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# Perimeter and Area of Composite Figures ## Presentation on theme: "Perimeter and Area of Composite Figures"— Presentation transcript: Perimeter and Area of Composite Figures Composite Figure: a figure that is made by combining triangles, quadrilaterals, semi-circles and other two-dimensional figures. Perimeter: Is the distance around a figure 7 cm 5 cm 3 cm ? 7 cm 4 cm P = 7 cm + 5 cm + 4 cm + 7 cm cm P = cm The perimeter of a circle has a special name, It is called the ______________, and is found using either of these formulas: circumference Alice needs to purchase 37 feet of border. Alice is putting a wallpaper border around her bedroom. She must find the __________ of her bedroom in order to purchase the correct amount of border. perimeter 12 ft 3 ft ? 4 ft 6.5 ft ? 3.5 ft 8 ft P = 8 ft + 6.5 ft + 12 ft + 3 ft + 4 ft + 3.5 ft Alice needs to purchase 37 feet of border. The area of a composite figure can be found by adding or subtracting the areas of the simpler figures that compose the composite figure. Area of Semi-Circle + Area of Rectangle + Area of Triangle Joe has to mow the lawn of the field below, how many square yards must he mow? 14 yd 3 yd 10 yd 10 yd 6 yd 6 yd 8 yd 8 yd Area of Semi-Circle + Area of Rectangle + Area of Triangle Is there another way of finding this area? Detective LeRue must investigate a crime committed at the local park. How many square meters of ground must he cover when looking for clues? Total area = 28 m m m2 = 65 square meters Is there another way of finding this area? Area = 4 m (7 m) Area = 4 m(4 m) 4 m 3 m 3 m Area =3 m(7 m) 4 m 28 m2 7 m 21 m2 7 m 16 m2 How can we use subtraction to find the area? Step 1: Find area of big rectangle Step 2: Find area of smaller rectangle Step 3: Subtract the smaller area from the larger area Total area = 77 m2 – 12 m2 = 65 square meters How can we use subtraction to find the area? 11 m 4 m 4 m 3 m 3 m Area = 3 m(4 m) 4 m 7 m 7 m Area =7 m (11 m) =12 m2 = 77 m2 Detective LaRue must isolate the crime scene, how much tape will he need? Perimeter = 7m + 4 m + 3 m + 4 m + 3 m + 3 m + 7 m + 11 m Detective LaRue needs 42 meters of crime scene tape. 4 m 3 m ? 3 m 4 m 7 m 7 m ? Area = 74 cm2 Perimeter = 40 cm 1. Find the area and perimeter 6 cm Area = 2300 in2 Perimeter = 240 in 2. Find the area and perimeter Area = 30 yd2 Perimeter = 29 yd 3. Find the area and perimeter 3 yd Area = 88.26 ft2 Perimeter = 38.84 ft 4. Find the area and perimeter. Use 3.14 for π Area = ft Perimeter = ft Area = 212.75 km2 Perimeter = 69.7 km 5. Find the area and perimeter 6. Find the area and perimeter Area = 97.5 m2 Perimeter= 36 m 3 m 12.5 m 7.5 m 13 m 7. Find the area of the arrow. Area = 90 ft2 Area = 437.5 mm2 8. Find the Area of the Shaded Region. Use 3.14 for π . Area = mm2 Similar presentations
# Square Roots and Cube Roots To find the square root of a number, you want to find some number that when multiplied by itself gives you the original number. In other words, to find the square root of 25, you want to find the number that when multiplied by itself gives you 25. The square root of 25, then, is 5. The symbol for square root is . Following is a list of the first eleven perfect (whole number) square roots. Special note: If no sign (or a positive sign) is placed in front of the square root, then the positive answer is required. Only if a negative sign is in front of the square root is the negative answer required. This notation is used in many texts and is adhered to in this book. Therefore, #### Cube roots To find the cube root of a number, you want to find some number that when multiplied by itself twice gives you the original number. In other words, to find the cube root of 8, you want to find the number that when multiplied by itself twice gives you 8. The cube root of 8, then, is 2, because 2 × 2 × 2 = 8. Notice that the symbol for cube root is the radical sign with a small three (called the index) above and to the left . Other roots are defined similarly and identified by the index given. (In square root, an index of two is understood and usually not written.) Following is a list of the first eleven perfect (whole number) cube roots. #### Approximating square roots To find the square root of a number that is not a perfect square, it will be necessary to find an approximate answer by using the procedure given in Example . ##### Example 1 Approximate . Since 62 = 36 and 72 = 49, then is between and . Therefore, is a value between 6 and 7. Since 42 is about halfway between 36 and 49, you can expect that will be close to halfway between 6 and 7, or about 6.5. To check this estimation, 6.5 × 6.5 = 42.25, or about 42. Square roots of nonperfect squares can be approximated, looked up in tables, or found by using a calculator. You may want to keep these two in mind: #### Simplifying square roots Sometimes you will have to simplify square roots, or write them in simplest form. In fractions, can be reduced to . In square roots, can be simplified to . There are two main methods to simplify a square root. Method 1: Factor the number under the into two factors, one of which is the largest possible perfect square. (Perfect squares are 1, 4, 9, 16, 25, 36, 49, …) Method 2: Completely factor the number under the into prime factors and then simplify by bringing out any factors that came in pairs. ##### Example 2 Simplify . In Example , the largest perfect square is easy to see, and Method 1 probably is a faster method. ##### Example 3 Simplify . In Example , it is not so obvious that the largest perfect square is 144, so Method 2 is probably the faster method. Many square roots cannot be simplified because they are already in simplest form, such as , , and .
# What is the slope intercept form for a line containing the points (10, 15) and (12, 20)? ##### 1 Answer Mar 24, 2018 $y = \frac{2}{5} \cdot x + 11$ #### Explanation: Given: Point 1 : (10,15) Point 2: (12,20) The Slope-Intercept form is y = mx + b; Slope (m) = $\frac{{x}_{2} - {x}_{1}}{{y}_{2} - {y}_{1}}$ m = $\frac{12 - 10}{20 - 15}$ = $\frac{2}{5}$ Therefore, y = $\frac{2}{5}$x + b. Now, plug any one of the above points in this equation to get the y-intercept. Using Point 1: (10,15); 15 = $\frac{2}{\cancel{5}} \cdot \cancel{10}$ + b 15 = 4 + b $\therefore$ b = 11 Therefore, the Slope-Intercept form for the above points is $\textcolor{red}{y = \frac{2}{5} \cdot x + 11}$
Courses Courses for Kids Free study material Offline Centres More Store Eye is most sensitive to $5500\overset{o}{\mathop{A}}\,$ and the diameter of the pupil is about 2 mm. Find the angular limit of the resolution of the eye. Last updated date: 17th Sep 2024 Total views: 431.1k Views today: 6.31k Verified 431.1k+ views Hint: We are given the wavelength of light to which the human eye is most sensitive and the diameter of the pupil. We are asked to find the angular limit of the resolution of the eye. We have an equation to find the angular limit of the resolution of the eye. By substituting the given values in the known equation we will get the required solution. Formula used: ${{D}_{\theta }}=\dfrac{1.22\lambda }{d}$ Complete step-by-step solution In the question, it is said that the human eye is most sensitive to light with wavelength $5500\overset{o}{\mathop{A}}\,$. We are also given the diameter of the pupil as 2 mm. Therefore we gave, Wavelength, $\lambda =5500\overset{o}{\mathop{A}}\,$ Since this is given in angstrom, we can convert it into meters. Thus we get, Wavelength, $\Rightarrow \lambda =5500\times {{10}^{-10}}m$ And the diameter of the pupil, $d=2mm$ By converting this into meters, we get $d=2\times {{10}^{-3}}m$ We are asked to calculate the angular limit of the resolution of the eye. We know that the equation for an angular limit of resolution for the eye is given as, ${{D}_{\theta }}=\dfrac{1.22\lambda }{d}$, were ‘$\lambda$’ is the wavelength and ‘d’ is the diameter By substituting the values for wavelength and diameter in the above equation, we will get ${{D}_{\theta }}=\dfrac{1.22\times 5500\times {{10}^{-10}}}{2\times {{10}^{-3}}}$ By solving this we will get, $\Rightarrow {{D}_{\theta }}=3.355\times {{10}^{-4}}rad$ Therefore the angular limit of resolution of eye is $3.355\times {{10}^{-4}}radians$. Note: Angular limit of resolution or resolving power is simply the ability of an eye or an optical instrument to differentiate between two objects or points. The sensitivity of the human eye to light varies over lights with its wavelength 380 – 800 nm. But in normal daylight conditions, the human eye is most sensitive to light of wavelength 555 nm.
Courses Courses for Kids Free study material Offline Centres More Store # How do you graph $y=-\dfrac{1}{2}\cos x$ ? Last updated date: 13th Jun 2024 Total views: 372.3k Views today: 7.72k Verified 372.3k+ views Hint: To graph the given function $y=-\dfrac{1}{2}\cos x$, first of all, we are going to draw the graph of $\cos x$ and then multiply the negative of 1 with $\cos x$. When we multiply the negative of 1 with $\cos x$ then we have to take the mirror image of the graph of $\cos x$ in the line mirror present on the x axis. Now, we will multiply one half to $-\cos x$ then we have to shrink the y value of the graph of $-\cos x$ by one half. The function given in the above problem which we have to graph is as follows: $y=-\dfrac{1}{2}\cos x$ To draw the above function on the graph paper, first of all, we are going to draw $\cos x$ on the graph. In the below, we have drawn the graph of $\cos x$. Now, we are going to draw the graph of $-\cos x$ by taking the mirror image of the above graph in the line mirror placed at x axis. The dotted line in the above graph is the mirror image of $\cos x$ in the line mirror of x axis so the dotted line pink graph above is of $-\cos x$. Now, we are drawing the graph of $-\dfrac{1}{2}\cos x$ by shrinking all the y values of the above graph by $\dfrac{1}{2}$. In the above graph, as you can see that the y value of the blue curve is lesser than that of the dotted pink curve so the blue curve is the graph of $-\dfrac{1}{2}\cos x$. Now, we are eliminating the pink dotted curve from the above graph then we get, Hence, we have drawn the graph of $y=-\dfrac{1}{2}\cos x$ which is shown below: Note: The mistake that could be possible in the drawing of the above graph is that when converting graph from $\cos x$ to $-\cos x$, you will take the mirror image of the graph in the line mirror placed on the y axis because you think that all the y values are changing so line mirror should present on the y axis. But this is the wrong way of conversion from $\cos x$ to $-\cos x$ so make sure you won’t make this mistake in the exam.
# Frustum Of A Cone: Definition, Volume, Surface Area, and Examples In this blog, we will discuss the frustum of a cone. first of all we will see what is a frustum. Let us take some clay and make a right circular cone. Cut it with a knife parallel to its base. Remove the smaller cone. What are we left with? We are left with a solid called a frustum of the cone. How can we find the surface area and volume of a frustum of a cone? For this, we will go with this example. 1. The radii of the ends of a frustum of a cone 45 cm high are 28 cm and 7 cm. Find its volume, the curved surface area, and the total surface area Solution: The frustum can be viewed as a difference of two right circular cones OAB and OCD. Let the height (in cm) of the cone OAB be h1 and its slant height l1,i.e., OP = h1 and OA = OB = l1. Let h2 be the height of cone OCD and l2 its slant height. We have: r1= 28 cm, r2= 7 cm and the height of frustum (h) = 45 cm. Also, h1 = 45 + h2 call this one. We first need to determine the respective heights h1 and h2 of the cone OAB and OCD. Since the triangles OPB and OQD are similar, we have h1 upon h2 = 28/7 = 4/1. Call this second. From first and second, we get h2 = 15 and h1 = 60. Now, the volume of the frustum = volume of the cone OAB – volume of the cone OCD Let h be the height, l the slant height and r1 and r2 the radii of the ends (r1 > r2) of the frustum of a cone. Then we can directly find the volume, the curved surface area and the total surface area of frustum by using these formulae. Let us apply these formulae in some examples. 1. Hanumappa and his wife Gangamma are busy making jaggery out of sugarcane juice. They have processed the sugarcane juice to make the molasses, which is poured into moulds in the shape of a frustum of a cone having the diameters of its two circular faces as 30 cm and 35 cm and the vertical height of the mould is 14 cm. If each cm3 of molasses has mass of about 1.2 g, find the mass of the molasses that can be poured into each mould. 2. An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6cm. Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also, find the volume of water the bucket can hold.
Best site for GRE, LSAT, SAT, GMAT, TOEFL, CCNA, CCSA and interview sample questions GMAT Sample Questions » GMAT Sample Problem Soving Ability # Quantitative Section : GMAT Sample Problem Solving Ability Solve the following questions and select the correct answer from the given five options. 1. If 3x = 243 then find out the value of 'x' to get this result. 1. 2 2. 3 3. 4 4. 5 5. None of the above Correct answer: d Explanation: According to the data given in the question, we know that the result is a multiple of 3 We have to find out the power such that when we raise 3 to that power the result is equal to 243 Therefore, we will try out each of the option given in the question to find out the correct answer Now, 3 raise to the power 2 = 33 = 9 Now, 3 raise to the power 3 = 333 = 27 Now, 3 raise to the power 4 = 3333 = 81 And, 3 raise to the power 5 = 33333 = 243 Hence, the correct answer is option d 2. In some arithmetic sequence, the sum of the digits of the first and the second term is the third term. The series starts from the whole number 0. Find the seventh term in this series. 1. 5 2. 7 3. 6 4. 8 5. None of the above Correct answer: d Explanation: According to the data given in the question, the numbers are in arithmetic sequence start from the number 0 Also, the sum of the first and the second term make up the third term in the sequence We have to find out the seventh term in this sequence. Hence, as per the description we will go on adding the numbers in the series until we get the seventh term of the series. Starting from 0, we get, the second term as 1, the third term as 1, the fourth term as 2, the fifth term as 3, the sixth term as 5 and lastly the seventh term as 8, in the following manner. 0+1 = 1 1+1 = 2 2+1 = 3 3+2 = 5 5+3 = 8 Hence, the correct answer is option d. 3. Adam got 5 academic text books and 15 note books. If the average cost of Adam's academic text books was \$ 105 and the average cost of his writing note books was \$ 150, then find out the average cost of the total number of books purchased by Adam. 1. 30.125 2. 31.875 3. 28.125 4. 55.550 5. None of the above Correct answer: b Explanation: According to the data given in the question, Adam got 5 academic text books and 15 writing note books respectively. ∴ Number of text books = 5 ……….i And, number of note books = 15 …………….ii Average cost of 5 academic text books = \$ 105 ……..iii ∴ Cost of 1 academic text book = \$ 105 / 5 = \$ 25 …………iv Also given that, average cost of 15 writing note books = \$ 150 ………..v ∴ Cost of 1 writing note book = \$ 150 / 15 = \$ 10 …………..vi Now, Total number of books purchased by Adam = 5 + 15 = 20 …………vii ∴ Average Cost of all the text books and note books = (\$ 105 + \$ 150) / 8 ∴ Average Cost of all the text books and note books = \$ 255 / 8 = \$ 31.875 ….viii Hence, the correct answer is option b. 4. If (a / b) = 0.12 then find the negative value of the reciprocal of the same fraction. 1. (b / a) 2. – (b / a) 3. 0.833 4. -8.33 5. None of the above Correct answer: d Explanation: According to the data given in the question the value of the fraction (a / b) = 0.12 We have to find the negative value of the reciprocal of this fraction which means we have to find the value of –(b / a) First we will find the value of (b / a) ∴ If (a / b) = 0.12, then (b / a) = (1 / 0.12) ∴ (b / a) = 8.33 Now, we will negate the sign of the result of (b/a) to get the value of –(b/a) as -8.33 Hence, the correct answer is option d. 5. If 4 - a = 8(1 – a), then find the value of 'a' according to this equation. 1. 1 / 5 2. 2 / 5 3. 4 / 7 4. 4 / 5 5. None of the above Correct answer: c Explanation: According to the data given in the question, 4 – a = 8(1 – a) …………..i ∴ 4 – a = 8 – 8a ∴ 8a – a = 8 – 4 ∴ 7a = 4 ∴ a = 4 / 7 Hence, the correct answer is option c. \|\| GMAT Sample Problem Solving Ability Question Number : 1-5 \| 6-10 \| 11-15 \| 16-20 \| 21-25 \| 26-30 \| 31-35 \| 36-40 \| 41-45 \| 46-50 \| 51-55 \| 56-60 \| 61-65 \| 66-70 \| 71-75 \| 76-80 \| 81-85 \| 86-90 \| 91-95 \| 96-100 \| 101-105 \| 106-110 \| 111-115 \| 116-120 \| 121-125 Sample Test Questions GRE Sample Questions CAT Sample Questions GMAT Sample Questions TOEFL Sample Questions ACT Sample Questions SAT Sample Questions LSAT Sample Questions PSAT Sample Questions MCAT Sample Questions PMP Sample Questions GED Sample Questions ECDL Sample Questions DMV Sample Questions CCNA Sample Questions MCSE Sample Questions Network+ Sample Questions A+ Sample Questions Technical Sample Questions WASL Sample Questions CISA Sample Questions Copyright © 2004-2013, Best BSQ. All Rights Reserved.
# Polynomial Division with a Box. Polynomial Multiplication: Area Method x + 5 x 2 x 3 5x25x2 -4x 2 -4x-4x -20x x +1 5 Multiply (x + 5)(x 2 – 4x + 1) x. ## Presentation on theme: "Polynomial Division with a Box. Polynomial Multiplication: Area Method x + 5 x 2 x 3 5x25x2 -4x 2 -4x-4x -20x x +1 5 Multiply (x + 5)(x 2 – 4x + 1) x."— Presentation transcript: Polynomial Division with a Box Polynomial Multiplication: Area Method x + 5 x 2 x 3 5x25x2 -4x 2 -4x-4x -20x x +1 5 Multiply (x + 5)(x 2 – 4x + 1) x 3 + x 2 – 19x+ 5 We will reverse this process to divide polynomials. Notice we just proved: Thus the following holds too: Dividend Divisor Quotient Polynomial Division: Area Method Divide x 4 – 10x 2 + 2x + 3 by x – 3 x 4 +0x 3 –10x 2 +2x + 3 x - 3 x 3 x 4 -3x 3 3x33x3 3x23x2 -9x 2 -x2-x2 -x-x 3x3x -x-x 3 x 3 + 3x 2 – x – 1 Divisor Dividend (make sure to include all powers of x) The sum of these boxes must be the dividend Needed Check Quotient Polynomial Division Divide 6x 3 + 7x 2 – 16x + 18 by 2x + 5 6x 3 + 7x 2 – 16x + 18 2x + 5 3x 2 6x36x3 15x 2 -8x 2 -4x -20x 4x4x 2 10 8 Rm Sometimes there is a Remainder. Download ppt "Polynomial Division with a Box. Polynomial Multiplication: Area Method x + 5 x 2 x 3 5x25x2 -4x 2 -4x-4x -20x x +1 5 Multiply (x + 5)(x 2 – 4x + 1) x." Similar presentations
### April 27, 2009 Math Homework Tuesday, April 28, 2009 Nicko's Stuff: Questions 9 to 12 #9 The distance to Grandma's house is 4/5 of the distance to Uncle Glen's. If Uncle Glen's house is 3 1/2 hours away, how long will it take to get to Grandma's house if you travel at the same speed? To multiply fractions, it would be way easier if there was no whole number. Let's turn 3 1/2 in to an improper fraction. To do this, you multiply the denominator by the whole number and add the numerator. 3 x 2 + 1 = 7 Now we have 4/5 x 7/2. Now we multiply the numerator then the denominator. 4 x 7 = 28 5 x 2 = 10 Now we have 28/10 as an answer. We then simplify the answer. 28/10 = 2 and 8/10. We can even simplify this more to 2 and 1/4. Now we know that it will take 2 and 1/4 hours to reach Grandma's house. #10 It takes 3/5 of a tank of gas to get to work and back each day. How much gas is used over 5 work days? In each 5 days, 3/5 is used. So all you have to do is multiply the numerator by 5. 5 x 3 = 15. 15/5 or 3 tanks In five days it takes 3 tanks of gas to get to work. #11 Owen is 2 1/4 times as old as Robin. When Robin celebrates his 8th b-day, how will old Owen be? 8 x 2 1/4 is the basic question. Again we should not multiply mixed fractions, so we convert. 2 x 4 + 1 = 9 The question becomes: 8 x 9/4 When we have a whole number being multiplied to a fraction, all you have to do is multiply the whole by the numerator and use the same denominator. 72/4 is what it will become but now we have to simplify. Both numerator are even so the can be divided by two. 36/2 again they are both even. 18/1 is the most simple form possible. It's basically just 18. Owen will then be 18 when Robin becomes 8. #12 The Karate club is arranging a grading for its members. It takes 3 1/4 hours to test a group of 4 candidates. How long will the club need the gym in order to process 3 groups of 4 candidates each? The question is basically asking: 3 x 3 1/4. 3 x 13/4 39/4 9 and 3/4 I got lazy and didn't show my work so I hope you figured out what I did. If not, check out the other question i did. It will take 9 and 45 minutes to test 3 groups of 4 candidates each. _Nicko---- ~_~" ----Lissa_
## Engage NY Eureka Math 3rd Grade Module 4 Lesson 13 Answer Key ### Eureka Math Grade 3 Module 4 Lesson 13 Problem Set Answer Key Question 1. Each of the following figures is made up of 2 rectangles. Find the total area of each figure. Figure 1: Area of A + Area of B: ____18_____ sq units + _____9____ sq units = _____27____ sq units Figure 2: Area of C + Area of D: ____18_____ sq units + ___15______ sq units = ___33______ sq units Figure 3: Area of E + Area of F: _____9____ sq units + ____21_____ sq units = ____30_____ sq units Figure 4: Area of G + Area of H: _____49____ sq units + ___6______ sq units = ____55_____sq units The total area of figure 1 = 27 sq units. the total area of figure 2 = 33 sq units. the total area of figure 3 = 30 sq units. the total area of figure 4 = 55 sq units. Explanation: In the above-given question, given that, the area of rectangle A + rectangle B = figure 1. area of rectangle A = 18 sq units. area of rectangle B = 9 sq units. so the area of figure 1 = 27 sq units. the area of rectangle C + rectangle D = figure 2. area of rectangle C = 18 sq units. area of rectangle D = 15 sq units. so the area of figure 2 = 33 sq units. the area of rectangle E + rectangle F = figure 3. area of rectangle E = 9 sq units. area of rectangle F = 21 sq units. so the area of figure 3 = 30 sq units. the area of rectangle G + rectangle H = figure 4. area of rectangle G = 49 sq units. area of rectangle H = 6 sq units. so the area of figure 4 = 55 sq units. Question 2. The figure shows a small rectangle cut out of a bigger rectangle. Find the area of the shaded figure. Area of the shaded figure: __90____ – __12____ = ___78___ Area of the shaded figure: ___78___ square centimeters The area of the shaded figure = 78 sq cm. Explanation: In the above-given question, given that, area of longer rectangle = 90 sq cm. area = l x b. area = 9 x 10. area = 90 sq cm. area of smaller rectangle = 12 sq cm. area = l x b. area = 3 x 4. area = 12 sq cm. area of shaded rectangle = area of longer rectangle – area of smaller rectangle. area of shaded rectangle = 90 – 12. area of shaded rectangle = 78 sq cm. Question 3. The figure shows a small rectangle cut out of a big rectangle. a. Label the unknown measurements. The unknown measurements = 4 cm and 5 cm. Explanation: In the above-given question, given that, the longer rectangle = 9 x 7 = 63 cm. 9 – 4 = 3. 7 – 3 = 4. so the unknown measurements = 4 cm and 5 cm. b. Area of the big rectangle: ___9__ cm × __7___ cm = __63___ sq cm The area of big rectangle = 63 sq cm. Explanation: In the above-given question, given that, area of the big rectangle = l x b. where l = length, and b = breadth. area = 9 x 7 = 63 sq cm. so the area of big rectangle = 63 sq cm. c. Area of the small rectangle: ___5__ cm × __4___ cm = ___20__ sq cm The area of small rectangle = 20 sq cm. Explanation: In the above-given question, given that, area of the big rectangle = l x b. where l = length, and b = breadth. area = 5 x 4 = 20 sq cm. so the area of big rectangle = 20 sq cm. d. Find the area of the shaded figure. The area of the shaded figure = 43 sq cm. Explanation: In the above-given question, given that, the area of the longer rectangle = 63 sq cm. area = l x b. area = 9 x 7. area = 63 sq cm. area of smaller rectangle = 20 sq cm. area = l x b. area = 5 x 4. area = 20 sq cm. area of shaded rectangle = area of the longer rectangle – the area of the smaller rectangle. area of shaded rectangle = 63 – 20. area of shaded rectangle = 43 sq cm. ### Eureka Math Grade 3 Module 4 Lesson 13 Exit Ticket Answer Key The following figure is made up of 2 rectangles. Find the total area of the figure. Area of A + Area of B: _____32___ sq units + ____20_____ sq units = ___52______ sq units The area of rectangle A and B = 52 sq units. Explanation: In the above-given question, given that, the area of rectangle A + rectangle B = figure 1. area of rectangle A = 32 sq units. area of rectangle B = 20 sq units. so the area of rectangle A and B = 52 sq units. ### Eureka Math Grade 3 Module 4 Lesson 13 Homework Answer Key Question 1. Each of the following figures is made up of 2 rectangles. Find the total area of each figure. Figure 1: Area of A + Area of B: ____15_____ sq units + ____9_____ sq units = ___24______ sq units Figure 2: Area of C + Area of D: ___3______ sq units + ____8_____ sq units = ____24_____ sq units Figure 3: Area of E + Area of F: ___12______ sq units + ____32_____ sq units = ____44_____ sq units Figure 4: Area of G + Area of H: _____15____ sq units + ____25_____ sq units = ___40______ sq units The total area of figure 1 = 24 sq units. the total area of figure 2 = 24 sq units. the total area of figure 3 = 44 sq units. the total area of figure 4 = 40 sq units. Explanation: In the above-given question, given that, the area of rectangle A + rectangle B = figure 1. area of rectangle A = 15 sq units. area of rectangle B = 9 sq units. so the area of figure 1 = 24 sq units. the area of rectangle C + rectangle D = figure 2. area of rectangle C = 3 sq units. area of rectangle D = 8 sq units. so the area of figure 2 = 24 sq units. the area of rectangle E + rectangle F = figure 3. area of rectangle E = 12 sq units. area of rectangle F = 32 sq units. so the area of figure 3 = 44 sq units. the area of rectangle G + rectangle H = figure 4. area of rectangle G = 15 sq units. area of rectangle H = 25 sq units. so the area of figure 4 = 40 sq units. Question 2. The figure shows a small rectangle cut out of a big rectangle. Find the area of the shaded figure. The area of the shaded figure = 45 sq cm. Explanation: In the above-given question, given that, the area of the longer rectangle = 56 sq cm. area = l x b. area = 7 x 8. area = 56 sq cm. area of smaller rectangle = 9 sq cm. area = l x b. area = 3 x 3. area = 9 sq cm. area of shaded rectangle = area of the longer rectangle – the area of the smaller rectangle. area of shaded rectangle = 56 – 9. area of shaded rectangle = 45 sq cm. Question 3. The figure shows a small rectangle cut out of a big rectangle. a. Label the unknown measurements. The unknown measurements = 3 cm and 4 cm. Explanation: In the above-given question, given that, the longer rectangle = 9 x 8 = 72 cm. 9 – 6 = 3. 8 – 4 = 4. so the unknown measurements = 3 cm and 4 cm. b. Area of the big rectangle: ____9__ cm × ___8___ cm = ___72___ sq cm The area of big rectangle = 72 sq cm. Explanation: In the above-given question, given that, area of the big rectangle = l x b. where l = length, and b = breadth. area = 9 x 8 = 72 sq cm. so the area of big rectangle = 72 sq cm. c. Area of the small rectangle: ___3___ cm × ___4___ cm = __12____ sq cm The area of small rectangle = 12 sq cm. Explanation: In the above-given question, given that, area of the big rectangle = l x b. where l = length, and b = breadth. area = 3 x 4 = 12 sq cm. so the area of small rectangle = 12 sq cm. d. Find the area of the shaded figure. The area of the shaded figure = 43 sq cm. Explanation: In the above-given question, given that, area of longer rectangle = 72 sq cm. area = l x b. area = 9 x 8. area = 72 sq cm. area of smaller rectangle = 12 sq cm. area = l x b. area = 3 x 4. area = 12 sq cm. area of shaded rectangle = area of longer rectangle – area of smaller rectangle. area of shaded rectangle = 72 – 12. area of shaded rectangle = 60 sq cm. ### Eureka Math Grade 3 Module 4 Lesson 13 Template Answer Key Area of A + Area of B: _____32___ sq units + ____20_____ sq units = ___52______ sq units
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ## MAP Recommended Practice ### Course: MAP Recommended Practice>Unit 34 Lesson 11: Subtracting decimals intro # Subtracting decimals We'll start with simple problems like 0.9 - 0.8 and build to more complex problems like 12.6 - 8.89. The problems go from easy to more difficult, and along the way there are examples and explanations in case you get stuck. If you get confused, just think of it as a chance to learn! Let's start by subtracting tenths. ## Problem set 1: Problem 1a $0.8-0.6=$ Beautiful! Let's move on to problems with whole numbers and tenths. ## Problem set 2: Problem 2a $1.5-1.2=$ Excellent! Let's move on to work with some larger numbers. ## Problem set 3: Problem 3a $12.3-2=$ Rockin'! Let's move on to some more difficult problems with whole numbers and tenths. ## Problem set 4: Problem 4a $9-4.9=$ Cool! Now let's subtract hundredths. ## Problem set 5: Problem 5a $0.64-0.41=$ Great! Let's move on to problems with whole numbers, tenths, and hundredths. ## Problem set 6: Problem 6a $0.53-0.4=$ Rockin'! Now we're ready to use bigger numbers. ## Problem set 7: Problem 7a $44.83-32=$ Sweet! Let's finish with some more challenging problems. ## Problem set 8: Problem 8a $9-6.35=$ ## Want to join the conversation? • this is confuseing • What part is confusing? • do we use math every day?😐 • yes, example: if you need 5 apples for a pie but you want to double the amount to make two pies... how many do you have to get? (that was just an easy example) • This is kinda easy for me but for some people it could be confusing and hard. • I’m glad this is easy for you. The likely cause of confusion for some people, when subtracting decimals, is failing to line up decimal points first and/or failing to put in zeros for missing digits. I've caught people saying "This is easy" Some people are bad at math and might have a unsuppoortive brain, please don't say that because someone might need help so instead of saying "this is easy" help them! So instead of an unsuppoortive brain turn it to a positive brain, I hope you understand. • i agree because we are old enough to know that... • I don't get that 1.0 - 0.59 is 0.41. How is it o.41? • Because the 1.0 can also be expressed as 1.00 If you view it that way, as 1.00 -0.59 It will equal 0.41 The steps are, First you subtract the 0 in the hundredth's place by the 9 in the other's hundredth's place. But 0 is less than 9, so you need to borrow from the tenth's place. Unfortunately, how do you borrow? The value there is 0 as well, and there isn't really anything for you to borrow. Fortunately, the one's place has a 1. This means the tenths place can borrow from that for a "10", and afterward the hundredth's place can borrow from the 10 in the tenth's place. 10 - 9 =1 9 - 5= 4 0 - 0 = 0 So you end up with 0.41 • what would happen if math wouldn't be in are life • are there negative decimals? • Yes, there are negative decimals. Every real positive number has an opposite that is negative. These are valid negative number: -8.32; -5 3/4; -0.05 • who else like turtles
# Quarter Wit, Quarter Wisdom: Exponents – The Bane of Your Preparation II Let me recap the rules we discussed in the last post: Rule 1: a^m × a^n = a^(m + n) Rule 2: a^m / a^n = a^(m – n) Rule 3: (a^m)^n = a^mn Rule 4: For any number a, a^0 = 1 In Rules 1 and 2 above, the bases were the same. What happens if the exponents are the same but the bases are different? 2^4 × 3^4 = (2 × 2 × 2 × 2) × (3 × 3 × 3 × 3) = (2 × 3) × (2 × 3) × (2 × 3) × (2 × 3) = 6^4 The bases got multiplied! Rule 5: a^m × b^m = (a × b)^m which also implies that (a × b)^m = a^m × b^m Example: 6^5 = 2^5 * 3^5 The treatment of division is very similar. 6^4/2^4 = (6 × 6 × 6 × 6)/(2 × 2 × 2 × 2) = 3 × 3 × 3 ×3 = 3^4 Essentially, the bases get divided and the exponent remains the same. Rule 6: a^m / b^m = (a / b)^m which also implies that (a / b)^m = a^m / b^m Example: (3/2)^5 = 3^5 / 2^5 Let’s take a quick look at negative exponents now. What is 3^-4? It is extremely easy to handle negative exponents. Just flip the number and the exponent becomes positive. So 3^-4 is 1/3^4. This also implies the following: • 1/3^-4 = 3^4 • 3^4 = 1/3^-4 • 1/3^4 = 3^-4 When you want to change the sign of the exponent, flip the number. When you want to flip the number, change the sign of the exponent! Example: What is the value of 3^2 / 3^(-3) This can be solved in two ways: 1. Flip the number to get rid of the negative exponent. 3^2 / 3^(-3) = 3^2 * 3^3 The base is the same and the two are multiplied so we add the exponents. 3^2 * 3^3 = 3^(2+3) = 3^5 OR 2. The bases are the same and the terms are divided so subtract the exponent of the divisor from the exponent of the dividend. 3^2 / 3^(-3) = 3^(2 – (-3)) = 3^5 Now that we have covered all the major rules of exponents, let’s work on the question we saw in the previous post. Question: If (2^a × 4 × 3^-4 × 3^b )/(3^4 × 2^2) = 8^-4 × 729, what is the value of a and b? (A) -10 and -14 (B) 10 and 12 (C) -10 and 12 (D) -12 and 14 (E) 12 and -14 Solution: First, we need to bring everything to prime number form. (2^a × 4 × 3^-4 × 3^b )/(3^4 × 2^2) = 8^-4 × 729 (2^a × 2^2 × 3^-4 × 3^b )/(3^4 × 2^2) = (2^3)^-4 × 3^6 If you do not remember that 729 is the sixth power of 3, you should know that it will be some power of 3 because the left hand side of the equation has only two prime numbers – 2 and 3. So the right hand side of the equation cannot have any prime other than 2 and 3 (if there is some other prime number, its exponent will be 0 to make the term 1). Since 729 is certainly not a power of 2 (since it is odd), it must be some power of 3. We just need to multiply 3 with itself a few times to figure out the power. Let’s work on the left hand side of the equation first. Get rid of negative exponent. (2^a × 2^2 × 3^b )/(3^4 × 3^4 × 2^2) Some terms have same bases and are multiplied. Add their exponents. (2^(a + 2) × 3^b )/(3^8 × 2^2) Some terms have same bases and are divided. Subtract their exponents. 2^a × 3^(b – 8) Let’s equate this to the right hand side now. 2^a × 3^(b – 8) = (2^3)^-4 × 3^6 Using Rule 3 on the right hand side, 2^a × 3^(b – 8) = 2^(-12) × 3^6 The exponent of 2 on left hand side is ‘a’ and on right hand side is -12. Therefore, a = -12. The exponent of 3 on left hand side is (b – 8) and on right hand side is 6. Therefore, b – 8 = 6 or b = 14 Answer (D) Note: We used the long method to solve this question since we wanted to discuss the application of various rules. You can use faster approaches once you are comfortable with these basic rules. Let me leave you with a couple of questions. I will discuss these in the next post. Question 1: Given (1/4)^18 × (1/5)^n = 1/(2 × 10^35), find the value of n. Question 2: Is 5^m < 1000? Statement 1: 5^(m + 1) > 3000 Statement 2: 5^(m – 1) = 5^m – 500 Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep in Detroit, and regularly participates in content development projects such as this blog!
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ### Course: High school geometry>Unit 1 Lesson 3: Translations # Properties of translations Experimentally verify the effect of geometric translations on segment length, angle measure, and parallel lines. When you translate something in geometry, you're simply moving it around. You don't distort it in any way. If you translate a segment, it remains a segment, and its length doesn't change. Similarly, if you translate an angle, the measure of the angle doesn't change. These properties may seem obvious, but they're important to keep in mind later on when we do proofs. To make sure we understand these properties, let's walk through a few examples. ## Property 1: Line segments are taken to line segments of the same length. Each square in the grid is $1$ unit long. Translate line segment $\stackrel{―}{ST}$ by $⟨2,-7⟩$. What is the length of the pre-image—the segment before the translation? units What is the length of the image—the segment after the translation? units As you can see for yourself, the pre-image and the image are both line segments with the same length. This is true for any line segment that goes under any translation. ## Property 2: Angles are taken to angles of the same measure. Translate $\mathrm{\angle }MNP$ by $⟨-5,-6⟩$. Has the measure of the angle changed after the translation? As you can see for yourself, the pre-image angle and the image angle have the same measure. This is true for any angle that goes under any translation. ## Property 3: Lines are taken to lines, and parallel lines are taken to parallel lines. Translate the parallel lines $\stackrel{↔}{FG}$ and $\stackrel{↔}{HJ}$ by $⟨-4,3⟩$. Are the two image lines parallel? As you can see for yourself, each line is taken to another line, and the image lines remain parallel to each other. This is true for any line—or lines—that go under any translation. ## Conclusion We found that translations have the following three properties: • line segments are taken to line segments of the same length; • angles are taken to angles of the same measure; and • lines are taken to lines and parallel lines are taken to parallel lines. This makes sense because a translation is simply like taking something and moving it up and down or left and right. You don't change the nature of it, you just change its location. It's like taking the elevator or going on a moving walkway: you start in one place and end in another, but you are the same as you were before, right? ## Want to join the conversation? • Ahhh yes! I need to know angles to do things in my everyday life! Sees pizza "WAIT! BEFORE YOU EAT THAT WE HAVE TO FIND OUT THE ANGLE OF THE PIZZA!!" -_- • I know right • Will this help me in the future • As a gardener -- no; as a computer scientist -- yes. • will this help me understand lam 1 • it helps best if you put thought in it and work toward it • i highly doubt this is highschool geometry... I am in 7th grade • The High School Geometry course gets harder as you get through it. But, from my personal experiences, as a 7th grader, I was able to get through the course quite easily. (1 vote) • Why is my dog sped? • you chose the dog. Make an inference • how do you do this i kind of get it and kind of dont so teach me the way • Hi Miguel, The key to understanding translations is that we are SLIDING a point or vertices of a figure LEFT or RIGHT along the x-axis and UP or DOWN along the y-axis. On our x-axis, if the translation number is positive, move that point right by the given number of units, and if the translation number is negative, move that point to its left. Use the same logic for y-axis; if the translation number is positive, move it up, and if the translation number is negative, move the point down. Let us have a look at an example. We are given a point A, and its position on the coordinate is (2, 5). The translation number is (-1, 3). So, we will move the point LEFT by 1 unit on the x-axis, as translation number is (-1). We will then move the point 3 units UP on the y-axis, as the translation number is (+3). The image will be A prime at (1, 8). We did this with a point, but the same logic is applicable when you have a line or any kind of figure. The other two points to remember in a translation are- 1. Lengths don't change 2. Angles don't change I hope this helped. Aiena. • hello how are y'all doing today • I'm doing pretty good myself (1 vote) • these comments are so old • The site defaults to showing the top rated questions & answers. Since the rating take time to accumulate, they are apt to be older. You should read them, they likely have good things to learn. If you want to see the more recent questions, just change the sort sequence to "recent". • this is differnt then what there teaching us in my class • The key to understanding translations is that we are SLIDING a point or vertices of a figure LEFT or RIGHT along the x-axis and UP or DOWN along the y-axis. On our x-axis, if the translation number is positive, move that point right by the given number of units, and if the translation number is negative, move that point to its left. Use the same logic for y-axis; if the translation number is positive, move it up, and if the translation number is negative, move the point down. Let us have a look at an example. We are given a point A, and its position on the coordinate is (2, 5). The translation number is (-1, 3). So, we will move the point LEFT by 1 unit on the x-axis, as translation number is (-1). We will then move the point 3 units UP on the y-axis, as the translation number is (+3). The image will be A prime at (1, 8). We did this with a point, but the same logic is applicable when you have a line or any kind of figure. The other two points to remember in a translation are- 1. Lengths don't change 2. Angles don't change I hope this helped.
# How do you solve 3x+4y=0 and x-4y= -8? Mar 5, 2018 $x = - 2$ and $y = \frac{3}{2}$ #### Explanation: Given [1]$\textcolor{w h i t e}{\text{XXX}} 3 x + 4 y = 0$ [2]$\textcolor{w h i t e}{\text{XXX}} x - 4 y = - 8$ Adding equations [1] and [2] together [3]$\textcolor{w h i t e}{\text{XXX}} 4 x = - 8$ Dividing both sides of [3] by $4$ [4]$\textcolor{w h i t e}{\text{XXX}} x = - 2$ Substituting $\left(- 2\right)$ for $x$ back in [2] [5]$\textcolor{w h i t e}{\text{XXX}} \left(- 2\right) - 4 y = - 8$ $\rightarrow$[6]$\textcolor{w h i t e}{\text{XXX}} - 4 y = - 6$ $r a r$r[7]$\textcolor{w h i t e}{\text{XXX}} y = \frac{3}{2}$ ...or if you prefer a mixed number $y = 1 \frac{1}{2}$
Question Video: Determining If a Graph Represents a Linear or a Nonlinear Function \| Nagwa Question Video: Determining If a Graph Represents a Linear or a Nonlinear Function \| Nagwa # Question Video: Determining If a Graph Represents a Linear or a Nonlinear Function Mathematics • Third Year of Preparatory School ## Join Nagwa Classes Is this the graph of a linear or a nonlinear function? 02:04 ### Video Transcript Is this the graph of a linear or a nonlinear function? In order to identify whether this is the graph of a linear or a nonlinear function, let’s begin by reminding ourselves what we mean when we describe an equation as a function. A function is a rule that relates an element in one set to exactly one element of a second set. We talk about functions as being one to one, in other words, one element in the first set maps to exactly one in the second, or many to one. That is, several elements in the first set could map to exactly one in the second. A one-to-many relationship is not a function. For this reason, a vertical line cannot represent a function. This would be an example of a one-to-many relationship. We substitute one value in, and we get a whole bunch out. So with that in mind, let’s remind ourselves what we mean by linear and nonlinear functions. A linear function is a function whose graph is a straight line. Therefore, a nonlinear function has a graph which is not a straight line. An example of this is something like 𝑦 equals 𝑥 cubed or 𝑦 equals 𝑒 to the power of 𝑥. And if we inspect our graph, we can see it is a nonvertical straight line. It must then be the graph of a linear function. In fact, we’re able to deduce the equation of this linear function. We might know this by heart, but let’s identify a few key points that lie on this line. We have the point with coordinates four, three; another point with coordinates negative two, three; and another with coordinates negative four, three. In fact, every single point that lies on this horizontal line has a 𝑦-coordinate of three. And so we can say that its equation is 𝑦 equals three. Or if we wish to choose function notation, we can write 𝑓 of 𝑥 equals three. This is the graph of a linear function. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
# What is 12/7 - 6? getcalc.com's two fractions subtraction calculator is an online basic math function tool to find what's the equivalent fraction for difference between fraction and a whole number, 12/7 minus 6. In mathematics, every integer is a rational number, hence a whole number 6 can be written as 6/1. 12/7 - 6 = -30/7 in fraction form 12/7 - 6 = -4.2857 in decimal form This calculator, formula, step by step calculation and associated information to find the difference between 12/7 and 6 may help students, teachers, parents or professionals to learn, teach, practice or verify such two fractions subtraction calculations efficiently. ## How to Find Equivalent Fraction for 12/7-6? The below workout with step by step calculation shows how to find the difference between fraction and a whole number by subtracting 6 from 12/7. use any one of the following methods to find the equivalent fraction for 12/7 minus 6. 1. LCM method 2. Cross multiplication method. Problem and Workout - LCM Method What is 12/7 minus 6 as a fraction by subtraction using LCM method? step 1 Address formula, input parameters and values. Input parameters and values: 12/7 & 6/1 12/7-6 = ? step 2 For unequal denominators, find the LCM (least common multiple) for both denominators The LCM for 7 and 1 is 7. step 3 To have the common denominator, multiply the LCM with both numerator and denominator =(12 x 7)/(7 x 7)-(6 x 7)/(1 x 7) =12/7-42/7 =(12 - 42)/7 step 4 Find difference between numerators and rewrite it in a single form. =-30/7 12/7-6=-30/7 By using LCM method, -30/7 is the equivalent fraction by subtracting 6 from 12/7. Problem and Workout - Cross Multiplication Method step 1 Address formula and input values. Input values: 12/7 & 6/1 12/7 - 6 = ? step 2 Cross multiply both numerator and denominator, multiply both denominators and rewrite as below = (12 x 1) - (6 x 7)/(7 x 1) step 3 Simplify and rewrite the fraction = (12 - 42)/7 12/7-6 = -30/7 By using cross multiplication method, -30/7 is the difference between two fractions 12/7 and 6.
### Archive Posts Tagged ‘sin’ ## The Chain Rule The chain rule allows you to differentiate composite functions (functions of other functions) ie) f(g(x)) such as sin(3x2) or (5x3+2x+3)2. The rule is as follows $\frac{d}{dx}(f(g(x)) = \frac{dg}{dx}\frac{df}{dg}(g(x)) = g'(x)f'(g(x))$ or to understand it more simply you differentiate the inner function and multiply it by the derivative of the outer function (leaving what’s inside alone). ## Differentiating brackets raised to a power The chain rule can be a great short cut to differentiating brackets raised to a power as it doesn’t require you to multiply them all out, it also enables you to differentiate brackets raised to an unknown power. Consider $\frac{d}{dx}((ax + b)^n)$ This is the composite of the functions ax+b and tn. So we differentiate them both to get a and ntn-1 and then apply the formula to get $\frac{d}{dx}((ax + b)^n) = an(ax+b)^{n-1}$ Notice how we multiplied the derivative of the inner function, a, by the derivative of the outer function ntn-1 but substituted ax+b back in for t. To generalise we can replace the ax+b with f(x) and by applying the above get $\frac{d}{dx}((f(x))^n) = f'(x)n(f(x))^{n-1}$ ## Differentiating Trigonometric functions We can also use the chain rule when differentiating sin(f(x)) and cos(f(x)) since we know how to differentiate sin(x) and cos(x). Using the chain rule we get $\frac{d}{dx}(sin(f(x)) = f'(x)cos(f(x))$ and $\frac{d}{dx}(cos(f(x)) = -f'(x)sin(f(x))$ ## Compound Angles – sin(A+B) = cosAsinB+sinAcosB Compound angles are angles made by adding two other angles together. When using trigonometry unfortunately you cant just “times out” the trig function but have to use an identity. This post will consider how we get the identity for sin(A+B): sin(A+B) = sinAcosB+sinBcosA Compound angle of A+B showing how they relate From the definition of sin=opp/hyp we find sin(A+B) = RT/OR But sinceRT comprises of RS+ST By David Woodford Categories: maths ## Sec, Cosec, Cot Sec, cosec and cot are all functions in trigonometry. They are simply equal to one over on of the other functions, ie cos, sin and tan. so Sec = 1/cos Cosec = 1/sin cot = 1/tan You can remember which is paired with which using the third letter rule. This is that the third letter is the first letter of the corresponding function ie) sec goes with cos cosec goes with sin cot goes with tan Categories: maths ## Differentiate Inverse Cos – Proof Now available from trevorpythag.blogspot.com How to differentiate cos-1x y=cos-1x Bring the cos across cosy = x Differentiate both sides, remember when differentiating y time by dy/dx -sin(y) dy/dx = 1 dy/dx = -1/siny However we want to get the differential in terms of x, to do this we can use the identity sin2t+cos2t = 1 so sint = √(1 – cos2t) putting this into our expression for dy/dx we get dy/dx = -1/√(1-cos2y) but cosy = x so dy/dx =- 1/√(1-x2) Categories: maths ## Trigonometry Identities There a number of “identities” in trigonometry that can be found from the basic ideas of sin, cos and tan as explained in my earlier post. These identities can help in solving equations involving trig functions, especially when there are 2 or more different functions as the often allow you to write the equation in terms of one function, eg sin, that you can then solve. One of the identities is: sin2 + cos2 = 1. To prove this consider a right angled triangle with side a,b and c as shown below From this we can use Pythagoras theorem to say: a2+b2=c2 now we know sin t = b/c so b = csin t cos t = a/c so a = ccos t substituting these values in the above equation we get (csint)2 +(ccost)2 = c2 canceling the c2 we get sint2 + cost2 = 1 There are trig functions that are equal to 1 over sin, cos and tan called cosec = 1/sin, sec = 1/cos and cot = 1/tan. These can be remembered using the third letter rule as the third letter of each of these corresponds to the the function it is one over. Using these a cos2 + sin2 = 1 we can calculate other identities tan2t + 1 = sec2t We can obtain this by dividing through by cos2 as we know sin/cos = tan, cos/cos = 1 and 1/cos = sec. Other similar identities can be obtained for cosec and cot. Categories: maths ## Tan = sin/cos this site is now at www.breakingwave.co.nr This is often useful when solving trig equations so i thought i’d include it basically: sin = opp/hyp and so so if we cancel the hyp’s we get tan = sin/cos Categories: maths Tags: , , , , ## Sine and Cos Graphs Differentiating sin and cos This is the basics of the sine cos and tan graphs and how sine and cos relate to give you tan. It also shows how to differentiate sin and cos. The output or range of both sine and cos is from -1 to 1 when given any angle. They can be shown on a graph where y = sin(x) and y = cos(x). In these graphs all the angles go along the x axis and you can see a wave type shape is formed Sine Graph Cosine Graph As you can see both the sin and cos graphs move periodically between -1 and 1 as the angles change, this pattern continues indefinitely because once you pass 360 degrees or 2 pi radians you will return back to the beginning. If you try to perform sin-1 of a value out side the range -1 to 1 you will get an error. Differentiate Sin and Cos also notice that the gradient of the sin graph is the value of the cos graph for the same angle and that the gradient of the cos graph is the -value of the sin graph for that angle. This means that we can differentiate the sin and cos graphs: if f(x) = sin(x) then f ‘ (x)=cos(x) and if f(x) = cos(x) then f ‘ (x) = -sin(x) however if we use ax instead of x we must differentiate it by bringing the a out, when its just x this doesn’t matter as the differential of x is 1. ie) let y = sin(f(x)) now let u = f(x) du/dx = f ‘ (x) also y=sin(u) as u = f(x) dy/du = cos(u) from the chain rule dy/dx = du/dx * dy/du therefore if y = sin(f(x)) dy/dx = f ‘ (x)cos(f(x)) and similarly for cos if y = cos(f(x)) dy/dx = -f ‘ (x)sin(f(x)) Categories: maths
Question Video: Finding and Evaluating the First Derivative of Polynomial Functions Using the Chain Rule and the Product Rule \| Nagwa Question Video: Finding and Evaluating the First Derivative of Polynomial Functions Using the Chain Rule and the Product Rule \| Nagwa # Question Video: Finding and Evaluating the First Derivative of Polynomial Functions Using the Chain Rule and the Product Rule Mathematics • Second Year of Secondary School ## Join Nagwa Classes Find the first derivative of the function π¦ = π₯β΄ (4π₯ + 9)βΉ at π₯ = β2. 04:24 ### Video Transcript Find the first derivative of the function π¦ equals π₯ to the power of four multiplied by four π₯ plus nine to the power of nine at π₯ equals negative two. Well, the first thing we need to do to actually solve this problem is actually differentiate our function. Well, we can see that our function is in the form π¦ equals π’π£. So therefore, weβre actually gonna use the product rule to help us differentiate it. And the product rule states that ππ¦ ππ₯ is equal to π’ ππ£ ππ₯ plus π£ ππ’ ππ₯. So first of all, we pick out whatβs gonna be our π’ and our π£. So our π’ is going to be π₯ to the power of four and our π£ is equal to four π₯ plus nine to the power of nine. So now, what we want to do is actually find out what ππ’ ππ₯ and ππ£ ππ₯ actually are. So Iβm gonna start with ππ’ ππ₯ and ππ’ ππ₯ is gonna be equal to four π₯ cubed. And thatβs because weβve obviously differentiated our π₯ to the power of four. And just to remind us how we actually did that, what we did is we actually multiplied the coefficient of our term by the exponent β so four multiplied by one. And then what we did is we actually subtracted one from our exponent β so four minus one. So that gave us four π₯ to the power of three. Okay, great, so now letβs move on and differentiate π£ to find ππ£ ππ₯. Well, now to find ππ£ ππ₯, what we need to do is we actually need to use a rule to help us actually differentiate four π₯ plus nine to the power of nine. And the rule weβre actually gonna use is the chain rule. And the chain rule actually tells us that ππ¦ ππ₯ is equal to ππ¦ ππ’ multiplied by ππ’ ππ₯. Okay, so now we have this rule, letβs look at our term and see if we can apply it. Well, first of all, weβre gonna have a look at π’ and say that itβs equal to four π₯ plus nine. So therefore, ππ’ ππ₯ is just gonna be four because if you differentiate four π₯, you get four and if you differentiate nine, you just get zero. Okay, so now we can move on to π¦. Well, π¦ would be equal to π’ to the power of nine. So therefore, ππ¦ ππ’ is equal to nine π’ to the power of eight. So then if we apply the chain rule, we can say that ππ¦ ππ₯ is equal to four multiplied by nine π’ to the power of eight, which is equal to 36π’ to the power of eight. So then, we just substitute back in that π’ is equal to four π₯ plus nine and we get that ππ¦ ππ₯ is equal to 36 multiplied by four π₯ plus nine to the power of eight. So therefore, we can say the ππ£ ππ₯ is equal to 36 multiplied by four π₯ plus nine to the power of eight. So therefore, now that we found our ππ£ ππ₯ and our ππ’ ππ₯, what we can actually do is apply our product rule to find ππ¦ ππ₯. So the first derivative of our function π₯ to the power of four multiplied by four π₯ plus nine to the power of nine. So therefore, we get ππ¦ ππ₯ is equal to π₯ to the power of four because thatβs our π’ and then multiplied by our ππ£ ππ₯ which is 36 multiplied by four π₯ plus nine to the power of eight plus our π£ which is four π₯ plus nine to the power of nine multiplied by our ππ’ ππ₯ which is four π₯ cubed. So great, weβre actually at the stage where we found the first derivative of our function. But what do we do now? So now, what we need to do is actually look at what the first derivative is going to be when π₯ is equal to negative two. And in order to do this, we need to substitute in π₯ is equal to negative two into our first derivative. So weβre gonna get that the first derivative with negative two substituted in for π₯ is equal to negative two to the power of four multiplied by 36 times four times negative two plus nine to the power of eight plus four multiplied by negative two plus nine to the power of nine multiplied by four multiplied by negative two cubed which is gonna be equal to 16 multiplied by 36 plus four multiplied by negative eight which is equal to 544. So therefore, we can say that the first derivative of the function π¦ equals π₯ to the power of four multiplied by four π₯ plus nine to the power of nine at π₯ equals negative two is equal to 544. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
# How do you find the square root of 203? Aug 31, 2016 Use a Newton Raphson method to find: $\sqrt{203} \approx \frac{6497}{456} \approx 14.247807$ #### Explanation: $203 = 7 \cdot 29$ has no square factors. So its square root cannot be simplified. It is an irrational number between $14$ and $15$, since: ${14}^{2} = 196 < 203 < 225 = {15}^{2}$ As such, it cannot be represented in the form $\frac{p}{q}$ for integers $p , q$. We can find rational approximations using a Newton Raphson method: To approximate the square root of $n$, start with an approximation ${a}_{0}$ then iterate using the formula: ${a}_{i + 1} = \frac{{a}_{i}^{2} + n}{2 {a}_{i}}$ I prefer to re-formulate this slightly using separate integers ${p}_{i} , {q}_{i}$ where ${p}_{i} / {q}_{i} = {a}_{i}$ and formulae: ${p}_{i + 1} = {p}_{i}^{2} + n {q}_{i}^{2}$ ${q}_{i + 1} = 2 {p}_{i} {q}_{i}$ If the resulting pair of integers have a common factor, then divide by that before the next iteration. So for our example, let $n = 203$, ${p}_{0} = 29$ and ${q}_{0} = 2$ (i.e. ${a}_{0} = 14.5$) ... $\left\{\begin{matrix}{p}_{0} = 29 \\ {q}_{0} = 2\end{matrix}\right.$ { (p_1 = p_0^2+ n q_0^2 = 29^2 + 2032^2 = 841+812 = 1653), (q_1 = 2 p_0 q_0 = 2292 = 116) :} Note that both ${p}_{1}$ and ${q}_{1}$ are divisible by $29$, so divide both by that: $\left\{\begin{matrix}{p}_{1 a} = \frac{1653}{29} = 57 \\ {q}_{1 a} = \frac{116}{29} = 4\end{matrix}\right.$ Next iteration: { (p_2 = p_(1a)^2 + n q_(1a)^2 = 57^2+2034^2 = 3249+3248 = 6497), (q_2 = 2 p_(1a) q_(1a) = 2574 = 456) :} If we stop here, we get: $\sqrt{203} \approx \frac{6497}{456} \approx 14.247807$ Each iteration roughly doubles the number of significant figures in the approximation.
# Multiplication of Rational Number: ## 1) Multiplication of rational number with whole number: Let us multiply the rational number (- 3)/5 "by 2, i.e., we find" (- 3)/5 xx 2. On the number line, it will mean two jumps of 3/5 to the left. We reach at (- 6)/5. Let us find it as we did in fractions. (- 3)/5 xx 2 = (- 3 xx 2)/5 = (- 6)/5. We arrive at the same rational number. So, we find that while multiplying a rational number by a positive integer, we multiply the numerator by that integer, keeping the denominator unchanged. Let us now multiply a rational number by a negative integer, (- 2)/9 xx (- 5) = (-2 xx (- 5))/9 = 10/9 ## 2) Multiplication of two rational number: We multiply two rational numbers in the following way: Step 1: Multiply the numerators of the two rational numbers. Step 2: Multiply the denominators of the two rational numbers. Step 3: Write the product as the product of "numerators"/"product of denominators". (-3)/5 xx 2/7 = (-3 xx 2)/(5 xx 7) = (-6)/35. (-5)/8 xx (-9)/7 = (- 5 xx (-9))/(8 xx 7) = 45/56. #### Example Multiply the following rational numbers. 9/13 xx 4/7 9/13 xx 4/7 = (9 xx 4)/(13 xx 7) = 36/91. #### Example Multiply the following rational numbers. 3/5 xx (- 4)/5 3/5 xx (- 4)/5 = (3 xx (- 4))/(5 xx 5) = (- 12)/25. #### Example Multiply the following rational numbers. 9/13 xx 26/3 9/13 xx 26/3 = (3 xx 2)/1 = 6/1. If you would like to contribute notes or other learning material, please submit them using the button below. ### Shaalaa.com Multiplication of Two Rational Numbers [00:09:30] S 0%
• 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 # Solution of equations by Numerical Methods. Extracts from this document... Introduction ## Pure 2 Coursework ### Solution of equations by Numerical Methods #### Method 1: The Change of Sign method The simplest method for solving an f(x) function is to use a change of sign method; these include the methods of bisection, decimal search and linear interpolations. Unfortunately, as well as being the simplest methods, they are also relatively cumbersome. The bisection was employed below for the function of f(x)=x5+3x2+x+2. This was also displayed graphically in the graph below. The actual calculations for this method are summarised below in this table. I placed these with relevant diagrams The error bounds for this result are conveniently provided by the use of this method, and are -1.485<x<-1.475. The maximum error is therefore ± 0.005. However, this method is clearly not particularly easy to work with. In addition, as with many methods there are some functions for x, which do not work. An example is the function f(x)=x3-2x2-x. This is shown below. Middle Using an estimate for (a.) as 0, the Newton-Raphson iterative formula was then used. This next diagram gives some indication as to how the Newton- Raphson method actually works. What it basically does is to ‘slide’ the tangent across and is indeed sometimes known as ‘tangent sliding’. For this method, the root can be calculated much more quickly then using the Change of Sign method. In accordance with the question the root was calculated to five significant figures, this means that the real value of x exists between 0.16655 and 0.16665, which give an error value of ± 0.00005. This method, despite its speed in calculation a root, does not work for every function; this is illustrated below. ##### Method 3: Fixed Point Estimation- Rearranging the function The third and final method for numerically solving a function of x is by rearranging an existing function of x and then using fixed-point iteration. This is illustrated for the function f(x)=x3+x-3 below. Diagram of f(x)=x3+x-3 As is displayed graphically, the function with regards to (g) Conclusion Rearranging the function is another useful technique, especially when a ‘cobweb’ effect is produced, as this gives natural error bounds for the root. However, it relies on an important rearrangement of the function, in conjunction with the plotting of at least two graphs. Once this has been done, the iterative formula is used, though within my calculations did not prove as fast as the Newton-Raphson method. As I have already mentioned, the rearrangement method, relies more heavily on suitable graphical generating software or hardware; this is therefore a clear disadvantage that it faces. However, the change of sign method requires an interval for the root and the Newton-Raphson method requires an approximation to the result. Though, as this investigation has proved clear, regardless of the method, it has been necessary to have some idea of the function before attempting calculations. This is particularly important in the case of turning points and roots close to each other. This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics 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 AS and A Level Core & Pure Mathematics essays 1. ## C3 Coursework - different methods of solving equations. 5 star(s) -1.96180 -0.004775254 -1.96179 -0.004280068 -1.96178 -0.003784896 -1.96177 -0.003289738 -1.96176 -0.002794593 -1.96175 -0.002299463 -1.96174 -0.001804347 -1.96173 -0.001309244 -1.96172 -0.000814155 -1.96171 -0.000319081 -1.96170 0.00017598 X f(x) -2.0 -2 -1.99 -1.45736 -1.98 -0.92928 -1.97 -0.41553 -1.96 0.084135 -1.95 0.569938 -1.94 1.042111 -1.93 1.500882 -1.92 1.946474 -1.91 2.37911 -1.90 2.79901 The root of the function f(x) 5 star(s) 1158.56201 1345.6201 4 1024 4.01 1036.86416 1286.416 4 1024 4.001 1025.28064 1280.6402 2 32 2.1 40.84101 88.4101 2 32 2.01 32.8080401 80.80401 2 32 2.001 32.08008004 80.08004 1 1 1.1 1.61051 6.1051 1 1 1.01 1.05101005 5.101005 1 1 1.001 1.00501001 5.01001 x x4 5x4 1 1 5 2 16 1. ## MEI numerical Methods must equal negative and f(b) must be positive. In the question we are provided with an interval which is 0 < x < ?/2. Lets assume K = 1 F(0) = -1 F(?/2) = a very large positive number. Our current interval is 0 < x < ?/2, in order to get a better approximation we must find the midpoint of the interval (h), and do the f(h). 2. ## This coursework is about finding the roots of equations by numerical methods. >1 then we would expect the iteration to diverge away from the root as a staircase. Comparison I am going to compare advantages and disadvantages of all three methods. y=-3x�-x�+x+2 There are three intervals containing roots: (-0, 1) To find the root in the interval (0, 1) 1. ## Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and ... =B3^3-5B3^2+4B3+2 =IF(G2>0,D2,F2) =D3^3-5D3^2+4D3+2 3 =IF(G3>0,F3,B3) =B4^3-5B4^2+4B4+2 =IF(G3>0,D3,F3) =D4^3-5D4^2+4D4+2 4 =IF(G4>0,F4,B4) =B5^3-5B5^2+4B5+2 =IF(G4>0,D4,F4) =D5^3-5D5^2+4D5+2 5 =IF(G5>0,F5,B5) =B6^3-5B6^2+4B6+2 =IF(G5>0,D5,F5) =D6^3-5D6^2+4D6+2 c f(c) Error =(B2+D2)/2 =F2^3-5F2^2+4F2+2 =(B2+D2)/2 =(B3+D3)/2 =F3^3-5F3^2+4F3+2 =H2/2 =(B4+D4)/2 =F4^3-5F4^2+4F4+2 =H3/2 =(B5+D5)/2 =F5^3-5F5^2+4F5+2 =H4/2 =(B6+D6)/2 =F6^3-5F6^2+4F6+2 =H5/2 This is a sample I select form my actual table to show how a spread sheet can help. 2. ## Numerical solution of equations, Interval bisection---change of sign methods, Fixed point iteration ---the Newton-Raphson ... 1 1.5 -1 1.25 -0.15625 0.25 3 1 1 1.25 -0.15625 1.125 0.39453125 0.125 4 1.125 0.39453125 1.25 -0.15625 1.188 0.11083984 0.0625 5 1.1875 0.110839844 1.25 -0.15625 1.219 -0.0249634 0.03125 6 1.1875 0.110839844 1.21875 -0.0249633 1.203 0.04239655 0.015625 7 1.203125 0.042396545 1.21875 -0.0249633 1.211 0.0085783 0.0078125 8 1.210938 0.0085783 1.21875 1. ## I am going to solve equations by using three different numerical methods in this ... -1.80065 0.001538(positive) Therefore, my answer is correct and it lies between this interval. Below shows the formulae for using in the Excel: A B f(a)<0 f(b)>0 (a+b)/2 y=2x�-3x�-8x+7 -2 -1 =2B4^3-3B4^2-8B4+7 =2C4^3-3C4^2-8C4+7 =(B4+C4)/2 =2F4^3-3F4^2-8F4+7 =IF(G4>0,B4,F4) =IF(G4>0,F4,C4) =2B4^3-3B4^2-8B4+8 =2C4^3-3C4^2-8C4+8 =(B4+C4)/3 =2F4^3-3F4^2-8F4+8 =IF(G4>0,B4,F5) 2. ## C3 COURSEWORK - comparing methods of solving functions 0.8xn3+2 xn2-1 1 0 -1 2 -1 0.2 3 0.2 -0.9136 4 -0.9136 0.05929 5 0.05929 -0.9928 6 -0.9928 0.188464 7 0.188464 -0.92361 From the table above, we can see that the iteration is diverging away from the root in interval [0, -1] This is illustrated graphically below: X3 y= 0.8x³+2x²-x–1 X1 y= x X2 We can conclude that g'(x) • Over 160,000 pieces of student written work • Annotated by experienced teachers • Ideas and feedback to
# CLASS-8DIVISION BY MONOMIAL BY MONOMIAL Division of a Monomial by a Monomial – To divide one monomial by another, divide the numerical coefficient of the dividend by the numerical coefficient of the divisor and the powers of variables in the dividend by the corresponding powers in the divisor. Then multiply all the quotients. Quotient of two monomials = (quotient of numerical factors) X (quotient of literal factors) There are some example are given below for your better understanding - Example.1) Divide 24x²y⁵ by 8xy Ans.) As per the given condition, we have to 24x²y⁵ ÷ 8xy 24x²y⁵ 24 So, 24x²y⁵ ÷ 8xy = ---------- = -------- x²⁻¹ y⁵⁻¹ = 3 xy⁴ (Ans.) 8xy 8 Example.2) Divide 36a⁴b⁸c²d⁶ by 9a⁵bᶟc²d² Ans.) As per the given condition, we have to 36a⁴b⁸c²d⁶ ÷ 9a⁵bᶟc²d² 36a⁴b⁸c²d⁶ ÷ 9a⁵bᶟc²d² 36a⁴b⁸c⁵d⁶ 36 a⁴ b⁸ c² d⁶ = ------------- = ------ X ------ X ------ X ------- X ------- 9a⁵bᶟc²d² 9 a⁵ bᶟ c² = 4 a⁴⁻⁵ b⁸⁻c²⁻² d⁶⁻² = 4 a⁻¹ b⁵ c⁰ d⁴ = 4 a⁻¹ b⁵ . 1 . d⁴ = 4 a⁻¹ b⁵ d⁴ 4 b⁵ d⁴ = ------------ (Ans. a Example.3) Divide 30x⁴y² + 15 x²y²- 20 x⁵y⁴ + 10 x⁵y⁶ - 25 x⁸y⁵ by 5 x⁴y⁶ Ans.) As per the given condition we have to 30x⁴y²+ 15 x²y²- 20 x⁵y⁴ + 10 x⁵y⁶ - 25 x⁸y⁵ ÷ 5 x⁴y⁶ 30x⁴y²+ 15 x²y² - 20 x⁵y⁴ + 10 x⁵y⁶ - 25 x⁸y⁵ ÷ 5 x⁴y⁶ 30x⁴y² + 15 x²y² - 20 x⁵y⁴ + 10 x⁵y⁶ - 25 x⁸y⁵ = ------------------------------------------------ 5 x⁴y⁶ 30x⁴y² 15 x²y² 20 x⁵y⁴ 10 x⁵y⁶ 25 x⁸y⁵ = --------- + --------- - ---------- + ---------- - ----------- 5 x⁴y⁶ 5 x⁴y⁶ 5 x⁴y⁶ 5 x⁴y⁶ 5 x⁴y⁶ = 6 x⁴⁻⁴. y²⁻⁶ + 3 x²⁻⁴. y²⁻⁶ - 4 x⁵⁻⁴ y⁴⁻⁶ + 2 x⁵⁻⁴ y⁶⁻⁶ - 5 x⁸⁻⁴ y⁵⁻⁶ = 6 x⁰ y⁻⁴ + 3 x⁻² y⁻⁴ - 4 x¹ y⁻² + 2x¹ y⁰ - 5x⁴ y⁻¹ = 6.1. y⁻⁴ + 3 x⁻² y⁻⁴ - 4 x y⁻² + 2x.1 - 5x⁴ y⁻¹ = 6 y⁻⁴ + 3 x⁻² y⁻⁴ - 4 x y⁻² + 2x - 5x⁴ y⁻¹ 6 3 4x 5x⁴ = -------- + -------- - --------- + 2x - --------- (Ans.) y⁴ x² y⁴ y² y
What is the polar form of ( -23,-3 )? Dec 5, 2015 The polar form of $\left(- 23 , - 3\right)$ is $\left(\sqrt{538} , {\tan}^{- 1} \left(\frac{3}{23}\right) + \pi\right) \approx \left(23.195 , 3.271\right)$ Explanation: This question has a list of equations used when converting between rectangular and polar coordinates. In this case, we will be using $\left\{\begin{matrix}{r}^{2} = {x}^{2} + {y}^{2} \\ \tan \left(\theta\right) = \frac{y}{x}\end{matrix}\right.$ $\implies \left\{\begin{matrix}r = \sqrt{{x}^{2} + {y}^{2}} \\ \theta = {\tan}^{- 1} \left(\frac{y}{x}\right)\end{matrix}\right.$ $r = \sqrt{{\left(- 23\right)}^{2} + {\left(- 3\right)}^{2}}$ theta = tan^(-1)((-3)/(-23)))^(color(red)("")) " "^color(red)("")(While calculating $\theta$, the $- 3$ and $- 23$ cancel negatives, causing the resulting angle puts us in quadrant $I$ when we want quadrant $I I I$. To fix this, all we need to do is add or subtract $\pi$ from the angle to put us in the correct quadrant.) $\implies \left\{\begin{matrix}r = \sqrt{538} \\ \theta = {\tan}^{- 1} \left(\frac{3}{23}\right) + \pi\end{matrix}\right.$ Thus we get the polar form of $\left(- 23 , - 3\right)$ to be $\left(\sqrt{538} , {\tan}^{- 1} \left(\frac{3}{23}\right) + \pi\right) \approx \left(23.195 , 3.271\right)$
# How do you solve abs(8x+8) = abs(9x-4)? Jul 11, 2015 Split into cases for intervals $\left(- \infty , - 1\right)$, $\left[- 1 , \frac{4}{9}\right)$ and $\left(\frac{4}{9} , \infty\right)$, solving the linear equations that result to find: $x = - \frac{4}{17}$ or $x = 12$ #### Explanation: Split into cases: Case 1: $x \in \left(- \infty , - 1\right)$ $8 x + 8 < 0$, so $\left\mid 8 x + 8 \right\mid = - \left(8 x + 8\right) = - 8 x - 8$ $9 x - 4 < 0$, so $\left\mid 9 x - 4 \right\mid = - \left(9 x - 4\right) = - 9 x + 4$ Equation becomes: $- 8 x - 8 = - 9 x + 4$ Add $9 x + 8$ to both sides to get: $x = 12$ This lies outside $\left(- \infty , - 1\right)$ so is not valid for this case. Case 2: $x \in \left[- 1 , \frac{4}{9}\right)$ $8 x + 8 \ge 0$, so $\left\mid 8 x + 8 \right\mid = 8 x + 8$ $9 x - 4 < 0$, so $\left\mid 9 x - 4 \right\mid = - \left(9 x - 4\right) = - 9 x + 4$ Equation becomes: $8 x + 8 = - 9 x + 4$ Add $9 x - 8$ to both sides to get: $17 x = - 4$ Divide both sides by $17$ to get: $x = - \frac{4}{17}$ This does lie in $\left[- 1 , \frac{4}{9}\right)$ so is a valid solution. Case 3: $x \in \left[\frac{4}{9} , \infty\right)$ $8 x + 8 \ge 0$, so $\left\mid 8 x + 8 \right\mid = 8 x + 8$ $9 x - 4 \ge 0$, so $\left\mid 9 x - 4 \right\mid = 9 x - 4$ Equation becomes: $8 x + 8 = 9 x - 4$ Add $- 8 x + 4$ to both sides to get: $x = 12$ This does lie in $\left[\frac{4}{9} , \infty\right)$ so is a valid solution. Jul 11, 2015 $\left\mid u \right\mid = \left\mid v \right\mid$ if and only if either $u = v$ or $u = - v$. #### Explanation: Two numbers have the same absolute value when the numbers are equal or they are opposites (negatives) of each other. $\left\mid 8 x + 8 \right\mid = \left\mid 9 x - 4 \right\mid$ if and only if: $8 x + 8 = 9 x - 4$ $\textcolor{w h i t e}{\text{xx}}$ or $\textcolor{w h i t e}{\text{xx}}$ $8 x + 8 = - \left(9 x - 4\right)$ Solving $8 x + 8 = 9 x - 4$ , we get $8 + 4 = 9 x - 9 x$ so $12 = x$, that is: $x = 12$ Solving $8 x + 8 = - \left(9 x - 4\right)$. we get $8 x + 8 = - \left(9 x - 4\right)$, so $8 x + 8 = - 9 x + 4$ and then $8 x + 9 x = 4 - 8$, so $17 x = - 4$, and finally $x = - \frac{4}{17}$ The solutions are: $12$ and $- \frac{4}{17}$
# DAV Class 5 Maths Chapter 7 Worksheet 1 Solutions The DAV Maths Book Class 5 Solutions and DAV Class 5 Maths Chapter 7 Worksheet 1 Solutions of Multiplication and Division of Decimal Numbers offer comprehensive answers to textbook questions. ## DAV Class 5 Maths Ch 7 Worksheet 1 Solutions Question 1. Find the product. (a) 0.3 × 3 Solution: 0.3 × 3 3 × 3 = 9 (Multiply ignoring decimal) 0.3 × 3 = 0.9 (Put decimal point 1 place to the left) (b) 0.3 × 4 Solution: 0.3 × 4 3 × 4 = 12(Multiply ignoring decimal point) 0.3 × 4 = 1.2 (Put decimal point 1 place to the left) (c) 0.412 × 2 Solution: 0.412 × 2 412 × 2 = 824 (Multiply ignoring decimal point) 0.412 × 2 = 0.824 (Put decimal point 3 places to the left) (d) 0.005 × 15 Solution: 0.005 × 15 5 × 15 = 75 (Multiply ignoring decimal point) 0.005 × 75 = .075 (Put decimal point 3 places to the left) (e) 2.4 × 23 Solution: 2.4 × 23 24 × 23 = 552 (Multiply ignoring decimal point) 2.4 × 23 = 55.2 (Put decimal point 1 place to the left) (f) 16.3 × 17 Solution: 16.3 × 17 So 163 × 17 = 2771 (Multiply ignoring decimal point) Now 16.3 × 17 = 277.1 (Put decimal point 1 place to the left) (g) 71.8 × 248 Solution: 71.8 × 248 718 × 248 = 178064 (Multiply ignoring decimal point) 71.8 × 248 = 17806.4 (Put decimal point 1 place to the left) (h) 7.37 × 56 Solution: 7.37 × 56 737 × 56 = 41272 (Multiply ignoring decimal point) 7.37 × 56 = 42.72 (Put decimal point 2 places to the left) (i) 1.001 × 96 Solution: 1.001 × 96 1001 × 19 = 96096 (Multiply ignoring decimal point) 1.001 × 96 = 96.096 (Put decimal point 3 places to the left) Question 2. If 3,485 × 16 = 55,760, find (a) 348.5 × 16 Solution: 348.5 × 16 = 5576.0 or 5576 (b) 34.85 × 16 Solution: 34.85 × 16 = 557.60 (c) 3.485 × 16 Solution: 3.485 × 16 = 55.760 (d) 0.3485 × 16 Solution: 0.3485 × 16 = 5.5760 DAV Class 5 Maths Chapter 7 Worksheet 1 Notes Whole numbers: Natural numbers along with zero are called whole numbers. Example: 0, 1, 2, 3, 4, 5, 6, ………. Decimal numbers: The fractions in which denominators are 10, 100, 1000 etc. are called decimal numbers. e.g. $$\frac{3}{100}$$ = 0.03 In order to multiply two decimal numbers, • Multiply the numbers ignoring the decimal points. • Make the decimal places in the product equal to the sum of decimal places in the multiplicand and multiplier. When we multiply a decimal number by 10,100 or 1000, we just shift the decimal point in the product to the right by as many places as there are zeros in the multiplier. e.g., 12.342 × 100 = 1234.2 When we divide a decimal number by 10, 100, or 1000 we just shift the decimal point in the quotient to the left by as many places as there are zeros in the divisor. e.g., 34.46 ÷ 1000 = 0.03446 The product remains the same if the order of two decimal number are changed. e.g., (a) 4.2 × 2.4 = 10.08 (b) 2.4 × 4.2 = 10.08 The product and division of a decimal number by one is the decimal number itself. e.g., (a) 4.1 × 1 = 4.1 (b) 3.2 ÷ 1 = 3.2 Product and division of decimal numbers by zero is always zero. e.g., (a) 3.3 × 0 = 0 (b) 0 ÷ 3.4 = 0 Multiplication of Decimal Numbers Example 1. Multiply 0.4 × 3 Solution: 4 × 3 = 12 (Multiply the no. ignoring decimal) so 0.4 × 3 = 1.2 (The number of decimal places in 0.4 is one. So we keep only one decimal place in the product) Example 2. Multiply 3.16 by 0.7 Solution: 316 × 0.7 = 2212 (Multiply ignoring the decimal point) 3.16 × 0.7 = 2.212 (Same number of decimal places in the product as in multiplicand and multiplier)
What is a radius? The geometry of a circle The radius of a circle is the distance from the center point to the edge of the circle. It’s the same distance anywhere on the circle, because the circle has radial symmetry. So it doesn’t matter where you measure the radius on the circle, and if you know one radius measurement for a circle, then you know all of them. Try it for yourself and see! Each radius is a line segment reaching from the center of the circle to the perimeter. ## How do I get the radius from the circumference? But suppose you don’t know the radius of this circle? Suppose all you know is that the circumference is 30 centimeters? Measuring the circumference of a tree. One example would be if you were trying to measure a big tree. It would be pretty easy to take a rope and wrap it around the tree, and then measure the length of the rope to find out how big around the tree was (the circumference), but you’d have to cut the tree down to measure the radius. ## How do you find the radius of a circle? We don’t want to cut this big tree down, so we’ll need to use math to figure out the radius instead. Luckily we know that the radius of any circle is always the same as half of the circumference divided by π (pi). So the radius of this tree is 30/π = 9.55 centimeters, divided by two is 4.77 centimeters. ## Prove the relationship between radius and circumference How can we be sure that this is true for every circle? We can do a simple geometrical proof that will convince anybody that the circumference of a circle is always, always, always going to be half of the circumference divided by pi. ## When did people know how to calculate the radius from the circumference of a circle? Mathematicians and engineers have known how to use approximate values of pi to calculate the radius of a circle certainly since about 2000 BC. In West Asia and in Egypt, mathematicians knew how to calculate the value of pi by around 1800 BC. They got it as accurate as 3.16. They used pi to figure out the area of circles and the volume of cylinders. ## What else is a radius good for? Once you have the radius of a circle, you can use it to figure out the circumference of the circle – the distance around the outside – and the area of the circle. If you know the radius of a cylinder, you can use that and the height of the cylinder to calculate the volume of the cylinder. You can also calculate the volume of a cone from the radius. And if you know the radius of a sphere – a ball – you can use it to calculate the volume of the sphere and the surface area of the sphere. So what is a radius? How do you find the radius of a circle? How do we know this is always true? Did you find out what you needed to know? Let us know in the comments!
# Angles of Inclination and of Intersection ## Angle of Inclination - inc Perpendicular and Parallel Lines DEFINITION The Angle of Inclination is the smallest angle (θ) between a line and where it intersects with the positive x-axis. If we recall, for a slope: m = Δ y Δ x We also know that the tangent of an angle, tan(θ) = the opposite line to the angle divided by the adjacent line to the angle. As Δ y = the opposite line of the angle and Δ x = the adjacent line of the angle We can say that tan(θ) = Δ y Δ x Therefore, m = tan(θ) ## Parallel Lines DEFINITION Two lines will be parallel if each of their angles of inclinations are equal. Remember, the angle of inclination is the angle the line makes with the x-axis in the positive direction Say we have two lines A and B, with angles of inclination θ1 and θ2 If θ1 = θ2 then the lines are parallel. Similarly, if the slope of A and the slope of B are equal, then they will also be parallel. We can say that, Two non-vertical lines are parallel if the slope of each line is equal to the other, m1 = m2. ## Example: Finding equation for a line when knowing a point on the line and the equation of another parallel line Question: A line of the equation 4x + 2y = 9 runs parallel to a line that has the point (2, 5). Determine the equation to the line with the point (2, 5). We know that the line of 4x + 2y = 9 runs parallel to another line that has the point (2, 5), Ask ourselves what can we work with from this statement? What do we know about parallel lines? If two lines are parallel then m1 = m2, meaning the slope of each line is the same. We want to know the equation to the line that passes through the point (2, 5). We can only work out an equation if we either know; two points on that line or one point on that line and the slope. We know the slope of the line we want the equation to is equal to the slope of the line 4x + 2y = 9, we can work out the slop of this line! Rearrange 4x + 2y = 9, to give us something like y =... Subtract 4x from both sides, 2y = 9 - 4x Divide both sides by 2, y = 9 2 - 4x 2 y = -2x + 9 2 We know that the slope of a line of the form y = mx + b, is equal to m, In this case, m = - 2, Now we have our slope of the line 4x + 2y = 9, we know that this slope is also equal to the slope of the line running parallel that goes through the point (2, 5) So the slope of the parallel line running through the point (2, 5) is -2, now we have these two values, we can calculate the equation for this line, Recall, our equation for building an equation of a line is, y - y1 = m(x - x1) Where m is the slope, and (x1, y1) is a point on that line, m = -2 x1 = 2 y1 = 5 So, y - 5 = -2(x - 2) y - 5 = -2x + 4 y = -2x + 4 + 5 y = -2x + 9 There we have our equation that goes through the point (2, 5) and runs parallel to the line of equation 4x + 2y = 9 ## Perpendicular Lines DEFINITION For two lines that are non-vertical, say A and B, with slopes m1 and m2 respectively, these lines will be perpendicular if, m1 -1 m2 This is called the negative inverse ## Example: Finding an equation of a line knowing a point on that line and the equation of a line that is perpendicular Question: Find the equation to the line that passes through the point (-3, 7) and runs perpendicular to the line of the equation 4y - 3x = 8 Similar to question 8, we use what we know about the relationship between the two lines, as well as what we know from the other line, to allow us to calculate the equation of the line we don't know. We have two lines that run perpendicular; One line of equation 4y - 3x = 8 And one line that passes through (-3, 7) If we find the slope of 4y - 3x = 8, we can find the slope of the line that passes through (-3, 7) as we know that for perpendicular lines the slope of one is equal to the negative inverse of the other, i.e. m1 -1 m2 Once we have the slope of the line with the point (-3, 7) we can work out the equation for that line. 4y = 8 + 3x Divide both sides by 4, y = 3x 4 + 8 4 y = (¾)x + 2 So, the slope, m, of 4y - 2x = 8 is (¾) If we take the negative inverse of this, it will equal the slope of the perpendicular line passing through the point (-3, 7) This slope becomes -(4/3) Now we know m = -(4/3) and a point along the line is (-3, 7) we can work out the equation using the formula, y - y1 = m(x - x1) with , m = -(4/3), x1 = -3, and y1 = 7 y - 7 = -4 3 (x - (-3)) y - 7 = -4 3 (x) - 4 3 (3) y - 7 = -4x 3 - 4 y = -4x 3 - 4 + 7 y = -4x 3 + 3 There we have our equation for the line that runs perpendicular to 4y - 3x = 8 and passes through the point (-3, 7) ## Angles Between Intersecting Lines DEFINITION If two non-vertical lines intersect, and that intersection is not perpendicular, then the angle, θ, between those two lines is, where m1 is the slope of the first line (L1) and m2 is the slope of the second line (L2), θ = arctan m2 - m1 1 + m2m1 ## Proof of equation for angle between lines Here is the proof: With θ1 as the angle L1 makes with the x-axis and θ2 is the angle L2 makes with the x-axis, Trigonometry laws state that in the visual above, θ2 = θ + θ1, therefore, θ = θ2 - θ1 We also know that tan(θ) = the slope of the line that makes the angle θ with the x-axis. So, applying tan to θ = θ2 - θ1 tan(θ) = tan(θ2 - θ1) (properties of tan give us) tan(θ2 - θ1) = tan(θ2) - tan(θ1) 1 + tan(θ2)tan(θ1) And remember, tan of an angle gives us the slope of the line making that angle, m2 - m1 1 + m2m1 For the arctan or inverse tan of an angle, the result will be between -pi ÷ 2 and pi ÷ 2 or -90° and 90°. So, with, tan(θ) = m2 - m1 1 + m2m1 θ = arctan m2 - m1 1 + m2m1 Arctan will always give us the smaller of the angles, we have to also account for the larger angle as well, which we get by taking pi - θ or 180° - θ So, we get our definition, The smaller angle formed by two non vertical, non perpendicular lines intersecting is, θ = arctan m2 - m1 1 + m2m1 ## Example: Finding point of intersection and angle between two lines Question: For the lines L1(y = x - 2) and L2(y = 2x - 5) a) Find the point of intersection b) Find the angle of the point of intersection in degrees to the nearest tenth of a decimal place a) This should be a review of work we have already done, but lets go over it in detail, The point of intersection is the point where it satisfies both of the equations i.e. it makes both equations true. We find this point by making the equations equal to the same variable (often y) and then equating the other part of the equation and solving it. Here we already have both of our equations equal to y, so we can make the “x-part” of each equation, y = x - 2 y = 2x - 5 Making them equal, x - 2 = 2x - 5 Subtract x from both sides, -2 = 2x - x - 5 -2 = x - 5 -2 + 5 = x x = 3 Now we have our x-value of the intersection point, find the y-value, we do this by subbing the x = 3 into either of the equations (as this is the point that is true for both equations) Sub x = 3 into y = x - 2, y = 3 - 2 y = 1 So, our point of intersection is (3,1) b) now we want to find the angle of intersection between two lines, We know the equation for an angle between two intersecting lines is, θ = arctan m2 - m1 1 + m2m1 Where m1 is the slope of the first line and m2 is the slope of the second line, We can calculate the slope of each line from each equation given, We know that an equation of the format y = mx + b, that m = slope So for L1: y = x -2, m1 = 1 And for L2: y = 2x - 5, m2 = 5 Now, sub these m1 and m2 values into, θ = arctan m2 - m1 1 + m2m1 θ = arctan 5 - 1 1 + (5)(1) θ = arctan 4 6 = arctan 2 3
# Graphically solving a system of linear equations solver Graphically solving a system of linear equations solver can support pupils to understand the material and improve their grades. Our website will give you answers to homework. ## The Best Graphically solving a system of linear equations solver In this blog post, we will be discussing about Graphically solving a system of linear equations solver. A right triangle is a triangle with two right angles. By definition, it has one leg that's longer than the other. A right triangle has three sides. A right triangle has three sides: the hypotenuse (the longest side) and two shorter sides. These are called legs. The legs are always equal in length. They have equal lengths to each other and to the hypotenuse. The hypotenuse is the longest side of a right triangle and is therefore the opposite side from the one with the highest angle. It is also called the altimeter or longer leg. Right triangles always have an altimeter (the longest side). It is opposite to the hypotenuse and is also called the longer leg or hypotenuse. The other two sides of a right triangle are called legs or short sides. These are always equal in length to each other and to the longer leg of the triangle, which is called the hypotenuse. The sum of any two angles in a right triangle must be 180 degrees, because this is one full turn in any direction around a vertical line from vertex to vertex of an angle-triangle intersection. An angle-triangle intersection occurs when two lines that intersect at a common point meet back together at another point on their way down from both vertexes to that point where they intersected at first! The system of equations is the mathematical representation of a set of related equations. It is an ordered list of equations with and without solutions. The solution of a system of equations is the set of values that satisfies all the given equations. To solve system of equations, first we need to identify all the variables involved in the given system. Then we need to add all unknowns and solve for them individually. Once all unknowns are known, we can add all knowns and solve for them individually. This way, we get a single solution from a set of individual solutions. We use algebra to find a solution or to solve a system of linear equations or inequalities. Algebra is used to simplify, manipulate and evaluate expressions and questions involving variables. Algebra is also used for solving more complicated problems such as quadratic equations, polynomial equations, rational expressions, exponential expressions etc. Algebra can be used to solve systems with several variables or when there are different types of questions (such as multiple choice, fill-in-the-blank). There are various methods one can use to solve system of linear equations like substitution method, elimination method and combination method etc. In this article, we will discuss several approaches on solving systems of linear equation i.e substitution method etc. The best way to learn math is by doing it. One of the easiest ways is by using a free step by step math solver like the one below. You can use the solver in just a few simple steps: 1) Click on the link above, or simply copy and paste it into your browser. 2) Enter some basic math facts (like 4 plus 6) or numbers (like 8). 3) Click "Answer" and see what you get. There are many different types of calculators out there, but this is one of the simplest and most reliable. It also has a built-in lesson plan that teaches you how to solve problems in math. So give it a try! Many times, however, inequalities are more complicated than linear equations and are better suited to coordinate geometry. The method of displacement gives you a way to accurately determine the location of a point on a line by measuring where it would move if you moved it up or down one unit in either direction. The method of variation proves that one line is longer or shorter than another by finding how much they change in length when rotated through an angle. Algebraic solutions can also be used to approximate values with interpolation, extrapolation, interpolation, or interpolation when solving for unknown values that are not perfect squares. For example, in order to estimate the value of x in an equation like x=1/2+5/4, we can approximate x with any value greater than 0 and less than 1 (e.g., x=1.5) and then use linear interpolation to estimate what value it should be closest to (e.g., x=1). Interpolation works well when dealing with large changes but may not be accurate enough for smaller changes ( Algebra is used to solve equations. Algebra equations can be written in the following ways: The three main types of algebra equations are linear, quadratic, and exponential. Linear equations involve one or two numbers. For example, 1x + 3 = 10. Quadratic equations have two unknown numbers and involve a squared number. For example, 4x2 + 2x + 5 = 25. Exponential equations have one number and involve an exponent (e) sign with a base number. For example for 4e-2x = 6. Algebra can be used to solve equations like the following: To solve the equation 5x - 8 = 7, we must first find the value of "a". To do this we use the formula: a = x - (5/8) br> br>Entering this in the formula above, we get: a = 7 - (1/8) br> br>Now that we know how to find "a", we can use it to find "b". To do this we use: b = a * x br> br>This gives us b = 1 * 7 br> br>The final result is that b = 9 br> br>To solve the equation y - 2 = 3, we must first find A super cool app. It helped me to solve my math easily the best part is that you also get detailed solutions. You all also have a special math keyboard But I am not finding the textbook by hands voted for. Pls look into that. And also include a conversion calculator like for area, temp, mass. Arabella Barnes This is a useful app and all, it helps me with math homework, but sometimes when I'm using the camera it goes blurry then not blurry then it misreads my problem that I have and it messes it up. Other than that, it’s perfect! Waneeta Harris Homework Math-solver Value Solve any math problems Instant math solver Math homework help How to solve matrices Mathematics tutors near me
Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis : # JEE Advanced Maths Circles Previous Year Questions with Solutions Students can find the JEE Advanced Maths circles previous year questions and solutions on this page. In the conic section, the topic circles have great importance. Students can expect questions from this topic. Revising previous year questions will help students to know about the exam pattern and difficulty levels of the exam. They can also download the solutions in PDF format for offline use. Question 1: A circle S passes through the point (0, 1) and is orthogonal to the circles (x-1)2 + y2 = 16 and x2 + y2 = 1. Then (a) radius of S is 8 (b) radius of S is 7 (c) centre of S is (-7, 1) (d) centre of S is (-8, 1) Solution: Let the equation of the circles be x2+y2+2gx+2fy+c = 0 ..(i) It passes through (0, 1) => 1 + 2f + c = 0 â€¦(ii) Since circle (i) is orthogonal to (x-1)2 + y2 = 16 => x2 + y2 – 2x – 15 = 0 and x2 + y2 – 1 = 0 2gÃ—(-1) + 2fÃ—0 = c-15 2g + c – 15 = 0 â€¦(iii) 2gÃ—0 + 2fÃ—0 = c-1 => c = 1 â€¦(iv) Solving (ii), (iii) and (iv) => c = 1, g = 7 and f = -1 The required circle is x2 + y2 + 14x – 2y + 1 = 0, with centre (-7, 1) and radius = 7. Hence option b and c are correct. Question 2: Let O be the centre of the circle x2 + y2 = r2, where r > âˆš5/2 . Suppose PQ is a chord of this circle and the equation of the line passing through P and Q is 2x + 4y = 5. If the centre of the circumcircle of the triangle OPQ lies on the line x + 2y = 4, then the value of r is (a) 1 (b) 4 (c) 6 (d) 2 Solution: S1: x2 + y2 = r2 where r > âˆš5/2 C1 = (0,0) let S2: x2 + y2 + ax + by = 0 C2 = (-a/2, -b/2) PQ: S1 -S2 = 0 PQ: ax + by + r2 = 0 â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (1) Given PQ: 2x + 4y – 5 = 0 â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (2) comparing equation (1) and (2) (a/2) = (b/4) = (r2/-5)â€¦â€¦â€¦â€¦..â€¦..(3) Also, centre of S2 lies on x + 2y = 4 (-a/2) – b = 4 â€¦(4) From equation (3) and (4) r = 2 Hence option d is the answer. Question 3: The circle passing through the point (-1, 0) and touching the y-axis at (0, 2) also passes through the point (a) (-3/2, 0) (b) (-5/2, 2) (c) (-3/2, 5/2) (d) (-4, 0) Solution: Equation of circle passing through a point (x1, y1) and touching the straight line L, is given by (x – x1)2 + (y – y1)2 + Î»L = 0 The circle is passing through the point (0, 2) => (x – 0)2 + (y – 2)2 + Î»L = 0 => x2 + (y – 2)2 + Î»x = 0 â€¦(i) Circle passes through (-1, 0). => 1 + 4 – Î» = 0 => Î» = 5 Put Î» = 5 in (i) x2 + (y – 2)2 + 5x = 0 => x2 + y2 – 4y + 4 + 5x = 0 => x2 + y2 + 5x – 4y + 4 = 0 Put y = 0 => x = -4, x = -1 Check the options. Hence option d is the answer. Question 4: The number of common tangents to the circles x2 + y2 = 4 and x2 + y2 – 6x – 8y = 24 is (a) 1 (b) 2 (c) 0 (d) 3 Solution: Given circle x2 + y2 = 4 with centre C1(0, 0) and R1 = 2 Also x2 + y2 – 6x – 8y – 24 = 0 with centre C2(3, 4) and R2 = 7 Distance between centres = C1C2 = 5 = R2 – R1 So the circles touch internally and they can have just one common tangent at the point of contact. Hence option a is the answer. Question 5: The points of intersection of the line 4x – 3y – 10 = 0 and the circle x2 + y2 – 2x + 4y – 20 = 0 are (a) (4, 2) (b) (-2, -6) (c) (2, 2) (d) (-2, -4) Solution: Given equation of the line 4x – 3y – 10 = 0 => x = (3y+10)/4 â€¦(i) Equation of the circle x2 + y2 – 2x + 4y – 20 = 0 ..(ii) Put (i) in (ii) => [(3y+10)2/42 ] + y2 – 2(3y+10)/4 + 4y – 20 = 0 => y2 + 4y – 12 = 0 => y = 2, -6 Put y in (i) => x = 4, -2 So the points are (4, 2) and (-2, -6). Hence option a and b are correct. Question 6: If the tangent at (1,7) to the curve x2 = y-6 touches the circle x2 + y2 + 16x + 12y + c = 0, then the value of c is (a) 185 (b) 85 (c) 195 (d) 95 Solution: Given curve is x2 = y – 6 Differentiate w.r.t.x 2x = dy/dx (dy/dx)(1,7) = 2 Equation of tangent at (1, 7) to x2 = y – 6 is y-y1 = m(x-x1) Here m = 2 (y-7) = 2 (x – 1) 2x – y + 5 = 0 The perpendicular from the centre (-8, -6) to 2x – y + 5 = 0 is equal to radius of circle. So \|(-16+6+5)/âˆš5 \| = âˆš(64 + 36 – c) => 5 = 100 – c => c = 95 Hence option d is the answer. Question 7: The circle passing through the intersection of the circles, x2 + y2 – 6x = 0 and x2 + y2 – 4xy = 0, having its centre on the line, 2x – 3y + 12 = 0, also passes through the point (a) (-1, 3) (b) (-3, 6) (c) (-3, 1) (d) (1, -3) Solution: Let the family of circles be S1 + Î»S2 = 0 x2 + y2 – 6x + Î»(x2 + y2 – 4y) = 0 => (1+Î»)x2 + (1+Î»)y2 – 6x – 4Î»y = 0 â€¦(i) Centre (-g, -f) = (3/(1+Î»), 2Î»/(1+Î»)) Centre lies on 2x – 3y + 12 = 0 Then (6/(Î»+1)) – (6Î»/(Î»+1)) + 12 = 0 => Î» = -3 Equation of circle (i) -2x2 – 2y2 – 6x + 12y = 0 => x2 + y2 + 3x – 6y = 0 Check options. (-3, 6) satisfy equation (ii). Hence option b is the answer. Question 8: Let the tangents drawn from the origin to the circle, x2 + y2 – 8x – 4y + 16 = 0 touch it at points A and B. Then (AB)2 is equal to (a) 52/5 (b) 56/5 (c) 64/5 (d) 32/5 Solution: Length of tangent, L = âˆšS1 = âˆš16 = 4 R = âˆš(16 + 4 – 16) = 2 Length of chord ofcontact = 2LR/âˆš(L2+R2) = 16/âˆš20 Square of length of chord of contact = 64/5 Hence option c is the answer. Question 9: The centre of the circle inscribed in the square formed by the lines x2 – 8x + 12 = 0 and y2 – 14y + 45 = 0 (a) (4, 7) (b) (7, 4) (c) (9, 4) (d) (4, 9) Solution: The centre is given by the intersection of the diagonals (the mid-point of a diagonal). x2 – 8x + 12 = 0 => (x – 6)(x-2) = 0 => x = 6 , 2 y2 – 14y + 45 = 0 => (y – 5)(y-9) = 0 => y = 5, 9 So A(2, 5), B(2, 9), C(6, 9), D(6, 5) form the square. Therefore the centre of circle inscribed in square will be ((2+6)/2, (5+9)/2) = (4, 7) Hence option a is the answer. Question 10: If the tangent at the point P on the circle is x2 + y2 + 6x + 6y = 2 meets a straight line 5x – 2y + 6 = 0 at a point Q on the y-axis, then the length of PQ is (a) 4 (b) 5 (c) 2âˆš5 (d) 3âˆš5 Solution: Given that the line 5x – 2y + 6 = 0 is intersected by tangent at P to the circle x2 + y2 + 6x + 6y – 2 = 0 on y ais at Q. On y axis, x = 0 => 2y = 6 => y = 3 So Q is (0, 3) Tangent passes throught (0, 3). PQ = length of tangent to the circle from (0, 3) = âˆš(0+9+0+18-2) = 5 Hence option b is the answer. Recommended videos ## Circles – Important and Previous Year JEE Questions Also Read JEE Circles and System of Circles Previous Year Questions With Solutions
# Computing the length of a path The rectangular grounds at Hillingham University is going to have a new path built which takes the curved shape of $y=2.2x^2$, starting from the south-west corner - (taken as the origin in a graph) over to the point on the path where $x=3.7$. The constructors of this path want to know the path's length. What is the exact length of this path? - The length of a curve is given by $$\int_a^b \sqrt{1+(f'(x))^2}\ dx.$$ Let $f(x)= y$, $f'(x) = \frac{dy}{dx}$, $a=0$ and $b=3.7$, and compute the integral in a straightforward manner. Where does this come from? Consider estimating the length of a segment of the curve by forming a right triangle whose vertices are $(x,f(x))$, $(x+\Delta x, f(x))$ and $(x+\Delta x, f(x+\Delta x))$. Let $\Delta y = f(x+\Delta x)-f(x)$. Then, the length of the hypotenuse, $s$, is given by the Pythagorean Theorem $s = (\Delta x)^2 + (\Delta y)^2$. Taking the limit as $\Delta x \to 0$, using the notation of infinitesimals, we see that $\Delta y \to dy$ and $\Delta x \to dx$. To get the total length of the curve, we add up all these infinitesimal lengths: $$\int_a^b ds.$$ But from the Pythagorean theorem we have that $ds^2 = dx^2+dy^2$, so $\frac{ds^2}{dx^2} = 1+\frac{dy^2}{dx^2} = 1+\left(f'(x)\right)^2$. Finally, multiplying $dx^2$ back over and taking the square root, we have $$ds = \sqrt{1+(f'(x))^2}\ dx.$$ The length of the curve is found by adding up all these infinitesimal hypotenuses, so $$L(f(x)) = \int_a^b ds = \int_a^b \sqrt{1+(f'(x))^2}\ dx.$$ - is this a standard result? how do you know this? – Anona anon Feb 20 '13 at 17:37 also, why does the function have to be differentiated first? – Anona anon Feb 20 '13 at 17:39 This is a standard result that comes from basic calculus. I'll expand the answer. – Emily Feb 20 '13 at 17:39 would you integrate this using trig substitution ? if so how? thank you – Anona anon Feb 21 '13 at 15:41 Well, $f(x) = 2.2x^2$, so $f'(x) = 4.4x$, so you integrate $\left[1+4.4x\right]^{1/2}$ through basic techniques. Substitution is a way; alternatively, you can recognize that multiplying the integral by 4.4 inside, and 1/4.4 outside gives you $\frac{1}{4.4}\int_0^{3.7} u^{1/2}\ du.$ – Emily Feb 21 '13 at 16:42
## Mar 18, 20150 comments Dedicated to my parents Prepared by O. MADHAVARAJ For most of the students, the subject of Mathematics is a bitter lemon, to make it as sweet lemon , the following some useful tricks on mathematical problems I have seen so far from the books of “Secret of mental maths “ and “Marvelous Maths tricks “ and the oldest script of “Vedic Maths “ were given below for the use of students. The secret of success in the mathematics is depends on the quickest way to get the answers for the problems. The object of this essay is to enlighten the easy way and tricks in the mathematics problem. 1.Eg: 1,2,3,4,5,6,7,8,9 S.C : Multiply last number with one more than that number and divide by 2 ie nx(n+1)/2 Multiply 9x9+1=90 Divide 90/2 =45 Result =45 2.Eg : 33,34,35,36,37,38,39,40,41 S.C : Add the lowest number with the highest, multilply with total number in group and then divide by 2. Multiply 74x9 =666 Divide 666/2 =333 Result =333 3.Find the sum of all odd numbers in between a series Eg: 1,3,5,7,9,11 …………99 S.C : Square the number in the series Here the total number in series is 50 Square 50^2 =2500 Result =2500 4.Find the sum of all even numbers in a series Eg: 2,4,6,8,10 ………..100 S.C : Multiply total numbers in series by more one Here the total number in series =50 Multilply 50x50+1=2550 Result =2550 5.Adding a series having a common difference Eg: 44,47,50,53 S.C: Add the lowest with highest number, divide b 2 then multiply with the total number in series. Divide 97/2 Multiply 97/2x4 =194 Result =194 6.Adding a series having a common ratio Eg: 44,88,176,352 S.C: Multiply the common ratio by itself as many times as the series having number, subtract one and multiply with first number in series.Divide result by less than one ratio Multiply the ratio 4 times 2x2x2x2 =16 Less one 16-1 =15 Multiply 15x44=660 Divide by 2-1 =1 ie 660/1=660 Result =660 7.Adding a sequence in the form 13+23+33…..103 S.C: Square the sum of the series Sum of the series =10x11/2=55 Square == 3025 8.Adding infinite series in the form a+a/b+a/b2 …… S.C: the first term divided by one subtract by the common term multiplied in series Eg: 6+4+8/3….. The common factor multiplied is 2/3 6 divided 1/3 =6x3/1=18 Visit : http://sapost.blogspot.in/ 9.Multiplying by 1001: 27848x1001 S.C: Write first 3 digits as it is ,for fourth add first digit with 4th digit and for 5th digit add second digit with last digit , then the last 3 digits as it is given. First 3 digit =278(multilier 4 digit hence first 3 digit as in the number) Add 7+8 =15(if more than 10 the remaining to be added with previous digit) Last three =848 Now answer 278615848 =27875848 Result 27875848 10.Multiplying by 100001: 423456x100001 S.C: Write first 5 digits as it is ,for sixth add first digit with last digit , then the last 5 digits as it is given. First 5 digit =42345(multilier 6 digit hence first 5 digit as in the number) Last 5 digits =23456 Now answer 423451023456 =42346023456 Result 42346023456 11. Eg: 234567x1000001 Multiplier is 7 digit hence the first six digit of answer is 234567 As there is no 7th digit to add the last digits were 234567 12.Multiplying 2 digit number both end in 5 and one starting with odd number S.C:To the product of the ten digits add one half of their sum(ignore fraction) affix 75 to the result Eg: 95x45 The product of ten digit 9x4 =36 Add one half of their sum =1/2(9+4) = 6 (ignore fraction) =36+6=42 Result 4275 13.Multiplying 2 digit number both end in 5 and both starting with even number S.C:To the product of the ten digits add one half of their sum(ignore fraction) affix 25 to the result Eg: 85x45 The product of ten digit 8x4 =32 Add one half of their sum =1/2(8+4) = 6 (ignore fraction) =32+6=38 Result 3825 14.Multiling two two digit numbers whose tens digits are both 5 and other digit both odd or both even Eg:52x58 S.C: Add one-half the sum of the tens digits to 25.Affix the product of the last digits to the result. If the product is less than 10 proceeds with 0 Add one half of he sum of digits 5+5 =10 =5+25=30 Product of the units =2x8 =16 Result =3016 15.Multipling two digit numbers whose tens digits are both 5 and other digit one odd and other even Eg:52x59 S.C: Add one-half the sum of the tens digits to 25.Affix 50 to the product of the last digits to the result. Add one half of he sum of digits 5+5 =10 =5+25=30 Product of the units =2x9+50 =68 Result =3068 Now from right to left 16.Multilying by 11 Eg: 3421634x11 S.C: Add the neighbor to the number Last 4 +neighbor 0 =4 Next 3+neighbor 4 =7 Next 6+neighbor 3 =9 Next 1+neighbor 6 =7 Next 2+neighbor 1 =3 Next 4+neighbor 2 =6 Next 3+neighbor 4 =7 Next 0+neighbor 3 =3 Result = 37637974 17.Eg: 377x11 Last 7 +neighbor 0 =7 Next 7+neighbor 7 =14 Next 3+neighbor 7 =10+1=11 Next 0+neighbor 3 =3+1 Result =4147 18.Multiplying by 12 Eg:3421634x12 S.C:Double the number add the neighbor Double the Last 4 +neighbor 0 =8 Next double the 3+neighbor 4 =10 Next double the 6+neighbor 3 =15+1 Next double the 1+neighbor 6 =8+1 Next double the 2+neighbor 1 =5 Next double the 4+neighbor 2 =10 Next double the 3+neighbor 4 =10+1 Next double the 0+neighbor 3 =3+1 Result = 41059608 19.Squaring any number ending with 5 Eg: 1952 S.C : Multiply the complete number left to 5 with one more and affix 25 to the result The number left to 5 is 19 Multiply with one more ie 19x(19+1) =380 Affix 25 ie 38025 Result =38025 20.Squaring any 3 digit number ending in 25 Eg: 6252 S.C: Add one half of hundredth digit with square of the hundredth digit (if it comes single digit then the result is 10 thousand digit else 100 thousand and 10 thousand digits),if the hundredth digit is odd number then 5 else 0 , suffix 625 First, square the hundredth digit ie 62 =36 Add one half of the hundredth digit ie 6/2=3+36=39(100 thousand and 10 thousand digits) The thousand digit number is zero (hundredth digit even ) Suffix 625 with above 390625 Result =390625 21.Eg: 5252 First, square the hundredth digit ie 52 =25 Add one half of the hundredth digit ie 5/2=2+25=27(100 thousand and 10 thousand digits) The thousand digit number is 5 (hundredth digit odd ) Suffix 625 with above 275625 Result =275625 22.Squaring any number ending in 9 Eg: 1492 S.C: Multiply the to the left of 9 by two more than itself,substract twice the number to the left of 9,affix 1 to the result Multiply the number to the left of 9 with two more than that number =14x14+2=224 Affix 8 =2248 Substract twice the number left side of 9 ie2248-( 2x14)=2220 Affix 1 to the above ie 22201 Result =22201 23.Squaring any number having 9 only Eg: 9992 S.C: Write one less 9 in the given number followed b 8 and then followed by as many zeros as nine with suffix 1 One 9 less =99 Followed b 8 =998 Zeros 99800 Suffix 1 998001 Result 998001 24.The trick with 37 Eg: 37x3 =111 say 37x3n =nnn Another Eg: 37x15 =37x(3x5)=555 25.The trick with 25 25x4=100 25x35 = 35/4=8 reminder 3x25 =875 26.Multiply by 99,999,9999 etc 56x99 56-1 =55 100-56 =44 Result =5544 567x999 567-1=566 1000-567=433 Result =566433 10245x99999 10245-1=10244 100000-10245=89755 Result =1024489755 27.Cross multiplication Eg : 35x35 3 5 3 5 a) first multiply RHS 5x5 =25 5 remaining(2) b) second multiply cross 3x5+3x5 =30 30+2 2 (3) c) thirdly multiply LHS 3x3 =9 9+3 12 Result : 1225 Splitting into two as one with multiple of 10 Eg: 962 Add the numbers 96+96 = 192 Split into two like 90,102 Multiply both splited 90x102 =9180 Add 6x6 =9180+63 =9216 Result =9216 28.Base method: base 10 Eg : 328x377 328 +28 377 +77 300 28x77 =2156 105 (28+77) ------- 405 x30 =12150 (30x10) 2156 ---------- 123656 ---------- Result =123656 29.Eg: 112x87 112 +12 87 -13 100 + 12x-13 = -156 97 (12-13) & -2 for substracting 156 ------ 97x10 =970 (10x10) 44(200-156) -------- 9744 -------- Result =9744 30.Eg: 87x88 87 -13 88 -12 100 -12x-13 =156 75(100-25) ie -13-12 75x10 =750 156 --------- 7656 Result =7656 31.Gelosia method: Eg: 89x67 8 9 4 8 5 6 5 4 6 3 First step 8x6 =48 4 in upper portion and 8 in lower portion Second 8x7 =56 5 and 6 Third 9x6 =54 5 and 4 and so on 3 = 3 4+6+6=18 =16 5+8+5=18 = 19 4 =5 Result =5963 32.WEIRD MULTIPLICATION: Eg : 34x23 3 4 2 3 The intersection points 6 =7 The intersection points 17 =18 The intersection points 12 =12 Result =782 33.Russian Multiplication Eg: 36X23 First half the LHS and Double RHS (ignore fraction) till LHS become 1 Say 18 46 9 92 4 184 2 368 1 736 Add RHS when LHS odd number ie 92+736=828 Result =828 34.Paper Strip Multiplication : Eg: 56x43 First reverse the multiplier Write the numbers in paper strip And kept as below and multily 6 5 4 3 6x3 =1 6 5 4 3 6x4 +5x3 =39+1 (reminder) =40 6 5 4 3 5x4 =20+4=24 Result =2408 35.DUPLEX METHOD: Eg: 45x57 Add zero before the numbers ie 045x057 Step 1 5x7 =35 (single multiplication) Step 2 5x5+4x7=56 (two multiplication) Step 3 5x0+4x5+0x7=20(three multiplication) Result = 2565 36.Eg: 35x24 Make it as a+b,c+d Say (30+5)x(20+4) = 600+120+100+20 =840 Result =840 37.Eg:35x24 Make it as a-b,c-d Say (40-5)x(30-6) = 1200-240-150+30 =840 Result =840 Squaring 38.Eg: 352 Surplus to base 10 3x10+5 = 5 Number+surplus 35+5 =40 Surplus square =25 LHS 40x3 =120 RHS 25 = 25 = 120 25 =1225 39.Eg: 352 deficit to base 10 4x10-5 = 5 Number-deficit 35-5 =30 deficit square =25 LHS 30x4 =120 RHS 25 = 25 = 120 25 =1225 40.Eg: 752 First step 52 =25 Second step 2x7x5 =70 Third step 72 =49 Answer is 56 72 25 =5625 THREE DIGITS MULTIPLICATION 41.Eg: 123x234 a) first multiply RHS 3x4 =12 2 remaining(1) b) second multiply cross 2x4+3x3 =17 17+1 8 (1) c) thirdly cross with straight 1x4+2x3+2x3 16+1 7 (1) d) second cross 1x3+2x2 7+1 8 e) final multiply LHS 1x2 2 Result 28782 42.Alternate 12 3x 23 4 a) first multiply RHS 3x4=12 2 remaining(1) b) second multiply cross 12x4+3x23 =117+1 8 (11) c) thirdly multiply LHS 12x23 =276+11 287 Result 28782 43.Eg: 123x117 Base 100 123 +23 117 +17 ------------------------ 140 23x17 = +391 14000 391 -------- 14391 -------- 44.Eg: 123x117 base 120 123 +3 117 -3 ------------------ 120 -9 (123-3 or 117+3) 120x12 =14400 - 9 --------- 14391 --------- Gelosia method: 45.Eg: 123x234 0 0 0 2 3 4 0 0 0 4 6 8 0 0 1 6 9 2 1 2 3 2 = 2 9+1+8 = 18 6+0+6+0+4=16+1 = 17 0+4+0+3+0=7+1 = 8 0+2+0 = 2 0 = 0 Result =28782 Duplex method: 46. Eg: 123x456 Add two zero on both number ie 00123x00456 a) single digit 3x6 =18 b) two digit 23x56 =2x6+3x5=27+1 =28 c) three digit 123x456 =3x4+2x5+1x6 =28 =30 d)four digit 0123x0456 =0x6+1x5+2x4+3x0=13 =16 e) Five digit 00123x00456=4+1 =5 Result =56088 Cube 47.Eg: 124 3 12 4 a b b/a=4/12=1/3 a 3 =12 =1728 1728 1728x1/3 (1728x1/3)x1/3 [(1728x1/3)x1/3]x13 (a3 a2b abb3 ) 1728 576 192 64 1152 384 (1 2 2 1) -------------------------------------------------------- 1728 1728 576 64 64 = 4 (6) 6+576 = 2 (58) 58+1728 = 6 (178) 1728+178 = 1906 Result =1906624 48.Eg: 233 A3=(A-D)(A+D)A+D2A here A=23 D=3 = 20x26x23+32*23 =12167 49.Eg: 63 4 6 3 a b b/a= 3/6=1/2 (a4 a3b a2b2 ab3 b4) a4= 64=1296 1296 648 324 162 81 1944 1620 486 (1 3 5 3 1 ) -------------------------------------------------------- 1296 2592 1944 648 81 -------------------------------------------------------- 81 = 1 (8) 648+8 = 6 (65) 1944+65 = 9(200) 2592+200=2(279) 1296+279 = 1575 Result 15752961 50.Eg: 1543 15 4 a b a3 =153 =3375 3375 3375x4/15 (3375x4/15)x4/15 [(3375x4/15)x4/15]x4/15 3375 900 240 64 1800 480 ----------------------------------------------------------------- 3375 2700 720 64 64 = 4 (6) 6+720 = 6 (72) 72+2700 = 2 (277) 277+3375 = 3652 Result =3652264 SQUARE ROOT 51.Eg: √169 From right mark two two digit First digit is 1 hence the square root having 1 as first digit Last digit =9 hence the digit ends with either 3 or 7 (square 9 or 49) Sum of the digits =1+6+9 =16=7 Either 13 or 17 1+3=7 or 1+49=5 Hence answer is 13 52.Eg : √9216 92 16 First two digit is 92 hence first digit be 9 The last digit is 16 ends in 6 hence either 4 or 6 Sum of the numbers 9+2+1+6 = 18=9 Answer should be either 94 or 96 9+4 =13=42 =16=7 not equal to 9 9+6 =15=62 =36=9 Result is 96 53.Eg: √15129 1 51 29 First digit be 1 Last digit be 3 or 7 Sum of the digit =1+5+1+2+9 =18=9 151 is between 122 hence the middle is 2 123 or 127 1+2+3 =6 or 1+2+7 =10=1 squaring 6 =9 hence last digit is 3 Result 123 54.Bakshali formula : Square root = a+b/2a √76 8 12 a b 8+12/16 = 8.75 but actual 8.71 Cube Root : 55.Eg : 3√ 1953125 1 2 5 \| 1 953 125 300(1)2 +30(1)(2) +22 =364 \| 1 -------- 300(12)2+30(12)(5)+52 =45025 364\| 953 728 ----- 45025\| 225125 225125 ---------- 0 ---------- 56.Eg: √1953125 1 953 125 First digit be 1 Last digit be 5 Sum of the numbers =1+9+5+3+1+2+5=26=8 hence middle digit 2 Result =125 DIVISION 57. A number divisible by 2 S.C: If the last digit is an even number Eg : 41256 is divisible by 2 Last digit is 6 which is divisible by 2 Hence the number is divisible. 58. A number divisible by 3 S.C: If the sum of the digits is divisible by 3 then divisible Eg : 41256 is divisible by 3 Sum of the digits 4+1+2+5+6 =18 which is divisible by 3 Hence the number is divisible. 59. A number divisible by 4 S.C: If the last two digit is an divisible by 4 Eg : 41256 is divisible by 4 Last two digit is 56 which is divisible by 4 Hence the number is divisible. 60. A number divisible by 5 S.C: If the last digit is either 0 or 5 is divisible Eg : 41256 is divisible by 5 Last digit is 6 Hence the number is not divisible. 61. A number divisible by 6 S.C: If the last digit is an even number and the sum of the number is divisible by 3 Eg : 41256 is divisible by 6 Last digit is 6 which is divisible by 2 Sum of the digits 4+1+2+5+6 = 18 is divisible by 3 Hence the number is divisible. 62.A number divisible by 8 S.C: If the last three digits is divisible by 8 Eg : 41256 is divisible by 8 Last two digit is 256 which is divisible by 8 Hence the number is divisible. 63.Eg: 8678991 divisible by 19 Osculator =2 2x1+9 = 11 (2x1)+1+9 = 12 (2x2)+1+8 = 13 (2x3)+1+7 = 14 (2x4)+1+6 = 15 (2x5)+1+8 = 19 19/19 hence it is divisible Number +osculator -osculator total 1 1 0 1 3 1 2 3 7 5 2 7 9 1 8 9 11 10 1 11 13 4 9 13 17 12 5 17 64.Eg: 17160384 divisible by 139 14x4+8 =64 14x4+6+3 =65 14x5+6+0 =76 14x6+7+6 =97 14x7+9+1 =108 14x8+10+7 =129 14x9+12+1 =139 Divisible by 139 65. Eg: 3245693 divisible by 11 Odd 3+4+ 6+3 = 16 Even 2+5+9 =16 both are equal hence divisible. 66.Eg:132101 / 9 132101 13210 1 1 1 1467 7 1+3 4 14677 8 1+3+2 6 1+3+2+1 7 1+3+2+1+0 7 14677 reminder 8 67.Eg: 23243 /9 2324 3 2 2 257 11 2+3 5 ----- 2+2+3 7 2581 14 2+2+3+4 11 2582 reminder 5 68. Eg : 2679502 / 43 6 2 3 1 4 ----------------------- 43\|26 7 9 5 0 2 24 -------- 2 7-(3x6) =9 8 ------ 19 –(3x2) = 13 12 ------- 1 5 –(3x3) =6 4 ----- 2 0 -3x1=17 16 ----- 12 -12 =0 Result 62314 69.Eg: 123456/69 1 7 8 9 --------------------- 7\|12 3 4 5 6 more than one ie 69+1=70 7 ---- 5 3+1 =54 49 ----- 54+7=61 56 ---- 5 5+8 =63 63 --- 0+9 +6 =15 Result 1789 reminder 15 70.Eg: 738704 divide by 79 9 3 5 0 ------------------- 8\| 73 8 7 0 4 72 ----- 18+9 =27 24 ----- 3 7+3 =40 40 ------ 54 ------ Result 9350 reminder 54 71.Eg: 113989/21 5 4 2 8 -------------- 2\| 11 3 9 8 9 10 ------ 13-5 =8 8 ------- 9 -4 =5 4 ------ 18-2=16 16 ----- 09-8 =1 Result 5428 reminder 1 72.Eg: 2042 LHS 22 (2x4x2) 42 RHS Result = 41616 73.Eg:3082 LHS32(3x8x2)82RHS Result = 94864 74.Find the value of 1/19 0.52631578947368 -------------------------------- 2\| 10 10 ------ 5 4 ------ 10+2=12 12 ----- 6 6 ---- 3 2 ---- 10+1=11 10 ------ 10+5 =15 14 ----- 10+7=17 16 ---- 18 18 ----- 9 8 ----- 10+4=14 14 ----- 7 6 ------ 10+3 =13 12 ----- 10+6=16 16 ------ Result =0.52631578947368 75. Find the value of 1/19 For a fraction of the form in whose denominator 9 is the last digit, we take the case of 1 / 19 as follows: For 1 / 19, 'previous' of 19 is 1. And one more than of it is 1 + 1 = 2. Therefore 2 is the multiplier for the conversion. We write the last digit in the numerator as 1 and follow the steps leftwards. Step. 2 : 21(multiply 1 by 2, put to left) Step. 3 : 421(multiply 2 by 2, put to left) Step. 4 : 8421(multiply 4 by 2, put to left) and so on 1 21 421 8421 168421 1368421 7368421 147368421 947368421 18947368421 178947368421 1578947368421 11578947368421 31578947368421 631578947368421 12631578947368421 52631578947368421 0. 052631578947368421 76.Eg: 113989 divided by 113 1 1 3 11 3 9 8 9 -1-3 -1 -3 0 0 -0 -0 -9 -27 ---------------------------------- 1009 -1 -18 1008 R 113-28=85 77.Eg: 124992 divided by 124 1 2 4 1 2 4 9 9 2 -2-4 -2 -4 0 0 0 0 -18 -36 ----------------------------------- 1 0 0 9 -9 -34 Q 1009 R-(34+90) =1 ie 1008 78.Eg: 1227352 divided by 9898 Divisor nearer to 10000 here deficit =10000-9898= 0102 Keep last four digits for reminder 0102 \| 1 2 2 7 3 5 2 0 1 0 2 0 2 0 4 0 3 0 6 ----------------------------- 1 2 3 9898 123+1 =124 Q=124 79.Eg: 111500 by 892 108\|1 1 1 5 0 0 1 0 8 2 0 16 3 0 24 ---------------------- 123 1784 =123+2=125 80.Eg: 97092 by 783 217\| 9 7 0 9 2 19 5 3 217x9 55 4 2 217x26 ------------------- 9 26 60 16 4 =4+16+6000=6164=783x8 90+26+8 =124 81. Karatsuba method of multiplication :two digits Eg: 78x21 axb= u x102+(u+w-v)x10+w Here u=7x2 =14 v=(8-7)x(1-2) = -1 w=8x1 =8 1400+230+8 =1638 82.Eg: 87x34 u= 8x3 =24 v =(8-7)x(4-3)=1 w=7x4=28 2400+530+28=2958 83.Karatsuba method of multiplication :four digits axb = ux104+(u+w-v)x102+w Eg: 5678x4321 u=56x43=2408 v=78-56x21-43=22x-22= -484 w=78x21=1638 24080000+453000+1638 =24534638 84.Fourier Techniques Eg:123x654 p(x) =3x0+2x+1x2 q(x) =4x0+5x+6x2 =3.4(x0)+x(3.5+4.2)+ x2(3.6+4.1+2.5)+ x3(2.6+5.1)+1.6x4 = 12+230+3200+17000+60000=80442 85.Eg: 54612 divided by 246 222 ---------- 24 6\|54 6 1 2 48 ------ 66-12=54 48 ------ 6 1-12=49 48 ----- 12-12 =0 Q=222 86.Eg: 28949025 divided by 2345 12345 ---------------- 234 5\|289 4 9 0 2 5 234 ------ 55 4-5=549 468 ------ 819 -10 =809 702 ------- 1070 -15 =1055 936 ------ 1192-20=1172 1170 ------- 25-25 =0 Q=12345 87. 14121 divided by 99 99 \| 1 41 21 1 \| 1+41\|41+21 1 42 62 Result Q= 142 R62 Another one Eg: 41089 divided by 33 33\| 4 10 89 4 4+10 4+10+89 414 103 414 103/33=3 R4 414x3+3 =1242+3 =1245 R4 Q=1245 R 4 88. Partial Quatients Method: 1220 divide by 16 16\| 1220 800 50 ------- 420 320 20 ------ 100 80 5 ----- 20 16 1 ----- 4 Result 50+20+5+1 =76 R 4 Some algebra problem 89.Eg: (a+2b)(3a+b) Cross multiplication method: a + 2b 3a + b 2bxb =2b2 ab+6ab =7ab ax3a =3a2 Result =3a2+7ab+2b2 90. Eg: (4x2+3)(5x+6) 4x2 +0x+3 0x2+5x+6 3x6 =18 (5x)(3)+(0x)(6) =15x (4x2)(6)+(0x2)(3)+(0x)(5x) =24x2 (4x2)(5x) +(ox2)(ox) =20x3 (4x2)(0x2) =0 Result 20x3+24x2+15x+18 91. x3+5x2+3x+7 divided by x-2 x+7x +17 ----------------------- x-2 \| x3 + 5x2 + 3x +7 x3 ---- 5x2 +2x2 =7x2 7x2 ----- 3x +14x=17x 17x ------ 7+34=41 Result =x2+7x+17 R 41 92. Eg :x3-3x2+10x-4 by x-5 1 +2 +20 ---------------- x-5\| 1 -3 +10 - 4 +5 --------- 10 ---- 100 ---------------------------------- 1 +2+20 96 Result = x2+2x+20 R 9 93. Four digit multiplication: Eg: 1188x1212 Cross multiplication 1. 8x2 =16 2. 8x2+1x8 =24+1 =25 3.1x2+2x8+1x8 =26+2=28 4.1x2+1x8+1x1+2x8 = 27+2=29 5.1x1+1x8+1x2 = 11+2=13 6.1x2+1x1 =3+1 =4 7.1x1 =1 Result =1439856 94. Squaring a four digit number Eg:12342 Surplus 1234-1200=34 base 1200 Surplus+number =1234+34=1268 Surplus square =34 =1156 1268 \|1156 x12 15216 1156 Result =1522756 Magic number 1/9 =0.111111 2/9 =0.222222 3/9=0.3333333 4/9=0.4444444 5/9=0.5555555 6/9=0.6666666 7/9=0.7777777 8/9=0.8888888 95.Eg: 600/9 =0.66666x100=66.66 1x1 = 1 11x11 = 121 111x111 = 12321 1111x1111 = 1234321 11111x11111 = 123454321 111111x111111 = 12345654321 1111111x1111111 = 1234567654321 11111111x11111111 =123456787654321 12345679x9 =111111111 12345679x18 =222222222 12345679x27 =333333333 135 =1+32+53 175 =1+72+53 518 =5+12+83 Fifth root : 96.Eg:8,58,73,40,257 x x5 1 100 thousands 2 3 million 3 24 million 4 100 million 5 300 million 6 777 million 7 1.6 billion or 160 crores 8 3 billion or 300 crores 9 6 billion or 600 crores In the above example number is more than 600 crores hence the first digit =9 The last digit is the last digit of the number =7 Result =97 97.Eg: 39135393 Having more than 24 million hence first digit is 3 Last digit is 3 Result =33 98. Fourth root: x x4 1 10000 2 160000 3 810000 4 2560000 5 6250000 6 12960000 7 24010000 8 40960000 9 65610000 99.Eg: 234256 The first digit is 2 The last digit is either 2 or 4 or 6 or 8 Sum of the digits =2+3+4+2+5+6 =22 =4 22=2+2 =44 =256 =13=4 244 =64=1296 =18=9 264 =84 =4096 =19=1 284=104=1 Hence 22 is the answer 100. Eg :37015056 The first digit is 7 The last digit is either 2 or 4 or 6 or 8 Sum of the digits =3+7+0+1+5+0+5+6 =27 =9 724=94=6561=18 =9 744=24=16=7 764=14 =1 784 =64=1296=18=9 Here 704=24010000 and 72 is nearer to 70 and will have 2800000 but our number is 3700000 hence 78 is the correct answer. Review : Multiplication Eg: 123x234 Method 1: 1 2 3 2 3 4 a) 3x4 =12 b) 2x4+3x3 =17+1 =18 c) 1x4+2x3+2x3 =16+1 =17 d) 1x3+2x2 =7=1 =8 e) 1x2 =2 Result =28782 Method 2: 12 3 23 4 a) 3x4 =12 b)12x4+23x3 =117+1 =118 c) 23x12=276+11 =287 Result =28782 Method 3: 123 -77 234 +34 (base 200) ------------------- 157 -77x34 157-15 (3000-2618) 15x200=3000 142x2 =28400 382 -------- 28782 ---------- Method 4: 0 0 0 2 3 4 0 0 0 4 6 8 0 0 1 6 9 2 1 2 3 2 = 2 9+1+8 = 18 6+0+6+0+4=16+1 = 17 0+4+0+3+0=7+1 = 8 0+2+0 = 2 0 = 0 Result =28782 Method 5: 123x234 00123x00234 a) single digit 3x4 =12 b) double digit 23x34=3x3+2x4 =17+1 =18 c) three digits 123x234=1x4+2x3+3x2 = 16+1 =17 d) four digits 0123x0234=0x4+1x3+2x2+3x0=7+1 =8 e) five digits 00123x00234=0x4+0x3+1x2+2x0+3x0=2 Result =28782 Method 6 : 123x234 100x234 =23400 20x234 = 4680 3x234= 702 -------- 28782 --------- Method 7: 123 234 61 468 30 936 15 1872 7 3754 3 7518 1 15036 ----------- 28782 Method 8: 1 2 3 Point of intersection = 12 = 17 = 16 = 7 =2 Result =28782 Method 9 : 123x234 3 2 1 2 3 4 3x4 =12 3 2 1 2 3 4 3x3+2x4 = 17+1=18 3 2 1 2 3 4 3x2+2x3+1x4 = 16+1 =17 3 2 1 2 3 4 2x2+1x3 =7+1 =8 3 2 1 2 3 4 1x2 =2 Result =28782 Method 10: p(x) =3x0+2x+1x2 q(x) =4x0+3x+2x2 =3.4(x0)+x(3.3+4.2)+ x2(1.4+2.3+2.3)+ x3(2.2+3.1)+1.2x4 = 12+170+1600+7000+20000=28782 Method 11: u=12x23= 276 v=(3-12)(4-23) =171 w=3x4 =12 u100+(u+w-v)10+w 276x100+117x10=12=28782 Result = 28782 a)Eg: 2716032 is divisible by 22? S.C : The last digit multiply by 1 and deduct from the remaining digits and so on till divisible by 22 or to become 0. 271603 2 - 2 --------- 27160 1 1 -------- 2715 9 9 ------ 270 6 6 ----- 26 4 4 ---- 22 22 ---- 0 --- b) Eg: 3936 is divisible by 32 S.C : The last digit multiply by 3/2 and deduct from the remaining digits and so on till divisible by 32 or to become 0 393 6 9 ----- 38 4 6 ---- 32 32 ---- 0 --- c)Eg: 191814 is divisible by 42 S.C: The last digit multiply by 2 and deduct from the remaining digits and so on till divisible by 42 or to become 0 19181 4 8 -------- 1917 3 6 ------ 191 1 2 ------- 18 9 18 --- 0 Q= 9134/2=4567 d) Eg: 42036 is divisible by 62 S.C: The last digit multiply by 3 and deduct from the remaining digits and so on till divisible by 62 or to become 0 4203 6 18 ------ 418 5 15 ---- 40 3 9 --- 3 1 3 -- 0 Q=1356/2=678 e)Eg: 1012290 is divisible by 82 S.C: The last digit multiply by 4 and deduct from the remaining digits and so on till divisible by 82 or to become 0 101229 0 0 -------- 10122 9 36 ------- 1008 6 24 ----- 98 4 16 ------- 82 82 ------- 0 Q= 12345 f) Eg: 125868 is divisible by 102 S.C: The last digit multiply by 5 and deduct from the remaining digits and so on till divisible by 102 or to become 0 12586 8 40 -------- 1254 6 30 ------ 122 4 20 ----- 102 102 ----- 0 ----- Q= 1 468/2=234=1234 g) Eg: 315614 is divisible by 122 S.C: The last digit multiply by 6 and deduct from the remaining digits and so on till divisible by 122 or to become 0 31561 4 24 -------- 3153 7 42 ------ 311 1 6 ----- 30 5 30 --- 0 5174/2=2587 i) Eg:14805 is divisible by 63 S.C: The last digit multiply by 2 and deduct from the remaining digits and so on till divisible by 63 or to become 0 1480 5 10 ------- 147 0 0 ------ 14 7 14 ---- 0 --- Q=705/3=235 j) Eg: 449934 is divisible by 123 S.C: The last digit multiply by 4 and deduct from the remaining digits and so on till divisible by 123 or to become 0 44993 4 16 -------- 4497 7 28 ------- 446 9 36 -------- 41 0 0 --------- 4 1 4 --- 0 Q =10974/3=3658 1 k) Eg:120717 divisible by 153 S.C: The last digit multiply by 5 and deduct from the remaining digits and so on till divisible by 153 or to become 0 12071 7 35 -------- 1203 6 30 ------ 117 3 15 ----- 10 2 10 ---- 0 Q=2367/3=789 l) Eg:8359074 is divisible by 183 S.C: The last digit multiply by 6 and deduct from the remaining digits and so on till divisible by 183 or to become 0 835907 4 24 ----------- 83588 3 18 -------- 8357 0 0 ------ 835 7 42 ---- 79 3 18 --- 6 1 6 -- 0 Q =137034/3=45678 m) Eg: 434544 divisible by 44 S.C: The last digit multiply by 1 and deduct from the remaining digits and so on till divisible by 44 or to become 0 43454 4 4 ------------- 4345 0 0 ------ 434 5 5 ----- 42 9 9 --- 3 3 3 --- 0 Q=39504/4=9876 n) Eg:5313672 is divisible by 84 S.C: The last digit multiply by 2 and deduct from the remaining digits and so on till divisible by 84 or to become 0 531367 2 4 --------- 53136 3 6 ------------ 5313 0 0 ------ 531 3 6 ------ 52 5 10 ----- 4 2 4 ------ 0 Q=253032/4=63258 o) Eg: 45756 is divisible by 124 S.C: The last digit multiply by 3 and deduct from the remaining digits and so on till divisible by 124 or to become 0 4575 6 18 ------ 455 7 21 ----- 43 4 12 --- 3 1 3 ---- 0 Q=1476/4=369 p) Eg : 5668988 is divisible by 164 S.C: The last digit multiply by 4 and deduct from the remaining digits and so on till divisible by 164 or to become 0 566898 8 32 ---------- 56686 6 24 --------- 5666 2 8 ------- 565 8 32 ----- 53 3 12 --- 4 1 4 --- 0 Q =138268/4 =34567 Eg: 315 is divisible by 7 S.C : A number is of form 100a+b is divisible by 7 if and only 2a+b is divisible by 7 (babylonian method) 3x100+15 a=3 b=15 2a+b = 6+15=21 which is divisible by 7 Hence divisible Eg: 168168 is divisible by 7 S.C: A six digit number of the form xyzxyz is divisible by 7 Eg: 1353 is divisible by 11 135\| 3 Multily last digit by -1 -3 ---- 13 \| 2 2x-1 -2 --- 1 \| 1 1x-1 1 -- 0 Hence divisible by 11 Q =123 Eg: 1968 is divisible by 16 196\| 8 Multiply by 1/6 195\| 18 8 is not divisible by 6 hence 1 borrowed 3 ----- 19\| 2 18\| 12 2 --- 1 6 1 -- 0 Hence divisible . Q = 6 12 18 divide by 6 Q=123 TABLE 1 Divisor the last digit multiply with 11 -1 12 - 1/2 13 - 1/3 14 - 1/4 16 - 1/6 17 - 1/7 18 - 1/8 19 - 1/9 21 -2 22 -1 23 - 2/3 24 - 2/4 26 - 2/6 27 - 2/7 28 - 2/8 29 - 2/9 31 -3 32 --3/2 33 -1 34 - 3/4 36 - 3/6 37 - 3/7 38 - 3/8 39 - 3/9 41 -4 42 -2 43 -4/3 44 -1 46 - 4/6 47 - 4/7 48 - 4/8 49 - 4/9 Sum of the digit tricks: Eg : 4567872 is divisible by 37 Sum of 3 digits from right ie 872+567+4 =1443 which is divisible by 37 Eg: 1222155 is divisible by 99 Sum of 2 digits 55+21+22+1 =99 which is divisible by 99 Divisor 3 9 11 27 33 37 99 101 Block to add 1 1 2 3 2 3 2 4 Sum of alternate digits : Eg: 9012345597 is divisible by 73 Sum of alternate 4 digits ie 5597+90 =5687 next 1234 5687-1234=4453 which is divisible by 73 Eg: 12469056 is divisible by 101 Sum of alternate 2 digits = 56+46 =102 next 90+12 =102 102-102=0 hence divisible Divisor 7 11 13 73 77 91 101 Block to add alternately 3 1 3 4 3 3 2 ------------------------------------------------------------------------------------------------ Reference : 1. 101 Shortcuts in maths an one can do by Gordon Rockmaker-Feredrick Fell publishers,Newyork 1965 2. Trenchenberg system of speed maths 3. Secret of mental maths by Arthur Benjamin and Michael Shermer 4. Vedic Maths 5. Mental Maths Tricks by Daryl Stephens 6. Mental arithmetic tricks by Andreas Klein 7. Simple Divisibility Rules by C.C.Briggs Penn State University 8. Stupid Divisibility Test by Marc Renault 9. Marvelous maths tricks Prepared by O. MADHAVARAJ
# Ex.2.4 Q2 Polynomials Solution - NCERT Maths Class 10 Go back to 'Ex.2.4' ## Question Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as $$2, –7, –14$$ respectively. Video Solution Polynomials Ex 2.4 \| Question 2 ## Text Solution What is known? Zeroes of a Cubic polynomials are $$2, –7, –14$$ respectively. What is unknown? A cubic polynomial with the sum, sum of the product of its zeroes taken two at a time. Reasoning: To solve this question, follow the steps below- We know that the general equation of the polynomial is $$a{x^3} + {{ }}b{x^2} + cx{{ }} + {{ }}d$$ and the zeroes are $$\alpha ,\;\beta$$ and $$\gamma .$$ Now we have to find the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes. For this we know that, \begin{align}{\alpha + \beta + \gamma}& = {\frac{{ - b}}{a}}\\{\alpha \beta + \beta \gamma + \gamma \alpha}& = {\frac{c}{a}}\\{\alpha \beta \gamma}& = \frac{{ - d}}{a}\end{align} Put the values of the known coefficients, you will get the value of unknown coefficient. Now put the values of coefficients in the general equation of the cubic polynomial $$a{{x}^{3}}+{ }b{{x}^{2}}+cx+d.~$$ Steps: Let the polynomial be $$a{x^3} + {{ }}b{x^2} + cx + d$$ and the zeroes are $$\alpha,\;\beta$$ and $$\gamma.$$ We know that \begin{align}\alpha + \beta + \gamma &= \frac{2}{1}\\ &= \frac{{ - b}}{a}\\\\\alpha \beta + \beta \gamma + \gamma \alpha &= \frac{{ - 7}}{1}\\ &= \frac{c}{a}\\\\\alpha \beta \gamma &= \frac{{ - 14}}{1}\\& = \frac{{ - d}}{a}\end{align} if $$a = 1$$ then $$b = - 2, \;c = - 7$$ and $$d = 14$$ Hence, the polynomial is $${x^3} - 2{x^2} - 7x + 14$$ Learn from the best math teachers and top your exams • Live one on one classroom and doubt clearing • Practice worksheets in and after class for conceptual clarity • Personalized curriculum to keep up with school
# How do you graph the function y=-x+6? Jun 22, 2018 See a solution process below: #### Explanation: First, solve for two points which solve the equation and plot these points: First Point: For $x = 6$ $y = - 6 + 6$ $y = 0$ or $\left(6 , 0\right)$ Second Point: For $y = 4$ $y = - 4 + 6$ $y = 2$ or $\left(4 , 2\right)$ We can next plot the two points on the coordinate plane: graph{((x-6)^2+y^2-0.035)((x-4)^2+(y-2)^2-0.035)=0 [-10, 10, -5, 5]} Now, we can draw a straight line through the two points to graph the line: graph{(y+x-6)((x-6)^2+y^2-0.035)((x-4)^2+(y-2)^2-0.035)=0 [-10, 10, -5, 5]} Jun 22, 2018 $\text{see explanation}$ #### Explanation: $\text{one way is to find the intercepts, that is where the graph}$ $\text{crosses the x and y axes}$ • " let x = 0, in the equation for y-intercept" • " let y = 0, in the equation for x-intercept" $x = 0 \Rightarrow y = 6 \leftarrow \textcolor{red}{\text{y-intercept}}$ $y = 0 \Rightarrow - x + 6 = 0 \Rightarrow x = 6 \leftarrow \textcolor{red}{\text{x-intercept}}$ $\text{plot the points "(0,6)" and } \left(6 , 0\right)$ $\text{draw a straight line through them for graph}$ graph{(y+x-6)((x-0)^2+(y-6)^2-0.04)((x-6)^2+(y-0)^2-0.04)=0 [-15.79, 15.81, -7.9, 7.9]}
0 like 0 dislike x-intercepts are (1,0) and (-3,0) The axis of symmetry is the x coordinate of the midpoint of x-intercepts:(1-3)/2=-1, x=-1 is the axis of symmetry. 0 like 0 dislike The equation f(x) = x² + 2x - 3 is an illustration of a quadratic function Calculate the left term The equation is f(x) = x² + 2x - 3 The left term is calculated using: $$x = -\frac{b}{2a}$$ This gives $$x = -\frac{2}{21}$$ x = -1 Hence, the left term is -1 Compare the vertex and the axis of symmetry The result above represents the axis of symmetry. Substitute x = -1 in f(x) = x² + 2x - 3 to calculate the y-coordinate of the vertex f(-1) = (-1)² + 2(-1) - 3 f(-1) = -4 This means that the vertex is (-1,-4) So, we can conclude that the axis of symmetry passes through the x-coordinate of the vertex and it divides the graph into two equal segments. Calculate the right term The formula is given as: $$x = \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$ So, we have: $$x = \pm \frac{\sqrt{2^2 - 4 1 * -3}}{2 * 1}$$ $$x = \pm \frac{\sqrt{16}}{2}$$ Evaluate $$x = \pm 2$$ Hence, the right term is ±2 The horizontal distance along the axis From the graph of the function (see attachment), we have the horizontal distance of the vertex between each x-intercept to be 2 Compare (d) and (e) In (d), we have: $$x = \pm 2$$ In (e), we have: horizontal distance = 2 This means that the horizontal distance is the absolute value of the right term. Summary of the findings The findings are: - The left term represents the axis of symmetry. - The right term represents the horizontal distance of the vertex between each x-intercept. - The axis of symmetry divides the graph into two equal segments.
# Solve Simple Equations (Mixed Decimals) In this worksheet, students solve simple one-stage equations with decimals. Key stage: KS 2 Curriculum topic: Maths and Numerical Reasoning Curriculum subtopic: Equations and Algebra Difficulty level: ### QUESTION 1 of 10 When we solve algebraic equations, our aim is to end up with one letter on one side of the equals sign and one number on the other. This is the solution. We do this by using inverse operations to undo things that get in the way, but remember that we must do the same thing to both sides. Example Solve for a 5.2 - a = 10 Add a to both sides. 5.2 - a + a = 10 + a Simplify 5.2 = 10 + a Subtract 10 from both sides. 5.2 -10 = 10 - 10 + a Simplify -4.8 = a a = -4.8 Example 5b = 24 Divide both sides by 5 5b ÷ 5 = 24 ÷ 5 Simplify b = 4.8 Example b ÷ 5 = 20.2 Multiply both sides by 5 b ÷ 5 x 5 = 20.2 x 5 Simplify b = 101 Solve for a: 7 + a = 15.4 a = 22.4 8.4 a = 8.4 Solve for a: a - 12.3 = 15 a = -27.3 a = 27.3 a = 2.7 Solve for a: 2a = 21 a = 42 a = 10.5 a = 19.0 Solve for a: a ÷ 3 = 21.5 a = 18.5 a = 7.17 a = 64.5 Solve for a: a - 28.5 = 21.5 a = -7 a = 49 a = 50 Solve for a: 18.4 - a = 24 a = -5.6 a = 6 a = -6.6 Solve for a: a ÷ 3.5 = 20 a = -23.5 a = 70 a = 16.5 Solve for a: 3a = 18.6 a = 15 a = 6.2 a = 55.8 Solve for a: 15.2 - a = 13.8 a = 1.4 a = -29 a = 2.6 Solve for a: 1.5a = -30 a = 15 a = -15 a = -20 • Question 1 Solve for a: 7 + a = 15.4 a = 8.4 • Question 2 Solve for a: a - 12.3 = 15 a = 27.3 • Question 3 Solve for a: 2a = 21 a = 10.5 • Question 4 Solve for a: a ÷ 3 = 21.5 a = 64.5 • Question 5 Solve for a: a - 28.5 = 21.5 a = 50 • Question 6 Solve for a: 18.4 - a = 24 a = -5.6 • Question 7 Solve for a: a ÷ 3.5 = 20 a = 70 • Question 8 Solve for a: 3a = 18.6 a = 6.2 • Question 9 Solve for a: 15.2 - a = 13.8 a = 1.4 • Question 10 Solve for a: 1.5a = -30 a = -20 ---- OR ---- Sign up for a £1 trial so you can track and measure your child's progress on this activity. ### What is EdPlace? We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them. Get started Start your £1 trial today. Subscribe from £10/month. • Tuition Partner
<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. # Words that Describe Patterns ## Use variable expressions to solve real-world problems. Estimated9 minsto complete % Progress Practice Words that Describe Patterns Progress Estimated9 minsto complete % Words that Describe Patterns What if you were given a word problem like "It took the Eagle Scouts one hour to wash 3 cars. How long did it take them to wash one car?" or "The distance from the East Coast to the West Coast is more than 2500 miles."? How could you write these sentences in algebraic form? After completing this Concept, you'll be able to write equations and inequalities for situations like these. ### Guidance In algebra, an equation is a mathematical expression that contains an equals sign. It tells us that two expressions represent the same number. For example, is an equation. An inequality is a mathematical expression that contains inequality signs. For example, is an inequality. Inequalities are used to tell us that an expression is either larger or smaller than another expression. Equations and inequalities can contain both variables and constants. Variables are usually given a letter and they are used to represent unknown values. These quantities can change because they depend on other numbers in the problem. Constants are quantities that remain unchanged. Ordinary numbers like and are constants. Equations and inequalities are used as a shorthand notation for situations that involve numerical data. They are very useful because most problems require several steps to arrive at a solution, and it becomes tedious to repeatedly write out the situation in words. Here are some examples of equations: To write an inequality, we use the following symbols: > greater than greater than or equal to < less than less than or equal to not equal to Here are some examples of inequalities: The most important skill in algebra is the ability to translate a word problem into the correct equation or inequality so you can find the solution easily. The first two steps are defining the variables and translating the word problem into a mathematical equation. Defining the variables means that we assign letters to any unknown quantities in the problem. Translating means that we change the word expression into a mathematical expression containing variables and mathematical operations with an equal sign or an inequality sign. #### Example A Define the variables and translate the following expressions into equations. a) A number plus 12 is 20. b) 9 less than twice a number is 33. c) $20 was one quarter of the money spent on the pizza. Solution a) Define Let the number we are seeking. Translate A number plus 12 is 20. b) Define Let the number we are seeking. Translate 9 less than twice a number is 33. This means that twice the number, minus 9, is 33. c) Define Let the money spent on the pizza. Translate$20 was one quarter of the money spent on the pizza. Often word problems need to be reworded before you can write an equation. #### Example B Find the solution to the following problems. a) Shyam worked for two hours and packed 24 boxes. How much time did he spend on packing one box? b) After a 20% discount, a book costs $12. How much was the book before the discount? Solution a) Define Let time it takes to pack one box. Translate Shyam worked for two hours and packed 24 boxes. This means that two hours is 24 times the time it takes to pack one box. Solve Answer Shyam takes 5 minutes to pack a box. b) Define Let the price of the book before the discount. Translate After a 20% discount, the book costs$12. This means that the price minus 20% of the price is $12. Solve Answer The price of the book before the discount was$15. Check If the original price was $15, then the book was discounted by 20% of$15, or $3. . The answer checks out. #### Example C Define the variables and translate the following expressions into inequalities. a) The sum of 5 and a number is less than or equal to 2. b) The distance from San Diego to Los Angeles is less than 150 miles. c) Diego needs to earn more than an 82 on his test to receive a in his algebra class. d) A child needs to be 42 inches or more to go on the roller coaster. Solution a) Define Let the unknown number. Translate b) Define Let the distance from San Diego to Los Angeles in miles. Translate c) Define Let Diego’s test grade. Translate d) Define Let the height of child in inches. Translate: Watch this video for help with the Examples above. ### Vocabulary • To write an inequality, we use the following symbols: > greater than greater than or equal to < less than less than or equal to not equal to ### Guided Practice Define the variables and translate the following expressions into inequalities. a) Jose took 3 train trips in a day, some of which cost$2.75 and some of which cost $3.95. His total cost was$9.45. b) The product of 3 and some number is more than the sum of 24 and that number. Solution: a) Let be the number of train rides that cost $2.75. Then is the number of train rides that cost$3.95. Then we get: b) Let be "some number." Then the product of 3 and is . The sume of 24 and is . Together we get: ### Practice For 1-10, define the variables and translate the following expressions into equations. 1. Peter’s Lawn Mowing Service charges $10 per job and$0.20 per square yard. Peter earns $25 for a job. 2. Renting the ice-skating rink for a birthday party costs$200 plus $4 per person. The rental costs$324 in total. 3. Renting a car costs $55 per day plus$0.45 per mile. The cost of the rental is $100. 4. Nadia gave Peter 4 more blocks than he already had. He already had 7 blocks. 5. A bus can seat 65 passengers or fewer. 6. The sum of two consecutive integers is less than 54. 7. The product of a number and 3 is greater than 30. 8. An amount of money is invested at 5% annual interest. The interest earned at the end of the year is greater than or equal to$250. 9. You buy hamburgers at a fast food restaurant. A hamburger costs $0.49. You have at most$3 to spend. Write an inequality for the number of hamburgers you can buy. 10. Mariel needs at least 7 extra credit points to improve her grade in English class. Additional book reports are worth 2 extra credit points each. Write an inequality for the number of book reports Mariel needs to do. ### Vocabulary Language: English $\ge$ $\ge$ The greater-than-or-equal-to symbol "$\ge$" indicates that the value on the left side of the symbol is greater than or equal to the value on the right. $\le$ $\le$ The less-than-or-equal-to symbol "$\le$" indicates that the value on the left side of the symbol is lesser than or equal to the value on the right. $\ne$ $\ne$ The not-equal-to symbol "$\ne$" indicates that the value on the left side of the symbol is not equal to the value on the right. constant constant A constant is a value that does not change. In Algebra, this is a number such as 3, 12, 342, etc., as opposed to a variable such as x, y or a. Equation Equation An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs. greater than greater than The greater than symbol, $>$, indicates that the value on the left side of the symbol is greater than the value on the right. greater than or equal to greater than or equal to The greater than or equal to symbol, $\ge$, indicates that the value on the left side of the symbol is greater than or equal to the value on the right. inequality inequality An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are $<$, $>$, $\le$, $\ge$ and $\ne$. less than less than The less-than symbol "<" indicates that the value on the left side of the symbol is lesser than the value on the right. less than or equal to less than or equal to The less-than-or-equal-to symbol "$\le$" indicates that the value on the left side of the symbol is lesser than or equal to the value on the right. not equal to not equal to The "not equal to" symbol, $\ne$, indicates that the value on the left side of the symbol is not equal to the value on the right. Variable Variable A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
1. ## Number sequence 5, 10, 17, 26, 37, ... I see the pattern, but I can't figure out the equation...any advice? 2. Originally Posted by jzellt 5, 10, 17, 26, 37, ... I see the pattern, but I can't figure out the equation...any advice? Hello jzellt: You might have noticed that difference between the numbers in sequence corresponds to the odd numbers sequence starting from 5. $5,7,9,11,.....$ Now we can write it as $5+0,5+5,5+5+7,5+5+7+9,...$ We can use this formula for the addition of the odd numbers: $\frac{n}{2}(2a+(n-1)d)$ Lets plug in the values: $\frac{n}{2}(2(5)+(n-1)2)$ $\frac{n-1}{2}(10+(n-2)*2)$ (because at n=1, there should be addition of zero i.e 5+0,5+5,....) Hence we have the equation: $5+\frac{n-1}{2}(8+2(n-1))$ Hope this helps 3. Hello, jzellt! I "eyeballed" the solution . . . $5, 10, 17, 26, 37, \hdots$ Take the differences of consecutive terms. Then take the differences of the differences ... and so on. . . $\begin{array}{cccccccccc}\text{Sequence} & 5 && 10 && 17 && 26 && 37 \\ \text{1st diff.} & & 5 && 7 && 9 && 11 \\ \text{2nd diff.} & & & 2 & & 2 & & 2 \end{array}$ We see that the second differences are constant. This indicates that the generating function is of the second degree, a quadratic. . . That is, $f(n)$ contains $n^2.$ With that in mind, I looked the sequence again, and I saw: . . $\begin{array}{c\|ccc} n & & a_n \\ \hline 1 & 5&=&2^2+1 \\ 2 & 10 &=&3^2+1 \\ 3 & 17 &=&4^2+1 \\ 4 & 26 &=&5^2+1 \\ 5 & 37 &=&6^2+1 \end{array}$ . . . $\begin{array}{c\|c}\vdots & \vdots \\ n & \;\;(n+1)^2+1 \end{array}$ Therefore: . $f(n) \:=\:(n+1)^2+1 \:=\:n^2+2n+2$ . . which verifies u2_wa's result.
# Maharashtra Board 9th Class Maths Part 2 Problem Set 6 Solutions Chapter 6 Circle Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 6 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 6 Circle. ## Problem Set 6 Geometry 9th Std Maths Part 2 Answers Chapter 6 Circle Question 1. Choose correct alternative answer and fill in the blanks. i. Radius of a circle is 10 cm and distance of a chord from the centre is 6 cm. Hence, the length of the chord is ____. (A) 16 cm (B) 8 cm (C) 12 cm (D) 32 cm ∴ OA2 = AC2 + OC2 ∴ 102 = AC2 + 62 ∴ AC2 = 64 ∴ AC = 8 cm ∴ AB = 2(AC)= 16 cm (A) 16 cm ii. The point of concurrence of all angle bisectors of a triangle is called the ____. (A) centroid (B) circumcentre (C) incentre (D) orthocentre (C) incentre iii. The circle which passes through all the vertices of a triangle is called ____. (A) circumcircle (B) incircle (C) congruent circle (D) concentric circle (A) circumcircle iv. Length of a chord of a circle is 24 cm. If distance of the chord from the centre is 5 cm, then the radius of that circle is ____. (A) 12 cm (B) 13 cm (C) 14 cm (D) 15 cm OA2 = AC2 + OC2 ∴ OA2 = 122 + 52 ∴ OA2 = 169 ∴ OA = 13 cm (B) 13 cm v. The length of the longest chord of the circle with radius 2.9 cm is ____. (A) 3.5 cm (B) 7 cm (C) 10 cm (D) 5.8 cm Longest chord of the circle = diameter = 2 x radius = 2 x 2.9 = 5.8 cm (D) 5.8 cm vi. Radius of a circle with centre O is 4 cm. If l(OP) = 4.2 cm, say where point P will lie ____. (A) on the centre (B) inside the circle (C) outside the circle (D) on the circle ∴Point P lies in the exterior of the circle. (C) outside the circle vii. The lengths of parallel chords which are on opposite sides of the centre of a circle are 6 cm and 8 cm. If radius of the circle is 5 cm, then the distance between these chords is _____. (A) 2 cm (B) 1 cm (C) 8 cm (D) 7 cm PQ = 8 cm, MN = 6 cm ∴ AQ = 4 cm, BN = 3 cm ∴ OQ2 = OA2 + AQ2 ∴ 52 = OA2 + 42 ∴ OA2 = 25 – 16 = 9 ∴ OA = 3 cm Also, ON2 = OB2 + BN2 ∴ 52 = OB2 + 32 ∴ OB = 4 cm Now, AB = OA + OB = 3 + 4 = 7 cm Question 2. Construct incircle and circumcircle of an equilateral ADSP with side 7.5 cm. Measure the radii of both the circles and find the ratio of radius of circumcircle to the radius of incircle. Solution: Steps of construction: i. Construct ∆DPS of the given measurement. ii. Draw the perpendicular bisectors of side DP and side PS of the triangle. iii. Name the point of intersection of the perpendicular bisectors as point C. iv. With C as centre and CM as radius, draw a circle which touches all the three sides of the triangle. v. With C as centre and CP as radius, draw a circle which passes through the three vertices of the triangle. Radius of incircle = 2.2 cm and Radius of circumcircle = 4.4 cm Question 3. Construct ∆NTS where NT = 5.7 cm. TS = 7.5 cm and ∠NTS = 110° and draw incircle and circumcircle of it. Solution: Steps of construction: For incircle: i. Construct ∆NTS of the given measurement. ii. Draw the bisectors of ∠T and ∠S. Let these bisectors intersect at point I. iii. Draw a perpendicular IM on side TS. Point M is the foot of the perpendicular. iv. With I as centre and IM as radius, draw a circle which touches all the three sides of the triangle. For circumcircle: i. Draw the perpendicular bisectors of side NT and side TS of the triangle. ii. Name the point of intersection of the perpendicular bisectors as point C. iii. Join seg CN iv. With C as centre and CN as radius, draw a circle which passes through the three vertices of the triangle. Question 4. In the adjoining figure, C is the centre of the circle, seg QT is a diameter, CT = 13, CP = 5. Find the length of chord RS. Given: In a circle with centre C, QT is a diameter, CT = 13 units, CP = 5 units To find: Length of chord RS Construction: Join points R and C. Solution: i. CR = CT= 13 units …..(i) [Radii of the same circle] In ∆CPR, ∠CPR = 90° ∴ CR2 = CP2 + RP2 [Pythagoras theorem] ∴ 132 = 52 + RP2 [From (i)] ∴ 169 = 25 + RP2 [From (i)] ∴ RP2 = 169 – 25 = 144 ∴ RP = $$\sqrt { 144 }$$ [Taking square root on both sides] ∴ RP = 12 cm ….(ii) ii. Now, seg CP _L chord RS [Given] ∴ RP = $$\frac { 1 }{ 2 }$$ RS [Perpendicular drawn from the centre of the circle to the chord bisects the chord.] ∴ 12 = $$\frac { 1 }{ 2 }$$ RS [From (ii)] ∴ RS = 2 x 12 = 24 ∴ The length of chord RS is 24 units. Question 5. In the adjoining figure, P is the centre of the circle. Chord AB and chord CD intersect on the diameter at the point E. If ∠AEP ≅ ∠DEP, then prove that AB = CD. Given: P is the centre of the circle. Chord AB and chord CD intersect on the diameter at the point E. ∠AEP ≅ ∠DEP To prove: AB = CD Construction: Draw seg PM ⊥ chord AB, A-M-B seg PN ⊥ chord CD, C-N-D Proof: ∠AEP ≅ ∠DEP [Given] ∴ Seg ES is the bisector of ∠AED. PoInt P is on the bisector of ∠AED. ∴ PM = PN [Every point on the bisector of an angle is equidistant from the sides of the angle.] ∴ chord AB ≅ chord CD [Chords which are equidistant from the centre are congruent.] ∴ AB = CD [Length of congruent segments] Question 6. In the adjoining figure, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that ∆ABC is an isosceles triangle. Given: O is the centre of the circle. diameter CD ⊥ chord AB, A-E-B To prove: ∆ABC is an isosceles triangle. Proof: diameter CD ⊥ chord AB [Given] ∴ seg OE ⊥ chord AB [C-O-E, O-E-D] ∴ seg AE ≅ seg BE ……(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord] In ∆CEA and ∆CEB, ∠CEA ≅ ∠CEB [Each is of 90°] seg AE ≅ seg BE [From (i)] seg CE ≅ seg CE [Common side] ∴ ∆CEA ≅ ∆CEB [SAS test] ∴ seg AC ≅ seg BC [c. s. c. t.] ∴ ∆ABC is an isosceles triangle. Maharashtra Board Class 9 Maths Chapter 6 Circle Problem Set 6 Intext Questions and Activities Question 1. Every student in the group should do this activity. Draw a circle in your notebook. Draw any chord of that circle. Draw perpendicular to the chord through the centre of the circle. Measure the lengths of the two parts of the chord. Group leader should prepare a table as shown below and ask other students to write their observations in it. Write the property which you have observed. (Textbook pg. no. 77) On completing the above table, you will observe that the perpendicular drawn from the centre of a circle on its chord bisects the chord. Question 2. Every student from the group should do this activity. Draw a circle in your notebook. Draw a chord of the circle. Join the midpoint of the chord and centre of the circle. Measure the angles made by the segment with the chord. Discuss about the measures of the angles with your friends. Which property do the observations suggest ? (Textbook pg. no. 77) The meausure of the angles made by the drawn segment with the chord is 90°. Thus, we can conclude that, the segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord. Question 3. Draw circles of convenient radii. Draw two equal chords in each circle. Draw perpendicular to each chord from the centre. Measure the distance of each chord from the centre. What do you observe? (Textbook pg. no. 79) Congruent chords of a circle are equidistant from the centre. Question 4. Measure the lengths of the perpendiculars on chords in the following figures. Did you find OL = OM in fig (i), PN = PT in fig (ii) and MA = MB in fig (iii)? Write the property which you have noticed from this activity. (Textbook pg. no. 80)
# Can you solve an expression In addition, Can you solve an expression can also help you to check your homework. Math can be a challenging subject for many students. ## The Best Can you solve an expression There are a lot of Can you solve an expression that are available online. The binomial solver can be used to solve linear equations, quadratic equations, and polynomial equations. The binomial solver is a versatile tool that can be used to solve many different types of equations. The binomial solver is a useful tool for solving equations that contain two variables. 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. A logarithmic equation solver is a tool that can be used to solve equations with Logarithms. Logarithmic equations often arise in settings where one is working with exponential functions. For example, if one were to take the natural log of both sides of the equation y = 2x, they would obtain the following equation: Log(y) = Log(2x). This equation can be difficult to solve without the use of a logarithmic equation solver. A logarithmic equation solver can be used to determine the value of x that satisfies this equation. In this way, a logarithmic equation solver can be a valuable tool for solving equations with Logarithms. The common factors of 3 and 4 are 1 and 3, so we can cancel out the 3 in both the numerator and denominator, leaving us with the simplified fraction 1/4. In general, it's helpful to start by finding any common factors in the numerator and denominator that are larger than 1. Once you've cancelled out as many factors as possible, you can then multiply both the numerator and denominator by any remaining factors in order to further simplify the fraction. Just be careful not to cancel out any essential parts of the fraction (like 2 in ¾). If you do, you'll end up with an incorrect answer! ## We cover all types of math issues This application is so good you should download it for your paper and if you are a student then what are you waiting for download it for the best examples of math and matrix papers students should download it. Odette Watson It's a good app. It really gives proper solutions and answers. You can input the numbers yourself and can also use the camera to scan you can also edit the scanned math problem. Very nice and helpful app thanks the app now I can solve all math equation. Thank you Haniya Lewis Helpful math apps Solve by completing the square calculator mathpapa Precalc help Input output rule solver Solving math gif
# Understanding Cube Roots Of 4: Definition, Calculation, And Properties // Thomas Gain a comprehensive understanding of the cube root of 4, including its , methods, , prime factorization, and simplification techniques. ## Understanding Cube Roots ### Definition of Cube Roots Have you ever wondered what cube roots are and how they work? Well, let’s dive in and explore! In mathematics, cube roots are a type of radical expression that involve finding the number which, when multiplied by itself three times, gives a given value. For example, the cube root of 8 is 2, because 2 multiplied by itself three times equals 8. Cube roots are denoted by the symbol ∛. ### How to Calculate Cube Roots Now that we understand what cube roots are, let’s learn how to calculate them. There are different methods to find the cube root of a number, but one common approach is using prime factorization. To do this, we break down the number into its prime factors and group them in sets of three. Then, we take one factor from each group and multiply them together. The resulting product is the cube root of the original number. For example, let’s calculate the cube root of 27. The prime factorization of 27 is 3 x 3 x 3. Grouping them in sets of three, we have (3 x 3) x (3 x 3) x (3 x 3). Taking one factor from each group, we get 3 x 3 x 3 = 27. Therefore, the cube root of 27 is 3. ### Properties of Cube Roots Cube roots have some interesting worth exploring. Here are a few of them: 1. Multiplication Property: The cube root of a product is equal to the product of the cube roots of the individual factors. In mathematical terms, ∛(a * b) = ∛a * ∛b. 2. Division Property: The cube root of a quotient is equal to the quotient of the cube roots of the numerator and denominator. In other words, ∛(a / b) = ∛a / ∛b. 3. Power Property: The cube root of a number raised to a power is equal to the number raised to the same power divided by 3. For example, ∛(a^3) = a. 4. Negative Numbers: Cube roots can also be calculated for negative numbers. The cube root of a negative number is equal to the negative of the cube root of its absolute value. For example, ∛(-8) = -2. Understanding these can help us calculations and solve problems involving cube roots more efficiently. Now that we have a solid understanding of cube roots, let’s move on to exploring the specific case of simplifying cube roots of 4. ## Simplifying Cube Roots of 4 ### Prime Factorization of 4 To simplify the cube root of 4, it is important to first understand the concept of prime factorization. Prime factorization involves breaking down a number into its prime factors, which are the smallest prime numbers that can divide the original number evenly. In the case of 4, the prime factorization is relatively simple. The number 4 can be expressed as 2 multiplied by 2, or 2^2. Since 2 is a prime number, we can write the prime factorization of 4 as 2^2. ### How to Simplify Cube Roots of 4 Now that we know the prime factorization of 4, we can simplify the cube root of 4 using this information. The cube root of a number represents the value that, when multiplied by itself three times, gives the original number. In the case of 4, the cube root can be simplified as follows: ∛4 = ∛(2^2) Since the cube root is asking for the number that, when multiplied by itself three times, gives 4, we can rewrite 4 as (2^2)^3: ∛4 = ∛((2^2)^3) Using the property of exponents, we can further: ∛4 = ∛(2^(2*3)) ∛4 = ∛(2^6) Now, the cube root of 4 can be expressed as the cube root of 2 raised to the power of 6. This simplifies to: ∛4 = 2^(6/3) ∛4 = 2^2 Therefore, the cube root of 4 is equal to 2. By understanding the of 4 and applying the concept of cube roots, we can easily the cube root of 4 to its simplest form. This knowledge can be helpful in various mathematical calculations and applications. ## Cube Root of 4 in Mathematics ### Cube Root of 4 as an Irrational Number The cube root of 4 is an interesting mathematical concept that falls under the category of irrational numbers. An irrational number is a number that cannot be expressed as a simple fraction and has an infinite number of non-repeating decimal places. In the case of the cube root of 4, it is an irrational number because it cannot be represented as a fraction or a terminating decimal. ### Decimal Representation of Cube Root of 4 When we calculate the cube root of 4, we find that it is approximately equal to 1.5874010519681994. This decimal representation is an approximation of the actual value and goes on indefinitely without repeating. It is important to note that the decimal representation of the cube root of 4 is an irrational number, meaning it cannot be expressed as a simple fraction or a finite decimal. Understanding the cube root of 4 as an irrational number and its decimal representation can be useful in various mathematical calculations and applications. It allows us to work with numbers that cannot be expressed precisely, and helps in solving equations and problems where the cube root of 4 is involved. In the next sections, we will explore the applications of the cube root of 4 in different fields, such as engineering and science, as well as its relevance in real-life scenarios. ## Applications of Cube Root of 4 ### Engineering Applications The cube root of 4 finds its applications in various engineering fields. Here are a few examples: 1. Structural Design: When designing structures such as bridges, buildings, or dams, engineers need to consider the cube root of 4 to determine the appropriate dimensions. This is crucial for ensuring the stability and safety of the structure. 2. Fluid Mechanics: In fluid mechanics, the cube root of 4 plays a role in calculating the flow rate of fluids through pipes or channels. Engineers use this information to design efficient systems for transporting liquids or gases. 3. Electrical Engineering: The cube root of 4 is relevant in electrical engineering when determining the voltage or current required for specific applications. It helps engineers in designing circuits and systems that meet the desired specifications. ### Scientific Applications Apart from engineering, the cube root of 4 also has scientific applications. Here are a few areas where it is utilized: 1. Physics: In certain physical calculations and equations, the cube root of 4 may arise. For example, when studying the behavior of particles or analyzing the of materials, scientists might encounter the cube root of 4 as a mathematical component. 2. Mathematical Modeling: Scientists often use mathematical models to simulate real-world phenomena. In some cases, the cube root of 4 may appear as a parameter in these models, helping scientists understand complex systems and make predictions. 3. Statistical Analysis: In statistical analysis, researchers sometimes need to manipulate data or perform transformations to achieve better results. The cube root of 4 can be applied to data sets to normalize them or reduce skewness, making them more suitable for analysis. By understanding and applying the cube root of 4 in engineering and scientific contexts, professionals in these fields can make accurate calculations, design efficient systems, and gain deeper insights into various phenomena. ## Cube Root of 4 in Real Life ### Examples of Objects with Volume 4 Have you ever wondered what objects in real life have a volume of 4? Well, let’s explore some examples. One such object is a Rubik’s Cube. This popular puzzle toy has a volume of 4 cubic units. Imagine holding a Rubik’s Cube in your hands and feeling its compactness. Another example is a small jewelry box. These boxes often have dimensions that result in a volume of 4 cubic units. Picture a beautiful box, just the right size to hold your favorite earrings or necklace. ### Practical Use of Cube Root of 4 You might be thinking, “Why is the cube root of 4 important in real life?” Well, let’s delve into its practical uses. One application is in the field of architecture. Architects use the cube root of 4 to determine the dimensions of small-scale models and prototypes. By understanding the cube root of 4, architects can accurately represent the volumes of their designs in miniature form. Another practical use of the cube root of 4 is in the field of cooking. Imagine you have a recipe that calls for a specific volume of an ingredient, and that volume is 4 cubic units. By knowing the cube root of 4, you can easily measure out the correct amount and ensure your dish turns out just right. In addition, the cube root of 4 has significance in the world of mathematics and geometry. It helps us understand the relationship between volume and dimensions, and it plays a role in various mathematical calculations. In summary, the cube root of 4 has both practical and mathematical significance in our everyday lives. From objects with a volume of 4 to its application in architecture and cooking, understanding the cube root of 4 can enhance our understanding of the world around us. Contact 3418 Emily Drive Charlotte, SC 28217 +1 803-820-9654
# How is 40percentage calculated? So, 40 percent would be 40 divided by 100. Once you have the decimal version of your percentage, simply multiply it by the given number (in this case, the amount of your paycheck). If your paycheck is \$750, you would multiply 750 by . 40. ## How do you calculate 40% of a total? To calculate the percentage, multiply this fraction by 100 and add a percent sign. 100 * numerator / denominator = percentage . In our example it’s 100 * 2/5 = 100 * 0.4 = 40 . Forty percent of the group are girls. ## What is 30% as a number? Once you have the decimal figure, multiply it by the number for which you seek to calculate the percentage; i.e., if you need to know 30 percent of 100, you convert 30 percent to a decimal (0.30) and multiply it by 100 (0.30 x 100, which equals 30). ## How do you find 20% of a number? First, convert the percentage discount to a decimal. A 20 percent discount is 0.20 in decimal format. Secondly, multiply the decimal discount by the price of the item to determine the savings in dollars. For example, if the original price of the item equals \$24, you would multiply 0.2 by \$24 to get \$4.80. ## How do you find 20% of an amount? As finding 10% of a number means to divide by 10, it is common to think that to find 20% of a number you should divide by 20 etc. Remember, to find 10% of a number means dividing by 10 because 10 goes into 100 ten times. Therefore, to find 20% of a number, divide by 5 because 20 goes into 100 five times. ## What percent is 3 out of 18? Now we can see that our fraction is 16.666666666667/100, which means that 3/18 as a percentage is 16.6667%. ## What does 20 percent off mean? A percent off of a product means that the price of the product is reduced by that percent. For example, given a product that costs \$279, 20% off of that product would mean subtracting 20% of the original price from the original price. For example: 20% of \$279 = 0.20 × 279 = \$55.80. ## How do you calculate 20% off in Excel? If you want to calculate a percentage of a number in Excel, simply multiply the percentage value by the number that you want the percentage of. For example, if you want to calculate 20% of 500, multiply 20% by 500. ## How do you take 20% off a price UK? To work out a price excluding the standard rate of VAT (20%) divide the price including VAT by 1.2. To work out a price excluding the reduced rate of VAT (5%) divide the price including VAT by 1.05. ## What percentage is 5 out of 60? Now we can see that our fraction is 8.3333333333333/100, which means that 5/60 as a percentage is 8.3333%. ## How do you find 60 percent of a number? To convert the fraction 60/100 to a percentage, you should first convert 60/100 to a decimal by dividing the numerator 60 by the denominator 100. This implies that 60/100 = 0.6. Then, multiply 0.6 by 100 = 60%. ## What percent is 80 out of 90? Now we can see that our fraction is 88.888888888889/100, which means that 80/90 as a percentage is 88.8889%. ## What is 3/8 as a percent? To convert to percentage To convert the fraction to decimal first convert to decimal and then multiply by 100. Therefore, the solution is 37.5%. ## What is a 25 out of 50 grade? Now we can see that our fraction is 50/100, which means that 25/50 as a percentage is 50%. And there you have it! Two different ways to convert 25/50 to a percentage. ## What percent is 5 out of 22? Now we can see that our fraction is 22.727272727273/100, which means that 5/22 as a percentage is 22.7273%.
Courses Courses for Kids Free study material Offline Centres More Store # If $\cos \left( {2{{\sin }^{ - 1}}x} \right) = \dfrac{1}{9}$, then $x$ is equal toA) Only $\dfrac{2}{3}$B) Only $- \dfrac{2}{3}$C) $\dfrac{2}{3}, - \dfrac{2}{3}$D) $\dfrac{1}{3}$ Last updated date: 12th Aug 2024 Total views: 410.7k Views today: 11.10k Verified 410.7k+ views Hint: First, assume ${\sin ^{ - 1}}x$ as another variable. Then use the fact that $\cos 2A = 1 - 2{\sin ^2}A$. Use the fact to convert the term on the left side in terms of the sign. After that, move the constant part to the other side and simplify them and cancel out the negative sign from both sides. Then take the square root. After that substitute back ${\sin ^{ - 1}}x$ in place of the variable and use the property $\sin \left( {{{\sin }^{ - 1}}x} \right) = x$ to remove the sine part to get the value of x. Complete step-by-step solution: Given:- $\cos \left( {2{{\sin }^{ - 1}}x} \right) = \dfrac{1}{9}$ The inverse trigonometric functions perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. We know that trig functions are especially applicable to the right-angle triangle. Let us assume $y = {\sin ^{ - 1}}x$. Then substitute in the expression, $\Rightarrow \cos 2y = \dfrac{1}{9}$ As we know that, $\cos 2A = 1 - 2{\sin ^2}A$ So, expand $\cos 2y$ using this formula, $\Rightarrow 1 - 2{\sin ^2}y = \dfrac{1}{9}$ Move the constant part on one side, $\Rightarrow - 2{\sin ^2}y = \dfrac{1}{9} - 1$ Take LCM on the right side, $\Rightarrow - 2{\sin ^2}y = \dfrac{{1 - 9}}{9}$ Subtract the value in the numerator, $\Rightarrow - 2{\sin ^2}y = - \dfrac{8}{9}$ Now, divide both sides by -2, $\Rightarrow {\sin ^2}y = \dfrac{4}{9}$ Take the square root on both sides, $\Rightarrow \sin y = \sqrt {\dfrac{4}{9}}$ As we know, the square root returns both positive and negative values. So, $\Rightarrow \sin y = \pm \dfrac{2}{3}$ Substitute back the value of y, $\Rightarrow \sin \left( {{{\sin }^{ - 1}}x} \right) = \pm \dfrac{2}{3}$ We know that, $\sin \left( {{{\sin }^{ - 1}}x} \right) = x$. Use this formula in the above equation, $\therefore x = \pm \dfrac{2}{3}$ Thus, the value of x is $\pm \dfrac{2}{3}$. Hence, option (C) is the correct answer. Note: Whenever such a type of question appears, then always try to write either sine in terms of cosine or vice- versa, so that the whole equation converts into similar terms, which is easier to solve. In the solution, cosine is written in terms of sine.
# How do you find the derivative of sqrt(x ln(x^4))? Mar 23, 2018 $\frac{\ln \left({x}^{4}\right) + 4}{2 \sqrt{x \ln \left({x}^{4}\right)}}$ #### Explanation: Let's rewrite it as: $\left[{\left(x \ln \left({x}^{4}\right)\right)}^{\frac{1}{2}}\right] '$ Now we have to derivate from the outside to the inside using the chain rule. $\frac{1}{2} {\left[x \ln \left({x}^{4}\right)\right]}^{- \frac{1}{2}} \cdot \left[x \ln \left({x}^{4}\right)\right] '$ Here we got a derivative of a product $\frac{1}{2} {\left(x \ln \left({x}^{4}\right)\right)}^{- \frac{1}{2}} \cdot \left[\left(x '\right) \ln \left({x}^{4}\right) + x \left(\ln \left({x}^{4}\right)\right) '\right]$ $\frac{1}{2} {\left(x \ln \left({x}^{4}\right)\right)}^{- \frac{1}{2}} \cdot \left[1 \cdot \ln \left({x}^{4}\right) + x \left(\frac{1}{x} ^ 4 \cdot 4 {x}^{3}\right)\right]$ Just using basic algebra to get a semplified version: $\frac{1}{2} {\left(x \ln \left({x}^{4}\right)\right)}^{- \frac{1}{2}} \cdot \left[\ln \left({x}^{4}\right) + 4\right]$ And we get the solution: $\frac{\ln \left({x}^{4}\right) + 4}{2 \sqrt{x \ln \left({x}^{4}\right)}}$ By the way you can even rewrite the inital problem to make it more simple: $\sqrt{4 x \ln \left(x\right)}$ $\setminus \sqrt{4} \sqrt{x \ln \left(x\right)}$ $2 \sqrt{x \ln \left(x\right)}$
### Theory: Let's see how to construct a quadrilateral when two adjacent sides and three angles are given. Example: Construct a quadrilateral $$CAKE$$ where $$CA = 7 \ cm$$, $$AK = 6.5 \ cm$$, $$\angle C = 90^\circ$$, $$\angle K = 110^\circ$$, $$\angle E = 100^\circ$$. Let us first draw the rough quadrilateral, which will help us to draw a fair quadrilateral. Rough diagram: Fair diagram: Step 1: Draw a line segment $$CA = 7 \ cm$$. Step 2: Construct $$\angle ACX = 90^\circ$$ using ruler and compass. Step 3: We know that sum of all the angles of a quadrilateral is $$360^\circ$$. $$\angle C + \angle A + \angle K + \angle E = 360^\circ$$ $$90^\circ + \angle A + 110^\circ + 100^\circ = 360^\circ$$ So, $$\angle A = 60^\circ$$ Construct $$\angle CAY = 60^\circ$$ using ruler and compass. Step 4: With $$A$$ as centre and $$6.5 \ cm$$ as radius, draw an arc, cutting $$AY$$ at $$K$$. Step 5: With $$K$$ as centre, make an angle of $$110^\circ$$, cutting $$CX$$ at $$E$$, then join $$EK$$. Thus, the quadrilateral $$CAKE$$ is obtained.
## อภิธานศัพท์ เลือกหนึ่งในคำหลักทางด้านซ้าย ... # Polygons and PolyhedraPolygons เวลาอ่านหนังสือ: ~30 min A polygon is a closed, flat shape that has only straight sides. Polygons can have any number of sides and angles, but the sides cannot be curved. Which of the shapes below are polygons? We give different names to polygons, depending on how many sides they have: Triangle 3 sides 4 sides Pentagon 5 sides Hexagon 6 sides Heptagon 7 sides Octagon 8 sides ## Angles in Polygons Every polygon with n sides also has n internal angles. We already know that the sum of the internal angles in a triangle is always ° but what about other polygons? \${a1[0]}° + \${a1[1]}° + \${a1[2]}° + \${360-a1[0]-a1[1]-a1[2]}° = \${a2[0]}° + \${a2[1]}° + \${a2[2]}° + \${a2[3]}° + \${540-a2[0]-a2[1]-a2[2]-a2[3]}° = It looks like the sum of internal angles in a quadrilateral is always ° – exactly the sum of angles in a triangle. This is no coincidence: every quadrilateral can be split into two triangles. The same also works for larger polygons. We can split a pentagon into triangles, so its internal angle sum is 3×180°= °. And we can split a hexagon into triangles, so its internal angle sum is 4×180°= °. A polygon with \${x} sides will have an internal angle sum of 180° × \${x-2} = \${(x-2)*180}°. More generally, a polygon with n sides can be split into triangles. Therefore, Sum of internal angles in an n-gon =n2×180°. ## Convex and Concave Polygons We say that a polygon is concave if it has a section that “points inwards”. You can imagine that this part has “caved in”. Polygons that are not concave are called convex. There are two ways you can easily identify concave polygons: they have at least one internal angle that is bigger than 180°. They also have at least one diagonal that lies outside the polygon. In convex polygons, on the other hand, all internal angles are less than °, and all diagonals lie the polygon. Which of these polygons are concave? ## Regular Polygons We say that a polygon is regular if all of its sides have the same length, and all of the angles have the same size. Which of these shapes are regular polygons? Regular polygons can come in many different sizes – but all regular polygons with the same number of sides ! We already know the sum of all internal angles in polygons. For regular polygons all these angles have , so we can work out the size of a single internal angle: angle= = 180°×x2x=180°360°x. If n=3 we get the size of the internal angles of an equilateral triangle – we already know that it must be °. In a regular polygon with \${x} sides, every internal angle is 180° – 360°\${x} = \${round(180-360/x)}°. ## The Area of Regular Polygons Here you can see a regular polygon with \${n} sides. Every side has length 1m. Let’s try to calculate its area! First, we can split the polygon into \${toWord(n)} congruent, triangles. We already know the of these triangles, but we also need the to be able to calculate its area. In regular polygons, this height is sometimes called the apothem. Notice that there is a right angled triangle formed by the apothem and half the base of the isosceles triangle. This means that we can use trigonometry! The base angles of the isosceles triangle (let’s call them α) are size of the internal angles of the polygon: α=12180°360°\${n}=\${round(90-180/n,2)} To find the apothem, we can use the definition of the function:
# What are the 4 squares called on a graph? ## What are the 4 squares called on a graph? A scatter-plot graph is divided into four quadrants due to the (0, 0) intersection point of the horizontal axis (x-axis) and vertical axis (y-axis). This intersection point is called the origin. ### What are the squares in graphs called? In general, graphs showing grids are sometimes called Cartesian graphs because the square can be used to map measurements onto a Cartesian (x vs. y) coordinate system. It is also available without lines but with dots at the positions where the lines would intersect. #### How many squares are in a 4×4 graph paper? This volume of Graph Paper 4×4 contains 100 pages printed with four squares per inch, each square measuring . 25″ x . 25″. It is frequently used for math or science for younger children. How do you find the number of squares? To get the total number of squares we need to find all the squares formed. 1. x 1: 8 * 8 = 64 squares. 2. x 2: 7 * 7 = 49 squares. 3. x 3: 6 * 6 = 36 squares. 4. x 4: 5 * 5 = 25 squares. 5. x 5: 4 * 4 = 16 squares. 6. x 6: 3 * 3 = 9 squares. 7. x 7: 2 * 2 = 4 squares. 8. x 8: 1 * 1 = 1 square. What is 4×4 vs 5×5 graph paper? 4×4 means 4 squares per inch and 5×5 means 5 squares per inch. 1 of 1 found this helpful. ## What are the 4 graphs? The four most common are probably line graphs, bar graphs and histograms, pie charts, and Cartesian graphs. They are generally used for, and are best for, quite different things. You would use: Bar graphs to show numbers that are independent of each other. ### How big are the squares on graph paper? Printable Grid Paper – Graphing Paper It is often called “quadrille paper” or “quad paper.” The squares usually have a specific size such as 1/2 inch, 1/4 inch or 1/8 inch – which gives the paper a name such as “1/2 Inch Grid Paper.” No axes are drawn on grid paper. #### How many types of graph paper are there? Let us know, if there is a type of graph paper you would like to see here. There are just over 100 different forms found here. How many squares are in a 7×7 grid? Rectangles Board Size # Squares # Rectangles 7×7 140 784 8×8 204 1,296 9×9 285 2,025 10×10 385 3,025 What is 5×5 graph paper? Pen Graph Paper, 5×5 (5 Squares per inch), 11″x8. ## How many squares are there in 5×5 grid? A 5×5 grid is made up of 25 individual squares, which can be combined to form rectangles. ### How many squares grids are there in the chess? 204 There are many more different-sized squares on the chessboard. Therefore, there are actually 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 squares on a chessboard! (in total 204).
# Division using Napier’s Bones On the previous post, we’ve learn the fun way of using Napier’s Bones to solve multiplication. Now, we will try to use Napier’s Bones to solve division problems. The process might be a little bit tedious for division problems involving one or two-digit divisors (I think long division might be faster for division problems involving one or two-digit divisors) but would be handy for problems involving large division. Let’s start with a two-digit divisor, so as we can clearly follow the process. Take for example: 8820 ÷ 98 To begin, you have to lay down the Napier Rods for 9 and 8 Next, you have to get the sums of each diagonal for each row We will be using our Napier’s Bones table to operate the long division From the Napier’s table of sums, we will use the sum that we can subtract to the first two digit of the dividend, that is 88. But, apparently, the smallest sum in the table is 98. We cannot subtract 98 to 88 and get a positive difference. We will then have to work on the first 3-digits of the dividend, that is 882. From the Napier’s table of sums, we’ve got 882 as the closest sum that we can subtract to 882, which in turn is in row 9 of the Napier’s Grid Subtracting, we get zero. Moving on, we bring down the next digit of the dividend, that is 0 We don’t have zero on the Napier’s Grid, we then use the value 0. Looking at steps, we get our respective quotient, that is 90. ### Example 2 Let’s take another example, this time a 3-digit divisor. 57268 ÷ 278 First, we lay down the Napier Rods for 2, 7 and 8 From the Napier’s table of sums, we will use the sum that we can subtract to the first two digit of the dividend, that is 572. From the sums, we’ve got 556 as the closest sum that we can subtract to 572, which in turn is in row 2 of the Napier’s Grid. Subtracting, we get 16. Moving on, we bring down the next digit of the dividend, that is 6. We don’t have a value on the Napier’s Grid that we can subtract to 166, we then use the value 0. Subtracting, we get 166. Moving on, we bring down the next digit of the dividend, that is 8. We’ve got 1668 as the closest sum that we can subtract to 1668, which in turn is in row 6 of the Napier’s Grid Looking at steps, we get our respective quotient, that is 206
# Expected Value How to find the expected value from the given probability condition: formula, examples, and their solutions. ## Formula The expected value literally means the expected value after an event. To find the expected value E(X), the product of the outcome (xi) and the probability (pi) for each case. So E(X) = x1p1 + x2p2 + x3p3 + ... + xnpn. Sigma notation ## Example 1 You get 3 points. So x1 = +3. The probability of getting a head is 1/2. So p1 = 1/2. Probability Case 2: Tail You lose 2 points. So x2 = -2. The probability of getting a tail is 1/2. So p2 = 1/2. x1 = +3, p1 = 1/2 x2 = -2, p2 = 1/2 So E(X) = 3⋅(1/2) + (-2)⋅(1/2). The numerators of both fractions are 2. So combine the fractions. 3 - 2 = 1 So E(X) = 1/2. E(X) = 1/2 means if a coin is tossed once, you expect to get 1/2 points. ## Example 2 For each toss, there are two outcomes: The coin is tossed 2 times. And each toss is an independent event. So (total) = 2⋅2 = 4 ways. These 4 ways are [H, H], [H, T], [T, H], and [T, T]. For each case, find the outcome xi and the probability pi. So x1 = 0. This is when [T, T] happens. There's 1 way to happen this. (total) = 4 ways So p1 = 1/4. So x2 = 1. This is when [H, T] or [T, H] happens. There are 2 ways to happen this. (total) = 4 ways So p2 = 2/4. So x3 = 2. This is when [H, H] happens. There's 1 way to happen this. (total) = 4 ways So p3 = 1/4. x1 = 0, p1 = 1/4 x2 = 1, p2 = 2/4 x3 = 2, p3 = 1/4 So E(X) = 0⋅(1/4) + 1⋅(2/4) + 2⋅(1/4). 0⋅(1/4) = 0 +1⋅(2/4) = +2/4 +2⋅(1/4) = +2/4 2/4 + 2/4 = 4/4 4/4 = 1 So E(X) = 1. E(X) = 1 means if a coin is tossed 2 times, you expect to see 1 head. ## Example 3 The total area of the dart, Atot, is Atot = π⋅32. So Atot = 9π. Area of a circle Case 1: 30 points So x1 = 30. The area of the blue part, A1, is A1 = π⋅12. So A1 = π. A1 = π Atot = 9π So p1 = π/9π. So p1 = 1/9. Case 2: 20 points So x2 = 20. The area of the green part, A2, is, the circle with r = 2, π⋅22 minus, the inner circle with r = 1, π⋅12. So A2 = 3π. A2 = 3π Atot = 9π So p2 = 3π/9π. So p2 = 3/9. Case 3: 10 points So x3 = 10. The area of the brown part, A3, is, the circle with r = 3, π⋅32 minus, the inner circle with r = 2, π⋅22. So A3 = 5π. A3 = 5π Atot = 9π So p3 = 5π/9π. So p3 = 5/9. x1 = 30, p1 = 1/9 x2 = 20, p2 = 3/9 x3 = 10, p3 = 5/9 So E(X) = 30⋅(1/9) + 20⋅(3/9) + 10⋅(5/9). The numerators of the fractions are all 9. So combine the fractions. 30 + 60 + 50 = 140 So E(X) = 140/9. E(X) = 140/9 means if you throw a dart once, you expect to get 140/9 points.
#### Please solve RD Sharma class 12 chapter Functions exercise 2.2 question 1 sub question (v) maths textbook solution. Answer : $f\; o\; g = 9x^{2}-18x+5$$f\; o\; g = 9x^{2}-18x+5\; \text {and}\; g\; o\; f=3x^{2}+6x-13$ Hint : $g\; o\; f$ means $f(x)$ function is in $g(x)$ function $f\; o\; g$ means $g(x)$ function is in $f(x)$ function Given : $f(x)=x^{2}+2x-3$ $g(x)= 3x-4$ Solution : Since $f:R\rightarrow R \text {and}g:R\rightarrow R$ $f\; o\; g:R\rightarrow R \; \text {and}\; g\; o\; f:R\rightarrow R$ Now, $g\; o\; f (x)=g(f(x))=g(x)$ $g\; o\; f (x)=g(f(x))=g\left ( x^{2}+2x -3\right )$ $g\; o\; f (x)= 3\left ( x^{2}+2x -3\right )-4$ $g\; o\; f (x)= 3x^{2}+6x -9-4$ $g\; o\; f (x)= 3x^{2}+6x -13$ $f\; o\; g (x)= f(g(x))=f(x)$ $f\; o\; g (x)= f(g(x))=f \left ( 3x-4 \right )$ $f\; o\; g (x)= \left ( 3x-4 \right )^{2}+2\left ( 3x-4 \right )-3$ $=9x^{2}+16-24x+6x-8-3$ $\because \left [ \left ( a-b \right )^{2}=a^{2}+b^{2} -2ab\right ]$ $f\; o\; g (x)=9x^{2}-18x+5$ Hence, $f\; o\; g = 9x^{2}-18x+5\; \text {and}\; g\; o\; f=3x^{2}+6x-13$
# RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21C In this chapter, we provide RS Aggarwal Solutions for Class 7 Chapter 21 Collection and Organisation of Data Ex 21C for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 7 Chapter 21 Collection and Organisation of Data Ex 21C Maths pdf, free RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21C Maths book pdf download. Now you will get step by step solution to each question. ### RS Aggarwal Solutions for Class 7 Chapter 21 Collection and Organisation of Data Ex 21C Download PDF Question 1. Solution: (i) Arranging in ascending order : 4, 6, 7, 8, 8, 8, 8, 10, 11, 15 We see that 8 occurs maximum times Mode = 8 (ii) Arranging in ascending order : 18, 21, 23, 27, 27, 27, 27, 27, 36, 39, 40 We see that 27 occurs maximum times Mode = 27 Question 2. Solution: Arranging in ascending order : 28, 31, 32, 32, 32, 32, 34, 36, 38, 40, 41. We see that 32 occurs maximum times Mode = 32 years Question 3. Solution: We prepare the table as given below: Here, number of terms = 45, which is odd Median = n+12 th term = 45+12 = 462 th term = 23th term = 450 Now, mode = 3(median) – 2(mean) = 3 x 450 – 2 x 470 = 1350 – 940 = 410 Question 4. Solution: We prepare the table as given below: Here, number of terms (N) = 41, which is odd Median = n+12 th term = 41+12 th term = 422 = 21 th term = 22 {value of 18 to 29 = 22} Mode = 3 (median) – 2 (mean) = 3 x 22 – 2 x 21.92 = 66 – 43.84 = 22.16 Question 5. Solution: We prepare the table as given below: All Chapter RS Aggarwal Solutions For Class 7 Maths —————————————————————————– All Subject NCERT Exemplar Problems Solutions For Class 7 All Subject NCERT Solutions For Class 7 ************************************************* I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam. If these solutions have helped you, you can also share rsaggarwalsolutions.in to your friends.
Q: # A college is currently accepting students that are both in-state and out-of-state. They plan to accept three times as many in-state students as out-of-state, and they only have space to accept 100 out-of-state students. Let x = the number of out-of-state students and y = the number in-state students. Write the constraints to represent the incoming students at the college. Accepted Solution A: Answer:$$0<x\leq 100\ and\ 0<y\leq 300$$Step-by-step explanation:Given:Maximum number of out-of-state students, $$x=100$$Maximum number of in-state students, $$y=3x$$Therefore, the range of out-of-state students can be anywhere between a number greater than 0 but less than or equal to 100 as this is maximum limit.$$0<x\leq 100$$In order to get the interval for $$y$$, we multiply the above by 3 throughout.$$3\times 0< 3\times x\leq 3\times 100\\0 < 3x\leq 300$$But $$3x = y$$. Therefore,$$0< y\leq 300$$
Está en la página 1de 16 # Form 2 [CHAPTER 4: FRACTION AND DECIMALS] Chapter 4: Fractions and Decimals 4.1 Understanding Fractions (Revision of equivalent fractions) A fraction describes part of a whole. Each fraction consists of a denominator (bottom) and a numerator (top), representing (respectively) the number of equal parts that an object is divided into. Numerator Denominator ## Are the following equivalent fraction? (i) 2 8 , 5 20 3 6 , 2 3 5 25 , 6 30 (ii) (iii) C.Camenzuli \| www.smcmaths.webs.com 1 ## Form 2 [CHAPTER 4: FRACTION AND DECIMALS] 4.2 " Adding and subtracting fractions and mixed numbers Example 1: Find: 2 1 + 3 5 Step 1: Find the lowest common multiple of both denominators. The multiples of 3 are 3, 6, 9, 12, 15, The multiples of 5 are 5, 10, 15, 15 is the lowest common multiple (LCM) Step 2: Change the denominators to make them both equal to the LCM by using equivalent fractions. ## 2 10 (Multiplying top and bottom by 3) = 3 15 1 3 (Multiplying top and bottom by 3) = 5 15 Step 3: Add or subtract both fractions with the same denominator. So 2 1 10 3 13 + = + = 3 5 15 15 15 ## Exercise 1: Work out the following: (i) 5 3 ! 6 4 (ii) 2 1 ! 3 5 C.Camenzuli \| www.smcmaths.webs.com 2 1 1 + 3 6 (iv) 3 1 + 4 3 ## Adding and Subtracting Mixed Numbers Example 2: Work out: 2 + 5 Step 1: Add the whole numbers: 2+5=7 Step 2: Work out 1 3 1 2 1 1 + 3 2 C.Camenzuli \| www.smcmaths.webs.com 3 1 2 7 !3 4 3 (ii) 3 6 2 +5 4 7 ## 4.3 Ordering fractions Example 1: Which fraction is bigger? a) 1 2 or 3 5 Step1: Find the LCM The multiples of 3 are 3, 6, 9, 12, 15, The multiples of 5 are 5, 10, 15, 15 is the lowest common multiple (LCM) C.Camenzuli \| www.smcmaths.webs.com 4 Form 2 [CHAPTER 4: FRACTION AND DECIMALS] Step 2: Multiply top and bottom so as to get the same denominator (Equivalent fractions). !!!!!! " !!!!!! ! " ## 1 5 2 6 = and = 5 15 3 15 !3 ! !5 " !!!!!! " !!!!!! Step 3: Compare the numerator. !5 !3 2 1 is bigger than . 5 3 3 4 or 4 5 ## Exercise 2: Order the following ! ! ! ! ! ! ! ! ! !" ! Step 1: Find the LCM of the denominators The multiples of 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, The multiples of 4 are 4, 8, 12, 16, 20, The multiples of 5 are 5, 10, 15, 20, The multiples of 10 are 10, 20, 20 is the lowest common multiple (LCM) Step 2: Get all fractions with a common denominator ! !" ! ! !" ! ! ! ! !" ! !" ! !" !" ! !" ! ! !" C.Camenzuli \| www.smcmaths.webs.com 5 Form 2 [CHAPTER 4: FRACTION AND DECIMALS] Step 2: Compare the numerators (in ascending order) ! ! ! ! ! ! ! ! ! !" ! ## Example 2: Arrange in descending order: 2 11 23 7 3 , , , , 3 15 30 10 5 ## 4.4 Multiplying fractions To multiply a fraction by an integer, multiply the numerator of the fraction by the integer. Do not change the denominator of the fraction. ## Example 1: Work out: 6 ! 6! 2 3 2 6! 2 = 3 3 6! 2 = 3 1 = 2! 2 1 =4 To multiply two fractions, multiply the numerators and then multiply the denominators. ## Example 2: Work out: = 3 2 ! 4 3 3! 2 4!3 1 1 3! 2 = 4! 3 2 1 C.Camenzuli \| www.smcmaths.webs.com 6 1! 1 2 !1 1 = 2 = 2 ! 33 3 2 2 ! 3 5 5 7 ! 14 10 ## Example 7: Work out: 4 12 of 3 18 When multiplying mixed numbers, first write the mixed numbers as improper fractions. ## 2 4 3 5 Step 1: Converting the mixed numbers as improper fractions: 2 8 (2 x 3 = 6 + 2 = 8) 2 = 3 3 Example 8: 2 !1 4 9 1 = 5 5 (1 x 5 = 5 + 4 = 9) C.Camenzuli \| www.smcmaths.webs.com 7 Form 2 [CHAPTER 4: FRACTION AND DECIMALS] Step 2: Carry out the multiplication: 2 4 8 9 Therefore, 2 !1 = ! 3 5 3 5 8! 9 = 3! 5 1 8! 3 = 1! 5 24 = 5 Step 3: Change the improper faction into a mixed number: 24 4 =4 5 5 1 2 3 5 ## Example 10: Work out ! 1 6 5 ! 3 15 7 C.Camenzuli \| www.smcmaths.webs.com 8 ## 4.5 Fraction of a quantity Example 1: Find 3 of 25 cm. 5 3 by 25cm. 5 ## To work out this problem we need to multiply To find a fraction of an amount, multiply by the numerator and divide by the denominator: 3 3 ! 25 ! 25 = 5 5 5 3 ! 25 = 5 1 = 15 cm 3! 5 = 1 Example 2: Find of 63 g. Example 3: Find of !30 C.Camenzuli \| www.smcmaths.webs.com 9 ## 4.6 Dividing fractions 4 3 5 Step 1: Find the reciprocal of the divisor. ## Example 1: Work out The reciprocal of 3 is 1 3 ## Step 2: Multiply the dividend by the reciprocal 4 1 ! 5 3 4 !1 = 5! 3 4 = 15 4 Answer: 15 5 3 6 4 Step 1: Find the reciprocal of the divisor. 3 4 The reciprocal of is 4 3 Step 2: Multiply the dividend by the reciprocal 5 4 ! 6 3 ## Example 2: Work out 5! 4 = 6! 3 3 5! 2 3! 3 10 = 9 C.Camenzuli \| www.smcmaths.webs.com 10 ## Form 2 [CHAPTER 4: FRACTION AND DECIMALS] Answer: 10 1 =1 9 9 5 15 8 32 ## Example 4: Work out 2 2 4 5 1 10 Step 1: Convert the mixed numbers to improper fractions. 4 14 2 = 5 5 1 21 2 = 10 10 4 1 14 21 Thus, 2 2 = 5 10 5 10 Step 2: Find the reciprocal of the divisor The reciprocal of 21 10 is 10 21 ## Step 3: Multiply the dividend by the reciprocal 14 10 ! 5 21 14! 10 = 5! 21 1 3 2! 2 = 1! 3 4 = 3 C.Camenzuli \| www.smcmaths.webs.com 11 ## Form 2 [CHAPTER 4: FRACTION AND DECIMALS] Answer: 4 1 =1 3 3 5 6 1 9 ## Example 6: Work out 2 1 1 !4 7 9 3 5 C.Camenzuli \| www.smcmaths.webs.com 12 ## 4.7 Changing from fraction to decimal All fractions can be changed back into a decimal ## Method 1: Using equivalent fractions Step 1: Check the denominator Step 2: Create an equivalent fraction with denominators 10, 100, 1000 etc. Step 3: Convert into a decimal Example1: Convert 1) 2) 2 into a decimal 5 ## Denominator is 5 Equivalent fraction 2 4 = 5 10 3) Convert into a decimal 0.4 ## Example 2: Convert the following fractions into decimals 4 5 (i) (ii) 7 20 C.Camenzuli \| www.smcmaths.webs.com 13 (iii) 3 25 (iv) 8 50 ## Method 2: Short division It is not always possible to create an equivalent fraction with denominators being multiples of 10. When this is not possible we can either use the calculator or perform a short division. ## Example 1: Use short division to change these fractions to decimals. 3 8 (i) (ii) 1 4 C.Camenzuli \| www.smcmaths.webs.com 14 (iii) 1 6 ## 4.8 Changing from decimal to fraction A terminating decimal is a decimal which ends. E.g. 0.26, 0.628 are terminating decimals All terminating decimals can be converted into fractions. Step 1: Observe the decimal Step 2: Find the place value of the digit further to the right Step 3: The place value of the last digit shows the number over which we have to express the fraction Step 4: Simplify the resulting fraction ## Example 1: Convert 0.24 into a fraction 1) 2) Observe the decimal 0.24 Finding the place value The place value of 4 is a hundredth( 3) Expressing the fraction 1 ) 100 0.24 = 4) 24 100 24 12 6 = = 100 50 25 C.Camenzuli \| www.smcmaths.webs.com 15 Simplify 0.24 = 5 1 =2 10 2 2.5 = 2 ## Example 3: Convert the following decimals into fraction (i) 5.45 (ii) 0.67 (iii) 56.42 (iv) 0.0003 C.Camenzuli \| www.smcmaths.webs.com 16
html5-img 1 / 30 # Geometric Sequences Geometric Sequences. Warm Up. Lesson Presentation. Lesson Quiz. Holt McDougal Algebra 1. Holt Algebra 1. Warm Up Find the value of each expression. 1. 2 5. 32. 2. 2 –5. 3. –3 4. –81. 4. (–3) 4. 81. 5. (0.2) 3. 6. 7(–4) 2. 0.008. 112. 7. 8. 12(–0.4) 3. –0.768. ## Geometric Sequences E N D ### Presentation Transcript 1. Geometric Sequences Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 1 Holt Algebra 1 2. Warm Up • Find the value of each expression. 1. 25 32 2. 2–5 3. –34 –81 4. (–3)4 81 5. (0.2)3 6. 7(–4)2 0.008 112 7. 8. 12(–0.4)3 –0.768 3. Objectives Recognize and extend geometric sequences. Find the nth term of a geometric sequence. 4. Vocabulary geometric sequence common ratio 5. The table shows the heights of a bungee jumper’s bounces. The height of the bounces shown in the table above form a geometric sequence. In a geometric sequence, the ratio of successive terms is the same number r, called the common ratio. 6. 1 2 3 4 Position 3 6 12 24 Term Geometric sequences can be thought of as functions. The term number, or position in the sequence, is the input, and the term itself is the output. a1a2a3a4 To find a term in a geometric sequence, multiply the previous term by r. 7. Writing Math The variable a is often used to represent terms in a sequence. The variable a4 (read “a sub 4”)is the fourth term in a sequence. 8. Example 1A: Extending Geometric Sequences Find the next three terms in the geometric sequence. 1, 4, 16, 64,… Step 1 Find the value of r by dividing each term by the one before it. 1 4 16 64 The value of r is 4. 9. 4 4 4 Example 1A Continued Find the next three terms in the geometric sequence. 1, 4, 16, 64,… Step 2 Multiply each term by 4 to find the next three terms. 64 256 1024 4096 The next three terms are 256, 1024, and 4096. 10. The value of r is . Example 1B: Extending Geometric Sequences Find the next three terms in the geometric sequence. Step 1 Find the value of r by dividing each term by the one before it. 11. Helpful Hint When the terms in a geometric sequence alternate between positive and negative, the value of r is negative. 12. Step 2 Multiply each term by to find the next three terms. The next three terms are Example 1B Continued Find the next three terms in the geometric sequence. 13. Check It Out! Example 1a Find the next three terms in the geometric sequence. 5, –10, 20,–40,… Step 1 Find the value of r by dividing each term by the one before it. 5 –10 20 –40 The value of r is –2. 14. Check It Out! Example 1a Continued Find the next three terms in the geometric sequence. 5, –10, 20,–40,… Step 2 Multiply each term by –2 to find the next three terms. –40 80 –160 320 (–2) (–2) (–2) The next three terms are 80, –160, and 320. 15. Check It Out! Example 1b Find the next three terms in the geometric sequence. 512, 384, 288,… Step 1 Find the value of r by dividing each term by the one before it. 512 384 288 The value of r is 0.75. 16. Check It Out! Example 1b Continued Find the next three terms in the geometric sequence. 512, 384, 288,… Step 2 Multiply each term by 0.75 to find the next three terms. 288 216 162 121.5 0.75 0.75 0.75 The next three terms are 216, 162, and 121.5. 17. To find the output an of a geometric sequence when n is a large number, you need an equation, or function rule. The pattern in the table shows that to get the nth term, multiply the first term by the common ratio raised to the power n – 1. 18. nth term 1st term Common ratio If the first term of a geometric sequence is a1, the nth term is an , and the common ratio is r, then an=a1rn–1 19. Example 2A: Finding the nth Term of a Geometric Sequence The first term of a geometric sequence is 500, and the common ratio is 0.2. What is the 7th term of the sequence? an = a1rn–1 Write the formula. a7 = 500(0.2)7–1 Substitute 500 for a1,7 for n, and 0.2 for r. Simplify the exponent. = 500(0.2)6 Use a calculator. = 0.032 The 7th term of the sequence is 0.032. 20. Example 2B: Finding the nth Term of a Geometric Sequence For a geometric sequence, a1 = 5, and r = 2. Find the 6th term of the sequence. an = a1rn–1 Write the formula. a6 = 5(2)6–1 Substitute 5 for a1,6 for n, and 2 for r. = 5(2)5 Simplify the exponent. = 160 The 6th term of the sequence is 160. 21. 2 –6 18 –54 Example 2C: Finding the nth Term of a Geometric Sequence What is the 9th term of the geometric sequence 2, –6, 18, –54, …? The value of r is –3. an = a1rn–1 Write the formula. Substitute 2 for a1,9 for n, and –3 for r. a9 = 2(–3)9–1 = 2(–3)8 Simplify the exponent. Use a calculator. = 13,122 The 9th term of the sequence is 13,122. 22. Caution When writing a function rule for a sequence with a negative common ratio, remember to enclose r in parentheses. –212 ≠ (–2)12 23. 1000 500 250 125 The value of r is . Substitute 1000 for a1,8 for n, and for r. a8 = 1000( )8–1 Check It Out! Example 2 What is the 8th term of the sequence 1000, 500, 250, 125, …? an = a1rn–1 Write the formula. Simplify the exponent. = 7.8125 Use a calculator. The 8th term of the sequence is 7.8125. 24. 300 150 75 Example 3: Application A ball is dropped from a tower. The table shows the heights of the balls bounces, which form a geometric sequence. What is the height of the 6th bounce? The value of r is 0.5. 25. Example 3 Continued an = a1rn–1 Write the formula. Substitute 300 for a1, 6 for n, and 0.5 for r. a6 = 300(0.5)6–1 = 300(0.5)5 Simplify the exponent. Use a calculator. = 9.375 The height of the 6th bounce is 9.375 cm. 26. Check It Out! Example 3 The table shows a car’s value for 3 years after it is purchased. The values form a geometric sequence. How much will the car be worth in the 10th year? 10,000 8,000 6,400 The value of r is 0.8. 27. Check It Out! Example 3 an = a1rn–1 Write the formula. Substitute 10,000 for a1,10 for n, and 0.8 for r. a6 = 10,000(0.8)10–1 = 10,000(0.8)9 Simplify the exponent. Use a calculator. = 1,342.18 In the 10th year, the car will be worth \$1342.18. 28. Lesson Quiz: Part I Find the next three terms in each geometric sequence. 1. 3, 15, 75, 375,… 2. 3. The first term of a geometric sequence is 300 and the common ratio is 0.6. What is the 7th term of the sequence? 4. What is the 15th term of the sequence 4, –8, 16, –32, 64…? 1875; 9375; 46,875 13.9968 65,536 29. Lesson Quiz: Part II Find the next three terms in each geometric sequence. 5. The table shows a car’s value for three years after it is purchased. The values form a geometric sequence. How much will the car be worth after 8 years? \$5570.39 More Related
Factor Theorem – Methods & Examples A polynomial is an algebraic expression with one or more terms in which an addition or a subtraction sign separates a constant and a variable. The general form of a polynomial is axn + bxn-1 + cxn-2 + …. + kx + l, where each variable has a constant accompanying it as its coefficient. Now that you understand how to use the Remainder Theorem to find the remainder of polynomials without actual division, the next theorem to look at in this article is called the Factor Theorem. We will study how the Factor Theorem is related to the Remainder Theorem and how to use the theorem to factor and find the roots of a polynomial equation. But, before jumping into this topic, let’s revisit what factors are. A factor is a number or expression that divides another number or expression to get a whole number with no remainder in mathematics. In other words, a factor divides another number or expression by leaving zero as a remainder. For example, 5 is a factor of 30 because when 30 is divided by 5, the quotient is 6, which a whole number and the remainder is zero. Consider another case where 30 is divided by 4 to get 7.5. In this case, 4 is not a factor of 30 because when 30 is divided by 4, we get a number that is not a whole number. 7.5 is the same as saying 7 and a remainder of 0.5. What is a Factor Theorem? Consider a polynomial f (x) of degree n ≥ 1. If the term ‘a’ is any real number, then we can state that; (x – a) is a factor of f (x), if f (a) = 0. Proof of the Factor Theorem Given that f (x) is a polynomial being divided by (x – c), if f (c) = 0 then, ⟹ f(x) = (x – c) q(x) + f(c) ⟹ f(x) = (x – c) q(x) + 0 ⟹ f(x) = (x – c) q(x) Hence, (x – c) is a factor of the polynomial f (x). Hence, the Factor Theorem is a special case of Remainder Theorem, which states that a polynomial f (x) has a factor xa, if and only if, a is a root i.e., f (a) = 0. How to use the Factor Theorem? Let’s see a few examples below to learn how to use the Factor Theorem. Example 1 Find the roots of the polynomial f(x)= x2 + 2x – 15 Solution f(x) = 0 x2 + 2x – 15 = 0 (x + 5) (x – 3) = 0 (x + 5) = 0 or (x – 3) = 0 x = -5 or x = 3 We can check if (x – 3) and (x + 5) are factors of the polynomial x2 + 2x – 15, by applying the Factor Theorem as follows: If x = 3 Substitute x = 3 in the polynomial equation/. f (x)= x2 + 2x – 15 ⟹ 32 + 2(3) – 15 ⟹ 9 + 6 – 15 ⟹ 15 – 15 f (3) = 0 And if x = -5 Substitute the values of x in the equation f(x)= x2 + 2x – 15 ⟹ (-5)2 + 2(-5) – 15 ⟹ 25 – 10 – 15 ⟹ 25 – 25 f (-5) = 0 Since the remainders are zero in the two cases, therefore (x – 3) and (x + 5) are factors of the polynomial x2 +2x -15 Example 2 Find the roots of the polynomial 2x2 – 7x + 6 = 0. Solution First factorize the equation. 2x2 – 7x + 6 = 0 ⟹ 2x2 – 4x – 3x + 6 = 0 ⟹ 2x (x – 2) – 3(x – 2) = 0 ⟹ (x – 2) (2x – 3) = 0 ⟹ x – 2 = 0 or 2x – 3 = 0 ⟹ x = 2 or x = 3/2 Hence, the roots are x = 2, 3/2. Example 3 Check whether x + 5 is a factor of 2x2 + 7x – 15. Solution x + 5= 0 x = -5 Now substitute the x= -5 into the polynomial equation. f (-5) = 2 (-5)2 + 7(-5) – 15 = 50 – 35 – 15 = 0 Hence, x + 5 is a factor of 2x2 + 7x – 15. Example 4 Determine whether x + 1 is a factor of the polynomial 3x4 + x3 – x2 + 3x + 2 Solution Given x + 1; x + 1 = 0 x = -1 Substitute x = -1 in the equation; 3x4 + x3 – x2 + 3x + 2. ⟹ 3(–1)4 + (–1)3 – (–1)2 +3(–1) + 2 = 3(1) + (–1) – 1 – 3 + 2 = 0 Therefore, x + 1 is a factor of 3x4 + x3 – x2 + 3x + 2 Example 5 Check whether 2x + 1 is a factor of the polynomial 4x3 + 4x2 – x – 1 Solution ⟹ 2x + 1 = 0 ∴ x = -1/2 Substitute x = -1/2 in the equation 4x3 + 4x2 – x – 1. ⟹ 4( -1/2)3 + 4(-1/2)2 – (-1/2) – 1 = -1/2 + 1 + ½ – 1 = 0 Since, the remainder = 0, then 2x + 1 is a factor of 4x3 + 4x2 – x – 1 Example 6 Check whether x + 1 is a factor of x6 + 2x (x – 1) – 4 Solution x + 1 = 0 x = -1 Now substitute x = -1 in the polynomial equation x6 + 2x (x – 1) – 4 ⟹ (–1)6 + 2(–1) (–2) –4 = 1 Therefore, x + 1 is not a factor of x6 + 2x (x – 1) – 4 Practice Questions 1. Using the factor theorem, we can confirm that $(x–4)$ is a factor of $x^3 – 9x^2 + 35x – 60$. 2. Which of the following shows the zeroes of $x^2 – 8 x – 9$? 3. Using the factor theorem, we can confirm that $(x + 2)$ is a factor of $x^3 + 4x^2 + x – 6$. 4. Using the factor theorem, we can confirm that $(x + 4)$ is a factor of $2x^3 – 3x^2 – 39x + 20$. 5. Which of the following shows the value of $k$ given that $x + 2$ is a factor of the equation $2x^3 -5x^2 + kx + k$.
Rd Sharma Xi 2020 2021 _volume 2 Solutions for Class 12 Humanities Maths Chapter 25 Parabola are provided here with simple step-by-step explanations. These solutions for Parabola are extremely popular among Class 12 Humanities students for Maths Parabola Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2020 2021 _volume 2 Book of Class 12 Humanities Maths Chapter 25 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2020 2021 _volume 2 Solutions. All Rd Sharma Xi 2020 2021 _volume 2 Solutions for class Class 12 Humanities Maths are prepared by experts and are 100% accurate. #### Question 1: Find the equation of the parabola whose: (i) focus is (3, 0) and the directrix is 3x + 4y = 1 (ii) focus is (1, 1) and the directrix is x + y + 1 = 0 (iii) focus is (0, 0) and the directrix 2xy − 1 = 0 (iv) focus is (2, 3) and the directrix x − 4y + 3 = 0. (i) Let P (x, y) be any point on the parabola whose focus is S (3, 0) and the directrix is 3x + 4y = 1. Draw PM perpendicular to 3x + 4y = 1. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-3\right)}^{2}+{\left(y-0\right)}^{2}={\left(\frac{3x+4y-1}{\sqrt{9+16}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-3\right)}^{2}+{y}^{2}={\left(\frac{3x+4y-1}{5}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒25\left\{{\left(x-3\right)}^{2}+{y}^{2}\right\}={\left(3x+4y-1\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\left(25{x}^{2}-150x+25{y}^{2}+225\right)=9{x}^{2}+16{y}^{2}+1+24xy-8y-6x\phantom{\rule{0ex}{0ex}}⇒16{x}^{2}+9{y}^{2}-24xy-144x+8y+224=0$ (ii) Let P (x, y) be any point on the parabola whose focus is S (1, 1) and the directrix is xy + 1 = 0. Draw PM perpendicular to xy + 1 = 0. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-1\right)}^{2}+{\left(y-1\right)}^{2}={\left\|\frac{x+y+1}{\sqrt{1+1}}\right\|}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-1\right)}^{2}+{\left(y-1\right)}^{2}={\left(\frac{x+y+1}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒2\left({x}^{2}+1-2x+{y}^{2}+1-2y\right)={x}^{2}+{y}^{2}+1+2xy+2y+2x\phantom{\rule{0ex}{0ex}}⇒\left(2{x}^{2}+2-4x+2{y}^{2}+2-4y\right)={x}^{2}+{y}^{2}+1+2xy+2y+2x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2xy-6x-6y+3=0$ (iii) Let P (x, y) be any point on the parabola whose focus is S (0, 0) and the directrix is 2x − y − 1 = 0. Draw PM perpendicular to 2x − y − 1 = 0. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}={\left\|\frac{2x-y-1}{\sqrt{4+1}}\right\|}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}={\left(\frac{2x-y-1}{\sqrt{5}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒5{x}^{2}+5{y}^{2}=4{x}^{2}+{y}^{2}+1-4xy+2y-4x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4{y}^{2}+4xy-2y+4x-1=0$ (iv) Let P (x, y) be any point on the parabola whose focus is S (2, 3) and the directrix is x − 4y + 3 = 0. Draw PM perpendicular to x − 4y + 3 = 0. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}={\left\|\frac{x-4y+3}{\sqrt{1+16}}\right\|}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}={\left(\frac{x-4y+3}{\sqrt{17}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒17\left({x}^{2}+4-4x+{y}^{2}-6y+9\right)={x}^{2}+16{y}^{2}+9-8xy-24y+6x\phantom{\rule{0ex}{0ex}}⇒\left(17{x}^{2}-68x-102y+17{y}^{2}+13×17\right)={x}^{2}+16{y}^{2}+9-8xy-24y+6x\phantom{\rule{0ex}{0ex}}⇒16{x}^{2}+{y}^{2}+8xy-74x-78y+212=0$ #### Question 2: Find the equation of the parabola whose focus is the point (2, 3) and directrix is the line x − 4y + 3 = 0. Also, find the length of its latus-rectum. Let P (x, y) be any point on the parabola whose focus is S (2, 3) and the directrix is x − 4y + 3 = 0. Draw PM perpendicular to x − 4y + 3=0. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}={\left(\frac{x-4y+3}{\sqrt{1+16}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}={\left(\frac{x-4y+3}{\sqrt{17}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒17\left({x}^{2}+4-4x+{y}^{2}-6y+9\right)={x}^{2}+16{y}^{2}+9-8xy-24y+6x\phantom{\rule{0ex}{0ex}}⇒\left(17{x}^{2}-68x+17{y}^{2}-102y+13×17\right)={x}^{2}+16{y}^{2}+9-8xy-24y+6x\phantom{\rule{0ex}{0ex}}⇒16{x}^{2}+{y}^{2}+8xy-74x-78y+212=0$ Length of the latus rectum = 2(Length of the perpendicular from the focus on the directrix) = 2(Length of the perpendicular from (2, 3) on the directrix) = $2\left\|\frac{2-12+3}{\sqrt{16+1}}\right\|=2\left\|\frac{-7}{\sqrt{17}}\right\|=2\left(\frac{7}{\sqrt{17}}\right)=\frac{14}{\sqrt{17}}$ units #### Question 3: Find the equation of the parabola if (i) the focus is at (−6, −6) and the vertex is at (−2, 2) (ii) the focus is at (0, −3) and the vertex is at (0, 0) (iii) the focus is at (0, −3) and the vertex is at (−1, −3) (iv) the focus is at (a, 0) and the vertex is at (a', 0) (v) the focus is at (0, 0) and vertex is at the intersection of the lines x + y = 1 and xy = 3. In a parabola, the vertex is the mid-point of the focus and the point of intersection of the axis and the directrix. Let (x1, y1) be the coordinates of the point of intersection of the axis and directrix. (i) It is given that the vertex and the focus of a parabola are (−2, 2) and (−6, −6), respectively. ∴ Slope of the axis of the parabola = $\frac{-6-2}{-6+2}=\frac{-8}{-4}=2$ Slope of the directrix = $\frac{-1}{2}$ Let the directrix intersect the axis at K (r, s). ∴ Required equation of the directrix: $y-10=\frac{-1}{2}\left(x-2\right)$ $2y+x-22=0$. Now, let P (x, y) be any point on the parabola whose focus is S (−6, −6), and the directrix is $2y+x-22=0$. Draw PM perpendicular to $2x+y+22=0$. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x+6\right)}^{2}+{\left(y+6\right)}^{2}={\left(\frac{2y+x-22}{\sqrt{5}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒5\left({x}^{2}+12x+36+{y}^{2}+12y+36\right)=4{y}^{2}+{x}^{2}+484+4xy-88y-44x\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}+{y}^{2}-4xy+104x+148y-124=0\phantom{\rule{0ex}{0ex}}⇒{\left(2x-y\right)}^{2}-4\left(26x+37y-31\right)=0$ (ii) It is given that the vertex and the focus of a parabola are (0, 0) and (0, −3), respectively. Thus, the slope of the axis of the parabola cannot be defined. Slope of the directrix = 0 Let the directrix intersect the axis at K (r, s). ∴ Required equation of directrix: $y=3$ Let P (x, y) be any point on the parabola whose focus is S (0, −3) and the directrix is $y=3$. Draw PM perpendicular to $y=3$. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(y+3\right)}^{2}={\left(\frac{y-3}{\sqrt{1}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}+6y+9={y}^{2}-6y+9\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=-12y$ (iii) It is given that the vertex and the focus of a parabola are (−1, −3) and (0, −3), respectively. Thus, the slope of the axis of the parabola is zero. And, the slope of the directrix cannot be defined. Let the directrix intersect the axis at K (r, s). ∴ Required equation of the directrix: $x+2=0$ Let P (x, y) be any point on the parabola whose focus is S (0, −3) and the directrix is $x+2=0$. Draw PM perpendicular to $x+2=0$. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(y+3\right)}^{2}={\left\|\frac{x+2}{\sqrt{1}}\right\|}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}+6y+9={x}^{2}+4x+4\phantom{\rule{0ex}{0ex}}⇒{y}^{2}+6y-4x+5=0$ (iv) It is given that the vertex and the focus of a parabola are (a', 0) and (a, 0), respectively. Thus, the slope of the axis of the parabola is zero. And, the slope of the directrix cannot be defined. Let the directrix intersect the axis at K (r, s). ∴ Required equation of the directrix is $x-2a\text{'}+a=0$. Let P (x, y) be any point on the parabola whose focus is S (a, 0), and the directrix is $x-2a\text{'}+a=0$. Draw PM perpendicular to $x-2a\text{'}+a=0$. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-a\right)}^{2}+{\left(y-0\right)}^{2}={\left(\frac{x-2a\text{'}+a}{\sqrt{1}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{y}^{2}={\left(x-2a\text{'}+a\right)}^{2}-{\left(x-a\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{y}^{2}={x}^{2}+4a{\text{'}}^{2}+{a}^{2}-4a\text{'}x-4aa\text{'}+2ax-{x}^{2}-{a}^{2}+2ax\phantom{\rule{0ex}{0ex}}⇒{y}^{2}=4a{\text{'}}^{2}-4a\text{'}x-4aa\text{'}+4ax\phantom{\rule{0ex}{0ex}}⇒{y}^{2}=-4\left(a\text{'}-a\right)\left(x-a\text{'}\right)$ (v) The point of intersection of is (2, −1). Thus, the vertex and the focus of the parabola are (2, −1) and (0, 0), respectively. ∴ Slope of the axis of the parabola = $\frac{0+1}{0-2}=\frac{-1}{2}$ The slope of the directrix is 2. Let the directrix intersect the axis at K (r, s). The required equation of the directrix is $y+2=2\left(x-4\right)$, which can be rewritten as $y-2x+10=0$. Let P (x, y) be any point on the parabola whose focus is S (0, 0) and directrix is $y-2x+10=0$. Draw PM perpendicular to $y-2x+10=0$. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}={\left(\frac{y-2x+10}{\sqrt{5}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒5{x}^{2}+5{y}^{2}={\left(y-2x+10\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4{y}^{2}+4xy+40x-20y-100=0\phantom{\rule{0ex}{0ex}}⇒{\left(x+2y\right)}^{2}+40x-20y-100=0$ #### Question 4: Find the vertex, focus, axis, directrix and latus-rectum of the following parabolas (i) y2 = 8x (ii) 4x2 + y = 0 (iii) y2 − 4y − 3x + 1 = 0 (iv) y2 − 4y + 4x = 0 (v) y2 + 4x + 4y − 3 = 0 (vi) y2 = 8x + 8y (vii) 4 (y − 1)2 = − 7 (x − 3) (viii) y2 = 5x − 4y − 9 (ix) x2 + y = 6x − 14 (i) Given: y2 = 8x On comparing the given equation with ${y}^{2}=4ax$: $4a=8⇒a=2$ ∴ Vertex = (0, 0) Focus = (a, 0) = (2, 0) Equation of the directrix: x = −a i.e. x = 2 Axis = y = 0 Length of the latus rectum = 4a = 8 units (ii) Given: 4x2 + y = 0 $⇒\frac{-y}{4}={x}^{2}$ On comparing the given equation with ${x}^{2}=-4ay$: $4a=\frac{1}{4}⇒a=\frac{1}{16}$ ∴ Vertex = (0, 0) Focus = (0, −a) = $\left(0,\frac{-1}{16}\right)$ Equation of the directrix: ya i.e. $y=\frac{1}{16}$ Axis = x = 0 Length of the latus rectum = 4a = $\frac{1}{4}$ units (iii) Given: y2 − 4y − 3x + 1 = 0 $⇒{\left(y-2\right)}^{2}-4-3x+1=0\phantom{\rule{0ex}{0ex}}⇒{\left(y-2\right)}^{2}=3\left(x+1\right)\phantom{\rule{0ex}{0ex}}⇒{\left(y-2\right)}^{2}=3\left(x-\left(-1\right)\right)$ Let $Y=y-2$, $X=x+1$ Then, we have: ${Y}^{2}=3X$ Comparing the given equation with ${Y}^{2}=4aX$: $4a=3⇒a=\frac{3}{4}$ ∴ Vertex = (X = 0, Y = 0) = Focus = (X = a, Y = 0) = Equation of the directrix: X = −a i.e$x+1=\frac{-3}{4}⇒x=\frac{-7}{4}$ Axis = Y = 0 i.e. $y-2=0⇒y=2$ Length of the latus rectum = 4a = 3 units (iv) Given: y2 − 4y + 4x = 0 $⇒{\left(y-2\right)}^{2}-4+4x=0\phantom{\rule{0ex}{0ex}}⇒{\left(y-2\right)}^{2}=-4\left(x-1\right)$ Let $Y=y-2$, $X=x-1$ Then, we have: ${Y}^{2}=-4X$ Comparing the given equation with ${Y}^{2}=-4aX$: $4a=4⇒a=1$ ∴ Vertex = (X = 0, Y = 0) = Focus = (X = −a, Y = 0) = Equation of the directrix: X = a i.e. $x-1=1⇒x=2$ Axis = Y = 0 i.e. $y-2=0⇒y=2$ Length of the latus rectum = 4a = 4 units (v) Given: y2 + 4y + 4x −3 = 0 $⇒{\left(y+2\right)}^{2}-4+4x-3=0\phantom{\rule{0ex}{0ex}}⇒{\left(y+2\right)}^{2}=-4\left(x-\frac{7}{4}\right)$ Let $Y=y+2$, $X=x-\frac{7}{4}$ Then, we have: ${Y}^{2}=-4X$ Comparing the given equation with ${Y}^{2}=-4aX$: $4a=4⇒a=1$ ∴ Vertex = (X = 0, Y = 0) = Focus = (X = −a, Y = 0) = Equation of the directrix: X = a i.e. $x-\frac{7}{4}=1⇒x=\frac{11}{4}$ Axis = Y = 0 i.e. $y+2=0⇒y=-2$ Length of the latus rectum = 4a = 4 units (vi) Given: y2 = 8x + 8y $⇒{\left(y-4\right)}^{2}=8\left(x+2\right)$ Putting $Y=y-4$, $X=x+2$: ${Y}^{2}=8X$ On comparing the given equation with ${Y}^{2}=4aX$: $4a=8⇒a=2$ ∴ Vertex = (X = 0, Y = 0) = Focus = (X = a, Y = 0) = Equation of the directrix: X = −a i.e. $x+2=-2⇒x+4=0$ Axis = Y = 0 i.e. $y-4=0⇒y=4$ Length of the latus rectum = 4a = 8 (vii) Given: 4(y − 1)2 = − 7 (x − 3) $⇒{\left(y-1\right)}^{2}=\frac{-7}{4}\left(x-3\right)$ Let $Y=y-1$, $X=x-3$ Then, we have: ${Y}^{2}=\frac{-7}{4}X$ Comparing the given equation with ${Y}^{2}=-4aX$: $4a=\frac{7}{4}⇒a=\frac{7}{16}$ ∴ Vertex = (X = 0, Y = 0) = Focus = (X = −a, Y = 0) = Equation of the directrix: X = a i.e. $x-3=\frac{7}{16}⇒x=\frac{55}{16}$ Axis = Y = 0 i.e. Length of the latus rectum = 4a = $\frac{7}{4}$ units (viii) Given: y 2 = 5x − 4y − 9 $⇒{y}^{2}+4y=5x-9\phantom{\rule{0ex}{0ex}}⇒{\left(y+2\right)}^{2}=5x-5=5\left(x-1\right)$ Putting $Y=y+2$, $X=x-1$: ${Y}^{2}=5X$ Comparing the given equation with ${Y}^{2}=4aX$: $4a=5⇒a=\frac{5}{4}$ ∴ Vertex = (X = 0, Y = 0) = Focus = (X = a, Y = 0) = Equation of the directrix: X = −a i.e. $x-1=\frac{-5}{4}⇒x=\frac{-1}{4}$ Axis = Y = 0 i.e. $y+2=0⇒y=-2$ Length of the latus rectum = 4a = 5 units (ix) Given: x2 = 6xy−14 $⇒{\left(x-3\right)}^{2}=-y-14+9\phantom{\rule{0ex}{0ex}}⇒{\left(x-3\right)}^{2}=-y-5=-\left(y+5\right)$ Let $Y=y+5$, $X=x-3$ Then, we have: ${X}^{2}=-Y$ Comparing the given equation with ${X}^{2}=-4aY$: $4a=1⇒a=\frac{1}{4}$ ∴ Vertex = (X = 0, Y = 0) = Focus = (X = 0, Y = −a) = Equation of the directrix: Y = a i.e. $y+5=\frac{1}{4}⇒y=\frac{-19}{4}$ Axis = X = 0 i.e. $x-3=0⇒x=3$ Length of the latus rectum = 4a = 1 units #### Question 5: For the parabola y2 = 4px find the extremities of a double ordinate of length 8 p. Prove that the lines from the vertex to its extremities are at right angles. The given equation of the parabola is y2 = 4px. Let PQ be the double ordinate of length 8p of the parabola ${y}^{2}=4px$. Then, we have: PR = RQ = 4p Let AR = x1 Then, the coordinates of P and Q are $\left({x}_{1},4p\right)$ and $\left({x}_{1},-4p\right)$, respectively. Now, P lies on ${y}^{2}=4px\phantom{\rule{0ex}{0ex}}$. ${\left(4p\right)}^{2}=4p{x}_{1}$ $⇒{x}_{1}=4p$ So, the coordinates of P and Q are and $\left(4p,-4p\right)$, respectively. The coordinates of A are (0, 0). Now, ${m}_{1}{m}_{2}=-1$ Thus, AP is perpendicular to AQ. Hence, the lines from the vertex to its extremities are at right angles. #### Question 6: Find the area of the triangle formed by the lines joining the vertex of the parabola ${x}^{2}=12y$ to the ends of its latus rectum. The given equation of the parabola is x2 = 12y. On comparing the given equation with ${x}^{2}=4ay$: a = 3 Required area = #### Question 7: Find the coordinates of the point of intersection of the axis and the directrix of the parabola whose focus is (3, 3) and directrix is 3x − 4y = 2. Find also the length of the latus-rectum. The given equation of the directrix is 3x − 4y = 2. (1) ∴ Slope of the directrix = $\frac{-3}{-4}=\frac{3}{4}$ Also, the axis is perpendicular to the directrix. ∴ Slope of the axis = $\frac{-4}{3}$ The focus lies on the axis of the parabola. ∴ Equation of the axis: $\left(y-3\right)=\frac{-4}{3}\left(x-3\right)$ $\left(3y-9\right)=-4x+12$ $3y+4x-21=0$ (2) Solving equations (1) and (2): Therefore, the intersection point of the axis and directrix is $\left(\frac{18}{5},\frac{11}{5}\right)$. Also, length of the latus rectum = 2 (Length of the perpendicular from the focus on the directrix) =$2\left\|\frac{3\left(3\right)+\left(-4\right)3-2}{\sqrt{16+9}}\right\|=2\left\|\frac{-5}{\sqrt{16+9}}\right\|=2$ square units #### Question 8: At what point of the parabola x2 = 9y is the abscissa three times that of ordinate? Putting x = 3y in the given equation of the parabola: $9{y}^{2}=9y\phantom{\rule{0ex}{0ex}}⇒9y\left(y-1\right)=0\phantom{\rule{0ex}{0ex}}⇒y=0,1$ At y = 0, x = 0 At y = 1, x = 3 Therefore, at (1, 3), the abscissa is three times that of the ordinate. #### Question 9: Find the equation of a parabola with vertex at the origin, the axis along x-axis and passing through (2, 3). CASE I: Let the equation of the required parabola be ${y}^{2}=4ax$ (1) Since (1) passes through (2, 3), we have: $9=4a\left(2\right)⇒a=\frac{9}{8}$ Thus, the required equation is ${y}^{2}=\frac{4\left(9\right)x}{8}$, i.e. $2{y}^{2}=9x$. CASE II: Let the equation of the required parabola be ${y}^{2}=-4ax$ (2) Since (2) passes through (2, 3), we have: $9=-4a\left(2\right)⇒a=\frac{-9}{8}$ Thus, the required equation is ${y}^{2}=\frac{-4\left(-9\right)x}{8}$, i.e. $2{y}^{2}=9x$. Hence, in either case, the required equation of the parabola is $2{y}^{2}=9x$. #### Question 10: Find the equation of a parabola with vertex at the origin and the directrix, y = 2. Let the equation of the directrix be y = a. Comparing with y = 2: a = 2 Equation of the parabola with directrix y =a is ${x}^{2}=-4ay$. Hence, the required equation of the parabola is ${x}^{2}=-8y$. #### Question 11: Find the equation of the parabola whose focus is (5, 2) and having vertex at (3, 2). Given: The vertex and the focus of the parabola are (3, 2) and (5, 2), respectively. ∴ Slope of the axis of the parabola = 0 Slope of the directrix cannot be defined. Let the directrix intersect the axis at K (r, s). Required equation of the directrix is $x-1=0$, which can be rewritten as x = 1. Let P (x, y) be any point on the parabola whose focus is S (5, 2) and the directrix is x =1. Draw PM perpendicular to x = 1. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-5\right)}^{2}+{\left(y-2\right)}^{2}={\left(\frac{x-1}{1}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+25-10x+{y}^{2}+4-4y={x}^{2}+1-2x\phantom{\rule{0ex}{0ex}}⇒25-10x+{y}^{2}+4-4y-1+2x=0\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-4y-8x+28=0$ #### Question 12: The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest wire being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle. Let X'OX be the bridge and PAQ be the suspension cable. The suspension cable forms a parabola with the vertex at (0, 6). Let the equation of the parabola formed by the suspension cable be ${\left(x-0\right)}^{2}=4a\left(y-6\right)$. (1) It passes through P (−50, 30) and Q (50, 30). $2500=4a\left(30-6\right)$ $4a=\frac{2500}{24}$ Putting the value of 4a in equation (1): ${x}^{2}=\frac{2500}{24}\left(y-6\right)$ (2) Let LM be the supporting wire attached at M, which is 18 m from the mid-point (O) of the bridge. Let the coordinates of L be (18, l). It lies on the parabola (2). ${18}^{2}=\frac{2500}{24}\left(l-6\right)$ Hence, the length of the supporting wire attached to the roadway 18 m from the middle is 9.11 m. #### Question 13: Find the equations of the lines joining the vertex of the parabola y2 = 6x to the point on it which have abscissa 24. [NCERT EXEMPLAR] Let A and B be points on the parabola y2 = 6x and OA, OB be the lines joining the vertex O to the points A and B whose abscissa are 24. Now, y2 = 6 × 24 = 144 y = ± 12 Therefore the coordinates of the points A and B are (24, 12) and (24, –12) respectively. Hence the lines are given by $y-0=±\frac{12-0}{24-0}\left(x-0\right)\phantom{\rule{0ex}{0ex}}⇒±2y=x\phantom{\rule{0ex}{0ex}}$ #### Question 14: Find the coordinates of points on the parabola y2 = 8x whose focal distance is 4. [NCERT EXEMPLAR] We have y2 = 8x ⇒ y2 = 4(2)x Comparing it with the general equation of parabola y2 = 4ax, we will get a = 2 Let the required point be (x1y1) Now, Focal distance = 4 ⇒ x1a = 4 ⇒ x1 + 2 = 4 ⇒ x1 = 2 Now, the point will satisfy the equation of parabola ∴ (y1)2 = 8(2) = 16 y1 = ± 4 Hence, the coordiantes of the points are (2, 4) and (2, −4). #### Question 15: Find the length of the line segment joining the vertex of the parabola y2 = 4ax and a point on the parabola where the line-segment makes an angle θ to the x-axis. [NCERT EXEMPLAR] Let the coordinates of the point on the parabola be B (x1y1). Let BO be the line segment In right triangle AOB ∴ x1 = OBcosθ and y1 = OBsinθ Now, the curve is passing through (x1y1) ∴ (y1)2 = 4a(x1) ⇒( OBsinθ)2 = 4a(OBcosθ) $⇒{\mathrm{OB}}^{2}{\mathrm{sin}}^{2}\mathrm{\theta }=4a\mathrm{OBcos\theta }\phantom{\rule{0ex}{0ex}}⇒\mathrm{OB}=\frac{4a\mathrm{cos\theta }}{{\mathrm{sin}}^{2}\mathrm{\theta }}=4a\mathrm{cosec}\theta .\mathrm{cot}\theta$ #### Question 16: If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola. [NCERT EXEMPLAR] As the vertex and focus lie on y-axis, so y-axis is the axis of the parabola. If the directrix meets the axis of the parabola at point Z, the AZ = AF = 2 OZ = OF + AZ + FA = 2 + 2 + 2 = 6 So, the equation of the directrix is y = 6 i.e., y − 6 = 0 Let P(xy) be any point in the plane of the focus and directrix and MP be the perpendicular distance from P to the directrix, then P lies on parabola iff FP = MP $⇒\sqrt{{\left(x-0\right)}^{2}+{\left(y-2\right)}^{2}}=\frac{\left\|y-6\right\|}{1}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-4y+4={y}^{2}-12y+36\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+8y=32$ #### Question 17: If the line y = mx + 1 is tangent to the parabola y2 = 4x, then find the value of m. [NCERT EXEMPLAR] We have y2 = 4x Substituting the value of y = mx + 1 in y2 = 4x, we get (mx + 1)2 = 4x m2x2 + 2mx + 1 = 4x m2x2 + (2m − 4)x + 1 = 0 .....(1) Since, a tangent touches the curve at a point, the roots of (1) must be equal. D = 0 ⇒ (2m − 4)2 − 4m2 = 0 ⇒ 4m−16m + 16 − 4m2 = 0 m = 1 #### Question 1: The coordinates of the focus of the parabola y2x − 2y + 2 = 0 are (a) (5/4, 1) (b) (1/4, 0) (c) (1, 1) (d) none of these (a) (5/4, 1) Given: The equation of the parabola is y2x − 2y + 2 = 0. $⇒{\left(y-1\right)}^{2}-1=\left(x-2\right)\phantom{\rule{0ex}{0ex}}⇒{\left(y-1\right)}^{2}=x-1$ Let ${Y}^{2}=X$ Comparing with ${Y}^{2}=4aX$: $a=\frac{1}{4}$ Focus = Hence, the focus is at (5/4, 1). #### Question 2: The vertex of the parabola (y + a)2 = 8a (xa) is (a) (−a, −a) (b) (a, −a) (c) (−a, a) (d) none of these (b) (a, −a) Given: The equation of the parabola is (y + a)2 = 8a (xa). Putting : ${Y}^{2}=8aX$ Vertex = Hence, the vertex is at (a, −a). #### Question 3: If the focus of a parabola is (−2, 1) and the directrix has the equation x + y = 3, then its vertex is (a) (0, 3) (b) (−1, 1/2) (c) (−1, 2) (d) (2, −1) (c) (−1, 2) Given: The focus S is at (−2, 1) and the directrix is the line x + y − 3 = 0. The slope of the line perpendicular to x + y − 3 = 0 is 1. The axis of the parabola is perpendicular to the directrix and passes through the focus. ∴ Equation of the axis of the parabola = $y-1=1\left(x+2\right)$ (1) Intersection point of the directrix and the axis is the intersection point of (1) and x + y − 3 = 0. Let the intersection point be K. Therefore, the coordinates of K will be (0, 3). Let (h, k) be the coordinates of the vertex, which is the mid-point of the segment joining K and the focus. Hence, the coordinates of the vertex are (−1, 2). #### Question 4: The equation of the parabola whose vertex is (a, 0) and the directrix has the equation x + y = 3a, is (a) x2 + y2 + 2xy + 6ax + 10ay + 7a2 = 0 (b) x2 − 2xy + y2 + 6ax + 10ay − 7a2 = 0 (c) x2 − 2xy + y2 − 6ax + 10ay − 7a2 = 0 (d) none of these (b) x2 − 2xy + y2 + 6ax + 10ay − 7a2 = 0 Given: The vertex is at (a, 0) and the directrix is the line x + y = 3a. The slope of the line perpendicular to x + y = 3a is 1. The axis of the parabola is perpendicular to the directrix and passes through the vertex. ∴ Equation of the axis of the parabola = $y-0=1\left(x-a\right)$ (1) Intersection point of the directrix and the axis is the intersection point of (1) and x + y = 3a. Let the intersection point be K. Therefore, the coordinates of K are . The vertex is the mid-point of the segment joining K and the focus (h, k). Let P (x, y) be any point on the parabola whose focus is S (h, k) and the directrix is xy = 3a. Draw PM perpendicular to xy = 3a. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(y+a\right)}^{2}={\left(\frac{x+y-3a}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{\left(y+a\right)}^{2}={\left(\frac{x+y-3a}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+2{y}^{2}+2{a}^{2}+4ay={x}^{2}+{y}^{2}+9{a}^{2}+2xy-6ax-6ay\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-7{a}^{2}+10ay+6ax-2xy=0$ #### Question 5: The parametric equations of a parabola are x = t2 + 1, y = 2t + 1. The cartesian equation of its directrix is (a) x = 0 (b) x + 1 = 0 (c) y = 0 (d) none of these (a) x = 0 Given: x = t2 + 1 (1) y = 2t + 1 (2) From (1) and (2): $x={\left(\frac{y-1}{2}\right)}^{2}+1$ On simplifying: ${\left(y-1\right)}^{2}=4\left(x-1\right)$ Let ${Y}^{2}=4X$ Comparing it with y2 = 4ax: a = 1 Therefore, the equation of the directrix is X = −a , i.e. $x-1=-1⇒x=0$. #### Question 6: If the coordinates of the vertex and the focus of a parabola are (−1, 1) and (2, 3) respectively, then the equation of its directrix is (a) 3x + 2y + 14 = 0 (b) 3x + 2y − 25 = 0 (c) 2x − 3y + 10 = 0 (d) none of these. (a) 3x + 2y + 14 = 0 Given: The vertex and the focus of a parabola are (−1, 1) and (2, 3), respectively. ∴ Slope of the axis of the parabola = $\frac{3-1}{2+1}=\frac{2}{3}$ Slope of the directrix = $\frac{-3}{2}$ Let the directrix intersect the axis at K (r, s). Equation of the directrix: $\left(y+1\right)=\frac{-3}{2}\left(x+4\right)$ $⇒3x+2y+14=0$ #### Question 7: The locus of the points of trisection of the double ordinates of a parabola is a (a) pair of lines (b) circle (c) parabola (d) straight line (c) parabola Suppose PQ is a double ordinate of the parabola ${y}^{2}=4ax$. Let R and S be the points of trisection of the double ordinates. Let $\left(h,k\right)$ be the coordinates of R. Then, we have: OL = and RL = k $\therefore RS=RL+LS=k+k=2k\phantom{\rule{0ex}{0ex}}⇒PR=RS=SQ=2k\phantom{\rule{0ex}{0ex}}⇒LP=LR+RP=k+2k=3k$ Thus, the coordinates of P are , which lie on ${y}^{2}=4ax$. $9{k}^{2}=4ah$ Hence, the locus of the point (h, k) is $9{y}^{2}=4ax$, i.e. ${y}^{2}=\left(\frac{4a}{9}\right)x$, which represents a parabola. #### Question 8: The equation of the directrix of the parabola whose vertex and focus are (1, 4) and (2, 6) respectively is (a) x + 2y = 4 (b) xy = 3 (c) 2x + y = 5 (d) x + 3y = 8 (a) x + 2y = 4 Given: The vertex and the focus of a parabola are (1, 4) and (2, 6), respectively. ∴ Slope of the axis of the parabola = $\frac{6-4}{2-1}=2$ Slope of the directrix = $\frac{-1}{2}$ Let the directrix intersect the axis at K (r, s). Equation of the directrix: $\left(y-2\right)=\frac{-1}{2}\left(x-0\right)$ $⇒$ x + 2y = 4 #### Question 9: If V and S are respectively the vertex and focus of the parabola y2 + 6y + 2x + 5 = 0, then SV = (a) 2 (b) 1/2 (c) 1 (d) none of these (b) 1/2 Given: The vertex and the focus of a parabola are V and S, respectively. The given equation of parabola can be rewritten as follows: ${\left(y+3\right)}^{2}-9+5+2x=0$ $⇒{\left(y+3\right)}^{2}+2x=4\phantom{\rule{0ex}{0ex}}⇒{\left(y+3\right)}^{2}=4-2x\phantom{\rule{0ex}{0ex}}⇒{\left(y+3\right)}^{2}=-2\left(x-2\right)$ Let Then, the equation of parabola becomes ${Y}^{2}=-2X$. Vertex = Comparing with y2 = 4ax: $4a=2⇒a=\frac{1}{2}$ Focus = SV = $\sqrt{{\left(2-\frac{3}{2}\right)}^{2}+{\left(-3+3\right)}^{2}}=\frac{1}{2}$ #### Question 10: The directrix of the parabola x2 − 4x − 8y + 12 = 0 is (a) y = 0 (b) x = 1 (c) y = − 1 (d) x = − 1 (c) y = −1 Given: x2 − 4x − 8y + 12 = 0 $⇒{\left(x-2\right)}^{2}-4-8y+12=0\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}=8y-8\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}=8\left(y-1\right)$ Putting X = x − 2, Y = y − 1: ${X}^{2}=8Y$ Comparing with ${X}^{2}=4aY$: a = 2 Equation of the directrix: $Y=-a$ $Y=-2\phantom{\rule{0ex}{0ex}}$ $⇒y-1=-2\phantom{\rule{0ex}{0ex}}⇒y=-2+1\phantom{\rule{0ex}{0ex}}⇒y=-1$ #### Question 11: The equation of the parabola with focus (0, 0) and directrix x + y = 4 is (a) x2 + y2 − 2xy + 8x + 8y − 16 = 0 (b) x2 + y2 − 2xy + 8x + 8y = 0 (c) x2 + y2 + 8x + 8y − 16 = 0 (d) x2y2 + 8x + 8y − 16 = 0 (a) x2 + y2 − 2xy + 8x + 8y − 16 = 0 Let P (x, y) be any point on the parabola whose focus is S (0, 0) and the directrix is xy = 4. Draw PM perpendicular to xy = 4. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}={\left(\frac{x+y-4}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}={\left(\frac{x+y-4}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+2{y}^{2}={x}^{2}+{y}^{2}+16+2xy-8x-8y\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2xy+8x+8y-16=0$ #### Question 12: The line 2xy + 4 = 0 cuts the parabola y2 = 8x in P and Q. The mid-point of PQ is (a) (1, 2) (b) (1, −2) (c) (−1, 2) (d) (−1, −2) (c) (−1, 2) Let the coordinates of P and Q be and , respectively. Slope of PQ = $\frac{2a{t}_{2}-2a{t}_{1}}{a{{t}_{2}}^{2}-a{{t}_{1}}^{2}}$ ......(1) But, the slope of PQ is equal to the slope of 2xy + 4 = 0. ∴ Slope of PQ = $\frac{-2}{-1}=2$ From (1), $\frac{2a{t}_{2}-2a{t}_{1}}{a{{t}_{2}}^{2}-a{{t}_{1}}^{2}}=2$ .....(2) Putting 4a = 8, a = 2 ∴ Focus of the given parabola = (a, 0) = . Using equation (2): $\frac{4\left({t}_{2}-{t}_{1}\right)}{2\left({{t}_{2}}^{2}-{{t}_{1}}^{2}\right)}=2$ $\frac{\left({t}_{2}-{t}_{1}\right)}{\left({{t}_{2}}^{2}-{{t}_{1}}^{2}\right)}=1$ $⇒{t}_{1}+{t}_{2}=1$ As, points P and Q lie on 2x-y+4=0 Let be the mid-point of PQ. Then, we have: ${y}_{1}=\frac{2a{t}_{2}+2a{t}_{1}}{2}=2\left({t}_{1}+{t}_{2}\right)=2$ And, ${x}_{1}=\frac{a{{t}_{1}}^{2}+a{{t}_{2}}^{2}}{2}={{t}_{1}}^{2}+{{t}_{2}}^{2}=-1$ #### Question 13: In the parabola y2 = 4ax, the length of the chord passing through the vertex and inclined to the axis at π/4 is (a) $4\sqrt{2}a$ (b) $2\sqrt{2}a$ (c) $\sqrt{2}a$ (d) none of these (a) $4\sqrt{2}a$ Let OP be the chord. Let the coordinates of P be . From the figure, we have: $O{P}^{2}={{x}_{1}}^{2}+{{y}_{1}}^{2}$ (1) And, $\mathrm{tan}\frac{\mathrm{\pi }}{4}=\frac{{y}_{1}}{{x}_{1}}$ $⇒{x}_{1}={y}_{1}$ (2) Also, lies on the parabola. ${{y}_{1}}^{2}=4a{x}_{1}$ (3) Using (2) and (3): ${{x}_{1}}^{2}=4a{x}_{1}⇒{x}_{1}=4a$ (4) ∴ From (4), (1) and (2), we have: $O{P}^{2}={\left(4a\right)}^{2}+{\left(4a\right)}^{2}=32{a}^{2}\phantom{\rule{0ex}{0ex}}⇒OP=4\sqrt{2}a$ Therefore, the length of the chord is . #### Question 14: The equation 16x2 + y2 + 8xy − 74x − 78y + 212 = 0 represents (a) a circle (b) a parabola (c) an ellipse (d) a hyperbola (b) a parabola Comparing the given equation with ax2 + by2 + 2hxy + 2gx+2fy + c =0, we get: We have: ${h}^{2}=16=ab$ Thus, the given equation represents a parabola. #### Question 15: The length of the latus-rectum of the parabola y2 + 8x − 2y + 17 = 0 is (a) 2 (b) 4 (c) 8 (d) 16 (c) 8 Given: y2 + 8x − 2y + 17 = 0 $⇒{\left(y-1\right)}^{2}-1+8x+17=0\phantom{\rule{0ex}{0ex}}⇒{\left(y-1\right)}^{2}+8x+16=0\phantom{\rule{0ex}{0ex}}⇒{\left(y-1\right)}^{2}=-8\left(x+2\right)$ Let ${Y}^{2}=-8X$ Comparing with y2=4ax: a = 2 Length of the latus rectum = 4a = 8 units #### Question 16: The vertex of the parabola x2 + 8x + 12y + 4 = 0 is (a) (−4, 1) (b) (4, −1) (c) (−4, −1) (d) (4, 1) (a) (−4, 1) Given: x2 + 8x + 12y + 4 = 0 $⇒{\left(x+4\right)}^{2}-16+12y+4=0\phantom{\rule{0ex}{0ex}}⇒{\left(x+4\right)}^{2}+12y-12=0\phantom{\rule{0ex}{0ex}}⇒{\left(x+4\right)}^{2}=-12\left(y-1\right)$ Let ${X}^{2}=-12Y$ Vertex = $\left(X=0,Y=0\right)=\left(x+4=0,y-1=0\right)=\left(x=-4,y=1\right)$ Hence, the vertex is at (−4, 1). #### Question 17: The vertex of the parabola (y − 2)2 = 16 (x − 1) is (a) (1, 2) (b) (−1, 2) (c) (1, −2) (d) (2, 1) (a) (1, 2) Given: (y − 2)2 = 16 (x − 1) Let ${Y}^{2}=16X$ Vertex = Hence, the vertex is at (1, 2). #### Question 18: The length of the latus-rectum of the parabola 4y2 + 2x − 20y + 17 = 0 is (a) 3 (b) 6 (c) 1/2 (d) 9 (c) 1/2 Given: 4y2 + 2x − 20y + 17 = 0 $⇒{y}^{2}+\frac{x}{2}-5y+\frac{17}{4}=0\phantom{\rule{0ex}{0ex}}⇒{\left(y-\frac{5}{2}\right)}^{2}+\frac{x}{2}-2=0\phantom{\rule{0ex}{0ex}}⇒{\left(y-\frac{5}{2}\right)}^{2}=-1\left(\frac{x}{2}-2\right)\phantom{\rule{0ex}{0ex}}⇒{\left(y-\frac{5}{2}\right)}^{2}=\frac{-1}{2}\left(x-4\right)$ Let ∴ Length of the latus rectum = 4a = $\frac{1}{2}$ units #### Question 19: The length of the latus-rectum of the parabola x2 − 4x − 8y + 12 = 0 is (a) 4 (b) 6 (c) 8 (d) 10 (c) 8 Given: x2 − 4x − 8y + 12 = 0 $⇒{\left(x-2\right)}^{2}-8y+8=0\phantom{\rule{0ex}{0ex}}⇒{\left(x-2\right)}^{2}=8y-8=8\left(y-1\right)$ Let ${X}^{2}=8Y$ ∴ Length of the latus rectum = 4a = 8 units #### Question 20: The focus of the parabola y = 2x2 + x is (a) (0, 0) (b) (1/2, 1/4) (c) (−1/4, 0) (d) (−1/4, 1/8) (c) (−1/4, 0) Given: Equation of the parabola = y = 2x2 + x $⇒{x}^{2}+\frac{x}{2}=\frac{y}{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x+\frac{1}{4}\right)}^{2}=\frac{y}{2}+\frac{1}{16}\phantom{\rule{0ex}{0ex}}⇒{\left(x+\frac{1}{4}\right)}^{2}=\frac{8y+1}{16}\phantom{\rule{0ex}{0ex}}⇒{\left(x+\frac{1}{4}\right)}^{2}=\frac{1}{2}\left(y+\frac{1}{8}\right)$ Let ${X}^{2}=\frac{1}{2}Y$ Comparing with ${X}^{2}=4aY$: $a=\frac{1}{8}$ Focus = Hence, the focus is at (−1/4, 0). #### Question 21: Which of the following points lie on the parabola x2 = 4ay? (a) x = at2, y = 2at (b) x = 2at, y = at2 (c) x = 2at2, y = at (d) x = 2at, y = at2 (d) x = 2at, y = at2 Substituting x = 2at, y = at2 in the given equation: ${\left(2at\right)}^{2}=4a\left(a{t}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒4{a}^{2}{t}^{2}=4{a}^{2}{t}^{2}$ Hence, (2at, at2) lies on the parabola x2 = 4ay. #### Question 22: The equation of the parabola whose focus is (1, −1) and the directrix is x + y + 7 = 0 is (a) x2 + y2 − 2xy − 18x − 10y = 0 (b) x2 − 18x − 10y − 45 = 0 (c) x2 + y2 − 18x − 10y − 45 = 0 (d) x2 + y2 − 2xy − 18x − 10y − 45 = 0 (d) x2 + y2 − 2xy − 18x − 10y − 45 = 0 Let P (x, y) be any point on the parabola whose focus is S (1, −1) and the directrix is xy + 7 = 0. Draw PM perpendicular to xy + 7 = 0. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-1\right)}^{2}+{\left(y+1\right)}^{2}={\left(\frac{x+y+7}{\sqrt{1+1}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-1\right)}^{2}+{\left(y+1\right)}^{2}={\left(\frac{x+y+7}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒2\left({x}^{2}+1-2x+{y}^{2}+1+2y\right)={x}^{2}+{y}^{2}+49+2xy+14y+14x\phantom{\rule{0ex}{0ex}}⇒\left(2{x}^{2}+2-4x+2{y}^{2}+2+4y\right)={x}^{2}+{y}^{2}+49+2xy+14y+14x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-45-10y-2xy-18x=0$ Hence, the required equation is x2 + y2 − 2xy − 18x − 10y − 45 = 0. #### Question 23: The area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latusrectum is (a) 12 sq. units (b) 16 sq. units (c) 18 sq. units (d) 24 sq. units The given equation of parabola is x= 12y ...(1) which is of the form x2 = 40y --(2) ∴ comparing (1) and (2), we get 12y = 40y i.e. a = 3 Let P be the focus for parabola, i.e. OP = 3 Length of lotus rectum is 4 = 4 × 3 = 12 i.e. Area of triangle formed = 18 square units. Hence, the correct answer is option C. #### Question 24: The equations of the lines joining the vertex of the parabola y2 = 6x to the points on it which have abscissa 24 are (a) y ± 2x = 0 (b) 2y ± x = 0 (c) x ± 2y = 0 (d) 2x ± y = 0 Given parabola is y2 = 6 Equation of line joining the vertex of parabola to the point on it which have abscissa 24 Hence, equation of live through (0, 0) and (24, $±$12) is y = mx Hence, the correct answer is option B. #### Question 25: The focus of the parabola is (0, –3) and and directrix is y = 3, then its equation is (a) x2 = –12y (b) x2 = 12y (c) y2 =–12x (d) y2 = 12x since focus of the parabola is (0, −3) i.e. y-axis. Equation of the parabola is either x2 = 4ay or x2 = −4ay Now since focus has negative co-ordinate ⇒ equation is of the form x2 = – 4ay ∴ Co-ordinate of focus (0, −a) = (0, −3) a = 3 for which, equation of parabola, x2 = −4ay i.e. x2 = −4(3) i.e. x2 = −12y Hence, the correct answer is option A. #### Question 26: If the vertex of the parabola is the of the point(–3, 0) and the directrix is the line x + 5 = 0, then its equation is (a) y2 = 8 (x + 3) (b) x2 = 8 (y + 3) (c) y2 = –8 (x + 3) (d) y2 = 8 (x + 5) Given that vertex is (−3, 0) and directrix is x + 5 = 0 Let P(x, y) be any point on parabola, Let S determine focus i.e. S(−1, 0) Since SP = PM Now, squaring both sides, Hence, the correct answer is option A. #### Question 27: If the parabola y2 = 4ax passes through the point (3, 2), then the length of its latusrectum is (a) $\frac{2}{3}$ (b) $\frac{4}{3}$ (c) $\frac{1}{3}$ (d) 4 Given parabola is y2 = 4ax and it passes through (3, 2) i.e. (2)2 = 4a(3) i.e. 4 = 4 × 3 × a i.e. a = $\frac{1}{3}$ ∴ Length of lotus rectum is 4a i.e. $4×\frac{1}{3}=\frac{4}{3}$ Hence, the correct answer is option B. #### Question 1: The coordinates of the points on the parabola y2 = 8x whose focal distance is 4 are __________. For y2 = 8defines a given parabola a = 2 [∵ y2 = 4ax Here A and B are the points an parabola whose focal distance is 4. i.e. A is determined by (2, 4) and B is determined by (2, −4) #### Question 2: If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then the length of its latusrectum is __________. For a parabola vertex is (0, 4) and focus is (0, 2) Since focus is (0, a), parabola is of the form x2 = 4ay. where a is 2 Since length of latus rectum is 4 Therefore, length of latus rectum is 4 × 2 = 8 #### Question 3: If the vertex of a parabola is at the origin and directrix is x + 5 = 0, then its latusrectum is __________. Given for a parabola vertex is (0, 0) and directrix is x + 5 = 0 i.e. x = −5 ⇒ parabola is of form y2 = 4ax for x = − a i.e. a = 5 ∴ Length of latus rectum is 4a = 4 × 5 = 20 #### Question 4: The latus rectum of the parabola whose directrix is x + y – 2 = 0 and the focus is (3, –4), is __________. For a given parabola vertex is given by (0, 0), focus is (3, 4) and directrix is x + y − 2 = 0 Distance between directrix and focus is 2a #### Question 5: The equation of the parabola with focus (3, 0) and the directrix x + 3 = 0 is __________. For a given parabola, focus is (3, 0) and directrix is x + 3 = 0 Since focus is (0, 0) a = 3 y2 = 4 × (3) x i.e. y2 = 12x is the required equation of parabola. #### Question 6: If a double ordinate of the parabola y2 = 4ax is a length 8a, then the angle between the lines joining the vertex of the parabola to the ends of this double ordinate is __________. Let us suppose that the ends of double ordinate are P(at2, 2at) and Q = (at2 , −2at Given distance PQ = 8a i.e. 4at = 8a i.e. t = 2 P is given by (4a, 4a) and Q is given by (4a , −4a) #### Question 7: The focal distances of points on the parabola y2 = 16x whose ordinate is twice the abscissa, is __________. Given parabola, y2 = 16x − 4ax y2 = 16x i.e. a = 4 Let us suppose co-ordinate of P are (t, 2t) [since ordinate is twice the abscissa.] Since P lies an y2 = 16x i.e. (2t)2 = 16(t i.e. 4t2 = 16t i.e. t(− 4) = 0 i.e. t = 0 or t = 4 Since t = 0 is not possible t = 4 Hence, co-ordinate of P are (4, 8) #### Question 8: The coordinates of the end-points of the latusrectum of the parabola x2 + 8y = 0 are __________. Given parabola is x2 + 8y = 0 i.e. x2 = −4(2)y i.e. a = 2 ∴ Focus point is (0, −2) Let us suppose co-ordinate of P be (p, 2) and Q be (p, −2) Since both points P and Q lie an x2 = −8y i.e. p2 = − 8 (−2) for (p, −2) i.e. p2 = 16 i.e. p = ± 4 ∴ co-ordinate of P are (4, −2) and Q are (−4, −2) #### Question 9: The coordinates of the point on the parabola y2 = 18x whose ordinate is three times the abscissa, are __________. For a given parabola, y2 = 18x. Let P be any point on parabola, i.e. P has co-ordinate (t, 3t ∴ (3t)2 = 18t i.e. 9t2 = 18t i.e. t(t − 2) = 0 Since t = 0 is not possible t = 2 i.e. co-ordinate of P are (2, 6) #### Question 10: The equations of the latus rectum and the tangent at the vertex of a parabola are x + y = 8 and x + y = 12 respectively. The length of the latus retum is __________. For a given parabola, equation of latus rectum is x + y = 8 and equation of tangent is x + y = 12 i.e. both the equations are parallel, Since distance between latus rectum and tangent is a Hence, length of latus rectum is 4a #### Question 11: If the vertex of the parabola y = x2 – 16x + k lies on x-axis, then k = __________. Given parabola is, y = x2 − 16x + k = x2 − 2x(8) + k = x2 − 2x(8) + 64 − 64 + k y = (x − 8)2 − 64 + k Since parabola has vertex an x - axis ⇒ −64 + k = 0 i.e. k = 64 #### Question 12: The equation of the axis of the parabola 2x2 + 5y – 3x + 4 = 0 is __________. Given parabola is, #### Question 13: The equation of the directrix of the parabola x2 + 8y – 2x – 7 = 0 is __________. Given equation of parabola is, x2 + 8y − 2− 7 = 0 i.e. x2 − 2x + 8y − 7 = 0 i.e. x2 − 2x + 1 + 8y − 7 − 1 = 0 i.e. (x − 1)2 = − 8y + 8 i.e. (− 1)2 = − 8(y − 1) = −4(2) (y − 1) ∴ equation of directrix is given by y − 1 = 2 i.e. y = 1 + 2 = 3 #### Question 14: The equation of the parabola whose focus is the point (2, 3) and directrix is the line x – 4y + 3 = 0, is __________. Let P(x, y) be any point an parabola, Given focus is (2, 3) ∴ Distance between focus and P is, ...(1) Also directrix is given by x − 4y + 3 = 0 ∴ The perpendicular distance from the point P to the line x − 4y + 3 = 0 is $\frac{\left\|x-4y+3\right\|}{\sqrt{1+16}}$ [since distance of any point an parabola plane focus is same as distance between point and directrix] By squaring both sides, is the required equation of parabola. #### Question 15: The equation of the parabola having focus at (–1, –2) and the directrix x – 2y + 3 = 0 is __________. Let P(xy) be any point in parabola The distance between focus and P is And distance (perpendicular distance) of directrix from point P is By equating (1) and (2), We get $\sqrt{{\left(x+1\right)}^{2}+{\left(y+2\right)}^{2}}=\left\|\frac{x-2y+3}{\sqrt{5}}\right\|$ By squaring both sides, we get is the required equation of parabola #### Question 1: Write the axis of symmetry of the parabola y2 = x. Clearly, the axis of symmetry of the given parabola is the x-axis. #### Question 2: Write the distance between the vertex and focus of the parabola y2 + 6y + 2x + 5 = 0. Given: ${y}^{2}+6y+2x+5=0$ Let Y = y+3, $X=x-2$ From (1), we have: ${Y}^{2}=-2X$ (2) Putting $4a=2$: $a=\frac{1}{2}$ Focus = Vertex = Thus, we have: Focus = $\left(\frac{3}{2},-3\right)$ Vertex = $\left(2,-3\right)$ Distance between the vertex and the focus: #### Question 3: Write the equation of the directrix of the parabola x2 − 4x − 8y + 12 = 0. Given: x2 − 4x − 8y + 12 = 0 Let Y = y−1, $X=x-2$ ∴ From (1), we have: ${X}^{2}=8Y$ (2) Comparing with ${x}^{2}=4ay$: $a=2$ Directrix = Y = −a y − 1 = −a ⇒y = −a + 1 = −2 + 1 = −1 Therefore, the required equation of the directrix is $y=-1$. #### Question 4: Write the equation of the parabola with focus (0, 0) and directrix x + y − 4 = 0. Let P (x, y) be any point on the parabola whose focus is S (0, 0) and the directrix is xy = 4. Draw PM perpendicular to xy = 4. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}={\left(\frac{x+y-4}{\sqrt{1+1}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}={\left(\frac{x+y-4}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+2{y}^{2}={x}^{2}+{y}^{2}+16+2xy-8y-8x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2xy+8x+8y-16=0$ #### Question 5: Write the length of the chord of the parabola y2 = 4ax which passes through the vertex and is inclined to the axis at $\frac{\mathrm{\pi }}{4}$. Let OP be the chord. Let the coordinates of P be . From the figure, we have: $O{P}^{2}={{x}_{1}}^{2}+{{y}_{1}}^{2}$ (1) And, $\mathrm{tan}\frac{\mathrm{\pi }}{4}=\frac{{y}_{1}}{{x}_{1}}$ $⇒{x}_{1}={y}_{1}$ (2) Also, lies on the parabola. ${{y}_{1}}^{2}=4a{x}_{1}$ (3) Using (2) and (3), we get: ${{x}_{1}}^{2}=4a{x}_{1}⇒{x}_{1}=4a$ ...(4) ∴ From (4), (1) and (2), we have: $O{P}^{2}={\left(4a\right)}^{2}+{\left(4a\right)}^{2}=32{a}^{2}\phantom{\rule{0ex}{0ex}}⇒OP=4\sqrt{2}a$ Therefore, the length of the chord is . #### Question 6: If b and c are lengths of the segments of any focal chord of the parabola y2 = 4ax, then write the length of its latus-rectum. Let S (a, 0) be the focus of the given parabola. Let the end points of the focal chord be . SP and SQ are segments of the focal chord with lengths b and c, respectively. SP = b, SQ = c Now, we have: $\frac{1}{SP}+\frac{1}{SQ}=\frac{1}{a\left(1+{t}^{2}\right)}+\frac{{t}^{2}}{a\left(1+{t}^{2}\right)}=\frac{1}{a}$ $⇒\frac{1}{b}+\frac{1}{c}=\frac{1}{a}\phantom{\rule{0ex}{0ex}}⇒\frac{b+c}{bc}=\frac{1}{a}\phantom{\rule{0ex}{0ex}}⇒a=\frac{bc}{b+c}$ ∴ Length of the latus rectum = 4a = $\frac{4bc}{b+c}$ #### Question 7: PSQ is a focal chord of the parabola y2 = 8x. If SP = 6, then write SQ. The coordinates of the focal chord are . Comparing y2 = 8x with ${y}^{2}=4ax$: a = 2 Therefore, the coordinates of the focus S is . Given: SP = 6 Thus, we have: SQ = $\sqrt{{\left(2-\frac{2}{{t}^{2}}\right)}^{2}+\left(\frac{4}{{t}^{2}}\right)}$=$\sqrt{{\left(2-\frac{2}{2}\right)}^{2}+\left(\frac{4}{2}\right)}$ = 3 #### Question 8: Write the coordinates of the vertex of the parabola whose focus is at (−2, 1) and directrix is the line x + y − 3 = 0. Given: The focus S is at (−2, 1) and the directrix is the line x + y − 3 = 0. The slope of the line perpendicular to x + y − 3 = 0 is 1. The axis of the parabola is perpendicular to the directrix and passes through the focus. ∴ Equation of the axis of the parabola = $y-1=1\left(x+2\right)$ (1) Intersection point of the directrix and axis is the intersection point of (1) and x + y − 3 = 0. Let the intersection point be K. Therefore, the coordinates of K are (0, 3). Let (h, k) be the coordinates of the vertex, which is the mid-point of the line segment joining K and the focus. Hence, the coordinates of the vertex are (−1, 2). #### Question 9: If the coordinates of the vertex and focus of a parabola are (−1, 1) and (2, 3) respectively, then write the equation of its directrix. Given: The vertex and the focus of a parabola are (−1, 1) and (2, 3), respectively. ∴ Slope of the axis of the parabola = $\frac{3-1}{2+1}=\frac{2}{3}$ Slope of the directrix = $\frac{-3}{2}$ Let the directrix intersect the axis at K (r, s). Now, required equation of the directrix: $\left(y+1\right)=\frac{-3}{2}\left(x+4\right)$ $⇒3x+2y+14=0$ #### Question 10: If the parabola y2 = 4ax passes through the point (3, 2), then find the length of its latus rectum. We have y2 = 4ax Since, the parabola is passing through the point (3, 2) Hence, it will satisfy the equation of the parabola. ∴ 22 = 4(a)(3) $⇒a=\frac{1}{3}$ Lenth of the latus ractum is given by $4a\phantom{\rule{0ex}{0ex}}=4×\frac{1}{3}\phantom{\rule{0ex}{0ex}}=\frac{4}{3}$ #### Question 11: Write the equation of the parabola whose vertex is at (−3,0) and the directrix is x + 5 = 0. The general equation of the parabola is (yk)2 = 4a(xh) Here, the (h, k) = (−3,0) Now, the directrix is given by x = h − a ⇒ −5 = −3 − a [∵ x + 5 = 0 ⇒ x = −5] a = 2 Hence, the equation is given by (y − 0)2 = 4(2)(x + 3) y2 = 8 (x + 3) View NCERT Solutions for all chapters of Class 16
# Into Math Grade 4 Module 3 Lesson 3 Answer Key Use Division to Solve Multiplicative Comparison Problems We included HMH Into Math Grade 4 Answer Key PDF Module 3 Lesson 3 Use Division to Solve Multiplicative Comparison Problems to make students experts in learning maths. ## HMH Into Math Grade 4 Module 3 Lesson 3 Answer Key Use Division to Solve Multiplicative Comparison Problems I Can solve a multiplicative comparison problem by using division with drawings and equations. Step It Out 1. Olivia is in the photography club. She takes 15 photos at a garden. That is 3 times as many photos as she takes at school. How many photos does Olivia take at school? A. Use a bar model to show the problem. Let n represent the unknown. Explanation: B. Write an equation to model the problem. Use n for the unknown. Problem: 3 times as many as n is 15. Equation: __ × n = ___ Explanation: 5 x 3 = 15 15 ÷ 3 = 15 Connect to Vocabulary Inverse operations, such as multiplication and division, undo each other. 3 × 26 6 ÷ 3 = 2 C. Use the inverse operation to solve the problem. ___ ÷ ___ = n ___ = n Explanation: 15 ÷ 3 = n 5 = n D. Olivia takes ___ photos at school. Explanation: Olivia takes 5 photos at school. Turn and Talk How did you represent 3 times as many in your bar model and equations? Explanation: By following the division equation. Step It Out 2. This year, 28 students are in the chess club. Last year, 7 students were in the dub. How many times as many students are in the chess club this year than last year? A. Write a multiplication equation to model the problem. Use n for the unknown. n × __ = __ Explanation: n x 4 = 28 7 x 4 = 28 B. Write a division equation to solve the problem. ___ ÷ ___ = ___ ___ = n Explanation: 28 ÷ 7 = 4 4 = n This year, 4 times as many students are in the chess club. Turn and Talk Why can you use division to solve a multiplicative comparison problem, but not an additive comparison problem? Explanation: One way you might describe the difference between the two is this. Additive comparisons. You focus on the difference between the two quantities. With multiplicative comparison, you focus on how many times larger or smaller one quantity is compared to the other. Check Understanding Math Board Write multiplication and division equations to model and solve the problem. Use n for the unknown. Question 1. Leslie collects 24 aluminum cans and 8 glass bottles for the recycling club. How many times as many cans as bottles does Leslie collect? Leslie collects ___ times as many cans as bottles. n x 8 = 24 4 x 8 = 24 24 ÷ 8 = 4 Explanation: Leslie collects 4 times as many cans as bottles. Question 2. The students in the cooking club bake 12 blueberry muffins. They bake twice as many blueberry muffins as corn muffins. How many corn muffins do the students bake? The students bake ___ corn muffins in the cooking club. n x 2 = 12 6 x 2 = 12 12 ÷ 2 = 6 Explanation: The students bake 6 corn muffins in the cooking club. Question 3. Use Tools The library club receives 56 fiction books and 8 nonfiction books. How many times as many fiction books as nonfiction books does the club receive? 56 ÷ 8 = 7 7 x 8 = 56 Explanation: 7 times as many fiction books as nonfiction books that the club receive Question 4. Use Structure Bob fills a basket with carnations and roses. He uses 18 roses and twice as many roses as carnations. How many carnations does Bob use? • Write an equation to model the problem. ____ n = ____ • Use the inverse operation to write a different equation and solve the problem. ____ ___ = n 2 x n = 18 9 x 2 = 18 18 ÷ 2 = 9 Explanation: 9 carnations that Bob use Question 5. Model with Mathematics A sweater costs $40. That is 5 times as much as a shirt. What is the price of the shirt ? Write multiplication and division equations to model and solve the problem. Use s for the unknown. Answer: n x 5 = 40 8 x 5 = 40 40 ÷ 5 = 8 Explanation:$8 is the price of the shirt. Question 6. Open Ended Write a multiplicative comparison word problem that can be modeled and solved with the equations 7 × m = 21 and 21 ÷ 7 = m. 21 ÷ 7 = m. m = 3 7 × m = 21 7 x 3 = 21 Explanation: In a Dance studio they played a song for 21 minutes. That is 7 times as long as the normal song. Question 7. Use Tools In the martial arts club, students spend 20 minutes working on blocking. That is 4 times as long as they spend meditating. How many minutes do the students spend meditating? • Complete the bar model to show the problem. • Use inverse operations to write equations to model and solve the problem. n x 4 = 20 5 x 4 = 20 20 ÷ 4 = 5 Explanation: 5 minutes that the students spend meditating Question 8. Students in the engineering club build two train tracks. Track A is 24 feet long. That is 3 times as long as the length of Track B. What is the length of Track B? Show your work. ___________________________ n x 3 = 24 8 x 3 = 24 24 ÷ 3 = 8 Explanation: length of Track B = 8 feet long. Question 9. Critique Reasoning Kevin walks 6 laps and jogs 2 laps. Kevin writes this equation to find how many times as many laps he walks as jogs. Is Kevin correct? Explain.
Courses Courses for Kids Free study material Offline Centres More Store # Solve the following. $\mathop {\lim }\limits_{n \to \infty } \sin \left[ {\pi \sqrt {{n^2} + 1} } \right]$ is equal toA) $\infty$ B) 0C) Does not existD) None of above Last updated date: 11th Aug 2024 Total views: 389.4k Views today: 10.89k Verified 389.4k+ views Hint: In this question, we will apply the trigonometry formula $\sin x = {\left( { - 1} \right)^{n + 1}}\sin \left[ {n\pi - x} \right]$. Because when we apply the limit over the given function, it will give $\sin \infty$. So, we have to check for two cases of$\sin (n\pi - x)$. The first case is n is even. When we put the even values of n they will lie in the fourth quadrant. In the fourth quadrant, the value of the sine function is negative. Therefore, $\sin (n\pi - x) = - \sin x$. The second case is odd. When we put the odd values of n they will lie in the second quadrant. In the second quadrant, the value of sin function is positive. Therefore, $\sin (n\pi - x) = \sin x$. Hence, we can write $\sin (n\pi - x) = {\left( { - 1} \right)^{n + 1}}\sin x$ in a generalized form. Then apply rationalization to the numerator. Here, we will apply rationalization because when we will apply the limit to the function we will get the value $\sin \left( {\infty - \infty } \right)$, which is not possible. The other algebraic formula that we will apply here during simplification is as stated below. $\left( {a - b} \right)\left( {a + b} \right) = \left( {{a^2} - {b^2}} \right)$ Complete step by step solution: In this question, it is given that $\Rightarrow \mathop {\lim }\limits_{n \to \infty } \sin \left[ {\pi \sqrt {{n^2} + 1} } \right]$ As we know the trigonometry formula $\sin x = {\left( { - 1} \right)^{n + 1}}\sin \left[ {n\pi - x} \right]$. Here, substitute the value of x is equal to $\pi \sqrt {{n^2} + 1}$. Therefore, we will get $\Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n + 1}}\sin \left( {n\pi - \pi \sqrt {{n^2} + 1} } \right)$ Take out $\pi$ as a common factor. Because it has a constant value. $\Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n + 1}}\sin \pi \left( {n - \sqrt {{n^2} + 1} } \right)$ When we substitute value $n = \infty$, we will get $\sin \left( {\infty - \infty } \right)$. That is not possible. Therefore, Let us rationalize the numerator of the above step. $\Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n + 1}}\sin \pi \left( {\dfrac{{\left( {n - \sqrt {{n^2} + 1} } \right)\left( {n + \sqrt {{n^2} + 1} } \right)}}{{n + \sqrt {{n^2} + 1} }}} \right)$ Let us apply the formula $\left( {a - b} \right)\left( {a + b} \right) = \left( {{a^2} - {b^2}} \right)$. Here, $a = n$ and $b = \sqrt {{n^2} + 1}$ So, $\Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n + 1}}\sin \pi \left( {\dfrac{{{n^2} - {{\left( {\sqrt {{n^2} + 1} } \right)}^2}}}{{n + \sqrt {{n^2} + 1} }}} \right)$ Let us simplify the above step. $\Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n + 1}}\sin \pi \left( {\dfrac{{{n^2} - ({n^2} + 1)}}{{n + \sqrt {{n^2} + 1} }}} \right)$ Now, open the brackets and multiply with -1 with bracket values. $\Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n + 1}}\sin \pi \left( {\dfrac{{{n^2} - {n^2} - 1}}{{n + \sqrt {{n^2} + 1} }}} \right)$ By applying subtraction, $\Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n + 1}}\sin \pi \left( {\dfrac{{ - 1}}{{n + \sqrt {{n^2} + 1} }}} \right)$ When we substitute value $n = \infty$, then we will get, $\Rightarrow {\left( { - 1} \right)^{\infty + 1}}\sin \pi \left( {\dfrac{{ - 1}}{\infty }} \right)$ As we know that the value of $\dfrac{1}{\infty }$ is 0. Therefore, $\Rightarrow {\left( { - 1} \right)^{\infty + 1}}\sin \pi \left( 0 \right)$ 0 multiply with any number the answer will be 0. $\Rightarrow {\left( { - 1} \right)^{\infty + 1}}\sin 0$ We know the value of $\sin 0$ is 0. $\Rightarrow 0$ $\Rightarrow \mathop {\lim }\limits_{n \to \infty } \sin \left[ {\pi \sqrt {{n^2} + 1} } \right] = 0$ Note: We have to remember the trigonometry formula and trigonometry ratio values at the angle $0^\circ ,30^\circ ,45^\circ ,60^\circ ,90^\circ$. And we have to remember the trigonometric function values in all quadrants.
# What is 12 out of 16 as a Percentage? 12 out of 16 is 75% when converted to a percentage. To get this answer, divide 12 by 16 and then multiply the answer by 100. Checkout this video: ## Introduction When you see a math problem that includes a fraction, such as “12 out of 16,” you may not know what to do with it. This is because most of us are more familiar with percentages, which are a way of expressing a number as a fraction of 100. So, when you want to know what 12 out of 16 equals as a percentage, you first need to convert 12/16 into an equivalent fraction with a denominator (bottom number) of 100. ## What is 12 out of 16 as a Percentage? To calculate 12 out of 16 as a percentage, divide 12 by 16 and multiply the result by 100. Here’s how that works in mathematical terms: 12 ÷ 16 × 100 = 75% Therefore, the answer is that 12 out of 16 as a percentage is 75%. ## Conclusion 100% of anything is the whole amount. If we divide 16 into 12, we get 0.75. Therefore, 12 out of 16 as a percentage is 75%.
# Triangle Medians and Centroids A centroid of a triangle is found at the intersection of the medians of the triangle. A median is a segment drawn from the vertex of a triangle to the midpoint of the opposite side. ## Finding a median To find the median of a triangle, you first need to find the midpoints of the sides of your triangle. To do this, we will use a simple tool in Geogebra, the Midpoint or Center tool . In the exercise below, select the Midpoint or Center tool , which is found under the Point tool . Then simple click on the segment AB. Answer the question below. What are the approximate coordinates of the midpoint of segment AB? In the exercise below, find the centroid of the triangle. First find the midpoint of all three sides as you did above. Then connect the midpoints with the opposite vertices. Do this by selecting the Segment tool , which is found under the Line menu . Select the vertex, and then the midpoint that is on the opposite side. After you have constructed all three medians, place a point at their intersection using the Intersect tool . Answer the question that follows. What are the approximate coordinates of the centroid? ## Special Property of the Centroid In the exercise below, the centroid (point C) has already been found for you. For this exercise, find the area of each portion of the overall triangle. Each portion (there are 6 total) is shaded a different color. First select the Area tool , which is found under the Angle menu . Then simply click on each of the individual portions. Answer the questions that follow. What do you notice about the area of each smaller triangle? Recall that the area of a shape is the amount of space inside that shape. Based on this, what could we say the centroid is the center of? Assume that the triangle portions all have a mass (similar to weight) that is distributed evenly across their entire area. Based on this, what could we now say the centroid is the center of? (*Hint, we could also call it the balancing point).
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ## Algebra I (2018 edition) ### Unit 2: Lesson 8 Linear equations with parentheses # Equations with parentheses Sal solves the equation -9 - (9x - 6) = 3(4x + 6) using the distributive property. Created by Sal Khan and Monterey Institute for Technology and Education. ## Video transcript We have the equation negative 9 minus this whole expression, 9x minus 6-- this whole thing is being subtracted from negative 9-- is equal to 3 times this whole expression, 4x plus 6. Now, a good place to start is to just get rid of these parentheses. And the best way to get rid of these parentheses is to kind of multiply them out. This has a negative 1-- you just see a minus here, but it's just really the same thing as having a negative 1-- times this quantity. And here you have a 3 times this quantity. So let's multiply it out using the distributive property. So the left-hand side of our equation, we have our negative 9. And then we want to multiply the negative 1 times each of these terms. So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to-- let's distribute the 3-- 3 times 4x is 12x. And then 3 times 6 is 18. Now what we want to do, let's combine our constant terms, if we can. We have a negative 9 and a 6 here, on this side, we've combined all of our like terms. We can't combine a 12x and an 18, so let's combine this. So let's combine the negative 9 and the 6, our two constant terms on the left-hand side of the equation. So we're going to have this negative 9x. So we're going to have negative 9x plus-- let's see, we have a negative 9 and then plus 6-- so negative 9 plus 6 is negative 3. So we're going to have a negative 9x, and then we have a negative 3, so minus 3 right here. That's the negative 9 plus the 6, and that is equal to 12x plus 18. Now, we want to group all the x terms on one side of the equation, and all of the constant terms-- the negative 3 and the positive 18 on the other side-- I like to always have my x terms on the left-hand side, if I can. You don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right. And the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. Now, on the left-hand side, I have negative 9x minus 12x. So negative 9 minus 12, that's negative 21. Negative 21x minus 3 is equal to-- 12x minus 12x, well, that's just nothing. That's 0. So I could write a 0 here, but I don't have to write anything. That was the whole point of subtracting the 12x from the left-hand side. And that is going to be equal to-- so on the right-hand side, we just are left with an 18. We are just left with that 18 here. These guys canceled out. Now, let's get rid of this negative 3 from the left-hand side. So on the left-hand side, we only have x terms, and on the right-hand side, we only have constant terms. So the best way to cancel out a negative 3 is to add 3. So it cancels out to 0. So we're going to add 3 to the left, let's add 3 to the right. And we get-- the left-hand side of the equation, we have the negative 21x, no other x term to add or subtract here, so we have negative 21x. The negative 3 and the plus 3, or the positive 3, cancel out-- that was the whole point-- equals-- what's 18 plus 3? 18 plus 3 is 21. So now we have negative 21x is equal to 21. And we want to solve for x. So if you have something times x, and you just want it to be an x, let's divide by that something. And in this case, that something is negative 21. So let's divide both sides of this equation by negative 21. Divide both sides by negative 21. The left-hand side, negative 21 divided by negative 21, you're just left with an x. That was the whole point behind dividing by negative 21. And we get x is equal to-- what's 21 divided by negative 21? Well, that's just negative 1. Right? You have the positive version divided by the negative version of itself, so it's just negative 1. So that is our answer. Now let's verify that this actually works for that original equation. So let's substitute negative 1 into that original equation. So we have negative 9-- I'll do it over here; I'll do it in a different color than we've been using-- we have negative 9 minus-- that 1 wasn't there originally, it's there implicitly-- minus 9 times negative 1. 9 times-- I'll put negative 1 in parentheses-- minus 6 is equal to-- well, actually, let me just solve for the left-hand side when I substitute a negative 1 there. So the left-hand side becomes negative 9, minus 9 times negative 1 is negative 9, minus 6. And so this is negative 9 minus-- in parentheses-- negative 9 minus 6 is negative 15. So this is equal to negative 15. And so we get negative 9-- let me make sure I did that-- negative 9 minus 6, yep, negative 15. So negative 9 minus negative 15, that's the same thing as negative 9 plus 15, which is 6. So that's what we get on the left-hand side of the equation when we substitute x is equal to negative 1. We get that it equals 6. So let's see what happens when we substitute negative 1 to the right-hand side of the equation. I'll do it in green. We get 3 times 4 times negative 1 plus 6. So that is 3 times negative 4 plus 6. Negative 4 plus 6 is 2. So it's 3 times 2, which is also 6. So when x is equal to negative 1, you substitute here, the left-hand side becomes 6, and the right-hand side becomes 6. So this definitely works out.
Document Sample ``` Business Calculus Math 1431 Unit 2.6 The Derivative Mathematics Department Louisiana State University Section 2.6: The Derivative Business Calculus - p. 1/44 Introduction Introduction The Tangent Problem The Slope? Introduction Secant lines Small h The main idea An animation Slope of Tangent Example Section 2.6: The Derivative Business Calculus - p. 2/44 Introduction Introduction In an earlier lecture we considered the idea of Introduction ﬁnding the average rate of change of some quan- The Tangent Problem tity Q(t) over a time interval [t1 , t2 ] and asked The Slope? Secant lines what would happen when the time interval became Small h smaller and smaller. This led to the notion of limits, The main idea An animation which we have examined in the previous two lec- Slope of Tangent Example erage rates of change and apply what we learned It turns out that the derivative has an equivalent (purely) mathematical formulation: that of ﬁnding the line tangent to a curve at some point. This is where we will start. Section 2.6: The Derivative Business Calculus - p. 3/44 The Tangent Problem Introduction Suppose y = f (x) is some given function and a is Introduction some ﬁxed point in the domain. We will explore the The Tangent Problem idea of ﬁnding the equation of the line tangent to The Slope? Secant lines the graph of f at the point (a, f (a)). Look at the Small h picture below: The main idea An animation Slope of Tangent Example f (a) a The tangent line at x = a is the unique line that goes through (a, f (a)) and just touches the graph there. Section 2.6: The Derivative Business Calculus - p. 4/44 The Slope? Introduction Recall the an equation of the line is determined Introduction once we know a point P and the slope m. If The Tangent Problem P = (a, f (a)) then the equation of the line tangent The Slope? Secant lines to the curve will take the form Small h The main idea y − f (a) = m(x − a) An animation Slope of Tangent Example for some slope m that we have to determine. How do we determine the slope of the tangent line? We will do so by a limiting process. Section 2.6: The Derivative Business Calculus - p. 5/44 Secant lines Given the graph of a function y = f (x) we call a Introduction secant line a line that connects two points on a Introduction The Tangent graph. Problem The Slope? f (a + h) Secant lines In the graph to the right a red Small h The main idea line joins the points (a, f (a)) and f (a) An animation (a + h, f (a + h). ( Think of h as a Slope of Tangent Example relatively small number.) aa+h Now, what is the slope of the secant line? The change in y is ∆y = f (a + h) − f (a) and the change in x is ∆x = a + h − a = h. So the slope of the secant line is: ∆y f (a + h) − f (a) = . ∆x h Section 2.6: The Derivative Business Calculus - p. 6/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 Small h Introduction The following graph illustrates what happens when Introduction we choose smaller values of h. Notice how the se- The Tangent Problem cant lines get closer to the tangent line. The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example f (a) a Section 2.6: The Derivative Business Calculus - p. 7/44 The main idea Introduction The main idea is to let h get smaller and smaller. Introduction Remember all we need is the slope of the tangent The Tangent Problem line so we will compute The Slope? Secant lines f (a + h) − f (a) Small h lim . The main idea h→0 h An animation Slope of Tangent Example In the following animation notice how the slopes of the secant lines approach the slope of the tangent line. Section 2.6: The Derivative Business Calculus - p. 8/44 Convergence of Secant Lines: An animation Introduction Introduction The Tangent Problem The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example Notice how the secant line and hence its slopes converge to the tangent line. Section 2.6: The Derivative Business Calculus - p. 9/44 Slope of the Tangent Line Introduction Since the slope of the secant lines are given by the Introduction formula The Tangent Problem f (a + h) − f (a) The Slope? , Secant lines h Small h the slope of the tangent line is given by The main idea An animation Slope of Tangent f (a + h) − f (a) Example lim , h→0 h when this exists. We will call this number the derivative of f at a and denote it f ′ (a). Thus The derivative of f at x = a is given by ′ f (a + h) − f (a) f (a) = lim . h→0 h Section 2.6: The Derivative Business Calculus - p. 10/44 Example To illustrate some of these ideas let’s consider the Introduction Introduction following example: The Tangent Problem The Slope? Example 1: Let f (x) = x2 and ﬁx a = 1. Compute Secant lines Small h the slope of the secant line that connects (a, f (a)) The main idea and (a + h, f (a + h) for An animation Slope of Tangent s h = 1 Example s h = .5 s h = .1 s h = .01 Compute the derivative of f at x = 1, i.e. f ′ (1). Finally, ﬁnd the equation of the line tangent to the graph of f (x) = x2 and x = 1. Section 2.6: The Derivative Business Calculus - p. 11/44 f (x) = x2 Let mh be the slope of the secant line. Then Introduction Introduction The Tangent f (1 + h) − f (1) (1 + h)2 − 1 Problem mh = = . The Slope? h h Secant lines Small h (1+1)2 −1 s For h = 1 m1 = 1 =4−1= 3. The main idea An animation (1.5)2 −1 Slope of Tangent s For h = .5 m.5 = .5 = 2.5. Example (1.1)2 −1 s For h = .1 m.1 = .1 = 2.1 (1.01)2 −a s For h = .01 m.01 = .01 = 2.01 Section 2.6: The Derivative Business Calculus - p. 12/44 f (x) = x2 Let mh be the slope of the secant line. Then Introduction Introduction The Tangent f (1 + h) − f (1) (1 + h)2 − 1 Problem mh = = . The Slope? h h Secant lines Small h (1+1)2 −1 s For h = 1 m1 = 1 =4−1= 3. The main idea An animation (1.5)2 −1 Slope of Tangent s For h = .5 m.5 = .5 = 2.5. Example (1.1)2 −1 s For h = .1 m.1 = .1 = 2.1 (1.01)2 −a s For h = .01 m.01 = .01 = 2.01 What do you think the limit will be as h goes to 0? Section 2.6: The Derivative Business Calculus - p. 12/44 f (x) = x2 The limit is the derivative. We compute Introduction Introduction The Tangent ′ f (a + h) − f (a) Problem f (1) = lim The Slope? h→0 h Secant lines (1 + h)2 − 1 Small h = lim The main idea h→0 h An animation Slope of Tangent 1 + 2h + h2 − 1 Example = lim h→0 h 2h + h2 = lim h→0 h = lim 2 + h = 2. h→0 Therefore the slope of the tangent line is m = 2. Section 2.6: The Derivative Business Calculus - p. 13/44 f (x) = x2 Introduction The equation of the tangent line is now an easy Introduction matter. The slope m is 2 and the point P that the The Tangent Problem line goes through is (1, f (1)) = (1, 1). Thus, we get The Slope? Secant lines Small h y − 1 = 2(x − 1) The main idea An animation or Slope of Tangent Example y = 2x − 1. On the next slide we give a graphical representa- tion of what we have just computed. Section 2.6: The Derivative Business Calculus - p. 14/44 f (x) = x2 Introduction Introduction The Tangent Problem The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example Section 2.6: The Derivative Business Calculus - p. 15/44 f (x) = x2 Introduction Introduction The Tangent Problem The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example Here h = 1 and the secant line goes through the points (1, 1) and (2, 4) Section 2.6: The Derivative Business Calculus - p. 15/44 f (x) = x2 Introduction Introduction The Tangent Problem The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example Here h = .5 and the secant line goes through the points (1, 1) and (1.5, 2.25) Section 2.6: The Derivative Business Calculus - p. 15/44 f (x) = x2 Introduction Introduction The Tangent Problem The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example Here h = .1 and the secant line goes through the points (1, 1) and (1.1, 1.21) Section 2.6: The Derivative Business Calculus - p. 15/44 f (x) = x2 Introduction Introduction The Tangent Problem The Slope? Secant lines Small h The main idea An animation Slope of Tangent Example Here h = .01 and the secant line goes through the points (1, 1) and (1.01, 1.0201) Section 2.6: The Derivative Business Calculus - p. 15/44 Rates of Change Rates of change Velocity Example Rates of Change Section 2.6: The Derivative Business Calculus - p. 16/44 Rates of change Rates of Change Given a function y = f (x) the difference quotient Rates of change Velocity f (x + h) − f (x) Example h measures the average rate of change of y with respect to x over the interval [x, x + h]. As the interval becomes smaller, i.e. as h goes to 0, we obtain the instantaneous rate of change f (x + h) − f (x) lim , h→0 h which is precisely our deﬁnition of the derivative f ′ (x). Thus, the derivative measures in an instant the rate of change of f (x) with respect to x. Section 2.6: The Derivative Business Calculus - p. 17/44 Velocity If s(t) is the distance travelled by an object (your Rates of Change car for instance) as a function of time t then the Rates of change Velocity quantity Example s(t + h) − s(t) h is the average rate of change of distance over the time interval [t, t + h]. This is none other than your average velocity. The quantity s(t + h) − s(t) lim h→0 h is the instantaneous rate of change: This is none thought of as a derivative machine. Section 2.6: The Derivative Business Calculus - p. 18/44 Example ice Rates of Change Example 2: Suppose the distance travelled by a Rates of change 1 car (in feet) is given by the function s(t) = 2 t2 + t Velocity Example where 0 ≤ t ≤ 20 is measured in seconds. s Find the average velocity over the time interval x [10, 11] x [10, 10.1] x [10, 10.01] s Find the instantaneous velocity at t = 10. s Compare the above results. Section 2.6: The Derivative Business Calculus - p. 19/44 s(t) = 1 t2 + t 2 s The average velocity over the given time intervals Rates of Change Rates of change are: Velocity Example s(11)−s(10) x 11−10 = 1 (11)2 + 11 − ( 1 (10)2 + 10) = 11.5 2 2 (ft/sec) s(10.1)−s(10) 1.105 x 10.1−10 = .1 = 11.05 (ft/sec) s(10.01)−s(10) 1.1005 x 10.01−10 = .01 = 11.005 (ft/sec) Section 2.6: The Derivative Business Calculus - p. 20/44 s(t) = 1 t2 + t 2 s We could probably guess that the instantaneous Rates of Change Rates of change velocity at t = 10 is 11 (ft/sec). But lets calculate Velocity Example this using the deﬁnition ′ s(t + h) − s(t) s (t) = lim h→0 h 1 (t + h)2 + (t + h) − ( 1 t2 + t) = lim 2 2 h→0 h 1 2 (t + 2th + h2 ) + t + h − 1 t2 − t = lim 2 2 h→0 h th + 1 h2 + h 2 = lim h→0 h 1 = lim t + h + 1 = t + 1. h→0 2 Section 2.6: The Derivative Business Calculus - p. 21/44 s(t) = 1 t2 + t 2 Rates of Change Notice that we have calculated the derivative at any Rates of change point t: Velocity Example s′ (t) = t + 1. We now evaluate at t = 10 to get s′ (10) = 11 just as we expected. The average velocity over the time intervals [10, 10 + h] for h = 1, h = .1 and h = .01 become closer to the instantaneous velocity at t = 10. This is as we should expect. Section 2.6: The Derivative Business Calculus - p. 22/44 Differeniation Notation An outline Example Example Finding the derivative of a function Do not despair! using the deﬁnition Section 2.6: The Derivative Business Calculus - p. 23/44 Notation Differeniation Differential calculus has various ways of denoting Notation the derivative, each with their own advantages. An outline Example We have used the prime notation, f ′ (x) (read: "f Example Do not despair! prime of x"), to denote the derivative of y = f (x). You will also see y ′ written when it is clear y = f (x). The prime notation is simple, quick to write, but not very inspiring. Section 2.6: The Derivative Business Calculus - p. 24/44 Differeniation Another notation is Notation An outline df dy Example or . Example dx dx Do not despair! This notation is much more suggestive. Recall that the derivative is the limit of the difference quotient ∆y ∆x : the change in y over the change in x. The notation "dy" or "df " is used to suggest the instan- taneous change in y after the limit is taken and like- wise for dx. One must not read too much into this df notation. dx is not a fraction but the limit of a frac- tion. There are other notations that are in use but these are the two most common. Section 2.6: The Derivative Business Calculus - p. 25/44 An outline df To compute the derivative dx = f ′ (x) of a function Differeniation Notation y = f (x) using the deﬁnition follow the steps: An outline Example Example Do not despair! Section 2.6: The Derivative Business Calculus - p. 26/44 An outline df To compute the derivative dx = f ′ (x) of a function Differeniation Notation y = f (x) using the deﬁnition follow the steps: An outline Example Example 1. Find the change in y: f (x + h) − f (x) Do not despair! Section 2.6: The Derivative Business Calculus - p. 26/44 An outline df To compute the derivative dx = f ′ (x) of a function Differeniation Notation y = f (x) using the deﬁnition follow the steps: An outline Example Example 1. Find the change in y: f (x + h) − f (x) Do not despair! f (x+h)−f (x) 2. Compute h Section 2.6: The Derivative Business Calculus - p. 26/44 An outline df To compute the derivative dx = f ′ (x) of a function Differeniation Notation y = f (x) using the deﬁnition follow the steps: An outline Example Example 1. Find the change in y: f (x + h) − f (x) Do not despair! f (x+h)−f (x) 2. Compute h 3. Determine limh→0 f (x+h)−f (x) . h Section 2.6: The Derivative Business Calculus - p. 26/44 Example ice Differeniation 3 Example 3: Find the derivative of y = x − x. Notation An outline Example Example Do not despair! Section 2.6: The Derivative Business Calculus - p. 27/44 Example ice Differeniation 3 Example 3: Find the derivative of y = x − x. Notation An outline Example Let f (x) = x3 − x. Then Example Do not despair! f (x + h) − f (x) = (x + h)3 − (x + h) − (x3 − x) = x3 + 3x2 h + 3xh2 + h3 −x − h − x3 + x = 3x2 h + 3xh2 + h3 − h Section 2.6: The Derivative Business Calculus - p. 27/44 Example ice Differeniation 3 Example 3: Find the derivative of y = x − x. Notation An outline Example Let f (x) = x3 − x. Then Example Do not despair! f (x + h) − f (x) = (x + h)3 − (x + h) − (x3 − x) = x3 + 3x2 h + 3xh2 + h3 −x − h − x3 + x = 3x2 h + 3xh2 + h3 − h Next we get f (x + h) − f (x) 3x2 h + 3xh2 + h3 − h = h h = 3x2 + 3xh + h2 − 1 Section 2.6: The Derivative Business Calculus - p. 27/44 Example ice Differeniation 3 Example 3: Find the derivative of y = x − x. Notation An outline Example Let f (x) = x3 − x. Then Example Do not despair! f (x + h) − f (x) = (x + h)3 − (x + h) − (x3 − x) = x3 + 3x2 h + 3xh2 + h3 −x − h − x3 + x = 3x2 h + 3xh2 + h3 − h Next we get f (x + h) − f (x) 3x2 h + 3xh2 + h3 − h = h h = 3x2 + 3xh + h2 − 1 dy Finally, dx = limh→0 3x2 + 3xh + h2 − 1 = 3x2 − 1. Section 2.6: The Derivative Business Calculus - p. 27/44 Example ice Differeniation Example 4: Find the equation of the line tangent Notation to √ An outline Example f (x) = x Example Do not despair! at the point (4, 2). Section 2.6: The Derivative Business Calculus - p. 28/44 √ f (x) = x Differeniation We need the slope of the tangent line at this point. Notation This is f ′ (4). An outline Example f (4 + h) − f (4) Example f ′ (4) = lim Do not despair! h→0 h √ 4+h−2 = lim h→0 h √ √ 4+h−2 4+h+2 = lim √ h→0 h 4+h+2 4+h−4 = lim √ h→0 h( 4 + h + 2) 1 1 = lim √ = . h→0 4+h+2 4 Section 2.6: The Derivative Business Calculus - p. 29/44 f ′ (4) = 1 4 and P = (4, 2) Given a point and a slope we compute the line: Differeniation Notation An outline 1 Example y − 2 = (x − 4) Example 4 Do not despair! or 1 y = x + 1. 4 Section 2.6: The Derivative Business Calculus - p. 30/44 Do not despair! Differeniation Admittedly, the calculation of a derivative using the Notation deﬁnition can be tedious. However, in the next An outline Example chapter we will discuss a set of rules for differenti- Example Do not despair! ation that will allow us to calculate the derivative of many commonly encountered functions very eas- ily. Nevertheless, it is important that you under- stand the deﬁnition and the underlying meaning of the derivative; at times, it will be necessary to come back to it. Section 2.6: The Derivative Business Calculus - p. 31/44 Continuity Reformulation Differentiability Proof Continuiity Differentiation and Continuity Section 2.6: The Derivative Business Calculus - p. 32/44 Reformulation of Continuity Continuity In the last section we discussed the meaning of Reformulation continuity. Recall a function y = f (x) is continu- Differentiability Proof ous at a point a if f (a) is deﬁned and Continuiity lim f (x) = f (a). x→a If we let x = a + h then x approaches a if h ap- proaches 0. This observations allows us the give an equivalent deﬁnition for continuity: f (a) is de- ﬁned and lim f (a + h) − f (a) = 0. h→0 Section 2.6: The Derivative Business Calculus - p. 33/44 Differentiable functions are Continuous Continuity A function is said to be differentiable at a point Reformulation x = a if f ′ (a) exists. This means that the limit Differentiability Proof limh→0 f (a+h)−f (a) exists. We say f is differentiable h Continuiity on an interval (a, b) if it is differentiable at every point in the interval. Notice the next theorem: Theorem: A function that is differentiable at a point x = a is continuous there. We have not been proving many theorems but this one is easy and short enough that we will do so on the next slide. Section 2.6: The Derivative Business Calculus - p. 34/44 Proof Proof: To say f is differentiable at x = a means Continuity Reformulation Differentiability f (a + h) − f (a) Proof lim Continuiity h→0 h exists and is a ﬁnite number, denoted f ′ (a). Thus f (a + h) − f (a) lim (f (a + h) − f (a)) = lim ·h h→0 h→0 h f (a + h) − f (a) = lim · lim (h) h→0 h h→0 = f ′ (a) · 0 = 0. This means that f is continuous at x = a. Section 2.6: The Derivative Business Calculus - p. 35/44 Continuity does not imply Differentiability We must not read something that is not in this theo- Continuity rem. Though a differentiable function is necessarily Reformulation Differentiability continuous a continuous function is not necessarily Proof Continuiity differentiable. Consider this classic example: y = \|x\| . At x = 0 there are several lines that just touch the graph at (0, 0); it is not unique. Section 2.6: The Derivative Business Calculus - p. 36/44 Continuity does not imply Differentiability We must not read something that is not in this theo- Continuity rem. Though a differentiable function is necessarily Reformulation Differentiability continuous a continuous function is not necessarily Proof Continuiity differentiable. Consider this classic example: y = \|x\| . At x = 0 there are several lines that just touch the graph at (0, 0); it is not unique. Section 2.6: The Derivative Business Calculus - p. 36/44 Continuity does not imply Differentiability We must not read something that is not in this theo- Continuity rem. Though a differentiable function is necessarily Reformulation Differentiability continuous a continuous function is not necessarily Proof Continuiity differentiable. Consider this classic example: y = \|x\| . At x = 0 there are several lines that just touch the graph at (0, 0); it is not unique. Section 2.6: The Derivative Business Calculus - p. 36/44 Continuity does not imply Differentiability We must not read something that is not in this theo- Continuity rem. Though a differentiable function is necessarily Reformulation Differentiability continuous a continuous function is not necessarily Proof Continuiity differentiable. Consider this classic example: y = \|x\| . At x = 0 there are several lines that just touch the graph at (0, 0); it is not unique. Section 2.6: The Derivative Business Calculus - p. 36/44 Continuity does not imply Differentiability We must not read something that is not in this theo- Continuity rem. Though a differentiable function is necessarily Reformulation Differentiability continuous a continuous function is not necessarily Proof Continuiity differentiable. Consider this classic example: y = \|x\| . At x = 0 there are several lines that just touch the graph at (0, 0); it is not unique. Section 2.6: The Derivative Business Calculus - p. 36/44 Continuity does not imply Differentiability We must not read something that is not in this theo- Continuity rem. Though a differentiable function is necessarily Reformulation Differentiability continuous a continuous function is not necessarily Proof Continuiity differentiable. Consider this classic example: y = \|x\| . At x = 0 there are several lines that just touch the graph at (0, 0); it is not unique. Section 2.6: The Derivative Business Calculus - p. 36/44 Continuity does not imply Differentiability We must not read something that is not in this theo- Continuity rem. Though a differentiable function is necessarily Reformulation Differentiability continuous a continuous function is not necessarily Proof Continuiity differentiable. Consider this classic example: y = \|x\| . At x = 0 there are several lines that just touch the graph at (0, 0); it is not unique. Section 2.6: The Derivative Business Calculus - p. 36/44 Continuity does not imply Differentiability We must not read something that is not in this theo- Continuity rem. Though a differentiable function is necessarily Reformulation Differentiability continuous a continuous function is not necessarily Proof Continuiity differentiable. Consider this classic example: y = \|x\| . At x = 0 there are several lines that just touch the graph at (0, 0); it is not unique. Remember, tangent lines are unique and since y = \|x\| has no unique tangent line it is not differ- entiable at x = 0. Section 2.6: The Derivative Business Calculus - p. 36/44 y = \|x\| at x = 0 Continuity Consider what happens here in terms of the deﬁni- Reformulation tion: Differentiability Proof ′ \|0 + h\| − 0 Continuiity y (0) = lim h→0 h \|h\| = lim . h→0 h Now, to compute this limit we will consider the left and right-hand limits. Section 2.6: The Derivative Business Calculus - p. 37/44 \|h\| Left and Right-hand limits of h Continuity If h is positive then \|h\| = h and Reformulation Differentiability ′ \|h\| h Proof y (0) = lim+ h = lim = 1. Continuiity h→0 h→0 h If h is negative then \|h\| = −h and ′ \|h\| −h y (0) = lim− h = lim = −1. h→0 h→0 h The left and right hand limits are not equal there- fore limh→0 \|h\| does not exist. h If y = \|x\| then y is continuous but not differentiable at x = 0. Section 2.6: The Derivative Business Calculus - p. 38/44 Summary Summary Summary Section 2.6: The Derivative Business Calculus - p. 39/44 Summary Summary This section is very important and likely new to Summary many students in this course. Here are some key concepts to master. Section 2.6: The Derivative Business Calculus - p. 40/44 Summary Summary This section is very important and likely new to Summary many students in this course. Here are some key concepts to master. s The deﬁnition of the derivative: f ′ (x) = limh→0 f (x+h)−f (x) . h Section 2.6: The Derivative Business Calculus - p. 40/44 Summary Summary This section is very important and likely new to Summary many students in this course. Here are some key concepts to master. s The deﬁnition of the derivative: f ′ (x) = limh→0 f (x+h)−f (x) . h s The meaning: The derivative of a function represents the instantaneous rate of change of f as a function of x. Section 2.6: The Derivative Business Calculus - p. 40/44 Summary Summary This section is very important and likely new to Summary many students in this course. Here are some key concepts to master. s The deﬁnition of the derivative: f ′ (x) = limh→0 f (x+h)−f (x) . h s The meaning: The derivative of a function represents the instantaneous rate of change of f as a function of x. s Primary Applications: Tangent lines, velocity Section 2.6: The Derivative Business Calculus - p. 40/44 Summary Summary This section is very important and likely new to Summary many students in this course. Here are some key concepts to master. s The deﬁnition of the derivative: f ′ (x) = limh→0 f (x+h)−f (x) . h s The meaning: The derivative of a function represents the instantaneous rate of change of f as a function of x. s Primary Applications: Tangent lines, velocity s Computation of the derivative. Section 2.6: The Derivative Business Calculus - p. 40/44 Summary Summary This section is very important and likely new to Summary many students in this course. Here are some key concepts to master. s The deﬁnition of the derivative: f ′ (x) = limh→0 f (x+h)−f (x) . h s The meaning: The derivative of a function represents the instantaneous rate of change of f as a function of x. s Primary Applications: Tangent lines, velocity s Computation of the derivative. s The connection between continuity and differentiation. Section 2.6: The Derivative Business Calculus - p. 40/44 Summary Summary This section is very important and likely new to Summary many students in this course. Here are some key concepts to master. s The deﬁnition of the derivative: f ′ (x) = limh→0 f (x+h)−f (x) . h s The meaning: The derivative of a function represents the instantaneous rate of change of f as a function of x. s Primary Applications: Tangent lines, velocity s Computation of the derivative. s The connection between continuity and differentiation. dy s Notation: y ′ or dx . Section 2.6: The Derivative Business Calculus - p. 40/44 In-Class Exercises ICE ICE ICE In-Class Exercises Section 2.6: The Derivative Business Calculus - p. 41/44 In-Class Exercise return In-Class Exercise 1: At a ﬁxed temperature the In-Class Exercises ICE volume V (in liters) of 1.33 g of a certain gas is ICE related to its pressure p (in atmospheres) by the ICE formula 1 V (p) = . p What is the average rate of change of V with re- spect to p as p increases from 5 to 6 ? 1. 1 1 2. 6 3. − 1 5 1 4. − 30 5. − 1 6 Section 2.6: The Derivative Business Calculus - p. 42/44 In-Class Exercise return In-Class Exercises In-Class Exercise 2: Use the deﬁnition of the ICE derivative to ﬁnd y ′ if ICE ICE y = 4x2 − x. 1. 4x2 − 1 2. 8x − 1 3. 8x 4. 4x2 − x 5. None of the above Section 2.6: The Derivative Business Calculus - p. 43/44 In-Class Exercise return In-Class Exercises In-Class Exercise 3: Find the equation of the line ICE tangent to ICE ICE y = x2 + x at the point (1, 2). 1. y = 3x − 1 2. y = 3x − 5 3. y = 2x 4. y = 2x − 3 5. None of the above Section 2.6: The Derivative Business Calculus - p. 44/44 ``` DOCUMENT INFO Shared By: Categories: Stats: views: 23 posted: 11/22/2010 language: English pages: 88
# Math Worksheets Land Math Worksheets For All Ages # Math Worksheets Land Math Worksheets For All Ages # Unknown Numbers in Addition And Subtraction Worksheets This is going to be the first time that students will be confronted with algebraic thinking. This can be very overwhelming for them, or they could embrace it. The best way to approach this new subject matter with them is to present it as a mystery. I usually call this my Math Detective unit. I tell students that now that we have a good on the basic math operations (addition and subtraction), we can now check the work of others. I tend to start this process visually as you will see in the progression of the worksheets. Using word problems that can easily be drawn is a good place to start. Once they get the basic concept of finding missing addends or subtrahends, you are ready to jump into working with actual equations. These worksheets will have students complete problems that undergo an addition or subtraction operation. This is an algebra primer for students. Again, this is most likely the first time that they have seen problems like this and it will require some patience on the part of the teacher. This is not a unit that you will fly through and it will need extra time set aside for it. We encourage you to give them as much practice as is possible within your time frame. ### Aligned Standard: Grade 1 Operations - 1.OA.8 • Answer Keys - These are for all the unlocked materials above. ### Guided Lessons I received many positive comments on this section. I hope you find it useful. ### Practice Worksheets I created this section as a primer section for word problems. ### Solving Unknowns in Addition Equations Example Problem: Find the unknown value in the equation: 5 + __ = 8 Math Based Solution Step 1) A good way to approach this is to rewrite the problem in what we call mountain format. In this format we write the sum or difference at the top and the parts that create it at the bottom. In this case, 8 at the top and 5 and the unknown (noted by the blank square) at the bottom. Step 2) Since the unknown is being added to 5. We can count up from 5 until we hit our sum. In this case, it requires us to count 3 times. Step 3)This tells us our missing addend is 3. ### Solving Unknowns in Addition Equations Example Problem: Find the unknown value in the equation: 5 + __ = 8 Model Based Solution Step 1)You can also model this problem with the use of simple shapes. In this case, I will use a circle. We know the sum is 8. As a result, we will draw 8 circles. Step 2) The sum indicates that you already have 5. We will model this by separating the first five circles. We then just need to count how many circles remain to make the total (8). We can see from the diagram that it is 3. Regardless of the method that you chose to use to solve the problem. The answer should follow as: 5 + __ = 8 \| 5 + 3 = 8 ### Solving Unknowns in Subtraction Equations Example Problem: Find the unknown value in the equation: 7 - __ = 5 Math Based Solution Step 1) We can once again apply the mountain format. In this format we write the sum or difference at the top and the parts that create it at the bottom. In this case, 5 at the top (difference) and 7 and the unknown (noted by the blank square) at the bottom. Step 2) Since the unknown is being subtracted from 7. We can count down from 7 until we hit our difference (5). In this case, it requires us to count down 2 times. Step 3)This tells us our missing addend is 2. ### Solving Unknowns in Subtraction Equations Example Problem: Find the unknown value in the equation: 7 - __ = 5 Model Based Solution Step 1)You can also model this problem with the use of simple shapes. In this case, I will use a circle. We know the starting value is 7. As a result, we will draw 7 circles. Step 2) The difference indicates a result of 5. We will model this by separating the last five circles. We then just need to count how many circles remain to make the total (7). We can see from the diagram that it is 2. Regardless of the method that you chose to use to solve the problem. The answer should follow as: 7 - __ = 5 \| 7 - 2 = 5 Unlock all the answers, worksheets, homework, tests and more! Save Tons of Time! Make My Life Easier Now ## Thanks and Don't Forget To Tell Your Friends! I would appreciate everyone letting me know if you find any errors. I'm getting a little older these days and my eyes are going. Please contact me, to let me know. I'll fix it ASAP.
# GSEB Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.3 Gujarat Board GSEB Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.3 Textbook Questions and Answers. ## Gujarat Board Textbook Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.3 Question 1. Prove that $$\sqrt { 5 }$$ is irrational. Solution: Let $$\sqrt { 5 }$$ be a rational number. Therefore we can find two integers a, b (b â‰ 0) such that $$\frac{a}{b}$$ = $$\sqrt { 5 }$$ Let a and b have a common factor other than 1 then we divide both by common factor. Then, we get $$\frac{p}{q}$$ = $$\sqrt { 5 }$$Â Â Â Â Â Â …(1) [Where p and q are co-prime] Squaring both sides $$\frac{p^{2}}{q^{2}}$$ = 5 p2 = 5q2Â Â Â Â Â Â Â Â Â Â Â Â Â Â …(2) therefore p2 is divisible by 5, then p will be divisible by 5. Let p = 5r Putting value of p in eqn (2) we get 25r2 = 5q2 5r2 = q2 This means q2 is divisible by 5 then q will be divisible by 5. Which means p and q have common factor 5. We reach at the contradiction as our supposition that p and q are co-prime is wrong. Hence $$\sqrt { 5 }$$ is irrational. Question 2. Prove that 3 + 2 $$\sqrt { 5 }$$ is irrational.(CBSE) Solution: Let us suppose 3 + 2$$\sqrt { 5 }$$ is a rational. Then $$\frac{a}{b}$$ = 3 + 2$$\sqrt { 5 }$$ [Where a and b are integers] â‡’ $$\frac{a}{b}$$ – 3 = 2$$\sqrt { 5 }$$ $$\frac{1}{2}$$[$$\frac{a}{b}$$ – 3] = $$\sqrt { 3 }$$ $$\frac{1}{2}$$[$$\frac{a}{b}$$ – 3] is rational as a and b are integers therefore $$\sqrt { 3 }$$ should be rational. This contradicts the fact that $$\sqrt { 3 }$$ is an irrational. Hence 3 + 2$$\sqrt { 5 }$$ is an irrational. Question 3. Prove that the following are irrationals: 1. $$\frac { 1 }{ \sqrt { 2 } }$$ 2. 7$$\sqrt { 5 }$$ 3. 6 + $$\sqrt { 2 }$$ Solution: 1. Let $$\frac { 1 }{ \sqrt { 2 } }$$ be a rational. Therefore we can find two integers a, b (b â‰ 0) Such that $$\frac { 1 }{ \sqrt { 2 } }$$ = $$\frac{a}{b}$$ $$\sqrt { 2 }$$ = $$\frac{b}{a}$$ $$\frac{b}{a}$$ is a rational. Therefore $$\sqrt { 2 }$$ will also be rational which contradicts to the fact that $$\sqrt { 2 }$$ is irrational. Hence our supposition is wrong and $$\frac { 1 }{ \sqrt { 2 } }$$ is irrational. 2. Let 7$$\sqrt { 5 }$$ be a rational Therefore 7$$\sqrt { 5 }$$ = $$\frac{a}{b}$$ [Where a and b are integers] $$\sqrt { 5 }$$ = $$\frac{a}{7b}$$ $$\frac{a}{7b}$$ is rational as a and b are integers Therefore $$\sqrt { 5 }$$ should be rational. This contradicts the facts that is $$\sqrt { 5 }$$ irrational therefore our supposition is wrong. Hence 7$$\sqrt { 5 }$$ is irrational. 3. Let 6 + $$\sqrt { 2 }$$ be rational Therefore 6 + $$\sqrt { 2 }$$ = $$\frac{a}{b}$$ [where a and b are integers] $$\sqrt { 2 }$$ = $$\frac{a}{b}$$ – 6 $$\frac{a}{b}$$ – 6 is rational as a and b are integers therefore, $$\sqrt { 2 }$$ should be rational. This contradicts the fact that $$\sqrt { 2 }$$ is irrational, therefore, our supposition is wrong. Hence 6 + $$\sqrt { 2 }$$ is irrational.
Tangent Circles of a Point on a Given Circle and a Given Line by Jiyoon Chun How to construct a tangent circle of an arbitrary point P on a given circle to a given line? Let the green circle and the blue line given. For an arbitrary point P on the green circle, there exists two tangent circle: The red tangent circle which is tangent to the green circle outside, and the orange circle which the green circle is tangent to inside of the circle. Construction and proof of the red tangent circle Construction and proof of the orange tangent circle The locus of the tangent circles The locus of the center of the tangent circle is a parabola of a focal point A and directrix k. A is a center of a given circle, and k is the line parallel to given line l and distance between l and k is the radius of the given circle A. What is the difference? When we constructed circle B, we first drew the line k. Therefore, the locus of B is the parabola of the center A and directrix k. WHen we constructed circle B', we first drew the line k'. Thus, the locus of B' is the parabola of the same center A and directrix k'. Since k' is closer to the focal point A than k, the locus of B' is narrower than the locus of B. The limiting case of tangent circles Let's think about the radius of the tangent circle when point Dāā 1) the Red tangent circle Intuitive thinking As we can see in the flash above, the slope of AD converges to 0 when Dāā. Then AD will be parallel to AC. Since B is on line m which is perpendicular to AD, B also goes to infinity when D goes infinity. Thus, when P is on the intersection of circle A and perpendicular line to the given line l passing through the center of the given circle A, than we have a tangent circle of radius infinity. Thus, the tangent circle will be a tangent line at a point P. See the picture below. Algebraic proof Suppose that A is at and . Then the locus of the center of tangent circles is . Thus, the coordinate of point B is . Since the tangent circle is tangent to the given circle A at point P, \|PB\| is the radius of B. \|PB\|=. Since P is at , . See the flash below. When I move the given circle so that point P is on the line given, we can observe that the radius of the red tangent circle is getting bigger and bigger and eventually it will be the line given. In addition, the line given is tangent to the circle given. The same intuitive thinking and algebraic proof can be applied to explain this. 2) the Orange tangent circle As in the flash above, we can observe that the tangent circle becomes the line given when point D goes to infinity. Since the construction is the same, the same reasoning of the red tangent circle also explains why the tangent circle becomes the line given. I am attaching another flash. RETURN
# Tangent and normal lines to an ellipse Find $\frac{dy}{dx}$ of $x^2 +y^2 +xy=27$ and use it to find the equations of the tangent and normal lines to the curve $(3\sqrt{3}, 0)$. - To find $\dfrac{dy}{dx}$: $$d(x^2 + y^2 + xy) = 2x dx + 2y dy + xdy + ydx = d(27) = 0\tag{1}$$ $$(x + 2y)\,dy = -(2x + y)\,dx$$ Then we can express $$\frac{dy}{dx} = -\frac{2x + y}{x+2y}\tag{2}$$ Now, using your given point $(3\sqrt 3, 0)$, evaluate $\large\frac{dy}{dx}$ at the given $x$ and $y$ coordinate in order to: • determine the slope of the line tangent to the ellipse at that point, and • then from that slope, determine the slope of the line normal to the ellipse$^+$ at that point. • Then, use the point-slope form of the equation of a line$^{++}$ to determine the equations of each of the line tangent to the ellipse at $(3\sqrt 3,0)$, and the line normal to the ellipse at that point. $^+$Recall: if slope of the tangent line is $\;\large\frac ab\;$, then the slope of the line normal at that point is $\;\large-\frac ba$. $^{++}$ Recall: the point-slope of the equation of a line, given point $\,(x_0, y_0)\,$ and slope $\,m,\,$ is given by $\;(y - y_0) = m(x - x_0)$ - Nice answer. I always set the relation as $F(x,y)=0$, then use the following $$y'=\frac{-F_x}{F_y}$$ instead. +1 for tomorrow. – Babak S. Feb 3 '13 at 20:05 user59633: if you're still struggling, let me know if you have any questions I can address. – amWhy Feb 3 '13 at 22:49 There's a much simpler way to find the tangent line at a point on an ellipse (or any conic section). If the conic has equation $ax^2 + 2bxy + cy^2 + 2dx + 2ey + f =0$, then the tangent at the point $(x_0, y_0)$ is $ax_0x + b(x_0y + xy_0) + cy_0y + d(x+x_0) + e(y+y_0) + f = 0$. Informally, you "half-way substitute" $(x_0, y_0)$ in place of $(x, y)$. Generally, the line you get this way is the "polar line" through the point $(x_0, y_0)$. But, if the point lies on the conic, then the polar line is a tangent. Once you have the tangent line, getting the normal line is easy, of course. No need for any differentiation. - $d(x^2 + y^2 + xy) = 2x dx + 2y dy + xdy + ydx = d(27) = 0$. Then $$\frac{dy}{dx} = -\frac{2x + y}{x+2y}$$ Then with this, you can find the slope of the tangent line at $(3\sqrt 3,0)$, then find the normal. - I am confused as to exactly how to find the tangent and normal lines? Could you provide a rough idea about how to do so? – user59633 Feb 3 '13 at 19:35
# What Is 33/49 as a Decimal + Solution With Free Steps The fraction 33/49 as a decimal is equal to 0.673. The division process consists of two numbers are dividend and a divisor. A divisor will basically split the dividend into equal parts. The usual process to find the resulting answer is done through a long division method. 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,Â which we will discuss in detail moving forward. So, letâ€™s go through the Solution of fraction 33/49. ## 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 = 33 Divisor = 49 Now, we introduce the most important quantity in our division process: theÂ 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 = 33 $\div$ 49 This is when we go through the Long Division solution to our problem. Given is the long division process in Figure 1: Figure 1 ## 33/49 Long Division Method We start solving a problem using the Long Division Method by first taking apart the divisionâ€™s components and comparing them. As we have 33Â and 49, we can see how 33 is Smaller than 49, and to solve this division, we require that 33 be Bigger than 49. 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 33, which after getting multiplied by 10 becomes 330. We take this 330 and divide it by 49; this can be done as follows: Â 330 $\div$ 49 $\approx$ 6 Where: 49 x 6 = 294 This will lead to the generation of a Remainder equal to 330 â€“ 294 = 36. Now this means we have to repeat the process by Converting the 36 into 360Â and solving for that: 360 $\div$ 49 $\approx$ 7Â Where: 49 x 7 = 343 This, therefore, produces another Remainder which is equal to 360 â€“ 343 = 17. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 170. 170 $\div$ 49 $\approx$ 3Â Where: 49 x 3 = 147 Finally, we have a Quotient generated after combining the three pieces of it as 0.673, with a Remainder equal to 23. Images/mathematical drawings are created with GeoGebra.
Arithmetic Of Polynomials Definition of a Polynomial (Vocabulary) Definition: A monomial is a number times a power of x: axn Examples 3x2, 1/2 x7, and 8 are all monomials. Definition: A polynomial is a sum or difference of monomials Examples: 4x5 - 3x2 - 1, 4x2, 2 are all polynomials. The degree of a polynomial is the largest power of x, the leading coefficient is the number in front of the term with the highest power of x, and the constant term is the number without any x's. Example: For the polynomial 4x5 - 3x2 - 1 the degree is 5, the leading coefficient is 4 and the constant term is -1. Notation: When we write P(x) = 3x3 - 2x2 +1 we say "P of x" To evaluate P(1), we find 3(1)3 - 2(1)2 + 1 = 2 To add or subtract polynomials, we just collect like terms: Example Let P(x) = x2 + 3x + 5 and Q(x) = 4x3 - 2x2 + 3x - 2 Then P(x) - Q(x) = (x2 + 3x + 5) - (4x3 - 2x2 + 3x - 2) = x2 + 3x + 5 - 4x3 + 2x2 - 3x + 2 Distributing the - sign = -4x3 + 3x2 + 7 Combining like terms Exercise Let P(x) = 3x2 + 4x - 2 and Q(x) = 5x2 - 3x - 5 Find P(x) + Q(x) Consider the multiplication of the following two first degree polynomials: (x + 3)(x + 4) = (x + 3)x + (x + 3)4 Distributing the x + 3 = x2 + 3x + 4x + 12 Distributing the x and the 4 = x2 + 7x + 12 Combining like terms Since this type of multiplication occurs so frequently, we have a systematic approach called FOIL- Firsts, Outers, Inners, Lasts. That is we multiply the first terms, the outer terms, the inner terms, and the last terms and add the four results together. Examples F O I L 1. (x + 2)(x + 5) = x2 + 5x + 2x + 10 = x2 + 7x + 10 2. (3x - 4)(5x + 2) = 15x2 + 6x - 20x - 8 = 15x2 - 14x - 8 Exercises: Evaluate the following 1. (x - 2)(3x + 1) 2. (5x + 4)(3x + 2) 3. (3x - y)(2x + 3y) 4. (x + y)(x - y) 5. (x + y)(x + y) 6. (x - y)(x - y) We will note the special products D, E and F as difference of squares, perfect square of sum, and perfect square of difference. Special Formulas (a + b)(a - b) = a2 - b2 Difference of Squares (a + b)(a + b) = a2 + 2ab + b2 Square of a Sum (a - b)(a - b) = a2 - 2ab + b2 Square of a Difference Examples: 1. (3 - x)(3 + x) = 9 - x2 2. (x + 3)2 = x2 + 6x + 9 3. (2x - 4)2 = 4x2 - 16x + 16 4. (x + 2)3 = (x + 2)2(x + 2) = (x2 + 4x + 4)(x + 2) = (x2 + 4x + 4)x + (x2 + 4x + 4)2 = x3 + 4x2 + 4x + 2x2 + 8x + 8 = x3 + 6x2 + 12x + 8. General Polynomial Multiplication When the polynomials have more than two terms, we must use the distributive property as follows: Example (x3 -3x +1) (x - 3) = (x3 -3x +1) (x) + (x3 -3x +1) (-3) = x4 - 3x2 + x - 3x3 + 9x - 3 = x4 - 3x3 -3x2 + 10x - 3 Back to the Math Department Home
# Sine law In plane and spherical trigonometry , the law of sines establishes a relationship between the angles of a general triangle and the opposite sides. ## Law of sines for plane triangles Sine law If , and the sides of a triangle with the area , the angles , and those of the associated side lie opposite and the radius of the circumference , then with the sine function : ${\ displaystyle a}$${\ displaystyle b}$${\ displaystyle c}$${\ displaystyle A}$${\ displaystyle \ alpha}$${\ displaystyle \ beta}$${\ displaystyle \ gamma}$ ${\ displaystyle R}$ ${\ displaystyle {\ frac {a} {\ sin \ alpha}} = {\ frac {b} {\ sin \ beta}} = {\ frac {c} {\ sin \ gamma}} = {\ frac {a \ cdot b \ cdot c} {2 \ cdot A}} = 2 \ cdot R}$ If angles in a triangle are to be calculated with the aid of the sine law , it must be ensured that there are generally two different angles with the same sine value in the interval [0 °; 180 °] . This ambiguity corresponds to that of the congruence theorem SSW. For the connection with the congruence theorems and the systematic of the triangle calculation, see the article on the cosine law . In spherical trigonometry there is a corresponding theorem, which is also referred to as the sine law. ### proof Triangle with height ${\ displaystyle h_ {c}}$ The height shown divides the triangle into two right-angled sub-triangles, in which the sine of and can be expressed as the quotient of the opposite cathetus and hypotenuse: ${\ displaystyle h_ {c}}$${\ displaystyle \ alpha}$${\ displaystyle \ beta}$ ${\ displaystyle \ sin \ alpha = {\ frac {h_ {c}} {b}}}$ ${\ displaystyle \ sin \ beta = {\ frac {h_ {c}} {a}}}$ Solving for gives: ${\ displaystyle h_ {c}}$ ${\ displaystyle h_ {c} = b \ cdot \ sin \ alpha}$ ${\ displaystyle h_ {c} = a \ cdot \ sin \ beta}$ So by equating one obtains ${\ displaystyle a \ cdot \ sin \ beta = b \ cdot \ sin \ alpha}$ If you now divide by , you get the first part of the claim: ${\ displaystyle \ sin \ alpha \ cdot \ sin \ beta}$ ${\ displaystyle {\ frac {a} {\ sin \ alpha}} = {\ frac {b} {\ sin \ beta}}}$ The equality with results from using the height or . In order to also show the correspondence with , which strictly speaking does not belong to the sine law, you need the known set of peripheral angles (circumferential angle) or the cosine law together with the peripheral / central angle set . ${\ displaystyle {\ tfrac {c} {\ sin \ gamma}}}$${\ displaystyle h_ {a}}$${\ displaystyle h_ {b}}$${\ displaystyle 2R}$ ### Connection with the periphery On the circumference of triangle ABC, D should be the point that, together with point A, forms a diameter so that the connection of A and D runs through the center of the circumference (see illustration). Then according to Thales's theorem , ABD is a right triangle and we have: ${\ displaystyle \ sin \ delta = {\ frac {c} {2 \ cdot R}}}$ According to the set of circumferential angles , the circumferential angles and over the side are equal, so the following applies: ${\ displaystyle \ gamma}$${\ displaystyle \ delta}$${\ displaystyle c}$ ${\ displaystyle \ sin \ delta = \ sin \ gamma = {\ frac {c} {2 \ cdot R}}}$ ${\ displaystyle {\ frac {c} {\ sin \ gamma}} = 2 \ cdot R}$ The same applies to and , i.e. in total ${\ displaystyle {\ frac {a} {\ sin \ alpha}} = 2 \ cdot R}$${\ displaystyle {\ frac {b} {\ sin \ beta}} = 2 \ cdot R}$ ${\ displaystyle {\ frac {a} {\ sin \ alpha}} = {\ frac {b} {\ sin \ beta}} = {\ frac {c} {\ sin \ gamma}} = 2 \ cdot R}$ ### Application example The following numerical values are rough approximations. In a triangle ABC the following side and angle sizes are known (designations as usual): ${\ displaystyle a = 5 {,} 4 \, \ mathrm {cm}; \ b = 3 {,} 8 \, \ mathrm {cm}; \ \ alpha = 73 ^ {\ circ}}$ Find the sizes of the remaining sides and angles. First you use the sine law to calculate . Thereafter applies ${\ displaystyle \ beta}$ ${\ displaystyle {\ frac {a} {\ sin \ alpha}} = {\ frac {b} {\ sin \ beta}}}$ what can be formed ${\ displaystyle \ sin \ beta = {\ frac {b \ cdot \ sin \ alpha} {a}} = {\ frac {3 {,} 8 \, \ mathrm {cm} \ cdot \ sin 73 ^ {\ circ }} {5 {,} 4 \, \ mathrm {cm}}} \ approx 0 {,} 67}$ from which, with the help of the arcsine , the inverse function of the sine, ${\ displaystyle \ beta \ approx \ arcsin (0 {,} 67) \ approx 42 ^ {\ circ}}$ can be calculated. There is actually a second angle with the same sine value, namely . However, this cannot be considered as a solution, since otherwise the sum of the angles of the triangle would exceed the prescribed values. ${\ displaystyle \ beta '= 180 ^ {\ circ} - \ beta \ approx 138 ^ {\ circ}}$${\ displaystyle 180 ^ {\ circ}}$ ${\ displaystyle \ gamma}$ is now obtained with the help of the sum of angles ${\ displaystyle \ gamma = 180 ^ {\ circ} - \ alpha - \ beta \ approx 180 ^ {\ circ} -73 ^ {\ circ} -42 ^ {\ circ} = 65 ^ {\ circ}}$ The side length should again be determined with the sine law. (The cosine law would also be possible here.) It applies ${\ displaystyle c}$ ${\ displaystyle {\ frac {a} {\ sin \ alpha}} = {\ frac {c} {\ sin \ gamma}}}$ The result is obtained by reshaping ${\ displaystyle c = {\ frac {a \ cdot \ sin \ gamma} {\ sin \ alpha}} \ approx {\ frac {5 {,} 4 \, \ mathrm {cm} \ cdot \ sin 65 ^ {\ circ}} {\ sin 73 ^ {\ circ}}} \ approx 5 {,} 1 \, \ mathrm {cm}}$ ## Sine law for spherical triangles The equations apply to spherical triangles ${\ displaystyle {\ frac {\ sin A} {\ sin a}} = {\ frac {\ sin B} {\ sin b}} = {\ frac {\ sin C} {\ sin c}}}$ Here are , and the sides ( arcs ) of the spherical triangle and , and the opposite angles on the spherical surface . ${\ displaystyle a}$${\ displaystyle b}$${\ displaystyle c}$${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle C}$ ### proof The radius of the unit sphere is given by ${\ displaystyle OA = OB = OC = 1}$ The point lies on the radius and the point lies on the radius , so . The point lies on the plane , so . It follows and . Because the perpendicular projection of is onto the plane , holds . According to the definition of the sine : ${\ displaystyle D}$${\ displaystyle OB}$${\ displaystyle E}$${\ displaystyle OC}$${\ displaystyle \ angle ADO = \ angle AEO = 90 ^ {\ circ}}$${\ displaystyle A '}$ ${\ displaystyle OBC}$${\ displaystyle \ angle A'DO = \ angle A'EO = 90 ^ {\ circ}}$${\ displaystyle \ angle ADA '= B}$${\ displaystyle \ angle AEA '= C}$${\ displaystyle A '}$${\ displaystyle A}$ ${\ displaystyle OBC}$${\ displaystyle \ angle AA'D = \ angle AA'E = 90 ^ {\ circ}}$ ${\ displaystyle \ sin c = {\ frac {AD} {OA}} = AD}$ ${\ displaystyle \ sin b = {\ frac {AE} {OA}} = AE}$ Also is . Insertion results ${\ displaystyle AA '= AD \ cdot \ sin B = AE \ cdot \ sin C}$ ${\ displaystyle \ sin c \ cdot \ sin B = \ sin b \ cdot \ sin C}$ Accordingly, one obtains , so in total ${\ displaystyle \ sin b \ cdot \ sin A = \ sin a \ cdot \ sin B}$ ${\ displaystyle {\ frac {\ sin A} {\ sin a}} = {\ frac {\ sin B} {\ sin b}} = {\ frac {\ sin C} {\ sin c}}}$
Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 Text Book Back Questions and Answers, Notes. ## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 Question 1. Find the combined equation of the straight lines whose separate equations are x – 2y – 3 = 0 and x + y + 5 = 0. Separate equations are x – 2y – 3 = 0; x + y + 5 = 0 So the combined equation is (x – 2y – 3) (x + y + 5) = 0 x2 + xy + 5x – 2y2 – 2xy – 10y – 3x – 3y – 15 = 0 (i.e) x2 – 2y2 – xy + 2x – 13y – 15 = 0 Question 2. Show that 4x2 + 4xy + y2 – 6x – 3y – 4 = 0 represents a pair of parallel lines. The equation of the given pair of straight lines is 4x2 + 4xy + y2 – 6x – 3y – 4 = 0 ………. (1) Compare this equation with the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ………. (2) a = 4, 2h = 4, b = 1, 2g = -6, 2f = -3, c = – 4 The condition for parallelism is h2 – ab = 0 22 – (4) (1) = 4 – 4 = 0 ∴ The given pair of straight lines represents a pair of parallel straight lines. Question 3. Show that 2x2 + 3xy – 2y2 + 3x + y + 1 = 0 represents a pair of perpendicular straight lines. Comparing the given equation with the general form a = 2,h = 3/2, b = -2,g= 3/2, f = 1/2 and c = 1 Condition for two lines to be perpendicular is a + b = 0. Here a + b = 2 – 2 = 0 ⇒ The given equation represents a pair of perpendicular lines. Question 4. Show that equation 2x2 – xy – 3y2 – 6x + 19y – 20 = 0 represents a pair of intersecting lines. Show further that the angle between them is tan-1(5) The equation of the given pair of lines is 2x2 – xy – 3y2 – 6x + 19y – 20 = 0 —- (1) Compare this equation with the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 —- (2) a = 2 , 2h = – 1, b = – 3, 2g = – 6, 2f = 19, c = – 20 h2 – ab = $$\left(-\frac{1}{2}\right)^{2}$$ – (2)(-3) = $$\frac{1}{4}$$ + 6 ≠ 0 ∴ The given line (1) is not parallel. ∴ They are intersecting lines. Let θ be the angle between the lines. Taking the acute angle θ = tan-1 (5) Question 5. Prove that the equation to the straight lines through the origin each of which makes an angle α with the straight line y = x is x2 – 2xy sec 2α + y2 = 0 Let OP be the given line y = x having slope m = 1 = tan 45° Given that OA is the line making angle α with the line y = x. Slope of the line OA = tan (45° – α) The equation of OA is the equation of the line passing through the point (0, 0) having slope tan (45° – α). y – 0 = tan(45°- α) (x – 0) y = x tan (45°- α) x tan (45° – α) – y = 0 Also given the line OB makes an angle α with the line y = x. Slope of the line OB = tan(45° + α) The equation of OB is the equation of the line passing through the point (0, 0) having slope tan(45° + α) y – 0 = tan (45° + α) (x – 0) y = x tan (45° + α) x tan (45° + α) – y = 0 ………. (2) The combined equation is (x tan(45°- α) – y) (x tan(45°+ α) – y) = 0 x2tan(45° – α) tan(45° + α) – xy tan (45° – α) – xy tan(45° + α) + yz = 0 Question 6. Find the equation of the pair of straight lines passing through the poInt (1,3) and perpendicular to the lines 2x -3y + 1 = 0 and 5x + y – 3 = 0. Equation of a line perpendicular to 2x – 3y + 1 = 0 is of the form 3x + 2y + k = 0. It passes through (1, 3) ⇒ 3 + 6 + k = 0 ⇒ k = – 9 So the line is 3x + 2y – 9 = 0 The equation of a line perpendicular to 5x + y – 3 = 0 will be of the form x – 5y + k = 0. It passes through (1, 3) ⇒ 1 – 15 + k = 0 ⇒ k = 14 So the line is x – 5y + 14 = 0. The equation of the lines is 3x + 2y – 9 = 0 and x – 5y + 14 = 0 Their combined equation is (3x + 2y – 9)(x – 5y + 14) = 0 (i.e) 3x2 – 15xy + 42x + 2xy – 10y2 + 28y – 9x + 45y – 126 = 0 (i.e) 3x2 – 13xy – 10y2 + 33x + 73y – 126 = 0 Question 7. Find the separate equation of the following pair of straight lines. (i) 3x2 + 2xy – y2 = 0 (ii) 6(x – 1)2 + 5(x – 1) (y – 2) – 4(y – 3)2 = 0 (iii) 2x2 – xy – 3y2 – 6x + 19y – 20 = 0 (i) Factorising 3x2 + 2xy – y2 we get 3x2 + 3xy – xy – y2 = 3x (x + y) – y (x + y) = (3 x – y)(x + y) So 3x2 + 2xy – y2 = 0 ⇒ (3x – y) (x + y) = 0 ⇒ 3x – y = 0 and x + y = 0 (ii) 6(x – 1)2 + 5 ( x – 1) (y – 2) – 4(y – 3 )2 = 0 Let X = x – 1 and Y = y – 2 ∴ The given equation becomes 6X2 + 5XY – 4Y2 = 0 6X2 + 8 XY – 3XY – 4Y2 = 0 2X(3X + 4Y) – Y (3X + 4Y) = 0 (2X – Y) (3X + 4Y) = 0 2X – Y = 0 and 3X + 4Y = 0 Substituting for X and Y , we have 2X – Y = 0 ⇒ 2(x – 1) – (y – 2) = 0 ⇒ 2x – 2 – y + 2 = 0 ⇒ 2x – y = 0 3X + 4Y = 0 ⇒ 3 (x – 1) + 4 ( y – 2 ) = 0 ⇒ 3x – 3 + 4y – 8 = 0 ⇒ 3x + 4y – 11 = 0 ∴ The separate equations are 2x – y = 0 and 3x + 4y – 11 = 0 (iii) 2x2 – xy – 3y2 – 6x + 19y – 20 = 0 The given equation is 2x2 – xy – 3y2 – 6x + 19y – 20 = 0 ……… (1) 2x2 – xy – 3y2 = 2x2 – 3xy + 2xy – 3y2 = x (2x – 3y) + y (2x – 3y ) = (x + y) (2x – 3y) Let the separate equation of the straight lines be x + y + 1 = 0 and 2x – 3y + m = 0 …….. (A) (1) ⇒ 2x2 – xy – 3y2 – 6x + 19y – 20 = (x + y + 1) (2x – 3y + m) Equating the coeffident of x , y and constant term on both sides, we have -6 = m + 2l ………. (2) 19 = m – 3l ………. (3) lm = – 20 ………. (4) Solving equations (2), (3) and (4) we have Substituting the values of l and m in equation (A) the required separate equations of the lines are x + y – 5 = 0 and 2x – 3y + 4 = 0 Question 8. The slope of one of the straight lines ax2 + 2hxy + by2 = 0 is twice that of the other, show that 8h2 = 9ab. The equation of the given straight line is ax2 + 2hxy + by2 = 0 ………… (1) Given that the slopes of the straight lines are m and 2m Question 9. The slope of one of the straight lines ax2 + 2h xy + by2 = 0 is three times the other, show that 3h2 = 4ab. The equation of the given straight line is ax2 + 2hxy + by2 = 0 ……….. (1) Given that the slopes of the lines are m and 3m. Question 10. A ∆ OPQ is formed by the pair of straight lines x2 – 4xy + y2 = 0 and the line PQ. The equation of PQ is x + y – 2 = 0, Find the equation of the median of the triangle ∆ OPQ drawn from the origin O The equation of the given pair of lines is x2 – 4xy + y2 = 0 ……….. (1) The equation of the line PQ is x + y – 2 = 0 y = 2 – x ……….. (2) To find the coordinates of P and Q, . Solve equations (1) and (2) (1) ⇒ x2 – 4x ( 2 – x) + ( 2 – x)2 = 0 x2 – 8x + 4x2 + 4 – 4x + x2 = 0 6x2 – 12x + 4 = 0 3x2 – 6x + 2 = 0 The midpoint of PQ is The equation of the median drawn from 0 is the equation of the line joining 0 (0, 0) and D (1, 1) ∴ The required equation is x = y Question 11. Find p and q, if the following equation represents a pair of perpendicular lines 6x2 + 5xy – py2 + 7x + qy – 5 = 0 The equation of the given pair of straight lines is 6x2 + 5xy – py2 + 7x + qy – 5 = 0 ……….. (1) Given that equation (1) represents a pair of perpendicular straight lines. ∴ Coefficient of x2 + coefficient of y2 = 0 6 – p = 0 ⇒ p = 6 6x2 + 5xy – 6y2 = 6x2 + 9xy – 4xy – 6y2 = 3x(2x + 3y) – 2y (2x + 3y) = (2x + 3y) (3x – 2y) Let the separate equation of the straight lines be 2x + 3y + 1 = 0 and 3x – 2y + m = 0 6x2 + 5xy – 6y2 + 7x + qy – 5 = (2x + 3y + 1)(3x – 2y + m) Comparing the coefficients of x , y and constant terms on both sides 2m + 3l = 7 ………… (2) 3m – 2l = q ……….. (3) lm = – 5 …….. (4) Equation (4) ⇒ l = 1, m = – 5 or l = – 1, m = 5 When l = 1, m = – 5 , equation (2) does not satisfy. ∴ l = – 1 , m = 5 Substituting in equation (3) 3 (5) – 2(-1) = q ⇒ q = 17 ∴ The required values are p = 6, q = 17 Question 12. Find the value of k, if the following equation represents a pair of straight lines. Further, find whether these lines are parallel or intersecting 12x2 + 7xy – 12y2 – x +7y + k = 0 The equation of the given pair of straight lines is 12x2 + 7xy – 12y2 – x + 7y + k = 0 ………. (1) Compare this equation with the equation ax2 + 2hxy + by2 + 2gx + 2f y + c = 0 ……….. (2) a = 12, 2h = 7 , b = – 12 , 2g = – 1, 2f = 7, k = c a = 12, h = $$\frac{7}{2}$$, b = – 12, g = $$-\frac{1}{2}$$, f = $$\frac{7}{2}$$, c = k The condition for a second degree equation in x and y to represent a pair of straight lines is abc + 2fgh – af2 – bg2 – ch2 = 0 Substituting the values Coefficient of x2 + coefficient of y2 = 12 – 12 = 0 ∴ The given pair of straight lines are perpendicular and hence they are intersecting lines. Question 13. For what values of k does the equation 12x2 + 2kxy + 2y2 +11x – 5y + 2 = 0 represent two straight lines. The given equation of the pair of straight line is 12x2 + 2kxy + 2y2 + 11x – 5y + 2 = 0 ……… (1) Compare this equation with the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ………. (2) a = 12, 2h = 2k, b = 2, 2g = 11, 2f = – 5 , c = 2 , a =12, h = k, b = 2, g = $$\frac{11}{2}$$, f = –$$\frac{5}{2}$$, c = 2, The condition for a second degree equation in x and y to represent a pair of straight lines is abc + 2fgh – af2 – bg2 – ch2 = 0 96 – 55k – 150 – 121 – 4k2 = o – 4k2 – 55k – 175 = 0 4k2 + 55k + 175 = 0 ∴ The given equation represents a pair of straight lines when k = – 5 and k = $$\frac{-35}{4}$$ Question 14. Show that the equation 9x2 – 24xy + 16y2 – 12x + 16y – 12 = 0 represents a pair of parallel lines. Find the distance between them. The given equation of the pair of straight line is 9x2 – 24xy + by2 – 12x + 16y – 12 = 0 ………… (1) 9x2 – 24xy + 16y2 = 9x2 – 12xy – 12xy + 16y2 = 3x (3x – 4y) – 4y (3x – 4y) = (3x – 4y) (3x – 4y) Let the separate equation of the straight lines be 3x – 4y + 1 = 0 and 3x – 4y + m = 0 9x2 – 24xy + 16y2 – 12x + 16y – 12 = (3x – 4y + l) ( 3x – 4y + m ) Comparing the coefficients of x , y and constant terms on both sides 3l + 3m = – 12 l + m = – 4 ……….. (2) – 4l – 4m = 16 l + m = – 4 ………… (3) lm = – 12 ……….. (4) (l – m)2 = (l + m)2 – 4lm = (- 4)2 – 4 × – 12 = 16 + 48 = 64 l – m = √64 = 8 l – m = 8 ………… (5) Solving equations (2) and (5 ), we have (2) ⇒ 2 + m = – 4 ⇒ m = – 6 ∴ l = 2 and m = – 6 ∴ The separate equation of the straight lines are 3x – 4y – 6 = 0 and 3x – 4y + 2 = 0 The distance between the parallel lines is given by ∴ The given pair of straight lines are parallel and the distance between them is $$\frac{8}{5}$$ units Question 15. Show that the equation 4x2 + 4xy + y2 – 6x – 3y – 4 = 0 represents a pair of parallel lines. Find the distance between them. The given equation of pair of straight lines is 4x2 + 4xy + y2 – 6x – 3y – 4 = 0 ………. (1) 4x2 + 4xy + y2 = (2x + y)2 Let the separate equation of the lines be 2x + y + l = 0 ……….. (2) 2x + y + m = 0 ………. (3) 4x2 + 4xy + y2 – 6x – 3y – 4 = (2x + y + l) (2x + y + m) Comparing the coefficients of x , y and constant terms on both sides we have 2l + 2m = – 6 l + m = – 3 ……… (4) l + m = – 3 ……… (5) l m = – 4 ……… (6) (l – m)2 = (l + m)2 – 4lm (l – m )2 = (- 3)2 – 4 × – 4 (l – m)2 = 9 + 16 = 25 l – m = 5 ………… (7) Solving equations (4) and (7) (4) ⇒ l + m = – 3 ⇒ m = – 4 ∴ The separate equation of the straight lines are 2x + y + 1 =0 and 2x + y – 4 = 0 The distance between the parallel lines is d = $$\frac{5}{\sqrt{5}}$$ = $$\sqrt{5}$$ ∴ The given equation represents a pair of parallel straight lines and the distance between the parallel lines is $$\sqrt{5}$$ units. Question 16. Prove that one of the straight lines given by ax2 + 2hxy + by2 = 0 will bisect the angle between the coordinate axes if (a + b)2 = 4h2 The equation of the given pair of straight lines is ax2 + 2hxy + by2 = 0 ……… (1) Let m1 and m2, be the slopes of the separate straight lines. Given that one of the straight lines of (1) bisects the angle between the coordinate axes. ∴ The angle made by that line with x-axis 45°. Slope of that line m1 = tan 45° m1 = 1 Squaring on both sides (a + b)2 = (- 2h)2 (a + b)2 = 4h2 Question 17. If the pair of straight lines x2 – 2kxy – y2 = 0 bisects the angle between the pair of straight lines x2 – 2lxy – y2 = 0. Show that the later pair also bisects the angle between the former. The equations of the given pair of straight lines are x2 – 2kxy – y2 = 0 ………… (1) x2 – 2lxy – y2 = 0 ………… (2) Given that the pair x2 – 2kxy – y2 = 0 bisects the angle between the pair x2 – 2lxy – y2 = 0 ∴ The equation of the bisector of the pair x2 – 2lxy – y2 = 0 is the pair x2 – 2kxy – y2 = 0 The equation of the bisector of x2 – 2lxy – y2 = 0 is Equation (3) and Equation (1) represents the same straight lines. ∴ The coefficients are proportional. To show that the pair x2 – 2lxy – y2 = 0 bisects the angle between the pair x2 – 2kxy – y2 = 0, it is enough to prove the equation of the bisector of x2 – 2kxy – y2 = 0 is x2 – 2lxy – y2 = 0 The equation of the bisector of x2 – 2kxy – y2 = 0 is x2 – y2 = 2lxy . x2 – 2lxy – y2 = 0 ∴ The pair x2 – 2lxy – y2 = 0 bisects the angle between the pair x2 – 2kxy – y2 = 0 Question 18. Prove that the straight lines joining the origin to the points of intersection of 3x2 + 5xy – 3y2 + 2x + 3y = 0 and 3x – 2y – 1 = 0 are at right angle.
Graph Linear Equations Related Topics: Math Worksheets In this lesson, we will learn • how to graph a linear equation when the equation is given in slope-intercept form. • how to graph a linear equation when the equation is given in general form. Other methods used to graph linear equations are by plotting points and using x-intercept and y-intercept. The following diagram shows how to graph a linear equation using slope-intercept. Scroll down the page for more examples and solutions. Graph Linear Equations in Slope-Intercept Form When we are given a slope-intercept form of a linear equation, we can use the slope and y-intercept to graph the equation. The slope-intercept form of an equation is y = mx + b, where m is the slope of the line and b is the y-intercept. Step 1: Find the y-intercept and plot the point. Step 2: From the y-intercept, use the slope to find the second point and plot it. Step 3: Draw a line to connect the two points. How to graph a line given in slope-intercept form? Slope intercept formula We look at what slope and intercept mean as well as how to graph the equation. Graph Linear Equations in General Form The equation y = mx+ b gives us information on the slope and the y-intercept. If the equation of a line is given in a different form, we rewrite it in the form y = mx+ b in order to get the slope and y-intercept of the line. * Example** :* Rewrite the following equation in slope-intercept form to determine the slope and y-intercept: 3x = 4y – 7 Solution: Rewrite 3x = 4y – 7 in the form y = mx+ c 3x = 4y – 7 4y = 3x + 7 y = The slope is and the y-intercept is How to convert a linear equation from standard form to slope-intercept form and graphing? How to linear equations given in general form? Slope and intercept of a linear equation Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
Instasolv IIT-JEE NEET CBSE NCERT Q&A 4.5/5 # NCERT Solutions for Class 7 Maths Chapter 4 – Simple Equations NCERT Solutions for Class 7 Maths Chapter 4 is about Simple Equations, the start of algebra. In this chapter, there are a total of 4 exercises and 14 questions. The Chapter 4 of NCERT Solution for Class 7 covers the concept of equations, formulation of Simple Equations and some practical applications Class 7 NCERT Solutions for Chapter 4 ‘Simple Equations’ is based on the latest syllabus approved by CBSE. At Instasolv, we provide the most reliable study material for Class 7 Maths. Chapter-wise Solutions of NCERT Class 7 Maths Book are also provided that help you in understanding the concepts of each exercise better. Chapter 4 Simple Equations, of CBSE NCERT Class 7 Maths Book introduces you to basic algebraic concepts that will be applied often in practical use. For the Maths subject in Class 7, we explain the important elements and summarise each part and subpart of the Chapter. The same sequence as in the book will be followed so that it becomes easy to understand and help as a reference. You will be able to strengthen your concepts of simple equations for class 7 NCERT and score high marks in the exams. ## NCERT Solutions for Class 7 Maths Chapter 4: Important Topics of Each Exercise In NCERT Solutions for Class 7 Maths Chapter 4 you will find exercise-wise solutions to all the 4 exercises that carries 19 questions in total. These exercises are based on the topics discussed below: Tell your friend to think of any number, multiply by 2 and then subtract 5 from the product. Now tell your friend to tell the result and you will tell him the number he had initially thought of. Does it not look like magic or as though you have read your friend’s mind? This game is the introduction of the chapter because the answer lies in Simple Equations. • Setting Up of an Equation Let us consider the same example we took above. Suppose the number your friend thought of was A. The first thing you did was tell him to multiply by 2 so the result is 2A. Now you said subtract 5 from the product so the result is 2A – 5. Suppose your friend says the result he got is 15. The equation then is 2A – 5 = 15. You can now tell him the number he had thought of is 10. Put 10 in the simple equation in place of A and check for yourself. • Review of What We Know From the above, we can infer that a simple equation is formed or set up when a condition on a variable is applied. The variable is not fixed and can take any different numerical value. We generally denote a variable with an alphabet. • What Equation is? In an equation, there is always an equality sign. The expression on the Left Hand Side (LHS) of the equality sign is equal to the expression on the Right Hand Side (RHS) of the equality sign. In totality LHS, equality sign and RHS together form a simple equation. Let us summarise: • An equation is a condition on a variable • The condition is that two expressions, LHS and RHS, should have equal value or in other words LHS = RHS • Since both sides are equal, interchanging the expressions will make no difference. • Note that the variable should be there in at least one of the expressions. • Solving an Equation – Since LHS is equal to RHS, the equation will still hold if we add the same number to both sides. In other words, we add the same number to both LHS and RHS. – By the same logic, the equation still holds if we subtract the same number from both sides. – This is true even if we multiply or divide by the same number on both sides provided it is a non-zero number. – Note that these are valid even if there is an unknown variable in the expressions LHS or RHS. • More Equations Let us take the same simple equation we took at the beginning 2A – 5 = 15 The equation still holds if we add 5 to both sides. 2A – 5 + 5 = 15 + 5 or 2A = 20 The equation still holds if we divide both sides by the non-zero number 2 2A/2 = 20/2 or A = 10 We can also simplify the process by transposing on either side. Let us use the same example for more clarity. 2A – 5 = 15 Instead of doing the addition function simply transpose so a negative number on one side becomes a positive number on the other side 2A = 15 + 5 or 2A = 20 Now transpose the multiplier of A which is 2. Check whether the multiplier is non-zero. 2 is a non-zero number so the condition holds. A multiplier on one side becomes a divider on the other side. 2A = 20 can now be written as A = 20/2 The answer is the same as before. A = 10. Note: that just as a negative number one side becomes a positive number on the other side, similarly a positive number on one side becomes a negative number of the other side. In the same manner, just as a multiplier on one side becomes a divider on the other side, a divider on one side becomes a multiplier on the other side. • From Solution to Equation Making the equation from the solution is how we played the mind-reading game. Knowing the solution as 15, we took the variable as A and made the equation just as we had given instructions. 2A – 5 = 15. Any number of equations can be created. Let us take the same example. We know A = 10 One equation we created was 2A – 5 = 15 We can create many more. Add the same divisor on both sides say 2. A/2 = 10/2. Then add the same number on both sides say 5 The new equation created now is A/2 +5 = 10/2 + 5 Solve and check A/2 + 5 = 5 + 5 or A/2 + 5 = 10 Transpose 5 from LHS to RHS, A/2 = 10 – 5 or A/2 = 5 Transpose 2 from LHS to RHS, A = 5 x 2 or A = 10. • Application of Simple Equations to Practical Situations There are several practical situations where Simple Equations can be used. NCERT Class 7 Maths Book has a self-practice exercise that we will use for our understanding. There are two types of boxes containing mangoes. Each box of the larger type contains 4 more mangoes than the number of mangoes contained in 8 boxes of the smaller type. Each larger box contains 100 mangoes. Find the number of mangoes contained in the smaller box? Given that each larger box contains 100 mangoes. Also given that the larger box contains four more mangoes than the smaller box. We have to find the number of mangoes in the smaller box which we will take as the variable and denote by A From the above statements, the simple equation is A = 100 – 4 = 96. Note: that there is extra information given here that there are 8 boxes of smaller type. When your concepts and understanding are clear you will not get confused by added information and use only what is relevant according to the requirement. That is how practical situations are. ## Study Tips for NCERT Class 7 Maths Chapter 4: Simple Equations Every student has an aim to get a high score in the exams. Instasolv recommends that you read chapter 4 Simple Equations of CBSE NCERT textbook for Class 7 Maths. Instasolv has provided a simple explanation and summary of each part. An example has been used at the beginning and the same example is used to explain all the parts and subparts. All of this can be used to develop a better understanding of all the concepts. • If any concept is still not understood, ask the experts on the Instasolv app which does not charge any student for this support. • Do all the 4 exercises, solved examples and self-practice questions on your own before you cross-check from the NCERT Solutions or ask the experts on the Instasolv app. A practical self-practice example from the CBSE NCERT textbook for Class 7 Maths has been used for explanation. In practical situations, data available can be much more than that required for the question asked. Being able to segregate the required data only is an important step. It is easy for students who have understood the concept and confusing for those who try to learn by rote. Learning by rote is never recommended by the subject matter experts of Instasolv. Follow the NCERT solutions for class 7 maths chapter 4 and tips provided by Instasolv and score high marks in your Maths exam. 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# Area of a Circular Ring or Annulus Annulus: A circular ring (annulus) is plane figure bounded by the circumference of two concentric circles of two different radii. The area of a circular ring is found by subtracting the area of the small circle from that of the large circle. An example of an annulus is the area of a washer and the area of a concrete pipe. If $A$ and $a$, $R$ and $r$ stand for the areas and the radii of two circles and ${A_r}$ for the area of the ring, the i.e. to find the area of a ring (or annulus), multiply the product of the sum and the difference of the two radii by $\pi$ in the first figure. Note: Rule holds good even when circles are not concentric as in second figure. Example: A path $14$cm wide surrounds a circular lawn with a diameter of $360$cm. Find the area of the path. Solution: Given that Radius of inner circle $= 180$cm Radius of outer circle $= (180 + 14) = 194$cm $\therefore$ the area of path $= \pi (R - r)(R + r)$ $= \frac{{22}}{7}(194 + 180)(194 - 180) = 16456$ Square cm Example: The areas of two concentric circles are $1386$square cm and $1886.5$ square cm respectively. Find the width of the ring. Solution: Let $R$ and $r$ be the radii of the outer and inner circles respectively. Let $d$ be the width of the ring $d = (R - r)$ $\therefore$ the area of the outer circle $= \pi {R^2} = 1886.5$ square cm or $R = 24.51$ cm $\therefore$ the area of the inner circle $= \pi {r^2} = 1386$ square cm $\therefore$ $\log \pi + 2\log r = \log 1386$ $2\log r = \log 1386 - \log (3.143)$ $= 3.1418 - 0.4973 = 2.6454$ $\log r = 1.3223$ $r = anti\log (1.3223) = 21$ Hence, the width of the ring $= R - r = 24.51 - 21 = 3.51$cm
# How do you find the coordinates of the center, foci, the length of the major and minor axis given 3x^2+9y^2=27? Dec 29, 2016 The center is $\left(0 , 0\right)$ The foci are $F = \left(\sqrt{6} , 0\right)$ and $F ' = \left(- \sqrt{6} , 0\right)$ The length of the major axis is $= 6$ The length of the minor axis is $= 2 \sqrt{3}$ #### Explanation: Let 's rearrange the equation $3 {x}^{2} + 9 {y}^{2} = 27$ Dividing by $27$ $\frac{3 {x}^{2}}{27} + \frac{9 {y}^{2}}{27} = 1$ ${x}^{2} / 9 + {y}^{2} / 3 = 1$ ${x}^{2} / {3}^{2} + {y}^{2} / {\left(\sqrt{3}\right)}^{2} = 1$ We compare this to the general equation of the ellipse ${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$ $a = 3$ and $b = \sqrt{3}$ The center is $\left(h . k\right) = \left(0 , 0\right)$ To calculate the foci, we need $c = \sqrt{{a}^{2} - {b}^{2}} = \sqrt{9 - 3} = \sqrt{6}$ The foci are F$= \left(h + c , k\right) = \left(\sqrt{6} , 0\right)$ and F'$= \left(h - c , k\right) = \left(- \sqrt{6} , 0\right)$ The length of the major axis is $= 2 a = 6$ The length of the minor axis is $= 2 b = 2 \sqrt{3}$ graph{3x^2+9y^2=27 [-5.546, 5.55, -2.773, 2.774]}
IMAT 2011 Q4 [Different colored balls] Three red balls, three yellow balls and one green ball are placed in a bag and the bag is shaken. I place my hand in the bag and pull out a red ball followed by a green ball. I do not replace either ball. Which one of the following statements is true? A. The next ball could be any one of red, yellow or green. B. The next ball will definitely be yellow. C. The next two balls cannot both be red. D. At least one of the next three balls must be yellow. E. At least one of the next three balls must be red. To solve the question above, we can follow a few methods, depending on your knowledge of mathematics. A. We can use probability formulas to understand the state of the bag in each turn. B. We can write down and draw the state of the bag in each turn. When it comes to this exam, I highly suggest using a combination of both. Understanding the mathematical idea of the question while also drawing and writing down the steps will allow you to create a clear picture of the question and compare it to the options provided. Step 1) Three red balls, three yellow balls, and one green ball are placed in a bag - There is \frac{3}{7} probability of pulling out a red ball, \frac{3}{7} of pulling out a yellow ball, and \frac{1}{7} of pulling out a green one, during the first round. You pull out one red ball and then one green ball Step 2) You have two red balls \frac{2}{5} , three yellow balls \frac{3}{5} , and no green ball \frac{0}{5} in your bag. A - The next ball could be any one of red, yellow, or green. We donâ€™t have any green balls left. B - The next ball will definitely be yellow - That cannot be true, as we have \frac{2}{5} chance of pulling out a red ball, as mentioned in the quick calculation we made. C - The next two balls cannot both be red. - This is a possibility, as we already calculated that we have two red balls left. D - At least one of the next three balls must be yellow. - It must be true as we only have two red balls left in the bag and all of the rest yellow. E - At least one of the next three balls must be red. - As we have three yellow balls in the bag, it is possible that we will pull all of them out in the next three rounds. Tip: This question might look straightforward when solving it at home, but you have to consider the fact that it will be a bit different on the exam. Some simple questions with a lot of info and steps might take a bit longer to solve, so I highly suggest always approaching these questions, knowing which actions you will follow while solving them.
# How do you use geometric solids? Place one of the more recognizable forms, such as a sphere, on the mat in front of you. Pick up and examine the solid by gently sliding your hands around it. "This is a spherical," explain to the youngster. Allow the youngster to have a turn repeating your activities and feeling the geometric solid. When finished, have him or her describe what they observed about the shape of the form. Now that you know how to use these solids effectively, it's time to move on to another type of solid: linear. To use a linear solid, start with a straightedge and compass. Have the child draw a line across the board. Explain that we can use this line to create other shapes by cutting along the edge of the line. Let him or her cut out a square, a rectangle, and a circle. Place each shape over the line on the board so that it fits exactly inside the corresponding shape cut from the solid. Help the child roll the solid in his or her mind's eye to see that it stays completely round or square regardless of which way it is rolled. These are only some examples of how to use geometrical solids. You may want to expand on these ideas by having your youngster explore other types of solids, such as triangular, quadrilateral, pentagonal, etc. The important thing is that he or she has fun learning about these interesting objects that make up our world around us. ## What are spherical objects? A sphere (from Greek sphairaâ€”sphaira, "globe, ball") is a three-dimensional geometrical entity that is the surface of a ball (viz., analogous to the circular objects in two dimensions, where a "circle" circumscribes its "disk"). These are also known as the radius and the center of the sphere, respectively. The term "sphere" can also be applied to other three-dimensional shapes, such as a cube or a dodecahedron; however, these are not generally considered solids for mathematical purposes. Spheres are important in mathematics and science because many problems involving spheres can be solved by using spheres instead. For example, if you want to know how much milk would be needed to fill a tank truck, you could use the formula PV = nRT, where V is volume, n is the number of gallons, R is the radius of the truck bed, and T is the temperature in degrees Fahrenheit. This equation shows that the volume of milk required is proportional to the square of the radius of the truck bed. To keep the load equal, we need to increase the radius by a factor of 7/5 every time we add milk to the truck. The problem with this approach is that it is extremely inefficient. It would require about a quart of milk for each gallon saved by this method. A better way to do this problem is to realize that the volume of milk required is constant, so we should just need to increase the amount of space we give it. ## How can one change the shape of the materials? The forms of solid objects made of certain materials may be modified by squashing, bending, twisting, and stretching. The way these shapes are modified is called deformation. In general, the more rigid a material is, the less it can be deformed without breaking; therefore, the more rigid its natural state is. A very rigid material, like glass, cannot be deformed at all in the natural state. All materials deform under pressure or force. Pressure from within, such as that produced by muscle tissue or blood cells, can cause a object to become thinner or longer. Pressure from without, such as that produced by hand or machine tools, can cause an object to become narrower or taller. When a person squeezes a ball of soft rubber, for example, they are merely modifying its form. When viewed microscopically, soft materials such as rubber appear smooth because there are no visible edges or corners. But at a larger scale, such as that of a human being, these same materials have rough surfaces caused by compression. If a person were to push on a corner of this rubber ball, it would deform into a slightly flattened ellipse. A sheet of paper can be bent without breaking if it is not too thick. ## How do you identify solids? The following qualities characterize solids: 1. Definite shape (rigid) 2. Definite volume. 3. Particles vibrate around fixed axes. ## How many solid shapes do we have? Solids, often known as three-dimensional things, have three dimensions: length, width, and height. Faces, edges, and vertices are all characteristics of solid forms. Learning about solid forms can benefit us in our daily lives because they are used in so many of our activities. As an example, Solid Shapes Cube 6 12 8 ## How do you describe a solid shape? Three-dimensional things are solid forms. They are triangular in shape and have three dimensions: length, breadth, and height. They take up space in the cosmos because they are three-dimensional. Two-dimensional things are flat, and they only have two dimensions: length and breadth. Three-dimensional objects contain areas without dimension; for example, the surface of a ball is two-dimensional but its interior is three-dimensional. Two-dimensional things can be shapes drawn with ink on paper, or they can be images stored on your computer's hard drive. When you view these images on your screen, you are viewing them visually but they are still just two-dimensional shapes. The term "solid" is used to describe shapes that have weight and take up space. A circle is an example of a two-dimensional solid shape while a cube is a three-dimensional solid shape. You can describe any two- or three-dimensional shape as solid. These could be shapes made from clay, wood, metal, or even ice! You can also describe a four-sided object as solid since there are no openings inside it where something could go through. A pyramid is an example of a four-sided solid shape. Shapes can also be described as hollow. Hollow shapes have spaces within them where things can go. ## Which is the most commonly used geometric shape? Geometric forms are most likely the most often utilized. They are the first thing that comes to mind. Circles, squares, triangles, pentagons, hexagons, and octagons are easily identified. They are so familiar to us since we began seeing and painting them as children. Geometric shapes are simple to draw yet provide the artist with a wide range of possibilities. Beyond their aesthetic value, these figures have important implications for physics and mathematics. Scientists use geometric shapes to model physical systems -- such as atoms, molecules, crystals, and metals -- that are too complex to study completely analytically. Mathematicians use them to prove theorems or devise new techniques. In both cases, geometric thinking is essential in order to achieve one's goals. There are many more types of figures than just those listed here. For example, there are convex and concave figures, orthographic and perspective views, empty and full figures. However, they are not as common as the ones mentioned above and thus do not get much attention in school geometry classes. It is difficult to say which type of figure is the most common since they all appear frequently in nature and art. However, based on what we see around us, circles seem to be the most popular choice followed by squares. Other shapes are less common but none of them are rare. Circle-shaped objects include rings, coins, and wheels. #### About Article Author ##### Donna Nease Donna Nease is an inspiration to many. She has overcome many hardships in her life, and she is now a successful businesswoman. She loves sharing her stories of struggle and victorious over-come because it shows people that no matter how bad things seem, they can overcome anything if they truly want it bad enough.
# How do you find lim_(xtooo)(Cos(x)/x)? Mar 28, 2017 Because the upper bound of cos(x) is 1 and the lower bound is -1: ${\lim}_{x \to \infty} - \frac{1}{x} \le {\lim}_{x \to \infty} \cos \frac{x}{x} \le {\lim}_{x \to \infty} \frac{1}{x}$ We know the limits of the two bounds to be 0: $0 \le {\lim}_{x \to \infty} \cos \frac{x}{x} \le 0$ $\therefore$ ${\lim}_{x \to \infty} \cos \frac{x}{x} = 0$ Mar 28, 2017 It should tend to zero. #### Explanation: I would say that as cos(x) oscilates between $1 \mathmr{and} - 1$ then $x$ will grow so much that the fraction will become very small tending to become zero. This is somewhat supported by the graph of our function: graph{(cos(x))/x [-10, 10, -5, 5]} Because $- 1 \le \cos x \le 1$ , we have that $- \frac{1}{x} \le \cos \frac{x}{x} \le \frac{1}{x}$ Hence ${\lim}_{x \to \infty} \left(- \frac{1}{x}\right) \le {\lim}_{x \to \infty} \cos \frac{x}{x} \le {\lim}_{x \to \infty} \frac{1}{x}$ $0 \le {\lim}_{x \to \infty} \cos \frac{x}{x} \le 0$ Finally ${\lim}_{x \to \infty} \cos \frac{x}{x} = 0$
# Drawing an Angle with a Protractor \| How to Construct Angle with Compass & Protractor Step by Step? Are you thinking more about how to draw an angle? So, without any postpone let’s go into this article and find your answers related to Drawing an angle. Drawing an angle is nothing but constructing an angle. Students who imagine that angles are very complex can refer to this article and find easy steps on the construction of an angle using a protractor. To draw an angle all you need are some basic tools like a protractor, ruler, compass, and a pencil. This article provides types of angles, how to draw an angle in different methods, and more examples to construct an angle. At the end of this page, you will be completely aware of constructing an angle using a compass & protractor. ## Types of Angles An Angle is nothing but when two lines intersect each other at a common vertex point. While drawing an angle, 4th Grade Math students & other higher grades kids should have a quick review of different angles that made them easy to grasp. The following are the five types of angles we have: 1. Acute Angle: An angle less than 90°. 2. Right Angle: An angle which is exactly 90°. 3. Obtuse Angle: An angle greater than 90°. 4. Straight Angle: An angle equals 180° and looks like a straight line. 5. Reflex Angle: An angle always greater than 180°. Do Refer the articles: ### Methods on How to Draw Angles Step-by-Step In geometry, drawing an angle can be possible with unknown and known measures by using some geometrical tools. Here, we have two methods to follow while constructing an angle. • Drawing an angle with a protractor • Drawing an angle using a compass ### How to Use a Protractor to Draw Angles? A protractor is a semi-circular tool used to draw and measure angles, from 0° to 180° in clockwise and anti-clockwise. If kids are familiar with drawing an angle with a protractor then they can easily draw an angle between 0° to 180°. Drawing Angles using a Protractor can be easy if you follow the below steps. To construct an angle of any given measure, children can follow these simple steps. Step 1: Draw a line segment of any length, and name both ends. Step 2: Place the protractor at the start point of a line segment drawn. Step 3: The protractor has two sets of markings. We examine the scale which has 0° near the endpoint of a line segment. Mark, another point next to the degrees given to measure or draw an angle and label the point. Step 4: Finally, join both the points, i.e., the starting point of the line segment and the point marked using a protractor by a line. By joining three points we get a required angle. ### How to Draw Angle with Compass? To draw an angle of any measure one more useful tool you need is a compass. Using compass children may feel difficult but by following these simple steps they can make an angle with a compass easily & effortlessly. Step 1: Draw a line segment of any line and mark both the points as AB. Step 2: Take the compass and place the pointer at point A, move the arm of the compass of some length. Draw an arc by the pencil which meets the line AB and point as P. Step 3: Move the compass without disturbing the length and place the pointer at point P and mark an arc that passes through point A and should intersect the previous arc at a point, name point as Q. Step 4: Draw a line from point A through point Q and name the line as point C. We get the required angle i.e., ∠CAB of degrees measured. Also Check: ### Examples on Drawing Angles Using Protractor & Compass Example 1: Draw an angle of 35° using a protractor. Solution: Draw a line segment YZ. Place the center of the protractor at point Y of the line segment. Mark the 45 degrees where the line YZ coincides with the 0° of the protractor. Label the point as X. Join the points Z and X by a line. We get a required angle ∠XYZ = 35°. Example 2: Draw an angle of 140° using a protractor. Solution: Draw a line segment QP. Place the protractor at straight line PQ and the midpoint of the protractor be at point P. Mark the 140° where the line coincides with the 0° and name the marked point as O. Join the points P and O by a line with the help of a pencil. We get a required angle ∠OPQ = 140°. Example 3: Draw an angle of 60° with a compass. Solution: Draw a line segment AB. Take a compass with A as a center of any width draw an arc that meets the line AB and label the point as O. Without changing the length of the compass, draw an arc from the point O as a center and name the point as Q that passes through A and intersects the previous arc. Now, draw a line from point A through Q and label the line endpoint as X. The required angle is ∠XAB = 60°. Example 4: Construct an obtuse angle of 120° by compass. Solution: Draw a line segment ST. Take a compass of any width. Place at point S as a center and draw an arc on line ST and name the arc as point A. Mark another arc of the same length of a compass and make a center as A. Label the arc as point B. Now, without disturbing the compass length, mark another arc taking point B as a center that intersects the previous arc A. Label the arc as point C. Draw a line from point S through C and extend the line and label it as R. We get a required obtuse angle ∠RST= 120°. Scroll to Top
Survey * Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project Document related concepts Bootstrapping (statistics) wikipedia, lookup Time series wikipedia, lookup Regression toward the mean wikipedia, lookup Transcript ```1.4 Defining Data Spread • An average alone doesn’t always describe a set of data effectively or completely. An average doesn’t indicate whether the data clusters, whether the set contains outliers, what the range is, how the data is spread etc. In general it does not tell about the set’s distribution. The various data distribution plots we have studied help to do that. • Investigate the following to discover a way to determine a single number that can indicate the spread and variation in a data set. • For the following data try to determine the average (mean) distance the values are from the mean of the set. Test Scores 35 44 56 58 62 67 70 72 76 88 90 94 • Step 1- Calculate the mean of the set mean = 67.7 approx. 68 Step 2 – Calculate the distance each value is from the mean ( mean – the value ). This is called the deviation from the mean. Data Value 35 44 56 58 62 67 70 72 76 88 90 94 Deviation(mean – value) 68-35= 33 24 12 10 6 1 -2 -4 -8 -20 -22 -26 • Step 3 – Square each deviation (to remove the negatives) Data Value Deviation Squared Deviation 35 33 1089 44 24 576 56 12 144 58 10 100 62 6 36 67 1 1 70 -2 4 72 -4 16 76 -8 64 88 -20 400 90 -22 484 94 -26 676 • Step 4 - Find the mean of the squared deviations. 3590 / 12 = 299.2 approx. 299 Step 5 – Find the square root of step 4 (the mean of the squared deviations) √ 299 = 17.3 • You just found what is called….. Standard deviation – a # that describes the spread/variation within a set of data. It represents the average distance the data values are from the mean of the set. • The greater the standard deviation…
## Precalculus: Sum and Difference Identities Practice Problems ... Precalculus: Sum and Difference Identities Practice Problems. Questions. 1. Find the value of sin. (. − π. 12. ) exactly. 2. Prove the identity cos. ( θ + π. 2. ). Precalculus: Sum and Difference Identities Practice Problems Questions π exactly. 1. Find the value of sin − 12 π = − sin θ. 2. Prove the identity cos θ + 2 3. Prove the identity sin 4x + sin 2x = 2 sin 3x cos x. 5π 4. Find the value of sin − exactly by using the sine of a sum identity. This problem shows you a method to 12 determine exactly the trig functions at angles other than the special angles on the unit circle. Page 1 of 4 Precalculus: Sum and Difference Identities Practice Problems Solutions π exactly. 1. Find the value of sin − 12 First, we need to figure out how to relate −π/12 to some of our special angles, since we are told to find this answer exactly. −π −2π 4π − 6π π π = = = − . 12 24 24 6 4 Therefore, π sin − 12 = = = = = √ hyp= 2 π − 6 4 π π π π sin cos − cos sin , use sin(u − v) = sin u cos v − cos u sin v 6 4 6 4 √ ! 1 1 1 3 √ √ − , using reference triangles below 2 2 2 2 √ 1 3 √ − √ 2 2 2 2 √ 1− 3 √ 2 2 sin opp=1 hyp=2 opp=1 π 2. Prove the identity cos θ + = − sin θ. 2 π cos θ + 2 π cos cos θ − sin π sin θ, using cos(u + v) = cos u cos v − sin u sin v 2 2 = (0) cos θ − (1) sin θ, using the unit circle to evaluate the trig functions of pi/2. = − sin θ = Page 2 of 4 Precalculus: Sum and Difference Identities Practice Problems 3. Prove the identity sin 4x + sin 2x = 2 sin 3x cos x. sin 4x + sin 2x = = = = = = = = = = sin(3x + x) + sin(x + x), (sin 3x cos x + cos 3x sin x) + (sin x cos x + cos x sin x) , use sin(u + v) = sin u cos v + cos u sin v twice. sin 3x cos x + (cos 3x) sin x + 2 sin x cos x, sin 3x cos x + (cos(2x + x)) sin x + 2 sin x cos x sin 3x cos x + (cos 2x cos x − sin 2x sin x) sin x + 2 sin x cos x, using cos(u + v) = cos u cos v − sin u sin v sin 3x cos x + cos 2x cos x sin x − sin 2x sin2 x + 2 sin x cos x sin 3x cos x + (cos 2x) cos x sin x − (sin 2x) sin2 x + 2 sin x cos x sin 3x cos x + (cos(x + x)) cos x sin x − (sin(x + x)) sin2 x + 2 sin x cos x sin 3x cos x + (cos x cos x − sin x sin x) cos x sin x − (sin x cos x + cos x sin x) sin2 x + 2 sin x cos x sin 3x cos x + cos3 x sin x − 3 sin3 x cos x + 2 sin x cos x We can see that this is very nasty looking. Let’s start over. sin 4x + sin 2x = sin(3x + x) + sin(3x − x), = (sin 3x cos x + cos 3x sin x) + (sin 3x cos x − cos 3x sin x) , = 2 sin 3x cos x use sin(u ± v) = sin u cos v ± cos u sin v That was certainly easier! In this case, we exploited some of the hidden symmetry in the problem by writing 2x = 3x − 1 rather than 2x = x + x. Page 3 of 4 Precalculus: Sum and Difference Identities Practice Problems 5π 4. Find the value of sin − exactly by using the sine of a sum identity. This problem shows you a method to 12 determine exactly the trig functions at angles other than the special angles on the unit circle. First, we need to figure out how to relate −5π/12 to some of our special angles, since we are told to find this answer exactly. We are told to use a sum formula, so the sum of two of our special angles should produce −5π/12. −5π −10π −6π − 4π π π = = =− − . 12 24 24 4 6 Therefore, 5π sin − 12 = = = = = =
## Engage NY Eureka Math 8th Grade Module 7 Lesson 5 Answer Key ### Eureka Math Grade 8 Module 7 Lesson 5 Example Answer Key Example 1. x3 + 9x = $$\frac{1}{2}$$ (18x + 54) Now that we know about square roots and cube roots, we can combine that knowledge with our knowledge of the properties of equality to begin solving nonlinear equations like x3 + 9x = $$\frac{1}{2}$$ (18x + 54). Transform the equation until you can determine the positive value of x that makes the equation true. Challenge students to solve the equation independently or in pairs. Have students share their strategy for solving the equation. Ask them to explain each step. x3 + 9x = $$\frac{1}{2}$$ (18x + 54) x3 + 9x = 9x + 27 x3 + 9x – 9x = 9x – 9x + 27 x3 = 27 $$\sqrt[3]{x^{3}}$$ = $$\sqrt [ 3 ]{ 27 }$$ x = $$\sqrt[3]{3^{3}}$$ x = 3 Now, we verify our solution is correct. 33 + 9(3) = $$\frac{1}{2}$$ (18(3) + 54) 27 + 27 = $$\frac{1}{2}$$ (54 + 54) 54 = $$\frac{1}{2}$$ (108) 54 = 54 Since the left side is the same as the right side, our solution is correct. Example 2. x(x – 3) – 51 = – 3x + 13 Let’s look at another nonlinear equation. Find the positive value of x that makes the equation true: x(x – 3) – 51 = – 3x + 13. Provide students with time to solve the equation independently or in pairs. Have students share their strategy for solving the equation. Ask them to explain each step. Sample response: x(x – 3) – 51 = – 3x + 13 x2 – 3x – 51 = – 3x + 13 x2 – 3x + 3x – 51 = – 3x + 3x + 13 x2 – 51 = 13 x2 – 51 + 51 = 13 + 51 x2 = 64 $$\sqrt{x^{2}}$$ = ±$$\sqrt{64}$$ x = ±$$\sqrt{64}$$ x = ±8 Now we verify our solution is correct. Provide students time to check their work. Let x = 8. 8(8 – 3) – 51 = – 3(8) + 13 8(5) – 51 = – 24 + 13 40 – 51 = – 11 – 11 = – 11 Let x = – 8. – 8( – 8 – 3) – 51 = – 3( – 8) + 13 – 8( – 11) – 51 = 24 + 13 88 – 51 = 37 37 = 37 Now it is clear that the left side is exactly the same as the right side, and our solution is correct. ### Eureka Math Grade 8 Module 7 Lesson 5 Exercise Answer Key Find the positive value of x that makes each equation true, and then verify your solution is correct. Exercise 1. a. Solve x2 – 14 = 5x + 67 – 5x. x2 – 14 = 5x + 67 – 5x x2 – 14 = 67 x2 – 14 + 14 = 67 + 14 x2 = 81 $$\sqrt{x^{2}}$$ = ±$$\sqrt{81}$$ x = ±$$\sqrt{81}$$ x = ±9 Check: 92 – 14 = 5(9) + 67 – 5(9) 81 – 14 = 45 + 67 – 45 67 = 67 ( – 9)2 – 14 = 5( – 9) + 67 – 5( – 9) 81 – 14 = – 45 + 67 + 45 67 = 67 b. Explain how you solved the equation. To solve the equation, I had to first use the properties of equality to transform the equation into the form of x2 = 81. Then, I had to take the square root of both sides of the equation to determine that x = 9 since the number x is being squared. Exercise 2. Solve and simplify: x(x – 1) = 121 – x. x(x – 1) = 121 – x x2 – x = 121 – x x2 – x + x = 121 – x + x x2 = 121 $$\sqrt{x^{2}}$$ = ±$$\sqrt{121}$$ x = ±$$\sqrt{121}$$ x = ±11 Check: 11(11 – 1) = 121 – 11 11(10) = 110 110 = 110 – 11( – 11 – 1) = 121 – ( – 11) – 11( – 12) = 121 + 11 132 = 132 Exercise 3. A square has a side length of 3x inches and an area of 324 in2. What is the value of x? (3x)2 = 324 32 x2 = 324 9x2 = 324 $$\frac{9 x^{2}}{9}$$ = $$\frac{324}{9}$$ x2 = 36 $$\sqrt{x^{2}}$$ = $$\sqrt{36}$$ x = 6 Check: (3(6)) 2 = 324 182 = 324 324 = 324 A negative number would not make sense as a length, so x = 6. Exercise 4. – 3x3 + 14 = – 67 – 3x3 + 14 = – 67 – 3x3 + 14 – 14 = – 67 – 14 – 3x3 = – 81 $$\frac{ – 3 x^{3}}{3}$$ = $$\frac{ – 81}{ – 3}$$ x3 = 27 $$\sqrt[3]{x^{3}}$$ = $$\sqrt [ 3 ]{ 27 }$$ x = 3 Check: – 3(3)3 + 14 = – 67 – 3(27) + 14 = – 67 – 81 + 14 = – 67 – 67 = – 67 Exercise 5. x(x + 4) – 3 = 4(x + 19.5) x(x + 4) – 3 = 4(x + 19.5) x2 + 4x – 3 = 4x + 78 x2 + 4x – 4x – 3 = 4x – 4x + 78 x2 – 3 = 78 x2 – 3 + 3 = 78 + 3 x2 = 81 $$\sqrt{x^{2}}$$ = ±$$\sqrt{81}$$ x = ±9 Check: 9(9 + 4) – 3 = 4(9 + 19.5) 9(13) – 3 = 4(28.5) 117 – 3 = 114 114 = 114 – 9( – 9 + 4) – 3 = 4( – 9 + 19.5) – 9( – 5) – 3 = 4(10.5) 45 – 3 = 42 42 = 42 Exercise 6. 216 + x = x(x2 – 5) + 6x 216 + x = x(x2 – 5) + 6x 216 + x = x3 – 5x + 6x 216 + x = x3 + x 216 + x – x = x3 + x – x 216 = x3 $$\sqrt [ 3 ]{ 216 }$$ = $$\sqrt[3]{x^{3}}$$ 6 = x Check: 216 + 6 = 6(62 – 5) + 6(6) 222 = 6(31) + 36 222 = 186 + 36 222 = 222 Exercise 7. a. What are we trying to determine in the diagram below? We need to determine the value of x so that its square root, multiplied by 4, satisfies the equation 52 + (4$$\sqrt{x}$$)2 = 112. 52 + (4$$\sqrt{x}$$)2 = 112 25 + 42 ($$\sqrt{x}$$)2 = 121 25 – 25 + 42 ($$\sqrt{x}$$)2 = 121 – 25 16x = 96 $$\frac{16x}{16}$$ = $$\frac{96}{16}$$ x = 6 The value of x is 6. Check: 52 + (4$$\sqrt{6}$$)2 = 112 25 + 16(6) = 121 25 + 96 = 121 121 = 121 ### Eureka Math Grade 8 Module 7 Lesson 5 Problem Set Answer Key Find the positive value of x that makes each equation true, and then verify your solution is correct. Question 1. x2 (x + 7) = $$\frac{1}{2}$$ (14x2 + 16) x2 (x + 7) = $$\frac{1}{2}$$ (14x2 + 16) x3 + 7x2 = 7x2 + 8 x3 + 7x2 – 7x2 = 7x2 – 7x2 + 8 x3 = 8 $$\sqrt[3]{x^{3}}$$ = $$\sqrt [ 3 ]{ 8 }$$ x = 2 Check: 22 (2 + 7) = $$\frac{1}{2}$$ (14(22 ) + 16) 4(9) = $$\frac{1}{2}$$ (56 + 16) 36 = $$\frac{1}{2}$$ (72) 36 = 36 Question 2. x3 = 1331 – 1 x3 = 1331 – 1 $$\sqrt[3]{x^{3}}$$ = $$\sqrt[3]{1331^{ – 1}}$$ x = $$\sqrt[3]{\frac{1}{1331}}$$ x = $$\sqrt[3]{\frac{1}{11^{3}}}$$ x = $$\frac{1}{11}$$ Check: ($$\left(\frac{1}{11}\right)^{3}$$)3 = 1331 – 1 $$\frac{1}{11^{3}}$$ = 1331 – 1 $$\frac{1}{1331}$$ = 1331 – 1 1331 – 1 = 1331 – 1 Question 3. Determine the positive value of x that makes the equation true, and then explain how you solved the equation. $$\frac{x^{9}}{x^{7}}$$ – 49 = 0 $$\frac{x^{9}}{x^{7}}$$ – 49 = 0 x2 – 49 = 0 x2 – 49 + 49 = 0 + 49 x2 = 49 $$\sqrt{x^{2}}$$ = $$\sqrt{49}$$ x = 7 Check: 72 – 49 = 0 49 – 49 = 0 0 = 0 To solve the equation, I first had to simplify the expression $$\frac{x^{9}}{x^{7}}$$ to x2. Next, I used the properties of equality to transform the equation into x2 = 49. Finally, I had to take the square root of both sides of the equation to solve for x. Question 4. Determine the positive value of x that makes the equation true. (8x)2 = 1 (8x)2 = 1 64x2 = 1 $$\sqrt{64^{2}}$$ = $$\sqrt{1}$$ 8x = 1 $$\frac{8x}{8}$$ = $$\frac{1}{8}$$ x = $$\frac{1}{8}$$ Check: (8($$\frac{1}{8}$$))2 = 1 12 = 1 1 = 1 Question 5. (9$$\sqrt{x}$$)2 – 43x = 76 (9$$\sqrt{x}$$)2 – 43x = 76 92 (√x)2 – 43x = 76 81x – 43x = 76 38x = 76 $$\frac{38x}{38}$$ = $$\frac{76}{38}$$ x = 2 Check: (9($$\sqrt{2}$$))2 – 43(2) = 76 92 ($$\sqrt{2}$$)2 – 86 = 76 81(2) – 86 = 76 162 – 86 = 76 76 = 76 Question 6. Determine the length of the hypotenuse of the right triangle below. 32 + 72 = x2 9 + 49 = x2 58 = x2 $$\sqrt{58}$$ = $$\sqrt{x^{2}}$$ $$\sqrt{58}$$ = x Check: 32 + 72 = ($$\sqrt{52}$$)2 9 + 49 = 58 58 = 58 Since x = $$\sqrt{58}$$, the length of the hypotenuse is $$\sqrt{58}$$ mm. Question 7. Determine the length of the legs in the right triangle below. x2 + x2 = (14$$\sqrt{2}$$)2 2x2 = 142 ($$\sqrt{2}$$)2 2x2 = 196(2) $$\frac{2 x^{2}}{2}$$ = $$\frac{196(2)}{2}$$ x2 = 196 $$\sqrt{x^{2}}$$ = $$\sqrt{196}$$ x = $$\sqrt{14^{2}}$$ x = 14 Check: 142 + 142 = (14$$\sqrt{2}$$)2 196 + 196 = 142 ($$\sqrt{2}$$)2 392 = 196(2) 392 = 392 Since x = 14, the length of each of the legs of the right triangle is 14 cm. Question 8. An equilateral triangle has side lengths of 6 cm. What is the height of the triangle? What is the area of the triangle? Note: This problem has two solutions, one with a simplified root and one without. Choose the appropriate solution for your classes based on how much simplifying you have taught them. Let h cm represent the height of the triangle. 32 + h2 = 62 9 + h2 = 36 9 – 9 + h2 = 36 – 9 h2 = 27 $$\sqrt{h^{2}}$$ = $$\sqrt{27}$$ h = $$\sqrt{27}$$ h = $$\sqrt{3^{3}}$$ h = $$\sqrt{3^{2}}$$×$$\sqrt{3}$$ h = 3$$\sqrt{3}$$ Let A represent the area of the triangle. A = $$\frac{6(3 \sqrt{3})}{2}$$) A = 3(3$$\sqrt{3}$$) A = 9$$\sqrt{3}$$ Simplified: The height of the triangle is 3$$\sqrt{3}$$ cm, and the area is 9$$\sqrt{3}$$ cm2. Unsimplified: The height of the triangle is $$\sqrt{27}$$ cm, and the area is 3$$\sqrt{27}$$ cm2 Question 9. Challenge: Find the positive value of x that makes the equation true. ($$\frac{1}{2}$$ x)2 – 3x = 7x + 8 – 10x ($$\frac{1}{2}$$ x)2 – 3x = 7x + 8 – 10x $$\frac{1}{4}$$ x2 – 3x = – 3x + 8 $$\frac{1}{4}$$ x2 – 3x + 3x = – 3x + 3x + 8 $$\frac{1}{4}$$ x2 = 8 4($$\frac{1}{4}$$) x2 = 8(4) x2 = 32 $$\sqrt{x^{2}}$$ = $$\sqrt{32}$$ x = $$\sqrt{2^{5}}$$ x = $$\sqrt{2^{2}}$$ ⋅ $$\sqrt{2^{2}}$$ ⋅ $$\sqrt{2}$$ x = 4$$\sqrt{2}$$ Check: ($$\frac{1}{2}$$ (4$$\sqrt{2}$$))2 – 3(4$$\sqrt{2}$$) = 7(4$$\sqrt{2}$$) + 8 – 10(4$$\sqrt{2}$$) $$\frac{1}{4}$$ (16)(2) – 3(4$$\sqrt{2}$$) = 7(4$$\sqrt{2}$$) – 10(4$$\sqrt{2}$$) + 8 $$\frac{32}{4}$$ – 3(4$$\sqrt{2}$$) = 7(4$$\sqrt{2}$$) – 10(4$$\sqrt{2}$$) + 8 8 – 3(4$$\sqrt{2}$$) = (7 – 10)(4$$\sqrt{2}$$) + 8 8 – 3(4$$\sqrt{2}$$) = – 3(4$$\sqrt{2}$$) + 8 8 – 8 – 3(4$$\sqrt{2}$$) = – 3(4$$\sqrt{2}$$) + 8 – 8 – 3(4$$\sqrt{2}$$) = – 3(4$$\sqrt{2}$$) Question 10. Challenge: Find the positive value of x that makes the equation true. 11x + x(x – 4) = 7(x + 9) 11x + x(x – 4) = 7(x + 9) 11x + x2 – 4x = 7x + 63 7x + x2 = 7x + 63 7x – 7x + x2 = 7x – 7x + 63 x2 = 63 $$\sqrt{x^{2}}$$) = $$\sqrt{63}$$ x = $$\sqrt{\left(3^{2}\right)(7)}$$ x = $$\sqrt{3^{2}}$$ ⋅ $$\sqrt{7}$$ x = 3$$\sqrt{7}$$ Check: 11(3$$\sqrt{7}$$) + 3$$\sqrt{7}$$ (3$$\sqrt{7}$$ – 4) = 7(3$$\sqrt{7}$$ + 9) 33$$\sqrt{7}$$ + 32 ($$\sqrt{7}$$)2 – 4(3$$\sqrt{7}$$) = 21$$\sqrt{7}$$ + 63 33$$\sqrt{7}$$ – 4(3$$\sqrt{7}$$) + 9(7) = 21$$\sqrt{7}$$ + 63 33$$\sqrt{7}$$ – 12$$\sqrt{7}$$ + 63 = 21$$\sqrt{7}$$ + 63 (33 – 12) $$\sqrt{7}$$ + 63 = 21$$\sqrt{7}$$ + 63 21$$\sqrt{7}$$ + 63 = 21$$\sqrt{7}$$ + 63 21$$\sqrt{7}$$ + 63 – 63 = 21$$\sqrt{7}$$ + 63 – 63 21$$\sqrt{7}$$ = 21$$\sqrt{7}$$ ### Eureka Math Grade 8 Module 7 Lesson 5 Exit Ticket Answer Key Question 1. Find the positive value of x that makes the equation true, and then verify your solution is correct. x2 + 4x = 4(x + 16) x2 + 4x = 4(x + 16) x2 + 4x = 4x + 64 x2 + 4x – 4x = 4x – 4x + 64 x2 = 64 $$\sqrt{x^{2}}$$ = $$\sqrt{64}$$ x = 8 Check: 82 + 4(8) = 4(8 + 16) 64 + 32 = 4(24) 96 = 96 Question 2. Find the positive value of x that makes the equation true, and then verify your solution is correct. (4x)3 = 1728 $$\frac{1}{64}$$(64x3) = (1728)$$\frac{1}{64}$$ $$\sqrt[3]{x^{3}}$$ = $$\sqrt [ 3 ]{ 27 }$$
Finding point when tangent is parallel/ perpendicular Chapter 6 Class 12 Application of Derivatives Concept wise Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month ### Transcript Ex 6.3, 8 Find a point on the curve π¦=(π₯β2)^2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).Given Curves is π¦=(π₯β2)^2 Let AB be the chord joining the Point (2 , 0) & (4 ,4) & CD be the tangent to the Curve π¦=(π₯β2)^2 Given that tangent is Parallel to the chord i.e. CD β₯ AB β΄ Slope of CD = Slope of AB If two lines are parallel, then their slopes are equal Slope of tangent CD Slope of tangent CD = ππ¦/ππ₯ =(π(π₯ β 2)^2)/ππ₯ = 2(π₯β2) (π (π₯ β 2))/ππ₯ = 2(π₯β2) (1β0) = 2(π₯β2) Slope of AB As AB is chord joining Points (2 , 0) & (4 , 4) Slope of AB =(4 β 0)/(4 β 2) =4/2 As slope of line joining point (π₯ , π¦) & (π₯2 , π¦2) ππ (π¦2 β π¦1)/(π₯2 β π₯1) =2 Now, Slope of CD = Slope of AB 2(π₯β2)=2 π₯β2=2/2 π₯β2=1 π₯=3 Finding y when π₯=3 π¦=(π₯β2)^2 π¦=(3β2)^2 π¦=(1)^2 π¦=1 Hence, Point is (π , π) Thus, the tangent is parallel to the chord at (3 ,1)
# Land Reform (an elaborated discussion) Topics: Measurement, Mathematics Pages: 3 (268 words) Published: September 21, 2013 Name: AnonymousDate: June 22, 2012 Year & Section:Weather: Fair Exercise No. 1 Title: DETERMINATION OF PACE FACTOR I. OBJECTIVE To let the student be familiar with the Pace factor its computation and uses by means of a field activity. II. PROCEDURES 1. Measure a horizontal distance of 55m. 2. Cover the measured distance by walking, count and record the number of paces in a trial. 3. Do the second procedure by seven trials. 4. Tabulate the gathered data accordingly and calculate the pace factor. III. EQUIPMENTS USED Steel Tape Chalk IV. FIGURE Figure 1 Determining the Pace Factor V. FIELD MEASUREMENT DATA Trial(n) Line No. of Paces(x) Taped Distance (TP) Average No. of Paces (AP) Pace Factor (PF) 1 AB 75 55 m 72.7 paces approx. 73paces 0.75m/pace 2 BA 77 3 AB 74 4 BA 72 5 AB 70 6 BA 71 7 AB 70 Total 509 VI. CALCULATION Given: *Tabulated Data To compute for the Average Pace: AP=(x1+ x2+ x3+ x4+ x5+ x6+ x7)/n =(75+77+74+72+70+71+70)/7 AP=72.7 paces approx. 73paces To compute for the Pace Factor: PF=TP / AP =55m/73paces PF=0.75m/pace VII. OBSERVATION Every individual varies in paces. The variation depends on the size of the individual or the way he/she walks. A certain individual can cover a certain distance in a few numbers of paces while some can do oppositely. VIII. CONCLUSION Several factors can affect the pace of a single individual, it could be either due to the size of the individual itself or maybe because of the way he/she walks. These factors do definitely affect the value of the pace factor. The lesser the pace the greater the pace factor, the greater the pace the lesser the pace factor. Please join StudyMode to read the full document
# Precalculus : Determine the domain of a trigonometric function ## Example Questions ### Example Question #1 : Determine The Domain Of A Trigonometric Function Which of the following is the correct domain of , where represents an integer? Explanation: The cotangent graph only has a period of intervals and is most similar to the tangent graph. The domain of cotangent exists everywhere except every value since an asymptote exists at those values in the domain. The y-intercept of 3 shifts the cotangent graph up by three units, so this does not affect the domain. Therefore, the graph exists everywhere except , where is an integer. ### Example Question #2 : Trigonometric Functions What is the domain of the following function? Explanation: All x values make the function work. Thus, making the domain . They're parentheses instead of brackets because parentheses are used when you can't actually use the specific value next to it. It is impossible to use infinity which makes parentheses appropriate. Brackets are used when you CAN use the specific value next to it. ### Example Question #2 : Determine The Domain Of A Trigonometric Function What is the domain of the following function: Explanation: All x values work for the function. Thus, making the domain all real numbers. Parentheses are required because you can never actually use the number infinity. ### Example Question #268 : Pre Calculus What is the domain of . Explanation: If you look at a graph of the function, you can see that every curve has a vertical asymptote that repeats every radians in the positive and negative x-direction, starting at radians. Also, the curve has a length that stretches radians which makes the domain . ### Example Question #3 : Determine The Domain Of A Trigonometric Function What is the restriction of the domain of the function given by: For all the answer choices below, is any integer. Explanation: has restrictions on its domain such that , where is any integer. To determine the domain for we equate the terms within the secant function and set them equal to the original domain restriction. Solving for The new domain restriction is: where is an integer
# Hawkes Learning Systems: College Algebra Save this PDF as: Size: px Start display at page: ## Transcription 1 Hawkes Learning Systems: College Algebra Section 1.2: The Arithmetic of Algebraic Expressions 2 Objectives o Components and terminology of algebraic expressions. o The field properties and their use in algebra. o The order of mathematical operations. o Basic set operations and Venn diagrams. 3 Components and Terminology of Algebraic Expressions o Algebraic Expressions are made of constants and variables combined by mathematical operations. o Constants are fixed numbers. o Variables are unspecified numbers. o Terms are the parts of an algebraic expression joined by addition (or subtraction). o Factors of a term are the parts of a term that are joined by multiplication (or division). o The coefficient of a term is the constant factor of the term while the remaining part of the term constitutes the variable factor. 4 Example: Components and Terminology of an Algebraic Expression Consider the algebraic expression What are its terms? What is the coefficient of each of the terms? 3, -2, 7, -12 What is the variable factor of each of the terms? (x y), none 5 Evaluating Algebraic Expressions o To evaluate an expression means to replace the variables with constants, perform the mathematical operations and simplify. 6 Example: Evaluate an Algebraic Expression Evaluate the following algebraic expression: 2 2x 5( x y) 2 2x 5( x y) for x 1and y 4 Replace x with 1 and y with 4. Simplify. 27 7 The Field Properties and Their Use in Algebra The set of real numbers forms what is known mathematically as a field and the properties below are called field properties. a, b and c represent arbitrary real numbers. Name of Property Closure Commutative Associative Identity Inverse Distributive a Additive Version b is real number Multiplicative Version is a real number a b b a ab ba a ( b c) ( a b) c a 0 0 a a a ( a) 0 ab a( b c) ab ac a( bc) ( ab) c a 1 1 a a 1 a 1(for a 0) a 8 Example: Field Properties Identify the property that justifies the statement. a) 4(y 3) = 4y 12 b) -3(4x 6 z) = (-3 4)(x 6 z) = -12x 6 z c) 9 The Field Properties and Their Use in Algebra Cancellation Properties Throughout this table, A, B and C represent algebraic expressions. The symbol can be read as if and only if or is equivalent to. Property A B A C B C For C 0, A B A C B C Description Additive Cancellation Adding the same quantity to both sides of an equation results in an equivalent equation Multiplicative Cancellation Multiplying both sides of an equation by the same non-zero quantity results in an equivalent equation 10 The Field Properties and Their Use in Algebra Zero-Factor Property Let A and B represent algebraic expressions. If the product of A and B is 0, then at least one of A and B is itself 0. Using the symbol AB 0 A 0 or B 0 for implies, we write 11 Example: Properties Identify the property that justifies the statement. (If it s a cancellation property, identify the quantity added or multiplied on both sides.) a) -14y = 7 y = -½ b) (x 3)(x + 2) = 0 x 3 = 0 or x + 2 = 0 c) 12 The Order of Mathematical Operations Order of Operations 1. If the expression is a fraction, simplify the numerator and denominator individually, according to the guidelines in the following steps. 2. Parentheses, braces and brackets are all used as grouping symbols. Simplify expressions within each set of grouping symbols, if any are present, working from the innermost outward. 3. Simplify all powers (exponents) and roots. 4. Perform all multiplications and divisions in the expression in the order they occur, working from left to right. 5. Perform all additions and subtractions in the expression in the order they occur, working from left to right. 13 Example: Order of Operations Evaluate following Order of Operations. a) b) 14 Basic Set Operations and Venn Diagrams o Set operations, union and intersection, combine two or more sets. o A Venn diagram is a pictorial representation of a set or sets and its aim is to indicate, through shading, the outcome of set operations such as union and intersection. 15 Basic Set Operations and Venn Diagrams A and B denote two sets and are represented in the Venn diagram by circles. The operation of union is demonstrated by shading. The symbol is read is an element of. The union of A and B, denoted A B, is the set x x A or x B. That is, an element x is in A B, if it is in the set A, A B the set B, or both. Note: the union of A and B contains both individual sets. 16 Basic Set Operations and Venn Diagrams A and B denote two sets and are represented in the Venn diagram by circles. The operation of intersection is demonstrated by shading. The symbol is read is an element of. The intersection of A and B, denoted A B, is the set x x A and x B. That is, an element x is in A B, A B if it is in both A and B. Note: the intersection of A and B is contained in each individual set. 17 Example: Set Operations Simplify each of the following set expressions. a. b. c. d. 3,7 1,11 3,11 3,7 1,11 1,7 2,5 5,7,2 17,, Since these two intervals overlap, their union is best described with a single interval. This intersection of two intervals can also be described with a single interval. These two intervals have no elements in common so their intersection is the empty set. The union of these two intervals constitutes the entire set of real numbers. 18 Example: Set Operations Simplify each of the following set expressions. a) b) ### COLLEGE ALGEBRA 10 TH EDITION LIAL HORNSBY SCHNEIDER 1.1-1 10 TH EDITION COLLEGE ALGEBRA LIAL HORNSBY SCHNEIDER 1.1-1 1.1 Linear Equations Basic Terminology of Equations Solving Linear Equations Identities 1.1-2 Equations An equation is a statement that two expressions ### Algebraic expressions are a combination of numbers and variables. Here are examples of some basic algebraic expressions. Page 1 of 13 Review of Linear Expressions and Equations Skills involving linear equations can be divided into the following groups: Simplifying algebraic expressions. Linear expressions. Solving linear ### Properties of Real Numbers 16 Chapter P Prerequisites P.2 Properties of Real Numbers What you should learn: Identify and use the basic properties of real numbers Develop and use additional properties of real numbers Why you should ### Chapter 2: Linear Equations and Inequalities Lecture notes Math 1010 Section 2.1: Linear Equations Definition of equation An equation is a statement that equates two algebraic expressions. Solving an equation involving a variable means finding all values of the variable ### Vocabulary Words and Definitions for Algebra Name: Period: Vocabulary Words and s for Algebra Absolute Value Additive Inverse Algebraic Expression Ascending Order Associative Property Axis of Symmetry Base Binomial Coefficient Combine Like Terms ### Linear Equations in One Variable Linear Equations in One Variable MATH 101 College Algebra J. Robert Buchanan Department of Mathematics Summer 2012 Objectives In this section we will learn how to: Recognize and combine like terms. Solve ### Click on the links below to jump directly to the relevant section Click on the links below to jump directly to the relevant section What is algebra? Operations with algebraic terms Mathematical properties of real numbers Order of operations What is Algebra? Algebra is ### Operations with positive and negative numbers - see first chapter below. Rules related to working with fractions - see second chapter below INTRODUCTION If you are uncomfortable with the math required to solve the word problems in this class, we strongly encourage you to take a day to look through the following links and notes. Some of them ### Solutions of Linear Equations in One Variable 2. Solutions of Linear Equations in One Variable 2. OBJECTIVES. Identify a linear equation 2. Combine like terms to solve an equation We begin this chapter by considering one of the most important tools ### Florida Department of Education/Office of Assessment January 2012. Algebra 1 End-of-Course Assessment Achievement Level Descriptions Florida Department of Education/Office of Assessment January 2012 Algebra 1 End-of-Course Assessment Achievement Level Descriptions Algebra 1 EOC Assessment Reporting Category Functions, Linear Equations, ### Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...} ### 2. Simplify. College Algebra Student Self-Assessment of Mathematics (SSAM) Answer Key. Use the distributive property to remove the parentheses College Algebra Student Self-Assessment of Mathematics (SSAM) Answer Key 1. Multiply 2 3 5 1 Use the distributive property to remove the parentheses 2 3 5 1 2 25 21 3 35 31 2 10 2 3 15 3 2 13 2 15 3 2 ### ModuMath Algebra Lessons ModuMath Algebra Lessons Program Title 1 Getting Acquainted With Algebra 2 Order of Operations 3 Adding & Subtracting Algebraic Expressions 4 Multiplying Polynomials 5 Laws of Algebra 6 Solving Equations ### 3.1. RATIONAL EXPRESSIONS 3.1. RATIONAL EXPRESSIONS RATIONAL NUMBERS In previous courses you have learned how to operate (do addition, subtraction, multiplication, and division) on rational numbers (fractions). Rational numbers ### 7. Solving Linear Inequalities and Compound Inequalities 7. Solving Linear Inequalities and Compound Inequalities Steps for solving linear inequalities are very similar to the steps for solving linear equations. The big differences are multiplying and dividing ### Mathematical Procedures CHAPTER 6 Mathematical Procedures 168 CHAPTER 6 Mathematical Procedures The multidisciplinary approach to medicine has incorporated a wide variety of mathematical procedures from the fields of physics, ### A Second Course in Mathematics Concepts for Elementary Teachers: Theory, Problems, and Solutions A Second Course in Mathematics Concepts for Elementary Teachers: Theory, Problems, and Solutions Marcel B. Finan Arkansas Tech University c All Rights Reserved First Draft February 8, 2006 1 Contents 25 ### Fractions. Cavendish Community Primary School Fractions Children in the Foundation Stage should be introduced to the concept of halves and quarters through play and practical activities in preparation for calculation at Key Stage One. Y Understand ### Basic Math Refresher A tutorial and assessment of basic math skills for students in PUBP704. 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For example, express the calculation Subtract y from 5 as 5 y. ### Accentuate the Negative: Homework Examples from ACE Accentuate the Negative: Homework Examples from ACE Investigation 1: Extending the Number System, ACE #6, 7, 12-15, 47, 49-52 Investigation 2: Adding and Subtracting Rational Numbers, ACE 18-22, 38(a), ### Order of Operations. 2 1 r + 1 s. average speed = where r is the average speed from A to B and s is the average speed from B to A. Order of Operations Section 1: Introduction You know from previous courses that if two quantities are added, it does not make a difference which quantity is added to which. For example, 5 + 6 = 6 + 5. ### 1.4 Variable Expressions 1.4 Variable Expressions Now that we can properly deal with all of our numbers and numbering systems, we need to turn our attention to actual algebra. Algebra consists of dealing with unknown values. These ### Maths Workshop for Parents 2. 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An Introduction: As is usually the case when learning a new concept in mathematics, the new concept is the reverse of the previous one. Remember how you first learned ### Arithmetic Operations. The real numbers have the following properties: In particular, putting a 1 in the Distributive Law, we get Review of Algebra REVIEW OF ALGEBRA Review of Algebra Here we review the basic rules and procedures of algebra that you need to know in order to be successful in calculus. Arithmetic Operations The real ### 1.3 Algebraic Expressions 1.3 Algebraic Expressions A polynomial is an expression of the form: a n x n + a n 1 x n 1 +... + a 2 x 2 + a 1 x + a 0 The numbers a 1, a 2,..., a n are called coefficients. Each of the separate parts, ### Negative Integer Exponents 7.7 Negative Integer Exponents 7.7 OBJECTIVES. Define the zero exponent 2. Use the definition of a negative exponent to simplify an expression 3. Use the properties of exponents to simplify expressions ### Students will be able to simplify and evaluate numerical and variable expressions using appropriate properties and order of operations. Outcome 1: (Introduction to Algebra) Skills/Content 1. Simplify numerical expressions: a). Use order of operations b). Use exponents Students will be able to simplify and evaluate numerical and variable ### Property: Rule: Example: Math 1 Unit 2, Lesson 4: Properties of Exponents Property: Rule: Example: Zero as an Exponent: a 0 = 1, this says that anything raised to the zero power is 1. 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To the applicant: KEY WORDS AND CONVERTING WORDS TO EQUATIONS To the applicant: The following information will help you review math that is included in the Paraprofessional written examination for the Conejo Valley Unified School District. The Education Code requires ### New York State Mathematics Content Strands, Grade 6, Correlated to Glencoe MathScape, Course 1 and Quick Review Math Handbook Book 1 New York State Mathematics Content Strands, Grade 6, Correlated to Glencoe MathScape, Course 1 and The lessons that address each Performance Indicator are listed, and those in which the Performance Indicator ### -126.7 87. 88. 89. 90. 13.2. Exponents, Roots, and Order of Operations. OBJECTIVE 1 Use exponents. In algebra, w e use exponents as a w ay of writing SECTION 1. Exponents, Roots, and Order of Operations 2-27.72-126.7-100 -50 87. 88. 89. 90. 1.2 6.2-0.01-0.05 Solve each problem. 91. The highest temperature ever recorded in Juneau, Alaska, was 90 F. The ### MATH 095, College Prep Mathematics: Unit Coverage Pre-algebra topics (arithmetic skills) offered through BSE (Basic Skills Education) MATH 095, College Prep Mathematics: Unit Coverage Pre-algebra topics (arithmetic skills) offered through BSE (Basic Skills Education) Accurately add, subtract, multiply, and divide whole numbers, integers, ### Order of Operations More Essential Practice Order of Operations More Essential Practice We will be simplifying expressions using the order of operations in this section. Automatic Skill: Order of operations needs to become an automatic skill. Failure ### The Properties of Signed Numbers Section 1.2 The Commutative Properties If a and b are any numbers, 1 Summary DEFINITION/PROCEDURE EXAMPLE REFERENCE From Arithmetic to Algebra Section 1.1 Addition x y means the sum of x and y or x plus y. Some other words The sum of x and 5 is x 5. indicating addition ### Multiplying and Dividing Fractions Multiplying and Dividing Fractions 1 Overview Fractions and Mixed Numbers Factors and Prime Factorization Simplest Form of a Fraction Multiplying Fractions and Mixed Numbers Dividing Fractions and Mixed ### MATH 90 CHAPTER 1 Name:. MATH 90 CHAPTER 1 Name:. 1.1 Introduction to Algebra Need To Know What are Algebraic Expressions? Translating Expressions Equations What is Algebra? They say the only thing that stays the same is change. 8. Radicals - Rationalize Denominators Objective: Rationalize the denominators of radical expressions. It is considered bad practice to have a radical in the denominator of a fraction. When this happens ### Set Theory: Shading Venn Diagrams Set Theory: Shading Venn Diagrams Venn diagrams are representations of sets that use pictures. We will work with Venn diagrams involving two sets (two-circle diagrams) and three sets (three-circle diagrams). ### Mathematics Review for MS Finance Students Mathematics Review for MS Finance Students Anthony M. Marino Department of Finance and Business Economics Marshall School of Business Lecture 1: Introductory Material Sets The Real Number System Functions, ### Algebra 2 Year-at-a-Glance Leander ISD 2007-08. 1st Six Weeks 2nd Six Weeks 3rd Six Weeks 4th Six Weeks 5th Six Weeks 6th Six Weeks Algebra 2 Year-at-a-Glance Leander ISD 2007-08 1st Six Weeks 2nd Six Weeks 3rd Six Weeks 4th Six Weeks 5th Six Weeks 6th Six Weeks Essential Unit of Study 6 weeks 3 weeks 3 weeks 6 weeks 3 weeks 3 weeks ### SOLVING EQUATIONS BY COMPLETING THE SQUARE SOLVING EQUATIONS BY COMPLETING THE SQUARE s There are several ways to solve an equation by completing the square. Two methods begin by dividing the equation by a number that will change the "lead coefficient" ### South Carolina College- and Career-Ready (SCCCR) Pre-Calculus South Carolina College- and Career-Ready (SCCCR) Pre-Calculus Key Concepts Arithmetic with Polynomials and Rational Expressions PC.AAPR.2 PC.AAPR.3 PC.AAPR.4 PC.AAPR.5 PC.AAPR.6 PC.AAPR.7 Standards Know ### 2 is the BASE 5 is the EXPONENT. Power Repeated Standard Multiplication. To evaluate a power means to find the answer in standard form. Grade 9 Mathematics Unit : Powers and Exponent Rules Sec.1 What is a Power 5 is the BASE 5 is the EXPONENT The entire 5 is called a POWER. 5 = written as repeated multiplication. 5 = 3 written in standard ### Integer Operations. Overview. Grade 7 Mathematics, Quarter 1, Unit 1.1. Number of Instructional Days: 15 (1 day = 45 minutes) Essential Questions Grade 7 Mathematics, Quarter 1, Unit 1.1 Integer Operations Overview Number of Instructional Days: 15 (1 day = 45 minutes) Content to Be Learned Describe situations in which opposites combine to make zero. ### MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS. + + x 2. x n. a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 a 31 a 32 a 3n b 3. a m1 a m2 a mn b m MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 1. SYSTEMS OF EQUATIONS AND MATRICES 1.1. Representation of a linear system. The general system of m equations in n unknowns can be written a 11 x 1 + a 12 x 2 + ### Licensed to: Printed in the United States of America 1 2 3 4 5 6 7 15 14 13 12 11 Licensed to: CengageBrain User This is an electronic version of the print textbook. Due to electronic rights restrictions, some third party content may be suppressed. Editorial review has deemed that any ### MATH 60 NOTEBOOK CERTIFICATIONS MATH 60 NOTEBOOK CERTIFICATIONS Chapter #1: Integers and Real Numbers 1.1a 1.1b 1.2 1.3 1.4 1.8 Chapter #2: Algebraic Expressions, Linear Equations, and Applications 2.1a 2.1b 2.1c 2.2 2.3a 2.3b 2.4 2.5 ### 3.1. Solving linear equations. Introduction. Prerequisites. Learning Outcomes. Learning Style Solving linear equations 3.1 Introduction Many problems in engineering reduce to the solution of an equation or a set of equations. An equation is a type of mathematical expression which contains one or ### 1.3 Polynomials and Factoring 1.3 Polynomials and Factoring Polynomials Constant: a number, such as 5 or 27 Variable: a letter or symbol that represents a value. Term: a constant, variable, or the product or a constant and variable. ### South Carolina College- and Career-Ready (SCCCR) Algebra 1 South Carolina College- and Career-Ready (SCCCR) Algebra 1 South Carolina College- and Career-Ready Mathematical Process Standards The South Carolina College- and Career-Ready (SCCCR) Mathematical Process ### This is a square root. The number under the radical is 9. (An asterisk * means multiply.) Page of Review of Radical Expressions and Equations Skills involving radicals can be divided into the following groups: Evaluate square roots or higher order roots. Simplify radical expressions. Rationalize ### STRAND: ALGEBRA Unit 3 Solving Equations CMM Subject Support Strand: ALGEBRA Unit Solving Equations: Tet STRAND: ALGEBRA Unit Solving Equations TEXT Contents Section. Algebraic Fractions. Algebraic Fractions and Quadratic Equations. Algebraic ### WORK SCHEDULE: MATHEMATICS 2007 , K WORK SCHEDULE: MATHEMATICS 00 GRADE MODULE TERM... LO NUMBERS, OPERATIONS AND RELATIONSHIPS able to recognise, represent numbers and their relationships, and to count, estimate, calculate and check ### 1. The algebra of exponents 1.1. Natural Number Powers. It is easy to say what is meant by a n a (raised to) to the (power) n if n N. CHAPTER 3: EXPONENTS AND POWER FUNCTIONS 1. The algebra of exponents 1.1. Natural Number Powers. It is easy to say what is meant by a n a (raised to) to the (power) n if n N. For example: In general, if ### No Solution Equations Let s look at the following equation: 2 +3=2 +7 5.4 Solving Equations with Infinite or No Solutions So far we have looked at equations where there is exactly one solution. It is possible to have more than solution in other types of equations that are
# RBSE Maths Class 11 Chapter 5: Important Questions and Solutions RBSE Maths Chapter 5 – Complex Numbers Class 11 Important questions and solutions is available here. The important questions and solutions of Chapter 5, available at BYJU’S, contain step by step explanations for the better understanding of the students. All these important questions are based on the new pattern prescribed by the RBSE. Students can also get the syllabus and textbooks on RBSE Class 11 solutions. Chapter 5 of the RBSE Class 11 Maths will help students to solve problems related to the set of complex numbers, theorems on complex numbers, operations on complex numbers and its properties, the conjugate and properties of a complex number, modulus and properties of a complex number, geometrical representation of complex numbers and their operations, cube roots of unity, quadratic equations. ### RBSE Maths Chapter 5: Exercise 5.1 Textbook Important Questions and Solutions Question 1: Write the following in the simplest form. [i] i52 [ii] √-2 * √-3 [iii] (1 + i)5 * (1 – i)5 Solution: [i] i52 abc = (ab)c i52 = (i2)26 i2 = -1 = (-1)26 = 126 = 1 [ii] √-2 * √-3 √-2 * √-3 = √2 * √-3i ab * ac = ab+c i * i = i1+1 = √2 i1+1 √3 = √2i2 √3 = i2 √2 * 3 = √6i2 = -√6 [iii] (1 + i)5 * (1 – i)5 (1 + i)5 = -4 – 4i (1 – i)5 = -4 + 4i (-4 – 4i) * (-4 + 4i) (a + b) (c + d) = ac + ad + bc + bd a = -4, b = -4i, c = -4, d = 4i = (-4) (-4) + (-4) (4i) + (-4i) (-4) + (-4i) (4i) = 4 * 4 – 4 * 4i + 4 * 4i – 4 * 4i2 = 16 – (-16) = 16 + 16 = 32 Question 2: Find the additive and multiplicative inverse of the following complex numbers. [i] 1 + 2i [ii] [1] / [3 + 4i] [iii] [3 + i]2 Solution: [i] 1 + 2i Let z = 1 + 2i Additive inverse = -z = – (1 + 2i) = -1 – 2i Multiplicative inverse = [1 / z] [1 / 1 + 2i] = [1 / 1 + 2i] * [(1 – 2i) / (1 – 2i)] = [(1 – 2i) / (12 – (2i)2)] = [(1 – 2i) / (1 – 4i2)] = (1 – 2i) / (1 + 4) = (1 – 2i) / 5 = (1 / 5) – i (2 / 5) [ii] [1] / [3 + 4i] Let z = [1] / [3 + 4i] = -1 / [3 + 4i] = [-1 / (3 + 4i)] * [(3 – 4i) / (3 – 4i)] = [-(3 – 4i) / (32 – (4i)2)] = [-(3 – 4i) / (9 – 16i2)] = -(3 – 4i) / (9 + 16) = -(3 – 4i) / 25 = -(3 / 25) + i (4 / 25) Multiplicative inverse = [1 / z] = (1) / [1] / [3 + 4i] = 3 + 4i [iii] [3 + i]2 Let z = [3 + i]2 = -[3 + i]2 [3 + i]2 = 32 + i2 + 6i = 9 + i2 + 6i = 9 – 1 + 6i = 8 + 6i Additive inverse = -(8 + 6i) = -8 – 6i Multiplicative inverse = [1 / z] = [1 / (3 + i)2)] = [1 / (8 + 6i)] * [(8 – 6i) / (8 – 6i)] = [(8 – 6i) / (64 – 36i2)] = [(8 – 6i) / (64 + 36)] = [(8 – 6i) / 100] = (8 / 100) – i (6 / 100) = (2 / 25) – i (3 / 50) Question 3: Find the conjugate of the complex number [(2 + i)3 / (3 + i)]. Solution: Let z = (2 + i)3 / [3 + i] Rewrite: (2 + i)3 / [3 + i] = (1 / 10) [(2 + i)3] / [(3 − i)] i2 = −1 (2 + i)3 = 2 + 11i [1 / 10] [(3 − i)] (2 + i)3 = [1 / 10] (3 − i) (2 + 11i) a (b + c) = ab + ac [1 / 10] (2 + 11i) = (1 / 10) (2) + (1 / 10) (11i) = [1 / 5] + [11i / 10] (3 − i) (1 / 10 (2 + 11i)) = (3 − i) (15 + 11i / 10) ((1 / 5 + 11i/ 10) (3 − i)) = (17 / 10 + 31i / 10) Hence the conjugate of z = (17 / 10) – (31i / 10) Question 4: Find the modulus of the following. [i] 4 + i [ii] -2 – 3i [iii] 1 / [3 – 2i] Solution: [i] 4 + i The modulus of a + ib is √a2 + b2 = √42 + 12 = √16 + 1 \|z\| = √17 [ii] -2 – 3i The modulus of a + ib is √a2 + b2 = √(-2)2 + (-3)2 = √4 + 9 \|z\| = √13 [iii] 1 / [3 – 2i] The modulus of a + ib is √a2 + b2 Let z = [1 / (3 – 2i)] = [1 / (3 – 2i)] * [(3 + 2i) / (3 + 2i)] = [3 + 2i] / [9 – 4i2] = [(3 + 2i) / (9 + 4)] = [(3 + 2i) / 13] = (3 / 13) + i (2 / 13) Modulus of z = √a2 + b2 = √(3 / 13)2 + (2 / 13)2 = √(9 / 169) + (4 / 169) = √13 / 169 \|z\| = 1 / √13 Question 5: If a2 + b2 = 1, then find the value of [1 + b + ia] / [1 + b – ia]. Solution: [1 + b + ia] / [1 + b – ia] = [1 + b + ia] / [1 + b – ia] * [1 + b + ia] / [1 + b + ia] = [(1 – b)2 – a2 + 2ia (1 + b)] / [1 + b2 + 2b + a2] = [(1 – a2) + 2b + b2 + 2ia (1 + b)] / [2(1 + b)] = [2b2 + 2b + 2ia (1 + b)] / [2 (1 + b)] = b + ia Question 6: If a = cos θ + i sin θ, then find the value of [1 + a] / [1 – a]. Solution: A = cosθ + isinθ [1 + a] / [1 – a] = (1 + cosθ + isinθ) / (1 – cosθ – isinθ) = (1 + cosθ + isinθ) (1 – cosθ + isinθ) / (1 – cosθ + isinθ) (1 – cosθ – isinθ) = (1 + cosθ + isinθ – cosθ – cos²θ – isinθ cosθ + isinθ + isinθ cosθ + i² sin²θ) / {(1 – cosθ)² – (isinθ)²} = (1 + 2 isinθ – cos²θ – sin²θ) / {1 – 2 cosθ + cos²θ – (i² sin²θ)} = {1 + 2 isinθ – (sin²θ + cos²θ)} / (1 – 2 cosθ + cos²θ + sin²θ) = (1 + 2isinθ – 1) / (1 – 2cosθ + 1) = 2 isinθ / (2 – 2 cosθ) = isinθ / (1 – cosθ) = i (2 sinθ / 2 cos θ/2) / (2 sin² θ/2) = (i cos θ/2) / (sin θ/2) = i cot θ/2 Question 7: Find the value of x and y which satisfy the equation {[(1 + i) x – 2i] / [3 + i]} + {[(2 – 3i) y + i] / [3 – i]} = i. Solution: {[(1 + i) x – 2i] / [3 + i]} + {[(2 – 3i) y + i] / [3 – i]} {[(1 + i) x – 2i] (3 – i) + [(2 – 3i) y + i]} / 32 – i2 = i [x + ix – 2i] (3 – i) + [2y – 3iy + i] (3 + i) = 10i 3x + 3ix – 6i – ix – i2x + 2i2 + 6y – 9iy + 3i + 2iy – 3i2y + i2 = 10i 3x + 3ix – 6i – ix + x – 2 + 6y – 9iy + 3i + 2iy + 3y – 1 = 10i 3x + 3ix – 6i – ix + x – 2 + 6y – 9iy + 3i + 2iy + 3y – 1 – 10i = 0 (4x + 9y – 3) + i (3x – 6 – x – 9y + 3 + 2y – 10) = 0 (4x + 9y – 3) + i (2x – 7y – 13) = 0 (4x + 9y – 3) —- (1) (2x – 7y – 13) —- (2) Equate the real and imaginary parts, Multiply the second equation by 2 and subtract it from the first equation. 4x + 9y – 4x + 14y = 3 – 26 23y = -23 y = -1 Substitute y = -1 in (1), we get x = 3 Hence x = 3 and y = -1. Question 8: If z1 and z2 are any two complex numbers, then prove that \|z1 + z2\|2 + \|z1 – z2\|2 = 2 \|z1\|2 + 2 \|z2\|2. Solution: LHS = \|z1 + z2\|2 + \|z1 – z2\|2 = \|z1\|2 + \|z2\|2 + 2 Re (z1 z2) + \|z1\|2 + \|z2\|2 – 2 Re (z1 z2) = 2 \|z1\|2 + 2 \|z2\|2 + 2 Re (z1 z2) – 2 Re (z1 z2) = 2 \|z1\|2 + 2 \|z2\|2 = RHS \|z1 + z2\|2 + \|z1 – z2\|2 = 2 \|z1\|2 + 2 \|z2\|2 Question 9: If a + ib = [c + i] / [c – i], where c is a real number, then prove that a2 + b2 = 1 and [b / a] = [2c] / [c2 – 1]. Solution: a + ib = [c + i] / [c – i] = (c + i) (c + i) / (c – i) (c + i) = (c² + 2ci + i²) / (c² – i²) = (c² + 2ci – 1) / {c² – (-1)} = {(c² – 1) + 2ci} / (c² + 1) a + ib = (c² – 1) / (c² + 1) + i (2c) / (c² + 1) Equating the real and imaginary part from both sides we get, a = c² – 1 / c ² + 1 and b = 2c / c² + 1 a² + b² = {(c² – 1) / (c² + 1)}² + {2c / (c² + 1)}² = {(c² – 1)² + 4c²} / (c² + 1)² = (c⁴ – 2c² + 1 + 4c²) / (c² + 1)² = (c⁴ + 2c² + 1) / (c² + 1)² = (c² + 1)² / (c² + 1)² = 1 Now, b / a = {2c / (c² + 1)} / {(c² – 1) / (c² + 1)} = 2c / (c² + 1) * (c² + 1) / (c² – 1) = 2c / (c² – 1) Question 10: If (x + iy) = a + ib, then prove that [x / a] + [y / b] = 4 (a2 – b2). Solution: (x + iy) = a + ib → x + iy = (a + ib)³ → x + iy = a³ + 3a²bi – 3ab² – ib³ → (x + iy) = (a³ – 3ab²) + (3a² – b³) i By the equality of complex numbers: x = a (a² – 3b²); y = b (3a² – b²) [x / a] = a2 − 3b2 ….. (1) [y / b] = 3a2 − b2 ….. (2) Adding (1) and (2) we get, [x / a] + [y / b] = 4 (a2 – b2) Question 11: If [x + i]2 / [3x + 2] = a + ib, then prove that [x2 + 1]2 / [3x + 2]2 = a2 + b2. Solution: ### RBSE Maths Chapter 5: Exercise 5.2 Textbook Important Questions and Solutions Question 1: Find the arguments of the following numbers. [i] [1 + i] / [1 – i] [ii] -1 + √3i [iii] [5 + i √3] / [4 – i 2√3] Solution: [i] z = [1 + i] / [1 – i] * [1 + i] / [1 + i] = [1 + i2 + 2i] / [1 + 1] = i = 0 + i In polar form, r (cosθ + i sinθ) = 0 + i ⇒ r cosθ = 0 and r sinθ = 1 ⇒ r2 (cos2θ + sin2θ) = 02 + 12 ⇒ r2 = 1 ⇒ r = 1 Now, cosθ = 1 ⇒ θ = π / 2 The argument is π / 2 and modulus is 1. [ii] -1 + √3i a = -1, b = √3 z = -1 + √3i = (-1, √3) lies in the second quadrant. Argument θ = π – tan-1 \|(b / a)\| = π – tan-1 \|(√3 / -1)\| = π – tan-1 (√3) = π – [π / 3] = 2π / 3 [iii] [5 + i √3] / [4 – i 2√3] 1 / [a + ib] = {1 / [(a – ib) (a + ib)]} * (a – ib) = [a – ib] / a2 + b2 On splitting, [a – ib] / a2 + b2 = [a / a2 + b2] – [ib / a2 + b2] Here a = 4, b = -2√3 [5 + i √3] * [1] / [4 – i 2√3] = [5 + i √3] ([1 / 7] + [i √3 / 2] ([1 / 7] + [i √3 / 2] * [5 + i √3] = ([1 / 2] + [i √3 / 2] For a complex number a + ib, the polar form is given by r [cos θ + i sin θ] where, r = √a2 + b2 and θ = a tan [b / a] a = [1 / 2] and b = √3 / 2 = [1 / 2] tan ([√3 / 2] / [1 / 2]) = π / 3 Question 2: Express the following complex numbers in polar form. [i] [1 + i] / √2 [ii] sin [π / 3] + i cos [π / 3] Solution: [i] [1 + i] / √2 For a complex number a + ib, the polar form is given by r [cos θ + i sin θ] where, r = √a2 + b2 and a = [√2 / 2] and b = √2 / 2 r = √[√2 / 2]2 + [√2 / 2]2 = 1 θ = a tan [b / a] = [√2 / 2] tan ([√2 / 2] / [√2 / 2]) = π / 4 [√2 / 2] + i [√2 / 2] = cos[π / 4] + i sin [π / 4] [ii] sin [π / 3] + i cos [π / 3] For a complex number a + ib, the polar form is given by r [cos θ + i sin θ] where, r = √a2 + b2 and a = [√3 / 2] and b = 1 / 2 r = √[√3 / 2]2 + [1 / 2]2 = 1 θ = a tan [b / a] = [√3 / 2] tan ([1 / 2] / [√3 / 2]) = π / 6 [√3 / 2] + i [1 / 2] = cos[π / 6] + i sin [π / 6] Question 3: If z1 and z2 are two non-zero complex numbers, then prove that arg z1 = arg z1 – arg z2. Solution: ### RBSE Maths Chapter 5: Exercise 5.3 Textbook Important Questions and Solutions Question 1: Find the square root of the following complex numbers. [i] -5 + 12i [ii] 8 – 6i [iii] -i Solution: [i] -5 + 12i If a + ib = -5 + 12i Here a = -5, b = 12 [ii] 8 – 6i If a + ib = 8 + 6i Here a = 8, b = 6 [iii] -i If a – ib = 0 – i Here a = 0, b = 1 Question 2: Find the value of √[4 + 3 √-20] + √[4 – 3 √-20]. Solution: Question 3: Find the cube root of the following. [i] -216 [ii] -512 Solution: (i) -216 = (-6) (1)1/3 = -6, -6ω, -6ω2 Hence, cube root of -216 = -6, -6ω, -6ω2 (ii) -512 = (-8) (1)1/3 = -8, -8ω, -8ω2 Hence, cube root of -512 = -8, -8ω, – 8ω2 Question 4: Prove that: (i) 1 + ωn + ω2n = 0, whereas n = 2, 4. (ii) 1 + ωn + ω2n = 3, whereas n is multiple of 3. Solution: [i] 1 + ωn + ω2n = 0, whereas n = 2, 4. When n = 2, 1 + ωn + ω2n = 1 + ω2 + ω22 = 1 + ω2 + ω4 = 1 + ω2 + ω ω3 = 1 + ω2 + ω = 1 + ω + ω2 = 0 When n = 4, 1 + ωn + ω2n = 1 + ω4 + ω24 = 1 + ω ω3 + ω3 * ω3 * ω2 = 1 + ω * 1 + 1 * 1 * ω2 = 1 + ω + ω2 = 0 [ii] 1 + ωn + ω2n = 3, whereas n is multiple of 3. n = 3k 1 + ωn + ω2n = 1 + ω3k + ω2 * 3k = 1 + (ω3)k + (ω3)2k = 1 + (1)k + (1)2k = 1 + 1 + 1 = 3 Question 5: Prove that: [i] [(-1 + √-3) / 2]29 + [(-1 – √-3) / 2]29 = -1 [ii] (1 + 5ω2 + ω) (1 + 5ω + ω2 ) (5 + ω + ω2 ) = 64 Solution: [i] [(-1 + √-3) / 2]29 + [(-1 – √-3) / 2]29 = -1 ω = [(-1 + i√3) / 2] and ω2 = [(-1 – i√3) / 2] LHS = [(-1 + √-3) / 2]29 + [(-1 – √-3) / 2]29 = [(-1 + i√3) / 2]29 + [(-1 – i√3) / 2]29 = (ω)29 + (ω2)29 = (ω3)9 ω2 + ω58 = ω2 + (1)19 ω = ω2 + ω = -1 [ii] (1 + 5ω2 + ω) (1 + 5ω + ω2 ) (5 + ω + ω2) = 64 LHS = (1 + 5ω2 + ω) (1 + 5ω + ω2 ) (5 + ω + ω2) = (1 + ω + ω2 + 4ω2) (1 + ω + ω2 + 4ω) (1 + ω + ω2 + 4) = (0 + 4ω2) (0 + 4ω) (0 + 4) = 4ω2 * 4ω * 4 = 64ω3 = 64 * 1 =64 ### RBSE Maths Chapter 5: Exercise 5.4 Textbook Important Questions and Solutions Question 1: Find the solution of the following equations by vedic method. (i) x2 + 4x + 13 = 0 (ii) 2x2 + 5x + 4 = 0 (iii) ix2 + 4x – 15 / 2 = 0 Solution: (i) x2 + 4x + 13 = 0 Comparing this equation with ax2 + bx + c = 0, we get a = 1, b = 4, c = 13 First derivative = Discriminant D1 = D 2x + 4 = ± √b2 – 4ac 2x + 4 = ± √42 – 4 * (1) * (13) 2x + 4 = ± √16 – 52 2x + 4 = ± √-36 2x = -4 ± 6i x = -2 ± 3i The solutions of x2 + 4x + 13 = 0 are -2 ± 3i. (ii) 2x2 + 5x + 4 = 0 Comparing this equation with ax2 + bx + c = 0, we get a = 2, b = 5, c = 4 First derivative = Discriminant D1 = D 4x + 5 = ± √b2 – 4ac 4x + 5 = ± √52 – 4 * (2) * (4) 4x + 5 = ± √25 – 32 4x = -5 ± √7i x = [-5 ± √7i] / 4 (iii) ix2 + 4x – 15 / 2 = 0 2ix2 + 8x – 15 = 0 Comparing this equation with ax2 + bx + c = 0, we get a = 2i, b = 8, c = (-15) First derivative = Discriminant Question 2: Find the quadratic equations which have the following roots: (i) 5 and -2 (ii) 1 + 2i Solution: (i) Root α = 5 and β = – 2 Then, sum of roots = α + β = 5 – 2 = 3 and Product of roots = αβ = 5 × (-2) = -10 Hence, the required equation whose roots are 5 and -2 is x2 – (sum of roots) x + product of roots = 0 ⇒ x2 – 3x + (-10) = 0 ⇒ x2 – 3x – 10 = 0 (ii) Roots α = 1 + 2i and β = 1 – 2i Then, sum of roots = α + β = 1 + 2i + 1 – 2i = 2 and Product of roots = αβ = (1 + 2i) * (1 – 2i) = 1 – 4i2 = 1 + 4 = 5 Hence, the required equation whose roots are 1 + 2i and 1 – 2i is x2 – (sum of roots) x + product of roots = 0 ⇒ x2 – 2x + 5 = 0. Question 3: If one root of equation x2 – px + q = 0 is twice the other then prove that 2p2 = 9q. Solution: The given equation is x2 – px + q = 0. If its roots are α and β, then according to question α = 2β then Sum of roots = α + β α + β = [-b / a] = – (-p / 1) 2β + β = p 3β = p —— (1) Product of roots = αβ αβ = [c / a] 2β * β = q 2 = q LHS = 2p2 = 2 * (3β)2 = 2 * 9β2 = 18β2 = 9 * 2β2 = 9q = RHS Question 4: Find that condition for which equation ax2 + bx + c = 0 has roots in the ratio m : n. Solution: The given equation is ax2 + bx + c = 0. If its roots are α and β, then according to question, α : β = m : n Question 5: If 2 + (2a + 5ib) = 8 + 10i, then find the values of a and b Solution: 2 + (2a + 5ib) = 8 + 10i ⇒ (2 + 2a) + 5bi = 8 + 10i On comparing, 2 + 2a = 8 ⇒ 2a = 6 ⇒ a = 3 and 5b = 10 ⇒ b = 2 So, a = 3 and b = 2.
Why find the domain? You can use domain to find the range of a function. It can be a square root, natural log, f, or ln(x-8) function. A domain is the set of all possible inputs. When you plot the domain against the x-axis, you will see that it excludes all negative numbers and all real numbers greater than 8. The domain can be a variable or a set of x-values that will generate y-values successfully. It is also a fraction with a radical sign. There are three common ways to find the domain. The smallest term of the interval (x-values) is written first. You can also write the domain with parentheses to exclude the endpoints. In the second case, you can use the natural log function instead of the fractional function. Domains can be useful when you are trying to determine the range of a rational function. Generally, a domain includes all values on an x-axis. It can be a good idea to avoid numbers like -3 and 5. The range of a function is the set of outputs. This can be a positive or negative number. The range of a function is the set of outputs for any given x. It is not uncommon for a function to have a range that is not defined in its domain. For example,you could look here a simple linear function has a domain of 4x + 3. Its range is a circle of three x-values. A graph is a graphical representation of a function. The domain contains all the possible inputs and outputs of the function. For example, if you write an equation y=2x+3, the graph of y=2x+7 will show all values of x. Its range is a red line, and it stops at y=-1. This is the graph’s domain. When you graph this, it can be useful to see its range and the x-axis coordinates. When writing a function, make sure to include the domain. The domain is the set of all x-values. Its range is the set of all y-values. It’s the set of x-values. If each x corresponds to one y-value, then it’s a function. If it doesn’t, then it’s not a function. So, you need to consider whether you need to know the range or the domain. There are many ways to identify the domain of a function. Using graphs, you can find out the range of the function. In a graph, the domain of a function is the set of all possible x-values. If the graph continues past the visible portion, then it has a greater domain. This way, you can see which variables are influencing the output of a function. You may want to check the domain of a function when you’re working with a piecewise function. Why Find the Domain?
What Are Decimals? ## What Are Decimals? The position of each digit in a number is important. Each position has a different place value. As you move left (👈), each place value is 10 times bigger than the one on its right. As you move right (👉), each place value gets 10 times smaller. ### Numbers Smaller Than 1 Are there any place values smaller than the Ones place? 🤔 Let's see. What happens when we divide 1 by 10? We get a fraction. Math people write this fraction as a decimal, 0.1! 0.1 = 1/10 = One tenth The decimal point . shows digits with values less than 1. If we divide 0.1 by 10 again, we get an even smaller decimal: 0.01 = One hundredth 0.01 is the same as the fraction 1/100! #### So, What Are Decimals Exactly? Decimals are numbers with parts smaller than 1. They use a decimal point, like 0.4, or 9.99. Tip: Decimals are just a convenient way to write fractions. For example, 0.4 is equal to 4/10. We can divide any decimal by 10 to get an even smaller decimal: 0.01 ÷ 10 = 0.001 = One thousandth Digits further to the right 👉 of decimal points have smaller and smaller values. 0.0001 = One ten-thousandth Whole numbers are positive numbers without decimal parts, like 1, 2, and 10. ### Tenths Place 72.4 The 4 above is in the Tenths place. The digit in the Tenths place is just right of the decimal point. Its value is a multiple of 1/10. Note that the Tenths place is different from the Tens place. Each part above is 1/10 of a whole, or 0.1. 0.4 is 4 of these parts, or 4/10. We can read decimals out loud like: seventy-two and four tenths or seventy-two point four ### Hundredths Place Let's add another digit to our number. 72.46 6 above is in the Hundredths place. Tip: This is different from the Hundreds place, no 'th'. Imagine one whole divided into 100 parts. 0.06 is equal to 6 of these tiny parts out of 100. We read the decimal 72.46 out loud as: "Seventy-two and forty-six hundredths" We don't say: "Seventy-two and four tenths and six hundredths" ❌ When reading decimals out loud, say all the parts after the decimal point as one number, and then say the place value of the rightmost (👉) digit after that. "0.01 = one hundredth 0. 11 = eleven hundredths 0. 211 = two hundred eleven thousandths" For short, you can also just say: 0.01 = "zero point zero one" 0.11 = "zero point one one" 0.211 = "zero point two one one"
# Integration By Substitution A LevelAQAEdexcelOCR ## Integration by Substitution Integration by substitution is another way to reverse the chain rule. In this one, we replace the integration variable $x$ with a different variable $u=f(x)$. We must also replace $dx$ with $du=f'(x)dx$ and replace the limits of the integral too. The aim is to end up with an integral that is easier to evaluate. A LevelAQAEdexcelOCR ## How to Integrate by Substitution Step 1: You will be presented with an integrand that is made up of two functions of $x$. Step 2: Substitute $u=f(x)$ where $f(x)$ is one of the functions of $x$. Step 3: Find $\dfrac{du}{dx}$ then rearrange to get $dx$ in terms of $du$. Step 4: Rewrite the original integral in terms of $u$ and $du$ and simplify it. Step 5: If you chose your substitution well, you will now be left with something much easier to integrate. Step 6: Integrate it. Step 7: Substitute $u$ for $f(x)$ in the answer to get the final answer in terms of $x$. A LevelAQAEdexcelOCR ## Changing the Limits For a definite integral of the form $\int^{b}_{a}$, if our substitution is $u=f(x)$, then rather than substitute $x$ back in at the end, we can change the limits to $\int^{f(b)}_{f(a)}$ and put those limits into our expression for $u$ to evaluate the integral. Example: Find $\int^{0.5}_{0}5x^{4}e^{x^{5}}dx$, using the substitution $u=x^{5}$. $u=f(x)=x^{5}$ So limits become: $0.5^{5}=0.03125$ and $0^{5}=0$ \begin{aligned}\dfrac{du}{dx}=5x^{4}\\[1.2em]du=5x^{4}dx\\[1.2em]dx=\dfrac{du}{5x^{4}}\end{aligned} Integral becomes: \begin{aligned}\int^{0.03125}_{0}5x^{4}e^{u}\dfrac{du}{5x^{4}}&=\int^{0.03125}_{0}e^{u}du\\[1.2em]&=[e^{u}]^{0.03125}_{0}\\[1.2em]&=e^{0.03125}-e^{0}=0.0317\end{aligned} A LevelAQAEdexcelOCR @mmerevise ## Integration by Substitution on Fractions When choosing a substitution for a fraction, the best thing to choose is almost always the denominator or part of the denominator. Example: Integrate ${\LARGE \int}\dfrac{4x^{3}}{(x^{4}-1)^{\frac{1}{6}}}dx$ with a suitable substitution. Choose $u=x^{4}-1$. \begin{aligned}\dfrac{du}{dx}=4x^{3}\\[1.2em]du=4x^{3}dx\\[1.2em]dx=\dfrac{1}{4x^{3}}du\end{aligned} Putting it in the integral: \begin{aligned}\int\dfrac{4x^{3}}{(x^{4}-1)^{\frac{1}{6}}}dx&=\int\dfrac{4x^{3}}{u^{\frac{1}{6}}}\dfrac{1}{4x^{3}}du\\[1.2em]&=\int\dfrac{1}{u^{\frac{1}{6}}}du\\[1.2em]&=\int u^{-\frac{1}{6}}du\\[1.2em]&=\dfrac{6}{5}u^{\frac{5}{6}}+c\\[1.2em]&=\dfrac{6}{5}(x^{4}-1)^{\frac{5}{6}}+c\end{aligned} A LevelAQAEdexcelOCR ## Trigonometric Integration by Substitution Integration by substitution questions involving trigonometry can be very difficult. They involve not only the skills on this page, but also a good knowledge of trigonometric integration and trigonometric identities is a must. Example: Integrate $\left(\dfrac{\sec(x)}{\tan(x)}\right)^{8}$ using the substitution $u=tan(x)$. $u=\tan(x)$ $\dfrac{du}{dx}=\sec^{2}(x)$ $du=\sec^{2}xdx$ $dx=\dfrac{1}{\sec^{2}(x)}du$ Put into integral: \begin{aligned}\int\left(\dfrac{\sec(x)}{\tan(x)}\right)^{8}dx&=\int\dfrac{\sec^{8}(x)}{\tan^{8}(x)}dx\\[1.2em]&=\dfrac{\sec^{8}(x)}{u^{8}}\dfrac{1}{\sec^{2}(x)}du\\[1.2em]&=\dfrac{\sec^{6}(x)}{u^{8}}du\end{aligned} How do we deal with the $\sec^{6}$ term? Recall: $\sec^{2}(x)=\tan^{2}(x)+1$ $\sec^{2}(x)=u^{2}+1$ $\sec^{6}(x)=(u^{2}+1)^{3}$ \begin{aligned}&\int\left(\dfrac{\sec(x)}{\tan(x)}\right)^{8}dx=\int\dfrac{(u^{2}+1)^{3}}{u^{8}}du\\[1.2em]&=\int\dfrac{u^{6}+3u^{4}+3u^{2}+1}{u^{8}}du\\[1.2em]&=\int \left( u^{-2}+3u^{-4}+3u^{-6}+u^{-8}du\right) \\[1.2em]&=-u^{-1}-\left( 3\times\dfrac{1}{3}u^{-3}\right) -\left( 3\times\dfrac{1}{5}u^{-5}\right) -\dfrac{1}{7}u^{-7}+c\\[1.2em]&=-u^{-1}-u^{-3}-\dfrac{3}{5}u^{-5}-\dfrac{1}{7}u^{-7}+c\\[1.2em]&=-\cot(x)-\cot^{3}(x)-\dfrac{3}{5}\cot^{5}(x)-\dfrac{1}{7}\cot^{7}(x)+c\end{aligned} A LevelAQAEdexcelOCR ## Integration By Substitution Example Questions Choose $u=x^{4}$ $\dfrac{du}{dx}=4x^{3}$ $du=4x^{3}dx$ $dx=\dfrac{1}{4x^{3}}du$ Put into integral: \begin{aligned}\int x^{3}e^{x^{4}}dx&=x^{3}e^{u}\dfrac{1}{4x^{3}}du\\[1.2em]&=\int\dfrac{1}{4}e^{u}du\\[1.2em]&=\dfrac{1}{4}e^{u}+c\\[1.2em]&=\dfrac{1}{4}e^{x^{4}}+c\end{aligned} Gold Standard Education $u=\sin(x)$ $\dfrac{du}{dx}=\cos(x)$ $du=\cos(x)dx$ $dx=\dfrac{1}{\cos(x)}du$ Lower limit $x=0$: $u=\sin(0)$ $u=0$ Upper limit $x=\dfrac{\pi}{2}$ $u=\sin(\dfrac{\pi}{2})$ $u=1$ Put into integral: \begin{aligned}\int^{\frac{\pi}{2}}_{0}\cos(x)sin^{2}(x)dx&=\int^{1}_{0}\cos(x)u^{2}\dfrac{1}{\cos(x)}du\\[1.2em]&=\int^{1}_{0}u^{2}du\\[1.2em]&=\left[\dfrac{1}{3}u^{3}\right]^{1}_{0}=\dfrac{1}{3}\times1^{3}-\dfrac{1}{3}\times0^{3}\\[1.2em]&=\dfrac{1}{3}\end{aligned} Gold Standard Education Choose $u=2x^{5}-16$ $\dfrac{du}{dx}=10x^{4}$ $du=10x^{4}dx$ $dx=\dfrac{1}{10x^{4}}du$ Put it into the integral: \begin{aligned}\int\dfrac{10x^{4}}{(2x^{5}-16)^{3}}dx&=\int\dfrac{10x^{4}}{u^{3}}\dfrac{1}{10x^{4}}du\\[1.2em]&=\int\dfrac{1}{u^{3}}du\\[1.2em]&=\int u^{-3}du\\[1.2em]&=-\dfrac{1}{2}u^{-2}+c\\[1.2em]&=-\dfrac{1}{2}(2x^{5}-16)^{-2}+c\end{aligned} Gold Standard Education $u=\cot(x)$ $\dfrac{du}{dx}=-\cosec^{2}(x)$ $du=-\cosec^{2}(x)dx$ $dx=\dfrac{-1}{\cosec^{2}(x)}du$ Put it into the integral: \begin{aligned}\int\dfrac{\cosec^{4}(x)}{\cot^{\frac{1}{3}}(x)}dx&=\int\dfrac{\cosec^{4}(x)}{u^{\frac{1}{3}}}\dfrac{(-1)}{\cosec^{2}(x)}du\\[1.2em]&=\int-\dfrac{\cosec^{2}(x)}{u^{\frac{1}{3}}}du\end{aligned} Recall: $\cosec^{2}(x)=1+\cot^{2}(x)$ $\cosec^{2}(x)=1+u^{2}$ \begin{aligned}\int\dfrac{\cosec^{4}(x)}{\cot^{\frac{1}{3}}(x)}dx&=\int-\dfrac{1+u^{2}}{u^{\frac{1}{3}}}du\\[1.2em]&=\int-u^{-\frac{1}{3}}-u^{\frac{5}{3}}du\\[1.2em]&=-\dfrac{3}{2}u^{\frac{2}{3}}-\dfrac{3}{8}u^{\frac{8}{3}}+c\\[1.2em]&=-\dfrac{3}{2}\cot^{\frac{2}{3}}(x)-\dfrac{3}{8}\cot^{\frac{8}{3}}(x)+c\end{aligned} Gold Standard Education A Level #### Formula Booklet A Level Gold Standard Education £19.99 /month. 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## Problem Given the value of integers $A, B$ and $C$ find a pair of integers $(x,y)$ such that it satisfies the equation $Ax + By = C$. For example, if $A = 2, B = 3$ and $C = 7$, then possible solution of $(x,y)$ for equation $2x + 3y = 7$ would be $(2,1)$ or $(5,-1)$. The problem above is a type of Diophantine problem. In the Diophantine problem, only integer solutions to an equation are required. Since $Ax + By = C$ is a linear equation, this problem is a Linear Diophantine Problem where we have to find a solution for a Linear Diophantine Equation. For now, let us assume that $A$ and $B$ are non-zero integers. ## Existence of Solution Before we jump in to find the solution for the equation, we need to determine whether it even has a solution. For example, is there any solution for $2x + 2y = 3$? On the left side we have $2x + 2y$ which is even no matter what integer value of $(x,y)$ is used and on the right side we have $3$ which is odd. This equation is impossible to satisfy using integer values. So how do we determine if the equation has a solution? Suppose $g = gcd(A,B)$. Then $Ax + By$ is a multiple of $g$. In order to have a valid solution, since left side of the equation is divisible by $g$, the right side too must be divisible by $g$. Therefore, if $g \not\| \: C$, then there is no solution. ## Simplifying the Equation Since both side of equation is divisible by $g$, i.e, $g \: \| \: \{ \: (Ax + By), C \: \}$, we can safely divide both side by $g$ resulting in a equivalent equation. Let $a = \frac{A}{g}$, $b = \frac{B}{g}$ and $c = \frac{C}{g}$. Then, $$(Ax + By = C) \: \equiv \: (ax + by = c)$$After simplification, $gcd(a,b)$ is either $1$ or $-1$. If it is $-1$, then we need multiply $-1$ with $a,b$ and $c$ so that $gcd(a,b)$ becomes $1$ and the equation remains unchanged. Why did we make the $gcd(a,b)$ positive? You will find the reason below. ## Using Extended Euclidean Algorithm Recall that in a previous post "Extended Euclidean Algorithm", we learned how to solve the Bezout's Identity $Ax + By = gcd(A, B)$. Can we apply that here in any way? Yes. Using $\text{ext_gcd()}$ function, we can find Bezout's coefficient for $ax + by = gcd(a,b)$. But we need to find solution for $ax + by = c$. Note that $gcd(a,b) = 1$, so when we use $\text{ext_gcd()}$ we find a solution for $ax + by = 1$. Let this solution be $(x_1,y_1)$. We can extend this solution to solve our original problem. Since we already have a solution where $ax_1 + by_1 = 1$, multiplying both sides with $c$ gives us $a(x_1c) + b(y_1c) = c$. So our result is $(x,y) = (x_1c, y_1c)$. This is why we had to make sure that $gcd(a,b)$ was $1$ and not $-1$. Otherwise, multiplying $c$ would have resulted $ax + by = -c$ instead. ## Summary of Solution Here is a quick summary of what I described above. We can find solution for Linear Diophantine Equation $Ax + By = C$ in 3 steps: 1. No Solution First check if solution exists for given equation. Let $g = gcd(A,B)$. If $g \not\| \: C$ then no solution exists. 2. Simplify Equation Let $a = \frac{A}{g}, b = \frac{B}{g}$ and $c = \frac{C}{g}$. Then finding solution for $Ax + By = C$ is same as finding solution for $ax + by = c$. In simplified equation, make sure $GCD(a,b)$ is $1$. If not, multiply $-1$ with $a,b,c$. 3. Extended Euclidean Algorithm Use $\text{ext_gcd()}$ to find solution $(x_1,y_1)$ for $ax + by = 1$. Then multiply the solution with $c$ to get solution for $ax + by = c$, where $x = x_1 \times c, y = y_1 \times c$. Let us try few examples. ### Example 1: $2x + 3y = 7$ Step $1$: $g = GCD(2,3) = 1$. Since $1$ divides $7$, solution exists. Step $2$: Since $g$ is already $1$ there is nothing to simplify. Step $3$: Using $\text{ext_gcd()}$ we get $(x,y) = (-1,1)$. But this is for $ax + by = 1$. We need to multiply $7$. So our solution is $(-7,7)$. $2 \times -7 + 3 \times 7 = -14 + 21 = 7$. The solution is correct. ### Example 2: $4x + 10y = 8$ Step $1$: $g = GCD(4,10) = 2$. Since $2$ divides $8$, solution exists. Step $2$: $a = \frac{4}{2}, b = \frac{10}{2}, c = \frac{8},{2}$. We will find solution of $2x + 5y = 4$. Step $3$: Using $\text{ext_gcd()}$ we get $(x,y) = (-2,1)$. But this is for $ax + by = 1$. We need to multiply $4$. So our solution is $(-8,4)$. $ax + by = 2 \times -8 + 5 \times 4 = -16 + 20 = 4 = c$. Also, $Ax + By = 4 \times -8 + 10 \times 4 = -32 + 40 = 8 = C$. The solution is correct. Both $ax + by = c$ and $Ax + By = C$ are satisfied. ## Finding More Solutions We can now find a possible solution for $Ax + By = C$, but what if we want to find more? How many solutions are there? Since the solution for $Ax + By = C$ is derived from Bezout's Identity, there are infinite solutions. Suppose we found a solution $(x,y)$ for $Ax + By = C$. Then we can find more solutions using the formula: $( x + k \frac{B}{g}, y - k \frac {A}{g})$, where $k$ is any integer. ## Code Let us convert our idea into code. bool linearDiophantine ( int A, int B, int C, int x, int y ) { int g = gcd ( A, B ); if ( C % g != 0 ) return false; //No Solution int a = A / g, b = B / g, c = C / g; ext_gcd( a, b, x, y ); //Solve ax + by = 1 if ( g < 0 ) { //Make Sure gcd(a,b) = 1 a = -1; b = -1; c = -1; } x = c; y = c; //ax + by = c return true; //Solution Exists } int main () { int x, y, A = 2, B = 3, C = 5; bool res = linearDiophantine ( A, B, C, &x, &y ); if ( res == false ) printf ( "No Solution\n" ); else { printf ( "One Possible Solution (%d %d) \n", x, y ); int g = gcd ( A, B ); int k = 1; //Use different value of k to get different solutions printf ( "Another Possible Solution (%d %d)\n", x + k ( B / g ), y - k * ( A / g ) ); } return 0; } $\text{linearDiophantine}()$ function finds a possible solution for equation $Ax + By = C$. It takes in $5$ parameters. $A,B,C$ defines the coefficients of equation and $x, y$ are two pointers that will carry our solution. The function will return $true$ if solution exists and $false$ if not. In line $2$ we calculate $gcd(A,B)$ and in line $3$ we check if $C$ is divisible by $g$ or not. If not, we return $false$. Next on line $5$ we define $a,b,c$ for simplified equation. On line $6$ we solve for $ax + by = 1$ using $\text{ext_gcd}$. Then we check if $g < 0$. If so, we multiply $-1$ with $a,b,c$ to make it positive. Then we multiply $c$ with $x,y$ so that our solution satisfies $ax + by = c$. A solution is found so we return true. In $\text{main}()$ function, we call $\text{linearDiophantine}()$ using $A=2,B=3,C=5$. In line $22$ we print a possible solution. In line $27$ we print another possible solution using formula for more solutions. ## $A$ and $B$ with Value $0$ Till now we assumed $\{A, B\}$ have non-zero values. What happens if they have value $0$? ### When Both $A = B = 0$ When both $A$ and $B$ are zero, the value of $Ax + By$ will always be $0$. Therefore, if $C \neq 0$ then there is no solution. Otherwise, any pair of value for $(x,y)$ will act as a solution for the equation. ### When $A$ or $B$ is $0$ Suppose only $A$ is $0$. Then equation $Ax + By = C$ becomes $0x + By = C \: \equiv \: By = C$. Therefore $y = \frac {C}{B}$. If $B$ does not divide $C$ then there is no solution. Else solution will be $(x,y) = (k, \frac{C}{B})$, where $k$ is any intger. Using same logic, when $B$ is $0$, solution will be $(x,y) = ( \frac{C}{A}, k )$. ## Coding Pitfalls When we use $gcd(a,b)$ in our code, we mean the result from Euclidean Algorithm, not what we understand mathematically. $gcd(4,-2)$ is $-2$ according to Euclidean Algorithm whereas it is $2$ in common sense.
# Addition of Polynomials – Definition, Rules, Examples \| How do you Solve Polynomials with Addition? The Addition of Polynomials is the process of adding different terms present in the polynomial. Check out the various problems and procedures on how do we add polynomials in this article. Know the different rules in addition of polynomials consisting of different exponents. Students of 6th Grade Math can get a strong grip on the Polynomial Addition by practicing the Addition of Polynomials Examples over here. The addition of two polynomials is the process of combining like terms present in the two polynomials. Like terms are the terms those having the same variable and same exponent. ## How to Solve Addition of Polynomials? \| Rules in Addition of Polynomials We have given the process to add polynomials below. Follow the procedure given here to solve all the addition of polynomial problems. (i) Arrange each polynomial along with their terms and also with the highest degree in decreasing order of degree. (ii) Next, group the like terms whose variables and exponents are the same. (iii) Finally, simplify by combining like terms. Go through the below procedure to do the Horizontal addition of polynomials. Let’s check the Addition of Polynomials horizontally with a simple example. Example: Find the sum of the following two polynomials: 12m4 + 3m3 + 8m – 5, 10m4 – 2m3 + 6m2 – 3m + 2. Solution: Given polynomials are 12m4 + 3m3 + 8m – 5 and 10m4 – 2m3 + 6m2 – 3m + 2. The first polynomial is 12m4 + 3m3 + 8m – 5 and the second is 10m4 – 2m3 + 6m2 – 3m + 2. Now, add the given polynomials horizontally. (12m4 + 3m3 + 8m – 5) + (10m4 – 2m3 + 6m2 – 3m + 2) Group the same variables with the same exponents. Terms that are not like terms cannot be added. 12m4 and 10m4 are the like terms. 3m3 and 2m3Â are the like terms. 8m and 3m are the like terms. 5 and 2 are constants. Now, add the given polynomials with the like terms. (12m4 + 3m3 + 8m – 5) + (10m4 – 2m3 + 6m2 – 3m + 2) = 12m4 + 3m3 + 8m – 5 + 10m4 – 2m3 + 6m2 – 3m + 2 12m4 + 10m4 + 3m3 – 2m3 + 6m2 + 8m – 3m – 5 + 2 = 22m4 + m3 + 6m2 + 5m – 3. Therefore, the addition of given polynomials is 22m4 + m3 + 6m2 + 5m – 3. The step-by-step process to Add Polynomials vertically is given here. Let’s check the Addition of Polynomials Vertically with a simple example. Example: Find the sum of the following two polynomials: 12m4 + 3m3 + 8m – 5, 10m4 – 2m3 + 6m2 – 3m + 2. Solution: Given polynomials are 12m4 + 3m3 + 8m – 5 and 10m4 – 2m3 + 6m2 – 3m + 2. The first polynomial is 12m4 + 3m3 + 8m – 5 and the second is 10m4 – 2m3 + 6m2 – 3m + 2. In the first term, we don’t have the m2 term. So, we can take it as 0. Now, add the given polynomials vertically. (12m4 + 3m3 + 0 + 8m – 5) + (10m4 – 2m3 + 6m2 – 3m + 2) ———————————— 22m4 + m3 + 6m2 + 5m – 3 Therefore, the addition of given polynomials is 22m4 + m3 + 6m2 + 8m – 3. Question 1. Add: 6x + 4y, 5x â€“ 5y + 2z and -2x + 6y + 3z Solution: Given polynomials are 6x + 4y, 5x â€“ 5y + 2z and -2x + 6y + 3z. The first polynomial is 6x + 4y and the second is 5x â€“ 5y + 2z and the third polynomial is -2x + 6y + 3z. Now, add the given polynomials horizontally. (6x + 4y) + (5x â€“ 5y + 2z) + (-2x + 6y + 3z) Group the same variables with the same exponents. Terms that are not like terms cannot be added. 6x, 5x, and 2x are the like terms. 4y, 5y, and 6y are the like terms. 2z and 3z are the like terms. Now, add the given polynomials with the like terms. (6x + 4y) + (5x â€“ 5y + 2z) + (-2x + 6y + 3z) = 6x + 4y + 5x â€“ 5y + 2z -2x + 6y + 3z 6x + 5x -2x+ 4y â€“ 5y + 6y + 2z + 3z = 9x + 5y + 8z. Therefore, the addition of given polynomials is 9x + 5y + 8z. Question 2. Add: 6a2 + 2ab â€“ 2b2, -2a2 + 4ab + 6b2 and 6a2 â€“ 20ab + 8b2 Solution: Given polynomials are 6a2 + 2ab â€“ 2b2, -2a2 + 4ab + 6b2 and 6a2 â€“ 20ab + 8b2. The first polynomial is 6a2 + 2ab â€“ 2b2, and the second is -2a2 + 4ab + 6b2 and the third polynomial is 6a2 â€“ 20ab + 8b2. Now, add the given polynomials horizontally. (6a2 + 2ab â€“ 2b2) + (-2a2 + 4ab + 6b2) + (6a2 â€“ 20ab + 8b2) Group the same variables with the same exponents. Terms that are not like terms cannot be added. 6a2, -2a2, and 6a2 are the like terms. 2ab, 4ab, and 20ab are the like terms. 2b2, 6b2, and 8b2Â are the like terms. Now, add the given polynomials with the like terms. (6a2 + 2ab â€“ 2b2) + (-2a2 + 4ab + 6b2) + (6a2 â€“ 20ab + 8b2) = 6a2 + 2ab â€“ 2b2 -2a2 + 4ab + 6b2 + 6a2 â€“ 20ab + 8b2 6a2 – 2a2 + 6a2 + 2ab + 4ab â€“ 20ab â€“ 2b2 + 6b2 + 8b2 = 10a2 – 18ab â€“ 12b2 Therefore, the addition of given polynomials is 10a2 – 18ab â€“ 12b2. Question 3. Add: 9a + 7b, 8a â€“ 8b + 5c and -7a + 9b + 6c Solution: Given polynomials are 9a + 7b, 8a â€“ 8b + 5c and -7a + 9b + 6c. The first polynomial is 9a + 7b and the second is 8a â€“ 8b + 5c and the third polynomial is -7a + 9b + 6c. In the first term, we don’t have the c term. So, we can take it as 0. Now, add the given polynomials vertically. 9a + 7b + 0 8a – 8b + 5c -7a + 9b + 6c —————— 10a +8b + 11c Therefore, the addition of given polynomials is 10a +8b + 11c. Question 4. Add: 5x3 â€“ 7x2 + 10x + 20, 17x3 â€“ 4x â€“ 25, 11x2 â€“ 2x + 17 and -10x3 + 4x2 â€“ 9x. Solution: Given polynomials are 5x3 â€“ 7x2 + 10x + 20, 17x3 â€“ 4x â€“ 25, 11x2 â€“ 2x + 17 and -10x3 + 4x2 â€“ 9x. The first polynomial is 5x3 â€“ 7x2 + 10x + 20 and the second is 17x3 â€“ 4x â€“ 25, and the third polynomial is 11x2 â€“ 2x + 17 also the fourth polynomial is -10x3 + 4x2 â€“ 9x. Where the variable is not present we can take it as 0. Now, add the given polynomials vertically. 5x3 â€“ 7x2 + 10x + 20 17x3 + 0Â Â â€“ 4x â€“ 25 0Â Â + 11x2 â€“ 2x + 17 -10x3 + 4x2 â€“ 9x + 0 ———————– 12x3 + 8x2 – 5x + 12 Therefore, the addition of given polynomials is 12x3 + 8x2 – 5x + 12. ### FAQs on Addition of Polynomials 1. Â What is a linear Polynomial? The linear polynomial is a polynomial that has the degree 1. Â 2. How do we add polynomials? We find similar terms first then we add the coefficients of the terms to add those polynomials. Â 3. Is it possible to add different terms with different exponents? No, generally we add similar terms having the same exponents. 4. Add 3x + 2y and 4x + 3y? By adding the given polynomials, we can get 7x + 5y. 5. Add 2a + 9b and 3a? By adding the given polynomials, we can get 5a. ### Conclusion The complete article will help you to learn the addition of polynomials easily. Without missing any part, read the complete concept and gain the knowledge. Scroll to Top Scroll to Top
Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 3 Integers Class 6 Practice Set 8 Answers Solutions. ## 6th Standard Maths Practice Set 8 Answers Chapter 3 Integers Question 1. Subtract the numbers in the top row from the numbers in the first column and write the proper number in each empty box: – 6 9 -4 -5 0 +7 -8 -3 3 3 – 6 = -3 8 8 – (-5) = 13 -3 -2 Solution: – 6 9 -4 -5 3 (+3) + (-6) = -3 (+3) + (-9) = -6 (+3) + (+4) = 7 (+3) + (+5) = 8 8 (+8) + (-6) = +2 (+8) + (-9) = -1 (+8) + (+4) = 12 (+8) + (+5) = 13 -3 (-3) + (-6) = -9 (-3) + (-9) = -12 (-3) + (+4) = 1 (-3) + (+5) = 2 -2 (-2) + (-6) = -8 (-2) + (-9) = -11 (-2) + (+4) = 2 (-2) + (+5) = 3 – 0 +7 -8 -3 3 (+3) – 0 = 3 (+3) + (-7) = -4 (+3) + (+8) = 11 (+3) + (+3) = 6 8 (+8) – 0 = 8 (+8) + (-7) = 1 (+8) + (+8) = 16 (+8) + (+3) = 11 -3 (-3) – 0 = -3 (-3) + (-7) = -10 (-3) + (+8) = 5 (-3) + (+3) = 0 -2 (-2) – 0 = -2 (-2) + (-7) = -9 (-2) + (+8) = 6 (-2) + (+3) = 1 #### Maharashtra Board Class 6 Maths Chapter 3 Integers Practice Set 8 Intext Questions and Activities Question 1. A Game of Integers. (Textbook pg. no. 20) The board for playing this game is given in the back cover of the textbook. Place your counters before the number 1. Throw the dice. Look at the number you get. It is a positive number. Count that many boxes and move your counter forward. If a problem is given in that box, solve it. If the answer is a positive number, move your counter that many boxes further. It it is negative, move back by that same number of boxes. Suppose we have reached the 18th box. Then the answer to the problem in it is -4 + 2 = -2. Now move your counter back by 2 boxes to 16. The one who reaches 100 first, is the winner. Solution: (Students should attempt this activity on their own)
Basic Laws for Differentiation # Basic Laws for Differentiation Before we look at some other methods of differentiation, let's first note some important properties of differentiation: Theorem 1 (Addition Law): If $f = g + j$ where $g$ and $j$ are differentiable, then $\frac{d}{dx} f(x) = \frac{d}{dx} g(x) + \frac{d}{dx} j(x)$. • Proof: From the definition of a derivative we know that $\frac{d}{dx} f(x) = \lim_{h\rightarrow 0} \frac{f(x + h) - f(x)}{h}$. Now suppose that $f = g + j$. Then: (1) \begin{align} \frac{d}{dx} f(x) &= \frac{d}{dx} (g(x) + j(x)) \\ \frac{d}{dx} f(x) &= \lim_{h \to 0} \left ( \frac{g(x + h) + j(x + h) - g(x) - j(x)}{h} \right ) \\ \frac{d}{dx} f(x) &= \lim_{h \to 0} \left ( \frac{g(x + h) - g(x)}{h} + \frac{j(x + h) - j(x)}{h} \right ) \\ \: \frac{d}{dx} f(x) &= \lim_{h \to 0} \frac{g(x + h) - g(x)}{h} + \lim_{h \to 0} \frac{j(x + h) - j(x)}{h} \\ \frac{d}{dx} f(x) &= \frac{d}{dx} g(x) + \frac{d}{dx} j(x) \quad \blacksquare \end{align} Theorem 2 (Subtraction Law): If $f = g - j$ where $g$ and $j$ are differentiable, then $\frac{d}{dx} f(x) = \frac{d}{dx} g(x) - \frac{d}{dx} j(x)$. • Proof: Like property 1, from the definition of a derivative we know that $\frac{d}{dx} f(x) = \lim_{h\rightarrow 0} \frac{f(x + h) - f(x)}{h}$, and suppose $f = g - j$, then: (2) \begin{align} \frac{d}{dx} f(x) &= \frac{d}{dx} (g(x) - j(x)) \\ \frac{d}{dx} f(x) &= \lim_{h \to 0} \left ( \frac{g(x + h) - j(x + h) - g(x) + j(x)}{h} \right ) \\ \frac{d}{dx} f(x) &= \lim_{h \to 0} \left ( \frac{g(x + h) - g(x)}{h} - \frac{j(x + h) - j(x)}{h} \right ) \\ \: \frac{d}{dx} f(x) &= \lim_{h \to 0} \frac{g(x + h) - g(x)}{h} - \lim_{h \to 0} \frac{j(x + h) - j(x)}{h} \\ \frac{d}{dx} f(x) &= \frac{d}{dx} g(x) - \frac{d}{dx} j(x) \quad \blacksquare \end{align} Theorem 3 (Constant Multiplication Law): If $f$ is differentiable and $k \in \mathbb{R}$, then $\frac{d}{dx} kf(x) = k\frac{d}{dx} f(x)$. • Proof: By the definition of a derivative we get that $\frac{d}{dx} f(x) = \lim_{h\rightarrow 0} \frac{f(x + h) - f(x)}{h}$. For some $k \in \mathbb{R}$, we will apply the definition of the derivative to get: (3) \begin{align} \frac{d}{dx} kf(x) &= \lim_{h \to 0} \left ( k \cdot \frac{f(x + h) - f(x)}{h} \right ) \\ \frac{d}{dx} kf(x) &= \lim_{h \to 0} k \cdot \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ \frac{d}{dx} kf(x) &= k \cdot \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ \frac{d}{dx} kf(x) &= k \frac{d}{dx} f(x) \quad \blacksquare \end{align}
# SSAT Upper Level Math Practice Test Questions Preparing your students for the SSAT Upper Level Math test? Try these free SSAT Upper Level Math Practice questions. Reviewing practice questions is the best way to brush up your student’s Math skills. Here, we walk you through solving 10 common SSAT Upper Level Math practice problems covering the most important math concepts on the SSAT Upper Level Math test. These SSAT Upper Level Math practice questions are designed to be similar to those found on the real SSAT Upper Level Math test. They will assess your student’s level of preparation and will give you a better idea of what students need to study on their exams. ## 10 Sample SSAT Upper Level Math Practice Questions 1- The sum of six different negative integers is $$-80$$. If the smallest of these integers is $$-16$$, what is the largest possible value of one of the other five integers? ☐A. $$-16$$ ☐B. $$-10$$ ☐C. $$-8$$ ☐D. $$-4$$ ☐E. $$-1$$ 2- What is the slope of a line that is perpendicular to the line $$2x-4y=24$$? ☐A. $$-2$$ ☐B. $$-\frac{1}{2}$$ ☐C. 6 ☐D. 12 ☐E. 14 3- The width of a box is one third of its length. The height of the box is one half of its width. If the length of the box is $$24$$ cm, what is the volume of the box? ☐A. $$91$$ cm$$^3$$ ☐B. $$172$$ cm$$^3$$ ☐C. $$254$$ cm$$^3$$ ☐D. $$768$$ cm$$^3$$ ☐E. $$2,990$$ cm$$^3$$ 4- A football team won exactly $$70\%$$ of the games it played during last session. Which of the following could be the total number of games the team played last season? ☐A. 59 ☐B. 45 ☐C. 72 ☐D. 20 ☐E. 11 5- The Jackson Library is ordering some bookshelves. If $$x$$ is the number of bookshelves the library wants to order, which each costs $$\200$$ and there is a one-time delivery charge of $$\600$$, which of the following represents the total cost, in dollar, per bookshelf? ☐A. $$\frac{200x+600}{x}$$ ☐B. $$\frac{200x+600}{600}$$ ☐C. $$200+600x$$ ☐D. $$200x+600$$ ☐E. $$200x-600$$ 6- There are 14 marbles in the bag A and $$18$$ marbles in the bag B. If the sum of the marbles in both bags will be shared equally between two children, how many marbles bag A has less than the marbles that each child will receive? ☐A. 2 ☐B. 3 ☐C. 4 ☐D. 5 ☐E. 6 7- If Jason’s mark is F more than Alex, and Jason’s mark is $$18$$, which of the following can be Alex’s mark? ☐A. $$18-F$$ ☐B. $$F-18$$ ☐C. $$\frac{F}{18}$$ ☐D. $$18F$$ ☐E. $$18+F$$ 8- When number $$81,602$$ is divided by $$240$$, the result is closest to? ☐A. 4 ☐B. 30 ☐C. 200 ☐D. 300 ☐E. 340 9- The price of a sofa is decreased by $$50\%$$ to $$\530$$. What was its original price? ☐A. $$\580$$ ☐B. $$\620$$ ☐C. $$\760$$ ☐D. $$\900$$ ☐E. $$\1,060$$ 10- If the perimeter of the following figure be $$33$$, what is the value of $$x$$? ☐A. 3 ☐B. 4 ☐C. 6 ☐D. 8 ☐E. 11 ## Best SSAT Upper Level Math Prep Resource for 2021 1- B The smallest number is $$-16$$. To find the largest possible value of one of the other five integers, we need to choose the smallest possible integers for four of them. Let $$x$$ be the largest number. Then: $$-80=(-16)+(-15)+(-14)+(-13)+(-12)+x→$$ $$-80=-70+x , →x=-80+70=-10$$ 2- A The equation of a line in slope intercept form is: $$y=mx+b$$, Solve for $$y$$. $$2x-4y=24⇒-4y=24-2x⇒y=(24-2x)÷(-4)⇒y=\frac{1}{2} x-6$$ The slope is $$\frac{1}{2}$$. The slope of the line perpendicular to this line is: $$m_{1}×m_{2}= -1⇒\frac{1}{2} ×m_{2}=-1⇒m_{2}=-2$$ 3- D If the length of the box is $$24$$, then the width of the box is one third of it, $$8$$, and the height of the box is $$4$$ (half of the width). The volume of the box is: V$$=$$(length)(width)(height)$$=(24)(8)(4)=768$$ cm$$^3$$ 4- D Choices A,B,C and E are incorrect because $$70\%$$ of each of the numbers is non-whole number. A.$$59, 70\%$$ of $$59 = 0.70×59=41.3$$ B.$$45, 70\%$$ of $$45=0.70×45=31.5$$ C.$$72, 70\%$$ of $$72=0.70×72=50.4$$ D.$$20, 70\%$$ of $$20=0.70×20=14$$ E. $$11, 70\%$$ of $$11=0.70×11=7.7$$ 5- A The amount of money for $$x$$ bookshelf is: $$200x$$, Then, the total cost of all bookshelves is equal to: $$200x+600$$, The total cost, in dollar, per bookshelf is: $$\frac{Total \ cost}{number \ of \ items}=\frac{200x+600}{x}$$ 6- A $$\frac{14+18}{2}=\frac{32}{2}=16$$ Then, $$16-14=2$$ 7- A Alex’s mark is F less than Jason’s mark. Then, from the choices provided Alex’s mark can only be $$18-F$$. 8- E $$\frac{8160}{240}≅340.0083≅340$$ 9- E Let $$x$$ be the original price. If the price of the sofa is decreased by $$50\%$$ to $$530$$, then: $$50\%$$ of $$x=530 ⇒ 0.50x=530 ⇒ x=530÷0.50=1,060$$ 10- E Let’s review the choices provided: A.$$x=3→$$ The perimeter of the figure is: $$3+5+3+3+3=17≠33$$ B.$$x=4→$$ The perimeter of the figure is: $$3+5+3+4+4=19≠33$$ C.$$x=6→$$ The perimeter of the figure is: $$3+5+3+6+6=23≠33$$ D.$$x=8→$$ The perimeter of the figure is: $$3+5+3+8+8=27≠33$$ E.$$x=11→$$ The perimeter of the figure is: $$3+5+3+11+11=33=33$$ Looking for the best resource to help you succeed on the SSAT Upper Level Math test? 52% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 3. Checkout Complete the quick and easy checkout process. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Help save the environment
# how to find cosine from sine cos(x) Note − This function is not accessible directly, so we need to import math module and then we need to call this function using math static object.. Parameters. Following is the syntax for cos() method −. The Pythagorean Identity is also useful for determining the sines and cosines of special angles. It is easy to memorise the values for these certain angles. When we find sin cos and tan values for a triangle, we usually consider these angles: 0°, 30°, 45°, 60° and 90°. The “length” of this interval of x … Introduction: In this lesson, the period and frequency of basic graphs of sine and cosine will be discussed and illustrated. I think I am a very visual learner and I always found that diagrams always made things clearer for my students. sin (x) = cos (90 -x) [within first quadrant] 0 0 Description. Teacher was saying that in right triangles the sine of one acute angle is the cosine of the other acute angle. When finding the equation for a trig function, try to identify if it is a sine or cosine graph. x − This must be a numeric value.. Return Value. You want to show that the sine function, slid 90 degrees to the left, is equal to the cosine function: Replace cos x with its cofunction identity. Understanding how to create and draw these functions is essential to these classes, and to nearly anyone working in a scientific field. The sine and cosine values are most directly determined when the corresponding point on the unit circle falls on an axis. Underneath the calculator, six most popular trig functions will appear - three basic ones: sine, cosine and tangent, and … Find $$\cos(20^\circ)$$ and $$\sin(20^\circ)\text{. See Example. We note that sin π/4=cos π/4=1/√2, and re-use cos θ=sin (π/2−θ) to obtain the required formula. The second one, y = cos( x 2 + 3) , means find the value ( x 2 + 3) first, then find the cosine of the result. Here’s how to prove this statement. To find the equation of sine waves given the graph: Find the amplitude which is half the distance between the maximum and minimum. The first one, y = cos x 2 + 3, or y = (cos x 2) + 3, means take the curve y = cos x 2 and move it up by 3 units. Trig calculator finding sin, cos, tan, cot, sec, csc To find the trigonometric functions of an angle, enter the chosen angle in degrees or radians. See Example. The sine and cosine functions appear all over math in trigonometry, pre-calculus, and even calculus. Sum Next, note that the range of the function is and that the function goes through the point . The Lesson: y = sin(x) and y = cos(x) are periodic functions because all possible y values repeat in the same sequence over a given set of x values. Example 26. From this information, we can find the amplitude: So our function must have a out in front. The shifted sine graph and the cosine graph are really equivalent — they become graphs of the same set of points. 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# Bar Model Method with Distributive Law For Chicken-Rabbit Approach ##### Ho Soo Thong Abstract This article illustrates a bar modelling problem solving approach to a classical chicken-rabbit problem which involves distributive property situations. ## 1. A Classic Chickens-Rabbits Problem The following classic word problem , chickens and rabbits in a cage, posed in the ancient Chinese book Sunzi Suanjing : There are chickens and rabbits in a cage. Look at the top of the cage – there are 35 heads. Look at the bottom of the cage – there are 94 legs. How many chickens and how many rabbits are there in the cage? The problem involves the Actual situation: There are 35 chickens and rabbits and there is a total of 94 legs where each rabbit has 4 legs and each chicken has 2 legs. The strategy here is to add the Supposed situation : There are 35 chickens and therefore a total of 35×2 = 70 legs. Figure 1 shows the bar models for the number of chickens and rabbits and the corresponding bar model for the number of legs. Therefore we have 23 chickens and 12 rabbits. ## 2. Variants of Chicken-Rabbit Problems Now, we simplify the presentation in the same approach to a counting problem. ### Example 1 There is a total of 33 triangles and squares, and they have a total of 113 vertices. How many triangles and how many squares are there? ### Solution Figure 2 shows the bar model for number of vertices in the Actual situation and the Supposed situation in which we suppose that the shapes are triangles. A square has 4 vertices and a triangle has 3 vertices. Therefore, there are 19 triangles and 14 squares. Next, we apply the Chickens-Rabbits problem solving problem to a simple finance problem. ### Example 2 Jennifer spent \$ 24.80 for a total of 30 apples and oranges. An apple costs 70₡ and an orange costs 90₡. How many apples and how many oranges did Jennifer buy? ### Solution Figure 3 shows the bar model for the amounts spent in the Actual situation and the Supposed situation in which we suppose that only apples are bought. An orange costs 20¢ more than an apple. Therefore, Jennifer bought 11 apples and 19 oranges ## 3. A Challenging Problem Finally, we post a complex problem which requires the chicken-rabbit problem solving approach to be applied twice. A total of 31 teachers and students plan for a two-nights excursion. Teachers prepare 96 kg of food and 43 kg of camping equipment. Each teacher will carry 4 kg of food. Among the students, each boy will carry 3 kg of food together with 2 kg of equipment and each girl will carry 3 kg of food together with 1 kg of equipment. How many boys and how many girls are there in the excursion? References [1] Ho Soo Thong, Ho Shuyuan, Bar Model Method for PSLE and Beyond – AceMath, 2011 [2] Ho Soo Thong, Ho Shuyuan and Leong Yu Kiang Problem Solving methods for Primary Olympiad Mathematics, AceMath, 2012.
Tamilnadu State Board New Syllabus Samcheer Kalvi 11th Business Maths Guide Pdf Chapter 9 Correlation and Regression Analysis Ex 9.1 Text Book Back Questions and Answers, Notes. ## Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 9 Correlation and Regression Analysis Ex 9.1 ### Samacheer Kalvi 11th Business Maths Correlation and Regression Analysis Ex 9.1 Text Book Back Questions and Answers Question 1. Calculate the correlation co-efficient for the following data: Solution: Coefficient of correlation Question 2. Find the coefficient of correlation for the following: Solution: Question 3. Calculate the coefficient of correlation for the ages of husbands and their respective wives: Solution: Without deviation: N = 10 Coefficient of correlation = $$\frac{1783}{45 \times 39.76}$$ = 0.9965 Note: We can do the above problem using deviations taken from arithmetic means of X and Y. i.e., using r = $$\frac{\Sigma \mathrm{XY}}{\sqrt{\Sigma \mathrm{X}^{2} \Sigma \mathrm{Y}^{2}}}$$ Question 4. Calculate the coefficient of correlation between X and Y series from the following data: The summation of product deviations of X and Y series from their respective arithmetic means is 122. Solution: N = 15, $$\overline{\mathrm{X}}$$ = 25, $$\overline{\mathrm{Y}}$$ = 18, x = X – $$\overline{\mathrm{X}}$$, y = Y – $$\overline{\mathrm{Y}}$$, Σx2 = 136, Σy2 = 138, Σxy = 122 Correlation coefficient Question 5. Calculate the correlation coefficient for the following data: Solution: N = 10, ΣX = 310, ΣY = 340, Σx2 = 870, Σy2 = 720, Σxy = 178 Correlation coefficient Question 6. Find the coefficient of correlation for the following: Solution: N = 8, ΣX = 600, ΣY = 724, Σdx2 = 1172, Σdy2 = 3372, Σdxdy = -146 Correlation coefficient Question 7. An examination of 11 applicants for an accountant post was taken by a finance company. The marks obtained by the applicants in the reasoning and aptitude tests are given below. Calculate Spearman’s rank correlation coefficient from the data given above. Solution: N = 11, Σd2 = 22 Rank correlation Question 8. The following are the ranks obtained by 10 students in commerce and accountancy are given below: To what extent is the knowledge of students in the two subjects related? Solution: N = 11, Σd2 = 128 Rank correlation Question 9. A random sample of recent repair jobs was selected and the estimated cost and actual cost were recorded. Calculate the value of spearman’s correlation coefficient. Solution: N = 8, Σd2 = 8 Rank correlation Question 10. The rank of 10 students of the same batch in two subjects A and B are given below. Calculate the rank correlation coefficient. Solution: N = 10, Σd2 = 226 Rank correlation
# Polar Equations A polar rose (Rhodonea Curve) This polar rose is created with the polar equation: $r = cos(\pi\theta)$. # Basic Description Polar equations are used to create interesting curves, and in most cases they are periodic like sine waves. Other types of curves can also be created using polar equations besides roses, such as Archimedean spirals and limaçons. See the Polar Coordinates page for some background information. # A More Mathematical Explanation Note: understanding of this explanation requires: *calculus, trigonometry ## Rose The general polar equations form to create a rose is UNIQ57e38ce34a3afc2f-math-00000001-Q [...] ## Rose The general polar equations form to create a rose is $r = a \sin(n \theta)$ or $r = a \cos(n \theta)$. Note that the difference between sine and cosine is $\sin(\theta) = \cos(\theta-\frac{\pi}{2})$, so choosing between sine and cosine affects where the curve starts and ends. $a$ represents the maximum value $r$ can be, i.e. the maximum radius of the rose. $n$ affects the number of petals on the graph: • If $n$ is an odd integer, then there would be $n$ petals, and the curve repeats itself every $\pi$. Examples: • If $n$ is an even integer, then there would be $2n$ petals, and the curve repeats itself every $2 \pi$. Examples: • If $n$ is a rational fraction ($p/q$ where $p$ and $q$ are integers), then the curve repeats at the $\theta = \pi q k$, where $k = 1$ if $pq$ is odd, and $k = 2$ if $pq$ is even. Examples: $r = \cos(\frac{1}{2}\theta)$ The angle coefficient is $\frac{1}{2} = 0.5$. $1 \times 2 = 2$, which is even. Therefore, the curve repeats itself every $\pi \times 2 \times 2 \approx 12.566.$ $r = \cos(\frac{1}{3}\theta)$ The angle coefficient is $\frac{1}{3} \approx 0.33333$. $1 \times 3 = 3$, which is odd. Therefore, the curve repeats itself every $\pi \times 3 \times 1 \approx 9.425.$ • If $n$ is irrational, then there are an infinite number of petals. Examples: $r = \cos(e \theta)$ $\theta \text{ from } 0 \text{ to...}$ ...$10$ ...$50$ ...$100$ $\text{Note: }e \approx 2.71828$ Below is an applet to graph polar roses, which is used to graph the examples above: If you can see this message, you do not have the Java software required to view the applet. Source code: Rose graphing applet ## Other Polar Curves Archimedean Spirals Archimedes' Spiral $r = a\theta$ The spiral can be used to square a circle and trisect an angle. Fermat's Spiral $r = \pm a\sqrt\theta$ This spiral's pattern can be seen in disc phyllotaxis. Hyperbolic spiral$r = \frac{a}{\theta}$ It begins at an infinite distance from the pole, and winds faster as it approaches closer to the pole. Lituus $r^2 \theta = a^2$It is asymptotic at the $x$ axis as the distance increases from the pole. Limaçon[1] The word "limaçon" derives from the Latin word "limax," meaning snail. The general equation for a limaçon is $r = b + a\cos(\theta)$. • If $b = a/2$, then it is a trisectrix (see figure 2). • If $b = a$, then it becomes a cardioid (see figure 3). • If $2a > b > a$, then it is dimpled (see figure 4). • If $b \geq 2a$, then the curve is convex (see figure 5). $r = \cos(\theta)$ 1 $r = 0.5 + \cos(\theta)$ 2 Cardioid $r = 1 + \cos(\theta)$3 $r = 1.5 + \cos(\theta)$4 $r = 2 + \cos(\theta)$5 ## Finding Derivatives[2] A derivative gives the slope of any point in a function. Consider the polar curve $r = f(\theta)$. If we turn it into parametric equations, we would get: • $x = r \cos(\theta) = f(\theta) \cos(\theta)$ • $y = r \sin(\theta) = f(\theta) \sin(\theta)$ Using the method of finding the derivative of parametric equations and the product rule, we would get: $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\frac{dr}{d\theta} \sin(\theta) + r \cos(\theta)}{\frac{dr}{d\theta} \cos(\theta) - r \sin(\theta)}$ Note: It is not necessary to turn the polar equation to parametric equations to find derivatives. You can simply use the formula above. Examples: Find the derivative of $r = 1 + \sin(\theta)$ at $\theta = \frac{\pi}{3}$. $\frac{dr}{d\theta} = \cos(\theta)$ $\frac{dy}{dx} = \frac{\frac{dr}{d\theta} \sin(\theta) + r \cos(\theta)}{\frac{dr}{d\theta} \cos(\theta) - r \sin(\theta)} = \frac{\cos(\theta) \sin(\theta) + (1 + \sin(\theta) ) \cos(\theta)}{\cos(\theta)\cos(\theta) - (1 + \sin(\theta) ) \sin(\theta)}$ $= \frac{\cos(\theta)\sin(\theta) + \cos(\theta) + \cos(\theta)\sin(\theta)}{\cos^2(\theta) - \sin(\theta) - \sin^2(\theta)}$ Note: Using the double-angle formula, we get $\cos^2(\theta) - \sin^2(\theta) = 1 - 2\sin^2(\theta)$ $= \frac{\cos(1+2\sin(\theta))}{1-2\sin^2(\theta)-\sin(\theta)} = \frac{\cos(1+2\sin(\theta))}{(1+\sin(\theta))(1-2\sin(\theta))}$ $\frac{dy}{dx} \Big \|_{\theta=\pi/3} = \frac{\cos(\pi/3)(1+2\sin(\pi/3))}{(1+\sin(\pi/3))(1-2\sin(\pi/3))}$ $= \frac{\frac{1}{2}(1+\sqrt{3})}{(1+\sqrt{3}/2)(1-\sqrt{3})} = \frac{1+\sqrt{3}}{(2+\sqrt{3})(1-\sqrt{3})} = \frac{1+\sqrt{3}}{-1-\sqrt{3}} = -1$ ## Finding Areas and Arc Lengths[2] Area of a sector of a circle. To find the area of a sector of a circle, where $r$ is the radius, you would use $A = \frac{1}{2} r^2 \theta$. $A = \int_{-\frac{\pi}{4}}^\frac{\pi}{4}\! \frac{1}{2} \cos^2(2\theta) d\theta$ Therefore, for $r = f(\theta)$, the formula for the area of a polar region is: $A = \int\limits_a^b\! \frac{1}{2} r^2 d\theta$ The formula to find the arc length for $r = f(\theta)$ and assuming $r$ is continuous is: $L = \int\limits_a^b\! \sqrt{r^2 + {\bigg(\frac{dr}{d\theta}\bigg)} ^2}$ $d\theta$ # Why It's Interesting $r = \sin^2(1.2\theta) + \cos^3(6\theta)$ The disc phyllotaxis of a sunflower.[3] Polar coordinates are often used in navigation, such as aircrafts. They are also used to plot gravitational fields and point sources. Furthermore, polar patterns are seen in the directionality of microphones, which is the direction at which the microphone picks up sound. A well-known pattern is the Cardioid. Archimedes' spiral can be used for compass and straightedge division of an angle into $n$ parts and circle squaring. [4] Fermat's spiral is a Archimedean spiral that is observed in nature. The pattern happens to appear in the mesh of mature disc phyllotaxis. Archimedean spirals can be seen in patterns of solar wind and Catherine's wheel. As you can see, these equations can create interesting curves and patterns. More complicated patterns can be created with more complicated equations, like the image on the right. Since intriguing patterns can be expressed mathematically, like these curves, they are often used for art and design. ## Possible Future work • More details can be written about the different curves, maybe they can get their own pages. • Applets can be made to draw these different curves, like the one on the page for roses. # About the Creator of this Image Polar Coordinates Cardioid Source code: Rose graphing applet # References Wolfram MathWorld: Rose, Limacon, Archimedean Spiral Wikipedia: Polar Coordinate System, Archimedean Spiral, Fermat's Spiral 1. Weisstein, Eric W. (2011). http://mathworld.wolfram.com/Limacon.html. Wolfram:MathWorld. 2. 2.0 2.1 Stewert, James. (2009). Calculus Early Transcendentals. Ohio:Cengage Learning. 3. Lauer, Christoph. http://www.christoph-lauer.de/Homepage/Blog/Eintrage/2009/10/6_Fermats_Spiral_Suflower_Generator.html. Christoph Lauer's Blog. 4. Weisstein, Eric W. (2011). http://mathworld.wolfram.com/ArchimedesSpiral.html. Wolfram:MathWorld.
# How to Find the Missing Number in an Equation Save In algebra, finding the missing number in an equation is known as “solving for x.” Because we don’t know what the missing number is, we represent it using the letter “x.” The letter "x" is also known as a variable. Equations can have one or more missing variables, depending how complex they are. Finding out a missing number in a simple equation is not too difficult, though it may seem confusing at first. ## Equations Involving Addition and Subtraction • Write out the equation and represent the unknown number with the letter "x." For instance: x+3=5. • Isolate the "x" to one side of the equation. To do that, you must get rid of the 3. Since you can’t just wipe it out, you will have to move it to the other side of the equation, which means the other side of the equal sign. When you move a number from one side of the equation to the other, you must change its sign. In the example, the 3 is positive on the left side of the equation. When you move it to the other side of the equation, it becomes negative, and the equation will look like this: x=5-3. Were this a subtraction problem (x-3=5), you would change the negative 3 to a positive when moving it the other side of the equation and the result would be: x=5+3. • Carry out the final step; in the original equation, that would mean subtracting 3 from 5. The result would give you x=2. You have solved the equation for x and found the unknown number. ## Equations Involving Multiplication and Division • Solve the equation 2x=6 by first isolating the x to the left side of the equation. Since division is the opposite of multiplication, you will need to divide 6 by 2. The new equation will look like this: x=6/2. You now know that x=3. • Isolate the "x" in a division problem by multiplying it by the number on the other side of the equation. Remember, division is the opposite of multiplication. Therefore, the equation x/2=6 would be rewritten as x=6(2). • Carry out the final step; multiply 6 by 2 and you have solved the equation for x. The result is x=12. You have found the unknown number in an equation. ## References • Photo Credit BananaStock/BananaStock/Getty Images Promoted By Zergnet ## Related Searches Check It Out ### Can You Take Advantage Of Student Loan Forgiveness? M Is DIY in your DNA? Become part of our maker community.
# Variable in Math – Definition with Examples Home » Math Vocabulary » Variable in Math – Definition with Examples ## What is Variables? In real-life there are things that remain constant like your date of birth. However, there are things that vary with time and place like temperature, age, height etc. Since these quantities change they are may be called variables. In algebra, a symbol (usually a letter) standing in for an unknown numerical value in an equation or an algebraic expression. In simple words, a variable is a quantity that can be changed and is not fixed. Variables are essential as they form a major algebra component. We usually use “x” and “y” to express an unknown integer. However, it isn’t necessary, and we can use any letter. Example- Let us take an example of the algebraic expression 2x + 6. Here, x is a variable and can take any value. If x = 1, the value of this algebraic expression will be 2(1) + 6 i.e. 8 and if x = 2, the value of the algebraic expression changes to 10. Hence, we can say that the value of the algebraic expression varies as the x varies. Now let us consider an equation 2x + 6 = 12. The variable x can take any value in an equation also. It may or may not satisfy the equation. If it does, it is called the solution of the equation. Here, x = 3 makes the equation true and is called the solution of this equation. ## Different Types of Variables There are two types of variables: Dependent Variables and Independent variables. ## Dependent Variables The dependent variable is a variable whose value is determined by the quantity another variable takes. For instance, in the equation y = 2x + 3, x can take any value, like 1, 2, 3. However, the value of y will depend on the value of x. So, if x = 1, y will become 5, and if x = 2, y will become 7, and so on. Therefore, y is called the dependent variable and x is called the independent variable. ## Independent Variables An independent variable in an algebraic equation is one whose values are unaffected by changes. If an algebraic equation has two variables, x, and y, and each value of x is related to any other value of y, then x is an independent variable, and y is a dependent variable. For instance, in the equation y = 2x, x can take any value. Hence, it is the independent variable in the equation. Conclusion Variables are a very important concept if you want to understand algebra. At SplashLearn, you will come across various resources, like games and worksheets, that will help you understand the concept of variables. The solved examples and practice problems give a step-by-step understanding, making it easier to solve sums on your own. ## Solved Examples: 1. Find one value of x that satisfies the equation 6x + 4 = 22 and one value that does not. Solution: x = 0 does not satisfy the equation as LHS is not equal to RHS. x = 3 satisfy the equation as LHS is equal to RHS. So, x = 3 is the solution for the given equation. 1. Is h = 3, the solution of equation 7h − 2 = 12? Solution: 7(3) − 2 = 21 − 2 = 19 19 $\neq$ 12 So, h = 3 is not a solution of a given equation. 1. What will be the value of the expression 3x + 2 when x = 4? Solution: 3(4) + 2 = 12 + 2 = 14
# What is the area of the largest isosceles triangle that can be inscribed in a circle of radius 4? Feb 25, 2016 $12 \sqrt{3} \cong 20.784$ #### Explanation: One could start by saying that the isosceles triangle with largest area inscribed in a triangle is also an equilateral triangle. However if you need a formal demonstration of this statement read the first part of this explanation. Suppose an isosceles ${\triangle}_{A B C}$ inscribed in a circle with center in $D$ and radius $r$, like the figure below. We can obtain the side $a$ in function of $r$ and $\alpha$ in this way (Law of Sines applied to ${\triangle}_{B C D}$): $\frac{a}{\sin} \left(2 \alpha\right) = \frac{r}{\sin} \beta$ Since $2 \alpha + \beta + \beta = {180}^{\circ}$ => $\beta = {90}^{\circ} - \alpha$ => $a = r \sin \frac{2 \alpha}{\sin} \left({90}^{\circ} - \alpha\right) = 2 r \frac{\sin \alpha \cdot \cancel{\cos \alpha}}{\cancel{\cos \alpha}}$ => $a = 2 r \sin \alpha = 4 r \sin \left(\frac{\alpha}{2}\right) \cos \left(\frac{\alpha}{2}\right)$ We can obtain the height $h$ in function of $r$ and $\alpha$ in this way: $\tan \left(\frac{\alpha}{2}\right) = \frac{\frac{a}{2}}{h}$ => $h = \frac{a}{2 \tan \left(\frac{\alpha}{2}\right)}$ Replacing $a$ for its value in function of $r$ and $\alpha$: $h = \frac{4 r \cancel{\sin \left(\frac{\alpha}{2}\right)} \cos \left(\frac{\alpha}{2}\right)}{2} \cdot \cos \frac{\frac{\alpha}{2}}{\cancel{\sin \left(\frac{\alpha}{2}\right)}}$ => $h = 2 r {\cos}^{2} \left(\frac{\alpha}{2}\right)$ Then we can obtain the area of the triangle in function of $r$ and $\alpha$: ${S}_{{\triangle}_{A B C}} = \frac{\text{base"*"height}}{2}$ ${S}_{{\triangle}_{A B C}} = \frac{\left(4 r \sin \left(\frac{\alpha}{2}\right) \cos \left(\frac{\alpha}{2}\right)\right) \left(\cancel{2} r {\cos}^{2} \left(\frac{\alpha}{2}\right)\right)}{\cancel{2}}$ ${S}_{{\triangle}_{A B C}} = 4 {r}^{2} \sin \left(\frac{\alpha}{2}\right) {\cos}^{3} \left(\frac{\alpha}{2}\right)$ Since in this problem $r$ is constant, we need to find the derivative relatively to $\alpha$ of ${S}_{{\triangle}_{A B C}}$ and equal it to zero to find the maximum or minimum of the area of the triangle. So $\frac{d}{\mathrm{da} l p h a} \left({S}_{{\triangle}_{A B C}}\right) = 4 {r}^{2} \cos \left(\frac{\alpha}{2}\right) \cdot \left(\frac{1}{2}\right) {\cos}^{3} \left(\frac{\alpha}{2}\right) + 4 {r}^{2} \sin \left(\frac{\alpha}{2}\right) \cdot 3 {\cos}^{2} \left(\frac{\alpha}{2}\right) \left(- \sin \left(\frac{\alpha}{2}\right)\right) \left(\frac{1}{2}\right)$ $= 2 {r}^{2} {\cos}^{4} \left(\frac{\alpha}{2}\right) - 6 {r}^{2} {\sin}^{2} \left(\frac{\alpha}{2}\right) {\cos}^{2} \left(\frac{\alpha}{2}\right)$ $= 2 {r}^{2} {\cos}^{2} \left(\frac{\alpha}{2}\right) \left({\cos}^{2} \left(\frac{\alpha}{2}\right) - 3 {\sin}^{2} \left(\frac{\alpha}{2}\right)\right)$ Equating the derivative to zero we get: ${\cos}^{2} \left(\frac{\alpha}{2}\right) = 0$ => $\cos \left(\frac{\alpha}{2}\right) = 0$ => $\frac{\alpha}{2} = {90}^{\circ}$ => $\alpha = {180}^{\circ}$ (this is the point of minimum area) And ${\cos}^{2} \left(\frac{\alpha}{2}\right) - 3 {\sin}^{2} \left(\frac{\alpha}{2}\right) = 0$ => $3 {\sin}^{2} \left(\frac{\alpha}{2}\right) = {\cos}^{2} \left(\frac{\alpha}{2}\right)$ => $\tan \left(\frac{\alpha}{2}\right) = \frac{1}{\sqrt{3}}$ => $\frac{\alpha}{2} = {30}^{\circ}$ => $\alpha = {60}^{\circ}$ (this is the point of maximum area) $\to$ But if $\alpha = {60}^{\circ}$ this means that $A \hat{B} C = A \hat{C} B = {60}^{\circ}$ and the isosceles triangle of maximum area is also an equilateral triangle. Area of the triangle of maximum area ${S}_{{\triangle}_{A B C}} = 4 \cdot {4}^{2} \sin {30}^{\circ} \cdot {\cos}^{3} {30}^{\circ} = 4 \cdot 16 \cdot \left(\frac{1}{2}\right) {\left(\frac{\sqrt{3}}{2}\right)}^{3} = 32 \cdot \frac{3 \sqrt{3}}{8} = 12 \sqrt{3} \cong 20.784$

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