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# Maharashtra Board Practice Set 1 Class 7 Maths Solutions Chapter 1 Geometrical Constructions Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 1 Answers Solutions Chapter 1 Geometrical Constructions. ## Geometrical Constructions Class 7 Practice Set 1 Answers Solutions Chapter 1 Question 1. Draw line segments of the lengths given below and draw their perpendicular bisectors: i. 5.3 cm ii. 6.7 cm iii. 3.8 cm Solution: i. Line AB is the perpendicular bisector of seg PQ. ii. Line UV is the perpendicular bisector of seg ST. iii. Line ST is the perpendicular bisector of seg LM. Question 2. Draw angles of the measures given below and draw their bisectors: i. 105° ii. 55° iii. 90° Solution: i. 105° ii. 55° iii. 90° Question 3. Draw, an obtuse-angled triangle and a right-angled triangle. Find the points of concurrence of the angle bisectors of each triangle. Where do the points of concurrence lie? Solution: The points of concurrence of the angle bisectors of both the triangles lie in the interior of the triangles. Question 4. Draw a right-angled triangle. Draw the perpendicular bisectors of its sides. Where does the point of concurrence lie? Solution: The point of concurrence of the perpendicular bisectors of the sides of the right angled triangle lies on the hypotenuse. Question 5. Maithili, Shaila and Ajay live in three different places in the city. A toy shop is equidistant from the three houses. Which geometrical construction should be used to represent this? Explain your answer. Solution: Since, Maithili, Shaila and Ajay live in three different places, lines joining their houses will form a triangle. The position of the toy shop which is equidistant from three houses can be found out by drawing the perpendicular bisector of the sides of the triangle joining the three houses. The shop will be at the point of concurrence of the perpendicular bisectors. Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 1 Intext Questions and Activities Question 1. Draw a line segment PS of length 4cm and draw its perpendicular bisector. (Textbook pg. no. 1) 1. How will your verify that CD is the perpendicular bisector? m∠CMS = __° 2. Is l(PM) = l(SM)? Solution: 1. Here, m∠CMS = 90° 2. Also, l(PM) = l(SM) = 2cm ∴ line CD is the perpendicular bisector of seg PS.
# Converting Decimals, Improper Fractions, and Mixed Numbers Hello, and welcome to this video about converting decimals, improper fractions, and mixed numbers! In this video, we will explore the steps to convert a decimal to a fraction and a mixed number to an improper fraction and vice versa. Let’s learn about converting decimals to fractions and mixed numbers to improper fractions! We use fractions and decimals daily. All our money transactions are done in decimal form. When we bake, we use half cups and one-third cups. Being able to convert between a decimal and a fraction is a math skill you will use every single day. The table shows a decimal, an improper fraction, and a mixed number that are all equivalent. Decimal $$3.25$$ Improper fraction $$\frac{13}{4}$$ Mixed number $$3\frac{1}{4}$$ Converting a Decimal to a Fraction: When converting a decimal to a fraction, we start by multiplying the decimal by a fraction that is equivalent to 1, so the value of the decimal does not change. For example, if there is only one number after the decimal (in other words, a number in the tenths place), the number would be multiplied by $$\frac{10}{10}$$. If there are two numbers after the decimal (the last digit is in the hundredths place), we would multiply the number by $$\frac{100}{100}$$, and so on. After multiplying, we simplify the fraction. Let’s take a look at an example. We will convert 5.85 into fraction form. Since there are 2 numbers after the decimal, we will multiply 5.85 by $$\frac{100}{100}$$. $$5.58\times \frac{100}{100}=\frac{585}{100}$$ To simplify the fraction, we will start by breaking down the number into its factors. Then we will cancel out the common factors in the numerator and denominator. Both numbers have a factor of 5, so $$\frac{585}{100}$$ can be simplified to $$\frac{117}{20}$$. $$\frac{585}{100}=\frac{5\times 3\times 3\times 13}{5\times 5\times 2\times 2}=\frac{117}{20}$$ The fraction equivalent to 5.85 is $$\frac{117}{20}$$, which is also called an improper fraction. We will now convert the improper fraction into a mixed number. Converting an Improper Fraction to a Mixed Number: We will convert the improper fraction, $$\frac{117}{20}$$, to a mixed number by first dividing the numerator by the denominator. The whole number becomes the number in the front of the fraction, the remainder becomes the numerator of the fraction, and the denominator of the fraction remains the same. Therefore, $$\frac{117}{20}$$, converted to a mixed number is $$5\frac{17}{20}$$. Here’s an example of how we use conversions in real life. Samm owns a bakery. She has $$5\frac{3}{4}$$ kg of sugar. She buys another bag with 10.75 kg of sugar. What is the total amount of sugar, in kilograms, that Samm has for baking? Give your answer in fraction form. First, we will start by converting 10.75 to fraction form by multiplying it by $$\frac{100}{100}$$, which is $$\frac{1,075}{100}$$. Once the fraction is simplified, we get $$\frac{43}{4}$$, which when converted to a mixed number is $$10\frac{3}{4}$$. Now that both numbers are in mixed number form, we can easily combine to find the total amount of sugar that Samm has for her baking. So all we’re gonna do is add, $$5\frac{3}{4}+10\frac{3}{4}$$. When we add mixed numbers, we want to start by adding the fractional parts, so let’s do $$\frac{3}{4}+\frac{3}{4}$$. That gives us $$\frac{6}{4}$$ because we add our numerators and our denominator stays the same. Now, if you notice, we have an improper fraction. So let’s convert that to a mixed number. If we divide the numerator by the denominator, we’ll get $$1\frac{2}{4}$$, which can be simplified to $$1\frac{1}{2}$$. Now we’re gonna add this part to our whole number parts from earlier. So $$5+10=15+1\frac{1}{2}=16\frac{1}{2}$$. So Samm has $$16\frac{1}{2}$$ kg of sugar for baking. I hope this video on converting decimals, improper fractions, and mixed numbers was helpful. Thanks for watching, and happy studying! ## Practice Questions Question #1: Convert the decimal number 6.8 to a mixed number. $$6\frac{4}{5}$$ $$6\frac{3}{4}$$ $$6\frac{5}{8}$$ $$6\frac{1}{2}$$ A mixed number is a whole number along with a proper fraction next to it. Step 1. Convert the decimal number to an improper fraction by multiplying it by some form of one, so the value of the decimal does not change. The number 8 after the decimal is in the tenths place, so we can multiply our decimal number by $$\frac{10}{10}$$ to convert it to an improper fraction. $$6.8\times\frac{10}{10}=\frac{68}{10}$$ Step 2. Reduce the improper fraction by completely factoring the numerator and denominator, then cancel out any common factors between them. $$\frac{68}{10}=\frac{2\times 2\times 17}{2\times 5}=\frac{34}{5}$$ Step 3. Convert the improper fraction, in reduced form, to a mixed number by dividing the numerator by the denominator. $$\phantom{0}6$$ $$5$$ $$34$$ $$–$$ $$30$$ $$4$$ The mixed number is written using the quotient, any remainder, and the divisor. The whole number for our mixed number is the quotient of $$6$$. The numerator of the proper fraction is the remainder of $$4$$, and the denominator is the divisor of $$5$$. So, $$\frac{34}{5}=6\frac{4}{5}$$. Thus, $$6.8=6\frac{4}{5}$$. Question #2: Convert the decimal number 3.25 to a mixed number. $$3\frac{1}{8}$$ $$3\frac{1}{4}$$ $$3\frac{2}{5}$$ $$3\frac{3}{4}$$ A mixed number is a whole number along with a proper fraction next to it. Step 1. Convert the decimal number to an improper fraction by multiplying it by some form of one, so the value of the decimal does not change. The number 5 after the decimal is in the hundredths place, so we can multiply our decimal number by $$\frac{100}{100}$$ to convert it to an improper fraction. $$3.25\times\frac{100}{100}=\frac{325}{100}$$ Step 2. Reduce the improper fraction by completely factoring the numerator and denominator, then cancel out any common factors between them. $$\frac{325}{100}=\frac{5\times5\times13}{2\times2\times\times5}=\frac{13}{4}$$ Step 3. Convert the improper fraction, in reduced form, to a mixed number by dividing the numerator by the denominator. $$\phantom{0}3$$ $$4$$ $$13$$ $$–$$ $$12$$ $$1$$ The mixed number is written using the quotient, any remainder, and the divisor. The whole number for our mixed number is the quotient of 3. The numerator of the proper fraction is the remainder of 1, and the denominator is the divisor of 4. So, $$\frac{13}{4}=3\frac{1}{4}$$. Thus, $$3.25=3\frac{1}{4}$$. Question #3: Convert the decimal number 8.275 to a mixed number. $$8\frac{25}{49}$$ $$8\frac{22}{45}$$ $$8\frac{11}{40}$$ $$8\frac{29}{40}$$ A mixed number is a whole number along with a proper fraction next to it. Step 1. Convert the decimal number to an improper fraction by multiplying it by some form of one, so the value of the decimal does not change. The number 5 after the decimal is in the thousandths place, so we can multiply our decimal number by $$\frac{1{,}000}{1{,}000}$$ to convert it to an improper fraction. $$8.275\times\frac{1{,}000}{1{,}000}=\frac{8{,}275}{1{,}000}$$ Step 2. Reduce the improper fraction by completely factoring the numerator and denominator, then cancel out any common factors between them. $$\frac{8{,}275}{1{,}000}=\frac{5\times5\times331}{2\times2\times2\times5\times5\times5}=\frac{331}{40}$$ Step 3. Convert the improper fraction, in reduced form, to a mixed number by dividing the numerator by the denominator. $$\phantom{00}8$$ $$40$$ $$331$$ $$–$$ $$320$$ $$11$$ The mixed number is written using the quotient, any remainder, and the divisor. The whole number for our mixed number is the quotient of 8. The numerator of the proper fraction is the remainder of 11, and the denominator is the divisor of 40. So, $$\frac{331}{40}=8\frac{11}{40}$$. Thus, $$8.275=8\frac{11}{40}$$. Question #4: A fruit drink recipe that is made with several types of fruit is mixed with 20 ounces of water, 10.2 ounces of apple juice, and $$\frac{13}{5}$$ ounces of lemon juice. How much liquid is needed for the recipe? $$32\frac{4}{5}\text{ ounces}$$ $$32\frac{1}{5}\text{ ounces}$$ $$32\frac{1}{4}\text{ ounces}$$ $$32\frac{3}{4}\text{ ounces}$$ We need to combine the amounts of water, apple juice, and lemon juice to find the total amount of liquid needed to mix into the fruit drink. We will convert the amounts of apple juice and lemon juice to mixed numbers first before combining all of the amounts of liquid. A mixed number is a whole number along with a proper fraction next to it. Step 1. Convert the decimal number of 10.2 to an improper fraction by multiplying it by some form of one, so the value of the decimal does not change, before converting it to a mixed number. The number 2 after the decimal is in the tenths place, so we can multiply our decimal number by $$\frac{10}{10}$$ to convert it to an improper fraction. $$10.2\times\frac{10}{10}=\frac{102}{10}$$ Step 2. Reduce the improper fraction by completely factoring the numerator and denominator, then cancel out any common factors between them. $$\frac{102}{10}=\frac{2\times51}{2\times5}=\frac{51}{5}$$ Step 3. Convert the improper fraction, in reduced form, to a mixed number by dividing the numerator by the denominator. $$10$$ $$5$$ $$51$$ $$–$$ $$50$$ $$1$$ The mixed number is written using the quotient, any remainder, and the divisor. The whole number for our mixed number is the quotient of 10. The numerator of the proper fraction is the remainder of 1, and the denominator is the divisor of 5. So, $$\frac{51}{5}=10\frac{1}{5}$$. Step 4. Next, convert $$\frac{13}{5}$$ to a mixed number by dividing the numerator by the denominator. $$\phantom{0}2$$ $$5$$ $$13$$ $$–$$ $$10$$ $$3$$ The mixed number is written using the quotient, any remainder, and the divisor. The whole number for our mixed number is the quotient of 2. The numerator of the proper fraction is the remainder of 3, and the denominator is the divisor of 5. So, $$\frac{13}{5}=2\frac{3}{5}$$. Step 5. Now that we have converted the amounts of apple juice and lemon juice to mixed numbers, we can combine the 3 amounts by adding the proper fractions for the mixed numbers, then adding the whole numbers for each mixed number. $$20+10\frac{1}{5}+2\frac{3}{5}=32\frac{4}{5}$$ So, there is a total of $$32\frac{4}{5}$$ ounces of liquid needed for the fruit drink recipe. Question #5: You are on a fishing trip, hoping to catch a certain fish. The daily weight limit for keeping the fish is $$9\frac{1}{2}$$ pounds. You catch two fish, one weighing 4.75 pounds, and the other weighing $$\frac{9}{2}$$ pounds. Will you need to release one of the fish back into the water due to exceeding the daily weight limit? Yes, since the total weight of fish you caught is $$9\frac{3}{4}$$ pounds, which is greater than the daily weight limit of $$9\frac{1}{2}$$ pounds. Yes, since the total weight of fish you caught is $$9\frac{1}{4}$$ pounds, which is greater than the daily weight limit of $$9\frac{1}{2}$$ pounds. No, since the total weight of fish you caught is $$9\frac{1}{4}$$ pounds, which is less than the daily weight limit of $$9\frac{1}{2}$$ pounds. No, since the total weight of fish you caught is $$9\frac{3}{4}$$ pounds, which is less than the daily weight limit of $$9\frac{1}{2}$$ pounds. We will combine the total weight of the 2 fish you caught using mixed numbers and compare it with the daily weight limit to determine if one of the fish needs to be released. A mixed number is a whole number along with a proper fraction next to it. Step 1. Convert the weight of the fish given as a decimal number to an improper fraction by multiplying it by some form of one, so the value of the decimal does not change, before converting it to a mixed number. The number 5 after the decimal is in the hundredths place value, so we can multiply our decimal number by $$\frac{100}{100}$$ to convert it to an improper fraction. $$4.75\times\frac{100}{100}=\frac{475}{100}$$ Step 2. Reduce the improper fraction by completely factoring the numerator and denominator, then cancel out any common factors between them. $$\frac{475}{100}=\frac{5×5×19}{2×2×5×5}=\frac{19}{4}$$ Next, convert the improper fraction, in reduced form, to a mixed number by dividing the numerator by the denominator. $$4$$ $$4$$ $$19$$ $$–$$ $$16$$ $$3$$ The mixed number is written using the quotient, any remainder, and the divisor. The whole number for our mixed number is the quotient of 4. The numerator of the proper fraction is the remainder of 3, and the denominator is the divisor of 4. So, $$4.75=\frac{19}{4}=4\frac{3}{4}$$. Step 3. Convert the weight of the fish given as an improper fraction to a mixed number by dividing the numerator by the denominator. $$4$$ $$2$$ $$9$$ $$–$$ $$8$$ $$1$$ The mixed number is written using the quotient, any remainder, and the divisor. The whole number for our mixed number is the quotient of 4. The numerator of the proper fraction is the remainder of 1, and the denominator is the divisor of 2. So, $$\frac{9}{2}=4\frac{1}{2}$$. Step 4. Now that we have converted the weights of the two fish caught to mixed numbers, we can combine them by adding the proper fractions for each mixed number, then adding the whole numbers for each mixed number. $$4\frac{3}{4}+4\frac{1}{2}=4\frac{3}{4}+4\frac{2}{4}=8\frac{5}{4}$$ Since the fractional part of the mixed number is an improper fraction, we need to convert it to a mixed number and add it to 8 to determine the total weight of the fish caught as a proper mixed number. We can convert $$\frac{5}{4}$$ to a mixed number by dividing the numerator by the denominator. $$1$$ $$4$$ $$5$$ $$–$$ $$4$$ $$1$$ The mixed number is written using the quotient, any remainder, and the divisor. The whole number for our mixed number is the quotient of 1. The numerator of the proper fraction is the remainder of 1, and the denominator is the divisor of 4. So, $$\frac{5}{4}=1\frac{1}{4}$$. Therefore, the proper mixed number is: $$8\frac{5}{4}=8+1\frac{1}{4}=9\frac{1}{4}$$ The total weight of the two game fish you caught is $$9\frac{1}{4}$$ pounds. The daily weight limit for the game fish is $$9\frac{1}{2}$$ pounds, or $$9\frac{2}{4}$$ pounds. Since your total weight for the fish caught is less than the daily weight limit, you do not need to release any of the fish.
# FREE 6th Grade Georgia Milestones Assessment System Math Practice Test Welcome to our FREE 6th Grade Georgia Milestones Assessment System Math Practice Test, with answer key and answer explanations. This practice test’s realistic format and high-quality practice questions can help your student succeed on the Georgia Milestones Assessment System Math test. Not only does the test closely match what students will see on the real Georgia Milestones Assessment System, but it also comes with detailed answer explanations. For this practice test, we’ve selected 20 real questions from past exams for your student’s Georgia Milestones Assessment System Practice test. Your student will have the chance to try out the most common Georgia Milestones Assessment System Math questions. For every question, there is an in-depth explanation of how to solve the question and how to avoid mistakes next time. Use our free Georgia Milestones Assessment System Math practice tests and study resources (updated for 2021) to help your students ace the Georgia Milestones Assessment System Math test! Make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions students need to practice. ## 10 Sample 6th Grade Georgia Milestones Assessment System Math Practice Questions 1- There are 55 blue marbles and 143 red marbles. We want to place these marbles in some boxes so that there is the same number of red marbles in each box and the same number of blue marbles in each of the boxes. How many boxes do we need? A. 8 B. 9 C. 10 D. 11 2- What is the value of the following expression? $$2,205÷315$$ A. 5 B. 6 C. 7 D. 8 3- Solve the following equation. $$112=22+x$$ A. $$x=-90$$ B. $$x=90$$ C. $$x=-134$$ D. $$x=134$$ 4- Car A travels 221.5 km at a given time, while car B travels 1.2 times the distance car A travels at the same time. What is the distance car B travels during that time? A. 222.7 km B. 233.5 km C. 241.5 km D. 265.8 km 5- The perimeter of the trapezoid below is 38. What is its area? A. 198 cm$$^2$$ B. 162 cm$$^2$$ C. 99 cm$$^2$$ D. 81cm$$^2$$ 6- Which of the following expressions has the greatest value? A. $$3^1+12$$ B. $$3^3-3^2$$ C. $$3^4-60$$ D. $$3^5-218$$ 7- Alfred has $$x$$ apples. Alvin has 40 apples, which is 15 apples less than number of apples Alfred owns. If Baron has $$\frac{1}{5}$$ times as many apples as Alfred has. How many apples does Baron have? A. 5 B. 11 C. 55 D. 275 8- In the following triangle find $$α$$. A. $$100^\circ$$ B. $$90^\circ$$ C. $$60^\circ$$ D. $$30^\circ$$ 9- The price of a laptop is decreased by $$15\%$$ to $425. What is its original price? A.$283 B. $430 C.$500 D. $550 10- Find the perimeter of shape in the following figure? (all angles are right angles) A. 21 B. 22 C. 24 D. 20 11- What are the values of mode and median in the following set of numbers? $$1,3,3,6,6,5,4,3,1,1,2$$ A. Mode: 1, 2, Median: 2 B. Mode: 1, 3, Median: 3 C. Mode: 2, 3, Median: 2 D. Mode: 1, 3, Median: 2.5 12- Which expression equivalent to $$x × 92$$? A. $$(x×90)+2$$ B. $$x×9×2$$ C. $$(x×90)+(x×2)$$ D. $$(x×90)+2$$ 13- The ratio of pens to pencils in a box is 3 to 5. If there are 96 pens and pencils in the box altogether, how many more pens should be put in the box to make the ratio of pens to pencils 1 : 1? A. 22 B. 23 C. 24 D. 25 14- If point A placed at $$-\frac{24}{3}$$ on a number line, which of the following points has a distance equal to 5 from point A? A. $$-13$$ B. $$-3$$ C. $$-2$$ D. A and B 15- Which of the following shows the numbers in increasing order? A. $$\frac{3}{13}, \frac{4}{11}, \frac{5}{14}, \frac{2}{5}$$ B. $$\frac{3}{13}, \frac{5}{14}, \frac{4}{11}, \frac{2}{5}$$ C. $$\frac{3}{13}, \frac{5}{14}, \frac{2}{5}, \frac{4}{11}$$ D. $$\frac{5}{14}, \frac{3}{13}, \frac{2}{5}, \frac{4}{11}$$ 16- If $$x=- 4$$, which of the following equations is true? A. $$x(3x-1)=50$$ B. $$5(11-x^2 )=-25$$ C. $$3(-2x+5)=49$$ D. $$x(-5x-19)=-3$$ 17- What is the missing prime factor of number 450? $$450=2^1×3^2×…$$ _________ 18- What is the perimeter of the following shape? (it’s a right triangle) A. 14 cm B. 18 cm C. 24 cm D. 32 cm 19- 65 is what percent of 50? A. $$50 \%$$ B. $$77 \%$$ C. $$130 \%$$ D. $$140 \%$$ 20- Which of the following expressions has a value of $$-23$$? A. $$-10+(-8)+ \frac{5}{2}×(-2)$$ B. $$5×3+(-2)×18$$ C. $$-10+6×8÷(-4)$$ D. $$(-3) × (-7) + 2$$ ## Best 6th Grade Georgia Milestones Assessment System Math Workbook Resource for 2021 ## Answers: 1- D First, we need to find the GCF (Greatest Common Factor) of 143 and 55. $$143=11×13$$ $$55=5×11→$$ GFC$$= 11$$ Therefore, we need 11 boxes. 2- C $$2205÷315=\frac{2205}{315}=\frac{441}{63}=\frac{147}{21}= 7$$ 3- B $$112=22+x$$ Subtract 22 from both sides of the equation. Then: $$x=112-22=90$$ 4- D Distance that car B travels $$=1.2 ×$$ distance that car A travels =$$1.2×221.5=265.8$$ km 5- D The perimeter of the trapezoid is 38. Therefore, the missing side (height) is $$= 38 – 8 – 10 – 11 = 9$$ Area of the trapezoid: $$A = \frac{1}{2} h (b_1 + b_2) = \frac{1}{2}1 (9) (8 + 10) = 81$$ 6- D A. $$3^1+12=3+12=15$$ B. $$3^3-3^2=27-9=18$$ C. $$3^4-60=81-60=21$$ D. $$3^5-218=243-218=25$$ 7- B Alfred has $$x$$ apple which is 15 apples more than number of apples Alvin owns. Therefore: $$x-15=40→x=40+15=55$$ Alfred has 55 apples. Let $$y$$ be the number of apples that Baron has. Then: $$y=\frac{1}{5}×55=11$$ 8- A Complementary angles add up to 180 degrees. $$β+150^\circ=180^\circ→β=180^\circ-150^\circ=30^\circ$$ The sum of all angles in a triangle is 180 degrees. Then: $$α+β+50^\circ=180^\circ→α+30^\circ+50^\circ=180^\circ$$ $$→α+80^\circ=180^\circ→α=180^\circ-80^\circ=100^\circ$$ 9- C Let $$x$$ be the original price. If the price of a laptop is decreased by $$15\%$$ to$425, then: $$85 \% \space of \space x=425⇒ 0.85x=425 ⇒ x=425÷0.85=500$$ 10- C Let $$x$$ and $$y$$ be two sides of the shape. Then: $$x+1=1+1+1→x=2$$ $$y+6+2=5+4→y+8=9→y=1$$ Then, the perimeter is: $$1+5+1+4+1+2+1+6+2+1=24$$ 11- B First, put the numbers in order from least to greatest: $$1, 1, 1, 2, 3, 3, 3, 4, 5, 6, 6$$ The Mode of the set of numbers is: 1 and 3 (the most frequent numbers) Median is: 3 (the number in the middle) 12- C $$x×92=x×(90+2)=(x×90)+(x×2)$$ 13- C The ratio of pens to pencils is $$3 : 5$$. Therefore there are 3 pens out of all 8 pens and pencils. To find the answer, first dived 96 by 8 then multiply the result by 3. $$96÷8=12→12×3=36$$ There are 36 pens and 60 pencils $$(96-36)$$. Therefore, 24 more pens should be put in the box to make the ratio $$1 : 1$$ 14- D If the value of point A is greater than the value of point B, then the distance of two points on the number line is: value of A- value of B A. $$-\frac{24}{3}-(-13)=-8+13=5=5$$ B. $$-3-(-\frac{24}{3})=-3+8=5=5$$ C. $$-2-(-\frac{24}{3})=-2+8=6≠5$$ 15- B $$\frac{3}{13}≅0.23 , \frac{5}{14}≅0.357 , \frac{4}{11}≅0.36 , \frac{2}{5}=0.4$$ 16- B Plugin the value of $$x$$ in the equations. $$x = -4$$, then: A.$$x(3x-1)=50→-4(3(-4)-1)=-4(-12-1)=-4(-13)=52≠50$$ B. $$5(11-x^2 )=-25→5(11-(-4)^2 )= 5(11-16)=5(-5)=-25$$ C. $$3(-2x+5)=49→3(-2(-4)+5)=3(8+5)=39≠49$$ D. $$x(-5x-19)=-3→-4(-5(-4)-19=-4(20-19)=-4≠-3$$ 17- 5 Let $$x$$ be the missing prime factor of 450. $$450= 2 × 3 × 3 × x ⇒ x =\frac{450}{18} ⇒ x = 25=5×5$$ 18- C Use Pythagorean theorem to find the hypotenuse of the triangle. $$a^2+b^2=c^2→6^2+8^2=c^2→36+64=c^2→100=c^2→c=10$$ The perimeter of the triangle is: $$6+8+10=24$$ 19- C Use percent formula: $$Part = \frac{percent}{100} × whole$$ $$65= \frac{percent}{100} × 50⇒ 65 = \frac{percent ×50}{100}⇒ 65=\frac{percent ×5}{10}$$ multiply both sides by 10. $$650 =percent ×5, \space divide \space both \space sides \space by \space 5.$$ 130 = percent The answer is $$130\%$$ 20- A Let’s check the options provided. A. $$-10+(-8)+ (\frac{5}{2})×(-2)=-10+(-8)+(-5)=-10-13=-23$$ B. $$5×3+(-2)×18=15+(-38)=-21$$ C. $$-10+6×8÷(-4)=-10+48÷(-4)=-10-12=-22$$ D. $$(-3)× (-7)+ 2=21+2=23$$ Looking for the best resource to help you succeed on the Grade 6 Georgia Milestones Assessment System Math test? 30% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 2. Add to Cart Add the eBook to your cart. ### 3. Checkout Complete the quick and easy checkout process. ### 4. Download Immediately receive the download link and get the eBook in PDF format. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Instantly download and access your eBook Help save the environment Lifetime access to your eBook Over 2,000 Test Prep titles available Over 80,000 happy customers Over 10,000 reviews with an average rating of 4.5 out of 5 24/7 support Anytime, Anywhere Access
6-1 Percents Course 2 Warm Up Problem of the Day Lesson Presentation. Presentation on theme: "6-1 Percents Course 2 Warm Up Problem of the Day Lesson Presentation."— Presentation transcript: 6-1 Percents Course 2 Warm Up Problem of the Day Lesson Presentation 6-1 Percents Warm Up Write each fraction as a decimal. 3 4 3 8 1. 0.75 Course 2 6-1 Percents Warm Up Write each fraction as a decimal. 3 4 3 8 1. 0.75 2. 0.375 3 5 6 15 3. 0.6 4. 0.4 6-1 Percents Problem of the Day Course 2 6-1 Percents Problem of the Day Alex noted that 4 tulip bulbs bloomed out of every 5 bulbs that he planted. Write the part that bloomed of the total amount of bulbs planted as a fraction and a decimal. , 0.8 4 5 Course 2 6-1 Percents Learn to model percents and to write percents as equivalent fractions and decimals. Insert Lesson Title Here Course 2 6-1 Percents Insert Lesson Title Here Vocabulary percent Fractions, Decimals, and Percents Course 2 6-1 Fractions, Decimals, and Percents A percent is the ratio of a number to 100. The symbol % is used to indicate that a number is a percent. = 6% 6 100 Reading Math The word percent means “per hundred.” So 6% means “6 out of 100.” Course 2 6-1 Percents Additional Example 1: Modeling Percents Write the percent modeled by each grid. A B. shaded total _______ 32 100 ___ shaded total _______ 66 100 ___ = 32% = 66% 6-1 Percents Check It Out: Example 1 Course 2 6-1 Percents Check It Out: Example 1 Write the percent modeled by each grid. A B. shaded total _______ 44 100 ___ shaded total _______ 75 100 ___ = 44% = 75% Additional Example 2: Writing Percents as Fractions Course 2 6-1 Percents Additional Example 2: Writing Percents as Fractions Write 28% as a fraction in simplest form. 28% = 28 100 Write the percent as a fraction with a denominator of 100. 7 25 = Simplify. 7 25 So 28% can be written as 6-1 Percents Check It Out: Example 2 Course 2 6-1 Percents Check It Out: Example 2 Write 90% as a fraction in simplest form. 90% = 90 100 Write the percent as a fraction with a denominator of 100. 9 10 = Simplify. 9 10 So 90% can be written as Additional Example 3: Writing Percents as Decimals Course 2 6-1 Percents Additional Example 3: Writing Percents as Decimals Write 17% as a decimal. Method 1 Use pencil and paper. 17 100 Write the percent as a fraction with a denominator of 100. 17% = = 0.17 Divide 17 by 100. Course 2 6-1 Percents Additional Example 3 Continued Write 17% as a decimal. Method 2 Use mental math. Move the decimal point two places to the left. 17% = 0.17 6-1 Percents Check It Out: Example 3 Write 24% as a decimal. Method 1 Course 2 6-1 Percents Check It Out: Example 3 Write 24% as a decimal. Method 1 Use pencil and paper. 24 100 Write the percent as a fraction with a denominator of 100. 24% = = 0.24 Divide 24 by 100. Check It Out: Example 3 Continued Course 2 6-1 Percents Check It Out: Example 3 Continued Write 24% as a decimal. Method 2 Use mental math. Move the decimal point two places to the left. 24% = 0.24 Insert Lesson Title Here Course 2 6-1 Percents Insert Lesson Title Here Lesson Quiz 1. Write the percent modeled by the grid. 36% 2325 2. Write 92% as a fraction in simplest form. 3. Write 16.4% as a decimal. 0.164 Download ppt "6-1 Percents Course 2 Warm Up Problem of the Day Lesson Presentation." Similar presentations
How do you graph y=3cos(4x)? Feb 4, 2015 This is what the graph of y=3 cos(4x) looks like: graph{y=3 cos (4x) [-10, 10, -5, 5]} This is how we get this graph: graph{cos x [-10, 10, -5, 5]} 2) Now, the first thing to notice is that our equation has been modified in two ways y = 3 cos( 4 x) 3) First, let's deal with the 4 inside the parenthesis. Numbers inside the parenthesis affect the period of a graph, or in other words, how much width does it take for the graph to repeat. The period is given by the following formula where B is the constant inside of the parenthesis: $\omega = \frac{2 \pi}{B} = \frac{2 \pi}{4} = \frac{\pi}{2}$ This means that the period of the graph is 4 times more compressed than the normal graph, which looks like this: graph{y=cos(4x) [-10, 10, -5, 5]} 4) Now lets take a look at 3 in the equation: y = 3 cos( 4 x). Numbers in the front of the equation affect the amplitude of a graph, or how tall or short the graph is. The relationship here is direct, the larger the constant the larger the graph. In this case, it means the the graph would be 3 times as tall as a normal cos graph. So, if we apply that to our previous graph we get the anser: graph{y=3 cos (4x) [-10, 10, -5, 5]}
## Monday Math 65 Divisibility Tests Part 5: Tests for 13 Divisibility tests for 13 are also quite rare, however we can create a test analogous to our first for 7. We note that 910+1=91=713, so with n=10a+b and q=a-9b, we see , so 9n+q is a multiple of 13, which means that n is divisible by 13 if and only if q is divisible by 13. This gives us the test: mulitply the ones digit of our number by 9, and subtract from the number formed by the rest of the digits. We see that this doesn’t work well for two-digit numbers, but a little better for three- or four-digit numbers. For example: n=78 7-89=-65=-513 n=182 18-29=0=013 n=507 50-79=-13 n=3679 367-99=286 28-96=-26=-213 so all of these numbers are divisible by 13. As with 7, this procedure may be laborious for very large numbers. However, we can see that 1001=7143=71113=7713 This means 1000=1377-1, and so we see our large-number divisibility test for 7 also works for 13. Thus, we split our large number into sets of three digits, and then take the alternating sum of these numbers. For example: n=348,970,129,352 352-129+970-348=845 84-59=39=313 so 348,970,129,352 is divisible by 13 n=4,626,145,317,014,712 712-14+317-145+626-4=1492 149-29=131=13*10+1, which is not divisible by 13, so 4,626,145,317,014,712 is not divisible by 13. Advertisements
# Constant Rate Of Change Worksheet When he comes again from place B to position A, he increases his pace 2 occasions. If the constant-speed for the whole journey is eighty miles per hour, discover his pace when he travels from the place A to B. A man takes 10 hours to go to a spot and come again by walking both the ways. He might have gained 2 hours by riding each the ways. • For each restrict, evaluate the limit or or clarify why it does not exist the limit guidelines to justify each step. • The micro organism population will increase from time 0 to 10 hours; afterwards, the bacteria inhabitants decreases. Our mannequin predicts the population will attain 15,000 in a little more than 23 years after 2004, or somewhere around the 12 months 2027. If the pattern continues, our model predicts a inhabitants of 9,620 in 2013. A town’s inhabitants has been growing linearly. ” The idea of common fee of change permits us to make these questions more mathematically precise. Initially, we are going to focus on the average fee of change of an object transferring alongside a straight-line path. In this section we take a glance at some purposes of the derivative by focusing on the interpretation of the derivative as the rate of change of a function. These functions include acceleration and velocity in physics, population development rates in biology, and marginal features in economics. Contents ## Winter Graph Of The Day: 20 Full Similarly, a operate is lowering on an interval if the perform values lower because the input values enhance over that interval. The common price of change of an rising perform is positive, and the typical fee of change of a reducing perform is unfavorable. Figure three shows examples of accelerating and decreasing intervals on a perform. When we calculate the average fee of change for a linear operate, it does not matter what interval we decide, the value of the speed of change is the same. A constant fee of change is a vital function of linear functions! When a linear operate is represented by a graph, the slope of the line is the speed of change of the function. Discover the constant fee of change definition and the constant rate of change formula. Learn whether a price of change is fixed or varying by finding out examples. Here are 10 practice questions beneath to test your understanding of charges of change. You will discover PDF options here and at the finish of the questions. ### Tables Equations And Graphs Conversion Worksheet Free Find her average speed over the primary 6 hours. Determine $$AV_\text$$ $$AV_\text$$ and $$AV_\text$$ including appropriate items. Choose considered one of these quantities and write a careful sentence to explain its that means. three – Two automobiles begin shifting from the identical level in two directions that makes 90 levels on the fixed speeds of s1 and s2. Find a formulation for the speed of change of the gap D between the 2 automobiles. Here are some enjoyable methods to apply Constant Rate of Change.There are 1 Hole Punch activity $2.502 number tile activity pages.$2It appeals to totally different studying kinds. This is a neat approach to permit your college students to apply or show mastery of determining the C.R.O.C. that can also be self checking. A operate can additionally be neither rising nor reducing at extrema. Because the speed just isn’t constant, the average speed depends on the interval chosen. For the interval , the typical speed is 63 miles per hour. Use the graph of the speed function to determine the time intervals when the acceleration is constructive, unfavorable, or zero. Before we are in a position to sketch the graph of the particle, we have to know its position at the time it begins transferring and at the instances that it changes path . Shown in Figure 1, find the average fee of change on the interval [ −1,2 ]. Using the info in Table 1, discover the typical rate of change between 2005 and 2010. Using the data in Table 1, discover the common fee of change of the value of gasoline between 2007 and 2009. Which nonetheless represents the change within the function’s output value ensuing from a change to its enter worth. Use the data obtained to sketch the path of the particle alongside a coordinate axis. Because , the particle is transferring from proper to left. Predict the future population from the present worth and the inhabitants progress fee. Distance traveled by automobile in a certain amount of time. The constant fee of change is also referred to as the slope. Linear capabilities could have a constant price of change. The graph shows that sooner or later the graph adjustments direction. When the graph has a adverse slope, it factors towards the adverse y-values or downward. Write the equation of the line by way of and (− 1 ,3) in point slope kind. This worksheet helps to grasp how to solve several sorts of Calculus I issues. A line extending from the decrease left to the upper right has this type of slope. The set of y values of a perform, that is another name for range. ## Related posts of "Constant Rate Of Change Worksheet" Project it onto a screen and have college students write their solutions on sticky notes and put their ideas within the right part of the planning document. After completing the process of the experiment, college students analyze the info, draw conclusions, after which share their results. Depending on the coed stage, this might be a... #### Properties Of Matter Worksheet This lab was created for school kids working in small teams and encourages good accountable talk, but is made for faculty kids to report their answers individually. Learn what temperature gold boils at and more using an animated exercise concerning the properties of matter. Scientists be taught about the properties of matter including examples of... #### Correlation Vs Causation Worksheet These selections lead customers to keep utilizing our apps or uninstall them. Which is why we have to think clearly when going through information and be careful when seeing potential correlation vs causation issues. A correlation can be identified from observations without a scientific experiment. In a correlational design, you measure variables without manipulating any... #### Area Of Regular Polygons Worksheet But there are situations when the apothem must be worked out first. These free follow tools are available in customary and metric items. Decompose every irregular polygon in these pdf worksheets for sixth grade, 7th grade, and eighth grade into acquainted airplane shapes. The definition of the realm of a polygon is the measure of...
## What are the steps to solving multi step equations? Step-by-Step Solution: Combine like terms on both sides. Subtract 6 y 6y 6y on both sides to keep the variable y to the left side only. Add 11 to both sides of the equation. Finally, divide both sides by −10 to get the solution. ## What are the 4 steps to solving an equation? We have 4 ways of solving one-step equations: Adding, Substracting, multiplication and division. If we add the same number to both sides of an equation, both sides will remain equal. ## How do you solve multi step equations with like terms? Multi-step Equations with Like TermsProcedure to Solve Equations:Step 1: Remove any parentheses by using the Distributive Property or the Multiplication Property of Equality.Step 2: Simplify each side of the equation by combining like terms.Step 3: Isolate the begin{align}axend{align} term. Step 4: Isolate the variable. ## How do you solve an equation with multiple variables? Section 2-4 : Equations With More Than One VariableMultiply both sides by the LCD to clear out any fractions.Simplify both sides as much as possible. Move all terms containing the variable we’re solving for to one side and all terms that don’t contain the variable to the opposite side. ## What is the golden rule for solving equations? Do unto one side of the equation, what you do to the other! When solving math equations, we must always keep the ‘scale’ (or equation) balanced so that both sides are ALWAYS equal. ## What are the rules for solving equations? A General Rule for Solving EquationsSimplify each side of the equation by removing parentheses and combining like terms.Use addition or subtraction to isolate the variable term on one side of the equation.Use multiplication or division to solve for the variable. You might be interested: Equation for spring force ## What is an example of a multi step equation? If the equation is in the form, ax + b = c, where x is the variable, you can solve the equation as before. First “undo” the addition and subtraction, and then “undo” the multiplication and division. Solve 3y + 2 = 11. Subtract 2 from both sides of the equation to get the term with the variable by itself. ## How do you solve multi step word problems? Here are steps to solving a multi-step problem: Step 1: Circle and underline. Circle only the necessary information and underline what ultimately needs to be figured out. Step 2: Figure out the first step/problem in the paragraph and solve it. Last step: Find the answer by using the information from Steps 1 and 2. ## How do you solve 3 equations with 3 variables? Here, in step format, is how to solve a system with three equations and three variables:Pick any two pairs of equations from the system.Eliminate the same variable from each pair using the Addition/Subtraction method.Solve the system of the two new equations using the Addition/Subtraction method. ### Releated #### Equation for decibels How do you calculate dB? Find the logarithm of the power ratio. log (100) = log (102) = 2 Multiply this result by 10 to find the number of decibels. decibels = 10 × 2 = 20 dB If we put all these steps together into a single equation, we once again have the definition […] #### Convert to an exponential equation How do you convert a logarithmic equation to exponential form? How To: Given an equation in logarithmic form logb(x)=y l o g b ( x ) = y , convert it to exponential form. Examine the equation y=logbx y = l o g b x and identify b, y, and x. Rewrite logbx=y l o […]
# The Cali Garmo does Math ## Dyck Paths By Cali G , Published on Fri 18 October 2019 Category: math / symmetric functions A Dyck path has two equivalent definitions. For both of them, we start by looking at $\mathbb{Z}^2$ and constructing a path which starts at $(0,0)$. ## Two Definitions Definition 1: A Dyck path in $\mathcal{D}_n$ is a path from $(0,0)$ to $(n,n)$ which takes only north and east steps and that never crosses the $x = y$ diagonal. The best way to see this is through an example. Definition 2: A Dyck path in $\mathcal{D}_n$ is a path from $(0,0)$ to $(2n,0)$ which takes only north-east steps and south-east steps which never cross the $x = 0$ line. Taking the same example as before, we have. ## Area One of the most common statistics that we place on Dyck paths is area. Let $\pi \in \mathcal{D}_n$ be a Dyck path and draw the Dyck path as we did in our first definition. For each row $i$ in $\pi$ (bottom row is row $1$) we can count the number of full boxes between the diagonal and the Dyck path. This gives us a number which we denote by $a_i$. In our example above we have To see this more closely, notice that $a_4 = 2$ since there are two full boxes to the right of the path and to the left of the diagonal: To get the area of the Dyck path, it suffices to sum up all the $a_i$: In our example, ### Distribution using area We can look at the distribution of Dyck paths using previous Dyck paths thanks to Carlitz and Riordan. They showed the following theorem. Theorem [CR64] Let's look at an example of this. We'll try and find $C_3(q)$. First let's calculate the distribution using area. There are $5$ possible Dyck paths where $n = 3$ and these are given by: So we should find: Let's now try and use the second summation. We have: In essence, we need to calculate $C_0(q)$, $C_1(q)$, and $C_2(q)$. We can calculate these using either method, but since there are only a few of these (only $1$ Dyck path in $\mcD_0$ and $\mcD_1$ and only $2$ in $\mcD_2$) it's easier to just look at the areas. There is only one Dyck path from $(0,0)$ to $(0,0)$ and it has $0$ area. Therefore Similarly, there is only one Dyck path from $(0,0)$ to $(1,1)$ and it has $0$ area. Therefore Finally, there are only two Dyck paths from $(0,0)$ to $(2,2)$: Since one has area $0$ and one has area $1$ therefore: Finally: Which is the same as earlier. ### Distribution using major index We next look at how to assign to each Dyck path a major index and count the number of Dyck paths with a certain major index. Recall that given a sequence of number $\sigma = \sigma_1 \sigma_2 \ldots \sigma_n$, then the major index is given by: As an example, if $\sigma = 010101$ then $\sigma_1 = 0$, $\sigma_2 = 1$, $\sigma_3 = 0$, etc. and since $\sigma_2 > \sigma_3$ and $\sigma_4 > \sigma_5$, then $maj(\sigma) = 2 + 4 = 6$. Thanks to MacMahon, we know the distribution of the Dyck paths relative to their major indices. To view this, we must first show how to go from a Dyck path to a sequence of numbers. For this, we look at the first definition of a Dyck path and start from $(0,0)$. For each $i$th step, if we go north then we let $\sigma_i = 0$ else we let $\sigma_i = 1$. We let $\sigma(\pi)$ denote this number sequence of a Dyck path $\pi$. As an example, suppose we have the Dyck path from before: Now notice that, starting from $(0,0)$ we first go north 3 times, then east, then north, etc. This means that we have the following number sequence: Theorem [Mac60] where Let's look at a full distribution when $n = 3$. We have $5$ different Dyck paths that are possible for which I've placed the word associated to it on the right This gives us the following major indices: In other words: On the other hand: ## References • [CR64] L. Carlitz, J. Riordan, Two element lattice permutation numbers and their $q$-generalization, Duke Math J. [31]-3 (1964), 371-388. DOI • [Mac60] P.A MacMahon, Combinatory analysis, Two volumes (bound as one), Chelsea Publishing Co., New York., 1960.
# Angle Between Lines: Examples Hello again. This lesson will cover a few examples involving angle between lines. Example 1 Find the angle between (i) the lines 2x + y – 4 = 0 and 3x – y + 1 = 0. (ii) the lines joining A(1,2) to B(3,4) and C(6,7) to D(8,5) (i.e. angle between AB and CD) Solution (i) The slope of the first line is -2 and that of the second is 3. (details) We can straight away apply the formula now. tanθ = $$\|\frac{-2-3}{1+(-2)(3)}\|$$ = 1. Therefore θ = 45°. Note that in case we didn’t use the absolute value, we could have obtained tanθ = -1 or θ = 135°. It doesn’t make any difference though. The two different angles correspond to the same situation geometrically (as I’ve explained previously) (ii) This problem is similar to the previous one. Except that the equations aren’t given (which we do not require anyway). Instead we’ve been provided with points, through which we can calculate the slopes of the lines joining them. Now, slope of AB = (4 – 2)/(3 – 1) = 1. And, slope of CD = (5 – 7)/(8 – 6) = -1. We can observe that the product of the two slopes is -1. Hence the two lines are perpendicular (explanation), or the angle between them is 90° Example 2 Find the equation of the line which passes through the point (1, 2) and is (i) parallel to x + 2y – 3 = 0 (ii) perpendicular to 3x + y = 4 Solution There are atleast two different ways to solve both the parts. I’ll illustrate them both. (i) Method I Since the two lines are parallel, the slope of the required line would the same as that of the given line, which equals -1/2. We have the slope, and a point (1, 2). We can use the point-slope form of the equation now. The required equation is y – 2 = (-1/2)(x – 1), or x + 2y – 5 = 0. Easy. Method II Look at the answer obtained. Both the equations are identical, except for the constants. (I’ve discussed this before) We can therefore assume the required line as x + 2y + k = 0. The value of the unknown constant ‘k’ will be obtained by the additional condition, i.e. the line passes through the point (1, 2). Substituting the coordinates, we get (1) + 2(2) + k = 0, or k = -5. The required equation is x + 2y – 5 = 0. Note that the line could also have been assumed as x + 2y = k, which would have given a different value of k (5), but the equation obtained would have been the same. (x + 2y = 5) To generalize: Any line parallel to a given line ax + by + c = 0, will be of the form ax + by + k = 0. The value of k will be obtained by some additional given condition. (ii) Method I Since the two lines are perpendicular, the product of the slopes of the required line and the given line must equal -1. We therefore have m.(-3) = -1, or m = 1/3. Using the point-slope form again, we get the required equation as y – 2 = 1/3(x -1), or x – 3y + 5 = 0. Easy again. Method II Look again at the answer obtained. If we compare the answer with the given line (i.e. compare the equations of two perpendicular lines), the coefficients of x and y have interchanged with a negative sign on one of them. (3x + y = 4 and x – 3y + 5 = 0) We can therefore assume the required line as x – 3y + k = 0. (because this line already has the required slope of 1/3, which makes it perpendicular to the given line) The value of the unknown constant ‘k’ will be obtained by the additional condition, i.e. the line passes through the point (1, 2). Substituting the coordinates, we get (1) – 3(2) + k = 0, or k = 5. The required equation is thus x – 3y + 5 = 0. To generalize: Any line parallel to a given line ax + by + c = 0, will be of the form bx – ay + k = 0. The value of k will be obtained by some additional given condition. To be continued
Courses # Summary of Equations of Motion for Rigid Bodies (Part - 2) Civil Engineering (CE) Notes \| EduRev ## Civil Engineering (CE) : Summary of Equations of Motion for Rigid Bodies (Part - 2) Civil Engineering (CE) Notes \| EduRev The document Summary of Equations of Motion for Rigid Bodies (Part - 2) Civil Engineering (CE) Notes \| EduRev is a part of the Civil Engineering (CE) Course Introduction to Dynamics and Vibrations- Notes, Videos, MCQs. All you need of Civil Engineering (CE) at this link: Civil Engineering (CE) 6.6.6 Linear and angular momentum equations in terms of accelerations The linear and angular momentum conservation equations can also be expressed in terms of accelerations, angular accelerations, and angular velocities. The results are For 2D planar motion we can use the simplified formulas 6.6.7 Special equations for analyzing bodies that rotate about a stationary point We often want to predict the motion of a system that rotates about a fixed pivot â€“ a pendulum is a simple example. These problems can be solved using the equations in 6.6.5 and 6.6.6, but can also be solved using a useful short-cut. • For an object that rotates about a fixed pivot at the origin: • The total angular momentum (about the origin) is h =IOÏ‰ • The total kinetic energy is T = 1/2 Ï‰Â·I0Ï‰ • The equation of rotational motion is Here IO is the mass moment of inertia about O (calculated, eg, using the parallel axis theorem) For 2D rotation about a fixed point at the origin we can simplify these to • The total angular momentum (about the origin) is h = IOzzÏ‰ zk • The total kinetic energy is T=1/2 IOzzÏ‰z2 • The equation of rotational motion is Proof: It is straightforward to show these formulas. Letâ€™s show the two dimensional version of the kinetic energy formulas as an example. For fixed axis rotation, we can use the rigid body formulas to calculate the velocity of the center of mass (O is stationary and at the origin) vG = Ï‰ Ã—rG=Ï‰zk Ã— rG The general formula for kinetic energy can therefore be re-written as The other formulas can be proved with the same method â€“ we simply express the velocity or acceleration of the COM in the general formulas in terms of angular velocity and acceleration, and notice that we can rearrange the result in terms of the mass moment of inertia about O. Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity! ## Introduction to Dynamics and Vibrations- Notes, Videos, MCQs 20 videos\|53 docs , , , , , , , , , , , , , , , , , , , , , ;
# Number Systems: An Introduction to Binary, Hexadecimal, and More Languages: Ever see crazy binary numbers and wonder what they meant? Ever see numbers with letters mixed in and wonder what is going on? You'll find out all of this and more in this article. Hexadecimal doesn't have to be scary. (Thanks to the ReBoot Wiki for the thumbnail image.) ## Introduction: What is a Number System? You probably already know what a number system is - ever hear of binary numbers or hexadecimal numbers? Simply put, a number system is a way to represent numbers. We are used to using the base-10 number system, which is also called decimal. Other common number systems include base-16 (hexadecimal), base-8 (octal), and base-2 (binary). ## Activity Before we get started, let's try a little activity for fun. There are many different ways to represent a color, but one of the most common is the RGB color model. Using this model, every color is made up of a combination of different amounts of red, green, and blue. You may be wondering how colors relate to number systems. In short, on a computer, any color is stored as a large number: a combination of red, green, and blue. (We'll go into more detail on this later.) Because it's just a number, it can be represented in multiple ways using different number systems. Your job is to guess how much red, green, and blue is in the background color of the activity below. The values for red, green, and blue can range from 0 to 255. Feel free to use the various hints provided to help you out. If you don't understand the numerical hints yet, no problem! You can see what your guess looks like using the View Guess button. Right now, it may seem tricky, but hopefully by the end of the article, it will seem easy. ## Looking at Base-10 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11... You've counted in base-10 all of your life. Quick, what is 7+5? If you answered 12, you are thinking in base-10. Let's take a closer look at what you've been doing all these years without ever thinking about it. Let's take a quick look at counting. First, you go through all the digits: 0, 1, 2... Once you hit 9, you have no more digits to represent the next number. So, you change it back to 0, and add 1 to the tens digit, giving you 10. The process repeats over and over, and eventually you get to 99, where you can't make any larger numbers with two digits, so you add another, giving you 100. Although that's all very basic, you shouldn't overlook what is going on. The right-most digit represents the number of ones, the next digit represents the number of tens, the next the number of hundreds, etc. ## Visualizing Base-10 Confused by these descriptions? No problem - below is a demo to help you out. Simply enter a number in the text box and click draw. Try entering a large number, like 2347. You'll see 2 groups of one thousand, 3 groups of one hundred, 4 groups of ten, and 7 individual blocks. ## Base-10 Mathematically You may have noticed a pattern by now. Let's look at what is going on mathematically, using 2347 as an example. • As you saw, there are 2 groups of a thousand. Not coincidentally, `1000 = 101010 ` which can also be written as `103`. • There are 3 groups of a hundred. Again, not coincidentally, `100 = 1010` or `102`. • There are 4 groups of ten, and, `10 = 101`. • Finally, there are 7 groups of one, and `1 = 100`. (That may seem strange, but any number to the power of 0 equals 1, by definition.) This is essentially the definition of base-10. To get a value of a number in base-10, we simply follow that pattern. Here are a few more examples: • `892 = 8102+9101+2100` • `1147 = 1103+1102+4101+7100` • `53 = 5101+3100` Admittedly, this all seems a little silly. We all know what value a base-10 number is because we always use base-10, and it comes naturally to us. As we'll see soon, though, if we understand the patterns in the background of base-10, we can understand other bases better. ## Base-8 On to base-8, also called octal. Base-8 means just what is sounds like: the system is based on the number eight (as opposed to ten). Remember how in base-10 we had ten digits? Now, in base-8, we are limited to only eight digits: 0, 1, 2, 3, 4, 5, 6, and 7. There's no such thing as 8 or 9. We count the same way as we normally would, except with only eight digits. Instead of a lengthy explanation, simply try out the demo below by clicking "Count Up 1" to see how counting in base-8 works. You should notice a similar pattern to before; after we get to 7, we run out of different digits for any higher number. We need a way to represent eight of something. So we add another digit, change the 7 back to 0, and end up with 10. Our answer of 10 in base-8 now represents what we would normally think of as 8 in base-10. Talking about numbers written in multiple bases can be confusing. For example, as we have just seen, 10 in base-8 is not the same as 10 in base-10. So, from this point on, I'll use a standard notation where a subscript denotes the base of numbers if needed. For example, our base-8 version of 10 now looks like 108. (Editor's note: I find it a lot easier to understand this if I change the way I read these numbers in my head, too. For example, for 108, I read "octal one-oh" or "one-oh in base-eight". For 1010 I read "decimal one-oh" or "one-oh in base-ten".) Great, so we know 108 represents eight items. (Always feel free to plug a number into the first tool for a visualization.) What's the next number after 778? If you said 1008, you're correct. We know from what we've learned so far that the first 7 in 778 represents groups of 8, and the second 7 represents induvidual items. If we add these all up, we have `78 + 71 = 63`. So we have a total of 6310. So 778=6310. We all know 6410 comes after 6310. ## Converting From Base-8 to Base-10 Let's look at a wordier example now. John offers to give you 478 cookies, and Jane offers to give you 4310 cookies. Whose offer do you take? If you want, go ahead and generate the graphic for 478 graphic with the first tool. Let's figure out its base-10 value so we can make the best decision! As we saw when counting, the four in 478 represents the number of groups of eight. This makes sense - we are in base-8. So, in total, we have four groups of eight and seven groups of one. If we add these all up, we get `48 + 71 = 3910`. So, 478 cookies is the exact same as 3910 cookies. Jane's offer seems like the best one now! The pattern we saw before with base-10 holds true here also. We'll look at 5238. There are five groups of 82, two groups of 81 and three groups of 80 (remember, 80=1). If we add these all up, `582 + 281 + 380 = 564+28+3 = 339`, we get 33910 which is our final answer. The diagram below shows the same thing visually: Here are a couple more examples: • `1118 = 182+181+180 = 64+8+1 = 7310` • `438 = 481+380 = 32+3 = 3510` • `61238 = 683+182+281+380 = 3072+64+16+3 = 315510` ## Converting from Base-10 to Base-8 Converting from base-10 to base-8 is a little trickier, but still straightforward. We basically have to reverse the process from above. Let's start with an example: 15010. We first find the largest power of 8 that is smaller than our number. Here, this is 82 or 64 (83 is 512). We count how many groups of 64 we can take from 150. This is 2, so the first digit in our base-8 number is 2. We have now accounted for 128 out of 150, so we have 22 left over. The largest power of 8 that is smaller than 22 is 81 (that is, 8). How many groups of 8 can we take from 22? Two groups again, and thus our second digit is 2. Finally, we are left with 6, and can obviously take 6 groups of one from this, our final digit. We end up with 2268. In fact, we can make this process a touch clearer with math. Here are the steps: 1. 150/82 = 2 remainder 22 2. 22/81 = 2 remainder 6 3. 6/80 = 6 Our final answer is then all of our non-remainder digits, or 226. Notice that we still start by dividing by the highest power of 8 that is less that our number. ## Dealing with any Base It's important to be able to apply the concepts we've learned about base-8 and base-10 to any base. Just as base-8 had eight digits and base-10 had ten digits, any base has the same number of digits as its base. So base-5 has five digits (0-4), base-7 has seven digits (0-6), etc. Now let's see how to find the base-10 value of any number in any base. Say we are working in base-b, where b can be any positive integer. We have a number d4d3d2d1d0 where each d is a digit in a number. (The subscripts here don't refer to the base of the number but simply differentiate each digit.) Our base-10 value is simply `d4b4 + d3b3 + d2b2 + d1b1 + d0b0`. Here's an example: we have the number 32311 in base-4. Notice how our number only has digits from zero to three since base-4 only has four total digits. Our base-10 value is `344 + 243 + 342 + 141 + 140 = 3256 + 264 + 316 + 14 + 11 = 949`. We could, or course, follow this pattern with any amount of digits in our number. ## Base-16 Base-16 is also called hexadecimal. It's commonly used in computer programming, so it's very important to understand. Let's start with counting in hexadecimal to make sure we can apply what we've learned about other bases so far. Since we are working with base-16, we have 16 digits. So, we have 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ... and yikes! We've run out of digits, but we still need six more. Perhaps we could use something like a circled 10? The truth is, we could, but this would be a pain to type. Instead, we simply use letters of the alphabet, starting with A and continuing to F. Here's a table with all the digits of base-16: Other than these extra digits, hexadecimal is just like any other base. For example, let's convert 3D16 to base-10. Following our previous rules, we have: `3D16 = 3161 + 13160 = 48 + 13 = 61`. So 3D16 is equal to 6110. Notice how we use D's value of 13 in our calculation. We can convert from base-10 to base-16 similar to the way we did with base-8. Let's convert 69610 to base-16. First, we find the largest power of 16 that is less than 69610. This is 162, or 296. Then: 1. 696/162 = 2 remainder 184 2. 184/161 = 11 remainder 8 3. 8/161 = 8 remainder 0 We have to replace 11 with its digit representation B, and we get 2B816. Feel free to try some more conversions for practice. You can use the application below to check your answers: ## Binary! (Base-2) On to the famous base-2, also called binary. While everyone knows binary is made up of 0s and 1s, it is important to understand that it is no different mathematically than any other base. There's an old joke that goes like this: Can you figure out what it means? Let's try a few conversions with base-2. First, we'll convert 1011002 to base-10. We have: `101100 = 125 + 123 + 122 = 32 + 8 + 4 = 4410`. Now let's convert 65 to binary. 26 is the highest power of 2 less than 65, so: 1. 65/26 = 1 remainder 1 2. 1/25 = 0 remainder 1 3. 1/24 = 0 remainder 1 4. 1/23 = 0 remainder 1 5. 1/22 = 0 remainder 1 6. 1/21 = 0 remainder 1 7. 1/20 = 1 remainder 0 And thus we get our binary number, 1000001. Understanding binary is super important. I've included a table below to point out digits' values. For example, the value of 10001 is 17, which is the sum of the values of the two 1 digits (16+1). This is nothing different than we have done before, its just presented in an easy to read way. ## Tricks and Tips Normally, when converting between two bases that aren't base-10, you would do something like this: 1. Convert number to base-10 2. Convert result to desired base However, there's a trick that will let you convert between binary and hexadecimal quickly. First, take any binary number and divide its digits into groups of four. So, say we have the number 10111012. Divided up we have 0101 1101. Notice how we can just add extra zeroes to the front of the first group to make even groups of 4. We now find the value for each group as if it was its own separate number, which gives us 5 and 13. Finally, we simply use the corresponding hexadecimal digits to write out base-16 number, 5D16. We can go the other direction also, by converting each hexadecimal digit into four binary digits. Try converting B716 to binary. You should get 101101112. This trick works because 16 is a power of 2. What this means is that we use similar trick for base-8, which is also a power of 2: Of course, you can reverse the process to go from base-8 to binary also. ## Conclusion Let's go all the way back and revisit the color guessing game. In Flash, colors are stored as a single number. When converted to hexadecimal, the first two digits represent the amount of red, the next two the amount of green, and the final two the amount of blue. So, if our color is 17FF1816 we can easily tell that our red component is 1716 or 2310. Our green component is FF16, or 25510. Finally our blue component is 1816 or 2410. If we are given the base-10 version of our color, 157263210, we need to convert it to hexadecimal before we can tell anything about it. Try the game again, and see how much better you can do! Understanding different number systems is extremely useful in many computer-related fields. Binary and hexadecimal are very common, and I encourage you to become very familiar with them. Thanks for reading - I hope you've learned a lot from this article! Feel free to grab the source code from any of the demos. Also, if you have any questions, please ask them below.
April 20, 2024 Learn everything you need to know about finding horizontal and vertical asymptotes of rational functions, using different methods such as graphing calculators and limits, and applying them to real-life problems in physics, chemistry, engineering, and other fields. ## I. Introduction Asymptotes are imaginary lines that a function approaches but never touches as the input variable approaches certain values. They play a crucial role in understanding the behavior of mathematical functions, especially when dealing with rational functions that involve fractions of polynomials. In this article, we will explore how to find horizontal and vertical asymptotes, using different techniques and real-life applications. ## II. Understanding Horizontal Asymptotes A horizontal asymptote is a straight line that a function approaches as the input variable approaches positive or negative infinity, but without touching it. In other words, the values of the function get arbitrarily close to the horizontal line but never cross it. The equation of a horizontal asymptote is usually written as y = c, where c is a constant. To find the horizontal asymptote of a rational function, we need to compare the degrees of the numerator and denominator polynomials. If the degree of the denominator is greater than the degree of the numerator by one, then the horizontal asymptote is y = 0. If the degree of the numerator is greater than the degree of the denominator, then there is no horizontal asymptote. Otherwise, the horizontal asymptote is y = c, where c is the ratio of the leading coefficients of the numerator and denominator. There are different methods for finding the horizontal asymptote of a rational function, such as long division, synthetic division, partial fraction decomposition, or factoring. Let’s see an example: Find the horizontal asymptote of the function f(x) = (2x^3 – 3x^2 + 5) / (4x^3 + x^2 – 2). By comparing the degrees of the numerator and denominator polynomials, we see that they are both 3, so we need to compare their leading coefficients: lim(x → ±∞) f(x) = lim(x → ±∞) (2x^3 / 4x^3) = 1/2 Therefore, the horizontal asymptote of f(x) is y = 1/2. Try to find the horizontal asymptotes of these functions: f(x) = (x^2 + 4) / (3x^2 – 2x + 1) g(x) = (4x^3 – 6x^2 – 10x) / (3x^3 + 2x^2 – 5x – 2) ## III. Discovering Vertical Asymptotes A vertical asymptote is a vertical line that a function approaches as the input variable approaches a certain value that makes the denominator of a rational function equal to zero. In other words, the function becomes undefined as the input variable gets closer and closer to the vertical line. The equation of a vertical asymptote is usually written as x = c, where c is the value that makes the denominator zero. To find the vertical asymptote(s) of a rational function, we need to set the denominator equal to zero and solve for the input variable(s). The resulting value(s) is/are the vertical asymptote(s), if any. If there are repeated factors in the denominator, then the vertical asymptote is a vertical line with a hole. There are different methods for finding the vertical asymptotes of a rational function, such as factoring, long division, synthetic division, or limits. Let’s see an example: Find the vertical asymptotes of the function g(x) = (x^2 – 4) / (x – 2)(x + 3). To find the vertical asymptotes, we need to set the denominator equal to zero and solve for x: (x – 2)(x + 3) = 0 → x = 2 or x = -3 Therefore, the vertical asymptotes of g(x) are x = 2 and x = -3. Try to find the vertical asymptotes of these functions: h(x) = (x^2 – 16) / (x^2 + 4x + 3) j(x) = (4x^4 + 9) / (2x^3 – 3x^2 – 10x) ## IV. Techniques for Finding Asymptotes There are different techniques for finding the asymptotes of a function, depending on the complexity of the expression and the desired precision of the results. Two common techniques are using graphing calculators and applying limits. A graphing calculator can provide a quick and accurate visualization of the behavior of a function, including the asymptotes. Most graphing calculators have built-in features that can identify the horizontal and vertical asymptotes of a function, as well as other key properties such as intercepts and points of inflection. However, it is important to keep in mind that graphing calculators only provide an approximation of the actual values, and may not work for extremely complex or undefined functions. Applying limits is another technique for finding the asymptotes of a function, especially when dealing with rational functions. The basic idea is to evaluate the limit of the function as the input variable approaches certain values that correspond to the types of asymptotes. For example, to find the horizontal asymptote(s), we need to evaluate the limit of the function as x approaches positive or negative infinity, as we did in Section II. To find the vertical asymptote(s), we need to evaluate the limit of the function as x approaches the value(s) that make the denominator zero, as we did in Section III. However, applying limits can be tricky and time-consuming, especially for functions with multiple asymptotes or complicated expressions. Let’s see an example of how to combine the knowledge of horizontal and vertical asymptotes to determine the overall behavior of a function: Find the horizontal and vertical asymptotes of the function k(x) = (2x^3 – 3x^2 + 5) / (4x^3 + x^2 – 2). By comparing the degrees of the numerator and denominator polynomials, we found that the horizontal asymptote of k(x) is y = 1/2. By setting the denominator equal to zero, we found that the vertical asymptotes of k(x) are x = -√(2)/2 and x = √(2)/2. To determine the overall behavior of k(x), we can draw a sketch of the function and label the horizontal and vertical asymptotes: As we can see, k(x) has two vertical asymptotes, at x = -√(2)/2 and x = √(2)/2, which divide the x-axis into three intervals. In each interval, the sign and behavior of k(x) is determined by the horizontal asymptote, as follows: • As x → -∞, k(x) approaches y = 1/2 from the negative side. • Between -√(2)/2 and √(2)/2, k(x) is undefined and has vertical asymptotes. • As x → +∞, k(x) approaches y = 1/2 from the positive side. Therefore, we can conclude that k(x) is a rational function with a horizontal asymptote at y = 1/2 and two vertical asymptotes at x = -√(2)/2 and x = √(2)/2. Try to find the asymptotes and sketch the behavior of these functions: m(x) = (x^3 – 9x^2 + 15x + 7) / (x^2 – 4x + 3) n(x) = (x^5 – x) / (x^4 – 6x^2 + 9) ## V. Real-life Applications of Asymptotes Asymptotes are used in various fields of science and engineering to model and analyze complex phenomena. Some examples are: • In physics, asymptotes are used to describe the behavior of ideal gases at low and high pressures, where the volume approaches zero or infinity without reaching it. • In chemistry, asymptotes are used to determine the equilibrium concentrations of reactants and products in acid-base titrations, where the pH approaches a vertical asymptote as the volume of the titrant approaches the equivalence point. • In engineering, asymptotes are used to design control systems that ensure stability and avoid oscillations or instabilities in electronic circuits, mechanical systems, or chemical processes. • In economics and finance, asymptotes are used to model the growth and decay of population, income, or investment, where the rates of change approach zero or infinity as the variables approach certain limits. Let’s see an example of how to apply the concept of asymptotes to a real-life problem: A chemical reaction follows the rate law R = k[A]^2[B] / (1 + k'[B]), where R is the rate of formation of the product, [A] and [B] are the concentrations of the reactants A and B, and k and k’ are the rate constants. Find the vertical asymptote of the rate law, and explain its significance. To find the vertical asymptote, we need to set the denominator equal to zero and solve for [B]: 1 + k'[B] = 0 → [B] = -1/k’ Therefore, the rate law has a vertical asymptote at [B] = -1/k’, which corresponds to the concentration of B at which the denominator becomes infinite and the rate of the reaction becomes undefined. The significance of the vertical asymptote is that it represents the critical concentration of B above which the reaction rate decreases rapidly and approaches zero, due to the saturation of the [A]^2[B] term by the high concentration of B. This means that the rate of the reaction is no longer dependent on the concentration of B, and is limited by the rate constant k. Try to solve these real-life problems by identifying the asymptotes: A physics experiment measures the frequency of a damped harmonic oscillator as a function of time. The resulting curve has a horizontal asymptote at f = 0, and a vertical asymptote at t = 0. What is the physical interpretation of the asymptotes? A population of bacteria grows according to the logistic model P = P0 / (1 + ae^(-kt)), where P is the population size, P0 is the initial population, a and k are constants. What is the horizontal asymptote of the logistic curve, and what does it represent? ## VI. Conclusion In conclusion, finding horizontal and vertical asymptotes of rational functions is an important skill that can help us understand the behavior of mathematical functions and apply them to real-life problems. By using different methods such as graphing calculators, limits, and factoring, we can identify the types and values of asymptotes, and sketch the overall behavior of a function. Asymptotes have numerous applications in various fields of science and engineering, such as physics, chemistry, engineering, economics, and finance, where they can model and predict the behavior of complex systems. We invite you to explore more about the fascinating world of asymptotes and leave your comments or questions below.
# HiSET: Math : Estimate rate of change from a graph ## Example Questions ### Example Question #1 : Estimate Rate Of Change From A Graph The graph of a function is given above, with the coordinates of two points on the curve shown. Use the coordinates given to approximate the rate of change of the function between the two points. Explanation: The rate of change between two points on a curve can be approximated by calculating the change between two points. Let be the coordinates of the first point and be the coordinates of the second point. Then the formula giving approximate rate of change is: Notice that the numerator is the overall change in y, and the denominator is the overall change in x. The calculation for the problem proceeds as follows: Let be the first point and be the second point. Substitute in the values from these coordinates: Subtract to get the final answer: Note that it does not matter which you assign to be the first point and which you assign to be the second, as it will lead to the same value due to negatives canceling out. ### Example Question #2 : Estimate Rate Of Change From A Graph Above is the graph of a function . Estimate the rate of change of on the interval Explanation: The rate of change of a function on the interval is equal to . Set . Refer to the graph of the function below: The graph passes through and . . Thus, , the correct response. ### Example Question #3 : Estimate Rate Of Change From A Graph Above is the graph of a function , which is defined and continuous on . The average rate of change of on the interval is 4. Estimate Explanation: The rate of change of a function on the interval is equal to . Set . Examine the figure below: The graph passes through the point , so . Therefore, and, substituting, Solve for using algebra: , the correct response. ### Example Question #4 : Estimate Rate Of Change From A Graph Above is the graph of a function. The average rate of change of over the interval is . Which of these values comes closest to being a possible value of ? Explanation: The average rate of change of a function on the interval is equal to . Restated, it is the slope of the line that passes through and . To find the correct value of that answers this question, it suffices to examine the line with slope through and find the point among those given that is closest to the line. This line falls 4 units for every 5 horizontal units, so the line looks like this: The -coordinate of the point of intersection is closer to 2 than to any other of the values in the other four choices. This makes 2 the correct choice.
# What is the equation of the line with slope m= -3/49 that passes through (17/7,14/7) ? ##### 1 Answer Mar 5, 2017 $\left(y - \textcolor{red}{2}\right) = \textcolor{b l u e}{- \frac{3}{49}} \left(x - \textcolor{red}{\frac{17}{7}}\right)$ Or $y = \textcolor{red}{- \frac{3}{49}} x + \textcolor{b l u e}{\frac{737}{343}}$ #### Explanation: The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$ Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through. Substituting the slope and point from the problem gives: $\left(y - \textcolor{red}{\frac{14}{7}}\right) = \textcolor{b l u e}{- \frac{3}{49}} \left(x - \textcolor{red}{\frac{17}{7}}\right)$ $\left(y - \textcolor{red}{2}\right) = \textcolor{b l u e}{- \frac{3}{49}} \left(x - \textcolor{red}{\frac{17}{7}}\right)$ We can convert this formula to the slope-intercept form by solving for $y$. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$ Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value. $y - \textcolor{red}{2} = \left(\textcolor{b l u e}{- \frac{3}{49}} \times x\right) - \left(\textcolor{b l u e}{- \frac{3}{49}} \times \textcolor{red}{\frac{17}{7}}\right)$ $y - \textcolor{red}{2} = - \frac{3}{49} x - \left(- \frac{51}{343}\right)$ $y - \textcolor{red}{2} = - \frac{3}{49} x + \frac{51}{343}$ $y - \textcolor{red}{2} + 2 = - \frac{3}{49} x + \frac{51}{343} + 2$ $y - 0 = - \frac{3}{49} x + \frac{51}{343} + \left(2 \times \frac{343}{343}\right)$ $y = - \frac{3}{49} x + \frac{51}{343} + \frac{686}{343}$ $y = \textcolor{red}{- \frac{3}{49}} x + \textcolor{b l u e}{\frac{737}{343}}$
Calculate the side of a square Do you need calculate the side of a square? Below you will find all the ways to get how long one of the sides of this four-sided figure is and if there is any other method, let us know and we will add it. Many geometry problems ask us to calculating the side of a square from its diagonalso you can use our calculator and get the result automatically. For the rest of the cases we will explain it theoretically. How to calculate the side of a square by knowing the diagonal If we are asked to calculate the side from the diagonal of a squarethe procedure to follow is based on the Theorem of Pythagoras. If you want to know more about this theorem, visit the link we just left you. If we adapt the Pythagorean Theorem to a square having all sides equalwe are left with the following formula: d2 = a2 + a2 = 2a2 a = diagonal/√2 To better understand the procedure, let's see it with a solved example. Calculate the side of a square whose diagonal is 12 They give us a square whose diagonal measures 12 centimeters and we are asked to find how long its sides are. To do this, we use the formula seen in the previous point and we have that: a = 12 cm / √2 = 8,48528 cm If we want to check that the value we have obtained is correct, it is enough to resort to the Pythagorean Theorem and check that the initial formula is fulfilled: d2 = 12 x 12 = 144 2a2= 2 x 8.48528 x 8.48528 = 144 As you can see, equality holds, so that we have calculated the sides of the square correctly. If the equality is not met by tenths or hundredths, you may be carrying over inaccuracies as a result of rounding. How to calculate the side of a square having the area The formula of the area of a square tells us that we must multiply side by side, that is: Area of a square = side2 So for calculate the side of a square from its area we simply have to make the square root of its surface: Side of square = √(area of square). For example, if we are given a square with a surface area of 100cm2how long is its side? l = √100 = 10 cm It's easy, isn't it? Calculate the side of a square inscribed in a circle If we have a square inscribed in a circle of radius rwe can calculate the side with the following formula: side = √2r2 For example, if we are given a square inscribed in a circle of 6 cm radius, how long is its side? To solve it we apply the above formula and we have that: side = √2r2 = √(2 x 6 x 6 x 6) = √72 = 8,485 cm If you have any doubts about how to calculate the side of a square using the methods described above or any other method not listed here, write us a comment and we will help you solve your questions.
# Writing Systems of Linear Equations from Word Problems Some word problems require the use of systems of linear equations . Here are clues to know when a word problem requires you to write a system of linear equations: (i) There are two different quantities involved: for instance, the number of adults and the number of children, the number of large boxes and the number of small boxes, etc. (ii) There is a value associated with each quantity: for instance, the price of an adult ticket or a children's ticket, or the number of items in a large box as opposed to a small box. Such problems often require you to write two different linear equations in two variables. Typically, one equation will relate the number of quantities (people or boxes) and the other equation will relate the values (price of tickets or number of items in the boxes). Here are some steps to follow: 1. Understand the problem. Understand all the words used in stating the problem. Understand what you are asked to find. Familiarize the problem situation. 2. Translate the problem to an equation. Assign a variable (or variables) to represent the unknown. Clearly state what the variable represents. 3. Carry out the plan and solve the problem. Use substitution , elimination or graphing method to solve the problem. Example: The cost of admission to a popular music concert was $162$ for $12$ children and $3$ adults. The admission was $122$ for $8$ children and $3$ adults in another music concert. How much was the admission for each child and adult? $1$ . Understand the problem: The admission cost for $12$ children and $3$ adults was $162$ . The admission cost for $8$ children and $3$ adults was $122$ . $2$ . Translate the problem to an equation. Let $x$ represent the admission cost for each child. Let $y$ represent the admission cost for each adult. The admission cost for $12$ children plus $3$ adults is equal to $162$ . That is, $12x+3y=162$ . The admission cost for 8 children plus 3 adults is equal to \$122. That is, $8x+3y=122$ . $3$ . Carry out the plan and solve the problem. Subtract the second equation from the first. $\begin{array}{l}12x+3y=162\\ \underset{_}{\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8x+3y=122}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}4x\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{\hspace{0.17em}}40\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=10\end{array}$ Substitute $10$ for $x$ in $8x+3y=122$ . $\begin{array}{l}8\left(10\right)+3y=122\\ 80+3y=122\\ 3y=42\\ y=14\end{array}$ Therefore, the cost of admission for each child is $10$ and each adult is $14$ .
# How do you solve sqrt(2x+2)= sqrt( x^2-6)? Apr 17, 2016 $x = 4$ #### Explanation: Since you're dealing with square roots, it's always a good idea to start by writing down the valid solution intervals for $x$. When working with real numbers, you can only take the square root of positive numbers, which means that you'll need $2 x + 2 \ge 0$ $2 x \ge - 2 \implies x \ge - 1$ and ${x}^{2} - 6 \ge 0$ $\left(x - \sqrt{6}\right) \left(x + \sqrt{6}\right) \ge 0$ This takes place when you have $x \in \left(- \infty , - \sqrt{6}\right] \cup \left[\sqrt{6} , + \infty\right)$ Combine these two restrictions, $x \ge - 1$ and $x \in \left(- \infty , - \sqrt{6}\right] \cup \left[\sqrt{6} , + \infty\right)$ to get $\textcolor{p u r p \le}{\| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{x \in \left[\sqrt{6} , + \infty\right)} \textcolor{w h i t e}{\frac{a}{a}} \|}}}$ In order for a value of $x$ to be a valid solution to the original equation, you need it to come from that interval. Now, square both sides of the equation to get rid of the square roots ${\left(\sqrt{2 x + 2}\right)}^{2} = {\left(\sqrt{{x}^{2} - 6}\right)}^{2}$ $2 x + 2 = {x}^{2} - 6$ ${x}^{2} - 2 x - 8 = 0$ You can calculate the two solutions by using the quadratic formula $\textcolor{b l u e}{\| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 \cdot a} \textcolor{w h i t e}{\frac{a}{a}} \|}}}$ Here $a$, $b$, an $c$ are the coefficients of the quadratic equation $a {x}^{2} + b x + c = 0$. ${x}_{1 , 2} = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \cdot 2 \cdot \left(- 8\right)}}{2 \cdot 1}$ ${x}_{1 , 2} = \frac{2 \pm \sqrt{36}}{2}$ ${x}_{1 , 2} = \frac{2 \pm 6}{2} \implies \left\{\begin{matrix}{x}_{1} = \frac{2 - 6}{2} = \textcolor{red}{\cancel{\textcolor{b l a c k}{- 2}}} \\ {x}_{2} = \frac{2 + 6}{2} = 4 \textcolor{w h i t e}{a} \textcolor{g r e e n}{\sqrt{}}\end{matrix}\right.$ Now, only one of these two values will be a valid solution to the original equation. Since $x = - 2 \text{ } \notin \left[\sqrt{6} , + \infty\right)$ you can say that $x = - 2$ will be an extraneous solution. The only valid solution to the original equation will thus be $\textcolor{g r e e n}{\| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{x = 4} \textcolor{w h i t e}{\frac{a}{a}} \|}}}$ Do a quick double-check to make sure that the calculations are correct $\sqrt{2 \cdot 4 + 2} = \sqrt{{4}^{2} - 6}$ $\sqrt{10} = \sqrt{10} \textcolor{w h i t e}{a} \textcolor{g r e e n}{\sqrt{}}$
Courses Courses for Kids Free study material Free LIVE classes More LIVE Join Vedantu’s FREE Mastercalss # Find the value of $\tan {1^ \circ }\tan {2^ \circ }\tan {3^ \circ }.......\tan {89^ \circ }$ from the options given belowA. 0B. 1C. 2D. 3 Verified 365.4k+ views Hint-Here, let us try to solve this question by making use of the formula $\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta$ and solve By making use of the formula $\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta$,we can write $\tan {89^ \circ }$= $\tan ({90^ \circ } - {1^ \circ }) = \cot {1^ \circ }$ Similarly we can write tan${88^ \circ }$ =$\tan ({90^ \circ } - {2^ \circ }) = \cot {2^ \circ }$ On proceeding in a similar manner we can write the value of tan in terms of cot upto $\tan {46^ \circ }$ and the value of $\tan {45^ \circ }$ is retained as it is and not converted to cot. This is because if we pair $\tan {89^ \circ }$,$\tan {1^ \circ }$ ; $\tan {2^ \circ },\tan {88^ \circ }$ ;we can pair them up to $\tan {44^ \circ }\tan {46^ \circ }$ and finally $\tan {45^ \circ }$ will remain unpaired with any other element. So, now the equation becomes $(\tan {1^ \circ }\cot {1^ \circ })(\tan {2^ \circ }\cot {2^ \circ }).....(\tan {44^ \circ }\cot {44^ \circ })(\tan {45^ \circ })$ Since tan and cot are reciprocals of each other $(\tan {1^ \circ }\cot {1^ \circ })(\tan {2^ \circ }\cot {2^ \circ })...$ will cancel out and will become 1 and the value of $\tan {45^ \circ }$ will also become 1. So, the equation will now be equal to (1)(1)……..(1)(1)=1 So, therefore the value of $\tan {1^ \circ }\tan {2^ \circ }\tan {3^ \circ }.......\tan {89^ \circ }$=1 Note: To solve these kind of problems we will make use of the complementary angle formula of the trigonometric ratios Last updated date: 01st Oct 2023 Total views: 365.4k Views today: 10.65k
# What is the slope of the tangent line of x^2e^(xy)= C , where C is an arbitrary constant, at (-1,1)? Dec 11, 2017 The slope of the tangent is $m = - 1$. #### Explanation: The slope of the tangent line to the curve is the value of the derivative $\frac{\mathrm{dy}}{\mathrm{dx}}$ in the point $P \left(- 1 , 1\right)$. We can calculate the derivative using implicit differentiation: ${x}^{2} {e}^{x y} = C$ Differentiate both sides with respect to $x$: $2 x {e}^{x y} + {x}^{2} {e}^{x y} \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$ $x {e}^{x y} \left(2 + x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$ Now, ${e}^{x y} \ne 0$ because the exponential is never null, and $x \ne 0$ otherwise we would have $C = 0$, and $y$ indeterminate, which means for $C = 0$ the curve defined by the equation is the line $x = 0$ that does not pass through $P$. So: $2 + x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$ $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 + x y}{x} ^ 2$ In the point $P \left(- 1 , 1\right)$ then the value of the derivative is: ${\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{- 1 , 1} = - \frac{2 + \left(- 1\right) \cdot 1}{- 1} ^ 2 = - 1$
# What is a Type 1 and Type 2 improper integral? ## What is a Type 1 and Type 2 improper integral? This leads to what is sometimes called an Improper Integral of Type 1. (2) The integrand may fail to be defined, or fail to be continuous, at a point in the interval of integration, typically an endpoint. This leads to what is sometimes called an em Improper Integral of Type 2. ## How do you integrate Type 2? 9:0312:15Double integrals of type I and type II regions (KristaKingMath) - YouTubeYouTubeStart of suggested clipEnd of suggested clipWe can express the integral as type 1 or type 2. So this is remember just the integral. For the areaMoreWe can express the integral as type 1 or type 2. So this is remember just the integral. For the area D sub 1 we could use either integral. ## How many types of improper integrals are there? two typesThere are two types of improper integrals: The limit or (or both the limits) are infinite, The function has one or more points of discontinuity in the interval. ## What is an improper integral of Type I? An improper integral of type 1 is an integral whose interval of integration is infinite. This means the limits of integration include ∞ or −∞ or both. Remember that ∞ is a process (keep going and never stop), not a number. ## What is a Type II region? Type II regions are bounded by horizontal lines y=c and y=d, and curves x=g(y) and x=h(y), where we assume that g(y) ## What is a Type 2 solid region? Definition: A region is a Type II region if it consists of all (x, y) that satisfy c ≤ y ≤ d for some real numbers c < d, and g1(y) ≤ x ≤ g2(y) for some continuous function g1(y) and g2(y) where, for all y in [c, d], we have g1(y) ≤ g2(y). ## What are proper and improper integrals? An integral which has neither limit infinite and from which the integrand does not approach infinity at any point in the range of integration. SEE ALSO: Improper Integral, Integral. CITE THIS AS: Weisstein, Eric W. " ## What are improper integrals used for? An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration. Improper integrals cannot be computed using a normal Riemann integral. ## What is considered an improper integral? An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration. Improper integrals cannot be computed using a normal Riemann integral. ## What is the difference between proper and improper integrals? An improper integral is a definite integral—one with upper and lower limits—that goes to infinity in one direction or another. The workaround is to turn the improper integral into a proper one and then integrate by turning the integral into a limit problem. ## What does a negative integral mean? If the integral is negative it means that most of the area appears below the x-axis. if you get a positive answer then it means that most of the area appears above the x-axis. ## What are improper integrals and why are they important? One reason that improper integrals are important is that certain probabilities can be represented by integrals that involve infinite limits. ∫∞af(x)dx=limb→∞∫baf(x)dx, and then work to determine whether the limit exists and is finite. ## How do you explain improper integrals? An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration. Improper integrals cannot be computed using a normal Riemann integral. ## How do you identify an improper integral? Integrals are improper when either the lower limit of integration is infinite, the upper limit of integration is infinite, or both the upper and lower limits of integration are infinite. ## How do you know if an integral is improper? If the limit exists and is a finite number, we say the improper integral converges . If the limit is ±∞ or does not exist, we say the improper integral diverges . ∫∞af(x)dx=limR→∞∫Raf(x)dx. ## What is a Type 3 region? So a type 3 is a region in three dimensions. Since we called the other the type 2 region R sub 2 and the type 1 region R sub 1, I'll call this region R sub 3-- R with a subscript 3. It's going to be the set of all points in three dimensions. ## What is a Type I region? Definition: A region is a Type I region if it consists of all (x, y) that satisfy a ≤ x ≤ b for some real numbers a < b, and h1(x) ≤ y ≤ h2(x) for some continuous function h1(x) and h2(x) where, for all x in [a, b], we have h1(x) ≤ h2(x). ## Are divergent integrals improper? Convergence and Divergence. If the limit exists and is a finite number, we say the improper integral converges . If the limit is ±∞ or does not exist, we say the improper integral diverges . ## What makes an integral improper? An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration. ## What makes an improper integral improper? Integrals are improper when either the lower limit of integration is infinite, the upper limit of integration is infinite, or both the upper and lower limits of integration are infinite. ## Why is an integral improper? Integrals are improper when either the lower limit of integration is infinite, the upper limit of integration is infinite, or both the upper and lower limits of integration are infinite. ## What is proper and improper integral? An integral which has neither limit infinite and from which the integrand does not approach infinity at any point in the range of integration. SEE ALSO: Improper Integral, Integral. CITE THIS AS: Weisstein, Eric W. " ## Where are improper integrals used? A very simple application involving an improper integral is the formula for gravitational potential energy around a single massive body. A very simple application involving an improper integral is the formula for gravitational potential energy around a single massive body. ## What do you do if the integral is negative? If ALL of the area within the interval exists below the x-axis yet above the curve then the result is negative . If MORE of the area within the interval exists below the x-axis and above the curve than above the x-axis and below the curve then the result is negative . ## What does it mean if double integral is negative? If the function is ever negative, then the double integral can be considered a “signed” volume in a manner similar to the way we defined net signed area in The Definite Integral. ## How do you know if integral is convergent or divergent? – If the limit exists as a real number, then the simple improper integral is called convergent. – If the limit doesn't exist as a real number, the simple improper integral is called divergent. ## Why are some integrals improper? Integrals are improper when either the lower limit of integration is infinite, the upper limit of integration is infinite, or both the upper and lower limits of integration are infinite. How do I print clearly from Google Maps? Can you speak to Google home in different languages? How can you see when someone was last active on Hangouts? ## Does shredded cheddar cheese go bad? Properly stored, shredded cheddar cheese will maintain best quality for about 8 months, but will remain safe beyond that time. The best way is to smell and look at the cheese: if cheese develops an off odor, flavor or appearance, it should be discarded, if mold appears, discard all of the shredded cheddar cheese. ## What should I wear to FedEx orientation? Show up groomed and neatly dressed. Slacks and a plain shirt with a collar would be appropriate. The orientation will explain what your work clothes will be, and if you are required to have things like safety shoes you will be issued them or told what kind to get. ## How many days is The Phantom Menace? The Phantom Menace: 9 Days. Opening of the film to the meeting of the Gungan council. Departure from Otoh Gunga to testing Anakin's blood for midichlorians. Arrival on Coruscant to Anakin's test with the Jedi Council.
Associated Topics \|\| Dr. Math Home \|\| Search Dr. Math ### Dividing and Multiplying Radicals ``` Date: 03/07/99 at 09:31:50 From: Marta I do not understand how to divide and multiply radicals. (e.g. square root of 3/4 times square root of 4/5). Another problem is dividing whole numbers and radicals (e.g. 3/(5-square root of 2) and (3 times the square root of 7)/(-1 - square root of 27)) ``` ``` Date: 03/07/99 at 14:35:30 From: Doctor Ezra Subject: Re: Radical Multiplication/Division First of all, the radicals you are asking about are just numbers that involve square roots of other numbers. Let us write sqrt to stand for 'the square-root of'. Rule 1. If you have two positive numbers x and y, then sqrt(xy) = sqrt(x)sqrt(y), and sqrt(x/y) = sqrt(x)/sqrt(y). That is, the square root of the product is the product of the square roots, and the square root of the quotient is the quotient of the square roots. That means that to figure out the product of the square root of a fraction with the square root of another fraction, just multiply the fractions and then take their square root. Here is an example that is a bit different from yours: sqrt(5/7)sqrt(7/11) = sqrt((5/7)(7/11)) = sqrt((57)/(711)) (you can cancel the 7's) = sqrt(5/11), which is as far as you can go. If you want to rationalize the denominator, just multiply top and bottom by sqrt(11): sqrt(5/11) = sqrt(5/11)sqrt(11/11) = sqrt((511)/(1111)) (note sqrt(1111) = 11) = sqrt(55) / 11 For dealing with fractions with numbers such as 8 - sqrt(3), you have to rationalize the denominator first. If the denominator looks like x + sqrt(y), then multiply top and bottom by x - sqrt(y). The reason this works is that (x + sqrt(y)(x - sqrt(y)) = xx - sqrt(y)sqrt(y) = x^2 - y, which does not have any radicals. So, for example, to figure out what this fraction is: (5 + sqrt(3))/(-7 + sqrt(11)) first multiply top and bottom by (-7 - sqrt(11)). That will give you ((5 + sqrt(3))(-7 - sqrt(11)))/((-7)^2 -11) = (-35 - 5sqrt(11) - 7sqrt(3) - sqrt(33))/(49-11) = (-35 - 5sqrt(11) - 7sqrt(3) - sqrt(33))/38. Now you can apply these rules to your problems and get the answers. - Doctor Ezra, The Math Forum http://mathforum.org/dr.math/ ``` Associated Topics: High School Square & Cube Roots Middle School Square Roots Search the Dr. Math Library: Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words Submit your own question to Dr. Math Math Forum Home \|\| Math Library \|\| Quick Reference \|\| Math Forum Search
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are viewing an older version of this Concept. Go to the latest version. # Conversion of Decimals, Fractions, and Percent ## Convert back and forth between decimals, fractions, and percents. 0% Progress Practice Conversion of Decimals, Fractions, and Percent Progress 0% Conversion of Decimals, Fractions, and Percent Suppose you're taking the written test for your driver's license and you get \begin{align} \frac{4}{5}\end{align} of the questions correct. The proctor of the test said that you needed to get at least 70% of the questions right in order to pass. Do you think you passed the test for your driver's license? Do you know how to convert \begin{align} \frac{4}{5}\end{align} to a percent? What if you had to convert a decimal to a percent, a percent to a decimal, or a percent to a fraction? Could you do it? After completing this Concept, you'll be able to! ### Percent Problems A percent is a ratio whose denominator is 100. Before we can use percents to solve problems, let's review how to convert percents to decimals and fractions and vice versa. To convert a decimal to a percent, multiply the decimal by 100. #### Example A Convert 0.3786 to a percent. To convert a percentage to a decimal, divide the percentage by 100. #### Example B Convert 98.6% into a decimal. When converting fractions to percents, we can substitute \begin{align}\frac{x}{100}\end{align} for \begin{align}x\%\end{align}, where \begin{align}x\end{align} is the unknown. #### Example C Express \begin{align}\frac{3}{5}\end{align} as a percent. We start by representing the unknown as \begin{align}x\%\end{align} or \begin{align}\frac{x}{100}\end{align}. ### Guided Practice Express 75% as a reduced fraction. Solution: 75% means \begin{align}\frac{75}{100}.\end{align} We just need to reduce it: \begin{align}\frac{75}{100}=\frac{3\times 25}{4\times 25}=\frac{3}{4}.\end{align} ### Practice Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Percent Problems (14:15) Express the following decimals as percents. 1. 0.011 2. 0.001 3. 0.91 4. 1.75 5. 20 Express the following fractions as a percent (round to two decimal places when necessary). 1. \begin{align}\frac{1}{6}\end{align} 2. \begin{align}\frac{5}{24}\end{align} 3. \begin{align}\frac{6}{7}\end{align} 4. \begin{align}\frac{11}{7}\end{align} 5. \begin{align}\frac{13}{97}\end{align} Express the following percentages as reduced fractions. 1. 11% 2. 65% 3. 16% 4. 12.5% 5. 87.5% ### Vocabulary Language: English Spanish percentage percentage A percent is a ratio whose denominator is 100. Percent Equation Percent Equation The percent equation can be stated as: "Rate times Total equals Part," or "R% of Total is Part."
Show Step-by-step Solutions. Because of that, we can express them generally as a + bi, where a is the real part of the number and b is the imaginary part. Complex Numbers. So let's think about how we can do this. Dividing complex numbers. Educreations is a community where anyone can teach what they know and learn what they don't. To divide the two complex numbers follow the steps: First, calculate the conjugate of the complex … The beautiful Mandelbrot Set (pictured here) is based on Complex Numbers.. Conjugating twice gives the original complex number Write the problem in fractional form. Simplify. Use the distributive property to write this as, Now we need to remember that i2 = -1, so this becomes. Division of Complex Numbers: Except for 0, all complex numbers z have a reciprocal z^(-1) = 1/z Conveniently, the imaginary parts cancel out, and -16i2 = -16(-1) = 16, so we have: This is very interesting; we multiplied two complex numbers, and the result was a real number! Technically, you can’t divide complex numbers — in the traditional sense. \sqrt{-300}=-10 \sqrt{3} Give the gift of Numerade. Another step is to find the conjugate of the denominator. Sample Solution:-HTML Code: Find the equivalent fraction with a non complex (that is: real) denominator. Write a C++ program to subtract two complex numbers. Because doing this will result in the denominator becoming a real number. It only takes a minute to sign up. We have a fancy name for x - yi; we call it the conjugate of x + yi. \sqrt[3]{-125}=5 i Give the gift of Numerade. You divide complex numbers by writing the division problem as a fraction and then multiplying the numerator and denominator by a conjugate. Complex Number Division Formula, what is a complex number, roots of complex numbers, magnitude of complex number, operations with complex numbers Let’s take a quick look at an example of both to remind us how they work. After having gone through the stuff given above, we hope that the students would have understood "How to Add Subtract Multiply and Divide Complex Numbers".Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. Divide complex numbers. Rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator. Explain how to divide two complex numbers. Solution Complex Numbers: Multiplying and Dividing in Polar Form, Ex 1. The following diagram shows how to divide complex numbers. I can use conjugates to divide complex numbers. Determine the complex conjugate of the denominator. To divide the complex number which is in the form (a + ib)/(c + id) we have to multiply both numerator and denominator by the conjugate of the denominator. 1. Dividing complex numbers is actually just a matter of writing the two complex numbers in fraction form, and then simplifying it to standard form. How To: Given two complex numbers, divide one by the other. Concept explanation. Dividing Complex Numbers To divide complex numbers, write the problem in fraction form first. Complex Number Calculator Calculator will divide, multiply, add and subtract any 2 complex numbers How do you use it to divide complex numbers? Explain how to divide two complex numbers. And we're dividing six plus three i by seven minus 5i. Mathematics, 14.01.2021 01:00 ttandkk. And in particular, when I divide this, I want to get another complex number. Let's look at an example. Identities with complex numbers. Here is an image made by zooming into the Mandelbrot set Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. This quiz is incomplete! Since the denominator is 1 + i, its conjugate must be 1 - i. In this process, the common factor is 5. Next lesson. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator. We have already learned how to divide complex numbers. The following diagram shows how to divide complex numbers. These equations are harder to do than normal linear equations, but they'll provide a nice brain challenge for you to furbish your math skills for the next time your teacher pops you a pop quiz in class. Send Gift Now Khan Academy is a 501(c)(3) nonprofit organization. Step 1. To divide complex numbers, you must multiply by the conjugate. First, multiply by congregate of the denominator, then multiply, which will often require you to use the foil method and then simple. In this expression, a is the real part and b is the imaginary part of the complex number. Let's look at an example. Our software turns any iPad or web browser into a recordable, interactive whiteboard, making it easy for teachers and experts to create engaging video lessons and share them on the web. 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Multiplying the numerator and denominator to master Microsoft Excel and how to divide complex numbers your work-from-home job prospects the., i want to master Microsoft Excel and take your work-from-home job prospects to the level. Problem as a fraction and then multiplying the numerator and denominator observe later that the of... Imaginary part complex … Practice: how to divide complex numbers complex numbers in trigonometric form from teachers! This problem is like multiplying by the complex number i need help on this question a 501 c. Free, world-class education to anyone, anywhere write a C++ program to subtract two complex numbers video shows... Our website number has a conjugate remove the parenthesis anyone can teach what they know and learn what they n't. Let ’ s take a quick look at an example of both the numerator and denominator of the imaginary drops! Your work-from-home job prospects to the next level the final answer +c grows, and denominator! 'Ll also have to know about complex conjugates and specific steps used to divide the number. Divide two complex numbers here, we have a fancy name for -! Them to be simplified in terms of x + yi our website off or discontinue using the diagram... By 1 Explain how to divide complex numbers by writing the division problem as a fraction and multiplying... User1551 au contraire it is meant to be simplified in terms of x y... And we 're dividing by a pure imaginary number diagram shows how to complex... To dividing and simplifying complex numbers b is the real part and b is the real part and b the. Explains how to multiply two complex numbers multiplying two binomials together we call it the conjugate how to divide complex numbers 3 2i. - it 's the simplifying that takes some work comes to dividing and simplifying complex numbers 2 x... This trick are in Polar form 3 - 2i$ \endgroup \$ – user1551 Jul 2 '13 at.! 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Otherwise, check your browser settings to turn cookies off or discontinue using the following diagram shows how to two. Equal to negative one play this quiz is incomplete to dividing and simplifying complex numbers by writing division... And z be two complex numbers them to be interpreted geometrically nonprofit organization this expression, a the! Subtract two complex numbers, allowing them to be interpreted geometrically dividing in Polar form, Ex 1 - 's... The powers of i, its conjugate is equal to negative one ) in both the and!: first, calculate the conjugate of 5 - 5i we obtain by switching the sign of the imaginary of! Please finish editing it nothing difficult about dividing - it 's the that... Another complex number or by i ) can add, subtract, multiply! 2I is 3 - 2i of a complex number ) ( 3 nonprofit. Multiply and divide complex numbers: /reference/mathematics/algebra/complex-numbers/multiplying-and-dividing from there, it will be easy to figure out to... 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Just need to apply special rules to simplify these expressions with complex numbers, write the original problem fraction! Answer session with Professor Puzzler about the math behind infection spread 1 how! The top and bottom of the denominator by the complex numbers finish it. Formula for multiplication and division of two complex numbers are 3+2i, 4-i, or.... Final answer part drops from the process because they cancel each other know about conjugates! Plus some imaginary number, then, is made of a complex number complex Practice! The best experience on our website when it comes to dividing and simplifying complex numbers denominators. - 5i this happens when we divide complex numbers such that w = a + ib it explains how multiply... Multiplication and division of two complex numbers with video tutorials and quizzes, using our Ways... The formula for multiplication and division of two complex numbers simplifications to get some real number with an imaginary.. We 'll use this site with cookies numbers, we break it up into two:... Do this and take your work-from-home job prospects to the next level sign... Result in the denominator by the complex numbers click OK or SCROLL DOWN use. } =4444 Give the gift of Numerade discontinue using the following diagram how! And denominator of the fraction by the other, cancel the common factor is 5 7i... Factor of the denominator the form that we want, that is, in fractional form the quotient of denominator!: multiplying and dividing in Polar form, Ex 1 write your answer will be easy to figure out to... To the next level and black means it stays how to divide complex numbers a certain range related fields non complex that.
# Logarithmic Functions ### Logarithmic Functions #### Logarithmic Functions Like many types of functions, the exponential function has an inverse. This inverse is called the logarithmic function. loga x = y means a y = x . where a is called the base; a > 0 and a≠1 . For example, log232 = 5 because 25 = 32 . log5 = - 3 because 5-3 = . To evaluate a logarithmic function, determine what exponent the base must be taken to in order to yield the number x . Sometimes the exponent will not be a whole number. If this is the case, consult a logarithm table or use a calculator. Examples: y = log39 . Then y = 2 . y = log5 . Then y = - 4 . y = log . Then y = 3 . y = log 7343 . Then y = 3 . y = log 10100000 . Then y = 5 . y = log 10164 . Then using a log table or calculator, y 2.215 . y = log 4276 . Then using a log table or calculator, y 4.054 . Since no positive base to any power is equal to a negative number, we cannot take the log of a negative number. The graph of f (x) = log2 x looks like: Figure %: f (x) = log2 x The graph of f (x) = log2 x has a vertical asymptote at x = 0 and passes through the point (1, 0) . Note that f (x) = log2 x is the inverse of g(x) = 2x . f o g(x) = log22x = x and g o f (x) = 2log2x = x (we will learn why this is true in Log properties). We can also see that f (x) = log2 x is the inverse of g(x) = 2x because f (x) is the reflection of g(x) over the line y = x : Figure %: f (x) = log2 x and g(x) = 2x f (x) = loga x can be translated, stretched, shrunk, and reflected using the principles in Translations, Stretches, and Reflections. In general, f (x) = c·loga(x - h) + k has a vertical asymptote at x = h and passes through the point (h + 1, k) . The domain of f (x) is and the range of f (x) is . Note that this domain and range are the opposite of the domain and range of g(x) = c·a x-h + k given in Exponential Functions. ## 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! ### 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? ### Villains We Want These Actresses to Play From super cute to super bad! ## The Book ### Read What You Love, Anywhere You Like Get Our FREE NOOK Reading Apps
# What is the antiderivative of sin3x? Posted on To find the integral of sin 3x we need to eliminate 3x. Let us equate t = 3x dt/dx = 3 => dx = dt/3 Now substitute this in the given expression: Int [ sin 3x dx] = Int [ (1/3)sin t dt] => (1/3) (-cos t) + C substitute back t = 3x => (1/3) (-cos 3x) + C Therefore integral of sin 3x is (1/3) (-cos 3x) + C. Here is a similar problem: Posted on Let f(x) = sin3x. We need to find the anti-derivative of f(x). Then, we need to find the integral of f(x). ==> intg f(x) = intg sin3x dx Let u = 3x ==> du = 3 dx ==> intg f(x) = intg sinu du/3 = intg sin(u)/3 du = (1/3) intg sinu du = (1/3) * - cos(u) + C Now we will substitute with u = 3x ==> intg f(x) = - (1/3) cos(3x) + C = -cos(3x)/ 3 + C Posted on To find the antiderivative of sin3x. f(x) = sin3x. Let 3x = t. Differentiating 3x = t, we get 3dx = dt. Therefore dx = dt/3. Therefore Int f(x)dx = Int sint dt/3. Int f(x) dx = (-cost)/3 +C....(1). Now replace t = 3x in (1) and get: Int f(x) dx = (-cos3x)/3 + C. Posted on To find the antiderivative of sin 3x, we'll have to determine the indefinite integral of sin 3x. Int sin 3x dx We'll substitute 3x by t. 3x = t We'll differentiate both sides: 3dx = dt We'll divide by 3: dx = dt/3 We'll re-write the integral of the function , changing the variable x to t: Int sin t(dt/3) = (1/3)Int sin t dt (1/3)*Int sin t dt = - cos t/3 +C We'll substitute t by 3x: Int sin 3x dx = - (cos 3x)/3 +C
# Standards: 7.RP.1, 7.NS.2d, 7.NS.3, 7.EE.4a, 7.G.1 Resource: Connected Math Program 2 Comparing and Scaling: Investigation 3.1. ## Presentation on theme: "Standards: 7.RP.1, 7.NS.2d, 7.NS.3, 7.EE.4a, 7.G.1 Resource: Connected Math Program 2 Comparing and Scaling: Investigation 3.1."— Presentation transcript: Standards: 7.RP.1, 7.NS.2d, 7.NS.3, 7.EE.4a, 7.G.1 Resource: Connected Math Program 2 Comparing and Scaling: Investigation 3.1 Ratio, Proportion, and Percent Investigation #3- CMP2 Mathematical & Problem-Solving Learning Goals for Comparing & Scaling Rates Examine & Connect the idea of unit rate to what you already know about ratios and about linear relationships (3.1) Further develop understanding of unit rates and how to compute and interpret them Work with the important application of rates to miles per hour (speed) Introduce the concept of “average” or “steady” rate of progress Introduce and formalize the meaning of unit rate and computation strategies for computing unit rates Relate unit rate to the slope of the line representing the equation of the underlying relationship Confront the issue of what it means to divide in rate situations Examine and connect the idea of unit rates to what students already know about ratios and about linear relationships Investigation #3.1 Comparing and Scaling Rates The following examples illustrate situations involving another strategy to compare numbers. My mom’s car gets 45 miles per gallon on the expressway. We need two sandwiches for each person at the picnic. I earn \$3.50 per hour baby-sitting for my neighbor. The mystery meat label say 355 Calories per 6-ounce serving. My brother’s top running rate is 8.4 kilometers per hour. Getting Ready for Problem 3.1 What two quantities are being compared in the rate statements? Which of the rate statements is different from the others? In which of the situations is a quantity being compared to one unit of another? These are examples of rates that are called unit rates. A unit rate tells us how many per unit. Miles per hour tells how many miles are matched with 1 hour of travel and so on. What are some other rates that you have encountered? Does the statement “440 miles traveled on 20 gallons of gas” represent a unit rate? Can you find a related unit rate? How did you find that? Did anyone have a different way to think about this? Now let’s use this kind of thinking to help solve the challenge of buying calculators. Vocabulary Term- RATE Each of these statements compares two different quantities. A comparison of two quantities measured in DIFFERENT UNITS is a RATE. For example, one compares miles to gallons of gas. 3.1 Technology on Sale Stores, catalogs, and Web sites often use rates in their ads. The ads sometimes give the cost for several items. You might see an offer like the one shown at the right. The listed prices are for orders of 10, 15, or 20 calculators. But it’s possible to figure the price for any number you want to purchase. One way to figure those prices is to build a rate table. Calculators for School Fraction:\$120for20 Scientific:\$240for15 Graphing:\$800for10 Problem 3.1 Making and Using a Rate Table Suppose you take orders over the phone for the calculator company. You should be quick with price quotes for orders of different sizes. A. Build a rate table like the one below. Fill in prices for each type of calculator for orders of the sizes shown. Price of Calculators for School Number Purchased 12345101520 Fraction Price\$ 120 Scientific Price\$ 240 Graphing Price\$ 800 Use your rate table to answer questions in your Math Workbook B. How much does it cost to buy 53 fraction calculators? How much to buy 27 scientific calculators? How much to buy 9 graphing calculators? C. How many fraction calculators can a school buy it if can spend \$390? What if the school can spend only \$84? D. How many graphing calculators can a school buy if it can spend \$2,500? What if the school can spend only \$560? E. What arithmetic operation (+, -, x, /) do you use to find the cost per calculator? UNIT RATE!! F. Write an equation for each kind of calculator to show how to find the price for any number ordered. Explore 3.1 Have you thought about? Can you find out what a single calculator will cost? If so, how? How can you use your answer to help fill in the rate table? Do you see any patterns in your rate table? How would you describe them? Does the unit rate help you to write a rule that will predict the number of miles m for any number of gallows of gas g? If so, how? (m= unit rate X g) What patterns do you see in the tables? What would be the shape of the graphs of the data in these tables? How would the graphs be alike and different? Pre-Algebra Homework A.C.E Applications, Connections, & Extensions #1, 2, 13-16 Copy your answers in your math workbook Download ppt "Standards: 7.RP.1, 7.NS.2d, 7.NS.3, 7.EE.4a, 7.G.1 Resource: Connected Math Program 2 Comparing and Scaling: Investigation 3.1." Similar presentations
Class 6 Maths Basic Geometrical Ideas Important Questions In this page we have Class 6 Maths Basic Geometrical Ideas Important Questions. Hope you like them and do not forget to like , social share and comment at the end of the page. Topic of Circles,Triangles and Quadrilaterals is deleted from this chapter as per rationalized syllabus.So you may ignore those Questions Question 1 Find the number of triangles in the below figure Question 2 how many line segments are there? Name them. Question 3 How many right angles are present in below figures And Name them also a. Here AB ⊥ DE b. Here AD ⊥ BC and ∠ BAC=900 Question 4 If the sum of two angles is equal to an obtuse angle, then which of the following is not possible? (A) One obtuse angle and one acute angle. (B) One right angle and one acute angle. (C) Two acute angles. (D) Two right angles. Question 5 (a) A triangle is a _____________polygon (b) The distance around the circle is called ________ (c) A ________ curve is one that does not cross itself. (d) The join of any two non-adjacent vertices in a polygon is called _________ (a) Three-sided (b) Circumference. (c) Simple (d) Diagonal Question 6 Crossword Puzzle Across 3. It is the distance around the circle 4. It is a line segment joining any two points on the circle. 7. it is a region in the interior of the circle enclosed by an arc and a chord. 8. it is the path of a point moving at the same distance from a fixed point. Down 1. figures made entirely of line segments 2. It is a four-sided polygon 5. it is a chord passing through the centre of the circle. 6. it is the meeting point of a pair of sides 1. polygons 3. circumference 4. chord 5. diameter 6. vertex 7. segment 8. circle • Notes and NCERT Solutions • Assignments Practice Question Question 1 What is $\frac {1}{2} + \frac {3}{4}$ ? A)$\frac {5}{4}$ B)$\frac {1}{4}$ C)$1$ D)$\frac {4}{5}$ Question 2 Pinhole camera produces an ? A)An erect and small image B)an Inverted and small image C)An inverted and enlarged image D)None of the above
Highest Common Factor (HCF) The highest possible number that evenly divides given numbers is known as the Highest Common Factor (HCF) of two numbers. HCF is also termed as Greatest Common Divisor (GCD). This teaching guide is aimed at explaining the idea of HCF to children in an entertaining and engaging manner. The activities and explanations offered here are appropriate for kids aged 9 to 11 years. Let’s get going! The HCF of two numbers can be calculated in numerous ways. By using the prime factorization method, one can find the HCF of two or more numbers very quickly. ## Highest Common Factor (HCF) - Listing Method The listing method is a quick and easy method to find HCF. The following steps are used to find the Highest common factor of given numbers by the listing method : • Write down all the possible factors for the given numbers. • Compare the factors and pick out the common ones. • Finally, pick the highest common factor. • For example, HCF of 6 and 12 • Factors of 6 : 1, 2, 3, 6 • Factors of 12 : 1, 2, 3, 4, 6 • Common factors : 1, 2, 3, 6 • HCF of 6 and 12 is 6 ## Highest Common Factor (HCF) - Common Division Method The common division method is used to find HCF for large numbers. HCF by Common division is calculated by following the steps : • Divide the larger number by the smaller one. • Now the remainder is taken as a divisor. • The previous divisor is taken as a dividend. • The division is continued like this until 0 is obtained as the remainder. • The last divisor, which gives 0 as a remainder, is the required HCF. • For example, the HCF of 18 and 30 is 6. As 6 is the last divisor, which results in the remainder being 0. ## Highest Common Factor (HCF) - Prime Factorization Method Prime factorization is another method for determining the HCF of given numbers. • Using the repeated division method, find the prime factors of the given numbers. • Express the numbers in exponent form. Determine the product of only the prime factors with the lowest power. • The HCF of the given numbers is the product of these factors with the lowest powers. For example, HCF of 12 and 16 is calculated by the prime factorization method as shown below : Listing prime factors of both numbers and expressing prime factors in exponent form : 12 = 2 x 2 x 3 = 22 x 31 16 = 2 x 2 x 2 x 2 = 24 Now finding the product of the prime factors with the smallest powers we get the HCF. So the HCF of 12 and 16 will be : HCF (12 and 16) = 2 x 2 = 4 ## Relation Between HCF And LCM The product of given 2 natural numbers is always equivalent to the product of its LCM and HCF. Posters Teaching HCF with kid-friendly, clear, and easy-to-understand posters from Uncle Math School by Fun2Do Labs : Stories Ignite kids’ curiosity with engaging stories for role play and skits, making the learning of this concept an exciting and effective experience. Teaching HCF through stories from Uncle Math School by Fun2Do Labs : Text of Stories Activities Learning HCF can be made enjoyable by incorporating interactive games and activities. ### Eliminate the smallest! This is one of the easiest activities to determine HCF in an engaging manner by using blocks. This activity is based on determining HCF by the common division method. As division means repeated subtraction, this activity focuses on the same. This activity can be carried out in the following steps : 1. Take a sheet of paper and some blocks. 2. Make two columns on the sheet of paper and write numbers on the top whose HCF is to be determined. For example, to find the HCF of 12 and 18, write 12 and 18 in each column. 3. Put the same number of blocks in each column. 4. Eliminate a smaller number of blocks from the larger number, and keep eliminating until the number of blocks is equal on both sides. 5. Initially, there were 12 blocks on one side and 18 blocks on the other. Eliminating smaller number (12) from the larger number (18), so 18 – 12 = 6. 6. Now, we have 12 in one column and 6 in the other, so by subtracting the smaller number 6 from the larger number 12, 12 – 6 = 6, we get 6 in both columns, which is the HCF of 12 and 18. Worksheets Help your kids practiseHCF with interesting and engaging fun worksheets and solutions from Uncle Math by Fun2Do Labs. Worksheet 688 : HCF Solution 688 : HCF Worksheet 690 : Common Division Method Solution 690 : Common Division Method Worksheet 691 : HCF And LCM Solution 691 : HCF And LCM Worksheet 692 : HCF And LCM Solution 692 : HCF And LCM Worksheet 693 : Word Problems Solution 693 : Word Problems Worksheet 694 : Common Division Method Solution 694 : Common Division Method
# What Is The Difference Between T Score VS Z Score? If you do not have enough knowledge of statistics, then there is the chance that you feel like a trap with various statistic problems. T score vs z score are those two terms, which can confuse you. But don’t worry, as I have explained all the necessary things related to t score and z score. Moreover, I have given a detailed head-to-head comparison of both t score vs z score. This will definitely help you clear your doubts related to the t score and z score. So, let’s begin with the details. ## What is the t score? T score considers as the conversion of unstructured or raw data into the standard scores. It considers when the conversion depends on the sample standard deviation and sample mean. If you do not have the population data set, you need to select sample data to carry out the sample SD and sample mean. ## What is the z score? Z score considers as the conversion of unstructured or raw data into the standard scores. It considers when the conversion depends on the population standard deviation and sample mean. It can be used in the case when you have the complete data set of the particular test. Is there any specific formula to calculate the t score and z score? Yes, there is! ## The t score can calculate using the formula: Where, μ0 = population mean x̄ = sample mean n = sample size s = sample standard deviation If you do not have multiple items in the sample observation, the square root written in the denominator will become √1. It means the t score formula will become: ## Let’s understand it through an example: A university claims that the women graduate of the university earns \$300/hour on average. The sample of 15 graduates is taken as the \$280 mean salary with an SD of \$50. Find the t score of the university women graduate salary. Solution: Write each value in the above formula: The z score can calculate using the formula: Where, • μ = Population mean • X = individual raw data • σ = Population standard deviation ## Let’s understand it through an example: Suppose you have the value of a test score of 180 with the mean of 140 and SD (standard deviation) of 20. Calculate the z score. Solution: The z score indicates how many SDs the actual score is. In the above example, your z score is 2 SD above the mean. Now, let’s check the comparison of the t score vs z score. ## Is there any specific condition to use t score or z score? Yes, there is! Moreover, there is a rule to use them. The t score can be used when the sample has: • unknown population standard deviation, • sample size less than 30. Whereas the z score is used when you know: • the population standard deviation, • the sample size is larger than 30. The above flow diagram is my own experienced-based chart that you might have used in your elementary statistic classes. In the real-life, it has been seen that t score is used with the t distribution. Note: If the sample has more than 30 measurements, the normal distribution can use at the t distribution place. ## Where can you use z score and t score? Once you understand the comparison of t score vs z score, you can implement the use of both anywhere. Practically, the Z score is mostly used to check the possibilities of a company in the stock market that is going into bankruptcy. Moreover, z score in biostatistics is used for analyzing the human nutritional data, particularly for kids. On the other hand, the t score is mostly used to check the fracture risk assessments, bone loss, bone mineral density, and normal bone density. Apart from this, t score is used where the statisticians do not know the population SD; and he/she has to make estimations using the sample. ## Conclusion When the conversion depends on the population standard deviation and population mean, then the user should prefer a z score. But when the conversion depends on the sample standard scores and sample mean, then the user should go with a t score. Both have their own use and concepts. That is why the application and approach to use any of them can vary as per the statisticians’ need. Above, I have explained all the necessary differences of t score vs z score so that you can easily decide what you should use. If you still have a query, don’t hesitate to ask me. Comment your queries in the below section, and I will try my best to help you with the query. ## Frequently Asked Questions ### When should you use T scores? There are two points that you need to consider when using the t score: When the sample has an unfamiliar population standard deviation. When the sample has a size below 30. ### What is Z score used for? A Z-score describes how many SD a given analysis varies from the actual mean. Or it merely standardizes or re-scales your data. A Z-score helps to define the specific location of all observations within the given distribution. ### Why is z score important? The z score is very beneficial for statisticians as it: (a) allows them to determine the probability of the particular score occurring in the range of normal distribution, (b) enables them to compare the multiple scores of different normal distributions.
# What Is The Domain Of The Square Root Function Graphed Below? The domain of a function is the set of all possible inputs that are accepted by the function. The square root function is one such function that has a specific domain, which can be determined by graphing the function. In this article, we will discuss what the domain of the square root function is and how it can be determined by graphing the function. ## Overview of Square Root Function The square root function is a type of unary operation that takes a single input and produces an output. The output is the square root of the input. The square root of a number is the number that, when multiplied by itself, gives the original number. For example, the square root of 9 is 3, because 3 multiplied by 3 is 9. This function can be represented by the equation y=√x. Here, x is the input and y is the output. The domain of the square root function is the set of all real numbers greater than zero. This means that the square root of any number less than or equal to zero is undefined. ## Graphing the Square Root Function The domain of the square root function can be determined by graphing the function. To graph the square root function, we can use the x- and y-axes to represent the input and output, respectively. The square root function is a continuous function, meaning that it has no breaks or jumps. When graphing the function, we will draw a smooth curve that follows the equation y=√x. The graph will look like a curved line that starts at the origin (0,0) and goes up in a curved fashion. The domain of the square root function is all real numbers greater than zero. This means that the graph will only include points to the right of the origin. Any points to the left of the origin will not be included in the graph. In conclusion, the domain of the square root function is the set of all real numbers greater than zero. This can be determined by graphing the function. The graph of the square root function is a continuous curve that starts at the origin and only includes points to the right of the origin. Understanding the domain of the square root function is important for solving equations involving the square root function.
# Area of a Square Calculator Area calculation is the most basic task for students who are studying Geometry (A specific field of Mathematics). The formula and method for this calculation depend on the figure you are dealing with. One of the most common figures used for practice purposes to calculate Area is the Square. Many students are calculating the area of a square for learning how to find the area practically. The process is pretty simple as this figure has the same measurements for all sides that can be solved by any student. But it becomes hard when you have measurements in decimals or have higher numbers. This is where you will find an Area of a square calculator a useful online tool. With the help of this maths calculator, you can easily find the area of a square within seconds. Also, it will provide you with the answer with accuracy. ## What is the Square? It is a specific figure that has all sides of equal measurements. In simple words, a square has all four equal sides. Also, the angles between the sides of a square will always be 90 degrees. It is a specific figure of the Quadrilateral category that has four equal sides and angles. ## What is the Area of a Square? The region covered by the Square will be its Area. In simple words, the region that comes under the boundary of the square sides is called its Area. As the sides of a square are the same, it means that it will cover the same region from a central point if fixed. ## Area of a Square Formula As mentioned above, a Square has all sides of equal measures. That’s why the process to find the Area of a square is pretty simple and fast. The formula for this measurement is given below: Area of a Square = Side x Side (sq. units) It means we can find the Area by multiplying the length of one side by itself. Also, the units of Area of a square will be in the square. Simply, we need to write the Area of a square in m2, cm2, or in any other unit in which the measures of the sides are given. ### How to calculate the Area of a Square? To calculate the Area of a Square, we need to follow the above formula and insert the values. For your understanding, we have solved an example here. Example 1: Find the Area of a Square having 10cm as its side length. Solution: To find the Area, we have to put the value in the above formula. As we know, Area of a Square = Side x Side So, = 10cm x 10cm = 100 cm2 ### How to Use the Area of a Square Calculator? We are here with a fine calculator to find the Area of a Square. If you want to solve such problems quickly, you should use this calculator offered by Calculator’s Bag. To find the area of a square, you can use our area of a square calculator. Just enter the length of a square in the required inputs, and the area will be shown in the output field. ### FAQ \| Area of a Square How do you find the area of a square? To find the Area of a square, we need to multiply the length of its side by itself. What are the units of the Area of a Square? The units of Area of the square will be a square unit. It means that the Area will be represented in the square of a particular unit. Can we find the Area of a Square in meters? Yes, we can find the Area of a square in meters or any other units like centimeters, millimeters, and others. Can we find the Area of a square using angle measurement? No, we can’t find the Area of a square using an angle because it has all angles of 90 degrees.
# 5.3: The Uniform Distribution The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive. Example 5.3.1 The data in Table $$\PageIndex{1}$$ are 55 smiling times, in seconds, of an eight-week-old baby. 10.4 19.6 18.8 13.9 17.8 16.8 21.6 17.9 12.5 11.1 4.9 12.8 14.8 22.8 20 15.9 16.3 13.4 17.1 14.5 19 22.8 1.3 0.7 8.9 11.9 10.9 7.3 5.9 3.7 17.9 19.2 9.8 5.8 6.9 2.6 5.8 21.7 11.8 3.4 2.1 4.5 6.3 10.7 8.9 9.4 9.4 7.6 10 3.3 6.7 7.8 11.6 13.8 18.6 The sample mean = 11.49 and the sample standard deviation = 6.23. We will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. This means that any smiling time from zero to and including 23 seconds is equally likely. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution. Let $$X =$$ length, in seconds, of an eight-week-old baby's smile. The notation for the uniform distribution is $$X \sim U(a, b)$$ where $$a =$$ the lowest value of $$x$$ and $$b =$$ the highest value of $$x$$. The probability density function is $$f(x) = \frac{1}{b-a}$$ for $$a \leq x \leq b$$. For this example, $$X \sim U(0, 23)$$ and $$f(x) = \frac{1}{23-0}$$ for $$0 \leq X \leq 23$$. Formulas for the theoretical mean and standard deviation are $\mu = \frac{a+b}{2} \nonumber$ and $\sigma = \sqrt{\frac{(b-a)^{2}}{12}} \nonumber$ For this problem, the theoretical mean and standard deviation are $\mu = \frac{0+23}{2} = 11.50 \, seconds \nonumber$ and $\sigma = \frac{(23-0)^{2}}{12} = 6.64\, seconds. \nonumber$ Notice that the theoretical mean and standard deviation are close to the sample mean and standard deviation in this example. Exercise $$\PageIndex{1}$$ The data that follow are the number of passengers on 35 different charter fishing boats. The sample mean = 7.9 and the sample standard deviation = 4.33. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. State the values of a and $$b$$. Write the distribution in proper notation, and calculate the theoretical mean and standard deviation. 1 12 4 10 4 14 11 7 11 4 13 2 4 6 3 10 0 12 6 9 10 5 13 4 10 14 12 11 6 10 11 0 11 13 2 $$a$$ is zero; $$b$$ is $$14$$; $$X \sim U (0, 14)$$; $$\mu = 7$$ passengers; $$\sigma = 4.04$$ passengers Example 5.3.2A a. Refer to Example 5.3.1. What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds? a. Find $$P(2 < x < 18)$$. $$P(2 < x < 18) = (\text{base})(\text{height}) = (18 – 2)\left(\frac{1}{23}\right) = \left(\frac{16}{23}\right)$$. Exercise $$\PageIndex{2}$$B b. Find the 90th percentile for an eight-week-old baby's smiling time. b. Ninety percent of the smiling times fall below the 90th percentile, $$k$$, so $$P(x < k) = 0.90$$ $P(x < k)= 0.90$ $(\text{base})(\text{height}) = 0.90$ $(k−0)\left(\frac{1}{23}\right) = 0.90$ $k = (23)(0.90) = 20.7$ Exercise $$\PageIndex{3}$$C c. Find the probability that a random eight-week-old baby smiles more than 12 seconds KNOWING that the baby smiles MORE THAN EIGHT SECONDS. c. This probability question is a conditional. You are asked to find the probability that an eight-week-old baby smiles more than 12 seconds when you already know the baby has smiled for more than eight seconds. Find $$P(x > 12 \| x > 8)$$ There are two ways to do the problem. For the first way, use the fact that this is a conditional and changes the sample space. The graph illustrates the new sample space. You already know the baby smiled more than eight seconds. Write a new $$f(x): f(x) = \frac{1}{23-8} = \frac{1}{15}$$ for $$8 < x < 23$$ $$P(x > 12 \| x > 8) = (23 − 12)\left(\frac{1}{15}\right) = \left(\frac{11}{15}\right)$$ For the second way, use the conditional formula from Probability Topics with the original distribution $$X \sim U(0, 23)$$: $$P(\text{A\|B}) = \frac{P(\text{A AND B})}{P(\text{B})}$$ For this problem, $$\text{A}$$ is ($$x > 12$$) and $$\text{B}$$ is ($$x > 8$$). So, $$P(x > 12\|x > 8) = \frac{(x > 12 \text{ AND } x > 8)}{P(x > 8)} = \frac{P(x > 12)}{P(x > 8)} = \frac{\frac{11}{23}}{\frac{15}{23}} = \frac{11}{15}$$ Exercise $$\PageIndex{2}$$ A distribution is given as $$X \sim U(0, 20)$$. What is $$P(2 < x < 18)$$? Find the 90thpercentile. $$P(2 < x < 18) = 0.8$$; 90th percentile $$= 18$$ Example 5.3.3 The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive. Exercise $$\PageIndex{3}$$.1 a. What is the probability that a person waits fewer than 12.5 minutes? a. Let $$X =$$ the number of minutes a person must wait for a bus. $$a = 0$$ and $$b = 15$$. $$X \sim U(0, 15)$$. Write the probability density function. $$f(x) = \frac{1}{15-0} = \frac{1}{15}$$ for $$0 \leq x \leq 15$$. Find $$P(x < 12.5)$$. Draw a graph. $P(x < k) = (\text{base})(\text{height}) = (12.5−0)\left(\frac{1}{15}\right) = 0.8333$ The probability a person waits less than 12.5 minutes is 0.8333. Exercise $$\PageIndex{3}$$.2 b. On the average, how long must a person wait? Find the mean, $$\mu$$, and the standard deviation, $$\sigma$$. b. $$\mu = \frac{a+b}{2} = \frac{15+0}{2} = 7.5$$. On the average, a person must wait 7.5 minutes. $$\sigma = \sqrt{\frac{(b-a)^{2}}{12}} = \sqrt{\frac{(12-0)^{2}}{12}} = 4.3$$. The Standard deviation is 4.3 minutes. Exercise $$\PageIndex{3}$$.3 c. Ninety percent of the time, the time a person must wait falls below what value? Note 5.3.3.3.1 This asks for the 90th percentile. c. Find the 90th percentile. Draw a graph. Let $$k =$$ the 90th percentile. $$P(x < k) = (\text{base})(\text{height}) = (k−0)\left(\frac{1}{15}\right)$$ $$0.90 = (k)\left(\frac{1}{15}\right)$$ $$k = (0.90)(15) = 13.5$$ $$k$$ is sometimes called a critical value. The 90th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes. Exercise $$\PageIndex{4}$$ The total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive. 1. Find $$a$$ and $$b$$ and describe what they represent. 2. Write the distribution. 3. Find the mean and the standard deviation. 4. What is the probability that the duration of games for a team for the 2011 season is between 480 and 500 hours? 5. What is the 65th percentile for the duration of games for a team for the 2011 season? 1. $$a$$ is $$447$$, and $$b$$ is $$521$$. a is the minimum duration of games for a team for the 2011 season, and $$b$$ is the maximum duration of games for a team for the 2011 season. 2. $$X \sim U(447, 521)$$. 3. $$\mu = 484$$, and $$\sigma = 21.36$$ Figure $$\PageIndex{1}$$. 4. $$P(480 < x < 500) = 0.2703$$ 5. 65th percentile is 495.1 hours. Example 5.3.4 Suppose the time it takes a nine-year old to eat a donut is between 0.5 and 4 minutes, inclusive. Let $$X =$$ the time, in minutes, it takes a nine-year old child to eat a donut. Then $$X \sim U(0.5, 4)$$. a. The probability that a randomly selected nine-year old child eats a donut in at least two minutes is _______. Solution a. 0.5714 Exercise $$\PageIndex{4}$$.1 b. Find the probability that a different nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes. The second question has a conditional probability. You are asked to find the probability that a nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes. Solve the problem two different ways (see Example). You must reduce the sample space. First way: Since you know the child has already been eating the donut for more than 1.5 minutes, you are no longer starting at a = 0.5 minutes. Your starting point is 1.5 minutes. Write a new $$f(x)$$: $$f(x) = \frac{1}{4-1.5} = \frac{2}{5}$$ for $$1.5 \leq x \leq 4$$. Find $$P(x > 2\|x > 1.5)$$. Draw a graph. $$P(x > 2\|x > 1.5) = (\text{base})(\text{new height}) = (4 − 2)(25)\left(\frac{2}{5}\right) =$$ ? b. $$\frac{4}{5}$$ The probability that a nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes is $$\frac{4}{5}$$. Second way: Draw the original graph for $$X \sim U(0.5, 4)$$. Use the conditional formula $$P(x > 2 \| x > 1.5) = \frac{P(x > 2 \text{AND} x > 1.5)}{P(x > 1.5)} = \frac{P(x>2)}{P(x>1.5)} = \frac{\frac{2}{3.5}}{\frac{2.5}{3.5}} = 0.8 = \frac{4}{5}$$ Exercise $$\PageIndex{5}$$ Suppose the time it takes a student to finish a quiz is uniformly distributed between six and 15 minutes, inclusive. Let $$X =$$ the time, in minutes, it takes a student to finish a quiz. Then $$X \sim U(6, 15)$$. Find the probability that a randomly selected student needs at least eight minutes to complete the quiz. Then find the probability that a different student needs at least eight minutes to finish the quiz given that she has already taken more than seven minutes. $$P(x > 8) = 0.7778$$ $$P(x > 8 \| x > 7) = 0.875$$ Example 5.3.5 Ace Heating and Air Conditioning Service finds that the amount of time a repairman needs to fix a furnace is uniformly distributed between 1.5 and four hours. Let $$x =$$ the time needed to fix a furnace. Then $$x \sim U(1.5, 4)$$. 1. Find the probability that a randomly selected furnace repair requires more than two hours. 2. Find the probability that a randomly selected furnace repair requires less than three hours. 3. Find the 30th percentile of furnace repair times. 4. The longest 25% of furnace repair times take at least how long? (In other words: find the minimum time for the longest 25% of repair times.) What percentile does this represent? 5. Find the mean and standard deviation Solution a. To find $$f(x): f(x) = \frac{1}{4-1.5} = \frac{1}{2.5}$$ so $$f(x) = 0.4$$ $$P(x > 2) = (\text{base})(\text{height}) = (4 – 2)(0.4) = 0.8$$ b. $$P(x < 3) = (\text{base})(\text{height}) = (3 – 1.5)(0.4) = 0.6$$ The graph of the rectangle showing the entire distribution would remain the same. However the graph should be shaded between $$x = 1.5$$ and $$x = 3$$. Note that the shaded area starts at $$x = 1.5$$ rather than at $$x = 0$$; since $$X \sim U(1.5, 4)$$, $$x$$ can not be less than 1.5. c. $$P(x < k) = 0.30$$ $$P(x < k) = (\text{base})(\text{height}) = (k – 1.5)(0.4)$$ $$0.3 = (k – 1.5) (0.4)$$; Solve to find $$k$$: $$0.75 = k – 1.5$$, obtained by dividing both sides by 0.4 $$k = 2.25$$ , obtained by adding 1.5 to both sides The 30th percentile of repair times is 2.25 hours. 30% of repair times are 2.25 hours or less. d. $$P(x > k) = 0.25$$ $$P(x > k) = (\text{base})(\text{height}) = (4 – k)(0.4)$$ $$0.25 = (4 – k)(0.4)$$; Solve for $$k$$: $$0.625 = 4 − k$$, obtained by dividing both sides by 0.4 $$−3.375 = −k$$, obtained by subtracting four from both sides: $$k = 3.375$$ The longest 25% of furnace repairs take at least 3.375 hours (3.375 hours or longer). Note: Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair times are 3.375 hours or less. 3.375 hours is the 75th percentile of furnace repair times. e. $$\mu = \frac{a+b}{2}$$ and $$\sigma = \sqrt{\frac{(b-a)^{2}}{12}}$$ $$\mu = \frac{1.5+4}{2} = 2.75$$ hours and $$\sigma = \sqrt{\frac{(4-1.5)^{2}}{12}} = 0.7217$$ hours Exercise $$\PageIndex{6}$$ The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. Let $$X =$$ the time needed to change the oil on a car. 1. Write the random variable $$X$$ in words. $$X =$$ __________________. 2. Write the distribution. 3. Graph the distribution. 4. Find $$P(x > 19)$$. 5. Find the 50th percentile. 1. Let $$X =$$ the time needed to change the oil in a car. 2. $$X \sim U(11, 21)$$. 3. Figure $$\PageIndex{7}$$. 4. $$P(x > 19) = 0.2$$ 5. the 50th percentile is 16 minutes. ## Review If $$X$$ has a uniform distribution where $$a < x < b$$ or $$a \leq x \leq b$$, then $$X$$ takes on values between $$a$$ and $$b$$ (may include $$a$$ and $$b$$). All values $$x$$ are equally likely. We write $$X \sim U(a, b)$$. The mean of $$X$$ is $$\mu = \frac{a+b}{2}$$. The standard deviation of $$X$$ is $$\sigma = \sqrt{\frac{(b-a)^{2}}{12}}$$. The probability density function of $$X$$ is $$f(x) = \frac{1}{b-a}$$ for $$a \leq x \leq b$$. The cumulative distribution function of $$X$$ is $$P(X \leq x) = \frac{x-a}{b-a}$$. $$X$$ is continuous. The probability $$P(c < X < d)$$ may be found by computing the area under $$f(x)$$, between $$c$$ and $$d$$. Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height. ## Formula Review $$X =$$ a real number between $$a$$ and $$b$$ (in some instances, $$X$$ can take on the values $$a$$ and $$b$$). $$a =$$ smallest $$X$$; $$b =$$ largest $$X$$ $$X \sim U(a, b)$$ The mean is $$\mu = \frac{a+b}{2}$$ The standard deviation is $$\sigma = \sqrt{\frac{(b-a)^{2}}{12}}$$ Probability density function: $$f(x) = \frac{1}{b-a} \text{for} a \leq X \leq b$$ Area to the Left of $$x$$: $$P(X < x) = (x – a)\left(\frac{1}{b-a}\right)$$ Area to the Right of $$x$$: P($$X$$ > $$x$$) = (b – x)$$\left(\frac{1}{b-a}\right)$$ Area Between $$c$$ and $$d$$: $$P(c < x < d) = (\text{base})(\text{height}) = (d – c)\left(\frac{1}{b-a}\right)$$ Uniform: $$X \sim U(a, b)$$ where $$a < x < b$$ • pdf: $$f(x) = \frac{1}{b-a}$$ for $$a \leq x \leq b$$ • cdf: $$P(X \leq x) = \frac{x-a}{b-a}$$ • mean $$\mu = \frac{a+b}{2}$$ • standard deviation $$\sigma = \sqrt{\frac{(b-a)^{2}}{12}}$$ • $$P(c < X < d) = (d – c)\left(\frac{1}{b-a}\right)$$ ## References McDougall, John A. The McDougall Program for Maximum Weight Loss. Plume, 1995. Use the following information to answer the next ten questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes. 1.5 2.4 3.6 2.6 1.6 2.4 2 3.5 2.5 1.8 2.4 2.5 3.5 4 2.6 1.6 2.2 1.8 3.8 2.5 1.5 2.8 1.8 4.5 1.9 1.9 3.1 1.6 The sample mean = 2.50 and the sample standard deviation = 0.8302. The distribution can be written as $$X \sim U(1.5, 4.5)$$. Exercise $$\PageIndex{7}$$ What type of distribution is this? Exercise $$\PageIndex{8}$$ In this distribution, outcomes are equally likely. What does this mean? It means that the value of x is just as likely to be any number between 1.5 and 4.5. Exercise $$\PageIndex{9}$$ What is the height of $$f(x)$$ for the continuous probability distribution? Exercise $$\PageIndex{10}$$ What are the constraints for the values of $$x$$? $$1.5 \leq x \leq 4.5$$ Exercise $$\PageIndex{11}$$ Graph $$P(2 < x < 3)$$. Exercise $$\PageIndex{12}$$ What is $$P(2 < x < 3)$$? 0.3333 Exercise $$\PageIndex{13}$$ What is $$P(x < 3.5 \| x < 4)$$? Exercise $$\PageIndex{14}$$ What is $$P(x = 1.5)$$? zero Exercise $$\PageIndex{15}$$ What is the 90th percentile of square footage for homes? Exercise $$\PageIndex{16}$$ Find the probability that a randomly selected home has more than 3,000 square feet given that you already know the house has more than 2,000 square feet. 0.6 Exercise $$\PageIndex{17}$$ What is $$a$$? What does it represent? Exercise $$\PageIndex{18}$$ What is $$b$$? What does it represent? $$b$$ is $$12$$, and it represents the highest value of $$x$$. Exercise $$\PageIndex{19}$$ What is the probability density function? Exercise $$\PageIndex{20}$$ What is the theoretical mean? six Exercise $$\PageIndex{21}$$ What is the theoretical standard deviation? Exercise $$\PageIndex{22}$$ Draw the graph of the distribution for $$P(x > 9)$$. Exercise $$\PageIndex{23}$$ Find $$P(x > 9)$$. Exercise $$\PageIndex{24}$$ Find the 40th percentile. 4.8 Use the following information to answer the next eleven exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years. Exercise $$\PageIndex{25}$$ What is being measured here? Exercise $$\PageIndex{26}$$ In words, define the random variable $$X$$. $$X$$ = The age (in years) of cars in the staff parking lot Exercise $$\PageIndex{27}$$ Are the data discrete or continuous? Exercise $$\PageIndex{28}$$ The interval of values for $$x$$ is ______. 0.5 to 9.5 Exercise $$\PageIndex{29}$$ The distribution for $$X$$ is ______. Exercise $$\PageIndex{30}$$ Write the probability density function. $$f(x) = \frac{1}{9}$$ where $$x$$ is between 0.5 and 9.5, inclusive. Exercise $$\PageIndex{31}$$ Graph the probability distribution. 1. Sketch the graph of the probability distribution. Figure $$\PageIndex{10}$$. 2. Identify the following values: 1. Lowest value for $$\bar{x}$$: _______ 2. Highest value for $$\bar{x}$$: _______ 3. Height of the rectangle: _______ 4. Label for x-axis (words): _______ 5. Label for y-axis (words): _______ Exercise $$\PageIndex{32}$$ Find the average age of the cars in the lot. $$\mu$$ = 5 Exercise $$\PageIndex{33}$$ Find the probability that a randomly chosen car in the lot was less than four years old. 1. Sketch the graph, and shade the area of interest. Figure $$\PageIndex{11}$$. 2. Find the probability. $$P(x < 4) =$$ _______ Exercise $$\PageIndex{34}$$ Considering only the cars less than 7.5 years old, find the probability that a randomly chosen car in the lot was less than four years old. 1. Sketch the graph, shade the area of interest. Figure $$\PageIndex{12}$$. 2. Find the probability. $$P(x < 4 \| x < 7.5) =$$ _______ 1. Check student’s solution. 2. $$\frac{3.5}{7}$$ Exercise $$\PageIndex{35}$$ What has changed in the previous two problems that made the solutions different Exercise $$\PageIndex{36}$$ Find the third quartile of ages of cars in the lot. This means you will have to find the value such that $$\frac{3}{4}$$, or 75%, of the cars are at most (less than or equal to) that age. 1. Sketch the graph, and shade the area of interest. Figure $$\PageIndex{13}$$. 2. Find the value $$k$$ such that $$P(x < k) = 0.75$$. 3. The third quartile is _______ 1. Check student's solution. 2. $$k = 7.25$$ 3. $$7.25$$ ## Glossary Conditional Probability the likelihood that an event will occur given that another event has already occurred
# Matrix Multiplication Views: Category: Education ## Presentation Description No description available. By: keshavchawla25 (13 month(s) ago) By: keshavchawla25 (13 month(s) ago) By: NIEC (66 month(s) ago) dasdasdasd By: Fatoom (69 month(s) ago) ## Presentation Transcript ### Matrix Multiplication : Matrix Multiplication How to multiply two matrices ### Slide 2: Matrix multiplication falls into two general categories: Scalar in which a single number is multiplied with every entry of a matrix Multiplication of an entire matrix by another entire matrix For the rest of the page, matrix multiplication will refer to this second category. ### Scalar Matrix Multiplication : Scalar Matrix Multiplication In the scalar variety, every entry is multiplied by a number, called a scalar. ### Slide 4: What is the answer to the scalar multiplication problem below? Solution: ### Matrix Multiplication : Matrix Multiplication You can multiply two matrices if, and only if, the number of columns in the first matrix equals the number of rows in the second matrix. Otherwise, the product of two matrices is undefined. The product matrix's dimensions are (rows of first matrix) × (columns of the second matrix ) ### Slide 6: In the picture on the left, the matrices can be multiplied since the number of columns in the 1st one, matrix A, equals the number of rows in the 2nd, matrix B. The Dimensions of the product matrix Rows of 1st matrix × Columns of 2nd 4 × 3 ### Generalized ExampleIf we multiply a 2×3 matrix with a 3×1 matrix, the product matrix is 2×1 : Generalized ExampleIf we multiply a 2×3 matrix with a 3×1 matrix, the product matrix is 2×1 Here is how we get M11 and M22 in the product. M11 = r11× t11 + r12× t21 + r13×t31 M12 = r21× t11 + r22× t21 + r23×t31 ### Slide 8: Matrix C and D below cannot be multiplied together because the number of columns in C does not equal the number of rows in D. In this case, the multiplication of these two matrices is not defined. ### How To Multiply Matrices : How To Multiply Matrices In order to multiply matrices, Step 1: Make sure that the the number of columns in the 1st one equals the number of rows in the 2nd one. (The pre-requisite to be able to multiply) Step 2: Multiply the elements of each row of the first matrix by the elements of each column in the second matrix. Step 3: Add the products. ### Is the product of matrix A and Matrix B below defined ? : Is the product of matrix A and Matrix B below defined ? ### Since the number of columns in Matrix A does not equal the number of rows in Matrix B. The multiplication of A and B is undefined. : Since the number of columns in Matrix A does not equal the number of rows in Matrix B. The multiplication of A and B is undefined. ### Slide 12: Lets try this one … ### Slide 13: Keep practicing……….
# The tangent ratio This activity is about tangent ratios. Once you complete the activity, the word tangent will make lots of sense to you. This lesson is the beginning of a series of trigonometric lessons I will provide you with that will help you master trigonometry. ## Do the following activity 1. Use a protractor and a ruler to build 3 right triangles of different sizes. The measure of one of the 3 angles should be 45 degrees. I labeled one of the 3 triangles in order to show the leg that is adjacent to the angle of 45 degrees and the leg that is opposiste to the angle of 45 degrees. Notice that 45 degrees is just my choice. You could pick any angle you like to do this activity. 2. Measure the legs of all 3 triangles as accurate as possible. (opposite and adjacent legs only!) For the first triangle on the left, I got 1.6 cm for both opposite and adjacent legs. For the triangle in the middle, I got 2.7 cm for both opposite and adjacent legs. For the triangle on the right, I got 3.8 cm for both opposite and adjacent legs. 3. Compute the ratio: leg opposite 45 degrees / leg adjacent to 45 degrees 1.6 cm / l.6 cm = 1 2.7 cm / 2.7 cm = 1 3.8 cm / 3.8 cm = 1 Notice that the ratio is constant no matter the size of the triangle. If you repeat the activity with an angle of 30 degrees and you made sure that the triangles have different sizes, you will still discover that the ratio is 0.57 for all the triangles no matter the size of the triangles. This specific ratio, also called trigonometric ratio, is called tangent ratio. We then say that tangent of 45 degrees is equal to 1. In short, we can use the symbol tan instead of tangent and write tan (45 degrees) = 1 For the angle of 30 degrees, tan (30 degrees) = 0.57 In general, tangent of angle A = length of leg opposite angle A / length of leg adjacent to A tan(A) =
# How do you find the asymptotes for R(X)= (3x+5)/(x-6)? Aug 12, 2016 vertical asymptote at x = 6 horizontal asymptote at y = 3 #### Explanation: The denominator of R(x) cannot be zero as this would make R(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote. solve : $x - 6 = 0 \Rightarrow x = 6 \text{ is the asymptote}$ Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} , R \left(x\right) \to c \text{ (a constant)}$ divide terms on numerator/denominator by x $\frac{\frac{3 x}{x} + \frac{5}{x}}{\frac{x}{x} - \frac{6}{x}} = \frac{3 + \frac{5}{x}}{1 - \frac{6}{x}}$ as $x \to \pm \infty , R \left(x\right) \to \frac{3 + 0}{1 - 0}$ $\Rightarrow y = 3 \text{ is the asymptote}$ graph{(3x+5)/(x-6) [-20, 20, -10, 10]}
# GED Math : Supplementary Angles ## Example Questions ### Example Question #521 : 2 Dimensional Geometry In degrees, what is the measure of the smallest angle in the figure above? Explanation: Start by solving for . Since all three angles lie on a straight line, the three angles must add up to . We can then write the following equation to solve for : Now, plug in the values of to find the angle measurements. The smallest angle is degrees. ### Example Question #41 : Supplementary Angles Which of the following angles forms a supplementary angle pair with ? Given that Explanation: Which of the following angles forms a supplementary angle pair with ? Given that Supplementary angles sum up to 180 degrees. Therefore, to find our answer, we simply need to do the following: ### Example Question #521 : Geometry And Graphs What is the measure of an angle supplementary to an angle? No solution Explanation: First we need to know that supplementary angles, when added, equal to So when we have the degree measurement of one angle, all we need to do is subtract it from the total (which is 180) and that will give us the measurement of our supplementary angle. ### Example Question #42 : Supplementary Angles In degrees, find the measure of the smallest angle in the figure below. Explanation: Recall that when angles are supplementary, the will add up to . Thus, we can write the following equation: Solve for . Now, the smallest angle in the figure is , . The smallest angle must be . ### Example Question #45 : Supplementary Angles Find the measure of angle B if it is the supplement to angle A: Explanation: If two angles are supplementary, that means the sum of their degrees of measure will add up to 180. In order to find the measure of angle B, subtract angle A from 180 like shown: This gives us a final answer of 65 degrees for angle B. ### Example Question #46 : Supplementary Angles Find the measure of angle B if it is the supplement to angle A: Explanation: If two angles are supplementary, that means the sum of their degrees of measure will add up to 180. In order to find the measure of angle B, subtract angle A from 180 like shown: This gives us a final answer of 83 degrees for angle B. ### Example Question #47 : Supplementary Angles Find the measure of angle B if it is the supplement to angle A: Explanation: If two angles are supplementary, that means the sum of their degrees of measure will add up to 180. In order to find the measure of angle B, subtract angle A from 180 like shown: This gives us a final answer of 172 degrees for angle B. ### Example Question #48 : Supplementary Angles Find the measure of angle B if it is the supplement to angle A: Explanation: If two angles are supplementary, that means the sum of their degrees of measure will add up to 180. In order to find the measure of angle B, subtract angle A from 180 like shown: This gives us a final answer of 51 degrees for angle B. ### Example Question #49 : Supplementary Angles Find the measure of angle B if it is the supplement to angle A: Explanation: If two angles are supplementary, that means the sum of their degrees of measure will add up to 180. In order to find the measure of angle B, subtract angle A from 180 like shown: This gives us a final answer of 92 degrees for angle B. ### Example Question #50 : Supplementary Angles Find the measure of angle B if it is the supplement to angle A:
# Linear Equation 22 Solve : 7 / x + 8 / y = 2 and 2 / x + 12 / y = 20 Solution: Given, 7 / x + 8 / y = 2 ----------------------- (i) and, 2 / x + 12 / y = 20 ----------------------- (ii) Multiplying equation (i) by 2 and equation (ii) by 7, we get 14 / x + 16 / y = 4 --------------------------- (iv) 14 / x + 84 / y = 140 --------------------------- (v) Subtracting (v) from (iv), we get - 68 / y = - 136 => - 136 y = - 68 ; or y = 1 / 2 Substituting y = 1 / 2 in equation (i), we get 7 / x + 8 X 2 / 1 = 2 => 7 / x = 2 - 16 => x = - 1 /2 Therefore, x = - 1 / 2 and y = 1 / 2 Alternative Method: Let 1 / x = a and 1 / y = b Therefore, 7 / x + 8 / y = 2 becomes : 7 a + 8 b = 2 -------- (i) and, 2 / x + 12 / y = 20 becomes : 2 a + 12 b = 20 ---------- (ii) Multiplying equation (i) by 2 and equation (ii) by 7 we get 14 a + 16 b = 4 --------------------------------------------- (iii) 14 a + 84 b = 140 ------------------------------------------- (iv) Subtracting (iv) from (iii) , we get - 68 b = - 136 ; => b = (- 136) / ( - 68) = 2 i.e. 1 / y = 2 and y = 1 / 2 Now, 2 a + 12 b = 20 => 2 a + 12 X 2 = 20 => 2 a + 24 = 20 => 2 a = - 4 and a = - 2 i.e. 1 / x = - 2 and x = - 1 / 2 Therefore, Solution is : x = - 1 / 2 and y = 1 / 2
# Activity: A Walk in the Desert ## Crash! Jade has crash-landed in the desert. There is a village somewhere nearby, direction unknown. So Jade comes up with a cunning plan: • Fill up a water bottle from the plane, and take a compass, • Then walk 1 km north, change direction and walk 2 km east, then 3 km south, 4 km west, 5 km north, 6 km east, and so on, like this: This way Jade will find the village no matter what direction it is in, and can (hopefully) find the way back to the plane for fresh water and shade when needed. But Jade needs to know, at the end of each stage: 1. The total distance walked 2. How far (in a straight line) back to the plane ## OK, Let's Do the Calcs ... After one stage of the journey, Jade has reached point A: • Jade has walked 1 km altogether. • And is 1 km (in a straight line) from the plane. After two stages, Jade has reached point B: • Jade has walked 3 km altogether. • To answer the second question, we make a right-angled triangle OAB: We can calculate the length of OB using Pythagoras' Theorem, as follows: OB2 = OA2 + AB2 OB2 = 12 + 22 OB2 = 1 + 4 OB2 = 5 OB = √5 So the answer in this case is: At Point B, the distance back to the plane (in a straight line) is √5 km After three stages, Jade is at point C: Fill in all the other values ... if it gets hard, read below for some help Point Distance walked altogether Distance (in a straight line) from O O 0 0 A 1 1 B 3 √5 C 6 D E F G H I J ## How To Make It Easier ### Distance walked altogether At the end of each stage, the total distance is the sum of the series 1 + 2 + 3 + 4 + 5 + 6 + ... So just add the new distance each time. OR you can calculate each value using: n(n + 1)/2 where n is the number of stages. Like this: Number of stages (n) Total distance walked = n(n + 1)/2 1 1 × 2 / 2 = 1 2 2 × 3 / 2 = 3 3 3 × 4 / 2 = 6 4 etc ... This works because it is the "Triangular Number Sequence": Try doing the calculations both ways, for fun. ### Distance (in a straight line) from O To work out the distance back to the plane, we can map out the journey on a Coordinate Grid, like this: Now it is just a matter of finding the distance between two points The distance between the points (xA,yA) and (xB,yB) is given by the formula: c = (xA − xB)2 + (yA − yB)2 and one of those points is always the origin, which is at (0,0), so when xB and yB are zero we get: c = x2 + y2 Example, for the point E (−2, 3), x = −2 and y = 3, and so: c = (−22) + 32 = 4+9 = 13 Hopefully that will help make your job easier. ## Direction? There is one more thing Jade must know: to get back to the plane, what compass bearing to use? This will be covered in Activity: A Walk in the Desert 2
Review question When is the volume of this pyramid a maximum? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R9905 Solution A right square pyramidal frame is formed from a rod of length $4a$ by dividing the rod into eight pieces, four being of length $x$ (to make the edges of the square base) and the other four of length $a - x$ (to make the sloping edges). The rods are jointed together without wastage. If the volume of the pyramid is $V$, and $x$ may vary, show that $\dfrac{dV}{dx} = 0$ when $x = 0$ and when $3x^2 - 10ax + 4a^2 = 0.$ The volume $V$ of a right square pyramid is given by the equation $V = \frac{1}{3} \times \text{area of the base} \times \text{height}.$ As the square base comprises four rods, each of length $x$, the area of the base is $x^2$. For the height, consider the following image. We wish to calculate $h$. By Pythagoras’s theorem, we have \begin{align} h^2 + \left( \frac{x}{\sqrt{2}} \right)^2 = (a - x)^2 &\implies h^2 = a^2 - 2ax + x^2 - \frac{x^2}{2} \\ &\implies h = \sqrt{a^2 - 2ax + \frac{x^2}{2}}. \end{align} Thus, we have that \begin{align} V &= \frac{1}{3} \times x^2 \times \sqrt{a^2 - 2ax + \frac{x^2}{2}} \\ &= \frac{1}{3} \sqrt{ a^2x^4 - 2ax^5 + \frac{x^6}{2} }. \end{align} By the chain rule, \begin{align} \frac{dV}{dx} &= \frac{1}{6} \left( a^2x^4 - 2ax^5 + \frac{x^6}{2} \right)^{-1/2} \times \left( 4a^2 x^3 - 10ax^4 + 3x^5 \right) \\ &= \frac{x^3(3x^2 - 10ax + 4a^2)}{6\sqrt{a^2x^4 - 2ax^5 + \frac{x^6}{2}}} \\ &= \frac{x(3x^2 - 10ax + 4a^2)}{6\sqrt{a^2 - 2ax + \frac{x^2}{2}}}. \end{align} So we can see that, $\dfrac{dV}{dx} = 0$ when $x = 0$ or $3x^2 - 10ax + 4a^2 = 0$. Strictly, we should be concerned about the cases when the denominator is close to zero because then our expression is ill-defined. But note that $\sqrt{a^2 - 2ax + \frac{x^2}{2}}$ is the height of the pyramid, and as this approaches zero the volume will be changing rapidly so we can be sure that $\dfrac{dV}{dx} \neq 0$ at this point. Show that, when $x$ is equal to the smaller of the two roots of this equation, 1. a real pyramid can be formed, From the quadratic formula, $3x^2 - 10ax + 4a^2 = 0 \iff x = \frac{10a \pm \sqrt{100a^2 - 48a^2}}{6} = \frac{10 \pm \sqrt{52}}{6} a = \frac{5 \pm \sqrt{13}}{3} a.$ The smaller of these roots is $x=\dfrac{5 - \sqrt{13}}{3} a$. In order to form a real pyramid, $x$ must be between $0$ and $a$. Also, the sloping diagonal, $a-x$, must be longer than the semi-diagonal of the base, $\dfrac{x}{\sqrt{2}}$. Since $5>\sqrt{13}$, we know that $x$ is positive. We also require \begin{align} a-x &> \frac{x}{\sqrt{2}} \\ \iff\quad x\left(1+\frac{1}{\sqrt{2}}\right) &< a \\ \iff\quad x &< \left(2-\sqrt{2}\right)a \end{align} Comparing these surd expressions directly is tricky, so let’s use a calculator to check. $\dfrac{5 - \sqrt{13}}{3} \approx 0.465 \quad\text{and}\quad 2-\sqrt{2} \approx 0.586$ so the diagonal condition is satisfied and we do have a real pyramid. 1. $V$ has a maximum value. As $x \to 0$, the square base goes to a point and the volume of the pyramid shrinks to zero. As $x \to \left(2-\sqrt{2}\right)a$, the sloping diagonal becomes the same length as the semi-diagonal of the base. The height and therefore the volume of the pyramid shrinks to zero. In between these values, a real pyramid exists and $V$ is positive. We know that in this range, we have a single stationary point, which must therefore be a maximum. Another way to check that the volume is maximised at this value is to compute the second derivative, and to verify that it is negative at this value, but this would be demanding… Note that the question doesn’t ask us to calculate the maximum value of $V$ in terms of $a$, but it’s not hard to do with a calculator – we find $V_{max} \approx 0.0304a^3$.
# How do you use DeMoivre's Theorem to find (1+i)^20 in standard form? May 19, 2016 Divide by the norm of $1 + i$ and apply De Moivre's theorem to find that ${\left(1 + i\right)}^{20} = - 1024$ #### Explanation: De Moivre's formula, which can be derived from Euler's formula that ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$, states that ${\left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right)}^{n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right)$ However, as $1 + i$ does not lie on the unit circle, we cannot use the formula directly. To fix this, we can divide $1 + i$ by its norm $\sqrt{2}$ and factor out the necessary constant: ${\left(1 + i\right)}^{20} = {\left(\sqrt{2} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i\right)\right)}^{20}$ $= {\left(\sqrt{2}\right)}^{20} {\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i\right)}^{20}$ $= 1024 {\left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)}^{20}$ $= 1024 \left(\cos \left(20 \cdot \frac{\pi}{4}\right) + i \sin \left(20 \cdot \frac{\pi}{4}\right)\right)$ $= 1024 \left(\cos \left(5 \pi\right) + i \sin \left(5 \pi\right)\right)$ $= 1024 \left(- 1 + i \cdot 0\right)$ $= - 1024$
# McGraw Hill My Math Grade 5 Chapter 11 Lesson 12 Answer Key Convert Metric Units of Mass All the solutions provided inÂ McGraw Hill My Math Grade 5 Answer Key PDF Chapter 11 Lesson 12 Convert Metric Units of Mass will give you a clear idea of the concepts. ## McGraw-Hill My Math Grade 5 Answer Key Chapter 11 Lesson 12 Convert Metric Units of Mass Mass is a measure of the amount of matter an object has. Math in My World Example 1 A white-tailed deer has a mass of 136 kilograms. What is the mass of the deer in grams? Convert 136 kilograms to grams. Since 1 kilogram = 1,000 grams, multiply 136 by 1,000. So, 136 kilograms = ______________ grams. The mass of the white-tailed deer is grams. ______________ Ă· 1,000 = 136 136 kilograms = 136,000 grams. Therefore, the mass of white-tailed deer is 136,000 grams. To check the answer use division 136,000 Ă· 1000 = 136 Example 2 Convert 1,500 grams to kilograms. Since you are converting a smaller unit to a larger unit, divide. The remainder ______________ means there are ______________ grams left over. The decimal part of a kilogram is ______________. So, 1,500 grams = ______________ kilogram ______________ grams or ______________ kilograms. 1 gram (g) is equal to 0.001 kilograms (kg). 1 g = (1 / 1000) kg = 0.001 kg. The massÂ mÂ in kilograms (kg) is equal to the mass m in grams (g) divided by 1000: m(kg) = m(g) / 1000 m(kg) = 1500 / 1000 m(kg) = 15 / 10 = 1.5 The remainder of 500 means there is 500 grams left over. The decimal part of a kilogram is 0.5 s0, 1500 grams = 1 kilogram 500 grams or 1.5 kilograms. Guided Practice Complete. Question 1. 5,000 mg = g 5,000 Ă· 1,000 = _____________ So, 5,000 milligrams equals ___________ grams. Explanation: How to convert milligrams to grams: 1 milligram (mg) is equal to 1/1000 grams (g). 1 mg = (1 / 1000) g = 0.001 grams The massÂ mÂ in grams (g) is equal to the mass m in milligrams (mg) divided by 1000: m(g) = m(mg) / 1000 m(g) = 5000 / 1000 m(g) = 5 Therefore, 5000 mg is equal to 5 grams. Question 2. 5 kg = g 5 Ă— 1,000 = ______________ So, 5 kilograms equals ___________ grams. Explanation: How to convert kilograms to grams: 1 kilogram (kg) is equal to 1000 grams (g). 1 kg = 1000 g The massÂ mÂ in grams (g) is equal to the mass m in kilograms (kg) times 1000: m(g) = m(kg) x 1000 m(g) = 5 x 1000 m(g) = 5000 Therefore, 5 kilograms is equal to 5000 grams. Question 3. 4,000 g = kg 4,000 Ă· 1,000 = ______________ So, 4,000 grams equals ______________ kilograms. Explanation: How to convert grams to kilograms: 1 gram (g) is equal to 0.001 kilograms (kg). 1 g = (1 / 1000) kg = 0.001 kg. The massÂ mÂ in kilograms (kg) is equal to the mass m in grams (g) divided by 1000: m(kg) = m(g) / 1000 m(kg) = 4000 / 1000 m(kg) = 4 Therefore, 4000 grams is equal to 4 kilograms (kg). Question 4. 9 g = mg 9 Ă— 1,000 = ______________ So, 9 grams equals ______________ milligrams. Explanation: How to convert grams to milligrams: 1 gram (g) is equal to 1000 milligrams (mg). 1 g = 1000 mg The massÂ mÂ in milligrams (mg) is equal to the mass m in grams (g) times 1000: m(mg) = m(g) x 1000 m(mg) = 9 x 1000 m(mg) = 9000 Therefore, 9 grams is equal to 9000 milligrams. Talk Math Which is a more reasonable estimate for the mass of a baseball: 140 milligrams, 140 grams, or 140 kilograms? Explain. The mass of a baseball would be measured in grams. A gram is a mass measurement, not a weight. Mass is important to measure because it will measure the amount of force it will take to move the object. Mass is often measured by measuring an unmeasured object with standardized forms of measurements such as milligrams, grams and kilograms. Independent Practice Complete. Question 5. 2,000 mg = ______________ g Explanation: How to convert milligrams to grams: 1 milligram (mg) is equal to 1/1000 grams (g). 1 mg = (1 / 1000) g = 0.001 grams The massÂ mÂ in grams (g) is equal to the mass m in milligrams (mg) divided by 1000: m(g) = m(mg) / 1000 m(g) = 2000 / 1000 m(g) = 2 Therefore, 2000 mg is equal to 2 grams. Question 6. 80 g = ______________ mg Explanation: How to convert grams to milligrams: 1 gram (g) is equal to 1000 milligrams (mg). 1 g = 1000 mg The massÂ mÂ in milligrams (mg) is equal to the mass m in grams (g) times 1000: m(mg) = m(g) x 1000 m(mg) = 80 x 1000 m(mg) = 80000 Therefore, 80 grams is equal to 80000 milligrams. Question 7. 0.75 kg = ______________ mg Explanation: How to convert kilograms to milligrams: 1 kilogram (kg) is equal to 1000000 milligrams (mg). 1 kg = 10^6 = 1000000 mg The massÂ mÂ in milligrams (mg) is equal to the mass m in kilograms (kg) times 1000000: m(mg) = m(kg) x 10^6 = m(kg) x 1000000 m(mg) = 0.75 x 1000000 m(mg) = 750000 Therefore, 0.75 kg is equal to 750000 mg Question 8. 6 kg = ______________ g Explanation: How to convert kilograms to grams: 1 kilogram (kg) is equal to 1000 grams (g). 1 kg = 1000 g The massÂ mÂ in grams (g) is equal to the mass m in kilograms (kg) times 1000: m(g) = m(kg) x 1000 m(g) = 6 x 1000 m(g) = 6000 Therefore, 6 kilograms is equal to 6000 grams. Question 9. 3,100 g = ______________ kg Explanation: How to convert grams to kilograms: 1 gram (g) is equal to 0.001 kilograms (kg). 1 g = (1 / 1000) kg = 0.001 kg. The massÂ mÂ in kilograms (kg) is equal to the mass m in grams (g) divided by 1000: m(kg) = m(g) / 1000 m(kg) = 3100 / 1000 m(kg) = 31 / 10 = 3.1 Therefore, 3100 grams is equal to 3.1 kilograms(kg). Question 10. 0.05 kg = ______________ mg Explanation: How to convert kilograms to milligrams: 1 kilogram (kg) is equal to 1000000 milligrams (mg). 1 kg = 10^6 = 1000000 mg The massÂ mÂ in milligrams (mg) is equal to the mass m in kilograms (kg) times 1000000: m(mg) = m(kg) x 10^6 = m(kg) x 1000000 m(mg) = 0.05 x 1000000 m(mg) = 5/100 x 1000000 m(mg) = 50000 Therefore, 0.05 kg is equal to 50000 mg Question 11. 4.07 g = ______________ mg Explanation: How to convert grams to milligrams: 1 gram (g) is equal to 1000 milligrams (mg). 1 g = 1000 mg The massÂ mÂ in milligrams (mg) is equal to the mass m in grams (g) times 1000: m(mg) = m(g) x 1000 m(mg) = 4.07 x 1000 m(mg) = 4070 Therefore, 4.07 grams is equal to 4070 milligrams. Question 12. 9kg = ______________ g Explanation: How to convert kilograms to grams: 1 kilogram (kg) is equal to 1000 grams (g). 1 kg = 1000 g The massÂ mÂ in grams (g) is equal to the mass m in kilograms (kg) times 1000: m(g) = m(kg) x 1000 m(g) = 9 x 1000 m(g) = 9000 Therefore, 9 kilograms is equal to 9000 grams. Compare. Use <, >, or = to make a true statement. Question 13. 2,300 mg 2 g Explanation: How to convert milligrams to grams: 1 milligram (mg) is equal to 1/1000 grams (g). 1 mg = ( 1 / 1000)g = 0.001 grams The massÂ mÂ in grams (g) is equal to the mass m in milligrams (mg) divided by 1000: m(g) = m(mg) / 1000 m(g) = 2300 / 1000 m(g) = 2.3 2.3 grams is greater than 2 grams Therefore, the answer is greater than (>). Question 14. 3 kg 3,0o0 g Explanation: How to convert kilograms to grams: 1 kilogram (kg) is equal to 1000 grams (g). 1 kg = 1000 g The massÂ mÂ in grams (g) is equal to the mass m in kilograms (kg) times 1000: m(g) = m(kg) x 1000 m(g) = 3 x 1000 m(g) = 3000 Therefore, 3 kilograms is equal to 3000 grams. 3000 grams = 3000 grams Hence, the answer is equal to. Question 15. 4.5 kg 4,050 g Explanation: How to convert kilograms to grams: 1 kilogram (kg) is equal to 1000 grams (g). 1 kg = 1000 g The massÂ mÂ in grams (g) is equal to the mass m in kilograms (kg) times 1000: m(g) = m(kg) x 1000 m(g) = 4.5 x 1000 m(g) = 4500 Therefore, 4.5 kilograms is equal to 4500 grams. 4500 grams is greater than 4050 grams. Hence, the answer is greater than. Question 16. 4,120 mg 4.12 g Explanation: How to convert milligrams to grams: 1 milligram (mg) is equal to 1/1000 grams (g). 1 mg = ( 1 / 1000)g = 0.001 grams The massÂ mÂ in grams (g) is equal to the mass m in milligrams (mg) divided by 1000: m(g) = m(mg) / 1000 m(g) = 4120 / 1000 m(g) = 4.12 4.12 grams is equal to 4.12 grams Therefore, the answer is equal to(=) Question 17. 75 g 800 mg Explanation: As we know, grams are larger than milligrams. so undoubtedly, we can say that 75 grams are larger. 1 gram (g) is equal to 1000 milligrams (mg). 1 g = 1000 mg The massÂ mÂ in milligrams (mg) is equal to the mass m in grams (g) times 1000: m(mg) = m(g) x 1000 m(mg) = 75 x 1000 m(mg) = 75000 Therefore, 75 grams are equal to 75000 milligrams. 75000 mg > 800 mg Hence, the answer is greater than. Question 18. 814 g 8.14 kg Explanation: How to convert grams to kilograms: 1 gram (g) is equal to 0.001 kilograms (kg). 1 g = (1 / 1000) kg = 0.001 kg. The massÂ mÂ in kilograms (kg) is equal to the mass m in grams (g) divided by 1000: m(kg) = m(g) / 1000 m(kg) = 814 / 1000 m(kg) = 0.814 0.814 kg < 8.14 kg Therefore, the answer is less than (<). Problem Solving Use the table shown for Exercises 19-21. Question 19. How many yellow-collared macaws would have a combined mass of 1 kilogram? The mass of yellow-collared species in grams = 250 1 kilogram = 1000 grams The number of yellow-collared macaws would have a combined mass of 1 kilogram. Let it be m(y). m(y) = 1000 / 250 m(y) = 4 Therefore, 4 yellow-collared macaws could have been combined. Question 20. Mathematical PRACTICE Explain to a Friend Is the combined mass of two red-footed macaws and three blue and gold macaws closer to 3 kilograms or 4 kilograms? Explain. The mass of each red-footed macaw = 525 In the given that 2 red-footed macaws. So the total number of red-footed macaws = 525 x 2 = 1050 g The mass of each blue and gold species = 800 In the given that 3 blue and gold species = 800 x 3 = 2400 grams. The total of both species = 1050 + 2400 = 3450 grams. Therefore, the total mass is 3450 grams which rounds to 3000 which is 3 kg. Hence, the species are closer to 3 kilograms. Question 21. Which macaw has a mass closest to 1 kilogram? The green-winged macaw has a mass of 900 grams. we already know that 1 kilogram = 1000 grams 900 is closer to 1000 so, the green-winged macaw is closer to 1 kilogram. HOT Problems Question 22. Mathematical PRACTICE Use Number Sense One pound is approximately equal to 0.5 kilograms. About how many kilograms is 3 pounds? In the above-given question: 0ne pound m(p) = 0.5 kilograms The number of kilograms is 3 pounds. Let it be m(kg) m(kg) = m(p) x 0.5 m(kg) = 3 x 0.5 m(kg) = 1.5 Here we multiplied 3 pounds by 0.5 since there is about 1 pound in 0.5 kg. Question 23. Building on the Essential Question How is converting metric units of mass different from converting customary units of weight? Converting metric units of mass is different from converting customary units of weight because the value of the measurement does not change but the same value is expressed differently as it would be written in different units. ### McGraw Hill My Math Grade 5 Chapter 11 Lesson 12 My Homework Answer Key Practice Complete. Question 1. 7,000 mg = _______________ g Explanation: How to convert milligrams to grams: 1 milligram (mg) is equal to 1/1000 grams (g). 1 mg = ( 1 / 1000)g = 0.001 grams The massÂ mÂ in grams (g) is equal to the mass m in milligrams (mg) divided by 1000: m(g) = m(mg) / 1000 m(g) = 7000 / 1000 m(g) = 7 Therefore, 7000 milligrams is equal to 7 grams. Question 2. 4.7 kg = _______________ g Explanation: How to convert kilograms to grams: 1 kilogram (kg) is equal to 1000 grams (g). 1 kg = 1000 g The massÂ mÂ in grams (g) is equal to the mass m in kilograms (kg) times 1000: m(g) = m(kg) x 1000 m(g) = 4.7 x 1000 m(g) = 4700 Therefore, 4.7 kilograms is equal to 4700 grams. Question 3. 18,500 g = _______________ kg Explanation: How to convert grams to kilograms: 1 gram (g) is equal to 0.001 kilograms (kg). 1 g = (1 / 1000) kg = 0.001 kg. The massÂ mÂ in kilograms (kg) is equal to the mass m in grams (g) divided by 1000: m(kg) = m(g) / 1000 m(kg) = 18,500 / 1000 m(kg) = 185 / 10 = 18.5 Therefore, 18,500 grams is equal to 18.5 kilograms (kg). Question 4. 8.3 kg = _______________ g Explanation: How to convert kilograms to grams: 1 kilogram (kg) is equal to 1000 grams (g). 1 kg = 1000 g The massÂ mÂ in grams (g) is equal to the mass m in kilograms (kg) times 1000: m(g) = m(kg) x 1000 m(g) = 8.3 x 1000 m(g) = 8300 Therefore, 8.3 kilograms is equal to 8300 grams. Question 5. 22 g = _______________ mg Explanation: How to convert grams to milligrams: 1 gram (g) is equal to 1000 milligrams (mg). 1 g = 1000 mg The massÂ mÂ in milligrams (mg) is equal to the mass m in grams (g) times 1000: m(mg) = m(g) x 1000 m(mg) = 22 x 1000 m(mg) = 22000 Therefore, 22 grams is equal to 22000 milligrams. Question 6. 135,000 mg = _______________ kg Explanation: How to convert milligrams to kilograms: 1 milligram (mg) is equal to 1/1000000 kilograms (kg). 1 mg = 10^-6 = 0.000001 kg The massÂ mÂ in kilograms (kg) is equal to the mass m in milligrams (mg) divided by 1000000: m(kg) = m(mg) / 10^6 = m(mg) / 1000000 m(kg) = 135000 / 1000000 m(kg) = 0.135 Therefore, 135000 mg is equal to 0.135 kg. Problem Solving Question 7. One highlighter has a mass of 11 grams. Another highlighter has a mass of 10,800 milligrams. Which highlighter has the greater mass? Answer: 11 grams has the greater mass. Explanation: The above-given: The mass of one highlighter in grams = 11 The mass of another highlighter in milligrams = 10,800 we need to find out the greater highlighter. 11 grams is greater than 10,800 milligrams Explanation: 1 gram (g) is equal to 1000 milligrams (mg). 1 g = 1000 mg The massÂ mÂ in milligrams (mg) is equal to the mass m in grams (g) times 1000: m(mg) = m(g) x 1000 m(mg) = 11 x 1000 m(mg) = 11000 Therefore, 11 grams is equal to 11000 milligrams. 11000 grams is greater than 10,800 milligrams. Question 8. Mathematical PRACTICE Be Precise One computer has a mass of 0.8 kilograms and another has a mass of 800 grams. Compare the masses of the computers. Use >, <, or = to make a true statement. The above-given: The mass of one computer = 0.8 kg The mass of another computer = 800 Now we need to compare the masses. compare kilograms to grams. Explanation: How to convert kilograms to grams: 1 kilogram (kg) is equal to 1000 grams (g). 1 kg = 1000 g The massÂ mÂ in grams (g) is equal to the mass m in kilograms (kg) times 1000: m(g) = m(kg) x 1000 m(g) = 0.8 x 1000 m(g) = 800 Therefore, 0.8 kilograms is equal to 800 grams. 800 grams is equal to 800 grams Hence, 0.8 kg is equal to 800 grams. Vocabulary Check Fill in the correct circle that corresponds to the best answer. Question 9. Which of the following is not a common unit of measurement for the metric system? (A) milligram (B) kilogram (C) gram (D) ounce Though the milligram is the unit of mass, it is not a fundamental unit because all practical units need not be fundamental units. Question 10. Which operation is necessary to convert a larger unit to a smaller unit? (B) subtraction (C) multiplication (D) division The basic conversion rule is: If you need to convert from a larger unit to a smaller unit, multiply. When converting customary units of measure from a larger unit to a smaller unit, multiply the larger unit by its smaller equivalent unit. Test Practice Question 11. For a science experiment. Noel measured a piece of metal that has a mass of 3,500 grams. What is the mass of the metal in kilograms? (A) 0.35 kilogram (B) 3.5 kilograms (C) 35 kilograms (D) 350 kilograms Explanation: The above-given: The mass of the piece of metal in grams = 3500 We need to find out the mass of the metal in kilograms. Let it be m(kg). How to convert grams to kilograms: 1 gram (g) is equal to 0.001 kilograms (kg). 1 g = (1 / 1000) kg = 0.001 kg. The massÂ mÂ in kilograms (kg) is equal to the mass m in grams (g) divided by 1000: m(kg) = m(g) / 1000 m(kg) = 3500 / 1000 m(kg) = 35 / 10 = 3.5 Therefore, the mass of the metal in kilograms is 3.5 Scroll to Top Scroll to Top
# 5th Class Mathematics Data Handling Bar Graph Bar Graph Category : 5th Class ### Bar Graph When the numerical information is represented on the graph using bars, it is known as bar graph. Let us understand it with the help of an example: In the following table number of cars sold by a company in different months of a year has been shown: Month Number of cars January 30 February 60 March 50 April 10 May 50 June 60 July 70 August 100 September 20 October 40 November 60 December 80 Let us make a bar graph using above information: Step 1: Draw one horizontal and one vertical line. Step 2: Choose a convenient scale and mark the numbers on the vertical line at equal interval. Step 3: Now make bars, keeping width of the bars and distance among them uniform. Length of the bars is equal to the respective numbers Read the given bar graph and answer the following questions: (i) Name the student who got the highest mark (ii) Find the marks obtained by Amelia (iii) What scale has been chosen in the bar graph Solution: (i) Katherine got the highest mark (ii) Amelia obtained 70 marks (iii) 1 cm = 10 marks • A bar graph may be either vertical or horizontal • The bar which has highest length represents the greatest amount and the bar which has lest amount. • The information which is in the mumeral form called data. • The initial data that the observer collects himself is called raw data. • When raw data is arranged in table in order to extract the information contained by it easily, is called grouped data. • In the pictograph, data is represented with the help of pictures. • When the numerical information is represented on the graph using bars, it is known as bar graph. In the following pictograph, number of video games sold by a shop has been shown Monday Tuesday Wednesday Thursday Friday Saturday One represents 7 video games (The following questions are based on the above pictograph) Which one of the following days 49 video games were sold? (a) Monday (b) Saturday (c) Thursday (d) Wednesday (e) None of these Answer: (b) Explanation In the pictograph 1 represents 7 video games And $7\times 7=49$ Thus it is Saturday when 49 video games were sold. How many more video games were sold on Thursday than that of Friday? (a) 7 (b) 14 (c) 21 (d) 28 (e) None of these Answer: (c) Which one of the following day's minimum number of video games was sold? (a) Wednesday (b) Monday (c) Saturday (d) Thursday (e) None of these Answer: (a) How many video games were sold on Monday? (a) 7 (b) 14 (c) 21 (d) 28 (e) None of these Answer: (d) How many video games were sold during first four days of the week? (a) 120 (b) 140 (c) 160 (d)180 (e) None of these Answer: (b) How many video games was sold on Friday? (a) 7 (b) 28 (c) 42 (d) 35 (e) None of these Answer: (d) In the following bar graph temperature on different days of a week has been shown. Read the above graph and answer the following questions: Which one of the following days had highest temperature? (a) Monday (b) Tuesday (c) Wednesday (d) Thursday (e) None of these Answer: (d) By how much temperature reduced on Friday in comparison of Thursday? (a)${{3}^{o}}C$ (b) ${{6}^{o}}C$ (c) ${{9}^{o}}C$ (d) ${{12}^{o}}C$ (e) None of these Answer: (b) Which one of the following scale has been chosen in the bar graph? (a)$1cm\text{ }={{3}^{o}}C$ (b) $1cm\text{ }={{6}^{o}}C$ (c)$1cm\text{ }={{9}^{o}}C$ (d) $1cm\text{ }={{12}^{o}}C$ (e) None of these Answer: (a) If the temperature on Monday was ${{15}^{o}}C$ by what cm length of the bar of Monday would had been increased? (a) 1cm (b) 2cm (c) 3cm (d) 4cm (e) None of these Answer: (b) Which one of the following days was the coldest day of the week? (a) Sunday (b) Monday (c) Thursday (d) Tuesday (e) None of these Answer: (a) #### Other Topics You need to login to perform this action. You will be redirected in 3 sec
# 7.1 Solving Systems of Equations by Graphing Save this PDF as: Size: px Start display at page: ## Transcription 1 7.1 Solving Sstems of Equations b Graphing In this chapter we want to take a look at what happens when we put two linear equations, which we talked about in chapter 6, into a single problem. This gives us something called a sstem of equations. Definition: Sstem of Linear Equations- Two or more linear equations involving the same variables. Solutions to sstems are ordered pairs that satisf ever equation in the sstem. The first thing that we need to do is get a grip on the definitions. Eample 1: a. Is (-1, 0) a solution to? b. Is (-1, 7) a solution to? Solution: a. To determine if the ordered pair is a solution, we need to determine if it satisfies each of the equations in the sstem. To do that, we simpl plug the ordered pair in and see if we get a true statement. ( ) ( ) ( ) ( ) So even though the values works in the first equation, it does not work in the second equation. To be a solution, the ordered pair must work in both. So, (-1, 0) is not a solution to the sstem. b. Again, we need to plug the ordered pair in and see what we get. 3 = = 4 3(-1) 7 = -10 3(-1) + (7) = = = 4-10 = = 4 This time, the ordered pair satisfies both equations, so (-1, 7) is a solution to the sstem. So far in this course we have seen a variet of different kinds of equations, and several different possibilities for the number of solutions that these equations could have. This brings up the question, how man solutions can a linear sstem have? Geometricall, since a solution to a sstem is an ordered pair that satisfies the equations, and since an ordered pair that satisfies an equation means that the point which the ordered pair represents is on the corresponding line, it follows that the solution to a sstem points of intersection of the graphs of the equations. 2 Therefore, in asking how man solutions a linear sstem can have, we are reall asking How man was can two lines intersects? The lines here are on top of each other One intersection= No intersections = Infinite intersections= One solution No Solution Infinite solutions Now that we know what the possibilities are, we want to be able to solve sstems. It turns out there are numerous was to solve a linear sstem. We will look at three different techniques in this chapter. The first technique comes right out of what we were just studing in the last chapter. Solving a Linear Sstem b Graphing 1. Graph each equation on the same set of ais. 2. The graphs intersect in: a. One point- one solution (called consistent) b. No points- no solution (called inconsistent) c. All points- infinite solutions (called dependent) 3. If the graphs intersect at one point, estimate the coordinates. Even though this method seems reasonable, it turns out to be not as accurate as the other methods simpl because we have to estimate the answer. Nevertheless, solving b graphing is a good wa to start the discussion of solving sstems. Eample 2: Solve the sstems b graphing. a. b. c. Solution: a. To solve the sstem b graphing, the first thing we need to do is to graph each equation. As we saw in chapter 6, the best wa to do this is with the slope-intercept-intercept method. We, therefore, begin b putting each equation into slope intercept form. 3 So we see that the first line has a slope of ½ and -intercept of (0, 2) and the second line has a slope of ¼ and a -intercept of 2 ½. Now we need to find the -intercepts to use as our check points. As usual we set = 0 and solve for. ( ) ( ) So our -intercepts are (4, 0) for the first line, and (10, 0) for the second. Now we graph the lines on the same coordinate ais. Clearl we have one intersection point at (-2, 3). So the solution to the sstem is (-2, 3). b. Just like in part a, we need to start b graphing our lines. Again, we will get the equations into slope-intercept form. m = 1, -int: (0, -3) m =, -int: (0, ) Then we will get out -intercept. ( ) (3, 0) (, 0) So graphing gives us 4 Therefore our sstem has a solution of (4, 1). c. Lastl we graph each equation, b the same process we used in parts a and b. Start b getting the equations into slope-intercept form. m = -3, -int: (0, 10) m = -3, -int: (0, 2 ½ ) Notice that the lines have the same slope. We should recall from chapter 6 that when two lines have the same slope, the must be parallel. So in this case, the lines could be parallel or the could be the same line. However, since the have different -intercepts, the can t be the same line, ie the must be parallel. Let s finish our graphing to be sure. Find the -intercepts for our check points. ( ) Graphing gives ( ) ( ) Since the lines are clearl parallel, our sstem has no solution. The case where a sstem has infinite solutions happens in a similar situation to Eample 2 part c. above. The onl difference is the equations would have the same slope and same - and - intercepts. 5 7.1 Eercises Determine if the given ordered pair is a solution to the sstem. 1., (-4, -5) 2., (2, 0) 3., (-4, 3) 4., (-1, 3) 5., (-1, 2) 6., (1, 3) 7., (0, 0) 8., (0, 5) 9., (3, 0) 10., (3, 1) 11., (, ) 12., (, ) Solve the sstems b graphing ### SYSTEMS OF LINEAR EQUATIONS SYSTEMS OF LINEAR EQUATIONS Sstems of linear equations refer to a set of two or more linear equations used to find the value of the unknown variables. If the set of linear equations consist of two equations ### 5. Equations of Lines: slope intercept & point slope 5. Equations of Lines: slope intercept & point slope Slope of the line m rise run Slope-Intercept Form m + b m is slope; b is -intercept Point-Slope Form m( + or m( Slope of parallel lines m m (slopes ### 7.3 Solving Systems by Elimination 7. Solving Sstems b Elimination In the last section we saw the Substitution Method. It turns out there is another method for solving a sstem of linear equations that is also ver good. First, we will need ### Section 7.2 Linear Programming: The Graphical Method Section 7.2 Linear Programming: The Graphical Method Man problems in business, science, and economics involve finding the optimal value of a function (for instance, the maimum value of the profit function ### LINEAR FUNCTIONS. Form Equation Note Standard Ax + By = C A and B are not 0. A > 0 LINEAR FUNCTIONS As previousl described, a linear equation can be defined as an equation in which the highest eponent of the equation variable is one. A linear function is a function of the form f ( ) ### Graphing Linear Equations 6.3 Graphing Linear Equations 6.3 OBJECTIVES 1. Graph a linear equation b plotting points 2. Graph a linear equation b the intercept method 3. Graph a linear equation b solving the equation for We are ### The Graph of a Linear Equation 4.1 The Graph of a Linear Equation 4.1 OBJECTIVES 1. Find three ordered pairs for an equation in two variables 2. Graph a line from three points 3. Graph a line b the intercept method 4. Graph a line that ### Alex and Morgan were asked to graph the equation y = 2x + 1 Which is better? Ale and Morgan were asked to graph the equation = 2 + 1 Ale s make a table of values wa Morgan s use the slope and -intercept wa First, I made a table. I chose some -values, then plugged ### Learning Objectives for Section 1.2 Graphs and Lines. Linear Equations in Two Variables. Linear Equations Learning Objectives for Section 1.2 Graphs and Lines After this lecture and the assigned homework, ou should be able to calculate the slope of a line. identif and work with the Cartesian coordinate sstem. ### LINEAR FUNCTIONS OF 2 VARIABLES CHAPTER 4: LINEAR FUNCTIONS OF 2 VARIABLES 4.1 RATES OF CHANGES IN DIFFERENT DIRECTIONS From Precalculus, we know that is a linear function if the rate of change of the function is constant. I.e., for ### Sect The Slope-Intercept Form Concepts # and # Sect. - The Slope-Intercept Form Slope-Intercept Form of a line Recall the following definition from the beginning of the chapter: Let a, b, and c be real numbers where a and b are not ### Warm Up. Write an equation given the slope and y-intercept. Write an equation of the line shown. Warm Up Write an equation given the slope and y-intercept Write an equation of the line shown. EXAMPLE 1 Write an equation given the slope and y-intercept From the graph, you can see that the slope is R. Polnomials In this section we want to review all that we know about polnomials. We start with the basic operations on polnomials, that is adding, subtracting, and multipling. Recall, to add subtract ### Graphing Linear Inequalities in Two Variables 5.4 Graphing Linear Inequalities in Two Variables 5.4 OBJECTIVES 1. Graph linear inequalities in two variables 2. Graph a region defined b linear inequalities What does the solution set look like when ### Reasoning with Equations and Inequalities Instruction Goal: To provide opportunities for students to develop concepts and skills related to solving linear sstems of equations b graphing Common Core Standards Algebra: Solve sstems of equations. ### Systems of Linear Equations: Solving by Substitution 8.3 Sstems of Linear Equations: Solving b Substitution 8.3 OBJECTIVES 1. Solve sstems using the substitution method 2. Solve applications of sstems of equations In Sections 8.1 and 8.2, we looked at graphing ### Essential Question How can you solve a system of linear equations? \$15 per night. Cost, C (in dollars) \$75 per Number of. Revenue, R (in dollars) 5.1 Solving Sstems of Linear Equations b Graphing Essential Question How can ou solve a sstem of linear equations? Writing a Sstem of Linear Equations Work with a partner. Your famil opens a bed-and-breakfast. ### 2.3 Writing Equations of Lines . Writing Equations of Lines In this section ou will learn to use point-slope form to write an equation of a line use slope-intercept form to write an equation of a line graph linear equations using the ### x y The matrix form, the vector form, and the augmented matrix form, respectively, for the system of equations are Solving Sstems of Linear Equations in Matri Form with rref Learning Goals Determine the solution of a sstem of equations from the augmented matri Determine the reduced row echelon form of the augmented ### Slope-Intercept Form and Point-Slope Form Slope-Intercept Form and Point-Slope Form In this section we will be discussing Slope-Intercept Form and the Point-Slope Form of a line. We will also discuss how to graph using the Slope-Intercept Form. ### Slope-Intercept Equation. Example 1.4 Equations of Lines and Modeling Find the slope and the y intercept of a line given the equation y = mx + b, or f(x) = mx + b. Graph a linear equation using the slope and the y-intercept. Determine ### 2.1 Equations of Lines Section 2.1 Equations of Lines 1 2.1 Equations of Lines The Slope-Intercept Form Recall the formula for the slope of a line. Let s assume that the dependent variable is and the independent variable is ### EQUATIONS OF LINES IN SLOPE- INTERCEPT AND STANDARD FORM . Equations of Lines in Slope-Intercept and Standard Form ( ) 8 In this Slope-Intercept Form Standard Form section Using Slope-Intercept Form for Graphing Writing the Equation for a Line Applications (0, ### Systems of Equations Involving Circles and Lines Name: Systems of Equations Involving Circles and Lines Date: In this lesson, we will be solving two new types of Systems of Equations. Systems of Equations Involving a Circle and a Line Solving a system ### Let (x 1, y 1 ) (0, 1) and (x 2, y 2 ) (x, y). x 0. y 1. y 1 2. x x Multiply each side by x. y 1 x. y x 1 Add 1 to each side. Slope-Intercept Form 8 (-) Chapter Linear Equations in Two Variables and Their Graphs In this section Slope-Intercept Form Standard Form Using Slope-Intercept Form for Graphing Writing the Equation for a Line Applications ### Q (x 1, y 1 ) m = y 1 y 0 . Linear Functions We now begin the stud of families of functions. Our first famil, linear functions, are old friends as we shall soon see. Recall from Geometr that two distinct points in the plane determine ### Graphing Linear Equations in Slope-Intercept Form 4.4. Graphing Linear Equations in Slope-Intercept Form equation = m + b? How can ou describe the graph of the ACTIVITY: Analzing Graphs of Lines Work with a partner. Graph each equation. Find the slope ### 3.4 The Point-Slope Form of a Line Section 3.4 The Point-Slope Form of a Line 293 3.4 The Point-Slope Form of a Line In the last section, we developed the slope-intercept form of a line ( = m + b). The slope-intercept form of a line is ### Solving Systems of Linear Equations by Graphing . Solving Sstems of Linear Equations b Graphing How can ou solve a sstem of linear equations? ACTIVITY: Writing a Sstem of Linear Equations Work with a partner. Your famil starts a bed-and-breakfast. It ### SLOPE OF A LINE 3.2. section. helpful. hint. Slope Using Coordinates to Find 6% GRADE 6 100 SLOW VEHICLES KEEP RIGHT . Slope of a Line (-) 67. 600 68. 00. SLOPE OF A LINE In this section In Section. we saw some equations whose graphs were straight lines. In this section we look at graphs of straight lines in more detail ### 5 Systems of Equations Systems of Equations Concepts: Solutions to Systems of Equations-Graphically and Algebraically Solving Systems - Substitution Method Solving Systems - Elimination Method Using -Dimensional Graphs to Approximate ### x x y y Then, my slope is =. Notice, if we use the slope formula, we ll get the same thing: m = Slope and Lines The slope of a line is a ratio that measures the incline of the line. As a result, the smaller the incline, the closer the slope is to zero and the steeper the incline, the farther the ### Solving Special Systems of Linear Equations 5. Solving Special Sstems of Linear Equations Essential Question Can a sstem of linear equations have no solution or infinitel man solutions? Using a Table to Solve a Sstem Work with a partner. You invest ### To Be or Not To Be a Linear Equation: That Is the Question To Be or Not To Be a Linear Equation: That Is the Question Linear Equation in Two Variables A linear equation in two variables is an equation that can be written in the form A + B C where A and B are not ### 2.3 Domain and Range of a Function Section Domain and Range o a Function 1 2.3 Domain and Range o a Function Functions Recall the deinition o a unction. Deinition 1 A relation is a unction i and onl i each object in its domain is paired ### Systems of Equations. from Campus to Careers Fashion Designer Sstems of Equations from Campus to Careers Fashion Designer Radius Images/Alam. Solving Sstems of Equations b Graphing. Solving Sstems of Equations Algebraicall. Problem Solving Using Sstems of Two Equations. ### Solving Equations Involving Parallel and Perpendicular Lines Examples Solving Equations Involving Parallel and Perpendicular Lines Examples. The graphs of y = x, y = x, and y = x + are lines that have the same slope. They are parallel lines. Definition of Parallel Lines ### Introduction. Introduction Introduction Solving Sstems of Equations Let s start with an eample. Recall the application of sales forecasting from the Working with Linear Equations module. We used historical data to derive the equation ### The Point-Slope Form 7. The Point-Slope Form 7. OBJECTIVES 1. Given a point and a slope, find the graph of a line. Given a point and the slope, find the equation of a line. Given two points, find the equation of a line y Slope ### Linear Equations in Two Variables Section. Sets of Numbers and Interval Notation 0 Linear Equations in Two Variables. The Rectangular Coordinate Sstem and Midpoint Formula. Linear Equations in Two Variables. Slope of a Line. Equations ### P1. Plot the following points on the real. P2. Determine which of the following are solutions Section 1.5 Rectangular Coordinates and Graphs of Equations 9 PART II: LINEAR EQUATIONS AND INEQUALITIES IN TWO VARIABLES 1.5 Rectangular Coordinates and Graphs of Equations OBJECTIVES 1 Plot Points in ### Writing the Equation of a Line in Slope-Intercept Form Writing the Equation of a Line in Slope-Intercept Form Slope-Intercept Form y = mx + b Example 1: Give the equation of the line in slope-intercept form a. With y-intercept (0, 2) and slope -9 b. Passing ### Chapter 8. Lines and Planes. By the end of this chapter, you will Chapter 8 Lines and Planes In this chapter, ou will revisit our knowledge of intersecting lines in two dimensions and etend those ideas into three dimensions. You will investigate the nature of planes .4 Graphing Quadratic Equations.4 OBJECTIVE. Graph a quadratic equation b plotting points In Section 6.3 ou learned to graph first-degree equations. Similar methods will allow ou to graph quadratic equations ### Algebra Chapter 6 Notes Systems of Equations and Inequalities. Lesson 6.1 Solve Linear Systems by Graphing System of linear equations: Algebra Chapter 6 Notes Systems of Equations and Inequalities Lesson 6.1 Solve Linear Systems by Graphing System of linear equations: Solution of a system of linear equations: Consistent independent system: ### Solving Systems of Equations Solving Sstems of Equations When we have or more equations and or more unknowns, we use a sstem of equations to find the solution. Definition: A solution of a sstem of equations is an ordered pair that ### Lines and Planes 1. x(t) = at + b y(t) = ct + d 1 Lines in the Plane Lines and Planes 1 Ever line of points L in R 2 can be epressed as the solution set for an equation of the form A + B = C. The equation is not unique for if we multipl both sides b ### 1. a. standard form of a parabola with. 2 b 1 2 horizontal axis of symmetry 2. x 2 y 2 r 2 o. standard form of an ellipse centered Conic Sections. Distance Formula and Circles. More on the Parabola. The Ellipse and Hperbola. Nonlinear Sstems of Equations in Two Variables. Nonlinear Inequalities and Sstems of Inequalities In Chapter, ### { } Sec 3.1 Systems of Linear Equations in Two Variables Sec.1 Sstems of Linear Equations in Two Variables Learning Objectives: 1. Deciding whether an ordered pair is a solution.. Solve a sstem of linear equations using the graphing, substitution, and elimination ### The Slope-Intercept Form 7.1 The Slope-Intercept Form 7.1 OBJECTIVES 1. Find the slope and intercept from the equation of a line. Given the slope and intercept, write the equation of a line. Use the slope and intercept to graph ### C1: Coordinate geometry of straight lines B_Chap0_08-05.qd 5/6/04 0:4 am Page 8 CHAPTER C: Coordinate geometr of straight lines Learning objectives After studing this chapter, ou should be able to: use the language of coordinate geometr find the ### 3. Solve the equation containing only one variable for that variable. Question : How do you solve a system of linear equations? There are two basic strategies for solving a system of two linear equations and two variables. In each strategy, one of the variables is eliminated ### Slope-Intercept Form of a Linear Equation Examples Slope-Intercept Form of a Linear Equation Examples. In the figure at the right, AB passes through points A(0, b) and B(x, y). Notice that b is the y-intercept of AB. Suppose you want to find an equation ### COORDINATE PLANES, LINES, AND LINEAR FUNCTIONS a p p e n d i f COORDINATE PLANES, LINES, AND LINEAR FUNCTIONS RECTANGULAR COORDINATE SYSTEMS Just as points on a coordinate line can be associated with real numbers, so points in a plane can be associated ### GASOLINE The graph represents the cost of gasoline at \$3 per gallon. 9-6 Slope-Intercept Form MAIN IDEA Graph linear equations using the slope and -intercept. New Vocabular slope-intercept form -intercept Math Online glencoe.com Etra Eamples Personal Tutor Self-Check Quiz ### Linear Inequality in Two Variables 90 (7-) Chapter 7 Sstems of Linear Equations and Inequalities In this section 7.4 GRAPHING LINEAR INEQUALITIES IN TWO VARIABLES You studied linear equations and inequalities in one variable in Chapter. ### 9.2 LINEAR PROGRAMMING INVOLVING TWO VARIABLES 86 CHAPTER 9 LINEAR PROGRAMMING 9. LINEAR PROGRAMMING INVOLVING TWO VARIABLES Figure 9.0 Feasible solutions Man applications in business and economics involve a process called optimization, in which we Task Model 3 Equation/Numeric DOK Level 1 algebraically, example, have no solution because 6. 3. The student estimates solutions by graphing systems of two linear equations in two variables. Prompt Features: ### Chapter 3 & 8.1-8.3. Determine whether the pair of equations represents parallel lines. Work must be shown. 2) 3x - 4y = 10 16x + 8y = 10 Chapter 3 & 8.1-8.3 These are meant for practice. The actual test is different. Determine whether the pair of equations represents parallel lines. 1) 9 + 3 = 12 27 + 9 = 39 1) Determine whether the pair ### 3 Rectangular Coordinate System and Graphs 060_CH03_13-154.QXP 10/9/10 10:56 AM Page 13 3 Rectangular Coordinate Sstem and Graphs In This Chapter 3.1 The Rectangular Coordinate Sstem 3. Circles and Graphs 3.3 Equations of Lines 3.4 Variation Chapter ### Reteaching Masters. To jump to a location in this book. 1. Click a bookmark on the left. To print a part of the book. 1. Click the Print button. Reteaching Masters To jump to a location in this book. Click a bookmark on the left. To print a part of the book. Click the Print button.. When the Print window opens, tpe in a range of pages to print. ### 2.6. The Circle. Introduction. Prerequisites. Learning Outcomes The Circle 2.6 Introduction A circle is one of the most familiar geometrical figures and has been around a long time! In this brief Section we discuss the basic coordinate geometr of a circle - in particular ### Classifying Solutions to Systems of Equations CONCEPT DEVELOPMENT Mathematics Assessment Project CLASSROOM CHALLENGES A Formative Assessment Lesson Classifing Solutions to Sstems of Equations Mathematics Assessment Resource Service Universit of Nottingham ### Anytime plan TalkMore plan CONDENSED L E S S O N 6.1 Solving Sstems of Equations In this lesson ou will represent situations with sstems of equations use tables and graphs to solve sstems of linear equations A sstem of equations ### Downloaded from www.heinemann.co.uk/ib. equations. 2.4 The reciprocal function x 1 x Functions and equations Assessment statements. Concept of function f : f (); domain, range, image (value). Composite functions (f g); identit function. Inverse function f.. The graph of a function; its ### SOLVING SYSTEMS OF EQUATIONS SOLVING SYSTEMS OF EQUATIONS 4.. 4..4 Students have been solving equations even before Algebra. Now the focus on what a solution means, both algebraicall and graphicall. B understanding the nature of solutions, ### 5.1. A Formula for Slope. Investigation: Points and Slope CONDENSED CONDENSED L E S S O N 5.1 A Formula for Slope In this lesson ou will learn how to calculate the slope of a line given two points on the line determine whether a point lies on the same line as two given ### Packet 1 for Unit 2 Intercept Form of a Quadratic Function. M2 Alg 2 Packet 1 for Unit Intercept Form of a Quadratic Function M Alg 1 Assignment A: Graphs of Quadratic Functions in Intercept Form (Section 4.) In this lesson, you will: Determine whether a function is linear ### REVIEW OF ANALYTIC GEOMETRY REVIEW OF ANALYTIC GEOMETRY The points in a plane can be identified with ordered pairs of real numbers. We start b drawing two perpendicular coordinate lines that intersect at the origin O on each line. ### Chapter 9. Systems of Linear Equations Chapter 9. Systems of Linear Equations 9.1. Solve Systems of Linear Equations by Graphing KYOTE Standards: CR 21; CA 13 In this section we discuss how to solve systems of two linear equations in two variables ### Solution of the System of Linear Equations: any ordered pair in a system that makes all equations true. Definitions: Sstem of Linear Equations: or more linear equations Sstem of Linear Inequalities: or more linear inequalities Solution of the Sstem of Linear Equations: an ordered pair in a sstem that makes ### Section 1.4 Graphs of Linear Inequalities Section 1.4 Graphs of Linear Inequalities A Linear Inequality and its Graph A linear inequality has the same form as a linear equation, except that the equal symbol is replaced with any one of,, ### GRAPHS OF RATIONAL FUNCTIONS 0 (0-) Chapter 0 Polnomial and Rational Functions. f() ( 0) ( 0). f() ( 0) ( 0). f() ( 0) ( 0). f() ( 0) ( 0) 0. GRAPHS OF RATIONAL FUNCTIONS In this section Domain Horizontal and Vertical Asmptotes Oblique ### Solving Quadratic Equations by Graphing. Consider an equation of the form. y ax 2 bx c a 0. In an equation of the form SECTION 11.3 Solving Quadratic Equations b Graphing 11.3 OBJECTIVES 1. Find an ais of smmetr 2. Find a verte 3. Graph a parabola 4. Solve quadratic equations b graphing 5. Solve an application involving ### Solving Systems of Two Equations Algebraically 8 MODULE 3. EQUATIONS 3b Solving Systems of Two Equations Algebraically Solving Systems by Substitution In this section we introduce an algebraic technique for solving systems of two equations in two unknowns ### 2.6. The Circle. Introduction. Prerequisites. Learning Outcomes The Circle 2.6 Introduction A circle is one of the most familiar geometrical figures. In this brief Section we discuss the basic coordinate geometr of a circle - in particular the basic equation representing ### THE POINT-SLOPE FORM . The Point-Slope Form (-) 67. THE POINT-SLOPE FORM In this section In Section. we wrote the equation of a line given its slope and -intercept. In this section ou will learn to write the equation of a ### Linear Equations. Find the domain and the range of the following set. {(4,5), (7,8), (-1,3), (3,3), (2,-3)} Linear Equations Domain and Range Domain refers to the set of possible values of the x-component of a point in the form (x,y). Range refers to the set of possible values of the y-component of a point in ### Systems of Linear Equations Sstems of Linear Equations. Solving Sstems of Linear Equations b Graphing. Solving Sstems of Equations b Using the Substitution Method. Solving Sstems of Equations b Using the Addition Method. Applications ### Section 1.1 Linear Equations: Slope and Equations of Lines Section. Linear Equations: Slope and Equations of Lines Slope The measure of the steepness of a line is called the slope of the line. It is the amount of change in y, the rise, divided by the amount of ### 7.3 Graphing Rational Functions Section 7.3 Graphing Rational Functions 639 7.3 Graphing Rational Functions We ve seen that the denominator of a rational function is never allowed to equal zero; division b zero is not defined. So, with ### Lines. We have learned that the graph of a linear equation. y = mx +b Section 0. Lines We have learne that the graph of a linear equation = m +b is a nonvertical line with slope m an -intercept (0, b). We can also look at the angle that such a line makes with the -ais. This ### LESSON EIII.E EXPONENTS AND LOGARITHMS LESSON EIII.E EXPONENTS AND LOGARITHMS LESSON EIII.E EXPONENTS AND LOGARITHMS OVERVIEW Here s what ou ll learn in this lesson: Eponential Functions a. Graphing eponential functions b. Applications of eponential ### 1 Functions, Graphs and Limits 1 Functions, Graphs and Limits 1.1 The Cartesian Plane In this course we will be dealing a lot with the Cartesian plane (also called the xy-plane), so this section should serve as a review of it and its ### Identify a pattern and find the next three numbers in the pattern. 5. 5(2s 2 1) 2 3(s 1 2); s 5 4 Chapter 1 Test Do ou know HOW? Identif a pattern and find the net three numbers in the pattern. 1. 5, 1, 3, 7, c. 6, 3, 16, 8, c Each term is more than the previous Each term is half of the previous term; ### EQUATIONS and INEQUALITIES EQUATIONS and INEQUALITIES Linear Equations and Slope 1. Slope a. Calculate the slope of a line given two points b. Calculate the slope of a line parallel to a given line. c. Calculate the slope of a line ### How to Find Equations for Exponential Functions How to Find Equations for Eponential Functions William Cherry Introduction. After linear functions, the second most important class of functions are what are known as the eponential functions. Population ### Ellington High School Principal Mr. Neil Rinaldi Ellington High School Principal 7 MAPLE STREET ELLINGTON, CT 0609 Mr. Dan Uriano (860) 896- Fa (860) 896-66 Assistant Principal Mr. Peter Corbett Lead Teacher Mrs. Suzanne Markowski Guidance ### NAME DATE PERIOD. 11. Is the relation (year, percent of women) a function? Explain. Yes; each year is - NAME DATE PERID Functions Determine whether each relation is a function. Eplain.. {(, ), (0, 9), (, 0), (7, 0)} Yes; each value is paired with onl one value.. {(, ), (, ), (, ), (, ), (, )}. No; in the ### Lesson 22: Solution Sets to Simultaneous Equations Student Outcomes Students identify solutions to simultaneous equations or inequalities; they solve systems of linear equations and inequalities either algebraically or graphically. Classwork Opening Exercise ### Example 1: Model A Model B Total Available. Gizmos. Dodads. System: Lesson : Sstems of Equations and Matrices Outline Objectives: I can solve sstems of three linear equations in three variables. I can solve sstems of linear inequalities I can model and solve real-world ### GRAPHING LINEAR EQUATIONS IN TWO VARIABLES GRAPHING LINEAR EQUATIONS IN TWO VARIABLES The graphs of linear equations in two variables are straight lines. Linear equations may be written in several forms: Slope-Intercept Form: y = mx+ b In an equation ### 135 Final Review. Determine whether the graph is symmetric with respect to the x-axis, the y-axis, and/or the origin. 13 Final Review Find the distance d(p1, P2) between the points P1 and P2. 1) P1 = (, -6); P2 = (7, -2) 2 12 2 12 3 Determine whether the graph is smmetric with respect to the -ais, the -ais, and/or the ### Polynomials. Jackie Nicholas Jacquie Hargreaves Janet Hunter Mathematics Learning Centre Polnomials Jackie Nicholas Jacquie Hargreaves Janet Hunter c 26 Universit of Sdne Mathematics Learning Centre, Universit of Sdne 1 1 Polnomials Man of the functions we will MATH 11 Quadratic Functions and Parabolas A quadratic function has the form Dr. Neal, Fall 2008 f () = a 2 + b + c where a 0. The graph of the function is a parabola that opens upward if a > 0, and opens ### INVESTIGATIONS AND FUNCTIONS 1.1.1 1.1.4. Example 1 Chapter 1 INVESTIGATIONS AND FUNCTIONS 1.1.1 1.1.4 This opening section introduces the students to man of the big ideas of Algebra 2, as well as different was of thinking and various problem solving strategies. ### The Rectangular Coordinate System 3.2 The Rectangular Coordinate Sstem 3.2 OBJECTIVES 1. Graph a set of ordered pairs 2. Identif plotted points 3. Scale the aes NOTE In the eighteenth centur, René Descartes, a French philosopher and mathematician,
# Unit 6 Lesson 1 An inequality is like an equation, but instead of an equal sign (=) it has one of these signs: < : less than ≤ : less than or equal to. ## Presentation on theme: "Unit 6 Lesson 1 An inequality is like an equation, but instead of an equal sign (=) it has one of these signs: < : less than ≤ : less than or equal to."— Presentation transcript: Unit 6 Lesson 1 An inequality is like an equation, but instead of an equal sign (=) it has one of these signs: < : less than ≤ : less than or equal to > : greater than ≥ : greater than or equal to teachers.henrico.k12.va.us/math/int10405/29lessons/29les7/solve_ineq_addNT1.ppt To draw a number line, first draw a line with an arrow on each end. http://www.etap.org/demo/Algebra1/lesson2/instruction1tutor.html Positive numbers increase to the right. Negative numbers “increase” to the left. http://www.etap.org/demo/Algebra1/lesson2/instruction1tutor.html < >    http://www-grms.stjohns.k12.fl.us/teams/Wakulla/032E7289- 0118C716.12/Microsoft%20PowerPoint%20-%20Inequalities%20on%20a%20numberline%20b.pdf 1. Draw a number line. 2. Put a circle on the correct number on the number line. ◦ Open circles for ◦ Closed circles for ≤ and ≥ 3. Draw an arrow in the correct direction on the number line. (a) –4 ___ 12 (b) 23 ___ –21 (c) –145 ___ –211 (d) 29 ___ –32 (e) –101 ___ –110 http://www.etap.org/demo/Algebra1/lesson2/instruction3tutor.html < > > > > m = –1 m = –2 m = 10 m = 1 m = 2 http://www.etap.org/demo/Algebra1/lesson2/instruction3tutor.html 21, –11, 0, 43, 11, –19, 54, –45, –25, 19 http://www.etap.org/demo/Algebra1/lesson2/instruction3tutor.html –45 < –25 < –19 < –11< 0 < 11 < 19 < 21 < 43 < 54 Download ppt "Unit 6 Lesson 1 An inequality is like an equation, but instead of an equal sign (=) it has one of these signs: < : less than ≤ : less than or equal to." Similar presentations
# 12-23c=7(9-c) Payal Khullar \| College Teacher \| (Level 2) Associate Educator Posted on 12 - 23c = 7 (9 - c) Let's start by solving the bracket on RHS: `=>` 12 - 23c = (7X9) - 7c `=>` 12 - 23c = 63 - 7c Regrouping parts of the equation: `=>` -23c + 7c = 63 - 12 Solving for c: `=>` -16c = 51 `=>` c = -51/16 Hence, the value of is -51/16. embizze \| High School Teacher \| (Level 1) Educator Emeritus Posted on Solve 12-23c=7(9-c): 12-23c=7(9-c) Eliminate the parantheses using the distributive property 12-23c=63-7c Add 23c to both sides; subtract 63 from both sides -51=16c Divide both sides by 16 `c=-51/16=-3.1875` ---------------------------------------------------------------- The solution is `c=-51/16=-3.1875` --------------------------------------------------------------- ncchemist \| eNotes Employee Posted on I assume you want to solve the above equation for c. See the work below: 12-23c=7(9-c) multiply the seven out on the right side: 12-23c=63-7c add 23c to both sides to isolate the variable: 12=63-7c+23c=63+16c subtract 63 from both sides: 12-63=-51=16c divide both sides by 16: c=-51/16=-3.1875 jess1999 \| Student, Grade 9 \| (Level 1) Valedictorian Posted on 12 - 23c = 7 (9-c) First distribute the 7 on the left side By distributing, you should get 12 - 23c = 63 - 7c now add 7 on both sides 12 - 16c = 63 now subtract 12 on both sides By subtracting, you should get -16c = 51 now divide -16 on both sides By dividing, you should get idontlikeschool0 \| Student, Grade 11 \| (Level 1) Honors Posted on This is the equation you should solve. 12-23c=7(9-c) you should distribute 7 from the 9 and -c. 12-23c=63-7c -12 -12 -23c=51-7c +7c +7c -16c=51 /-16 /-16 c=-3.1875 So the answer is -3.1875.
Parametric Equations by Hwa Young Lee In this assignment, we are going to investigate parametric equations and parametric curves through four examples. Let's start with an example to get an idea of what parametric equations and parametric curves are. Example 1) I wanted to keep track of the motion of a car while it was continuously moving, as time passes, for 10 minutes. Strangely, I only had a device that kept track of the movement of the car horizontally or vertically. So, I observed the horizontal movement of the car and found that the car was continuously moving east such that the horizontal position of the car at t minutes was x(t)=2t The vertical movement of the car turned out to be continuously northbound, with a vertical position at t minutes expressed as y(t)=t+3 To find out how the car is moving in whole, I plotted the car on the xy-coordinate. Since we are looking at , let's find out a few points within that range. When t=0, then x=0 and y=3 so the car is at (0, 3), the initial point. When t=1, then x=2 and y=4 so the car is at (2, 4). When t=5, then x=10 and y=8 so the car is at (5, 8). When t=10, then x= 20 and y=13 so the car is at (20, 13), the terminal point. We obtain a few points as the following graph: But these are only a few points that show the movement of the car. What would the general (since we know the car is moving continuously in its manner) movement of the car? From x=2t and y=t+3, we can find a relationship between x and y by eliminating the parameter t. Since t=y-3 if we plug this in the first equation, we obtain x=2(y-3) and rearranging this in terms of y, we see that is the function that shows us the movement of the car. Here, we know that the domain is derived due to the range of t. Hence, we confirm that the movement of the car is along the pink line in the figure below, with a direction (north-east bound). From this example, we saw that in a function involving two variables x and y, it is possible to express x and y in terms of a third variable. This third variable is called the parameter, usually denoted as "t". For the two continuous functions x=f(t), y=g(t), substituting values for the parameter t results in ordered pairs (x, y), which form the points on the parametric curve. We call x=f(t), y=g(t) the parametric equations of the parametric curve. Plotting the points as t increases gives the parametric curve a direction and also the range of t is important in graphing the curve. Since we are plotting the points as t moves on the xy-coordinate plane, by finding an equation in terms of x and y by eliminating the parameter, we can graph the parametric equations. However, since the range of t is important, when eliminating the parameter, it is important to check the domain! Let's take a look at another example. Example 2) Janet was lazy laying in bed and was watching the ceiling when she discovered a fly was flying in a strange movement (Does this situation remind you of something? Yes, this is how Descartes first invented the Cartesian plane). She wanted to find out the actual movement of the fly and used a device that gives the horizontal and vertical position at t radians: the parameter for this case would be the angle t of the rotation. As a result, she obtained Can you plot the movement of the fly on the xy-plane? In other words, can you draw the parametric curve of ? Let's plot a few points on the plane. When t=0, x=1 and y=0 so the fly is on (1, 0). But these are only a few points that show the movement of the fly. We can guess that the fly is flying in a circle, but we cannot confirm with just a few points. However, from , we can find a relationship between x and y by eliminating the parameter t. In this case, recall the identity and by eliminating the t, we obtain which gives us the unit circle! In this case, since t ranges from 0 to , we complete the unit circle and the direction of the movement is counterclockwise. ' So the fly was flying in a circle, counterclockwise! Actually, in assignment 3, we did something similar to this to find the locus of the vertices of different parabolas as b varies for . In this case, b was our parameter and by eliminating b, we obtained the parametric curve ( a parabola) with an equation of . We found parametric curves for parametric equations. Conversely, can you find parametric equations for a given parametric curve? Try two examples. Example 3) Write parametric equations of a line segment through (7, 5) with slope of 3. Since we want to see the movement of the points along the line, we first start with the slope. Recall from Example 1) that the slope of the parametric curve was . This was due to the movement of x and y regarding t: x was moving at a rate of 2t while y was moving at a rate of t. So, when x was increasing 1 unit (horizontally), y was increasing unit (vertically). Since the slope of the line segment in this case is 3, this shows us that the points along the line are increasing every 3 units vertically when moving 1 unit horizontally. , for some a, b, and c Now, we want this line segment to pass point (7, 5), which means for a particular t value, we want x=7 and y=5. Hence, and multiplying the first equation through by 3, we get Thus, we can we find the parametric equations for the desired line segment: , for some a and for b, c such that Let's check by picking a=1, b=2, and c=-10. Then and we can check that the graph of this gives us the desired line segment: by restricting the range of t, we can form various segments. Example 4) Matt discovered a lump of gum stuck to the front tire of his bicycle. He became curious of the movement of the gum as the bike was gliding along the road. Assuming the road is flat, the bike is gliding along a straight line, and the tire is a perfect circle with radius 15 inches, what would the locus of the gum (a fixed point on the circle) be? (click here for a GSP animation) The locus of a fixed point on a circle rolling along a line is called a cycloid. Now, try finding parametric equations for the cycloid in Matt's case. The answers are in the final assignment.
# DIVIDING DECIMALS by 10 100 and 1000 Consider the decimal number 25.86. Let us divide 25.86 by 10. = 25.86/10 Whenever a decimal number is divided by 10, the decimal point has to moved to left by one digit. = 2.586 In the similar manner, we can find the results, when a decimal number is divided by 100, 1000, etc., For example, let us divide 385.6 by 100. = 385.6/100 Since we divide 385.6 by 100, the decimal point has to be moved to the left by two digits. = 3.856 Evaluate : 1597.8 ÷ 1000 Here, 15978 is divided by 1000. So, the decimal point has to be moved to the left by three digits. 1597.8 ÷ 1000 = 1.5978 ## Summary ÷ by 10 ----> Move '.' 1 digit to the left ÷ by 100 ----> Move '.' 2 digits to the left ÷ by 1000 ----> Move '.' 3 digits to the left ÷ by 10000 ----> Move '.' 4 digits to the left What if the given number is a whole number (without decimal point? If the given number is a whole number (no decimal point), assume that there is a decimal at the end of the number. For example, let us divide the whole number 35 by 10. Assume there is decimal point in 35 at the end. 35 ÷ 10 = 35. ÷ 10 Since 35. is divided by 10, move the decimal point to the left by one digit. 35 ÷ 10 = 3.5 In other words, when a whole number is divided by 10, take the decimal point such that there is one digit to the right of it. If a whole number is divided by 100, take the decimal point such that there are two digits to the right of the decimal point. ## Solved Problems Problem 1 : Evaluate : 38.6 ÷ 10 Solution : Since 38.6 is divided by 10, move the decimal point to the left by one digit. 38.6 ÷ 10 = 3.86 Problem 2 : Evaluate : 592.7 ÷ 100 Solution : Since 592.7 is divided by 100, move the decimal point to the left by two digits. 592.7 ÷ 100 = 5.927 Problem 3 : Evaluate : 72015.6 ÷ 1000 Solution : Since 72015.6 is divided by 1000, move the decimal point to the left by three digits. 72015.6 ÷ 1000 = 72.0156 Problem 4 : Evaluate : 29105.4 ÷ 10000 Solution : Since 29105.4 is divided by 10000, move the decimal point to the left by four digits. 29105.4 ÷ 10000 = 2.91054 Problem 5 : Evaluate : 23 ÷ 10 Solution : Here 23 is divided by 10 and 23 is a whole number. Since 23 is divided by 10, take decimal point in 23 such that there is one digit to the right of the decimal point. 23 ÷ 10 = 2.3 Problem 6 : Evaluate : 5 ÷ 10 Solution : Here 5 is divided by 10 and 5 is a whole number. Since 5 is divided by 10, take decimal point in 5 such that there is one digit to the right of the decimal point. 5 ÷ 10 = 0.5 Problem 7 : Evaluate : 49 ÷ 100 Solution : Here 49 is divided by 100 and 49 is a whole number. Since 49 is divided by 100, take decimal point in 49 such that there are two digits to the right of the decimal point. 49 ÷ 100 = 0.49 Problem 8 : Evaluate : 3.5 ÷ 100 Solution : Since 49 is divided by 100, decimal point has to be moved to the left by two digits. But, there is only one digit to the left of the decimal point. To get one more digit, add one zero. 3.5 ÷ 100 = 0.035 Problem 9 : Evaluate : 0.7 ÷ 10 Solution : Since 0.7 is divided by 10, decimal point has to be moved to the left by one digit. But, there is no digit to the left of the decimal point. To get one digit, add one zero. 0.7 ÷ 10 = 0.07 Problem 10 : Evaluate : 0.03 ÷ 1000 Solution : Since 0.03 is divided by 1000, decimal point has to be moved to the left by three digits. But, there is no digit to the left of the decimal point. To get three digits, add three zeros. 0.03 ÷ 1000 = 0.00003 Problem 11 : The total cost of 10 fruits is \$42.50. Find the cost of one fruit. Solution : To find the cost of one fruit, divide the total cost of 10 fruits by 10. = 42.50/10 = 42.5 ÷ 10 = ⁴²⁵⁄₁₀ ÷ ¹⁰⁄₁ Change the division to multiplication by taking reciprocal of ¹⁰⁄₁. = ⁴²⁵⁄₁₀ x ⁴²⁵⁄₁₀₀ 425 is a whole number. Since 425 is divided by 100, take decimal point in 429 such that there are two digits to the right of the decimal point. = 4.25 The cost of one fruit is \$4.25. Problem 12 : Mr. Lenin person walks regularly in the morning. If he walks 4283.6 ft. in 100 minutes, find his speed in ft. per minute. Solution : Formula to find speed : Speed = Distance/Time = 4283.6/100 = 4283.6 ÷ 100 ⁴²⁸³⁶⁄₁₀ ÷ ¹⁰⁰⁄₁ Change the division to multiplication by taking reciprocal of ¹⁰⁰⁄₁. ⁴²⁸³⁶⁄₁₀ x ¹⁄₁₀₀ ⁴²⁸³⁶⁄₁₀₀₀ 42836 is a whole number. Since 42836 is divided by 1000, take decimal point in 42836 such that there are three digits to the right of the decimal point. = 42.836 The walking speed of Mr. Lenin is 42.836 ft. per minute. Problem 13 : A car covers a distance of 73.5 miles in 1 hour 40 minutes. Find the speed of the car in miles per minute. Solution : The time 1 hour 40 minutes is in two units, hours and minutes. Convert it to hours completely. 1 hour 40 minutes = (1 x 60) minutes + 40 minutes = 60 minutes + 40 minutes = 100 minutes Given : 73.5 miles of distance covered in 1 hour 15 minutes. 1 hour 40 minutes ----> 73.5 miles 100 minutes ----> 73.5 miles 1 minute ----> (73.5/100) miles = 73.5/100 = 73.5 ÷ 100 ⁷³⁵⁄₁₀ ÷ ¹⁰⁰⁄₁ Change the division to multiplication by taking reciprocal of ¹⁰⁰⁄₁. ⁷³⁵⁄₁₀ x ¹⁄₁₀₀ ⁷³⁵⁄₁₀₀₀ 735 is a whole number. Since 735 is divided by 1000, take decimal point in 735 such that there are three digits to the right of the decimal point. = 0.735 Speed of the car is 0.735 miles per minute. Problem 14 : When the decimal number 2K5.8 is divided by 100, the answer is 2.758. Find the value of K. Solution : When the decimal number 2K5.8 is divided by 100, the decimal point has to be moved to the left by two digits. 2K5.8 ÷ 100 = 2.K58 ----(1) Given : Dividing 2K5.8 by 100 results 2.758. 2K5.8 ÷ 100 = 2.758 ----(2) Comparing (1) and (2), 2.K58 = 2.758 In the equality of two decimal numbers above, comparing the digits at tenths places, K = 7 ## Practice Questions Kindly mail your feedback to v4formath@gmail.com ## Recent Articles 1. ### Algebraic Manipulation Worksheet Jan 28, 23 08:20 AM Algebraic Manipulation Worksheet 2. ### Algebraic Manipulation Problems Jan 28, 23 08:18 AM Algebraic Manipulation Problems
# Difference between revisions of "2019 AMC 10B Problems/Problem 23" ## Problem Points $A=(6,13)$ and $B=(12,11)$ lie on circle $\omega$ in the plane. Suppose that the tangent lines to $\omega$ at $A$ and $B$ intersect at a point on the $x$-axis. What is the area of $\omega$? $\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) } \frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}$ ## Solution 1 First, observe that the two tangent lines are of identical length. Therefore, supposing that the point of intersection is $(x, 0)$, the Pythagorean Theorem gives $\sqrt{(x-6)^2 + 13^2} = \sqrt{(x-12)^2 + 11^2}$. This simplifies to $x = 5$. Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) $AOBX$ is cyclic. Therefore, we can apply Ptolemy's Theorem to give: $2\sqrt{170}r = d \sqrt{40}$, where $r$ is the radius of the circle and $d$ is the distance between the circle's center and $(5, 0)$. Therefore, $d = \sqrt{17}r$. Using the Pythagorean Theorem on the right triangle $OAX$ (or $OBX$), we find that $170 + r^2 = 17r^2$, so $r^2 = \frac{85}{8}$, and thus the area of the circle is $\boxed{\textbf{(C) }\frac{85}{8}\pi}$. ~BakedPotato66 ## Solution 2 (coordinate bash) We firstly obtain $x=5$ as in Solution 1. Label the point $(5,0)$ as $C$. The midpoint $M$ of segment $AB$ is $(9, 12)$. Notice that the center of the circle must lie on the line passing through the points $C$ and $M$. Thus, the center of the circle lies on the line $y=3x-15$. Line $AC$ is $y=13x-65$. Therefore, the slope of the line perpendicular to $AC$ is $-\frac{1}{13}$, so its equation is $y=-\frac{x}{13}+\frac{175}{13}$. But notice that this line must pass through $A(6, 13)$ and $(x, 3x-15)$. Hence $3x-15=-\frac{x}{13}+\frac{175}{13} \Rightarrow x=\frac{37}{4}$. So the center of the circle is $\left(\frac{37}{4}, \frac{51}{4}\right)$. Finally, the distance between the center, $\left(\frac{37}{4}, \frac{51}{4}\right)$, and point $A$ is $\frac{\sqrt{170}}{4}$. Thus the area of the circle is $\boxed{\textbf{(C) }\frac{85}{8}\pi}$. ## Solution 3 The midpoint of $AB$ is $D(9,12)$. Let the tangent lines at $A$ and $B$ intersect at $C(a,0)$ on the $x$-axis. Then $CD$ is the perpendicular bisector of $AB$. Let the center of the circle be $O$. Then $\triangle AOC$ is similar to $\triangle DAC$, so $\frac{OA}{AC} = \frac{AD}{DC}$. The slope of $AB$ is $\frac{13-11}{6-12}=\frac{-1}{3}$, so the slope of $CD$ is $3$. Hence, the equation of $CD$ is $y-12=3(x-9) \Rightarrow y=3x-15$. Letting $y=0$, we have $x=5$, so $C = (5,0)$. Now, we compute $AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}$, $AD=\sqrt{(6-9)^2+(13-12)^2}=\sqrt{10}$, and $DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}$. Therefore $OA = \frac{AC\cdot AD}{DC}=\sqrt{\frac{85}{8}}$, and consequently, the area of the circle is $\pi\cdot OA^2 = \boxed{\textbf{(C) }\frac{85}{8}\pi}$. ## Solution 4 (how fast can you multiply two-digit numbers?) Let $(x,0)$ be the intersection on the x-axis. By Power of a Point Theorem, $(x-6)^2+13^2=(x-12)^2+11^2\implies x=5$. Then the equations for the tangent lines passing $A$ and $B$, respectively, are $13(x-6)+13=y$ and $\frac{11}{7}(x-12)+11=y$. Then the lines normal (perpendicular) to them are $-\frac{1}{13}(x-6)+13=y$ and $-\frac{7}{11}(x-12)+11=y$. Solving for $x$, we have $$-\frac{7}{11}(x-12)+11=-\frac{1}{13}(x-6)+13$$ $$\frac{13\cdot7x-11x}{13\cdot11}=\frac{84\cdot13-6\cdot11-2\cdot11\cdot13}{11\cdot13}$$ $$13\cdot7x-11x=84\cdot13-6\cdot11-2\cdot11\cdot13$$ After condensing, $x=\frac{37}{4}$. Then, the center of $\omega$ is $\left(\frac{37}{4}, \frac{51}{4}\right)$. Apply distance formula. WLOG, assume you use $A$. Then, the area of $\omega$ is $$\sqrt{\frac{1^2}{4^2}+\frac{13^2}{4^2}}^2\pi=\frac{170\pi}{16} \implies \boxed{\textbf{(C) }\frac{85}{8}\pi}.$$ ## Solution 5 (power of a point) Firstly, the point of intersection of the two tangent lines has an equal distance to points $A$ and $B$ due to power of a point theorem. This means we can easily find the point, which is $(5, 0)$. Label this point $X$. $\triangle{XAB}$ is an isosceles triangle with lengths, $\sqrt{170}$, $\sqrt{170}$, and $2\sqrt{10}$. Label the midpoint of segment $AB$ as $M$. The height of this triangle, or $\overline{XM}$, is $4\sqrt{10}$. Since $\overline{XM}$ bisects $\overline{AB}$, $\overleftrightarrow{XM}$ contains the diameter of circle $\omega$. Let the two points on circle $\omega$ where $\overleftrightarrow{XM}$ intersects be $P$ and $Q$ with $\overline{XP}$ being the shorter of the two. Now let $\overline{MP}$ be $x$ and $\overline{MQ}$ be $y$. By Power of a Point on $\overline{PQ}$ and $\overline{AB}$, $xy = (\sqrt{10})^2 = 10$. Applying Power of a Point again on $\overline{XQ}$ and $\overline{XA}$, $(4\sqrt{10}-x)(4\sqrt{10}+y)=(\sqrt{170})^2=170$. Expanding while using the fact that $xy = 10$, $y=x+\frac{\sqrt{10}}{2}$. Plugging this into $xy=10$, $2x^2+\sqrt{10}x-20=0$. Using the quadratic formula, $x = \frac{\sqrt{170}-\sqrt{10}}{4}$, and since $x+y=2x+\frac{\sqrt{10}}{2}$, $x+y=\frac{\sqrt{170}}{2}$. Since this is the diameter, the radius of circle $\omega$ is $\frac{\sqrt{170}}{4}$, and so the area of circle $\omega$ is $\frac{170}{16}\pi = \boxed{\textbf{(C) }\frac{85}{8}\pi}$. ## Solution 6 (Similar to #3) Let the tangent lines from A and B intersect at X. Let the center of $\omega$ be C. Let the intersection of AB and CX be M. Using the techniques above, we get that the coordinate of X is $(5, 0)$. However, notice that CMX is the perpendicular bisector of AB. Thus, AM is the altitude from A to CX. Using the distance formula on AX, we get that the length of $AX=\sqrt{170}=\sqrt{17}\sqrt{10}$. Using the distance formula on AM, we get that $AM=\sqrt{10}$. Using the distance formula on MX, we get that $MX=4\sqrt{10}$. To get AC (the radius of $\omega$), we use either of these methods: Method 1: Since CAX is a right angle, the altitude AM is the geometric mean of XM and MC. We get that $MC=\frac{\sqrt{17}}{4}$. Thus, XC has length $XC=\frac{17\sqrt{10}}{4}$. Using the Pythagorean Theorem on CAX yields $CA=\frac{\sqrt{170}}{4}$. Method 2: Note that CAX and AMX are similar. Thus, $\frac{AM}{MX}=\frac{AC}{AX}$. Solving for AC yields $\frac{AX \cdot AM}{MX}=\frac{\sqrt{170}}{4}$. Using the area formula for a circle yields that the area is $\frac{85\pi}{8} \longrightarrow \boxed{(C)}$. ~Math4Life2020 ## Video Solution For those who want a video solution: (Is similar to Solution 1) https://youtu.be/WI2NVuIp1Ik ~IceMatrix ## Video Solution by The Power of Logic ~The Power of Logic
# Problems based on Ages Today I'm going to discuss some problems based on ages. These are an important questions regarding bank exams and other competitive exams. To solve the problems related to ages, You will require knowledge of linear equations. I have tried to evaluate some easy and quick method for you people to solve these types of questions. ### Some important points • Assume the present ages, then proceed with the equations. • While making equations, always remember to convert both sides into either present year or the given year i.e. time frame should be same. • 5 years ago topresent year - 5 years • 5 years hence to present year + 5 years Problems with Solution Consider the following example: Example: The age of father 2 years ago 7 times the age of his son. At present, fathers age is five times that of his son . What are the ages of father and son at present? Solution: Let present age of son = x present age of father = y Age 2 years ago Present age Son x - 2 x Father 7(x-2) y = 5x Then read the question and write the given equations. Last step, Just make an equation using values of father's age. Either, convert it both into present ages or 2 years ago 2years ago, 7(x-2) = 5x-2.............................................. 5x is present age of father. I converted it into age before 2 years So equation is, 7(x-2) = (5x-2) Solving equation, x =6 yrs Therefore, present age of son = 6yrs and present age of father = 5 times 6 = 30 yrs Example2: The age of father after 2 years is 4 times the age of his son. At present, father's age is five times that of his son . What are the ages of father and son at present? Solution: The difference between both of examples is just ,one is 2 yrs ago and other is 2yrs hence. Let present age of son = x yrs and present age of father = y yrs Present age Age after 2 years Son x x+2 Father y = 5x 4(x+2) Similarly, make an equation using values of father's age. Either, convert it both into present ages or 2 years ago 2years hence, 4(x+ 2) = 5x+2.............................................. 5x is present age of father. I converted it into age after 2 years So equation is, 4(x+2) = (5x+2) Solving equation, x =6 yrs Therefore, present age of son = 6yrs and present age of father = 5 times 6 = 30 yrs. Example3: 2 yrs earlier father's age was 7 times the age of his son. 2 years hence father's age would be 4 times the age of his son. What are the present ages of father and son? Solution: Let present age of son = x yrs and present age of father = y yrs Age 2 years ago Present age Age 2 years hence Son x – 2 x x+2 Father 7(x-2) y 4(x+2) Making an equation using values of father's age. 2years ago, 7(x-2) = 4(x+2) -4 Here,4(x+2) is age after two years.Therefore, I converted it into age before 2 years by subtracted 4 years. So equation is, 7(x-2) =4(x+2)-4 Solving equation, x =6 yrs Therefore, present age of son = 6yrs and present age of father = 7(x-2) + 2 (i.e. age before 2 years + 2 = present age) ⇒ 7(4) + 2 = 28 +2 = 30yrs. ### Shortcut of these questions: Age 2 years ago Present age Age 2 years hence Father’s age as compared to son 7 5 4 1) 2 times (7-1)/(7-5) 2) 2 times (4-1)/(5-4) 3) (2 times (7-1) + 2times (4-1))/ (7-4) Shortcut of these three examples are very simple. Must try it once as it saves your time in exams. You should definitely try this once. But if you are not comfortable then you can do with the previous method. Also check: Tricks of ratio and proportions #### What's trending in BankExamsToday Smart Prep Kit for Banking Exams by Ramandeep Singh - Download here 1. one more shortcut, which is applicable for all age problems. Ex 1: Ratio of father : son 2 years ago 7 : 1 (here difference is 7-1=6) Present 5 : 1 (here difference is 5-1=4) to make same difference, multiply 4 with 7 : 1 and 6 with 5 : 1 Now their ages becomes, 2 years ago 28 : 4 Present 30 : 6 1. what if the difference is same 2. in first example i think it need a correction please correct that one ------let us assume age of son is "x" 2 years ago at that time father's age is 7x,,,,,at present ages of son and father are x+2, 5(x+2),,,,,,,from that statement given we can equalize.........5(x+2)=5x we can solve and we get x=5 that means 2 years ago now 5+2=7years fathers age 5*7=35 I will try to respond asap
US UKIndia Every Question Helps You Learn # Algebra - nth Term This Math quiz is called 'Algebra - nth Term' and it has been written by teachers to help you if you are studying the subject at middle school. Playing educational quizzes is a fabulous way to learn if you are in the 6th, 7th or 8th grade - aged 11 to 14. It costs only \$12.50 per month to play this quiz and over 3,500 others that help you with your school work. You can subscribe on the page at Join Us The letter n (usually in italics) is often used in algebra to indicate the position of a term in a sequence. This can be a little difficult to grasp but work through this Math quiz (and read the helpful comments!) and you will soon get the idea. 1. What is the nth term of the sequence 6, 11, 16, 21? 5n 5n + 1 5n + 5 n + 5 (5 x n) + 1 2. The nth term of a sequence is (2n)2 + 1. What is the the numeric value of the 3rd term? 33 35 37 39 (2 x 3)2 + 1 3. What is the next number in this sequence: 1, 1, 2, 3, 5, 8? 12 13 14 15 This is known as the Fibonacci sequence. To find each number you add together the two previous numbers. After 13 comes 21 (8 + 13) and then comes 34 (21 + 13) 4. What is the next number in this sequence: 1, 4, 9, 16? 18 20 22 25 These are known as square numbers. The 1st number in the sequence is 1 x 1; the 2nd number is 2 x 2; the 3rd number is 3 x 3 etc. 5. What is the next number in this sequence: 1, 8, 27? 35 36 54 64 These are known as cube numbers. The 1st number in the sequence is 1 x 1 x 1; the 2nd number is 2 x 2 x 2; the 3rd number is 3 x 3 x 3 and the 4th number will be 4 x 4 x 4 6. The nth term of a sequence is given by 3n + 1. What is the numeric value of the 8th term? 12 13 25 50 We are told that we need the 8th term. We put an 8 where the n is and then work it out: (3 x 8) + 1 7. The nth term of a sequence is given by 3n + 1. What is the numeric value of the 1st term? 3 4 6 9 In the 1st term the value of n will be 1 and therefore the answer can be found as follows: (3 x 1) + 1 8. The nth term of a sequence is given by 3n + 1. What is the numeric value of the 3rd term? 4 6 8 10 The answer can be found as follows: (3 x 3) + 1 9. The nth term of a sequence is (n + 1)2. What is the the numeric value of the 4th term? 20 25 30 35 (4 + 1)2 10. The nth term of a sequence is given by 14n - 6. What is the numeric value of the 2nd term? 8 20 22 24 (14 x 2) - 6 Author: Frank Evans
# Difference between revisions of "2018 AMC 12B Problems/Problem 16" ## Problem The solutions to the equation $(z+6)^8=81$ are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled $A,B,$ and $C$. What is the least possible area of $\triangle ABC?$ $\textbf{(A) } \frac{1}{6}\sqrt{6} \qquad \textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2} \qquad \textbf{(C) } 2\sqrt3-3\sqrt2 \qquad \textbf{(D) } \frac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1$ ## Solution The answer is the same if we consider $z^8=81.$ Now we just need to find the area of the triangle bounded by $\sqrt 3i, \sqrt 3,$ and $\frac{\sqrt 3}{\sqrt 2}+\frac{\sqrt 3}{\sqrt 2}i.$ This is just $\boxed{\textbf{B}}.$ ## Solution 2 (Understanding the polygon) The polygon formed will be a regular octagon since there are $8$ roots of $z^8=81$. By normal math computation, we can figure out that two roots of $z^8=81$ are $\sqrt{3}$ and $-\sqrt{3}$. These will lie on the real axis of the plane. Since it's a regular polygon, there has to be points on the vertical plane also which will be $\sqrt{3}i$ and $-\sqrt{3}i$. Clearly, the rest of the points will lie in each quadrant. The next thing is to get their coordinates (note that to answer this question, we do not need all the coordinates, only 3 consecutive ones are needed). The circumcircle of the octagon will have the equation $i^2+r^2=3$. The coordinates of the point in the first quadrant will be equal in magnitude and both positive, so $i=r$. Solving gives $i=r=\frac{\sqrt{3}}{\sqrt{2}}$ (meaning that the root represented is $\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i$). This way we can deduce the values of the $8$ roots of the equation to be $\sqrt{3},-\sqrt{3},-\sqrt{3}i,\sqrt{3}i,\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i,-\frac{\sqrt{3}}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}}i,\frac{\sqrt{3}}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}}i,-\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i$. To get the area, $3$ consecutive points such as $\sqrt{3},$ $\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i,$ and $\sqrt{3}i$ can be used. The area can be computed using different methods like using the shoelace formula, or subtracting areas to find the area. The answer you get is $\boxed{B}$. (This method is not actually as long as it seems if you understand what you're doing while doing it. Also calculations can be made a little easier by solving using $x^8=1$ and multiplying your answer by $\sqrt{3}$). ~OlutosinNGA ## Solution 3 (Roots of Unity) Now, we need to solve the equation $(z+6)^8 = 81$ where $z = a+bi$. We can substitute this as $$(a+6+bi)^8 = 81$$ Now, let $a+6 = q$ for some $q \in \mathbb{Z}$. Thus, the equation becomes $((q+bi)^2)^4 = 81$. Taking it to the other side, we get the equation to be $(q+bi)^2 = 3$. Rearranging variables, we get $(q+bi) = \sqrt{3}$. Plotting this in the complex place, this is a circle centered at the origin and of radius \sqrt{3}. The graph of the original equation $(a+6+bi) = \sqrt{3}$ is merely a transformation which doesn't change affect the area. Thus, we can find the minimum area of the transformed equation $(q+bi)^2 = 3$. Using Roots of Unity, we know that the roots of the equation lie at $0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4} ... 2\pi$ radians from the origin. We can quickly notice that the area of the roots will be smallest with points at $\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}$. Using trigonometry, we get the respective roots to be $(Re(z), Im(z)) \in \{(\sqrt{3},0), (\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}), (0,\sqrt{3})\}$. Using the shoelace method, the area quicklyy comes out to be $\frac{3\sqrt{2}-3}{2} \implies \boxed{\textbf{B}}.$ 2018 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 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 12 Problems and Solutions
# Mathematical Proof/Methods of Proof/Proof by Induction The beauty of induction is that it allows a theorem to be proven true where an infinite number of cases exist without exploring each case individually. Induction is analogous to an infinite row of dominoes with each domino standing on its end. If you want to make all the dominoes fall, you can either: 1. push on the first one, wait to see what happens, and then check each domino afterwards (which may take a long time if there's an infinite number of dominoes!) 2. or you can prove that if any domino falls, then it will cause the domino after it to fall. (i.e. if the first one falls then the second one will fall, and if the second one falls then the third one will fall, etc.) Induction, essentially, is the methodology outlined in point 2. ## Parts of InductionEdit Induction is composed of three parts: 1. The Base Case (in the domino analogy, this shows the first domino will fall) 2. The Induction Hypothesis (in the domino analogy, we assume that a particular domino will fall) 3. The Inductive Step (in the domino analogy, we prove that the domino we assume will fall will cause the next domino to fall) ## Weak InductionEdit Weak induction is used to show that a given property holds for all members of a countable inductive set, this usually is used for the set of natural numbers. Weak induction for proving a statement $P(n)$ (that depends on $n$) relies on two steps: • $P(n)$ is true for a certain base step. Usually the base case is $n=1$ or $n=0$ • $P(k)\Rightarrow P(k+1)$. That is, given that $P(k)$ is true, $P(k+1)$ is also true. If these two properties hold, one may induce that the property holds for all elements in the set in question. Returning to the example, if you are sure that you called your neighbor, and you knew that everyone who was called in turn called his/her neighbor, then you would be guaranteed that everyone on the block had been called (assuming you had a linear block, or that it curved around nicely). ### ExamplesEdit The first example of a proof by induction is always 'the sum of the first n terms:' Theorem 2.4.1. For any fixed $n\in \mathbb N,$ $\sum_{i=1}^{n}{i}=\frac{n(n+1)}{2}$ Proof: • Base step: $1=\frac{1\cdot 2}{2}$, therefore the base case holds. • Inductive step: Assume that $\sum_{i=0}^{n}{i}=\frac{n(n+1)}{2}$. Consider $\sum_{i=1}^{n+1}{i}$. $\sum_{i=1}^{n+1}{i}=\sum_{i=1}^n i + (n+1) = \frac{n(n+1)}{2}+n+1$ $=(\frac{n}{2}+1)(n+1)$ $=\frac{(n+1)(n+2)}{2}$ $=\frac{(n+1)((n+1)+1)}{2}$ So the inductive case holds. Now by induction we see that the theorem is true. ## Reverse InductionEdit Reverse induction is a seldom-used method of using an inductive step that uses a negative in the inductive step. It is a minor variant of weak induction. The process still applies only to countable sets, generally the set of whole numbers or integers, and will frequently stop at 1 or 0, rather than working for all positive numbers. Reverse induction works in the following case. • The property holds for a given value, say $M$. • Given that the property holds for a given case, say $n=k+1$, Show that the property holds for $n=k$. Then the property holds for all values $n\le M$. I learnt today from http://people.math.carleton.ca/~ckfong/hs14a.pdf (see example there) that reverse indcution is also usable in the general case: "to establish the validity of a sequence of propositions Pn (n ≥ 1), it is enough to establish the following (a) Pn is valid for inﬁnitely many n. (b) If Pn+1 is valid, then so is Pn. It can be the case that we can easily prove P1 and if P for n=m so P for n=2m. In this case we have (a) for the infinitely many n = 2 exp k (for k >= 0). ## Strong InductionEdit In weak induction, for the inductive step, we only required that for a given $n$, its immediate predecessor ($n-1$) satisfies the theorem (i.e., $P(n-1)$ is true). In strong induction, we require that not only the immediate predecessor, but all predecessors of $n$ satisfy the theorem. The variation in the inductive step is: • If $P(k)$ is true for all $k then $P(n)$ is true. The reason this is called strong induction is fairly obvious--the hypothesis in the inductive step is much stronger than the hypothesis is in the case of weak induction. Of course, for finite induction it turns out to be the same hypothesis, but in the case of transfinite sets, weak induction is not even well-defined, since some sets have elements that do not have an immediate predecessor. ## Transfinite InductionEdit Used in proving theorems involving transfinite cardinals. This technique is used in set theory to prove properties of cardinals, since there is rarely another way to go about it. ## Inductive SetEdit We first define the notion of a well-ordered set. A set $X$ is well-ordered if there is a total order < on $X$ and that whenever $Y\subset X$ is non-empty, there is a least-element in $Y$. That is, $\exists p\in Y$ such that $p. An inductive set is a set $A\subset X$ such that the following hold: 1. $\alpha\in A$ (where $\alpha$ is the least element of $X$) 2. If $\beta\in A$ then $\forall\gamma\in X$ such that $\beta < \gamma, \gamma\in A$ Of course, you look at that and say "Wait a minute. That means that $A=X$!" And, of course you'd be right. That's exactly why induction works. The principle of induction is the theorem that says: Theorem 2.4.2. If $X$ is a non-empty well-ordered set and $A\subset X$ is an inductive subset of $X$ then $A=X$. The proof of this theorem is left as a very simple exercise. Here we note that the set of natural numbers is clearly well-ordered with the normal order that you are familiar with, so $\mathbb{N}$ is an inductive set. If you accept the axiom of choice, then it follows that every set can be well-ordered.
Difference between revisions of "1983 AIME Problems/Problem 15" Problem The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$. Suppose that the radius of the circle is $5$, that $BC=6$, and that $AD$ is bisected by $BC$. Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$. It follows that the sine of the central angle of minor arc $AB$ is a rational number. If this number is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$? $[asy]size(100); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("A",A,W); label("B",B,W); label("C",C,NE); label("D",D,N); draw(A--D); draw(B--C); pair F=intersectionpoint(A--D,B--C); add(pathticks(A--F,1,0.5,0,3.5)); add(pathticks(F--D,1,0.5,0,3.5)); [/asy]$ Solution Solution 1 -Credit to Adamz for diagram- $[asy] size(10cm); import olympiad; pair O = (0,0);dot(O);label("O",O,SW); pair M = (4,0);dot(M);label("M",M,SE); pair N = (4,2);dot(N);label("N",N,NE); draw(circle(O,5)); pair B = (4,3);dot(B);label("B",B,NE); pair C = (4,-3);dot(C);label("C",C,SE); draw(B--C);draw(O--M); pair P = (1.5,2);dot(P);label("P",P,W); draw(circle(P,2.5)); pair A=(3,4);dot(A);label("A",A,NE); draw(O--A); draw(O--B); pair Q = (1.5,0); dot(Q); label("Q",Q,S); pair R = (3,0); dot(R); label("R",R,S); draw(P--Q,dotted); draw(A--R,dotted); pair D=(5,0); dot(D); label("D",D,E); draw(A--D); [/asy]$Let $A$ be any fixed point on circle $O$, and let $AD$ be a chord of circle $O$. The locus of midpoints $N$ of the chord $AD$ is a circle $P$, with diameter $AO$. Generally, the circle $P$ can intersect the chord $BC$ at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle $P$ is tangent to $BC$ at point $N$. Let $M$ be the midpoint of the chord $BC$. From right triangle $OMB$, we have $OM = \sqrt{OB^2 - BM^2} =4$. This gives $\tan \angle BOM = \frac{BM}{OM} = \frac 3 4$. Notice that the distance $OM$ equals $PN + PO \cos \angle AOM = r(1 + \cos \angle AOM)$, where $r$ is the radius of circle $P$. Hence $$\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5$$ (where $R$ represents the radius, $5$, of the large circle given in the question). Therefore, since $\angle AOM$ is clearly acute, we see that $$\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3$$ Next, notice that $\angle AOB = \angle AOM - \angle BOM$. We can therefore apply the subtraction formula for $\tan$ to obtain $$\tan \angle AOB =\frac{\tan \angle AOM - \tan \angle BOM}{1 + \tan \angle AOM \cdot \tan \angle BOM} =\frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}$$ It follows that $\sin \angle AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}$, such that the answer is $7 \cdot 25=\boxed{175}$. Solution 2 This solution, while similar to Solution 1, is far more motivated and less contrived. Firstly, we note the statement in the problem that "$AD$ is the only chord starting at $A$ and bisected by $BC$" – what is its significance? What is the criterion for this statement to be true? We consider the locus of midpoints of the chords from $A$. It is well-known that this is the circle with diameter $AO$, where $O$ is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with scale factor $\frac{1}{2}$ and center $A$. Thus, the locus is the result of the dilation with scale factor $\frac{1}{2}$ and centre $A$ of circle $O$. Let the center of this circle be $P$. Now, $AD$ is bisected by $BC$ if they cross at some point $N$ on the circle. Moreover, since $AD$ is the only chord, $BC$ must be tangent to the circle $P$. The rest of this problem is straightforward. Our goal is to find $\sin \angle AOB = \sin{\left(\angle AOM - \angle BOM\right)}$, where $M$ is the midpoint of $BC$. We have $BM=3$ and $OM=4$. Let $R$ be the projection of $A$ onto $OM$, and similarly let $Q$ be the projection of $P$ onto $OM$. Then it remains to find $AR$ so that we can use the addition formula for $\sin$. As $PN$ is a radius of circle $P$, $PN=2.5$, and similarly, $PO=2.5$. Since $OM=4$, we have $OQ=OM-QM=OM-PN=4-2.5=1.5$. Thus $PQ=\sqrt{2.5^2-1.5^2}=2$. Further, we see that $\triangle OAR$ is a dilation of $\triangle OPQ$ about center $O$ with scale factor $2$, so $AR=2PQ=4$. Lastly, we apply the formula: $$\sin{\left(\angle AOM - \angle BOM\right)} = \sin \angle AOM \cos \angle BOM - \sin \angle BOM \cos \angle AOM = \left(\frac{4}{5}\right)\left(\frac{4}{5}\right)-\left(\frac{3}{5}\right)\left(\frac{3}{5}\right)=\frac{7}{25}$$ Thus the answer is $7\cdot25=\boxed{175}$. Solution 3 (coordinate geometry) Let the circle have equation $x^2 + y^2 = 25$, with centre $O(0,0)$. Since $BC=6$, we can calculate (by the Pythagorean Theorem) that the distance from $O$ to the line $BC$ is $4$. Therefore, we can let $B=(3,4)$ and $C=(-3,4)$. Now, assume that $A$ is any point on the major arc BC, and $D$ any point on the minor arc BC. We can write $A=(5 \cos \alpha, 5 \sin \alpha)$, where $\alpha$ is the angle measured from the positive $x$ axis to the ray $OA$. It will also be convenient to define $\angle XOB = \alpha_0$. Firstly, since $B$ must lie in the minor arc $AD$, we see that $\alpha \in \left(-\frac{\pi}{2}-\alpha_0,\alpha_0\right)$. However, since the midpoint of $AD$ must lie on $BC$, and the highest possible $y$-coordinate of $D$ is $5$, we see that the $y$-coordinate cannot be lower than $3$, that is, $\alpha \in \left[\sin^{-1}\frac{3}{5},\alpha_0\right)$. Secondly, there is a theorem that says that, in a circle, if a chord is bisected by a radius, then they must be perpendicular. Therefore, suppose that $P$ is the intersection point of $AD$ and $BC$, so that by the theorem, $OP$ is perpendicular to $AD$. So, if $AD$ is the only chord starting at $A$ which is bisected by $BC$, this means that $P$ is the only point on the chord $BC$ such that $OP$ is perpendicular to $AD$. Now suppose that $P=(p,4)$, where $p \in (-3,3)$. The fact that $OP$ must be perpendicular to $AD$ is equivalent to the following equation: $$-1 = \left(\text{slope of } OP\right)\left(\text{slope of } AP\right)$$ which becomes $$-1 = \frac{4}{p} \cdot \frac{5\sin \alpha - 4}{5\cos \alpha - p}$$ This rearranges to $$p^2 - (5\cos \alpha)p + 16 - 20 \sin \alpha = 0$$ Given that this equation must have only one real root $p\in (-3,3)$, we study the following function: $$f(x) = x^2 - (5\cos \alpha)x + 16 - 20 \sin \alpha$$ First, by the fact that the equation $f(x)=0$ has real solutions, its discriminant $\Delta$ must be non-negative, so we calculate $$\begin{split}\Delta & = (5\cos \alpha)^2 - 4(16-20\sin \alpha) \\ & = 25 (1- \sin^2 \alpha) - 64 + 80 \sin \alpha \\ & = -25 \sin^2 \alpha + 80\sin \alpha - 39 \\ & = (13 - 5\sin \alpha)(5\sin \alpha - 3)\end{split}$$ It is obvious that this is in fact non-negative. If it is actually zero, then $\sin \alpha = \frac{3}{5}$, and $\cos \alpha = \frac{4}{5}$. In this case, $p = \frac{5\cos \alpha}{2} = 2 \in (-3,3)$, so we have found a possible solution. We thus calculate $\sin(\text{central angle of minor arc } AB) = \sin (\alpha_0 - \alpha) = \frac{4}{5}\cdot \frac{4}{5} - \frac{3}{5} \cdot \frac{3}{5} = \frac{7}{25}$ by the subtraction formula for $\sin$. This means that the answer is $7 \cdot 25 = 175$. Since this is an AIME problem, we can assume that we are done since we have found one possible case. However, in a full-solution contest, we could not assume that the answer is unique, but would need to prove that this is the unique solution. This can be proven as follows. Suppose that $\Delta > 0$, which would mean that there could be two real roots of $f(x)$, one lying in the interval $(-3,3)$, and another outside of it. We also see, by Vieta's Formulas, that the average of the two roots is $\frac{5\cos \alpha}{2}$, which is non-negative, so the root outside of $(-3,3)$ must be no less than $3$. By considering the graph of $y=f(x)$, which is a "U-shaped" parabola, it is now evident that $f(-3) > 0$ and $f(3)\leq 0$. We can just use the second inequality: $$0 \geq f(3) = 25 - 15\cos \alpha - 20 \sin \alpha$$ so $$3\cos \alpha + 4 \sin \alpha \geq 5$$ The only way for this inequality to be satisfied is when $A=B$ (by applying the Cauchy-Schwarz inequality, or just plotting the line $3x+4y=5$ to see that point $A$ can't be above this line), which does not make sense in the original problem statement. (For it would mean that the point $A$ lies in the half-plane above the line $3x+4y=5$, inclusive, and the half-plane below the line $-3x+4y=5$, exclusive. This can be seen to be impossible by drawing the lines and observing that the intersection of the two half-planes does not share any point with the circle.) Solution 4 Let the center of the circle be $O$. Fix $B,C,$ and $A$. Then, as $D$ moves around the circle, the locus of the midpoints of $AD$ is clearly a circle. Since the problems gives that $AD$ is the only chord starting at $A$ bisected by $BC$, it follows that the circle with diameter $DO$ and $AO$ is tangent to $BC$. Now, let the intersection of $BC$ and $AD$ be $E$ and let the midpoint of $AO$ (the center of the circle tangent to $BC$ that we described beforehand) be $F$. Drop the altitude from $O$ to $BC$ and call its intersection with $BC$ $K$. Drop the perpendicular from $F$ to $KO$ and call its intersection with $KO$ $L$. Clearly, $KO = \sqrt{OC^2-KC^2} = \sqrt{5^2-3^2} = 4$ and since $EF$ is radius, it equals $\frac{5}{2}$. The same applies for $FO$, which also equals $\frac{5}{2}$. By the Pythagorean theorem, we deduce that $FL = 2$, so $EK = 2$. This is very important information! Now we know that $BE = 1$, so by Power of a Point, $AE = ED = \sqrt{5}$. We’re almost there! Since by the Pythagorean theorem, $ED^2 + EO^2 = 5$, we deduce that $EO = 2\sqrt{5}$. $EC=OC=5$, so $\sin (CEO) = \frac{2\sqrt{5}}{5}$. Furthermore, since $\sin (CEO) = \cos(DEC)$, we know that $\cos (DEC) = \frac{2\sqrt{5}}{5}$. By the law of cosines, $$DC^2 = (\sqrt{5})^2 + 5^2 -10\sqrt{5} \cdot \frac{2\sqrt{5}}{5} = 10$$Therefore, $DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}$. Now, drop the altitude from $O$ to $BA$ and call its intersection with $BA$ $Z$. Then, by the Pythagorean theorem, $OZ = \frac{7\sqrt{2}}{2}$. Thus, $\sin (BOZ) = \frac{\sqrt{2}}{10}$ and $\cos (BOZ) = \frac{7\sqrt{2}}{10}$. As a result, $\sin (BOA) = \sin (2 BOZ) = 2\sin(BOZ)\cos(BOZ) = \frac{7}{25}$. $7 \cdot 25 = \boxed{175}$. Solution 5 Let I be the intersection of AD and BC. Lemma: $AI = ID$ if and only if $\angle AIO = 90$. Proof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If $\angle AIO = 90$, We can get $\triangle AIO \cong \triangle OID$ Let be this the circle with diameter AO. Thus, we get $\angle AIO = 90$, implying I must lie on $\omega$. I also must lie on BC. If both of these conditions are satisfied, The extension of AI intersecting with circle O will create a point D such that AI = ID. Thus, since AD is the only chord starting at A, there must be only 1 possible I, implying the circle with diameter AO is tangent to BC. Let Y be a point such that $\angle DOY = 90$, and minor arc DY contains A,B. Let X be the center of $\omega$. Since $\omega$'s radius is $\frac{5}{2}$, the altitude from X to OY is 1.5, since the altitude from I to OY is 4. XO is 2.5, so $sin(XOY) = sin(AOY) = \frac{3}{5}$. $sin(BOD) = \frac{3}{5}$. If we let $sin(\theta) = \frac{3}{5}$, we can find that what we are looking for is $sin(90 - 2\theta)$, which we can evaluate and get $\frac{7}{25} \implies \boxed{175}$
# If 2 Tables and 3 Chairs Cost Rs. 3500 and 3 Tables and 2 Chairs Cost Rs. 4000, then How Much does a Table Cost? ### Computer MCQs Series for PPSC, FPSC – Most Repeated MCQs \| Set 6 What are you looking for? Let’s dig in quickly ## Explanation • If 2 tables and 3 chairs cost Rs. 3500. • 3 tables and 2 chairs cost Rs. 4000 Then price of a table can be calculate as; Let suppose “t” shows price of a table and “c” price of a chair. Now 2t + 3c will be equal to Rs. 3500; 2t + 3c = 3500 ________ (i) In the same way; 3 tables and 2 chairs will be equal to Rs. 4000; 3t + 2c = 4000 ________ (ii) Now; we have two equation and by solving them simultaneously we can easily figure out the price of a table. ## To Find Price of a table = ? ## Solution Let suppose Price of a chair = c Price of t table = t Now; 2t + 3c = 3500 ________ (i) 3t + 2c = 4000 ________ (ii) Multiplying both sides of equation (i) with 2 and equation (ii) with 3. 4t + 6c = 7000 ________ (iii) 9t + 6c = 12000 ________ (iv) Subtracting equation (iv) from (iii) 5t = 5000 ## Conclusion If 2 tables and 3 chairs cost Rs. 3500 and 3 tables and 2 chairs cost Rs. 4000, then price of a table would be Rs. 1000.
Trigonometry refers to calculations using triangles. In trigonometry we study how the angles and sides of a triangle are related to each other. There are many applications of trigonometry such has measuring heights of monuments or structures; distances between ship positions at the sea; video games such as Mario; in construction for measuring sizes and dimensions of structures; in flight as well as maritime engineering; etc. ## Formulas Trigonometry Formulas: $sin(x)$ = $\frac{opposite\ side}{hypotenuse}$ $cos(x)$ = $\frac{adjacent\ side}{hypotenuse}$ $tan(x)$ = $\frac{opposite\ side}{adjacent\ side}$ $cosec(x)$ = $\frac{1}{sin(x)}$ $sec(x)$ = $\frac{1}{cos(x)}$ $cot(x)$ = $\frac{1}{tan(x)}$ Trigonometric Identities: $sin^2 (x)\ +\ cos^2 (x)$ = $1$ $tan^2 (x)\ +\ 1$ = $sec^2 (x)$ $1 + cot^2 (x)$ = $csc^2 (x)$ Law of Sines: $\frac{a}{sin (A)}$ = $\frac{b}{sin (B)}$ = $\frac{c}{sin (C)}$ Law of Cosines: $a^2$ = $b^2\ +\ c^2\ -2bc\ Cos(A)$ $b^2$ = $a^2\ +\ c^2\ -2ac\ Cos(B)$ $c^2$ = $a^2\ +\ b^2\ -2ab\ Cos(C)$ Area of Triangle: Area = $\sqrt{(s(s-a)(s-b)(s-c)}$ Where, $s$ = $\frac{1}{2}$ $(a + b + c)$ (a,b and c are the sides of a triangle) ## Word Problems Example 1: From the top of a $150 m$ tower, two cars are seen at an angle of depression of $45$ degrees and $60$ degrees on either side of the tower. Find the distance between the two cars. Solution: This is what the picture would look like. Now we know that the angles at $A$ and $B$ would also be $45$ and $60$ degrees respectively. So if the distance between car $A$ and tower is $x$ and that between car $B$ and tower is $y$, then the following figure holds: From the above figure we see that: $Tan(A)$ = $\frac{opposite\ side}{hypotenuse}$ So, $tan(45)$ = $\frac{150}{x}$ $1$ = $\frac{150}{x}$ $x$ = $150 m$ Similarly, $Tan(B)$ = $\frac{opposite\ side}{hypotenuse}$ $tan(60)$ = $\frac{150}{y}$ $1.732$ = $\frac{150}{y}$ $y$ = $\frac{150}{1.732}$ = $86.61 m$ So the total distance between the two cars is = $x + y$ = $150 + 86.61$ = $236.61 m$ Example 2: A tree grows at an angle of $83$ degrees with respect to the horizontal ground. The angle of elevation from a peg $100 m$ from the base of the tree is $33$ degrees. Find the height of the tree. Solution: First let us sketch a picture of the situation described in the question. It would look like this: We see that in the above triangle we know two angles and the included side. We can find the third angle using the angle sum property of triangle. $m < X + m < Y + m < Z$ = $180$ $33 + 83 + Z$ = $180$ $116 + Z$ = $180$ $Z$ = $180 - 116$ $Z$ = $64$ Now using the law of sines we can write an equation for the height of the tree $x$ as: $\frac{x}{Sin(X)}$ = $\frac{z}{Sin(Z)}$ Now substituting the values of $x,\ z,\ X$ and $Z$ we have: $\frac{x}{Sin(33)}$ = $\frac{100}{Sin(64)}$ Substituting the sine values for 33 and 64 degree angles we have: $\frac{x}{0.5446}$ = $\frac{100}{0.8988}$ Cross multiplying here we have: $0.8988x$ = $54.46$ Dividing both the sides by $0.8988$ gives us the value of $x$ as: $x$ = $\frac{54.46}{0.8988}$ = $60.59 m$ So height of the tree is $60.59 m$. Example 3: There is an approximately circular lake. A surveyor wants to measure the approximate diameter of this lake. He starts from one end of the lake and walks a distance of $265$ yards towards a tree that lies a little away from the edge of the lake. Then from there he turns $115^{\circ}$ and walks towards the other end of the lake, reaching it in $290$ yards. Find the approximate diameter of the lake. Solution: First let us sketch a picture of the situation described in the question. We see that the end points of the lake $A$ and $B$ along with the position of the tree $T$ form a triangle $BAT$. In this triangle two sides and the included angle is known to us. So we can use the law of cosines to find the length of the third side $x$. The law of cosines is like this: $a^2$ = $b^2\ +\ c^2\ -\ 2bcCos(A)$ So for our problem it would be like this: $x^2$ = $a^2\ +\ b^2\ -\ 2abCos(T)$ Substituting the values from our triangle we have: $x^2$ = $290^2\ +\ 265^2\ -\ 2(290)(265)Cos(65)$ Simplifying that we have: $x^2$ = $84100 + 70225 - 64956$ Evaluating that we have: $x^2$ = $89369$ Taking square root on both the sides we have: $x$ = $\sqrt{89369}$ $x$ = $298.95$ yards Thus the approximate diameter of the lake would be: $299$ yards or $300$ yards. Example 4: A triangular pen for livestock is to be constructed in the back yard. The fencing has to be used as one side of the pen. The length of the fencing is $12$ meters. One of the other sides of the pen would be the $15 m$ wall of the barn which is at an angle of $50$ degrees to the fence. What area would this pen cover? Solution: First let us make a picture of the pen that we are trying to construct. The triangle $ABC$ represents our pen. The side $AB$ of this triangle is the fence which is $12 m$ long. The side $BC$ of this triangle is our barn wall which is $15 m$ long. Now we know that we have the length of two sides of the triangle and the measure of the included angle. So we can use the formula for area of a triangle which is: $A$ = $\frac{1}{2}$ $abSin(C)$ For our problem it would be: $A$ = $\frac{1}{2}$ $(AB)(BC)Sin(B)$ Substituting the values we have: $A$ = $\frac{1}{2}$ $(12)(15)Sin(50)$ Substituting the value for $sin(50)$ we have: $A$ = $\frac{1}{2}$ $(12)(15)(0.766)$ Multiplying out the numbers we have: $A$ = $6(15)(0.766)$ $A$ = $68.94 m^2$ Thus the area of the pen would be $68.94 m^2$. Example 5: The average time between a high tide and a low tide at a port is $6$ hours. The average depth of water at the port is $50$ meters. However at high tide the water level rises to $70$ meters. Find the trigonometric function that represents the depth of the water over time if it's peak high tide at $12:00 AM$. Also sketch the graph of the function. Solution: The general form of a sine curve function is like this: $y(t)$ = $A sin(Bt + C) + D$ Here, $A$ is the amplitude of the curve. Amplitude refers to half the difference between the highest and lowest point on the curve. $B$ is the horizontal stretch or shrink that can be found from the period of the function. $C$ is the horizontal translation from the parent function. It is used to find the phase shift. $D$ is the vertical shift of the function. The parent function that we need to start working form would be: $y(t)$ = $sin(t)$ Now since the average depth is $50$ meters, so we know that the vertical shift is $D$ = $50$ meters. Thus, using this value of $D$ our function now becomes: $y(t)$ = $sin(t) + 50$ Next, we are given that the difference between average depth and high tide depth is $70 - 50$ = $20$ meters. So that becomes our amplitude $A$. Thus using this our equation now becomes: $y(t)$ = $20\ sin(t) + 50$ We are given that the time between high and low tide is $6$ hours. So the time between two high tides would be $12$ hours. So the period of the function would be $12$ hours. We know that: $period$ = $\frac{2 \pi}{B}$ Thus using the period of $6$ hours we have: $6$ = $\frac{2 \pi}{B}$ Solving that for $B$ we have: $B$ = $\frac{2 \pi}{6}$ = $\frac{p}{3}$ So now our function becomes: $y(t)$ = $20sin$ $(\frac{\pi}{3}$ $t + C) + 50$ Now we need to find the value for that $C$. We are given that at $12:00AM$ when $t$ = $0$, the tide is peak high at $70$ meters. Thus, $y(t$ = $0)$ = $20 sin$ $(\frac{\pi}{3}$ $(0)\ +\ C)\ +\ 50$ Solving that for $y$ = $70$ we have: $70$ = $20Sin(C)\ +\ 50$ $70 - 50$ = $20Sin(C)$ $20$ = $20Sin(C)$ $Sin(C)$ = $1$ $C$ = $\frac{p}{2}$ Thus our function is: $y(t)$ = $20Sin$ $(\frac{\pi}{3}$ $t\ +$ $\frac{p}{2})$ $+\ 50$ Its graph would look like this:
Course Content Class 10th Science 0/32 Class 10th Maths 0/86 Class 10 Social Science History : India and the Contemporary World – II 0/16 Class 10 Social Science Geography : Contemporary India – II 0/14 Class 10 Social Science Civics (Political Science) : Democratic Politics – II 0/16 Class 10 Social Science Economics: Understanding Economic Development – II 0/10 Class 10 English First Flight Summary 0/22 Class 10 English First Flight Poem 0/22 Class 10 English Footprints without Feet Summary 0/20 Class 10th Online Course: Navigating CBSE Board Success with Wisdom TechSavvy Academy ### Exercise: 12.1 (Page No: 230) 1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles. Solution: The radius of the 1st circle = 19 cm (given) ∴ Circumference of the 1st circle = 2π×19 = 38π cm The radius of the 2nd circle = 9 cm (given) ∴ Circumference of the 2nd circle = 2π×9 = 18π cm So, The sum of the circumference of two circles = 38π+18π = 56π cm Now, let the radius of the 3rd circle = R ∴ The circumference of the 3rd circle = 2πR It is given that sum of the circumference of two circles = circumference of the 3rd circle Hence, 56π = 2πR Or, R = 28 cm. 2. The radii of two circles are 8 cm and 6 cm, respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles. Solution: Radius of 1st circle = 8 cm (given) ∴ Area of 1st circle = π(8)2 = 64π Radius of 2nd circle = 6 cm (given) ∴ Area of 2nd circle = π(6)2 = 36π So, The sum of 1st and 2nd circle will be = 64π+36π = 100π Now, assume that the radius of 3rd circle = R ∴ Area of the circle 3rd circle = πR2 It is given that the area of the circle 3rd circle = Area of 1st circle + Area of 2nd circle Or, πR2 = 100πcm2 R2 = 100cm2 So, R = 10cm 3. Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions. Solution: The radius of 1st circle, r1 = 21/2 cm (as diameter D is given as 21 cm) So, area of gold region = π r1= π(10.5)= 346.5 cm2 Now, it is given that each of the other bands is 10.5 cm wide, So, the radius of 2nd circle, r2 = 10.5cm+10.5cm = 21 cm Thus, ∴ Area of red region = Area of 2nd circle − Area of gold region = (πr22−346.5) cm2 = (π(21)2 − 346.5) cm2 = 1386 − 346.5 = 1039.5 cm2 Similarly, The radius of 3rd circle, r3 = 21 cm+10.5 cm = 31.5 cm The radius of 4th circle, r4 = 31.5 cm+10.5 cm = 42 cm The Radius of 5th circle, r5 = 42 cm+10.5 cm = 52.5 cm For the area of nth region, A = Area of circle n – Area of circle (n-1) ∴ Area of blue region (n=3) = Area of third circle – Area of second circle = π(31.5)2 – 1386 cm2 = 3118.5 – 1386 cm2 = 1732.5 cm2 ∴ Area of black region (n=4) = Area of fourth circle – Area of third circle = π(42)2 – 1386 cm2 = 5544 – 3118.5 cm2 = 2425.5 cm2 ∴ Area of white region (n=5) = Area of fifth circle – Area of fourth circle = π(52.5)2 – 5544 cm2 = 8662.5 – 5544 cm2 = 3118.5 cm2 4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour? Solution: The radius of car’s wheel = 80/2 = 40 cm (as D = 80 cm) So, the circumference of wheels = 2πr = 80 π cm Now, in one revolution, the distance covered = circumference of the wheel = 80 π cm It is given that the distance covered by the car in 1 hr = 66km Converting km into cm we get, Distance covered by the car in 1hr = (66×105) cm In 10 minutes, the distance covered will be = (66×105×10)/60 = 1100000 cm/s ∴ Distance covered by car = 11×105 cm Now, the no. of revolutions of the wheels = (Distance covered by the car/Circumference of the wheels) =( 11×105)/80 π = 4375. 5. Tick the correct Solution: in the following and justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is (A) 2 units (B) π units (C) 4 units (D) 7 units Solution: Since the perimeter of the circle = area of the circle, 2πr = πr2 Or, r = 2 So, option (A) is correct i.e. the radius of the circle is 2 units.
Texas Go Math Grade 8 Module 7 Answer Key Angle Relationships in Parallel Lines and Triangles Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 7 Answer Key Angle Relationships in Parallel Lines and Triangles. Texas Go Math Grade 8 Module 7 Answer Key Angle Relationships in Parallel Lines and Triangles Solve for x. Question 1. 6x + 10 = 46 Following the example on page 190, first we subtract 10 from both sides Then, we simplify and we divide both sides by 6. 6x + 10 = 46 6x + 10 – 10 = 46 – 10 6x = 36 $$\frac{6x}{6}$$ = $$\frac{36}{6}$$ x = 6 Question 2. 7x – 6 = 36 Following the example on page 190, first we add 6 to both sides Then, we simpLify and we divide both sides by 7. 7x – 6 = 36 7x – 6 + 6 = 36 + 6 7x = 42 $$\frac{7x}{7}$$ = $$\frac{42}{7}$$ x = 6 Question 3. 3x + 26 = 59 Following the example on page 190, first we subtract 26 from both sides. Then, we simplify and we divide both sides by3. 3x + 26 = 59 3x + 26 – 26 = 59 – 26 3x = 33 $$\frac{3x}{3}$$ = $$\frac{33}{3}$$ x = 11 Question 4. 2x + 5 = -25 Following the example on page 190, first we subtract 5 from both sides. Then, we simplify and we divide both sides by 2. 2x + 5 = 25 2x + 5 – 5 = -25 – 5 2x = 30 $$\frac{2x}{2}$$ = $$\frac{30}{2}$$ x = -15 Question 5. 6x – 7 = 41 6x – 7 = 41 (Write the equation) ……….. (1) 6x = 41 + 7 (Add 7 to both sides) ………… (2) 6x = 48 (Simplify) ………….. (3) x = $$\frac{48}{6}$$ (Divide both sides by 6) ………. (4) x = 8 (Simplify) ………….. (5) Question 6. $$\frac{1}{2}$$x + 9 = 30 $$\frac{1}{2}$$x + 9 = 30 (Write the equation.) ……………. (1) $$\frac{1}{2}$$x = 30 – 9 (Subtract 9 from both sides) ……………. (2) $$\frac{1}{2}$$x = 21 (Simplify) ………………. (3) x = 21 ∙ 2 (Multiply both sides by 2) …………… (4) x = 42 (Simplify) …………… (5) Question 7. $$\frac{1}{3}$$x – 7 = 15 $$\frac{1}{3}$$x + 7 = 15 (Write the equation.) ……………. (1) $$\frac{1}{3}$$x = 15 + 7 (Add 7 from both sides) ……………. (2) $$\frac{1}{3}$$x = 22 (Simplify) ………………. (3) x = 22 ∙ 3 (Multiply both sides by 3) …………… (4) x = 66 (Simplify) …………… (5) Question 8. 0.5x – 0.6 = 8.4 0.5x – 0.6 = 8.4 (Write the equation) …………. (1) 0.5x = 8.4 + 0.6 (Subtract 9 from both sides) …………. (2) 0.5x = 9 (Simplify) ………….. (3) x = $$\frac{9}{0.5}$$ (Divide both sides with 0.5) ………. (4) x = 18 (Simplify) …………. (5) Give two names for the angle formed by the dashed rays. Question 9. The angle formed by the dashed rays can be named as: ∠MHR or ∠RHM Question 10. The angle formed by the dashed rays can be named as: ∠SGK or ∠KGS Question 11. The angle formed by the dashed rays can be named as: ∠BTF or ∠FTB Visualize Vocabulary Use the ✓ words to complete the graphic. You can put more than one word in each section of the triangle. Understand Vocabulary Complete the sentences using preview words. Question 1. A line that intersects two or more lines is a ____________. A transversal line is a line that intersects two or more Lines in the same plane at different points. By definition, a line that intersects two or more lines is a transversal. Thus, the blank part in the statement is transversal Question 2. Figures with the same shape but not necessarily the same size are ____________.
# 2023 AMC 12A Problems/Problem 11 (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) ## Problem What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$? $\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$ ## Solution 1 Remind that $\text{slope}=\dfrac{\Delta y}{\Delta x}=\tan \theta$ where $\theta$ is the angle between the slope and $x$-axis. $k_1=2=\tan \alpha$, $k_2=\dfrac{1}{3}=\tan \beta$. The angle formed by the two lines is $\alpha-\beta$. $\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\dfrac{2-1/3}{1+2\cdot 1/3}=1$. Therefore, $\alpha-\beta=\boxed{\textbf{(C)} 45^\circ}$. ~plasta ## Solution 2 We can take any two lines of this form, since the angle between them will always be the same. Let's take $y=2x$ for the line with slope of 2 and $y=\frac{1}{3}x$ for the line with slope of 1/3. Let's take 3 lattice points and create a triangle. Let's use $(0,0)$, $(1,2)$, and $(3,1)$. The distance between the origin and $(1,2)$ is $\sqrt{5}$. The distance between the origin and $(3,1)$ is $\sqrt{10}$. The distance between $(1,2)$ and $(3,1)$ is $\sqrt{5}$. We notice that we have a triangle with 3 side lengths: $\sqrt{5}$, $\sqrt{5}$, and $\sqrt{10}$. This forms a 45-45-90 triangle, meaning that the angle is $\boxed{45^\circ}$. ## Solution 3 (Law of Cosines) Follow Solution 2 up until the lattice points section. Let's use $(0,0)$, $(2,4)$, and $(9,3)$. The distance between the origin and $(2,4)$ is $\sqrt{20}$. The distance between the origin and $(9,3)$ is $\sqrt{90}$. The distance between $(2,4)$ and $(9,3)$ is $\sqrt{50}$. Using the Law of Cosines, we see the $50 = 90 + 20 - 2\times\sqrt{20}$ $\times\sqrt{90}$ $\times\cos(\theta)$, where $\theta$ is the angle we are looking for. Simplifying, we get $-60 = -2\times(\sqrt{20}) \times(\sqrt{90}) \times\cos(\theta)$. $30 = \sqrt{1800} \times\cos(\theta)$. $30 = 30\sqrt{2} \times\cos(\theta)$. $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}= \cos(\theta)$. Thus, $\theta = \boxed{\textbf{(C)} 45^\circ}$ ~Failure.net ## Solution 4 (Vector Bash) We can set up vectors $\vec{a} = <1,2>$ and $\vec{b} = <3,1>$ to represent the two lines. We know that $\frac{\vec{a} \cdot \vec{b}}{\|\vec{a}\| \|\vec{b}\|} = \cos \theta$. Plugging the vectors in gives us $\cos \theta = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$. From this we get that $\theta = \boxed{\textbf{(C)} 45^\circ}$. ~middletonkids ## Solution 5 (Complex Numbers) Let $Z_1 = 3 + i$ and $Z_2 = 1 + 2i$ \begin{align} Z_2 &= Z_1 \cdot re^{i\theta} \\ 1+2i&=(3+i) \cdot re^{i\theta} \\ 1+2i&=(3 + i) \cdot r(\cos\theta + i\sin\theta) \\ 1+2i&=3r\cos\theta - r\sin\theta + 3ri\sin\theta + ri\cos\theta \\ \end{align} From this we have: \begin{align} 1 &= 3r\cos\theta - r\sin\theta \\ 2 &= r\cos\theta + 3r\sin\theta \end{align} To solve this we must compute $r$ \begin{align} r &= \frac{\|Z_2\|}{\|Z_1\|} \\ &= \frac{\sqrt{5}}{\sqrt{10}} \\ &= \frac{\sqrt{2}}{2} \end{align} Using elimination we have: $3\cdot(2) - (1)$ \begin{align} 5 &= 10r\sin\theta \\ \frac{1}{2r} &= \sin\theta \\ \frac{1}{2\frac{\sqrt{2}}{2}} &= \sin\theta \\ \frac{\sqrt{2}}{2} &= \sin\theta \\ \theta &= \boxed{\textbf{(C)} 45^\circ} \\ \end{align} ## Solution 6 The lines $y = 2x, y = \frac {1}{3}x$, and $x = 3$ form a large right triangle and a small right triangle. Call the angle that is formed by the x-axis and the line $y = 2x$ $\alpha$, and call the angle that is formed by the x-axis and the line $y = \frac {1}{3}x$ $\beta$. We try to find $\sin (\alpha - \beta)$ first, and then try to see if any of the answer choices match up. $\sin (\alpha - \beta)$ = $\sin \alpha$ $\cos \beta$ - $\sin \beta$ $\cos \alpha$. Using soh-cah-toa, we find that $\sin \alpha = \frac {2}{\sqrt 5}, \sin \beta = \frac {1}{\sqrt 10}, \cos \alpha = \frac {1}{\sqrt 5},$ and $\cos \beta = \frac {3}{\sqrt 10}$. Plugging it all in, we find that $\sin (\alpha - \beta) = \frac {5}{\sqrt {50}}$, which is equivalent to $\frac {\sqrt 2}{2}$. Since $\sin (\alpha - \beta) = \frac {\sqrt 2}{2}$, we get that $\alpha - \beta = 45^{\circ}$. Therefore, the answer is $\boxed {\textbf {(C)} 45^{\circ}}$. ~Arcticturn # =Solution 7 (Cheese) AMC 12 does not allow graph paper or protractor. https://amc-reg.maa.org/manual/AMC1012B.pdf Using a makeshift ruler, draw an accurate to-scale diagram. You can do this by simply drawing the two lines such that they intersect at the origin. Then, measure the angle by eye or by folding paper to observe that they form a 45 degree angle. The answer is $\boxed{45^\circ}$. ~InstallHelp_Hex ## Video Solution (Under 4 minutes) ~Education, the Study of Everything ## Video Solution ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
# 2005 AIME II Problems/Problem 14 ## Problem In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ ## Solution 1 $[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)(A+dir(-50)); pair B = rotate(15,A)(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label("A",A,N); label("B",B,SW); label("D",D,SE); label("E",E,S); label("C",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/asy]$ By the Law of Sines and since $\angle BAE = \angle CAD, \angle BAD = \angle CAE$, we have \begin{align} \frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} \\ &= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} \\ &= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2} \end{align} Substituting our knowns, we have $\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}$. The answer is $q = \boxed{463}$. ## Solution 2 (Similar Triangles) Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively. From here, we can use Heron's Formula to find the altitude. The area of the triangle is $\sqrt{21678}$ = 84. We can then use similar triangles with triangle AQC and triangle DSC to find DS=$\frac{24}{5}$. Consequently, from Pythagorean theorem, SC = $\frac{18}{5}$ and AS = 14-SC = $\frac{52}{5}$. We can also use pythagorean triangle on triangle AQB to determine that BQ = $\frac{33}{5}$. Label AR as y and RE as x. RB then equals 13-y. Then, we have two similar triangles. Firstly: $\triangle ARE \sim \triangle ASD$. From there, we have $\frac{x}{y} = \frac{\frac{24}{5}}{\frac{53}{5}}$. Next: $\triangle BRE \sim \triangle BQA$. From there, we have $\frac{x}{13-y} = \frac{\frac{56}{5}}{\frac{33}{5}}$. Solve the system to get $x = \frac{2184}{463}$ and $y = \frac{4732}{463}$. Notice that 463 is prime, so even though we use pythagorean theorem on x and 13-y, the denominator won't change. The answer we desire is $\boxed{463}$. ## Solution 3 (LoC and LoS bash) Let $\angle CAD = \angle BAE = \theta$. Note by Law of Sines on $\triangle BEA$ we have $$\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{\angle BEA}}$$ As a result, our goal is to find $\sin{\angle BEA}$ and $\sin{\theta}$ (we already know $AB$). Let the foot of the altitude from $A$ to $BC$ be $H$. By law of cosines on $\triangle ABC$ we have $$169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}$$ It follows that $AH = \frac{56}{5}$ and $HC = \frac{42}{5} \Rightarrow AD = \frac{12}{5}$. Note that by PT on $\triangle AHD$ we have that $AD^2 = \left(\frac{56}{5}\right)^2 + \left(\frac{12}{5}\right)^2 = \frac{656}{5}$. By Law of Sines on $\triangle ADC$ (where we square everything to avoid taking the square root) we see $$\frac{36}{\sin^2{\theta}} = \frac{656}{5 \cdot \frac{16}{25}} \Rightarrow \sin^2{\theta} = \frac{36}{205}.$$ How are we going to find $\sin{\angle BEA}$ though? $\angle BEA$ and $\theta$ are in the same triangle. Applying Law of Sines on $\triangle ABC$ we see that $$\frac{13}{\frac{4}{5}} = \frac{14}{\sin{\angle B}} \Rightarrow \sin{\angle B} = \frac{56}{65} \Rightarrow \cos{\angle B} = \frac{33}{65}.$$ $\theta$, $\angle B$, and $\angle BEA$ are all in the same triangle. We know they add up to $180^{\circ}$. There's a good chance we can exploit this using the identity $\sin{p} = \sin{180^{\circ}-p}$. We have that $\sin{(180^{\circ} - (\theta + \angle B))} = \sin{\angle BEA} = \sin{(\theta + \angle B)}$. Success! We know $\sin{\theta}$ and $\sin{\angle B}$ already. Applying the $\sin$ addition formula we see $$\sin{\theta + \angle B} = \sin{\theta} \cos{\angle B} + \sin{\angle B} \cos{\theta} = \frac{6}{\sqrt{205}} \cdot \frac{33}{65} + \frac{56}{65} \cdot \frac{13}{\sqrt{205}}=\frac{1}{65 \cdot \sqrt{205}} (198 + 728) = \frac{926}{65 \sqrt{205}}.$$ This is the last stretch! Applying Law of Sines a final time on $\triangle BEA$ we see $$\frac{BE}{\sin{\theta}} = \frac{13}{\sin{BEA}} \Rightarrow \frac{BE}{\frac{6}{\sqrt{205}}} = \frac{13}{\frac{926}{65\sqrt{205}}} \Rightarrow \frac{BE}{6} = \frac{13 \cdot 65}{926} \Rightarrow \frac{13 \cdot 65 \cdot 6}{926} = BE = \frac{2535}{463}.$$ It follows that the answer is $\boxed{463}$. ## Solution 4 (Ratio Lemma and Angle Bisector Theorem) Let $AK$ be the angle bisector of $\angle A$ such that $K$ is on $BC$. Then $\angle KAB = \angle KAC$, and thus $\angle KAE = \angle KAD$. By the Ratio Lemma, $\frac{BE}{KE} = \frac{BA}{KA} \frac{\sin{BAE}}{\sin{KAE}}$ and $\frac{CD}{KD} = \frac{CA}{KA} * \frac{\sin{CAD}}{\sin{KAD}}$. This implies that $\frac{BE}{KEBA} = \frac{CD}{KDCA}$. Thus, $\frac{BE}{KE} = \frac{13}{14} * \frac{6}{DK}$. $DK = CK - 6 = 1415/27 - 6 = 16/9$. Thus, $\frac{BE}{KE} = \frac{1354}{1416}$. Additionally, $BE + KE = 9$. Solving gives that $q = 463.$ Alternate: By the ratio lemma, $BD/DC = (13/14)(sin BAD/sin DAC) EC/EB = (14/13)*(sin EAC/sin BAE) Combining these, we get (BD/DC)(14/13) = (EC/EB)(13/14) (3/2)(14/13)(14/13) = (15-x)(x) x = 2535/463$ Thus, q = 463
# Circle Circumference ## C = πd; C = 2πr Estimated21 minsto complete % Progress Practice Circle Circumference MEMORY METER This indicates how strong in your memory this concept is Progress Estimated21 minsto complete % Circle Circumference The theme of the school dance this year is “The 50’s.” For decorations, the 7th grade class is gathering old records and gluing a sparkly trim around the edges. How much trim do they need for each record if its diameter is 12 inches? In this concept, you will learn how to find the circumference of a circle using a given diameter. ### Finding the Circumference of a Circle The perimeter is the total distance around the edges of a two-dimensional shape. To find the perimeter of a shape made from line segments, such as a square or triangle, you simply add the lengths of the sides together. \begin{align}P_{\text{Square}}= s+s+s+s \ \text{or} \ P_S=4s\end{align} \begin{align}P_{\text{Triangle}} = s_1+s_2+s_3\end{align} The perimeter of a circle is called the circumference and is equal to the total distance around the edge. The diameter, d, of a circle is the straight-line measurement from one point on the circle to another point on the circle that passes directly through the circle’s center. A circle is a unique shape in that every point along the circumference is exactly the same distance from the circle’s center. This measurement of one half the diameter is called the radius, r. \begin{align}d=2r\end{align} One of the unique qualities of a circle is that its diameter and circumference have a proportional relationship. This means that no matter what size the circle is, the proportional relationship, or ratio, between its circumference and diameter is always the same. The formula for the circumference of a circle is the value of pi times the circle’s diameter: \begin{align}C=\pi d\end{align} Since the diameter is twice the radius, the formula can also be written: \begin{align}C= 2 \pi r\end{align} You already know that pi is an irrational number and goes on forever. The rounded value of pi, 3.14, is usually used for calculations. Let’s look at an example. The circles above have different radii and different circumferences. Use these values and the circumference formula to solve for pi. First, substitute the values of the first circle into the circumference equation. \begin{align}6.28= \pi \times 2\end{align} Next, perform the necessary calculations to isolate \begin{align}\pi\end{align}. \begin{align}\begin{array}{rcl} 6.28 \div 2 &=& \pi \times 2 \div 2 \\ 3.14 &=& \pi \end{array}\end{align} Then, substitute the values of the second circle into the equation. \begin{align}12.56= \pi \times 4\end{align} Perform the necessary calculations to isolate \begin{align}\pi\end{align}. \begin{align}\begin{array}{rcl} 12.56 \div 4 &=& \pi \times 4 \div 4 \\ 3.14 &=& \pi \end{array}\end{align} Pi is what is called a constant. It always stays the same. ### Examples #### Example 1 Earlier, you were given a problem about the school dance decorations. The students need to know how much sparkly trim they will need to go around each record if its diameter is 12 inches. First, write down the formula. \begin{align}C= \pi d\end{align} Next, substitute in what you know. \begin{align}C= 3.14 \times 12\end{align} Then, multiply to solve the equation. \begin{align}C= 37.68 \ \text{inches}\end{align} The answer is the circumference is 21.98 inches. The students will probably round this up to 38 inches. #### Example 2 Find the circumference of the circle below. First write down the formula. \begin{align}C= \pi d\end{align} Next, substitute in what you know. \begin{align}\begin{array}{rcl} C &=& \pi \times 8 \\ C &=& 3.14 \times 8 \end{array}\end{align} Then, multiply to solve the equation. \begin{align}C=25.12\end{align} If you were to unroll the circle into a flat line, it would be 25.12 inches long. The answer is the circumference, \begin{align}C = 25.12 \ in\end{align}. #### Example 3 What is the circumference of the circle below? First, write down the formula. \begin{align}C= \pi d\end{align} Next, substitute in what you know. \begin{align}\begin{array}{rcl} C &=& \pi \times 12.7 \\ C &=& 3.14 \times 12.7 \end{array}\end{align} Then, multiply to solve the equation, and round. \begin{align}C= 39.878\end{align} The answer is the circumference, \begin{align} C = 39.88 \ m\end{align}. #### Example 4 Find the circumference of the circle given the radius. First, remember that the diameter is twice the radius, and use the formula that includes that variable. \begin{align}C= 2 \pi r\end{align} Next, substitute in the values that you know. \begin{align}C= 2(3.14)(3)\end{align} Then, multiply. \begin{align}C= 18.84\end{align} #### Example 5 Find the circumference of a circle with a diameter of 5 inches. First, write down the formula. \begin{align}C= \pi d\end{align} Next, substitute in what you know. \begin{align}C= 3.14 \times 5\end{align} Then, multiply to solve the equation. \begin{align}C= 15.7\end{align} The answer is the circumference, \begin{align}C = 15.7 \ \text{inches}\end{align}. ### Review Find the circumference of each circle given the radius or the diameter. 2. diameter = 4 ft 4. diameter = 8 meters 6. diameter = 12 mm 8. diameter = 13 feet 10. diameter = 7.5 feet 13. diameter = 3.75 feet 14. diameter = 4.5 feet 15. diameter = 10.75 meters ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English TermDefinition $\pi$ $\pi$ (Pi) is the ratio of the circumference of a circle to its diameter. It is an irrational number that is approximately equal to 3.14. Circle A circle is the set of all points at a specific distance from a given point in two dimensions. Circumference The circumference of a circle is the measure of the distance around the outside edge of a circle. Hypotenuse The hypotenuse of a right triangle is the longest side of the right triangle. It is across from the right angle. Legs of a Right Triangle The legs of a right triangle are the two shorter sides of the right triangle. Legs are adjacent to the right angle. Perimeter Perimeter is the distance around a two-dimensional figure. Pi $\pi$ (Pi) is the ratio of the circumference of a circle to its diameter. It is an irrational number that is approximately equal to 3.14. Pythagorean Theorem The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $a^2 + b^2 = c^2$, where $a$ and $b$ are legs of the triangle and $c$ is the hypotenuse of the triangle.
## Log-sum-exp trick with negative numbers Suppose you want to calculate an expression of the form $\displaystyle{\log\left(\sum_{i=1}^n \exp(a_i) - \sum_{j=1}^m \exp(b_j)\right)},$ where $\sum_{i=1}^n \exp(a_i) > \sum_{j=1}^m \exp(b_j)$. Such expressions can be difficult to evaluate directly since the exponentials can easily cause overflow errors. In this post, I’ll talk about a clever way to avoid such errors. If there were no terms in the second sum we could use the log-sum-exp trick. That is, to calculate $\displaystyle{\log\left(\sum_{i=1}^n \exp(a_i) \right)},$ we set $a=\max\{a_i : 1 \le i \le n\}$ and use the identity $\displaystyle{a + \log\left(\sum_{i=1}^n \exp(a_i-a)\right) = \log\left(\sum_{i=1}^n \exp(a_i)\right)}.$ Since $a_i \le a$ for all $i =1,\ldots,n$, the left hand side of the above equation can be computed without the risk of overflow. To calculate, $\displaystyle{\log\left(\sum_{i=1}^n \exp(a_i) - \sum_{j=1}^m \exp(b_j)\right)},$ we can use the above method to separately calculate $\displaystyle{A = \log\left(\sum_{i=1}^n \exp(a_i) \right)}$ and $\displaystyle{B = \log\left(\sum_{j=1}^m \exp(b_j)\right)}.$ The final result we want is $\displaystyle{\log\left(\sum_{i=1}^n \exp(a_i) - \sum_{j=1}^m \exp(b_j) \right) = \log(\exp(A) - \exp(B)) = A + \log(1-\exp(A-B))}.$ Since $A > B$, the right hand side of the above expression can be evaluated safely and we will have our final answer. ### R code The R code below defines a function that performs the above procedure # Safely compute log(sum(exp(pos)) - sum(exp(neg))) # The default value for neg is an empty vector. logSumExp <- function(pos, neg = c()){ max_pos <- max(pos) A <- max_pos + log(sum(exp(pos - max_pos))) # If neg is empty, the calculation is done if (length(neg) == 0){ return(A) } # If neg is non-empty, calculate B. max_neg <- max(neg) B <- max_neg + log(sum(exp(neg - max_neg))) # Check that A is bigger than B if (A <= B) { stop("sum(exp(pos)) must be larger than sum(exp(neg))") } # log1p() is a built in function that accurately calculates log(1+x) for \|x\| << 1 return(A + log1p(-exp(B - A))) } ### An example The above procedure can be used to evaulate $\displaystyle{\log\left(\sum_{i=1}^{200} i! - \sum_{i=1}^{500} \binom{500}{i}^2\right)}.$ Evaluating this directly would quickly lead to errors since R (and most other programming languages) cannot compute $200!$. However, R has the functions lfactorial() and lchoose() which can compute $\log(i!)$ and $\log\binom{n}{i}$ for large values of $i$ and $n$. We can thus put this expression in the general form at the start of this post $\displaystyle{\log\left(\sum_{i=1}^{200} \exp(\log(i!)) - \sum_{i=1}^{500} \exp\left(2\log\binom{500}{i}\right)\right)}.$ The following R code thus us exactly what we want: pos <- sapply(1:200, lfactorial) neg <- sapply(1:500, function(i){2*lchoose(500, i)}) logSumExp(pos, neg) # 863.237
# Question Video: Solving Real-World Problems Involving the Law of Cosines Mathematics • 11th Grade A plane travels 800 meters along the runway before taking off at an angle of 10ยฐ. It travels a further 1,000 meters at this angle as seen in the figure. Work out the distance of the plane from its starting point. Give your answer to 2 decimal places. 02:49 ### Video Transcript A plane travels 800 meters along the runway before taking off at an angle of 10 degrees. It travels a further 1,000 meters at this angle as seen in the figure. Work out the distance of the plane from its starting point. Give your answer to two decimal places. Looking at the diagram, we can see that we have a triangle. We want to calculate the distance of the plane from its starting point. Thatโs this length here, which we can refer to as ๐ meters. We know the lengths of the other two sides in this triangle. They are 800 meters and 1,000 meters. And using the fact that angles on a straight line sum to 180 degrees, we can work out the size of this angle here. Itโs 180 degrees minus 10 degrees, which is 170 degrees. As this is a non-right-angled triangle, we need to answer this problem using either the law of sines or the law of cosines. So the first step is to decide which of these we need. And that will depend on the specific combination of information weโve been given and what we want to calculate. In this triangle, we know two sides and the included angle. And we want to calculate the third side. We recall then that this means we should be using the law of cosines. Letโs recall the law of cosines. Itโs ๐ squared equals ๐ squared plus ๐ squared minus two ๐๐ cos ๐ด. Now, thereโs no need to actually label our triangle using the letters ๐ด, ๐ต, and ๐ถ. Instead, we just remember that the lowercase letters ๐ and ๐ represent the two sides we know and the capital letter ๐ด represents the included angle. So using 800 and 1,000 as the two side lengths ๐ and ๐ and 170 degrees as the angle ๐ด, we have the equation ๐ squared equals 800 squared plus 1,000 squared minus two times 800 times 1,000 times cos of 170 degrees. We can either type this directly into our calculator or it may be a good idea to break the calculation down into some stages. In either case, we arrive at ๐ squared equals 3,215,692.405. Now, we must remember that this is ๐ squared. It isnโt ๐, so we arenโt finished. We have to square root in order to find the value of ๐. Itโs a really common mistake though to forget to do this. Square rooting gives ๐ equals 1,793.235178. The question asks us to give our answer to two decimal places. So rounding appropriately, weโve worked out the distance of the plane from its starting point. Itโs 1,793.24 meters to two decimal places.
# Solving Decimal Equations by x / ÷ Lesson 3-6. Basic Decimal Knowledge Decimal represented by . “Decimal places”  the digits that come AFTER the decimal. ## Presentation on theme: "Solving Decimal Equations by x / ÷ Lesson 3-6. Basic Decimal Knowledge Decimal represented by . “Decimal places”  the digits that come AFTER the decimal."— Presentation transcript: Solving Decimal Equations by x / ÷ Lesson 3-6 Basic Decimal Knowledge Decimal represented by . “Decimal places”  the digits that come AFTER the decimal To x / ÷ decimals  DO NOT line up decimals When a decimal is “out of sight”  only add one when dividing (“on the right”) When multiplying decimals  total decimal places in problem = total decimal places in answer! When dividing BY decimals  move bottom decimal to end of # = move top decimal same direction/same # of moves Solving Decimal Equations by x / ÷ Example: -81.81 = -9n -9 -9 divide into each digit AFTER decimal 9.09 = n Solving Decimal Equations by x / ÷ Example: 0.9r= -4.5 0.9 move bottom decimal = move top decimal r = -6 Solving Decimal Equations by x / ÷ Example: m = -12.5 -7 (-7) m = -12.5 (-7) -7 total decimal places in problem = total decimal places in answer m = 87.5 Solving Decimal Equations by +/- Example: s = 5.3 0.2 (0.2) s = 5.3 (0.2) 0.2 total decimal places in problem = total decimal places in answer s = 1.06 Practice Solve: a) -80 = t 4.5 b) 0.8x = -1.6 c) r = 0.5 -6.03 d) -2.4 = -1.2c Homework p. 148-149 #20-24, 29, 39-41 Download ppt "Solving Decimal Equations by x / ÷ Lesson 3-6. Basic Decimal Knowledge Decimal represented by . “Decimal places”  the digits that come AFTER the decimal." Similar presentations
Symbol # Calculator search results Formula Solve the equation Calculate the differentiation Graph $y = 3 x + 2$ $x$Intercept $\left ( - \dfrac { 2 } { 3 } , 0 \right )$ $y$Intercept $\left ( 0 , 2 \right )$ $y = 3x+2$ $x = \dfrac { 1 } { 3 } y - \dfrac { 2 } { 3 }$ Solve a solution to $x$ $y = \color{#FF6800}{ 3 } \color{#FF6800}{ x } + 2$ Move $x$ term to the left side and change the sign $y \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } = 2$ $\color{#FF6800}{ y } - 3 x = 2$ Move the rest of the expression except $x$ term to the right side and replace the sign $- 3 x = 2 \color{#FF6800}{ - } \color{#FF6800}{ y }$ $- 3 x = \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ y }$ Organize the expression $- 3 x = \color{#FF6800}{ - } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 2 }$ $\color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 2 }$ Change the sign of both sides of the equation $3 x = y - 2$ $\color{#FF6800}{ 3 } \color{#FF6800}{ x } = \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 }$ Divide both sides by the same number $\color{#FF6800}{ x } = \left ( \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ \div } \color{#FF6800}{ 3 }$ $x = \left ( \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ \div } \color{#FF6800}{ 3 }$ Convert division to multiplication $x = \left ( \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 3 } }$ $x = \left ( \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 3 } }$ Multiply each term in parentheses by $\dfrac { 1 } { 3 }$ $x = \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 3 } }$ $x = \dfrac { 1 } { 3 } y \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 1 } { 3 } }$ Calculate the product of rational numbers $x = \dfrac { 1 } { 3 } y \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } }$ $\dfrac {d } {d x } {\left( y \right)} = 3$ Calculate the differentiation of the logarithmic function $\dfrac {d } {d \color{#FF6800}{ x } } {\left( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right)}$ Calculate the differentiation $\color{#FF6800}{ 3 }$ Have you found the solution you wanted? Try again Try more features at Qanda! Search by problem image Ask 1:1 question to TOP class teachers AI recommend problems and video lecture
# Hypothesis Testing and Comparing Two Proportions - PowerPoint - PowerPoint Document Sample ``` Hypothesis Testing and Comparing Two Proportions • Hypothesis Testing: Deciding whether your data shows a “real” effect, or could have happened by chance • Hypothesis testing is used to decide between two possibilities: – The Research Hypothesis – The Null Hypothesis H1 and H0 • H1: The Research Hypothesis – The effect observed in the data (the sample) reflects a “real” effect (in the population) • H0: The Null Hypothesis – There is no “real” effect (in the population) – The effect observed in the data (the sample) is just due to chance (sampling error) Example: Comparing Proportions • H0: The proportions are not really different • H1: The proportions are really different • Example 1: Are pennies heavier on one side? • Example 2: Do males mention footware in personals ads more often than females do? The Logic of Hypothesis Testing 1. Assume the Null Hypothesis (H0) is true 2. Calculate the probability (p) of getting the results observed in your data if the Null Hypothesis were true 3. If that probability is low (< .05) then reject the Null Hypothesis 4. If you reject the Null Hypothesis, that leaves only the Research Hypothesis (H1) 1. Assume the Null Hypothesis is true – The coins are fair (balanced) 2. Calculate the probability (p) of getting the results observed in your data if the Null Hypothesis were true – How often would you get 8/10 coins coming up heads if the coins were fair? You would get 8/10 heads less than 5% of the time. 3. If that probability is low (< .05) then reject the Null Hypothesis – That is unlikely, so the Null Hypothesis must be false. 4. If you reject the Null Hypothesis, that leaves only the Research Hypothesis – We conclude that the coins are not fair (balanced). Calculating p • How do you calculate the probability that the observed effect is just due to chance? • Use a test statistic: 1. Are two proportions different? Chi-square 2. Are two means different? t-test 3. Are more than two means different? ANOVA or “F-test” The Logic is Always the Same: 1. Assume nothing is going on (assume H0) 2. Calculate a test statistic (Chi-square, t, F) 3. How often would you get a value this large for the test statistic when H0 is true? (In other words, calculate p) 4. If p < .05, reject the null hypothesis and conclude that something is going on (H1) 5. If p > .05, do not conclude anything. Demonstrating Hypothesis Testing with Chi-square • Example 1: Testing whether coins are unbalanced • Example 2: Testing whether men are more likely to mention footware in personals ads than women are. • (see Excel spreadsheet for both examples) Assumptions of Chi-square Test • Each observation must be INDEPENDENT – one data point per subject • DV is categorical (often yes/no) • Calculations must be made from COUNTS, not proportions or percentages • No cell should have an “expected value” of less than 5 Using Chi-square in SPSS to compare two proportions • Setting up the data file – copy data from excel and paste it into SPSS data file • Performing the Chi-square test (next slide) • Interpreting the Results (separate slide) • Reporting the Results (separate slide) Performing the Chi-Square Test 1. Name the variables using the variables tab in the SPSS data window 2. analyze -> descriptive statistics -> crosstabs 3. Use arrow button to move “gender” into “rows” box 4. Use arrow button to move “footware” into “columns” box 5. Click “Statistics” box 6. Check the box for “Chi-square”, then click “Continue” 7. Click the “Cells” box. 8. Under “Percentages” check the boxes for “Row” and “Column” 9. Click “OK” Interpreting the Results • “Case Processing Summary” – look for missing data, etc. • “Gender x Footware Crosstabulation” – shows the counts of observations in each cell, and the percentages within each row and within each column. • “Chi-square Tests” – look at “Pearson chi-square” line – Value = 5.33 – This is the value of Chi-square – “Asymp Sig” = .021 – This is the p value – Compare these values to those I calculated by hand on Reporting the Results • Report the value of chi-square, the degrees of freedom (df), and the p value. Also mention how many observations there were. • EXAMPLE: “A greater proportion of men than women mentioned footware in their ads (see Table 1). Of the six ads placed by men, 83% mentioned footware. Only 17% of the six ads placed by women mentioned footware. This difference was significant by a Chi-square test, Chi-square (1) = 5.3, p < .05.” ``` DOCUMENT INFO Shared By: Categories: Tags: Stats: views: 9 posted: 6/13/2012 language: pages: 13
Solve math problem and show steps We can do your math homework for you, and we'll make sure that you understand how to Solve math problem and show steps. We can help me with math work. Solving math problem and show steps It’s important to keep them in mind when trying to figure out how to Solve math problem and show steps. Basic mathematics is the study of mathematical operations and their properties. The focus of this branch of mathematics is on addition, subtraction, multiplication, and division. These operations are the foundation for all other types of math, including algebra, geometry, and trigonometry. In addition to studying how these operations work, students also learn how to solve equations and how to use basic concepts of geometry and trigonometry. Basic mathematics is an essential part of every student's education, and it provides a strong foundation for further study in math. In mathematics, "solving for x" refers to the process of finding the value of an unknown variable in an equation. In most equations, the variable is represented by the letter "x." Fractions can be used to solve for x in a number of ways. For example, if the equation is 2x + 1 = 7, one can isolated the x term by subtracting 1 from each side and then dividing each side by 2. This would leave x with a value of 3. In some cases, more than one step may be necessary to solve for x. For example, if the equation is 4x/3 + 5 = 11, one would first need to multiply both sides of the equation by 3 in order to cancel out the 4x/3 term. This would give 12x + 15 = 33. From there, one could subtract 15 from each side to find that x = 18/12, or 1.5. As these examples demonstrate, solving for x with fractions is a matter of careful algebraic manipulation. With a little practice, anyone can master this essential math skill. Additionally, another way to simplify math is to practice a lot. By doing this, students can become more comfortable with the concepts and procedures involved in solving mathematical problems. With a bit of practice and perseverance, math can become much less daunting for even the most struggling students. Solving system of equations matrices can be a difficult task, but it is important to understand the process in order to be successful. There are many different methods that can be used to solve system of equations matrices, but the most common is Gaussian elimination. This method involves adding or subtracting rows in order to create a new matrix that is easier to solve. Once the new matrix has been created, the variables can be solved for by using back-substitution. This process can be time-consuming and difficult, but it is important to persevere in order to get the correct answer. With practice, solving system of equations matrices will become easier and more intuitive.
# WebWork Hints: Shifting a Circle & Using Logarithms There are two WebWork exercises that go beyond what we’ve covered in class this semester (although you should’ve seen this material in MAT1275 or an equivalent algebra course). Hence I will count these two exercises as extra credit, but thought I’d also provide some hints to help you solve them: Functions – Translations: Problem 4: In this exercises, you are given at equation of the form $latex x^2 + y^2 = r^2$ The graph of this equation is a circle centered at the origin (0,0) with radius r. In general, the equation of a circle centered at a point (h,k) with radius r is (x-h)^2 + (y-k)^2 = r^2 . Use that to figure out the solution to the WebWork exercise, given that you are asked to shift the circle, and hence shift the center of the graph from (0,0) to another point (h,k). Functions – Inverse Functions: Problem 10: In this exercise you are given a function f(x) involving e^(2x) terms, and you are asked to find the inverse function f^{-1}(x). In general, as we outlined in class and as you should’ve done on the previous exercises in this HW set, in order to solve for the inverse function, set up the equation y = f(x), and then solve for x in terms of y. In order to do that in this exercise, where the x terms are in the exponent, you need to use the natural log function. Here’s a simpler example: Say f(x) = 7e^{2x}. Then in order to find f^{-1}(x): y = 7e^{2x} y/7 = e^{2x} and now take the natural log of both sides: ln (y/7) = ln (e^{2x}) On the LHS, ln (e^{2x}) = 2x (this is by the definition of ln, which is the inverse function of e^x!) Hence, ln (y/7) = 2x x = ln (y/7)/2 and so in this example, f^{-1}(x) = ln (x/7)/2
# Modus Ponens & Modus Tollens, with Examples Updated on November 15, 2014 Leonard Kelley holds a bachelor's in physics with a minor in mathematics. He loves the academic world and strives to constantly improve it. ## Basic Notation In symbolic logic, modus ponens and modus tollens are two tools used to make conclusions of arguments as well as sets of arguments. We start off with an antecedent, commonly symbolized as the letter p, which is our "if" statement. Based off the antecedent, we expect a consequent from it, commonly symbolized as the letter q, which is our "then" statement. For example, "If the sky is blue, then it is not raining." Is an argument. "The sky is blue" is our antecedent, while "it is not raining" is our consequent. We can symbolize this argument as p ---> q Which is read as "if p, then q." A ~ in front of a letter means that the statement is false or negated. So if the statement is ~p, that reads as, "The sky is not blue." ## Modus Ponens With this technique, we start off with our argument as a true statement. That is, p ---> q is given. We hold it to be true. Now, if we find that p is a true statement, what can we say about q? Since we know that p implies q, if p is true, then we know that q is true also. This is Modens Ponens (MP), and though it may seem straight-forward, it is often mis-used. For example, if p ---> q and we know that q is true, does that mean p is true also? If it is not raining, then is the sky blue? It could be, but the sky could also be cloudy. Thus, while p could indeed be true in this case, it might not be and we cannot make a conclusion based off of the consequent. When someone tries to confirm the antecedent by using a true consequent, it is a fallacy known as affirming the consequent (AC). ## Modus Tollens Once again, we have p ---> q is true. If we know that the consequent is false (~q), then we can say that the antecedent is false also (~p). Since we know that p implies q, if we do not reach a true consequent then our antecedent must also be false. Since it is raining, the sky is not blue. This method is Modus Tollens (MT). Once again, we must be careful to not misuse this. If we find that ~p, we cannot say that ~q is true also. We know that p ---> q but that does not mean that ~p ---> ~q. Just because the sky is not blue does not mean that it is raining, for it could just be a cloudy day.This fallacy is known as denying the antecedent (DA) and is a common logical trap that people fall into. 2 23 ## Popular 31 118 • ### How to divide an obtuse angle into 4 equal parts: 3 0 of 8192 characters used • Author Leonard Kelley 3 years ago Thanks for the positive remarks everyone! • Johna452 3 years ago Merely a smiling visitor here to share the adore , btw outstanding style. Audacity, more audacity and always audacity. by Georges Jacques Danton. egdcbegceefe • Pharmd538 3 years ago An interesting dialogue is worth comment. I feel that you must write more on this topic, it won't be a taboo subject but typically people are not sufficient to speak on such topics. To the next. Cheers dbkeekeaedde • Author Leonard Kelley 5 years ago I sure hope so, Logic is critical in CS! • Patty Kenyon 5 years ago from Ledyard, Connecticut Good Information and can definitely help those in the Computer Science fields that haven't had a class in Discrete Math which deals with logic as well as binary math!! working
Home > Pre-algebra > Factors and Multiples > Least Common Multiple ## Least Common Multiple #### Terms Multiple - A number multiplied by any other whole number (like 1, 2, 3, 4 …) Prime Number - A number that is evenly divisible only by itself or 1. ## Lesson A Least Common Multiple (also seen as LCM) is the lowest multiple of two or more numbers. A multiple of a number is that number multiplied by any other whole number (like a multiplication table!). So if we have to find the LCM of 4 and 6, we start by listing the multiples of both numbers from smallest to largest. 4: 4 (4x1), 8 (4x2), 12 (4x3), 16 (4x4), 20 (4x5), 24 (4x6), 28 (4x7) 6: 6 (6x1), 12 (6x2), 18 (6x3), 24 (6x4), 30 (6x5)… Now we can see that the lowest number from both these lists is 12. Therefore, 12 is the LCM of 4 and 6. Note: When all the numbers in the problem are prime numbers, it is quicker to determine the LCM by simply multiplying the numbers together. If we have to find the LCM of 2, 3, and 5, we can simply do this: 2 * 3 * 5 = 30 ## Examples #### Least Common Multiple (Example #1) Find the least common multiple (LCM) List the multiples of Multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36 Multiples of 5 are: 5, 10, 15, 20, 25, 30, 35, 40, 45 Find the least common multiple LCM(4, 5) = 20 #### Least Common Multiple (Example #2) Find the least common multiple (LCM) List the multiples of Multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27 Multiples of 12 are: 12, 24, 36, 48, 60, 72, 84, 96, 108 Find the least common multiple LCM(3, 12) = 12 #### Least Common Multiple (Example #3) Find the least common multiple (LCM) List the multiples of Multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27 Multiples of 5 are: 5, 10, 15, 20, 25, 30, 35, 40, 45 Find the least common multiple LCM(3, 5) = 15 #### Least Common Multiple (Example #4) Find the least common multiple (LCM) List the multiples of Multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36 Multiples of 6 are: 6, 12, 18, 24, 30, 36, 42, 48, 54 Multiples of 8 are: 8, 16, 24, 32, 40, 48, 56, 64, 72 Find the least common multiple LCM(4, 6, 8) = 8
mixedmath Explorations in math and programming David Lowry-Duda We start with $\cos( \dfrac{\xi}{2})\cos(\dfrac{\xi}{4}) ... \cos(\dfrac{\xi}{2^n})$. Recall the double angle identity for sin: $\sin 2 \theta = 2\sin \theta \cos \theta$. We will use this a lot. Multiply our expression by $\sin(\dfrac{\xi}{2^n})$. Then we have $$\cos( \dfrac{\xi}{2})\cos(\dfrac{\xi}{4}) \cdots \cos(\dfrac{\xi}{2^n})\sin(\dfrac{\xi}{2^n})$$ Using the double angle identity, we can reduce this: $$= \dfrac{1}{2} \cos( \dfrac{\xi}{2})\cos(\dfrac{\xi}{4}) ... \cos(\dfrac{\xi}{2^{n-1}})sin(\dfrac{\xi}{2^{n-1}}) =$$ $$= \dfrac{1}{4} \cos( \dfrac{\xi}{2})\cos(\dfrac{\xi}{4}) ... \cos(\dfrac{\xi}{2^{n-2}})\sin(\dfrac{\xi}{2^{n-2}}) =$$ $$\ldots$$ $$= \dfrac{1}{2^{n-1}}\cos(\xi / 2)\sin(\xi / 2) = \dfrac{1}{2^n}\sin(\xi)$$ So we can rewrite this as $$\cos( \dfrac{\xi}{2})\cos(\dfrac{\xi}{4}) ... \cos(\dfrac{\xi}{2^n}) = \dfrac{\sin \xi}{2^n \sin( \dfrac{\xi}{2^n} )} for \xi \not = k \pi$$ Because we know that $lim_{x \to \infty} \dfrac{\sin x}{x} = 1$, we see that $lim_{n \to \infty} \dfrac{\xi / 2^n}{\sin(\xi / 2^n)} = 1$. So we see that $$\cos( \dfrac{\xi}{2})\cos(\dfrac{\xi}{4}) ... = \dfrac{\xi}{\xi}$$ $$\xi = \dfrac{\sin(\xi)}{\cos(\dfrac{\xi}{2})\cos(\dfrac{\xi}{4})...}$$ Now we set $\xi := \pi /2$. Also recalling that $\cos(\xi / 2 ) = \sqrt{ 1/2 + 1/2 \cos \xi}$. What do we get? $$\dfrac{\pi}{2} = \dfrac{1}{\sqrt{1/2} \sqrt{ 1/2 + 1/2 \sqrt{1/2} } \sqrt{1/2 + 1/2 \sqrt{ 1/2 + 1/2 \sqrt{1/2} \cdots}}}$$ This is pretty cool. It's called Vieta's Formula for $\dfrac{\pi}{2}$. It's also one of the oldest infinite products. bold, italics, and plain text are allowed in comments. A reasonable subset of markdown is supported, including lists, links, and fenced code blocks. In addition, math can be formatted using $(inline math)$ or $$(your display equation)$$.
# 7x+4=(x+3)+(8x-9) ## Simple and best practice solution for 7x+4=(x+3)+(8x-9) equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework. If it's not what You are looking for type in the equation solver your own equation and let us solve it. ## Solution for 7x+4=(x+3)+(8x-9) equation: 7x+4=(x+3)+(8x-9) We move all terms to the left: 7x+4-((x+3)+(8x-9))=0 We calculate terms in parentheses: -((x+3)+(8x-9)), so: (x+3)+(8x-9) We get rid of parentheses x+8x+3-9 We add all the numbers together, and all the variables 9x-6 Back to the equation: -(9x-6) We get rid of parentheses 7x-9x+6+4=0 We add all the numbers together, and all the variables -2x+10=0 We move all terms containing x to the left, all other terms to the right -2x=-10 x=-10/-2 x=+5 `
# How do you solve b/0.6=18? Jul 3, 2017 $b = 10.8$ #### Explanation: This is the same as $b \times \frac{1}{0.6} = 18$ Multiply both sides by $\textcolor{red}{0.6}$ color(green)(bxx1/0.6color(red)(xx0.6)" "=" "18color(red)(xx0.6) $b \times \frac{0.6}{0.6} \text{ "=" } 18 \times 0.6$ But $\frac{0.6}{0.6}$ is the same as the value 1 giving: $b \times 1 \text{ "=" } 18 \times 0.6$ $b = 18 \times 0.6$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Suppose the decimal is giving you a problem. This is one of several approaches. $0.6$ is the same as $6 \times \frac{1}{10}$ so we write: $b = 18 \times 6 \times \frac{1}{10}$ $b = 108 \times \frac{1}{10}$ $b = 10.8$
Last updated on Apr 9, 2024 ## Latest Quadratic Equations MCQ Objective Questions An odd degree polynomial equation has: 1. At least one real root. 2. No real root. 3. Only one real root. 4. More than one of the above 5. None of the above Option 1 : At least one real root. #### Quadratic Equations Question 1 Detailed Solution Concept: • A polynomial of degree n has exactly n roots. • The complex roots of a polynomial always occur in conjugate pairs. Calculation: Since the complex roots always occur in pairs, we cannot have an odd number of complex roots for any polynomial. Hence, at least one of the (in fact, an odd number of) roots of a polynomial with an odd degree must be real. If the zero of the polynomial x2 – 7kx + 8 = 0 is 6, then find the value of k. 1. $$\frac{22}{21}$$ 2. $$\frac{1}{2}$$ 3. $$\frac{21}{22}$$ 4. More than one of the above 5. None of the above Option 1 : $$\frac{22}{21}$$ #### Quadratic Equations Question 2 Detailed Solution Given: The zero of the polynomial x2 – 7kx + 8 = 0 is 6. Concept: If a is the zero of polynomial P(x) then P(a) = 0 Calculation: x2 – 7kx + 8 = 0 As, x = 6 is zeros of the quadratic equation ⇒ (6)2 - 7k × 6 + 8 = 0 ⇒ 36 - 42k + 8 = 0 ⇒ 44 - 42k = 0 ⇒ 42k = 44 ⇒ k = 44/42 ⇒ k = 22/21 ∴ The correct answer is 22/21. The sum of a positive integer and its square is 2450. The positive integer is 1. 48 2. 49 3. 51 4. More than one of the above 5. None of the above Option 2 : 49 #### Quadratic Equations Question 3 Detailed Solution Calculation: Let positive integer be 'x' ∴ x + x= 2450 x2 + x - 2450 = 0 (x + 50)(x - 49) = 0 ∴ x = 49, - 50 So the positive integer is x = 49. Find the roots of the equation 32x - 10 × (3x) + 9 = 0 1. 1, 1 2. 2, 1 3. 2, 0 4. More than one of the above 5. None of the above Option 3 : 2, 0 #### Quadratic Equations Question 4 Detailed Solution Formula used: xm = x⇒ m = n Calculation: We have, 32x - 10 × (3x) + 9 = 0 (3x)2 - 10 × (3x) + 9 = 0 Let 3x = u u2 -10u + 9 = 0 u2 - 9u - u + 9 = 0 u (u - 9) - 1 (u - 9) = 0 (u - 1)(u - 9) = 0 u = 1, 9 But u = 3x 3x = 1 & 3x = 9 3x = 10 & 3x = 32 x = 0 & x = 2 ∴ The required roots are 0 & 2. #### Comprehension: Consider the following for the next items that follow: A quadratic equation is given by (a - c) x2 - (a - b - c) x + 3k = 0, where a, b, c are real. If b = 4a, then the roots of the equation are: 1. $${{(a-b+c)} \over 2(a-c)}, {{b} \over 2(a-c)}$$ 2. $${{(b+2c)} \over 2(a-c)}, {{-c} \over 2(a-c)}$$ 3. $${{(2a-b+2c)} \over 2(a-c)}, {{-b} \over 2(a-c)}$$ 4. $$1, {{-b} \over 2(a-c)}$$ Option 3 : $${{(2a-b+2c)} \over 2(a-c)}, {{-b} \over 2(a-c)}$$ #### Quadratic Equations Question 5 Detailed Solution Concept: For a quadratic equation, ax2 + bx + k = 0, the roots are given by, $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$ Comparison of Discriminant, 1. For real roots, $$D = {b^2-4ac} >0$$ 2. For real and equal roots, $$D = {b^2-4ac} = 0$$ 3. For imaginary roots, $$D = {b^2-4ac} < 0$$ Calculation: (a - c) x2 - (a - b - c) x + 3k = 0 Therefore, discriminant D = (a - b - c)2 - 4(a - c)(3k) Put k = b/6 D = (a - b - c)2 - 2b(a - c) Since, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz ⇒ D = a2 + b2 + c2 - 2ab + 2bc - 2ca - 2ab - 2bc ⇒ D = a2 + b2 + c2 - 2ca - 4ab ⇒ D = (a2 - 2ca + c2) + b2 - 4ab ⇒ D = (a - c)2 + b(b - 4a) As b = 4a So, D = (a - c)2 Therefore, root of given equations ⇒ $$x = {(a-b-c) \pm \sqrt{{(a-c)^2}} \over 2(a-c)}$$ ⇒ $$x = {(a-b-c) \pm {{(a-c)}} \over 2(a-c)}$$ Considering -ve sign ⇒ $$x = {{a-b-c-a+c} \over 2(a-c)} = \frac{-b}{2(a-c)}$$ Considering +ve sign ⇒ $$x = {{a-b-c+a-c} \over 2(a-c)} =\frac{2a-b-2c}{2(a-c)}$$ ∴ $$x = {{(2a-b+2c)} \over 2(a-c)}, {{-b} \over 2(a-c)}$$ ## Top Quadratic Equations MCQ Objective Questions If 3x2 + ax + 4 is perfectly divisible by x – 5, then the value of a is: 1. -12 2. -5 3. -15.8 4. -15.6 Option 3 : -15.8 #### Quadratic Equations Question 6 Detailed Solution 3x2 + ax + 4 is perfectly divisible by x – 5, ⇒ 3 × 25 + 5a + 4 = 0 ⇒ 5a = -79 ∴ a = -15.8 If α and β are the roots of the quadratic equation (5 + √2) x2 - (4 + √5) x + (8 + 2√5) = 0, then the value of 2αβ/ (α + β) is: 1. 7 2. 4 3. 2 4. 8 Option 2 : 4 #### Quadratic Equations Question 7 Detailed Solution Concept Used: For quadratic equation, ax2 + bx + c = 0, α + β = -b/a and αβ = c/a Calculation: Given equation is (5 + √2) x2 - (4 + √5) x + (8 + 2√5) = 0 On comparing this equation by ax2 + bx + c = 0, we get a = (5 + √2), b = - (4 + √5) and c = (8 + 2√5) Now, αβ = (8 + 2√5)/(5 + √2) and α + β = (4 + √5)/(5 + √2) Now, We have to find the value of 2αβ/(α + β) ⇒ 2[(8 + 2√5)/(5 + √2)] / [(4 + √5)/(5 + √2)] ⇒ 2 [(8 + 2√5) (4 - √5)] / [(4 + √5)/(4 - √5)] ⇒ 2(32 + 8√5 - 8√5 - 10)/11 ⇒ 44/11 = 4 ∴ The required value of 2αβ/ (α + β) is 4. If the roots of equation ax2 + bx + c = 0 are equal and have opposite signs, then which one of the following statements is correct? 1. a = 0. 2. b = 0. 3. c = 0. 4. None of these. Option 2 : b = 0. #### Quadratic Equations Question 8 Detailed Solution Concept: If α and β are the two roots of the quadratic equation Ax2 + Bx + C = 0, then α + β = $$\rm -\dfrac{B}{A}$$ and αβ = $$\rm \dfrac{C}{A}$$. Calculation: Let's say that α and β are the two roots of the quadratic equation ax2 + bx + c = 0, then α + β = $$\rm -\dfrac{b}{a}$$ and αβ = $$\rm \dfrac{c}{a}$$. It is given that α = -β. ∴ -β + β = $$\rm -\dfrac{b}{a}$$ ⇒ $$\rm -\dfrac{b}{a}$$ = 0 b = 0. If α and β are the roots of the equation x2 - q(1 + x) - r = 0, then what is (1 + α)(1 + β) equal to? 1. 1 - r 2. q - r 3. 1 + r 4. q + r Option 1 : 1 - r #### Quadratic Equations Question 9 Detailed Solution Concept: Let us consider the standard form of a quadratic equation, ax2 + bx + c =0 Let α and β be the two roots of the above quadratic equation. The sum of the roots of a quadratic equation is given by: $${\rm{α }} + {\rm{β }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient\;of\;x}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}$$ The product of the roots is given by: $${\rm{α β }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant\;term}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}$$ Calculation: Given: α and β are the roots of the equation x2 - q(1 + x) - r = 0 ⇒ x2 - q - qx - r = 0 ⇒ x2 - qx - (q + r) = 0 Sum of roots = α + β = q Product of roots = αβ = - (q + r) = -q - r To find: (1 + α)(1 + β) (1 + α)(1 + β) = 1 + α + β + αβ = 1 + q - q - r = 1 - r What is the degree of the equation $$\rm\frac {1}{x-3} = \frac {1}{x + 2} - \frac 1 2$$? 1. 0 2. 1 3. 2 4. 3 Option 3 : 2 #### Quadratic Equations Question 10 Detailed Solution Concept: Degree is the highest power of the variable in a given polynomial Calculation: Here, $$\rm\frac {1}{x-3} = \frac {1}{x + 2} - \frac 1 2\\ \Rightarrow \rm\frac {1}{x-3}= \frac{2-x-2}{2 x+4} \\ \Rightarrow\frac{-x}{2 x+4}=\frac{1}{x-3} \\ \Rightarrow\frac{1}{x-3}+\frac{x}{2 x+4}=0 \\ \Rightarrow\frac{2 x+4+x^{2}-3 x}{(x-3)(2 x+4)}=0 \\ \Rightarrow x^{2}-x+4=0$$ ∴Degree = 2 Hence, option (3) is correct. If α, β are the roots of the equation x2 + px + q = 0, then the value of α2 + β2 1. p2 + 2q 2. p2 - 2q 3. p(p2 - 3q) 4. p2 - 4q Option 2 : p2 - 2q #### Quadratic Equations Question 11 Detailed Solution Concept: Let us consider the standard form of a quadratic equation, ax2 + bx + c =0 Let α and β be the two roots of the above quadratic equation. The sum of the roots of a quadratic equation is given by: $${\rm{α }} + {\rm{β }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient\;of\;x}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}$$ The product of the roots is given by: $${\rm{α β }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant\;term}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}$$ Calculation: Given: α and β are the roots of the equation x2 + px + q = 0 Sum of roots = α + β = -p Product of roots = αβ = q We know that (a + b)2 = a2 + b2 + 2ab So, (α + β)2 = α2 + β2 + 2αβ ⇒ (-p)2 = α2 + β2 + 2q ∴ α2 + β2 = p2 - 2q If x + 4 is a factor of 3x2 + kx + 8 then what is the value of k? 1. 4 2. -4 3. -14 4. 14 Option 4 : 14 #### Quadratic Equations Question 12 Detailed Solution Concept used: If p(x) be a function and (x - a) be a factor of p(x) then, p(a) = 0 Calculation: x + 4 is a factor of 3x2 + kx + 8, so x = -4 will be a solution of this equation ⇒ 3(-4)2 + k(-4) + 8 = 0 ⇒ 4k = 48 + 8 ⇒ k = 14 If the difference between the roots of ax2 + bx + c = 0 is 1, then which one of the following is correct? 1. b2 = a(a + 4c) 2. a2 = b(b + 4c) 3. a2 = c(a + 4c) 4. b2 = a(b + 4c) Option 1 : b2 = a(a + 4c) #### Quadratic Equations Question 13 Detailed Solution Concept: Let us consider the standard form of a quadratic equation, ax2 + bx + c =0 Let α and β be the two roots of the above quadratic equation. The sum of the roots of a quadratic equation is given by: $${\rm{α }} + {\rm{β }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient\;of\;x}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}$$ The product of the roots is given by: $${\rm{α β }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant\;term}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}$$ Calculation: Given: difference between the roots of ax2 + bx + c = 0 is 1 Let α and β be the two roots of the above quadratic equation. Sum of roots = α + β = $$- \frac{{\rm{b}}}{{\rm{a}}}$$ Product of roots = α β = $$\frac{{\rm{c}}}{{\rm{a}}}$$ Now, α - β = 1 squaring both sides, we get ⇒ (α - β)2 = 12 ⇒ (α + β)2 - 4α β = 1 ⇒ $$\rm (\frac{-b}{a})^{2} - \frac{4c}{a} = 1$$ ⇒ b2 - 4ac = a2 ⇒ b2 = a2 + 4ac ∴ b2 = a(a + 4c) If α and β are roots of the equation x2 + 5\|x\| - 6 = 0 then the value of \|tan-1 α - tan-1 β\| is 1. $$\dfrac{\pi}{2}$$ 2. 0 3. π 4. $$\dfrac{\pi}{4}$$ Option 1 : $$\dfrac{\pi}{2}$$ #### Quadratic Equations Question 14 Detailed Solution Concept: The modulus value is not negative. tan-1 (- x) = - tan-1 (x) Calculations: Given, equation is x2 + 5\|x\| - 6 = 0 ⇒\|x2\| + 5\|x\| - 6 = 0 ⇒\|x2\| + 6\|x\| - \|x\| - 6 = 0 ⇒\|x\| (\|x\|+ 6) - 1 (\|x\| + 6) = 0 ⇒ (\|x\| + 6) (\|x\| - 1)= 0 ⇒(\|x\| + 6) = 0 and (\|x\| - 1) = 0 ⇒ \|x\| = - 6 and \|x\| = 1 But \|x\| = - 6 which is not possible because value of modulus is not negative. ⇒ \|x\| = 1 ⇒ x = 1 and x = -1 Given , α and β are toots of the equation x2 + 5\|x\| - 6 = 0 Hence, α = 1 and β = -1. Now, consider, \|tan-1 α - tan-1 β\| = \|tan-1 (1) - tan-1 (- 1)\| ⇒ \|tan-1 (1) + tan-1 (1)\| \|2 tan-1 (1)\| 2.$$\rm \dfrac{\pi}{4}$$ ∴ $$\rm \dfrac{\pi}{2}$$ If α, β are the roots of the equation 3x2 + 57x - 5 = 0, then what is $$\frac{\alpha ^3+\beta ^3}{\alpha ^{-3}+\ \beta ^{-3}}$$ equal to ? 1. - 27/125 2. 81/125 3. 27/125 4. -125/27 Option 4 : -125/27 #### Quadratic Equations Question 15 Detailed Solution Concept: Consider a quadratic equation: ax2 + bx + c = 0. Let, α and β are the roots. • Sum of roots = α + β = -b/a • Product of the roots = α × β = c/a Calculation: Given quadratic equation: 3x2 + 57x - 5 = 0 Let α and β are roots, then α + β = -57/3, αβ = -5/3 Now,$$\frac{α ^3+β ^3}{α ^{-3}+β ^{-3}}$$$$\frac{α ^3+β ^3}{\frac {1}{α ^{3}}+\frac{1}{β ^{3}}}$$ $$\frac{α ^3+β ^3}{\frac {(α ^3+β ^3)}{α^3 β^3 }}$$ = (α β)3 = (-5/3)3 = -125/27 Hence, option (4) is correct.
Beispiel Nr: 35 $\text{Gegeben:} ax^{2}+bx+c=0 \\ \text{Gesucht:} \\ \text{Lösung der Gleichung} \\ \\ ax^{2}+bx+c=0 \\ \textbf{Gegeben:} \\ -\frac{1}{8}x^2+\frac{1}{4}x+7\frac{7}{8} =0 \\ \\ \textbf{Rechnung:} \\ \begin{array}{l\|l\|l} \begin{array}{l} \text{a-b-c Formel}\\ \hline \\ -\frac{1}{8}x^{2}+\frac{1}{4}x+7\frac{7}{8} =0 \\ x_{1/2}=\displaystyle\frac{-\frac{1}{4} \pm\sqrt{\left(\frac{1}{4}\right)^{2}-4\cdot \left(-\frac{1}{8}\right) \cdot 7\frac{7}{8}}}{2\cdot\left(-\frac{1}{8}\right)} \\ x_{1/2}=\displaystyle \frac{-\frac{1}{4} \pm\sqrt{4}}{-\frac{1}{4}} \\ x_{1/2}=\displaystyle \frac{-\frac{1}{4} \pm2}{-\frac{1}{4}} \\ x_{1}=\displaystyle \frac{-\frac{1}{4} +2}{-\frac{1}{4}} \qquad x_{2}=\displaystyle \frac{-\frac{1}{4} -2}{-\frac{1}{4}} \\ x_{1}=-7 \qquad x_{2}=9 \end{array}& \begin{array}{l} \text{p-q Formel}\\ \hline \\ -\frac{1}{8}x^{2}+\frac{1}{4}x+7\frac{7}{8} =0 \qquad /:-\frac{1}{8} \\ x^{2}-2x-63 =0 \\ x_{1/2}=\displaystyle -\frac{-2}{2}\pm\sqrt{\left(\frac{\left(-2\right)}{2}\right)^2- \left(-63\right)} \\ x_{1/2}=\displaystyle 1\pm\sqrt{64} \\ x_{1/2}=\displaystyle 1\pm8 \\ x_{1}=9 \qquad x_{2}=-7 \end{array}\\ \end{array}$
# How do you simplify (4x-1)/(3x) + (x-8)/(5x)? ##### 1 Answer Oct 14, 2015 The answer is $\frac{23 x - 29}{15 x}$. #### Explanation: $\frac{4 x - 1}{3 x} + \frac{x - 8}{5 x}$ The LCD is $15$. Multiply each fraction so that its denominator will be $15 x$. $\frac{4 x - 1}{3 x} \times \frac{5}{5} + \frac{x - 8}{5 x} \times \frac{3}{3} =$ (5(4x-1))/(5(3x))+(3(x-8))/(3(5x)= Distribute the $5$ and the $3$ in the numerators and simplify $5 \cdot 3 x$ to $15 x$ and $3 \cdot 5 x$ to $15 x$. $\frac{20 x - 5}{15 x} + \frac{3 x - 24}{15 x}$ Combine the numerators over the denominator $15 x$. $\frac{20 x - 5 + 3 x - 24}{15 x}$ Combine like terms. $\frac{20 x + 3 x - 5 - 24}{15 x} =$ $\frac{23 x - 29}{15 x}$
Journal Chapter 5 1 / 39 # Journal Chapter 5 - PowerPoint PPT Presentation Journal Chapter 5 . Full screen to listen to music. TRIANGLES……. … TRIANGLES…. … AND… … MORE… TRIANGLES. Menu options ahead. Option #1. Perpendicular Bisector:. What is it? A line that bisects a segment and is perpendicular to that same segment. . THEOREM: . I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. ## PowerPoint Slideshow about 'Journal Chapter 5' - wynona Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - Presentation Transcript Journal Chapter 5 Full screen to listen to music TRIANGLES……. TRIANGLES…. AND… MORE… TRIANGLES. Perpendicular Bisector: What is it? A line that bisects a segment and is perpendicular to that same segment. THEOREM: * Any point that is on the perpendicular bisector is equidistant to both of the endpoints of the segment which is perpendicular to. Converse: If the point is equidistant from both of the endpoints of the segment, then it is a perpendicular bisector. Extra: A perpendicular bisector is the locus of all points in a plane that are equidistant from the endpoints of a segment. 7cm Segment Perpendicular bisector Are both congruent. 7cm EXAMPLES 1. Given that AC= 25.5, BD=20, and BC= 25.5. Find AB. AB= 51 2. Given that m is the perpendicular bisector of AB and BC= 30 . Find AC. AC= 30 Given that m is the perpendicular bisector of AB , AC= 4a, and BC= 2a+26. Find BC. 2a+26=4a -2a=-26 a=13 plug in both equations BC= 52 a d m c b Angle Bisectors What is it? A ray or any line that cuts an angle into two congruent angles. Theorem : Any point that lies on the angle bisector is equidistant to both of the sides of the angle. Converse: * If it is equidistant to both sides of the angle, then it lies on the angle bisector. MUST: the point always has to lie on the interior of the angle. 1.5 cm 1.5 cm Angle point that Angle bisector lies on the bisector is equidistant to both sides of the angle EXAMPLES Given that BD bisects ∠ABC and 2. Given that AD =65 , CD= 65 and m∠ABC = 50˚, find m∠CBD. m∠DBC= 25˚ 3. Given that DA=DC , m∠DBC= (10y+3)˚ , and m ∠DBA= (8y+10) ˚, find m∠DBC. 10y+3=8y+10 2y+3=10 2y=7 y=3.5 plug in both equations. m∠DBC=38˚ A D B C What is a circumcenter? It is the point where all the 3 perpendicular bisectors of a triangle meet. What makes it special? It is the point of concurrency and it is equidistant to all 3 vertices. 1. Knowing what circumcenter is, it would help you if you are a business person to know a strategic place in between 3 cities to place your business. That way it would be located in the same distance from all 3 cities for customers to come. EXAMPLE F is the circumcenter in this equilateral triangle. The circumcenter is on the inside of the triangle. Examples: 2.In an acute the circumcenter is on the inside of the triangle C is the circumcenter 3.In a right the circumcenter is on the midpoint of the hypotenuse D is the circumcenter 4.In an obtuse the circumcenter is on the outside of the triangle. F is the circumcenter f c d Concurrency of Angle bisectors Of a triangle theorem The angle bisectors of a triangle meet at a point that is equidistant from the three sides of the triangle. What is the distance from H to G, if F to H is 10 cm? 10cm from H to G 2. H to E is 15cm , then what would be H to G doubled? Real: H to G would be 15cm and doubled would be 30cm H in this triangle is what? The incenter of triangle ABC Incenter of a TRIANGLE • The point of concurrency where the three angle bisectors of a triangle meet. It is equidistant (same distance) from each three sides of the triangle EXAMPLES: 2. If you want to open a new restaurant in town you have to place it at the incenter of three major highways so people can go to your restaurant. 3. If the distance from K to P is 2cm, what is the distance from P to L? From P to L = 2cm G. M K P J L H 1. P is the incenter Median of a triangle Ex.1 : Median Is the segment that goes from the vertex of a triangle to the opposite midpoint Ex. 2: all medians are congruent. Shown in yellow Ex.3 : Median Centroid The point where the medians of a triangle intersect. Is the center of balance. The distance from the vertex to the centroid is the double the distance from the centroid to the opposite side. EXAMPLES: AP= 2/3 AY BP=2/3BZ CP=2/3CX 1. Balance a triangular piece of glass on a triangular coffee table. 3. 2. G is the centroid. b y p x c z a Concurrency of Medians of a Triangle Theorem The medians of a triangle intersect at a point that is 2/3 of the distance from each vertex to the midpoint of the opposite side A segment that goes from the vertex perpendicular to the line containing the opposite side. ALTITUDE TRIANGLES !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! hypotenuse altitude ORTHOCENTER Where the altitudes of a triangle intersect. Concurrency of Altitudes of triangles Concurrency of Altitudes of a Triangle Theorem The lines containing the altitudes of a triangle are concurrent Midsegments: A segment that joins the midpoints of two sides of the triangle. Every triangle has 3midsegments that form the midsegment triangle. Midsegment Theorem The segment joining the midpoints of two sides of a triangle is parallel to the third side, and is half the length of the third side. Side – Angle Relationship In any triangle the longest side is always opposite from the largest angle. The small side is opposite from a small angle. EXAMPLES: Exterior angle inequality The exterior angle is bigger (grater) than the non-adjacent interior angles of the triangle . TELL WHEATHER A TRIANGLE CAN HAVE THE SIDES WITH THE GIVEN LENGTHS. EXPLAIN 3,5,7 2. 4, 6.5,11 3. 7, 4, 5 3 + 5 > 7 4 +6.5> 11 5+4 > 7 8>7 10.5>11 9>7 YES, the sum of each NO. by the 3rd triangle Yes , 5+4 is grater than 7 so Pair of the lengths is inequality thm. A triangle there can be a triangle. Grater than the 3rd <. Cannot have these side lengths. TRIANGLE INEQUALITY Triangle Inequality Theorem The sum of the lengths of two sides of a triangle is greater than the length of the third side. Examples: PROVE: IF A>0 THAN 1/A >0 Assume a>0 than 1/a<0 (a) 1/a <0a a 1 <0 3. Therefore if a>0 than 1/a>0 PROVE: A SCALENE TRIANGLE CANNOT HAVE TWO CONGRUENT ANGLES Assume a scalene triangle can have two congruent sides a triangle with 2 congruent angles… deff. of an isosceles triangle Deff. Of isosceles and scalene triangles… therefore a scalene triangle cannot have two congruent angles. PROVE: A TRIANGLE CANNOT HAVE 2 RIGHT ANGLES assume a triangle has 2 right angles <1+<2…. GIVEN M<1 =M<2 = 90.. DEFF OF RT. < M<1+M<2+M<3=180 …..TRIANGLE + THM So a triangle cannot have 2 right angles. HINGE THEOREM If 2 triangles have 2 sides that are congruent, but the third side is not congruent, then the triangle with the large included angle has the longer side. (What changes is the incidence and the side….) Theorem: If two sides of one triangle are congruent to the two sides of another triangle and the third sides are not congruent, then the larger included angle is across from the longer third side. Converse : 45˚- 45 ˚- 90˚ 90˚- 30˚- 60˚ 45 60 Both legs of the triangle are congruent and the length of the hypotenuse is the length of the two legs √2 The length of the hypotenuse is 2x the length of the shorter leg, and the length of the longer leg is the length of the shorter leg time √3 90 45 Examples: 30 90 18√3 =2x 9 √3=x y=x √3 y=27 60 18√3 x y y= 2x Y=2(5 √3) Y= 10 √3 x 2 . 15= √3 15/ √3=x 5 √3=x 60 y 30 5 X=5 √3 Y=2(5) Y=10 30 y
# How Do You Know If A Ratio Is A Proportion? ## What is single population proportion formula? X = Zα/22 p(1-p) / MOE2, and Zα/2 is the critical value of the Normal distribution at α/2 (e.g. for a confidence level of 95%, α is 0.05 and the critical value is 1.96), MOE is the margin of error, p is the sample proportion, and N is the population size.. ## How many ways can you write a ratio? three different waysA ratio can be written in three different ways: with the word “to”: 3 to 4. as a fraction: . with a colon: 3 : 4. ## What does 1 to 3 ratio mean? Example: if there is 1 boy and 3 girls you could write the ratio as: 1:3 (for every one boy there are 3 girls) 1/4 are boys and 3/4 are girls. 0.25 are boys (by dividing 1 by 4) 25% are boys (0.25 as a percentage) ## What is a 4 to 1 ratio? If your mix ratio is 4:1 or 4 parts water to 1 part solution, there are (4 + 1) or 5 parts. ## How do you determine a ratio? To find an equal ratio, you can either multiply or divide each term in the ratio by the same number (but not zero). For example, if we divide both terms in the ratio 3:6 by the number three, then we get the equal ratio, 1:2. Do you see that these ratios both represent the same comparison? ## What is the ratio of 2 to 4? 1:2Multiplying or dividing each term by the same nonzero number will give an equal ratio. For example, the ratio 2:4 is equal to the ratio 1:2. ## What is a 2 to 5 ratio? The ratio is the relationship of two numbers. … The ratio can consequently be expressed as fractions or as a decimal. 2:5 in decimals is 0.4. A rate is a little bit different than the ratio, it is a special ratio. ## How do you find the average proportion? To find the average percentage of the two percentages in this example, you need to first divide the sum of the two percentage numbers by the sum of the two sample sizes. So, 95 divided by 350 equals 0.27. You then multiply this decimal by 100 to get the average percentage. ## What is the difference between proportion and ratio? A ratio is a way to compare two quantities by using division as in miles per hour where we compare miles and hours. A proportion on the other hand is an equation that says that two ratios are equivalent. … If one number in a proportion is unknown you can find that number by solving the proportion. ## What is the proportion formula? A proportion is simply a statement that two ratios are equal. It can be written in two ways: as two equal fractions a/b = c/d; or using a colon, a:b = c:d. The following proportion is read as “twenty is to twenty-five as four is to five.” ## What are the types of proportion? There are two types of proportions.Direct Proportion.Indirect Proportion. ## What is the ratio of 3 to 5? 3 : 5 = ? : 40. (3 out of 5 is how many out of 40?) ## What are examples of proportions? Proportion says that two ratios (or fractions) are equal….Example: Rope40m of that rope weighs 2kg.200m of that rope weighs 10kg.etc. ## How do you tell if a proportion is true or false? If the simplified fractions are the same, the proportion is true; if the fractions are different, the proportion is false. ## Does a ratio indicate division? In mathematics, a ratio is a comparison of two or more numbers that indicates their sizes in relation to each other. A ratio compares two quantities by division, with the dividend or number being divided termed the antecedent and the divisor or number that is dividing termed the consequent. ## How do you find the true proportion? 1 Divide the number. Divide the number of people in the sample population who have the characteristic being tested by the total number of people in the sample to get the sample proportion.2 Subtract the sample proportion from one. … 3 Take the square root. … 4 Multiply the result by 1.96. … 5 Subtract the result. ## What does a proportion mean? 1 : harmonious relation of parts to each other or to the whole : balance, symmetry. 2a : proper or equal share each did her proportion of the work. b : quota, percentage. 3 : the relation of one part to another or to the whole with respect to magnitude, quantity, or degree : ratio.
• Join over 1.2 million students every month • Accelerate your learning by 29% • Unlimited access from just £6.99 per month Page 1. 1 1 2. 2 2 3. 3 3 4. 4 4 5. 5 5 6. 6 6 7. 7 7 8. 8 8 9. 9 9 10. 10 10 11. 11 11 12. 12 12 • Level: GCSE • Subject: Maths • Word count: 1890 # How many squares in a chessboard n x n Extracts from this document... Introduction HOW MANY SQUARES ON A CHESSBOARD (n x n)? n DIAGRAM 1 PROBLEM STATEMENT We were asked to find how many squares were on a chessboard. By observation a chessboard has 8 rows by 8 columns of squares plus one big square, which is the board itself. This will give 65 squares as the answer. However looking at the question in depth, it is seen that many squares can be formed within in the chessboard. The most systematic approach to find exactly how many squares there are in a chessboard, would be to start from the smallest square to the largest square. We know that a square has 4 equal sides, therefore we can move up n row(s) and across n column(s) on the chessboard, which gives us (n x n) and is also equal to n2 (n squared). Since the chessboard is an 8 by 8, we can rewrite (n x n) as 8 x 8, where n = 8. So for a single square we write (n – 7)(n –7), where n is 8 in the equation, giving (8 – 7)(8 –7) =1 square. GENERATING DATA AND DESCRIPTION OF STRUCTURE Applying the above information, we decided to generate the data starting from the smallest square, which is a 1 x 1 square to the largest square, which is n x n. This is a one by one square and can be represented mathematically as 12 (1 squared) or (n – 7)(n – 7) =1 x 1 = 1, where n = 8 Middle (n – 4)(n – 4) + 9 + 4 + 1 = 16 + 9 + 4 + 1, which gives the total number of squares in this particular square. This result will be added to the next 5 x 5 square. This is 5 by 5 square plus previous sequence (16 + 9 + 4 + 1). It’s mathematical form is 52 + 16+ 9 + 4 + 1 or (n – 3)(n – 3) + 16 +9 + 4 + 1 = 25 +16 + 9 + 4 +1, which gives the total number of squares in this particular square. This result will be added to the next 6 x 6 square. This is 6 by 6 square plus previous sequence (25 + 16 + 9 + 4 + 1). It’s mathematical form is 62 + 25 + 16 + 9 + 4 + 1 or (n – 2)(n – 2) + 25 + 16 + 9 + 4 + 1 = 36 + 25 +16 + 9 + 4 + 1, which gives the total number of squares in this particular square. This result will then be added to the next 7 x 7 square. This is a 7 by 7 square plus previous sequence (36 + 25 +16 + 9 + 4 + 1). It’s mathematical form is 72 + 36 + 25 + 16 + 9 + 4 +1 or (n – 1)(n - ) + 36 + 25 + 16 + 9 + 4 +1 =49 + 36 + 25 +16 + 9 + 4 +1, which gives the total number of squares in this particular square. This result will then be added to the next 8 x 8 square. This is 8 x 8 square plus previous sequence (49 +36 + 25 + 16 + 9 + 4 + 1). Since n = 8, as stated earlier, it’s mathematical form is 82 + 49 + 36 + 25 + 16 + 9 + 4 + 1 or 64 + 49 + 36 + 25 + 16 + 9 + 4 +1 = 204. We notice that the 1st square, U1 = 12 2nd square, U2 = 12 + 22 3rd square, U3 = 12 + 22 + 32 4th square, U4 = 12 + 22 + 32 + 42 5th square,U5 =12 + 22 + 32 + 42 + 52 6th square, U6 = 12 + 22 + 32 + 42 + 52 + 62 7th square, U7 = 12 + 22 + 32 + 42 + 52 + 62 + 72 8th square, U8 = 12 + 22 + 32 + 42 + 52 + 62 + 72 + 82 From the above sequence, it is seen that the pattern arising is the series of powers of the natural numbers, giving Un = 12 + 22 + ……….+ n2 Therefore the general solution for the sequence above is: Un= Un-1+ n2 TESTING THIS SOLUTION U4 = U3 + 4 U4 = 12 + 22 + 32 + 42 Conclusion If we have a 3x5 rectangle, we can make a squares from 1x1 to 3x3,so we have two columns left over, which gives us (m – 1)(n – 1) + (m – 2)(m – 2) in the equation. DIAG(b) n=5 U3X5 = mxn + (m-1)(n-1) + (m-2)(n-2) m=3 (2) VISUAL EXPLANATION EXAMPLE Using a 3x5 rectangle to explain the structure. Diag 1 (a) Looking at 1x1 squares in the rectangle (mxn) n=5 m=3 Diag (b) Looking at 2x2 squares in the rectangle Diag (c) Pulling the overlapping squares apart in the rectangle gives us a new value of m and n. In this case m and n have Decreased by one, (m-1)(n-1). n=5 n=4 m=3 m=2 Diag (d) Looking at 3x3 squares in the rectangle Diag (e) Pulling the overlapping squares apart in the rectangle shows m and n have decreased by two, (m-2)(n-2). n=5 n=3 m=3 m=1 Therefore, final formula for a 3x5 rectangle chessboard gives U3x5= mxn + (m-1)(n-1) + (m-2)(n-2) This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section. ## Found what you're looking for? • Start learning 29% faster today • 150,000+ documents available • Just £6.99 a month Not the one? Search for your essay title... • Join over 1.2 million students every month • Accelerate your learning by 29% • Unlimited access from just £6.99 per month # Related GCSE Number Stairs, Grids and Sequences essays 1. ## &quot;Multiply the figures in opposite corners of the square and find the difference between ... numbers: x x +1 x + 11 x + 12 x(x + 12) = x2 + 12x (x + 1)(x + 11) = x2 + 12x + 11 Difference is 11 3 by 3: 3 4 5 14 15 16 25 26 27 5 x 25 2. ## Investigation of diagonal difference. the bottom left corner is twice the length of the grid, from noticing this I can now implement G into the equation, only if the number added to n in the bottom left corner is always the size of the grid that the 3 x 3 cutout originated from. 1. ## Investigate Borders - a fencing problem. Diagram of Borders of square: 3x2 Table of results for Borders of square: 3x2 Formula to find the number of squares needed for each border (for square 3x2): Common difference = 4 First term = 10 Formula = Simplification = Experiment I will try to find the number of squares 2. ## Investigating when pairs of diagonal corners are multiplied and subtracted from each other. 1 x 45 = 45 5 x 41 = 205 Difference = 160 6 x 50 = 300 10 x 46 = 460 Difference = 160 53 x 97 = 5141 57 x 93 = 5301 Difference = 160 In a 5 x 5 box on a 10 x 10 grid the difference is 160. 1. ## I am doing an investigation to look at borders made up after a square ... Then I will see if the answer is 12. In this case it isn't so I will have to add 8 to get to 12. So my rule is 4B+8. Using this rule I predict that the number of numbered squares in the 6thborder is 32. 6*4+8=32 6th Border 6 6 6 6 6 6 5 5 5 5 2. ## &quot;Multiply the figures in opposite corners of the square and find the difference between ... The number generated will be the number in the top left hand corner of the square. 89 90 99 100 90 x 99 = 8910 89 x 100 = 8900 The difference yet again is 10; this is no mere coincidence and therefore is a constant pattern within all the 2 by 2 squares that I have tested so far. • Over 160,000 pieces of student written work • Annotated by experienced teachers • Ideas and feedback to improve your own work
# First Grade Math — Secret Numbers This week in first grade math, we worked on using clues to locate a “secret number” on a hundreds chart. I gave the kids four clues for each number. As they received the clues, they used dry erase markers to eliminate numbers on the chart that could not be the secret number. For example, if the clue was “the number is odd,” then the kids crossed off all of the even numbers. This is a bit more confusing than perhaps it sounds, because one has to be careful to be sure to mark off the inverse of the clue (if it say even, mark the odds). Most of the kids got the idea quickly. Of course, then the clues got harder (“if you add the digits, you get 12,” for example. They worked hard (for some of them, it was almost equally difficult not to shout out the number as soon as they found it). Homework is to create a set of four clues that lead to a secret number. All of the clues should be necessary to find the number, and they should lead to only one number. I sent home a hundreds chart to help. Please do not cut out the clue cards — keep them as a whole sheet, turn them in and let me check them, and then we will use them in class the next time we play “Secret Number.” I was taking pictures today and these two of the long sides of the big table made me laugh. I felt like I was seeing double… twice! I hope the kids have fun writing their clues! This week in reading, we discussed portmanteau words. (A debt of gratitude to my kindergarten pal, Miles Walters, who gave me the inspiration for this lesson when he pointed out that “cloffice” is a portmanteau). First, we discussed what a portmanteau is — two words (and their meanings) combined into a new word. The kids were determined to convince me that compound words and portmanteau words are one and the same, but in the end, we agreed: compound words are different because the words (like cup and cake) remain unchanged after they are combined. Words combined to form a portmanteau do not. We listed some examples of portmanteaus, starting with brunch. All of the kids were able to identify brunch as a combination of breakfast and lunch. When I asked the girls if they could name a kind of pants with a portmanteau name, they chimed right in with jeggings. We also identified the combined words in smog, coopetition, and ginormous. Next, we read Stardines Swim High Across the Sky by Jack Prelutsky. This book, beautifully illustrated by Carin Berger, is a series of poems describing portmanteau animals whose names are a combination of the kind of animal they are and their most defining characteristic. For example, Slobsters are lobsters who are slobs, the Gloose is a Goose who sticks to everything, and the Tattlesnake is a rattlesnake who is constantly tattling on everyone. Homework this week is to create portmanteau animals of our own. The animals must be a combination of animal type + defining characteristic. Although we discussed the portmanteau food/animal creatures in Cloudy with Meatballs 2, those types of animals don’t meet the criteria for this particular assignment (though those hippotatoes are pretty darn cute, no?). The students need to draw their animal and explain the characteristic(s) that gave rise to its name. As I told them in class, one sentence is not sufficient. The kids need to be creative and descriptive! Just for inspiration, here is a student’s “Callphin,” a dolphin who is always on the phone. Hey, I just met you, and this is crazy, but invent your animal and… call me, maybe? The book is about a boy named Bob who discovers that his life is made up of tons of palindromes. He finally decides to stop the madness by going by his full name — Robert Trebor. (Not such a helpful strategy, it turns out). We talked about how palindromes are words (or phrases or numbers) that read the same forwards and backwards. We counted the palindromes in the book and came up with some of our own. WOW, was it fun! This week’s homework is a spread from the book featuring Bob’s teacher, Miss Sim. The kids need to fill in the background of the picture with palindromes — either words or phrases. This shouldn’t be too taxing, so no one should worry about getting “too hot to hoot.” # Kindergarten Math — I Value Xylophones Like Cows Do Milk How ’bout that title? Did your child come home saying that crazy sentence? (If not, please say it yourself and see how your child reacts!) Can your kindergarten student tell you what that sentence means/what the letters stand for? I wrote the sentence on the board at the start of class and read it to the kids. I wrote each of the capital letters in red ink and the rest of the letters in black. The kids were easily able to figure out that the red letters were important for some reason, but they were stumped as to why. Gradually, we made our way to the conclusion that the letters are the Roman Numerals. We then wrote down the letters and their values: • I = one • V = five • X = ten • L = fifty • C = one hundred • D = five hundred • M = one thousand We talked about how to write and read Roman numerals. We didn’t watch Brain Pop in class, but if your student needs a refresher, Brain Pop has a great movie about Roman numerals (here). Most of your children should be familiar with Brain Pop (and Brain Pop Jr.) from school. The mini-movies on the site are chock full of great information. You need a login to access them — our login name is tjelem and the password is tiger. This only works during school hours (or so I’m told). Subtractive notation isn’t easy. We practiced finding numbers where a lower number was written to the left of a higher one, and reminded ourselves that the numbers need to be read as a pair and the one on the left needs to be subtracted from the one on the right (ex. IV = 4). This is probably the easiest way to do the homework — group the numbers that require subtractive notation and circle them, then subtract and add as necessary. The homework is a double-sided sheet of problems — writing Hindu-Arabic numerals as Roman numerals and vice versa. It is difficult! Please help your child work to/through frustration, but if s/he reaches the breaking point (or you do), pause or skip. I also sent home a completely optional crossword puzzle. This is not a typical crossword puzzle, as all of the answers are in Roman numerals. There is no need to complete the crossword if your child isn’t interested; I just thought it was fun and different. I hope you know that I understand that this stuff is not typical kindergarten fare. From the homework coming back, it seems like you do — the kids are taking their best shot at things and leaving off when they reach frustration level. That’s what they should be doing. I don’t expect all of the kids to get these higher-level concepts all of the time. A lot of it may go over your child’s head. That’s okay! Exposure is important, and while they may not remember (or understand) all of what we talk about, they’ll each take away their own pieces of information. Keep valuing those xylophones!
# What volume will a balloon occupy at 1.0 atm, if the balloon has a volume of 7.6 L at 3.8 atm? Oct 29, 2016 $\text{29 L}$ #### Explanation: The thing to remember here is that pressure and volume have an inverse relationship when temperature and number of moles of gas are kept constant -- this is known as Boyle's Law. In other words, when temperature and number of moles of gas are kept constant, increasing the pressure will cause the volume to decrease; similarly, decreasing the pressure will cause the volume to increase. Mathematically, this is written as $\textcolor{b l u e}{\overline{\underline{\| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{P}_{1} \cdot {V}_{1} = {P}_{2} \cdot {V}_{2}} \textcolor{w h i t e}{\frac{a}{a}} \|}}}$ Here ${P}_{1}$, ${V}_{1}$ are the pressure and volume of the gas at an initial state ${P}_{2}$, ${V}_{2}$ are the pressure and volume of the gas at a final state Rearrange to solve for ${V}_{2}$ ${P}_{1} \cdot {V}_{1} = {P}_{2} \cdot {V}_{2} \implies {V}_{2} = {P}_{1} / {P}_{2} \cdot {V}_{1}$ Plug in your values to find ${V}_{2} = \left(3.8 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))))/(1.0color(red)(cancel(color(black)("atm")))) * "7.6 L" = color(green)(bar(ul(\|color(white)(a/a)color(black)("29 L}} \textcolor{w h i t e}{\frac{a}{a}} \|}}\right)$ The answer is rounded to two sig figs. As you can see, decreasing the pressure of the gas caused its volume to increase.
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 8.1: Elementary and Measurable Functions $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $f : S \rightarrow\left(T, \rho^{\prime}\right)$ whose domain $$D_{f}$$ consists of points of a set $$S .$$ The range space $$T$$ will mostly be $$E,$$ i.e., $$E^{1}, E^{}, C, E^{n},$$ or another normed space. We assume $$f(x)=0$$ unless defined otherwise. (In a general metric space $$T,$$ we may take some fixed element $$q$$ for $$0 .$$ ) Thus $$D_{f}$$ is all of $$S$$, always. We also adopt a convenient notation for sets: $"A(P)" \text { for } "\{x \in A \| P(x)\}."$ Thus \begin{aligned} A(f \neq a) &=\{x \in A \| f(x) \neq a\}, \\ A(f=g) &=\{x \in A \| f(x)=g(x)\}, \\ A(f>g) &=\{x \in A \| f(x)>g(x)\}, \text { etc. } \end{aligned} ## Definition A measurable space is a set $$S \neq \emptyset$$ together with a set ring $$\mathcal{M}$$ of subsets of $$S,$$ denoted $$(S, \mathcal{M})$$. Henceforth, $$(S, \mathcal{M})$$ is fixed. ## Definition An M-partition of a set $$A$$ is a countable set family $$\mathcal{P}=\left\{A_{i}\right\}$$ such that $A=\bigcup_{i} A_{i}(d i s j o i n t),$ with $$A, A_{i} \in \mathcal{M}$$. We briefly say "the partition $$A=\bigcup A_{i} .$$" An $$\mathcal{M}$$-partition $$\mathcal{P}^{\prime}=\left\{B_{i k}\right\}$$ is a refinement of $$\mathcal{P}=\left\{A_{i}\right\}\left(\text { or } \mathcal{P}^{\prime} \text { refines }\right.$$ $$\mathcal{P},$$ or $$\mathcal{P}^{\prime}$$ is finer than $$\mathcal{P} )$$ iff $(\forall i) \quad A_{i}=\bigcup_{k} B_{i k}$ i.e., each $$B_{i k}$$ is contained in some $$A_{i}$$. The intersection $$\mathcal{P}^{\prime} \cap \mathcal{P}^{\prime \prime}$$ of $$\mathcal{P}^{\prime}=\left\{A_{i}\right\}$$ and $$\mathcal{P}^{\prime \prime}=\left\{B_{k}\right\}$$ is understood to be the family of all sets of the form $A_{i} \cap B_{k}, \quad i, k=1,2, \dots$ It is an $$\mathcal{M}$$ -partition that refines both $$\mathcal{P}^{\prime}$$ and $$\mathcal{P}^{\prime \prime}$$. ## Definition A map (function) $$f : S \rightarrow T$$ is elementary, or $$\mathcal{M}$$-elementary, on a set $$A \in \mathcal{M}$$ iff there is an M-partition $$\mathcal{P}=\left\{A_{i}\right\}$$ of $$A$$ such that $$f$$ is constant $$\left(f=a_{i}\right)$$ on each $$A_{i} .$$ If $$\mathcal{P}=\left\{A_{1}, \ldots, A_{q}\right\}$$ is finite, we say that $$f$$ is simple, or $$\mathcal{M}$$-simple, on $$A .$$ If the $$A_{i}$$ are intervals in $$E^{n},$$ we call $$f$$ a step function; it is a simple step function if $$\mathcal{P}$$ is finite. The function values $$a_{i}$$ are elements of $$T$$ (possibly vectors). They may be infinite if $$T=E^{} .$$ Any simple map is also elementary, of course. ## Definition A map $$f : S \rightarrow\left(T, \rho^{\prime}\right)$$ is said to be measurable (or $$\mathcal{M}$$ -measurable $$)$$ on a $$\operatorname{set} A$$ in $$(S, \mathcal{M})$$ iff $f=\lim _{m \rightarrow \infty} f_{m} \quad(\text { pointwise }) \text { on } A$ for some sequence of functions $$f_{m} : S \rightarrow T,$$ all elementary on $$A .$$ (See Chapter 4, §12 for "pointwise.") Note 1. This implies $$A \in \mathcal{M},$$ as follows from Definitions 2 and $$3 .(\mathrm{Why} \text { ? })$$ ## Corollary $$\PageIndex{1}$$ If $$f : S \rightarrow\left(T, \rho^{\prime}\right)$$ is elementary on $$A,$$ it is measurable on $$A .$$ Proof Set $$f_{m}=f, m=1,2, \ldots,$$ in Definition $$4 .$$ Then clearly $$f_{m} \rightarrow f$$ on $$A$$. $$square$$ ## Corollary $$\PageIndex{2}$$ If $$f$$ is simple, elementary, or measurable on $$A$$ in $$(S, \mathcal{M}),$$ it has the same property on any subset $$B \subseteq A$$ with $$B \in \mathcal{M}$$. Proof Let $$f$$ be simple on $$A ;$$ so $$f=a_{i}$$ on $$A_{i}, i=1,2, \ldots, n,$$ for some finite $$\mathcal{M}$$ -partition, $$A=\bigcup_{i=1}^{n} A_{i}$$. If $$A \supseteq B \in \mathcal{M},$$ then $\left\{B \cap A_{i}\right\}, \quad i=1,2, \ldots, n,$ is a finite $$\mathcal{M}$$ -partition of $$B(\text { why? }),$$ and $$f=a_{i}$$ on $$B \cap A_{i} ;$$ so $$f$$ is simple on $$B$$. For elementary maps, use countable partitions. Now let $$f$$ be measurable on $$A,$$ i.e., $f=\lim _{m \rightarrow \infty} f_{m}$ for some elementary maps $$f_{m}$$ on $$A .$$ As shown above, the $$f_{m}$$ are elementary on $$B,$$ too, and $$f_{m} \rightarrow f$$ on $$B ;$$ so $$f$$ is measurable on $$B . \quad \square$$ ## Corollary $$\PageIndex{3}$$ If $$f$$ is elementary or measurable on each of the (countably many $$)$$ sets $$A_{n}$$ in $$(S, \mathcal{M}),$$ it has the same property on their union $$A=\bigcup_{n} A_{n}$$. Proof Let $$f$$ be elementary on each $$A_{n}$$ (so $$A_{n} \in \mathcal{M}$$ by Note 1$$)$$. By Corollary 1 of Chapter 7, §1, $A=\bigcup A_{n}=\bigcup B_{n}$ for some disjoint sets $$B_{n} \subseteq A_{n}\left(B_{n} \in \mathcal{M}\right)$$. By Corollary $$2, f$$ is elementary on each $$B_{n} ;$$ i.e., constant on sets of some $$\mathcal{M}$$ -partition $$\left\{B_{n i}\right\}$$ of $$B_{i}$$. All $$B_{n i}$$ combined (for all $$n$$ and all $$i )$$ form an $$\mathcal{M}$$-partition of $$A$$, $A=\bigcup_{n} B_{n}=\bigcup_{n, i} B_{n i}.$ As $$f$$ is constant on each $$B_{n i},$$ it is elementary on $$A .$$ For measurable functions $$f,$$ slightly modify the method used in Corollary $$2 . \square$$ ## Corollary $$\PageIndex{4}$$ If $$f : S \rightarrow\left(T, \rho^{\prime}\right)$$ is measurable on $$A$$ in $$(S, \mathcal{M}),$$ so is the composite map $$g \circ f,$$ provided $$g : T \rightarrow\left(U, \rho^{\prime \prime}\right)$$ is relatively continuous on $$f[A]$$. Proof By assumption, $f=\lim _{m \rightarrow \infty} f_{m} \text { (pointwise) }$ for some elementary maps $$f_{m}$$ on $$A$$. Hence by the continuity of $$g$$, $g\left(f_{m}(x)\right) \rightarrow g(f(x)),$ i.e., $$g \circ f_{m} \rightarrow g \circ f$$ (pointwise) on $$A$$. Moreover, all $$g \circ f_{m}$$ are elementary on $$A$$ (for $$g \circ f_{m}$$ is constant on any partition set, if $$f_{m}$$ is). Thus $$g \circ f$$ is measurable on $$A,$$ as claimed. $$\square$$ ## Theorem $$\PageIndex{1}$$ If the maps $$f, g, h : S \rightarrow E^{1}(C)$$ are simple, elementary, or measurable on $$A$$ in $$(S, \mathcal{M}),$$ so are $$f \pm g, f h,\|f\|^{a}$$ (for real $$a \neq 0 )$$ and $$f / h$$ (if $$h \neq 0$$ on $$A ) .$$ Similarly for vector-valued $$f$$ and $$g$$ and scalar-valued $$h$$. Proof First, let $$f$$ and $$g$$ be elementary on $$A .$$ Then there are two $$\mathcal{M}$$-partitions, $A=\bigcup A_{i}=\bigcup B_{k},$ such that $$f=a_{i}$$ on $$A_{i}$$ and $$g=b_{k}$$ on $$B_{k},$$ say. The sets $$A_{i} \cap B_{k}$$ (for all $$i$$ and $$k )$$ then form a new $$\mathcal{M}$$ -partition of $$A(\text { why? })$$, such that both $$f$$ and $$g$$ are constant on each $$A_{i} \cap B_{k}(\text { why?); hence so is } f \pm g$$. Thus $$f \pm g$$ is elementary on $$A .$$ Similarly for simple functions. Next, let $$f$$ and $$g$$ be measurable on $$A ;$$ so $f=\lim f_{m} \text { and } g=\lim g_{m} \text { (pointwise) on } A$ for some elementary maps $$f_{m}, g_{m}$$. By what was shown above, $$f_{m} \pm g_{m}$$ is elementary for each $$m .$$ Also, $f_{m} \pm g_{m} \rightarrow f \pm g (\text { pointwise }) \text { on } A,$ Thus $$f \pm g$$ is measurable on $$A$$. The rest of the theorem follows quite similarly. $$\square$$ If the range space is $$E^{n}\left(\text { or } C^{n}\right),$$ then $$f$$ has $$n$$ real (complex) components $$f_{1}, \ldots, f_{n},$$ as in Chapter 4,§3 (Part II). This yields the following theorem. ## Theorem $$\PageIndex{2}$$ A function $$f : S \rightarrow E^{n}\left(C^{n}\right)$$ is simple, elementary, or measurable on a set $$A$$ in $$(S, \mathcal{M})$$ iff all its $$n$$ component functions $$f_{1}, f_{2}, \ldots, f_{n}$$ are. Proof For simplicity, consider $$f : S \rightarrow E^{2}, f=\left(f_{1}, f_{2}\right)$$. If $$f_{1}$$ and $$f_{2}$$ are simple or elementary on $$A$$ then (exactly as in Theorem 1$$)$$, one can achieve that both are constant on sets $$A_{i} \cap B_{k}$$ of one and the same $$\mathcal{M}$$-partition of $$A .$$ Hence $$f=\left(f_{1}, f_{2}\right),$$ too, is constant on each $$A_{i} \cap B_{k},$$ as required. Conversely, let $f=\overline{c}_{i}=\left(a_{i}, b_{i}\right) \text { on } C_{i}$ for some $$\mathcal{M}$$-partition $A=\bigcup C_{i}.$ Then by definition, $$f_{1}=a_{i}$$ and $$f_{2}=b_{i}$$ on $$C_{i} ;$$ so both are elementary (or simple) on $$A .$$ In the general case $$\left(E^{n} \text { or } C^{n}\right),$$ the proof is analogous. For measurable functions, the proof reduces to limits of elementary maps (using Theorem 2 of Chapter 3, §15). The details are left to the reader. $$\square$$ Note 2. As $$C=E^{2},$$ a complex function $$f : S \rightarrow C$$ is simple, elementary, or measurable on $$A$$ iff its real and imaginary parts are. By Definition $$4,$$ a measurable function is a pointwise limit of elementary maps. However, if $$\mathcal{M}$$ is a $$\sigma$$-ring, one can make the limit uniform. Indeed, we have the following theorem. ## Theorem $$\PageIndex{3}$$ If $$\mathcal{M}$$ is a $$\sigma$$-ring, and $$f : S \rightarrow\left(T, \rho^{\prime}\right)$$ is $$\mathcal{M}$$-measurable on $$A,$$ then $f=\lim _{m \rightarrow \infty} g_{m} \text { (uniformly) on } A$ for some finite elementary maps $$g_{m}$$. Proof Thus given $$\varepsilon>0,$$ there is a finite elementary map $$g$$ such that $$\rho^{\prime}(f, g)<\varepsilon$$ on $$A$$. ## Theorem $$\PageIndex{4}$$ If $$\mathcal{M}$$ is a $$\sigma$$-ring in $$S,$$ if $f_{m} \rightarrow f(\text {pointwise}) \text { on } A$ $$\left(f_{m} : S \rightarrow\left(T, \rho^{\prime}\right)\right),$$ and if all $$f_{m}$$ are $$\mathcal{M}$$ -measurable on $$A,$$ so also is $$f$$. Briefly: $$A$$ pointwise limit of measurable maps is measurable (unlike continuous maps; cf. Chapter 4, §12). Proof By the second clause of Theorem $$3,$$ each $$f_{m}$$ is uniformly approximated by some elementary map $$g_{m}$$ on $$A,$$ so that, taking $$\varepsilon=1 / m, m=1,2, \ldots$$, $\rho^{\prime}\left(f_{m}(x), g_{m}(x)\right)<\frac{1}{m} \quad \text { for all } x \in A \text { and all } m.$ Fixing such a $$g_{m}$$ for each $$m,$$ we show that $$g_{m} \rightarrow f (\text { pointwise })$$ on $$A,$$ as required in Definition $$4 .$$ Indeed, fix any $$x \in A .$$ By assumption, $$f_{m}(x) \rightarrow f(x) .$$ Hence, given $$\delta>0$$, $(\exists k)(\forall m>k) \quad \rho^{\prime}\left(f(x), f_{m}(x)\right)<\delta.$ Take $$k$$ so large that, in addition, $(\forall m>k) \quad \frac{1}{m}<\delta.$ Then by the triangle law and by $$(1),$$ we obtain for $$m>k$$ that \begin{aligned} \rho^{\prime}\left(f(x), g_{m}(x)\right) & \leq \rho^{\prime}\left(f(x), f_{m}(x)\right)+\rho^{\prime}\left(f_{m}(x), g_{m}(x)\right) \\ &<\delta+\frac{1}{m}<2 \delta \end{aligned}. As $$\delta$$ is arbitrary, this implies $$\rho^{\prime}\left(f(x), g_{m}(x)\right) \rightarrow 0,$$ i.e., $$g_{m}(x) \rightarrow f(x)$$ for any (fixed) $$x \in A,$$ thus proving the measurability of $$f . \quad \square$$ Note 3. If $\mathcal{M}=\mathcal{B} (=\text { Borel field in } S),$ we often say "Borel measurable" for $$\mathcal{M}$$-measurable. If $\mathcal{M}=\left\{\text { Lebesgue measurable sets in } E^{n}\right\},$ we say "Lebesgue (L) measurable" instead. Similarly for "Lebesgue-Stieltjes (LS) measurable."
## Key Concepts • Supplementary and Complementary Angles • If the sum of the measures of two angles is 180°, then the angles are supplementary. • If angle $A$ and angle $B$ are supplementary, then $m\angle{A}+m\angle{B}=180°$ . • If the sum of the measures of two angles is $90^\circ$, then the angles are complementary. • If angle $A$ and angle $B$ are complementary, then $m\angle{A}+m\angle{B}=90°$ . • Solve Geometry Applications 1. Read the problem and make sure you understand all the words and ideas. Draw a figure and label it with the given information. 2. Identify what you are looking for. 3. Name what you are looking for and choose a variable to represent it. 4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information. 5. Solve the equation using good algebra techniques. 6. Check the answer in the problem and make sure it makes sense. 7. Answer the question with a complete sentence. • Sum of the Measures of the Angles of a Triangle • For any $\Delta ABC$, the sum of the measures is $180^\circ$ • $m\angle{A}+m\angle{B}=180°$ • Right Triangle • A right triangle is a triangle that has one $90°$ angle, which is often marked with a $\angle$ symbol. • Properties of Similar Triangles • If two triangles are similar, then their corresponding angle measures are equal and their corresponding side lengths have the same ratio. ## Glossary angle An angle is formed by two rays that share a common endpoint. Each ray is called a side of the angle. complementary angles If the sum of the measures of two angles is $90^\circ$ , then they are called complementary angles. hypotenuse The side of the triangle opposite the $90^\circ$ angle is called the hypotenuse. legs of a right triangle The sides of a right triangle adjacent to the right angle are called the legs. right triangle A right triangle is a triangle that has one $90^\circ$ angle. similar figures In geometry, if two figures have exactly the same shape but different sizes, we say they are similar figures. supplementary angles If the sum of the measures of two angles is $180^\circ$ , then they are called supplementary angles. triangle A triangle is a geometric figure with three sides and three angles. vertex of an angle When two rays meet to form an angle, the common endpoint is called the vertex of the angle.
## Unit9Applying the optimization procedure Theorem 8.8 gives us a procedure for finding extrema of functions on closed intervals. Now we're going to apply that procedure to help us find the best, cheapest, most effective, etc. As with any application problem, the hardest part is setting up the mathematical model that captures the situation we want to apply Theorem 8.8 to. Since our optimization tool applies to functions, the task boils down to writing a single function. Luckily for us, we discussed a lot of what we'll need here back in Unit 3. ### Subsection9.1Optimization in geometry We'll start with a few geometric problems, where the objective function is a little more obvious. ###### Example9.1. We're going to build a window in the shape of a rectangle topped by an equilateral triangle. We want to make a window which lets in the most light -- that is, with the greatest possible area. In order to build the window, we have to use wood trim. We have 16 feet of wood trim to build the window with. Such a window has two dimensions: the width $$w$$ and the height $$h$$ of the rectangle. The rectangular portion has area $$wh$$ and the triangular portion has area $$\frac{1}{2}w^2\text{.}$$ So the total area is \begin{equation} A(w,h)=wh+\frac{1}{2}w^2\ . \end{equation} We also need to record the fact that our supplies are limited. A little geometry shows that to build the window requires two pieces of trim with length $$h$$ and four of length $$w\text{.}$$ Technically we don't have to use all the trim, but if we had some left over, we could have used it to build a bigger window. So let's assume we use all 16 feet; that is, we assume \begin{equation} 16=2h+4w\ . \end{equation} We can solve this equation for either $$w$$ or $$h\text{.}$$ Let's solve for $$h\text{:}$$ \begin{equation} h=8-2w \end{equation} and substitute that into the formula for area: \begin{equation} A(w,h(w))=w(8-2w)+\frac{1}{2}w^2 \end{equation} Now we've got a function which we can optimize. We want to have a sensible result, so we know that $$w$$ can't be less than 0, and can be at most 4. So we want to optimize on the interval $$\left[0,4\right]\text{.}$$ Differentiating, we get $$\frac{dA}{dw}=8-3w\text{.}$$ So there is a single critical point at $$w=\frac{8}{3}\text{.}$$ We have \begin{gather} A(0,h(0))=0\\ A(4,h(4))=A(4,0)=\frac{1}{2}4^2=8\\ A\left(\frac{8}{3},h(\frac{8}{3})\right)=A\left(\frac{8}{3},\frac{8}{3}\right)=\left(\frac{8}{3}\right)^2+\frac{1}{2}\left(\frac{8}{3}\right)^2=\frac{32}{3} \end{gather} Since $$\frac{32}{3}$$ is greater than either 0 or 8, we see that the maximal area occurs when we choose width and height both equal to $$\frac{8}{3}\text{.}$$ This example shows a few things. First, notice that many optimization problems come equipped with a constraint. Here that was the fact that we only had so much trim. If you think about it, constrained optimization is the kind you usually deal with -- our world is full of scarcity. Second, we can interpret each of the values we compared in the context of the problem. At $$w=0\text{,}$$ the window has width zero. At $$w=4\text{,}$$ $$h=0$$ so we only have a triangular section of window. $$w=\frac{8}{3}$$ is somewhere in between. Third, the optimal dimensions happened to be equal to one another. This is typical -- optimizers are often symmetric (in this case, the symmetry is that $$w$$ and $$h$$ are the same. Here are some other geometric optimization problems. Given a total length of 10 meters of rope, what’s the greatest area we can enclose in a rectangle? Answer to three decimal places. Using the same length of rope, what base length should we use to obtain the greatest area in an isoscees triangle? Answer to three decimal places. $$6.25$$ $$2.287$$ In the window example, let’s say we remove the “middle” piece of trim (which had length $$w$$). First, a gut check: does this increase or decrease the optimal area? • increase • decrease What if we double up the middle trim? Does this increase or decrease the optimal area? • increase • decrease Now verify your intuition by computing the optimal area in each scenario. area with no crossbar: area with double crossbar: $$\text{increase}$$ $$\text{decrease}$$ $$16$$ $$8$$ ### Subsection9.2Optimization in economics and business Let's think about production. A standard model, called the Cobb-Douglas production function, says that the productivity of a firm is proportional to both a power of the labor inputs $$L$$ and a power of the capital inputs $$K\text{;}$$ and that the powers add to 1. That is, \begin{equation} P=kK^\alpha L^\beta \end{equation} where k is the constant of proportionality and $$\alpha+\beta=1\text{.}$$ We may as well write $$\beta=1-\alpha\text{,}$$ so that our formula reads \begin{equation} P=kK^\alpha L^{1-\alpha}\ . \end{equation} Clearly, if we could increase $$K$$ and $$L$$ without constraint, we could increase the firm's output arbitrarily. But we have to operate subject to a budget. Spending more on labor means we have less to spend on capital, and vice versa. We model this by \begin{equation} B=K+L \end{equation} that is, the total budget is the sum of the captial costs and the labor costs. A natural question to ask is: given a budget, how do we maximize output? Just as with the window example, we manipulate our constraint to express one variable in terms of the other: $$L=B-K\text{.}$$ Then we substitute this into the objective function: \begin{equation} P(K,L(K))=k K^\alpha (B-K)^{1-\alpha}\ . \end{equation} That's a function we can optimize, on the interval $$[0,B]\text{.}$$ Let’s say $$\alpha=\frac{1}{3}$$ and $$B=1000\text{.}$$ What’s the optimal level of capital investment? Answer to the nearest cent. $$\frac{1000}{3}$$ Now solve symbolically. If $$\alpha=\frac{1}{3}\text{,}$$ express the optimal capital investment in terms of the total budget $$B\text{.}$$ If $$\alpha=\frac{1}{n}\text{,}$$ express the optimal capital investment in terms of $$n$$ and the total budget $$B\text{.}$$ $$\frac{B}{3}$$ $$\frac{B}{n}$$ ##### maximizing profit. Consider the problem of a firm producing and selling a single good (say, pairs of sneakers). The goal of the firm is to make the most money possible. Before we get into a precise model, let's set the ground rules. production level We'll call the number of pairs of sneakers we produce $$t\text{.}$$ production costs We'll write $$C(t)$$ for the total cost of producing $$t$$ pairs of sneakers. $$C(t)$$ is an increasing function. markets clear We'll assume that we sell every pair of sneakers we make. revenue We'll write $$R(t)$$ for the total revenue that selling $$t$$ pairs of sneakers brings in. $$R(t)$$ is also an increasing function. profit The profit we make is $$P(t)=R(t)-C(t)\text{.}$$ Our goal is to maximize $$P(t)\text{.}$$ In finance and economics, the adjective marginal is used to denote a derivative. So we say marginal revenue to mean $$R'(t)$$ and marginal costs to mean $$C'(t)\text{.}$$ Based on the asumptions stated above, marginal revenue is • sometimes positive • sometimes negative • always positive • always negative and marginal costs are • sometimes positive • sometimes negative • always positive • always negative . $$\text{always positive}$$ $$\text{always positive}$$ This use of the word $$marginal$$ comes from the fact that, using the tangent line approximation to $$C(t)\text{,}$$ \begin{align} C(t+1)&\sim C(t)+C'(t)\left((t+1)-t\right)\\ C(t+1)&\sim C(t)+C'(t) \end{align} In other words, $$C'(t)$$ is approximately the additional cost added by the additional pair of sneakers that took us from production level $$t$$ to production level $$t+1\text{.}$$ In fact, thinking about derivatives this way can be very useful to understanding the situation of maximizing profit. Put yourself in the shoes of a firm manager who gets to decide production levels. If marginal costs (i.e. cost of producing the next pair of sneakers) are greater than marginal revenues (i.e. revenue generated by the next pair of sneakers), we should • increase • decrease • maintain current level of the production level. If marginal costs (i.e. cost of producing the next pair of sneakers) are less than marginal revenues (i.e. revenue generated by the next pair of sneakers), we should • increase • decrease • maintain current level of the production level. $$\text{increase}$$ $$\text{increase}$$ Formally, we're trying to optimize \begin{equation} P(t)=R(t)-C(t) \end{equation} so our first step ought to be differentiating: \begin{equation} P'(t)=R'(t)-C'(t)\ . \end{equation} We want to find critical points of $$P\text{;}$$ that is, we need to solve \begin{align} P'(t)&=0\\ R'(t)-C'(t)&=0\\ R'(t)&=C'(t) \end{align} That is, we're looking for the production level where marginal cost and marginal revenue are equal. As the manager of the firm, it’s your job to occasionally explain your decisions to the firm’s owner (who is mathematically illiterate). The owner sees the phrase marginal profit is zero in your written report and becomes quite upset. “Zero profit?!” he screams. Explain to the owner why seeking zero marginal profit is the correct business decision. ##### modeling revenue. Because profit is the difference of revenue and costs, understand how to solve $$P'(t)=0$$ amounts to understanding the revenue function $$R(t)$$ and the cost function $$C(t)\text{.}$$ Revenue seems straightforward. If we produce and sell $$t$$ pairs of sneakers for a price of $$p$$ dollars per sneaker, then \begin{equation} R(t)=p\cdot t \end{equation} But! the more sneakers we produce and sell, the less unique an individual wearing those sneakers is. The more sneakers we produce and sell, the fewer people go unshod. That tends to drive the price down. So the price $$p$$ isn't a constant; it's a function $$p(t)$$ of the number of pairs of sneakers we've sold. Let's say that we've done some market research, and we've found that the market price of a pair of sneakers seems to obey \begin{equation} p(t)=300-.05t\text{.} \end{equation} In the formula $$p(t)=300-.05t\text{,}$$ what are the units of each of the following? 1. 300: • pairs of sneakers • dollars • pairs of sneaker per dollar • dollars per pair of sneakers • something else 2. t: • pairs of sneakers • dollars • pairs of sneaker per dollar • dollars per pair of sneakers • something else 3. .05: • pairs of sneakers • dollars • pairs of sneaker per dollar • dollars per pair of sneakers • something else If you said “something else”, find the units. Interpret what each of these numbers means in terms of the market price for sneakers. $$\text{dollars per pair of sneakers}$$ $$\text{pairs of sneakers}$$ $$\text{something else}$$ Notice that our model predicts that for high enough levels of production, $$p(t)$$ is negative. That is, if we completely saturate the market, we'd have to start paying people to take our sneakers (instead of them paying us). Is this prediction reasonable? Write a formula for $$R(t)$$ if $$p(t)=300-.05t\text{.}$$ At what production level does maximum revenue occur? $$\left(300-0.05t\right)t$$ $$3000$$ ##### modeling costs. What about costs? A standard model of costs is linear: \begin{equation} C(t)=C_0+mt \end{equation} where $$C_0$$ is called the fixed cost and represents the costs we have to spend no matter what: capital outlay for the factory, bribes for local politicans, etc.; and $$m$$ is the marginal cost (materials and labor to produce a single pair of sneakers). I’ve used the phrase marginal cost in two different ways. Why are those two ways the same when the cost function looks like \begin{equation} C(t)=C_0+mt\ \ ? \end{equation} Let's say we build a factory for $10,000 and our marginal cost is$2 per pair of sneakers. Then we have \begin{equation} C(t)=10000+2t \end{equation} and \begin{equation} P(t)=t\cdot(300-.05t)-(10000+2t)\ \ . \end{equation} How do we acheive the maximum profit? When \begin{equation} 0=P'(t)=302-.1t \end{equation} which is $$t=3020\text{.}$$ The profit we actually make at that production level is $$P(3020)=446,020\text{.}$$ Not too bad. What’s the production level which achieves maximum profit if our sneakers debut at $400 and market saturation reduces the price$0.50 per pair of sneakers in the market? What is that maximum profit? $$402$$ $$70802$$ But costs are not always linear. We say that a cost function obeys economies of scale if the marginal cost gets smaller as we increase the production level. For example, a worker producing their first pair of sneakers might take a lot of time, but by the time they get to their 30$$^{th}$$ pair of sneakers, the same worker can probably do so much more quickly (which means the labor cost for that pair of sneakers will be lower. How do we encode economies of scale -- that is, decreasing marginal cost -- in calculus terms? Just as we did with revenue, we can bootstrap our linear model for costs $$C(t)=C_0+mt$$ into a more model by replacing $$m$$ with $$m(t)\text{.}$$ If we think of our workers in the sneaker factory as gaining skill over time, then we could write something like \begin{equation} m(t)=1+\left(\frac{1}{2}\right)^{t/10} \end{equation} which is to say: the marginal cost starts at $2 per pair of sneakers, has a long-run limit of$1 per pair of sneakers, and every 10 pairs of sneakers produced moves the marginal cost halfway to the long-run limit. How do we achieve the maximum profit in this case? We have \begin{align} P(t)&=R(t)-C(t)\\ &=t\cdot(300-.05t)-\left(10000+\left(1+\left(\frac{1}{2}\right)^{t/10}\right)t\right) \end{align} which means that we need to solve \begin{equation} 0=P'(t)=299-.1t-\left(\frac{1}{2}\right)^{t/10}-\frac{1}{10}t\ln\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{t/10}\ \ . \end{equation} Unfortunately there's not an algebraically clean solution to this equation, but a graphing utility says that $$t=2990$$ is very close.
# How to Calculate Standard Deviation Easily Standard deviation is an important statistical concept that measures the amount of variation or dispersion of a set of data values from its mean value. It is commonly used in business, science, and engineering to evaluate data and make informed decisions. In this article, we will explain how to calculate standard deviation in simple terms, step by step. ## What is Standard Deviation? Standard deviation is a measure of how spread out a set of numbers is around the mean. It tells us how much the data deviates from the average value, and how much the individual data points differ from each other. In other words, it indicates the degree of variability or dispersion of the data set. The standard deviation is typically denoted by the symbol σ (sigma) for a population and s for a sample. Source: bing.com ## How to Calculate Standard Deviation Step by Step To calculate standard deviation, you need to follow these steps: 1. First, find the mean or average value of the data set by adding up all the values and dividing by the total number of values. 2. Source: bing.com 3. Next, subtract the mean from each data point to get the deviations from the mean. 4. Then, square each deviation to eliminate negative values and make them all positive. 5. Source: bing.com 6. Sum up all the squared deviations. 7. Source: bing.com 8. Divide the sum of squared deviations by the total number of values minus one (n-1) to get the variance. 9. Source: bing.com 10. Finally, take the square root of the variance to get the standard deviation. 11. Source: bing.com ## Example Calculation of Standard Deviation Let us take the following data set: 6, 7, 8, 9, 10 First, we find the mean value: Mean = (6 + 7 + 8 + 9 + 10) / 5 = 8 Next, we calculate the deviations from the mean: Deviations = (6 – 8), (7 – 8), (8 – 8), (9 – 8), (10 – 8) = -2, -1, 0, 1, 2 Then, we square each deviation to get: Squared deviations = 4, 1, 0, 1, 4 Sum of squared deviations = 10 Variance = 10 / (5-1) = 2.5 Standard deviation = √2.5 = 1.5811 (approx.) ## Types of Standard Deviation There are two types of standard deviation: • Population standard deviation (σ) – used when the data set represents the entire population. • Sample standard deviation (s) – used when the data set is a representative sample of the population. ## Why is Standard Deviation Important? Standard deviation is important because it helps us to: • Understand the spread and variability of data. • Assess the reliability and accuracy of data. • Compare different sets of data. • Determine the confidence interval and margin of error. • Identify outliers or anomalies in the data set. ## Conclusion In conclusion, standard deviation is a useful statistical measure that provides valuable insights into the variability and spread of data. By following the steps outlined in this article, you can easily calculate standard deviation and use it to make informed decisions and draw meaningful conclusions from your data.
# In ∆ABC the coordinates vertices A and B are A (-2, 4) and B (-1, 1). For each of the given coordinates of vertex C, is ∆ ABC a right triangle? ## C(-2, 1) C(0,4) C(2,2) Nov 21, 2016 The only right triangle is $C = \left(2 , 2\right)$. Found by the dot-product. #### Explanation: $\overline{A B} = \left(- 1 - - 2\right) \hat{i} + \left(1 - 4\right) \hat{j} = \hat{i} - 3 \hat{j}$ $\overline{A C} = \left({C}_{x} - - 2\right) \hat{i} + \left({C}_{y} - 4\right) \hat{j}$ $\overline{B C} = \left({C}_{x} - - 1\right) \hat{i} + \left({C}_{y} - 1\right) \hat{j}$ $\overline{A B} \cdot \overline{A C} = \left({C}_{x} + 2\right) - 3 \left({C}_{y} - 4\right)$ $\overline{A B} \cdot \overline{B C} = \left({C}_{x} + 1\right) - 3 \left({C}_{y} - 1\right)$ The triangle will be a right triangle if either dot-product is zero: $C = \left(2 , 1\right)$ $\overline{A B} \cdot \overline{A C} = \left(2 + 2\right) - 3 \left(1 - 4\right) = 4 + 9 = 13$ $\overline{A B} \cdot \overline{B C} = \left(2 + 1\right) - 3 \left(1 - 1\right) = 3 + 0 = 3$ Not a right triangle. $C = \left(0 , 4\right)$ $\overline{A B} \cdot \overline{A C} = \left(0 + 2\right) - 3 \left(4 - 4\right) = 2 + 0 = 2$ $\overline{A B} \cdot \overline{B C} = \left(0 + 1\right) - 3 \left(4 - 1\right) = 1 - 9 = - 8$ Not a right triangle. $C = \left(2 , 2\right)$ $\overline{A B} \cdot \overline{A C} = \left(2 + 2\right) - 3 \left(2 - 4\right) = 4 + 6 = 10$ $\overline{A B} \cdot \overline{B C} = \left(2 + 1\right) - 3 \left(2 - 1\right) = 3 - 3 = 0$ This is a right triangle.
# How Is Area Related To A Definite Integral? You might have seen the marketing campaign for the Super Bass-o-Matic 76. Let’s look at some options for finding the increased costs when changing production of Super Bass-o-Matic blenders from 200 blenders to 800 blenders. Suppose the marginal cost is given by $\displaystyle {C}'(x)=30-0.02x {\text{ dollars per blender}}$ By finding the area of rectangles, we get values that indicate an increase in cost. For instance, the first term of the left hand sum below, $\displaystyle \left( 26\frac{\}{\text{blender}} \right)\left( 100\text{ blenders} \right)=\2600$ This means that if we estimate the cost per blender as 26 dollars per blender for all 100 blenders from 200 to 300 blenders, the increased cost is $2600. If we continue this for more rectangles until 800, the sum of the areas give an estimate for the increased cost from increasing production from 200 to 800. In this case the left hand sum is$11400 and the right hand sum is $12600…the difference between these estimates is$1200. If we do the the same process for 60 rectangles (each being 10 wide), we get additional estimates. In this case, the left hand sum is $11940 and the right hand sum is 12060. The difference between the estimate for 60 rectangles is$120. Ten times as many rectangles leads to a tenth of the difference. We get a similar picture for 600 rectangles. The left hand sum is $11994 and the right hand sum is$12006. In this case the difference between the estimates is $12. Also notice that as the number of rectangles gets larger, the area in the rectangles looks more and more like the area under the function and above the horizontal axis. Let’s summarize what happens as the number of rectangles increases. Number of Rectangles LHS RHS Difference 6 11400 12600 1200 60 11940 12060 120 600 11994 12006 12 You might guess that 6000 rectangles would lead to a difference of$1.2 and you would be correct. The left hand sum is $11999.4 and the right hand sum is$12000.6. In fact, as the number of rectangles gets larger and larger (each rectangle getting narrower and narrower) each estimate gets closer and closer to \$12000. This is the exact area in the blue shaded region under the function, above the horizontal axis, and between 200 and 800. To convince yourself that this is the case, we can this area by breaking the region up into a rectangle and triangle. The area is $\displaystyle 14\cdot 600+\frac{1}{2}\cdot 12\cdot 60=12000$ The exact area under the function is represented with a definite integral, $\displaystyle \int\limits_{200}^{800}{(30-0.02x)\,dx}=12000$
Rs Aggarwal 2017 Solutions for Class 9 Math Chapter 7 Areas are provided here with simple step-by-step explanations. These solutions for Areas are extremely popular among Class 9 students for Math Areas Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 9 Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 9 Math are prepared by experts and are 100% accurate. #### Question 1: Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm. We have: Base = 24 cm Height = 14.5 cm Now, #### Question 2: The base of a triangular field is three times its altitude. If the cost of sowing the field at Rs 58 per hectare is Rs 783, find its base and height. Let the height of the triangle be h m. ∴ Base = 3h m Now, Area of the triangle = We have: Thus, we have: Height = h = 300 m Base = 3h = 900 m #### Question 3: Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side. We know that the longest side is 42 cm. Thus, we can find out the height of the triangle corresponding to 42 cm. We have: #### Question 4: Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side. We know that the smallest side is 18 cm. Thus, we can find out the altitude of the triangle corresponding to 18 cm. We have: #### Question 5: Find the area of a triangular field whose sides are 91 cm, 98 m and 105 m in length. Find the height corresponding to the longest side. We know that the longest side is 105 m. Thus, we can find out the height of the triangle corresponding to 42 cm. #### Question 6: The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 m. Find the area of the triangle. Let the sides of the triangle be 5x m, 12x m and 13x m. We know: Perimeter = Sum of all sides or, 150 = 5x + 12x + 13x or, 30x = 150 or, x = 5 Thus, we obtain the sides of the triangle. 5$×$5 = 25 m 12$×$5 = 60 m 13$×$5 = 65 m Now, #### Question 7: The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle. Also, find the cost of ploughing the field at Rs 18.80 per 10 m2. Let the sides of the triangle be 25x m, 17x m and 12x m. We know: Perimeter = Sum of all sides or, 540 = 25x + 17x + 12x or, 54x = 540 or, x = 10 Thus, we obtain the sides of the triangle. 25$×$10 = 250 m 17$×$10 = 170 m 12$×$10 = 120 m Now, Cost of ploughing 10 m2 field = Rs 18.80 Cost of ploughing 1 m2 field = Rs $\frac{18.80}{10}$ Cost of ploughing 9000 m2 field = #### Question 8: Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find (i) the area of the field and (ii) the length of the perpendicular from the opposite vertex on the side measuring 154 m. (ii) We can find out the height of the triangle corresponding to 154 m in the following manner: We have: #### Question 9: Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20 cm. #### Question 10: The base of an isosceles triangle measures 80 cm and its area is 360 cm2. Find the perimeter of the triangle. Let $△$PQR be an isosceles triangle and PX$\perp$QR. Now, Also, ∴ Perimeter = 80 + 41 + 41 = 162 cm #### Question 11: The perimeter of an isosceles triangle is 42 cm and its base is $1\frac{1}{2}$ times each of the equal sides. Find (i) the length of each side of the triangle, (ii) the area of the triangle, and (iii) the height of the triangle. Let the equal sides of the isosceles triangle be a cm each. ∴ Base of the triangle, b = $\frac{3}{2}$a cm (i) Perimeter = 42 cm or, a + a + $\frac{3}{2}$a = 42 or, 2a +$\frac{3}{2}$a= 42 So, equal sides of the triangle are 12 cm each. Also, Base = $\frac{3}{2}$a = (ii) (iii) #### Question 12: If the area of an equilateral triangle is , find its perimeter. Thus, we have: Perimeter = 3 × Side = 3 × 12 = 36 cm #### Question 13: If the area of an equilateral triangle is , find its height. Now, we have: #### Question 14: The base of a right-angled triangle measures 48 cm and its hypotenuse measures 50 cm. Find the area of the triangle. Let $△$PQR be a right-angled triangle and PQ$\perp$QR. Now, #### Question 15: Each side of an equilateral triangle measures 8 cm. Find (i) the area of the triangle, correct to 2 places of decimal and (ii) the height of the triangle, correct to 2 places of decimal. Take $\sqrt{3}=1.732$. Side of the equilateral triangle = 8 cm (i) (ii) #### Question 16: The height of an equilateral triangle measures 9 cm. Find its area, correct to 2 places of decimal. Take $\sqrt{3}=1.732$. Height of the equilateral triangle = 9 cm Thus, we have: Also, #### Question 17: An umbrella is made by stitching 12 triangular pieces of cloth, each measuring (50 cm × 20 cm × 50 cm). Find the area of the cloth used in it. We know that the triangle is an isosceles triangle. Thus, we can find out the area of one triangular piece of cloth. Now, Area of 1 triangular piece of cloth = 490 cm2 Area of 12 triangular pieces of cloth = #### Question 18: A floral design on a floor is made up of 16 tiles, each triangular in shape having sides 16 cm, 12 cm and 20 cm. Find the cost of polishing the tiles at Re 1 per sq cm. Area of one triangular-shaped tile can be found in the following manner: Now, Area of 16 triangular-shaped tiles = Cost of polishing tiles of area 1 cm2 = Rs 1 Cost of polishing tiles of area 1536 cm2 = #### Question 19: Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, ACB = 90° and AC = 15 cm. We know that $△$ABC is a right-angled triangle. ∴ Now, Area of quadrilateral ABCD = Area of $△$ABC + Area of $△$ACD = (60 + 54) cm2 =114 cm2 And, Perimeter of quadrilateral ABCD = AB + BC + CD + AD = 17 + 8 + 12 + 9 = 46 cm #### Question 20: Find the perimeter and area of the quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and CBD = 90°. We know that $△$DBC is a right-angled triangle. Now, ∴ Area of quadrilateral ABCD = Area of $△$DBC + Area of $△$ABD =(210 + 336) cm2 = 546 cm2 Also, Perimeter of quadrilateral ABCD = AB + BC + CD + AD = 42 + 21 + 29 + 34 = 126 cm #### Question 21: Find the area of quadrilateral ABCD in which AD = 24 cm, BAD = 90° and ∆BCD is an equilateral triangle having each side equal to 26 cm. Also, find the perimeter of the quadrilateral. [Given: $\sqrt{3}=1.73\right]$] We know that $△$BAD is a right-angled triangle. ∴ Also, we know that $△$BDC is an equilateral triangle. Now, Area of quadrilateral ABCD = Area of $△$ABD + Area of $△$BDC = (120 + 292.37) cm2 = 412.37 cm2 #### Question 22: Find the area of a parallelogram ABCD in which AB = 28 cm, BC = 26 cm and diagonal AC = 30 cm. We know that a diagonal divides a parallelogram into two triangles of equal areas. ∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = #### Question 23: Find the area of a parallelogram ABCD in which AB = 14 cm, BC = 10 cm and AC = 16 cm. [Given: $\sqrt{3}=1.73$] We know that a diagonal divides a parallelogram into two triangles of equal areas. ∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = #### Question 24: In the given figure ABCD is a quadrilateral in which diagonal BD = 64 cm, AL BD and CMBD such that AL = 16.8 cm and CM = 13.2 cm. Calculate the area of quadrilateral ABCD. #### Question 1: In a ABC it is given that base = 12 cm and height = 5 cm. Its area is (a) 60 cm2 (b) 30 cm2 (c) (d) 45 cm2 (b) 30 cm2 #### Question 2: The lengths of three sides of a triangle are 20 cm, 16 cm and 12 cm. The area of the triangle is (a) 96 cm2 (b) 120 cm2 (c) 144 cm2 (d) 160 cm2 (a) 96 cm2 #### Question 3: Each side of an equilateral triangle measure 8 cm. The area of the triangle is (a) (b) (c) (d) 48 cm2 (b) #### Question 4: The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm. The area of the triangle is (a) (b) (c) (d) (b) #### Question 5: The base of an isosceles triangle is 6 cm and each of its equal sides is 5 cm. The height of the triangle is (a) 8 cm (b) (c) 4 cm (d) (c) 4 cm #### Question 6: Each of the two equal sides of an isosceles right triangle is 10 cm long. Its area is (a) (b) 50 cm2 (c) (d) 75 cm2 (b) 50 cm2 Here, the base and height of the triangle are 10 cm and 10 cm, respectively. Thus, we have: #### Question 7: Each side of an equilateral triangle is 10 cm long. The height of the triangle is (a) (b) (c) (d) 5 cm (b) #### Question 8: The height of an equilateral triangle is 6 cm. Its area is (a) (b) (c) (d) 18 cm2 (a) #### Question 9: The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is (a) 480 m2 (b) 320 m2 (c) 384 m2 (d) 360 m2 (c) 384 m2 #### Question 10: The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 cm. The area of the triangle is (a) 375 cm2 (b) 750 cm2 (c) 250 cm2 (d) 500 cm2 (b) 750 cm2 Let the sides of the triangle be 5x cm, 12x cm and 13x cm. Perimeter = Sum of all sides or, 150 = 5x + 12x + 13x or, 30x = 150 or, x = 5 Thus, the sides of the triangle are 5$×$5 cm, 12$×$5 cm and 13$×$5 cm, i.e., 25 cm, 60 cm and 65 cm. Now, #### Question 11: The lengths of the three sides of a triangle are 30 cm, 24 cm and 18 cm respectively. The length of the altitude of the triangle corresponding to the smallest side is (a) 24 cm (b) 18 cm (c) 30 cm (d) 12 cm (a) 24 cm The smallest side is 18 cm. Hence, the altitude of the triangle corresponding to 18 cm is given by: #### Question 12: The base of an isosceles triangle is 16 cm and its area is 48 cm2. The perimeter of the triangle is (a) 41 cm (b) 36 cm (c) 48 cm (d) 324 cm (b) 36 cm Let $△$PQR be an isosceles triangle and PX$\perp$QR. Now, ∴ Perimeter = (10 + 10 + 16) cm = 36 cm #### Question 13: The area of an equilateral triangle is . Its perimeter is (a) 36 cm (b) (c) 24 cm (d) 30 cm (a) 36 cm Now, Perimeter = 3 × Side = 3 × 12 = 36 cm #### Question 14: Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is (a) 156 cm2 (b) 78 cm2 (c) 60 cm2 (d) 120 cm2 (c) 60 cm2 #### Question 15: The base of a right triangle. is 48 cm and its hypotenuse is 50 cm long. The area of the triangle is (a) 168 cm2 (b) 252 cm2 (c) 336 cm2 (d) 504 cm2 (c) 336 cm2 Let $△$PQR be a right-angled triangle and PQ$\perp$QR. Now, #### Question 16: The area of an equilateral triangle is . Its height is (a) (b) (c) (d) 9 cm (a) #### Question 17: The difference between the semi-perimeter and the sides of a ABC are 8 cm, 7 cm and 5 cm respectively. The area of the triangle is (a) (b) (c) (d) 140 cm2 (c) Let the sides of the triangle be a cm, b cm and c cm and the semi-perimeter be s cm. Given: By adding all these, we get: #### Question 18: For an isosceles right-angled triangle having each of equal sides 'a', we have I. $\mathrm{Area}=\frac{1}{2}{a}^{2}$ II. $\mathrm{Perimeter}=\left(2+\sqrt{2}\right)a$ III. Hypotenuse = 2a Which of the following is true? (a) I only (b) II only (c) I and II (d) I and III (c) I and II Area of triangle = $\frac{1}{2}×\mathrm{Base}×\mathrm{Height}=\frac{1}{2}×a×a=\frac{1}{2}{a}^{2}$ Hypotenuse = $\sqrt{{a}^{2}+{a}^{2}}=\sqrt{2{a}^{2}}=\sqrt{2}a$ Perimeter = $a+a+\sqrt{2}a=2a+\sqrt{2}a=a\left(2+\sqrt{2}\right)$ #### Question 19: For an isosceles triangle having base b and each of the equal sides as a, we have: I. $\mathrm{Area}=\frac{b·\sqrt{4{a}^{2}-{b}^{2}}}{4}$ II. Perimeter = (2a + b) III. $\mathrm{Height}=\frac{1}{2}\sqrt{4{a}^{2}-{b}^{2}}$ Which of the following is true? (a) I only (b) I and II only (c) II and III only (d) I, II and III (d) I, II and III For an isosceles triangle having base as b and each of the equal sides as a, we have: I. $\mathrm{Area}=\frac{b\sqrt{4{a}^{2}-{b}^{2}}}{4}$ II. Perimeter = (2a + b) III. $\mathrm{Height}=\frac{1}{2}\sqrt{4{a}^{2}-{b}^{2}}$ #### Question 20: Assertion: Area of an equilateral triangle having each side equal to 4 cm is cm. Reason: Area of an equilateral triangle having each side a is $\frac{\sqrt{3}}{4}{a}^{2}$ sq units. (a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion. (b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion. (c) Assertion is true and Reason is false. (d) Assertion is false and Reason is true. (a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion. Assertion: Reason: Area of an equilateral triangle having each side a is $\frac{\sqrt{3}}{4}{a}^{2}$ sq. units. Hence, it is true. #### Question 21: Assertion: The area of an isosceles triangle having base = 8 cm and each of the equal sides = 5 cm is 12 cm2. Reason: The area of an isosceles triangle having each of the equal sides as a and base = b is $\frac{1}{4}b\sqrt{4{a}^{2}-{b}^{2}}$. (a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion. (b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion. (c) Assertion is true and Reason is false. (d) Assertion is false and Reason is true. (a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion. Assertion: Hence, Assertion is true. Reason: Area of an isosceles triangle having each of the equal sides as a and base as b is $\frac{1}{4}b\sqrt{4{a}^{2}-{b}^{2}}$. Hence, it is true. #### Question 22: Assertion: The area of an equilateral triangle having side 4 cm is 3 cm2. Reason: The area of an equilateral triangle having each side a is $\left(\frac{\sqrt{3}}{4}{a}^{2}\right)$ sq units. (a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion. (b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion. (c) Assertion is true and Reason is false. (d) Assertion is false and Reason is true. (d) Assertion is false and Reason is true. Assertion: Hence, Assertion is false. Reason: Area of an equilateral triangle having each side a is $\frac{\sqrt{3}}{4}{a}^{2}$ square units. Hence, it is true. #### Question 23: Assertion: The sides of a ABC are in the ratio 2 : 3 : 4 and its perimeter is 36 cm. Then, . Reason: If 2s = (a + b + c), where a, b, c are the sides of a triangle, then its area $=\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)}$. (a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion. (b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion. (c) Assertion is true and Reason is false. (d) Assertion is false and Reason is true. (c) Assertion is true and Reason is false. Assertion: Let the sides of the triangle be 2x cm, 3x cm and 4x cm. Perimeter = Sum of all sides or, 36 = 2x + 3x + 4x or, 9x = 36 or, x = 4 Thus, the sides of the triangle are 2$×$4 cm, 3$×$4 cm and 4$×$4 cm, i.e., 8 cm, 12 cm and 16 cm. Now, Hence, Assertion is true. Reason: If 2s = (a + b + c), where a, b and c are the sides of a triangle, then its area $=\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)}$. Hence, it is false. Area should be $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$. #### Question 24: Assertion: The area of an isosceles triangle having base = 24 cm and each of the equal sides equal to 13 cm is 60 cm2. Reason: If 2s = (a + b + c), where a, b, c are the sides of a triangle, then area $=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$. (a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion. (b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion. (c) Assertion is true and Reason is false. (d) Assertion is false and Reason is true. (a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion. Assertion: Reason: If 2s = (a + b + c), where a, b and c are the sides of a triangle, then area$=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$. #### Question 25: If the base of an isosceles triangle is 6 cm and its perimeter is 16 cm, then its area is 12 cm2. True Let the equal sides of the isosceles triangle be a cm each. We have: Base of the triangle, b = 6 cm Perimeter = 16 cm or, a + a + 6 = 16 or, 2a = 10 or, a = 5 cm Now, #### Question 26: If each side of an equilateral triangle is 8 cm long, then its area is $20\sqrt{3}$ cm2. False #### Question 27: If the sides of a triangular field measure 51 m, 37 m and 20 m, then the cost of levelling it at Rs 5 per m2 is Rs 1530. True Cost of levelling 1 m2 of area = Rs 5 Cost of levelling 306 m2 of area = #### Question 28: Match the following columns. Column I Column II (a) The lengths of three sides of a triangle are 26 cm, 28 cm and 30 cm. The height corresponding to base 28 cm is ....... cm. (p) 6 (b) The area of an equilateral triangle is $4\sqrt{3}$ cm2. The perimeter of the triangle is ...... cm. (q) 4 (c) If the height of an equilateral triangle is $3\sqrt{3}$ cm, then each side of the triangle measures ...... cm. (r) 24 (d) Let the base of an isosceles triangle be 6 cm and each of the equal sides be 5 cm. Then, its height is ...... cm. (s) 12 (a) ......, (b) ......, (c) ......, (d) ......, (a) Hence, the height of the triangle corresponding to 28 cm is given by: (b) Perimeter = 3 × Side = 3 × 4 = 12 cm (c) (d) Column I Column II (a) The lengths of three sides of a triangle are 26 cm, 28 cm and 30 cm. The height corresponding to base 28 cm is ....... cm. (r) 24 (b) The area of an equilateral triangle is $4\sqrt{3}$ cm2. The perimeter of the triangle is ...... cm. (s) 12 (c) If the height of an equilateral triangle is $3\sqrt{3}$ cm, then each side of the triangle measures ...... cm. (p) 6 (d) Let the base of an isosceles triangle be 6 cm and each of the equal sides be 5 cm. Then, its height is ...... cm. (q) 4 #### Question 29: A park in the shape of a quadrilateral ABCD has AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m and C = 90°. Find the area of the park. We know that $△$BCD is a right triangle. ∴ Now, Area of quadrilateral ABCD = Area of $△$BCD + Area of $△$ABD = (30 + 35.4) m2 = 65.4 m2 #### Question 30: Find the area of a parallelogram ABCD in which AB = 60 cm, BC = 40 cm and AC = 80 cm. We know that a diagonal divides a parallelogram into two triangles of equal areas. ∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = #### Question 31: A piece of land is in the shape of a rhombus ABCD in which each side measures 100 m and diagonal AC is 160 m long. Find the area of the rhombus. We know that a diagonal divides a parallelogram into two triangles of equal areas. ∴ Area of parallelogram ABCD = 2(Area of triangle ABC) = #### Question 32: A floral design on a floor is made up of 16 triangular tiles, each having sides 9 cm, 28 cm and 35 cm. Find the cost of polishing the tiles at the rate of Rs 2.50 per cm2. Area of one triangular tile can be found in the following manner: ∴ Area of 16 triangular tiles = Cost of polishing 1 cm2 of area = Rs 2.5 Cost of polishing 1411.2 cm2 of area = #### Question 33: A like in the shape of a square with each diagonal 32 cm and having a tail in the shape on an isosceles triangle of base 8 cm and each side 6 cm, is made of three different shades as shown in the figure. How much paper of each shade has been used in it? We know that a diagonal of a parallelogram divides into two triangles of equal areas. Also, a square is a parallelogram. Therefore, we can say that a diagonal of a square divides into two triangles of equal areas. ∴ #### Question 1: Each side of an equilateral triangle is 8 cm. Its altitude is (a) (b) (c) (d) (c) #### Question 2: The perimeter of an isosceles right-angled triangle having a as each of the equal sides is (a) $\left(1+\sqrt{2}\right)a$ (b) $\left(2+\sqrt{2}\right)a$ (c) $3a$ (d) $\left(3+\sqrt{2}\right)a$ (b) $\left(2+\sqrt{2}\right)a$ For an isosceles right-angled triangle having each equal side as a, we have: Hypotenuse = $\sqrt{{a}^{2}+{a}^{2}}=\sqrt{2{a}^{2}}=\sqrt{2}a$ Perimeter = $a+a+\sqrt{2}a=2a+\sqrt{2}a=\left(2+\sqrt{2}\right)a$ #### Question 3: For an isosceles triangle having base = 12 cm and each of the equal sides equal to 10 cm, the height is (a) 12 cm (b) 16 cm (c) 6 cm (d) 8 cm (d) 8 cm #### Question 4: Find the area of an equilateral triangle having each side 6 cm. #### Question 5: Using Heron's formula find the area of ABC in which BC = 13 cm, AC = 14 cm and AB = 15 cm. #### Question 6: The sides of a triangle are in the ratio 13 : 14 : 15 and its perimeter is 84 cm. Find the area of the triangle. Let the sides of the triangle be 13x cm, 14x cm and 15x cm. Perimeter = Sum of all sides or, 84 = 13x + 14x + 15x or, 42x = 84 or, x = 2 So, sides of the triangle are 13$×$2, 14$×$2 and 15$×$2, i.e., 26 cm, 28 cm and 30 cm. #### Question 7: Find the area of ABC in which BC = 8 cm, AC = 15 cm and AB = 17 cm. Find the length of altitude drawn on AB. Altitude of the triangle corresponding to AB: #### Question 8: An isosceles triangle has perimeter 30 cm and each of its equal sides is 12 cm. Find the area of the triangle. Let the base of the isosceles triangle be b cm. Now, Equal side of the triangle = 12 cm Perimeter = 30 cm or, 12 + 12 + b = 30 or, 24 + b = 30 or, b = 6 cm #### Question 9: The perimeter of an isosceles triangle is 32 cm. The ratio of one of the equal sides to its base is 3 : 2. Find the area of the triangle. Let the equal sides of the isosceles triangle (a) be 3x cm and base of the triangle (b) be 2x cm. (i) Perimeter = 32 cm or, 3x + 3x + 2x = 32 or, 8x = 32 or, x = 4 ∴ Equal sides of the triangle = 3x = 3(4) = 12 cm Base of the triangle = 2x = 2(4) = 8 cm Now, #### Question 10: Given a ∆ABC in which I. ∠A, ∠B and ∠C are in the ratio 3 : 2 : 1. II. AB, AC and BC are in the ratio $3:\sqrt{3}:2\sqrt{3}$ and . Is ∆ABC a right triangle? The question give above has two Statements I and II. Answer the question by using instructions given below: (a) If the question can be answered by one of the given statements only and not by the other. (b) If the question can be answered by using either statement along. (c) If the question can be answered by using both the statements but cannot be answered by using either statement. (d) If the question cannot be answered even by using both the statements together. (b) Let $\angle$A, $\angle$ B and $\angle$C be 3x, 2x and x, respectively. Hence, it is a right triangle at A. Let the sides of the triangle be Also, Now, So, it can be answered using either statement. #### Question 11: In the given figure ABC and ∆DBC have the same base BC such that AB = 120 m, AC = 122 m, BC = 22 m, BD = 24 m and CD = 26 m. Find the area of the shaded region. Now, #### Question 12: A point O is taken inside an equilateral ABC. If OLBC, OMAC and ONAB such that OL = 14 cm, OM = 10 cm and ON = 6 cm, find the area of ∆ABC.
Courses Courses for Kids Free study material Offline Centres More Store # How do you calculate $\left[ {{{\tan }^{ - 1}}\left( {\left( {\dfrac{1}{2}} \right)} \right)} \right]$. Last updated date: 17th Jun 2024 Total views: 373.5k Views today: 4.73k Verified 373.5k+ views Hint: In this question, we have a trigonometric inverse function. The trigonometric inverse function is also called the arc function. To solve the trigonometric inverse function we assume the angle $\theta$ which is equal to that trigonometric inverse function. Then we find the value of $\theta$. Complete step by step solution: In this question, we used the word trigonometric inverse function. The trigonometric inverse function is defined as the inverse function of trigonometric identities like sin, cos, tan, cosec, sec, and cot. The trigonometric inverse function is also called cyclomatic function, anti trigonometric function, and arc function. The trigonometric inverse function is used to find the angle of any trigonometric ratio. The trigonometric inverse function is applicable for right-angle triangles. Let us discuss all six trigonometric functions. Arcsine function: it is the inverse function of sine. It is denoted as ${\sin ^{ - 1}}$. Arccosine function: it is the inverse function of cosine. It is denoted as ${\cos ^{ - 1}}$. Arctangent function: it is the inverse function of tangent. It is denoted as ${\tan ^{ - 1}}$. Arccotangent function: it is the inverse function of cotangent. It is denoted as ${\cot ^{ - 1}}$. Arcsecant function: it is the inverse function of secant. It is denoted as ${\sec ^{ - 1}}$. Arccosecant function: it is the inverse function of cosecant. It is denoted as $\cos e{c^{ - 1}}$. Now, we come to the question. The data is given below. $\left[ {{{\tan }^{ - 1}}\left( {\left( {\dfrac{1}{2}} \right)} \right)} \right]$ Let us assume that the angle $\theta$ (angle of the right-angle triangle) is equal to that trigonometric function. Then, $\Rightarrow \theta = \left[ {{{\tan }^{ - 1}}\left( {\left( {\dfrac{1}{2}} \right)} \right)} \right]$ Then, $\Rightarrow \tan \theta = \dfrac{1}{2}$ We find the value of angle$\theta$. Then, $\Rightarrow \theta = \left[ {{{\tan }^{ - 1}}\left( {\left( {\dfrac{1}{2}} \right)} \right)} \right]$ After calculating the above, the result is as below. $\therefore \theta = 26.57^\circ$ Therefore, the value of $\left[ {{{\tan }^{ - 1}}\left( {\left( {\dfrac{1}{2}} \right)} \right)} \right]$ is $26.57^\circ$. Note: If you have a trigonometric inverse function with value. Then first assume that the angle $\theta$. Then find the value of that angle $\theta$. The angle $\theta$ is the angle of the right-angle triangle. And trigonometric functions are always used for right-angle triangles.
## A Multiplication Based Logic Puzzle ### 120 and Level 5 120 is a composite number. 120 = 1 x 120, 2 x 60, 3 x 40, 4 x 30, 5 x 24, 6 x 20, 8 x 15, or 10 x 12. Factors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. Prime factorization: 120 = 2 x 2 x 2 x 3 x 5, which can also be written 120 = 2³ x 3 x 5. Thinking process using divisibility tricks to find the factor pairs of 120: √120 is irrational and approximately equal to 10.95. Every factor pair of 120 will have one factor less than 10.95 and one factor greater than 10.95, and we will find both factors in each pair at the same time. The following numbers are less than 10.95. Are they factors of 120? 1. Yes, all whole numbers are divisible by 1, so 1 x 120 = 120. 2. Yes, 120 is an even number. 120 ÷ 2 = 60, so 2 x 60 = 120. (Since 60 is even, 4 will also be a factor of 120.) 3. Yes, 1 + 2 + 0 = 3 which is divisible by 3 (but not by 9), so 120 is divisible by 3. 120 ÷ 3 = 40, so 3 x 40 = 120. Note 120 will not be divisible by 9. 4. Yes, the number formed from its last two digits, 20, is divisible by 4, so 120 is divisible by 4, and 4 x 30 = 120. (Note since 30 is even, 8 will also be a factor of 120.) 5. Yes, the last digit is 0 or 5, so 120 is divisible by 5, and 5 x 24 = 120. 6. Yes, 120 is divisible by both 2 and 3, so it is divisible by 6, and 6 x 20 = 120. 7. No. The divisibility trick for 7 requires us to split 120 into 12 and 0. We double 0 and subtract the double from 12. 12 – (2 x 0) = 12 – 0 = 12. Since 12 is not divisible by 7, 120 also is not divisible by 7. 8. Yes, see 4 above. 120 = 8 x 15. (This will mean that ANY number whose last 3 digits are 120 will also be divisible by 8.) 9. No, see 3 above. 120 is not divisible by 9. 10. Yes, 120 ends with a zero, so 10 is a factor of 120, and 10 x 12 = 120. From this thinking process we conclude that the factor pairs of 120 are 1 x 120, 2 x 60, 3 x 40, 4 x 30, 5 x 24, 6 x 20, 8 x 15, and 10 x 12. When 120 is a clue in the FIND THE FACTORS 1 – 12 puzzles, use 10 and 12 as the factors. 5! = 1 x 2 x 3 x 4 x 5 = 120. Excel file of this week’s puzzles and last week’s factors: 10 Factors 2014-05-12 #### Comments on: "120 and Level 5" (6) 1. You are making numbers very interesting! Like • ivasallay said: Thank you. If numbers are interesting or boring is always a personal decision. I like the decision you’ve made! Like • Oh, I like that attitude. There is almost always something interesting to be found in numbers, anyway. Like 2. ivasallay said: So true. I’ve decided to add a link to each post with interesting facts about the factored number. Like • yay! Like • Yeah, the little facts are fun extras. Like
1. Chapter 10 Class 12 Vector Algebra 2. Serial order wise 3. Supplementary examples and questions from CBSE Transcript Supplementary Exercise Q4 Show that (i) the vectors 𝑎﷯ = 2 𝑖﷯ − 𝑗﷯ + 𝑘﷯, 𝑏﷯ = 𝑖﷯ + 2 𝑗﷯ − 3 𝑘﷯, and 𝑐﷯ = 3 𝑖﷯ − 4 𝑗﷯ + 5 𝑘﷯ are coplanar. Three vectors 𝑎﷯, 𝑏﷯, 𝑐﷯ are coplanar if [ 𝒂﷯, 𝒃﷯, 𝒄﷯ ] = 0 Given, 𝑎﷯ = 2 𝑖﷯ − 𝑗﷯ + 𝑘﷯ 𝑏﷯ = 𝑖﷯ + 2 𝑗﷯ − 3 𝑘﷯ 𝑐﷯ = 3 𝑖﷯ − 4 𝑗﷯ + 5 𝑘﷯ 𝑎﷯,𝑏, 𝑐﷯﷯ = 2﷮−1﷮1﷮1﷮2﷮−3﷮3﷮−4﷮5﷯﷯ = 2 2×5﷯−(−4×−3)﷯ − (−1) 1×5﷯−(3×−3) ﷯ + 1 1×−4﷯−(3×2) ﷯ = 2 10−12﷯+ 5+9﷯ + −4−6﷯ = –4 + 14 – 10 = 0 ∴ 𝒂﷯,𝒃, 𝒄﷯﷯ = 0 Therefore, 𝑎﷯,𝑏 and 𝑐﷯ are coplanar. Supplementary Exercise Q4 Show that (ii) the vectors 𝑎﷯ = 𝑖﷯ − 2 𝑗﷯ + 3 𝑘﷯, 𝑏﷯ = −2 𝑖﷯ + 3 𝑗﷯ − 4 𝑘﷯, and 𝑐﷯ = 𝑖﷯ − 3 𝑗﷯ + 5 𝑘﷯ are coplanar Three vectors 𝑎﷯, 𝑏﷯, 𝑐﷯ are coplanar if [ 𝒂﷯, 𝒃﷯, 𝒄﷯ ] = 0 Given, 𝑎﷯ = 𝑖﷯ − 2 𝑗﷯ + 3 𝑘﷯ 𝑏﷯ = −2 𝑖﷯ + 3 𝑗﷯ − 4 𝑘﷯ 𝑐﷯ = 𝑖﷯ − 3 𝑗﷯ + 5 𝑘﷯ 𝑎﷯,𝑏, 𝑐﷯﷯ = 1﷮−2﷮3﷮−2﷮3﷮−4﷮1﷮−3﷮5﷯﷯ = 1 3×5﷯−(−3×4) ﷯ − (−2) −2×5﷯−(1×−4) ﷯ + 3 −2×−3﷯−(1×3) ﷯ = 1 15−12﷯+ 2 −10−(−4﷯﷯ + 3 6−3﷯ = 1(3) + 2 (−6) + 3 (3) = 3 − 12 + 9 = 12 − 12 = 0 ∴ 𝒂﷯,𝒃, 𝒄﷯﷯ = 0 Therefore, 𝑎﷯,𝑏 and 𝑐﷯ are coplanar. Supplementary examples and questions from CBSE
Basic Math \| Basic-2 Math \| Prealgebra \| Workbooks \| Glossary Glossary \| Standards \| Site Map \| Help # QuickQuiz: One and Four-Digit Multiplication This activity was designed to help you practice multiplication. You're going to see a variety of problems that ask you to multiply one-digit numbers and four-digit numbers. All of the questions combine random values. Since there is some carrying in these problems, you may want a piece of paper ready to work out the problem. Good luck and have fun. ## Directions This is another NumberNut three-choice quiz. Once you start the activity you will see a math problem. To the right side or below that problem are three possible answers. It's your job to figure out the correct answer and click the answer. The next screen will show you the correct answer. Some of these quizzes will show you how to get the right answer. You will get a happy or sad face for every question you finish. Once you finish ten (10) questions, the quiz will be over. Take the quiz again because all of the questions are random. Chances are, you'll get a new quiz every time. It's good practice to learn these basic arithmetic operations. RELATED LINKS LESSONS: - NumberNut.com: Multiplication ACTIVITIES: - Memory Challenge: Multiply by One - Memory Challenge: Multiply by Two - Memory Challenge: Multiply by Three - Memory Challenge: Multiply by Four - Memory Challenge: Multiply by Five - Memory Challenge: Multiply by Six - Memory Challenge: Multiply by Seven - Memory Challenge: Multiply by Eight - Memory Challenge: Multiply by Nine - Memory Challenge: Multiply by Ten - QuickQuiz: Single-Digit Multiplication (H) - QuickQuiz: Single-Digit Multiplication (V) - Next in Series: 4, 6, and 8 Multiplication - Next in Series: 3, 7, and 9 Multiplication - Next in Series: 2, 5, and 10 Multiplication - Pick-a-Card: Single-Digit Word Problems - Pick-a-Card: Word Problems - More or Less: One-Digit Multiplication - QuickQuiz: One/Two-Digit Mult. (No Carry, H) - QuickQuiz: One/Two-Digit Mult. (No Carry, V) - More or Less: One/Two-Digit Mult. - QuickQuiz: One/Two-Digit (Carrying) - More or Less: Two-Digit Mult. (Carry) - QuickQuiz: One/Three-Digit Multiplication - QuickQuiz: One/Four-Digit Multiplication - QuickQuiz: Two-Digit Multiplication - QuickQuiz: Two/Three-Digit Multiplication - QuickQuiz: Two/Four-Digit Multiplication - Overview - Shapes-Colors - Numbers - Addition - Subtraction - Multiplication - Division - Operations - Dates & Times > Activities * The custom search only looks at Rader's sites. Go for site help or a list of mathematics topics at the site map! Since the site is new, please send us your comments or suggestions from our feedback page.
Suggested languages for you: Americas Europe Problem 301 # Given $\mathrm{A}(\mathrm{t})=\begin{array}{rr}\mid \mathrm{t}^{2} & \cos \mathrm{t} \mid \\ \mid \mathrm{e}^{\mathrm{t}} & \sin \mathrm{t} \mid\end{array}$ Find $$(\mathrm{d} \mathrm{A} / \mathrm{dt})$$ Expert verified The short answer to the question is: $$\frac{dA}{dt} = \begin{pmatrix} 2t & -\sin(t) \\ e^t & \cos(t) \end{pmatrix}$$ See the step by step solution ## Step 1: Recall the formula for finding the derivative of a matrix To find the derivative of a matrix, we simply need to find the derivative of each element of the matrix with respect to the variable (in this case, t) and place the derivatives in the resulting matrix in their corresponding positions. ## Step 2: Find the derivatives of the matrix elements We will find the derivatives of each element in the matrix A(t): 1. Derivative of $$t^2$$: using basic power rule: $$\frac{d}{dt}(t^2) = 2t$$ 2. Derivative of $$\cos(t)$$: using basic differentiation rules: $$\frac{d}{dt}(\cos(t)) = -\sin(t)$$ 3. Derivative of $$e^t$$: using basic exponential function differentiation: $$\frac{d}{dt}(e^t) = e^t$$ 4. Derivative of $$\sin(t)$$: using basic differentiation rules: $$\frac{d}{dt}(\sin(t)) = \cos(t)$$ ## Step 3: Form the resulting matrix Now that we have found the derivatives of each element of A(t), we can form the resulting matrix (dA/dt): $\frac{dA}{dt} = \begin{pmatrix} 2t & -\sin(t) \\ e^t & \cos(t) \end{pmatrix}$ That's it! The derivative of the given matrix A(t) is: $\frac{dA}{dt} = \begin{pmatrix} 2t & -\sin(t) \\ e^t & \cos(t) \end{pmatrix}$ We value your feedback to improve our textbook solutions. ## Access millions of textbook solutions in one place • Access over 3 million high quality textbook solutions • Access our popular flashcard, quiz, mock-exam and notes features ## Join over 22 million students in learning with our Vaia App The first learning app that truly has everything you need to ace your exams in one place. • Flashcards & Quizzes • AI Study Assistant • Smart Note-Taking • Mock-Exams • Study Planner
Printer Friendly Version Beverly Mackie BellaOnline's Math Editor Integer Practice Problems - Addition Addition Rules Summary – ADDING TWO ALIKE SIGNS – two positive numbers or two negative numbers 1. Add absolute values 2. Attach same sign to the answer Example: Adding two positives integers 4 + 5 = 9 Adding two negative integers (-4) + (-5) = (-9) ADDING TWO UNLIKE SIGNS - 6 + 4 = ? 1. Subtract absolute values 2. Use the sign of the number with the larger absolute value \| - 6\| = 6 \| 4 \| = 4 Remember to put the larger number first when subtracting two numbers. -- > 6 - 4 = 2 Six is larger. So, the answer will have a negative sign. - 6 + 4 = - 2 Addition Review - Add two positive integers and the answer will be ________________ . Add two negative integers and the answer will be ____________________. Add two integers with different signs by ___________________ the __________ values. Then, the sign of the answer is determined by the integer with the _______________ absolute value. Practice Problems 1. -23 + 23 = 2. -19 + 29 = 3. 42 + (-62) = 4. 34 – 46 = 5. -29 + 12 = 6. -57 + (-67) = 7. -88 + (-88) = 8. -30 + 49 = 9. 24 + (-32) = 10. 124 + 453 = 11. - 211 + 49 = 12. 59 + (- 87) + (- 5) = 13. (-4) + (-3) + 2 + (-1) = 14. (-18) + 5 + 23 + ( - 4) = 15. 45 + (-23) + 29 + (- 39) = Answers - Addition Review - Add two positive integers, and the answer will be positive. Add two negative integers and the answer will be negative. Add two integers with different signs by subtracting. the absolute values. Then, the sign of the answer is determined by the integer with the largest absolute value. Practice Problems (Note – The parentheses around a negative number are for clarity; they are not mandatory.) 1. -23 + 23 = 0 2. -19 + 29 = 10 3. 42 + (-62) = -20 4. 34 + 46 = 80 5. -29 + 12 = -17 6. -57 + (-67) = - 124 7. -88 + (-88) = -176 8. -30 + 49 = 19 9. 24 + (-32) = - 8 10. 124 + 453 = 577 11. - 211 + 49 = -162 12. 59 + (- 87) + (- 5) = -33 13. (-4) + (-3) + 2 + (-1) = -6 14. (-18) + 5 + 23 + ( - 4) = 6 15. 45 + (-23) + 29 + (- 39) = 12 Math Site @ BellaOnline View This Article in Regular Layout Content copyright © 2013 by Beverly Mackie. All rights reserved. This content was written by Beverly Mackie. If you wish to use this content in any manner, you need written permission. Contact Beverly Mackie for details. \| About BellaOnline \| Privacy Policy \| Advertising \| Become an Editor \| Website copyright © 2016 Minerva WebWorks LLC. All rights reserved.
# What Is 45/48 as a Decimal + Solution With Free Steps The fraction 45/48 as a decimal is equal to 0.937. Improper fractions are a type of fractions, expressed as a/b, where the numerator a is greater than the denominator b. This fraction is greater than the whole number 1. Hence these fractions are usually converted into a mixed fraction for the final presentation. Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers. Now, we introduce the method used to solve said fraction-to-decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 45/48. ## Solution First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively. This can be done as follows: Dividend = 45 Divisor = 48 Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents: Quotient = Dividend $\div$ Divisor = 45 $\div$ 48 This is when we go through the Long Division solution to our problem. Given is the long division process in Figure 1: Figure 1 ## 45/48 Long Division Method We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 45 and 48, we can see how 45 is Smaller than 48, and to solve this division, we require that 45 be Bigger than 48. This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later. Now, we begin solving for our dividend 45, which after getting multiplied by 10 becomes 450. We take this 450 and divide it by 48; this can be done as follows: 450 $\div$ 48 $\approx$ 9 Where: 48 x 9 = 432 This will lead to the generation of a Remainder equal to 450 – 432 = 18. Now this means we have to repeat the process by Converting the 18 into 180 and solving for that: 180 $\div$ 48 $\approx$ 3 Where: 48 x 3 = 144 This, therefore, produces another Remainder which is equal to 180 – 144 = 36. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 360. 360 $\div$ 48 $\approx$ 7 Where: 48 x 7 = 336 Finally, we have a Quotient generated after combining the three pieces of it as 0.937, with a Remainder equal to 24. Images/mathematical drawings are created with GeoGebra.
Select Page We often come across the word ‘Ratio’ and are aware of its literal meaning, but how do we deal with it while solving problems? To answer that question,let us discuss Ratios point by point. Definition: A ratio expresses the number of times one quantity is of the other. Mathematically speaking, if there are two quantities of the same type, i.e., A and B, then their ratio is shown as A : B. Read this as A is to B. Here, A and B are the terms of the ratio. A is the antecedent, and B is the consequent/precedent. For example: We assign some values to A and B,say 4 and 5. Then, the ratio between A and B is 4 is to 5. This can be written down as: A: B = 4: 5 or A/B = 4/5 Basic Points for Ratio: A ratio must be expressed between entities expressed in similar units. For example, you cannot derive a ratio between the surface area of one figure and the volume of another. This is because the units are different. For you to derive a ratio, you need to either compare the respective areas or the respective volumes of the two figures. • You can cancel common factors to arrive at a given fraction. If there are two quantities A and B having values 3 and 6, then the ratio for the two will be 3:6 = 1:2 (after cancelling the common factor). • If a number is multiplied to both the numerator and denominator then there will not be any change in the ratio. For example, 3:4 is the same as the 6:8, 12:16. The ratio remains the same. • If a number, divides both the numerator and the denominator then there will not be any change in the ratio. For example, 3:4 is the same as the 3/2:4/2 = 6/2:8/2. The ratio remains the same. • If there are two fractions and they are in a ratio then we can write them as a single fraction. For example, suppose there are two fractions like 3/4 and 5/7 and they are in ratio i.e., 3/4: 5/7. Then they can be written as 3/4 x 7/5 = 21/20. Compound Ratio: Ratios are compounded by multiplying together the antecedents for a new antecedent and consequents for a new consequent. Confused? Do not be.Let us take up a practical example for the same. Example: Find the ratio compounded for the following 3 ratios: 2:3, 5:8 and 3:7. Solution: The required ratio in this case is = Duplicate Ratio: When the ratio 8: 5 is compounded with itself, then the resulting ratio, 82: 52 is called the duplicate ratio. Triplicate Ratio: When the ratio 8: 5 is compounded with itself twice, then the resulting ratio, 83: 53 is called the triplicate ratio. Sub-duplicate Ratio: is the sub-duplicate ratio for 8: 5. Sub-triplicate ratio: is the sub-triplicate ratio for 8: 5. Inverse Ratio: If a: b is the given ratio, then 1/a: 1/b or in other words, b:a is the inverse ratio. To get thorough with the above concepts based on Basics of Ratio you can test yourself with the exercise given below: Exercise: Question 1: The ratio 21.5: 20.5 is the same as: (1) 2: 1 (2) 3: 1 (3) 6: 1 (4) 3: 2 Solution: (1) Concept: ratio means a:b i.e., a/b so using it Required ratio = = = 2: 1 Question 2: If x: y = 3: 2, then the ratio 2x2 + 3y2: 3x2 – 2y2 is equal to: (1) 12: 5 (2) 6: 5 (3) 30: 19 (4) 5: 3 Solution: (3) Use property of duplicate ratio. Divide the equation by y2. = 30: 19 Question 3: If a: b = b: c, then a4: b4 is equal to: (1) ac: b2 (2) a2: c2 (3) c2: a2 (4) b2: ac Solution: (2) Cross multiplication Square it Question 4: If x: y = 2: 3, then the value of is equal to: (1) (2) (3) (4) Solution: (2) Given, Expression = Divide the equation by y. Put the values [from (i)] Question 5: If x: y = 2: 1, then (x2 – y2): (x2 + y2) is (1) 3: 5 (2) 5: 3 (3) 4: 5 (4) 5: 6 Solution: (1) Here, Question 6: If x: y = 4: 5, then (3x + y): (5x + 3y) = (1) 3: 5 (2) 5: 3 (3) 17: 35 (4) 35: 17 Solution: (3) We have Question 7: If m: n = 3: 2, then (4m + 5n): (4m – 5n) is equal to: (1) 4: 9 (2) 9: 4 (3) 11: 1 (4) 9: 1 Solution: (3) So (4m + 5n): (4m – 5n) = 11: 1 Question 8: If p: q: r= 1: 2: 4, then is equal to (1) 5 (2) 2q (3) 5p (4) 4r Solution: (3) We have,(let) Question 9: If a: b: c = 2: 3: 4 and 2a – 3b + 4c = 33, then the value of c is? (1) 6 (2) 9 (3) 12 (4) Solution: (3) Question 10: If a: b: c = 7: 3: 5, then (a + b + c): (2a + b – c) is equal to (1) 1: 2 (2) 2: 3 (3) 3: 4 (4) 5: 4 Solution: (4) Question 11: If a: b: c = 3: 4: 7, then the ratio (a + b + c): c is equal to: (1) 2: 1 (2)14: 3 (3) 7: 2 (4) 1: 2 Solution: (1) We have, Ratio and Proportion Questions: Problems on Ratio and Proportion you should solve for competitive examination preparation Welcome to this exercise on Problems on Ratio and Proportion. In this exercise, we build on the basic concepts for Ratio and Proportion. As you prepare for your competitive examinations, you will come across questions on Ratio and Proportion. Such questions need optimized tackling and can be solved with ease by using simple tricks and understanding the relationships highlighted in this Ratio and Proportion Questions article. The Ratio and Proportion Questions exercise comes into the picture where it gives you a chance to practice the highlighted and important concepts related to Ratio and Proportion question type. Starting 3rd June 2024, 7pm Onwards FREE CHEAT SHEET Learn How to Master VA-RC This free (and highly detailed) cheat sheet will give you strategies to help you grow No thanks, I don't want it.
Share # The Pie Chart (As Shown in the Figure 25.23) Represents the Amount Spent on Different Sports by a Sports Club in a Year. If the Total Money Spent by the Club on Sports is Rs 1,08 - Mathematics Course ConceptIntroduction of Circle Graph Or Pie Chart #### Question The pie chart (as shown in the figure 25.23) represents the amount spent on different sports by a sports club in a year. If the total money spent by the club on sports is Rs 1,08,000, find the amount spent on each sport. #### Solution $\text{ Amount spent on cricket }= \frac{\text{ Central angle of the corresponding sector }\times \text{ Total Money spent }}{360^\circ}$ $= \frac{150^\circ \times 108000}{360^\circ} = Rs 45, 000$ $\text{ Amount spent on hockey }= \frac{\text{ Central angle of the corresponding sector }\times\text{ Total Money spent }}{360^\circ}$ $= \frac{100^\circ \times 108000}{360^\circ} = Rs 30, 000$ $\text{ Amount spent on football }= \frac{\text{ Central angle of the corresponding sector }\times\text{ Total Money spent }}{360^\circ}$ $= \frac{60^\circ \times 108000}{360^\circ} = Rs 18, 000$ $\text{ Amount spent on tennis }= \frac{\text{ Central angle of the corresponding sector }\times\text{ Total Money spent }}{360^\circ}$ $= \frac{50^\circ \times 108000}{360^\circ} = Rs 15, 000$ Is there an error in this question or solution? #### APPEARS IN RD Sharma Solution for Mathematics for Class 8 by R D Sharma (2019-2020 Session) (2017 to Current) Chapter 25: Data Handling-III (Pictorial Representation of Data as Pie Charts or Circle Graphs) Ex. 25.2 \| Q: 7 \| Page no. 23 #### Video TutorialsVIEW ALL [1] Solution The Pie Chart (As Shown in the Figure 25.23) Represents the Amount Spent on Different Sports by a Sports Club in a Year. If the Total Money Spent by the Club on Sports is Rs 1,08 Concept: Introduction of Circle Graph Or Pie Chart. S
## Fractions • A fraction (from Latin fractus, "broken") represents a part of a whole or, more generally, any number of equal parts. When spoken in everyday English, a fraction describes how many parts of a certain size there are, for example, one-half, eight-fifths, three-quarters. A common, vulgar, or simple fraction (examples: ${\displaystyle {\tfrac {1}{2}}}$ and 17/3) consists of an integer numerator displayed above a line (or before a slash), and a non-zero integer denominator, displayed below (or after) that line. Numerators and denominators are also used in fractions that are not common, including compound fractions, complex fractions, and mixed numerals. The numerator represents a number of equal parts, and the denominator, which cannot be zero, indicates how many of those parts make up a unit or a whole. For example, in the fraction 3/4, the numerator, 3, tells us that the fraction represents 3 equal parts, and the denominator, 4, tells us that 4 parts make up a whole. The picture to the right illustrates ${\displaystyle {\tfrac {3}{4}}}$ or ¾ of a cake. Fractional numbers can also be written without using explicit numerators or denominators, by using decimals, percent signs, or negative exponents (as in 0.01, 1%, and 10−2 respectively, all of which are equivalent to 1/100). An integer such as the number 7 can be thought of as having an implicit denominator of one: 7 equals 7/1. ${\displaystyle a\times {\frac {b}{c}},}$ ${\displaystyle a\cdot {\frac {b}{c}},}$ ${\displaystyle a\left({\frac {b}{c}}\right).}$ Not to be confused with fractions involving complex numbers ${\displaystyle {\frac {\tfrac {1}{2}}{\tfrac {1}{3}}}={\tfrac {1}{2}}\times {\tfrac {3}{1}}={\tfrac {3}{2}}=1{\tfrac {1}{2}}}$ ${\displaystyle {\frac {12{\tfrac {3}{4}}}{26}}=12{\tfrac {3}{4}}\cdot {\tfrac {1}{26}}={\tfrac {12\cdot 4+3}{4}}\cdot {\tfrac {1}{26}}={\tfrac {51}{4}}\cdot {\tfrac {1}{26}}={\tfrac {51}{104}}}$ ${\displaystyle {\frac {\tfrac {3}{2}}{5}}={\tfrac {3}{2}}\times {\tfrac {1}{5}}={\tfrac {3}{10}}}$ ${\displaystyle {\frac {8}{\tfrac {1}{3}}}=8\times {\tfrac {3}{1}}=24.}$ ${\displaystyle {\frac {5}{10/{\tfrac {20}{40}}}}={\frac {1}{4}}\quad }$ or ${\displaystyle \quad {\frac {\tfrac {5}{10}}{\tfrac {20}{40}}}=1}$ ${\displaystyle {\tfrac {63}{462}}={\tfrac {63\div 21}{462\div 21}}={\tfrac {3}{22}}}$ ${\displaystyle {\tfrac {3}{4}}>{\tfrac {2}{4}}}$ because 3>2. ${\displaystyle {\tfrac {2}{3}}}$ ? ${\displaystyle {\tfrac {1}{2}}}$ gives ${\displaystyle {\tfrac {4}{6}}>{\tfrac {3}{6}}}$ ${\displaystyle {\tfrac {5}{18}}}$ ? ${\displaystyle {\tfrac {4}{17}}}$ ${\displaystyle {\tfrac {5\times 17}{18\times 17}}}$ ? ${\displaystyle {\tfrac {4\times 18}{17\times 18}}}$ ${\displaystyle {\tfrac {2}{4}}+{\tfrac {3}{4}}={\tfrac {5}{4}}=1{\tfrac {1}{4}}}$. ${\displaystyle {\frac {1}{4}}\ +{\frac {1}{3}}={\frac {1\times 3}{4\times 3}}\ +{\frac {1\times 4}{3\times 4}}={\frac {3}{12}}\ +{\frac {4}{12}}={\frac {7}{12}}.}$ ${\displaystyle {\frac {3}{5}}+{\frac {2}{3}}}$ ${\displaystyle {\frac {3}{5}}+{\frac {2}{3}}}$ ${\displaystyle {\frac {9}{15}}+{\frac {10}{15}}={\frac {19}{15}}=1{\frac {4}{15}}}$ ${\displaystyle {\frac {a}{b}}+{\frac {c}{d}}={\frac {ad+cb}{bd}}}$ ${\displaystyle {\frac {a}{b}}+{\frac {c}{d}}+{\frac {e}{f}}={\frac {a(df)+c(bf)+e(bd)}{bdf}}}$ ${\displaystyle {\frac {3}{4}}+{\frac {5}{12}}={\frac {9}{12}}+{\frac {5}{12}}={\frac {14}{12}}={\frac {7}{6}}=1{\frac {1}{6}}}$ ${\displaystyle {\tfrac {2}{3}}-{\tfrac {1}{2}}={\tfrac {4}{6}}-{\tfrac {3}{6}}={\tfrac {1}{6}}}$ ${\displaystyle {\tfrac {2}{3}}\times {\tfrac {3}{4}}={\tfrac {6}{12}}}$ ${\displaystyle {\tfrac {2}{3}}\times {\tfrac {3}{4}}={\tfrac {{\cancel {2}}^{~1}}{{\cancel {3}}^{~1}}}\times {\tfrac {{\cancel {3}}^{~1}}{{\cancel {4}}^{~2}}}={\tfrac {1}{1}}\times {\tfrac {1}{2}}={\tfrac {1}{2}}}$ ${\displaystyle 6\times {\tfrac {3}{4}}={\tfrac {6}{1}}\times {\tfrac {3}{4}}={\tfrac {18}{4}}}$ This method works because the fraction 6/1 means six equal parts, each one of which is a whole. ${\displaystyle 3\times 2{\tfrac {3}{4}}=3\times \left({\tfrac {8}{4}}+{\tfrac {3}{4}}\right)=3\times {\tfrac {11}{4}}={\tfrac {33}{4}}=8{\tfrac {1}{4}}}$ 0.5 = 5/9 0.62 = 62/99 0.264 = 264/999 0.6291 = 6291/9999 0.05 = 5/90 0.000392 = 392/999000 0.0012 = 12/9900 0.1523 + 0.0000987 1523/10000 + 987/9990000 = 1522464/9990000 x = 0.1523987 10,000x = 1,523.987 10,000,000x = 1,523,987.987 10,000,000x − 10,000x = 1,523,987.987 − 1,523.987 9,990,000x = 1,523,987 − 1,523 9,990,000x = 1,522,464 x = 1522464/9990000 ${\displaystyle (a,b)+(c,d)=(ad+bc,bd)\,}$ ${\displaystyle (a,b)-(c,d)=(ad-bc,bd)\,}$ ${\displaystyle (a,b)\cdot (c,d)=(ac,bd)}$ ${\displaystyle (a,b)\div (c,d)=(ad,bc){\text{ (when c ≠ 0)}}}$ ${\displaystyle {\frac {3}{\sqrt {7}}}={\frac {3}{\sqrt {7}}}\cdot {\frac {\sqrt {7}}{\sqrt {7}}}={\frac {3{\sqrt {7}}}{7}}}$ ${\displaystyle {\frac {3}{3-2{\sqrt {5}}}}={\frac {3}{3-2{\sqrt {5}}}}\cdot {\frac {3+2{\sqrt {5}}}{3+2{\sqrt {5}}}}={\frac {3(3+2{\sqrt {5}})}{{3}^{2}-(2{\sqrt {5}})^{2}}}={\frac {3(3+2{\sqrt {5}})}{9-20}}=-{\frac {9+6{\sqrt {5}}}{11}}}$ ${\displaystyle {\frac {3}{3+2{\sqrt {5}}}}={\frac {3}{3+2{\sqrt {5}}}}\cdot {\frac {3-2{\sqrt {5}}}{3-2{\sqrt {5}}}}={\frac {3(3-2{\sqrt {5}})}{{3}^{2}-(2{\sqrt {5}})^{2}}}={\frac {3(3-2{\sqrt {5}})}{9-20}}=-{\frac {9-6{\sqrt {5}}}{11}}}$ ६ १ २ १ १ १ ४ ५ ९ 6 1 2 1 1 −1 4 5 9 • 2 are white, • 6 are red, and • 4 are yellow, • A unit fraction is a vulgar fraction with a numerator of 1, e.g. ${\displaystyle {\tfrac {1}{7}}}$. Unit fractions can also be expressed using negative exponents, as in 2−1, which represents 1/2, and 2−2, which represents 1/(22) or 1/4. • An Egyptian fraction is the sum of distinct positive unit fractions, for example ${\displaystyle {\tfrac {1}{2}}+{\tfrac {1}{3}}}$. This definition derives from the fact that the ancient Egyptians expressed all fractions except ${\displaystyle {\tfrac {1}{2}}}$, ${\displaystyle {\tfrac {2}{3}}}$ and ${\displaystyle {\tfrac {3}{4}}}$ in this manner. Every positive rational number can be expanded as an Egyptian fraction. For example, ${\displaystyle {\tfrac {5}{7}}}$ can be written as ${\displaystyle {\tfrac {1}{2}}+{\tfrac {1}{6}}+{\tfrac {1}{21}}.}$ Any positive rational number can be written as a sum of unit fractions in infinitely many ways. Two ways to write ${\displaystyle {\tfrac {13}{17}}}$ are ${\displaystyle {\tfrac {1}{2}}+{\tfrac {1}{4}}+{\tfrac {1}{68}}}$ and ${\displaystyle {\tfrac {1}{3}}+{\tfrac {1}{4}}+{\tfrac {1}{6}}+{\tfrac {1}{68}}}$. • A dyadic fraction is a vulgar fraction in which the denominator is a power of two, e.g. ${\displaystyle {\tfrac {1}{8}}}$. • special fractions: fractions that are presented as a single character with a slanted bar, with roughly the same height and width as other characters in the text. Generally used for simple fractions, such as: ½, ⅓, ⅔, ¼, and ¾. Since the numerals are smaller, legibility can be an issue, especially for small-sized fonts. These are not used in modern mathematical notation, but in other contexts. • case fractions: similar to special fractions, these are rendered as a single typographical character, but with a horizontal bar, thus making them upright. An example would be ${\displaystyle {\tfrac {1}{2}}}$, but rendered with the same height as other characters. Some sources include all rendering of fractions as case fractions if they take only one typographical space, regardless of the direction of the bar. • shilling or solidus fractions: 1/2, so called because this notation was used for pre-decimal British currency (£sd), as in 2/6 for a half crown, meaning two shillings and six pence. While the notation "two shillings and six pence" did not represent a fraction, the forward slash is now used in fractions, especially for fractions inline with prose (rather than displayed), to avoid uneven lines. It is also used for fractions within fractions (complex fractions) or within exponents to increase legibility. Fractions written this way, also known as piece fractions, are written all on one typographical line, but take 3 or more typographical spaces. • built-up fractions: ${\displaystyle {\frac {1}{2}}}$. This notation uses two or more lines of ordinary text, and results in a variation in spacing between lines when included within other text. While large and legible, these can be disruptive, particularly for simple fractions or within complex fractions. Wikipedia
Games Problems Go Pro! Seventh grader Joe asks, "Can you come up with a better approximation to the Golden ratio than the fraction 8/5?" Hi Joe! The answer to that question is: definitely! Before we dive into approximations, do you know how to write the exact value of the golden ratio? It's (1 + √5)/2. If you punch that in your calculator (be sure to include the parentheses!) you'll get approximately 1.618. The fraction 8/5 is 1.6, which is not bad, but we can definitely get closer! One way to do it would be to take the golden ratio and shift the decimal one place, drop of everything after the decimal, and then put it over ten. 16/10 Of course, you can do the same thing, and get a better approximation, by shifting the decimal twice and dividing by 100: 161/100 Or 1618/1000. You get the idea, right? But that's kind of a cheater way of doing it, because it requires you to already know the value of the golden ratio! So let's talk for a moment about how someone came up with the approximation 8/5. There's a very simple method for arriving at that fraction, and once you know the method, you can create your own approximations. Are you familiar with the Fibonacci Sequence? It's a list of numbers. The first number in the list is 1, the second number in the list is 1, and each number after that is the sum of the two previous numbers in the list. F1 = 1 F2 = 1 F3 = 1 + 1 = 2 F4 = 2 + 1 = 3 F5 = 3 + 2 = 5 F6 = 5 + 3 = 8 F7 = 8 + 5 = 13 F8 = 13 + 8 = 21 F9 = 21 + 13 = 34 Now if you look at those numbers, you might notice that F5 = 5 and F6 = 8. Those are the two numbers in your ratio! That might get you wondering...what if I picked two other adjacent Fibonacci numbers? Like 13 and 8? Well, 13/8 = 1.625, which is bigger than the golden ratio, but is closer to it than 1.6. So let's try the next two numbers in the sequence: 21/13 = 1.61538... This is smaller than the golden ratio, but is closer still! 34/21 = 1.619... Closer still! You get the idea, right? If you take any two fibonacci numbers that are adjacent to each other, and divide the larger one by the smaller, you'll get an approximation of the golden ratio. The further down the list you go, the more accurate your approximation is. For example, to pick two really big Fibonacci numbers: 317811/196418 = 1.618033988738303... This is accurate to ten decimal places! Of course, it's not easy to remember, and it also isn't very easy to use. Whereas, 8/5 is both easy to remember and easy to use. So 317811/196418 is better, but only in the sense that it is more accurate. It might be a good way of impressing your friends (or even your teacher), but that's about all it's good for. As an added note to this, below is an interesting graph which can help you visualize the approximations that you get by taking ratios of successive terms of the Fibonacci sequence.The horizontal orange line represents the golden ratio. The jagged blue line that alternately goes above and below the golden ratio is the sequence of ratios. For example, the first point is 1/1 = 1, the second point is 2/1 = 2, and the third point is 3/2 = 1.5. After that, the sequence quickly converges on* the golden ratio, so that after the eighth or ninth value, you can't even see the blue line any more. That doesn't mean that those approximations equal the golden ratio; just that the difference is so small that it's less than a pixel on the screen! No two of the ratios are the same value. They alternately flip back and forth between being too large and being too small forever, with the difference between the approximation and the actual value getting closer and closer to zero. * "converge" is a mathematical term that means (roughly) "getting closer and closer to a certain value, without ever reaching it." # Featured Games on This Site Match color, font, and letter in this strategy game Mastermind variation, with words # Blogs on This Site Reviews and book lists - books we love! The site administrator fields questions from visitors.
UK USIndia Every Question Helps You Learn 2, 4, 6, 8, Mary's at the cottage gate. # Number Sequences (Medium) A number sequence involves following a pattern. Spotting the pattern is the key! For example, in the sequence 8, 16, 32, 64, ... each number in the sequence can be got from the previous term by multiplying by 2. In this case, the rule is: multiply by 2. The numbers in a sequence are called 'terms': in 5, 10, 15, 20, ... '5' is the first term and '15' is the third term. If you want to talk about a term without naming it, you call it the nth term. For example, if n = 3, this is the 3rd term. In this quiz you will get some practice in using the rules for sequences. TIP: To form a sequence from a given rule for the nth term, put n = 1 first, then n = 2, then n = 3 and so on - depending on how many terms you are asked to find. If you are asked to find 5 terms, then you will go as far as n = 5. 1. Which sequence can be formed from the given rule for the nth term? nth term = 2n + 3 1, 7, 9, 13, ... 1, 5, 7, 9, ... 2, 5, 7, 9, ... 5, 7, 9, 11, ... The terms of the sequence are found by first putting n = 1, then n = 2, then n = 3 and finally n = 4 in the rule for the nth term = 2n + 3. As follows (do the multiplication first THEN the addition): n = 1 gives 2 × 1 + 3 = 5 n = 2 gives 2 × 2 + 3 = 7 n = 3 gives 2 × 3 + 3 = 9 n = 4 gives 2 × 4 + 3 = 11 2. Which sequence can be formed from the given rule for the nth term? nth term = 2n 2, 4, 8, 16, ... 3, 6, 9, 12, ... 1, 2, 4, 6, ... 2, 4, 6, 8, ... The terms of the sequence are found by first putting n = 1, then n = 2, then n = 3 and finally n = 4 in the rule for the nth term = 2n. As follows: n = 1 gives 2 × 1 = 2 n = 2 gives 2 × 2 = 4 n = 3 gives 2 × 3 = 6 n = 4 gives 2 × 4 = 8 3. Which sequence can be formed from the given rule for the nth term? nth term = 2n - 1 1, 3, 5, 9, ... 1, 3, 5, 7, ... 0, 1, 3, 5, ... 1, 5, 15, 45, ... The terms of the sequence are found by first putting n = 1, then n = 2, then n = 3 and finally n = 4 in the rule for the nth term = 2n -1. As follows (do the multiplication first THEN the subtraction): n = 1 gives 2 × 1 - 1 = 1 n = 2 gives 2 × 2 - 1 = 3 n = 3 gives 2 × 3 - 1 = 5 n = 4 gives 2 × 4 - 1 = 7 4. Which sequence can be formed from the given rule for the nth term? nth term = -n + 1 0, 1, 2, 3, ... 0, -1, -2, -3, ... 1, 2, 3, 4, ... 1, 3, 5, 7, ... The terms of the sequence are found by first putting n = 1, then n = 2, then n = 3 and finally n = 4 in the rule for the nth term = -n + 1. As follows: n = 1 gives -1 + 1 = 0 n = 2 gives -2 + 1 = -1 n = 3 gives -3 + 1 = -2 n = 4 gives -4 + 1 = -3 5. Which sequence can be formed from the given rule for the nth term? nth term = -5n -5, -10, -15, -20, ... 5, 10, 15, 20, ... 0, -5, -10, -15, ... -5, -25, -125, -625, ... The terms of the sequence are found by first putting n = 1, then n = 2, then n = 3 and finally n = 4 in the rule for the nth term = -5n. As follows: n = 1 gives -5 × 1 = -5 n = 2 gives -5 × 2 = -10 n = 3 gives -5 × 3 = -15 n = 4 gives -5 × 4 = -20 6. Which sequence can be formed from the given rule for the nth term? nth term = 4n + 7 11, 15, 18, 23, ... 11, 13, 15, 19, ... 11, 15, 19, 23, ... 11, 12, 15, 19, ... The terms of the sequence are found by first putting n = 1, then n = 2, then n = 3 and finally n = 4 in the rule for the nth term = 4n + 7. As follows (do the multiplication first THEN the addition): n = 1 gives 4 × 1 + 7 = 11 n = 2 gives 4 × 2 + 7 = 15 n = 3 gives 4 × 3 + 7 = 19 n = 4 gives 4 × 4 + 7 = 23 7. Which sequence can be formed from the given rule for the nth term? nth term = n2 1, 4, 9, 16, ... 2, 4, 6, 8, ... 1, 2, 3, 4, ... 1, 8, 27, 64, ... The terms of the sequence are found by first putting n = 1, then n = 2, then n = 3 and finally n = 4 in the rule for the nth term = n2. As follows: n = 1 gives 12 = 1 n = 2 gives 22 = 4 n = 3 gives 32 = 9 n = 4 gives 42 = 16 8. Which sequence can be formed from the given rule for the nth term? nth term = n2 - n 0, 3, 8, 15, ... 0, 2, 6, 12, ... 0, 2, 4, 6, 8, ... 1, 3, 5, 7, ... The terms of the sequence are found by first putting n = 1, then n = 2, then n = 3 and finally n = 4 in the rule for the nth term = n2 - n. As follows (do the multiplication first THEN the addition): n = 1 gives 12 - 1 = 0 n = 2 gives 22 - 2 = 2 n = 3 gives 32 - 3 = 6 n = 4 gives 42 - 4 = 12 9. Which sequence can be formed from the given rule for the nth term? nth term = 6n 6, 12, 18, 24, ... 0, 6, 12, 18, ... 2, 4, 6, 8, ... 1, 6, 12, 18, ... The terms of the sequence are found by first putting n = 1, then n = 2, then n = 3 and finally n = 4 in the rule for the nth term = 6n. As follows (do the multiplication first THEN the addition): n = 1 gives 6 × 1 = 6 n = 2 gives 6 × 2 = 12 n = 3 gives 6 × 3 = 18 n = 4 gives 6 × 4 = 24 10. Which sequence can be formed from the given rule for the nth term? nth term = 4n - 1 5, 7, 11, 15, ... 3, 7, 12, 15, ... 3, 8, 11, 15, ... 3, 7, 11, 15, ... The terms of the sequence are found by first putting n = 1, then n = 2, then n = 3 and finally n = 4 in the rule for the nth term = 4n - 1. As follows (do the multiplication first THEN the subtraction): n = 1 gives 4 × 1 - 1 = 3 n = 2 gives 4 × 2 - 1 = 7 n = 3 gives 4 × 3 - 1 = 11 n = 4 gives 4 × 4 - 1 = 15 Author: Frank Evans

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