3. Write the result as a simplified fraction: x + 2 As you can see, factoring the polynomials first made the process much simpler and helped to avoid any potential mistakes. In summary, multiplying polynomials with fractions is not too difficult once you understand the process. The key is to multiply the numerators and denominators separately and simplify the resulting fraction if possible. Factoring the polynomials first can also make the process easier and help you avoid mistakes. ## How to Multiply Polynomials FAQ ### What is the box method for multiplying polynomials? The box method is a visual method for multiplying two polynomials. It involves drawing a box and placing the two polynomials in the box, with each term of one polynomial being multiplied by each term of the other polynomial. The resulting products are then added together to obtain the final answer. ### What is the formula for multiplying polynomials? The formula for multiplying two polynomials is to multiply each term of one polynomial by each term of the other polynomial, and then add the resulting products together. The distributive property of multiplication is used to simplify the process. ### What are some examples of multiplying polynomials with answers? • (x + 2)(x – 3) = x^2 – x – 6 • (2x – 3)(3x + 4) = 6x^2 + 5x – 12 • (x^2 + 2x + 1)(x – 1) = x^3 + x^2 – x – 1 ### What are the steps to multiplying polynomials by monomials? To multiply a polynomial by a monomial, you simply multiply each term of the polynomial by the monomial. This can be done using the distributive property of multiplication. ### What are the three ways to multiply polynomials? The three ways to multiply polynomials are: 1. Box method 2. Distributive property 3. FOIL method ### What is the most common mistake in multiplying polynomials? The most common mistake in multiplying polynomials is forgetting to distribute the terms properly. It is important to multiply each term of one polynomial by each term of the other polynomial, and then add the resulting products together. ### What are the 3 ways to multiply polynomials? As mentioned earlier, the three ways to multiply polynomials are: 1. Box method 2. Distributive property 3. FOIL method ### What is the common mistake in multiplying polynomials? The most common mistake in multiplying polynomials is forgetting to distribute the terms properly. It is important to multiply each term of one polynomial by each term of the other polynomial, and then add the resulting products together. ## Free Multiplying Polynomials Worksheet Download Enter your email to download the free Subtracting Polynomials worksheet ### Worksheet Downloads ###### Practice makes Perfect. 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# 2014 AMC 10B Problems/Problem 20 ## Problem For how many integers $x$ is the number $x^4-51x^2+50$ negative? $\textbf {(A) } 8 \qquad \textbf {(B) } 10 \qquad \textbf {(C) } 12 \qquad \textbf {(D) } 14 \qquad \textbf {(E) } 16$ ## Solution 1 First, note that $50+1=51$, which motivates us to factor the polynomial as $(x^2-50)(x^2-1)$. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so $x^2-50<0. Solving this inequality, we find $1. There are exactly $12$ integers $x$ that satisfy this inequality, $\pm \{2,3,4,5,6,7\}$. Thus our answer is $\boxed{\textbf {(C) } 12}$. ## Solution 2 Since the $x^4-51x^2$ part of $x^4-51x^2+50$ has to be less than $-50$ (because we want $x^4-51x^2+50$ to be negative), we have the inequality $x^4-51x^2<-50 \rightarrow x^2(x^2-51) <-50$. $x^2$ has to be positive, so $(x^2-51)$ is negative. Then we have $x^2<51$. We know that if we find a positive number that works, it's parallel negative will work. Therefore, we just have to find how many positive numbers work, then multiply that by $2$. If we try $1$, we get $1^4-51(1)^4+50 = -50+50 = 0$, and $0$ therefore doesn't work. Test two on your own, and then proceed. Since two works, all numbers above $2$ that satisfy $x^2<51$ work, that is the set {${2,3,4,5,6,7}$}. That equates to $6$ numbers. Since each numbers' negative counterparts work, $6\cdot2=\boxed{\textbf{(C) }12}$. ## Solution 3 (Graph) As with Solution $1$, note that the quartic factors to $(x^2-50)\cdot(x^2-1)$, which means that it has roots at $-5\sqrt{2}$, $-1$, $1$, and $5\sqrt{2}$. Now, because the original equation is of an even degree and has a positive leading coefficient, both ends of the graph point upwards, meaning that the graph dips below the $x$-axis between $-5\sqrt{2}$ and $-1$ as well as $1$ and $5\sqrt{2}$. $5\sqrt{2}$ is a bit more than $7$ ($1.4\cdot 5=7$) and therefore means that $-7,-6,-5,-4,-3,-2,2,3,4,5,6,7$ all give negative values. ## Solution 4 Let $x^{2}=u$. Then the expression becomes $u^{2}-51u+50$ which can be factored as $\left(u-1\right)\left(u-50\right)$. Since the expression is negative, one of $\left(u-1\right)$ and $\left(u-50\right)$ need to be negative. $u-1>u-50$, so $u-1>0$ and $u-50<0$, which yields the inequality $50>u>1$. Remember, since $u=x^{2}$ where $x$ is an integer, this means that $u$ is a perfect square between $1$ and $50$. There are $6$ values of $u$ that satisfy this constraint, namely $4$, $9$, $16$, $25$, $36$, and $49$. Solving each of these values for $x$ yields $12$ values (as $x$ can be negative or positive) $\Longrightarrow \boxed{\textbf {(C) } 12}$. ~JH. L
1. Chapter 10 Class 11 Straight Lines 2. Serial order wise 3. Miscellaneous Transcript Misc 11 Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x – 2y = 3. Let the equation line AB be x − 2y = 3 Let line CD pass through point (3, 2) & make an angle of 45° with line AB i.e. angle between CD & AB is 45° Let m1 be the slope of line AB & m2 be the slope of line CD Given angle between CD & AB is 45° i.e. θ = 45° Now, angle between two lines is given by tan θ =\|(𝑚_2 − 𝑚_1)/(1 + 𝑚_2 𝑚_1 )\| Here, m1 = Slope of line AB m2 = Slope of line CD & θ = angle between AB & CD = 45° Putting values tan 45° =\|(𝑚_2 − 𝑚_1)/(1 + 𝑚_2 𝑚_1 )\| 1 = \|(𝑚_2 − 𝑚_1)/(1 + 𝑚_2 𝑚_1 )\| Finding slope of line AB Given equation of line AB is x − 2y =3 − 2y = 3 − x y = (3 − 𝑥)/( − 2) y = 3/( − 2) + (( − 𝑥)/( − 2)) y = (−3)/2 + 𝑥/3 y = 𝑥/3 + ((−3)/2) y = 1/3 x – 3/2 The above equation is of the form y = mx + c Where m is slope of line Slope of line AB = 1/3 i.e. m1 = 1/3 Putting m1 = 1/3 in (1) 1 = \|(𝑚_2 − 1/2)/(1 + 𝑚_2 × 1/2)\| 1 = \|( (2𝑚_2 − 1)/2)/( (2 + 𝑚_2)/2)\| 1 = \|(2𝑚_2 − 1)/(2 + 𝑚_2 )\| \|(2𝑚_2 − 1)/(2 + 𝑚_2 )\| = 1 Hence, (2𝑚_2 − 1)/(2 + 𝑚_2 ) = 1 or (2𝑚_2 − 1)/(2 + 𝑚_2 ) = − 1 So, m2 = 3 or ( − 1)/3 ∴ Slope of line CD is 3 or ( − 1)/3 Now we need to calculate equation of line CD Equation of line passes through (x1, y1) & having slope m is (y – y1) = m(x – x1) Equation of line CD passing through (3, 2) & having slope m2 is (y – 2) = m2(x – 3) Hence required equation of line is 3x − y = 7 or x + 3y = 9 Miscellaneous
How do you factor 4x^2-12x+9? Apr 25, 2018 Answer: $\left(2 x - 3\right) \left(2 x - 3\right)$ or ${\left(2 x - 3\right)}^{2}$ Explanation: Use the rainbow method. Multiply the two outer coefficients. $4 \cdot 9 = 36$ Find two numbers, that when multiplied equal 36, and when added equal -12. -6 and -6. $- 6 + - 6 = 12$ $- 6 \cdot - 6 = 36$ Rewrite the equation with the x value replaced by the two new values. $4 {x}^{2} - 6 x - 6 x + 9$ Seperate the equation into two parts. $4 {x}^{2} - 6 x$ and $- 6 x + 9$ Find the GCF of the two parts. $2 x \left(2 x - 3\right)$ and $- 3 \left(2 x - 3\right)$ Take the GCF as your first factor, and the two remaining values as the second. 2x-3 and 2x-3 Apr 25, 2018 Answer: $4 {x}^{2} - 12 x + 9 = 4 {\left(x - \frac{3}{2}\right)}^{2}$ Explanation: First, factor x^2 $4 {x}^{2} - 12 x + 9 = 4 \left({x}^{2} - 3 x + \frac{9}{4}\right)$ Then, identify the y element in the form ${\left(x + y\right)}^{2} = {x}^{2} + 2 x y + {y}^{2}$ $4 \left({x}^{2} - 3 x + \frac{9}{4}\right) = 4 \left({x}^{2} - 2 \cdot \frac{3}{2} x + {\left(\frac{3}{2}\right)}^{2}\right)$ Last, factor the perfect square trinomial $4 \left({x}^{2} - 2 \cdot \frac{3}{2} x + {\left(\frac{3}{2}\right)}^{2}\right) = 4 {\left(x - \frac{3}{2}\right)}^{2}$
# Secant of an Angle – Formulas and Examples The secant of an angle can be calculated by relating the sides of a right triangle. The secant is defined as the reciprocal function of the cosine, so it is equal to the length of the hypotenuse over the length of the adjacent side. The secant of the most important angles are obtained by using the proportions of the known special triangles. Here, we will learn about secant using diagrams. We will derive a formula and apply it to solve some practice problems. ##### TRIGONOMETRY Relevant for Learning about the secant of an angle with examples. See examples ##### TRIGONOMETRY Relevant for Learning about the secant of an angle with examples. See examples ## Definition of secant of an angle The secant of an angle in a right triangle is defined as the length of the hypotenuse divided by the length of the side adjacent to the angle. Additionally, we can also define the secant of an angle as the reciprocal of the cosine. This is because the cosine is defined as the adjacent side on the hypotenuse, so by taking its reciprocal, we obtain the secant. $latex \sec (\theta)=\frac{1}{\cos}=\frac{H}{A}$ where H is the hypotenuse and A is the adjacent side. ## Formula for the secant in right triangles We are going to use the following right triangle ABC that has a right angle at C to find the ratio of the secant of an angle. We use lowercase letters to represent the lengths of the sides and we use capital letters to represent the angles of the triangle. For example, the letter a represents the side that is opposite angle A, the letter b represents the side that is opposite angle B, and the letter c represents the side that is opposite angle C. The secant of an angle in a triangle rectangle is equal to the hypotenuse divided by the adjacent side: In the triangle above, we have $latex \sec(A)=\frac{c}{b}$ and also $latex \sec(B)=\frac{c}{a}$. ## Secant for common special angles We can use the proportions of the sides of special triangles to find the values of the secants of important angles. For example, we can consider the right isosceles triangle, which has angles 45°-45°-90°. We find its proportions using the Pythagorean theorem: $latex {{c}^2}={{a}^2}+{{b}^2}$, but in this case, $latex a=b$, so we have $latex {{c}^2}=2{{a}^2}$. We conclude that $latex c = a \sqrt{2}$. Therefore, the secant of 45° is equal to $latex \sqrt{2}$. Additionally, we also use the right triangle with angles 30°-60°-90°. The ratios of the sides of this triangle are 1: $latex \sqrt{3}$: 2. Using these proportions, we have $latex \sec(30^{\circ}) = \frac{2}{\sqrt{3}}$, which is equivalent to $latex \frac{2 \sqrt{3}}{3}$. We also have $latex \sec(60^{\circ})= 2$. Finally, we can consider the angles 0 and 90°. When the angle is 0, the adjacent side like the hypotenuse is equal to 1 in the unit circle, so the secant of 0 is equal to 1. On the other hand, when the angle is 90°, the adjacent side is equal to 0 and the hypotenuse is equal to 1. However, we cannot divide by 0, so the secant of 90° is undefined. ## Secant of an angle – Examples with answers The following examples are solved using the secant formula seen above. Each of the following examples refers to the right triangle shown above, so we use the same notation for the sides and angles. ### EXAMPLE 1 What is the value of b if we have $latex \sec(A)= 1.7$ and $latex c = 5$? We use the right triangle above and look at the relationship $latex \sec(A) = \frac{c}{b}$. Therefore, we use this relation together with the given values to find the value of b: $latex \sec(A)=\frac{c}{b}$ $latex 1.7=\frac{5}{b}$ $latex b=\frac{5}{1.7}$ $latex b=2.94$ The value of the b side is equal to 2.94. ### EXAMPLE 2 If we have $latex a=8$ and $latex \sec(B) = 1.44$, determine the value of c. Using the right triangle above as a reference, we have $latex \sec(B) = \frac{c}{a}$. We use this relationship in conjunction with the given values to determine the value of c: $latex \sec(B)=\frac{c}{a}$ $latex 1.44=\frac{c}{8}$ $latex c=1.44(8)$ $latex c=11.52$ The value of the hypotenuse is 11.52. ### EXAMPLE 3 If we have $latex c = 2$ and $latex b = \sqrt{3}$, what is the value of angle A? Using the right triangle above as a reference, we can form the following relation $latex \sec(A) = \frac{c}{b}$. Therefore, using the given values, we have: $latex \sec(A)=\frac{c}{b}$ $latex \sec(A)=\frac{2}{\sqrt{3}}$ $latex \sec(A)=\frac{2\sqrt{3}}{3}$ Using a calculator with the function $latex {{\sec}^{-1}}$ or using the table above, we know that we have: $latex A=30$° Angle A measures 30°. ## Secant of an angle – Practice problems Use what you have learned about the secant of an angle to solve the following problems. These problems refer to the right triangle seen above, so they use the same notation for sides and angles. Choose an answer Choose an answer Choose an answer ## See also Interested in learning more about secant, cosecant, and cotangent? Take a look at these pages: ### Jefferson Huera Guzman Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students. LEARN MORE
# Drill #37 Factor the following polynomials Simplify. ## Presentation on theme: "Drill #37 Factor the following polynomials Simplify."— Presentation transcript: Drill #37 Factor the following polynomials Simplify. Find the value of r: Drill #38 Factor the following polynomials Simplify. Drill #60 Factor the following polynomial Evaluate the following roots: 7-4 Nth Roots Objective: To simplify radicals having various indices, and to use a calculator to estimate the roots of numbers. Square Roots What power is a square root? A square is the inverse of a square root… Square Root* Definition: For any real numbers a and b, if then a is a square root of b or We can also write square roots using the ½ power. Cube Root* Definition: For any real numbers a and b, if then a is a cube root of b or We can also write cube roots using the 1/3 power. nth Root* Definition: For any real numbers a and b, if then a is a nth root of b or We can also write nth roots using the power. How many ways can you … Multiply two numbers to get a positive #? Multiply two numbers to get a negative #? Examples: Roots (of powers of 2) Even Roots: Odd Roots: Principal Root* Definition: When there is more than one real root of a number (even numbered roots), the non-negative root is the principal root  Principal Root  Negative Root  Both Examples: Roots (of powers of 2) Both Principal Negative Roots: Roots: Roots: Roots of negative numbers* Even roots: Negative numbers have no even roots. (undefined) Odd Roots: Negative numbers have negative roots. Examples: Roots (of powers of 2) Even Roots: Odd Roots: Roots: Number and Types Even Roots Odd Roots Positive 2 (one positive, one negative) 1 (positive) Negative 0 (undefined) 1 (negative) Even Roots (of variable expressions)* When evaluating even roots (n is even) use absolute values (if resulting power is odd). Odd Roots (of variable expressions)* When evaluating odd roots (n is odd) do not use absolute values. Evaluating Roots of Monomials To evaluate nth roots of monomials: (where c is the coefficient, and x, y and z are variable expressions) or Simplify coefficients (if possible) For variables, evaluate each variable separately Evaluating Roots of Monomials* To find a root of a monomial Split the monomial into a product of the factors, and evaluate the root of each factor. Variables: divide the power by the root Coefficients: re-write the number as a product of prime numbers with powers, then divide the powers by the root. Use a calculator to find the roots* Find the following roots using a calculator (round to the 3 decimal places): Examples: Simplify Examples: Simplify
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 1.6: Polar Equations and Complex Numbers Difficulty Level: At Grade Created by: CK-12 ## Polar Coordinates Polar Coordinates Introduction On example 1, you may need to stress that we are moving clockwise because the \begin{align}\theta\end{align}-coordinate is negative, and that otherwise we would be moving counterclockwise. Sinusoids of One Revolution Introduction An important thing to explain about graphing in polar coordinates is that \begin{align}r\end{align} is always the dependent variable and \begin{align}\theta\end{align} is the independent variable—so the equations we graph in polar coordinates will take forms like \begin{align}r = \sin \theta\end{align}, or more complex expressions based on \begin{align}\theta\end{align}, where \begin{align}\theta\end{align} is always an angle measure. The fact that \begin{align}r\end{align} is always written first in the ordered-pair representation of a point is a little counterintuitive, because we have previously been used to seeing the independent variable (which is usually \begin{align}x\end{align}) written first. Why do we write \begin{align}r\end{align} first if \begin{align}r\end{align} is the dependent variable? One reason is that graphing points is easier, or at least more intuitive, if we look at the \begin{align}r-\end{align}coordinate first; we can think of ourselves as starting at the origin, moving along the \begin{align}x-\end{align}axis to the given \begin{align}r-\end{align}value, and then moving around the circle to the given \begin{align}\theta-\end{align}value. (Another reason has to do with the conventions for representing complex numbers in polar form, which will be covered later in this chapter.) Example 3 contains an interesting optical illusion: the diagonal lines passing through the graph make it look slightly warped and may prevent students from realizing that it is in fact a perfect circle. Also worth pointing out about this example is that all of the points on the graph have in fact been traced twice over. The \begin{align}\theta-\end{align}values from \begin{align}0^\circ\end{align} to \begin{align}180^\circ\end{align} traced out the circle; then the \begin{align}\theta-\end{align}values from \begin{align}180^\circ\end{align} to \begin{align}360^\circ\end{align} produced the same set of \begin{align}r-\end{align}values all over again, but negative. Since \begin{align}(-r, \theta + 180^\circ)\end{align} always represents the same point as \begin{align}(r, \theta)\end{align}, the second set of \begin{align}r-\end{align}values correspond to the same set of points as the first set. The important thing to point out about cardioids is that they are dimpled limaçons whose “dimple” passes directly through the pole. The depth of the dimple in a limaçon depends on the ratio \begin{align}\frac{a}{b}\end{align} (the smaller the ratio, the deeper the dimple), and the limaçon becomes a cardioid when the ratio equals \begin{align}1\end{align}. If the ratio got any smaller, the limaçon would dimple so far that it would develop a loop. Applications, Trigonometric Tools Introduction The constraints on example 1 should read “\begin{align}-2\pi \le \theta \le 2 \pi\text{''}\end{align} rather than \begin{align}0 \le \theta \le 2 \pi\text{''}\end{align}. ## Polar-Cartesian Transformations Graphs of Polar Equations Note that when we graph a basic cosine equation in polar coordinates, the domain only needs to go from to \begin{align}\pi\end{align} rather than \begin{align}2 \pi\end{align}. As noted in the previous lesson, the sine or cosine graph is traced out twice over on an interval of \begin{align}2 \pi\;\mathrm{units}\end{align}. In a sense, we can almost say that when we are using polar coordinates, the sine and cosine functions have a period of \begin{align}\pi\end{align} instead of \begin{align}2 \pi\end{align}. Conic Section Transformations Introduction An ellipse is actually the result of the intersection of a cone on one side by a plane that may or may not be parallel to the base of the cone. When the plane is parallel to the base, the ellipse is a circle. The plane that creates a parabola cannot just be non-parallel to the base; it must be parallel to the slanted line that forms the edge of the cone. The plane that creates a hyperbola does not actually have to be perpendicular to the base, as long as it intersects both halves of the cone. Also, you may need to clarify that the definition of “cone” used here differs from the one students encountered in geometry: a “cone” here is really two cones lined up tip to tip, and the two cones actually extend infinitely far outward from the point where they meet. Parabolas can theoretically be stretched horizontally as well as vertically, but stretching them horizontally is in a sense the same as shrinking (or “un-stretching”) them vertically, so it can be expressed in the same way, with a little adjusting of arbitrary constants. The focal axis of an ellipse is also called the major axis, and its length is denoted as 2a. The perpendicular line passing through the center of the ellipse is the minor axis and its length is 2b. This will be important later. Points to Consider Circles centered at the origin are certainly easier to express in polar coordinates, but those that have been shifted away from the origin may be a little harder. Parabolas tend to be easier to represent in rectangular coordinates. In general, taking a familiar equation and shifting or stretching it in one direction or another tends to be easier when the equation is expressed in rectangular coordinates. (These are just a few examples; students may provide others.) Polar curves may of course intersect, as we saw during this lesson. This question prepares students for the next lesson, where they will learn to find the intersection points of such curves. Since two different sets (in fact, infinitely many sets) of coordinates can be used to represent the same set of points, it makes sense that two different equations could be used to represent the same polar curve. For example, \begin{align}r = \sin \theta\end{align} would produce the same graph as \begin{align}r = \sin (\theta + 2\pi)\end{align}. One important difference between rectangular and polar representation is that polar graphs are more likely to stay within a finite viewing space. When we graph a function on a rectangular grid, if the function’s domain is unlimited, then the graph extends infinitely far to each side, so we can’t ever really draw the entire graph. Polar graphs, on the other hand, can extend infinitely far outward if the range is unlimited, but if the range is limited, then the domain can be unlimited and the graph will still be conveniently compact. (Then again, this can be an inconvenience at times, as it makes it harder to show when we are deliberately only graphing part of a function instead of the whole thing. Since the whole thing should fit in the visible part of the graph, viewers will expect the visible part of the graph to contain the whole function unless we include a note specifying otherwise. With rectangular coordinates, we can simply narrow the graphing window to show only the part we want to graph, and viewers don’t need to be told that there’s really more to the graph than just that part.) ## Systems of Polar Equations Graph and Calculate Intersections of Polar Curves Introduction In the solution to example 2, the notation \begin{align}k \epsilon I\end{align} may be unfamiliar to students; it means “\begin{align}k\end{align} is an element (a member) of the set of all integers,” or simply “\begin{align}k\end{align} is an integer.” So saying that the solution set includes \begin{align}\left (1, \frac{5 \pi}{3}\right ) + 2 \pi k, k \epsilon I\end{align} is simply another way of saying that when we add any integer multiple of \begin{align}2 \pi\end{align} to the \begin{align}\theta-\end{align}coordinate in the solution \begin{align}\left (1, \frac{5 \pi}{3}\right )\end{align}, we get another valid solution—and of course the same holds true for the solution \begin{align}\left (1, \frac{\pi}{3}\right )\end{align}. The solution to Example 3 may be confusing at first—how can \begin{align}(0,0)\end{align} and \begin{align}\left (0, \frac{\pi}{2}\right )\end{align} represent the same point? Students should grasp by now that adding a multiple of \begin{align}2 \pi\end{align} to the \begin{align}\theta-\end{align}coordinate of a point yields another representation of the same point, but in this case we’ve added \begin{align}\frac{\pi}{2}\end{align}, which isn’t a multiple of \begin{align}2 \pi\end{align}—so what’s going on here? The important thing to explain here is that the pole (as we call the origin when we are using polar coordinates) is a very special point. Normally, any given \begin{align}r-\end{align}coordinate designates a circle centered at the pole with radius \begin{align}r\end{align}, and we use different \begin{align}\theta-\end{align}coordinates to pick out specific points on the circle. But a circle with radius is just a single point—the pole itself—so no matter what \begin{align}\theta-\end{align}coordinate we choose, we always end up at that same point. \begin{align}(0,0)\end{align} represents the same point as \begin{align}(0,3), \left (0, \frac{\pi}{2}\right ), (0, 4\pi),\end{align} or any ordered pair whatsoever that has as the \begin{align}r-\end{align}coordinate. Points to Consider A very simple example of two polar curves that do not intersect is the pair \begin{align}r = 1\end{align} and \begin{align}r = 2\end{align}. And we have seen again in this lesson how the same point can be expressed in more than one way in polar coordinates; in the next lesson we will see how the same curve can too. Equivalent Polar Curves It is important that students do not get the mistaken idea that expressions with equivalent graphs are necessarily equivalent expressions. In example 1b, the two equations graphed are indeed equivalent, as they are both simply different ways of expressing \begin{align}r = 5\end{align}. But in example 1a, although the two equations trace out the same graph, they do not actually have the same \begin{align}r-\end{align}value for any given \begin{align}\theta-\end{align}value, and so are not equivalent equations. If they were equivalent, plugging the same \begin{align}\theta-\end{align}value into both of them would always yield the same \begin{align}r-\end{align}value. This is especially confusing because it only happens in polar coordinates, where the \begin{align}\theta\end{align}-values overlap and repeat themselves. Here’s an analogy that may help you to explain it: Suppose you ride the same bus to work or school every day, and suppose the bus maintains a very strict schedule, so it always reaches the same stop on its route at exactly the same time each day. (Of course no real bus could manage this, but let’s assume it does in order to simplify the problem.) Now suppose you draw a graph each day representing the route the bus travels, and when you compare two consecutive graphs, you see that they look exactly the same—if you plotted them on the same axes, they would look like just one graph. But does this mean they are describing the exact same trip? No—they represent two different trips taken on two different days, and when you plot them on the same axis you are simply leaving out the “two different days” part. Really, the time-values of the second graph are the time-values of the first graph “plus \begin{align}24\;\mathrm{hours}\end{align},” and if you choose to plot them on the same graph to save space, you must still remember that the times on the two graphs aren’t “really” the same. And that’s what happens when we use polar coordinates—we sometimes end up graphing the \begin{align}\theta-\end{align}values for two or more different \begin{align}r-\end{align}values in what looks like the same spot, but we must remember that just because they are sharing space, that doesn’t mean they are really the same coordinates. Even when a whole graph looks the same as another, sometimes it simply consists of a different set of values that happen to be graphed in the same places. ## Imaginary and Complex Numbers Recognize Introduction Here’s another example of why the rule \begin{align}\sqrt{ab} = \sqrt{a} \sqrt{b}\end{align} only applies if \begin{align}a\end{align} and \begin{align}b\end{align} are not both negative: Without that exception, we could apply the rule to \begin{align}\sqrt{36}\end{align} and express it as \begin{align}\sqrt{-4} \sqrt{-9}\end{align}, which would equal \begin{align}2i\;\mathrm{times}\ 3i\end{align}, or \begin{align}-6\end{align}. Technically, \begin{align}(-6)^2\end{align} is of course \begin{align}36\end{align}, but officially -6 is not the square root of 36, so that answer would be incorrect. The last line of the solution to example 2b should have a \begin{align}5\end{align}, rather than a \begin{align}3\end{align}, under the radical sign. Points to Consider Students needn’t know the answers to these questions; they are simply a preview of the next section. Standard Form of Complex Numbers (a + bi) Introduction Students may be a little confused by the statement that a and b are both real numbers in the standard form \begin{align}a + bi\end{align}. Clarify if necessary that the imaginary part of a complex number is \begin{align}bi\end{align}, not just \begin{align}b\end{align}; \begin{align}bi\end{align} is a pure imaginary number because it is a real number multiplied by \begin{align}i\end{align}. Students may not quite see why the answer to example 1c is expressed as it is. The answer is indeed in standard rectangular form, with \begin{align}a = 3\sqrt{2}\end{align} and \begin{align}b = -2\sqrt{2}\end{align}, but we traditionally put the \begin{align}i\end{align} in front of the radical sign so that it doesn’t look like it is included under the radical sign, and that makes it harder to see that the whole expression has the form \begin{align}a + bi\end{align}. After reviewing example 3, you might also want to challenge students to find the conjugate of a real number, like \begin{align}5\end{align}. (Answer: \begin{align}5\end{align} is really \begin{align}5 + 0i\end{align}, so its conjugate is \begin{align}5 - 0i\end{align}, or simply \begin{align}5\end{align} again. In other words, the conjugate of any real number is simply itself.) Points to Consider We will see in the next lesson what operations can be performed on complex numbers and with what results. Complex Number Plane If you’ve covered vectors recently, you might point out here that the absolute value of the complex number \begin{align}a + bi\end{align} is the same as the magnitude of the vector represented by the point \begin{align}(a,b)\end{align}. You may need to skip over the problem about two students walking home, since the original formulation of the problem is missing from the lesson. ## Operations on Complex Numbers Points to Consider When the roots of an equation are complex, we know that the graph of the equation does not intersect the \begin{align}x-\end{align}axis. Conversely, when the graph does not intersect the \begin{align}x-\end{align}axis, we know the roots are complex, and when it does, we know there are either two real roots or one real root repeated twice. Sums and Differences of Complex Numbers Points to Consider We will see in the next lesson how complex numbers can be expressed in polar form. Products and Quotients of Complex Numbers (conjugates) Introduction There is a typographical error in the formula for multiplying complex numbers: the term that reads \begin{align}(ad - bd)\end{align} should read \begin{align}(ac - bd)\end{align}. Students may not quite see where the \begin{align}-bd\text{''}\end{align} term comes from. Explain that multiplying \begin{align}bi\end{align} and \begin{align}di\end{align} yields \begin{align}bdi^2\end{align}, and \begin{align}i^2\end{align} is simply \begin{align}-1\end{align}, leaving us with \begin{align}-bd\end{align}. The procedure for dividing complex numbers may make more sense if you remind students that \begin{align}i\end{align} is equal to \begin{align}\sqrt{-1}\end{align}. When we express the quotient of two complex numbers as a fraction, substituting \begin{align}\sqrt{-1}\end{align} for \begin{align}i\end{align} shows that this fraction essentially has a radical in the denominator which we must rationalize. Multiplying the numerator and denominator by the conjugate of the denominator, then, is clearly the way to get the imaginary part out of the denominator; and once the denominator is a real number, we can divide both parts of the numerator by that real number and thus express the answer in standard form. Points to Consider Once again, we will see in the next lesson that the answer to both of these questions is “yes.” Applications, Trigonometric Tools Operations on Complex Numbers Example 2 provides an excellent illustration of the relationship between operations on complex numbers and operations on vectors. You may want to point out to students that when they solve a problem like this by working with the real and imaginary parts of complex numbers separately, they are really just resolving vectors into horizontal and vertical components and working with each component separately. But instead of using \begin{align}\hat i\end{align} and \begin{align}\hat j\end{align} to represent those components, they are using \begin{align}1\end{align} and \begin{align}i\end{align}, because the horizontal component is a multiple of the unit vector in the real direction or “\begin{align}1-\end{align}direction” and the vertical component is a multiple of the unit vector in the imaginary or “\begin{align}i-\end{align}direction.” ## Trigonometric Form of Complex Numbers Trigonometric Form of Complex Numbers: Relationships among \begin{align}x, y, r,\end{align} and \begin{align}\theta\end{align} In case students don’t immediately see why \begin{align}x = r \cos \theta\end{align} and \begin{align}y = r \sin \theta\end{align}, you can show them fairly easily on the diagram that \begin{align}\sin \theta = \frac{y}{r}\end{align}, and then solving for \begin{align}y\end{align} yields \begin{align}y = r \sin \theta\end{align}. Similar reasoning, of course, holds for \begin{align}x\end{align}. The Trigonometric or Polar Form of a Complex Number \begin{align}(r cis \theta)\end{align} The term \begin{align}\mathrm{cis} \theta\text{''}\end{align} is easier to remember if you point out that it is somewhat like an acronym, derived from \begin{align}\cos \theta + i \sin \theta.\text{''}\end{align} The term “argument” is also worth explaining, as it often appears elsewhere in mathematics. Generally it refers to the “input” of a given function, so for example, in the expression \begin{align}\cos \theta,\text{''}\end{align} \begin{align}\theta\end{align} is called the argument of the cosine function. In the polar form of a complex number, of course, \begin{align}\theta\end{align} appears as the argument of both the sine and cosine functions, so it makes a kind of sense to call it the “argument” of the complex number as a whole. Thinking of r as the “absolute value” of a complex number may be counterintuitive for students, but it is really just an extension of the idea of absolute value of real numbers. The absolute value of a real number is its distance from ; the absolute value of a complex number is its distance from the point \begin{align}(0,0)\end{align}. And in the complex plane, the distance of a real number from becomes the same thing as its distance from \begin{align}(0,0)\end{align}. Trigonometric Form of Complex Numbers: Steps for Conversion Introduction The very first table in this section is the most useful for students to know, and it is most helpful for them to think of the left half and the right half separately. Emphasize that if they know the polar coordinates \begin{align}r\end{align} and \begin{align}\theta\end{align}, they should use the two equations on the left to find the rectangular coordinates \begin{align}x\end{align} and \begin{align}y\end{align}, whereas if they know \begin{align}x\end{align} and \begin{align}y\end{align} they should use the two equations on the right to find \begin{align}r\end{align} and \begin{align}\theta\end{align}. (They could use the equations on the right to get \begin{align}x\end{align} and \begin{align}y\end{align} from \begin{align}r\end{align} and \begin{align}\theta\end{align}, or use the equations on the left to get \begin{align}r\end{align} and \begin{align}\theta\end{align} from \begin{align}x\end{align} and \begin{align}y\end{align}, but that would be a much messier process.) The last step of Example 3 is important: finding the inverse tangent just tells us the reference angle for \begin{align}\theta,\end{align} not \begin{align}\theta\end{align} itself. We must then apply our knowledge of what quadrant the complex number is in to figure out what angle \begin{align}\theta\end{align} really is. We can find out what quadrant the number is in by graphing the rectangular coordinates we started out with, or by simply noting the signs of those coordinates: \begin{align}x\end{align} is only positive in the first two quadrants, and \begin{align}y\end{align} is only positive in the first and fourth. Points to Consider In the next lesson, we will see how to perform basic operations such as multiplication and division on complex numbers in polar form. ## Product and Quotient Theorems Product Theorem The first line of the derivation here uses the FOIL method for multiplying binomials; in the second line, we group together the terms with \begin{align}i\end{align} in them and factor out the \begin{align}i\end{align}; and in the third line, we apply the angle sum rules for sine and cosine in reverse. The Product Theorem may be easier for students to remember if summarized in words: “To multiply complex numbers in polar form, multiply the \begin{align}r-\end{align}coordinates and add the \begin{align}\theta-\end{align}coordinates.” Quotient Theorem Similarly, the Quotient Theorem can be summarized: “To divide complex numbers in polar form, divide the \begin{align}r-\end{align}coordinates and subtract the \begin{align}\theta-\end{align}coordinates.” The derivation follows much the same procedure as the one for the product theorem; note that the denominator of the final fraction equals \begin{align}1\end{align}, so we can cancel it out. Incidentally, some students may notice that we haven’t discussed how to add and subtract complex numbers in polar form. It turns out that there is no handy formula for doing so; the only good way to add and subtract complex numbers is to convert them to rectangular form first. Using the Quotient and Product Theorem Introduction Students may be confused by example 3, or may be tempted to do the division on the numbers as given, in rectangular form. Doing it as the book suggests, however, will give them practice in converting from rectangular to polar form as well as in performing division on numbers in polar form. Meanwhile, the other three examples will give them practice in working with complex numbers expressed in polar form in more than one way. For extra practice, you might have them find both the product and the quotient of the two numbers in each example, instead of just the product or the quotient. Points to Consider So far, we’ve actually just barely touched on squares and square roots of complex numbers, and we still don’t know how to find them for most numbers. The next lesson will cover these as well as other powers and roots of complex numbers. Applications and Trigonometric Tools: Real-Life Problem You may need to skip these problems, as they contain terms that students are not likely to know . ## Powers and Roots of Complex Numbers De Moivre’s Theorem A more intuitive way to express De Moivre’s Theorem is “To raise a complex number to the \begin{align}n^{\mathrm{th}}\end{align} power, raise the \begin{align}r-\end{align}coordinate to the \begin{align}n^{\mathrm{th}}\end{align} power and multiply the \begin{align}\theta-\end{align}coordinate by \begin{align}n\end{align}.” In example 2, the expression \begin{align}\left (\frac{-1}{2} + \frac{\sqrt{3}} {2}\right )\end{align} should read \begin{align}\left (\frac{-1}{2} + \frac{\sqrt{3}} {2} i \right )\end{align}. nth Root Theorem Here’s a much more intuitive way to explain the \begin{align}n^{\mathrm{th}}\end{align} Root Theorem: Every complex number has exactly \begin{align}n \ n^{\mathrm{th}}\end{align} roots, which are evenly spaced around a circle in the complex plane. If the original number has coordinates \begin{align}(r, \theta)\end{align}, then the first of the \begin{align}n^{\mathrm{th}}\end{align} roots (which, incidentally, is known as the principal root) has coordinates \begin{align}\left (\sqrt[n]{r}, \frac{\theta}{n}\right )\end{align}. The rest of the \begin{align}n^{\mathrm{th}}\end{align} roots all have the same \begin{align}r-\end{align}coordinate, and their \begin{align}\theta-\end{align}coordinates are each \begin{align}\frac{\theta}{n}\end{align} plus some multiple of \begin{align}\frac{2 \pi}{n}\end{align}; in other words, each of them is \begin{align}\frac{1}{n}\end{align} of the way around the circle from the one before it. For example, the fourth roots of \begin{align}(16, 60^\circ)\end{align} are \begin{align}(2, 15^\circ), (2, 105^\circ), (2, 195^\circ),\end{align} and \begin{align}(2, 285^\circ)\end{align}. Plotting these points shows that they are evenly spaced around a circle of radius \begin{align}2\end{align}, and a little thought will show why. Raising \begin{align}2\end{align} to the fourth power of course gives us \begin{align}16\end{align}, and multiplying an angle of \begin{align}15^\circ\end{align} by \begin{align}4\end{align} gives us \begin{align}60^\circ\end{align} —but multiplying an angle of \begin{align}15^\circ\end{align}-plus-some-multiple-of-\begin{align}90^\circ\end{align} by \begin{align}4\end{align} gives us \begin{align}60^\circ\end{align}-plus-some-multiple-of-\begin{align}360^\circ\end{align}, which is equivalent to \begin{align}60^\circ\end{align}. So that’s why, if \begin{align}(2, 15^\circ)\end{align} is one fourth root of \begin{align}(16, 60^\circ)\end{align}, all the other fourth roots are of the form \begin{align}(2, 15^\circ + k90^\circ)\end{align} for some integer \begin{align}k\end{align}. (If they were fifth roots, they’d be \begin{align}\frac{360^\circ}{5}\end{align} or \begin{align}72^\circ\end{align} apart; if they were sixth roots they’d be \begin{align}60^\circ\end{align} apart, and so on.) Incidentally, we could just keep going, adding \begin{align}90^\circ\end{align} to each previous \begin{align}\theta-\end{align}coordinate to get yet another one. But of course, after we’ve added \begin{align}90^\circ\end{align} three times to the first solution to get three more solutions, adding \begin{align}90^\circ\end{align} one more time would just give us the first solution plus \begin{align}360^\circ\end{align}, or in other words the first solution all over again, and then we’d start cycling through all the solutions over again. So the first four solutions (the first \begin{align}n\end{align} solutions, if we’re finding \begin{align}n^{\mathrm{th}}\end{align} roots) are the only unique ones, and we can stop after finding them. Solve Equations It should become clear here that we can use polar coordinates to determine the \begin{align}n^{\mathrm{th}}\end{align*} roots of a pure real number or pure imaginary number just as easily as a complex number. However, the roots are usually complex and tend to be somewhat messy to express in rectangular coordinates. For example, in the problem shown in the text, we can see that the roots can be expressed precisely in polar coordinates, and that we can use just one expression to summarize them all, whereas in rectangular coordinates we must list all the roots separately and can only express them in decimal approximations. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Please to create your own Highlights / Notes Show Hide Details Description Tags: Subjects:
# Question 9e045 Aug 6, 2017 ${f}^{- 1} \left(27\right) = 9$ #### Explanation: The trick here is to realize that for a given value, the inverse function has its output as the input of the original function for the same value. This means that if you plug something into the original function (the red apples), you will get an output (the yellow pears). If you then plug that output (the yellow pears) into the inverse function, you will get the something (the red apples) that you plugged into the original function. The problem wants you to determine the output of the inverse function of $f \left(x\right)$, which we label as ${f}^{- 1} \left(x\right)$, when the input of the inverse function is equal to $\textcolor{b l u e}{27}$. f^(-1)(color(blue)(27)) = ? In other words, you need to find the input of $f \left(x\right)$ when its output is equal to $\textcolor{b l u e}{27}$. So you need f(?) = 3 * ? = color(blue)(27) This will give you ? = color(blue)(27)/3 = color(red)(9)# Therefore, ${f}^{- 1} \left(\textcolor{b l u e}{27}\right) = \textcolor{red}{9}$. So, if you plug $\textcolor{red}{9}$ into $f \left(x\right)$ you will get an output of $\textcolor{b l u e}{27}$. If you then plug $\textcolor{b l u e}{27}$ into the inverse function, you will get $\textcolor{red}{9}$. $f \left(\textcolor{red}{9}\right) = 3 \cdot \textcolor{red}{9} = \textcolor{b l u e}{27}$ ${f}^{- 1} \left(\textcolor{b l u e}{27}\right) = \frac{\textcolor{b l u e}{27}}{3} = \textcolor{red}{9}$ This shows that the inverse function takes the form ${f}^{- 1} \left(x\right) = \frac{x}{3}$
# Ex.14.1 Q1 Factorization - NCERT Maths Class 8 Go back to 'Ex.14.1' ## Question Find the common factors of the terms (i) $$12x,\;\,36$$ (ii) $$2y,\;\,22xy$$ (iii) $$14pq,\;\,28{p^2}{q^2}$$ (iv) $$2x,\;\,3{x^2},\;\,4$$ (v) $$6abc,\;\,24a{b^2},\;\,12{a^2}b$$ (vi) $$16{x^3},\; - 4{x^2},\;32x$$ (vii) $$10pq,\;20qr,\;30rp$$ (viii) $$3{x^2}{y^3},\;\,10{x^3}{y^2},\;\,6{x^2}{y^2}z$$ Video Solution Factorisation Ex 14.1 \| Question 1 ## Text Solution What is known: Terms. What is unknown: Common factors of given terms. Reasoning: First we will find factors of each terms then find out which factors are common in each term. Steps: \begin{align}({\rm{i}})\quad 12 x &= 2 \times 2 \times 3 \times x\\36 &= 2 \times 2 \times 3 \times 3\end{align} The common factors are $$2, 2, 3.$$ And, $$~2\times 2\times 3=12$$ \begin{align}({\rm{ii}})\quad 2y &= 2 \times y\\22xy &= 2 \times 11 \times x \times y\end{align} The common factors are $$2, y.$$ And, $$2 \times y = 2y$$ \begin{align}({\rm{iii}})\quad {\rm{14 }}pq &= {\rm{2}} \times {\rm{7}} \times p \times q\\{\rm{28}}{p^{\rm{2}}}{q^{\rm{2}}} &= 2 \times 2 \times 7 \times p \times p \times q \times q \end{align} The common factors are $$2, 7, p, q.$$ And, $$2 \times 7 \times p \times q = 14pq$$ \begin{align}({\rm{iv}})\quad 2x &= {\rm{2}} \times x\\3x^2 &= 3 \times x \times x\\4 &= 2 \times 2\end{align} The common factor is $$1.$$ \begin{align}({\rm{v}}) \quad 6abc &= 2 \times 3 \times a \times b \times c\\24a{b^2} &= 2 \times 2 \times 2 \times 3 \times a \times b \times b \\12{a^2}b &= 2 \times 2 \times 3 \times a \times a \times b \end{align} The common factors are $$2,\, 3,\, a,\, b.$$ And, $$2 \times 3 \times a \times b = 6ab$$ \begin{align}({\rm{vi}})\quad16{x^3} &= 2 \times 2 \times 2 \times 2 \times x \times x \times x \\ - 4{x^2} &= - 1 \times 2 \times 2 \times x \times x\\32x &= 2 \times 2 \times 2 \times 2 \times 2 \times x \end{align} The common factors are $$2,\, 2,\, x.$$ And,$$2\times 2\times x=4x$$ \begin{align}({\rm{vii}})\quad 10pq &= 2 \times 5 \times p \times q\\20\,qr &= 2 \times 2 \times 5 \times q \times r\\30\,rp &= 2 \times 3 \times 5 \times r \times p\end{align} The common factors are $$2, \;5.$$ And, $$2 \times 5 = 10$$ \begin{align}({\rm{viii}})\quad 3{x^2}{y^3} &= 3 \times x \times x \times y \times y \times y \\ 10{x^3}{y^2} &= 2 \times 5 \times x \times x \times x \times y \times y \\6{x^2}{y^2}z &= 2 \times 3 \times x \times x \times y \times y \times z \end{align} The common factors are $$x, \,x,\, y, \,y.$$ And, $$x \times x \times y \times y = {x^2}{y^2}$$ 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
Courses Courses for Kids Free study material Offline Centres More Last updated date: 05th Dec 2023 Total views: 279.3k Views today: 5.79k # You are given several identical resistors each of value of $5\,\Omega$ and each capable of carrying a maximum current of $2\,A$. It is required to make a suitable combination of these resistances to produce a resistance of $\,2.5\,\Omega$ which can carry current of $4\,A$. the minimum number of resistances required for this job is? Verified 279.3k+ views Hint: In series current remains the same and equivalent resistance is greater than the individual resistances. In parallel combination the current gets divided and the equivalent resistance is lesser that the individual resistances. Series combination: -The resistors are in series combination when same current passes through each of the resistors present in the combination. -Voltage across each resistance is directly proportional to its value Hence, ${V_1}\, = \,I \times {R_1}$ ,${V_2}\, = \,I \times {R_2}$, ${V_3}\, = \,I \times {R_3}$ -Sum of voltage across each individual resistor is equal to the Voltage applied across the whole combination. $V\, = \,{V_1} + {V_2} + {V_3}\,$ $\Rightarrow I \times R\,\, = \,I \times {R_1} + I \times {R_2} + I \times {R_3}$ $\Rightarrow \,R\, = \,{R_1} + {R_2} + {R_3}$ where $R$ is equivalent resistance of the combination. Parallel combination: -The resistors are in parallel when each of the resistors present in the combination has the same voltage difference across them. -Current in each resistance is inversely proportional to the value of resistance. ${I_{1\,}}\, = \,\dfrac{V}{{{R_1}}}$, ${I_2}\, = \,\dfrac{V}{{{R_2}}}$, ${I_3}\, = \,\dfrac{V}{{{R_3}}}$. Current flowing in the circuit is sum of the currents in individual resistance. $I\, = \,\,{I_1} + {I_2} + {I_3}\,$ $\Rightarrow \dfrac{V}{R} = \,\dfrac{V}{{{R_1}}}\, + \,\,\dfrac{V}{{{R_2}}}\, + \,\,\dfrac{V}{{{R_3}}}$ $\Rightarrow \dfrac{1}{R} = \,\dfrac{1}{{{R_1}}}\, + \,\,\dfrac{1}{{{R_2}}}\, + \,\,\dfrac{1}{{{R_3}}}$. To get maximum resistance, resistance must be connected in series and in series the resultant is greater than the individual. To get minimum resistance, resistance must be connected in parallel and the equivalent resistance of parallel combination is lower than the value of lowest resistance in the combination. As the equivalent resistance of the combination is lower than the individual resistances taken so from above two statements it is clear that the combination is a parallel combination. Also, as the current has increased (doubled) so the combination can’t be series as the same current flows through each of the resistance in series combination.Hence, it is clear that the combination of resistances in parallel combination. Let the number of resistances used be $x$. Given resistance of the resistors = $5\,\Omega$ $\therefore \,\,\dfrac{1}{{{R_{eq}}}}\, = \,\dfrac{1}{{{R_1}}}\, + \,\dfrac{1}{{{R_2}}}\, + \,\,\dfrac{1}{{{R_3}}}.......$ Here all resistors are of same resistance ($5\,\Omega$) $\therefore \,\,\dfrac{1}{{{R_{eq}}}}\, = \,\dfrac{1}{5}\, + \,\dfrac{1}{5}\, + \,\,\dfrac{1}{5}.......\,x\,times$ $\Rightarrow \,\,\dfrac{1}{{2.5}}\, = \,\dfrac{x}{5}\,$ $\Rightarrow \,\,x\,\, = \,\,2$ Hence two resistances in parallel make the desired combination. Verifying this combination with current. As we know that current gets divided in the inverse ratio of current. Hence, here the current will get halved. So, if $4\,A$ is passed through the combination. Each resistance will have $2\,A$ of current through them, which is inside their capacity. The required combination is two $5\,\Omega$ resistance in parallel as shown in diagram. Note: Current is constant in series combination whereas voltage is constant in parallel combination. In our home parallel combinations are there to avoid short circuits and to not operate all the appliances simultaneously.
# How do you divide (-x^5-4x^3+x-12)/(x^2-x+3)? Jul 10, 2018 The remainder is $= \left(8 x - 15\right)$ and the quotient is $= \left(- {x}^{3} - {x}^{2} - 2 x + 1\right)$ #### Explanation: Perform a long division $\textcolor{w h i t e}{a a}$$- {x}^{5} + 0 {x}^{4} - 4 {x}^{3} + 0 {x}^{2} + x - 12$$\textcolor{w h i t e}{a a}$$\|$${x}^{2} - x + 3$ $\textcolor{w h i t e}{a a}$$- {x}^{5} + {x}^{4} - 3 {x}^{3}$$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a}$$\|$$- {x}^{3} - {x}^{2} - 2 x + 1$ $\textcolor{w h i t e}{a a a a}$$0 - 1 {x}^{4} - 1 {x}^{3} + 0 {x}^{2}$ $\textcolor{w h i t e}{a a a a a a}$$- 1 {x}^{4} + 1 {x}^{3} - 3 {x}^{2}$ $\textcolor{w h i t e}{a a a a a a a a a}$$0 - 2 {x}^{3} + 3 {x}^{2} + x$ $\textcolor{w h i t e}{a a a a a a a a a a a}$$- 2 {x}^{3} + 2 {x}^{2} - 6 x$ $\textcolor{w h i t e}{a a a a a a a a a a a a a a}$$0 + {x}^{2} + 7 x - 12$ $\textcolor{w h i t e}{a a a a a a a a a a a a a a a a}$$+ {x}^{2} - 1 x + 3$ $\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a}$$- 0 + 8 x - 15$ Therefore, $\frac{- {x}^{5} + 0 {x}^{4} - 4 {x}^{3} + 0 {x}^{2} + x - 12}{{x}^{2} - x + 3}$ $= \left(- {x}^{3} - {x}^{2} - 2 x + 1\right) + \frac{8 x - 15}{{x}^{2} - x + 3}$ The remainder is $= \left(8 x - 15\right)$ and the quotient is $= \left(- {x}^{3} - {x}^{2} - 2 x + 1\right)$
# 0.2 Motion in one dimension (Page 2/16) Page 2 / 16 For this chapter, we will only use frames of reference in the $x$ -direction. Frames of reference will be covered in more detail in Grade 12. ## Position Position Position is a measurement of a location, with reference to an origin. A position is a measurement of a location, with reference to an origin. Positions can therefore be negative or positive. The symbol $x$ is used to indicate position. $x$ has units of length for example cm, m or km. [link] shows the position of a school. Depending on what reference point we choose, we can say that the school is $300\phantom{\rule{2pt}{0ex}}\mathrm{m}$ from Joan's house (with Joan's house as the reference point or origin) or $500\phantom{\rule{2pt}{0ex}}\mathrm{m}$ from Joel's house (with Joel's house as the reference point or origin). The shop is also $300\phantom{\rule{2pt}{0ex}}m$ from Joan's house, but in the opposite direction as the school. When we choose a reference point, we have a positive direction and a negative direction. If we choose the direction towards the school as positive, then the direction towards the shop is negative. A negative direction is always opposite to the direction chosen as positive. ## Discussion : reference points Divide into groups of 5 for this activity. On a straight line, choose a reference point. Since position can have both positive and negative values, discuss the advantages and disadvantages of choosing 1. either end of the line, 2. the middle of the line. This reference point can also be called “the origin". ## Position 1. Write down the positions for objects at A, B, D and E. Do not forget the units. 2. Write down the positions for objects at F, G, H and J. Do not forget the units. 3. There are 5 houses on Newton Street, A, B, C, D and E. For all cases, assume that positions to the right are positive. 1. Draw a frame of reference with house A as the origin and write down the positions of houses B, C, D and E. 2. You live in house C. What is your position relative to house E? 3. What are the positions of houses A, B and D, if house B is taken as the reference point? ## Displacement and distance Displacement Displacement is the change in an object's position. The displacement of an object is defined as its change in position (final position minus initial position). Displacement has a magnitude and direction and is therefore a vector. For example, if the initial position of a car is ${x}_{i}$ and it moves to a final position of ${x}_{f}$ , then the displacement is: ${x}_{f}-{x}_{i}$ However, subtracting an initial quantity from a final quantity happens often in Physics, so we use the shortcut $\Delta$ to mean final - initial . Therefore, displacement can be written: $\Delta x={x}_{f}-{x}_{i}$ The symbol $\Delta$ is read out as delta . $\Delta$ is a letter of the Greek alphabet and is used in Mathematics and Science to indicate a change in a certain quantity, or a final value minus an initial value. For example, $\Delta x$ means change in $x$ while $\Delta t$ means change in $t$ . where we get a research paper on Nano chemistry....? what are the products of Nano chemistry? There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others.. learn Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level learn da no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts Bhagvanji hey Giriraj Preparation and Applications of Nanomaterial for Drug Delivery revolt da Application of nanotechnology in medicine what is variations in raman spectra for nanomaterials I only see partial conversation and what's the question here! what about nanotechnology for water purification please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment. Damian yes that's correct Professor I think Professor Nasa has use it in the 60's, copper as water purification in the moon travel. Alexandre nanocopper obvius Alexandre what is the stm is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.? Rafiq industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong Damian How we are making nano material? what is a peer What is meant by 'nano scale'? What is STMs full form? LITNING scanning tunneling microscope Sahil how nano science is used for hydrophobicity Santosh Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq Rafiq what is differents between GO and RGO? Mahi what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq Rafiq if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION Anam analytical skills graphene is prepared to kill any type viruses . Anam Any one who tell me about Preparation and application of Nanomaterial for drug Delivery Hafiz what is Nano technology ? write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? why we need to study biomolecules, molecular biology in nanotechnology? ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. why? what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good Got questions? Join the online conversation and get instant answers!
# C Program to Find the Sum of First N Natural Numbers devquora Posted On: Dec 23, 2020 ### Overview: In this article, We will discuss the C Program to Find the Sum of First N Natural Numbers. First of all I would like to explain what a natural number is? Then, I will demonstrate how to calculate the nth term of the natural number. Moreover, I would explain the logic to Find the Sum of First N Natural Numbers. Furthermore, I would write the logic and C Program to Find the Sum of First N Natural Numbers with output. ### What is Number? Number is an object that uses digits to perform mathematical tasks. Calculus that is a branch of mathematics included many numbers such as integers, whole numbers, real numbers, and imaginary numbers. Here, I will discuss numbers to perform arithmetic addition operations. Moreover, Number is a combination of digits, some symbols and decimal points. For instance, 127.23 is a rational number. There are many numbers such as whole numbers, natural numbers(positive integers), negative integers, rational numbers etc. ### What is Natural Number? Natural numbers are the numbers which are positive integers and include numbers from 1 till infinity(∞). These numbers are countable and are generally used for calculation purposes. The set of natural numbers is represented by the letter “N”. N = {1,2,3,4,5,6,7,8,9,10……,∞} ### What is the formula to find the Sum of First N Natural Numbers? Sums of the First n Natural Numbers The natural numbers are 1,2,3,4,... so on The sum of the first n natural numbers, Sn is as follows: 1 + 2 + 3 + 4 + ... + n Sn = 1 + 2 + 3 + 4 + ... + n As we know, that Sn=n[2a+(n−1)d]/2 According to 'n' natural numbers, The first term = a = 1 The common difference = d = 1 ∴ Sn=n[2×1+(n−1)×1]/2 Sn=n[2+n−1]/2 Sn=n[n+1]/2 ### Demonstration to Find the Sum of First N Natural Numbers Let's look at this problem for n=1, 2, 3, 4, and 5 and calculate the sum: 1=1 1+2=3 1+2+3=6 1+2+3+4=10 1+2+3+4+5=15 What is the formula to calculate this sum of n natural number ? The sum of n Natural numbers is calculated by using following formula: ### Logic to Find the Sum of First N Natural Numbers • Step 1:Declare and initialize variable, i,num, s=0. • Step 2:Enter the value of num i.e. number upto which sum is to be calculated Use Loop to add number. • Step 3:Any of while, do-while or for loop can be used. • Step 4:Print the sum ### C Program to Find the Sum of First N Natural Numbers ```#include <stdio.h> int main() { int n, count, sum = 0; printf("Enter the value of n(positive integer): "); scanf("%d",&n); for(count=1; count <= n; count++) { sum = sum + count; } printf("Sum of first %d natural numbers is: %d",n, sum); return 0; } ``` The output of the Program with code: When the above code is executed, it produces the following results: Enter the value of n(positive integer): 6 Sum of first 6 natural numbers is: 21 Conclusion: This C program to add first N natural numbers includes a single header file: stdio.h to control standard input and output function. As the program is executed, it asks for the value of the number to which the sum of natural numbers is to be found out. Then, the numbers are added using for loop. It can also be done with for and while loop as well. Finally, it displays the sum of numbers stored in integer variable ‘s’. Never Miss an Articles from us. ## Related Articles #### C Program to convert Decimal to Binary Number A Decimal Number is constructed with any digit from 0 to 9. For instance, 24 is a Decimal Number constructed from the digits 2 and 4. A Binary Number is constructed with digits 0 and 1. For instance, .. #### C program to reverse a string This program reversed the entered string that means opposite of the previous string sequence... #### C program to reverse an Integer number Are you want a demonstration of The C Program that reverses an Integer number? This Program reversing the sequence of digits in an Integer. After the successful compilation of the program, a message..
Home » Math Theory » Geometry » Ordered Pairs # Ordered Pairs ## Introduction A pair of items that have a specific significance for the order of their placements is called an ordered pair. Ordered pairs are typically used to represent a point on a coordinate plane in coordinate geometry. They can also be used to indicate relationships between items. You’ve probably already encountered ordered pairs without recognizing them if you have used a video game or consulted a map. Let us explore ordered pairs further, including their definition, significance, attributes, and more. Understanding ordered pairs is generally introduced to students around the 4th grade. But don’t worry if you’re younger or older; the beauty of learning is that it has no age limit. You are in the right place whether you’re 8, 15, or even a little beyond! ## Math Domain Ordered pairs belong to the domain of mathematics called “Coordinate Geometry.” Coordinate Geometry is like a bridge combining algebra and geometry to help us understand shapes, sizes, and positions. ## Applicable Common Core Standards CCSS.MATH.CONTENT.5.G.A.1: Use a pair of perpendicular number lines, called axes, to define a coordinate system, with the intersection of the lines (the origin) arranged to coincide with the 0 on each line and a given point in the plane located by using an ordered pair of numbers, called its coordinates. CCSS.MATH.CONTENT.6.NS.C.6.C: Find and position integers and other rational numbers on a horizontal or vertical number line diagram; find and position pairs of integers and other rational numbers on a coordinate plane. ## Definition of the Topic An ordered pair is a set of two numbers or objects in which the order of the elements matters. It is often written in parentheses (like this: (4,5)). The first number is the x-coordinate, and the second is the y-coordinate. ## Key Concepts Ordered Pair: An ordered pair (x, y) consists of two numbers: the x-coordinate and the y-coordinate. Coordinate Plane: A plane spanned by two perpendicular lines called axes. The horizontal line is called the x-axis, and the vertical line is the y-axis. The point where these two lines intersect is called the origin, represented by the ordered pair (0,0). Plotting Ordered Pairs: Ordered pairs are used to denote a point’s location on the coordinate plane. The x-coordinate denotes the horizontal position (left or right), and the y-coordinate denotes the vertical position (up or down) from the origin. ## Discussion with Illustrative Examples Ordered pairs are composed of the x-coordinate and y-coordinate written inside the parentheses and separated by a comma. The x-coordinate is the horizontal location in the cartesian plane, called the abscissa. The y-coordinate is the vertical location in the cartesian plane, called the ordinate. Let us imagine you’re playing a treasure hunt game. You get a hint: “Walk three steps to the right and two steps forward.” In our coordinated world, that is like saying the treasure is at the point (3,2). Here, 3 is the x-coordinate, and 2 is the y-coordinate. When we plot this on the coordinate plane, we start at the origin (0,0), move three steps to the right along the x-axis, and then two steps upwards along the y-axis. That’s where we find our treasure! Quadrants are quarter areas inside the coordinate plane. It can be identified by numbers and is divided into half axes. The center of the cartesian plane is called the origin and has an ordered pair of (0, 0). A cartesian plane consists of four quadrants, as shown in the figure below. You can determine the quadrant of an ordered pair or point by checking if the coordinates are negative or positive. ### Steps to Graph Ordered Pairs The following are the steps to follow when graphing ordered pairs: Step 1: Start from the origin (0,0) and move horizontally x units to the right if the first coordinate is positive and to the left if it is negative. Step 2: From where you have stopped, move y units vertically upward if positive and downward if negative. Step 3: Place a dot when you have stopped to represent the ordered pair. ## Examples with solutions Example 1 Plot the point (2,4) on the coordinate plane. Solution Start at the origin (0,0), then move two units to the right along the x-axis and four upwards along the y-axis. Mark the point. Example 2 Plot the ordered pair (-3,-1) on the coordinate plane. Solution Start at the origin (0,0), then move three units to the left along the x-axis and 1 unit downward along the y-axis. Label the point. Example 3 What is the ordered pair for a point with five units left of the origin and three units below? Solution The x-coordinate is -5 (to the left of the origin), and the y-coordinate is -3 (below the y-axis). So, the ordered pair is (-5,-3). Example 4 Identify the quadrant where each of the given ordered pair is located. a. (1,4) b. (-2,6) c. (7,-4) d. (-8,-3) Solution a. Since the x and y-coordinates are positive, (1,4) is in Quadrant I. b. The ordered pair (-2,6) is in Quadrant II since the x-coordinate is negative while the y-coordinate is positive. c. The ordered pair (7,-4) is in Quadrant IV since the x-coordinate is positive while the y-coordinate is negative. d. Since both the x and y-coordinates are negative, (-8,-3) is in Quadrant III ## Real-life Application with Solution Your school is planning a picnic at a park. The park is divided into a grid for different activities. The entrance is considered the origin (0,0). The playground is located four units to the right and three units up from the entrance. In terms of an ordered pair, where is the playground located? Solution The playground is at the point (4,3). ## Practice Test 1. Plot these points on a coordinate plane: (3,2), (-1,4), (0,-3), (-4,-2), and (5,0). 2. What is the ordered pair for a point with two units right of the origin and three units below? 3. You’re playing a game. You move three spaces to the left and then four spaces up. What is your position as an ordered pair? 4. What ordered pair represents the origin? 5. To get to the point (2, 2), how would you move from the origin? 1. 2. The ordered pair is 2,-3. 3. You are located at -3,4. 4. The ordered pair that represents the origin is 0,0. 5. To get to the ordered pair (2,2), move two units to the right of the x-axis, then two units upward along the y-axis. ### What happens if we switch the order of the numbers in an ordered pair? Switching the order of the numbers will denote a different point in the coordinate plane. ### Can the ordered pair be (0,0)? Yes, that is called the origin. It is where the x-axis and the y-axis intersect. ### Do ordered pairs have negative numbers? Absolutely! Negative numbers indicate a location left of the origin for the x-coordinate or below the origin for the y-coordinate. ### Are the coordinates always whole numbers? No, they can be fractions or even irrational numbers. ### What is the use of ordered pairs in real life? Ordered pairs are used in navigation, computer graphics, architecture, and more! Remember, like in a fun game, every point in the world of ordered pairs is an adventure. With your new knowledge, you are ready to explore the world of ordered pairs. Happy learning! We spend a lot of time researching and compiling the information on this site. If you find this useful in your research, please use the tool below to properly link to or reference Helping with Math as the source. We appreciate your support! • "Ordered Pairs". Helping with Math. Accessed on July 15, 2024. https://helpingwithmath.com/ordered-pairs/. • "Ordered Pairs". Helping with Math, https://helpingwithmath.com/ordered-pairs/. Accessed 15 July, 2024.
# How Many Prime Numbers Are There? (3 Key Concepts) Prime numbers are the building blocks of integers, and they are important in number theory and cryptography. There are lots of prime numbers that we know of, but this still leaves the question of how numerous they are. So, how many prime numbers are there? There are infinitely many prime numbers, and we can never run out of prime numbers. Also, there is no largest prime number – they grow without bound. There are 4 primes between 1 and 10, 8 primes between 1 and 20, 25 primes between 1 and 100, and 168 primes between 1 and 1000. Of course, there are also twin primes (such as the pair 3 and 5) to consider – there may be infinitely many such pairs, but this is still unproven. In this article, we’ll talk about prime numbers and how to prove that there are infinitely many. We’ll also look at some examples of how to find all of the primes between 1 and N. Let’s get started. ## How Many Prime Numbers Are There? There are infinitely many prime numbers. That is, no matter how many prime numbers we write out in a list, we can always find more to add to the list. To show this, we will use proof by contradiction. Assume that there are only finitely many prime numbers, and write them in a list, labeling them with subscripts from 1 to N: • p1, p2, … , pN Then, take the product P of all N prime numbers on our list: • P = p1 x p2 x … x pN Finally, add 1 to this number to get P + 1. Since P is a product of every prime number in our list, then P + 1 is larger than any number on the list. Thus, P + 1 is not on the list, and so it is not prime. Since P + 1 is not prime, it must be composite. Therefore, it is the product of prime numbers from our list. This means that there must be some prime number pi in our list so that P + 1 divided by pi has a zero remainder. However, P + 1 divided by any prime in our list will have a remainder of 1, since: • (P + 1) / p1 = (p1 x p2 x … x pN + 1) / p1 = p2 x … x pN + 1 / p1 • (P + 1) / p2 = (p1 x p2 x … x pN + 1) / p2 = p1 x p3 x … x pN + 1 / p2 • (P + 1) / pN = (p1 x p2 x … x pN + 1) / pN = p1 x p2 x … x pN-1 + 1 / pN No matter which prime pi we divide by, we always get a remainder of 1. So, P + 1 is not divisible by any prime on our list. This implies that either: • P + 1 is prime, which is a contradiction (we said P + 1 is not prime), or • P + 1 is composite and is a product of other primes that were not on our list, which is also a contradiction (we said our list contained all prime numbers). So, there must be infinitely many prime numbers. Thus, we can never run out of prime numbers. Note: due to the Fundamental Theorem of Arithmetic, it is also true that every integer greater than 1 is either prime or has a unique prime factorization. That means that there is only one combination of prime numbers (raised to the proper powers) that will result in a given integer product. ### How Many Prime Numbers Are Between 1 & N? To find out how many prime numbers are between 1 and N, we can use a method called the Sieve of Eratosthenes. Basically, we are using the process of elimination to remove composite numbers from our list until only prime numbers remain. Here are the steps for using the Sieve of Eratosthenes: • First, list out the numbers from 2 to N in a grid. • Next, cross out all of the multiples of 2 (that is, even numbers, or those divisible by 2 with no remainder – anything ending in 0, 2, 4, 6, or 8). • Then, cross out all of the multiples of 3 (that is, numbers divisible by 3 with no remainder). • Continue in this way by crossing out all of the multiples of 5, 7, etc. • Only the prime numbers will remain in the final list. Let’s look at an example of how to use the Sieve. ### How Many Prime Numbers Are There Between 1 & 10? Using the Sieve of Eratosthenes, we list out the numbers 1 to 10 in a grid: Next, we cross out the numbers that are multiples of 2 (except 2, which is prime): Next, we cross out the numbers that are multiples of 3 (except 3, which is prime): The remaining numbers in the list are prime. So, there are 4 prime numbers between 1 and 10: 2, 3, 5, and 7. The other numbers between 1 and 10 are composite, since: • 4 = 22 = 22 • 6 = 23 • 8 = 222 = 23 • 9 = 33 = 32 • 10 = 25 ### How Many Prime Numbers Are There Between 1 & 20? There are 8 prime numbers between 1 and 20: 2, 3, 5, 7, 11, 13, 17, and 19. The other numbers between 1 and 20 are composite, since: • 12 = 213 = 223 • 14 = 27 • 15 = 35 • 16 = 2222 = 24 • 18 = 233 = 232 • 20 = 225 = 225 ### How Many Prime Numbers Are There Between 1 & 100? There are 25 prime numbers between 1 and 100: • 2 • 3 • 5 • 7 • 11 • 13 • 17 • 19 • 23 • 29 • 31 • 37 • 41 • 43 • 47 • 53 • 59 • 61 • 67 • 71 • 73 • 79 • 83 • 89 • 97 ### How Many Prime Numbers Are There Between 1 & 1000? There are 168 prime numbers between 1 and 1000. The largest prime number less than 1000 is 997. ### How Many Prime Numbers Are Even? There is only one even prime number: 2. We can prove this as follows: • 2 is even, since we can write it as 2 = 21. 2 is also prime, since it only has 1 and itself as factors. • Any other positive even number E has the form E = 2N, where N is an integer greater than 1. Since E has factors of 1, 2, and N for N > 1, then E is composite (not prime). So, no other even number besides 2 is prime. Thus, 2 is the only even prime number. Another way to say this is that 2 is the only prime number divisible by 2. Every prime number other than 2 is an odd prime. Similarly, 3 is the only prime number divisible by 3, and 5 is the only prime number divisible by 5, and so forth. ### What Are Twin Primes? A pair of twin primes P and P + 2 are both prime and are only 2 units apart (that is, one is 2 greater than the other). Some examples of twin primes are: • 3 and 5 • 5 and 7 • 11 and 13 • 17 and 19 • 29 and 31 • 41 and 43 The twin prime conjecture suggests that there are infinitely many pairs of twin primes. However, this remains to be proven. ### How Many Numbers Between 1 & N Are Relatively Prime To N? Two integers M and N are relatively prime (also called coprime or mutually prime) if the only common factor of M and N is 1. In other words, M and N are relatively prime if they share no common prime factor. For example, 10 and 21 are relatively prime, since 10 = 25 and 21 = 3*7. Since there is no overlap between {2, 5} and {3, 7}, 10 and 21 are relatively prime. Note that two distinct prime numbers are automatically relatively prime. Also, every number from 1 to P – 1 is relatively prime to a prime number P. If we want to find out how many numbers are relatively prime to N, we can use Euler’s totient function (or Euler’s phi function). For a prime number P, there are P – 1 positive integers that are relatively prime to P (the numbers 1, 2, … , P – 1). ## Conclusion Now you know how many prime numbers there are (infinitely many!) and a little more about how to come to this conclusion.
IBPS Clerk 2021 Notification IBPS (Institute of Banking Personnel Selection) conducts a common recruitment process (CRP) every year to the r... # Quantitative Aptitude Basic Maths Tricks and shortcuts ## Basic Maths Tricks and shortcuts for competitive exams:Multiplication and Division shortcut Basic Maths Tricks and shortcuts for competitive exams:In every competitive exam, if you want to succeed you should manage your time well.To crack a bank exam , time management is a crucial factor.Understanding some shortcuts and tricks will definitely save your time during common written examinations.Here we are sharing some simple but powerful time saving shortcut tricks for division and multiplication. ### 1. Division-shortcuts In division instead of direct division, use factoring method Example:1848/264=(234711)/(23411)=7 ### 2.Multiplication-shortcuts SUM – 10 METHOD: Example: 78 and 72. These two numbers, if we add the numbers in the unit's place, the resultant is 10 and thenumbers in the ten's place are both the same. In such cases, we can have a simple solution. Step1: multiply the numbers in the unit's place and write down the resultant. (82 = 16) Step2: say, the number in the ten's digit is a, then multi a(a+1) and write down the resultant. => (7(7+1) = 56) Step3: write the final result: 5616 82 = 16; and 11(11+1) = 1112 = 132. And hence the result is: 13216. In short: abac = (a(a+1))(bc) ### Base Method: Base numbers, in general, are nothing but multiples of 10. If the given numbers are nearer to base numbers, then you can follow this method to multiply them. Example: 9895 =? Here 98 is ,2 less than the base number 100 and 95 is ,5 less than 100. We can write them like this: 98 -2 95 -5 The first step will be deducting/subtracting the resultant of the diff between the base number and the given number with the given number in a cross-way! That is, you need to subtract 98 and 5 (which is the resultant of difference between the base number and 95) or you can also cross-subtract 95 and 2, the result will be same. This result forms the 1st part of the resultant at the start. The last part of the resultant will be multiplication of the differences from base numbers (i.e., 2 5 = 10) 98 -2 95 -5 (98 – 5) (-2 * -5) Hence, the answer will be: 9310 Example: 998997 =? 998 -2 997 -3 Observe carefully, in the second part, the multiplication of difference yield in a single digit number, but no. of zeroes in the base number, here 1000, is three. Hence add two zeroes before the result. Therefore, the answer will be: (998-3) \| (-2 -3) = 995006 What if the numbers we get are like this? I mean, the base is 50 here. We will follow the same procedure as above but a small difference that the resultant in the first part will be halved. And if the base is 200, then the number will be doubled and so on based on the base number. ### Multiplication with 5,25,50 etc... Substitute 5 by 10/2,25 by 100/4 and 50 by 100/2. Examples: 1. 518=1810/2=180/2=90 2. 2425=24100/4=2400/4=600 3. 7350=73100/2=7300/2=3650 ### Multiplication with 9, 99,999 etc.. Examples: 1. 139=13(10-1)=130-13=117 2. 2699=26(100-1)=2600-26=2574 3. 350999=350(1000-1)=350000-350=349650 Hope this shortcuts will help you do better.Share it if you find this shortcuts useful. [Tags:quantitative aptitude tricks and shortcuts pdf,quantitative aptitude tricks and shortcuts pdf free download,quantitative aptitude tricks and shortcuts books,quantitative aptitude tricks and shortcuts for bank exams,quantitative aptitude tricks and shortcuts for cat,quantitative aptitude tricks and shortcuts videos,quantitative aptitude shortcuts and tricks for bank po pdf,quantitative aptitude shortcut methods pdf-tricks to solve download] 1. i have found the tricks so helpful . My heartiest thanks to the bankaspire online community. 1. you are welcome..keep supporting bankaspire..if you want any specific topic or shortcuts,feel free to contact us ,we will present it asap 2. please provide trick to remember Cubes upto 50 3. Thanks for providing aptitude shortcuts and basic maths. It was really useful at the time of bank exam. with regards Bank Exam Coaching Centres in Chennai 4. Candidates who are searching IBPS PO Previous Year Papers (2011 - 2017). Here is the best result for you. Our team is providing the direct link to IBPS PO Old Paper pdf, which is mention this page. More details about IBPS PO Previous Year Question Paper, We advise to visit our portal. 5. Best SSC Coaching in Bangalore http://www.globalias.in/ssc-coaching-in-bangalore/ ## Contact Form Name Email * Message * © 2015 bankAspire. WP Theme-junkie converted by Bloggertheme9
# APEX Calculus ## Section8.4Modeling with Differential Equations In the first three sections of this chapter, we focused on the basic ideas behind differential equations and the mechanics of solving certain types of differential equations. We have only hinted at their practical use. In this section, we use differential equations for mathematical modeling, the process of using equations to describe real world processes. We explore a few different mathematical models with the goal of gaining an introduction to this large field of applied mathematics. ### Subsection8.4.1Models Involving Proportional Change Some of the simplest differential equation models involve one quantity that changes at a rate proportional to another quantity. In the introduction to this chapter, we considered a population that grows at a rate proportional to the current population. The words in this assumption can be directly translated into a differential equation as shown below. There are some key ideas that can be helpful when translating words into a differential equation. Any time we see something about rates or changes, we should think about derivatives. The word “is” usually corresponds to an equal sign in the equation. The words “proportional to” mean we have a constant multiplied by something. The differential equation in Figure 8.4.1 is easily solved using separation of variables. We find \begin{equation} p = Ce^{kt}\text{.} \end{equation} Notice that we need values for both $$C$$ and $$k$$ before we can use this formula to predict population size. We require information about the population at two different times in order to fully determine the population model. #### Example8.4.2.Bacterial Growth. Suppose a population of e-coli bacteria grows at a rate proportional to the current population. If an initial popluation of 200 bacteria has grown to 1600 three hours later, find a function for the size of the population at time $$t\text{,}$$ and use it to predict when the population size will reach 10,000. Solution 1. We already know that the population at time $$t$$ is given by $$p = Ce^{kt}$$ for some $$C$$ and $$k\text{.}$$ The information about the initial size of the population means that $$p(0)=200\text{.}$$ Thus $$C=200\text{.}$$ Our knowledge of the population size after three hours allows us to solve for $$k$$ via the equation \begin{equation} 1600 = 200e^{3k}\text{.} \end{equation} Solving this exponential equation yields $$k =\ln(8)/3 \approx 0.6931\text{.}$$ The popluation at time $$t$$ is given by \begin{equation} p = 200 e^{(\ln(8)/3)t}\text{.} \end{equation} Solving \begin{equation} 10000 = 200e^{(\ln(8)/3)t} \end{equation} yields $$t =(3\ln(50))/\ln(8) \approx 5.644\text{.}$$ The population is predicted to reach 10,000 bacteria in slightly more than five and a half hours. Solution 2. Video solution Another example of porportional change is Newton’s Law of Cooling. The laws of thermodynamics state that heat flows from areas of higher temperature to areas of lower temperature. A simple example is a hot object that cools down when placed in a cool room. Newton’s Law of Cooling is the simple assumption that the temperature of the object changes at a rate proportional to the difference between the temperature of the object and the ambient temperature of the room. If $$T$$ is the temperature of the object and $$A$$ is the constant ambient temperature, Newton’s Law of Cooling can be expressed as the differential equation \begin{equation} \frac{dT}{dt} = k(A - T)\text{.} \end{equation} This differential equation is both linear and separable. The separated form is \begin{equation} \frac{1}{A-T}\,dT = k\,dt\text{.} \end{equation} Then an implicit definition of the temperature is given by \begin{equation} -\ln\abs{A-T} = kt + C\text{.} \end{equation} If we solve for $$T\text{,}$$ we find the explicit temperature \begin{equation} T = A-Ce^{-kt}\text{.} \end{equation} Though we didn’t show the steps, the explicit solution involves the typical process of renaming the constant $$\pm e^{-C}$$ as $$C\text{,}$$ and allowing $$C$$ to be positive, negative, or zero to account for both cases of the absolution value and to catch the constant solution $$T=A\text{.}$$ Notice that the temperature of the object approaches the ambient temperature in the limit as $$t\to\infty\text{.}$$ #### Example8.4.3.Hot Coffee. A freshly brewed cup of coffee is set on the counter and has a temperature of 200$$^\circ$$ Fahrenheit. After 3 minutes, it has cooled to 190$$^\circ\text{,}$$ but is still too hot to drink. If the room is 72$$^\circ$$ and the coffee cools according to Newton’s Law of Cooling, how long will the impatient coffee drinker have to wait until the coffee has cooled to 165$$^\circ\text{?}$$ Solution 1. Since we have already solved the differential equation for Newton’s Law of Cooling, we can immediately use the function \begin{equation} T = A - Ce^{-kt}\text{.} \end{equation} Since the room is 72$$^\circ\text{,}$$ we know $$A = 72\text{.}$$ The initial temperature is 200$$^\circ\text{,}$$ which means $$C = -128\text{.}$$ At this point, we have \begin{equation} T = 72 + 128e^{-kt} \end{equation} The information about the coffee cooling to 190$$^\circ$$ in 3 minutes leads to the equation \begin{equation} 190 = 72 + 128e^{-3k}\text{.} \end{equation} Solving the exponential equation for $$k\text{,}$$ we have \begin{equation} k = -\frac{1}{3}\ln \left(\frac{59}{64}\right) \approx 0.0271\text{.} \end{equation} Finally, we finish the problem by solving the exponential equation \begin{equation} 165 = 72 + 128e^{\frac{1}{3}\ln \left(\frac{59}{64}\right)t}\text{.} \end{equation} The coffee drinker must wait $$\displaystyle t = \frac{3 \ln \left(\frac{93}{128}\right)}{\ln \left(\frac{59}{64}\right)} \approx 11.78$$ minutes. Solution 2. Video solution We finish our discussion of models of proportional change by exploring three different models of disease spread through a population. In all of the models, we let $$y$$ denote the proportion of the population that is sick ($$0 \leq y \leq 1$$). We assume a proportion of $$0.05$$ is initially sick and that a proportion of $$0.1$$ is sick 1 week later. Suppose a disease spreads through a population at a rate proportional to the number of individuals who are sick. If 5% of the population is sick initially and 10% of the population is sick one week later, find a formula for the proportion of the popoulation that is sick at time $$t\text{.}$$ Solution. The assumption here seems to have some merit because it matches our intuition that a disease should spread more rapidly when more individuals are sick. The differential equation is simply \begin{equation} \frac{dy}{dt} = ky\text{,} \end{equation} with solution \begin{equation} y = Ce^{kt}\text{.} \end{equation} The conditions $$y(0)=0.05$$ and $$y(1) = 0.1$$ lead to $$C = 0.05$$ a and $$k = \ln(2)\text{,}$$ so the function is \begin{equation} y = 0.05e^{(\ln(2)t}\text{.} \end{equation} We should point out a glaring problem with this model. The variable $$y$$ is a proportion and should take on values between 0 and 1, but the function $$y = 0.05e^{2t}$$ grows without bound. After $$t \approx 4.32$$ weeks, $$y$$ exceeds 1, and the model ceases to make physical sense. Suppose a disease spreads through a population at a rate proportional to the number of individuals who are not sick. If 5% of the population is sick initially and 10% of the population is sick one week later, find a formula for the proportion of the popoulation that is sick at time $$t\text{.}$$ Solution. The intuition behind the assumption here is that a disease can only spread if there are individuals who are susceptible to the infection. As fewer and fewer people are able to be infected, the disease spread should slow down. Since $$y$$ is proportion of the population that is sick, $$1-y$$ is the proportion who are not sick, and the differential equation is \begin{equation} \frac{dy}{dt} = k(1-y)\text{.} \end{equation} Though the context is quite different, the differential equation is identical to the differential equation for Newton’s Law of Cooling, with $$A=1\text{.}$$ The solution is \begin{equation} y = 1 - Ce^{-kt}\text{.} \end{equation} The conditions $$y(0)=0.05$$ and $$y(1) = 0.1$$ yield $$C = 0.95$$ and $$k = -\ln\left(\frac{18}{19}\right) \approx 0.0541\text{,}$$ so the final function is \begin{equation} y = 1-.95e^{\ln\left(\frac{18}{19}\right)t}\text{.} \end{equation} Notice that this function approaches $$y=1$$ in the limit as $$t \to \infty\text{,}$$ and does not suffer from the non-physical behavior described in Example 8.4.4. In Example 8.4.4, we assumed disease spread depends on the number of infected individuals. In Example 8.4.6, we assumed disease spread depends on the number of susceptible individuals who are able to become infected. In reality, we would expect many diseases to require the interaction of both infected and susceptible individuals in order to spread. One of the simplest ways to model this required interaction is to assume disease spread depends on the product of the proportions of infected and uninfected individuals. This assumption (regularly seen in the context of chemical reactions) is often called the law of mass action. Suppose a disease spreads through a population at a rate proportional to the product of the number of infected and uninfected individuals. If 5% of the population is sick initially and 10% of the population is sick one week later, find a formula for the proportion of the population that is sick at time $$t\text{.}$$ Solution 1. The differential equation is \begin{equation} \frac{dy}{dt} = ky(1-y)\text{.} \end{equation} This is exactly the logistic equation with $$M = 1\text{.}$$ We solved this differential equation in Example 8.2.8, and found \begin{equation} y = \frac{1}{1 + be^{-kt}}\text{.} \end{equation} The conditions $$y(0)=0.05$$ and $$y(1) = 0.1$$ yield $$b = 19$$ and $$k = -\ln\left(\frac{9}{19}\right) \approx 0.7472\text{.}$$ The final function is \begin{equation} y = \frac{1}{1+19e^{\ln\left(\frac{9}{19}\right)t}}\text{.} \end{equation} Based on the three different assumptions about the rate of disease spread explored in the last three examples, we now have three different functions giving the proportion of a population that is sick at time $$t\text{.}$$ Each of the three functions meets the conditions $$y(0)=0.05$$ and $$y(1) = 0.1\text{.}$$ The three functions are shown in Figure 8.4.8. Notice that the logistic function mimics specific parts of the functions from Examples 8.4.4 and 8.4.6. We see in Figure 8.4.8.(a) that the logistic and exponential functions are virtually indistinguishable for small $$t$$ values. When there are few infected individuals and lots of susceptible individuals, the spread of a disease is largely determined by the number of sick people. The logistic curve captures this feature, and is “almost exponential” early on. In Figure 8.4.8.(b), we see that the logistic curve leaves the exponential curve from Example 8.4.4 and approaches the curve from Example 8.4.6. This result implies that when most of the population is sick, the spread of the disease is largely dependent on the number of susceptible individuals. Though there are much more sophisticated mathematical models describing the spread of infections, we could argue that the logistic model presented in this example is the “best” of the three. Solution 2. Video solution ### Subsection8.4.2Rate-in Rate-out Problems One of the classic ways to build a mathematical model involves tracking the way the amount of something can change. We sometimes say these models are based on conservation laws. Consider a box with some amount of a specific type of material inside. (Some type of chemical, for example.) The amount of material of the specific type in the box can only change in four ways; we can add more to the box, we can remove some from the box, some of the material can change into material of a different type, or some other type of material can turn into the type we’re tracking. In the examples that follow, we assume material doesn’t change type, so we only need to keep track of material coming into the box and material leaving the box. To derive a differential equation, we track rates: \begin{equation} \text{ rate of change of some quantity } = \text{ rate in } - \text{ rate out }\text{.} \end{equation} Though we stick to relatively simple examples, this basic idea can be used to derive some very important differential equations in mathematics and physics. The examples to follow involve tracking the amount of a chemical in solution. We assume liquid containing some chemical flows into a container at some rate. That liquid mixes instantaneously with the liquid already in the container. Then the liquid from the container flows out at some (potentially different) rate. #### Example8.4.10.Equal Flow Rates. Suppose a 10 liter tank has 5 liters of salt solution in it. The initial concentration of the salt solution is 1 gram per liter. A salt solution with concentration 3 gL flows into the tank at a rate of 2 Lmin. Suppose the salt solution mixes instantaneously with the solution already in the tank, and that the mixed solution from the tank flows out at a rate of 2 Lmin. Find a function that gives the amount of salt in the tank at time $$t\text{.}$$ Solution 1. We use the rate in - rate out setup described above. The quantity here is the amount (in grams) of salt in the tank at time $$t\text{.}$$ Let $$y$$ denote the amount of salt. In words, the differential equation is given by \begin{equation} \frac{dy}{dt} = \text{ rate in } - \text{ rate out }\text{.} \end{equation} Thinking in terms of units can help fill in the details of the differential equation. Since $$y$$ has units of grams, the left hand side of the equation has units g/min. Both terms on the right hand side must have these same units. Notice that the product of a concentration (with units g/L) and a flow rate (with units L/min) results in a quantity with units g/min. Both terms on the right hand side of the equation will include a concentration multiplied by a flow rate. For the rate in, we multiply the inflow concentration by the rate that fluid is flowing into the bucket. This is $$\displaystyle \left(3 \frac{\text{g} }{\text{L} }\right)\left(2 \frac{\text{L} }{\text{ min } }\right) = 6$$ g/min. The rate out is more complicated. The flow rate is still 2 Lmin, meaning that the overall volume of the fluid in the bucket is the constant 5 L. The salt concentration in the bucket is not constant though, meaning that the outflow concentration is not constant. In particular, the outflow concentration is not the constant 1 Lmin. This is simply the initial concentration. To find the concentration at any time, we need the amount of salt in the bucket at that time and the volume of liquid in the bucket at that time. The volume of liquid is the constant 5 L, and the amount of salt is given by the dependent variable $$y\text{.}$$ Thus, the outflow concentration is $$\displaystyle \frac{y}{5}$$ g/L, yielding a rate out given by \begin{equation} \left(\frac{y}{5}\frac{\text{ g } }{\text{ L } }\right)\left(2 \frac{\text{ L } }{\text{ min } }\right) = \frac{2y}{5}\text{ g/min }\text{.} \end{equation} The differential equation we wish to solve is given by \begin{equation} \frac{dy}{dt} = 6 - \frac{2y}{5}\text{.} \end{equation} To furnish an initial condition, we must convert the initial salt concentration into an initial amount of salt. This is $$\left(1\displaystyle \frac{\text{ g } }{\text{ L } }\right)(5 \text{ L } ) = 5$$ g, so $$y(0) = 5$$ is our initial condition. Our differential equation is both separable and linear. We solve using separation of variables. The separated form of the differential equation is \begin{equation} \frac{5}{30 - 2y}\,dy = dt\text{.} \end{equation} Integration yields the implicit solution \begin{equation} -\frac{5}{2}\ln\abs{30 - 2y} = t+C\text{.} \end{equation} Solving for $$y$$ (and redefining the arbitrary constant $$C$$ as necessary) yields the explicit solution \begin{equation} y = 15 + Ce^{-\frac{2}{5}t}\text{.} \end{equation} The initial condition $$y(0) = 5$$ means that $$C = -10$$ so that \begin{equation} y = 15 - 10e^{-\frac{2}{5}t} \end{equation} is the particular solution to our initial value problem. This function is plotted in Figure 8.4.11. Notice that in the limit as $$t\to\infty\text{,}$$ $$y$$ approaches $$15\text{.}$$ This corresponds to a bucket concentration of $$15/5 = 3$$ g/L. It should not be surprising that salt concentration inside the tank will move to match the inflow salt concentration. Solution 2. Video solution #### Example8.4.12.Unequal Flow Rates. Suppose the setup is identical to the setup in Example 8.4.10 except that now liquid flows out of the bucket at a rate of 1 L/min. Find a function that gives the amount of salt in the bucket at time $$t\text{.}$$ What is the salt concentration when the solution ceases to be valid? Solution 1. Because the inflow and outflow rates no longer match, the volume of liquid in the bucket is not the constant 5 L. In general, we can find the volume of liquid via the equation \begin{equation} \text{ volume } = \text{ initial volume } + \text{ (inflow rate - outflow rate) } t\text{.} \end{equation} In this example, the volume at time $$t$$ is $$5 + t$$ liters. Because the total volume of the bucket is only 10 L, it follows that our solution will only be valid for $$0 \leq t \leq 5\text{.}$$ At that point it is no longer possible to have liquid flow into a the bucket at a rate of 2 L/min and out of the bucket at a rate of 1 L/min. To update the differential equation, we must modify the rate out. Since the volume is $$5 + t\text{,}$$ the concentration at time $$t$$ is given by $$\frac{y}{5+t}$$ g/L. Thus for rate out, we must use $$\left( \frac{y}{5+t}\right)(1)$$ g/min. The initial value problem is \begin{equation} \frac{dy}{dt} = 6 - \frac{y}{5+t}, \text{ with } y(0)=5\text{.} \end{equation} Unlike Example 8.4.10, where we had equal flow rates, this differential equation is no longer separable. We must proceed with an integrating factor. Writing the differential equation in the form \begin{equation} \frac{dy}{dt} + \frac{1}{5+t}y = 6\text{,} \end{equation} we identify the integrating factor \begin{equation} \mu(t) = e^{\int \frac{1}{5+t}\,dt} = e^{\ln(5+t)} = 5+t\text{.} \end{equation} Then \begin{equation} \frac{d}{dt}\big((5+t)y\big) = 6(5+t)\text{,} \end{equation} yielding the implicit solution \begin{equation} (5+t)y = 30t + 3t^2 + C\text{.} \end{equation} The initial condition $$y(0) = 5$$ implies $$C = 25\text{,}$$ so the explicit solution to our initial value problem is given by \begin{equation} y = \frac{3t^2 + 30t + 25}{5+t}\text{.} \end{equation} This solution ceases to be valid at $$t=5\text{.}$$ At that time, there are 25 g of salt in the tank. The volume of liquid is 10 L, resulting in a salt concentration of $$2.5$$ g/L. Solution 2. Video solution Differential equations are powerful tools that can be used to help describe the world around us. Though relatively simple in concept, the ideas of proportional change and matching rates can serve as building blocks in the development of more sophisticated mathematical models. As we saw in this section, some simple mathematical models can be solved analytically using the techniques developed in this chapter. Most sophisticated mathematical models don’t allow for analytic solutions. Even so, there are an array of graphical and numerical techniques that can be used to analyze the model to make predictions and infer information about real world phenomena. ### Exercises8.4.3Exercises #### Problems ##### Exercise Group. In the following exercises, use the tools in the section to answer the questions presented. ###### 1. Suppose the rate of change of $$y$$ with respect to $$x$$ is proportional to $$10 - y\text{.}$$ Write down and solve a differential equation for $$y\text{.}$$ ###### 2. A rumor is spreading through a middle school with 250 students. Suppose the rumor spreads at a rate proportional to the number of students who haven’t heard the rumor yet. If 1 person starts the rumor, and 75 students have heard the rumor 3 days later, how many days will it take until 80% of the students in the school have heard the rumor? ###### 3. A rumor is spreading through a middle school with 250 students. Suppose the rumor spreads at a rate proportional to the product of number of students who have heard the rumor and the number who haven’t heard the rumor. If 1 person starts the rumor, and 75 students have heard the rumor 3 days later, how many days will it take until 80% of the students in the school have heard the rumor? ###### 4. A feature of radioactive decay is that the amount of a radioactive substance decreases at a rate proportional to the current amount of the substance. The half life of a substance is the amount of time it takes for half of a given amount of substance to decay. The half life of carbon-14 is approximately 5730 years. If an ancient object has a carbon-14 amount that is 20% of the original amount, how old is the object? ###### 5. Consider a chemical reaction where molecules of type A combine with molecules of type B to form molecules of type C. Suppose one molecule of type A combines with one molecule of type B to form one molecule of type C, and that type C is produced at a rate proportional the product of the remaining number of molecules of types A and B. Let $$x$$ denote moles of molecules of type $$C\text{.}$$ Find a function giving the number of moles of type C at time $$t$$ if there are originally $$a$$ moles of type A, $$b$$ moles of type B, and zero moles of type C. ###### 6. Suppose an object with a temperature of 100$$^\circ$$ is introduced into a room with an ambient temperature of 70$$^\circ\text{.}$$ Suppose the temperature of the object changes at a rate proportional to the difference between the temperature of the object and the temperature of the room (Newton’s Law of Cooling). If the object has cooled to 92$$^\circ$$ in 10 minutes, how long until the object has cooled to 84$$^\circ?$$ ###### 7. Suppose an object with a temperature of 100$$^\circ$$ is introduced into a room with an ambient temperature given by $$60 + 20e^{-\frac{1}{4}t}$$ degrees. Suppose the temperature of the object changes at a rate proportional to the difference between the temperature of the object and the temperature of the room (Newton’s Law of Cooling). If the object is 80$$^\circ$$ after 20 minutes, find a formula giving the temperature of the object at time $$t\text{.}$$ (Note: This problem requires a numerical technique to solve for the unknown constants.) ###### 8. A tank contains 5 gallons of salt solution with concentration 0.5 g/gal. Pure water flows into the tank at a rate of 1 gallon per minute. Salt solution flows out of the tank at a rate of 1 gallon per minute. (Assume instantaneous mixing.) Find the concentration of the salt solution at 10 minutes. ###### 9. Dead leaves accumulate on the ground at a rate of 4 grams per square centimeter per year. The dead leaves on the ground decompose at a rate of 50% per year. Find a formula giving grams per square centimeter on the ground if there are no leaves on the ground at time $$t=0\text{.}$$ ###### 10. A pond initially contains 10 million gallons of fresh water. Water containing an undesirable chemical flows into the pond at a rate of 5 million gallons per year, and fluid from the pond flows out at the same rate. (Assume instantaneous mixing.) If the concentration (in grams per million gallons) of the incoming chemical varies periodically according to the expression $$2 + \sin(2t)\text{,}$$ find a formula giving the amount of chemical in the pond at time $$t\text{.}$$ ###### 11. A large tank contains 1 gallon of a salt solution with concentration 2 g/gal. A salt solution with concentration 1 g/gal flows into the tank at a rate of 4 gal/min. Salt solution flows out of the tank at a rate of 3 gal/min. (Assume instantaneous mixing.) Find the amount of salt in the tank at 10 minutes. ###### 12. A stream flows into a pond containing 2 million gallons of fresh water at a rate of 1 million gallons per day. The stream flows out of the first pond and into a second pond containing 3 million gallons of fresh water. The stream then flows out of the second pond. Suppose the inflow and outflow rates are the same so that both ponds maintain their volumes. A factory upstream of the first pond starts polluting the stream. Directly below the factory, pollutant has a concentration of 55 grams per million gallons, and this concentration starts to flow into the first pond. Find the concentration of pollutant in the first and second ponds at 5 days.
# Communications of the ACM BLOG@CACM ## How Does One Divide with Napier's Rods? View as: Print Mobile App Share: Napier's rods, also called Napier's bones (see Figure 1), were invented at the beginning of the 17th century. They have been used for multiplications and divisions until the 19th century. There were various forms, e.g. rotating cylinders. Operating Instructions for Napier's Bones Napier's multiplication and division rods, deriving from the basic multiplication table, simplify calculations considerably. They were also built into several mechanical calculating aids as rotating drums. The intermediate results are read from the diagonal grid and added by hand, taking account of the tens carry. According to the model, the diagonals run from the lower left to the upper right or from the upper left to the lower right. The tens are at the upper left or the lower left, respectively, and the ones at the lower right or the upper right, respectively. Division Example 3: 204,383: 347 = 589 (see Fig. 2) 1. Lay the rods with the numerals of the divisor next to each other. Place the basic rod with the numerical column 1 to 9 to the left of these. 2. Read off how many times 347 goes into the first numbers 2043 of the dividend (5x). The next lower value with a whole number quotient is 1735, and the next higher value 2082 is too high. 3. Then subtract 1735 (5 × 347) from 2043. Add the next digit (8) of the dividend to the remainder (308). 4. Control how often 347 goes into 3088 (8x). The next lower value with a whole number quotient is 2776. 5. Then subtract 2776 (8 × 347) from 3088. Add the last digit (3) of the dividend to the remainder (312). 6. For the division 347 into 3123, this gives the value 9. 7. The final result is 589. Napier's bones Example: 204,383: 347 = 589 A few selected examples of rotating drums (see Figures 3 and 4). Source Bruderer, Herbert: Milestones in Analog and Digital Computing, Springer Nature Switzerland AG, Cham, 3rd edition 2020, 2 volumes, 2113 pages, 715 illustrations, 151 tables https://www.springer.com/de/book/9783030409739 (translated by Dr. John McMinn) Herbert Bruderer is a retired lecturer in didactics of computer science at ETH Zurich. More recently, he has been an historian of technology. bruderer@retired.ethz.ch, herbert.bruderer@bluewin. No entries found
## Engage NY Eureka Math 5th Grade Module 6 Lesson 32 Answer Key ### Eureka Math Grade 5 Module 6 Lesson 32 Problem Set Answer Key Question 1. Ashley decides to save money, but she wants to build it up over a year. She starts with $1.00 and adds 1 more dollar each week. Complete the table to show how much she will have saved after a year. Answer: Explanation : Amount with which ashley started =$1 Each week she adds $1 more that the previous week So, Week she has$1 in her account Week 2 she adds one more dollar than previous week so, that is $2 after adding her account balance is$2 +$1 =$3 Week 3 she adds one more than than previous week so, that is $3 after adding her account balance is$3 + $3 =$ 6 and so on …. till Week 52 the balance amount is calculated . Question 2. Carly wants to save money, too, but she has to start with the smaller denomination of quarters. Complete the second chart to show how much she will have saved by the end of the year if she adds a quarter more each week. Try it yourself, if you can and want to! Explanation : Amount with which Carlie started = $0.25 (Quarters) Each week she adds$0.25 more that the previous week So, Week 1 she starts with $0.25 in her account Week 2 she adds one more Quarter dollar than previous week so, that is$0.50 after adding her account balance is $0.25 +$0.50 = $0.75 Week 3 she adds one more Quarter dollar than previous week so, that is$0.75 after adding her account balance is $0.75 +$0.75 = $1.50 and so on …. till week 52 amount is calculated . Question 3. David decides he wants to save even more money than Ashley did. He does so by adding the next Fibonacci number instead of adding$1.00 each week. Use your calculator to fill in the chart and find out how much money he will have saved by the end of the year. Is this realistic for most people? Explain your answer. Explanation : By end of the year he will save $86,267,571,271 . It is not realistic because In Fibonacci number the rule is followed So we can write the rule: The Rule is xn = xn−1 + xn−2 where: xn is term number “n” xn−1 is the previous term (n−1) xn−2 is the term before that (n−2) In Fibonacci rule the previous term and the term before that is added to get next fibonacci number , That means the David should starts with$1 . for week he should add $1 for week1 later, he should add$1 the previous money and before that week in week 2 total is $2, In week 3 he should add$2 that double of previous then amount becomes $4. and so on till week 52 for every week he should add nearly 2 weeks amount which will be difficulty to add. so this not realistic to follow this rule as Savings are done with smaller amounts not with huge amounts . ### Eureka Math Grade 5 Module 6 Lesson 32 Reflection Answer Key Today, we watched how savings can grow over time, but we did not discuss how the money saved was earned. Have you ever thought about how math skills might help you to earn money? If so, what are some jobs that might require strong math skills? If not, think about it now. How might you make a living using math skills? Answer: Money math involves many of the math skills you learn in school, such as addition, subtraction, multiplication, division, fractions, decimals, and percentages. … Handling money can help you learn how to count it. Making purchases in a store can help you become comfortable counting money. Some jobs that might require strong math skills are in Auditor, Statistician, Actuary, Mathematician, Operations Research Analyst, Math Professor, Banking , Cashier , Math Matters in Everyday Life Managing money$ Balancing the checkbook. Shopping for the best price. Preparing food. Figuring out distance, time and cost for travel. Understanding loans for cars, trucks, homes, schooling or other purposes. ### Eureka Math Grade 5 Module 6 Lesson 32 Homework Answer Key Question 1. Jonas played with the Fibonacci sequence he learned in class. Complete the table he started. 1 2 3 4 5 6 7 8 9 10 1 1 2 3 5 8 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 1 1 2 3 5 8 13 21 34 55 11 12 13 14 15 16 17 18 19 20 89 144 233 377 610 987 1597 2584 4181 6765 Explanation : The Fibonacci Sequence can be written as a “Rule” First, the terms are numbered from 0 onwards like this: So we can write the rule: The Rule is xn = xn−1 + xn−2 where: xn is term number “n” xn−1 is the previous term (n−1) xn−2 is the term before that (n−2) x2= x2−1 + x2−2 = x1 + x0 = 1 + 0 = 1 x3= x3−1 + x3−2 = x2 + x1 = 1 + 1 = 2 With this rule all the terms are calculated . Question 2. As he looked at the numbers, Jonas realized he could play with them. He took two consecutive numbers in the pattern and multiplied them by themselves and then added them together. He found they made another number in the pattern. For example, (3 × 3) + (2 × 2) = 13, another number in the pattern. Jonas said this was true for any two consecutive Fibonacci numbers. Was Jonas correct? Show your reasoning by giving at least two examples of why he was or was not correct. Yes he is write . 1 2 3 4 5 6 7 8 9 10 1 1 2 3 5 8 13 21 34 55 11 12 13 14 15 16 17 18 19 20 89 144 233 377 610 987 1597 2584 4181 6765 Explanation : The Fibonacci number of 6 is 8 The Fibonacci number of 7 is 5 Taking two consecutive numbers and squaring and adding them we get, ( 3 × 3)+(5 × 5 ) = 9 + 25 = 34 34 is the Fibonacci number of 9 (4 + 5 = 9) It is true The Fibonacci number of 6 is 8 The Fibonacci number of 7 is 13 Taking two consecutive numbers and squaring and adding them we get, ( 8 × 8 )+(13 × 13 ) = 64 + 169 = 233 233 is the Fibonacci number of 13 (6 + 7 = 13) It is true Question 3. Fibonacci numbers can be found in many places in nature, for example, the number of petals in a daisy, the number of spirals in a pine cone or a pineapple, and even the way branches grow on a tree. Find an example of something natural where you can see a Fibonacci number in action, and sketch it here.
# Lesson 14 Equivalent Linear Expressions The 14th lesson of Algebra focuses on equivalent linear expressions – a major part of understanding algebraic equations. It is essential for students to understand what makes two expressions equivalent. Equivalent linear expressions are important to understand in order to solve various algebraic problems. ## What are Equivalent Linear Expressions? Equivalent linear expressions refer to two different expressions that have the same value. For example, “2x + 5” and “2x + 3 + 2” are two different expressions but both have the same value. This means that they are equivalent linear expressions. In order to determine if two expressions are equivalent, they must be simplified. This means that all the terms must be combined and reduced to the same value in order to determine if the expressions are equivalent. ## How to Simplify Expressions? In order to simplify expressions, the terms must be added, subtracted, multiplied, or divided in order to get the same value for both expressions. For example, “2x + 5” and “2x + 3 + 2” can be simplified by adding the two terms on the right side of the equation. This would result in a new expression of “2x + 5”. This is the same value as the original expression, meaning that the two expressions are equivalent. ## Examples of Equivalent Linear Expressions There are many examples of equivalent linear expressions. One example is “4x – 9” and “4x + 1 – 10”. These two expressions can be simplified by adding the terms on the right side of the equation. This will result in a new expression of “4x – 9”, which is the same value as the original expression. Another example is “2x + 5” and “2x – 3 + 8”. These two expressions can be simplified by subtracting the terms on the right side of the equation. This will result in a new expression of “2x + 5”, which is the same value as the original expression. ## Conclusion Equivalent linear expressions are an important concept in algebra. Understanding what makes two expressions equivalent is essential in order to solve various algebraic problems. Equivalent linear expressions can be determined by simplifying the terms in the expression. There are many examples of equivalent linear expressions that can be used to help students understand this concept.
# Lesson video In progress... Loading... Hi everyone, I'm Mrs Crane, and welcome to today's lesson. How are you today? I hope you're well, and I hope you're ready for some maths learning. In today's lesson, we're going to be looking at the place value of each digit in a four digit number. What I'd like you to do for me is make sure that you've turned off all of your apps or anything else that might distract you on your devices. And then if you can, try and find somewhere nice and quiet, so that you're not going to be distracted in our learning today. I'll run through the agenda with you in a moment so you know exactly what we're learning, and I'll explain any equipment that you might need. So, when you're ready, let's get started. Okay then. We're going to run through the agenda of today's lesson. So we're going to start off by looking at the value of a digit. Then we're going to be exploring and recognising the different place value of different columns. Then we're going to be looking at when we can write numbers in words, and finally, it will be time for your independent task, but I'll be representing place value. And of course, we'll go through the answers together at the end of that. So let's get started. Today, you're going to need a pencil, maybe a rubber and a piece of paper, but then four digit numbers. So what can you say, or what do you notice on your screen? Have a little think. I can see that there are four columns here. One, two, three, four. And I can see each column is labelled with thousands, hundreds, tens, and ones. And if I look down this side of my screen here, I can see some representations. Now these are representations of Dienes. So pictures of Dienes, this is a ones Dienes. This is a tens Dienes. This is a hundreds Dienes, and this is a thousands Dienes. What we're going to do, first of all, is think about how we can make a four digit number. How does our column, how do our columns work for it? So we're going to start off in the ones column. So if I were to put nine ones in my ones column, that's straightforward. I can put them in here. I've just drawn that line in here so that I can show you with the written numbers underneath it, the number that's in my column. So here I've got nine ones. Ready to count with me? One, two, three, four, five, six, seven, eight, nine. Now, if I wanted to add another one to this column, how would it work kind of thing? Can I put another one in this column here? Let's try it. I've put another one here, that gives me 10 ones. I need to write that in a different way. I have to do something different and I have to regroup my 10 ones for one ten Dienes, because I can't put 10 ones here. When I go to my to here, it won't work because I have two digits and one column which doesn't work. So I have regrouped to my 10 ones or one group of ten, and then I can place it in-- in my work--in my writing--in my numbers--sorry-- at the bottom. So I can put in 10 here, like that. Now this time, I want to put in nine tens here. So let's check one, two, three, four, five, six, seven, eight, nine. And I want to put in one more ten. Can I do that? I've done it, but can I write that in here? Is that correct? Nope, it's not correct. I cannot put 10 groups of 10 in my tens column. I have to regroup 10 groups of 10 for one group of a hundred. So I'm going to put my one group of a hundred and in here. Then I can write it in here. I can put in one, so I have one hundred, and I have zero tens and zero ones. So these zeros kind of become what we call a place holder. We'll look at that word even more so later on in the lesson. Okay. Now, this time I want to put nine hundreds in my hundreds column. Let's count them. One, two, three, four, five, six, seven, eight, nine. Yup! I've got nine hundreds in that. And I want to put one more hundred in that. Oh no, I've done it again. I've put too many numbers, too many hundreds in my column. I need to do something called regrouping. So I need to cross those out and I need to regroup my 10 hundreds for thousand. Here it is, there's my 1000. Now, if I wanted to write this in here, I'm going to have a one here, and zeros here, because I'm showing I have one thousand, zero hundreds, zero tens, and zero ones. Okay then. Now, what I want you to do is have a look at these four numbers here: Four thousand, eight hundred two. Three thousand, two hundred eighty-four. Five thousand, four hundred twenty-six. And, two thousand, forty-six. What I want you to do is just have a think. How does the value of my two, the number two, change in each of those four numbers? How does the value of the number two change in each of the four numbers? Have a little think. Okay. Let's start off by looking at this number here: Four thousand, eight hundred two. The two here, represents two ones. I've got two ones Dienes here to show you. I'm going to go to our next number then: Three thousand, two hundred eighty-four. The two here represents two hundreds, Five thousand, four hundred twenty-six. The two here represents two tens And, two thousand, and forty-six. The two here represents two thousands. So even though there's a two in each number that two, and the value of that two changes depending on which column that it's in and which number, which value it represents. So here is the two representing the ones column. Here, my two represented two hundreds. Here, my two represents two tens. And here, my two represented two thousands. The same number can represent different amounts depending on which column it's in, within our numbers. And that's going to be really important as we go on with our learning today. Okay. So, now we're going to look at the value of each digit, and I've been given my number here. I've got a blank representation of my number here, and I've got my number partitioned here. So my number is four thousand, and fifty-six. So, when I'm looking at how many thousands, hundreds, and tens and ones, there are in my number, I can see there's four thousands, zero hundreds, this five tens, and this six ones, Why is it so important that I fill in this block, this box--sorry, if there's zero there? Why is that important? Well done to those of you that thought actually, if Mrs. Crane didn't fill in this box here, she's going to get confused when it comes to writing in her four-- she might put it into the wrong column. That zero is there as a place holder, that word that we looked at earlier in the session. The word was placeholder, so that's going to come in really, really helpful when it comes to us filling in and completing different representations. So, what I'm going to do now is represent this using Dienes. So, I'm going to start in my ones column. I've got six ones, so let's check: one, two, three, four, five, six. Fantastic! I should have five tens, lets see: one, two, three, four, five tens. Absolutely. Now. Oh no. I know that I should have nothing in this hundreds column. What have I accidentally done in my hundreds column? Well done those of you who are saying: you've put your thousands in the wrong column, miss! Silly me! So my hundreds column should have nothing in it. These four thousands here, need to be in the correct column. So I need to get rid of them in my hundreds column, like I've just done. I need to put them into the correct column. So on my Dienes model, it's just going to be nothing in my hundreds column. That's okay. It shows that it's zero, here. And in my thousands column, there will be four thousands. Okay. If you're feeling really confident now, what I'd like you to do is pause the screen, and have a go at drawing the Dienes that would go in the correct place value columns, and filling in the partitioning of our place value in the four columns: thousands, hundreds, tens, and ones. If you're not feeling so confident, that's okay, because we're going to go through this example together before I give you an example to prepare for it, okay? So, pause the screen now if you want to have go on your own, if you don't, don't worry. The number is three thousand, five hundred and two. When I first look at this, the first thing I've noticed is that there's a zero there. So I know I'm going to need a placeholder in one of my columns. I can see it, that zero is in the tens column, so I know that I'm going to need my zero here and I can make sure that that makes sense in a moment. So, let's have a go at filling it in then. So I know I've got two ones, so I'm going to put this here. I know I've got no tens, so I've not filled in that column. I've gone straight to my hundreds column to fill in my five hundred--one, two, three, four, five hundred. Then I'm going to look at my thousands. I know I've got three thousand, so I can fill in one, three thousands. Then if I'm filling this in I know that there are three thousands, five hundred zero tens and two ones in my number. Fantastic! So, today we're going to do a 'Let's Explore', and we're going to be recognising place value. So, for your 'Let's Explore' today, what I would like you to do is choose one of these Dienes representations, here. So these are three separate numbers, here. Choose the one that you want to represent, and you're going to have a go at drawing it or writing it in the place value chart like we've just been doing. My challenge for you today is can you write it into the sentence? And here's the sentence model here for you to have a go at. Pause the screen now, and have a go at today's 'Lets Explore'. Okay. Welcome back. What we're going to do now is going to discuss one of the examples from the left 'Let's Explore'. So, I've taken this example here, this number here, and now I'm choosing to use this example, and to write it into here, okay? So, I'm going to write it rather than draw it. You might have drawn it. That's absolutely fine. It doesn't matter. So here I've put zero in my ones column. Why have I put a zero in my ones column? Well done. I don't have any ones, and the ones Dienes in this number here, do I? I do have one ten, so I need one ten here. I do have one, two, three, four hundreds, so I need to put four hundreds here, and I do have one, two thousands, so I need to put my thousands here. Then when I write it into my frame I can say there are two thousands, four hundreds, one ten, and zero ones, making sure I put that zero there to show that there's nothing there, okay? Fantastic! Let's move on then and have a look at how we can identify four digit numbers when that, in words. So at the moment you can't see them in words. That's okay. We've got four different representations here. We've got it written in a place value chart. We're going to have Dienes in a place value chart, we're going to have it partitioned here. Here. And we're going to write our number in words here, okay? So, our number, let's start off by reading it all together is two thousand, seven hundreds and thirty six. Brilliant! Now we're going to have a go at putting that into our Dienes. So you can see here, I've put in six ones, one, two, three, four, five, six. I spend three tens on that. Fantastic! I've put in one, two, three, four, five, six, seven hundreds. And I've put in two thousands. So that's my number represented with Dienes. Now I can quite straightforward--I can do this bit quite eloquent, straightforward way, because I know there's two thousands, seven hundreds three tens and six ones. Now, this part we haven't looked at just yet. So I'm partitioning it into it's thousands, hundreds, tens, and ones, okay? So here's my whole number here: two thousand, seven hundreds and thirty six partitioned into two thousand, seven hundred, thirty, and six. Because if I added those all up, that would give me my two thousand, seven hundred and thirty-six, okay? Now this part is the new part that we haven't looked at yet, which is why we write it in word. So, as we say it, we would say two thousand, seven hundred and thirty-six. So we're going to write that how we would say it. So we would say two thousand, seven hundred and thirty six. So that's it written, okay? Now, if you're feeling really confident, you can use, but it's written to have a go at filling out the other different representations. If you're feeling confident, pause the screen now, and you can have a go at that. If you're not feeling so confident, don't worry. We're going to go through it together, okay? So, this time, as you've spotted, it's different because we've been given our number in words first. You've got to use that to work backwards to find out what our a number looks like when we write it in numbers with our Dienes. So, I know it's fifty-three. So under this three ones, fifty tells me that there are five tens, one, two, three, four, five, two hundred. How many hundreds are going to go in my hundreds column? Hold on. It's two hundred, so there's going to be two. And last, but not least, four thousand, I've got to have four thousands in my thousand columns to show my number three ones here. Now I can write it partitioned because I can say my number is two thous-- four thousand, sorry--two hundred and fifty-three partitioned is four thousand, two hundred, fifty, and three. And then I can write it in my place value grid with my four two, five, and three, representing the number on the column above. So four thousands, two hundreds, five tens, and three ones. Right, then. It's now time for your independent task today. I'm going to read through the different questions for you today, so you know what you're doing before you have a go. You're going to be representing the place value of four digit numbers. So, just like we've already looked at, you've got three questions and I'd like you to fill in the missing representations for the following four digit numbers. Have a look at where the number is and the representations. Use that to work through the other representations to help you. You've got question one, two, and three. What I'd like you to do is pause your video to complete your task. Okay. So we're going to go through the answers then. So here is our number here. We've got five thousand, and seventy-nine, so I need to fill in the other representations for this number. So I'm going to put in my nine ones Dienes. My seven tens Dienes. I'm going to leave my hundreds column blank because I've got zero hundreds. And I'm going to put my five thousands here. There are five thousands, zero hundreds, seven tens and nine ones. So if I partition my number again in a different way, I can show here, it's five thousand and seventy-nine, which is equal to five thousand, zero hundreds, seventy, and nine. And my number is five thousand, and seventy-nine. Question two then, has the number here, and we have to fill out the other representations here. There in my number, I'm told there are six thousands, two hundreds, five tens, and zero ones. So I'm going to fill out this part first. I've got six thousand here. I've got my two hundreds here, my five tens here, and my zero ones. My number is six thousand, two hundred and fifty. I can put that into my place value here, and I can write it here: six thousand, two hundred and fifty. Then I can show it in my Dienes here. I've got nothing in my ones column because I've got fifty. No ones. I've got five tens, my two hundreds, and I've got my six thousands. And my final number then, I've got it written in two different--represented, sorry-- in two different ways. I have it written here and I put it represented in a different way here. So my number is four thousand, two hundred and eighty-three. There were four thousands, two hundreds, eight tens, and three ones. So I could show that, here with my three ones, my eight tens, my two hundreds, and my four thousands. I can write that straight into my place value grid, here: four, two, eight, and three to represent Four thousand, two hundred, eighty--okay, so eight tens, and three ones. Then I can write it in here, so I can write my four thousand, two hundred and eighty three partitioned into four thousand, plus two hundred, plus eighty, plus three. If you'd like to, please ask your parent or carer to share your work today on Twitter, by tagging, @OakNational and #LearnwithOak. I've been really impressed with your work today. Now what I'd like you to do is using everything you've learned today, have a go at the final quiz. Thank you and hopefully we'll see you again soon. Bye!.
# How do you find the limit of (3x^2-x-10)/(x^2+5x-14) as x approaches 2? May 30, 2018 $\frac{11}{9}$ #### Explanation: Observe that $3 {x}^{2} - x - 10 = \left(3 x + 5\right) \left(x - 2\right)$ ${x}^{2} + 5 x - 14 = \left(x + 7\right) \left(x - 2\right)$ So we get ${\lim}_{x \to 2} \frac{3 {x}^{2} - x - 10}{{x}^{2} + 5 x - 1} = {\lim}_{x \to 2} \frac{3 x + 5}{x + 7} = \frac{11}{9}$ May 30, 2018 ${\lim}_{x \to 2} \frac{3 {x}^{2} - x - 10}{{x}^{2} + 5 x - 14} = \frac{11}{9}$ #### Explanation: Method 1): factorize the polynomials As both numerator and denominator vanish for $x = 2$ they can be divided by $\left(x - 2\right)$: $3 {x}^{2} - x - 10 = \left(3 x + 5\right) \left(x - 2\right)$ ${x}^{2} + 5 x - 14 = \left(x + 7\right) \left(x - 2\right)$ Then: ${\lim}_{x \to 2} \frac{3 {x}^{2} - x - 10}{{x}^{2} + 5 x - 14} = {\lim}_{x \to 2} \frac{\left(3 x + 5\right) \left(x - 2\right)}{\left(x + 7\right) \left(x - 2\right)}$ ${\lim}_{x \to 2} \frac{3 {x}^{2} - x - 10}{{x}^{2} + 5 x - 14} = {\lim}_{x \to 2} \frac{3 x + 5}{x + 7}$ ${\lim}_{x \to 2} \frac{3 {x}^{2} - x - 10}{{x}^{2} + 5 x - 14} = \frac{11}{9}$ Method 2): L'Hospital's rule: As both numerator and denominator vanish for $x = 2$: ${\lim}_{x \to 2} \frac{3 {x}^{2} - x - 10}{{x}^{2} + 5 x - 14} = {\lim}_{x \to 2} \frac{\frac{d}{\mathrm{dx}} \left(3 {x}^{2} - x - 10\right)}{\frac{d}{\mathrm{dx}} \left({x}^{2} + 5 x - 14\right)}$ ${\lim}_{x \to 2} \frac{3 {x}^{2} - x - 10}{{x}^{2} + 5 x - 14} = {\lim}_{x \to 2} \frac{6 x - 1}{2 x + 5} = \frac{11}{9}$
# If 2+sqrt 3 is a Polynomial Root If 2+sqrt 3 is a Polynomial Root. Contents: § 2.4 (p. 200) of the text. Suggested Issues from Text: p. 212 #7, viii, xi, 15, 17, 18, 23, 26, 35, 38, 41, 43, 46, 47, 51, 54, 57, 60, 63, 66, 71, 72, 75, 76, 81, 87, 88, 95, 97 ### Equations Involving Fractional Expressions or Absolute Values A quadratic equation is one of the form axtwo + bx + c = 0, where a, b, and c are numbers, and a is not equal to 0. ### Factoring This approach to solving equations is based on the fact that if the product of two quantities is zero, then at to the lowest degree one of the quantities must be nix. In other words, if ab = 0, so either a = 0, or b = 0, or both. For more on factoring polynomials, see the review department P.three (p.26) of the text. Instance ane. 2xtwo – 5x – 12 = 0. (2x + 3)(x – four) = 0. 2x + 3 = 0 or 10 – 4 = 0. 10 = -3/two, or 10 = 4. ### Square Root Principle If x2 = k, then x = ± sqrt(chiliad). Example two. tentwo – nine = 0. xtwo = ix. ten = 3, or 10 = -3. Example 3. Instance 4. x2 + 7 = 0. xii = -7. x = ± . Note that = = , and then the solutions are x = ± , ii complex numbers. ### Completing the Square The idea behind completing the square is to rewrite the equation in a form that allows u.s.a. to employ the foursquare root principle. Example 5. x2 +6x – i = 0. x2 +6x = ane. x2 +6x + 9 = 1 + 9. The nine added to both sides came from squaring half the coefficient of 10, (6/two)two = 9. The reason for choosing this value is that now the left hand side of the equation is the square of a binomial (two term polynomial). That is why this procedure is called completing the square. [ The interested reader can see that this is true by considering (x + a)2 = x2 + 2ax + aii. To get “a” one need only divide the x-coefficient by 2. Thus, to complete the square for ten2 + 2ax, one has to add aii.] (ten + three)two = 10. At present we may utilize the foursquare root principle and so solve for x. 10 = -3 ± sqrt(10). Example 6. 2x2 + 6x – 5 = 0. 2x2 + 6x = five. The method of completing the square demonstrated in the previous case simply works if the leading coefficient (coefficient of x2) is 1. In this example the leading coefficient is ii, merely nosotros tin change that by dividing both sides of the equation by ii. xii + 3x = 5/2. Now that the leading coefficient is 1, nosotros take the coefficient of x, which is now three, divide it past 2 and foursquare, (3/2)two = 9/4. This is the constant that we add together to both sides to consummate the foursquare. x2 + 3x + 9/iv = 5/2 + nine/iv. The left hand side is the foursquare of (x + iii/2). [ Verify this!] (x + iii/2)2 = 19/four. Now nosotros use the square root principle and solve for x. x + iii/2 = ± sqrt(xix/4) = ± sqrt(19)/2. x = -three/2 ± sqrt(19)/two = (-iii ± sqrt(19))/ii So far nosotros accept discussed three techniques for solving quadratic equations. Which is best? That depends on the problem and your personal preference. An equation that is in the right class to use the foursquare root principle may be rearranged and solved by factoring every bit we run into in the side by side example. Instance 7. 102 = 16. tentwo – 16 = 0. (10 + 4)(x – 4) = 0. 10 = -iv, or x = 4. In some cases the equation can be solved by factoring, merely the factorization is not obvious. The method of completing the square will always piece of work, even if the solutions are complex numbers, in which instance nosotros will take the square root of a negative number. Furthermore, the steps necessary to consummate the square are always the same, so they can exist practical to the full general quadratic equation axii + bx + c = 0. The effect of completing the square on this full general equation is a formula for the solutions of the equation chosen the Quadratic Formula. The solutions for the equation ax2 + bx + c = 0 are We are maxim that completing the square e’er works, and we have completed the square in the full general case, where nosotros take a,b, and c instead of numbers. And then, to notice the solutions for any quadratic equation, nosotros write information technology in the standard course to detect the values of a, b, and c, then substitute these values into the Quadratic Formula. One consequence is that y’all never accept to complete the square to observe the solutions for a quadratic equation. However, the process of completing the square is important for other reasons, so y’all still demand to know how to do it! Examples using the Quadratic Formula: Example 8. 2x2 + 6x – 5 = 0. In this case, a = two, b = 6, c = -5. Substituting these values in the Quadratic Formula yields Notice that we solved this equation earlier by completing the foursquare. Note: In that location are two real solutions. In terms of graphs, at that place are 2 intercepts for the graph of the role f(x) = 2x2 + 6x – 5. Case 9. 4x2 + 4x + one = 0 In this example a = four, b = 4, and c = 1. • There is just one solution. In terms of graphs, this means there is only one x-intercept. • The solution simplified so that at that place is no foursquare root involved. This ways that the equation could have been solved by factoring. (All quadratic equations tin can be solved past factoring! What I mean is information technology could take been solved hands by factoring.) 4xtwo + 4x + one = 0. (2x + one)2 = 0. x = -i/ii. Example ten. xtwo + x + 1 = 0 a = 1, b = i, c = i Notation: There are no real solutions. In terms of graphs, there are no intercepts for the graph of the part f(10) = x2 + x + 1. Thus, the solutions are complex because the graph of y = xii + 10 + 1 has no x-intercepts. The expression under the radical in the Quadratic Formula, b2 – 4ac, is chosen the discriminant of the equation. The terminal 3 examples illustrate the iii possibilities for quadratic equations. one. Discriminant > 0. Two existent solutions. 2. Discriminant = 0. I real solution. 3. Discriminant < 0. Ii circuitous solutions. Notes on checking solutions None of the techniques introduced so far in this section can introduce extraneous solutions. (Run across case three from the Linear Equations and Modeling department.) However, it is still a good thought to check your solutions, because information technology is very easy to brand careless errors while solving equations. The algebraic method, which consists of substituting the number back into the equation and checking that the resulting statement is true, works well when the solution is “uncomplicated”, merely it is not very applied when the solution involves a radical. For example, in our adjacent to last case, 4xii + 4x + 1 = 0, nosotros constitute one solution ten = -one/ii. The algebraic bank check looks like 4(-ane/ii)two +four(-one/2) + ane = 0. four(1/iv) – 2 + i = 0. 1 – 2 + ane = 0. 0 = 0. The solution checks. In the instance earlier that, 2xii + 6x – 5 = 0, we found two real solutions, 10 = (-3 ± sqrt(19))/2. It is certainly possible to check this algebraically, simply it is not very easy. In this case either a graphical check, or using a calculator for the algebraic check are faster. First, discover decimal approximations for the 2 proposed solutions. (-iii + sqrt(xix))/2 = 0.679449. (-three – sqrt(nineteen))/2 = -3.679449. Now use a graphing utility to graph y = 2x2 + 6x – five, and trace the graph to observe approximately where the x-intercepts are. If they are close to the values above, then yous can be pretty sure you have the correct solutions. You can also insert the judge solution into the equation to see if both sides of the equation give approximately the same values. However, you lot nevertheless need to be conscientious in your claim that your solution is correct, since it is not the exact solution. Note that if you had started with the equation 2x2 + 6x – 5 = 0 and gone directly to the graphing utility to solve it, then you would non become the verbal solutions, considering they are irrational. Nevertheless, having found (algebraically) 2 numbers that you think are solutions, if the graphing utility shows that intercepts are very shut to the numbers you found, and so you are probably right! Exercise 1: Solve the following quadratic equations. (a) 3xtwo -5x – 2 = 0. Respond (b) (x + 1)2 (c) x2 = 3x + 2. Respond Render to Contents ## Equations Involving Radicals Equations with radicals tin often be simplified past raising to the advisable power, squaring if the radical is a foursquare root, cubing for a cube root, etc. This operation can introduce extraneous roots, so all solutions must be checked. If there is only one radical in the equation, and so earlier raising to a power, you lot should arrange to accept the radical term past itself on one side of the equation. Case 11. Now that we take isolated the radical term on the right side, we foursquare both sides and solve the resulting equation for ten. Bank check: x = 0 When nosotros substitute 10 = 0 into the original equation nosotros get the statement 0 = two, which is not true! So, x = 0 is not a solution. x = 3 When we substitute x = 3 into the original equation, nosotros become the statement 3 = 3. This is true, so x = 3 is a solution. Solution: ten = iii. Note: The solution is the x-coordinate of the intersection point of the graphs of y = x and y = sqrt(x+1)+1. Look at what would have happened if we had squared both sides of the equation before isolating the radical term. This is worse than what we started with! If there is more than than one radical term in the equation, then in general, nosotros cannot eliminate all radicals past raising to a power one fourth dimension. Yet, nosotros can decrease the number of radical terms by raising to a power. If the equation involves more than ane radical term, then nosotros yet want to isolate one radical on i side and raise to a power. Then we repeat that process. Case 12. At present foursquare both sides of the equation. This equation has just 1 radical term, so we have made progress! Now isolate the radical term and so foursquare both sides over again. Check: Substituting 10 = 5/4 into the original equation yields sqrt(9/4) + sqrt(i/four) = 2. 3/two + 1/ii = 2. This statement is true, so x = five/4 is a solution. Notation on checking solutions: The algebraic check was easy to do in this instance. However, the graphical check has the advantage of showing that there are no solutions that nosotros accept not found, at to the lowest degree within the scope of the viewing rectangle. The solution is the 10-coordinate of the intersection point of the graphs of y = two and y = sqrt(10+1)+sqrt(10-i). Exercise 2: Solve the equation sqrt(x+ii) + ii = 2x. Answer Render to Contents ## Polynomial Equations of College Degree We have seen that any caste 2 polynomial equation (quadratic equation) in 1 variable can be solved with the Quadratic Formula. Polynomial equations of degree greater than two are more complicated. When we encounter such a trouble, and then either the polynomial is of a special grade which allows us to factor it, or nosotros must approximate the solutions with a graphing utility. ### Zero Constant One common special instance is where there is no constant term. In this case we may factor out one or more than powers of x to begin the problem. Case 13. 2xthree + 3xii -5x = 0. x (2xtwo + 3x -five) = 0. Now we have a product of x and a quadratic polynomial equal to 0, so we have two simpler equations. x = 0, or 2x2 + 3x -5 = 0. The first equation is lilliputian to solve. x = 0 is the just solution. The second equation may exist solved by factoring. Note: If nosotros were unable to factor the quadratic in the second equation, then we could have resorted to using the Quadratic Formula. [Verify that yous get the same results as below.] x = 0, or (2x + 5)(ten – one) = 0. So there are three solutions: x = 0, x = -five/2, 10 = i. Note: The solution is constitute from the intercepts of the graphs of f(ten) = 2xiii + 3x2 -5x. ### Cistron by Grouping Example 14. x3 -2xii -9x +18 = 0. The coefficient of xtwo is -2 times that of x3, and the aforementioned relationship exists betwixt the coefficients of the third and 4th terms. Group terms one and ii, and likewise terms iii and four. x2 (x – 2) – 9 (x – 2) = 0. These groups share the common cistron (x – 2), so we tin can factor the left hand side of the equation. (ten – 2)(x2 – 9) = 0. Whenever nosotros find a product equal to nil, we obtain two simpler equations. ten – 2 = 0, or x2 – 9 = 0. 10 = 2, or (ten + 3)(x – 3) = 0. So, there are 3 solutions, x = ii, ten = -3, x = iii. Annotation: These solutions are found from the intercepts of the graph of f(x) = x3 -2x2 -9x +18. ### Quadratic in Form Example fifteen. x4 – tenii – 12 = 0. This polynomial is not quadratic, it has degree four. All the same, it can be idea of as quadratic in x2. (102) 2 -(xii) – 12 = 0. It might assistance you lot to actually substitute z for x2. z2 – z – 12 = 0 This is a quadratic equation in z. (z – 4)(z + 3) = 0. z = 4 or z = -3. We are not done, because we need to find values of 10 that brand the original equation true. Now supersede z past ten2 and solve the resulting equations. xtwo = 4. x = ii, x = -2. x2 = -iii. 10 = i , or x = –i. So there are iv solutions, two real and two complex. Note: These solutions are institute from the intercepts of the graph of f(x) = xiv – x2 – 12. A graph of f(x) = xiv – x2 – 12 and a zoom showing its local extrema. Practice 3: Solve the equation xiv – 5x2 + iv = 0. Answer Render to Contents ## Equations Involving Partial Expressions or Absolute Values Example xvi. The least mutual denominator is x(ten + 2), and then we multiply both sides by this product. This equation is quadratic. The Quadratic Formula yields the solutions Checking is necessary because nosotros multiplied both sides by a variable expression. Using a graphing utility we see that both of these solutions check. The solution is the x-coordinate of the intersection indicate of the graphs of y = 1 and y = 2/x-1/(x+ii). Case 17. 5 \| 10 – 1 \| = x + 11. The primal to solving an equation with absolute values is to retrieve that the quantity inside the absolute value bars could be positive or negative. Nosotros will have two separate equations representing the different possibilities, and all solutions must exist checked. Case 1. Suppose 10 – 1 >= 0. And so \| x – ane \| = x – 1, so we accept the equation 5(x – ane) = ten + 11. 5x – v = x + 11. 4x = sixteen. x = 4, and this solution checks because v3 = iv + 11. Case 2. Suppose x – 1 < 0. Then ten – 1 is negative, then \| x – i \| = -(x – ane). This bespeak oftentimes confuses students, because it looks as if nosotros are maxim that the absolute value of an expression is negative, only nosotros are not. The expression (x – ane) is already negative, so -(10 – i) is positive. Now our equation becomes -5(x – 1) = ten + xi. -5x + five = 10 + xi. -6x = 6. ten = -1, and this solution checks because fiveii = -ane + 11. If yous employ the Java Grapher to check graphically, notation that abs() is accented value, then you would graph 5abs(ten - 1) - x - 11 and look at ten-intercepts, or you can find the solution equally the x-coordinates of the intersection points of the graphs of y = x+11 and y = v*abs(10-i). Exercise iv: (a) Solve the equation
Tags Definition A function is an equation between two quantities x and y such that one quantity can be defined in terms of the other quantity. Example y = mx + c – eq. 1 Where x and y are two quantities, m is some number, a multiple of x and c is a constant number. This equation may also be written as: x = (y – c) / m – eq. 2 Equation eq.1 defines y in yerms of x. Equation eq.2 defines x in yerms of y. The function defines a line on a graph. Let m=2 and c=–3, then the equation can be written as y = 2x – 3 This equation can be plotted on a graph with real values for x and y Intercept When the value of x is set to 0, the line cuts the y-axis of the graph at y = –3. This is called the x-intercept. On the graph, the coordinates are (0, –3). When the value of y is set to 0, the line cuts the x-axis at 3/2 or 1.5. This is called the y-intercept. On the graph, the coordinates are (3/2, 0). A straight line can be drawn on the graph with all possible values of x and y between the two intercepts and beyond. Slope In the generic function above, m is the called the slope of the line represented by the equation. The slope can be expressed in terms of x, y and the constant c. m = (y – c) / x The slope is a property of the line and is a constant for all values of x and y. It is calculated by the formula (y1 – y2) / (x1 – x2), where (x1, y1) and (x2, y2) are two points on the line. For the function y = 2x – 3, the slope obtained is 2 by substituting the intercept points (0, –3) and (3/2, 0) in the slope formula. Slope of a line may also be expressed in a trigonometric form. The line on the graph subtends an angle with the x-axis. The tangent of this angle is the slope. If theta is the angle of the line with x-axis, then tan(theta) = opp. Side/adjacent Side. Therefore, –3/(3/2) , which is 2 below the axis, since the negative sign implies that the angle is below the axis line in the fourth quadrant). See graph below. Graph Graph for the function y = 2x – 3 looks as shown below (from WolframAlpha): Domain The domain of a function is all values of x. Range The range of a function is all values of y corresponding to the values of x. In the function y = 2x – 3, the domain and range are both real numbers. Note Consider an equation of the form ax+by+c=0, where a,b,c are real numbers and x and y are unknown quantities. This equation can be expressed in the function form as shown below: Dividing the equation by b, it becomes y = -ax/b-c/b Where slope m = -ax/b and the constant is -c/b
### Complex Numbers - Hinchingbrooke ```Lesson Objective Understand what Complex Number are and how they fit into the mathematical landscape. Be able to do arithmetic with complex numbers Solve the equations: x2 +4x + 1 = 0 x2 +4x + 4 = 0 x2 +4x + 6 = 0 What can we say about the graph of: y = x2 + 4x + c? What are the conditions for each case? Arithmetic of complex numbers: Let z = 3 + 4j w = 2 – 5j Find: a) w + z b) z – w c) z2 d) zw e) Complex Conjugates, Uniqueness and Real and Imaginary Parts: If z = 3 + 4j we say that z* = 3 – 4j is the Complex Conjugate of z we say that Re(z) = 3 is the real part of z we say that Im(z) = -4 is the imaginary part of z Two complex numbers are identical if the imaginary and real parts are the same; In other words if 3 - 2j = a + bj a must be equal to 3 b must be equal to -2 Complex Conjugates, Uniqueness and Real and Imaginary Parts: If z = 3 + 4j we say that z = 3 – 4j is the Complex Conjugate of z we say that Re(z) = 3 is the real part of z we say that Im(z) = -4 is the imaginary part of z Let w = 2 – 5j Find: a) w b) z + w* c) w - z* e) wz f) (zw)* g) 1 w∗ d) (z + w)* 1 ∗ h) ∗ Complex Conjugates, Uniqueness and Real and Imaginary Parts: If z = 3 + 4j we say that z* = 3 – 4j is the Complex Conjugate of z Let w = 2 – 5j Find: a) w* b) z + w* c) w - z* e) wz f) (zw)* g) 1) If y is a complex number y = a + bj x is a complex number x = c + dj y + y * = 2Re(y) y - y* = 2Im(y) (xy)* = xy (x) = x 1 1 2Re(y) e) + ∗= 2+2 Prove that: a) b) c) d) From FP1 page 53 1 w∗ d) (z + w)* 1 ∗ h) ∗ 2) Find real numbers a and b (with a>0) such that a) (a + bj)2 = 21 + 20j b) (a + bj)2 = -40 - 42j 3) Find real numbers ‘a’ and ‘b’ such that + =1− 3+ 1+2 4) Find real numbers z for which z2 = 2z* 5) Solve z + jw = 13 3z – 4w = 2j for complex numbers z and w Complex Conjugates, Uniqueness and Real and Imaginary Parts: If z = 3 + 4j we say that z* = 3 – 4j is the Complex Conjugate of z Let w = 2 – 5j Find: a) w* 2+5j e) wz 26+7j b) z + w* 5+9j f) (zw)* 26+7j g) 1) If y is a complex number y = a + bj x is a complex number x = c + dj Re(y + y ) = 2a Im(y - y) = 2b (xy)* = xy (x) = x 1 1 2Re(y) e) + ∗= 2+2 Prove that: a) b) c) d) From FP1 page 53 c) w - z* 1 w∗ -1-j (2-5j)/29 d) (z + w)* 5+j 1 ∗ h) ∗ (3-4j)/25 2) Find real numbers a and b (with a>0) such that a) (a + bj)2 = 21 + 20j b) (a + bj)2 = -40 - 42j 3) Find real numbers ‘a’ and ‘b’ such that + =1− 3+ 1+2 4) Find real numbers z for which z2 = 2z* 5) Solve z + jw = 13 3z – 4w = 2j for complex numbers z and w Lesson Objective Be able to display complex numbers on an Argand Diagram Understand how to find the modulus and argument of a complex number Consider z = 3 +4j Re(z) = Im(z) = = Arg(z) = and w = -2 – 5j Re(w) = Im(w) = = Arg(w) = How can we represent the following on the Argand diagram: a) z + w ? b) z – w ? What is − ? What does it represent? To generate a mandelbrot set: Solve the equation z2 + 1 =0 By using an iterative approach. (Like C3 coursework) Count the number of iterations required to get within an acceptable margin of the solution. Plot the starting value on an Argand diagram, with a colour that corresponds to the number of steps until convergence. Eg Rearrange to make z = -1/z Choose 1+i as starting value Keep iterating until within a small 1) Write down the modulus and the argument of these complex numbers: a) 2 + 2j b) -3 +4j c) -3-4j d) 3j 2) What can you say about the modulus and argument of z* compared to z? 3) Let z be a complex number On the Argand diagram, show all the complex numbers ‘z’, such that: a) =6 b) Arg(z) = 3 c) − 3 =5 d) − 3 − 4 =5 e) Arg(z-3-4j) = 3 f) − 3 = − 3 g) − 2 =2 − 3 e) -2 Let ‘z’ be a complex number On the Argand diagram, show all the complex numbers ‘z’, such that: − 3 − 4 =5 Let ‘w’ be a complex number On the Argand diagram, show all the complex numbers ‘w’, such that: − 3 − 4 = − 3 + 2 How does this differ from: − 3 − 4 > − 3 + 2 − 3 − 4 < − 3 + 2 ? ? z 3  2 z  2 Note that: z a represents the distance from the complex number a to the complex number z Arg(z – a) represents the angle to the complex number z from the complex number a measured from a line parallel to the +ve part of the real axis If z is a complex number x + iy then: Re(z) = x Im(z) = y z* = The complex conjugate of z= x - iy z 3j  4 z2 6 z 42 j  6 z 1  z  j z  2  2 z 1 z2j 3z Re( z )  Im(z ) Im(z )  0 2 2 Arg ( z  1) 3 Arg ( z  2  j )  Arg ( 2 z ) Arg ( z )    4 3 4  Arg ( z  1)  Arg ( z 1)  4  Arg ( z  j )   Arg ( z j ) 6 Re(z )  2 1 Re( z  * )  0 z  z2 Re 0  z  Lesson Objective Modulus and Argument Form and the beginnings of powerful arithmetic Write down the modulus and the argument of these complex numbers: a) a = 1 + 2j b) b = 2j c) c = -j d) d = -1-j Lesson Objective Modulus and Argument Form and the beginnings of powerful arithmetic Write down the modulus and the argument of these complex numbers: a) a = 1 + 2j b) b = 2j c) c = -j d) d = -1-j Section A Find ab ab2 ab3 ab4 ……. In each case plot the new complex number on the Argand diagram and find its modulus and argument Generalise for abn Section B What can you say about the modulus and Generalised results: Let w be a complex number in the form a + ib Then: The modulus of w = = 2 + 2 The argument of w = Arg(w) = the angle that w makes with the +ve section of the real axes (usually given between 180 and -180 degrees or We can write w as: w = (∅ + jsin∅) this is called modulus/argument form (or polar form) When two complex numbers are multiplied together, the resulting complex number: Will have an argument = to the sum of the two original arguments and a modulus =to the product of the two original moduli. Puzzle 1 The points 1 + j and 3 + 4j are two adjacent corners of a square. Where are the other corners? Can you solve this problem? More importantly, how can you solve it using complex numbers? Puzzle 2 Suppose that 1 + j and 3 + 4j were two adjacent corners of an equilateral triangle, where would the final vertex be? This is much easier to solve using complex numbers! Puzzle 3 Can we generalise this method to a regular ‘n’ sided shape? This really demonstrates just how cool complex numbers are!!!!!!!! Lesson Objective Understand the Fundametal Thm of Algebra be able to solve cubic/quartic equations with Real coefficients even when some roots are complex Write down any quadratic equation with complex roots. Solve it. What do you notice about the solutions? Is it true that (wn)* (w)n ? Try it with (wn) How can we prove this key result? Lesson Objective Understand the Fundamental Thm of Algebra and be able to solve cubic/quartic equations with Real coefficients even when some roots are complex Starter: Is it true that: (w)n = (wn) for all n ≥ 0 where w is a complex number? Try it with (w)3 How can we prove this key result? Lesson Objective Understand the Fundamental Thm of Algebra be able to solve cubic/quartic equations with Real coefficients even when some roots are complex Write down any quadratic equation with complex roots? Solve it. What do you notice about the solutions? Lesson Objective Understand the Fundamental Thm of Algebra be able to solve cubic/quartic equations with Real coefficients even when some roots are complex Write down any quadratic equation with complex roots? Solve it. What do you notice about the solutions? The Fundamental Thm of algebra states that any polynomial equation of degree ‘n’ will have exactly ‘n’ solutions (if you count repeated roots). The Complex Conjugate Root Thm goes further, and states that, as long as the coefficients are real, the solutions will come in complex conjugate pairs. Proof of Fundamental Thm of Algebra (Go to University – it took Gauss years) Proof of the Complex Conjugate Root Thm Consider: αzn + βzn-1 + γzn-2 + …….. + δz2 + εz + ζ = 0 Then ‘z’ is a solution to the equation Now consider: (αzn + βzn-1 + γzn-2 + …….. + δz2 + εz + ζ) = (0)* (αzn)* + (βzn-1)* + (γzn-2)* + …….. + (δz2)* + (εz)* + (ζ)* = (0)* α(zn) + β* (zn-1)* + γ* (zn-2)* + …….. + δ* (z2)* + ε* (z)* + (ζ)* = 0* α(z)n + β* (z)n-1 + γ (z)n-2 + …….. + δ (z)2 + ε (z) + (ζ) = 0* α(z)n + β (z)n-1 + γ (z)n-2 + …….. + δ (z)2 + ε (z) + ζ = 0 Example Shows the 2 + j is a solution to z3 – z2 -7z + 15 = 0 Hence find all the other roots. Example Give than z = -2 + j is a solution to z4 + az3 +bz2 +10z + 25 = 0 Find the values of a and b Hence factorise the cubic and find all 4 solutions Example Give than z = -2 + j is a solution to z4 + az3 +bz2 +10z + 25 = 0 Find the values of a and b Hence factorise the cubic and find all 4 solutions a=2b=2 Other two solutions are 1 + 2j and 1 – 2j ```
# Angle Measure and Side Length Relationship of a Triangle In the previous post, we proved that if the two sides of a triangle are not congruent, then the angles opposite to these sides are also not congruent. In addition, we have shown that the angle opposite to the longer side is the larger angle. In this post, we continue with the converse of this theorem. We show that if two angles of a triangle are not congruent, then the sides opposite to these angles are not congruent. Also, the side opposite to the larger angle is the longer side. This is evident as shown in the figure above. As we can see, $latex \angle B$ is the largest angle and $latex \overline{AB}$, the side opposite to it is also the longest side. Therefore, we show that if $latex \angle C > \angle B$ then, $latex \overline{AB} > \overline{AC}$. Theorem In any triangle $latex ABC$, if $latex \angle C < \angle B$ then, $latex \overline{AB} > \overline{AC}$. Proof Assume $latex \overline{AB} \ngtr \overline{AC}$. Then $latex \overline{AB} \cong \overline{AC}$ or $latex \overline{AB} < \overline{AC}$. If $latex \overline{AB} \cong \overline {AC}$, then $latex \triangle ABC$ is isosceles and $latex \angle C \cong \angle B$ by the Isosceles Triangle Theorem. If $latex \overline{AB} < \angle \overline{BC}$, by the previous theorem, $latex \angle C < \angle B$. Now, both of these statements contradict the given that $latex \angle C > \angle B$. Therefore, the assumption that $latex \overline{AB} \ngtr \overline{AC}$ must be false. So, in any triangle $latex ABC$, if $latex \angle C < \angle B$ then, $latex \overline{AB} > \overline{AC}$.
# Finding $b(a+c)$ for Real Roots of $\sqrt{2014}x^3-4029x^2+2=0$ • MHB • anemone In summary, to find the value of $b(a+c)$ in the given equation, use the quadratic formula with the values of $a=\sqrt{2014}$, $b=-4029$, and $c=2$. There are various methods, such as factoring, completing the square, and using the quadratic formula, that can be used to solve this type of equation. To check if the solution for $b(a+c)$ is correct, plug the value back into the original equation. The equation can only have one solution for $b(a+c)$ and there are restrictions on the values of $a$, $b$, and $c$ in order for the equation to have real roots. anemone Gold Member MHB POTW Director Let $a>b>c$ be the real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$. Find $b(a+c)$. To avoid radicals let $\sqrt{2014}=p$ So we get $px^3-(2p^2+1)x^2 +2 = 0$ Or factoring we get $(px-1)(x^2-2px-2)$ = 0 So one root is $x= \frac{1}{p}$ and other two roots are roots of $x^2-2px-2=0$ For the equation $x^2-2px-2=0$ sum of the roots is 2p and product is -2. so one root has to be -ve and the postiive root shall be above 2p So $b=\frac{1}{p}\cdots(1)$ And c is the -ve root and $a> 2p$ a,c are roots of $x^2-2px-2=0$ so $a+c = 2p\cdots(2)$ Hence $b(a+c) = 2$ using (1) and (2) ## 1. What is the formula for finding $b(a+c)$? The formula for finding $b(a+c)$ in the equation $\sqrt{2014}x^3-4029x^2+2=0$ is $b(a+c) = \frac{-b}{a}$, where $a$ and $b$ are the coefficients of the quadratic term and the linear term, respectively. ## 2. How do you know if a given equation has real roots? A given equation has real roots if the discriminant, $b^2-4ac$, is greater than or equal to 0. If the discriminant is less than 0, the equation will have complex roots. ## 3. Can the value of $b(a+c)$ be negative? Yes, the value of $b(a+c)$ can be negative. This would occur if the coefficient of the linear term, $b$, is negative and the coefficient of the quadratic term, $a$, is positive. ## 4. What is the significance of finding $b(a+c)$ for real roots? Finding $b(a+c)$ for real roots is significant because it allows us to determine the sum of the roots of the equation. This can help us in solving the equation and understanding the behavior of the graph of the equation. ## 5. Can the value of $b(a+c)$ be 0? Yes, the value of $b(a+c)$ can be 0. This would occur if the coefficient of the linear term, $b$, is 0. In this case, the equation would simplify to $ax^2+c=0$ and the value of $b(a+c)$ would be 0 regardless of the value of $a$ and $c$. • General Math Replies 1 Views 1K • General Math Replies 4 Views 1K • General Math Replies 1 Views 1K • General Math Replies 2 Views 885 • General Math Replies 6 Views 986 • General Math Replies 19 Views 2K • General Math Replies 1 Views 952 • General Math Replies 2 Views 986 • General Math Replies 1 Views 856 • General Math Replies 1 Views 825
# Fraction math solver Fraction math solver is a mathematical tool that helps to solve math equations. We can solve math problems for you. ## The Best Fraction math solver This Fraction math solver provides step-by-step instructions for solving all math problems. Pros and cons of probability PROS: Probability is a great tool for beginners and people who are unfamiliar with statistics. It’s straightforward to understand, which makes it an ideal way to learn the basics of statistics. There are many different types of probability questions that can be used in a variety of applications. This makes probability a versatile tool that can help solve a wide range of problems. CONS: Probability questions may be challenging for some students. They have to keep in mind both the probabilities for each outcome and the overall likelihood of each outcome occurring. Probability questions also require understanding of how to interpret data and how to identify patterns in data. When calculating a circle’s radius, you need to take into account both the radius of the circle’s circumference and the radius of its diameter. You can use this formula to solve for either or both: With these formulas, all you have to do is find the radius of each side in relation to the other one. You should also remember that the radius increases as your circle gets larger. If a circle has a radius of 1 unit, then its radius will double (or triple) as it grows from 1 unit in size. Once you know how much bigger a circle is than another one, you can calculate its diameter. Divide the first circle’s circumference by the second one’s diameter and multiply by pi to get the answer. Once you have the roots, you can use them to determine which values of x satisfy the inequality. If the roots are real, you will need to use the sign of the quadratic equation to determine which values of x satisfy the inequality. If the roots are complex, you will need to use the conjugate roots to determine which values of x satisfy the inequality. Algebra is the branch of mathematics that deals with the rules of operations and relations, and the study of quantities which may be either constant or variable. Factoring is a technique used to simplify algebraic expressions. When an expression is factored, it is rewritten as a product of simpler factors. This can be helpful in solving equations and graphing functions. In general, factoring is the process of multiplying two or more numbers to get a product. For example, 6 can be factored as 2 times 3, since 2 times 3 equals 6. In algebra, factoring is often used to simplify equations or to find solutions. For example, the equation x^2+5x+6 can be simplified by factoring it as (x+3)(x+2). This can be helpful in solving the equation, since now it can be seen that the solution is x=-3 or x=-2. Factoring can also be used to find zeroes of polynomials, which are important in graphing functions. In general, polynomials can be factored into linear factors, which correspond to zeroes of the function. For example, the function f(x)=x^2-4x+4 has zeroes at x=2 and x=4. These zeroes can be found by factoring the polynomial as (x-2)(x-4). As a result,factoring is a powerful tool that can be used to simplify expressions and solve equations. ## Instant assistance with all types of math Ok so it may seem like a malicious piece of software made to cheat homework. But I have used it a few times to walk me through problems that troubled me or things the teacher had not taught yet. I know it is used to cheat homework and I have seen people use it in this way, but this provides a thorough explanation to each problem that goes in my way. It has a very impressive text reader and can somehow read my [awful] handwriting. When it does mess up though, you can always correct it, too. Sylvia Morgan This app is godly. It automatically calculates the answer to the problem, shows its work, and graphs it. Recently, I had been struggling with a topic in math and just couldn’t figure it out. After using this app to complete a practice problem, I looked through the work it did and I immediately understood what to do. This app is very handy for students in any grade and I 100% recommend it. Belinda Reed

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