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Lesson 6 # Lesson 6 - GCD(3537 8739-2 3537 = GCD(3537 8739-7074 =... This preview shows pages 1–2. Sign up to view the full content. Greatest Common Divisor This algorithm has existed for the last 2000 years. When two things have GCD equal to 1, we say that they are relatively prime. Algorithm steps: 1. Write out GCD( a, b ) where a is bigger than b. 2. Divide a by b, getting remainder r. 3. Now the problem is to ﬁnd GCD( b, r ). 4. Repeat steps 2 and 3 until r is 1 or 0. 5. If r is 1, then 1 is the greatest common divisor, and if r is 0 than the previous remainder is the greatest common divisor. NOTE: For poly- nomials, the algorithm stops for any constant remainder. Also, we can always rewrite any polynomial by multiplying it by any non-zero number. Examples Example 1: Find the greatest common divisor of 29753 and 8739. GCD(29754 , 8739) = GCD(8739 , 29754 - 3 * 8739) = GCD(8739 , 29754 - 26217) = GCD(8739 , 3537) = This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: GCD(3537 , 8739-2 * 3537) = GCD(3537 , 8739-7074) = GCD(3537 , 1665) = GCD(1665 , 3537-2 * 1665) = GCD(1665 , 3537-3330) = GCD(1665 , 207) = GCD(202 , 1665-8 * 207) = GCD(202 , 1665-1656) = GCD(207 , 9) = GCD(9 , 207-23 * 9) = GCD(9 , 0) = 9 So the GCD of 29753 and 8739 is 9. Example 2: Find the greatest common divisor of-2 x 2 + 5 x-12 x + 2 x + 1 and x 2-3 x + 5 GCD(-2 x 2 + 5 x-12 x + 2 x + 1 , x 2-3 x + 5) 3-5-2 5-12 2 1 ↓-6-3-15 ↓ 10 5 25-2-1-5-8 26 GCD( x 2-3 x + 5 ,-8 x + 26) GCD( x 2-3 x + 5 , x-13 / 4) 13 / 4 1-3 5 ↓ 13 / 4 13 / 16 1 1 / 4 93 / 16 GCD( x-13 / 4 , 93 / 16) Since 93/16 is a constant, we say that the GCD is 1. So the two polyno-mials are relatively prime.... View Full Document ## This note was uploaded on 03/22/2010 for the course MATH 1104 taught by Professor Unknown during the Fall '10 term at Carleton. ### Page1 / 2 Lesson 6 - GCD(3537 8739-2 3537 = GCD(3537 8739-7074 =... This preview shows document pages 1 - 2. Sign up to view the full document. View Full Document Ask a homework question - tutors are online
Ignore the bases, and simply set the exponents equal to each other $$x + 1 = 9$$ Step 2 Get help with your Exponential function homework. See the chapter on Exponential and Logarithmic Functions if you need a refresher on exponential functions before starting this section.] Finish solving the problem by subtracting 7 from each side and then dividing each side by 3. Questions on Logarithm and exponential with solutions, at the bottom of the page, are presented with detailed explanations.. Southern MD's Original Stone Fabricator Serving the DMV Area for Over 30 Years We need to make the bases equal before attempting to solve for .Since we can rewrite our equation as Remember: the exponent rule . If we have an exponential function with some base b, we have the following derivative: (d(b^u))/(dx)=b^u ln b(du)/(dx) [These formulas are derived using first principles concepts. answer as appropriate, these answers will use 6 decima l places. Solve the equation (1/2) 2x + 1 = 1 Solve x y m = y x 3 for m.; Given: log 8 (5) = b. Whenever an exponential function is decreasing, this is often referred to as exponential decay. In an exponential function, the variable is in the exponent and the base is a positive constant (other than the Other examples of exponential functions include: $$y=3^x$$ $$f(x)=4.5^x$$ $$y=2^{x+1}$$ The general exponential function looks like this: $$\large y=b^x$$, where the base b is any positive constant. Solve: $$4^{x+1} = 4^9$$ Step 1. Q. The base b could be 1, but remember that 1 to any power is just 1, so it's a particularly boring exponential function! The amount of ants in a colony, f, that is decaying can be modeled by f(x) = 800(.87) x, where x is the number of days since the decay started.Suppose f(20) = 49. Exponential Functions We have already discussed power functions, such as ( )= 3 ( )=5 4 In a power function the base is the variable and the exponent is a real number. Access the answers to hundreds of Exponential function questions that are explained in a … We need to be very careful with the evaluation of exponential functions. https://www.onlinemathlearning.com/exponential-functions.html Now that our bases are equal, we can set the exponents equal to each other and solve for . Therefore, the solution to the problem 5 3x + 7 = 311 is x ≈ –1.144555. Solving Exponential Equations with Different Bases The concepts of logarithm and exponential are used throughout mathematics. Exponential functions are used to model relationships with exponential growth or decay. This lesson covers exponential functions. Exponential Function. In more general terms, we have an exponential function, in which a constant base is raised to a variable exponent.To differentiate between linear and exponential functions, let’s consider two companies, A and B. 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# using the difference formula • Nov 6th 2009, 10:28 PM maryanna91 using the difference formula So Im supposed to use the difference formula to figure out cos(2x). This is what I have so far: cos(2x)=cos(3x-x) cos(3x-x)=(cos3x)(cosx)+(sin3x)(sinx) Im just not sure how to finish solving! (my math homework is done online so it's supposed to be as simplified as possible). • Nov 6th 2009, 10:33 PM Gusbob Are you sure you can use only the difference formula? Most people derive it with the sum formula using cos(2x) = cos(x+x) • Nov 6th 2009, 10:36 PM maryanna91 yep we're supposed to use the difference formula. • Nov 6th 2009, 10:49 PM Gusbob In that case you will need these identities: $\sin(3x) = 3\sin(x) - 4\sin^3(x)$ $\cos(3x) = 4\cos^3(x) - 3\cos(x)$ • Nov 6th 2009, 11:05 PM maryanna91 Hmmm...tried plugging them in and I'm still still stuck (Headbang). Would you mind showing me what you'd do after they've been plugged in? • Nov 6th 2009, 11:56 PM Gusbob $\cos(3x)\cos(x) + \sin(3x)\sin(x)$ = $[4\cos^3(x) - 3\cos(x)] \cos(x) + [3\sin(x) - 4\sin^3(x)] \sin(x)$ = $4\cos^4(x) - 3\cos^2(x) + 3\sin^2(x) - 4\sin^4(x)$ = $4[\cos^4(x) - \sin^4(x)] - 3[ \cos^2(x) - \sin^2(x)]$ Now notice that $\cos^4(x) - \sin^4(x)$ can be written as a difference of two squares to $(\cos^2(x) + \sin^2(x) ) (\cos^2(x) - \sin^2(x))$. Do you see where to go from here? • Nov 7th 2009, 12:16 AM maryanna91 yep got it!!! thanks sooo much for all the help!
ENCYCLOPEDIA 4U .com Web Encyclopedia4u.com # Series (mathematics) In mathematics, a series is a sum of a sequence of terms. Examples of simple series include arithmetic series which is a sum of a arithmetic progression which can be written as: and geometric series which is a sum of a geometric progression which can be written as: An infinite series is a sum of infinitely many terms. Such a sum can have a finite value; if it has, it is said to converge; if it does not, it is said to diverge. The fact that infinite series can converge resolves several of Zeno's paradoxes. The simplest convergent infinite series is perhaps It is possible to "visualize" its convergence on the real number line: we can imagine a line of length 2, with successive segments marked off of lengths 1, 1/2, 14, etc. There is always room to mark the next segment, because the amount of line remaining is always the same as the last segment marked: when we have marked off 1/2, we still have a piece of length 1/2 unmarked, so we can certainly mark the next 1/4. This argument does not prove that the sum is equal to 2, but it does prove that it is at most 2 -- in other words, the series has an upper bound. This series is a geometric series and mathematicians usually write it as: Formally, if an infinite series is given with real (or complex) numbers an, we say that the series converges towards S or that its value is S if the limit exists and is equal to S. If there is no such number, then the series is said to diverge. Table of contents 1 Some types of infinite series 2 Convergence criteria 3 Examples 4 Absolute convergence 5 Power series 6 Generalizations ### Some types of infinite series • A geometric series is one where each successive term is produced by multiplying the previous term by a constant number. Example: 1 + 1/2 + 1/4 + 1/8 + 1/16... • The harmonic series is the series 1 + 1/2 + 1/3 + 1/4 + 1/5... • An alternating series is a series where terms alternate signs. Example: 1 - 1/2 + 1/3 + 1/4 - 1/5... ### Convergence criteria 1) If the series ∑ an converges, then the sequence (an) converges to 0 for n→∞; the converse is in general not true. 2) If all the numbers an are positive and ∑ bn is a convergent series such that anbn for all n, then ∑ an converges as well. Conversely, if all the bn are positive, anbn for all n and ∑ bn diverges, then ∑ an diverges as well. 3) If the an are positive and there exists a constant C < 1 such that an+1/anC, then ∑ an converges. 4) If the an are positive and there exists a constant C < 1 such that (an)1/nC, then ∑ an converges. 5) If f(x) is a positive monotone decreasing function defined on the interval [1, ∞) with f(n) = an for all n, then ∑ an converges if and only if the integral1 f(x) dx exists. 6) A series of the form ∑ (-1)n an (with an ≥ 0) is called alternating. Such a series converges if the sequence an is monotone decreasing and converges towards 0. The converse is in general not true. ### Examples The series converges if r > 1 and diverges for r ≤ 1, which can be shown with the integral criterion 5) from above. As a function of r, the sum of this series is Riemann's zeta function. converges if and only if \|z\| < 1. The telescoping series converges if the sequence bn converges to a limit L as n goes to infinity. The value of the series is then b1 - L. ### Absolute convergence The sum is said to converge absolutely if the series of absolute values converges. In this case, the original series, and all reorderings of it, converge, and converge towards the same sum. If a series converges, but not absolutely, then one can always find a reordering of the terms so that the reordered series diverges. Even more: if the an are real and S is any real number, one can find a reordering so that the reordered series converges with limit S (Riemann). ### Power series Several important functions can be represented as Taylor series; these are infinite series involving powers of the independent variable and are also called power series. Historically, mathematicians such as Leonhard Euler operated liberally with infinite series, even if they were not convergent. When calculus was put on a sound and correct foundation in the nineteenth century, rigorous proofs of the convergence of series were always required. However, the formal operation with non-convergent series has been retained in rings of formal power series which are studied in abstract algebra. Formal power series are also used in combinatorics to describe and study sequences that are otherwise difficult to handle; this is the method of generating functions. ### Generalizations The notion of series can be defined in every abelian topological group; the most commonly encountered case is that of series in a Banach space. Content on this web site is provided for informational purposes only. We accept no responsibility for any loss, injury or inconvenience sustained by any person resulting from information published on this site. We encourage you to verify any critical information with the relevant authorities.
Lesson 4 Dilating Lines and Angles • Let’s dilate lines and angles. 4.1: Angle Articulation Triangle $$A’B’C’$$ is a dilation of triangle $$ABC$$ using center $$P$$ and scale factor 2. 1. What do you think is true about the angles in $$A’B’C’$$ compared to the angles in $$ABC$$? 2. Use the tools available to figure out if what you thought was true is definitely true for these triangles. 3. Do you think it would be true for angles in any dilation? 4.2: Dilating Lines 1. Dilate point $$A$$ using center $$C$$ and scale factor $$\frac{3}{4}$$. 2. Dilate point $$B$$ using center $$C$$ and scale factor $$\frac13$$. 3. Dilate point $$D$$ using center $$C$$ and scale factor $$\frac32$$. 4. Dilate line $$CE$$ using center $$C$$ and scale factor 2. 5. What happens when the center of dilation is on a line and then you dilate the line? • $$X$$ is the midpoint of $$AB$$. • $$B'$$ is the image of $$B$$ after being dilated by a scale factor of 0.5 using center $$C$$. • $$A'$$ is the image of $$A$$ after being dilated by a scale factor of 0.5 using center $$C$$. Call the intersection of $$CX$$ and $$A'B'$$ point $$X'$$. Is point $$X'$$ a dilation of point $$X$$? Explain or show your reasoning. 4.3: Proof in Parallel Jada dilated triangle $$ABC$$ using center $$P$$ and scale factor 2. 1. Jada claims that all the segments in $$ABC$$ are parallel to the corresponding segments in $$A’B’C’$$. Write Jada's claim as a conjecture. 3. In Jada’s diagram the scale factor was greater than one. Would your proof have to change if the scale factor was less than one? Summary When one figure is a dilation of the other, we know that corresponding side lengths of the original figure and dilated image are in the same proportion, and are all related by the same scale factor, $$k$$. What is the relationship of corresponding angles in the original figure and dilated image? For example, if triangle $$ABC$$ is dilated using center $$P$$ with scale factor 2, we can verify experimentally that each angle in triangle $$ABC$$ is congruent to its corresponding angle in triangle $$A’B’C’$$. $$\angle A \cong \angle A’, \angle B \cong \angle B’, \angle C \cong \angle C’$$. What is the image of a line not passing through the center of dilation? For example, what will be the image of line $$BC$$ when it is dilated with center $$P$$ and scale factor 2? We can use congruent corresponding angles to show that line $$BC$$ is taken to parallel line $$B’C'$$. What is the image of a line passing through the center of dilation? For example, what will be the image of line $$GH$$ when it is dilated with center $$C$$ and scale factor $$\frac12$$? When line $$GH$$ is dilated with center $$C$$ and scale factor $$\frac12$$, line $$GH$$ is unchanged, because dilations take points on a line through the center of a dilation to points on the same line, by definition. So, a dilation takes a line not passing through the center of the dilation to a parallel line, and leaves a line passing through the center unchanged. Glossary Entries • dilation A dilation with center $$P$$ and positive scale factor $$k$$ takes a point $$A$$ along the ray $$PA$$ to another point whose distance is $$k$$ times farther away from $$P$$ than $$A$$ is. Triangle $$A'B'C'$$ is the result of applying a dilation with center $$P$$ and scale factor 3 to triangle $$ABC$$. • scale factor The factor by which every length in an original figure is increased or decreased when you make a scaled copy. For example, if you draw a copy of a figure in which every length is magnified by 2, then you have a scaled copy with a scale factor of 2.
Ncert Science Class 10 Solutions Chapter 11 Ncert Science Class 10 Solutions Chapter 11 Electricity Welcome to NCTB Solutions. Here with this post we are going to help 10th class students for the Solutions of NCERT Class 10 Science Book Chapter 11, Electricity. Here students can easily find step by step solutions of all the questions in Electricity. Also our Expert Science Teacher’s solved all the problems with easily understandable language with proper guidance so that all the students can understand easily. Here in this post students will get chapter 11 solutions. Here all Question Answer are based on NCERT latest syllabus. Electricity Exercise question Solutions : (1) A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is The correct option is – (d) 25 As given R is resistance cut in to five equal parts, So the each resistance Is R/5 Now if resistance is connected in parallel. 1/Rsum = 1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5 Now putting each value of resistance, 1/Rsum = 5/R1 + 5/R2 + 5/R3 + 5/R4 + 5/R5 1/Rsum = 25/R So, Rsum = R/25 And now ratio of R/Rsum = 25 (2) Which of the following terms does not represent electrical power in a circuit? The appropriate answer is option – (b) IR2 As we know P = VI……………..c And From ohm’s law R = V/I – V = RI – I = V/R P = (RI)I P = RI2…………………………..a And also, P = VI Putting value of I = V/R P = V2/R…………………………d (3) An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be The answer is alternative – (d) 25 W Power consumed is given by P = VI P = V2/R We Have Voltage and Power So we can use P = V2/R R = V2/P R = 220 × 220/100 R = 22 × 22 R = 284 ohm Now when use 110 volt P = V2/R P = 110 × 110/484 P = 25 Watt (4) Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be The correct option is – (c) 1 : 4 Heat produced is given by H = Pt H = t × V2/R Now heat produced in series Hseries = t × V2/RS Now heat produced in Parallel Hparallel = t × V2/RP Now removing Common, Ratio = Hseries/Hparallel = 1/RS/1/RP = RP/RS ……eq 1 Now, Rs = R + R Rs = 2R and Rp = 1/1/(r+r) RP = 1/1/2r RP = R/2 Putting value in eq 1 Ratio = R/2/2R Ratio = 1/4 There fore, the answer is 1/4 (5) How is a voltmeter connected in the circuit to measure the potential difference between two points? Voltmeter should be always connected in parallel between the two points whose potential difference has to measure. (6) A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 W m. What will be the length of this wire to make its resistance 10 W? How much does the resistance change if the diameter is doubled? We have formulae, R = ρ 1/a A is area of conductor, A = πr^2 A = π〖0.25〗^2 A = 0.1964 mm2 Now, l = R × A/ρ l = 10 × 0.1964/1.6 × 10–8 l = 122.71 meter Now, if diameter is doubled A = πr^2 A = π〖0.5〗^2 A = 0.7854 mm2 Now, R = ρ l/A R = 1.6 × 10-8 × 122.71/0.7854 R = 2.5 Ohm (7) The values of current (I) flowing through a given resistor of resistance (R), for the corresponding values of potential difference (V) across the resistor are as given below V (volt) 0.5 1 1.5 2 2.5 3 4 5 I (amperes) 0.1 0.2 0.3 0.4 0.5 0.6 0.8 0.9 Plot a graph between current (I) and potential difference (V) and determine the resistance (R) of the resistor. (8) When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor. As per the given question, V = 12 v I = 2.5 mA R = ? R = V/I R = 12/2.5 × 10-3 R = 4.8 × 103 ohm R = 4.8 KΩ Therefore, the value of the resistance of the resistor will be 4.8 KΩ (9) A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor? V = 9v R = 0.2,0.3,0.4,0.5,12 Ω Resister are connected in series, Rsum = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω R = V/I I = V/R I = 9/13.4 I = 0.671 Ampere. Hence, 0.671 Ampere current will flow through the 12 Ω resistor. (10) How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? V = 220 v I = 5 A R = 176 Ω Actual resistance Ra = V/I = 220/5 = 44 Now these resister are connected in parallel so, No of resister = R/Ra = 176/44 = 4 Therefore, the number of resister will be 4. (11) Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω Lets take first 9 Ω, If we connect resister in series it will be 6 + 6 + 6 = 18 Ω And if we connect resister in parallel it will be, 1/R = 1/6 + 1/6 + 1/6 = 3/6 R = 2 So we have to make combination of series and parallel connection, one series and two parallel would give, R = 6 + 1/1/6 + 1/6 = 6 + 3 = 9 Ω Now lets take second case, Two series and one parallel resister would give, Two series resister = 6 + 6 = 12 Now connect one parallel resister to it, R = 1/1/12 + 1/6 = 72/18 = 4 Ω (12) Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A? As per the given question, V = 220 v P = 10 W I = 5 A R = V/I R = 220/5 R = 44 Ω this is maximum resistance we can design, Now resistance of each bulb P = V2/R R = 220 × 220/10 R = 4840 Ω Now no of bulb we can connect parallel can be calculated be 1/R = 1/R1 + 1/R2 up to N numbers 1/R = 1/R × N 1/44 = 1/4840 × N N = 4840/44 N = 110 Thus, we can connect maximum 110 bulbs. (13) A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases? As per the given data, V = 220 v R = 24 Ω (1) When using separately, I = V/R I = 220/24 I = 9.166 A (2) when using in series, Resistance in series R = 24 + 24 = 48 I = V/R I = 220/48 I = 4.58 A (3) when using in parallel, Resistance in parallel, 1/R = 1/1/24 + 1/24 1/R = 1/12 R = 12 Ω I = V/R I = 220/12 I = 18.33 A (14) Compare the power used in the 2 Ωresistor in each of the following circuits : (i) a 6 V battery in series with 1 Ωand 2 Ωresistors, (ii) a 4 V battery in parallel with 12 Ωand 2 Ω resistors Lets take first case, V = 6 v R = 1 + 2 R = 3 Ω Now, I = V/R I = 6/3 I = 2 The power across 2 Ω resister can be calculated as, P = I2R P = 22 × 2 P = 8 W Lets take second case, V = 4 v And the resister are connected parallel so the voltage across two resister is same, So the power across 2 Ω resister can be calculated as, P = V2/R P = 42/2 P = 16/2 P = 8 W (15) Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V? According to the given data, V = 220 v P1 = 100 W P2 = 60 W Since both bulbs are connected parallelly Now , P = V × I I = P/V I = 100/220 + 60/220 I = 160/220 I = 0.727 Ampere (10) Which uses more energy, a 250 W TV set in 1 hr or a 1200 W toaster in 10 minutes? According to the question, P1 = 250 W T1 = 1 Hour = 3600 seconds P2 = 1200 W T2 = 10 Minute = 600 seconds Now, H = P × t H is the energy consumed by appliances. H1 = 250 × 3600 H1 = 900 × 103 joule H2 = 1200 × 600 H2 = 720 × 103 joule From comparing both value the TV set uses more energy. (17) An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater. As per the question, I = 15 A R = 8 Ω T = 2 Hour Now Applying Formula, P = I2R P = (15)2 × 8 P = 1800 W or 1800 J/s The rate of the heater is 1800 J/s. (18) Explain the following. (a) Why is the tungsten used almost exclusively for filament of electric lamps? = The tungsten does not burn or break by electricity because, It has high Melting point and resistance. That’s why tungsten is used almost exclusively. (b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal? = Pure metal have low resistance so other metal is used to form alloy of high resistance, Which can generate high heat when used as conductor. (c) Why is the series arrangement not used for domestic circuits? = Because in series arrangement addition of resistance take place, so the voltage gets dropped in next device and device may burn due to handling of large current. (d) How does the resistance of a wire vary with its area of cross-section? = Resistance is directly proportional to the cross sectional area of conductor, that’s why the resistance increases when the area of conductor increased. (e) Why are copper and aluminium wires usually employed for electricity transmission? = The conductivity of both metal is high and have low resistance so the copper and aluminum is usually employed for electricity transmission. More Solutions : 👉 Light – Reflection and Refraction 👉 Our Environment Updated: June 30, 2023 — 7:23 am
Search IntMath Close 450+ Math Lessons written by Math Professors and Teachers 5 Million+ Students Helped Each Year 1200+ Articles Written by Math Educators and Enthusiasts Simplifying and Teaching Math for Over 23 Years # 6. Products and Quotients of Complex Numbers by M. Bourne When performing addition and subtraction of complex numbers, use rectangular form. (This is because we just add real parts then add imaginary parts; or subtract real parts, subtract imaginary parts.) When performing multiplication or finding powers and roots of complex numbers, use polar and exponential forms. (This is because it is a lot easier than using rectangular form.) We start with an example using exponential form, and then generalise it for polar and rectangular forms. ### Example 1 Find (3e4j)(2e1.7j), where j=sqrt(-1). Here we are multiplying two complex numbers in exponential form. It is no different to multiplying whenever indices are involved. (3e4j)(2e1.7j) = (3)(2)e4j+1.7j = 6e5.7j ## Multiplying Complex Numbers in Polar Form We can generalise the example we just did, as follows: (r_1\ e^(\ theta_1j))(r_2\ e^(\ theta_2j))=r_1r_2\ e^((theta_1+\ theta_2)j From this, we can develop a formula for multiplying using polar form: r_1(cos\ theta_1+j\ sin\ theta_1) xxr_2(cos\ theta_2+j\ sin\ theta_2) =r_1r_2(cos[theta_1+theta_2] {:+j\ sin[theta_1+theta_2]) or with equivalent meaning: r_1/_theta_1xxr_2/_theta_2=r_1r_2/_[theta_1+theta_2] In words, all this confusing-looking algebra simply means... To multiply complex numbers in polar form, Multiply the r parts Continues below ### Example 2 Find 3(cos 120° + j sin 120°) × 5(cos 45° + j sin 45°) 3(cos 120° + j sin 120°) × 5(cos 45° + j sin 45°) = (3)(5)(cos(120° + 45°) +j sin(120° + 45°) = 15 [cos(165°) +j sin(165°)] In this example, the r parts are 3 and 5, so we multiplied them. The angle parts are 120° and 45°, and we added them. ## Division As we did before, we do an example in exponential form first, then generalise it for polar form. ### Example in Exponential Form: 8\ e^(\ 3.6j)-:2\ e^(\ 1.2j) =4\ e^(\ 3.6j-1.2j) =4\ e^(\ 2.4j) [We divided the number parts, and subtracted the indices, just using normal algebra.] From the above example, we can conclude the following: (r_1(costheta_1+j\ sintheta_1))/(r_2(costheta_2+j\ sintheta_2)) =r_1/r_2(cos[theta_1-theta_2]+j\ sin[theta_1-theta_2]) or (r_1/_theta_1)/(r_2/_theta_2)=r_1/r_2/_[theta_1-theta_2] In words, this simply means... To divide complex numbers in polar form, Divide the r parts Subtract the angle parts ### Example 3 Find (2/_90^"o")/(4/_75^"o") Here, the r parts are 2 and 4, so we divide them, and the angle parts are 90° and 75°, so we subtract them. (2 angle 90^text(o))/(4 angle 75^text(o))=2/4angle(90^text(o)-75^text(o)) =1/2angle15^text(o) ### Example 4 Find (3/_20^"o")/(9/_60^"o") (3 angle 20^text(o))/(9 angle 60^text(o)) =3/9angle(20^text(o)-60^text(o))=1/3angle-40^text(o) In this section, we normally leave our answer as a positive angle, so since -40° = 360° − 40° = 320°, we write : (3 angle 20^text(o))/(9 angle 60^text(o))=1/3angle320^text(o) ### Example 5 Find (8j)/(7+2j) using polar form. First, we express 8j and 7 + 2j in polar form. 8j = 8\ ∠\ 90° and 7 + 2j = 7.28\ ∠\ 15.95° So (8j)/(7+2j) =(8angle90^text(o))/(7.28 angle 15.95^text(o)) =1.10angle74.05^text(o) ### Exercises: 1. Evaluate: (0.5 ∠ 140^"o")(6 ∠ 110^"o") 0.5xx6=3 and 140^"o"+110^"o"=250^"o" So (0.5 ∠ 140^"o")(6 ∠ 110^"o")=3 ∠ 250^"o" 2. Evaluate: (12/_320^"o")/(5/_210^"o") 12/5=2.4; and 320^"o"-120^"o"=110^"o" So (12/_320^"o")/(5/_210^"o") =12/5angle(320^text(o)-210^text(o)) =2.4angle110^text(o) 3. (i) Evaluate the following by first converting numerator and denominator into polar form. (ii) Then check your answer by multiplying numerator and denominator by the conjugate of the denominator. (-2+5j)/(-1-j) #### Solution, Part (i) (-2+5j)/(-1-j) = (5.39 angle 112^text(o))/(1.41 angle 225^text(o)) =5.39/1.41angle(112^text(o)-225^text(o)) =3.82angle247^text(o) =-1.49-3.52j Note: 112^"o" - 225^"o" = -113^"o" is equivalent to positive 247^"o". #### Part (ii) CHECK: (-2+5j)/(-1-j) =((-2+5j)(-1+j))/((-1-j)(-1+j)) =(-3-7j)/2 =-1.5-3.5j Here's an explanation of what happened in the expansion of terms in the above answer. The top of the fraction (the numerator) is: (−2 + 5j)(−1 + j) This gives us −2(−1 + j) + 5j(−1 + j) = 2 − 2j − 5j + 5j2 = 2 − 2j − 5j − 5 = −3 − 7j The bottom part of the fraction is (−1 − j)(−1 + j) = (−1)2 − (j)2 = 1 − (−1) = 1 + 1 = 2 [There is a rounding error in Part (i) since we used decimal approximations throughout. The answer in Part (ii) is exact.]
# Question: What Is One’S Complement In Binary? ## Why 2s complement is so important? Two’s complement allows negative and positive numbers to be added together without any special logic. This means that subtraction and addition of both positive and negative numbers can all be done by the same circuit in the cpu. ## How do you find complement? 1) If A = { 1, 2, 3, 4} and U = { 1, 2, 3, 4, 5, 6, 7, 8} then find A complement ( A’). Complement of set A contains the elements present in universal set but not in set A. Elements are 5, 6, 7, 8. ∴ A complement = A’ = { 5, 6, 7, 8}. ## How do you find complements? A mutually exclusive pair of events are complements to each other. For example: If the desired outcome is heads on a flipped coin, the complement is tails. The Complement Rule states that the sum of the probabilities of an event and its complement must equal 1, or for the event A, P(A) + P(A’) = 1. ## What is 1s and 2s complement? To get 1’s complement of a binary number, simply invert the given number. To get 2’s complement of a binary number, simply invert the given number and add 1 to the least significant bit (LSB) of given result. … end-around-carry-bit addition occurs in 1’s complement arithmetic operations. It added to the LSB of result. ## How do you find the 9’s complement of a binary number? The 9’s complement of a number is calculated by subtracting each digit of the number by 9. For example, suppose we have a number 1423, and we want to find the 9’s complement of the number. For this, we subtract each digit of the number 1423 by 9. So, the 9’s complement of the number 1423 is 9999-1423= 8576. ## What is 2’s complement? It is a system in which the negative numbers are represented by the two’s complement of the absolute value. For example : -9 converts to 11110111 (to 8 bits), which is -9 in two’s complement. … The two’s complement is a method for representing positive and negative integer values in the decimal number system. ## What is 2’s complement with example? To get 2’s complement of binary number is 1’s complement of given number plus 1 to the least significant bit (LSB). For example 2’s complement of binary number 10010 is (01101) + 1 = 01110. ## Which is not a binary number? Which of the following is not a binary number? Explanation: A binary number can have only two possible digits, 0 and 1. In the third option, there is an alphabet E present which makes it an invalid binary number. Alphabets are only allowed in the hexadecimal number system. ## What is R and r1 complement? The r’s complement is also known as Radix complement (r-1)’s complement, is known as Diminished Radix complement. … If the base of the number is 2, then we can find 1’s and 2’s complement of the number. ## Why are complements used in binary arithmetic? Signed-complement forms of binary numbers can use either 1’s complement or 2’s complement. The 1’s complement and the 2’s complement of a binary number are important because they permit the representation of negative numbers. ## How do you do 2’s complement? To get the two’s complement negative notation of an integer, you write out the number in binary. You then invert the digits, and add one to the result. ## What is the use of 2’s complement? Two’s complement is a mathematical operation on binary numbers, and is an example of a radix complement. It is used in computing as a method of signed number representation. ## Why is it called 2s complement? According to Wikipedia, the name itself comes from mathematics and is based on ways of making subtraction simpler when you have limited number places. The system is actually a “radix complement” and since binary is base two, this becomes “two’s complement”. ## How do you convert binary to one’s complement? To get 1’s complement of a binary number, simply invert the given number. For example, 1’s complement of binary number 110010 is 001101. To get 2’s complement of binary number is 1’s complement of given number plus 1 to the least significant bit (LSB)….One’s Complement.Binary number1’s complement1110007 more rows•Feb 21, 2019 ## What is one complement of a binary number? 1’s complement of a binary number is another binary number obtained by toggling all bits in it, i.e., transforming the 0 bit to 1 and the 1 bit to 0. 2’s complement of a binary number is 1 added to the 1’s complement of the binary number. These representations are used for signed numbers.
How To Divide Complex Numbers In Polar Form 2021. To find θ, we have to consider cases. Multiplication and division of complex numbers in polar form. Furthermore, since $$a=2$$ and $$b=3$$, we have $$\tan(\theta)=\dfrac{3}{2}$$. Multiplication and division of complex numbers in polar form. (2.2.4) r e i θ ⋅ s e i β = ( r s) e i ( θ + β). ### Finding Roots Of Complex Numbers In Polar Form. Writing a complex number in polar form involves the following conversion formulas: We use the definition of the complex exponential and some trigonometric identities. = +𝑖 ∈ℂ, for some , ∈ℝ ### Again, To Convert The Resulting Complex Number In Polar Form, We Need To Find The Modulus And Argument Of The Number. This is an advantage of using the polar form. Dividing complex numbers in polar form. To multiply/divide complex numbers in polar form, multiply/divide the two moduli and add/subtract the arguments. ### There Are Several Ways To Represent A Formula For Finding $$N^{Th}$$ Roots Of Complex Numbers In Polar Form. Furthermore, since $$a=2$$ and $$b=3$$, we have $$\tan(\theta)=\dfrac{3}{2}$$. Let 𝑖2=−බ ∴𝑖=√−බ just like how ℝ denotes the real number system, (the set of all real numbers) we use ℂ to denote the set of complex numbers. To find the $$n^{th}$$ root of a complex number in polar form, we use the $$n^{th}$$ root theorem or de moivre’s theorem and raise the complex number to a power with a rational exponent. ### Read More On Complex Numbers: A vector emanating from the zero point can also be used as a pointer. The conjugate of the complex z = a+ib is a−ib. C 1 ⋅ c 2 = r 1 ⋅ r 2 ∠ (θ 1 + θ 2 ). ### Find More Mathematics Widgets In Wolfram\|Alpha. Learn the process of converting complex numbers to polar form from rectangular form, and how de moivre's formula can isolate the power of complex numbers. More specifically, for any t wo complex numbers, $$z_1=r_1(cos(\theta_1)+isin(\theta_1))$$ and $$z_2=r_2(cos(\theta_2)+isin(\theta_2))$$, we have: Complex numbers are often denoted by z. Categories: How to
# Basic Mathematical Operations with Polynomials In Mathematics, a polynomial is an expression that consists of variables, constant, and mathematical operators. The exponents of the variables are positive integers, whereas the constants are real numbers. The polynomial cannot have a negative value exponent, variable in the denominator, variable inside the radical sign. The three primary terminologies used in the polynomials are coefficients, degrees, and the variable. The important process in polynomials is factorization. While factoring polynomials, we can get the zeros of the polynomials. Let us discuss these terms with the help of an example. Consider an example, 4x2 + 10y – 3 The variables in the above-given examples are x and y. The coefficients are the constant values present before the variables. Here, the coefficients are 4, 10, -3. The highest power of the variable in the expression is called the degree. In this case, the degree of the given polynomial is 2. Now, let us have a look at the basic operations of polynomials. The basic operations are: • Subtraction of Polynomials • Multiplication of Polynomials • Division of Polynomials ## Addition and Subtraction of Polynomials In addition and the subtraction of polynomials, first, arrange the like terms, and the like terms are added together. For example, 4x +10y and 6x + 4y are the two polynomials. Now, we have to add polynomials. Now, arrange the like terms together , (4x +10y) + ( 6x + 4y ) Here 4x and 6x are the like terms, 10y and 4y are the like terms = 4x + 6x + 10y+ 4y Now, add the like terms, we get 10x + 14y, which is the result of the given polynomial ## Multiplication In the process of polynomial multiplication, multiply each term in one polynomial by each term in the other polynomial. Now, add those answers, and simplify the answer, if it is needed. Let us take an example (4x+ 2y+1)(2x + 2) are the two polynomials. Follow the steps mentioned above, to get the product. (4x+ 2y+1)(2x + 2) =8x2 + 8x + 4xy + 4y + 2x +2 Now, perform addition operation, we get 8x2 + 4xy +10x + 4y+ 2. ## Division The polynomial long division method is used for dividing polynomials, which is the generalized version of the arithmetic method called the long division method. In this method, the division of one polynomial by another polynomial is performed, which should have the same or the lower degree. For more examples and related concepts, subscribe to BYJU’S YouTube Channel. ### 11,336 thoughts on “Basic Mathematical Operations with Polynomials” 1. 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# SAT II Physics : Motion in Two Dimensions ## Example Questions ### Example Question #1 : Motion In Two Dimensions Sam throws a rock off the edge of a tall building at an angle of from the horizontal. The rock has an initial speed of . How long is the rock in the air? Explanation: We first need to find the vertical component of the velocity. We can plug in the given values for the angle and the initial velocity to find the vertical component. Now we need to solve for the time that the rock travels upward. We can then add the upward travel time to the downward travel time to find the total time in the air. Remember that the vertical velocity at the highest point of a parabola is zero. We can use that to find the time for the rock to travel upward. Now let's find the time for the downward travel. We don't know the final velocity for the rock, but we CAN use the information we have been given to find the height it travels upward. Remember, only tells us the vertical CHANGE. Since the rock started at the top of a building, if it rose an extra , then at its highest point it is above the ground. This means that our will be as it will be traveling down from the highest point. Using this distance, we can find the downward travel time. Add together the time for upward travel and downward travel to find the total flight time. ### Example Question #2 : Motion In Two Dimensions Sam throws a rock off the edge of a tall building at an angle of from the horizontal. The rock has an initial speed of . What is the horizontal distance that the rock travels? Explanation: We first need to find the horizontal component of the initial velocity. We can plug in the given values for the angle and initial velocity and solve. The only force acting on the rock during flight is gravity; there are no forces in the horizontal direction, meaning that the horizontal velocity will remain constant. We can set up a simple equation to find the relationship between distance traveled and the velocity. We know , but now we need to find the time the rock is in the air. We need to solve for the time that the rock travels upward. We can then add the upward travel time to the downward travel time to find the total time in the air. Remember that the vertical velocity at the highest point of a parabola is zero. We can use that to find the time for the rock to travel upward. Now let's find the time for the downward travel. We don't know the final velocity for the rock, but we CAN use the information we have been given to find the height it travels upward. Remember, only tells us the vertical CHANGE. Since the rock started at the top of a building, if it rose an extra , then at its highest point it is above the ground. This means that our will be as it will be traveling down from the highest point. Using this distance, we can find the downward travel time. Add together the time for upward travel and downward travel to find the total flight time. Now that we've finally found our time, we can plug that back into the equation from the beginning of the problem, along with our horizontal velocity, to solve for the final distance. ### Example Question #5 : Mechanics Sam throws a rock off the edge of a tall building at an angle of from the horizontal. The rock has an initial speed of . At what angle to the horizontal will the rock impact the ground? Explanation: The question gives the total initial velocity, but we will need to find the horizontal and vertical components. To find the horizontal velocity we use the equation . We can plug in the given values for the angle and initial velocity to solve. We can find the vertical velocity using the equation . The horizontal velocity will not change during flight because there are no forces in the horizontal direction. The vertical velocity, however, will be affected. We need to solve for the final vertical velocity, then combine the vertical and horizontal vectors to find the total final velocity. We know that the rock is going to travel a net distance of , as that is the distance between where the rock's initial and final positions. We now know the displacement, initial velocity, and acceleration, which will allow us to solve for the final velocity. Because the rock is traveling downward, our velocity will be negative: . Now that we know our final velocities in both the horizontal and vertical directions, we can find the angle created between the two trajectories. The horizontal and vertical velocities can be compared using trigonometry. , Plug in our values and solve for the angle. ### Example Question #4 : Sat Subject Test In Physics If air resistance is negligible, 8 seconds after it is released, what would be the velocity of a stone dropped from a helicopter that has a horizontal velocity of 60 meters per second? Explanation: We are looking for total velocity, which in this case has both a horizontal and vertical component. Because the helicopter is flying horizontally we know We can assume that this takes place near the surface of the earth so We can plug this into the equation: Next we must find the horizontal velocity. Because there are no additional forces, the horizontal velocity is the same as the initial horizontal velocity of the helicopter, so: Next we must use vector addition to add the horizontal and vertical components of velocity. Because this horizontal and vertical velocities are perpendicular the sum will be the hypotenuse of a right triangle:
# Class 12 Maths NCERT Solutions for Chapter 10 Vector Algebra Miscellaneous Exercise ### Vector Algebra Miscellaneous Exercise Solutions 1. Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis. Solution If r is a unit vector in the XY-plane, then r = cosÎ¸i + sinÎ¸j ^ . Here, Î¸ is the angle made by the unit vector with the positive direction of the x-axis. Therefore, for Î¸ = 30° . r = cosÎ¸i + sinÎ¸j ^ = (√3/2)i + (1/2)j ^ Hence, the required unit vector is (√3/2) i + (1/2)j ^. 2. Find the scalar components and magnitude of the vector joining the points P(x1, y1, z1 ) and Q(x2, y2, z2) . Solution The vector joining the points P(x1, y1, z1) and Q(x2, y2, z2) can be obtained by, Hence, the scalar components and the magnitude of the vector joining the given points are respectively {(x2 - x1), (y2 - y1), (z2 - z1)} and 3. A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure. Solution Let O and B be the initial and final positions of the girl respectively. Then, the girl's position can be shown as : Now, we have: By the triangle law of vector addition, we have : Hence, the girl's displacement from her initial point of departure is (-5/2) i ^ + (3√3/2) j ^ 4. If a⃗ = b⃗ + c, then is it true that \|a\| = \|b\| + \|c\| ? Justify your answer. Solution Now, by the triangle law of vector addition, we have a = b⃗ + c It is clearly known that \| a\|, \|b\|, and \|c\| represent the sides of Î”ABC. Also, it is known that the sum of the lengths of any two sides of a triangle is greater than the third side. ∴ \| a\| < \|b\| + \|c\| Hence, it is not true that \|a\| = \|b\| + \|c\| . 5. Find the value of x for which x(i ^+ j ^+ k ^) is a unit vector. Solution 6. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors a = 2i ^+3j ^-k^ and b = i ^- 2j ^+ k^ . Solution We have, a = 2i + 3j ^ - k^ and b = i - 2j ^ + k^ Let c be the resultant of a and b Then, Hence, the vector of magnitude 5 units and parallel to the resultant of vectors a and b is 7. If a = i ^+ j ^+ k^, b = 2i ^- j ^+3k^ and c = i ^- 2j ^+k^ find a unit vector parallel to the vector 2a - b +3c . Solution We have, 8. Show that the points A (1, –2, –8), B (5, 0, –2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC. Solution The given points are A (1, –2, –8), B (5, 0, –2) and C (11, 3, 7) . Thus, the given points A, B, and C are collinear. Now, let point B divide AC in the ratio Î» : 1. Then, we have : On equating the corresponding components, we get : 5(Î» + 1) = 11Î» + 1 ⇒ 5Î» + 5 =11Î» + 1 ⇒ 6Î» = 4 ⇒ Î» = 4/6 = 2/3 Hence, point B divides AC in the ratio 2 : 3. 9. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are P(2a + b) and Q(a - 3b) externally in the ratio 1 : 2. Also, show that P is the mid point of the line segment RQ. Solution It is given that It is given that point R divides a line segment joining two points P and Q externally in the ratio 1 : 2. Then, on using the section formula, we get : Therefore, the position vector of point R is Position vector of the mid - point of RQ = Hence, P is the mid - point of the line segment RQ. 10 . The two adjacent sides of a parallelogram are 2i - 4j ^ + 5k^, and i ^-2j ^-3k^ . Find the unit vector parallel to its diagonal. Also, find its area. Solution Adjacent sides of a parallelogram are given as: a = 2i -4j ^+5k^ and b = i ^-2j ^+3k^ . Then, the diagonal of a parallelogram is given by a + b . ∴ Area of parallelogram ABCD = \| a + b \| Hence, the area of the parallelogram is 11√5 square units. 11. Show that the direction cosines of a vector equally inclined to the axes OX, OY, and OZ are 1/√3, 1/√3, 1/√3 . Solution Let a vector be equally inclined to axes OX, OY, and OZ at angle Î±. Then, the direction cosines of the vector are cos Î±, cos Î± and cos Î±. Now, cos2 Î± + cos2 Î± + cos2 Î± = 1 ⇒ 3cos2 Î± = 1 ⇒ cos Î± = 1/√3 Hence, the direction cosines of the vector which are equally inclined to the axes are 1/√3, 1/√3, 1/√3. 12. Let a = i ^+4j ^+2k^ , b=3i ^-2j ^+7k^ and c = 2i ^- j ^+4k. Find a vector d which is perpendicular to both a and b, and c . d = 15. Solution Let Since d is perpendicular to both a and b, we have : 13. The scalar product of the vector i ^+ j ^+ k^, with a unit vector along the sum of vectors 2i ^+4j ^-5k^ and Î»i ^+2j ^+3k^ is equal to one. Find the value of Î» . Solution ⇒ Î»2 + 4 Î» + 44 = (Î» + 6)2 ⇒ Î»2 + 4Î» + 44 = Î»2 + 12Î» + 36 ⇒ 8Î» = 8 ⇒ Î» = 1 Hence, the value of Î» is 1. 14. If a, b, c are mutually perpendicular vectors of equal magnitudes, show that the vector a + b + c is equally inclined to a, b and c . Solution Since, a , b and c are mutually perpendicular vectors, we have Let vector a+b +c be inclined to a, b and c at angles Î¸1, Î¸2, Î¸3 respectively . Then , we have : Now, as \|a\|+\|b\|+ \|c\|, cos Î¸1 = cos Î¸2 = cosÎ¸3. ∴ Î¸1 = Î¸2 = Î¸3 Hence, the vector (a +b + c) is equally inclined to a, b, and c 15. Prove that (a +b) . (a +b) = \|a\|2 +\|b\|2 if and only if a⃗ . b are perpendicular, given a ≠ 0 , b ≠ 0 Solution 16. If Î¸ is the angle between two vectors a and b, then a . b ≥ 0 only when (A) 0 < Î¸ < Ï€/2 (B) 0 ≤ Î¸ ≤ Ï€/2 (C) 0 < Î¸ < Ï€ (D) 0 ≤ Î¸ ≤ Ï€ Solution Let Î¸ be the angle between two vectors a and b . Then, without loss of generality, a and b are non-zero vectors so that \|a\| and \|b\| are positive. Hence, a . b ≥ 0 when 0 ≤ Î¸ ≤ Ï€/2 . 17. Let a and b be two unit vectors and Î¸ is the angle between them. Then a + b is a unit vector if ____ (A) Î¸ = Ï€/4 (B) Î¸ = Ï€/3 (C) Î¸ = Ï€/2 (D) Î¸ = 2Ï€/3 Solution Let a and b be two unit vectors and Î¸ be the angle between them. Then, \| a\| = \|b\| = 1. Now, a + b is a unit vector if \|a +b\| = 1. Hence, a + b is a unit vector if Î¸ = 2Ï€/3. 18. The value of is (A) 0 (B) -1 (C) 1 (D) 3 Solution
Basic Math \| Basic-2 Math \| Prealgebra \| Workbooks \| Glossary \| Standards \| Site Map \| Help By the end of grade two, students understand place value and number relationships in addition and subtraction, and they use simple concepts of multiplication. They measure quantities with appropriate units. They classify shapes and see relationships among them by paying attention to their geometric attributes. They collect and analyze data and verify the answers. NUMBER SENSE 1.0 Students understand the relationship between numbers, quantities, and place value in whole numbers up to 1,000: 1.1 Count, read, and write whole numbers to 1,000 and identify the place value for each digit. - Recognizing Numbers 1-1,000 Card Quiz - "Before and After" 1-1,000 - Roman Numerals 1-1,000 Card Quiz 1.2 Use words, models, and expanded forms (e.g., 45 = 4 tens + 5) to represent numbers (to 1,000). 1.3 Order and compare whole numbers to 1,000 by using the symbols <, =, >. - "More or Less" 1-1,000 2.0 Students estimate, calculate, and solve problems involving addition and subtraction of two-and three-digit numbers: 2.1 Understand and use the inverse relationship between addition and subtraction (e.g., an opposite number sentence for 8 + 6 = 14 is 14 - 6 = 8) to solve problems and check solutions. 2.2 Find the sum or difference of two whole numbers up to three digits long. - Three-Digit Numbers (No Carrying) - Three-Digit Numbers (Carrying) - Three-Digit Number Quiz (No Borrowing) - Three-Digit Number Quiz (Borrowing) 2.3 Use mental arithmetic to find the sum or difference of two two-digit numbers. 3.0 Students model and solve simple problems involving multiplication and division: 3.1 Use repeated addition, arrays, and counting by multiples to do multiplication. 3.2 Use repeated subtraction, equal sharing, and forming equal groups with remainders to do division. 3.3 Know the multiplication tables of 2s, 5s, and 10s (to "times 10") and commit them to memory. - Twos Memory Game - Fives Memory Game - Tens Memory Game - 2, 5, and 10 Multiplication Quiz 4.0 Students understand that fractions and decimals may refer to parts of a set and parts of a whole: 4.1 Recognize, name, and compare unit fractions from 1/12 to 1/2. 4.2 Recognize fractions of a whole and parts of a group (e.g., one-fourth of a pie, two-thirds of 15 balls). - Identifying Fractions of Shapes - Identifying Fractions of Groups 4.3 Know that when all fractional parts are included, such as four-fourths, the result is equal to the whole and to one. 5.0 Students model and solve problems by representing, adding, and subtracting amounts of money: 5.1 Solve problems using combinations of coins and bills. - Adding Amounts Under One Dollar 5.2 Know and use the decimal notation and the dollar and cent symbols for money. - Adding Amounts Under Ten Dollars - Subtracting Amounts Under Ten Dollars 6.0 Students use estimation strategies in computation and problem solving that involve numbers that use the ones, tens, hundreds, and thousands places: 6.1 Recognize when an estimate is reasonable in measurements (e.g., closest inch). ALGEBRA AND FUNCTIONS 1.0 Students model, represent, and interpret number relationships to create and solve problems involving addition and subtraction: 1.1 Use the commutative and associative rules to simplify mental calculations and to check results. 1.2 Relate problem situations to number sentences involving addition and subtraction. - One and Two-Digit Addition Word Problems - Single-Digit Word Subtraction Problems - One and Two-Digit Subtraction Word Problems 1.3 Solve addition and subtraction problems by using data from simple charts, picture graphs, and number sentences. MEASUREMENT AND GEOMETRY 1.0 Students understand that measurement is accomplished by identifying a unit of measure, iterating (repeating) that unit, and comparing it to the item to be measured: 1.1 Measure the length of objects by iterating (repeating) a nonstandard or standard unit. 1.2 Use different units to measure the same object and predict whether the measure will be greater or smaller when a different unit is used. 1.3 Measure the length of an object to the nearest inch and/ or centimeter. 1.4 Tell time to the nearest quarter hour and know relationships of time (e.g., minutes in an hour, days in a month, weeks in a year). - Time Passed: Fifteen Minute Amounts - Time Passed: Five Minute Amounts - Converting Days to Weeks - Converting Hours to Days 1.5 Determine the duration of intervals of time in hours (e.g., 11:00 a.m. to 4:00 p.m.). - Time Passed: Thirty Minute Amounts 2.0 Students identify and describe the attributes of common figures in the plane and of common objects in space: 2.1 Describe and classify plane and solid geometric shapes (e.g., circle, triangle, square, rectangle, sphere, pyramid, cube, rectangular prism) according to the number and shape of faces, edges, and vertices. 2.2 Put shapes together and take them apart to form other shapes (e.g., two congruent right triangles can be arranged to form a rectangle). STATISTICS, DATA ANALYSIS, AND PROBABILITY 1.0 Students collect numerical data and record, organize, display, and interpret the data on bar graphs and other representations: 1.1 Record numerical data in systematic ways, keeping track of what has been counted. 1.2 Represent the same data set in more than one way (e.g., bar graphs and charts with tallies). 1.3 Identify features of data sets (range and mode). 2.0 Students demonstrate an understanding of patterns and how patterns grow and describe them in general ways: 2.1 Recognize, describe, and extend patterns and determine a next term in linear patterns (e.g., 4, 8, 12 ...; the number of ears on one horse, two horses, three horses, four horses). 2.2 Solve problems involving simple number patterns. MATHEMATICAL REASONING 1.0 Students make decisions about how to set up a problem: 1.1 Determine the approach, materials, and strategies to be used. 1.2 Use tools, such as manipulatives or sketches, to model problems. 2.0 Students solve problems and justify their reasoning: 2.1 Defend the reasoning used and justify the procedures selected. 2.2 Make precise calculations and check the validity of the results in the context of the problem. 3.0 Students note connections between one problem and another. * The custom search only looks at Rader's sites. Go for site help or a list of mathematics topics at the site map!
Skip to main content $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 8.3: Trigonometric Substitutions So far we have seen that it sometimes helps to replace a subexpression of a function by a single variable. Occasionally it can help to replace the original variable by something more complicated. This seems like a "reverse'' substitution, but it is really no different in principle than ordinary substitution. Example 8.3.1 Evaluate $$\int \sqrt{1-x^2}\,dx$$. Solution Let $$x=\sin u$$ so $$dx=\cos u\,du$$. Then $$\int \sqrt{1-x^2}\,dx=\int\sqrt{1-\sin^2 u}\cos u\,du= \int\sqrt{\cos^2 u}\cos u\,du.$$ We would like to replace $$\sqrt{\cos^2 u}$$ by $$\cos u$$, but this is valid only if $$\cos u$$ is positive, since $$\sqrt{\cos^2 u}$$ is positive. Consider again the substitution $$x=\sin u$$. We could just as well think of this as $$u=\arcsin x$$. If we do, then by the definition of the arcsine, $$-\pi/2\le u\le\pi/2$$, so $$\cos u\ge0$$. Then we continue: \eqalign{ \int\sqrt{\cos^2 u}\cos u\,du&=\int\cos^2u\,du=\int {1+\cos 2u\over2}\,du = {u\over 2}+{\sin 2u\over4}+C\cr &={\arcsin x\over2}+{\sin(2\arcsin x)\over4}+C.\cr } This is a perfectly good answer, though the term $$\sin(2\arcsin x)$$ is a bit unpleasant. It is possible to simplify this. Using the identity $$\sin 2x=2\sin x\cos x$$, we can write $$\sin 2u=2\sin u\cos u=2\sin(\arcsin x)\sqrt{1-\sin^2 u}= 2x\sqrt{1-\sin^2(\arcsin x)}=2x\sqrt{1-x^2}.$$ Then the full antiderivative is $${\arcsin x\over2}+{2x\sqrt{1-x^2}\over4}= {\arcsin x\over2}+{x\sqrt{1-x^2}\over2}+C.$$ This type of substitution is usually indicated when the function you wish to integrate contains a polynomial expression that might allow you to use the fundamental identity $$\sin^2x+\cos^2x=1$$ in one of three forms: $$\cos^2 x=1-\sin^2x \qquad \sec^2x=1+\tan^2x \qquad \tan^2x=\sec^2x-1.$$ If your function contains $$1-x^2$$, as in the example above, try $$x=\sin u$$; if it contains $$1+x^2$$ try $$x=\tan u$$; and if it contains $$x^2-1$$, try $$x=\sec u$$. Sometimes you will need to try something a bit different to handle constants other than one. Example 8.3.2 Evaluate $$\int\sqrt{4-9x^2}\,dx$$. Solution We start by rewriting this so that it looks more like the previous example: $$\int\sqrt{4-9x^2}\,dx=\int\sqrt{4(1-(3x/2)^2)}\,dx =\int 2\sqrt{1-(3x/2)^2}\,dx.$$ Now let $$3x/2=\sin u$$ so $$(3/2)\,dx=\cos u \,du$$ or $$dx=(2/3)\cos u\,du$$. Then \eqalign{ \int 2\sqrt{1-(3x/2)^2}\,dx&=\int 2\sqrt{1-\sin^2u}\,(2/3)\cos u\,du ={4\over3}\int \cos^2u\,du\cr &={4u\over 6}+{4\sin 2u\over12}+C\cr &={2\arcsin(3x/2)\over3}+{2\sin u \cos u\over3}+C\cr &={2\arcsin(3x/2)\over3}+{2\sin(\arcsin(3x/2))\cos(\arcsin(3x/2))\over3}+C\cr &={2\arcsin(3x/2)\over3}+{2(3x/2)\sqrt{1-(3x/2)^2}\over3}+C\cr &={2\arcsin(3x/2)\over3}+{x\sqrt{4-9x^2}\over2}+C,\cr } using some of the work from example 8.3.1. Example 8.3.3 Evaluate $$\int\sqrt{1+x^2}\,dx$$. Solution Let $$x=\tan u$$, $$dx=\sec^2 u\,du$$, so $$\int\sqrt{1+x^2}\,dx=\int \sqrt{1+\tan^2 u}\sec^2u\,du= \int\sqrt{\sec^2u}\sec^2u\,du.$$ Since $$u=\arctan(x)$$, $$-\pi/2\le u\le\pi/2$$ and $$\sec u\ge0$$, so $$\sqrt{\sec^2u}=\sec u$$. Then $$\int\sqrt{\sec^2u}\sec^2u\,du=\int \sec^3 u \,du.$$ In problems of this type, two integrals come up frequently: $$\int\sec^3u\,du$$ and $$\int\sec u\,du$$. Both have relatively nice expressions but they are a bit tricky to discover. First we do $$\int\sec u\,du$$, which we will need to compute $$\int\sec^3u\,du$$: \eqalign{ \int\sec u\,du&=\int\sec u\,{\sec u +\tan u\over \sec u +\tan u}\,du\cr &=\int{\sec^2 u +\sec u\tan u\over \sec u +\tan u}\,du.\cr } Now let $$w=\sec u +\tan u$$, $$dw=\sec u \tan u + \sec^2u\,du$$, exactly the numerator of the function we are integrating. Thus \eqalign{ \int\sec u\,du=\int{\sec^2 u +\sec u\tan u\over \sec u +\tan u}\,du&= \int{1\over w}\,dw=\ln \|w\|+C\cr &=\ln\|\sec u +\tan u\|+C.\cr } Now for $$\int\sec^3 u\,du$$: \eqalign{ \sec^3u&={\sec^3u\over2}+{\sec^3u\over2}={\sec^3u\over2}+{(\tan^2u+1)\sec u\over 2}\cr &={\sec^3u\over2}+{\sec u \tan^2 u\over2}+{\sec u\over 2}= {\sec^3u+\sec u \tan^2u\over 2}+{\sec u\over 2}.\cr } We already know how to integrate $$\sec u$$, so we just need the first quotient. This is "simply'' a matter of recognizing the product rule in action: $$\int \sec^3u+\sec u \tan^2u\,du=\sec u \tan u.$$ So putting these together we get $$\int\sec^3u\,du={\sec u \tan u\over2}+{\ln\|\sec u +\tan u\| \over2}+C,$$ and reverting to the original variable $$x$$: \eqalign{ \int\sqrt{1+x^2}\,dx&={\sec u \tan u\over2}+{\ln\|\sec u +\tan u\|\over2}+C\cr &={\sec(\arctan x) \tan(\arctan x)\over2} +{\ln\|\sec(\arctan x) +\tan(\arctan x)\|\over2}+C\cr &={ x\sqrt{1+x^2}\over2} +{\ln\|\sqrt{1+x^2} +x\|\over2}+C,\cr } using $$\tan(\arctan x)=x$$ and $$\sec(\arctan x)=\sqrt{1+\tan^2(\arctan x)}=\sqrt{1+x^2}$$. ### Contributors • Integrated by Justin Marshall.
# How To Calculate Winning Odds in California Lottery Ever wonder how to calculate winning odds of lottery games? The winning odds of the top prize of Fantasy 5 in California Lottery are 1 in 575,757. The winnings odds of the top prize of SuperLOTTO plus are 1 in 41,416,353. The winnings odds of the top prize of Mega Millions are 1 in 175,711,534. In this post, we show how to calculate the odds for these games in the California Lottery. The calculation is an excellent combinatorial exercise as well as in calculating hypergeometric probability. All figures and data are obtained from the California Lottery. ____________________________________________________ Fantasy 5 The following figures show a playslip and a sample ticket for the game of Fantasy 5. Figure 1 Figure 2 In the game of Fantasy 5, the player chooses 5 numbers from 1 to 39. If all 5 chosen numbers match the 5 winning numbers, the player wins the top prize which starts at $50,000 and can go up to$500,000 or more. The odds of winning the top prize are 1 in 575,757. There are lower tier prizes that are easier to win but with much lower winning amounts. The following figure shows the prize categories and the winning odds of Fantasy 5. Figure 3 All 5 of 5 In matching the player’s chosen numbers with the winning numbers, the order of the numbers do not matter. Thus in the calculation of odds, we use combination rather than permutation. Thus we have: $\displaystyle (1) \ \ \ \ \ \binom{39}{5}=\frac{39!}{5! \ (39-5)!}=575757$ Based on $(1)$, the odds of matching all 5 winning numbers is 1 in 575,757 (the odds of winning the top prize). Any 4 of 5 To match 4 out of 5 winning numbers, 4 of the player’s chosen numbers are winning numbers and 1 of the player’s chosen numbers is from the non-winning numbers (34 of them). Thus the probability of matching 4 out of 5 winning numbers is: $\displaystyle (2) \ \ \ \ \ \frac{\displaystyle \binom{5}{4} \ \binom{34}{1}}{\displaystyle \binom{39}{5}}=\frac{5 \times 34}{575757}=\frac{1}{3386.8} \ \ \text{(1 out of 3,387)}$ Any 3 of 5 To find the odds for matching 3 out of 5 winning numbers, we need to find the probability that 3 of the player’s chosen numbers are from the 5 winning numbers and 2 of the selected numbers are from the 34 non-winning numbers. Thus we have: $\displaystyle (3) \ \ \ \ \ \frac{\displaystyle \binom{5}{3} \ \binom{34}{2}}{\displaystyle \binom{39}{5}}=\frac{10 \times 561}{575757}=\frac{1}{102.63} \ \ \text{(1 out of 103)}$ Any 2 of 5 Similarly, the following shows how to calculate the odds of matching 2 out of 5 winning numbers: $\displaystyle (4) \ \ \ \ \ \frac{\displaystyle \binom{5}{2} \ \binom{34}{3}}{\displaystyle \binom{39}{5}}=\frac{10 \times 5984}{575757}=\frac{1}{9.6216} \ \ \text{(1 out of 10)}$ ____________________________________________________ SuperLOTTO Plus Here are the pictures of a playslip and a sample ticket of the game of SuperLOTTO Plus. Figure 4 Figure 5 Based on the playslip (Figure 4), the player chooses 5 numbers from 1 to 47. The player also chooses an additional number called a Mega number from 1 to 27. To win the top prize, there must be a match between the player’s 5 selections and the 5 winning numbers as well as a match between the player’s Mega number and the winning Mega number (All 5 of 5 and Mega in Figure 6 below). Figure 6 All 5 of 5 and Mega To find the odds of the match of “All 5 of 5 and Mega”, the total number of possibilities is obtained by choosing 5 numbers from 27 numbers and choose 1 number from 27 numbers. We have: $\displaystyle (5) \ \ \ \ \ \binom{47}{5} \times \binom{27}{1}=41,416,353$ Thus the odds of matching “All 5 of 5 and Mega” are 1 in 41,416,353. Any 5 of 5 To find the odds of matching “All 5 of 5″ (i.e. the player’s 5 selections match the 5 winning numbers but no match with the Mega winning number), we need to choose 5 numbers from the 5 winning numbers, choose 0 numbers from the 42 non-winning numbers, choose 0 numbers from the 1 Mega winning number and choose 1 number from the 26 non-Mega winning numbers. This may seem overly precise, but will make it easier to the subsequent derivations. We have: \displaystyle \begin{aligned}(6) \ \ \ \ \ \frac{\displaystyle \binom{5}{5} \ \binom{42}{0} \ \binom{1}{0} \ \binom{26}{1}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{1 \times 1 \times 1 \times 26}{41416353} \\&=\frac{1}{1592936.654} \\&\text{ } \\&=\text{1 out of 1,592,937} \end{aligned} Any 4 of 5 and Mega To calculate the odds for matching “any 4 of 5 and Mega”, we need to choose 4 out of 5 winning numbers, choose 1 out of 42 non-winning numbers, choose 1 out of 1 Mega winning number, and choose 0 out of 26 non-winning Mega numbers. We have: \displaystyle \begin{aligned}(7) \ \ \ \ \ \frac{\displaystyle \binom{5}{4} \ \binom{42}{1} \ \binom{1}{1} \ \binom{26}{0}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{5 \times 42 \times 1 \times 1}{41416353} \\&=\frac{1}{197220.7286} \\&\text{ } \\&=\text{1 out of 197,221} \end{aligned} Any 4 of 5 To calculate the odds for matching “any 4 of 5″ (no match for Mega number), we need to choose 4 out of 5 winning numbers, choose 1 out of 42 non-winning numbers, choose 0 out of 1 Mega winning number, and choose 1 out of 26 non-winning Mega numbers. We have: \displaystyle \begin{aligned}(8) \ \ \ \ \ \frac{\displaystyle \binom{5}{4} \ \binom{42}{1} \ \binom{1}{0} \ \binom{26}{1}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{5 \times 42 \times 1 \times 26}{41416353} \\&=\frac{1}{7585.412637} \\&\text{ } \\&=\text{1 out of 7,585} \end{aligned} Any 3 of 5 and Mega To calculate the odds for matching “any 3 of 5 and Mega”, we need to choose 3 out of 5 winning numbers, choose 2 out of 42 non-winning numbers, choose 1 out of 1 Mega winning number, and choose 0 out of 26 non-winning Mega numbers. We have: \displaystyle \begin{aligned}(9) \ \ \ \ \ \frac{\displaystyle \binom{5}{3} \ \binom{42}{2} \ \binom{1}{1} \ \binom{26}{0}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{10 \times 861 \times 1 \times 1}{41416353} \\&=\frac{1}{4810.261672} \\&\text{ } \\&=\text{1 out of 4,810} \end{aligned} The rest of the calculations for SuperLOTTO Plus should be routine. It is a matter to deciding how many to choose from the 5 winning numbers, how many to choose from the 42 non-winning numbers as well as how many to choose from the 1 winning Mega number and how many to choose from the 26 non-winning Mega numbers. Any 3 of 5 \displaystyle \begin{aligned}(10) \ \ \ \ \ \frac{\displaystyle \binom{5}{3} \ \binom{42}{2} \ \binom{1}{0} \ \binom{26}{1}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{10 \times 861 \times 1 \times 26}{41416353} \\&=\frac{1}{185.0100643} \\&\text{ } \\&=\text{1 out of 185} \end{aligned} Any 2 of 5 and Mega \displaystyle \begin{aligned}(11) \ \ \ \ \ \frac{\displaystyle \binom{5}{2} \ \binom{42}{3} \ \binom{1}{1} \ \binom{26}{0}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{10 \times 11480 \times 1 \times 1}{41416353} \\&=\frac{1}{360.7696254} \\&\text{ } \\&=\text{1 out of 361} \end{aligned} Any 1 of 5 and Mega \displaystyle \begin{aligned}(12) \ \ \ \ \ \frac{\displaystyle \binom{5}{1} \ \binom{42}{4} \ \binom{1}{1} \ \binom{26}{0}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{5 \times 111930 \times 1 \times 1}{41416353} \\&=\frac{1}{74.00402573} \\&\text{ } \\&=\text{1 out of 74} \end{aligned} None of 5 only Mega \displaystyle \begin{aligned}(13) \ \ \ \ \ \frac{\displaystyle \binom{5}{0} \ \binom{42}{5} \ \binom{1}{1} \ \binom{26}{0}}{\displaystyle \binom{47}{5} \times \binom{27}{1}}&=\frac{1 \times 850668 \times 1 \times 1}{41416353} \\&=\frac{1}{48.68685903} \\&\text{ } \\&=\text{1 out of 49} \end{aligned} ____________________________________________________ Mega Millions The following are a playslip, a sample ticket and the winning odds of the game of Mega Millions. Figure 7 Figure 8 Figure 9 Based on the playslip (Figure 7), the player chooses 5 numbers from 1 to 56. The player also chooses an additional number called a Mega number from 1 to 46. To win the top prize, there must be a match between the player’s 5 selections and the 5 winning numbers as well as a match between the player’s Mega number and the winning Mega number. The calculation of the odds indicated in Figure 9 are left as exercises.
# Prove that the sum of the three angle of triangle is two right angle in a right angle triangle an acute is one fourth the other,find the acute angle. 1 by s111 2015-10-13T20:14:39+05:30 Let ABC is a triangle To prove: to prove all the sum of side is 180° construction: draw a line DE through the point A such that DE is parallel to BC proof DE is parallel to BC and AC & BA are transversal ∴ ∠DAB=∠ABC and ∠EAC =∠ACB we know ∠DAB +∠BAC+∠EAC=180° [straight line] we can also write that ∠ABC+∠BAC+∠ACB= 180° Hence proved which is equl to 2 right angle 90° + 90° = 180° A/\q In a right angled triangle one angle is x second angle is x third angle 90° again 90°+x + x =180° x + + 90°= 180° x = 75° the second angle (acute angle)= =18.5
# Determinants and Matrices - PowerPoint PPT Presentation 1 / 102 Title: ## Determinants and Matrices Description: ### (A) A square array of real (or complex) numbers arranged in n rows ... (7) The Idempotent Matrix : A square matrix is called idempotent if and only if A2 = A. ... – PowerPoint PPT presentation Number of Views:1027 Avg rating:3.0/5.0 Slides: 103 Provided by: tangt1 Category: Tags: Transcript and Presenter's Notes Title: Determinants and Matrices 1 Unit 5 • Determinants and Matrices 2 5.1 Definition of Determinants • (A) A square array of real (or complex) numbers arranged in n rows and n columns is called a determinant of the nth order . 3 5.1 Definition of Determinants • The number in the determinant is called an element or entry of the determinant. There are n2 elements in a determinant of the nth order. Usually, we use small letters to denote elements. The symbol aij denotes the element located in the ith row, the jth column. e.g. a34 located in 3th row 4th column 4 5.1 Definition of Determinants • (B) The value of a determinant of order 2 is defined by The expression a11a22 - a12a21 on the right hand side of (1) is called the expansion of the second order determinant. 5 5.1 Definition of Determinants • (C) The value of a determinant of order 3 is defined by The expression on the right hand side of (2) is called the expansion of the third order determinant. 6 5.2 Properties of Determinants • (1) The value of a determinant remains unchanged if all the rows and columns are correspondingly interchanged. 7 5.2 Properties of Determinants • (2) The determinant changes sign only but its absolute value remains unchanged if any two rows (or any two columns) are interchanged. 8 5.2 Properties of Determinants • (3) If the elements of a row (resp. a column) are proportional to those of another row (resp. another column), then the value of the determinant is zero. 9 5.2 Properties of Determinants • (4) If the elements of a row (resp. a column) are identical to those of another row (resp. another column), then the value of the determinant is zero. 10 5.2 Properties of Determinants • (5) If the elements of any row (resp. a column) of a determinant are multiplied by the same factor, the resulting determinant is equal to the product of that factor and the original determinant. 11 5.2 Properties of Determinants • (5 ext) If the elements of a row (resp. a column) are zero, the value of the determinant is zero. 12 5.2 Properties of Determinants • (6) If the elements of any row (resp. a column) are added or subtracted by equimultiples of the corresponding elements of another row (resp. another column), the value of the determinant is altered. 13 5.2 Properties of Determinants • (6) If the elements of any row (resp. a column) are added or subtracted by equimultiples of the corresponding elements of another row (resp. another column), the value of the determinant is altered. 14 5.2 Properties of Determinants • (7) If the elements of a row (resp. a column) of a determinant consists of an algebraic sum of the terms, the determinant is equal to the sum of two other determinants in each of which the elements consist of single term. 15 5.2 Properties of Determinants • (7) If the elements of a row (resp. a column) of a determinant consists of an algebraic sum of the terms, the determinant is equal to the sum of two other determinants in each of which the elements consist of single term. 16 P.133 Ex.5A 17 5.3 Minors and Cofactors • Let aij be an element of the determinant A located in the ith row, the jth column. The minor of aij , is denoted by ?ij, is defined as the determinant formed by deleting the ith row and jth column of A. 18 5.3 Minors and Cofactors • Let A be a determinant, aij be an element of A in the ith row, the jth column and ?ij be the minor of aij. The cofactor of aij is defined by (-1)ij ?ij and is usually denoted by Aij. 19 5.3 Minors and Cofactors 20 5.3 Minors and Cofactors The sum of the products of the elements of any row (or any column) and their own cofactors is equal to the value of the determinant. 21 5.3 Minors and Cofactors The sum of the products of the elements of any row (or any column) and their own cofactors is equal to the value of the determinant. 22 P.145 Ex.5B 23 5.4 Factorization of Determinants To factorize a given determinant, the following two methods are usually employed (1) Apply properties of a determinant to transform the entries in a row (or a column) until there is a common factor among the entries. (2) If the determinant is a polynomial in x and if the determinant vanishes when x a, then (x - a) is a factor. 24 5.4 Factorization of Determinants 25 5.4 Factorization of Determinants • Method II • remains unchanged if a, b, c are respectively replaced by b, c, a.Thus it is a cyclic symmetric about a, b, c and is homogeneous. • If we put a b or b c or c a, ? 0. • Hence ? has factors a b, b c and c a. • As ? is of degree 3, we let ? ? k(a-b)(b-c)(c-a). • By comparing coefficients of bc2, we get k 1. • ? ? ? (a-b)(b-c)(c-a) 26 P.149 Ex.5C 27 5.5 Definition and Basic Operation of Matrices Matrix A rectangular array of real (or complex) numbers arranged in m rows and n column is called a m x n matrix. A m x n matrix is usually represented in the form 28 5.5 Definition and Basic Operation of Matrices The number aij in the ith row and the jth column of a matrix is called an element or entry. Hence a m x n matrix contains mn elements. We call m x n the order of a matrix, and we usually use capital letters to denote matrices. Matrix is a mathematical tool, it is not a number. 29 5.5 Definition and Basic Operation of Matrices Equality of matrices Two matrices A (aij) and B (bij) are equal if and only if they are of the same orders and aij bij, for all i 1, 2 , .., m j 1, 2,..., n. 30 5.5 Definition and Basic Operation of Matrices Sum of Matrices The sum of two m x n matrices A (aij) and B (bij) is the m x n matrix C (cij), where cij aij bij, for all i 1, 2, ., m j 1, 2, , n. 31 5.5 Definition and Basic Operation of Matrices Scalar Multiplication The scalar multiplication of a m x n matrix A (aij) by a scalar ? (here ? is a real or a complex number) is the m x n matrix C (cij), where cij ? aij, for every i 1, 2, , m j 1, 2,, n. We usually write C ?A. This defines scalar multiplication of a matrix A by a scalar ?. 32 5.5 Definition and Basic Operation of Matrices Negative Matrix Let A be any matrix, the symbol A denotes (-1)A. 33 5.5 Definition and Basic Operation of Matrices Difference of Matrices Let A, B be two matrices of the same order. The difference A - B is defined by A - B A (-1)B. 34 5.5 Definition and Basic Operation of Matrices For any three matrices A, B, C of the same order and any scalars ?, ?, we have • (A B) C A (B C) (2) Addition is commutative A B B A (3) Scalar multiplication is distributive over 35 5.5 Definition and Basic Operation of Matrices Multiplication of Matrices Let A (aik) be a m x n matrix and B (bkj) be a n x p matrix, the product AB is the m x p matrix C (cij), where for all i 1, 2,., m j 1, 2, , p. The product AB defines multiplication of matrices. 36 5.5 Definition and Basic Operation of Matrices Multiplication of Matrices 37 5.5 Definition and Basic Operation of Matrices Multiplication of Matrices 38 5.5 Definition and Basic Operation of Matrices Multiplication of Matrices 39 5.5 Definition and Basic Operation of Matrices Multiplication of Matrices Multiplication of matrices is non-commutative. 40 5.5 Definition and Basic Operation of Matrices Multiplicative Properties of Matrices • Let A, B, C be three matrices of order m x n, n x p, p x q respectively, then (AB)C A(BC) (2) Let A, B be two matrices of order m x n and C, D be matrices of order n x p, q x m respectively, then (3) Let A, B be two matrices of order m x n and n x p respectively. For any scalar ?, we have ?(AB) (?A)B A(?B). 41 5.6 Special Types of Matrices (1) The square Matrix A n x n matrix is called a square matrix of order n. (2) The identity Matrix A square matrix A (aij) of order n is called an identity matrix or a unit matrix of order n if and only if and is denoted by In. 42 5.6 Special Types of Matrices (3) The scalar Matrix A square matrix (bij) is called a scalar matrix if and only if 43 5.6 Special Types of Matrices 44 5.6 Special Types of Matrices (5) The Zero/ Null of Matrix A matrix (not necessarily square) is called a zero matrix or a null matrix if and only if all its elements are zero. A zero matrix of order m x n is usually denoted by 0mxn. (i) For any matrix A and a zero matrix, both are of the same order, A 0 0 A A (ii) For any square matrix A and a square zero matrix 0, both are of the same order, A x 0 0 x A 0 45 5.6 Special Types of Matrices 46 5.6 Special Types of Matrices (7) The Idempotent Matrix A square matrix is called idempotent if and only if A2 A. 47 5.6 Special Types of Matrices (8) The transpose of Matrix The transpose of a m x n matrix A (aij), denoted by AT or At or A, is the n x m matrix AT (bij), where bij aji. 48 5.6 Special Types of Matrices (8 ext.) Properties of the transpose of matrix 49 5.6 Special Types of Matrices (9) The Symmetric Matrix A square matrix A is said to be symmetric if and only if AT A. 50 5.6 Special Types of Matrices (10) The skew-symmetric Matrix A square matrix a is said to be skew-symmetric if and only if AT - A 51 5.6 Special Types of Matrices (11) The Row Vector and the Column Vector A 1 x n matrix is called a row vector. A m x 1 matrix is called column vector. 52 P.156 Ex.5D 53 5.7 Multiplicative Inverse of a Square Matrix The determinant of A, denoted by det A or A, is defined as the determinant 54 5.7 Multiplicative Inverse of a Square Matrix General properties between determinant and matrix • For any square matrix A of order n and any scalar ?, det (?A) ?n det A or Why is there power n? 55 5.7 Multiplicative Inverse of a Square Matrix Verification 56 5.7 Multiplicative Inverse of a Square Matrix General properties between determinant and matrix (2) For any two square matrices A and B of the same order, det (AB) det A x det B or How can we verify this property? 57 5.7 Multiplicative Inverse of a Square Matrix Verification 58 5.7 Multiplicative Inverse of a Square Matrix Verification By direct expansion of the determinants, the identity holds. 59 5.7 Multiplicative Inverse of a Square Matrix General properties between determinant and matrix (3) For any square matrix A, det (AT) det A or How can we prove this property? 60 5.7 Multiplicative Inverse of a Square Matrix 61 5.7 Multiplicative Inverse of a Square Matrix The Cofactor Matrix Let A (aij)nxn be a square matrix. The cofactor matrix of A, denoted by cof A, is defined by cof A (Aij)nxn, where Aij is the cofactor of aij, for every i, j 1, 2,,n. 62 5.7 Multiplicative Inverse of a Square Matrix The Adjoint Matrix Let A be a square matrix. The transpose of the cofactor matrix of A, i.e. (cof A)T, is called the adjoint matrix of A, 63 5.7 Multiplicative Inverse of a Square Matrix The Non-singular/Invertible Matrix A square matrix A of order n is said to be non-singular or invertible if and only if there exists a square matrix B such that AB BA In, where In is an identity matrix of order n, and the matrix B is called the multiplicative inverse or simply the inverse of A, which is denoted by A-1, i.e. AA-1 A-1A In. The inverse of a non-singular matrix is unique. 64 5.7 Multiplicative Inverse of a Square Matrix Lemma For any square matrix of order n, identity matrix of order n. ? 65 5.7 Multiplicative Inverse of a Square Matrix Verification Why do the elements arrange in this way? 66 5.7 Multiplicative Inverse of a Square Matrix How can we simplify these three elements? Then how about the other elements? Why does each row equal zero? And 67 5.7 Multiplicative Inverse of a Square Matrix 68 5.7 Multiplicative Inverse of a Square Matrix Verification 69 5.7 Multiplicative Inverse of a Square Matrix Lemma For any square matrix of order n, identity matrix of order n. 70 5.7 Multiplicative Inverse of a Square Matrix 71 5.7 Multiplicative Inverse of a Square Matrix On the other hand, A square matrix A is said to be singular or not invertible if and only if the inverse, A-1, of A does not exist. A square matrix A is singular if and only if det A 0. 72 5.8 Properties of Inverses If A and B are non-singular matrices of the same order, and any scalar ? we have (1) If AC1 AC2 or C1A C2A, then C1 C2 (2) A-1 is non-singular and (A-1)-1 A (3) AB is non-singular and (AB)-1 B-1A-1 Quite similar to (AB)T BTAT We will show the identity (AB)-1 B-1A-1 later. (4) For any positive integer n, An is non-singular, and (An)-1 (A-1)n. 73 5.8 Properties of Inverses If A and B are non-singular matrices of the same order, and any scalar ? we have (6) AT is non-singular and (AT)-1 (A-1)T. (7) If AC 0, then C 0. 74 5.8 Properties of Inverses If A and B are non-singular matrices of the same order, and any scalar ? we have Prove (AB)-1 B-1A-1 5E 75 P.164 Ex.5E 76 5.9 Some Illustrative Examples 77 5.9 Some Illustrative Examples 78 P.176 Ex.5F 79 Transformations of Points on the Coordinate Plane 80 5.10 Linear Transformations on the Rectangular Cartesian Plane 81 5.10 Linear Transformations on the Rectangular Cartesian Plane 82 5.10 Linear Transformations on the Rectangular Cartesian Plane 83 5.10 Linear Transformations on the Rectangular Cartesian Plane 84 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 85 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 86 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 87 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 88 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 89 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 90 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 91 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 92 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 93 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane (IV) Reflection 94 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane (IV) Reflection The matrix representing the reflection about 95 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 96 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 97 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 98 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 99 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 100 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 101 5.11 Some special Linear Transformations on the Rectangular Cartesian Plane 102 P.195 Ex.5G
# Here is an arithmetic sequece find the 20 th term Find the 20th term of this sequence. $1, 5, 4, 8, 7, 11, 10, 14.........$ it is like add 4 to the st term then subtract 1 i need to find a recursive formula that can help me get the 20th term. Help please. • Do you need a recursive formula, or can you just write it out? Feb 10, 2015 at 20:44 • oeis.org/A166517 Feb 10, 2015 at 20:45 There are two arithmetic series in the given series. The numbers at odd position constitute the following series $1,4,7,10,\cdots$ The numbers at even position constitute the following series: $5,8,11,14,\cdots$ Since $20$ is an even number, the $20^{th}$ term will be the $10^{th}$ term of the second series. And it is $32$. Here's a single formula that works for all positive $n$: $$a_n = \lfloor (3n/2) + 1 \rfloor + (-1)^n.$$ Then, $a_{20} = \lfloor 31 \rfloor + 1 = 32.$ I know you didn't ask for an explicit rule but, you can also find that too... $\\a_{2n}=5+3(n-1) \text{ or } a_{2n+1}=1+3n$ one of these gives you the even entries and the other gives you the odd entries. Every two steps, you add $3$. So after $20$ steps you'll have added $3$ ten times. This gets you the $21$st term. Now you can work out the $20$-th. Note that $$a_{2n}=a_{2n-1}+1\\ a_{2n-1}=a_{2n-2}+4,\quad n\ge 1$$ The list: 0 1 2 3 4 5 6 7 8 9 10 ... 19 20 1 5 4 8 7 11 10 14 13 17 16 ... 32 31 The series: (4.5+3n-2.5(-1)^n)/2 To write a recursive formula you need two parts: a starting point, and a rule for finding each term from the previous term. For this sequence, a recursive formula could look like: $$\begin{array}{ll} a_1 = 1 \\ a_n = \left\{ \begin{array}{ll} a_{n-1} + 4 & \text{if } n \text{ is even} \\ a_{n-1} - 1 & \text{if } n \text{ is odd} \\ \end{array} \right. \end{array}$$ To use a recursive formula to find a specific term in a sequence you must find all of the terms before the specific term. I would just write out the sequence to the 20th term. As other answers have noted, there are more efficient ways of finding the 20th term than using a recursive formula. For example using one of the explicit formulas given in other answers will give you the 20th term without needing to find any of the previous terms. In Mathematica: FindSequenceFunction[{1, 5, 4, 8, 7, 11, 10, 14}, 20] (* 32 *) and the function derived is: $\frac{1}{4} (-1)^n \left(3 (-1)^n (2 n+1)+5\right)$
## How do you calculate the speed of a crash? How Do I Calculate Speed and Distance in a Car Accident Case? The formula for speed and distance is the same for a car as any other object: distance ÷ time. So if you want to calculate the speed of a car at sixty miles an hour, the math is (60 x 5280) ÷ (60 x 60) = 88 feet per second. ## How skid marks can be used to determine the speed of a moving vehicle before colliding with other vehicle? Measuring the length of skid marks is another way that investigators can determine how fast the vehicles were traveling. When the skid distance, the drag factor due to road surface friction, and the braking efficiency are determined, the minimum speed of a car as it started skidding can be found. Can you tell how fast a car was going after an accident? The severity of the damage can tell investigators important information. A minor dent may indicate that a driver was traveling at low speed or that the driver had nearly enough time to finish braking. Severe damage can tell investigators how fast a vehicle may have been traveling or how hard the car was hit. How can speed be calculated? The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time. ### How is skid distance calculated? dis the distance the car skidded (in feet). fis a special number (called the coefficient of friction) that depends on the road surface and road conditions. S = ~30d(I.O) (dry tar road). For a wet tar road, f is about 0.5, so the formula is S = ~ 30d(0.5) (wet tar road). ### What is the skid distance formula? What is the difference between skid marks and yaw marks? Characteristics. Skid marks are divided into “acceleration marks” created on acceleration, if the engine provides more power than the tire can transmit; “braking marks,” if the brakes “lock up” and cause the tire to slide; or “yaw marks”, if the tire slides sideways. What are the formulas used in finding the estimated speed? The formula for speed is speed = distance ÷ time. #### How can skid marks determine speed? How to Calculate Skid Speed Measure the distance of the skid marks in feet and inches. If there are multiple skid marks, measure all of them and take the average. Convert the measurement to a decimal by dividing the number of inches by 12 and adding the result to the number of feet. Estimate the drag factor of the road that you were traveling on. #### What is the skid to stop formula? The over-used SKID TO STOP formula is only appropriate when called upon to calculate the initial (i.e., beginning) speed of a vehicle found to have skidded to a stop. In its most basic form, this equation (Eq. 1) takes the form: = {30Df}1/2 (Eq. 1) How do you calculate stopping distance? To determine the stopping distance, you calculate: Perception Distance (71 feet) + Reaction Distance (71 feet) + Braking Distance (525 feet) = Stopping Distance (667 feet) Common sense also says when conditions change, times and distances change too.
1 / 75 # Chapter Four - PowerPoint PPT Presentation Chapter Four. Elementary Probability Theory. 4.1 What Is Probability?. Focus Points Assign probabilities to events. Explain how to law of large numbers relates to relative frequencies. Apply basic rules of probability in everyday life. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. ## PowerPoint Slideshow about ' Chapter Four' - julian-snow Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - Presentation Transcript ### Chapter Four Elementary Probability Theory Focus Points • Assign probabilities to events. • Explain how to law of large numbers relates to relative frequencies. • Apply basic rules of probability in everyday life. • Explain the relationship between statistics and probability. • Probability is a numerical measurement of likelihood of an event. • The probability of any event is a number between zero and one. • Events with probability close to one are more likely to occur. • Events with probability close to zero are less likely to occur. If A represents an event, P(A) represents the probability of A. If P(A) = 1 Event A is certain to occur If P(A) = 0 Event A is certain not to occur • Intuition • Relative frequency • Equally likely outcomes • Incorporates past experience, judgment, or opinion. • Is based upon level of confidence in the result • Example: “I am 95% sure that I will attend the party.” Probability of an event = the fraction of the time that the event occurred in the past = f n where f = frequency of an event n = sample size as Relative Frequency If you note that 57 of the last 100 applicants for a job have been female, the probability that the next applicant is female would be: • No one result is expected to occur more frequently than any other. When tossing a coin, the probability of getting head= In the long run, as the sample size increases and increases, the relative frequencies of outcomes get closer and closer to the theoretical (or actual) probability value. Statistical Experiment or Statistical Observation • Any random activity that results in a definite outcome • A collection of one or more outcomes of a statistical experiment or observation • An outcome of a statistical experiment that consists of one and only one of the outcomes of the experiment • The set of all possible distinct outcomes of an experiment • The sum of all probabilities of all simple events in a sample space must equal one. Sample Space for the rolling of an ordinary die: 1, 2, 3, 4, 5, 6 For the experiment of rolling an ordinary die: • P(even number) = • P(result less than five) = • P(not getting a two) = • the event that A does not occur • Notation for the complement of event A: Event A and its complement Ac • P(event A does not occur) = P(Ac) = 1 – P(A) • So, P(A) + P(Ac) = 1 If the probability that it will snow today is 30%, P(It will not snow) = 1 – P(snow) = 1 – 0.30 = 0.70 • Probability makes statements about what will occur when samples are drawn from a known population. • Statistics describes how samples are to be obtained and how inferences are to be made about unknown populations. • The occurrence (or non-occurrence) of one event does not change the probability that the other event will occur. If events A and B are independent, • P(A and B) = P(A) •P(B) • If events are dependent, the fact that one occurs affects the probability of the other. • P(A, givenB) equals the probability that event A occurs, assuming that B has already occurred. • P(A and B) = P(A) •P(B, given A) • P(Aand B) = P(B) •P(A, given B) • For independent events: P(A and B) = P(A) •P(B) • For any events: P(A and B) = P(A) •P(B, given A) P(A and B) = P(B) •P(A, given B) For independent events:P(A and B) = P(A) •P(B) When choosing two cards from two separate decks of cards, find the probability of getting two fives. P(two fives) = P(5 from first deck and 5 from second) = For dependent events:P(A and B) = P(A) · P(B, given A) When choosing two cards from a deck without replacement, find the probability of getting two fives. P(two fives) = P(5 on first draw and 5 on second) = And” versus “or” • And means both events occur together. • Or means that at least one of the events occur. The Event A and B The Event A or B For any eventsA and B, P(A or B) = P(A) + P(B) – P(A and B) When choosing a card from an ordinary deck, the probability of getting a five or a red card: P(5 ) + P(red) – P(5 and red) = When choosing a card from an ordinary deck, the probability of getting a five or a six: P(5 ) + P(6) – P(5 and 6) = Mutually Exclusive Events of getting a • Events that are disjoint, cannot happen together. Addition Rule for Mutually Exclusive Events of getting a For any mutually exclusive events A and B, P(A or B) = P(A) + P(B) When rolling an ordinary die: of getting a P(4 or 6) = Survey results: of getting a P(male and college grad) = ? Survey results: of getting a Survey results: of getting a P(male or college grad) = ? Survey results: of getting a Survey results: of getting a P(male, given college grad) = ? Survey results: of getting a Counting Techniques of getting a • Tree Diagram • Multiplication Rule of Counting • Permutations • Combinations Tree Diagram of getting a • a visual display of the total number of outcomes of an experiment consisting of a series of events Experiment of rolling two dice, one after the other and observing any of the six possible outcomes each time . Number of paths = 6 x 6 = 36 Multiplication Rule of Counting diagram: For any series of events, if there are n1possible outcomes for event E1 and n2 possible outcomes for event E2, etc. then there are n1X n2X···X nm possible outcomes for the series of events E1 through Em. Area Code Example diagram: In the past, an area code consisted of 3 digits, the first of which could be any digit from 2 through 9. The second digit could be only a 0 or 1. The last could be any digit. How many different such area codes were possible? 8·2 ·10 = 160 Ordered Arrangements diagram: In how many different ways could four items be arranged in order from first to last? 4 ·3 ·2·1 = 24 Factorial Notation diagram: • n! is read "n factorial" • n! is applied only when n is a whole number. • n! is a product of n with each positive counting number less than n Calculating Factorials diagram: 120 • 5! = 5 • 4 • 3 • 2 • 1 = • 3! = 3 • 2 • 1 = 6 Definitions diagram: • 1! = 1 • 0! = 1 Complete the Factorials: diagram: 24 4! = 10! = 6! = 15! = 3,628,800 720 1.3077 x 1012 Permutations diagram: • A Permutation is an arrangement in a particular order of a group of items. • There are to be no repetitions of items within a permutation. Listing Permutations diagram: How many different permutations of the letters a, b, c are possible? Solution: There are six different permutations: abc, acb, bac, bca, cab, cba. Listing Permutations diagram: How many different two-letter permutations of the letters a, b, c, d are possible? Solution: There are twelve different permutations: ab, ac, ad, ba, ca, da, bc, bd, cb, db, cd, dc. Permutation Formula diagram: The number of ways to arrange in order n distinct objects, taking them r at a time, is: where n and r are whole numbers and n>r. Find diagram:P7, 3 Applying the Permutation Formula diagram: P3, 3 = _______ P4, 2 = _______ P6, 2 = __________ P8, 3 = _______ P15, 2 = _______ 6 12 30 336 210 Application of Permutations diagram: A teacher has chosen eight possible questions for an upcoming quiz. In how many different ways can five of these questions be chosen and arranged in order from #1 to #5? Solution: P8,5 = = 8• 7 • 6 • 5 • 4 = 6720 Combinations diagram: A combination is a grouping in no particular order of items. Combination Formula diagram: The number of combinations of n objects taken r at a time is: where n and r are whole numbers and n>r. Find diagram:C9, 3 Applying the Combination Formula diagram: 10 35 C5, 3 = C7, 3 = C3, 3 = C15, 2 = C6, 2 = 1 105 15 Application of Combinations diagram: A teacher has chosen eight possible questions for an upcoming quiz. In how many different ways can five of these questions be chosen if order makes no difference? Solution: C8,5 = = 56 Determining the Number of diagram:Outcomes of an Experiment • If the experiment consists of a series of stage with various outcomes, use the multiplication rule or a tree diagram. Determining the Number of diagram:Outcomes of an Experiment If the outcomes consist of ordered subgroups of r outcomes taken from a group of n outcomes use the permutation rule. Determining the Number of diagram:Outcomes of an Experiment If the outcomes consist of non-ordered subgroups of r items taken from a group of n items use the combination rule.
# Question #3d423 Jan 16, 2018 $641.67 \frac{m}{m t s}$ #### Explanation: Suppose,the speed of the hawk in still air is $v \frac{m}{m t s}$ So,while going against wind flow its net velocity becomes $\left(v - u\right) \frac{m}{m t s}$ where, $u$ is the velocity of wind. So,if in this process it goes for a distance of $x$ meters, time taken $\left(t\right)$ will be, $\frac{x}{v - u}$ Again when it goes in the direction of wind,its net velocity becomes $\left(v + u\right) \frac{m}{m t s}$ Hence, here time required $\left(t 1\right)$ to travel the same distance will become $\frac{x}{v + u}$ Now,given that $x = 24 , t = 45 , t 1 = 32$ Solving it we get, $v = 641.67 \frac{m}{m t s}$ Jan 16, 2018 Let the speed of the hawk in still air be ${v}_{h}$km/hr and the speed of wind be ${v}_{w}$km/hr Hence time taken by the hawk to cover a distance of 24 km flying against the wind will be $\frac{24}{{v}_{h} - {v}_{w}}$hr By the problem $\frac{24}{{v}_{h} - {v}_{w}} = \frac{45}{60} = \frac{3}{4}$hr $\implies {v}_{h} - {v}_{w} = 32. \ldots \ldots . . \left(1\right)$ Again in return journey the time taken by the hawk to cover same distance flying with the wind is given by $\frac{24}{{v}_{h} + {v}_{w}} = \frac{32}{60} = \frac{8}{15}$hr $\implies {v}_{h} + {v}_{w} = 45. \ldots \ldots \ldots \left(2\right)$ Adding (1) and (2) we get $2 {v}_{h} = 77$ $\implies {v}_{h} = \frac{77}{2} = 38.5$km/hr
# How can you use addition or subtraction to solve an equation? Save this PDF as: Size: px Start display at page: ## Transcription 1 7.2 Solving Equations Using Addition or Subtraction How can you use addition or subtraction to solve an equation? When two sides of a scale weigh the same, the scale will balance. When you add or subtract the same amount on each side of the scale, it will still balance. 1 ACTIVITY: Solving an Equation Work with a partner. a. Use a model to solve n + 3 = 7. Explain how the model represents the equation n + 3 = 7. How much does one weigh? How do you know? The solution is n =. COMMON CORE Solving Equations In this lesson, you will use addition or subtraction to solve equations. use substitution to check answers. solve real-life problems. Learning Standards 6.EE.5 6.EE.7 b. Describe how you could check your answer in part (a). c. Which model below represents the solution of n + 1 = 9? How do you know? 300 Chapter 7 Equations and Inequalities 3 7.2 Lesson Lesson Tutorials Key Vocabulary solution, p. 302 inverse operations, p. 303 Equations may be true for some values and false for others. A solution of an equation is a value that makes the equation true. Value of x x + 3 = 7 Are both sides equal? =? no Reading The symbol means is not equal to =? 7 7 = 7 yes =? no So, the value x = 4 is a solution of the equation x + 3 = 7. EXAMPLE 1 Checking Solutions Tell whether the given value is a solution of the equation. a. p + 10 = 38; p = =? 38 Substitute 18 for p Sides are not equal So, p = 18 is not a solution. b. 4y = 56; y = 14 4(14) =? 56 Substitute 14 for y. 56 = 56 Sides are equal So, y = 14 is a solution. Exercises 6 11 Tell whether the given value is a solution of the equation. 1. a + 6 = 17; a = g = 5; g = = 7n; n = 5 4. q = 28; q = Chapter 7 Equations and Inequalities 4 You can use inverse operations to solve equations. Inverse operations undo each other. Addition and subtraction are inverse operations. Addition Property of Equality Words When you add the same number to each side of an equation, the two sides remain equal. Numbers 8 = 8 Algebra x 4 = = 13 x = 9 Subtraction Property of Equality Words When you subtract the same number from each side of an equation, the two sides remain equal. Numbers 8 = 8 Algebra x + 4 = = 3 x = 1 EXAMPLE 2 Solving Equations Using Addition a. Solve x 2 = 6. x 2 = 6 Write the equation. Undo the subtraction Addition Property of Equality x = 8 Simplify. The solution is x = 8. Check x 2 = =? 6 6 = 6 Study Tip You can check your solution by substituting it for the variable in the original equation. b. Solve 18 = x = x 7 Write the equation Addition Property of Equality 25 = x Simplify. The solution is x = 25. Check 18 = x 7 18 =? = 18 Exercises Solve the equation. Check your solution. 5. k 3 = 1 6. n 10 = = r 6 Section 7.2 Solving Equations Using Addition or Subtraction 303 6 7.2 Exercises Help with Homework 1. WRITING How can you check the solution of an equation? Name the inverse operation you can use to solve the equation. 2. x 8 = n + 3 = b + 14 = WRITING When solving x + 5 = 16, why do you subtract 5 from the left side of the equation? Why do you subtract 5 from the right side of the equation? 9+(-6)=3 3+(-3)= 4+(-9)= 9+(-1)= 1 Tell whether the given value is a solution of the equation. 6. x + 42 = 85; x = b = 48; b = g = 7; g = m 4 = 16; m = w + 23 = 41; w = s 68 = 11; s = 79 Use a scale to model and solve the equation. 12. n + 7 = t + 4 = c + 2 = 8 Write a question that represents the equation. Use mental math to answer the question. Then check your solution. 15. a + 5 = v + 9 = = d 6 Solve the equation. Check your solution y 7 = z 3 = = r p + 5 = k + 6 = = h f 27 = = q = j x = = m a = 17.3 ERROR ANALYSIS Describe and correct the error in solving the equation. 30. x + 7 = x = = y = y Section 7.2 Solving Equations Using Addition or Subtraction 305 7 24 in. 32. PENGUINS An emperor penguin is 45 inches tall. It is 24 inches taller than a rockhopper penguin. Write and solve an equation to find the height of a rockhopper penguin. Is your answer reasonable? Explain. 33. ELEVATOR You get in an elevator and go down 8 floors. You exit on the 16th floor. Write and solve an equation to find what floor you got on the elevator. 34. AREA The area of Jamaica is 6460 square miles less than the area of Haiti. Write and solve an equation to find the area of Haiti. JAMAICA Kingston Area = 4181 mi 2 HAITI Port-au- Prince DOMINICAN REPUBLIC Santo Domingo 35. REASONING The solution of the equation x + 3 = 12 is shown. Explain each step. Use a property, if possible. x + 3 = 12 x = 12 3 x + 0 = 9 x = 9 Write the equation. Write the word sentence as an equation. Then solve the equation subtracted from a number w is A number k increased by 7 is is the difference of a number n and is the sum of a number g and 58. Solve the equation. Check your solution. 40. b = y = m = v 7 = = 2 + r = d 17 GEOMETRY Write and solve an addition equation to find x. 46. Perimeter = 48 ft 47. Perimeter = 132 in. 48. Perimeter = 93 ft 16 in. 18 ft 18 ft x 20 ft 34 in. 34 in. 15 ft 15 ft 12 ft x x 49. REASONING Explain why the equations x + 4 = 13 and 4 + x = 13 have the same solution. 50. REASONING Explain why the equations x 13 = 4 and 13 x = 4 do not have the same solution. 306 Chapter 7 Equations and Inequalities 8 51. SIMPLIFYING AND SOLVING Compare and contrast the two problems. Simplify the expression 2(x + 3) 4. 2(x + 3) 4 = 2x = 2x + 2 Solve the equation x + 3 = 4. x + 3 = x = PUZZLE In a magic square, the sum of the numbers in each row, column, and diagonal is the same. Write and solve equations to find the values of a, b, and c. 53. FUNDRAISER You participate in a dance-a-thon fundraiser. After your parents pledge \$15.50 and your neighbor pledges \$8.75, you have \$ Write and solve an equation to find how much money you had before your parents and neighbor pledged. a b 34 c MONEY On Saturday, you spend \$33, give \$15 to a friend, and receive \$20 for mowing your neighbor s lawn. You have \$21 left. Use two methods to find how much money you started with that day. Bumper Cars: Roller Coaster: Giant Slide: Ferris Wheel: \$1.75 \$1.25 more than Ferris Wheel \$0.50 less than Bumper Cars \$1.50 more than Giant Slide 55. AMUSEMENT PARK You have \$15. a. How much money do you have left if you ride each ride once? b. Do you have enough money to ride each ride twice? Explain. 56. Consider the equation x + y = 15. The value of x increases by 3. What has to happen to the value of y so that x + y = 15 remains true? Find the value of the expression. Use estimation to check your answer. (Section 1.1) MULTIPLE CHOICE What is the area of the parallelogram? (Section 4.1) A 25 in. 2 B 30 in. 2 C 50 in. 2 D 100 in. 2 5 in. 10 in. Section 7.2 Solving Equations Using Addition or Subtraction 307 ### Writing and Graphing Inequalities 4.1 Writing and Graphing Inequalities solutions of an inequality? How can you use a number line to represent 1 ACTIVITY: Understanding Inequality Statements Work with a partner. Read the statement. Circle ### Writing and Graphing Inequalities .1 Writing and Graphing Inequalities solutions of an inequality? How can you use a number line to represent 1 ACTIVITY: Understanding Inequality Statements Work with a partner. Read the statement. Circle ### Inequalities. and Graphing Inequalities 4.3 Solving Inequalities Using. What would you have? Inequalities.1 Writing ii in and Graphing Inequalities. Solving Inequalities Using Addition or Subtraction. Solving Inequalities Using Multiplication or Division. Solving Two-Step Inequalities If you reached ### Is the sum of two integers positive, negative, or zero? How can you tell? ACTIVITY: Adding Integers with the Same Sign .2 Adding Integers Is the sum of two integers positive, negative, or zero? How can you tell? ACTIVITY: Adding Integers with the Same Sign Work with a partner. Use integer counters to find 4 + ( ). Combine ### Solving Inequalities Using Addition or Subtraction 7.6. ACTIVITY: Writing an Inequality. ACTIVITY: Writing an Inequality 7.6 Solving Inequalities Using Addition or Subtraction How can you use addition or subtraction to solve an inequality? 1 ACTIVITY: Writing an Inequality Work with a partner. In 3 years, your friend will ### Words to Review. Give an example of the vocabulary word. Numerical expression. Variable. Evaluate a variable expression. Variable expression 1 Words to Review Give an example of the vocabulary word. Numerical expression 5 12 Variable x Variable expression 3x 1 Verbal model Distance Rate p Time Evaluate a variable expression Evaluate the expression ### How can you use multiplication or division to solve an equation? ACTIVITY: Finding Missing Dimensions 7.3 Solving Equations Using Multiplication or Division How can you use multiplication or division to solve an equation? 1 ACTIVITY: Finding Missing Dimensions Work with a partner. Describe how you would ### Words to Review. Give an example of the vocabulary word. Numerical expression. Variable. Variable expression. Evaluate a variable expression 1 Words to Review Give an example of the vocabulary word. Numerical expression 5 1 Variable x Variable expression 3x 1 Verbal model Distance Rate p Time Evaluate a variable expression Evaluate the expression ### How can you write and evaluate an expression that represents a real-life problem? ACTIVITY: Reading and Re-Reading .1 Algebraic Expressions How can you write and evaluate an expression that represents a real-life problem? 1 ACTIVITY: Reading and Re-Reading Work with a partner. a. You babysit for hours. You receive ### How can you use multiplication or division to solve an inequality? ACTIVITY: Using a Table to Solve an Inequality . Solving Inequalities Using Multiplication or Division How can you use multiplication or division to solve an inequality? 1 ACTIVITY: Using a Table to Solve an Inequality Work with a partner. Copy and ### Think of three math problems using the four operations where order matters and three where order doesn t matter. ACTIVITY: Commutative Properties 3.3 Properties of Addition and Multiplication operation matter? Does the order in which you perform an ACTIVITY: Does Order Matter? Work with a partner. Place each statement in the correct oval. a. Fasten ### Study Guide For use with pages 63 68 2.1 For use with pages 63 68 GOAL Use properties of addition and multiplication. VOCABULARY Lesson 2.1 Commutative Property of Addition: In a sum, you can add the numbers in any order. Associative Property ### ACTIVITY: Simplifying Algebraic Expressions . Algebraic Expressions How can you simplify an algebraic expression? ACTIVITY: Simplifying Algebraic Expressions Work with a partner. a. Evaluate each algebraic expression when x = 0 and when x =. Use ### Writing Equations in One Variable 7.1 Writing Equations in One Variable solve the word problem? How does rewriting a word problem help you 1 ACTIVITY: Rewriting a Word Problem Work with a partner. Read the problem several times. Think ### Unit 4, Lesson 1: Number Puzzles Unit 4, Lesson 1: Number Puzzles Let s solve some puzzles! 1.1: Notice and Wonder: A Number Line What do you notice? What do you wonder? 1.2: Telling Temperatures Solve each puzzle. Show your thinking. ### Algebra I Solving & Graphing Inequalities Slide 1 / 182 Slide 2 / 182 Algebra I Solving & Graphing Inequalities 2016-01-11 www.njctl.org Slide 3 / 182 Table of Contents Simple Inequalities Addition/Subtraction click on the topic to go to that ### ACTIVITY: Quarterback Passing Efficiency 3. Solving Inequalities Using Addition or Subtraction solve an inequality? How can you use addition or subtraction to 1 ACTIVITY: Quarterback Passing Efficiency Work with a partner. The National Collegiate ### MathB65 Ch 1 I,II,III, IV.notebook. August 23, 2017 Chapter 1: Expressions, Equations and Inequalities I. Variables & Constants II. Algebraic Expressions III. Translations IV. Introduction to Equations V. The Additive/Multiplication Principles VI. Strategy ### Solving and Graphing Linear Inequalities Chapter Questions. 2. Explain the steps to graphing an inequality on a number line. Solving and Graphing Linear Inequalities Chapter Questions 1. How do we translate a statement into an inequality? 2. Explain the steps to graphing an inequality on a number line. 3. How is solving an inequality ### How can you find decimal approximations of square roots that are not rational? ACTIVITY: Approximating Square Roots . Approximating Square Roots How can you find decimal approximations of square roots that are not rational? ACTIVITY: Approximating Square Roots Work with a partner. Archimedes was a Greek mathematician, ### Essential Question How can you solve an absolute value inequality? Work with a partner. Consider the absolute value inequality x Learning Standards HSA-CED.A.1 HSA-REI.B.3.6 Essential Question How can you solve an absolute value inequality? COMMON CORE Solving an Absolute Value Inequality Algebraically MAKING SENSE OF PROBLEMS To ### Investigating Inequalities: Investigating Inequalities: Choose roles: Record each group member s name next to their role: Anuncer: Recorder: Walker A: Walker B: Set-up: use the number cards to construct a number line on the floor. ### Systems of Linear Inequalities . Sstems of Linear Inequalities sstem of linear inequalities? How can ou sketch the graph of a ACTIVITY: Graphing Linear Inequalities Work with a partner. Match the linear inequalit with its graph. + Inequalit ### How can you use inductive reasoning to observe patterns and write general rules involving properties of exponents? 0. Product of Powers Property How can you use inductive reasoning to observe patterns and write general rules involving properties of exponents? ACTIVITY: Finding Products of Powers Work with a partner. ### Quiz For use after Section 4.2 Name Date Quiz For use after Section.2 Write the word sentence as an inequality. 1. A number b subtracted from 9.8 is greater than. 2. The quotient of a number y and 3.6 is less than 6.5. Tell whether ### Regular Algebra 1 Fall Final Exam Review Regular Algebra 1 Fall Final Exam Review Name: 1. A home store is having a 10%- off sale on all in-stock bathroom floor tile. What is the relationship between the sale price of the tile and the original ### Unit 5. Linear equations and inequalities OUTLINE. Topic 13: Solving linear equations. Topic 14: Problem solving with slope triangles Unit 5 Linear equations and inequalities In this unit, you will build your understanding of the connection between linear functions and linear equations and inequalities that can be used to represent and ### SOLVING LINEAR INEQUALITIES Topic 15: Solving linear inequalities 65 SOLVING LINEAR INEQUALITIES Lesson 15.1 Inequalities on the number line 15.1 OPENER Consider the inequality x > 7. 1. List five numbers that make the inequality ### ACTIVITY: Areas and Perimeters of Figures 4.4 Solving Two-Step Inequalities the dimensions of a figure? How can you use an inequality to describe 1 ACTIVITY: Areas and Perimeters of Figures Work with a partner. Use the given condition to choose ### Solving Equations with Variables on Both Sides . Solving Equations with Variables on Both Sides variables on both sides? How can you solve an equation that has ACTIVITY: Perimeter and Area Work with a partner. Each figure has the unusual property that ### Solving Systems of Linear Equations by Elimination 5.3 Solving Systems of Linear Equations by Elimination system of linear equations? How can you use elimination to solve a ACTIVITY: Using Elimination to Solve a System Work with a partner. Solve each system ### 6th Grade. Equations & Inequalities. 1 6th Grade Equations & Inequalities 2015 12 01 www.njctl.org 2 Table of Contents Equations and Identities Tables Determining Solutions of Equations Solving an Equation for a Variable Click on a topic ### Solving Equations by Multiplying or Dividing Practice A Solving Equations by Multiplying or Dividing Solve and check. 1. 3 = 12 2. 2y = 18 3. 8z = 32 4. a 3 = 2 5. b 5 = 6 6. c 9 = 3 7. 5n = 35 8. 3k = 24 9. 7p = 63 10. m 4 = 4 11. v 2 = 6 12. s 7.2 Adding and Subtracting Polynomials subtract polynomials? How can you add polynomials? How can you 1 EXAMPLE: Adding Polynomials Using Algebra Tiles Work with a partner. Six different algebra tiles ### ACTIVITY: Factoring Special Products. Work with a partner. Six different algebra tiles are shown below. 7.9 Factoring Special Products special products? How can you recognize and factor 1 ACTIVITY: Factoring Special Products Work with a partner. Six different algebra tiles are shown below. 1 1 x x x 2 x ### UNIT 5 INEQUALITIES CCM6+/7+ Name: Math Teacher: UNIT 5 INEQUALITIES 2015-2016 CCM6+/7+ Name: Math Teacher: Topic(s) Page(s) Unit 5 Vocabulary 2 Writing and Graphing Inequalities 3 8 Solving One-Step Inequalities 9 15 Solving Multi-Step Inequalities ### Algebra 1 Unit 6: Linear Inequalities and Absolute Value Guided Notes Section 6.1: Solving Inequalities by Addition and Subtraction How do we solve the equation: x 12 = 65? How do we solve the equation: x 12 < 65? Graph the solution: Example 1: 12 y 9 Example 2: q + 23 ### Algebraic Expressions and Properties Algebraic Expressions and Properties. Algebraic Expressions. Writing Expressions. Properties of Addition and Multiplication.4 The Distributive Property 4 56 4 5 6 4 6 8 0 6 9 5 8 4 5 6 4 5 6 8 0 5 8 6 ### Eureka Math Module 4 Topic G Solving Equations Eureka Math Module 4 Topic G Solving Equations Lessons 23-27 6.EE.B.5 Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make ### Unit Essential Questions. Can equations that appear to be different be equivalent? How can you solve equations? Unit Essential Questions Can equations that appear to be different be equivalent? How can you solve equations? What kinds of relationships can proportions represent? Williams Math Lessons TARGET ONE-STEP ### How can you use algebra tiles to solve addition or subtraction equations? . Solving Equations Using Addition or Subtraction How can you use algebra tiles to solve addition or subtraction equations? ACTIVITY: Solving Equations Work with a partner. Use algebra tiles to model and ### Fair Game Review. 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Use algebra ### Lesson 1 Reteach. Equations. Example 1. Example 2. Exercises Lesson 1 Reteach Equations An equation is a mathematical sentence showing two expressions are equal. An equation contains an equals sign, =. Some equations contain variables. When you replace a variable ### GRADE 7 MATH LEARNING GUIDE. Lesson 26: Solving Linear Equations and Inequalities in One Variable Using GRADE 7 MATH LEARNING GUIDE Lesson 26: Solving Linear Equations and Inequalities in One Variable Using Guess and Check Time: 1 hour Prerequisite Concepts: Evaluation of algebraic expressions given values ### UNIT 7 CCM6 and CCM6+ UNIT 7. Equations and Inequalities. CCM6 and CCM Unit 7 Vocabulary...2 UNIT 7 Equations and Inequalities CCM6 and CCM6+ 2015-16 Name: Math Teacher: Projected Test Date: Unit 7 Vocabulary.......2 Identify Equations/Expressions/Inequalities. 3-6 Identify Solutions to Equations ### 2-4. 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I can write an algebraic expression for addition, subtraction, multiplication, ### 3rd Grade Emergency Sub Plans 3rd Grade Emergency Sub Plans created by The Curriculum Corner Measurement in a Bag Pick a shape from the bag. Trace the shape. Measure and label the sides. Find the perimeter. The perimeter is: Measurement ### What You ll Learn. or irrational. New Vocabulary perfect square, square root, irrational numbers, real numbers. Why Learn This? -. Plan - Exploring Square Roots and Irrational Numbers Objective To find and estimate square roots and to classify numbers as rational or irrational Examples Finding Square Roots of Perfect Squares Estimating ### Fair Game Review. Chapter inches. Your friend s height is 5.6 feet. Who is taller? Explain. Name Date Chapter 8 Fair Game Review Complete the number sentence with , or =. 1. 3 0.. 7 0.7 10 3. 0.6. 3 1.75 3 5. 1 6 6. 31 1.8 16 7. Your height is 5 feet and 1 5 8 inches. Your friend s height ### CCGPS Coordinate Algebra. EOCT Review Units 1 and 2 CCGPS Coordinate Algebra EOCT Review Units 1 and 2 Unit 1: Relationships Among Quantities Key Ideas Unit Conversions A quantity is a an exact amount or measurement. A quantity can be exact or approximate ### ALGEBRA 1 SUMMER ASSIGNMENT Pablo Muñoz Superintendent of Schools Amira Presto Mathematics Instructional Chair Summer Math Assignment: The Passaic High School Mathematics Department requests all students to complete the summer assignment. ### Chapter 1 ( )? Chapter 1 Opener. Section 1.1. Worked-Out Solutions. 2π π = π. Try It Yourself (p. 1) So, x = 95.3. Chapter Chapter Opener Try It Yourself (p. ). + ( ) 7.. + 8. ( ) +. 7. ( 7) + 7 7. 8 () 0 + 8. 7. ( 7) 8 0.. 8. Section.. Activity (pp. ). Triangle Angle A (degrees) Angle B (degrees). a. The sum of the ### Pre-Algebra Notes Unit Two: Solving Equations Pre-Algebra Notes Unit Two: Solving Equations Properties of Real Numbers Syllabus Objective: (.1) The student will evaluate expressions using properties of addition and multiplication, and the distributive ### Solving Systems of Linear and Quadratic Equations 9.5 Solving Systems of Linear and Quadratic Equations How can you solve a system of two equations when one is linear and the other is quadratic? ACTIVITY: Solving a System of Equations Work with a partner. ### CHAPTER 2: INTRODUCTION TO VARIABLES AND PROPERTIES OF ALGEBRA CHAPTER 2: INTRODUCTION TO VARIABLES AND PROPERTIES OF ALGEBRA Chapter Objectives By the end of this chapter, students should be able to: Interpret different meanings of variables Evaluate algebraic expressions ### Evaluate algebraic expressions and use exponents. Translate verbal phrases into expressions. Algebra 1 Notes Section 1.1: Evaluate Expressions Section 1.3: Write Expressions Name: Hour: Objectives: Section 1.1: (The "NOW" green box) Section 1.3: Evaluate algebraic expressions and use exponents. ### Unit 3. Expressions. Unit 3 Calendar Unit 3 Expressions Exponents Order of Operations Evaluating Algebraic Expressions Translating Words to Math Identifying Parts of Exprsessions Evaluating Formulas Algebraic Properties Simplifying Expressions ### THANK YOU FOR YOUR PURCHASE! THANK YOU FOR YOUR PURCHASE! The resources included in this purchase were designed and created by me. I hope that you find this resource helpful in your classroom. Please feel free to contact me with any ### 1-5 Study Guide and Intervention 1-5 Study Guide and Intervention One-Step Inequalities The following properties can be used to solve inequalities. Addition and Subtraction Properties for Inequalities For any real numbers a, b, and c: ### Why? 0.10d 2x z _ Variables and Expressions Then You performed operations on integers. (Lesson 0-3) Now 1Write verbal expressions for algebraic expressions. 2Write algebraic expressions for verbal expressions. Why? Cassie ### Name Period Date DRAFT Name Period Date Equations and Inequalities Student Packet 4: Inequalities EQ4.1 EQ4.2 EQ4.3 Linear Inequalities in One Variable Add, subtract, multiply, and divide integers. Write expressions, equations, ### Fair Game Review. Chapter 4. x > Name Date. Graph the inequality. 5. x 2.3 Name Date Chapter 4 Fair Game Review Graph the inequality. 1. x < 3 2. x 5 3. x 2 4. x > 7 5. x 2.3 6. x > 2 5 7. The deepest free dive by a human in the ocean is 417 feet. The depth humans have been in ### UNIT 2 SOLVING EQUATIONS UNIT 2 SOLVING EQUATIONS NAME: GRADE: TEACHER: Ms. Schmidt _ Solving One and Two Step Equations The goal of solving equations is to. We do so by using. *Remember, whatever you to do one side of an equation. ### Lesson 22: Solving Equations Using Algebra Student Outcomes Students use algebra to solve equations (of the form px + q = r and p(x + q) = r, where p, q, and r are specific rational numbers); using techniques of making zero (adding the additive ### Can a system of linear equations have no solution? Can a system of linear equations have many solutions? 5. Solving Special Sstems of Linear Equations Can a sstem of linear equations have no solution? Can a sstem of linear equations have man solutions? ACTIVITY: Writing a Sstem of Linear Equations Work with ### Solving Systems of Linear Equations by Substitution 5.2 Solving Systems of Linear Equations by Substitution a system of linear equations? How can you use substitution to solve 1 ACTIVITY: Using Substitution to Solve a System Work with a partner. Solve each ### 2-2. Warm Up. Simplify each expression. 1. ( 7)(2.8) ( 9)( 9) Warm Up Simplify each expression. 1. ( 7)(2.8) 2. 0.96 6 3. ( 9)( 9) 4. 5. 6. Learning Goals 1. Students will solve and check equations using multiplication 2. Students will solve and check equations using ### Writing and Solving Equations Writing and Solving Equations Melody s Music Solution Lesson 6-1 Modeling and Writing Two-Step Equations ACTIVITY 6 Learning Targets: Use variables to represent quantities in real-world problems. Model ### Algebraic expression is formed from variables and constants using different operations. NCERT UNIT 10 ALGEBRAIC EXPRESSIONS (A) Main Concepts and Results Algebraic expression is formed from variables and constants using different operations. Expressions are made up of terms. A term is the product ### Lesson 3-7: Absolute Value Equations Name: Lesson 3-7: Absolute Value Equations Name: In this activity, we will learn to solve absolute value equations. An absolute value equation is any equation that contains an absolute value symbol. To start, ### MATH ALGEBRA AND FUNCTIONS Students: 1. Use letters, boxes, or other symbols to stand for any number in simple expressions or equations. 1. Students use and interpret variables, mathematical symbols and properties to write and simplify
Question of Exercise 3 (True and False) # Question The age of father is 3 times the age of the son. If sum of their ages is 48 years, then the age of father and son are 35 and 12 years respectively Factorize x2+x-6 Solution: Explanation: On factoring the equation, we get, x2+x-6 x2+3x-2x-6 x(x+3)-2(x+3) (x+3)(x-2) The simplified form of the equation x2+x-6 is (x+3)(x-2). Factorize x^2-2x-8 Solution: Explanation: We have; x2-2x-8 x2-4x+2x-8 x(x-4)+2(x-4) (x-4) (x+2) x=4,x=-2 Hence, we factorized the given expression as, x=4,x=-2 The opposite angles of a parallelogram are are (3x - 2) and (x + 48) Find the measure of each angle of the parallelogram. Solution: Explanation: Let the parallelogram be ABCD & the angles of parallelogram be <A,<B,<C & <D From the given question, opposite angles of a parallelogram are (3x-2) & (x+48). Let <A=3x-2 & <B=x+48 As we know that, “the opposite angles of a parallelogram are always equal”. Therefore, we can write; (3x-2)=(x+48) ⇒3x-x=48+2 ⇒2x=50 ⇒x=50/2 ⇒x=25 Substituting the value of x in 3x-2, we get; 3(25)-2 =75-2 =73º ⇒<A=<C=73 Finding the measure of other two angles: We know that, the sum of adjacent angles of a parallelogram is equal to 180. ∴ <A+<B=180º ⇒73º+<B=180º ⇒<B=180º-73º ⇒<B=107º ∴ <D=107º Therefore, the measure of each angle of the parallelogram is <A=73º,<B=107º,<C=73º,<D=107º. Hence, we measured all the angles of parallelogram as;<A=73º,<B=107º,<C=73º,<D=107º. In the given figure the value of x is Solution: Explanation:- The value of x is 125º. Estimate the value of square root square root :√22 Solution: Final Answer: The estimated square root of √22 is 4.690.
# Critical Points ### Definition Let $$f\left(x\right)$$ be a function and let $$c$$ be a point in the domain of the function. The point $$c$$ is called a critical point of $$f$$ if either $$f’\left( c \right) = 0$$ or $$f’\left( c \right)$$ does not exist. ### Examples of Critical Points A critical point $$x = c$$ is a local minimum if the function changes from decreasing to increasing at that point. 1. The function $$f\left( x \right) = x + {e^{ – x}}$$ has a critical point (local minimum) at $$c = 0.$$ The derivative is zero at this point. $f\left( x \right) = x + {e^{ – x}}.$ ${f^\prime\left( x \right) = \left( {x + {e^{ – x}}} \right)^\prime }={ 1 – {e^{ – x}}.}$ ${f^\prime\left( c \right) = 0,\;\;} \Rightarrow {1 – {e^{ – c}} = 0,\;\;} \Rightarrow {{e^{ – c}} = 1,\;\;} \Rightarrow {{e^{ – c}} = {e^0},\;\;} \Rightarrow {c = 0.}$ 2. The function $$f\left( x \right) = \left\| {x – 3} \right\|$$ has a critical point (local minimum) at $$c = 3.$$ The derivative does not exist at this point. 3. A critical point $$x = c$$ is a local maximum if the function changes from increasing to decreasing at that point. 4. The function $$f\left( x \right) = 2x – {x^2}$$ has a critical point (local maximum) at $$c = 1.$$ The derivative is zero at this point. $f\left( x \right) = 2x – {x^2}.$ ${f^\prime\left( x \right) = \left( {2x – {x^2}} \right)^\prime }={ 2 – 2x.}$ ${f^\prime\left( c \right) = 0,\;\;} \Rightarrow {2 – 2c = 0,\;\;} \Rightarrow {c = 1.}$ 5. The function $$f\left( x \right) = 1 – \left\| {x + 2} \right\|$$ has a critical point (local maximum) at $$c = -2.$$ The derivative does not exist at this point. 6. A critical point $$x = c$$ is an inflection point if the function changes concavity at that point. 7. The function $$f\left( x \right) = {x^3}$$ has a critical point (inflection point) at $$c = 0.$$ The first and second derivatives are zero at $$c = 0.$$ $f\left( x \right) = {x^3}.$ $f^\prime\left( x \right) = \left( {{x^3}} \right)^\prime = 3{x^2}.$ ${f^\prime\left( c \right) = 0,\;\;} \Rightarrow {3{c^2} = 0,\;\;} \Rightarrow {c = 0.}$ 8. Trivial case: Each point of a constant function is critical. For example, any point $$c \gt 0$$ of the function $$f\left( x \right) = \left\{ {\begin{array}{{20}{l}} {2 – x,\;x \le 0}\\ {2,\;x \gt 0} \end{array}} \right.$$ is a critical point since $$f^\prime\left( c \right) = 0.$$ $f\left( x \right) = \left\{ {\begin{array}{{20}{l}} {2 – x,\;x \le 0}\\ {2,\;x \gt 0} \end{array}} \right..$ $f^\prime\left( x \right) = \left\{ {\begin{array}{{20}{l}} { – 1,\;x \le 0}\\ {0,\;x \gt 0} \end{array}} \right..$ ${f^\prime\left( c \right) = 0,\;\;} \Rightarrow {c \gt 0.}$ ## Solved Problems Click or tap a problem to see the solution. ### Example 1 Find the critical points of the function $$f\left( x \right) = {x^2}\ln x$$. ### Example 2 Find the critical points of the function $$f\left( x \right) = 8{x^3} – {x^4}.$$ ### Example 3 Find the critical points of the function $$f\left( x \right) = {x^4} – 5{x^4} + 5{x^3} – 1.$$ ### Example 4 Find the critical points of the function $$f\left( x \right) = \large{\frac{{{x^2} – 4x + 3}}{{x – 2}}}\normalsize.$$ ### Example 5 Find all critical points of the function $$f\left( x \right) = x\sqrt {1 – {x^2}}.$$ ### Example 6 Indicate all critical points of the function $$f\left( x \right) = \large{\frac{{{e^x}}}{x}}\normalsize.$$ ### Example 7 Indicate all critical points of the function $$f\left( x \right) = \left\| {{x^2} – 5} \right\|.$$ ### Example 8 Find all critical points of the function $$f\left( x \right) = \large{\frac{x}{{\ln x}}}\normalsize.$$ ### Example 9 Determine critical points of the function $$f\left( x \right) = \left\| {{x^2} – 4x + 3} \right\|.$$ ### Example 10 Find the critical points of the function $$f\left( x \right) = {\sin ^2}x – \cos x$$ on the interval $$\left( {0,2\pi } \right).$$ ### Example 11 How many critical points does the function $$f\left( x \right) = \left\| {{x^3} – 12x} \right\|$$ have? ### Example 1. Find the critical points of the function $$f\left( x \right) = {x^2}\ln x$$. Solution. Take the derivative using the product rule: ${f^\prime\left( x \right) = \left( {{x^2}\ln x} \right)^\prime }={ 2x \cdot \ln x + {x^2} \cdot \frac{1}{x} }={ 2x\ln x + x }={ x\left( {2\ln x + 1} \right).}$ Determine the points where the derivative is zero: ${f^\prime\left( c \right) = 0,\;\;} \Rightarrow \cssId{element14}{c\left( {2\ln c + 1} \right) = 0.}$ The first root $${c_1} = 0$$ is not a critical point because the function is defined only for $$x \gt 0.$$ Consider the second root: ${2\ln c + 1 = 0,\;\;} \Rightarrow {\ln c = – \frac{1}{2},\;\;} \Rightarrow {{c_2} = {e^{ – \frac{1}{2}}} = \frac{1}{{\sqrt e }}.}$ Hence $${c_2} = \large{\frac{1}{{\sqrt e }}}\normalsize$$ is a critical point of the given function. ### Example 2. Find the critical points of the function $$f\left( x \right) = 8{x^3} – {x^4}.$$ Solution. The function is defined and differentiable over the entire set of real numbers. Therefore, we just differentiate it to determine where the derivative is zero. ${f^\prime\left( x \right) = \left( {8{x^3} – {x^4}} \right)^\prime }={ 24{x^2} – 4{x^3}.}$ ${f^\prime\left( c \right) = 0,}\;\; \Rightarrow {24{c^2} – 4{c^3} = 0,}\;\; \Rightarrow {4{c^2}\left( {6 – c} \right) = 0,}\;\; \Rightarrow {{c_1} = 0,{c_2} = 6.}$ Hence, the function has 2 critical points $${c_1} = 0,{c_2} = 6.$$ ### Example 3. Find the critical points of the function $$f\left( x \right) = {x^4} – 5{x^4} + 5{x^3} – 1.$$ Solution. The function is defined and differentiable for all $$x$$. Calculate the derivative: ${f^\prime\left( x \right) = \left( {{x^4} – 5{x^4} + 5{x^3} – 1} \right)^\prime }={ 5{x^4} – 20{x^3} + 15{x^2}.}$ By equating the derivative to zero, we get the critical points: ${f^\prime\left( c \right) = 0,}\;\; \Rightarrow {5{c^4} – 20{c^3} + 15{c^2} = 0,}\;\; \Rightarrow {5{c^2}\left( {{c^2} – 4c + 3} \right) = 0.}$ We can factor the quadratic expression: ${c^2} – 4c + 3 = \left( {c – 1} \right)\left( {c – 3} \right),$ so the latter equation is written as $5{c^2}\left( {c – 1} \right)\left( {c – 3} \right) = 0.$ Thus, the function has the following critical points: ${c_1} = 0,\,{c_2} = 1,\,{c_3} = 3.$ ### Example 4. Find the critical points of the function $$f\left( x \right) = \large{\frac{{{x^2} – 4x + 3}}{{x – 2}}}\normalsize.$$ Solution. Take the derivative by the quotient rule: ${f^\prime\left( x \right) = \left( {\frac{{{x^2} – 4x + 3}}{{x – 2}}} \right)^\prime }={ \frac{{\left( {2x – 4} \right)\left( {x – 2} \right) – \left( {{x^2} – 4x + 3} \right) \cdot 1}}{{{{\left( {x – 2} \right)}^2}}} = \frac{{{x^2} – 4x + 5}}{{{{\left( {x – 2} \right)}^2}}}.}$ Solve the equation $$f^\prime\left( c \right) = 0:$$ ${\frac{{{c^2} – 4c + 5}}{{{{\left( {c – 2} \right)}^2}}} = 0,}\;\; \Rightarrow {\left\{ {\begin{array}{{20}{l}} {{c^2} – 4c + 5 = 0}\\ {c – 2 \ne 0} \end{array}} \right..}$ The quadratic equation has no roots as the discriminant $$D = 16 – 20 = – 4 \lt 0.$$ Note that $$x = 2$$ is a not a critical point as the function is not defined at this point. Thus, the given function has no critical points. ### Example 5. Find all critical points of the function $$f\left( x \right) = x\sqrt {1 – {x^2}}.$$ Solution. First we determine the domain of the function: ${1 – {x^2} \ge 0,}\;\; \Rightarrow {{x^2} \le 1,}\;\; \Rightarrow {- 1 \le x \le 1.}$ Find the derivative by the product rule: ${f^\prime\left( x \right) = \left( {x\sqrt {1 – {x^2}} } \right)^\prime }={ x^\prime\sqrt {1 – {x^2}} + x\left( {\sqrt {1 – {x^2}} } \right)^\prime }={ \sqrt {1 – {x^2}} + x \cdot \frac{{\left( { – 2x} \right)}}{{2\sqrt {1 – {x^2}} }} }={ \frac{{1 – {x^2} – {x^2}}}{{\sqrt {1 – {x^2}} }} }={ \frac{{1 – 2{x^2}}}{{\sqrt {1 – {x^2}} }}.}$ Determine the points at which the derivative is zero: ${f^\prime\left( c \right) = 0,}\;\; \Rightarrow {\frac{{1 – 2{c^2}}}{{\sqrt {1 – {c^2}} }} = 0,}\;\; \Rightarrow {\left\{ {\begin{array}{{20}{l}} {1 – 2{c^2} = 0}\\ {\sqrt {1 – {c^2}} \ne 0} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{{20}{l}} {{c^2} = \frac{1}{2}}\\ {{c^2} \ne 1} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{{20}{l}} {{c_{1,2}} = \pm \frac{{\sqrt 2 }}{2}}\\ {c \ne \pm 1} \end{array}} \right.,}\;\; \Rightarrow {{c_{1,2}} = \pm \frac{{\sqrt 2 }}{2}.}$ So we have two points in the domain of the function where the derivative is zero. Hence, these points are critical, by definition. ${{c_1} = – \frac{{\sqrt 2 }}{2},}\;{{c_2} = \frac{{\sqrt 2 }}{2}.}$ ### Example 6. Indicate all critical points of the function $$f\left( x \right) = \large{\frac{{{e^x}}}{x}}\normalsize.$$ Solution. The function is defined over all $$x$$ except $$x = 0$$ where it has a discontinuity. Differentiate the function using the quotient rule: ${f^\prime\left( x \right) = \left( {\frac{{{e^x}}}{x}} \right)^\prime }={ \frac{{\left( {{e^x}} \right)^\prime \cdot x – {e^x} \cdot x^\prime}}{{{x^2}}} }={ \frac{{{e^x} \cdot x – {e^x} \cdot 1}}{{{x^2}}} }={ \frac{{\left( {x – 1} \right){e^x}}}{{{x^2}}}.}$ Solve the equation $$f^\prime\left( c \right) = 0:$$ ${f^\prime\left( c \right) = 0,}\;\;\Rightarrow {\frac{{\left( {c – 1} \right){e^c}}}{{{c^2}}} = 0,}\;\; \Rightarrow {c = 1.}$ Note that $$c =0$$ is not a critical point since the function itself is not defined here. Therefore, the function has one critical point $$c = 1.$$ ### Example 7. Indicate all critical points of the function $$f\left( x \right) = \left\| {{x^2} – 5} \right\|.$$ Solution. Find the roots of the function: ${f\left( x \right) = 0,}\;\; \Rightarrow {\left\| {{x^2} – 5} \right\| = 0,}\;\; \Rightarrow {{x_{1,2}} = \pm \sqrt 5 .}$ The derivative does not exist at the corner points $$x = – \sqrt 5$$ and $$x = \sqrt 5 ,$$ i.e. these points are critical. In the interval $$\left[ { – \sqrt 5 ,\sqrt 5 } \right],$$ the function is written as ${f\left( x \right) = – \left( {{x^2} – 5} \right) }={ – {x^2} + 5.}$ Solving the equation $$f^\prime\left( c \right) = 0$$ on this interval, we get one more critical point: ${f^\prime\left( c \right) = 0,}\;\; \Rightarrow {- 2c = 0,}\;\; \Rightarrow {c = 0.}$ Hence, the function has three critical points: ${{c_1} = – \sqrt 5,}\;{{c_2} = 0,}\;{{c_3} = \sqrt 5 .}$ ### Example 8. Find all critical points of the function $$f\left( x \right) = \large{\frac{x}{{\ln x}}}\normalsize.$$ Solution. The domain of $$f\left( x \right)$$ is determined by the conditions: $\left\{ \begin{array}{l} x \gt 0\\ \ln x \ne 0 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} x \gt 0\\ x \ne 1 \end{array} \right..$ Take the derivative using the quotient rule: ${f^\prime\left( x \right) = \left( {\frac{x}{{\ln x}}} \right)^\prime }={ \frac{{x^\prime \cdot \ln x – x \cdot \left( {\ln x} \right)^\prime}}{{{{\ln }^2}x}} }={ \frac{{1 \cdot \ln x – x \cdot \frac{1}{x}}}{{{{\ln }^2}x}} }={ \frac{{\ln x – 1}}{{{{\ln }^2}x}}.}$ Equating the derivative to zero, we find the critical points $$c:$$ ${f^\prime\left( c \right) = 0,}\;\; \Rightarrow {\frac{{\ln c – 1}}{{{{\ln }^2}c}} = 0,}\;\; \Rightarrow {\left\{ {\begin{array}{{20}{l}} {\ln c = 1}\\ {{{\ln }^2}c \ne 0} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} {c = e}\\ {c \ne 1} \end{array}} \right..}$ Note that the derivative does not exist at $$c = 1$$ (where the denominator of the derivative approaches zero). But the function itself is also undefined at this point. Therefore, $$c = 1$$ is not a critical point. Hence, the function has one critical point $$c = e.$$ ### Example 9. Determine critical points of the function $$f\left( x \right) = \left\| {{x^2} – 4x + 3} \right\|.$$ Solution. First, we find the roots of the function and sketch its graph: ${f\left( x \right) = 0,}\;\; \Rightarrow {\left\| {{x^2} – 4x + 3} \right\| = 0.}$ ${D = {\left( { – 4} \right)^2} – 4 \cdot 3 = 4,}\;\; \Rightarrow {{x_{1,2}} = \frac{{4 \pm \sqrt 4 }}{2} = 1,3.}$ We see that the function has two corner points (or V-points): $$c = 1$$ and $$c = 3,$$ where the derivative does not exist. Therefore, $$c = 1$$ and $$c = 3$$ are critical points of the function. Besides that, the function has one more critical point at which the derivative is zero. Indeed, in the interval $$1 \le x \le 3,$$ the function is written as ${f\left( x \right) = – \left( {{x^2} – 4x + 3} \right) }={ – {x^2} + 4x – 3.}$ Differentiating and equating to zero, we get ${f^\prime\left( x \right) = \left( { – {x^2} + 4x – 3} \right)^\prime }={ – 2x + 4.}$ ${f^\prime\left( c \right) = 0,}\;\; \Rightarrow {- 2c + 4 = 0,}\;\; \Rightarrow {c = 2.}$ Thus, the function has three critical points: ${{c_1} = 1,}\;{{c_2} = 2,}\;{{c_3} = 3.}$ (a pretty nice answer, isn’t it?) ### Example 10. Find the critical points of the function $$f\left( x \right) = {\sin ^2}x – \cos x$$ on the interval $$\left( {0,2\pi } \right).$$ Solution. Determine the derivative of $$f\left( x \right)$$ using the chain rule and trig derivatives: ${f^\prime\left( x \right) = \left( {{{\sin }^2}x – \cos x} \right)^\prime }={ 2\sin x\cos x – \left( { – \sin x} \right) }={ 2\sin x\cos x + \sin x }={ \sin x\left( {2\cos x + 1} \right).}$ Solving the equation $$f^\prime\left( c \right) = 0,$$ we obtain two solutions: ${f^\prime\left( c \right) = 0,}\;\; \Rightarrow {\sin c\left( {2\cos c + 1} \right) = 0.}$ ${1.\;\sin c = 0,}\;\; \Rightarrow {c = \pi n,\;n \in Z.}$ The equation $$\sin c = 0$$ has one root $$c = \pi$$ in the open interval $$\left( {0,2\pi } \right).$$ ${2.\;2\cos c + 1 = 0,}\;\; \Rightarrow {2\cos x = – 1,}\;\; \Rightarrow {\cos c = – \frac{1}{2},}\;\; \Rightarrow {c = \pm \arccos \left( { – \frac{1}{2}} \right) + 2\pi n,}\;\; \Rightarrow {c = \pm \frac{{2\pi }}{3} + 2\pi n,\,n \in Z.}$ Only one solution $$c = \large{\frac{{2\pi }}{3}}\normalsize$$ belongs to the open interval $$\left( {0,2\pi } \right).$$ So, the function has two critical points: ${{c_1} = \pi ,}\;{{c_2} = \frac{{2\pi }}{3}.}$ ### Example 11. How many critical points does the function $$f\left( x \right) = \left\| {{x^3} – 12x} \right\|$$ have? Solution. Determine the roots of the function: ${f\left( x \right) = 0,}\;\; \Rightarrow {\left\| {{x^3} – 12x} \right\| = 0,}\;\; \Rightarrow {x\left( {{x^2} – 12} \right) = 0,}\;\; \Rightarrow {{x_1} = 0,\,{x_{2,3}} = \pm 2\sqrt 3 .}$ We see that the function has 3 corner points (or V-points) at $$x = – 2\sqrt 3 ,$$ $$x = 0$$ and $$x = 2\sqrt 3 .$$ Since the derivative does not exist at these points, we have 3 critical points here. Consider other critical points which can occur at local extrema. In the interval $$\left[ { – 2\sqrt 3 ,0} \right],$$ the function has the form $f\left( x \right) = {x^3} – 12x.$ Its derivative is given by ${f^\prime\left( x \right) = \left( {{x^3} – 12x} \right)^\prime }={ 3{x^2} – 12.}$ By equating the derivative to zero, we get ${f^\prime\left( x \right) = 0,}\;\; \Rightarrow {3{x^2} – 12 = 0,}\;\; \Rightarrow {x = \pm 2.}$ Only one point $$x = -2$$ lies in the interval under consideration. So $$x = -2$$ is also a critical point. Similarly, we find that the function has one more critical point $$x = 2$$ in the interval $$\left[ {0,2\sqrt 3 } \right]$$. Hence, the function has $$5$$ critical points ($$3$$ V-points and $$2$$ local extrema points).
# Area and Perimeter of the Triangle Here we will discuss about the area and perimeter of the triangle. If a, b, c are the sides of the triangle, then the perimeter of triangle = (a + b + c) units. Area of the triangle = √(s(s - a) (s - h) (s - c)) The semi-perimeter of the triangle, s = (a + b + c)/2 In a triangle if 'b' is the base and h is the height of the triangle then Area of triangle = 1/2 × base × height Similarly, 1/2 × AC × BD 1/2 × BC × AD Base of the triangle = (2 Area)/height Height of the triangle = (2 Area)/base Area of right angled triangle If a represents the side of an equilateral triangle, then its area = (a²√3)/4 Area of right angled triangle A = 1/2 × BC × AB = 1/2 × b × h Worked-out examples on area and perimeter of the triangle: 1. Find the area and height of an equilateral triangle of side 12 cm. (√3 = 1.73). Solution: Area of the triangle = $$\frac{√3}{4}$$ a² square units = $$\frac{√3}{4}$$ × 12 × 12 = 36√3 cm² = 36 × 1.732 cm² = 62.28 cm² Height of the triangle = $$\frac{√3}{2}$$ a units = $$\frac{√3}{2}$$ × 12 cm = 1.73 × 6 cm = 10.38 cm 2. Find the area of right angled triangle whose hypotenuse is 15 cm and one of the sides is 12 cm. Solution: AB² = AC² - BC² = 15² - 12² = 225 - 144 = 81 Therefore, AB = 9 Therefore, area of the triangle = ¹/₂ × base × height = ¹/₂ × 12 × 9 = 54 cm² 3. The base and height of the triangle are in the ratio 3 : 2. If the area of the triangle is 243 cm² find the base and height of the triangle. Solution: Let the common ratio be x Then height of triangle = 2x And the base of triangle = 3x Area of triangle = 243 cm² Area of triangle = 1/2 × b × h 243 = 1/2 × 3x × 2x ⇒ 3x² = 243 ⇒ x² = 243/3 ⇒ x = √81 ⇒ x = √(9 × 9) ⇒ x = √9 Therefore, height of triangle = 2 × 9 = 18 cm Base of triangle = 3x = 3 × 9 = 27 cm 4. Find the area of a triangle whose sides are 41 cm, 28 cm, 15 cm. Also, find the length of the altitude corresponding to the largest side of the triangle. Solution: Semi-perimeter of the triangle = (a + b + c)/2 = (41 + 28 + 15)/2 = 84/2 = 42 cm Therefore, area of the triangle = √(s(s - a) (s - b) (s - c)) = √(42 (42 - 41) (42 - 28) (42 - 15)) cm² = √(42 × 1 × 27 × 14) cm² = √(3 × 3 × 3 × 3 × 2 × 2 × 7 × 7) cm² = 3 × 3 × 2 × 7 cm² = 126 cm² Now, area of triangle = 1/2 × b × h Therefore, h = 2A/b = (2 × 126)/41 = 252/41 = 6.1 cm More solved examples on area and perimeter of the triangle: 5. Find the area of a triangle, two sides of which are 40 cm and 24 cm and the perimeter is 96 cm. Solution: Since, the perimeter = 96 cm a = 40 cm, b = 24 cm Therefore, C = P - (a + b) = 96 - (40 + 24) = 96 - 64 = 32 cm Therefore, S = (a + b + c)/2 = (32 + 24 + 40)/2 = 96/2 = 48 cm Therefore, area of triangle = √(s(s - a) (s - b) (s - c)) = √(48 (48 - 40) (48 - 24) (48 - 32)) = √(48 × 8 × 24 × 16 ) = √(2 × 2 × 2 × 2 × 3 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 2 × 2 × 2 × 2) = 3 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 384 cm² 6. The sides of the triangular plot are in the ratio 2 : 3 : 4 and the perimeter is 180 m. Find its area. Solution: Let the common ratio be x, then the three sides of triangle are 2x, 3x, 4x Now, perimeter = 180 m Therefore, 2x + 3x + 4x = 180 ⇒ 9x = 180 ⇒ x = 180/9 ⇒ x = 20 Therefore, 2x = 2 × 20 = 40 3x = 3 × 20 = 60 4x = 4 × 20 = 80 Area of triangle = √(s(s - a) (s - b) (s - c)) = √(90(90 - 80) (90 - 60) (90 - 40)) = √(90 × 10 × 30 × 50)) = √(3 × 3 × 2 × 5 × 2 × 5 × 3 × 2 × 5 × 5 × 5 × 2) = 3 × 2 × 5 × 2 × 5 √(3 × 5) = 300 √15 m² = 300 × 3.872 m² = 1161.600 m² = 1161.6 m² The above explanation on area and perimeter of the triangle are explained using step-by-step solution. ● Mensuration Area and Perimeter Perimeter and Area of Rectangle Perimeter and Area of Square Area of the Path Area and Perimeter of the Triangle Area and Perimeter of the Parallelogram Area and Perimeter of Rhombus Area of Trapezium Circumference and Area of Circle Units of Area Conversion Practice Test on Area and Perimeter of Rectangle Mensuration - Worksheets Worksheet on Area and Perimeter of Rectangles Worksheet on Area and Perimeter of Squares Worksheet on Area of the Path Worksheet on Circumference and Area of Circle Worksheet on Area and Perimeter of Triangle
# Scales, Maps and Ratios Lesson ## Scales A scale compares the distance on a map to the actual distance on the ground. Scales can be depicted in a few different ways: ## Graphic scale A graphic scale represents a scale by using a small line with markings similar to a ruler. One side of the line represents the distance on the map, while the other side represents the true distances of objects in real life. By measuring the distance between two points on a map and then referring to the graphic scale, we can calculate the actual distance between those points. In this picture, you can see that one centimetre on the scale represents $250$250 kilometres in real life. ## Verbal scale A verbal scale uses words to describe the ratio between the map's scale and the real world. For example, we could say "One centimetre equals fifteen kilometres" or we could write it as $1$1cm = $15$15km. This means that one centimetre on the map is equivalent to $15$15 kilometres in the real world. ## Scale Ratio Some maps use a representative fraction to describe the ratio between the map and the real world. If you need a refresher about ratios, see Looking at Relationships between Different Groups. We write scale ratios for maps just like other ratios with the colon in the middle. For example $1:100000$1:100000. #### Examples ##### Question 1 Convert the following description to a proper scale ratio: $5$5cm on the map = $25$25m in real life. Remember we need to have our two quantities in the same unit of measurement in a ratio. I'm going to convert everything to centimetres. $25$25m $=$= $25\times100$25×100 cm $=$= $2500$2500 cm Once we have equivalent quantities, we can write it as a scale ratio. $5:2500$5:2500 $=$= $1:500$1:500 ##### Question 2 Question: Given that the scale on a map is $1$1:$50000$50000, find the actual distance between two points that are $8$8 cm apart on the map. Think: This means that $1$1 cm on the map represents $50000$50000cm (or $500$500m) in real life. Do: So to work out how far $8$8cm represents, we need to multiply $8$8 by $50000$50000. Then convert to km. $8\times50000$8×50000 $=$= $400000$400000 cm $=$= $4000$4000 m $=$= $4$4 km Now let's look at how we can do this process in reverse. ##### Question 3 Question: Given that the scale on a map of a garden is $1$1:$2000$2000 , how far apart should two fountains be drawn on the map if the actual distance between the fountains is $100$100 m? ### Outcomes #### GM5-8 Interpret points and lines on co-ordinate planes, including scales and bearings on maps
# 30+ Numerical Aptitude Questions For All Competitive Exams \| Quantitative Aptitude Cracking Indian competitive exams is a dream for many, One of the most obvious hurdles that students often face can be the tough numerical aptitude questions. But fear not, with dedication and the right approach, you can get a high score in this section. This article lists 30+ multiple-choice numerical aptitude questions for practice so that you may get high scores in the exams. We understand that the vast syllabus and diverse question types can be overwhelming. But worry not, we’ll break it down into manageable chunks, covering all the important topics first. ## Tips and Tricks to Solve Numerical Aptitude Questions We have listed certain tips and tricks to Identify the question type (percent, ratio, etc.). Your first step is to break down the question and estimate an answer to rule out some options that don’t go well with the question. Backsolving with answer choices can save time. Read the following tips and tricks to solve numerical aptitude questions easily:: 1. Master the Basics: Sharpen your skills with addition, subtraction, multiplication, and division. 2. Practice Makes Perfect: Solve many practice problems to build speed and accuracy. 3. Focus on the Relevant: Read questions carefully and identify the important information to solve them. 4. Befriend Estimation: Estimate answers to check your calculations and save time. 5. Identify Shortcuts: Learn time-saving tricks like percentages of 100 or quick fraction conversion etc. ## Practice 30+ Numerical Aptitude Questions 1. What is the sum of the solutions of the equation x² – 5x + 6 = 0? a) 3 b) 4 c) 5 d) 6 1. If a triangle has sides of lengths 10 cm, 15 cm, and 20 cm, what type of triangle is it? a) Scalene b) Isosceles c) Equilateral d) Right-angled 1. What is the value of tan(π/4)? a) 0 b) 1 c) √2 d) Undefined 1. If log(x) – log(3) = log(5), what is the value of x? a) 3 b) 5 c) 8 d) 15 1. Solve: e^(2x) = 7 a) x = ln(7) b) x = ln(3.5) c) x = ln(2.5) d) x = ln(2) 1. What is the result of √(12^2 + 16^2)? a) 20 b) 25 c) 28 d) 30 1. If a train travels at a speed of 120 km/h for 2.5 hours, how far does it travel? a) 250 km b) 300 km c) 325 km d) 350 km 1. Solve: 2^(3x – 1) = 64 a) x = 1 b) x = 2 c) x = 3 d) x = 4 1. What is the volume of a cylinder with a radius of 5 cm and a height of 10 cm? a) 100π cm³ b) 125π cm³ c) 150π cm³ d) 200π cm³ 1. If the principal amount is Rs. 5000, the interest rate is 8%, and the time period is 2 years, what is the compound interest? a) Rs. 800 b) Rs. 820 c) Rs. 840 d) Rs. 860 1. Solve: log₂(x + 5) = 3 a) x = 8 b) x = 10 c) x = 13 d) x = 16 1. What is the sum of the first 20 terms of the arithmetic progression 3, 7, 11, …? a) 440 b) 450 c) 460 d) 470 1. If sin θ = 3/5 and cos φ = 5/13, what is sin(θ + φ)? a) 56/65 b) 57/65 c) 58/65 d) 59/65 1. If the perimeter of a rectangle is 40 cm and its length is 3 times its width, what is the area of the rectangle? a) 60 cm² b) 70 cm² c) 80 cm² d) 90 cm² 1. What is the value of (2 + i) * (3 – 2i), where i is the imaginary unit? a) 4 + 7i b) 5 – 4i c) 6 – i d) 7 + 2i 1. If f(x) = 2x^3 – 5x^2 + 3x – 7, what is f'(x)? a) 6x^2 – 10x + 3 b) 6x^2 – 10x + 7 c) 6x^2 – 5x + 3 d) 6x^2 – 5x + 7 Answer: b) 6x^2 – 10x + 7 1. What is the value of ∫(2x + 3) dx from x = 1 to x = 4? a) 19 b) 21 c) 23 d) 25 1. Solve: 2^(x – 3) = 8^(x + 1) a) x = 1 b) x = 2 c) x = 3 d) x = 4 1. If a coin is flipped 5 times, what is the probability of getting exactly 3 heads? a) 1/8 b) 5/16 c) 3/8 d) 5/8 1. What is the value of cos(π/3)? a) 1/2 b) √3/2 c) √2/2 d) 1/√2 1. If the roots of the quadratic equation x² – 5x + k = 0 are equal, what is the value of k? a) 4 b) 5 c) 6 d) 7 1. Solve: ∫(2sinx + 3cosx) dx from x = 0 to x = π/2 a) 2 b) 3 c) 4 d) 5 1. What is the coefficient of x² in the expansion of (3x – 2)^4? a) -16 b) -32 c) 16 d) 32 1. If a circle has a radius of 10 cm and is inscribed in a square, what is the area of the square? a) 100 cm² b) 200 cm² c) 300 cm² d) 400 cm² 1. The population of a city increases by 5% annually. If the population is 100,000 now, what will it be after 3 years? a) 115,762 b) 116,155 c) 116,550 d) 116,955 1. Solve: 5 log(x) = 10 a) x = 10 b) x = 20 c) x = 100 d) x = 1000 1. The nth term of an arithmetic progression is given by tn = 3n – 1. What is the sum of the first 10 terms? a) 145 b) 150 c) 155 d) 160 1. If a cube has a volume of 512 cm³, what is the length of one edge? a) 6 cm b) 7 cm c) 8 cm d) 9 cm 1. What is the remainder when 17^23 is divided by 7? a) 1 b) 2 c) 3 d) 4 1. If log₃(x) = 4, what is x? a) 64 b) 81 c) 256 d) 729 29. The nth term of an arithmetic progression is given by tn = 3n – 1. What is the sum of the first 10 terms? a) 145 b) 150 c) 155 d) 160 30. If a cube has a volume of 512 cm³, what is the length of one edge? a) 6 cm b) 7 cm c) 8 cm d) 9 cm Answer: c) 8 cm ## FAQs How do you solve aptitude questions on permutation and combination? Understand the problem, identify if it’s a permutation or combination problem, apply relevant formulas, and ensure clarity in counting elements. What are aptitude questions on permutation and combination in competitive exams? Permutation involves arranging elements in a particular order, while combination involves selecting elements without considering the order. Both are frequently tested in competitive exams to assess quantitative aptitude skills. What is permutation and combination in quants? In quantitative aptitude, permutation refers to arrangements where the order matters, while combination refers to selections where the order doesn’t matter. They’re important for solving problems involving arrangements and selections. What are aptitude questions on permutation and combination? Permutation questions involve arranging elements in different orders, while combination questions involve selecting elements without regard to order. Both types of questions are common in quantitative aptitude assessments. This was all about ”numerical aptitude questions”. For more such informative blogs, check out our Study Material Section, or you can learn more about us by visiting our Indian exams page.
• Written By Prince # Permutation and Combination: Definition, Formula, Solved Questions Permutation and Combination: Permutation and Combination are one of the most important concepts in Mathematics. It is used in the practical applications in science, engineering and research, and many other fields. The difference between permutation and combination can be defined as; when a set of data is selected from a certain group, it is known as permutation; while the order in which the data is arranged is known as combination. Thus, we can say that permutation is an ordered combination. Grouping of data and its interpretation is possible with the help of permutation and combination. Mathematics students should be well versed in the concepts because it finds applications in all advanced research and data analysis. This article provides insight on the permutation and combination questions, solutions, permutation and combination formulas and more. ## Permutation and Combination Definition Here we have given the mathematical definition of permutation and combination: What is Permutation? A permutation is an arrangement in a definite order of several objects taken, some or all at a time, with permutations, every tiny detail matters. It means the order in which elements are arranged is significant. There are two types of permutations: • Repetition is Allowed: For the number lock example provided above, it could be “2-2-2”. • No Repetition Allowed: For example, the first three people in a race. You can’t be first and second at the same time. What is Combination? The combination is a way of selecting elements from a set so that the order of selection doesn’t matter. With the combination, only choosing elements matters. It means the order in which elements are chosen is not essential. There are two types of combinations: • Repetition is Allowed: For example, coins in your pocket (2, 5, 5, 10, 10) • No Repetition Allowed: For example, lottery numbers (2, 14, 18, 25, 30, 38) ### Permutation and Combination Formula There are many formulas that are used to solve permutation and combination problems. We have provided the complete permutation formula list here: • When repetition is not allowed: P is a permutation or arrangement of r things from a set of n things without replacement. • When repetition is allowed: P is a permutation or arrangement of r things from a set of n things when duplication is allowed. Derivation of Permutation Formula: Let us assume that there are r boxes, and each of them can hold one thing. There will be as many permutations as there are ways of filling in vacant boxes by n objects. • No. of ways the first box can be filled: n • No. of ways the second box can be filled: (n – 1) • No. of ways the third box can be filled: (n – 2) • No. of ways the fourth box can be filled: (n – 3) • No. of ways rth box can be filled: [n – (r – 1)] • The number of permutations of different objects taken at a time, where 0 < r ≤ n and the objects do not repeat is: n(n – 1)(n – 2)(n – 3) . . . (n – r + 1) We have provided the complete combination formula list here: • When repetition is not allowed: C is a combination of n distinct things taking r at a time (order is not important). • When repetition is allowed: C is a combination of n distinct things taking r at a time (order is not important) with repetition. We define C as: Derivation of Combination Formula: Let us assume that there are r boxes, and each of them can hold one thing. • No. of ways to select the first object from n distinct objects: n • No. of ways to select the second object from (n-1) distinct objects: (n-1) • No. of ways to select the third object from (n-2) distinct objects: (n-2) • No. of ways to select rth object from [n-(r-1)] distinct objects: [n-(r-1)] Completing the selection of r things from the original set of n things creates an ordered subset of r elements. ∴ The number of ways to make a selection of r elements of the original set of elements is: (– 1) (– (n-3) . . . (– (– 1)) or (– 1) (– 2) … (– + 1) Let us consider the ordered subset of r elements and all their permutations. The total number of permutations of this subset equals r! because objects in every combination can be rearranged in r! ways. Hence, the total number of permutations of different things taken at a time is (nCr×r!). It is nothing but nPr. We can summarize the permutation combination formula in the table below: ### Difference Between Permutation and Combination We have provided the permutation and combination differences in the table below: Check: ### Solved Examples of Permutation and Combination We have provided some permutation and combination examples with detailed solutions. Get Permutation and Combination Class 11 NCERT Solutions for on Embibe. Q.1: Find the number of permutations and combinations, if n = 15 and r = 3. Ans: n = 15, r = 3 (Given) Using the formulas for permutation and combination, we get: Permutation, P = n!/(n – r)! = 15!/(15 – 3)! = 15!/12! = (15 x 14 x 13 x 12!)/12! = 15 x 14 x 13 = 2730 Also, Combination, C = n!/(n – r)!r! = 15!/(15 – 3)!3! = 15!/12!3! = (15 x 14 x 13 x 12! )/12!3! = 15 x 14 x 13/6 = 2730/6 = 455 Q.2: In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together? Ans: The word ‘OPTICAL’ has 7 letters. It has the vowels’ O’, ‘I’, and ‘A’ in it and these 3 vowels should always come together. Hence these three vowels can be grouped and considered as a single letter. That is, PTCL(OIA). Hence we can assume the total letters as 5 and all these letters are different. Number of ways to arrange these letters = 5! = 5×4×3×2×1 = 120 All the 3 vowels (OIA) are different. Number of ways to arrange these vowels among themselves = 3! = 3×2×1 = 6 Hence, the required number of ways: = 120×6 = 720 Q.3: How many 3-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’ if repetition of letters is not allowed? Ans: The word ‘LOGARITHMS’ has 10 different letters. Hence, the number of 3-letter words (with or without meaning) formed by using these letters 10P3 = 10×9×8 = 720 Q.4: There are 8 men and 10 women and you need to form a committee of 5 men and 6 women. In how many ways can the committee be formed? Ans: We need to select 5 men from 8 men and 6 women from 10 women. Number of ways to do this 8C5 × 10C6 8C3 × 10C4 [∵ nCr = nC(n-r)] = [(8 x 7 x 6)/(3 x 2 x 1)] x [(10 x 9 x 8 x 7)/(4 x 3 x 2 x 1)] = 56×210 = 11760 Q.5: A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag if at least one black ball is to be included in the draw? Ans: From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that at least one black ball should be there. Hence we have 3 choices as given below. Choice 1: We can select 3 black balls. Choice 2: We can select 2 black balls and 1 non-black ball. Choice 3: We can select 1 black ball and 2 non-black balls. Number of ways to select 3 black balls = 3C3 Number of ways to select 2 black balls and 1 non-black ball = 3C2 × 6C1 Number of ways to select 1 black ball and 2 non-black balls = 3C1 × 6C2 Total number of ways = 3C3 + 3C2 × 6C1 + 3C1 × 6C2 = 3C3 + 3C1 × 6C1 + 3C1 × 6C2[∵ nCr = nC(n-r)] = 1 + (3×6) + [3 x (6×5)/(2×1)] = 1 + 18 + 45 = 64 Q.6: An event manager has ten patterns of chairs and eight patterns of tables. In how many ways can he make a pair of tables and chairs? Ans: The event manager has 10 patterns of chairs and 8 pattern tables. A chair can be selected in 10 ways. A table can be selected in 8 ways. Hence one chair and one table can be selected in 10 × 8 ways = 80 ways Q.7: In how many ways can three boys can be seated on five chairs? Ans: There are 3 boys. The 1st boy can sit in any of the five chairs (5 ways). Now there are 4 chairs remaining. The second boy can sit in any of the four chairs (4 ways). Now there are 3 chairs remaining. The third boy can sit in any of the three chairs (3 ways). Hence, the total number of ways in which 3 boys can be seated on 5 chairs = 5 × 4 × 3 = 60 Q.8: In how many ways can a team of 5 persons be formed out of a total of 10 persons such that two particular persons should be included in each team? Answer: Two particular persons should be included in each team. Therefore we have to select the remaining (5 – 2) = 3 persons from (10 – 2) = 8 persons. Hence, the required number of ways 8C3 = (8×7×6)/(3×2×1) = 8×7 = 56 Other important Maths articles: ### Permutation and Combination Practice Questions Here are some practice questions on permutation and combination concepts for you to practice: You can also check, ### FAQs on Permutation and Combination Unleash Your True Potential With Personalised Learning on EMBIBE
# Warm Up Simplify each expression. 1. 6² ## Presentation on theme: "Warm Up Simplify each expression. 1. 6²"— Presentation transcript: Warm Up Simplify each expression. 1. 6² 36 2. 112 121 25 36 3. (–9)(–9) 81 4. Write each fraction as a decimal. 2 5 5 9 5. 0.4 6. 0.5 5 3 8 –1 5 6 7. 5.375 8. –1.83 Square Roots Vocabulary & Notes Vocabulary Objectives square root rational numbers perfect square irrational numbers real numbers repeating decimal natural numbers terminating decimal whole numbers Integers Objectives Evaluate expressions containing square roots. Classify numbers within the real number system. Square Roots A number that is multiplied by itself to form a product is called a square root of that product. The operations of squaring and finding a square root are inverse operations. The radical symbol , is used to represent square roots. Positive real numbers have two square roots. 4  4 = 42 = 16 = 4 Positive square root of 16 (–4)(–4) = (–4)2 = 16 Negative square root of 16 = –4 Perfect Squares The nonnegative square root is represented by The negative square root is represented by – A perfect square is a number whose positive square root is a whole number. Some examples of perfect squares are shown in the table. 1 4 9 16 25 36 49 64 81 100 02 12 22 32 42 52 62 72 82 92 102 The expression does not represent a real number because there is no real number that can be multiplied by itself to form a product of –36. Reading Math Finding Square Roots of Perfect Squares Find each square root. A. Think: What number squared equals 16? 42 = 16 Positive square root positive 4. = 4 B. Think: What is the opposite of the square root of 9? 32 = 9 = –3 Negative square root negative 3. Finding Square Roots of Perfect Squares Find the square root. Think: What number squared equals ? 25 81 Positive square root positive 5 9 Find the square root. 1a. 22 = 4 = 2 1b. 52 = 25 Try This! Find the square root. 1a. 22 = 4 Think: What number squared equals 4? = 2 Positive square root positive 2. 1b. Think: What is the opposite of the square root of 25? 52 = 25 Negative square root negative 5. Download ppt "Warm Up Simplify each expression. 1. 6²" Similar presentations
# Term A term is an algebraic expression which can form a separable part of another expression such as an algebraic equation or a sequence. Terms are a specific part of the symbolic language of algebra. The symbols of this language were primarily developed during the sixteenth and seventeenth centuries and are used to represent otherwise lengthy expressions. They can be as simple as using the single character, +, to mean addition, or as complicated as y = 4x2 + 2x - 3 to represent an algebraic polynomial equation. In general, there are three types of algebraic expressions which can be classified as terms. These include expressions made up of a single variable or constant, ones that are the product or quotient of two or more variables and/or constants, and those that are the product or quotient of other expressions. For example, the number 4 and the variable x are both terms because they consist of a single symbol. The expression 2z is also a term because it represents the product of two symbols. It should be noted that terms like 2z, in which a number and a variable are written together, are indicated products because multiplication is implied. Therefore, the symbol 2z means 2 × z. Finally, an expression like 2pq(a + 5)n is a term because it represents a quotient (the result of division) of two expressions. The symbols that make up a term are known as coefficients. In the term 4x, the number 4 is known as a numerical coefficient and the letter x is known as the literal coefficient. For this expression, we could say that 4 is the coefficient of x or x is the coefficient of 4. Terms should be thought of as a single unit that represents the value of a particular number. This is particularly useful when discussing the terms of a larger expression such as an equation. In the expression 5x3 + 2x2 + 4x - 7, there are four terms. Numbering them from left to right, the first term is 5x3, the second is 2x2, the third is 4x, and the fourth is -7. Notice that the sign in front of a term is actually part of it. Some expressions contain terms which can be combined to form a single term. These "like terms" contain the same variable raised to the same power. For example, the like terms in the expression 3x + 2x can be added and the equation simplifies to 5x. Similarly, the expression 7y2 - 3y2 can be simplified to 4y2. Expressions containing unlike terms can not be simplified. Therefore, 4x2 2x is in its simplest form because the differences in the power of x prevents these terms from being combined.
## Time and Work Questions and Answers Part-10 1. Three taps A, B and C together can fill an empty cistern in 10 minutes. The tap A alone can fill it in 30 minutes and the tap B alone in 40 minutes. How long will the tap C alone take to fill it? a) 16 minutes b) 24 minutes c) 32 minutes d) 40 minutes Explanation: A, B and C together can fill 100% empty tank in 10 minutes Work rate of (A + B + C) = $$\frac{{100}}{{10}}$$ = 10% per minute A alone can fill the tank in 30 minutes Work rate of A = $$\frac{{100}}{{30}}$$ = 3.33% per minute B alone can fill the tank in 40 minutes Work rate of B = $$\frac{{100}}{{40}}$$ = 2.5% Work rate of (A + B) = 3.33 + 2.5 = 5.83% per minute Work rate of C, = Work rate of (A + B + C) - (A + B) = 10 - 5.83 = 4.17% per minute C takes = $$\frac{{100}}{{4.17}}$$ ≈ 24 minutes to fill the tank 2. A and B working separately can do a piece of work in 9 and 15 days respectively. If they work for a day alternately, with A beginning, then the work will be completed in: a) 9 days b) 10 days c) 11 days d) 12 days Explanation: Work rate of A = $$\frac{{100}}{9}$$ = 11.11% work per day Work rate of B = $$\frac{{100}}{{15}}$$ = 6.66% work per day They together can do (A + B) = 11.11 + 6.66 ≈ 18% work per day They are working in alternate day, so we take 2 days = 1 unit of day Therefore, in one unit of day they can complete 18% of work (A + B) can complete 90% of work in 5 units of days. i.e. (5 × 18) And the rest 10% work will be completed by A in Next day Total number of day = 5 Unit of days + 1 day of A = 2 × 5 + 1 = 11 days 3. Two pipes A and B can fill a tank in 36 min. and 45 min. respectively. Another pipe C can empty the tank in 30 min. First A and B are opened. After 7 minutes, C is also opened. The tank filled up in: a) 39 min. b) 46 min c) 40 min. d) 45 min. Explanation: Pipe A can fill empty tank in 36 min. Pipe A can fill the tank = $$\frac{{100}}{{36}}$$ = 2.77% per minute Pipe B can fill empty tank in 45 min. Pipe B can fill the tank = $$\frac{{100}}{{45}}$$ = 2.22% per min. A and B can together fill the tank = (2.77 + 2.22) ≈ 5% per minute So, A and B can fill the tank in 7 min. = 7 × 5 = 35% of the tank Rest tank to be filled = 100 - 35 = 65% C can empty the full tank in 30 min. C can empty the tank = $$\frac{{100}}{{30}}$$ = 3.33% per min. C is doing negative work i.e. emptying the tank A, B and C can together fill the tank, = 2.77% + 2.22% - 3.33% = 1.67% tank per minute A, B and C will take time to fill 65% empty tank, = $$\frac{{65}}{{1.67}}$$ = 39 min. (Approx) 4. Three men A, B, C working together can do a job in 6 hours less time than A alone, in one hour less time than B alone and in one half the time needed by C when working alone. Then A and B together can do the job in: a) $$\frac{2}{3}$$ hours b) $$\frac{3}{4}$$ hours c) $$\frac{3}{2}$$ hours d) $$\frac{4}{3}$$ hours Explanation: Time taken by A =x hours. Therefore taken by A, B and C together = (x - 6) Time taken by B = (x - 5) Time taken by C = 2(x - 6) Now, rate of work of A + Rate of work of B + Rate of work of C = Rate of work of ABC. $$\frac{1}{x} + \frac{1}{{x - 5}} + \frac{1}{{2\left( {x - 6} \right)}} = \frac{1}{{x - 6}}$$ On solving above equation, x = 3, $$\frac{{40}}{6}$$ When x = 3, the expression (x - 6) becomes negative, thus it's not possible. $$x = \frac{{40}}{6}$$ Time taken by A & B together = $$\frac{1}{{\frac{3}{{20}} + \frac{3}{5}}}$$ = $$\frac{4}{3}$$ hours 5. A does half as much work as B in one -sixth of the time.If together they take 10 days to complete a work, how much time shall B take to do it alone? a) 13.33 days b) 20 days c) 30 days d) 40 days Explanation: Given, $${\text{A}} \times \frac{1}{6} = {\text{B}} \times \frac{1}{2}$$ A = 3B Given they together complete the work in 10 days So, One Day's work of, (A + B) = $$\frac{{100}}{{10}}$$ = 10% (3B + B) = 10% 4B = 10% one day work of B = $$\frac{{10}}{4}$$ = 2.5% B can complete 100% work in = $$\frac{100}{2.5}$$ = 40 days 6. An employee pays Rs. 26 for each day a worker and forfeits Rs. 7 for each day he idle. At the end of 56 days, if the worker got Rs. 829, for how many days did the worker remain idle? a) 21 days b) 15 days c) 19 days d) 13 days Explanation: His Per day pay = Rs. 26 Total pay employee got = Rs. 829 Total pay he gets if he did not remain idle a single day, = 26 × 56 = Rs. 1456 He Forfeits or fined = 1456 - 829 = Rs. 627 Per day he Forfeits Rs. 7 Means per idle day he loses = 26 + 7 = Rs. 33 Total idle days = $$\frac{{627}}{{33}}$$ = 19 days 7. A is 60% more efficient than B. In how many days will A and B working together complete a piece of work which A alone takes 15 days to finish? a) $$\frac{{124}}{{13}}$$ days b) $$\frac{{113}}{{13}}$$ days c) $$\frac{{108}}{{13}}$$ days d) $$\frac{{120}}{{13}}$$ days Explanation: Given, A is 60% more efficient of B Means, \eqalign{ & {\text{A}} = {\text{B}} + 60\% \,{\text{of B}} \cr & {\text{A}} = {\text{B}} + \frac{{60{\text{B}}}}{{100}} \cr & {\text{A}} = \frac{{100{\text{B}} + 60{\text{B}}}}{{100}} \cr & {\text{A}} = \frac{{160{\text{B}}}}{{100}} \cr & {\text{A}} = \frac{{8{\text{B}}}}{5} \cr} A can complete whole work in 15 days. So, One day work of A = $$\frac{1}{{15}}$$ One day work of A = $$\frac{{8{\text{B}}}}{5}$$ = $$\frac{1}{{15}}$$ One day work of B = $$\frac{5}{{120}}$$ = $$\frac{1}{{24}}$$ One day work, (A + B) = $$\frac{1}{{15}} + \frac{1}{{24}}$$ One day work, (A + B) = $$\frac{{24 + 15}}{{360}}$$ = $$\frac{{39}}{{360}}$$ Time taken to finish the work by A and B together = $$\frac{{360}}{{39}}$$ = $$\frac{{120}}{{13}}$$ days 8. A pipe can fill a tank in 0.9 hours and another pipe can empty in 0.7 hours. If tank is completely filled and both pipes are opened simultaneously then 450 liters of water is removed from the tank is 2.5 hours. What is the capacity of the tank? a) 200 liters b) 350 liters c) 456 liters d) 567 liters Explanation: Pipe A can fill the empty tank in = 0.9 hours So work rate of the Pipe A = $$\frac{{100}}{{0.9}}$$ % per hour Pipe B can empty the tank in = 0.7 hours Negative Work rate of B = $$\frac{{100}}{{0.7}}$$ % per hour. (B is removing water, so, taken as negative work) Tank fill per hour = $$\frac{{100}}{{0.7}} - \frac{{100}}{{0.9}}$$ = 31.75% per hour Time Taken to empty the tank = $$\frac{{100}}{{31.75}}$$ ≈ 3.15 hours Capacity of the tank = 3.15 × 180 = 567 liters 9. A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is : a) $$\frac{1}{4}$$ b) $$\frac{1}{{10}}$$ c) $$\frac{7}{{15}}$$ d) $$\frac{8}{{15}}$$ \eqalign{ & {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{15}} \cr & {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{20}} \cr & \left( {{\text{A + B}}} \right){\text{'s}}\,{\text{1day's}}\,{\text{work}} \cr & = {\frac{1}{{15}} + \frac{1}{{20}}} = \frac{7}{{60}} \cr & \left( {{\text{A + B}}} \right){\text{'s}}\,{\text{4}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{7}{{60}} \times 4} = \frac{7}{{15}} \cr & {\text{Remaining}}\,{\text{work}}\, = {1 - \frac{7}{{15}}} = \frac{8}{{15}} \cr} a) $$9\frac{1}{5}$$ days b) $$9\frac{2}{7}$$ days c) $$9\frac{3}{5}$$ days \eqalign{ & \left( {{\text{A + B + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{4} \cr & {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{16}} \cr & {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{12}} \cr & {\text{C's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = \frac{1}{4} - \left( {\frac{1}{{16}} + \frac{1}{{12}}} \right) = {\frac{1}{4} - \frac{7}{{48}}} = \frac{5}{{48}} \cr & {\text{C}}\,\,{\text{alone}}\,{\text{can}}\,{\text{do}}\,{\text{the}}\,{\text{work}}\,{\text{in}} \cr & \frac{{48}}{5} = 9\frac{3}{5}\,{\text{days}} \cr}
# Multiplication Table of 21 Repeated addition by 21’s means the multiplication table of 21. (i) When 6 rows of twenty-one soldiers each. By repeated addition we can show 21 + 21 + 21 + 21 + 21 + 21 = 126 Then, twenty-one 6 times or 6 twenty-one’s 6 × 21 = 126 Therefore, there are 126 soldiers. (ii) When 9 bunches each having 21 flowers. By repeated addition we can show 21 + 21 + 21 + 21 + 21 + 21 + 21 + 21 + 21 = 189 Then, twenty-one 9 times or 9 twenty-one’s 9 × 21 = 189 Therefore, there are 189 flowers. We will learn how to use the number line for counting the multiplication table of 21. (i) Start at 0. Hop 21, four times. Stop at 84. 4 twenty-one’s are 84 4 × 21 = 84 (ii) Start at 0. Hop 21, seven times. Stop at ____. Thus, it will be 147 7 twenty-one’s are 147 7 × 21 = 147 (iii) Start at 0. Hop 21, eleven times. Stop at ____. Thus, it will be 231 11 twenty-one’s are 231 11 × 21 = 231 How to read and write the table of 21? The above chart will help us to read and write the 21 times table. Read 1 twenty-one is 21 2 twenty-one’s are 42 3 twenty-one’s are 63 4 twenty-one’s are 84 5 twenty-one’s are 105 6 twenty-one’s are 126 7 twenty-one’s are 147 8 twenty-one’s are 168 9 twenty-one’s are 189 10 twenty-one’s are 210 11 twenty-one’s are 231 12 twenty-one’s are 252 Write 1 × 21 = 21 2 × 21 = 42 3 × 21 = 63 4 × 21 = 84 5 × 21 = 105 6 × 21 = 126 7 × 21 = 147 8 × 21 = 168 9 × 21 = 189 10 × 21 = 210 11 × 21 = 231 12 × 21 = 252 Now we will learn how to do forward counting and backward counting by 21’s. Forward counting by 21’s: 0, 21, 42, 63, 84, 105, 126, 147, 168, 189, 210, 231, 252, 273, 294, 315, 336, 357, 378, 399, 420, 441, 462, 483, 504, 525, …… Backward counting by 21’s: ……, 525, 504, 483, 462, 441, 420, 399, 378, 357, 336, 315, 294, 273, 252, 231, 210, 189, 168, 147, 126, 105, 84, 63, 42, 21, 0 Multiplication Table Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. ## Recent Articles 1. ### Addition and Subtraction of Fractions \| Solved Examples \| Worksheet Jul 18, 24 03:08 PM Addition and subtraction of fractions are discussed here with examples. To add or subtract two or more fractions, proceed as under: (i) Convert the mixed fractions (if any.) or natural numbers 2. ### Worksheet on Simplification \| Simplify Expressions \| BODMAS Questions Jul 18, 24 01:19 AM In worksheet on simplification, the questions are based in order to simplify expressions involving more than one bracket by using the steps of removal of brackets. This exercise sheet 3. ### Fractions in Descending Order \|Arranging Fractions an Descending Order Jul 18, 24 01:15 AM We will discuss here how to arrange the fractions in descending order. Solved examples for arranging in descending order: 1. Arrange the following fractions 5/6, 7/10, 11/20 in descending order. First… 4. ### Fractions in Ascending Order \| Arranging Fractions \| Worksheet \|Answer Jul 18, 24 01:02 AM We will discuss here how to arrange the fractions in ascending order. Solved examples for arranging in ascending order: 1. Arrange the following fractions 5/6, 8/9, 2/3 in ascending order. First we fi… 5. ### Worksheet on Comparison of Like Fractions \| Greater & Smaller Fraction Jul 18, 24 12:45 AM In worksheet on comparison of like fractions, all grade students can practice the questions on comparison of like fractions. This exercise sheet on comparison of like fractions can be practiced Worksheet on Multiplication Table of 0 Worksheet on Multiplication Table of 2 Worksheet on Multiplication Table of 3 Worksheet on Multiplication Table of 4 Worksheet on Multiplication Table of 5 Worksheet on Multiplication Table of 6 Worksheet on Multiplication Table of 7 Worksheet on Multiplication Table of 8 Worksheet on Multiplication Table of 10 Worksheet on Multiplication Table of 11 Worksheet on Multiplication Table of 12 Worksheet on Multiplication Table of 13 Worksheet on Multiplication Table of 14 Worksheet on Multiplication Table of 17 Worksheet on Multiplication Table of 18 Worksheet on Multiplication Table of 19 Worksheet on Multiplication Table of 20 Worksheet on Multiplication Table of 21 Worksheet on Multiplication Table of 23 Worksheet on Multiplication Table of Worksheet on Multiplication Table of 25
# Table of 28 In this, we will read and memorize the table of 28. You can also download the 28 times table image from here for further reference. Multiplication tables help us to solve the mathematical problems easily and quickly. Let’s look at the table here. ## Table of 28 Charts (28 Times Table) Tables from 1 to 20 are important to study the further numbers tables. 28 X 1 = 28 28 X 2 = 56 28 X 3 = 84 28 X 4 = 112 28 X 5 = 140 28 X 6 = 168 28 X 7 = 196 28 X 8 = 224 28 X 9 = 252 28 X 10 = 280 28 X 11 = 308 28 X 12 = 336 28 X 13 = 364 28 X 14 = 392 28 X 15 = 420 28 X 16 = 448 28 X 17 = 476 28 X 18 = 504 28 X 19 = 532 28 X 20 = 560 ### 28 Times Table Repeatedly adding by 28’s gives the table of 28. (i) When 2 packs each having 28 pastel colours. By repeated addition we can show 28 + 28 = 56 Then, twenty-eight times 2 is 2 × 28 = 56 Therefore, there are 56 pastel colours. (ii) When 3 bunches of 28 bananas each. By repeated addition we can show 28 + 28 + 28 =84 Then, twenty-eight times 3 is 3 × 28 = 84 Therefore, there are 84bananas. (iii) When 4 bunches of 28 bananas each. By repeated addition we can show 28 + 28 + 28 +28 = 112 Then, twenty-eight times 4 is 4 × 28 = 112 Therefore, there are 112 bananas. (iv) When 5 bunches of 28 bananas each. By repeated addition we can show 28 + 28 + 28 + 28 +28 = 140 Then, twenty-eight times 5 is 5 × 28 = 140 Therefore, there are 140 bananas. (v) When 6 bunches of 28 bananas each. By repeated addition we can show 28 + 28 + 28 + 28 + 28 + 28 = 168 Then, twenty-eight times 6 is 6 × 28 = 168 Therefore, there are 168 bananas. (vi)When 7 bunches of 28 bananas each. By repeated addition we can show 28 + 28 + 28 + 28 + 28 + 28 +28 = 196 Then, twenty-eight times 7 is 7 × 28 = 196 Therefore, there are 196 bananas. (vii) When 8 bunches of 28 bananas each. By repeated addition we can show 28 + 28 + 28 + 28 + 28 + 28 + 28 + 28 = 224 Then, twenty-eight times 8 is 8 × 28 = 224 Therefore, there are 224 bananas. (viii) When 9 bunches of 28 bananas each. By repeated addition we can show 28 + 28 + 28 + 28 + 28 + 28 + 28 + 28 +28 = 252 Then, twenty-eight times 9 is 9 × 28 = 252 Therefore, there are 252 bananas. (ix) When 10 bunches of 28 bananas each. By repeated addition we can show 28 + 28 + 28 + 28 + 28 + 28 + 28 + 28 +28 +28 = 280 Then, twenty-eight times 10 is 10 × 28 = 280 Therefore, there are 280 bananas. Register to BYJU’S to get all the tables from 1 to 30. Explore more multiplication Tables Table of 12 Table of 13 Table of 15 Table of 16 Table of 17 Table of 18 Table of 19 Table of 20
# Chapter 1 A quick introduction to R ## 1.1 Using R as a big calculator ### 1.1.1 Basic arithmetic The end of the Get up and running with R and RStudio chapter demonstrated that R can handle familiar arithmetic operations: addition, subtraction, multiplication, division. If we want to add or subtract a pair of numbers just place the + or - symbol in between two numbers, hit Enter, and R will read the expression, evaluate it, and print the result to the Console. This works exactly as we expect it to: 3 + 2 ## [1] 5 5 - 1 ## [1] 4 Multiplication and division are no different, though we don’t use x or ÷ for these operations. Instead, we use * and / to multiply and divide: 7 * 2 ## [1] 14 3 / 2 ## [1] 1.5 We can also exponentiate a numbers: raise one number to the power of another. We use the ^ operator to do this: 4^2 ## [1] 16 This raises 4 to the power of 2 (i.e. we squared it). In general, we can raise a number x to the power of y using x^y. Neither x or y need to be a whole numbers either. Arithmetic operations can also be combined into one expression. Assume we want to subtract 6 from 23. The expression to perform this calculation is: 2^3 - 6 ## [1] 2 $$2^3=8$$ and $$8-6=2$$. Simple enough, but what if we had wanted to carry out a slightly longer calculation that required the last answer to then be divided by 2? This is the wrong the way to do it: 2^3 - 6 / 2 ## [1] 5 The answer we were looking for is $$1$$. So what happened? R evaluated $$6/2$$ first and then subtracted this answer from $$2^3$$. If that’s obvious, great. If not, it’s time to learn a bit about the order of precendence used by R. R uses a standard set of rules to decide the order in which arithmetic calculations feed into one another so that it can unambiguously evaluate any expression. It uses the same order as every other computer language, which thankfully is the same one we all learned in mathematics class at school. The order of precedence used is: 1. exponents and roots (“taking powers”) 2. multiplication and division #### BODMAS and friends If you find it difficult to remember order of precedence used by R, there are a load of mnemonics that can to help. Pick one you like and remember that instead. In order to get the answer we were looking for we need to take control of the order of evaluation. We do this by enclosing grouping the necessary bits of the calculation inside parentheses (“round brackets”). That is, we place ( and ) either side of them. The order in which expressions inside different pairs of parentheses are evaluated follows the rules we all had to learn at school. The R expression we should have used is therefore: (2^3 - 6) / 2 ## [1] 1 We can use more than one pair of parentheses to control the order of evaluation in more complex calculations. For example, if we want to find the cube root of 2 (i.e. 21/3) rather than 23 in that last calculation we would instead write: (2^(1/3) - 6) / 2 ## [1] -2.370039 The parentheses around the 1/3 in the exponent are needed to ensure this is evaluated prior to being used as the exponent. ### 1.1.2 Problematic calculations Now is a good time to highlight how R handles certain kinds of awkward numerical calculations. One of these involves division of a number by 0. Some programming languages will respond to an attempt to do this with an error. R is a bit more forgiving: 1/0 ## [1] Inf Mathematically, division of a finite number by 0 equals A Very Large Number: infinity. R has a special built in data value that allows it to handle this kind of thing. This is Inf, which of course stands for “infinity”. The other special kind of value we sometimes run into can be generated by numerical calculations that don’t have a well-defined result. For example, it arises when we try to divide 0 or infinity by themselves: 0/0 ## [1] NaN The NaN in this result stands for Not a Number. R produces NaN because $$0/0$$ is not defined mathematically: it produces something that is Not a Number. The reason we are pointing out Inf and NaN is not because we expect to use them. It’s important to know what they represent because they often arise as a result of a mistake somewhere in a program. It’s hard to track down such mistakes if we don’t know how Inf and NaN arise. That is enough about using R as a calculator for now. What we’ve seen—even though we haven’t said it yet—is that R functions as a REPL: a read-eval-print loop (there’s no need to remember this term). R takes user input, evaluates it, prints the results, and then waits for the next input. This is handy, because it means we can use it interactively, working through an analysis line-by-line. However, to use R to solve for complex problems we need to learn how to store and reuse results. We’ll look at this in the next section. #### Working efficiently at the Console Working at the Console soon gets tedious if we have to retype similar things over and over again. There is no need to do this though. Place the cursor at the prompt and hit the up arrow. What happens? This brings back the last expression sent to R’s interpreter. Hit the up arrow again to see the last-but-one expression, and so on. We go back down the list using the down arrow. Once we’re at the line we need, we use the left and right arrows to move around the expression and the delete key to remove the parts we want to change. Once an expression has been edited like this we hit Enter to send it to R again. Try it! ## 1.2 Storing and reusing results So far we’ve not tried to do anything remotely complicated or interesting, though we now know how to construct longer calculations using parentheses to control the order of evaluation. This approach is fine if the calculation is very simple. It quickly becomes unwieldy for dealing with anything more. The best way to see what we mean is by working through a simple example—solving a quadratic equation. Quadratic equations looks like this: $$a + bx + cx^2 = 0$$. If we know the values of $$a$$, $$b$$ and $$c$$ then we can solve this equation to find the values of $$x$$ that ensure the left hand side equals the right hand side. Here’s the well-known formula for these solutions: $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ Let’s use R to calculate these solutions for us. Say that we want to find the solutions to the quadratic equation when $$a=1$$, $$b=6$$ and $$c=5$$. We just have to turn the above equation into a pair of R expressions: (-6 + (6^2 -4 * 1 * 5)^(1/2)) / (2 * 1) ## [1] -1 (-6 - (6^2 -4 * 1 * 5)^(1/2)) / (2 * 1) ## [1] -5 The output tells us that the two values of $$x$$ that satisfy this particular quadratic equation are -1 and -5. What should we do if we now need to solve a different quadratic equation? Working at the Console, we could bring up the expressions we typed (using the up arrow) and then go through each of these, changing the numbers to match the new values of $$a$$, $$b$$ and $$c$$. Editing individual expressions like this is fairly tedious, and more importantly, it’s fairly error prone because we have to make sure we substitute the new numbers at exactly the right positions. A partial solution to this problem is to store the values of $$a$$, $$b$$ and $$c$$. We’ll see precisely why this is useful in a moment. First, we need to learn how to store results in R. The key to this is to use the assigment operator, written as a left arrow <-. Sticking with our original example, we need to store the numbers 1, 6 and 5. We do this using three expressions, one after the another: a <- 1 b <- 6 c <- 5 Notice that we don’t put a space between < and -—R won’t like it if we try to add one. R didn’t print anything to screen, so what actually happened? We asked R to first evaluate the expression on the right hand side of each <- (just a number in this case) and then assign the result of that evaluation instead of printing it. Each result has a name associated with it, which appears on the left hand side of the <-. #### RStudio shortcut We use the assignment operator <- all the time when working with R, and because it’s inefficient to have to type the < and - characters over and over again, RStudio has a built in shortcut for typing the assignment operator: Alt + - . Try it. Move the curser to the Console, hold down the Alt key (‘Option’ on a Mac), and press the - sign key. RStudio will auto-magically add insert <-. The net result of all this is that we have stored the numbers 1, 6 and 5 somewhere in R, associating them with the letters a, b and c, respectively. What does this mean? Here’s what happens if we type the letter a into the Console and hit Enter: a ## [1] 1 It looks the same as if we had typed the number 1 directly into the Console. The result of typing b or c is hopefully obvious. What we just did was to store the output that results from evaluating three separate R expressions, associating each a name so that we can access them again3. Whenever we use the assignment operator <- we are telling R to keep whatever kind of value results from the calculation on the right hand side of <-, giving it the name on the left hand side so that we can access it later. Why is this useful? Let’s imagine we want to do more than one thing with our three numbers. If we want to know their sum or their product we can now use: a + b + c ## [1] 12 a * b * c ## [1] 30 So once we’ve stored a result and associated it with a name we can reuse it wherever it’s needed. Returning to our motivating example, we can now calculate the solutions to the quadratic equation by typing these two expressions into the Console: (-b + (b^2 -4 * a * c)^(1/2)) / (2 * a) ## [1] -1 (-b - (b^2 -4 * a * c)^(1/2)) / (2 * a) ## [1] -5 Imagine we’d like to find the solutions to a different quadratic equation where $$a=1$$, $$b=5$$ and $$c=5$$. We just changed the value of $$b$$ here to keep things simple. To find our new solutions we have to do two things. First we change the value of the number associated with b b <- 5 …then we bring up those lines that calculate the solutions to the quadratic equation and run them, one after the other: (-b + (b^2 -4 * a * c)^(1/2)) / (2 * a) ## [1] -1.381966 (-b - (b^2 -4 * a * c)^(1/2)) / (2 * a) ## [1] -3.618034 We didn’t have to retype those two expressions. We could just use the up arrow to bring each one back to the prompt and hit Enter. This is much simpler than editing the expressions. More importantly, we are beginning to see the benefits of using something like R: we can break down complex calculations into a series of steps, storing and reusing intermediate results as required. ## 1.3 How does assignment work? It’s important to understand, at least roughly, how assignment works. The first thing to note is that when we use the assignment operator <- to associate names and values, we informally refer to this as creating (or modifying) a variable. This is much less tedious than using words like “bind”, “associate”, value“, and”name" all the time. Why is it called a variable? What happens when we run these lines: myvar <- 1 myvar <- 7 The first time we used <- with myvar on the left hand side we created a variable myvar associated with the value 1. The second line myvar <- 7 modified the value of myvar to be 7. This is why we refer to myvar as a variable: we can change the its value as we please. What happened to the old value associated with myvar? In short, it is gone, kaput, lost… forever. The moment we assign a new value to myvar the old one is destroyed and can no longer be accessed. Remember this. Keep in mind that the expression on the right hand side of <- can be any kind of calculation, not just just a number. For example, if I want to store the number 1, associating it with answer, I could do this: answer <- (1 + 2^3) / (2 + 7) That is a strange way to assign the number 1, but it illustrates the point. More generally, as along as the expression on the right hand side generates an output it can be used with the assignment operator. For example, we can create new variables from old variables: newvar <- 2 * answer What happened here? Start at the right hand side of <-. The expression on this side contained the variable answer so R went to see if answer actually exists in the global environment. It does, so it then substituted the value associated with answer into the requested calculation, and then assigned the resulting value of 2 to newvar. We created a new variable newvar using information associated with answer. Now look at what happens if we just copy a variable using the assignment operator: myvar <- 7 mycopy <- myvar At this point we have two variables, myvar and mycopy, each associated with the number 7. There is something very important going on here: each of these is associated with a different copy of this number. If we change the value associated with one of these variables it does not change the value of the other, as this shows: myvar <- 10 myvar ## [1] 10 mycopy ## [1] 7 R always behaves like this unless we work hard to alter this behaviour (we never do this in this book). So remember, every time we assign one variable to another, we actually make a completely new, independent copy of its associated value. For our purposes this is a good thing because it makes it much easier to understand what a long sequence of R expressions will do. That probably doesn’t seem like an obvious or important point, but trust us, it is. ## 1.4 Global environment Whenever we associate a name with a value we create a copy of both these things somewhere in the computer’s memory. In R the “somewhere” is called an environment. We aren’t going to get into a discussion of R’s many different kinds of environments—that’s an advanced topic well beyond the scope of this book. The one environment we do need to be aware of though is the Global Environment. Whenever we perform an assignment in the Console the name-value pair we create (i.e. the variable) is placed into the Global Environment. The current set of variables are all listed in the Environment tab in RStudio. Take a look. Assuming that at least one variable has been made, there will be two columns in the Environment tab. The first shows us the names of all the variables, while the second summarises their values. #### The Global Environment is temporary By default, R will save the Global Environment whenever we close it down and then restore it in the next R session. It does this by writing a copy of the Global Environment to disk. In theory this means we can close down R, reopen it, and pick things up from where we left off. Don’t do this—it only increases the risk of making a serious mistake. Assume that when R and RStudio are shut down, everything in Global Environment will be lost. ## 1.5 Naming rules and conventions We don’t have to use a single letter to name things in R. The words tom, dick and harry could be used in place of a, b and c. It might be confusing to use them, but tom, dick and harry are all legal names as far as to R is concerned: • A legal name in R is any sequence of letters, numbers, ., or _, but the sequence of characters we use must begin with a letter. Both upper and lower case letters are allowed. For example, num_1, num.1, num1, NUM1, myNum1 are all legal names, but 1num and _num1 are not because they begin with 1 and _. • R is case sensitive—it treats upper and lower case letters as different characters. This means that num and Num are treated as distinct names. Forgetting about case sensitivity is a good way to create errors when using R. Try to remember that. #### Don’t begin a name with . We are allowed to begin a name with a ., but this usually is A Bad Idea. Why? Because variable names that begin with . are hidden from view in the Global Environment—the value it refers to exists but it’s invisible. This behaviour exists to allow R to create invisible variables that control how it behaves. This is useful, but it isn’t really meant to be used by the average user. 1. Technically, this is called binding the name to a value. You really don’t need to remember this though.
# IIT JEE, UPSEE & WBJEE: Detailed Notes on Circular Permutation Mar 29, 2017 15:03 IST Circular arrangement is one of the interesting topics of Permutations. Before understanding circular arrangement let us review some important concepts of linear permutation. • In linear arrangement n objects can be arranged in n! ways taken all at time. For circular permutation let us consider four objects A, B, C and D. If we shift A, B, C, D one position in anticlockwise direction, then we get following arrangements. Now, wait for a moment. Can you see that in above all four arrangements the relative position of none of the four objects A, B, C, D is changed? So, in circular arrangement they are not considered as different arrangements instead they are considered one. Now suppose that if the above four objects A, B, C, D are to be arranged in linear, then the number of ways it can be done is 4!. Also, in case of linear arrangements the above four circular arrangements can be shown as: Let us generalize the above example. Let x be the number of ways in which n different objects have to arranged in circle, taken all at a time. For every unique circular arrangement we have n different circular arrangements in which the relative position of none of the objects is changed. In other words, for one circular permutation there are n linear arrangements. So, for x circular arrangements there are nx linear arrangements. But from linear permutation n different things can be arranged in n! ways. So, n! = nx x = (n-1)! So, if clock wise and anti clock wise arrangements are considered different, then the number of circular permutations of n different things taken all at a time is (n - 1)!. Ring arrangement (Formation of necklace/garland using bead/flower) Now consider the above circular arrangements, in the above discussion they were treated different because of clockwise or anti clockwise order but if no distinction is made then again the relative position of none of the objects is changed and they can be considered as same arrangement. The above figure (i) on being flipped to the right gives figure (ii). In the above discussion figure (i) and (ii) are counted as two different arrangements which should be counted one when no distinction is allowed and hence the actual number of arrangements is (n - 1)!/2. How to guess the correct option in IIT JEE 2017 Examination: Must know these 7 tips So, if clockwise and anti clockwise arrangement is considered same then, the number of circular permutation of n different things taken all at a time is (n-1)!/2 Now, extending the above discussion to more general: Number of circular permutation of n different things taken r at a time • If clockwise and anti clockwise arrangements are taken as different, then the required number of circular permutation is (nPr)/r • If clockwise and anti clockwise arrangements are taken as same, then the required number of circular permutation is (nPr)/2r Example 1: In how many ways can 15 persons be seated around a table for dinner if there are 9 chairs? Solution: In case of circular table the clockwise and anticlockwise arrangements will be treated as different arrangements. Hence the total number of ways is 15P9/9 Example 2: How many different necklaces can be formed from 20 beads of different colors if 10 beads are required to make a necklace? Solution: In case of necklace there is no distinction between the clockwise and anticlockwise arrangements. So, the required number of circular permutation is 20P10/(2 × 10) IIT JEE Main Question Papers JEE Main Practice Papers JEE Main Sample Papers ## Register to get FREE updates All Fields Mandatory • (Ex:9123456789)
## The Vault: Powers of two My new book of maths puzzles is on its way! It’s packed full of interesting problems to sink your teeth into. I’ll post an update as the launch approaches. In the meantime, here is a fantastic problem from the USA: It is given that $2^{2004}$ is a 604-digit number beginning with a 1. How many of the numbers $2^0, 2^1, 2^2, 2^3, ..., 2^{2003}$ begin with a 4? On the face of it, the fact that this can even be solved in a reasonable length of time is astonishing. There are so many numbers in that list! And yet, the solution is remarkably simple. The first idea is to consider how the first digit of a number changes when you double the number. After some thought, we arrive at the following possible ‘first digit chains’: 1,2,4,8,1… 1,2,4,9,1,… 1,2,5,1,… 1,3,6,1,… 1,3,7,1,… Notice that we keep coming back to a first digit of 1, whatever other first digits appear in the chain. But the crucial observation is that the digit 4 appears precisely in the long chains, and never in the short chains. So, in moving through the powers of 2 from $2^0$ to $2^{2003}$, the first digits appear in blocks of length four (1,2,4,8,… or 1,2,4,9,…) or length three (1,2,5,… or 1,3,6, … or 1,3,7,…). Furthermore, the number of times the first digit 4 appears is equal to the number of chains of length four. But all this information can be summarised in one pair of equations! If we let $x$ and $y$ be the number of four-chains and three-chains respectively, then we have $x+y=603$ (since every chain represents an increase of one digit in the length of the number ) and $4x+3y=2004$ (since there are 2004 terms from $1$ to \$2^{2003}\$. Solving simultaneously yields $x=195$, so the number of terms beginning with the digit 4 is 195. The problem is utterly brilliant. I’ve seen many questions in which one has to form their own simultaneous equations, but rarely have I seen one of such creative invention.
5.6 ## National STEM Learning Centre Skip to 0 minutes and 7 secondsSPEAKER 1: So let's have a look now at how we divide one fraction by another. We're going to have a look at 2/5 divided by 2/3. Skip to 0 minutes and 15 secondsSPEAKER 2: So I've come across a method for doing this, and I've heard it being called the KFC method. So K means that you keep this one the same. F, we're going to flip the second one, and then C, we're going to change this from a division to a multiplication. And that seems to work every time. Skip to 0 minutes and 33 secondsSPEAKER 1: OK, so you're saying we're going to keep this the same, so 2/5. We're going to change to a multiplication and flip this one to be 3/2. Skip to 0 minutes and 45 secondsSPEAKER 2: And I can do multiplication, so 2 times 3 gives us 6 on the top. 5 times 2 is 10, so it's going to be 6/10. Skip to 0 minutes and 53 secondsSPEAKER 1: OK, which we can simplify to be 3/5. OK, and do you know why that works? Where does it come from? Skip to 1 minute and 2 secondsSPEAKER 1: So maths, it's quite poor practise if we don't understand the reason behind some things. So let's have a look into that. So if we have 2/5 divided by 2/3-- so you put 2/5. And we could write this quite crudely as 2/5 divided by 2/3. Skip to 1 minute and 17 secondsSPEAKER 2: So the fraction and then another fraction underneath? Skip to 1 minute and 20 secondsSPEAKER 1: Yes, it isn't ideal, but just to show the reasoning where it comes from. Now, we know if we want to get rid of a denominator, if we multiply by the reciprocal, that would equal 1. So you would have 3/2. Because these are fractions and equivalent, do the same thing top and bottom, so 3/2. Skip to 1 minute and 43 secondsSPEAKER 2: So this thing that you're multiplying by on the right-hand side is 3/2 over 3/2. So that's just 1, so it's not going to change the value? Skip to 1 minute and 51 secondsSPEAKER 1: Absolutely, perfect, good, and the reason we're doing this is to have our reciprocal to get rid of our denominator. So with that in mind, our denominator now becomes 1. Our numerator becomes 2/5 times 3/2. So we have our 2/5 multiplied by 3/2. So can we see now that's where our reasoning comes from? Skip to 2 minutes and 15 secondsSPEAKER 2: Ah, because that's 1, we can almost ignore it. And then we just do the multiplication? Skip to 2 minutes and 20 secondsSPEAKER 1: Exactly, yes, that's exactly where it comes from. So we have our 2 multiplied by 3. As we solved earlier, it's our 6. 5 multiplied by 2, product 10. So finally, same answer but with the more reasoning, it's 3/5. # Dividing fractions How would you explain what happens when we divide one fraction by another? Before you watch the video, have a go at this task: ## Describe In the comments below, before you watch the video or continue, just have a go at writing out an explanation of what happens when you divide one fraction by another. Explaining the exact meaning of what is meant when we divide one fraction by another, for example dividing two fifths by three halves, might be difficult as it is not easy to represent this using manipulatives. In these cases, we might decide to show pictorially that we are attempting to find how many three halves go into two fifths. However, this too can be very confusing especially, when students know that three halves is a bigger fraction than two fifths. This may explain why students are often taught ‘how’ to divide fractions by using a simple algorithm. We can, however, if we understand the properties of reciprocals, explain quite simply why the algorithm for dividing fractions works. Watch the video in which Paula and Michael explain. How would you amend your description? ## Problem worksheet Complete questions 7 and 8 from this week’s worksheet. ## Teaching resource This teaching resource directly addresses one of the common misconceptions to do with dividing whole numbers by fractions: Dividing whole numbers by fractions.
# Properties of multiplication (10x10) Lesson ## Any order is okay Before we get too far into this one, have a play with the applet below. 1. Create an array by using the sliders 2. Then rotate the array around to a different orientation 3. Does the size of the array change? 4. Why do we write the multiplication differently? 5. Are the answers to the multiplication the same? With multiplication, it doesn't matter which way two numbers are multiplied together, we get the same answer each time. This is called the commutative property of multiplication or turnarounds. You might like to refresh your memory on using arrays for multiplication (and division), as this is what we are going to use to prove that we can multiply numbers in any order. ## Using multiplication turnarounds Now that we know we can multiply numbers in any order, let's look at how we can use them to solve some multiplication questions. We will look at multiplying by 0 and multiplying by 1 Did you know? Not only can you multiply two numbers in any order, you can multiply any quantity of numbers, in any order. Why don't you see if you can prove that this works? Perhaps you can solve $3\times4\times5$3×4×5 and see if you get the same answer as you do when you solve $5\times4\times3$5×4×3. Can you write this problem in a different order again, and solve it? #### Worked example ##### Question 1 The commutative property of multiplication says that the order of numbers doesn't matter in multiplication. For example, $3\times7=7\times3$3×7=7×3. Use the commutative property to complete the following statements: 1. If $5\times3=15$5×3=15, then $3\times5=\editable{}$3×5=. 2. If $5\times4=20$5×4=20, then $4\times5=\editable{}$4×5=. 3. If $5\times\editable{}=35$5×=35, then $\editable{}\times5=35$×5=35. ##### Question 2 1. When $16$16 is multiplied by $1$1, the answer is: $1$1 A $16$16 B $0$0 C $1$1 A $16$16 B $0$0 C 2. When $1195$1195 is multiplied by $1$1, the answer is: $1$1 A $1195$1195 B $0$0 C $1$1 A $1195$1195 B $0$0 C 3. When any number (that isn't zero) is multiplied by $1$1, the answer is: $0$0 A $1$1 B Itself C $0$0 A $1$1 B Itself C ##### Question 3 Let's see if we can work out $4\times5\times3$4×5×3 by multiplying any pair of the numbers first. 1. First find $4\times5=\editable{}$4×5=. Then multiply the result by $3$3, to get $\editable{}\times3=\editable{}$×3=. 2. Now find $5\times3=\editable{}$5×3=. Then multiply the result by $4$4, to get $4\times\editable{}=\editable{}$4×=. 3. This time, find $4\times3=\editable{}$4×3=. Then multiply the result by $5$5, to get $\editable{}\times5=\editable{}$×5=. 4. Does the order of multiplication matter? No A Yes B No A Yes B ### Outcomes #### NA3-2 Know basic multiplication and division facts. #### NA3-7 Generalise the properties of addition and subtraction with whole numbers
# Know about different parts of the circle In the geometry field, students come across several shapes. One of the popular shapes that we all are familiar with is the circle. In layman terms, it is the round in shape like coin, bangles, bottle, tyres and others. Most of you may argue that earth is also in shape, but it is not. The earth is spherical. If we talk about the circle, there are different parts of circle. Whether it is a sequence and seriesof a circle, or only the circle, terminologies of them will be the same. It means that there is a radius, diameter, pie and much more included in a circle. If you have read about the circle in the past, you might have known its different parts. In this guide also, we have mentioned different parts of the circle. Just keep reading the post. ## What are different circle parts? You might think that the circle is only about a round figure and it has no parts then you are wrong. There are different parts of the circle. • Diameter • Centre • Circumference of a circle • Chord • Arc • Segment • Sector If you are reading such terminologies first time, no need to worry because we have brought explanation also about them The central point and the outer rim of a circle is there and the distance between them is called a radius. ### Diameter To put it simply, diameter is twice of the radius that divides the circle into two parts. And both parts should be equal in measurements. Basically there is a chord that segregates the circle into two different parts. ### Centre The centre of a circle is a point from where the radius is drawn. It means that there is a fixed point from which a closed figure is drawn that we call a circle. If you have drawn a circle with the help of a compass you might understand what we want to convey to you. ### Circumference of a Circle You can understand this part of a circle by the length of a circle. We are making this statement simpler by sharing one example. For instance, if you have a wire of 15 cm, bend it and form a circle. The circular wire has a circumference of 15 cm. ### Chord A line segment that joins any two points of the circle can be considered as a chord. If you have a doubt whether the diameter is a chord or not, then you are right. Even diameter is also considered as a chord, because it joins two points. In a circle, it is the longest chord. ### Arc Arc is a piece of a circle that can be drawn the circumference of a circle. A continuous piece of a Suppose A,B,C,D,E,F be the points on the circumference of the circle. AB, BC, CD, DE, EF etc pieces will be counted as arcs. ### Segment To understand it, draw a diagram circle and one chord named as AB. The divided region AB is enclosed in a circle. The segment of a circle that contains the minor arc is called the minor segment and the major arc covers the major segment. Basically, it differentiates between chords of a circle. ### Sector It is a circular disc region that lies between an ac and two radius. And these two radii join extreme arc points and the centre of a circle. You can call it a pie shape. If you want to know how to draw a circle, take the help of a compass and draw with it. Drawing circles with the help of a compass is easier and will give you an accurate figure. ## FAQ’s ### What are the 3 main parts of a circle? The three main parts of a circle are the radius, chord, and diameter. The radius is the distance from the center of the circle to its edge, the chord is a line segment that connects two points on the circumference of the circle, and the diameter is a line segment that runs through the center of the circle and connects its two edges. ### What is circle and parts of circle? A circle is defined as a set of all points in a plane that are at the same distance from a given point, called the center of the circle. The radius is the distance from the center to any point on the circle. The circumference of a circle is the distance around it. The area of a circle is pi times the radius squared. ### What is half circle called? A semicircle is a half circle. The word “semicircle” comes from the Latin words semi meaning “half” and circulus meaning “circle”. So literally, a semicircle is a half circle. ### What is a slice of a circle called? A slice of a circle is typically referred to as either a Sector or a Segment. A Sector is a portion of the circumference of the circle, formed by two radii and their intercepted arc; it looks like you would imagine cutting a pizza slice. A Segment is also made up of two radii, but this time they are connected by a chord – an imaginary line between two points on the circle. The Segment differs from the Sector in that the intercepted arc does not form part of it; instead, it consists only of the area between the chord and the circumference. Both Sectors and Segments are useful when measuring angles, calculating areas or deriving other elements of geometry related to circles. In mathematics, these two shapes are vital components in finding solutions to problems involving circles and can be used to calculate areas and other measurements with great precision and accuracy.
# NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables Written by Team Trustudies Updated at 2021-05-07 ## NCERT solutions for class 9 Maths Chapter 4 Linear Equations In Two Variables Exercise 4.1 Q1 ) The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be Rs. x and that of a pen to be Rs. y) NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables Let the cost of a notebook = Rs. x and the cost of a pen = Rs. y According to question, Cost of a Notebook = ( Cost of a 2 Pen) x = 2y Therefore the equation, x - 2y = 0 represents this statement. Q2 ) Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case. i) $$2x + 3y = 9\overline{.35}$$ ii)$$x - \frac{y}{5} - 10 = 0$$ iii) -2x + 3y = 6 iv) x = 3y v) 2x = -5y vi) 3x + 2 = 0 vii) y - 2 = 0 viii)5 = 2x NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables i)$$2x + 3y = 9\overline{.35}$$ Rearranging the terms we get, $$2x + 3y - 9\overline{.35} = 0$$ On comparing with ax + by + c = 0, then the values of a = 2, b = 3 and c = -$$9\overline{.35}$$ ii)$$x - \frac{y}{5} - 10 = 0$$ Rearranging the terms we get, $$x - \frac{y}{5} - 10 = 0$$ On comparing with ax + by + c = 0, then the values of a = 1, b = $$\frac{-1}{5}$$ and c = -10 iii)-2x + 3y = 6 Rearranging the terms we get, -2x + 3y - 6 = 0 On comparing with ax + by + c = 0, then the values of a = -2, b = 3 and c = -6 iv)x = 3y Rearranging the terms we get, x - 3y + 0= 0 On comparing with ax + by + c = 0, then the values of a = 1, b = -3 and c = 0 v)2x = -5y Rearranging the terms we get, 2x + 5y + 0 = 0 On comparing with ax + by + c = 0, then the values of a = 2, b = 5 and c = 0 vi)3x + 2 = 0 Rearranging the terms we get, 3x + (0)y + 2 = 0 On comparing with ax + by + c = 0, then the values of a = 3, b = 0 and c = 2 vii)y - 2 = 0 Rearranging the terms we get, (0)x + y - 2 = 0 On comparing with ax + by + c = 0, then the values of a = 0, b = 1 and c = -2 viii)5 = 2x Rearranging the terms we get, 2x + (0)y - 5 = 0 On comparing with ax + by + c = 0, then the values of a = 2, b = 0 and c = -5 ## NCERT solutions for class 9 Maths Chapter 4 Linear Equations In Two Variables Exercise 4.2 Q1 ) Which one of the following options is true and why? y = 3x + 5 has i) A unique solution. ii) Only two solutions. iii) Infinitely many solutions. NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables The iii) option is true. A linear equation in two variables has infinitely many solutions. Q2 ) Write four solutions for each of the following equations : i) 2x + y = 7 ii) $$\pi$$x + y = 9 iii) x = 4y NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables i)By inspection, x = 2 and y = 3 is a solution because for these values of x and y, we get,2(2) + 3 = 4 + 3 = 7 Now, if we put x = 0, then the equation reduces to an unique solution of y = 7. So, one solution of 2x + y = 7 is (0, 7). Similarly, if we put y = 0, then the equation reduces to an unique solution of x = $$\frac{7}{2}$$. So, another solution of 2x + y = 7 will be ($$\frac{7}{2}$$ , 0). Finally, let us put x = 1, then, we get, 2(1) + y = 7 => 2 + y = 7 => y = 5 Hence, (1, 5) is also a solution of given equation. Therefore, four of the infinitely many solutions of the equation 2x + y = 7 are : (2, 3) , (0, 7) , ($$\frac{7}{2}$$ , 0) and (1, 5). ii)By inspection, x = $$\frac{1}{\pi }$$ and y = 8 is a solution because for these values of x and y, we get,$$\pi (\frac{1}{\pi}$$) + 8 = 1 + 8 = 9 Now, if we put x = 0, then the equation reduces to an unique solution of y = 9. So, one solution of $$\pi$$x + y = 9 is (0, 9)). Similarly, if we put y = 0, then the equation reduces to an unique solution of x = $$\frac{9}{\pi }$$. So, another solution of 2x + y = 7 will be ($$\frac{9}{\pi }$$, 0). Finally, let us put x = 1, then, we get, $$\pi$$(1) + y = 9 => $$\pi$$ + y = 9 => y = 9 - $$\pi$$ Hence, (1, 9 - $$\pi$$) is also a solution of given equation. Therefore, four of the infinitely many solutions of the equation 2x + y = 7 are : ($$\frac{1}{ \pi }$$, 8) , (0, 9) , ($$\frac{9}{\pi }$$, 0) and (1, 9 - $$\pi$$). iii)By inspection, x = 0 and y = 0 is a solution because for these values of x and y, we get,0 = 4(0) => 0 = 0 Now, if we put x = 4, then the equation reduces to solution of y = 1. So, one solution of x = 4y is (4, 1). Similarly, if we put y = -1, then the equation reduces to solution of x = -4. So, another solution of x = 4y will be (-4, -1). Finally, let us put y = $$\frac{1}{2}$$ , then, we get, x = 4($$\frac{1}{2}$$) => x = 2 Hence, (2, $$\frac{1}{2}$$ ) is also a solution of given equation. Therefore, four of the infinitely many solutions of the equation x = 4y are : (0, 0) , (4, 1) , (-4, -1) and (2, $$\frac{1}{2}$$ ). Q3 ) Check of the following are solution of the equation x – 2y = 4 and which are not? i) (0, 2) ii) (2, 0) iii) (4, 0) iv) ($$\sqrt{2}$$, 4$$\sqrt{2}$$) v) (1, 1) NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables If (x, y) are the solutions of the equation x – 2y = 4 then, they satisfies that equation. i) Let's put x = 0 and y = 2, we get, L.H.S. = (0) - 2(2) = -4 so, since, L.H.S. $$\ne$$ R.H.S. (0, 2) are not the solutions of given equation. ii) Let's put x = 2 and y = 0, we get, L.H.S. = (2) - 2(0) = 2 so, since, L.H.S. $$\ne$$ R.H.S. (2, 0) are not the solutions of given equation. iii) Let's put x = 4 and y = 0, we get, L.H.S. = (4) - 2(0) = 4 so, since, L.H.S. = R.H.S. (4, 0) are the solutions of given equation. iv) Let's put x = $$\sqrt{2}$$ and y = 4$$\sqrt{2}$$, we get, L.H.S. = ($$\sqrt{2}$$) - 2(4$$\sqrt{2}$$) = -7$$\sqrt{2}$$ so, since, L.H.S. $$\ne$$ R.H.S. ($$\sqrt{2}$$, 4$$\sqrt{2}$$) are not the solutions of given equation. v) Let's put x = 1 and y = 1, we get, L.H.S. = (1) - 2(1) = -1 so, since, L.H.S. $$\ne$$ R.H.S. (1, 1) are not the solutions of given equation. Q4 ) Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k. NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables As x = 2 and y = 1 is the solution of the equation 2x + 3y = k, so it satisfies the equation. put x = 2 and y = 1, we get, k = 2(2) + 3(1) = 4 + 3 = 7 Therefore, the value of k is 7. ## NCERT solutions for class 9 Maths Chapter 4 Linear Equations In Two Variables Exercise 4.3 Q1 ) Draw the graph of each of the following linear equations in two variables i) x + y = 4 ii) x - y = 2 iii) y = 3x iv) 3 = 2x + y NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables i) To draw the graph, we need atleast two solutions of the equation. x04 y40 (x, y)(0, 4)(4, 0) Thus, we draw the graph by plotting the two points from table and then joining by a line. ii) To draw the graph, we need atleast two solutions of the equation. x02 y-20 (x, y)(0, -2)(2, 0) Thus, we draw the graph by plotting the two points from table and then joining by a line. iii)To draw the graph, we need atleast two solutions of the equation. x01-1 y03-3 (x, y)(0, -2)(2, 0)(-1, -3) Thus, we draw the graph by plotting the two points from table and then joining by a line. iv)To draw the graph, we need atleast two solutions of the equation. x01-1 y315 (x, y)(0, 3)(1, 1)(-1, 5) Thus, we draw the graph by plotting the two points from table and then joining by a line. Q2 ) Give the equations of two lines passing through (2, 14). How many more such lines are there, and why? NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables Here, (2, 14) is a solution of a linear equation. One example of such a linear equation is y = 6x + 2 and another is x + y = 16. There are infinitely many lines because there are infinitely many linear equations which are satisfied by the coordinates of the point (2, 14). Q3 ) If the point (3, 4) lies on the graph of the equation 3y - ax - 7 , find the value of a . NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables If the point (3, 4) lies on the graph, then it satisfies the equation. $$\because$$ 3y - ax - 7 = 0 $$\Rightarrow$$ 3(4) - a(3) - 7 = 0 $$\Rightarrow$$ 12 - 3a = 7 $$\Rightarrow$$ 12 - 7 = 3a $$\Rightarrow$$ 5 = 3a $$\Rightarrow$$ a = $$\frac{5}{3}$$ Q4 ) The taxi fare in a city is as follows: For the first kilometer, the fare is Rs. 8 and for the subsequent distance it is Rs. 5 per km. Taking the distance covered as x km and the total fare as Rs. y, write a linear equation for this information and draw its graph. NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables Distance covered = x km = 1 + (x – 1) km The fare for the first kilometer is Rs. 8. Fare for next (x - 1)km = (x - 1) × 5 = 5(x - 1). According to question, Total fare = y. $$\therefore$$ 8 + 5(x - 1) = y $$\Rightarrow$$ 8 + 5x - 5 = y So, the equation becomes 5x - y + 3 = 0 x01 y38 (x, y)(0, 3)(1, 8) Now, plot the points (0, 3) and (1, 8) on a graph paper and joining them, to form a line . Q5 ) From the choices given below, choose the equation whose graphs are given in fig. (a) and fig. (b). For fig. (a)For fig. (b) i) y = x (i) y = x + 2 ii)x + y = 0y = x - 2 iii)y = 2xy = -x + 2 iv)2 + 3y = 7xiv)x + 2y = 6 NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables In fig. (a), we can observe that the points(1, -1) and (-1, 1) lies on the line of equation x + y = 0. Because, for (-1, 1), (1) + (-1) = 0 and for (1, -1), (-1) + (1) = 0 Similarly, in fig. (b), we can observe that the points(2, 0),(0, 2) and (-1, 3) lies on the line of equation x + y = 2. Because, for (2, 0), (2) + (0) = 2, for(0, 2), (0) + (2) = 2 And for (-1, 3), (-1) + (3) = 2 Q6 ) If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also, read from the graph the work done when the distance travelled by the body is (i) 2 units (ii) 0 units. NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables Given that, work done by a body on application of a constant force is directly proportional to the distance travelled by the body. i.e., (W) work done in distance (s) = F.s Where,(F is an arbitrary constant which take the value 5 units) Because, W = 5s.....(i) sWPoint 00O(0, 0) 15A(1, 5) 210B(2, 10) Now, plot the points on graph paper and join all the points to get a line. i) From point B(2, 10), draw a line parallel to OY to intersect the x-axis at (2, 10) and draw a line parallel to OY to the x-axis intersect at C(0, 10). work done by a body when distance travelled is 2 units = 10 units. ii) Clearly y = 0, when x = 0, so the work done by a body when distance travelled is 0 units Q7 ) Yamini and Fatima, two students of class IX of a school, together contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs. x and Rs. y.) Draw the graph of the same. NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables Let the contributions of Yamini and Fatima together towards the Prime Minster’s Relief Fund to help the earthquake victms are Rs. x and Rs. y, respectively. Then, by given condition, x + y = 100 … (i) x0100 y = 100 - x1000 (x, y)B (0, 100)A (100, 0) Here, we plot the points B (0, 100) and A (100, 0) on graph paper and join all these points to form a line AB. Q8 ) In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is linear equation that converts Fahrenheit to Celsius. F = ($$\frac{9}{5}$$ )C + 32 (i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis. (ii) If the temperature is $$30^{\circ}$$C , what is the temperature in Fahrenheit? (iii) If the temperature is 95F , what is the temperature in Celsius? (iv) If the temperature is $$0^{\circ}$$C , what is the temperature in Fahrenheit and if the temperature is 0F , what is the temperature in Celsius? (v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it. NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables The linear equation that converts Fahrenheit to Celsius is F = ($$\frac{9}{5}$$)C + 32 $$\Rightarrow$$ 5F - 9C = 160....(i) For plotting points, we get the following table : C0-160/9-40 F = (9/5)C + 32320-40 PointsA(0, 32)B(-160/9, 0)c(-40, -40) Here, we plot the given points on graph paper and join all these points to form a line. ii) ii)If the temperature is $$30^{\circ}$$C i.e.,C = $$30^{\circ}$$C Then we get, F = ($$\frac{9}{5}$$ )×(30) + 32 = 9 × 6 + 32 = 54 + 32 = 86 Therefore, the temperature in Fahrenheit is 86F iii)If the temperature is 95F i.e.,F = 95 Then we get, from eq.(i) 5(95) - 9C = 160 $$\Rightarrow$$ 9C = 475 - 160 = 315 $$\therefore$$ C = $$35^{\circ}$$C. Therefore, the temperature in Celsius is $$35^{\circ}$$C iv)If the temperature is $$0^{\circ}$$C i.e.,C = $$0^{\circ}$$ Then we get, F = ($$\frac{9}{5}$$ )×(0) + 32 = 32 Therefore, the temperature in Fahrenheit is 32F. If the temperature is 0F i.e.,F = 0 Then we get, from eq.(i) 5(0) - 9C = 160 $$\Rightarrow$$ C = $$\frac{-160}{9} ^{\circ}$$C Therefore, the temperature in celsius is $$\frac{-160}{9} ^{\circ}$$C. V)For this, let us take C = F, Thus from eq.(i) we get, 5F - 9F = 160 $$\Rightarrow$$ -4F = 160 $$\Rightarrow$$ F = -40 Therefore, C = F = $$-40^{\circ}$$ ## NCERT solutions for class 9 Maths Chapter 4 Linear Equations In Two Variables Exercise 4.4 Q1 ) Give the geometric representations of y = 3 as an equation (i) In one variable (ii) In two variables NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables The given linear equation is y = 3 ...(i) (i) The representation of the solution on the number line is shown in the figure below, where y = 3 is treated as an equation in one variable. ii)The representation of the solution on the number line is shown in the figure below, where y = 3 can be written as an equation in of two variable i.e., (0).x + y = 3 Here, if we observe, all the values of x are permissible as (0).x is always going to be 0. However, y always has to be equal to 3. Hence, graph of y = 3 will be a line parallel to x-axis and at a distance of 3 units along positive y- axis. Q2 ) Give the geometric representations of 2x + 9 = 0 as an equation (i) In one variable (ii) In two variables NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables The given linear equation is $$\because$$ 2x + 9 = 0 $$\Rightarrow x = \frac{-9}{2}$$ ...(i) (i) The representation of the solution on the number line is shown in the figure below, where x = $$\frac{-9}{2}$$ is treated as an equation in one variable. ii)The representation of the solution on the number line is shown in the figure below, where 2x + 9 = 0 can be written as an equation in of two variable i.e., 2x + (0).y = -9 Here, if we observe, all the values of y are permissible as (0).y is always going to be 0. However, x always has to satisfy the equation to 2x + 9 = 0. Hence, graph of x = $$\frac{-9}{2}$$ will be a line parallel to y-axis and at a distance of $$\frac{-9}{2}$$ units along negative x- axis. ##### FAQs Related to NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables There are total 16 questions present in ncert solutions for class 9 maths chapter 4 linear equations in two variables There are total 3 long question/answers in ncert solutions for class 9 maths chapter 4 linear equations in two variables There are total 4 exercise present in ncert solutions for class 9 maths chapter 4 linear equations in two variables
# Set Theory- Basics, Definition, Representation of Sets Set Theory is a branch of mathematics and is a collection of objects known as numbers or elements of the set. Set theory is a vital topic and lays stronger basics for the rest of the Mathematics. You can learn about the axioms that are essential for learning the concepts of mathematics that are built with it. For instance, Element a belongs to Set A can be denoted by a ∈ A and a ∉ A represents the element a doesn’t belong to Set A. { 3, 4, 5} is an Example of Set. In this article of Introduction to Set Theory, you will find Representation of Sets in different forms such as Statement Form, Roster Form, and Set Builder Form, Types of Sets, Cardinal Number of a Set, Subsets, Operations on Sets, etc. • Representation of a Set • Types of Sets • Finite Sets and Infinite Sets • Power Set • Problems with the Union of Sets • Problems with Intersection of Sets • Difference between two Sets • Complement of a Set • Problems with the Complement of a Set • Problems with Operation on Sets • Word Problems on Sets • Venn Diagrams in Different Situations • Relationship in Sets using Venn Diagram • Union of Sets using Venn Diagram • Intersection of Sets using Venn Diagram • Disjoint of Sets using the Venn Diagram • Difference of Sets using Venn Diagram • Examples of Venn Diagrams ### Set Definition Set can be defined as a collection of elements enclosed within curly brackets. In other words, we can describe the Set as a Collection of Distinct Objects or Elements. These Elements of the Set can be organized into smaller sets and they are called the Subsets. Order isn’t that important in Sets and { 1, 2, 4} is the same as { 4,2, 1}. ### Examples of Sets • Odd Numbers less than 20, i.e., 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 • Prime Factors of 15 are 3, 5 • Types of Triangles depending on Sides: Equilateral, Isosceles, Scalene • Top two surgeons in India • 10 Famous Engineers of the Society. Among the Examples listed the first three are well-defined collections of elements whereas the rest aren’t. ### Important Sets used in Mathematics N: Set of all natural numbers = {1, 2, 3, 4, …..} Q: Set of all rational numbers R: Set of all real numbers W: Set of all whole numbers Z: Set of all integers = {….., -3, -2, -1, 0, 1, 2, 3, …..} Z+: Set of all positive integers ### Representation of a Set Set can be denoted using three common forms. They are given along the following lines by taking enough examples • Statement Form • Roaster Form or Tabular Form • Set Builder Form Statement Form: In this representation, elements of the set are given with a well-defined description. You can see the following examples for an idea Example: Consonants of the Alphabet Set of Natural Numbers less than 20 and more than 5. Roaster Form or Tabular Form: In Roaster Form, elements of the set are enclosed within a pair of brackets and separated by commas. Example: N is a set of Natural Numbers less than 7 { 1, 2, 3, 4, 5, 6} Set of Vowels in Alphabet = { a, e, i, o, u} Set Builder Form: In this representation, Set is given by a Property that the members need to satisfy. {x: x is an odd number divisible by 3 and less than 10} {x: x is a whole number less than 5} ### Size of a Set At times, we are curious to know the number of elements in the set. This is called cardinality or size of the set. In general, the Cardinality of the Set A is given by \|A\| and can be either finite or infinite. ### Types of Sets Sets are classified into many kinds. Some of them are Finite Set, Infinite Set, Subset, Proper Set, Universal Set, Empty Set, Singleton Set, etc. Finite Set: A Set containing a finite number of elements is called Finite Set. Empty Sets come under the Category of Finite Sets. If at all the Finite Set is Non-Empty then they are called Non- Empty Finite Sets. Example: A = {x: x is the first month in a year}; Set A will have 31 elements. Infinite Set: In Contrast to the finite set if the set has infinite elements then it is called Infinite Set. Example: A = {x : x is an integer}; There are infinite integers. Hence, A is an infinite set. Power Set: Power Set of A is the set that contains all the subsets of Set A. It is represented as P(A). Example: If set A = {-5,7,6}, then power set of A will be: P(A)={ϕ, {-5}, {7}, {6}, {-5,7}, {7,6}, {6,-5}, {-5,7,6}} Sub Set: If Set A contains the elements that are in Set B as well then Set A is said to be the Subset of Set B. Example: If set A = {-5,7,6}, then Sub Set of A will be: P(A)={ϕ, {-5}, {7}, {6}, {-5,7}, {7,6}, {6,-5}, {-5,7,6}} Universal Set: This is the base for all the other sets formed. Based on the Context universal set is decided and it can be either finite or infinite. All the other Sets are Subsets of the Universal Set and are given by U. Example: Set of Natural Numbers is a Universal Set of Integers, Real Numbers. Empty Set: There will be no elements in the set and is represented by the symbol ϕ or {}. The other names of Empty Set are Null Set or Void Set. Example: S = { x \| x ∈ N and 9 < x < 10 } = ∅ Singleton Set: If a Set contains only one element then it is called a Singleton Set. Example: A = {x : x is an odd prime number} ### Operations on Sets Consider Two different sets A and B, they are several operations that are frequently used Union: Union Operation is given by the symbol U. Set A U B denotes the union between Sets A and B. It is read as A Union B or Union of A and B. It is defined as the Set that contains all the elements belonging to either of the Sets. Intersection: Intersection Operation is represented by the symbol ∩. Set A ∩ B is read as A Intersection B or Intersection of A and B. A ∩ B is defined in general as a set that contains all the elements that belong to both A and B. Complement: Usually, the Complement of Set A is represented as Ac or A or ~A. The Complement of Set A contains all the elements that are not in Set A. Power Set: The power set is the set of all possible subsets of S. It is denoted by P(S). Remember that the Empty Set and the Set itself also come under the Power Set. The Cardinality of the Power Set is 2n in which n is the number of elements of the set. Cartesian Product: Consider A and B to be Two Sets. The Cartesian Product of the two sets is given by AxB i.e. the set containing all the ordered pairs (a, b) where a belongs to Set A, and b belongs to Set B. Representation of Cartesian Product A × B = {(a, b) \| a ∈ A ∧ b ∈ B}. The cardinality of AxB is N*M where N is the cardinality of A and M is the Cardinality of B. Remember that AxB is not the same as BxA.
Albert Teen YOU ARE LEARNING: Circle Theorems: Angle in a semi-circle # Circle Theorems: Angle in a semi-circle ### Circle Theorems: Angle in a semi-circle We can apply universal rules to solve geometrical problems involving circles, where there are tangents, secants or chords present. The properties of circles have been been studied for centuries, and there are a range of generalised results which are true for circles of any size. First, let's define some lines! 1 This is a secant A secant moves through two points on the circumference of a circle, and keeps going 2 This is a tangent A tangent touches the circumference of a circle at a single point 3 This is a chord A chord is a straight line which connects any two points on the circumference of a circle 4 What type of line is this? Let's go through the first two circle theorems! 1 The angle in a semicircle is $90\degree$ The line moves through the centre of the circle. No matter where you move the yellow point, the angle is always $90\degree$! 2 What is the sum of the other two angles in this triangle? 1 The angle in the centre is twice the angle at the circumference By drawing two radii and two chords, we form a situation where there are two angles. Try dragging around the green points - the angle in the centre is always twice the angle at the circumference. 2 What would the angle at the circumference be here? Write the number of degrees. 3 The angle at the circumference is $51\degree$ This is because the angle would be half the angle at the centre, so $102\degree\div2=51\degree$ 4 What is the angle at the centre here? Write the number of degrees. 5 The angle is $156\degree$ This is because the angle in the centre is double the angle at the circumference, so $78\degree\times2=156\degree$
#### What do the interior angles of a quadrilateral add up to A quadrilateral is a shape with 4 sides. For any quadrilateral , we can draw a diagonal line to divide it into two triangles. Each triangle has an angle sum of 180 degrees. Therefore the total angle sum of the quadrilateral is 360 degrees. ## Do angles in a quadrilateral equal 360? The sum of interior angles in a quadrilateral is 360 °. Another way to calculate the sum of the interior angles of a polygon is to see how many triangles the shape is composed of. Quadrilaterals are composed of two triangles. ## What is the formula for interior angles? An interior angle is located within the boundary of a polygon. The sum of all of the interior angles can be found using the formula S = (n – 2)*180. It is also possible to calculate the measure of each angle if the polygon is regular by dividing the sum by the number of sides. ## What are the 7 types of angles? Types of Angles – Acute, Right, Obtuse , Straight and Reflex Acute angle. Right angle. Obtuse angle. Straight angle. Reflex angle. ## How do you prove a quadrilateral has 360 degrees? A quadrilateral is a polygon which has 4 vertices and 4 sides enclosing 4 angles and the sum of all the angles is 360 °. When we draw a draw the diagonals to the quadrilateral , it forms two triangles. Both these triangles have an angle sum of 180°. Therefore, the total angle sum of the quadrilateral is 360 °. ## Do all interior angles add up to 360? All sides are the same length (congruent) and all interior angles are the same size (congruent). To find the measure of the interior angles , we know that the sum of all the angles is 360 degrees (from above) And there are four angles So, the measure of the interior angle of a square is 90 degrees. You might be interested: How to fix interior roof of car ## How do you figure angles? Find the angle of elevation of the plane from point A on the ground. Step 1 The two sides we know are Opposite (300) and Adjacent (400). Step 2 SOHCAHTOA tells us we must use Tangent. Step 3 Calculate Opposite/Adjacent = 300/400 = 0.75. Step 4 Find the angle from your calculator using tan1 ## Which of the following can be four interior angles of a quadrilateral? 38 Which of the following can be four interior angles of a quadrilateral ? Solution. (a) We know that, the sum of interior angles of a quadrilateral is 360°. Thus, the angles in option (a) can be four interior angles of a quadrilateral as their sum is 360°. ## How do you find three missing angles in a quadrilateral? Subtract the sum of the angles from 180 degrees to get the missing angle . For example if a triangle in a quadrilateral had the angles of 30 and 50 degrees, you would have a third angle equal to 100 degrees (180 – 80 = 100). ## What are f angles called? Corresponding angles These are sometimes known as ‘ F ‘ angles . The two angles marked in this diagram are called corresponding angles and are equal to each other. The two angles marked in each diagram below are called alternate angles or Z angles . They are equal. ## What angles add up to 90? Two angles are called complementary when their measures add to 90 degrees . Two angles are called supplementary when their measures add up to 180 degrees . ## Do co interior angles add up to 180? Co – interior angles lie between two lines and on the same side of a transversal. In each diagram the two marked angles are called co – interior angles . If the two lines are parallel, then co – interior angles add to give 180 o and so are supplementary. ### Releated #### How to paint your home interior What supplies do I need to paint the interior of my house? Important items for your supply list include: Primer. Paint . Stir sticks. Paint rollers. Roller covers (a .-inch nap is best for most projects) Small paint brushes, for cutting in or touch-ups. Paint trays, and if needed a sturdy holder for your paint […] #### How to window trim interior How do you measure for interior window trim? Measure from inside corner to inside corner across the bottom of the window . Add 1/2-inch to the measurement for the bottom piece of trim . The extra 1/2 inch provides a 1/4-inch gap — also known as the reveal — on all sides between the casing […]
# How do you find the amplitude, period, and shift for y=1/3 sin (2x - pi/3)? Dec 21, 2017 See below. #### Explanation: If we write a trigonometric equation in the form: $y = a \sin \left(b x + c\right) + d$ Then: Amplitude is a Period is $\frac{2 \pi}{b}$ Phase shift is $\frac{- c}{b}$ Vertical shift is d From example: Amplitude $= a = \frac{1}{3}$ Period $= \frac{2 \pi}{b} = \pi$ Phase shift $= \frac{- c}{b} = \frac{\frac{\pi}{3}}{2} = \frac{\pi}{6}$ Vertical shift $= d = 0$ no vertical shift. Dec 21, 2017 $\frac{1}{3} , \pi , \frac{\pi}{6} , 0$ #### Explanation: $\text{the standard form of the sine function is}$ $\textcolor{red}{\overline{\underline{\| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a \sin \left(b x + c\right) + d} \textcolor{w h i t e}{\frac{2}{2}} \|}}}$ $\text{where amplitude "=\|a\|," period } = \frac{2 \pi}{b}$ $\text{phase shift "=-c/b" and vertical shift } = d$ $\text{here } a = \frac{1}{3} , b = 2 , c = - \frac{\pi}{3} , d = 0$ $\Rightarrow \text{amplitude "=\|1/3\|=1/3," period } = \frac{2 \pi}{2} = \pi$ $\text{phase shift } = - \frac{- \frac{\pi}{3}}{2} = \frac{\pi}{6}$ $\text{there is no vertical shift}$
# Fun Math Activities for the 6th Grade Classroom By Keren Perles Learning about fractions, perimeter and area, and percentages might seem fascinating to you, but your students may think otherwise. Catch their interest with some of these fun 6th grade math activities. ## Percentages Percentages tend to bore 6th grade students. After all, what do they care about earning interest and calculating statistics? There are percentages, however, that they can relate to: sales. After all, if they see a pair of designer jeans on sale for 60% off, they have to know whether they can afford them, right? Tap into this interest by making your own “store." Over the course of a week or so, generously hand out fake money to students who are working well, complete assignments, or take on extra work. At the end of the week, set out different prizes along a table in the classroom. Provide various cheap objects that students might like, such as decorative school supplies, snack foods, or gift certificates for class privileges, and label them with both prices and sale signs. For example, a small notebook might have the price “\$5.00" on it, with a for sale sign that reads “20% off." Have students make their selections on paper and hand them in to you at the end of the period. That night, check their papers to make sure that each student has calculated correctly. The next day, give each student the objects that she calculated correctly. ## Fractions Are your students struggling with fractions? Helping them visualize the abstract concepts can help. Have them fold a paper into quarters. Then show them how to fold a second paper into eighths. Have them color in one quarter of the first paper, and then ask them to color in an equivalent value on the second paper. It should be easy for them to see that they’ll need to color in two sections of the paper, or 2/8, to get the same value. You can also teach addition and subtraction of fractions this way. ## Perimeter and Area Figuring out the perimeter and area of a shape can be boring for 6th grade students. Help students use their newfound skills to solve some interesting puzzles with these 6th grade math activities. Provide them with graph paper and encourage them to use it to test out their answers. Have students work in groups to answer some of the following questions: · What is the largest perimeter a rectangle can have, if its area is 24 cm? · What is the largest area a rectangle can have, if its perimeter is 20 cm? · Draw a circle and a square that have approximately the same area. How are the radius of the circle and the length of the square’s side related?
# Equation of tangent line to following curve. • Nov 4th 2007, 09:24 AM Paige05 Equation of tangent line to following curve. Find the equation of the tangent lines to the following curve at the indicated point. y= x / y+a at (0,0) Thanks. :) • Nov 4th 2007, 06:55 PM topsquark Quote: Originally Posted by Paige05 Find the equation of the tangent lines to the following curve at the indicated point. y= x / y+a at (0,0) Thanks. :) Is this $y = \frac{x}{y + a}$ at (0, 0)? The solution method is to solve for the derivative at the point x = 0. This then is the slope of your tangent line, and you may use some version of the line equation and the given point to find the equation for the line. In this case we would probably want to use implicit differentiation: $\frac{dy}{dx} = \frac{1(y + a) - x \cdot \frac{dy}{dx} }{(y + a)^2}$ So $(y + a)^2 \frac{dy}{dx} = y + a - x \cdot \frac{dy}{dx}$ $[(y + a)^2 + x] \frac{dy}{dx} = y + a$ $\frac{dy}{dx} = \frac{y + a}{(y + a)^2 + x}$ At the point (0, 0) the value of the derivative is $\frac{dy}{dx} = \frac{0 + a}{(0 + a)^2 + 0} = \frac{1}{a}$ So the slope of the tangent line at (0, 0) is 1/a. Thus the tangent line has the form: $y = \frac{1}{a} \cdot x + b$ We know this line passes through the point (0, 0), so $0 = \frac{1}{a} \cdot 0 + b$ Thus b = 0 and the tangent line is $y = \frac{1}{a} \cdot x$ -Dan • Nov 5th 2007, 10:23 AM Paige05 Thank you Dan :) • Nov 5th 2007, 10:45 AM Soroban Hello, Paige05! Quote: Find the equation of the tangent lines (?) to the following curve at the indicated point. . . $y \:= \:\frac{x}{y+a}$ at $(0,0)$ We have: . $y^2 + ay \:=\:x$ Differentiate implicitly: . $2yy' + ay' \:=\:1\quad\Rightarrow\quad y'(2y+a) \:=\:1\quad\Rightarrow\quad y' \:=\:\frac{1}{2y + a}$ At $(0,0)\!:\;\;y' \:=\:\frac{1}{a}$ The equation of the tangent is: . $y - 0 \:=\:\frac{1}{a}(x-0)\quad\Rightarrow\quad\boxed{ y \:=\:\frac{1}{a}x}$
# Difference between revisions of "2004 AMC 10B Problems/Problem 24" In triangle $ABC$ we have $AB=7$, $AC=8$, $BC=9$. Point $D$ is on the circumscribed circle of the triangle so that $AD$ bisects angle $BAC$. What is the value of $AD/CD$? $\text{(A) } \dfrac{9}{8} \quad \text{(B) } \dfrac{5}{3} \quad \text{(C) } 2 \quad \text{(D) } \dfrac{17}{7} \quad \text{(E) } \dfrac{5}{2}$ ## Solution Set $\overline{BD}$'s length as $x$. $CD$'s length must also be $x$ since $\angle BAD$ and $\angle DAC$ intercept arcs of equal length. Using Ptolemy's Theorem, $7x+8x=9(AD)$. The ratio is $\boxed{\frac{5}{3}}\implies(B)$ ## Solution 2 Let $P = \overline{BC}\cap \overline{AD}$. Observe that $\angle ABC = \angle ADC$ because they subtend the same arc. Furthermore, $\angle BAP = \angle PAC$, so $\triangle ABP$ is similar to $\triangle ADC$ by AAA similarity. Then $\dfrac{AD}{AB} = \dfrac{CD}{BP}$. By angle bisector theorem, $\dfrac{7}{BP} = \dfrac{8}{CP}$ so $\dfrac{7}{BP} = \dfrac{8}{9-BP}$ which gives $BP = \dfrac{21}{5}$. Plugging this into the similarity proportion gives: $\dfrac{AD}{7} = \dfrac{CD}{\dfrac{21}{5}} \implies \dfrac{AD}{CD} = \boxed{\dfrac{5}{3}} = \textbf{B}$. ## See Also 2004 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 25 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username Login to AoPS
## Key Concepts • If $a,b$, and $c$ are numbers where $c\ne 0$ , then ${\Large\frac{a}{c}}+{\Large\frac{b}{c}}={\Large\frac{a+c}{c}}$ . • To add fractions, add the numerators and place the sum over the common denominator. • Fraction Subtraction • If $a,b$, and $c$ are numbers where $c\ne 0$ , then ${\Large\frac{a}{c}}-{\Large\frac{b}{c}}={\Large\frac{a-b}{c}}$ . • To subtract fractions, subtract the numerators and place the difference over the common denominator. • Find the prime factorization of a composite number using the tree method. 1. Find any factor pair of the given number, and use these numbers to create two branches. 2. If a factor is prime, that branch is complete. Circle the prime. 3. If a factor is not prime, write it as the product of a factor pair and continue the process. 4. Write the composite number as the product of all the circled primes. • Find the prime factorization of a composite number using the ladder method. 1. Divide the number by the smallest prime. 2. Continue dividing by that prime until it no longer divides evenly. 3. Divide by the next prime until it no longer divides evenly. 4. Continue until the quotient is a prime. 5. Write the composite number as the product of all the primes on the sides and top of the ladder. • Find the LCM using the prime factors method. 1. Find the prime factorization of each number. 2. Write each number as a product of primes, matching primes vertically when possible. 3. Bring down the primes in each column. 4. Multiply the factors to get the LCM. • Find the LCM using the prime factors method. 1. Find the prime factorization of each number. 2. Write each number as a product of primes, matching primes vertically when possible. 3. Bring down the primes in each column. 4. Multiply the factors to get the LCM. • Find the least common denominator (LCD) of two fractions. 1. Factor each denominator into its primes. 2. List the primes, matching primes in columns when possible. 3. Bring down the columns. 4. Multiply the factors. The product is the LCM of the denominators. 5. The LCM of the denominators is the LCD of the fractions. • Equivalent Fractions Property • If $a,b$ , and $c$ are whole numbers where $b\ne 0$ , $c\ne 0$ then$\Large\frac{a}{b}=\Large\frac{a\cdot c}{b\cdot c}$ and $\Large\frac{a\cdot c}{b\cdot c}=\Large\frac{a}{b}$ • Convert two fractions to equivalent fractions with their LCD as the common denominator. 1. Find the LCD. 2. For each fraction, determine the number needed to multiply the denominator to get the LCD. 3. Use the Equivalent Fractions Property to multiply the numerator and denominator by the number from Step 2. 4. Simplify the numerator and denominator. • Add or subtract fractions with different denominators. 1. Find the LCD. 2. Convert each fraction to an equivalent form with the LCD as the denominator. 3. Add or subtract the fractions. 4. Write the result in simplified form. ## Glossary composite number A composite number is a counting number that is not prime divisibility If a number $m$ is a multiple of $n$ , then we say that $m$ is divisible by $n$ least common denominator (LCD) The least common denominator (LCD) of two fractions is the least common multiple (LCM) of their denominators multiple of a number A number is a multiple of $n$ if it is the product of a counting number and $n$ ratio A ratio compares two numbers or two quantities that are measured with the same unit. The ratio of $a$ to $b$ is written $a$ to $b$ , $\Large\frac{a}{b}$ , or $a:b$ prime number A prime number is a counting number greater than 1 whose only factors are 1 and itself
SEMATHS.ORG Asked by: Carroll Abbott Updated: 21 July 2021 08:27:00 PM # What is an example of direct variation? where k is the constant of variation. For example, if y varies directly as x, and y = 6 when x = 2, the constant of variation is k = = 3. Thus, the equation describing this direct variation is y = 3x. ## In addition, you may be interested in what are direct variations? Direct variation describes a simple relationship between two variables . We say y varies directly with x (or as x , in some textbooks) if: y=kx. for some constant k , called the constant of variation or constant of proportionality . ## In the same way how do you know if an equation is a direct variation? A direct variation is when x and y (or f(x) and x) are directly proportional to each other For example, if you have a chart that says x and y, and in the x column is 1, 2 and 3, and the y column says 2, 4 and 6 then you know it's proportional because for each x, y increases by 2 ## Subsequently, question is, what are the characteristics of direct variation? Direct variation is a relationship between two variables, x and y. When two variables vary directly, their ratio (the fraction ) always equals the same number. That number is called the Constant of Variation. ## Do you have your own answer or clarification? ### Related questions and answers #### How do you tell if a graph is a direct variation? 1 Answer. A graph shows direct variation if it goes through the origin, (0,0) . The equation is y=kx , where k is a constant, which is apparent when we write the equation as yx=k . #### How do you answer joint variation? Example 1 – If y varies jointly as x and z, and y = 12 when x = 9 and z = 3, find z when y = 6 and x = 15. Step 1: Write the correct equation. Joint variation problems are solved using the equation y = kxz. Step 2: Use the information given in the problem to find the value of k. #### What is constant variation? The constant of variation is the number that relates two variables that are directly proportional or inversely proportional to one another. #### How do you solve joint and combined variation? Joint variation is just like direct variation, but it involves two or more variables: y=k(xz). Combined variation is a combination of direct and inverse variation: y=kx/z. #### How do you find the constant variation? Since k is constant (the same for every point), we can find k when given any point by dividing the y-coordinate by the x-coordinate. For example, if y varies directly as x, and y = 6 when x = 2, the constant of variation is k = = 3. #### What is the meaning of variation? 1a : the act or process of varying : the state or fact of being varied. b : an instance of varying. c : the extent to which or the range in which a thing varies. #### How do you teach direct variation? Solving a Direct Variation Problem 1. Write the variation equation: y = kx or k = y/x. 2. Substitute in for the given values and find the value of k. 3. Rewrite the variation equation: y = kx with the known value of k. 4. Substitute the remaining values and find the unknown. #### What is the formula for indirect variation? An inverse variation can be represented by the equation xy=k or y=kx . That is, y varies inversely as x if there is some nonzero constant k such that, xy=k or y=kx where x≠0,y≠0 . #### What is the formula for joint variation? Equation for a joint variation is X = KYZ where K is constant. One variable quantity is said to vary jointly as a number of other variable quantities, when it varies directly as their product. If the variable A varies directly as the product of the variables B, C and D, i.e., if. #### What are 3 characteristics of direct variations? The graph is an example of a direct variation. 1) The rate of change is constant (\$\$ k = 1/1 = 1), so the graph is linear. 2) The line passes through the origin (0, 0). 3) The equation of the direct variation is \$\$ y =1 x or simply \$\$ y = x . #### Does direct variation have to go through the origin? The graph of a direct variation always passes through the origin, and always has a slope that is equal to the constant of proportionality, k. #### Which linear function shows a direct variation? The equation for a linear direct variation is y = kx, where k is the slope of the line y = mx + b, and the y-intercept, or b, equals zero. #### What is a direct variation function? 1 : mathematical relationship between two variables that can be expressed by an equation in which one variable is equal to a constant times the other. 2 : an equation or function expressing direct variation — compare inverse variation. #### How do you find the constant variation? Since k is constant (the same for every point), we can find k when given any point by dividing the y-coordinate by the x-coordinate. For example, if y varies directly as x, and y = 6 when x = 2, the constant of variation is k = = 3. #### What are the 4 types of variation? Examples of types of variation include direct, inverse, joint, and combined variation. #### How do you identify direct and indirect variation? Direct & Inverse Variation 1. Direct variation describes a simple relationship between two variables . 2. This means that as x increases, y increases and as x decreases, y decreases—and that the ratio between them always stays the same. 3. Inverse variation describes another kind of relationship. #### Why is direct variation important? Direct variation is a critical topic in Algebra 1. A direct variation represents a specific case of linear function, and it can be used to model a number of real-world situations. #### What is a direct variation table? Tables can be used to write direct variation equations. You divide y by x and get the same value, which will become k. For example, in the direct variation table above, each time you divide y by x, you should get 8. This means that 8 is the constant and that each time x increases by 1, y increases by 8. #### What is the initial step in solving joint and combined variation? Write the general variation formula of the problem. Find the constant of variation k. Rewrite the formula with the value of k. Solve the problem by inputting known information. #### Does direct variation have to be positive? 1 Answer. Depends. When your direct variation is linear (i.e. y=kx ), you have a line with a positive slope. #### What is indirect variation examples? We gave an example of inverse proportion above, namely speed and time for a particular journey. In this case, if you double the speed, you halve the time. So the product, speed x time = constant. In general, if x and y are inversely proportional, then the product xy will be constant. #### What is direct square variation? Direct Square Variation Word Problem: Again, a Direct Square Variation is when y is proportional to the square of x, or y=k{{x}^{2}}. #### What is direct and indirect variation? Direct variation means when one quantity changes, the other quantity also changes in direct proportion. Inverse variation is exactly opposite to this. As the bill at the shopping centre increases, the amount to be paid also increases. #### Which is an example of a joint variation? When a variable is dependent on the product or quotient of two or more variables, this is called joint variation. For example, the cost of busing students for each school trip varies with the number of students attending and the distance from the school. #### What does a joint variation look like? Joint variation is just like direct variation, but involves more than one other variable. All the variables are directly proportional, taken one at a time. Suppose x varies jointly with y and the square root of z. When x=-18 and y=2, then z=9. #### What is direct variation class 8? Therefore, if the ratio between two variables remains constant, it is said to be in direct variation.
Collection of recommendations and tips # What is the multiple of 30? ## What is the multiple of 30? Multiples of 30: 30, 60, 90, 120, 150, 180, 210, 240, 270, 300, 330… Common multiples of 25 and 30 include 150 and 300. The lowest common multiple or least common multiple is the lowest multiple two numbers have in common. What are multiples of 19? The first 10 multiples of 19 are 19, 38, 57, 76, 95, 114, 133, 152, 171, and 190. What are the first 5 multiples of 31? The first 5 multiples of 31 are 31, 62, 93, 124, 155. The sum of the first 5 multiples of 31 is 465 and the average of the first 5 multiples of 31 is 93. Multiples of 31: 31, 62, 93, 124, 155, 186, 217, 248, 279, 310 and so on. ### How do you find the multiples of 30? The successive multiples of 30 are obtained as a product of 30 and the successive natural numbers. 30 × 1 = 30, 30 × 2 = 60, 30 × 3 = 90, 30 × 4 = 120 and so on. This shows that the first multiple of 30 is itself. The second multiple of 30 is 60, and the third multiple of 30 is 90 and so on. Which is the 30th non zero multiple of 30? Answer: 1,2,3,5,6,15. these are the non zero multiples of 30. What are the multiples of 32? First 5 multiples of 32 are 32, 64, 96, 128, and 160. #### What are multiples 23? The first 5 multiples of 23 are 23, 46, 69, 92, and 115. What are multiples numbers? A multiple of a number is the product of the number and a counting number. So a multiple of 3 would be the product of a counting number and 3 . What are the multiples of 28? The first five multiples of 28 are 28, 56, 84, 112, and 140. ## What is a multiple of 39? Multiples of 39: 39, 78, 117, 156, 195, 234, 273, 312, 351, 390 and so on. What is the least common multiple of 29? To find the Least Common Multiple of 5 and 29, you can also make a list of multiples of 5 and a list of multiples of 29. Multiples of 5 would be 5, 10, 15, 20 and so on. Multiples of 29 would be 29, 58, 87, 116 and so on. What is 29 as a percent? To write .29 as a percent have to remember that 1 equal 100% and that what you need to do is just to multiply the number by 100 and add at the end symbol % . .29 * 100 = 29%. And finally we have: .29 as a percent equals 29%. ### What can multiply to 29? There are at least two combinations of two numbers that you can multiply together to get 29. For your convenience, we have made a list of all the combinations of two numbers multiplied by each other that will make 29: 1 x 29 = 29 29 x 1 = 29 What is the meaning of the number 29? In numerology, number 29 is a combination of the vibrational energies of the numbers 2 and 9. Number 2 signifies cooperation, partnerships, diplomacy, and teamwork. It also symbolizes humanitarianism to an extent.
# How do you write 13/4 as a decimal? Since it is −134 , the decimal is −1.75 . The percentage is −175% . Convert fraction (ratio) 13 / 4 1. 13/4 = 2. 13 ÷ 4 = 3. 3.25 = 4. 3.25 × 100/100 = 5. 3.25 × 100% = Also Know, what is 5 4 as a decimal? Fraction to decimal conversion table Fraction Decimal 1/5 0.2 2/5 0.4 3/5 0.6 4/5 0.8 Subsequently, one may also ask, what is 1 and 3/4 as a decimal? Since it is −134 , the decimal is −1.75 . The percentage is −175% . What is 13/4 as a decimal? Fraction Decimal Percentage 15/4 3.75 375% 14/4 3.5 350% 13/4 3.25 325% 12/4 3 300% ### What is in a decimal? What is a Decimal? In algebra, a decimal number can be defined as a number whose whole number part and the fractional part is separated by a decimal point. The dot in a decimal number is called a decimal point. The digits following the decimal point show a value smaller than one. ### What is as a fraction? A fraction simply tells us how many parts of a whole we have. You can recognize a fraction by the slash that is written between the two numbers. We have a top number, the numerator, and a bottom number, the denominator. For example, 1/2 is a fraction. ### How do u turn a number into a fraction? Step 1: Write down the decimal divided by 1, like this: decimal 1. Step 2: Multiply both top and bottom by 10 for every number after the decimal point. (For example, if there are two numbers after the decimal point, then use 100, if there are three then use 1000, etc.) Step 3: Simplify (or reduce) the fraction. ### What is 5/8 as a decimal? Common Fractions with Decimal and Percent Equivalents Fraction Decimal Percent 5/6 0.8333… 83.333…% 1/8 0.125 12.5% 3/8 0.375 37.5% 5/8 0.625 62.5% ### How do you write 4/12 as a decimal? 0.3333 is a decimal and 33.33/100 or 33.33% is the percentage for 4/12. ### What is 2.4 as a fraction? Since there are 1 digits in 4, the very last digit is the “10th” decimal place. So we can just say that . 4 is the same as 4/10. terms by dividing both the numerator and denominator by 2. ### What does a decimal fraction look like? What is a decimal fraction? In algebra, a decimal fraction is a fraction whose denominator is 10 or a multiple of 10 like 100, 1,000, 10,000, etc. ### What is 1 11 as a decimal? 1/11 = 0.0909090909 You don’t need a calculator with this idea. What is 5 tenths as a decimal? ### What is 0.25 as a fraction? The decimal 0.25 represents the fraction 25/100. Decimal fractions always have a denominator based on a power of 10. We know that 5/10 is equivalent to 1/2 since 1/2 times 5/5 is 5/10. Therefore, the decimal 0.5 is equivalent to 1/2 or 2/4, etc. ### What is 0.75 as a fraction? Example Values Percent Decimal Fraction 75% 0.75 3/4 80% 0.8 4/5 90% 0.9 9/10 99% 0.99 99/100 ### What is 1.5 as a fraction? 1.5 in fraction form is 3/2. ### What is 1 over 4 as a decimal? The result of 1 divided by 4 is 0.25.
# What Do The Stars Say About Nick Lachey? (12/04/2019) How will Nick Lachey perform on 12/04/2019 and the days ahead? Let’s use astrology to undertake a simple analysis. Note this is of questionable accuracy – don’t get too worked up about the result. I will first calculate the destiny number for Nick Lachey, and then something similar to the life path number, which we will calculate for today (12/04/2019). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology experts. PATH NUMBER FOR 12/04/2019: We will analyze the month (12), the day (04) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. This is how it’s calculated. First, for the month, we take the current month of 12 and add the digits together: 1 + 2 = 3 (super simple). Then do the day: from 04 we do 0 + 4 = 4. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 3 + 4 + 12 = 19. This still isn’t a single-digit number, so we will add its digits together again: 1 + 9 = 10. This still isn’t a single-digit number, so we will add its digits together again: 1 + 0 = 1. Now we have a single-digit number: 1 is the path number for 12/04/2019. DESTINY NUMBER FOR Nick Lachey: The destiny number will take the sum of all the letters in a name. Each letter is assigned a number per the below chart: So for Nick Lachey we have the letters N (5), i (9), c (3), k (2), L (3), a (1), c (3), h (8), e (5) and y (7). Adding all of that up (yes, this can get tiring) gives 46. This still isn’t a single-digit number, so we will add its digits together again: 4 + 6 = 10. This still isn’t a single-digit number, so we will add its digits together again: 1 + 0 = 1. Now we have a single-digit number: 1 is the destiny number for Nick Lachey. CONCLUSION: The difference between the path number for today (1) and destiny number for Nick Lachey (1) is 0. That is less than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t go jumping for joy yet! As mentioned earlier, this is of questionable accuracy. If you want really means something, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind. ### Abigale Lormen Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. #### Latest posts by Abigale Lormen (see all) Abigale Lormen Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
# How to Do Conic Sections Conic sections are an interesting branch of mathematics involving the cutting of a double-napped cone. By cutting the cone in different ways, you can create a shape as simple as a point or as complex as a hyperbola. ### Part 1 Part 1 of 5:Conic Sections: General Information 1. 1 Understand what is special about a conic section. Unlike regular coordinate equations, conic sections are general equations and don't necessarily have to be functions. For instance, ${\displaystyle x=5}$, while an equation, is not a function. 2. 2 Know the difference between a degenerate case and a conic section. The degenerate cases are those where the cutting plane passes through the intersection, or apex of the double-napped cone. Some examples of degenerates are lines, intersecting lines, and points. The four conic sections are circles, parabolas, ellipses and hyperbolas.[1] 3. 3 Realize the idea that conic sections rely on. A conic section on a coordinate plane is just a collection of points which follow a certain rule which relates them all to the direction and focal points of the conic. ### Part 2 Part 2 of 5:Conic Section 1: Circles 1. 1 Know what part of the cone you are looking at. A circle is defined as "the collection of points equidistant from a fixed point."[2] 2. 2 Find the coordinates of the center of the circle. For formula sake, we will call the center ${\displaystyle (h,k)}$ as is custom when writing the general equation of a conic section. 3. 3 Find the radius of the circle. The circle is defined as a collection of points which are the same distance away from a set center point ${\displaystyle (h,k)}$. That distance is the radius. 4. 4 Plug them into the equation of a circle. The equation of a circle is one of the easiest to remember of all the conic sections. Given a center of ${\displaystyle (h,k)}$ and a radius of length ${\displaystyle r}$, a circle is defined by ${\displaystyle (x-h)^{2}+(y-k)^{2}}$. Be sure to realize that this isn't a function. If you are trying to graph a circle on your graphing calculator, you will have to do some algebra to separate it into two equations which can be graphed using a calculator or use the "draw" feature. 5. 5 Graph the circle, if necessary. If the graph is not given to you, graphing can help give you a better idea of what the circle should look like. Plot the point of the center, extend a line the length of the radius from each side, and draw the circle. ### Part 3 Part 3 of 5:Conic Section 2: Parabolas 1. 1 Understand what a parabola is. By definition, a parabola is "the set of all points equidistant from a line (the directrix) and a fixed point not on the line (the focus)."[3] 2. 2 Find the coordinates of the vertex. The vertex, ${\displaystyle (h,k)}$, is the point where the graph has its axis of symmetry. Drawing this point will help you graph the parabola. 3. 3 Find the focus. The equation for the focus is ${\displaystyle (h,k+p)}$, ${\displaystyle p}$ being the distance between the vertex and the focus. 4. 4 Plug in to find the directrix. The directrix has an equation of ${\displaystyle y=k-p}$. By using the vertex and focus to create a system of two equations, solve for the variables and plug them into the directrix formula. 5. 5 Solve for the axis of symmetry. The parabola's axis of symmetry is defined as ${\displaystyle x=h}$. This line shows how the parabola is symmetrical and should cross through the vertex. 6. 6 Find the equation of the parabola. The formula for the equation of the parabola is ${\displaystyle (y-k)={\frac {1}{4p}}(x-h)^{2}}$. Plug in the variables ${\displaystyle k}$, ${\displaystyle h}$, and ${\displaystyle p}$ to find the equation. 7. 7 Graph the parabola if the graph is not given to you. This will show how the parabola appears. Plot the point of the vertex and focus, and draw the directrix and axis of symmetry. Draw the parabola either upwards or downwards, depending on if ${\displaystyle p}$ is positive or negative, respectively. ### Part 4 Part 4 of 5:Conic Section 3: Ellipses 1. 1 Know what an ellipse is. An ellipse is defined as "the set of points such that the sum of the distances from any point on the ellipse to two other fixed points is constant."[4] 2. 2 Find the center. The center of the ellipse is defined as ${\displaystyle (h,k)}$. 3. 3 Find the major axis. The equation for an ellipse is ${\displaystyle {\frac {(x-h)^{2}}{a^{2}}}+{\frac {(y-k)^{2}}{b^{2}}}=1}$ or ${\displaystyle {\frac {(x-h)^{2}}{b^{2}}}+{\frac {(y-k)^{2}}{a^{2}}}=1}$, where ${\displaystyle a>b}$. Whichever denominator has the larger number, the variable in the numerator's (either ${\displaystyle x}$ or ${\displaystyle y}$) corresponding axis is the major axis. The other is the minor axis. 4. 4 Solve for the vertices. An ellipse has four vertices. To solve for the vertices, let ${\displaystyle x}$ and ${\displaystyle y=0}$ and solve for the two variables. These will give you the points on your graph where the ellipse intersects. 5. 5 Graph the ellipse, if neccesary. Plot the points of the vertices and connect the dots to graph the ellipse. The major axis should appear longer than the minor axis. ### Part 5 Part 5 of 5:Conic Section 4: Hyperbolas 1. 1 Understand what a hyperbola is. By definition, a hyperbola is "the set of all points such that the difference of the distances between any point on the hyperbola and two fixed points is constant."[5] This is similar to the ellipse; however, the hyperbola is the difference of the distances, while the ellipse is the sum. 2. 2 Find the hyperbola's center. The center is defined as ${\displaystyle (h,k)}$ and will be the point between the two curves. 3. 3 Find the transverse axis. The equation of a hyperbola is ${\displaystyle {\frac {(x-h)^{2}}{a^{2}}}-{\frac {(y-k)^{2}}{b^{2}}}=1}$ or ${\displaystyle {\frac {(y-k)^{2}}{a^{2}}}-{\frac {(x-h)^{2}}{b^{2}}}=1}$, where ${\displaystyle a>b}$. Whichever variable is first in the equation and is greater (either ${\displaystyle x}$ or ${\displaystyle y}$) is the transverse axis. 4. 4 Solve for the vertices. Unlike the ellipse, a hyperbola only has two vertices. To solve for them, let ${\displaystyle x}$ and ${\displaystyle y=0}$ and solve for the two variables. The solutions for the variable corresponding with the transverse axis will give you the points on your graph where the hyperbola intersects. • The other two solutions will not be real numbers but eliminating the imaginary component (${\displaystyle i}$) will give you two other coordinates on the real plane. These points, called the covertices, can help you graph the hyperbola. 5. 5 Find the asymptotes. The asymptotes are two lines that the hyperbola will never touch but continuously get closer to. You can simply use the slope formula (${\displaystyle m={\frac {rise}{run}}}$) or solve by factoring to find the asymptotes. 6. 6 Graph the hyperbola if it isn't given to you. Construct a box using the four points (the two vertices and the two other points found) as the vertices of the box. From here, draw the asymptotes coming out of the corners of the box. Then, draw the two curves coming out of the box, touching the two vertices. Erase the box if you desire. ## Community Q&A 200 characters left
The square root of 4096 is expressed as √4096 in the radical kind and together (4096)½ or (4096)0.5 in the exponent form. The square source of 4096 is 64. It is the positive solution the the equation x2 = 4096. The number 4096 is a perfect square. You are watching: Square root of 4,096 Square root of 4096: 64Square source of 4096 in exponential form: (4096)½ or (4096)0.5Square source of 4096 in radical form: √4096 1 What is the Square source of 4096? 2 How to uncover the Square root of 4096? 3 Is the Square root of 4096 Rational? 4 FAQs ## What is the Square source of 4096? The square source of 4096, (or root 4096), is the number which as soon as multiplied through itself provides the product together 4096. Therefore, the square source of 4096 = √4096 = 64. ☛ Check: Square source Calculator ## How to find Square root of 4096? ### Value of √4096 through Long division Method Explanation: Forming pairs: 40 and also 96Find a number Y (6) such that whose square is bring down the next pair 96, to the appropriate of the remainder 4. The new dividend is now 496.Add the critical digit that the quotient (6) to the divisor (6) i.e. 6 + 6 = 12. Come the right of 12, uncover a number Z (which is 4) such that 12Z × Z divide 496 by 124 through the quotient as 4, offering the remainder = 496 - 124 × 4 = 496 - 496 = 0.We protect against the procedure since the remainder is currently 0 and there are no much more digits that can be brought down. Therefore, the square root of 4096 by long division method is 64. ## Is Square source of 4096 Rational? The worth of √4096 is 64. Hence, the square root of 4096 is a rational number. ☛ also Check: ## Square root of 4096 fixed Examples Example 1: deal with the equation x2 − 4096 = 0 Solution: x2 - 4096 = 0 i.e. X2 = 4096x = ±√4096Since the value of the square source of 4096 is 64,⇒ x = +√4096 or -√4096 = 64 or -64. Show equipment > go come slidego come slidego to slide Ready to see the world through math’s eyes? Math is in ~ the core of whatever we do. Gain solving real-world math problems in live classes and become an experienced at everything. Book a totally free Trial Class ## FAQs ~ above the Square root of 4096 ### What is the value of the Square root of 4096? The square source of 4096 is 64. ### Why is the Square root of 4096 a reasonable Number? Upon prime factorizing 4096 i.e. 212, we find that every the prime determinants are in even power. This implies that the square root of 4096 is a hopeful integer. Therefore, the square source of 4096 is rational. ### Evaluate 2 to add 9 square root 4096 The given expression is 2 + 9 √4096. We recognize that the square source of 4096 is 64. Therefore, 2 + 9 √4096 = 2 + 9 × 64 = 2 + 576 = 578 ### What is the Square of the Square root of 4096? The square that the square root of 4096 is the number 4096 chin i.e. (√4096)2 = (4096)2/2 = 4096. ### What is the value of 2 square source 4096? The square source of 4096 is 64. Therefore, 2 √4096 = 2 × 64 = 128. See more: What Body Covering Do Amphibians Have ? Types Of Amphibians ### If the Square source of 4096 is 64. Find the value of the Square source of 40.96. Let us represent √40.96 in p/q type i.e. √(4096/100) = 40.96/10 = 6. Hence, the worth of √40.96 = 6
THE BINOMIAL EXPANSION - Number Explorations - Numbers: Their Tales, Types, and Treasures ## Chapter 6: Number Explorations ### 6.5.THE BINOMIAL EXPANSION The Pascal triangle contains many unexpected number relationships beyond the characteristics that Pascal intended in his original use of the triangle. Pascal's original use was to exhibit the coefficients of successive terms of a binomial expansion. We shall now consider the binomial expansion—that is, taking a binomial such as (a + b) to successively higher powers. By now, you ought to be able to recognize the pattern formed by the coefficients of the terms shown in figure 6.9. The coefficients of each of the binomial-expansion lines is also represented as a row of the Pascal triangle. Figure 6.9: The binomial expansion. This allows us to expand a binomial without actually multiplying it by itself many times to get the end result. There is a pattern also among the variables’ exponents: one descends while the other ascends in value—each time keeping the sum of the exponents constant—that is, the sum is equal to the exponent of the power to which the original binomial was taken. The formulas in figure 6.9 can be written in compact form in a single line, which we provide here because of its beauty. This formula is called the binomial theorem. The symbol is pronounced “n over k.” This is the modern notation for the numbers in the Pascal triangle, the so-called binomial coefficients, which we denoted by B(n,k) in chapter 5. Actually, there is a formula that allows us to compute any binomial coefficient given its position n,k in the Pascal triangle, without having to evaluate the binomial coefficients above. This formula uses the notation n! = 1 × 2 × 3 ×…× (n – 1) × n for the product of all natural numbers from 1 to n. The expression n! is pronounced “n-factorial.” With this abbreviation, the binomial coefficient is given by For example, gives the binomial coefficient B(7,3) = 35. According to chapter 5, it describes in how many ways we can choose three elements out of a set of seven elements. It also tells us how many ways seven coin tosses come up with three heads. For a = 1 and b = 1, the expressions in figure 6.9 can be simplified, as all products of powers of a and b can be replaced by 1. We obtain, for example, (1 + 1)6 = 26 = 1 + 6 + 15 + 20 + 15 + 6 + 1, which is just the sum of the entries in the corresponding row of the Pascal triangle. In that way, we can reproduce the result of figure 5.12: The sums across the lines of the Pascal triangle are the powers of 2. For a = 10 and b = 1, all factors an bk in figure 6.9 simply become 10n. This leads to an interesting observation. For example, (10 + 1)3 = 113 = 1 × 103 + 3 × 102 + 3 × 101 + 1 = 1,331. The binomial formula, in this case, provides the representation of a number in our decimal numeral system. And the digits of this number are the entries of the Pascal triangle in the corresponding row. This shows us that if one reads a row of the Pascal triangle as a single number whose digits are the elements of that row, we get a power of 11. Indeed, 121 = 112, and the numbers 1, 4, 6, 4, 1 in the fourth row lead to 14,641 = 114. However, beginning with the fifth row, we would have to regroup the digits: 115 = 1 × 105 + 5 × 104 + 10 × 103 + 10 × 102 + 5 × 101 + 1 = 161,051. In this case, some of the elements in the Pascal triangle have more than one digit, and one has to carry the leading digits over to the next order. So, for example, the sixth power of 11 has to be computed as follows: Figure 6.10: How to determine a power of 11 from the Pascal triangle. The occurrence of Fibonacci numbers, powers of 2, and powers of 11 in the Pascal triangle are once again illustrated in figure 6.11. Figure 6.11: Amazing number relationships in the Pascal triangle There are many number relationships present in the Pascal triangle. The turf is fertile. The opportunity to find more gems in this triangular arrangement of numbers is practically boundless! We encourage the motivated reader to search for more hidden treasures embedded in this number arrangement.
Note 1 ##### Take Note: Take a note while surfing. ##### Note With Ink Give your Note a Colorful Tag. ##### Easy to Access Stay on same information and in Sync wherever you are. Note 2 ##### Take Note: Organize your information,It may take Shape. ##### Easy to Access Easy to pull up your content from anywhere anytime. Note 3 ##### Take Note: Don't Let information to miss,Because it take shape ##### Note With Ink Simple an Easy Way to take a note. ##### Easy to Access Get the same in next visit. # Arithmetic Aptitude :: Time and Work Home > Arithmetic Aptitude > Time and Work > General Questions Overview #### Co. Cloud 11. A and B together can do a piece of work in 12 days, which B and C together can do in 16 days. After A has been working at it for 5 days and B for 7 days, C finishes it in 13 days. In how many days C alone will do the work ? \| \| \| \| Explanation: (A and B) 1 day work = 1/12 -----(1) (B and C) 1 day work = 1/16 -----(2) Given A's 5 days' work + B's 7 days' work + C's 13 days' work = 1 simplify the above .. => (A + B)'s 5 days' work + (B + C)'s 2 days' work + C's 11 days' work = 1 Put the values from equation (1) & (2) => (51/12)+(21/16) + C's 11 days' work = 1 => C's 11 days' work = (1-5/12+2/16)=11/24 => C's 1 day's work = (11/24?1/11)=1/24 So C alone can finish the work = 24 days. Workspace Tags: No Tags on this question yet! 12. A and B can do a piece of work in 45 days and 40 days respectively. They began to do the work together but A leaves after some days and then B completed the remaining work in 23 days. The number of days after which A left the work was : \| \| \| \| Explanation: A's 1 day work = 1/45 B's 1 day work = 1/40 so (A + B)'s 1 day's work = (1/45+1/40)=17/360 Work done by B in 23 days = (140?23)=2340 Remaining work =(1-23/40)=17/40 Now, 17/360 work was done by (A + B) = 1 day. 17/40 part of work was done by (A + B) = (1?360/17?17/40) = 9 days. A left after 9 days. Workspace Tags: No Tags on this question yet! 13. A, B and C together earn Rs. 300 per day, while A and C together earn Rs. 188 and B and C together earn Rs. 152. The daily earning of C is : \| \| \| \| Explanation: B's daily earning = Rs. (300 - 188) = Rs. 112. A's daily earning = Rs. (300 - 152) = Rs. 148. C's daily earning = Rs. [300 - (112 + 148)] = Rs. 40. Workspace Tags: No Tags on this question yet! 14. Kim can do a work in 3 days while David can do the same work in 2 days. Both of them finish the work together and get Rs. 150. What is the share of Kim ? \| \| \| \| Explanation: Kim 1 day work = 1/3 David 1 day work = 1/2 Kim's 1 day's work : David's 1 day's work=(1/3)/(1/2) = 2 : 3. Kim's share = Rs. (2/5?150)=Rs. 60 Workspace Tags: No Tags on this question yet! 15. A can do a piece of work in 14 days which B can do in 21 days. They begin together but 3 days before the completion of the work, A leaves off. The total number of days to complete the work is: \| \| \| \| Explanation: A's 1 day work = 1/14 B' 1 day work = 1/21 A left 3 day before the completion of work. B's 3 days' work = (1/21?3)=1/7 Remaining work = (1?1/7)=6/7 (A + B)'s 1 day's work = (1/14+1/21)=5/42 Now, 5/42 work is done by A and B = 1 day. or whole work done by A&B=42/5 6/7 part of work done by A and B = 42/5?6/7=36/5 days. Hence, total time taken = (3+36/5)days = 10 1/5 days. Workspace Tags: No Tags on this question yet! 16. A, B and C can complete a work separately in 24, 36 and 48 days respectively. They started together but C left after 4 days of start and A left 3 days before the completion of the work. In how many days will the work be completed ? \| \| \| \| Explanation: (A + B + C)'s 1 day's work = (1/24+1/36+1/48)=13/144 Work done by (A + B + C) in 4 days = 4*13/144=13/36 Work done by B in 3 days = 1/36?3=1/12 Left work = [1?(13/36+1/12)]=5/9 (A + B)'s 1 day's work = (1/24+1/36)=5/72 5/72 part of work is done by A and B = 1 day 5/9 work done by A & B = (72/5?5/9) = 8 days. So, total time taken = (4 + 3 + 8) days = 15 days. Workspace Tags: No Tags on this question yet! 17. A, B and C are employed to do a piece of work for Rs. 529. A and B together are supposed to do 19/23 of the work and B and C together 8/23 of the work. What amount should A be paid ? \| \| \| \| Explanation: B and C together done = 8/23 part So Work done by A = (1?8/23)=15/23 and work done by C = (1-19/23) = 4/23, so work done by B = 4/23 A : B : C = 15/23:4/23:4/23 = 15:4:4 So, A's share = Rs. (15/23?529) = Rs. 345. Workspace Tags: No Tags on this question yet! 18. A tyre has two punctures. The first puncture alone would have made the tyre flat in 9 minutes and the second alone would have done it in 6 minutes. If air leaks out at a constant rate, how long does it take both the punctures together to make it flat ? \| \| \| \| Explanation: 1 minute's Work by both punctures = (1/9+1/6) = (2/18+3/18) = 5/18 So both punctures can make tyre flat = 18/5 = 3 3/5 Workspace Tags: No Tags on this question yet! 19. A man can do a job in 15 days. His father takes 20 days and his son finishes it in 25 days. How long will they take to complete the job if they all work together? \| \| \| \| Explanation: Find the 1 day work for all three 1 day's work for all three = (1/15 + 1/20 + 1/25) = (20/300 + 15/300 + 12/300) = 47/300 So all together can do the work in 300/47 days. => 6.4 days. Workspace Tags: No Tags on this question yet! 20. A, B and C can complete a piece of work in 24, 6 and 12 days respectively. Working together, they will complete the same work in \| \| \| \|
# How do you differentiate (cos x) / (1-sinx)? Jun 13, 2016 Quotient Rule:- If $u$ and $v$ are two differentiable functions at $x$ with $v \ne 0$, then $y = \frac{u}{v}$ is differentiable at $x$ and $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v \cdot \mathrm{du} - u \cdot \mathrm{dv}}{v} ^ 2$ Let $y = \frac{\cos x}{1 - \sin x}$ Differentiate w.r.t. 'x' using quotient rule $\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 - \sin x\right) \frac{d}{\mathrm{dx}} \left(\cos x\right) - \cos x \frac{d}{\mathrm{dx}} \left(1 - \sin x\right)}{1 - \sin x} ^ 2$ Since $\frac{d}{\mathrm{dx}} \left(\cos x\right) = - \sin x$ and $\frac{d}{\mathrm{dx}} \left(1 - \sin x\right) = - \cos x$ Therefore $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 - \sin x\right) \left(- \sin x\right) - \cos x \left(- \cos x\right)}{1 - \sin x} ^ 2$ $\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \sin x + {\sin}^{2} x + {\cos}^{2} x}{1 - \sin x} ^ 2$ Since $S {\in}^{2} x + C o {s}^{2} x = 1$ Therefore $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - \sin x}{1 - \sin x} ^ 2 = \frac{1}{1 - S \in x}$ Hence, derivative of the given expression is $\frac{1}{1 - \sin x} .$
# Eureka Math Precalculus Module 2 Lesson 14 Answer Key ## Engage NY Eureka Math Precalculus Module 2 Lesson 14 Answer Key ### Eureka Math Precalculus Module 2 Lesson 14 Opening Exercise Answer Key Opening Exercise: Ahmad says the matrix $$\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right]$$ applied to the point $$\left[\begin{array}{l} 4 \\ 1 \end{array}\right]$$ will reflect the point to $$\left[\begin{array}{l} 1 \\ 4 \end{array}\right]$$ Randelle says that applying the matrix to the given point will produce a rotation of 1800 about the origin. Who is correct? Explain your answer and verify the result. Randelle is correct. Applying the matrix $$\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right]$$ to the given point produces a rotation of 180Â° about the origin of the $$\left[\begin{array}{l} 4 \\ 1 \end{array}\right]$$ point to the image point $$\left[\begin{array}{l} -4 \\ -1 \end{array}\right]$$. Applying the matrix $$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$$ to the given point would produce a reflection to the image point $$\left[\begin{array}{l} 1 \\ 4 \end{array}\right]$$. ### Eureka Math Precalculus Module 2 Lesson 14 Example Answer Key Example: a. Describe a transformation not already discussed that results in an image point of $$\left[\begin{array}{l} 4 \\ 1 \end{array}\right]$$ and represent the transformation using a 2 Ã— 2. Answers will :ary. An example of an appropriate response is as follows: A rotation of the point $$\left[\begin{array}{c} -1 \\ 4 \end{array}\right]$$ 90Â° to the point $$\left[\begin{array}{l} 4 \\ 1 \end{array}\right]$$ can be represented with the matrix $$\left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right]$$. b. Determine whether any of the matrices listed represent linear transformations that can produce the image point $$\left[\begin{array}{l} 4 \\ 1 \end{array}\right]$$. Justify your answers by describing the transformations represented by the matrices. i. $$\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$$ This matrix represents a collapse to the origin, so it cannot produce the ima ge point $$\left[\begin{array}{l} 4 \\ 1 \end{array}\right]$$. ii. $$\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]$$ This matrix represents a transformation to the diagonal defined by y = x, so it cannot produce the image point $$\left[\begin{array}{l} 4 \\ 1 \end{array}\right]$$. iii. $$\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]$$ This matrix represents a transformation to the y-axis, so it cannot produce the image point $$\left[\begin{array}{l} 4 \\ 1 \end{array}\right]$$. c. Suppose a linear transformation L is represented by the matrix $$\left[\begin{array}{cc} 2 & -1 \\ 3 & 1 \end{array}\right]$$. Find a point L$$\left[\begin{array}{l} x \\ y \end{array}\right]$$so that L$$\left[\begin{array}{l} x \\ y \end{array}\right]$$ = $$\left[\begin{array}{l} 4 \\ 1 \end{array}\right]$$. $$\left[\begin{array}{l} x \\ y \end{array}\right]$$ = $$\left[\begin{array}{c} 1 \\ -2 \end{array}\right]$$ ### Eureka Math Precalculus Module 2 Lesson 14 Exercise Answer Key Exercises: Exercise 1. Given the system of equations 2x + 5y = 4 3x – 8y = -25 a. Show how this system can be written as a statement about a linear transformation of the form Lx = b, with x = $$\left[\begin{array}{l} x \\ y \end{array}\right]$$ and b = $$\left[\begin{array}{c} 4 \\ -25 \end{array}\right]$$ $$\left[\begin{array}{cc} 2 & 5 \\ 3 & -8 \end{array}\right]$$ $$\left[\begin{array}{l} x \\ y \end{array}\right]$$ = $$\left[\begin{array}{c} 4 \\ -25 \end{array}\right]$$ b. Determine whether L has an inverse. If it does, compute L-1b, and verify that the coordinates represent the solution to the system of equations. L-1b = $$\frac{1}{2(-8)-(5)(3)}$$$$\left[\begin{array}{cc} -8 & -5 \\ -3 & 2 \end{array}\right]\left[\begin{array}{c} 4 \\ -25 \end{array}\right]$$ L-1b = $$\frac{1}{-31}$$$$\left[\begin{array}{cc} -8 & -5 \\ -3 & 2 \end{array}\right]\left[\begin{array}{c} 4 \\ -25 \end{array}\right]$$ = $$\frac{1}{-31}$$$$\left[\begin{array}{c} 93 \\ -62 \end{array}\right]$$ = $$\left[\begin{array}{c} -3 \\ 2 \end{array}\right]$$ Verification using back substitution: 2(-3) + 5(2) = 4 3(-3) – 8(2) = -25 Exercise 2. The path of a piece of paper carried by the wind into a tree can be modeled with a linear transformation, where L = $$\left[\begin{array}{cc} 3 & -4 \\ 5 & 3 \end{array}\right]$$ and b = $$\left[\begin{array}{c} 6 \\ 10 \end{array}\right]$$. a. Write an equation that represents the linear transformation of the piece of paper. $$\left[\begin{array}{cc} 3 & -4 \\ 5 & 3 \end{array}\right]$$ $$\left[\begin{array}{l} x \\ y \end{array}\right]$$ = $$\left[\begin{array}{c} 6 \\ 10 \end{array}\right]$$. b. Solve the equation from part (a). $$\left[\begin{array}{l} x \\ y \end{array}\right]$$ = $$\frac{1}{29}$$$$\left[\begin{array}{cc} 3 & 4 \\ -5 & 3 \end{array}\right]\left[\begin{array}{c} 6 \\ 10 \end{array}\right]$$ = $$\frac{1}{29}$$$$\left[\begin{array}{c} 58 \\ 0 \end{array}\right]$$ = $$\left[\begin{array}{l} 2 \\ 0 \end{array}\right]$$ c. Use your solution to provide a reasonable interpretation of the path of the piece of paper under the transformation by the wind. Answers will vary. An example of an appropriate response would be that the piece of paper started on the ground 2 feet to the right of the location defined as the origin, and it was moved by the wind to a spot 6 feet to the right of the origin and 10 feet above the ground (in the tree). Exercise 3. For each system of equations, write the system as a linear transformation represented by a matrix and apply inverse matrix operations to find the solution, or explain why this procedure cannot be performed. a. 6x + 2y = 1 y = 3x + 1 b. 4x – 6y = 10 2x – 3y = 1 This system cannot be represented as a linear transformation because the transformation matrix L has a determinant of 0. The system represents parallel lines, so there is no solution. Exercise 4. In a two-dimensional plane, A represents a rotation of 30Â° counterclockwise about the origin, B represents a reflection over the line y = x, and C represents a rotation of 60Â° counterclockwise about the origin. a. Write matrices A, B, and C. A = $$\left[\begin{array}{ll} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{array}\right]$$ B = $$\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$$ C = $$\left[\begin{array}{cc} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{array}\right]$$ b. Transformations A, B, and C are applied to Point $$\left[\begin{array}{l} x \\ y \end{array}\right]$$ successively and produce the image point $$\left[\begin{array}{c} 1+2 \sqrt{3} \\ 2-\sqrt{3} \end{array}\right]$$. Use inverse matrix operations to find $$\left[\begin{array}{l} x \\ y \end{array}\right]$$. We must apply the inverse transformations in the reverse order. The inverse of matrix C C-1 = 1$$\left[\begin{array}{cc} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{array}\right]$$ Applied to the image ### Eureka Math Precalculus Module 2 Lesson 14 Problem Set Answer Key Question 1. In a two-dimensional plane, a transformation represented by L = $$\left[\begin{array}{cc} 1 & 5 \\ 2 & -4 \end{array}\right]$$ is applied to point x, resulting in an image point $$\left[\begin{array}{l} 0 \\ 5 \end{array}\right]$$. Find the location of the point before it was transformed. a. Write an equation to represent the linear transformation of point x. $$\left[\begin{array}{cc} 1 & 5 \\ 2 & -4 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 0 \\ 5 \end{array}\right]$$ b. Solve the equation to find the coordinates of the pre-image point. Question 2. Find the location of the point $$\left[\begin{array}{l} x \\ y \end{array}\right]$$ before it was transformed when given: a. The transformation L = $$\left[\begin{array}{ll} 3 & 5 \\ 1 & 2 \end{array}\right]$$, and the resultant is $$\left[\begin{array}{l} 1 \\ 2 \end{array}\right]$$. Verify your answer. b. The transformation L = $$\left[\begin{array}{cc} 4 & 7 \\ -1 & -2 \end{array}\right]$$ and the resultant is $$\left[\begin{array}{c} 2 \\ -1 \end{array}\right]$$. Verify your answer. c. The transformation L = $$\left[\begin{array}{cc} 0 & -1 \\ 2 & 1 \end{array}\right]$$, and the reasultant is $$\left[\begin{array}{l} 1 \\ 3 \end{array}\right]$$. Verify your answer. d. The transformation L = $$\left[\begin{array}{cc} 2 & 3 \\ 0 & -1 \end{array}\right]$$, and the resultant is $$\left[\begin{array}{l} 3 \\ 0 \end{array}\right]$$. Verify your answer. e. The transformation L = $$\left[\begin{array}{cc} 2 & -1 \\ 1 & 2 \end{array}\right]$$, and the reasultant is $$\left[\begin{array}{l} 3 \\ 2 \end{array}\right]$$. Verify your answer. Question 3. On a computer assembly line, a robot is placing a CPU onto a motherboard. The robotâ€™s arm is carried out by the transformation L = $$\left[\begin{array}{cc} 2 & 3 \\ 1 & 2 \end{array}\right]$$. a. If the cu is attached to the motherboard at point $$\left[\begin{array}{l} -2 \\ 3 \end{array}\right]$$, at what location does the robot pick up the CPU? $$\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} -2 \\ 3 \end{array}\right]$$ $$\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{1}\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right]\left[\begin{array}{c} -2 \\ 3 \end{array}\right]=\left[\begin{array}{c} -13 \\ 8 \end{array}\right]$$ b. If the CPU is attached to the motherboard at point $$\left[\begin{array}{l} 3 \\ 2 \end{array}\right]$$, at what location does the robot pick up the CPU? $$\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \end{array}\right]$$ $$\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{1}\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \end{array}\right]=\left[\begin{array}{l} 0 \\ 1 \end{array}\right]$$ c. Find the transformation L = $$\left[\begin{array}{cc} -1 & c \\ b & 3 \end{array}\right]$$ that places the CPU starting at $$\left[\begin{array}{c} 2 \\ -3 \end{array}\right]$$ onto the motherboard at the location $$\left[\begin{array}{c} -8 \\ 3 \end{array}\right]$$ $$\left[\begin{array}{cc} -1 & c \\ b & 3 \end{array}\right]$$ $$\left[\begin{array}{c} 2 \\ -3 \end{array}\right]$$ = $$\left[\begin{array}{c} -8 \\ 3 \end{array}\right]$$ -2 – 3c = -8, c = 2 2b – 9 = 3, b = 6 $$\left[\begin{array}{cc} -1 & 2 \\ 6 & 3 \end{array}\right]$$ Question 4. On a construction site, a crane is moving steel beams from a truck bed to workers. The crane is programmed to perform the transformation L = $$\left[\begin{array}{ll} 1 & 1 \\ 2 & 3 \end{array}\right]$$. a. If the workers are at location $$\left[\begin{array}{l} 2 \\ 5 \end{array}\right]$$, where does the truck driver need to unload the steel beams so that the crane can pick them up and bring them to the workers? $$\left[\begin{array}{ll} 1 & 1 \\ 2 & 3 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 2 \\ 5 \end{array}\right]$$ $$\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{1}\left[\begin{array}{cc} 3 & -1 \\ -2 & 1 \end{array}\right]\left[\begin{array}{l} 2 \\ 5 \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \end{array}\right]$$ b. If the workers move to another location $$\left[\begin{array}{c} -3 \\ 1 \end{array}\right]$$, where does the truck driver need to unload the steel beams so that the crane can pick them up and bring them to the workers? $$\left[\begin{array}{ll} 1 & 1 \\ 2 & 3 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} -3 \\ 1 \end{array}\right]$$ $$\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{1}\left[\begin{array}{cc} 3 & -1 \\ -2 & 1 \end{array}\right]\left[\begin{array}{c} -3 \\ 1 \end{array}\right]=\left[\begin{array}{c} -10 \\ 7 \end{array}\right]$$ Question 5. A video game soccer player is positioned at $$\left[\begin{array}{l} \mathbf{0} \\ 2 \end{array}\right]$$, where he kicks the ball. The ball goes into the goal, which is at point $$\left[\begin{array}{c} 10 \\ 0 \end{array}\right]$$. When the player moves to point $$\left[\begin{array}{l} 1 \\ 1 \end{array}\right]$$ and kicks the ball, he misses the goal. The ball lands at point $$\left[\begin{array}{l} 10 \\ -1 \end{array}\right]$$.What is the program/transformation L = $$\left[\begin{array}{ll} a & c \\ b & d \end{array}\right]$$ that this video soccer player uses? $$\left[\begin{array}{ll} a & c \\ b & d \end{array}\right]\left[\begin{array}{l} 0 \\ 2 \end{array}\right]=\left[\begin{array}{c} 10 \\ 0 \end{array}\right]$$, 2c = 10, c = 5. 2d = 0, d = 0 $$\left[\begin{array}{ll} a & 5 \\ b & 0 \end{array}\right]\left[\begin{array}{l} 1 \\ 1 \end{array}\right]=\left[\begin{array}{l} 10 \\ -1 \end{array}\right]$$, a + 5 = 10, a = 5, b = -1 $$\left[\begin{array}{cc} 5 & 5 \\ -1 & 0 \end{array}\right]$$ Question 6. Tim bought 5 shirts and 3 pairs of pants, and it cost him $250. Scott bought 3 shirts and 2 pairs of pants, and It cost him$160. All the shirts have the same cost, and all the pants have the same cost. a. Write a system of linear equations to find the cost of the shirts and pants. 5S + 3P = 250 3S + 2P = 160 b. Show how this system can be written as a statement about a linear transformation of the form Lx = b with x = $$\left[\begin{array}{l} S \\ P \end{array}\right]$$ and b = $$\left[\begin{array}{l} 250 \\ 160 \end{array}\right]$$ $$\left[\begin{array}{ll} 5 & 3 \\ 3 & 2 \end{array}\right]\left[\begin{array}{l} S \\ P \end{array}\right]=\left[\begin{array}{l} 250 \\ 160 \end{array}\right]$$ c. Determine whether L has an inverse. If it does, compute L-1b, and verify your answer to the system of equations. The determinant of $$\left[\begin{array}{ll} 5 & 3 \\ 3 & 2 \end{array}\right]$$ is 1. L-1 = $$\frac{1}{1}\left[\begin{array}{cc} 2 & -3 \\ -3 & 5 \end{array}\right]$$ = $$\left[\begin{array}{cc} 2 & -3 \\ -3 & 5 \end{array}\right]$$ $$\left[\begin{array}{l} S \\ P \end{array}\right]=\left[\begin{array}{cc} 2 & -3 \\ -3 & 5 \end{array}\right]\left[\begin{array}{l} 250 \\ 160 \end{array}\right]=\left[\begin{array}{c} 20 \\ 50 \end{array}\right]$$ Verification using back substitution: 5(20) + 3(50) = 250, 3(20) + 2(50) = 160. Question 7. In a two-dimensional plane, A represents a reflection over the x-axis, B represents a reflection over the y-axis, and C represents a reflection over the line y = x. a. Write matrices A, B, and C. A = $$\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right]$$ B = $$\left[\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right]$$ C = $$\left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right]$$ b. Write an equation for each linear transformation, assuming that each one produces an image point of $$\left[\begin{array}{l} -2 \\ -3 \end{array}\right]$$. For transformation A, $$\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} -2 \\ -3 \end{array}\right]$$ For transformation B, $$\left[\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} -2 \\ -3 \end{array}\right]$$ For transformation C, $$\left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} -2 \\ -3 \end{array}\right]$$ c. Use inverse matrix operations to find the pre-image point for each equation. Explain how your solutions make sense based on your understanding of the effect of each geometric transformation on the coordinates of the pre-image points. For transformation A, $$\left[\begin{array}{l} x \\ y \end{array}\right]$$ = -1 $$\left[\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{l} -2 \\ -3 \end{array}\right]=\left[\begin{array}{c} -2 \\ 3 \end{array}\right]$$ When a point is reflected over the x-axis, the x-coordinate remains unchanged, and the y-coordinare changes signs. For transformation B, $$\left[\begin{array}{l} x \\ y \end{array}\right]$$ = -1 $$\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right]\left[\begin{array}{l} -2 \\ -3 \end{array}\right]=\left[\begin{array}{c} 2 \\ -3 \end{array}\right]$$. When a point is reflected over the y-axis, the y-coordinate remains unchanged, and the x-coordinate changes signs. For transformation C, $$\left[\begin{array}{l} x \\ y \end{array}\right]$$ = $$\left[\begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array}\right]\left[\begin{array}{l} -2 \\ -3 \end{array}\right]=\left[\begin{array}{l} -3 \\ -2 \end{array}\right]$$ When a point is reflected over the line y = x, the coordinates of the pre-image point are interchanged (x and y are switched). Question 8. A system of equations is shown: 2x + 5y + z = 3 4x + y – z = 5 3x + 2y + 4z = 1 a. Represent this system as a linear transformation in three-dimensional space represented by a matrix equation in the form of Lx = b. $$\left[\begin{array}{ccc} 2 & 5 & 1 \\ 4 & 1 & -1 \\ 3 & 2 & 4 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3 \\ 5 \\ 1 \end{array}\right]$$ b. What assumption needs to be made to solve the equation in part (a) for x. To use inverse operations, we need to assume that L has an inverse. c. Use algebraic methods to solve the system. Adding equations 1 and 2 gives 6x + 6y = 8. Adding 4 times equation 2 and equation 3 gives 19x + 6y = 21. Subtracting (6x + 6y = 8) from (19x + 6y = 21) gives 13x = 13, so x = 1. Back substituting into 6x + 6y = 8 gives 6y = 2, or y = $$\frac{1}{3}$$ Back substituting for y and x into the first equation gives 2(1) + 5($$\frac{1}{3}$$) + z = 3, so z = –$$\frac{2}{3}$$ x = $$\left[\begin{array}{c} 1 \\ \frac{1}{3} \\ \frac{-2}{3} \end{array}\right]$$ Question 9. Assume L-1 = $$\frac{1}{78}\left[\begin{array}{ccc} -6 & 18 & 6 \\ 19 & -5 & -6 \\ -5 & -11 & 18 \end{array}\right]$$ Use inverse matrix operations to solve the equation from Problem 8, part (a) for x. Verify that your solution is the same as the one you found in Problem 8, part (c). $$\left.\begin{array}{l} x \\ y \\ z \end{array}\right]$$ = $$\frac{1}{78}\left[\begin{array}{ccc} -6 & 18 & 6 \\ 19 & -5 & -6 \\ -5 & -11 & 18 \end{array}\right]$$$$\left[\begin{array}{l} 3 \\ 5 \\ 1 \end{array}\right]$$ = $$\left[\begin{array}{c} 1 \\ \frac{1}{3} \\ \frac{-2}{3} \end{array}\right]$$, which is the same solution found in part (c). ### Eureka Math Precalculus Module 2 Lesson 14 Exit Ticket Answer Key Question 1. In two-dimensional space, point x is rotated 180Â° to the point $$\left[\begin{array}{l} 3 \\ 4 \end{array}\right]$$. a. Represent the transformation of point x using an equation in the format Lx = b. $$\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 3 \\ 4 \end{array}\right]$$ b. Use inverse matrix operations to find the coordinates of x. $$\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{1}\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right]\left[\begin{array}{l} 3 \\ 4 \end{array}\right]=\left[\begin{array}{l} -3 \\ -4 \end{array}\right]$$ c. Verify that this solution makes sense geometrically. When a point is rotated 180Â° about the origin in the coordinate plane, the x- and y-coordinates of the image point are the opposite of those of the pre-image point. Scroll to Top Scroll to Top
# What is the equation of the normal line of f(x)=ln2x-x at x=4 ? Dec 6, 2016 $y = \frac{4}{3} x + 3 \ln 2 - \frac{28}{3}$ #### Explanation: The equation of the line normal to the curve $y = f \left(x\right)$ in the point $x = \overline{x}$ is given by: $y = f \left(\overline{x}\right) - \frac{1}{f ' \left(\overline{x}\right)} \left(x - \overline{x}\right)$ Given $\overline{x} = 4$: $f \left(x\right) = \ln 2 x - x$ $f \left(4\right) = \ln 8 - 4 = 3 \ln 2 - 4$ $f ' \left(x\right) = \frac{1}{x} - 1$ $f ' \left(4\right) = - \frac{3}{4}$ The equation of the normal line is: $y = \frac{4}{3} \left(x - 4\right) + 3 \ln 2 - 4 = \frac{4}{3} x + 3 \ln 2 - \frac{28}{3}$
# What is meant by injective function? In mathematics, an injective function (also known as injection, or one-to-one function) is a function f that maps distinct elements to distinct elements; that is, f(x1) = f(x2) implies x1 = x2. (Equivalently, x1 ≠ x2 implies f(x1) ≠ f(x2) in the equivalent contrapositive statement.) ## What is injective and Surjective function? If the codomain of a function is also its range, then the function is onto or surjective. If a function does not map two different elements in the domain to the same element in the range, it is one-to-one or injective. ## What is meant by Surjective function? In mathematics, a surjective function (also known as surjection, or onto function) is a function f that maps an element x to every element y; that is, for every y, there is an x such that f(x) = y. In other words, every element of the function's codomain is the image of at least one element of its domain. ## What is an injective function Class 12? The injective function is defined as a function in which for every element in the codomain there is an image of exactly one in the domain. ## What is an injective functions and give three 3 examples? Examples of Injective Function The identity function X → X is always injective. If function f: R→ R, then f(x) = 2x is injective. If function f: R→ R, then f(x) = 2x+1 is injective. If function f: R→ R, then f(x) = x2 is not an injective function, because here if x = -1, then f(-1) = 1 = f(1). ## How do you know if a function is Injective? To show that a function is injective, we assume that there are elements a1 and a2 of A with f(a1) = f(a2) and then show that a1 = a2. Graphically speaking, if a horizontal line cuts the curve representing the function at most once then the function is injective. ## What is meant by Bijective function? A function is said to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. ## Why is x3 injective? As we all know, this cannot be a surjective function, since the range consists of all real values, but f(x) can only produce cubic values. Also from observing a graph, this function produces unique values; hence it is injective. ## What is Bijective function with example? A function f: X→Y is said to be bijective if f is both one-one and onto. Example: For A = {1,−1,2,3} and B = {1,4,9}, f: A→B defined as f(x) = x2 is surjective. Example: Example: For A = {−1,2,3} and B = {1,4,9}, f: A→B defined as f(x) = x2 is bijective. ## Is square root function injective? If you intend the domain and codomain as "the non-negative real numbers" then, yes, the square root function is bijective. To show that you show it is "injective" ("one to one"): if then x= y. ## Is Sinx injective? The wiki page tells you that in the case of R→[−1,1] that sin(x) is a surjection but is not an injection. It is a surjection because every possible output has a preimage. ## How do you prove two sets are injective? So how do we prove whether or not a function is injective? To prove a function is injective we must either: Assume f(x) = f(y) and then show that x = y. Assume x doesn't equal y and show that f(x) doesn't equal f(x). ## What is the difference between bijective and injective functions? The function is bijective (one-to-one and onto, one-to-one correspondence, or invertible) if each element of the codomain is mapped to by exactly one element of the domain. That is, the function is both injective and surjective. A bijective function is also called a bijection. ## What is subjective and injective? Injective is also called "One-to-One" Surjective means that every "B" has at least one matching "A" (maybe more than one). There won't be a "B" left out. Bijective means both Injective and Surjective together. Think of it as a "perfect pairing" between the sets: every one has a partner and no one is left out. ## How many injective functions are there? The composition of two injective functions is injective. ## How do you find the number of injective functions? Number of Injective Functions (One to One) If set A has n elements and set B has m elements, m≥n, then the number of injective functions or one to one function is given by m!/(m-n)!. ## How do you prove surjective and injective? To prove a function, f : A → B is surjective, or onto, we must show f(A) = B. In other words, we must show the two sets, f(A) and B, are equal. ## Why is E X not surjective? Why is it not surjective? The solution says: not surjective, because the Value 0 ∈ R≥0 has no Urbild (inverse image / preimage?). But e^0 = 1 which is in ∈ R≥0. ## Are all inverse function bijective? Then, ∀ y∈Y,f(x)=11y=y. So f is surjective. Show activity on this post. The claim that every function with an inverse is bijective is false. ## Is TANX injective? The function is injective because it is a monotonically increasing function. This means that it is impossible for two different (real) values to have the same arctangent, and this is the definition of injective (given that the domain is the real numbers). ## What is empty relation? An empty relation (or void relation) is one in which there is no relation between any elements of a set. For example, if set A = {1, 2, 3} then, one of the void relations can be R = {x, y} where, \|x – y\| = 8. ## Can a function be injective but not surjective? An example of an injective function R→R that is not surjective is h(x)=ex. This "hits" all of the positive reals, but misses zero and all of the negative reals. Previous question Who is the poorest superhero?
7638 Riddle #1 from Fall 2010 Newsletter What number comes next in the sequence: 61, 691, 163, 487, 4201, ? Solution: 9631. The sequence consists of the prime numbers which, when their digits are reversed, are perfect squares. Riddle #1 from Spring 2010 Newsletter Bob and John form a team together. Bob is as old as John will be when Bob is twice as old as John was when Bob was half as old as the sum of their current ages. John is as old as Bob was when John was half as old as he will become over ten years. How old are Bob and John? Solution: Let us call the current age of Bob B and of John J. The first fact is that Bob is as old as John will be when Bob is twice as old as John was when Bob was half as old as the sum of their current ages. When Bob was half as old as the sum of their current ages, he had reached the age of (B + J) / 2 years. This is now B – (B + J) / 2 = B/2 – J/2 years ago. At that moment, John was J – (B/2 – J/2) = 3/2 J – B/2 years old. If Bob is twice as old as John at that moment, then Bob is (3/2 J – B/2) * 2 = 3 J – B years old. That moment is 3 J – 2 B years from now. Then John will be J + 3 J – 2 B = 4 J – 2 B years old. And that is exactly the age of Bob, so B = 4 J – 2 B which gives that 3 B = 4 J. The second fact is that John is as old as Bob was when John was half as old as he will become over ten years. When John had half the age as he will have over ten years, he was (J+10)/2 years old. This is now J – (J + 10) / 2 = J/2 – 5 years ago. At that moment Bob was B – (J/2 – 5) years old. According to the second fact, John is now as old as Bob was at that moment, so J = B – (J/2 – 5). It now follows that 3 J / 2 = B + 5. Summarizing, we end up with two equations. 3 B = 4 J B + 5 = 3 J / 2 Multiplying the bottom equation with -3 gives after adding the top equation -15 = (4 – 9/2) J Solving this equation gives J = 30. And now it easily follows that B = 40. So John is 30 years old and Bob 40. Riddle #1 from Fall 2009 Newsletter What is the integral of “one over cabin” with respect to “cabin”? http://www.onlinemathlearning.com/calculus-riddles.html Natural log cabin + c = houseboat. Sudoku Solution (pdf) Riddle #1 from Spring 2009 Newsletter A man was found murdered on Sunday morning. His wife immediately called the police. The police questioned the wife and staff and got these alibis: The Wife said she was sleeping. The Cook was cooking breakfast. The Gardener was picking vegetables. The Maid was getting the mail. The Butler was cleaning the closet. The police instantly arrested the murdered. Who did it and how did they know? It was the Maid. She said she was getting the mail. There is no mail on Sunday! (e-mail does not count) Riddle #1 from Spring 2008 Newsletter Driving along, Terry notices that the last four digits on the odometer are palindromic. A mile later, the last five digits are palindromic. Two miles later, all six are palindromic. What was the odometer reading when Terry first looked at it? 1-9-8, 8-8-8. That’s what he first saw when he looked. One mile later it was 1-9-8, 8-8-9, the last five are palindromic. One mile later, 1-9-8, 8-9-0, the middle four are palindromic. You drive one more mile and it’s 1-9-8-, 8-9-1, all six are palindromic. Riddle #1 from Fall 2007 Newsletter A boat has a ladder that has six rungs, each rung is one foot apart. The bottom rung is one foot from the water. The tide rises at 12 inches every 15 minutes. High tide peaks in one hour. When the tide is at it’s highest, how many rungs are under water? Answer: No rungs will be under water; the boat floats. Riddle # 2 from Fall 2007 Newsletter There are 4 men who want to cross a bridge. They all begin on the same side. You have 17 minutes to get all of them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either 1 or 2 people, must have the flashlight with them. The flashlight must be walked back and forth, it cannot be thrown, etc. Each man walks at a different speed. A pair must walk together at the rate of the slower man. Man 1: 1 minute to cross Man 2: 2 minutes to cross Man 3: 5 minutes to cross Man 4: 10 minutes to cross Man 1 and Man 2 cross 2 min Man 2 comes back 2min Man 3 and Man 4 cross 10 min Man 1 comes back 1 min Man 1 and Man 2 cross 2 min Total 17 min
# 2014 AMC 10A Problems/Problem 3 ## Problem Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for $\textdollar 2.50$ each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf costs $\textdollar 0.75$ for her to make. In dollars, what is her profit for the day? $\textbf{(A)}\ 24\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}}\ 48\qquad\textbf{(E)}\ 52$ (Error compiling LaTeX. ) ## Solution $\dfrac{1}{2}$ of the bread is $24$ loaves. $24\times\textdollar 2.50=\textdollar60$. This leaves $24$ loaves left. $\dfrac{2}{3}\times 24=16$ The new price will be $\dfrac{1}{2}\textdollar2.50=\textdollar1.25$. So $\textdollar1.25\times16=\textdollar20.00$. This leaves 8 loaves remaining. We have $\textdollar 1\times 8=\textdollar 8.00$. The total amount of money she made for the day is the sum of these amounts, which is $60+20+8=\textdollar 88$. The total amount it cost her to make all of the loaves is $\textdollar 0.75*48=\textdollar 36$. Therefore, her total profit is the amount of money she spent subtracted from the amount of money she made. $88-36=52\implies\boxed{\textbf{(E)}52}}$ (Error compiling LaTeX. ).
What are the applications of differential equations Differential equations and their applications Chapter 1. First order differential equations This book deals with differential equations and their applications. A differential equation establishes a relationship between a function and its derivatives. The equations \$\$ \ frac {dy} {dt} = 3 {{y} ^ {2}} \ sin \ left (t + y \ right) \$\$ and \$\$ \ frac {{{d} ^ {3}} y} {d {{t} ^ {3}}} = {{e} ^ {- y}} + t + \ frac {{{d} ^ { 2}} y} {d {{t} ^ {2}}} \$\$ are examples of differential equations. The order of a differential equation is the order of the highest derivative of the function y that occurs in the equation. So (i) is a first-order differential equation, (ii) a third-order differential equation. A solution of a differential equation is a continuous function y (t) which, together with its derivatives, satisfies the given relationship. For example, the function solves \$\$ y \ left (t \ right) = 2 \ sin t - \ frac {1} {3} \ cos 2t \$\$ the differential equation of the 2nd order \$\$ \ frac {{{d} ^ {2}} y} {d {{t} ^ {2}}} + y = \ cos2t \$\$ , because simple recalculation results \$\$ \ frac {{{d ^ 2}}} {{d {t ^ 2}}} \ left ({2 \, \ sin \; t - \ frac {1} {3} \ cos 2t} \ right ) + \ left ({2 \, \ sin \; t - \ frac {1} {3} \ cos 2t} \ right) = \ left ({- 2 \, \ sin \; t + \ frac {4} {3} \ cos 2t} \ right) + 2 \, \ sin \; t - \ frac {1} {3} \ cos 2t = \ cos 2t \$\$ . Chapter 2. Linear differential equations of the second order A relationship of form \$\$ \ frac {{{d} ^ {2}} y} {d {{t} ^ {2}}} = f \ left (t, y, \ frac {dy} {dt} \ right) \$\$ is called the second order differential equation. Chapter 3. Systems of differential equations In this chapter we want to investigate systems of first-order differential equations. Such a system consists of simultaneous first-order differential equations in several variables and has the form \$\$ \ begin {gathered} \ frac {{{\ text {d}} {{\ text {x}} _ {\ text {1}}}}}} {{{\ text {dt}}}} {\ text {=}} {{\ text {f}} _ {\ text {1}}} \ left ({{\ text {t,}} {{\ text {x}} _ {\ text {1}} } {\ text {, \ ldots,}} {{\ text {x}} _ {\ text {n}}}} \ right) {\ text {,}} \ hfill \ \ frac {{{\ text {d}} {{\ text {x}} _ {\ text {2}}}}} {{{\ text {dt}}}} {\ text {=}} {{\ text {f}} _ {\ text {2}}} \ left ({{\ text {t,}} {{\ text {x}} _ {\ text {1}}} {\ text {\ ldots,}} {{\ text {x}} _ {\ text {n}}}} \ right) {\ text {, \ ldots,}} \ frac {{{\ text {d}} {{\ text {x}} _ {\ text {n}}}}} {{{\ text {dt}}}} {\ text {=}} {{\ text {f}} _ {\ text {n}}} \ left ({{\ text { t,}} {{\ text {x}} _ {\ text {1}}} {\ text {, \ ldots,}} {{\ text {x}} _ {\ text {n}}}} \ right) \ hfill \ \ end {gathered} \$\$ Chapter 4. Qualitative theory of differential equations In this chapter we consider the differential equation \$\$ \ overset {\ centerdot} {\ mathop {x}} \, = f (t, x), \$\$ in the \ (x = \ left (\ begin {matrix} {{x} _ {1}} (t) \ \ vdots \ {{x} _ {n}} (t) \ end {matrix } \ right) \) and \ (f (t, x) = \ left (\ begin {matrix} {{f} _ {1}} (t, {{x} _ {1}}, \ cdots, {{x} _ {n}}) \ \ vdots \ {{f} _ {n}} (t, {{x} _ {1}}, \ cdots, {{x} _ {n}} ) \ \ end {matrix} \ right) \) a nonlinear function of x1, ..., xn designated. Unfortunately, no methods of solving equation (1) are known. This is of course very regrettable, but in most applications it is not necessary to find the solutions of (1) explicitly. For example, let x1(t) and x2(t) the populations of two species at time t, which are fighting against each other for the food and habitat that is only available to a limited extent in their microcosm. Then the growth rates of x1(t) and x2(t) is described by the differential equation (1), in this case we are less attached to the values of x1(t) and x2(t) at any point in time t, rather than the qualitative properties of x1(t) and x2(t) interested. Chapter 5. Separation of Variables and Fourier Series In the applications that we want to examine more closely in this chapter, we are faced with the following problem: Problem: For which values of λ can we find nontrivial functions y (x) that \$\$ \ frac {{{d ^ 2} y}} {{d {x ^ 2}}} + \ lambda y = o; ay \ left (o \ right) + by '(o) = o, cy \ left (\ ell \ right) + dy '\ left (\ ell \ right) = o \$\$ fulfill? Equation (1) is called the boundary value problem because, in contrast to the initial value problem, in which we have the value of y and y 'at a point x = xO specify the values of the solution y (x) and its derivation y '(x) in two different points x = o and x = ℓ.
Upcoming SlideShare × # Equivalence, simplifying and ordering keynote 556 Published on Published in: Technology, Self Improvement 0 Likes Statistics Notes • Full Name Comment goes here. Are you sure you want to Yes No Your message goes here • Be the first to comment • Be the first to like this Views Total Views 556 On Slideshare 0 From Embeds 0 Number of Embeds 1 Actions Shares 0 2 0 Likes 0 Embeds 0 No embeds No notes for slide • \n • \n • \n • \n • \n • \n • \n • \n • \n • \n • \n • \n • \n • \n • \n • \n • ### Equivalence, simplifying and ordering keynote 1. 1. Equivalence, Simplifying and A presentation to help you learnBy St a rg ir l 5A 2. 2. IntroductionIn this presentation, you will learn:What are equivalent fractions?How to get equivalent fractionsHow to simplify and order fractionsEnjoy! 3. 3. What are equivalent fractions?Equivalent fractions are fractions that all havethe same value, for example:2/3=4/6=8/12Equivalence means equal, so all the equivalentfractions have to be equal.Let me explain it a bit more...... 4. 4. What are Equivalent Fractions? 2 5. 5. What are Equivalent Fractions? 2 6. 6. What are Equivalent Fractions? 2 7. 7. What are Equivalent Fractions? 2 8. 8. What are Equivalent Fractions? 2 9. 9. What are Equivalent Fractions? 2 10. 10. What are Equivalent Fractions? 2If I have a pie cut in to half, then theoutlined section is 1/2. 11. 11. What are Equivalent Fractions? 2If I have a pie cut in to half, then theoutlined section is 1/2.If I cut it into 4, the outlined sectionwould be 2/4, which is still half if you simplifyit. 12. 12. What are Equivalent Fractions? 2If I have a pie cut in to half, then theoutlined section is 1/2.If I cut it into 4, the outlined sectionwould be 2/4, which is still half if you simplifyit.2/4 is the equivalent fraction of 1/2! 13. 13. How to Get Equivalent FractionsIf 3/4 is our main fraction we are going to use,then just multiply the numerator and thedenominator by the same number. Then you willget the equivalent fraction! Lets try:3x2/4x2 so an equivalent fraction of 3/4 is6/8! You can keep on doing that to keep ongetting equivalent fractions. 14. 14. How to simplify fractionsHere’s how to simplify fractions:If the fraction you want to simplify is 5/10, thenyou have to ﬁnd the common factor of 5 and10.The common factor is 5, so divide each one by5You’re simpliﬁed fraction is 1/2! 15. 15. How to Order FractionsYour fractions are 3/7, 1/2, 4/5You need to put them in order from lowest tohighest. How do you do that?First, change all the denominators to the LCM of 7,2 and 5So your denominator is 70!Now, the fractions are 30/70, 35/70 and 56/70We’ll carry on on the next page...... 16. 16. How to Order Fractions 2So, just put them in order: 30/70, 35/70 and56/70Simplify them.You have just learnt how to order fractions! 17. 17. We are Done!Thanks for watching my presentation!Hope you enjoyed it!Bye! 18. 18. We are Done!Thanks for watching my presentation!Hope you enjoyed it!Bye! 1. #### A particular slide catching your eye? Clipping is a handy way to collect important slides you want to go back to later.
# Can variance and mean be zero? ## Can variance and mean be zero? A large variance indicates that numbers in the set are far from the mean and far from each other. A variance value of zero, though, indicates that all values within a set of numbers are identical. Every variance that isn’t zero is a positive number. A variance cannot be negative. ### What happens to the variance of the sampling distribution of the sample means when the sample size increases? As sample sizes increase, the sampling distributions approach a normal distribution. As the sample sizes increase, the variability of each sampling distribution decreases so that they become increasingly more leptokurtic. How does the variance of the sample mean and the variance of the population differ? Summary: Population variance refers to the value of variance that is calculated from population data, and sample variance is the variance calculated from sample data. Due to this value of denominator in the formula for variance in case of sample data is ‘n-1’, and it is ‘n’ for population data. READ: Why did we move from C to C++? What can you say about the variance of the sample means and the variance of the population? The mean of the sample means is the same as the population mean, but the variance of the sample means is not the same as the population variance. ## Can a random variable have 0 variance? Zero variance means all observations are equal. For example, the variance of the observations say, 5, 5, 5, 5 is zero. If the variance of a random variable is zero, then that random variable must be a constant. ### Can variance of a random variable be 0? By definition, the variance of X is the average value of (X−μX)2. Since (X−μX)2≥0, the variance is always larger than or equal to zero. How does mean affect variance? As the draws spread out from the mean (both above and below), the variance increases. Since some observations are above the mean and others below, we square the difference between a single observation (k i) and the mean (μ) when calculating the variance. What is the variance of the sample mean? The variance of the sampling distribution of the mean is computed as follows: That is, the variance of the sampling distribution of the mean is the population variance divided by N, the sample size (the number of scores used to compute a mean). The variance of the sum would be σ2 + σ2 + σ2. READ: Is it wrong to date a siblings ex? ## Is sample variance always smaller than population variance? Given a sample from a normal (or asymptotic normal) distribution, the sample variance is more often less than the population variance due to the skewed nature of the distribution of the unbiased sample estimate. ### What does sample variance mean in statistics? Sample variance (s2) is a measure of the degree to which the numbers in a list are spread out. If the numbers in a list are all close to the expected values, the variance will be small. If they are far away, the variance will be large. Why is sample variance important? When you collect data from a sample from a population, the sample variance is used to make estimates about the population variance. So, uneven variances between samples result in biased and skewed test results. That’s why we need homogeneity or similar variances when comparing samples. Why is the variance of a constant 0? The variance of a constant is zero. Adding a constant value, c, to a random variable does not change the variance, because the expectation (mean) increases by the same amount. Rule 3. Multiplying a random variable by a constant increases the variance by the square of the constant. ## What is simple random sampling in statistics? 1 Simple Random Sampling. The goal is to estimate the mean and the variance of a variable of interest in a nite population by collecting a random sample from it. Suppose there are N members of the population, numbered 1 through N and let the values assumed by the variable of interest be x. ### What is the difference between sample mean and sample variance? The sample mean \\ (m\\) is simply the expected value of the empirical distribution. Similarly, if we were to divide by \\ (n\\) rather than \\ (n – 1\\), the sample variance would be the variance of the empirical distribution. What is the square root of the sample variance? In any event, the square root \\ (s\\) of the sample variance \\ (s^2\\) is the sample standard deviation. It is the root mean square deviation and is also a measure of the spread of the data with respect to the mean. What is the relationship between sample size and sample mean? When drawing a single random sample, the larger the sample is the closer the sample mean will be to the population mean (in the above quote, think of “number of trials” as “sample size”, so each “trial” is an observation).
# Precalculus : Find the Distance Between Two Parallel Lines ## Example Questions ### Example Question #1 : Find The Distance Between Two Parallel Lines Find the distance between the two lines. Explanation: Since the slope of the two lines are equivalent, we know that the lines are parallel. Therefore, they are separated by a constant distance. We can then find the distance between the two lines by using the formula for the distance from a point to a nonvertical line: First, we need to take one of the line and convert it to standard form. where Now we can substitute A, B, and C into our distance equation along with a point, , from the other line. We can pick any point we want, as long as it is on line . Just plug in a number for x, and solve for y. I will use the y-intercept, where x = 0, because it is easy to calculate: Now we have a point, , that is on the line . So let's plug our values for : ### Example Question #2 : Find The Distance Between Two Parallel Lines Find the distance between and Explanation: To find the distance, choose any point on one of the lines. Plugging in 2 into the first equation can generate our first point: this gives us the point We can find the distance between this point and the other line by putting the second line into the form : subtract the whole right side from both sides now we see that We can plug the coefficients and the point into the formula where represents the point. ### Example Question #3 : Find The Distance Between Two Parallel Lines Find the distance between and Explanation: To find the distance, choose any point on one of the lines. Plugging in into the second equation can generate our first point: this gives us the point We can find the distance between this point and the other line by putting the second line into the form : subtract the whole right side from both sides multiply both sides by now we see that We can plug the coefficients and the point into the formula where represents the point. ### Example Question #4 : Find The Distance Between Two Parallel Lines How far apart are the lines and ? Explanation: To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point: this gives us the point We can find the distance between this point and the other line by putting the second line into the form : subtract the whole right side from both sides multiply both sides by now we see that We can plug the coefficients and the point into the formula where represents the point. ### Example Question #5 : Find The Distance Between Two Parallel Lines Find the distance between and Explanation: To find the distance, choose any point on one of the lines. Plugging in into the second equation can generate our first point: this gives us the point We can find the distance between this point and the other line by putting the second line into the form : subtract the whole right side from both sides multiply both sides by now we see that We can plug the coefficients and the point into the formula where represents the point. ### Example Question #6 : Find The Distance Between Two Parallel Lines Find the distance between and Explanation: To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point: this gives us the point We can find the distance between this point and the other line by putting the second line into the form : subtract the whole right side from both sides multiply both sides by now we see that We can plug the coefficients and the point into the formula where represents the point. ### Example Question #7 : Find The Distance Between Two Parallel Lines Find the distance between the lines and Explanation: To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point: this gives us the point We can find the distance between this point and the other line by putting the second line into the form : subtract the whole right side from both sides multiply both sides by now we see that We can plug the coefficients and the point into the formula where represents the point.
# NCERT Exemplar Solutions for Class 11 Maths Chapter 2 Relations and Functions NCERT Exemplar Solutions for Class 11 Maths Chapter 2 Relations and Functions are created by the subject experts at BYJUâ€™S. These Solutions of NCERT Exemplar Maths help students in solving problems efficiently. They also focus on cracking the solutions of Maths and making it easy for students. NCERT Exemplar Solutions for Class 11 aim at equipping them with detailed, step-wise explanations for all answers to questions given in the exercises of this chapter. Chapter 2 Relations and Functions of NCERT Exemplar Solutions for Class 11 Maths explains the concepts related to relations and functions. A relation can be defined as a set of outputs as well as inputs. These are also written as ordered pairs by representing a relation of a graph or by a mapping diagram. The graph of the relationship is one of the best visual indicators between the inputs and the outputs. Functions, on the other hand, are the relation with a single input and output. These Exemplar Solutions can also be used as a reference tool while preparing for the final exam. Click here to get exemplar solutions for all chapters. In Chapter 2, students will learn and solve exemplar problems based on topics, like • Cartesian product of sets • Relations between sets • Functions on sets • Identity function • Constant function • Polynomial function • Rational function • The Modulus function • Signum function • Greatest integer function • Algebra of functions • Addition of two real functions • Subtraction of a real function from another • Multiplication by a Scalar • Multiplication of two real functions • Quotient of two real function • Graphical representation of functions • Algebraic representation of real functions ## Download the PDF of the NCERT Exemplar Class 11 Maths Solutions Chapter 2 Relations and Functions ### Access Answers to the NCERT Exemplar Class 11 Maths Solutions Chapter 2 Relations and Functions Exercise Page No: 27 Short Answer Type 1. Let A = {â€“1, 2, 3} and B = {1, 3}. Determine (i) A Ã— B (ii) B Ã— A (iii) B Ã— B (iv) A Ã— A Solution: According to the question, A = {â€“1, 2, 3} and B = {1, 3} (i) A Ã— B {â€“1, 2, 3} Ã— {1, 3} So, A Ã— B = {(â€“1, 1), (â€“1, 3), (2, 1), (2, 3), (3, 1), (3, 3)} Hence, the Cartesian product = {(â€“1, 1), (â€“1, 3), (2, 1), (2, 3), (3, 1), (3, 3)} (ii) B Ã— A. {1, 3} Ã— {â€“1, 2, 3} So, B Ã— A = {(1, â€“1), (1, 2), (1, 3), (3, â€“1), (3, 2), (3, 3)} Hence, the Cartesian product = {(1, â€“1), (1, 2), (1, 3), (3, â€“1), (3, 2), (3, 3)} (iii) B Ã— B {1, 3} Ã—{1, 3} So, B Ã— B = {(1, 1), (1, 3), (3, 1), (3, 3)} Hence, the Cartesian product = {(1, 1), (1, 3), (3, 1), (3, 3)} (iv) A Ã— A {â€“1, 2, 3} Ã— {â€“1, 2, 3} So, A Ã— A = {(â€“1, â€“1), (â€“1, 2), (â€“1, 3), (2, â€“1), (2, 2), (2, 3), (3, â€“1), (3, 2), (3, 3)} Hence, the Cartesian product ={(â€“1, â€“1), (â€“1, 2), (â€“1, 3), (2, â€“1), (2, 2), (2, 3), (3, â€“1), (3, 2), (3, 3)} 2. If P = {x : x < 3, xÂ âˆˆÂ N}, Q = {x : x â‰¤ 2, xÂ âˆˆÂ W}. Find (PÂ âˆªÂ Q) Ã— (PÂ âˆ©Â Q), whereÂ WÂ is the set of whole numbers. Solution: According to the question, P = {x: x < 3, xÂ âˆˆN}, Q = {x : x â‰¤ 2, xÂ âˆˆW} whereÂ WÂ is the set of whole numbers P = {1, 2} Q = {0, 1, 2} Now (PâˆªQ) = {1, 2}âˆª{0, 1, 2} = {0, 1, 2} And, (Pâˆ©Q) = {1, 2}âˆ©{0, 1, 2} = {1, 2} We need to find the Cartesian product of (PâˆªQ) = {0, 1, 2} and (Pâˆ©Q) = {1, 2} So, (PâˆªQ) Ã— (Pâˆ©Q) = {0, 1, 2} Ã— {1, 2} = {(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)} Hence, the Cartesian product = {(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)} 3. If A = {x : xÂ âˆˆÂ W, x < 2}, B = {x : xÂ âˆˆÂ N, 1 < x < 5}, C = {3, 5} find (i) A Ã— (BÂ âˆ©Â C) (ii) A Ã— (BÂ âˆªÂ C) Solution: According to the question, A = {x: xÂ âˆˆ W, x < 2}, B = {x : xÂ âˆˆN, 1 < x < 5} C = {3, 5};Â WÂ is the set of whole numbers A = {x: xÂ âˆˆ W, x < 2} = {0, 1} B = {x : xÂ âˆˆN, 1 < x < 5} = {2, 3, 4} (i) (Bâˆ©C) = {2, 3, 4} âˆ© {3, 5} (Bâˆ©C) = {3} A Ã— (Bâˆ©C) = {0, 1} Ã— {3} = {(0, 3), (1, 3)} Hence, the Cartesian product = {(0, 3), (1, 3)} (ii) (BâˆªC) = {2, 3, 4} âˆª {3, 5} (BâˆªC) = {2, 3, 4, 5} A Ã— (BâˆªC) = {0, 1} Ã— {2, 3, 4, 5} = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)} Hence, the Cartesian product = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)} 4. In each of the following cases, find a and b. (i) (2a + b, a â€“ b) = (8, 3) (ii) (a/4 , a â€“ 2b) = (0, 6 + b) Solution: (i) According to the question, (2a + b, a â€“ b) = (8, 3) Given the ordered pairs are equal, so corresponding elements will be equal. Hence, 2a + b = 8 and aâ€“b = 3 Now aâ€“b = 3 â‡’a = 3 + b Substituting the value of a in the equation 2a + b = 8, We get, 2(3 + b) + b = 8 â‡’Â 6 + 2b + b = 8 â‡’Â 3b = 8â€“6 = 2 â‡’ b = 2/3 Substituting the value of b in equation (aâ€“b = 3), We get, â‡’ a â€“ 2/3 = 3 â‡’ a = 3 + 2/3 â‡’ a = (9 + 2)/3 â‡’ a = 11/3 Hence the value of a = 11/3 and b =Â 2/3Â respectively. (ii) According to the question, Given the ordered pairs are equal, so corresponding elements will be equal. a/4 = 0Â and a â€“ 2b = 6 + b NowÂ a/4 = 0 â‡’a = 0 Substituting the value of a in the equation (aâ€“2b = 6 + b), We get, 0 â€“ 2b = 6 + b â‡’Â â€“ 2b â€“ b = 6 â‡’Â â€“ 3b = 6 â‡’ b = â€“ 6/3 â‡’ b = â€“ 2 Hence, the value of a = 0 and b = â€“ 2 respectively 5. Given A = {1, 2, 3, 4, 5}, S = {(x, y) : xÂ âˆˆÂ A, yÂ âˆˆÂ A}. Find the ordered pairs which satisfy the conditions given below: (i) x + y = 5 (ii) x + y < 5 (iii) x + y > 8 Solution: According to the question, A = {1, 2, 3, 4, 5}, S = {(x, y) : xÂ âˆˆA, yÂ âˆˆA} (i) x + y = 5 So, we find the ordered pair such that x + y = 5, where x and y belongs to set A = {1, 2, 3, 4, 5}, 1 + 1 = 2â‰ 5 1 + 2 = 3â‰ 5 1 + 3 = 4â‰ 5 1 + 4 = 5â‡’Â the ordered pair is (1, 4) 1 + 5 = 6â‰ 5 2 + 1 = 3â‰ 5 2 + 2 = 4â‰ 5 2 + 3 = 5â‡’Â the ordered pair is (2, 3) 2 + 4 = 6â‰ 5 2 + 5 = 7â‰ 5 3 + 1 = 4â‰ 5 3 + 2 = 5â‡’Â the ordered pair is (3, 2) 3 + 3 = 6â‰ 5 3 + 4 = 7â‰ 5 3 + 5 = 8â‰ 5 4 + 1 = 5â‡’Â the ordered pair is (4, 1) 4 + 2 = 6â‰ 5 4 + 3 = 7â‰ 5 4 + 4 = 8â‰ 5 4 + 5 = 9â‰ 5 5 + 1 = 6â‰ 5 5 + 2 = 7â‰ 5 5 + 3 = 8â‰ 5 5 + 4 = 9â‰ 5 5 + 5 = 10â‰ 5 Therefore, the set of ordered pairs satisfying x + y = 5 = {(1,4), (2,3), (3,2), (4,1)}. (ii) x + y < 5 So, we find the ordered pair such that x + y<5, where x and y belongs to set A = {1, 2, 3, 4, 5} 1 + 1 = 2<5 â‡’ the ordered pairs is (1, 1) 1 + 2 = 3<5 â‡’ the ordered pairs is (1, 2) 1 + 3 = 4<5 â‡’ the ordered pairs is (1, 3) 1 + 4 = 5 1 + 5 = 6>5 2 + 1 = 3<5 â‡’ the ordered pairs is (2, 1) 2 + 2 = 4<5 â‡’ the ordered pairs is (2, 2) 2 + 3 = 5 2 + 4 = 6>5 2 + 5 = 7>5 3 + 1 = 4<5 â‡’ the ordered pairs is (3, 1) 3 + 2 = 5 3 + 3 = 6>5 3 + 4 = 7>5 3 + 5 = 8>5 4 + 1 = 5 4 + 2 = 6>5 4 + 3 = 7>5 4 + 4 = 8>5 4 + 5 = 9>5 5 + 1 = 6>5 5 + 2 = 7>5 5 + 3 = 8>5 5 + 4 = 9>5 5 + 5 = 10>5 Therefore, the set of ordered pairs satisfying x + y< 5 = {(1,1), (1,2), (1,3), (2, 1), (2,2), (3,1)}. (iii) x + y > 8 So, we find the ordered pair such that x + y>8, where x and y belongs to set A = {1, 2, 3, 4, 5} 1 + 1 = 2<8 1 + 2 = 3<8 1 + 3 = 4<8 1 + 4 = 5<8 1 + 5 = 6<8 2 + 1 = 3<8 2 + 2 = 4<8 2 + 3 = 5<8 2 + 4 = 6<8 2 + 5 = 7<8 3 + 1 = 4<8 3 + 2 = 5<8 3 + 3 = 6<8 3 + 4 = 7<8 3 + 5 = 8 4 + 1 = <8 4 + 2 = 6<8 4 + 3 = 7<8 4 + 4 = 8 4 + 5 = 9>8, so one of the ordered pairs is (4, 5) 5 + 1 = 6<8 5 + 2 = 7<8 5 + 3 = 8 5 + 4 = 9>8, so one of the ordered pairs is (5, 4) 5 + 5 = 10>8, so one of the ordered pairs is (5, 5) Therefore, the set of ordered pairs satisfying x + y > 8 = {(4, 5), (5, 4), (5,5)}. 6. Given R = {(x, y) : x, yÂ âˆˆÂ W, x2 + y2 = 25}. Find the domain and Range of R. Solution: According to the question, R = {(x, y) : x, yÂ âˆˆW, x2Â + y2Â = 25} R = {(0,5), (3,4), (4, 3), (5,0)} The domain of R consists of all the first elements of all the ordered pairs of R. Domain of R = {0, 3, 4, 5} The range of R contains all the second elements of all the ordered pairs of R. Range of R = {5, 4, 3, 0} 7. If R1Â = {(x, y) \| y = 2x + 7, where xÂ âˆˆÂ RÂ and â€“ 5 â‰¤ x â‰¤ 5} is a relation. Then find the domain and Range of R1. Solution: According to the question, R1Â = {(x, y) \| y = 2x + 7, where xÂ âˆˆRÂ and â€“ 5 â‰¤ x â‰¤ 5} is a relation The domain of R1Â consists of all the first elements of all the ordered pairs of R1, i.e., x, It is also given â€“ 5 â‰¤ x â‰¤ 5. Therefore, Domain of R1Â = [â€“5, 5] The range of R contains all the second elements of all the ordered pairs of R1, i.e., y It is also given y = 2x + 7 Now xÂ âˆˆÂ [â€“5,5] Multiply LHS and RHS by 2, We get, 2x âˆˆ [â€“10, 10] Adding LHS and RHS with 7, We get, 2x + 7 âˆˆ [â€“3, 17] Or, y âˆˆ [â€“3, 17] So, Range of R1Â = [â€“3, 17] 8. If R2Â = {(x, y) \| x and y are integers and x2Â + y2Â = 64} is a relation. Then find R2. Solution: We have, R2Â = {(x, y) \|Â x and y are integers and x2Â + y2Â â€“ 64} So, we get, x2Â = 0 and y2Â = 64 or x2Â = 64 and y2Â = 0 x = 0 and y = Â±8 or x = Â±8 and y = 0 Therefore, R2Â = {(0, 8), (0, â€“8), (8,0), (â€“8,0)} 9. If R3Â = {(x, \|x\| ) \|x is a real number} is a relation. Then find domain and range of R3. Solution: According to the question, R3Â = {(x, \|x\|) \|x is a real number} is a relation Domain of R3Â consists of all the first elements of all the ordered pairs of R3, i.e., x, It is also given that x is a real number, So, Domain of R3Â = R Range of R contains all the second elements of all the ordered pairs of R3, i.e., \|x\| It is also given that x is a real number, So, \|x\| = \|R\| â‡’Â \|x\|â‰¥0, i.e., \|x\| has all positive real numbers including 0 Hence, Range of R3Â = [0, âˆž) 10. Is the given relation a function? Give reasons for your answer. (i) h = {(4, 6), (3, 9), (â€“ 11, 6), (3, 11)} (ii) f = {(x, x) \| x is a real number} (iii) g = n, (1/n) \|n is a positive integer (iv) s = {(n, n2) \| n is a positive integer} (v) t = {(x, 3) \| x is a real number. Solution: (i) According to the question, h = {(4, 6), (3, 9), (â€“ 11, 6), (3, 11)} Therefore, element 3 has two images, namely, 9 and 11. A relation is said to be function if every element of one set has one and only one image in other set. Hence, h is not a function. (ii) According to the question, f = {(x, x) \| x is a real number} This means the relation f has elements which are real number. Therefore, for every xÂ âˆˆÂ R there will be unique image. A relation is said to be function if every element of one set has one and only one image in other set. Hence, f is a function. (iii) According to the question, g = n, (1/n) \|nÂ is a positive integer Therefore, the element n is a positive integer and the corresponding 1/nÂ will be a unique and distinct number. Therefore, every element in the domain has unique image. A relation is said to be function if every element of one set has one and only one image in other set. Hence, g is a function. (iv) According to the question, s = {(n, n2) \| n is a positive integer} Therefore, element n is a positive integer and the corresponding n2Â will be a unique and distinct number, as square of any positive integer is unique. Therefore, every element in the domain has unique image. A relation is said to be function if every element of one set has one and only one image in other set. Hence, s is a function. (v) According to the question, t = {(x, 3) \| x is a real number. Therefore, the domain element x is a real number. Also, range has one number i.e., 3 in it. Therefore, for every element in the domain has the image 3, it is a constant function. A relation is said to be function if every element of one set has one and only one image in other set. Hence, t is a function. 11. If f and g are real functions defined by f (x) = x2Â + 7 and g (x) = 3x + 5, find each of the following (a) f (3) + g (â€“ 5) (b) f(Â½) Ã— g(14) (c) f (â€“ 2) + g (â€“ 1) (d) f (t) â€“ f (â€“ 2) (e) (f(t) â€“ f(5))/ (t â€“ 5), if t â‰ 5 Solution: According to the question, f and g are real functions such that f (x) = x2Â + 7 and g (x) = 3x + 5 (a) f (3) + g (â€“ 5) f (x) = x2Â + 7 Substituting x = 3 in f(x), we get f (3) = 32Â + 7 = 9 + 7 = 16 â€¦(i) And, g (x) = 3x + 5 Substituting x = â€“5 in g(x), we get g (â€“5) = 3(â€“5) + 5 = â€“15 + 5 = â€“10â€¦â€¦â€¦â€¦(ii) Adding equations (i) and (ii), We get, f (3) + g (â€“ 5) = 16â€“10 = 6 (b) f(Â½) Ã— g(14) f (x) = x2Â + 7 Substituting x = Â½ in f(x), we get f(Â½) = (Â½)2 + 7 = Â¼ + 7 = 29/4 â€¦(i) And, g (x) = 3x + 5 Substituting x = 14 in g(x), we get g (14) = 3(14) + 5 = 42 + 5 = 47â€¦â€¦â€¦â€¦(ii) Multiplying equation (i) and (ii), We get, f(Â½) Ã— g(14) = (29/4) Ã— 47 = 1363/4 (c) f (â€“ 2) + g (â€“ 1) f (x) = x2Â + 7 Substituting x = â€“2 in f(x), we get f (â€“2) = (â€“2)2Â + 7 = 4 + 7 = 11â€¦â€¦..(i) And, g (x) = 3x + 5 Substituting x = â€“1 in g(x), we get g (â€“1) = 3(â€“1) + 5 = â€“3 + 5 = 2â€¦â€¦â€¦â€¦(ii) Adding equation (i) and (ii), We get, f (â€“ 2) + g (â€“ 1) = 11 + 2 = 13 (d) f (t) â€“ f (â€“ 2) f (x) = x2Â + 7 Substituting x = t in f(x), we get f (t) = t2Â + 7â€¦â€¦..(i) Considering the same function, f (x) = x2Â + 7 Substituting x = â€“2 in f(x), we get f (â€“2) = (â€“2)2Â + 7 = 4 + 7 = 11â€¦â€¦.(ii) Subtracting equation (i) with (ii), We get, f (t) â€“ f (â€“ 2) = t2Â + 7 â€“ 11= t2 â€“ 4 (e)Â (f(t) â€“ f(5))/ (t â€“ 5), ifÂ t â‰ 5 f (x) = x2Â + 7 Substituting x = t in f(x), we get f (t) = t2Â + 7â€¦â€¦..(i) Considering the same function, f (x) = x2Â + 7 Substituting x = 5 in f(x), we get f (5) = (5)2Â + 7 = 25 + 7 = 32â€¦â€¦..(ii) From equation (i) and (ii), we get 12. Let f and g be real functions defined by f (x) = 2x + 1 and g (x) = 4x â€“ 7. (a) For what real numbers x, f (x) = g (x)? (b) For what real numbers x, f (x) < g (x)? Solution: According to the question, f and g be real functions defined by f(x) = 2x + 1 and g(x) = 4x â€“ 7 (a) For what real numbers x, f (x) = g (x) To satisfy the condition f(x) = g(x), Should also satisfy, 2x + 1 = 4xâ€“7 â‡’Â 7 + 1 = 4xâ€“2x â‡’Â 8 = 2x Or, 2x = 8 â‡’Â x = 4 Hence, we get, For x = 4, f (x) = g (x) (b) For what real numbers x, f (x) < g (x) To satisfy the condition f(x) < g(x), Should also satisfy, 2x + 1 < 4xâ€“7 â‡’Â 7 + 1 < 4xâ€“2x â‡’Â 8 < 2x Or, 2x > 8 â‡’Â x > 4 Hence, we get, For x > 4, f (x) > g (x) 13. If f and g are two real valued functions defined as f (x) = 2x + 1, g (x) = x2 + 1, then find. (i) f + g (ii) f â€“ g (iii) fg (iv)f/g Solution: According to the question, f and g be real valued functions defined as f (x) = 2x + 1, g (x) = x2Â + 1, (i) f + g â‡’Â f + g = f(x) + g(x) = 2x + 1 + x2Â + 1 = x2Â + 2x + 2 (ii) f â€“ g â‡’Â f â€“ g = f(x) â€“ g(x) = 2x + 1 â€“ (x2Â + 1) = 2x â€“ x2 (iii) fg â‡’Â fg = f(x) g(x) = (2x + 1)( x2Â + 1) = 2x(x2Â ) + 2x(1) + 1(x2) + 1(1) = 2x3Â + 2x + x2Â + 1 = 2x3Â + x2Â + 2x + 1 (iv) f/g f/g = f(x)/g(x) 14. Express the following functions as set of ordered pairs and determine their range. f: XÂ â†’Â R, f (x) = x3Â + 1, where X = {â€“1, 0, 3, 9, 7} Solution: According to the question, A function f: XÂ â†’R, f (x) = x3Â + 1, where X = {â€“1, 0, 3, 9, 7} Domain = f is a function such that the first elements of all the ordered pair belong to the set X = {â€“1, 0, 3, 9, 7}. The second element of all the ordered pair are such that they satisfy the condition f (x) = x3Â + 1 When x = â€“ 1, f (x) = x3Â + 1 f (â€“ 1) = (â€“ 1)3Â + 1 = â€“ 1 + 1 = 0 â‡’ ordered pair = (â€“1, 0) When x = 0, f (x) = x3Â + 1 f (0) = (0)3Â + 1 = 0 + 1 = 1â‡’ ordered pair = (0, 1) When x = 3, f (x) = x3Â + 1 f (3) = (3)3Â + 1 = 27 + 1 = 28â‡’ ordered pair = (3, 28) When x = 9, f (x) = x3Â + 1 f (9) = (9)3Â + 1 = 729 + 1 = 730â‡’ ordered pair = (9, 730) When x = 7, f (x) = x3Â + 1 f (7) = (7)3Â + 1 = 343 + 1 = 344â‡’ ordered pair = (7, 344) Therefore, the given function as a set of ordered pairs is f = {(â€“1, 0), (0, 1), (3, 28), (7, 344), (9, 730)} And, Range of f = {0, 1, 28, 730, 344} 15. Find the values of x for which the functions f (x) = 3x2Â â€“ 1 and g (x) = 3 + x are equal Solution: According to the question, f and g functions defined by f (x) = 3x2Â â€“ 1 and g (x) = 3 + x For what real numbers x, f (x) = g (x) To satisfy the condition f(x) = g(x), Should also satisfy, 3x2Â â€“ 1 = 3 + x â‡’Â 3x2Â â€“ x â€“ 3 â€“ 1 = 0 â‡’Â 3x2Â â€“ x â€“ 4 = 0 Splitting the middle term, We get, â‡’Â 3x2Â + 3x â€“ 4xâ€“4 = 0 â‡’Â 3x(x + 1) â€“ 4(x + 1) = 0 â‡’Â (3x â€“ 4)(x + 1) = 0 â‡’Â 3x â€“ 4 = 0 or x + 1 = 0 â‡’Â 3x = 4 or x = â€“1 â‡’ x = 4/3, â€“1 Hence, forÂ x = 4/3, â€“1, f (x) = g (x), i.e., given functions are equal. Long Answer Type 16. Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? Justify. If this is described by the relation, g (x) = Î±x + Î², then what values should be assigned to Î± and Î²? Solution: According to the question, g = {(1, 1), (2, 3), (3, 5), (4, 7)}, and is described by relation g (x) = Î±x + Î² Now, given the relation, g = {(1, 1), (2, 3), (3, 5), (4, 7)} g (x) = Î±x + Î² For ordered pair (1,1), g (x) = Î±x + Î², becomes g (1) = Î±(1) + Î² = 1 â‡’Â Î± + Î² = 1 â‡’Â Î± = 1 â€“ Î² â€¦(i) Considering ordered pair (2, 3), g (x) = Î±x + Î², becomes g (2) = Î±(2) + Î² = 3 â‡’Â 2Î± + Î² = 3 Substituting value of Î± from equation (i), we get â‡’Â 2(2) + Î² = 3 â‡’Â Î² = 3 â€“ 4 = â€“ 1 Substituting value of Î² in equation (i), we get Î± = 1â€“Î² = 1â€“(â€“1) = 2 Now, the given equation becomes, i.e., g (x) = 2xâ€“1 17. Find the domain of each of the following functions given by Solution: (i) According to the question, We know the value of cos x lies between â€“1, 1, â€“1 â‰¤ cos x â‰¤ 1 Multiplying by negative sign, we get Or 1 â‰¥ â€“ cos x â‰¥ â€“1 Adding 1, we get 2 â‰¥ 1â€“ cos x â‰¥ 0 â€¦(i) Now, 1â€“ cosÂ x â‰ 0 â‡’Â cos x â‰ 1 Or, x â‰ 2nÏ€Â âˆ€ n âˆˆ Z Therefore, the domain of f = Râ€“{2nÏ€:nâˆˆZ} (ii) According to the question, For real value of f, x + \|x\| > 0 When x > 0, x + \|x\| > 0â‡’Â x + x > 0 â‡’Â 2x > 0â‡’Â x > 0 When x < 0, x + \|x\| > 0â‡’Â x â€“ x > 0 â‡’Â 2x > 0â‡’Â x > 0 So, x > 0, to satisfy the given equation. Therefore, the domain of f = R+ (iii) f(x) = x\|x\| According to the question, We know x and \|x\| are defined for all real values. Therefore, the domain of f = R (iv) According to the question, For real value of x2â€“1â‰ 0 â‡’Â (xâ€“1)(x + 1)â‰ 0 â‡’Â xâ€“1â‰ 0 or x + 1â‰ 0 â‡’Â xâ‰ 1 or xâ‰ â€“1 Therefore, the domain of f = Râ€“{â€“1, 1} (v) According to the question, For real value of 28 â€“ x â‰ 0 â‡’Â xâ‰ 28 Therefore, the domain of f = Râ€“{28} 18. Find the range of the following functions given by Solution: (i) â‡’Â y>0 and 2yâ€“3â‰¥0 â‡’Â y>0 and 2yâ‰¥3 â‡’Â y>0 andÂ y â‰¥ 3/2 Or f(x)>0 andÂ f(x) â‰¥ 3/2 f(x) âˆˆ ( â€“ âˆž, 0) âˆª [ 3/2 , âˆž) Therefore, the range of f =Â ( â€“ âˆž, 0) âˆª [ 3/2 , âˆž) (ii) f(x) = 1â€“\|xâ€“2\| According to the question, For real value of f, \|xâ€“2\|â‰¥ 0 Adding negative sign, we get Or â€“\|xâ€“2\|â‰¤ 0 Adding 1 we get â‡’Â 1â€“\|xâ€“2\|â‰¤ 1 Or f(x)â‰¤1 â‡’Â f(x)âˆˆÂ (â€“âˆž, 1] Therefore, the range of f = (â€“âˆž, 1] (iii) f(x) = \|xâ€“3\| According to the question, We know \|x\| are defined for all real values. And \|xâ€“3\| will always be greater than or equal to 0. i.e., f(x) â‰¥ 0 Therefore, the range of f = [0, âˆž) (iv) f (x) = 1 + 3 cos2x According to the question, We know the value of cos 2x lies between â€“1, 1, so â€“1â‰¤ cos 2xâ‰¤ 1 Multiplying by 3, we get â€“3â‰¤ 3cos 2xâ‰¤ 3 Adding 1, we get â€“2â‰¤ 1 + 3cos 2xâ‰¤ 4 Or, â€“2â‰¤ f(x)â‰¤ 4 Hence f(x)âˆˆÂ [â€“2, 4] Therefore, the range of f = [â€“2, 4] 19. Redefine the functionÂ f(x) = \|x â€“ 2\| + \|2 + x\|, â€“ 3 â‰¤ x â‰¤ 3 Solution: According to the question, function f(x) = \|xâ€“2\| + \|2 + x\|, â€“3â‰¤ xâ‰¤ 3 We know that, when x>0, \|x â€“ 2\| is (xâ€“2), xâ‰¥2 \|2 + x\| is (2 + x), xâ‰¥â€“2 when x>0 \|x â€“ 2\| is â€“(xâ€“2), x<2 \|2 + x\| is â€“(2 + x), x<â€“2 Given that, f(x) = \|xâ€“2\| + \|2 + x\|, â€“3â‰¤ xâ‰¤ 3 It can be rewritten as, 20. IfÂ , then show that: Solution: (i) Hence proved (ii) Substituting x by â€“ 1/x, we get Hence proved 21. Let f(x) = âˆšxÂ and g (x) = x be two functions defined in the domain R+âˆªÂ {0}. Find (i) (f + g) (x) (ii) (f â€“ g) (x) (iii) (fg) (x) (iv) (f/g) (x) Solution: (i) (f + g)(x) â‡’Â (f + g)(x) = f(x) + g(x) â‡’Â f(x) + g(x) = âˆšx + x (ii) (f â€“ g)(x) â‡’Â (f â€“ g)(x) = f(x) â€“ g(x) â‡’Â f(x) â€“ g(x) = âˆšxâ€“x (iii) (fg)(x) â‡’Â (fg)(x) = f(x) g(x) â‡’Â (fg)(x) = (âˆšx)(x) â‡’Â f(x)g(x)= xâˆšx (iv) (f/q)(x) = f(x)/g(x)
## Semi-Bluff Shove Poker Math There are very few things in poker that are more fun than shoving. And if you are considering doing more 5bet bluff shoving preflop or semi-bluff jamming postflop, then understanding the math behind it is crucial. In this video I explain when and how to expand the math to make sure you are solving the spot correctly. If you’d rather read the script of this video, read on below. Otherwise, turn the video to 720p and enjoy easy-to-digest complicated poker math! (Switch to 720p and subscribe for new poker vids) Hello, and welcome to today’s Quick Plays video on advanced EV in poker. We’ve done another video on basic EV, but there are many situations in poker when a basic EV formula just doesn’t quite cut it. So in this video ill show you a more complex EV formula and how to use it with an example. The basic EV formula we worked with in the past was EV = (%W\$W)-(%L\$L). So essentially what we stand the win multiplied by how often we’ll win…minus what we stand to lose multiplied by how often we’ll lose. If this seems confusing at all, please first watch the basic EV poker video and then come back to this one. But there are times when we need a more complex version of this formula. So let’s look at an example to get us started… In this hand it folds to the cutoff who opens, we semi-bluff 3bet to \$10 with 8♥ 6♥, the cutoff 4bets to \$23 and we decide to 5bet shove. Like most plays we can proof this using some simple poker math, so let’s pull out our EV equation. You may notice that this basic formula doesn’t account for all possible outcomes. Once we shove there are 3 things that can happen: 1. He calls and we lose 2. He calls and we win 3. He folds and we win So at this point the basic EV formula needs to be expanded to account for each outcome. The expanded formula would then look like this: EV = F(\$Pot) + C(%W\$W) – C(%L\$L) Where “F” stands for “times villain folds” and “C” stands for “times villain calls”. And if you only know one of them, you can always figure out the other since their sum is always 100%. If you know F is 20%, then you take 100% – 20%, and get 80% for C. Now we can just start plugging numbers in. Since most of these numbers are related (F+C=100% and %W+%L=100%), it makes life even easier. Let’s review how to get each number quickly: %F/%C = These are estimations based upon how often you think villain will call your shove. If you think he was bluff 4betting a ton and thus wouldn’t be able to call your shove often…then F would be a very large percentage. Conversely, if you think villain were 4betting a strong range and would call your shove often, then F would be a small percentage and C would be quite large. %W/%L = These are based upon equity, which we can calculate using a free program like Equilab. The W% is your equity against the range villain would open, 4bet, and call your shove with. \$Pot = the size of the pot BEFORE you shove \$W = what you would win the times you get called and win \$L = what you would lose the times you get called and villain wins Now we just have to make some assumptions on his range and frequencies, plug in some numbers, and proof the validity of this shove. Let’s assume villain would call our shove with TT+/AK. In that case we would have 27% equity…so %W is .27 and %L is .73 Let’s also assume that villain does bluff 4bet sometimes, so we assume he will 4bet/fold 25% of the time. This means F is 25% and C is 75%. Now for the dollar amounts and we can solve! The Pot before we shove is \$34.50, so that’s easy. If we shove and villain wins we lose \$96. Because villain has the shortest stack we can only lose \$83 plus the \$13 to match his 4bet. Our \$10 3bet no longer belongs to us, and thus we cannot lose it once we shove. If we shove and villain calls we can win \$117.5. Because we have the largest stack the shortcut is just current pot + villain’s stack. Now we have all of the necessary inputs! We see that at this point our shove has a -\$20.14 expected value. Given the parameters and assumptions we’ve used, this is a bad shove and we should avoid making it. But since we are analyzing this hand away from the table, and have this extra time, let’s do some experimenting… Assume for a moment that villain 4bet bluffs a LOT more often, and thus we can expect a fold from him 60% of the time. That changes F to 60% and C to 40%. Let’s also change his calling range from TT+/AK to TT+/AQ+. This increases our equity up to 30%, and thus changes both %W and %L. Now if we plug everything in we notice our EV jumps up to +\$7.92. All of the sudden our shove is looking pretty good! With this formula the money won and lost will remain constant, but changes in ranges and frequencies can alter the outcome a ton. Essentially, the more villain folds and we pick up the pot outright the better for us. The more equity we have when called the better since we’ll pick up the all-in pot more often and lose less often. And if we can ever increase both our equity when called AND the times villain folds preflop…the better and better our semi-bluff will be. Knowing how and when to expand the basic EV formula can greatly benefit you on and off the table. Now in real time you won’t be able to plug-n-play with the formula…but with enough off-table practice things will get ingrained and you will be able to more correctly estimate the math at the tables. And to be honest, there are times when you could expand the formula even further…but just knowing how to do this gives you a nice mathematical leg-up on your opponents! Same as always, if you have any questions please don’t hesitate to let me know…otherwise…good luck and happy grinding! ### SplitSuit My name is James "SplitSuit" Sweeney and I'm a poker player, coach, and author. I've released 300+ videos, coached 500+ players, and co-founded the training site Red Chip Poker. Contact me if you need any help improving your poker game! HAVE A POKER QUESTION?
# 7-2 PROBLEM SOLVING FACTORING BY GCF In this tutorial we are going to look at two ways to factor polynomial expressions, factoring out the greatest common factor and factoring by grouping. If you want to test this, go ahead and divide both and by 42—they are both evenly divisible by this number! The fully factored polynomial will be the product of two binomials. Factor out the common factor, 2 x — 3 , from both terms. And then you have minus 2 divided by 2 is 1. To factor a number is to rewrite it as a product. When we divide it out of the second term, we are left with Notice that when you factor two terms, the result is a monomial times a polynomial. Now, if you were to undistribute 2x squared out of the expression, you’d essentially get 2x squared times this term, minus this term, minus this term. Well 4 divided by 2 is 2. Identify the GCF of the polynomial. Introduction Factoring is to write an expression as a product of factors. So the GCF of our variable part is xy. That simplifies to 1, maybe I fachoring write it below. Notice that in the example below, the first term is x 2and x is the only variable present. And y divided by 1, you can imagine, is just y. The polynomial is now factored. And to figure that something else we can fcf undistribute the 2x squared, say this is the same thing, or even before we undistribute the 2x squared, we could say look, 4x to the fourth y is the same thing as 2x squared, times 4x to the fourth y, over 2x squared. LESSON 8-8 PROBLEM SOLVING SOLVING RADICAL EQUATIONS AND INEQUALITIES Rewrite each term as the product of the GCF and the remaining terms. # Factoring by grouping (article) \| Khan Academy Just as any integer can be written as the product of factors, ffactoring too can any monomial or polynomial be expressed as a product of factors. Factor 45 c 2 d 2. And then you have minus 8 divided by 2 is 4. Well, the biggest coefficient that divides all of these is a 2, so let me put that 2, let me factor 2 out. You might say OK, let me look at each of these. Well 4 divided by 2 is 2. problemm Take the numbers 50 and Similarly, you could say that 8x to the third y– I’ll put the negative out front– is the same thing as 2x squared, our greatest common factor, times 8x to the third y, over 2x squared. Find the GCF of the second pair of terms. Likewise to factor a polynomialyou rewrite it as a product. In this case, it does check out. ## Greatest Common Factor (GCF) Calculator Let me factor an x ffactoring out. Note that if we multiply our answer out that we do get the original polynomial. It involves organizing the polynomial in groups. Factor the GCF, 4 aout of the first group. Product of a number and a sum: The largest monomial that we can factor out of each term is. MY DOG ATE MY HOMEWORK POEM KENN NESBITT Example Problem Find the greatest common factor of 81 c 3 d and 45 c 2 d 2. Others will be asking you for help with factoring. Notice that when you factor two terms, the result is a monomial times a polynomial. The entire term xy 3 is not a factor of either monomial. Note that this is not in factored form because of the minus sign we have before the 7 in the problem. We can also do this with polynomial dolving. # Greatest Common Factor But the factored form of a four-term polynomial is the product of two binomials. And the reason why I kind of of went through great pains to show you exactly what we’re doing is so you know exactly what we’re doing. So it’s 2x squared times 2x squared y, and then you have minus 2x squared probem, 8 divided by 2 is 4. Pulling out common factors, you find: Find the greatest common factor of 56 xy and 16 y 3.
# Find Domain and Range of Relations Given by Graphs Examples and Questions With Solutions Find the domain and range of a relation given by its graph. Examples are presented along with detailed Solutions and explanations and also more questions with detailed solutions. ## Examples with Solutions Example 1 a) Find the domain and b) the range of the relation given by its graph shown below and c) state whether the relation is a function or not. Solution: a) Domain: We first find the 2 points on the graph of the given relation with the smallest and the largest x-coordinate. In this example the 2 points are A(-2,-4) and B(4,-6) (see graph above). The domain is the set of all x values from the smallest x-coordinate (that of A) to the largest x-coordinate (that of B) and is written as: -2 ≤ x ≤ 4 The double inequality above has the inequality symbol ≤ at both sides because the closed circles at points A and B indicate that the relation is defined at these values of x. b) Range: We need to find the coordinates of the 2 points on the graph with the lowest and the largest values of the y coordinate. In this example, these points are B(4,-6) and C(2,2). The range is the set of all y values between the smallest and the largest y coordinates and given by the double inequality: -6 ≤ y ≤ 2 The inequality symbol ≤ is used at both sides because the closed circles at points B and C indicates the relation is defined at these values. c) The relation graphed above is a function because no vertical line can intersect the given graph at more than one point. Example 2 Find the a) domain and b) range of the relation given by its graph shown below and c) state whether the relation is a function or not. Solution: a) Domain: In this example points A(-3,-5) and B(8,4) have the smallest and the largest x-coordinates respectively, hence the domain is given by: -3 ≤ x ≤ 8 The use of the symbol ≤ at both sides is due to the fact that the relation is defined at points A and B (closed circles at both points). b) Range: Points A and B have the smallest and the largest values of the y-coordinate respectively. The range is given by the inequality: - 5≤ y ≤ 4 The use of the symbol ≤ at both sides is due to the fact that the relation is defined at points A and B. c) No vertical line can cut the given graph at more than one point and therefore the relation graphed above is a function. Example 3 Find the a) domain and b) range of the relation given by its graph shown below and c) state whether the relation is a function or not. Solution: a) Domain: Points A(-3,-2) and B(1,-2) have the smallest and the largest x-coordinates respectively, hence the domain: -3 ≤ x ≤ 1 The use of the symbol ≤ at both sides is due to the fact that the relation is defined at points A and B (closed circles at both points). b) Range: Points C(-1,-5) and D(-1,1) have the smallest and the largest y-coordinate respectively. The range is given by the double inequality: - 5≤ y ≤ 1 The relation is defined at points C and D (closed circles), hence the use of the inequality symbol ≤. c) There is at least one vertical line that cuts the given graph at two points (see graph below) and therefore the relation graphed above is NOT a function. Example 4 Find the a) domain and b) range of the relation given by its graph shown below and c) state whether the relation is a function or not. Solution: a) Domain: Points A(-3,0) has the smallest x-coordinate. The arrow at the top right of the graph indicates that the graph continues to the left as x increases. Hence there is no limit to the largest x-coordinate of points on the graph. The domain is given by all values greater than or equal to the smallest values x = -3 and is written as: x ≥ -3 The use of the symbol ≥ at because the relation is defined at points A (closed circle at point A). b) Range: Points B and C have equal and smallest y-coordinates equal to -2. The arrow at the top right of the graph indicates that the y coordinate increases as x increases. Hence there is no limit to the y-coordinate and therefore the range is given by all values greater than or equal to the smallest value y = -2 and is written as: y ≥ -2 The use of the inequality symbol ≥ is due to the fact that the relation is defined at y = -2 (closed circle at B and C). c) There is no vertical line that cuts the given graph at more than one point (see graph below) and therefore the relation graphed above is a function. Example 5 Find the a) domain and b) range of the relation given by its graph shown below and c) state whether the relation is a function or not. Solution: a) Domain: Points A(-2,-3) has the smallest x-coordinate. The arrow at the top right of the graph indicates that the graph continues to the left as x increases. Hence there is no limit to the largest x-coordinate of points on the graph. The domain is given by all values greater than the smallest values x = - 2 and is written as: x > -2 We use of the inequality symbol > (with no equal) because the relation is not defined at points A (open circle at point A). b) Range: Points A(-2,-3) has the smallest y-coordinate equal to - 3. The arrow at the top right of the graph indicates that the y coordinate increases as x increases. Therefore there is no limit to the y-coordinate. Hence the range is given by all values greater than the smallest value y = - 3 and is written as: y > - 3 The inequality symbol > is used because the relation is not defined at y = - 3 (open circle at point A). c) The graph represents a function because there is no vertical line that cuts the given graph at more than one point. ## More Questions and their detailed solutions For each relation below, find the domain and range and state whether the relation is a function. 1) 2) 3 4) 5) ## More References and links Domain and Range of Functions Online tutorials on how to find the domain and range of functions. Middle School Maths (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers Home Page {ezoic-ad-1} {ez_footer_ads}
Indeterminate Limits---Sine Forms. How to Solve, Examples, practice problems with work # Indeterminate Limits---Sine Forms ### Quick Overview • $$\displaystyle \lim_{\theta\to0} \frac{\sin(\theta)} \theta = 1$$ • This limit was derived in the lesson on the Squeeze Theorem • The denominator must be the same as the argument of the sine, and both must approach zero in the limit. ### Examples ##### Example 1 Evaluate $$\displaystyle \lim_{\theta\to0}\frac{\sin(4\theta)} \theta$$ Step 1 Multiply by $$\frac 4 4$$ so the denominator matches the argument. \begin{align} \lim_{\theta\to0}\,\frac{\sin(4\theta)} \theta % & = \lim_{\theta\to0} \left(% \frac{\red 4}{\red 4} \cdot \frac{\sin(4\theta)} \theta \right)\\[6pt] % & = \lim_{\theta\to0}\left(% \frac{\red 4} 1\cdot \frac{\sin(4\theta)}{\red 4\theta} \right)\\[6pt] % & = \red{4}\,\displaystyle\lim_{\theta\to0}\, \frac{\sin(4\theta)}{\red 4\theta} \end{align} Step 2 Evaluate the limit. Since the denominator is the same as the argument of the sine function, and both are going to 0, the limit is equal to 1. $$4\,\blue{\displaystyle\lim_{\theta\to0}\, \frac {\sin(4\theta)} {4\theta}} = 4(\blue 1) = 4$$ Answer: $$\displaystyle \lim_{\theta\to0}\,\frac{\sin(4\theta)} \theta = 4$$ ##### Example 2 Evaluate $$\displaystyle \lim_{x\to0}\,\frac{\sin(5x)}{2x}$$ Step 1 Factor out the 2 from the denominator. $$\displaystyle\lim_{x\to0}\,\frac{\sin(5x)}{\blue{2}x} % = \displaystyle\lim_{x\to0}\left(% \frac 1 {\blue{2}}\cdot \frac{\sin(5x)}{x} \right) % = \frac 1 {\blue 2}\cdot \displaystyle\lim_{x\to0}\,\frac{\sin(5x)} x$$ Step 2 Multiply by $$\frac 5 5$$ so the denominator matches the argument of the sine function. \begin{align} \frac 1 2\cdot\lim_{x\to0}\,\frac{\sin(5x)}{x} % & = \frac 1 2\cdot\lim_{x\to0}\left(% \frac{\red 5}{\red 5}\cdot \frac{\sin(5x)}{x} \right)\\[6pt] % & = \frac 1 2\cdot\displaystyle\lim_{x\to0}\left(% \frac{\red 5} 1\cdot \frac{\sin(5x)}{\red 5 x} \right)\\[6pt] % & =\frac{\red 5} 2 \cdot \lim_{x\to0}\,\frac{\sin(5x)}{\red 5 x} \end{align} Step 3 Evaluate the limit. $$\frac 5 2\cdot \blue{\displaystyle\lim_{x\to0}\, \frac{\sin(5x)}{5x}} = \frac 5 2 (\blue 1) = \frac 5 2$$ Answer: $$\displaystyle \displaystyle\lim_{x\to0}\,\frac{\sin(5x)}{2x} = \frac 5 2$$ ##### Example 3 Evaluate $$\displaystyle \lim_{x\to0} \,\frac{6x}{\sin 3x}$$ Step 1 Factor a 2 out of the numerator. $$\displaystyle\lim_{x\to0} \,\frac{\blue 6 x}{\sin 3x} % = \displaystyle\lim_{x\to0} \left(% \frac{\blue 2} 1 \cdot \frac{\blue 3 x}{\sin 3x} \right) % = \blue 2\cdot \displaystyle\lim_{x\to0}\,\frac{\blue 3 x}{\sin 3x}$$ Step 2 Rewrite the fraction as its reciprocal to the -1 power. $$2\cdot \displaystyle\lim_{x\to0}\,\frac{3x}{\sin 3x} % = 2\cdot \displaystyle\lim_{x\to0}\,\left(\frac{\sin 3x}{3x}\right)^{-1}$$ Step 3 Pass the limit inside the exponent (see the page on Limit Laws), and evaluate. $$2\cdot \displaystyle\lim_{x\to0}\,\left(\frac{\sin 3x}{3x}\right)^{-1} % = 2\,\left(\blue{\displaystyle\lim_{x\to0}\frac{\sin 3x}{3x}}\right)^{-1} % = 2(\blue 1)^{-1} % = 2$$ Answer: $$\displaystyle\, \lim_{x\to0} \,\frac{6x}{\sin 3x} = 2$$ Important! Example 3 shows us that $$\displaystyle \lim_{x\to 0}\, \frac{\theta}{\sin \theta} = 1.$$ ##### Example 4 Evaluate $$\displaystyle \lim_{t\to0} \,\frac{\sin 3t}{\sin 8t}$$ Step 1 Rewrite as separate fractions. $$\displaystyle\lim_{t\to0}\, \frac {\sin 3t} {\sin 8t} = \displaystyle\lim_{t\to0}\, \left(% \frac{\sin 3t } 1 \cdot \frac 1 {\sin 8t } \right)$$ Step 2 Multiply by $$\frac{3t}{3t}$$ and $$\frac{8t}{8t}$$. \begin{align} \lim_{t\to0}\, \left(% \frac{\sin 3t } 1 \cdot \frac 1 {\sin 8t } \right) % & = \lim_{t\to0}\, \left(% \frac{\blue{3t}}{\blue{3t}} \cdot \frac{\sin 3t } 1 \cdot \frac{\red{8t}}{\red{8t}} \cdot \frac 1 {\sin 8t } \right)\\[6pt] % & = \lim_{t\to0}\, \left(% \frac{\blue{3t}}{1} \cdot \frac{\sin 3t }{\blue{3t}} \cdot \frac{1}{\red{8t}} \cdot \frac{\red {8t}} {\sin 8t } \right) \end{align} Step 3 Simplify the non-trigonometric fractions \begin{align} \lim_{t\to0} \left(% \blue{\frac{3t}{1}} \cdot \frac{\sin 3t}{3t}\cdot \red{\frac{1}{8t}} \cdot \frac{8t} {\sin 8t} \right) % & = \lim_{t\to0} \left(% \blue{\frac{3t}{1}} \cdot \red{\frac{1}{8t}} \cdot \frac{\sin 3t}{3t}\cdot \frac{8t} {\sin 8t} \right)\\[6pt] % & = \lim_{t\to0} \left(% \frac{\blue{3t}}{\red{8t}} \cdot \frac{\sin 3t}{3t}\cdot \frac{8t} {\sin 8t} \right)\\[6pt] % & = \lim_{t\to0} \left(% \frac{\blue{3}}{\red{8}} \cdot \frac{\sin 3t}{3t}\cdot \frac{8t} {\sin 8t} \right)\\[6pt] % & = \frac{\blue{3}}{\red{8}} \cdot \lim_{t\to0} \left(% \frac{\sin 3t}{3t}\cdot \frac{8t} {\sin 8t} \right) \end{align} Step 4 Evaluate each of the limits. $$\frac 3 8 \left(% \blue{\displaystyle\lim_{t\to0} \frac{\sin 3t}{3t}}\right) \left(\red{\displaystyle\lim_{t\to0} \frac{8t} {\sin 8t}} \right) % = \frac 3 8 (\blue 1)(\red 1) % = \frac 3 8$$ Answer: $$\displaystyle \lim_{t\to0} \frac{\sin 3t}{\sin 8t} = \frac 3 8$$ ### Practice Problems Step 1 Multiply by $$\frac 7 7$$. \begin{align} \lim_{x\to0} \frac{\sin 7x}x % & = \lim_{x\to0} \left(% \blue{\frac 7 7} \cdot \frac{\sin 7x} x \right)\\[6pt] % & = \lim_{x\to0}\left(% \frac{\blue 7} 1 \cdot \frac{\sin 7x}{\blue 7 x} \right)\\[6pt] % & = \blue 7\cdot\lim_{x\to0}\,\frac{\sin 7x}{\blue 7 x} \end{align} Step 2 Evaluate the limit. $$7\cdot\blue{\displaystyle\lim_{x\to0} \frac{\sin 7x}{7 x}} = 7(\blue 1) = 7$$ Answer: $$\displaystyle \lim_{x\to0}\, \frac{\sin 7x}x =7$$ Step 1 Factor a 2 out of the numerator. $$\displaystyle\lim_{x\to0} \,\frac{\blue{18}x}{\sin 9x} % = \displaystyle\lim_{x\to0}\left(% \blue{2}\cdot \frac{\blue{9}x}{\sin 9x} \right) % = \blue 2\cdot\displaystyle\lim_{x\to0}\, \frac{\blue{9}x}{\sin 9x}$$ Step 2 Evaluate the limit. $$2\cdot\blue{\displaystyle\lim_{x\to0}\, \frac{9x}{\sin 9x}} = 2(\blue 1) = 2$$ Answer: $$\displaystyle \lim_{x\to0}\, \frac{18x}{\sin 9x} = 2$$ Step 1 Factor the 8 out of the denominator. $$\displaystyle\lim_{x\to0}\, \frac{\sin 6x}{\blue 8 x} % = \displaystyle\lim_{x\to0}\left(% \frac 1 {\blue 8} \cdot \frac{\sin 6x} x \right) % = \frac 1 {\blue 8}\cdot\displaystyle\lim_{x\to0}\, \frac{\sin 6x} x$$ Step 2 Multiply by $$\frac 6 6$$. \begin{align} \frac 1 8\cdot\lim_{x\to0}\, \frac{\sin 6x} x % & = \frac 1 8\cdot\lim_{x\to0}\left(% \blue{\frac 6 6}\cdot \frac{\sin 6x} x \right) \\[6pt] % & = \frac 1 8\cdot\lim_{x\to0}\left(% \frac{\blue 6} 1\cdot \frac{\sin 6x}{\blue 6 x} \right)\\[6pt] % & = \frac{\blue 6} 8\cdot\lim_{x\to0}\, \frac{\sin 6x}{\blue 6 x}\\[6pt] % & = \frac 3 4\cdot\lim_{x\to0}\, \frac{\sin 6x}{6 x} \end{align} Step 3 Evaluate the limit. $$\frac 3 4 \cdot\blue{\displaystyle\lim_{x\to0}\, \frac{\sin 6x}{6x}} % = \frac 3 4(\blue 1) % = \frac 3 4$$ Answer: $$\displaystyle \lim_{x\to0} \,\frac{\sin 6x}{8x} = \frac 3 4$$ Step 1 Factor the 5 out of the numerator. $$\displaystyle\lim_{x\to0}\, \frac{\blue 5 x}{\sin 9x} % = \displaystyle\lim_{x\to0} \left(% \frac{\blue 5} 1 \cdot \frac x {\sin 9x} \right) % =\blue5\cdot\displaystyle\lim_{x\to0} \,\frac{x}{\sin 9x}$$ Step 2 Multiply by $$\frac 9 9$$. \begin{align} 5\cdot\lim_{x\to0}\,\frac x {\sin 9x} % & = 5\cdot\lim_{x\to0}\left(% \blue{\frac 9 9} \cdot \frac x {\sin 9x} \right)\\[6pt] % & = 5\cdot\lim_{x\to0}\left(% \frac 1 {\blue 9}\cdot \frac{\blue 9 x}{\sin 9x} \right)\\[6pt] % & = \frac 5 {\blue 9}\cdot\lim_{x\to0}\,\frac{\blue 9 x}{\sin 9x} \end{align} Step 3 Evaluate the limit. $$\frac 5 9\cdot\blue{\displaystyle\lim_{x\to0}\,\frac{9 x}{\sin 9x}} % = \frac 5 9 (\blue 1) % = \frac 5 9$$ Answer: $$\displaystyle \lim_{x\to0}\, \frac{5x}{\sin 9x} = \frac 5 9$$ Step 1 Rewrite in separate fractions. $$\displaystyle\lim_{x\to0}\,\frac{\sin 4x}{\sin 11x} % = \displaystyle\lim_{x\to0} \left(% \frac{\sin 4x} 1 \cdot \frac 1 {\sin 11x} \right)$$ Step 2 Multiply by $$\frac{4x}{4x}$$ and $$\frac{11x}{11x}$$. \begin{align} \lim_{x\to0} \left(% \frac{\sin 4x} 1 \cdot \frac 1 {\sin 11x} \right) % & = \lim_{x\to0} \left(% \blue{\frac{4x}{4x}}\cdot \frac{\sin 4x} 1 \cdot \red{\frac{11x}{11x}}\cdot \frac 1 {\sin 11x} \right)\\[6pt] % & = \lim_{x\to0} \left(% \frac{\blue{4x}}{1}\cdot \frac{\sin 4x}{\blue{4x}} \cdot \frac 1 {\red{11x}}\cdot \frac{\red{11x}}{\sin 11x} \right)\\[6pt] \end{align} Step 3 Simplify the non-trigonometric fractions. \begin{align} \lim_{x\to0} \left(% \blue{\frac{4x} 1}\cdot \frac{\sin 4x}{4x} \cdot \red{\frac 1 {11x}}\cdot \frac{11x}{\sin 11x} \right) % & =\lim_{x\to0} \left(% \frac{\blue{4x}}{\red{11x}}\cdot \frac{\sin 4x}{4x}\cdot \frac{11x}{\sin 11x} \right)\\[6pt] % & = \lim_{x\to0}\left(% \frac{\blue{4}}{\red{11}}\cdot \frac{\sin 4x}{4x} \cdot \frac{11x}{\sin 11x} \right)\\[6pt] % & = \frac{\blue 4}{\red{11}}\cdot \lim_{x\to0}\left(% \frac{\sin 4x}{4x}\cdot \frac{11x}{\sin 11x} \right) \end{align} Step 4 Find the limit of each factor. $$\frac 4 {11}\cdot\displaystyle\lim_{x\to0}\left(% \blue{\frac{\sin 4x}{4x}}\cdot \red{\frac{11x}{\sin 11x}} \right) % = \left(% \blue{\displaystyle\lim_{x\to0}\,\frac{\sin 4x}{4x}} \right) % \left(% \red{\displaystyle\lim_{x\to0}\,\frac{11x}{\sin 11x}} \right) % = \frac 4 {11}(\blue 1)(\red 1) = \frac 4 {11}$$ Answer: $$\displaystyle \lim_{x\to0}\, \frac{\sin 4x}{11x} = \frac 4 {11}$$ Step 1 Write the numerator and denominator in separate fractions. $$\displaystyle\lim_{x\to0}\,\frac{\sin 5x}{\sin 2x} % = \displaystyle\lim_{x\to0}\left(% \frac{\sin 5x} 1 \cdot \frac 1 {\sin 2x} \right)$$ Step 2 Multiply by $$\frac{5x}{5x}$$ and $$\frac{2x}{2x}$$. \begin{align} \lim_{x\to0}\left(% \frac{\sin 5x} 1 \cdot \frac 1 {\sin 2x} \right) % & = \lim_{x\to0}\left(% \blue{\frac{5x}{5x}}\cdot \frac{\sin 5x} 1 \cdot \red{\frac{2x}{2x}}\cdot \frac 1 {\sin 2x} \right)\\[6pt] % & = \lim_{x\to0}\left(% \frac{\blue{5x}} 1\cdot \frac{\sin 5x}{\blue{5x}} \cdot \frac 1 {\red{2x}}\cdot \frac{\red{2x}}{\sin 2x} \right) \end{align} Step 3 Simplify the non-trigonometric fractions. \begin{align} \lim_{x\to0}\left(% \blue{\frac{5x} 1}\cdot \frac{\sin 5x}{5x} \cdot \red{\frac 1 {2x}}\cdot \frac{2x}{\sin 2x} \right) % & = \lim_{x\to0}\left(% \frac{\blue{5x}}{\red{2x}}\cdot \frac{\sin 5x}{5x} \cdot \frac{2x}{\sin 2x} \right)\\[6pt] % & = \lim_{x\to0}\left(% \frac{\blue{5}}{\red{2}}\cdot \frac{\sin 5x}{5x} \cdot \frac{2x}{\sin 2x} \right)\\[6pt] % & = \frac{\blue{5}}{\red{2}}\cdot\lim_{x\to0}\left(% \frac{\sin 5x}{5x} \cdot \frac{2x}{\sin 2x} \right) \end{align} Step 4 Evaluate the limits of each factor. $$\frac 5 2 \cdot\displaystyle\lim_{x\to0}\left(% \blue{\frac{\sin 5x}{5x}} \cdot \red{\frac{2x}{\sin 2x}} \right) % =\frac 5 2\cdot\left(% \blue{\displaystyle\lim_{x\to0}\frac{\sin 5x}{5x}} \right) % \left(% \red{\displaystyle\lim_{x\to0}\frac{2x}{\sin 2x}} \right) % = \frac 5 2 (\blue 1)(\red 1) % = \frac 5 2$$ Answer: $$\displaystyle \lim_{x\to0}\, \frac{\sin 5x}{\sin 2x} = \frac 5 2$$ ### Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there!
# Numerical Integration We have seen that definite integrals arise in many different areas and that the Fundamental Theorem of Calculus is a powerful tool for evaluating definite integrals. However, it cannot always be applied: there are some functions which do not have an antiderivative which can be expressed in terms of familiar functions such as polynomials, exponentials and trigonometric functions. One such example is ; of course, this is an important function since it is the probability density function for the normal distribution. In this section, we will demonstrate two tools for approximating a definite integral to any degree of accuracy. These tools can be used to evaluate definite integrals when the integrand has no simple antiderivative. Moreover, we sometimes only have information about a function by making observations at a certain number of points. In that case, we do not have a nice formula for the function we are integrating, but only some data points. Our methods can help us to evaluate a definite integral in this case as well. The Trapezoidal Rule In our first method, we will approximate the definite integral by approximating the graph of the function by a straight line. That means that we approximate the area under the graph by the trapezoid formed below. The area of the trapezoid is just the area of the rectangle plus the area of the triangle. That means our approximation is Of course, we cannot expect this to be a good approximation. However, we can break the region into many smaller pieces and apply the approximation on each piece. On the smaller pieces, the graph looks more and more like a straight line so the approximation should improve. Let's choose some positive integer n and break the interval into n equal pieces. The width of each piece is . We will label the points defined by the sub-intervals by and call . If we the approximate the area under the graph by the area of the trapezoids, we have Example: We will consider the integral . Of course, we can evaluate this integral directly with the Fundamental Theorem of Calculus: . We now will build the trapezoidal approximation so that we might see explicitly how much of an error we are making. First, we will consider . In this case, . There are only two points and . Then we have and . Then the trapezoidal rule produces This means that the Trapezoidal Rule with produces an approximation of to the integral which we know is . Now let's see what happens when . In this case, we have . The points for us to consider are and . This produces and . Then we have Clearly, this is a better approximation. The following demonstration will show what happens when we increase n further. Notice that the approximation becomes quite good. The reason for this is clear from the picture: as the size of the intervals becomes very small, the graph is better approximated by a straight line on each interval. Simpson's Rule In the example above, you can see that the Trapezoidal Rule provides a reasonable approximation to a definite integral if we take a large number of steps. Notice that the error in the approximation originates in the fact that general graphs are curved and we are approximating them by straight lines. We will now form an approximation which takes into account the curvature of the graph: the result is a more efficient approximation called Simpson's Rule. Simpson's Rule is formed by approximating a general curve by a parabola. In this picture, the red graph is a parabola which approximates the blue graph. Remember that a parabola is the graph of a quadratic function and so three pieces of information are required to determine the coefficients , and . This means that we must use three data points and to fix the parabola. We won't show you how to determine the coefficients for the parabola, but it is fairly straightforward. As before, the width of each of the two intervals is . With a little bit of work, you would find the approximation This is Simpson's Rule with one step. More generally, we can break the interval into several pieces and apply Simpson's Rule on each interval. For instance, to use n steps, break the interval into pieces, each of width . Call the x coordinates and let . Then we have This is the same idea as the Trapezoidal Rule but you see that the algorithm is slightly different. Inside the sum, the endpoints are weighted once, while the odd values of are weigted four times and the even values of in the middle are weighted twice. Examples: 1. First, we'll consider the same example as we studied with the Trapezoidal Rule: . Let's apply Simpson's Rule with one term: so that . Then and . This means that and . Now applying Simpson's Rule gives This means that, with one step, Simpson's Rule gives the correct answer. This shouldn't be too surprising since the Rule uses a parabola to approximate the graph. In this case, the graph we are interested in is already a parabola and so Simpson's Rule will produce the correct answer. In comparing to the Trapezoidal Rule, you can see the savings in effort: there we needed many steps to get a good approximation to the integral. 2. Consider the integral . You have probably been told for most of your mathematical career that but you have probably never seen where that number comes from. We can use Simpson's Rule to approximate the integral and then use the fact that to obtain an approximation for . With , we have and and . This means that and . Our approximation is then This gives an approximation for . So with just one step, we have a pretty good approximation for . With , we have This produces an approximation With just two steps of Simpson's Rule, we have computed to four decimal places! The demonstration below will provide approximations for this integral using both the Trapezoidal Rule and Simpson's Rule. You can see how much better Simpson's Rule really is. 3. Of course, the real value of Simpson's Rule is in computing integrals of functions which cannot be antidifferentiated using familiar functions. One such example is where the approximation is obtained using Simpson's Rule. This integral is needed to determine the probability that a typical sample lies within one standard deviation of the mean in the normal distribution.
Question Video: Writing a Complex Number in Polar Form given Its Modulus and Principal Argument Mathematics • 12th Grade Given that \|π\| = 5 and the argument of π is π = 2π + 2ππ, where π β β€, find π, giving your answer in trigonometric form. 01:40 Video Transcript Given that the modulus of π is equal to five and the argument of π is π equals two π plus two ππ, where π is an integer, find π, giving your answer in trigonometric form. When we write a complex number in trigonometric or polar form, we write it as π equals π multiplied by cos π plus π sin π, where π is known as the modulus of the complex number π and π is its argument. In polar form, π can be in degrees or radians whereas in exponential form, it does need to be in radians. Letβs substitute what we know about the complex number π into this formula. The modulus of π was five and the argument was two π plus two ππ. So we get π equals five multiplied by cos of two π plus two ππ plus π sin of two π plus two ππ. Next, we can recall what we know about the cosine and sine functions. They are periodic. That is to say, they repeat. And their period is two π radians. That means that for any value of π, sin of π plus some multiple of two π is equal to sin π. And cos of π plus some multiple of two π is equal to cos of π. This means that cos of two π plus two ππ is simply cos of two π. And sin of two π plus two ππ is equal to sin of two π. Our complex number can therefore be written as five multiplied by cos of two π plus π sin of two π, in trigonometrical polar form.
# Word Problems on Area of a Rectangle In word problems on area of a rectangle we will find the area whose length and breadth are given. For finding the area of a given rectangle, we should make sure that the sides (length or breadth) are in the same unit of length. If they are given in different units, change them to the same unit. 10 mm = 1 cm 100 cm = 1 m 1000 m = 1 km Word problems on area of a rectangle: 1. Find the area of a rectangle of length 78 mm and breadth 26 mm. Length of a rectangle = 78 mm. Breadth of a rectangle = 26 mm. Area of a rectangle = length × breadth = 78 × 26 sq mm. = 2028 sq mm. 2. Finding the area of a rectangle whose length is 12 cm and breadth 9 cm. Length of the rectangle = 12 cm Breadth of the rectangle = 9 cm Area of the rectangle = 12 × 9 cm2 = 108 cm2 3. Find the area of a rectangle of length 43 m and breadth 13 m. Length of a rectangle = 43 m. Breadth of a rectangle = 13 m. Area of a rectangle = l × b = 43 × 13 sq m. = 559 sq m. 4. The area of a rectangle is 48 cm2. If its breadth is 6 cm then find its length. Area of rectangle = 48 cm2 Breadth of rectangle = 6 cm Length of rectangle = Area of the rectangle/Breadth of the rectangle = 48 cm2/6 cm = 8 cm Therefore, the length of the rectangle is 8 cm. 5. How many pieces of stone slabs each 26 cm long and 10 cm broad will be require to lay a path 260 m long and 15 m wide. For each slab, length = 26 cm, breadth = 10 cm Therefore, the area of a slab = length × breadth = 26 cm × 10 cm = 260 cm2 For the path, length = 26 cm, breadth = 15 m Therefore, the area of the path = 260 × 15 × 10000 cm2 Therefore, required number of slabs = Area of path/Are of each slab = 260 × 15 × 10000/260 = 150000 Therefore, 150000 slabs will be required. Area. Area of a Rectangle. Area of a Square. To find Area of a Rectangle when Length and Breadth are of Different Units. To find Length or Breadth when Area of a Rectangle is given. Areas of Irregular Figures. To find Cost of Painting or Tilling when Area and Cost per Unit is given. To find the Number of Bricks or Tiles when Area of Path and Brick is given. Worksheet on Area. Worksheet on Area of a Square and Rectangle Practice Test on Area.
You’ll find lots of triangles on the PSAT/NMSQT, especially right triangles. The Greeks weren’t the only mathematicians in the ancient world, but they managed to place their “brand” on geometry, a word which, by the way, comes from the Greek words for “earth measure.” Specifically, a mathematician named Pythagoras wrote the Pythagorean Theorem: a2 + b2 = c2 You can use this formula to find the sides of any right triangle, in which a and b are defined as the two legs of the triangle and c is the hypotenuse, a fancy word for the side opposite the 90̊ angle. Note: This formula — the Pythagorean Theorem — appears in the information box on the exam. A few common right-triangle ratios are frequent fliers on the PSAT/NMSQT, so it’s worth memorizing them: • The 3:4:5 triangle: The sides can be any multiple of these numbers (for example, 15:20:25, with each side multiplied by 5, or 21:28:35, with each side multiplied by 7). • The 5:12:13 triangle: Strange numbers, huh? But this ratio behaves like any other, so you can multiply each side by 2 and get a 10:24:26 triangle, or multiply by 5 and get a 25:60:65 triangle. • The triangle: The s stands for a side, and because you have two sides that are equal (both are s), this is an isosceles right triangle, and the interior angles are 45°, 45°, and 90°. Note: This formula appears in the information box on the exam. You can use the information in the preceding bullet to calculate the diagonal (a line connecting opposite corners) of a square. If the sides of a square are 65 meters long, the diagonal is You can easily see why this formula works: A square is just two isosceles right triangles glued together, because each side of a square is the same length. • The triangle: This one has 30°-60°-90° angles, and for some reason, the exam-writers love it. The hypotenuse (the long side) is double the length of the side opposite the 30º angle. Note: This formula appears in the information box on the exam. If you cut an equilateral triangle (one with equal sides) in half, you get two 30°-60°-90° triangles. So if you see a question on the exam about an equilateral triangle, drag out this formula and you’ll find the answer in a flash. Stretch those triangular muscles! Try these problems: 1. An equilateral triangle has a side with a length of x. What is the area of the triangle, in terms of x? 2. What is the area of the following figure? (A) 6 (B) 15 (C) 32 (D) 36 (E) 42 3. In the following square, the product of diagonals AC and BD is 18. What is the perimeter of triangle ABC? (C) 18 (D) 24 (E) 36 1. B. Always draw a picture if you’re having any trouble visualizing a problem: Remember that you can transform all equilateral triangles into a pair of 30º-60º-90º triangles by cutting them in half. That lets you see that the base of one of the smaller triangles is x/2, and the height is making the area of the whole triangle equal 2. D. 36 How well do you know your Pythagorean triples? Either way, you can use the Pythagorean Theorem to help you solve this problem, or any problem with right triangles. First, look at the small triangle. Its area is Next, you can quickly solve for the hypotenuse using a2 + b2 = c2, b = 12 and determine that it’s 5 units long. Pythagorean theorem to the rescue again: 52 + b2 = 132, b = 12. That means that the area of the larger triangle is Add those two areas together: 6 + 30 = 36, and you can see that Choice (D) is correct. 3. A. Diagonals AC and BD must be the same length, so they’re each in length. Now you can ignore the square and just pay attention to triangle ABC, which is a 45̊-45̊-90̊ triangle, with a hypotenuse of Using your knowledge of special triangles (or the formula box), you know that the legs of the triangle must each be 3 units long. Therefore, the perimeter of the triangle is
Divisibility tests allow us to calculate whether a number can be divided by another number. For example, can 354 be divided by 3? Can 247,742 be divided by 11? So what are the rules behind divisibility tests, and more interestingly, how can we prove them? Divisibility rule for 3 The most well known divisibility rule is that for dividing by 3. All you need to do is add the digits of the number and if you get a number that is itself a multiple of 3, then the original number is divisible by 3. For example, 354 is divisible by 3 because 3+5+4 = 12 and 12 can be divided by 3. We can prove this using the modulo function. This allows us to calculate the remainder when any number is divided by another. For example, 21 ≡ 3 (mod 6). This means that the remainder when 21 is divided by 6 is 3. So we first start with our number that we want to divide by 3: n = a + 10b + 100c + 1000d + ……. Now, dividing by 3 is the same as working in mod 3, so we can rewrite this n in terms of mod 3: n = a + 1b + 1c + 1d + ……. (mod 3) (this is because 10 ≡ 1 mod 3, 100 ≡ 1 mod 3, 1000 ≡ 1 mod 3 etc) Now, for a number to be divisible by 3 this sum needs to add to a multiple of 3. Therefore if a + b + c + d + …. ≡ 0 (mod 3) then the original number is also divisible by 3. Divisibility rule for 11 This rule is much less well known, but it’s quite a nice one. Basically you take the digits of any number and alternately subtract and add them. If the answer is a multiple of 11 (or 0) then the original number is divisible by 11. For example, 121 is a multiple of 11 because 1-2+1 = 0. 247,742 is because 2-4+7-7+4-2 = 0. Once again we can prove this using the modulo operator. n = a + 10b +100c + 1000d + ….. and this time work in mod 11: n = a + -1b +1c + -1d + ….. (mod 11) this is because for ease of calculation we can write 10 ≡ -1 (mod 11). This is because -1 ≡ 10 ≡ 21 ≡ 32 ≡ 43 (mod 11). All numbers 11 apart are the same in mod 11. Meanwhile 100 ≡ 1 (mod 11). This alternating pattern will continue. Therefore if we alternately subtract and add digits, then if the answer is divisible by 11, then the original number will be as well. Palindromic Numbers Palindromic numbers are numbers which can be read the same forwards as backwards. For example, 247,742 is a palindromic number, as is 123,321. Any palindromic number which is an even number of digits is also divisible by 11. We can see this by considering (for example) the number: n = a + 10b + 100c + 1000c + 10,000b + 100,000a Working in mod 11 we will then get the same pattern as previously: n = a – b +c – c +b – a (mod 11) so n = 0 (mod 11). Therefore n is divisible by 11. This only works for even palindromic numbers as when the numbers are symmetric they cancel out.
In this explainer, we will learn how to approximate the nth root of a number and then use it to solve real-world problems. A perfect square is the square root of a square number; for example, . The square root of a fraction made of perfect squares is a fraction; for instance, . But the square root of a number that is neither a perfect square nor a fraction made of perfect squares is an irrational number. ### Definition: Irrational Numbers An irrational number is a real number that cannot be expressed as a simple fraction (i.e., whose numerator and denominator are integers). Recall, when seeking to find the square root of a number, that it is equivalent to finding the length of the side of the square whose area is equal to this number. In this explainer, we will develop a method to estimate the value of an irrational square root. Let us take , for example. A square with side length has an area of 3. To estimate the square root, we find the square of closest size, whose area is a perfect square. The number 3 is between the square numbers 1 and 4. As 3 is closer to 4 than 1, the closest square in size to the square of area 3 is the square of area 4 and side 2. Therefore, the nearest integer to is 2. ### How To: Estimating the Value of the Square Root of an Irrational Square Root to the Nearest Integer Let us consider a square of area . Its side is . To estimate the value of , we need to find the nearest square number to , which we call SN. SN is either larger or smaller than . The value of can then be estimated as . Remember that since SN is a square number, its square root is an integer. In the first example, we will look at how we can identify the relative positions of square roots on a number line. ### Example 1: Placing Square Roots on a Number Line The positions of the numbers , , and have been identified on the number line. By considering their size, decide which number is represented by . The numbers , , and are the sides of three squares. We have ; therefore, the corresponding areas are in the same order: . Since , we deduce that ### Example 2: Finding Which Irrational Square Root Is Closest to Five Which of the following numbers is closest to 5? All the proposed answers are in the form . Therefore, we are looking for the square root of number so that is close to 5. If we imagine two squares of sides and 5, their areas are and 25. Among the proposed numbers in the form , we need to identify where is closest to 25. ### Example 3: Estimating the Value of an Irrational Square Root to the Nearest Integer The formula for the area of a square is , where is the side length. Estimate the side length of a square whose area is 74 square inches. We need here to estimate . We therefore need to find the closest square number to 74. Considering and , we find that the square number closest to 74 is 81 since . The nearest integer to is thus 9. Our estimate of the length of the square of area 74 square inches is 9 inches. Let us now consider the two consecutive square numbers, and , between which a number lies. We have and thus It means that we can always find the two consecutive integers ( and are necessarily consecutive integers) that a square root lies between. ### Example 4: Finding the Irrational Number That Lies between Two and Three Which of the following is an irrational number that lies between 2 and 3? 1. 2.6 We are looking for an irrational square root that lies between 2 and 3, that is, the square root of number so that . If we think of these three numbers as being the lengths of the sides of three squares, we can write about their respective areas that . This is of course equivalent to simply squaring the three numbers. Among the proposed answers, only answers B, C, and D are irrational numbers. And B is the only option whose square is between 4 and 9. So far, we have found how to find the two consecutive integers between which a square root lies. We are going to see how we can find more accurate estimates. Let us consider, for instance, . We have so Remember that and 5 represent the side and the area of a square. We can find the areas of the squares with sides 2.1, 2.2, 2.3, and so on, up to 3. We see that , which means that We can apply exactly the same reasoning to be able to frame between two consecutive two-decimal-place numbers. However, we can avoid working out all the areas of the squares of sides 2.21, 2.22, 2.23, and so on, up to 2.3. Let us see how. The idea is to work out the area of one square, possibly close to where we believe the square of side lies. Since 5 is closer to 4.89 than 5.29, we can expect to be closer to 2.2 than to 2.3. So, we can work out, for instance, the area of the square of side 2.23, that is, . We find 4.9729. This value of 4.9729 is less than 5, which means that 2.23 is less than . Let us work out ; it gives 5.0176. Since we have Here, we go from one inequality (about areas) to the other (about squares’ sides) by taking the square root of the three numbers. We can of course apply the same reasoning again if we want to find between which two consecutive three-decimal-place numbers lies, and so forth. Let us now apply this method in the following example. ### Example 5: Finding between Which Two Consecutive One-Decimal-Place Numbers a Square Root Lies Find the two consecutive one-decimal-place numbers that lies between. We want to find between which two consecutive one-decimal-place numbers lies. Remember that is the side of the square of area 151. We need to find first the two closest square numbers between which 151 lies. We find that which can be interpreted as the area of the square of side that lies between the areas of the squares of sides and . Therefore, we have Now, to find between which two consecutive one-decimal-place numbers lies, we are going to work out the area of a square whose side is a one-decimal-place number between 12 and 13. As 151 is closer to 144 than 169, we can pick a one-decimal-place number closer to 12 than to 13, for instance, 12.4. We find that This value is larger than 151. Therefore, we need to check the area of the square of side 12.3. We have We see that this area is closer to 151 but still higher than 151. Let us work out the area of the square of side 12.2; this is Now, we can write that and therefore ### Key Points • The square root of a square number, also called a perfect square, is an integer; for instance, . The square root of a fraction made of square numbers is a fraction, for instance, . But the square root of a number that is neither a square number nor a fraction made of square numbers is an irrational number. • An irrational number is a real number that cannot be expressed as a simple fraction (i.e., whose numerator and denominator are integers). • The value of an irrational square root can be approximated first by finding the two nearest square numbers to , and . We have and thus We can approximate the value of more accurately by finding between which two consecutive one-decimal-place numbers lies. For this, we consider the 9 squares whose side lengths are comprised between the two consecutive integers and , where each side is 0.1 larger than the side of the previous square. The area of each square can be calculated by squaring its side. Then, we identify between which two square areas lies, from which we deduce that lies between their two side lengths. • For nonnegative numbers , , and , if we can conclude that Furthermore, we can go from the bottom expression to the top. Hence, we can either take the square of or the square root of the three numbers or expressions in the inequality.
HomeTren&dHow Many Edges Does a Cube Have? How Many Edges Does a Cube Have? A cube is a three-dimensional geometric shape that is composed of six square faces, twelve edges, and eight vertices. In this article, we will focus on exploring the number of edges a cube has and delve into the properties and characteristics of this fascinating shape. The Definition of a Cube Before we dive into the number of edges a cube possesses, let’s first establish a clear definition of what a cube is. A cube is a regular polyhedron, which means it has congruent faces and identical angles between faces. In the case of a cube, all six faces are squares, and each face meets at a right angle with the adjacent faces. The Anatomy of a Cube To understand the number of edges a cube has, it is essential to familiarize ourselves with the different components of this shape: • Faces: A cube has six faces, and each face is a square. All the faces are congruent, meaning they have the same size and shape. • Edges: A cube has twelve edges. Each edge is a line segment where two faces meet. • Vertices: A cube has eight vertices, which are the points where three edges meet. Calculating the Number of Edges Now that we understand the components of a cube, let’s calculate the number of edges it has. Since each edge is formed by the intersection of two faces, we can determine the number of edges by counting the number of intersections. For a cube, each face shares an edge with four other faces. Since there are six faces in total, we can calculate the number of edges by multiplying the number of faces by the number of shared edges: Number of edges = Number of faces × Number of shared edges Number of edges = 6 faces × 4 shared edges Number of edges = 24 edges Therefore, a cube has 24 edges in total. Visualizing the Edges of a Cube It can be helpful to visualize the edges of a cube to gain a better understanding of their arrangement. Imagine a cube placed on a flat surface, with one face at the bottom. The edges can be seen as the lines connecting the corners of the cube. Each edge connects two vertices and forms a straight line segment. By following the edges, we can trace the outline of the cube and see how the faces are connected. Real-World Examples Cubes are not only mathematical abstractions but also appear in various real-world objects and structures. Let’s explore a few examples: • Dice: A traditional die used in board games is a cube with each face displaying a different number of dots. • Rubik’s Cube: This popular puzzle toy consists of smaller cubes arranged in a 3×3 grid, forming a larger cube. The edges of the smaller cubes align to create the edges of the entire Rubik’s Cube. • Building Blocks: Children often play with building blocks that are shaped like cubes. These blocks can be stacked and connected using their edges. Q&A 1. Can a cube have curved edges? No, a cube cannot have curved edges. By definition, a cube has straight edges that form right angles with the adjacent faces. 2. How many edges does a rectangular prism have? A rectangular prism has twelve edges, just like a cube. However, unlike a cube, a rectangular prism has rectangular faces instead of square faces. 3. What is the difference between an edge and a face? An edge is a line segment where two faces meet, while a face is a flat surface of a three-dimensional shape. 4. Can a cube have more than six faces? No, a cube cannot have more than six faces. A cube is a specific type of polyhedron with six congruent square faces. 5. How many edges does a triangular pyramid have? A triangular pyramid, also known as a tetrahedron, has six edges. Each edge connects two vertices of the pyramid. Summary In conclusion, a cube has a total of 24 edges. Each edge is a line segment formed by the intersection of two faces. Understanding the components and properties of a cube, such as its faces, edges, and vertices, allows us to visualize and comprehend this three-dimensional shape more effectively. Cubes can be found in various real-world objects and structures, from dice to Rubik’s Cubes, showcasing their practical applications beyond mathematics. By exploring the number of edges a cube has, we have gained valuable insights into the fundamental characteristics of this geometric shape. Whether you encounter a cube in a mathematical problem or in everyday life, you now have a solid understanding of its edge count and can appreciate its unique properties. Riya Sharma Riya Sharma is a tеch bloggеr and UX/UI dеsignеr spеcializing in usеr еxpеriеncе dеsign and usability tеsting. With еxpеrtisе in usеr-cеntric dеsign principlеs, Riya has contributеd to crafting intuitivе and visually appеaling intеrfacеs. RELATED ARTICLES
# Baseball: finding trajectory of ball • Naeem In summary, the conversation discusses a baseball hit by Ted Williams, who claims he could have hit it out of the stadium if there was no air resistance. The question posed asks for the maximum height of the stadium at its back wall, assuming Ted's statement is true. The necessary equations and units are also provided. Naeem Q.Ted Williams hits a baseball with an initial velocity of 120 miles per hour (176 ft/s) at an angle of q = 35 degrees to the horizontal. The ball is struck 3 feet above home plate. You watch as the ball goes over the outfield wall 420 feet away and lands in the bleachers. After you congratulate Ted on his hit he tells you, 'You think that was something, if there was no air resistance I could have hit that ball clear out of the stadium!' Assuming Ted is correct, what is the maximum height of the stadium at its back wall x = 565 feet from home plate, such that the ball would just pass over it? You may need: 9.8 m/s2 = 32.2 ft/s2 1 mile = 5280 ft h = feet Can somebody tell me what have to find step by step in this problem. The help provided with this problem isn't very helpful.. Thanks #### Attachments • Baseball.gif 3.9 KB · Views: 601 x(t) = ? y(t) = ? x(t1) = 565, t1 = ? y(t1) = ? for your question! Let's break down the steps needed to solve this problem: Step 1: Understand the given information We are given the initial velocity of the ball (120 miles per hour or 176 ft/s), the angle at which it was hit (35 degrees), and the distance it traveled (420 feet). We are also told that the ball was struck 3 feet above home plate. Step 2: Convert units Since the given velocity is in miles per hour and the distance is in feet, we need to convert the velocity to feet per second. We can do this by multiplying 120 miles per hour by 5280 feet per mile and dividing by 3600 seconds per hour. This gives us 176 ft/s, which is the same as the given velocity. Step 3: Find the time of flight Using the given initial velocity and angle, we can use the kinematic equations to find the time of flight of the ball. The equation we will use is t = (2vsin(q))/g, where v is the initial velocity, q is the angle, and g is the acceleration due to gravity (32.2 ft/s^2). Plugging in the given values, we get t = (2176sin(35))/32.2 = 3.35 seconds. Step 4: Find the maximum height To find the maximum height, we can use the equation h = v^2sin^2(q)/(2g), where v is the initial velocity, q is the angle, and g is the acceleration due to gravity. Plugging in the given values, we get h = (176^2sin^2(35))/(232.2) = 89.19 feet. This means that the maximum height of the ball's trajectory is 89.19 feet. Step 5: Find the distance traveled by the ball at the maximum height To find the distance traveled by the ball at the maximum height, we can use the equation d = v^2sin(2q)/g, where v is the initial velocity, q is the angle, and g is the acceleration due to gravity. Plugging in the given values, we get d = (176^2sin(2*35))/32.2 = 563.56 feet. This means that the ball traveled 563.56 feet horizontally at the maximum height. Step 6: Find the height of the stadium ## 1. How do you calculate the trajectory of a baseball? The trajectory of a baseball can be calculated using the equations of motion, which take into account the initial velocity and angle of the ball, as well as the effects of gravity and air resistance. ## 2. What factors affect the trajectory of a baseball? The trajectory of a baseball is affected by the initial velocity and angle of the ball, as well as the effects of gravity, air resistance, and any external forces such as wind or spin on the ball. ## 3. Can the trajectory of a baseball be predicted accurately? While the trajectory of a baseball can be calculated using mathematical equations, it is difficult to predict with complete accuracy due to variables such as air resistance, wind, and the spin of the ball. However, advanced technology and modeling techniques can greatly improve the accuracy of trajectory predictions. ## 4. How does the surface of the baseball affect its trajectory? The surface of the baseball, specifically the stitches and seams, can affect its trajectory by creating turbulence and altering the air resistance on the ball. This is why pitchers often manipulate the grip and spin of the ball to create different trajectories. ## 5. Can the trajectory of a baseball be altered during its flight? Yes, the trajectory of a baseball can be altered during its flight through external forces such as wind or spin, or through the intentional manipulation of the ball's grip and release by the pitcher. Additionally, the trajectory can also be affected by collisions with other objects, such as a player's glove or the ground. ### Similar threads • Introductory Physics Homework Help Replies 4 Views 18K • General Math Replies 5 Views 2K • Introductory Physics Homework Help Replies 4 Views 2K • Introductory Physics Homework Help Replies 6 Views 4K • Introductory Physics Homework Help Replies 2 Views 2K • Introductory Physics Homework Help Replies 25 Views 7K • Introductory Physics Homework Help Replies 4 Views 2K • Introductory Physics Homework Help Replies 7 Views 5K • Introductory Physics Homework Help Replies 1 Views 3K • Introductory Physics Homework Help Replies 2 Views 3K
Love to Cut and Paste A Twitter post we want to share! We love @RightStartMath because we get to cut and glue Thanks for sharing with us @ServingFromHome Valentine’s Day Applied Math In a small school district in North Dakota, elementary students make Valentine’s Day card boxes to collect cards from their friends. Some students just write their name on their boxes, others go all out. Here’s what Kylee created. Notice how she has the planets proportionate in distance and size? Hmmmm. Wonder if she used math? Good job, Kylee! What are Check Numbers? Check numbers are a method of checking addition. Sometimes, this is called Casting out Nines. Check numbers also works with subtraction, multiplication, and division. I like to think of check numbers as a cool tool for my math toolbox. Some people use check numbers frequently, others, not so much. However, if we don’t let people know about these cool things, we’ll never know who might use them! Let’s look at how to find check numbers, then how to apply them. We also have a presentation on check numbers for you to review. Check numbers are first taught in RightStart™ Mathematics Level D, starting in lesson 47, and Level E, lesson 4, and Math Card Games, game #A63. Finding Simple Check Numbers Check numbers are one digit numbers from 0 to 8. We will designate the check numbers by using parenthesis. Add the digits together: 1 + 7 = 8 Check number of 17 is (8). Now let’s try another: 49 Add the digits together: 4 + 9 = 13 Remember check numbers are only one digit, so we’ll need to take the 13 found above, and continue to add the digits together: 1 + 3 = 4 Check number of 49 is (4). Another: 99 Add the digits together: 9 + 9 = 18 And again: 1 + 8 = 9 However, remember we said that check numbers are from 0 to 8? There are no 9s. Now what? Well, all 9s are 0s. So, on this example, we have 1 + 8 = 9, and 9 = 0. Check number of 99 is (0). Because all 9s = 0s, we have a quick shortcut to help find check number. Let’s back up to our second example: 49 If 9 = 0, then it looks like this: 4 + 0 = (4), which is what we had the “long” way. Neat, right? Let’s reapply to our third example: 99 Well, that’d be: 0 + 0 = (0) Remember the other name for Check Numbers is Casting Out Nines. If we “cast” the 9s out, which is the same as 0, our work is simplified! Finding More Check Numbers Let’s find check numbers with a four-digit number: 4639 Add the digits together: 4 + 6 + 3 + 9 = 22 And again: 2 + 2 = 4 Check number of 4639 is (4). Let’s try this again using some of our newly discovered shortcuts. Remember, 9 = 0. 4639 We can “cast out” the 9, so now we have: 4 + 6 + 3 + 0 But 6 + 3 = 9, so let’s “cast” that out too! 4 + 0 + 0 + 0 = 4 Well! That was easy! Check number of 4639 is quickly found as (4). Another one: 7326 See anything to “cast”? How about 7 & 2 and 3 & 6? Check number is (0). Applying Check Numbers So now that we can find check numbers, let’s use them! Consider the following equation: . 4639 + 7326 . 11965 If you’re like me, you wonder if you added it correctly and often will double check either by recalculating and/or checking on a calculator. We can check accuracy by using check numbers! So, figure the check numbers: . 4639 (4) + 7326 (0) . 11965 (4) Look at the check numbers! (4)+(0)=(4)! Let’s do another: . 364 + 4426 Calculate the answer then calculate the check numbers. Did you do it right? Should look like this: . 364 (4) + 4426 (7) . 4790 (2) and the check numbers are correct too. Now, let’s assume you came up with a wrong sum (which happens) and it looked like this: . 364 (4) + 4426 (7) . 4780 (1) ERROR Notice now how the check numbers don’t add up. (4) + (7) does not equal (1). This becomes our check! We now know something is wrong and needs to be corrected. More Applying Check Numbers As we can see, check numbers are a method of checking and verifying addition calculations. If the check numbers are not adding up, the answer is probably wrong. Remember that check numbers work with subtraction, multiplication, and division? We’re going to save that for another post. Meanwhile, play around and see what you discover! Stay tuned……
# How to Find Distance if You Have Velocity and Time ### TheHOUSERule Think of a house with a desk $d$ in the attic, and a sofa $s$ and a table $t$ in the living room. Hold your finger over the thing you want to find, for example the time $t$. Then there is a $d$ above an $s$, giving you that $t=\frac{\text{d}}{\text{s}}$. A house that illustrates the relationship between distance, speed and time. Another word for speed is velocity. Rule ### Distance,Speed,Time–theHOUSERule $d=s\cdot t\phantom{\rule{2em}{0ex}}t=\frac{d}{s}\phantom{\rule{2em}{0ex}}s=\frac{d}{t}$ Rule ### UnitsofDistance,SpeedandTime Distance: meters m or kilometers km, Time: seconds s or hours h Then the units of speed are: ### Converting from m/s to km/h and Vice Versa $\begin{array}{llll}\hfill \text{km/h}& =\text{m/s}\cdot 3.6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{m/s}& =\text{km/h}÷3.6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ $\text{km/h}=\text{m/s}\cdot 3.6\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\text{m/s}=\text{km/h}:3.6$ Remember that the distance is how far you have moved. Example 1 A car is driving at a speed of $\text{}80\text{}\phantom{\rule{0.17em}{0ex}}\text{km/h}$. How long does it take to drive $\text{}470\text{}\phantom{\rule{0.17em}{0ex}}\text{km}$? $\begin{array}{lll}\hfill t=\frac{d}{s}=\frac{470\phantom{\rule{0.17em}{0ex}}\text{km}}{80\phantom{\rule{0.17em}{0ex}}\text{km/h}}=5.875\phantom{\rule{0.17em}{0ex}}\text{h}& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$ Then you need to convert the decimals to find the minutes and seconds. From the answer you can see that it takes $5$ full hours. To find the number of minutes, multiply the decimal by $60$. That gives you $0.875\cdot 60=52.5$ Next, you find the number of seconds: $0.5\cdot 60=30$ This means it takes $5$ hours, $52$ minutes and $30$ seconds to drive $470$ km at $80$ km/h. Example 2 A plane is flying at a speed of $\text{}250\text{}\phantom{\rule{0.17em}{0ex}}\text{m/s}$. How far will it fly in 3 hours? $\begin{array}{llll}\hfill 250\phantom{\rule{0.17em}{0ex}}\text{m/s}\cdot 3.6& =900\phantom{\rule{0.17em}{0ex}}\text{km/h}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill d& =s\cdot t\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =900\phantom{\rule{0.17em}{0ex}}\text{km/h}\cdot 3\phantom{\rule{0.17em}{0ex}}\text{h}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2700\phantom{\rule{0.17em}{0ex}}\text{km}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ The plane will fly $2700$ km in $3$ hours. Example 3 In 2009, Usain Bolt ran 100 meters in $\text{}9.58\text{}$ seconds. How fast did he run on average? Give the answer in both $\text{m/s}$ and $\text{km/h}$. $\begin{array}{llll}\hfill s& =\frac{d}{t}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{100\phantom{\rule{0.17em}{0ex}}\text{m}}{9.58\phantom{\rule{0.17em}{0ex}}\text{s}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 10.44\phantom{\rule{0.17em}{0ex}}\text{m/s}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(10.44\cdot 3.6\right)\phantom{\rule{0.33em}{0ex}}\text{km/h}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 37.58\phantom{\rule{0.17em}{0ex}}\text{km/h}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ He ran at an average speed of $10.44$ m/s, which is pretty close to $37.58$ km/h!
```Statistics Notes   In the previous section, you learned how to find the probability of two events, A and B, occurring in sequence. Such probabilities are denoted by P(A and B). (ex – flipping a heads and rolling a 3) In this section, you will learn how to find the probability that at least one of two events will occur. Probabilities such as these are denoted by P(A or B) and depend on whether the events are mutually exclusive. Two events A and B are mutually exclusive if A and B cannot occur at the same time.  The Venn diagrams show the relationship between events that are mutually exclusive and events that are not mutually exclusive.  1. Event A: Roll a 3 on a die. Event B: Roll a 4 on a die. 2. Event A: Randomly select a male student. E vent B: Randomly select a nursing major. 3. Event A: Randomly select a blood donor with type “0” blood. Event B: Randomly select a female blood donor. 1. Event A: Randomly select a jack from a standard deck of cards Event B: Randomly select a face card from a standard deck of cards 2. Event A: Randomly select a 20 -year.- old student. Event B: Randomly select a student with blue eyes. 3. Event A: Randomly select a vehicle that is a Ford. Event B. Randomly select a vehicle that is a Toyota.   If events A and B are mutually exclusive, then the rule can be simplified to  P(A or B) = P(A) + P(B)   This simplified rule can be extended to any number of mutually exclusive events.  The probability that events A or B will occur P(A or B) is given by  P(A or B) = P(A) + P(B) - P(A and B).    A box contains 3 glazed doughnuts, 4 jelly doughnuts, and 5 chocolate doughnuts. If a person selects a doughnut at random, find the probability that it is either a glazed doughnut or a chocolate doughnut. P(glazed or chocolate)= P(glazed) + P (Chocolate)= 3/12 + 5/12 = 8/12 = 2/3 The events are Mutually Exclusive.    At a political rally, there are 20 Republicans, 13 Democrats, and 6 independents. If a person is selected at random, find the probability that he or she is either Democrat or Republican. P(Democrat or Independent)= P(Democrat) + P(Independent) =13/39 + 6/39 =19/39    A day of the week is selected at random. Find the probability that it is a weekend day. P(Saturday or Sunday) = P(Saturday) + P(Sunday) P(Saturday or Sunday) = 1/7 + 1/7 = 2/7    A single card is drawn at random from an ordinary deck of cards. Find the probability that it is either an ace or a black card. P(ace or black card) = P(ace) + P(black card) P(ace or black card) = 4/52 + 26/32 – 2/52 = 28/52 = 7/13    In a hospital units there are 8 nurses and 5 physicians; 7 nurses and 3 physicians are female. If a staff person is selected, find the probability that the subject is a nurse or a male. P(nurse or male) = P(nurse) + P(male) P(nurse or male) = 8/13 + 3/13 – 1/13= 10/13 On New Year’s Eve, the probability of a person driving while intoxicated is 0.32 the probability of a person having a driving accident is 0.09, and the probability of a person having a driving accident while intoxicated is 0.06. What is the probability of a person driving while intoxicated or having a driving accident?  P(intoxicated or accident) = P(intoxicated) + P ( accident) – P(intoxicated and accident)  =.32 + 0.09 – 0.06= 0.35   The frequency distribution shows the volume of sales (in dollars) and the number of months a sales representative reached each sales level during the past three years. If this sales pattern continues, what is the probability that the sales representative will sell between \$75,000 and \$124,999 next month?  Find the probability that the sales representative will sell between \$0 and \$49,999.  A blood bank catalogs the types of blood, including positive or negative Rh-factor, given by donors during the last five days. The number of donors who gave each blood type is shown in the table. A donor is selected at random.  Find the probability that the donor has type 0 or type A blood.  Find the probability that the donor has type B blood or is Rhnegative.  Find the probability that the donor has type B or type AB blood.  Find the probability that the donor has type O blood or is Rh-positive.  Use the graph at the right to find the probability that a randomly selected draft pick is not a running back or a wide
# 0.8 Exponential functions and graphs (Page 2/2) Page 2 / 2 Therefore, if $a<0$ , then the range is $\left(-\infty ,q\right)$ , meaning that $f\left(x\right)$ can be any real number less than $q$ . Equivalently, one could write that the range is $\left\{y\in \mathbb{R}:y . For example, the domain of $g\left(x\right)=3·{2}^{x+1}+2$ is $\left\{x:x\in \mathbb{R}\right\}$ . For the range, $\begin{array}{ccc}\hfill {2}^{x+1}& >& 0\hfill \\ \hfill 3·{2}^{x+1}& >& 0\hfill \\ \hfill 3·{2}^{x+1}+2& >& 2\hfill \end{array}$ Therefore the range is $\left\{g\left(x\right):g\left(x\right)\in \left[2,\infty \right)\right\}$ . ## Domain and range 1. Give the domain of $y={3}^{x}$ . 2. What is the domain and range of $f\left(x\right)={2}^{x}$ ? 3. Determine the domain and range of $y={\left(1,5\right)}^{x+3}$ . ## Intercepts For functions of the form, $y=a{b}^{\left(x+p\right)}+q$ , the intercepts with the $x$ - and $y$ -axis are calculated by setting $x=0$ for the $y$ -intercept and by setting $y=0$ for the $x$ -intercept. The $y$ -intercept is calculated as follows: $\begin{array}{ccc}\hfill y& =& a{b}^{\left(x+p\right)}+q\hfill \\ \hfill {y}_{int}& =& a{b}^{\left(0+p\right)}+q\hfill \\ & =& a{b}^{p}+q\hfill \end{array}$ For example, the $y$ -intercept of $g\left(x\right)=3·{2}^{x+1}+2$ is given by setting $x=0$ to get: $\begin{array}{ccc}\hfill y& =& 3·{2}^{x+1}+2\hfill \\ \hfill {y}_{int}& =& 3·{2}^{0+1}+2\hfill \\ & =& 3·{2}^{1}+2\hfill \\ & =& 3·2+2\hfill \\ & =& 8\hfill \end{array}$ The $x$ -intercepts are calculated by setting $y=0$ as follows: $\begin{array}{ccc}\hfill y& =& a{b}^{\left(x+p\right)}+q\hfill \\ \hfill 0& =& a{b}^{\left({x}_{int}+p\right)}+q\hfill \\ \hfill a{b}^{\left({x}_{int}+p\right)}& =& -q\hfill \\ \hfill {b}^{\left({x}_{int}+p\right)}& =& -\frac{q}{a}\hfill \end{array}$ Since $b>0$ (this is a requirement in the original definition) and a positive number raised to any power is always positive, the last equation above only has a real solution if either $a<0$ or $q<0$ (but not both). Additionally, $a$ must not be zero for the division to be valid. If these conditions are not satisfied, the graph of the function of the form $y=a{b}^{\left(x+p\right)}+q$ does not have any $x$ -intercepts. For example, the $x$ -intercept of $g\left(x\right)=3·{2}^{x+1}+2$ is given by setting $y=0$ to get: $\begin{array}{ccc}\hfill y& =& 3·{2}^{x+1}+2\hfill \\ \hfill 0& =& 3·{2}^{{x}_{int}+1}+2\hfill \\ \hfill -2& =& 3·{2}^{{x}_{int}+1}\hfill \\ \hfill {2}^{{x}_{int}+1}& =& \frac{-2}{2}\hfill \end{array}$ which has no real solution. Therefore, the graph of $g\left(x\right)=3·{2}^{x+1}+2$ does not have a $x$ -intercept. You will notice that calculating $g\left(x\right)$ for any value of $x$ will always give a positive number, meaning that $y$ will never be zero and so the graph will never intersect the $x$ -axis. ## Intercepts 1. Give the y-intercept of the graph of $y={b}^{x}+2$ . 2. Give the x- and y-intercepts of the graph of $y=\frac{1}{2}{\left(1,5\right)}^{x+3}-0,75$ . ## Asymptotes Functions of the form $y=a{b}^{\left(x+p\right)}+q$ always have exactly one horizontal asymptote. When examining the range of these functions, we saw that we always have either $y or $y>q$ for all input values of $x$ . Therefore the line $y=q$ is an asymptote. For example, we saw earlier that the range of $g\left(x\right)=3·{2}^{x+1}+2$ is $\left(2,\infty \right)$ because $g\left(x\right)$ is always greater than 2. However, the value of $g\left(x\right)$ can get extremely close to 2, even though it never reaches it. For example, if you calculate $g\left(-20\right)$ , the value is approximately 2.000006. Using larger negative values of $x$ will make $g\left(x\right)$ even closer to 2: the value of $g\left(-100\right)$ is so close to 2 that the calculator is not precise enough to know the difference, and will (incorrectly) show you that it is equal to exactly 2. From this we deduce that the line $y=2$ is an asymptote. ## Asymptotes 1. Give the equation of the asymptote of the graph of $y={3}^{x}-2$ . 2. What is the equation of the horizontal asymptote of the graph of $y=3{\left(0,8\right)}^{x-1}-3$ ? ## Sketching graphs of the form $f\left(x\right)=a{b}^{\left(x+p\right)}+q$ In order to sketch graphs of functions of the form, $f\left(x\right)=a{b}^{\left(x+p\right)}+q$ , we need to determine four characteristics: 1. domain and range 2. $y$ -intercept 3. $x$ -intercept For example, sketch the graph of $g\left(x\right)=3·{2}^{x+1}+2$ . Mark the intercepts. We have determined the domain to be $\left\{x:x\in \mathbb{R}\right\}$ and the range to be $\left\{g\left(x\right):g\left(x\right)\in \left(2,\infty \right)\right\}$ . The $y$ -intercept is ${y}_{int}=8$ and there is no $x$ -intercept. ## Sketching graphs 1. Draw the graphs of the following on the same set of axes. Label the horizontal asymptotes and y-intercepts clearly. 1. $y={b}^{x}+2$ 2. $y={b}^{x+2}$ 3. $y=2{b}^{x}$ 4. $y=2{b}^{x+2}+2$ 1. Draw the graph of $f\left(x\right)={3}^{x}$ . 2. Explain where a solution of ${3}^{x}=5$ can be read off the graph. ## End of chapter exercises 1. The following table of values has columns giving the $y$ -values for the graph $y={a}^{x}$ , $y={a}^{x+1}$ and $y={a}^{x}+1$ . Match a graph to a column. $x$ A B C -2 7,25 6,25 2,5 -1 3,5 2,5 1 0 2 1 0,4 1 1,4 0,4 0,16 2 1,16 0,16 0,064 2. The graph of $f\left(x\right)=1+a.{2}^{x}$ (a is a constant) passes through the origin. 1. Determine the value of $a$ . 2. Determine the value of $f\left(-15\right)$ correct to FIVE decimal places. 3. Determine the value of $x$ , if $P\left(x;0,5\right)$ lies on the graph of $f$ . 4. If the graph of $f$ is shifted 2 units to the right to give the function $h$ , write down the equation of $h$ . 3. The graph of $f\left(x\right)=a.{b}^{x}\phantom{\rule{3.33333pt}{0ex}}\left(a\ne 0\right)$ has the point P(2;144) on $f$ . 1. If $b=0,75$ , calculate the value of $a$ . 2. Hence write down the equation of $f$ . 3. Determine, correct to TWO decimal places, the value of $f\left(13\right)$ . 4. Describe the transformation of the curve of $f$ to $h$ if $h\left(x\right)=f\left(-x\right)$ . Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? why we need to study biomolecules, molecular biology in nanotechnology? ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. why? what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe anyone know any internet site where one can find nanotechnology papers? research.net kanaga sciencedirect big data base Ernesto Introduction about quantum dots in nanotechnology what does nano mean? nano basically means 10^(-9). nanometer is a unit to measure length. Bharti do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment? absolutely yes Daniel how to know photocatalytic properties of tio2 nanoparticles...what to do now it is a goid question and i want to know the answer as well Maciej Abigail for teaching engĺish at school how nano technology help us Anassong Do somebody tell me a best nano engineering book for beginners? there is no specific books for beginners but there is book called principle of nanotechnology NANO what is fullerene does it is used to make bukky balls are you nano engineer ? s. fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball. Tarell what is the actual application of fullerenes nowadays? Damian That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes. Tarell what is the Synthesis, properties,and applications of carbon nano chemistry Mostly, they use nano carbon for electronics and for materials to be strengthened. Virgil is Bucky paper clear? CYNTHIA carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc NANO so some one know about replacing silicon atom with phosphorous in semiconductors device? Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure. Harper Do you know which machine is used to that process? s. how to fabricate graphene ink ? for screen printed electrodes ? SUYASH What is lattice structure? of graphene you mean? Ebrahim or in general Ebrahim in general s. Graphene has a hexagonal structure tahir On having this app for quite a bit time, Haven't realised there's a chat room in it. Cied what is biological synthesis of nanoparticles how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good Other chapter Q/A we can ask
# Find the x that makes <2,x> and <4,-6> parallel? Aug 9, 2018 $x = - 8$ #### Explanation: Recall.. For $x$ to be parallel it means the gradients must be equal to $1$ $m = 1$ ${m}_{1} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$ ${y}_{2} = - 6$ ${y}_{1} = x$ ${x}_{2} = 4$ ${x}_{1} = 2$ Plugging in the values.. ${m}_{1} = \frac{\left(- 6\right) - x}{4 - 2} = \frac{- 6 - x}{2}$ Therefore; $\frac{- 6 - x}{2} = 1$ $\frac{- 6 - x}{2} = \frac{1}{1}$ Cross multiply; $\left(- 6 - x\right) \times 1 = 1 \times 2$ $- 6 - x = 2$ $- x = 2 + 6$ $- x = 8$ $x = - 8$ Aug 9, 2018 $x = - 3$. #### Explanation: By definition, two non-null vectors vecu, &, vecv" are parallel "iff vecu=kvecv" for some "k in RR-{0}. $\therefore \left(2 , x\right) = k \left(4 , - 6\right) \Rightarrow 4 k = 2 , - 6 k = x$. $\therefore k = \frac{2}{4} = \frac{1}{2}$. $\therefore x = - 6 k = - 6 \cdot \frac{1}{2} = - 3$.
# Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.1 ## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.1 Question 1. Which of the following are sets? (i) The Collection of prime numbers upto 100. (ii) The Collection of rich people in India. (iii) The Collection of all rivers in India. (iv) The Collection of good Hockey players. Solution: (i) A = {2, 3, 5, 7,11, 13,17,19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97} As the collection of prime numbers upto 100 is known and can be counted (well defined). Hence this is a set. (ii) The collection of rich people in India. Rich people has no definition. Hence, it is not a set. (iii) A = {Cauvery, Sindhu, Ganga, } Hence, it is a set. (iv) The collection of good hockey players is not a well – defined collection because the criteria for determining a hockey player’s talent may vary from person to person. Hence, this collection is not a set. Question 2. Listthe set of letters of the following words in Roster form. (i) INDIA (ii) PARALLELOGRAM (iii) MISSISSIPPI (iv) CZECHOSLOVAKIA Solution: (i) A = {I, N, D, A} (ii) B = {P, A, R, L, E, O, G, M} (iii) C = {M, I, S, P} (iv) D = {C, Z, E, H, O, S, L, V, A, K, I}. Question 3. Consider the following sets A = {0, 3, 5, 8} B = {2, 4, 6, 10} C = {12, 14, 18, 20} (a) State whether True or false. (i) 18 ∈ C (ii) 6 ∉A (iii) 14 ∉ C (iv) 10 ∈ B (v) 5 ∈ B (vi) 0 ∈ B (b) Fill in the blanks? (i) 3 ∈ ___ (ii) 14 e ___ (iii) 18 ___ B (iv) 4 ___ B Solution: (a) (i) True (ii) True (iii) False (iv) True (v) False (vi) False, (b) (i) A (ii) C (iii) ∉ (iv) ∈ Question 4. Represent the following sets in Roster form. (i) A = The set of all even natural numbers less than 20. (ii) B = {y : y = $$\frac{1}{2 n}$$, n ∈ N, n ≤ 5} (iii) C = (x : x is perfect cube, 27 < x < 216} (iv) D = {x : x ∈ Z, -5 < x ≤ 2} Solution: (i) A= {2,4, 6, 8,10, 12, 14,16, 18} (ii) N = { 1, 2, 3, 4, 5} (iii) C = {64, 125} (iv) D = {-4,-3, -2, -1,0, 1, 2} Question 5. Represent the following sets in set builder form. (i) B = The set of all Cricket players in India who scored double centuries in One Day Internationals. (ii) C = { $$\frac{1}{2}, \frac{2}{3}, \frac{3}{4} \ldots . .$$}. (iii) D = The set of all tamil months in a year. (iv) E = The set of odd Whole numbers less than 9. Solution: (i) B = {x : x is an Indian player who scored double centuries in one day internationals} (ii) C = {x : x = $$\frac{n}{n+1}$$, n ∈ N} (iii) D = {x : x is a tamil month in a year} (iv) E = {x : x is odd number, x ∈ W, x < 9, where W is the set of whole numbers}. Question 6. Represent the following sets in descriptive form. (i) P = {January, June, July} (ii) Q = {7, 11, 13, 17, 19, 23, 29} (iii) R= {x : x ∈ N,x< 5} (iv) S = {x : x is a consonant in English alphabets} Solution: (i) P is the set of English Months begining with J. (ii) Q is the set of all prime numbers between 5 and 31. (iii) R is the set of all natural numbers less than 5. (iv) S is the set of all English consonants.
## Video and Examples There are many ways to solve a proportion. This page shows you three methods. $\frac{6}{9}\,=\,\frac{x}{12}$ Finding the missing value in a proportion is much like finding the missing value for two equal fractions. There are three main methods for determining whether two fractions (or ratios) are equivalent. • Vertical • Horizontal • Diagonal (often called "cross-products") Method 1: Vertical Are these two ratios equivalent? Since the numerator and denominator are related (by multiplying or dividing by 2), we know these two ratios are equivalent. Method 2: Horizontal Are these two ratios equivalent? Since the numerators are related (by multiplying or dividing by 3) to each other and the denominators are related to each other, we know these two ratios are equivalent. Method 3: Cross-products Are these two ratios equivalent? Since the cross-products are equal to each other, the two ratios are equivalent. All three methods work for every problem, but often one method is easier to use than the others. Always watch for the easiest method to use! Example 1 The following two ratios are equivalent. Find the missing value. Answer: Let's use the vertical method on this. Since 3 x 4 = 12, then 5 x 4 = 20. So the missing value is 20. Example 2 The following two ratios are equivalent. Find the missing value. Answer: We will use the horizontal method on this one. Since 5 x 2 = 10, then 4 x 2 = 8. So, the missing value is 8. Example 3 The following two ratios are equivalent. Find the missing value. Answer: In this problem it is easiest to look at the cross-products. Directions: 1. Look at the proportion on the right. 2. Solve it. (You may use the calculator below.) 3. Then enter your answer and press the "Enter" or "Return" key on your keyboard. 4. Press the "New problem" button to get a new problem. (Duh.) This free script provided by JavaScript Kit # Self-Check Q1: Find the missing value. $\,\,\,\frac{9}{12}=\frac{12}{m}\,\,\,$[show answer] Q2: Find the missing value. $\,\,\,\frac{18}{12}=\frac{k}{4}\,\,\,$[show answer] Q3: Find the missing value. $\,\,\,\frac{a}{24}=\frac{5}{20}\,\,\,$[show answer] Last modified: Friday, 17 April 2020, 10:23 AM
## Plus Blog February 19, 2013 Solving equations often involves taking square roots of numbers and if you're not careful you might accidentally take a square root of something that's negative. That isn't allowed of course, but if you hold your breath and just carry on, then you might eventually square the illegal entity again and end up with a negative number that's a perfectly valid solution to your equation. People first noticed this fact in the 15th century. A lot later on, in the 19th century, William Rowan Hamilton noticed that the illegal numbers you come across in this way can always be written as where and are ordinary numbers and stands for the square root of The number itself can be represented in this way with and Numbers of this form are called complex numbers. You can add two complex numbers like this: And you multiply them like this: The complex number 1+2i. But how can we visualise these numbers and their addition and multiplication? The and components are normal numbers so we can associate to them the point with coordinates on the plane, which is where you get to if you walk a distance in the horizontal direction and a distance in the vertical direction. So the complex number which is the sum of and corresponds to the point you get to by walking a distance in the horizontal direction and a distance in the vertical direction. Makes sense. What about multiplication? Think of the numbers that lie on your horizontal axis with coordinates Multiplying them by flips them over to the other side of the point : goes to goes to and so on. In fact, you can think of multiplication by as a rotation: you rotate the whole plane through 180 degrees about the point Multiplying by i. What about multiplication by the square root of ? Multiplying twice by is the same as multiplying by So if the latter corresponds to a rotation through 180 degrees, the former should correspond to rotation by 90 degrees. And this works. Try multiplying any complex number, say by and you will see that the result corresponds to the point you get to by rotating through 90 degrees (counter-clockwise) about And what about multiplying not just by but by a more difficult complex number Well, multiplying by an ordinary positive number corresponds to stretching or shrinking the plane: multiplication by 2 takes a point to which is further away from (that’s stretching) and multiplication by 1/2 takes it to which is closer to (shrinking). Multiplying by 2 is stretching. It turns out that multiplication by a complex number corresponds to a combination of rotation and shrinking/stretching. For example, multiplication by is rotation through 120 degrees followed by stretching by a factor of 2. So complex numbers are not just weird figments of the imagination designed to help you solve equations, they’ve got a geometric existence in their own right. You can find out more about complex numbers and things you can do with them in the Plus articles Curious quaternions, Unveiling the Mandelbrot set, Non-Euclidean geometry and Indra's pearls and Maths goes to the movies. February 17, 2013 Science advisors to government are an embattled lot. Remember the l'Aquila earthquake debacle or David Nutt's stance on drugs that cost him his job. Bridging the gap between politics and science isn't easy. Politicians like clear messages but science, and the reality it tries to describe, is rarely clear-cut. Full marks for Obama. So how do you advise a politician about science, its uncertainties and about risk? What better person to ask than John P. Holdren, Assistant to President Obama for Science and Technology. As he told a packed auditorium at the annual AAAS meeting in Boston, Holdren is actually quite happy with his own boss. Obama, he says, always wants to know the level of confidence scientists have in a specific result. And Holdren's first memo for Obama, which stuck to the traditional two pages, came back with "where's the rest?" scrawled over it. That's reassuring! Anne Glover, Chief Scientific Advisor to the European Commission, on the other hand, has come across leading European politicians who prefer to do without science advisors altogether, since "scientists never agree". Glover says that it's important to emphasise consent rather than disagreement. Scientists may be unsure or disagree about the details of something, say the exact relationship between CO2 emissions and global climate, but they may be certain about the big picture — that climate change is happening. Holdren points out that it's important to know where uncertainties are coming from — can they be sorted out with a bit more time and effort, or are they down to deeper gaps in or understanding or to processes we just can't pin down with better accuracy. Advisors should make sure uncertainties aren't exaggerated or understated, or simply ignored because they are too difficult to deal with. Both Glover and Holdren agree that it's important to speak plainly to politicians. Advisors should use examples and visualisations of uncertainties (see this Plus article for some ideas) and when there's a range of possible outcomes of something, say an epidemic, use scenarios to examine the possibilities. Holdren advises to look for a policy that remains robust in the face of all of them. It's important to be prudent, since new evidence may always come along. But when there is a large, coherent and consistent body of evidence, as there is with climate change, it's safe to talk in terms of certainty. For more on risk and uncertainty see our understanding uncertainty section. February 16, 2013 Sequences of numbers can have limits. For example, the sequence 1, 1/2, 1/3, 1/4, ... has the limit 0 and the sequence 0, 1/2, 2/3, 3/4, 4/5, ... has the limit 1. But not all number sequences behave so nicely. For example, the sequence 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 4/5, ... keeps jumping up and down, rather than getting closer and closer to one particular number. We can, however, discern some sort of limiting behaviour as we move along the sequence: the numbers never become larger than 1 or smaller than 0. And what's more, moving far enough along the sequence, you can find numbers that get as close as you like to both 1 and 0. So both 0 and 1 have some right to be considered limits of the sequence — and indeed they are: 1 is the limit superior and 0 is the limit inferior, so-called for obvious reasons. But can you define these limits superior and inferior for a general sequence , for example the one shown in the picture? Here’s how to do it for the limit superior. First look at the whole sequence and find its least upper bound: that’s the smallest number that’s bigger than all the numbers in the sequence. Then chop off the first number in the sequence, and again find the least upper bound for the new sequence. This might be smaller than the previous least upper bound (if that was equal to ), but not bigger. Then chop off the first two numbers and again find the least upper bound. Keep going, chopping off the first three, four, five, etc numbers, to get a sequence of least upper bounds (indicated by the red curve in the picture). In this sequence every number is either equal to or smaller than the number before. The limit superior is defined to be the limit of these least upper bounds. It always exists: since the sequence of least upper bounds is either constant or decreasing, it will either approach minus infinity or some other finite limit. The limit superior could also be equal to plus infinity, if there are numbers in the sequence that get arbitrarily large. The limit inferior is defined in a similar way, only that you look at the sequence of greatest lower bounds and then take the limit of that. You can read more about the limits inferior and superior in the Plus article The Abel Prize 2012. February 13, 2013 An infinite set is called countable if you can count it. In other words, it's called countable if you can put its members into one-to-one correspondence with the natural numbers 1, 2, 3, ... . For example, a bag with infinitely many apples would be a countable infinity because (given an infinite amount of time) you can label the apples 1, 2, 3, etc. Two countably infinite sets A and B are considered to have the same "size" (or cardinality) because you can pair each element in A with one and only one element in B so that no elements in either set are left over. This idea seems to make sense, but it has some funny consequences. For example, the even numbers are a countable infinity because you can link the number 2 to the number 1, the number 4 to 2, the number 6 to 3 and so on. So if you consider the totality of even numbers (not just a finite collection) then there are just as many of them as natural numbers, even though intuitively you'd think there should only be half as many. Something similar goes for the rational numbers (all the numbers you can write as fractions). You can list them as follows: first write down all the fractions whose denominator and numerator add up to 2, then list all the ones where the sum comes to 3, then 4, etc. This is an unfailing recipe to list all the rationals, and once they are listed you can label them by the natural numbers 1, 2, 3, ... . So there are just as many rationals as natural numbers, which again seems a bit odd because you'd think that there should be a lot more of them. It was Galileo who first noticed these funny results and they put him off thinking about infinity. Later on the mathematician Georg Cantor revisited the idea. In fact, Cantor came up with a whole hierarchy of infinities, one "bigger" than the other, of which the countable infinity is the smallest. His ideas were controversial at first, but have now become an accepted part of pure mathematics. You can find out more about all this in our collection of articles on infinity. February 11, 2013 A buckyball in Madison Square. Yesterday we opened the Plus New York office, amidst snow covered streets at the foot of the Empire State Building! The day started with a trip to MoMath, the recently opened maths museum in central New York. It was filled with a fascinating array of interactive exhibits demonstrating the beauty and playfulness of mathematics. And as one of the volunteers told us, playfulness is what it's all about. There were musical spheres demonstrating the maths of music, a fractal machine with cameras creating fractals from their surroundings, and a chance to discover the paths of mathematical rolling stones. It was full of children and the young at heart discovering the joy of maths for themselves. We also discovered an illuminated buckyball in the park just across from our hotel and the arithmetic of relationships in the High Line park. Maths is everywhere in NYC! The Institute for Advanced Study in Princeton. Today we had a very early start, taking the train from New York Penn Station to Princeton to visit the Institute for Advanced Studies. We were very lucky to speak with Freeman Dyson and Edward Witten about quantum field theory (QFT), the mathematical framework that has made much of the advancements in physics possible in the last century. This is the main reason for our trip to the States and we are looking forward to more interviews this week with other luminaries of theoretical physics to continue our series telling the story of QFT. You can read our first articles here. We'd like to thank Jeremy Butterfield and Nazim Boutta, our gurus in QFT for all their help in preparing for the trip! After a Manhattan or two tonight we're heading to Boston tomorrow to continue our quantum adventure! The arithmetic of relationships on the High Line. February 7, 2013 On 14 March at 1.59pm GMT, Marcus du Sautoy will host Pi Day Live, an interactive exploration of the number which has fascinated mathematicians throughout the ages. He wants to rediscover pi using ancient and intriguing techniques, and he needs your help! Get a slice of the action... (Image from www.freeimages.co.uk) Everyone at Pi Day Live will be using marbles, pins, maps and other household items to discover pi using methods that range from 3500 to around 250 years old. It’s not all low-tech, though, as they will be using the web to gather everyone’s results live, combining them to find out if they can collectively calculate a more accurate approximation of pi. Will it be possible to derive pi to one, two, three or more, decimal places? Can we do better than the ancient Greeks or have we lost the ability to rediscover this amazing number without using computers? Mathematicians (and the American House of Representatives) have christened 14 March Pi Day because the date, when written in the US date format, is 3.14. Add the 1.59pm time of the Pi Day Live experiment and you get 3.14159, or pi at around the accuracy Archimedes calculated it over 2000 years ago using simple geometry. Pi has obsessed generations of mathematicians for millennia because it is integral to one of the most important and elegant geometric objects in nature, the circle. Attempting to calculate an accurate value for this never-ending transcendental number has been one of the big themes running throughout the history of mathematics. Even though you only need to know pi to 39 decimal places to calculate a circumference the size of the observable universe to the precision comparable to the size of a hydrogen atom, mathematicians have pushed the limits of computing technology to calculate the number to over one trillion digits. How close can Pi Day Live get to this accuracy using ancient techniques? You can connect with Marcus and Pi Day Live via an Online Lecture Theatre or by watching online on the ‘Big Screen’. If your computer can run YouTube videos then you have what you need to get involved. The event will be recorded and will be available on YouTube afterwards for anyone who can’t take part on the day. Just go to Pi Day Live website to find out more. And you can get live updates and all the pi facts you could ever want on Twitter at and Facebook. And you can read more about pi on Plus:
# Difference between revisions of "1991 AIME Problems/Problem 2" ## Problem Rectangle $ABCD_{}^{}$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$, and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$. For $1_{}^{} \le k \le 167$, draw the segments $\overline {P_kQ_k}$. Repeat this construction on the sides $\overline {AD}$ and $\overline {CD}$, and then draw the diagonal $\overline {AC}$. Find the sum of the lengths of the 335 parallel segments drawn. ## Solution An image is supposed to go here. You can help us out by creating one and editing it in. Thanks. The length of the diagonal is $\sqrt{3^2 + 4^2} = 5$ (a 3-4-5 right triangle). For each $k$, $\overline{P_kQ_k}$ is the hypotenuse of a 3-4-5 right triangle with sides of $3 \cdot \frac{k}{168}, 4 \cdot \frac{k}{168}$. Thus, its length is $5 \cdot \frac{k}{168}$. The sum we are looking for is $2 \cdot \left(\sum_{k = 1}^{167} 5 \cdot \frac{k}{168}\right) + 5 = \frac{5}{84}\sum_{k=1}^{167}k + 5$. Using the formula for the sum of the first n numbers, we find that the solution is $\frac{5}{84} \cdot \frac{168 \cdot 167}{2} + 5 = 5 \cdot 167 + 5 = 840$.
# Eigenvalues and their Algebraic Multiplicities of a Matrix with a Variable ## Problem 206 Determine all eigenvalues and their algebraic multiplicities of the matrix $A=\begin{bmatrix} 1 & a & 1 \\ a &1 &a \\ 1 & a & 1 \end{bmatrix},$ where $a$ is a real number. ## Proof. To find eigenvalues we first compute the characteristic polynomial of the matrix $A$ as follows. \begin{align} \det(A-tI)&=\begin{vmatrix} 1-t & a & 1 \\ a &1-t &a \\ 1 & a & 1-t \end{vmatrix}\\ &=(1-t)\begin{vmatrix} 1-t & a\\ a& 1-t \end{vmatrix}-a\begin{vmatrix} a & a\\ 1& 1-t \end{vmatrix}+\begin{vmatrix} a & 1-t\\ 1& a \end{vmatrix} \end{align} We used the first row cofactor expansion in the second equality. After we compute three $2 \times 2$ determinants and simply, we obtain $\det(A-tI)=-t(t^2-3t+2-2a^2).$ The eigenvalues of $A$ are roots of this characteristic polynomial. Thus, eigenvalues are $0, \quad \frac{3\pm \sqrt{1+8a^2}}{2}$ by the quadratic formula. Now the only possible way to obtain a multiplicity $2$ eigenvalue is when $\frac{3- \sqrt{1+8a^2}}{2}=0$ and it is straightforward to check that this happens if and only if $a=1$. Therefore, when $a=1$ eigenvalues of $A$ are $0$ with algebraic multiplicity $2$ and $3$ with algebraic multiplicity $1$. When $a \neq 1$, eigenvalues are $0, \quad \frac{3\pm \sqrt{1+8a^2}}{2}$ and each of them has algebraic multiplicity $1$. Show that eigenvalues of a Hermitian matrix $A$ are real numbers. (The Ohio State University Linear Algebra Exam Problem)
# Parallelogram Definition ## Parallelogram ### Properties of parallelogram are as follows: Opposite sides of a parallelogram are equal. If ABCD is a parallelogram then AD = BC, AB = DC. Opposite angles of a parallelogram are equal. If ABCD is a parallelogram ∠A = ∠Cand ∠B = ∠D. The diagonals of a parallelogram bisect each other. If ABCD is a parallelogram then AO = OC and BO = OD. Area of Parallelogram (A) : Base (b) . Height (h) Theorem: In a parallelogram, prove that • the opposite sides are equal; • the opposite angles are equal; • diagonals bisect each other. Proof: Let ABCD be a parallelogram. Draw its diagonal AC. In triangles ABC and CDA, we have ∠1 = ∠3 [Alternate angles] ∠2 = ∠4 [Alternate angles] and, AC = AC [Common side] So, by ASA congruence criterion, we have ΔABC≅ΔCDA ⇒ AB = CD, BC = AD and ∠B = ∠D Corresponding parts of congruent triangle are equal] Similarly, by drawing the diagonal BD, we can prove that ΔABD ≅ΔCDB ⇒ AD = BCand∠A = ∠C Hence, in parallelogram ABCD, we have AB = CD, AD = BC, ∠A = ∠C, ∠B = ∠D This proves (i) and (ii) In order to prove (iii), consider a parallelogram ABCD. Draw its diagonals AC and BD, intersecting each other at O. In Δs AOB and COD, we have AB = CD [Opposite sides of a parallelogram are equal] ∠AOB = ∠COD [Vertically opposite angles] ∠OAB = ∠DCO [Alternate angles] ∴ ΔOAB≅ΔOCD ⇒ OA = OC and OB = OD This proves (iii). ### Types of Parallelogram: #### Rectangle Rectangle is parallelogram with a right angle. It has the following properties. Each of the angles is a right angle. Diagonals are equal. #### Rhombus Rhombus is a parallelogram whose adjacent sides and adjacent angles are equal. So, rhombus has all the properties of a parallelogram. Hence, from the properties of a parallelogram, we have PQ = SR, PS = RQ ∠P = ∠R and ∠S = ∠Q Diagonals PR and SQ bisect each other at right angle Diagonals are unequal in length. #### Square A square is an equilateral rectangle. This means a square has all the properties of a rectangle with an additional requirement that all the sides have equal length. In square, the diagonals • bisect one another • are of equal length • are perpendicular to one another. Do follow NCERT Solutions for Class 8 Maths prepared by expert faculty of Physics Wallah. For additional information related to the subject you can check the Maths Formula section.
# Visualizing the factorial Often in basic mathematics, we can visualize things very easily, which I believe helps understanding (instead of just working out a number theoretical proof). For example: $$(n+1)^2 - n^2 = (n+1) +n$$ can be visualized by squares. Remove a square with sided $n$ from a square with sides $n+1$ leaves the top row ($n+1$) and the right row without the top ($n$) (done here with diamonds and bullets for $n = 4$). $$\diamond \diamond \diamond \diamond \diamond \\ \bullet \bullet \bullet \bullet \diamond \\ \bullet \bullet \bullet \bullet \diamond \\ \bullet \bullet \bullet \bullet \diamond \\ \bullet \bullet \bullet \bullet \diamond$$ Another example is proving that $$\sum_{i = 1}^n 2\cdot i = n^2 + n$$ which can be done in the following way (for $n = 4$): $$\diamond \diamond \diamond \diamond \\ \diamond \diamond \diamond \bullet \\ \diamond \diamond \bullet \bullet \\ \diamond \bullet \bullet \bullet \\ \bullet \bullet \bullet \bullet$$ Here, we see two triangles, the one with diamonds with row lengths from $1$ to $n$ and the one with bullets going from $1$ to $n$, which represents the sum. We also see a $(n+1) \times n$ rectangle, which represents the right hand side. This proves the theorem. I was working through same basic number theory proofs and induction proof because I like to visualize these. It is easy enough to visuale $n^a$ as an $a$-dimensional cube with sides $n$. The problem is that I have often difficulty to visualize the factorial: $n!$ Does anybody know of a nice way to visualise the factorial? The best I could come up with is the following: See $2!$ as just two dots $\bullet \bullet$. See $3!$ as a triangle with the sides made with $2!$, e.g. $$\cdot \\ \bullet \quad \bullet \\ \bullet \quad \quad \bullet \\ \cdot \space \space \bullet \bullet \space \space \cdot$$ Now see $n!$ as an $n$-gon with the sides made of the $(n-1)$-gon. (So $4!$ would be a square with a $3!$-triangle on its sides.) This visualization is not very easy to work with when you want to visualize proofs. Are there better ways to visualize $n!$? EDIT: I should emphasis that I would like to visualize $n!$ using dots or lines or so, not so much with concepts ( it is definitely easier to understand the factorial using permutations, just as it is easier to prove some statements using algebra, however the point is that I am trying to prove these things using these very concrete and real visualizations.) • Maybe the number of ways to order $n$ books? Aug 3, 2015 at 15:47 • Visualizing $n!$ can't be easy as it grows very fast. – user65203 Aug 3, 2015 at 15:49 • If I recall correctly, it's not just an $(n-1)$-gon, it's actually the next-higher dimension's "three-sided" shape, i.e., the progression is "point, line, triangle, tetrahedron, $4$th-dimension tetrahedron, ..." In particular, it directly follows the progression of Pascal's Triangle for the $n$ number of terms in $(a_1+ a_2+a_3+\dots+a_n)^k$. Aug 3, 2015 at 15:49 • Im looking for a more concrete visualization, like with dots. Aug 3, 2015 at 15:49 • Perhaps something like this — a tree with $n$ "children" coming out of the first vertex, $n-1$ "children" coming out of the second, etc. (Or the reverse order.) Aug 3, 2015 at 17:13 One way is the total number of leaves of a (single) rooted tree in which each leaf is minimally linked to the root by exactly $n-1$ edges, and which has the following property: the root has $2$ children, each child of the root has $3$ children, each child of each child of the root has $4$ children, and so on until the leaves are reached. A natural term for this is factorial tree, but I don't know if this phrase is in general use for this notion. For example, for $n = 4$: • Excellent! Closely related to the comment by@columbus8myhw, although in reverse. Aug 3, 2015 at 18:05 • I didn't see @columbus8myhw's comment until just now, or I might not have bothered! Apparently the comment wasn't available just when my lunch hour began, a little over an hour ago when I thought about posting (and eventually decided to post) an answer, and I didn't look back over the comments to see if any new ones had appeared. Aug 3, 2015 at 18:10 • @DaveL.Renfro: I think you mean "child" rather than "sibling"? Aug 3, 2015 at 20:00 • @Dave L. Renfro, I have created the factorial tree for $n = 4$, here. Perhaps it would be useful to display it in your answer. Aug 3, 2015 at 20:21 • I don't know how to paste pictures into a stackexchange answer. That was what I meant by "how to do such things in a stackexchange post". Aug 3, 2015 at 20:25 The way I see $n!$ is a hybrid of avid19's and Dave L. Renfro's visualizations: I imagine $n$ people lining up one by one. I think it really helps to imagine people or animals or fruits or something, rather than boring symbols: that's the way it's done in Burns and Weston's Math For Smarty Pants, and it seems to have made quite an impression on me. My keyboard has no fruits on it, unfortunately, so maybe try to imagine the digits below pinned to some hockey players. • The first person doesn't have any choice about where they join the line, since there is no line yet. 1 • The second person can join in two places: the front or the back. 21 12 • The third person can join in three places: the front, the middle, or the back. 321 231 213 312 132 123 • The fourth person can join in four places. 4321 3421 3241 3214 4231 2431 2341 2314 4213 2413 2143 2134 4312 3412 3142 3124 4132 1432 1342 1324 4123 1423 1243 1234 • The fifth person can join in five places... Here's a geometric visualization in higher dimensions. You can take a hyper-cube in dimension $d$ (basically the Cartesian product of $n$ copies of the interval $[0,c]$ for any $c > 0$ that you want), and then you triangulate (i.e. partition) into equal volume simplices (a simplex in $d$ dimensions is a full dimensional convex hulls of $d+1$ points, i.e. higher dimensional analog of triangles for $d = 2$) by first drawing the edge from the origin to the opposite corner of the hypercube (so the opposite corners are vertices included in each simplex), and then move along one edge of the cube incident to the origin to get your next vertex, then move closer to the opposite corner by taking one edge incident to that vertex to get the next vertex, and so on until you reach the opposite corner. You can traverse the dimension-aligned edges in any order you want to get $d+1$ vertices of a distinct simplex, and the interiors of the simplexes are disjoint, and the number of congruent simplexes you get in this partition is equal to the number of ways you can order the dimensions, which is $d!$. Thus, if $c = 1$, then each simplex in this partition has volume $1/d!$ and they are all congruent. A related construction is to consider the volume of the simplex whose vertices are the origin along with the endpoints of $d$ linearly independent vectors $v_i$ extending from the origin. This solid has the description $\{ \sum_i c_i v_i \, \| \, \sum_i c_i \leq 1, c_i \geq 0$ } where the $v_i$ are your vectors. The parallelepiped (analog of hypercube) spanned by these vectors $v_i$ on the other hand has the description $\{ \sum_i c_i v_i \, \| \, 0 \leq c_i \leq 1 \}$. It is a geometric fact that the volume of the parallelepiped is $d!$ times the volume of the simplex, and the volume of the parallelepiped is $\|\det V\|$ where $V$ is the matrix of the vectors that span the parallelepiped. • This is quite a difficult visualization. I do like them (expecially the second one) but was looking for a simpler one, one that would still make sense for a high school student or a first year undergraduate. Aug 3, 2015 at 16:29 • @Krijn Fairly early in undergraduate (e.g. in the 1st or 2nd year) one usually has the opportunity to take basic linear algebra and then, at least if they take a geometric approach at times, this should hopefully all be explained. I apologize I couldn't give you an answer closer to what you were looking for. Aug 3, 2015 at 16:34 • Don't apologize! I certainly liked the answer. I however was looking for an answer that would easily explain for example why $n!$ grows faster then $n^a$ or that could prove $\sum_{i=1}^n\frac{i-1}{i!} = \frac{n!-1}{n!}$ using a visual approach. Aug 3, 2015 at 16:37 • @Krijn I don't know if I would exactly call it "visual", but just based on the definition of factorial, you can easily show the pattern that gives you your second equation for example, just by adding $1/n!$ to both sides and seeing the pattern of what happens in the computations if you start with $i=n$ and work your way down. That basically has to do with the fact that the factorials can be used as a base system for writing down numbers, which is an interesting fact about them. Aug 3, 2015 at 16:46 • @Krijn Also for your first claim, you can write $(n!)^2$ and imagine visually taking pairwise products of terms with $n!$ written forward as a product in one copy and backwards in the other. Then it's not too hard to see that each pairwise product of terms is greater than or equal to $n$, which shows $n! \geq n^{n/2}$. Visual, just maybe not how you meant visual. Aug 3, 2015 at 16:48 This might not be what you're looking for, but I visualize a factorial as a process. $5!$ is how many ways you can arrange 5 things. I visualize arranging 5 things. Not a representation as dots but personally it's powerful. • You can visualize the number of ways one can order $5$ things? I'm impressed. I can hardly do $3$ without trying unnaturally in my head. – MT_ Aug 4, 2015 at 21:07 • @soke Not the individual ways of course. But the process of it. "5 things here, then 4, then ..." which is basically the definition of the factorial – user223391 Aug 6, 2015 at 23:00
# One-sided Limit Example Q: Find the one-sided limit (if it exists): limx→-1– (x+1)/(x4-1) A: So we need to find the limit of this function (x+1)/(x4-1) as x approaches -1 from the left. Remember, from the left means as x gets closer and closer to -1, but is still smaller. The concept: What is happening to this function as x = -2, x = -1.5, x = -1.1, x = -1.0001, etc… We test first and plug -1 into the function: (-1+1)/((-1)4-1) = 0/0 Whenever you get 0/0, that is your clue that maybe you need to do “more work” before just plugging in or jumping to conclusions. So, let’s try “more work” — usually that means simplifying. I see that the denominator can factor. We have: (x+1) / (x4-1) = (x+1) / [(x2-1)(x2+1)] Let’s keep factoring the denominator: (x+1) / [(x-1)(x+1)(x2+1)] Now, it appears there is a “removable hole” in the function. This means, we can remove this hole by reducing the matching term in the numerator with the matching term in the denominator: (x+1) / [(x-1)(x+1)(x2+1)] = 1 / [(x-1)(x2+1)] Notice that hole exists when x = -1 (and it was removable! This is good news for us since we are concerned with the nature of the function as x approaches -1) Now that we have removed that hole, let’s once again try to plug in -1 to see what we get. 1 / [(-1-1)((-1)2+1)] 1 / [(-2)(2)] -1/4 So, after removing the hole at (x+1), we found the function value when x=-1 is -1/4. Due to the nature of this function, this means: limx→-1(x+1)/(x4-1) = -1/4 Since the limit exists as both the right-handed and left-handed limit, it follows that limx→-1– (x+1)/(x4-1) must also be -1/4. It ended up not being necessary that we only do a “one-handed limit analysis.”
# Circle A has a center at (2 ,3 ) and an area of 8 pi. Circle B has a center at (11 ,7 ) and an area of 54 pi. Do the circles overlap? Jul 28, 2018 $\text{circles overlap}$ #### Explanation: $\text{What we have to do here is compare the distance (d)}$ $\text{between the centres to the sum of the radii}$ • " if sum of radii">d" then circles overlap" • " if sum of radii"< d" then no overlap" $\text{To calculate the radii use the area formula } A = \pi {r}^{2}$ ${r}_{A} = \sqrt{8} = 2 \sqrt{2} \text{ and } {r}_{B} = \sqrt{54} = 3 \sqrt{6}$ $\text{to calculate d use the "color(blue)"distance formula}$ •color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2) $\text{let "(x_1,y_1)=(2,3)" and } \left({x}_{2} , {y}_{2}\right) = \left(11 , 7\right)$ $d = \sqrt{{\left(11 - 2\right)}^{2} + {\left(7 - 3\right)}^{2}} = \sqrt{81 + 16} = \sqrt{97} \approx 9.85$ $\text{sum of radii } = 2 \sqrt{2} + 3 \sqrt{6} \approx 10.18$ $\text{Since sum of radii">d" then circles overlap}$ graph{((x-2)^2+(y-3)^2-(2sqrt2)^2)((x-11)^2+(y-7)^2-(3sqrt6)^2)=0 [-20, 20, -10, 10]}
# Difference between revisions of "2003 AMC 12A Problems/Problem 18" ## Problem Let $n$ be a $5$-digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$. For how many values of $n$ is $q+r$ divisible by $11$? $\mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090$ ## Solution 1 When a $5$-digit number is divided by $100$, the first $3$ digits become the quotient, $q$, and the last $2$ digits become the remainder, $r$. Therefore, $q$ can be any integer from $100$ to $999$ inclusive, and $r$ can be any integer from $0$ to $99$ inclusive. For each of the $9\cdot10\cdot10=900$ possible values of $q$, there are at least $\left\lfloor \frac{100}{11} \right\rfloor = 9$ possible values of $r$ such that $q+r \equiv 0\pmod{11}$. Since there is $1$ "extra" possible value of $r$ that is congruent to $0\pmod{11}$, each of the $\left\lfloor \frac{900}{11} \right\rfloor = 81$ values of $q$ that are congruent to $0\pmod{11}$ have $1$ more possible value of $r$ such that $q+r \equiv 0\pmod{11}$. Therefore, the number of possible values of $n$ such that $q+r \equiv 0\pmod{11}$ is $900\cdot9+81\cdot1=8181 \Rightarrow\boxed{(B)}$. ## Solution 2 Notice that $q+r\equiv0\pmod{11}\Rightarrow100q+r\equiv0\pmod{11}$. This means that any number whose quotient and remainder sum is divisible by 11 must also be divisible by 11. Therefore, there are $\frac{99990-10010}{11}+1=8181$ possible values. The answer is $\boxed{\textbf{(B) }8181}$. ## Solution 3 Let $abcde$ be the five digits of $n$. Then $q = abc$ and $r = de$. By the divisibility rules of $11$, $q = a - b + c \pmod{11}$ and $r = -d + e \pmod{11}$, so $q + r = a - b + c - d + e = abcde = n \pmod{11}$. Thus, $n$ must be divisble by $11$. There are $\frac{99990 - 10010}{11} + 1 = 8181$ five-digit multiples of $11$, so the answer is $\boxed{\textbf{(B) }8181}$.
# What is 1/90 as a decimal? ## Solution and how to convert 1 / 90 into a decimal 1 / 90 = 0.011 Fraction conversions explained: • 1 divided by 90 • Numerator: 1 • Denominator: 90 • Decimal: 0.011 • Percentage: 0.011% 1/90 or 0.011 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Both are used to handle numbers less than one or between whole numbers, known as integers. But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. If we need to convert a fraction quickly, let's find out how and when we should. 1 / 90 as a percentage 1 / 90 as a fraction 1 / 90 as a decimal 0.011% - Convert percentages 1 / 90 1 / 90 = 0.011 ## 1/90 is 1 divided by 90 The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 1 is being divided into 90. Think of this as our directions and now we just need to be able to assemble the project! Fractions have two parts: Numerators and Denominators. This creates an equation. Now we divide 1 (the numerator) into 90 (the denominator) to discover how many whole parts we have. This is our equation: ### Numerator: 1 • Numerators are the portion of total parts, showed at the top of the fraction. Comparatively, 1 is a small number meaning you will have less parts to your equation. 1 is an odd number so it might be harder to convert without a calculator. Ultimately, having a small value may not make your fraction easier to convert. Let's look at the fraction's denominator 90. ### Denominator: 90 • Denominators represent the total parts, located at the bottom of the fraction. 90 is a large number which means you should probably use a calculator. And it is nice having an even denominator like 90. It simplifies some equations for us. Have no fear, large two-digit denominators are all bark no bite. Let's start converting! ## How to convert 1/90 to 0.011 ### Step 1: Set your long division bracket: denominator / numerator $$\require{enclose} 90 \enclose{longdiv}{ 1 }$$ Use long division to solve step one. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well. ### Step 2: Extend your division problem $$\require{enclose} 00. \\ 90 \enclose{longdiv}{ 1.0 }$$ Uh oh. 90 cannot be divided into 1. So we will have to extend our division problem. Add a decimal point to 1, your numerator, and add an additional zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 90 into 1 + 0 or 10. ### Step 3: Solve for how many whole groups you can divide 90 into 10 $$\require{enclose} 00.0 \\ 90 \enclose{longdiv}{ 1.0 }$$ How many whole groups of 90 can you pull from 10? 0 Multiply this number by 90, the denominator to get the first part of your answer! ### Step 4: Subtract the remainder $$\require{enclose} 00.0 \\ 90 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$ If there is no remainder, you’re done! If there is a remainder, extend 90 again and pull down the zero ### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit. In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value. ### Why should you convert between fractions, decimals, and percentages? Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. Same goes for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But 1/90 and 0.011 bring clarity and value to numbers in every day life. Without them, we’re stuck rounding and guessing. Here are real life examples: ### When you should convert 1/90 into a decimal Dollars & Cents - It would be silly to use 1/90 of a dollar, but it makes sense to have $0.1. USD is exclusively decimal format and not fractions. (Yes, yes, there was a 'half dollar' but the value is still$0.50) ### When to convert 0.011 to 1/90 as a fraction Distance - Any type of travel, running, walking will leverage fractions. Distance is usually measured by the quarter mile and car travel is usually spoken the same. ### Practice Decimal Conversion with your Classroom • If 1/90 = 0.011 what would it be as a percentage? • What is 1 + 1/90 in decimal form? • What is 1 - 1/90 in decimal form? • If we switched the numerator and denominator, what would be our new fraction? • What is 0.011 + 1/2? ### Convert more fractions to decimals From 1 Numerator From 90 Denominator What is 1/91 as a decimal? What is 2/90 as a decimal? What is 1/92 as a decimal? What is 3/90 as a decimal? What is 1/93 as a decimal? What is 4/90 as a decimal? What is 1/94 as a decimal? What is 5/90 as a decimal? What is 1/95 as a decimal? What is 6/90 as a decimal? What is 1/96 as a decimal? What is 7/90 as a decimal? What is 1/97 as a decimal? What is 8/90 as a decimal? What is 1/98 as a decimal? What is 9/90 as a decimal? What is 1/99 as a decimal? What is 10/90 as a decimal? What is 1/100 as a decimal? What is 11/90 as a decimal? What is 1/101 as a decimal? What is 12/90 as a decimal? What is 1/102 as a decimal? What is 13/90 as a decimal? What is 1/103 as a decimal? What is 14/90 as a decimal? What is 1/104 as a decimal? What is 15/90 as a decimal? What is 1/105 as a decimal? What is 16/90 as a decimal? What is 1/106 as a decimal? What is 17/90 as a decimal? What is 1/107 as a decimal? What is 18/90 as a decimal? What is 1/108 as a decimal? What is 19/90 as a decimal? What is 1/109 as a decimal? What is 20/90 as a decimal? What is 1/110 as a decimal? What is 21/90 as a decimal? ### Convert similar fractions to percentages From 1 Numerator From 90 Denominator 2/90 as a percentage 1/91 as a percentage 3/90 as a percentage 1/92 as a percentage 4/90 as a percentage 1/93 as a percentage 5/90 as a percentage 1/94 as a percentage 6/90 as a percentage 1/95 as a percentage 7/90 as a percentage 1/96 as a percentage 8/90 as a percentage 1/97 as a percentage 9/90 as a percentage 1/98 as a percentage 10/90 as a percentage 1/99 as a percentage 11/90 as a percentage 1/100 as a percentage
# How do you find the value of cos(pi/4)? ##### 2 Answers Mar 7, 2018 You would look on the unit circle. #### Explanation: $\cos \left(\frac{\pi}{4}\right)$= $\left(\frac{1}{\sqrt{2}}\right) = \frac{\sqrt{2}}{2}$ !unit circle Mar 7, 2018 $\cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, refer to the explanation below for how to find the exact value without a calculator. #### Explanation: It is possible to find the exact value of $\cos \left(\frac{\pi}{4}\right)$ by constructing a right triangle with one angle set to $\frac{\pi}{4}$ radians. First, let's convert radians into degrees $\frac{\pi}{4} \text{ rad"=pi/4 " rad" * 180^"o"/(pi)* "rad"^-1=45^"o}$ Now let's draw a right triangle with one of the acute angles set to $45$ degrees. Remeber that these two angles would be supplementary, meaning that their sum would be ${90}^{\text{o}}$. As a result, the other angle in this triangle would also be ${45}^{\text{o}}$, making an isosceles right triangle. By setting the length of one of the sides adjacent to the right angle to $1$ and applying the Pythagorean theorem, you'll find the length of the hypotenuse $\sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}$. Thus $\cos \left(\frac{\pi}{4}\right) = \cos \left({45}^{\text{o")=("adj.")/("hyp.}}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.
# What is the slope of a line that has an What is the slope of a line that has an X-Intercept of 8 and a Y-Intercept of 11? You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it If we know the X-Intercept and the Y-Intercept, then we have two points. With two points, we can define the slope of the line and, indeed, an equation for the line through those two points. Using your example, suppose the X-Intercept is 8 and Y-Intercept is 11. Then we know that $\left(8,0\right)$ and $\left(0,11\right)$ are points on this line. Therefore, the slope of the line is $\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}$, or the change in $y$ over the change in $x$. Therefore, $slope=\frac{0-11}{8-0}=\frac{-11}{8}$. Now we can use point-slope form for a line through a point to give a formula for the line. Point-slope form is defined by $y-{y}_{1}=m\left(x-{x}_{1}\right)$) where ${x}_{1},{y}_{1}$ are the x and y values from one of your points, and m is the slope. I will use the point $\left(8,0\right)$, although we can very easily choose the other point and get the same formula, so that the formula for this line is $y-0=\frac{-11}{8}\left(x-8\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}y=\frac{-11}{8}x+11$ ###### Did you like this example? gnatopoditw The intercepts can be treated as special points. In your case you have $\left(8,0\right)$ and $\left(0,11\right)$. $m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{11-0}{0-8}=-\frac{11}{8}$
# Thread: Integration by trigonometric substitution 1. ## Integration by trigonometric substitution I'm new to this type of question and I can only get so far into the question before I don't know what to do - I haven't been taught more but I very much would like to know. Use the substitution $x$ = $3\sin\theta$ to show that $\int_{1.5}^{3} \sqrt{9 - x^2} dx$ = $\int_{a}^{b} k \cos^2\theta d\theta$ where the values of a, b and k are to be found. Hence evaluate $\int_{1.5}^{3} \sqrt{9 - x^2} dx$. So far I can say that $x$ = $3\sin\theta$ and $dx$ = $3\cos\theta d\theta$ $\sqrt{9 - (3\sin\theta)^2}$ = $\sqrt{9 - 9\sin^2\theta}$ = $\sqrt{9 (1 - \sin^2\theta)}$ = $\sqrt{9\cos^2\theta}$ = $3\cos\theta$ And this is as far as I can get. Could someone please help me finish the question and understand how it works from this point onwards? Any help is greatly appreciated 2. $\int_{1.5}^{3} \sqrt{9 - x^2} dx$ = $\int_{a}^{b} k \cos^2\theta d\theta$ where the values of a, b and k are to be found. As you have done we let: $x$ = $3\sin\theta$ and $dx$ = $3\cos\theta d\theta$ so $\int_{1.5}^{3} \sqrt{9 - x^2} dx$ = $\int_{1.5}^{3}\sqrt{9 - (3\sin\theta)^2} 3\cos\theta d\theta$ = $\int_{1.5}^{3} 3\cos \theta \cdot 3\cos\theta d\theta$ = $\int_{1.5}^{3} 9 \cos^2\theta d\theta$ so a=1.5 b=3 k=9 now to evaluate this integral you can use the identity: $\cos^2 x = \frac{1+cos(2x)}{2}$
Review question Ref: R6429 ## Solution Solve the simultaneous equations \begin{align} 2x + 4y &= 9, \\ 4x^2 + 16y^2 &= 20x + 4y - 19. \end{align} We will solve the equations by substitution, rearranging the first (linear) equation and substituting it into the second (quadratic) equation. We have two choices: we could either write $2x$ in terms of $y$ or write $4y$ in terms of $x$. Looking at the second equation, we notice that the $x$ terms ($4x^2=(2x)^2$ and $20x=10(2x)$) can be easily written in terms of $2x$, though the coefficients involved are somewhat large, especially if we do not happen to have a calculator handy. On the other hand, the $y$ terms ($16y^2=(4y)^2$ and $4y$) are far more straightforward, so we will use this substitution. We can rearrange the first equation to get $4y = 9 - 2x$. Substituting this into the second equation, we obtain $4x^2 + (9-2x)^2 = 20x + (9-2x) - 19.$ Expanding this gives $4x^2 + (81 - 36x + 4x^2) = 18x - 10,$ which, by bringing everything over to the left-hand side, becomes $8x^2 - 54x + 91 = 0.$ We could either work to discover that this factorises as $(4x-13)(2x-7)=0$ so that $x=\frac{13}{4}$ or $x=\frac{7}{2}$, or we could use the quadratic formula to reach the same conclusion. When $x=\frac{13}{4}$, $y = \frac{9 - 2x}{4} = \frac{9 - \tfrac{13}{2}}{4} = \frac{5}{8},$ and when $x = \tfrac{7}{2}$, $y = \frac{9 - 2x}{4} = \frac{9 - 7}{4} = \frac{1}{2}.$ So the two solutions are $\left( \frac{13}{4}, \frac{5}{8} \right) \quad\text{and}\quad \left( \frac{7}{2}, \frac{1}{2} \right).$ Another way of thinking about these equations is as a pair of curves in the $(x,y)$-plane. The first equation, $2x + 4y = 9,$ is that of a straight line. The second equation, $4x^2 + 16y^2 = 20x + 4y - 19,$ can be rearranged into $\left( x - \frac{5}{2} \right)^2 + 4\left( y - \frac{1}{8} \right)^2 = \frac{29}{16},$ which is an ellipse centred at $(\frac{5}{2},\frac{1}{8})$. In the sketch below, the lines $y = \tfrac{1}{8}$ and $x = \tfrac{5}{2}$ are shown in grey. The sketch shows why there are at most two (real) solutions: a straight line and an ellipse cannot intersect more than twice. We cannot solve this problem accurately by just sketching the graph, but it tells us something about how many solutions there could be.
Geometry Module 5, Topic B, Lesson 10: Teacher Version ```Lesson 10 NYS COMMON CORE MATHEMATICS CURRICULUM M5 GEOMETRY Lesson 10: Unknown Length and Area Problems Student Outcomes  Students apply their understanding of arc length and area of sectors to solve problems of unknown area and length. Lesson Notes This lesson continues the work started in Lesson 9 as students solve problems on arc length and area of sectors. The lesson is intended to be 45 minutes of problem solving with a partner. Problems vary in level of difficulty and can be assigned specifically based on student understanding. The problem set can be used in class for some students or assigned as homework. Students who need to focus on a small number of problems could finish the other problems at home. Teachers may choose to model two or three problems with the entire class. Exercise 4 is a modeling problem highlighting G-MG.A.1 and MP.4. Scaffolding: Classwork Begin with a quick whole class discussion of an annulus. Project the figure on the right on the board. Opening Exercise (3 minutes) Opening Exercise In the following figure, a cylinder is carved out from within another cylinder of the same height; the bases of both cylinders share the same center. a. Sketch a cross section of the figure parallel to the base.  Post area of sector and arc length formulas for easy reference.  A review of compound figures may be required before this lesson.  Scaffold the task by asking students to compute the area of the circle with radius , then the circle with radius , and then ask how the shaded region is related to the two circles.  Use an example with numerical values for and on the coordinate plane, and ask students to estimate the area first (see example below). Confirm that students’ sketch is correct before allowing them to proceed to part (b). Lesson 10: Date: Unknown Length and Area Problems 2/6/16 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 119 Lesson 10 NYS COMMON CORE MATHEMATICS CURRICULUM M5 GEOMETRY b. MP.2 & MP.7 Mark and label the shorter of the two radii as and the longer of the two radii . Show how to calculate the area of the shaded region and explain the parts of the expression. () = ( − ) = radius of outer circle = radius of inner circle The figure you sketched in part (b) is called an annulus; it is a ring shaped region or the region lying between two concentric circles. In Latin, annulus means “little ring.” Exercises (35 minutes) 1. Find the area of the following annulus. () = ( − . ) () = . The area of the annulus is . . 2. The larger circle of an annulus has a diameter of cm, and the smaller circle has a diameter of . cm. What is the area of the annulus? The radius of the larger circle is cm, and the radius of the smaller circle is . cm. () = ( − . ) () = . The area of the annulus is . . 3. In the following annulus, the radius of the larger circle is twice the radius of the smaller circle. If the area of the following annulus is , what is the radius of the larger circle? (): (() − ) = = = The radius of the larger circle is twice the radius of the smaller circle or () = ; the radius of the larger circle is . Lesson 10: Date: Unknown Length and Area Problems 2/6/16 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 120 Lesson 10 NYS COMMON CORE MATHEMATICS CURRICULUM M5 GEOMETRY MP.4 4. An ice cream shop wants to design a super straw to serve with their extra thick milkshakes that is double the width and thickness of a standard straw. A standard straw is mm in diameter and . mm thick. a. What is the cross-sectional (parallel to the base) area of the new straw (round to the nearest hundredth)? . mm2 b. If the new straw is mm long, what is the maximum volume of milkshake that can be in the straw at one time (round to the nearest hundredth)? , . mm3 c. A large milkshake is ounces (approximately mL). If Corbin withdraws the full capacity of a straw times a minute, what is the minimum amount of time that it will take him to drink the milkshake (round to the nearest minute)? minutes 5. In the circle given, ̅̅̅̅ is the diameter and is perpendicular to chord ̅̅̅̅ . ̂, = cm and = cm. Find , , ∠, the arc length of ̂ (round to the nearest hundredth, if necessary). and the area of sector = cm, = cm, ∠ = (. ) = . °, arc length = . cm, area = . cm2 6. Given circle with ∠ ≅ ∠, find the following (round to the nearest hundredth, if necessary): a. ̂ ˚ b. ̂ ˚ c. ̂ . ˚ d. ̂ Arc length . yds e. ̂ Arc length . yds Lesson 10: Date: Unknown Length and Area Problems 2/6/16 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 121 Lesson 10 NYS COMMON CORE MATHEMATICS CURRICULUM M5 GEOMETRY f. ̂ Arc length . yds g. ̂ Area of sector . yds2 h. ̂ Area of sector . yds2 i. ̂ Area of sector . yds2 7. Given circle , find the following (round to the nearest hundredth, if necessary): a. Circumference of circle yds b. . yds c. ̂ Area of sector . yds2 8. Given circle , find the following (round to the nearest hundredth, if necessary): a. ∠ . ° b. ̂ Area of sector Lesson 10: Date: Unknown Length and Area Problems 2/6/16 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 122 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 10 M5 GEOMETRY 9. Find the area of the shaded region (round to the nearest hundredth). . 10. Many large cities are building or have built mega Ferris wheels. One is feet in diameter and has cars each seating up to people. Each time the Ferris wheel turns Ө degrees, a car is in a position to load. a. How far does a car move with each rotation of Ө degrees (round to the nearest whole number)? feet b. What is the value of Ө in degrees? . ˚ 11. ∆ is an equilateral triangle with edge length cm. , , and are midpoints of the sides. The vertices of the triangle are the centers of the circles creating the arcs shown. Find the following (round to the nearest hundredth): a. The area of the sector with center . . cm2 b. The area of triangle . . cm2 c. The area of the shaded region. . cm2 d. The perimeter of the shaded region. . cm Lesson 10: Date: Unknown Length and Area Problems 2/6/16 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 123 Lesson 10 NYS COMMON CORE MATHEMATICS CURRICULUM M5 GEOMETRY 12. In the figure shown, = = cm, = cm, and ∠ = °. Find the area in the rectangle, but outside of the circles (round to the nearest hundredth). . cm2 13. This is a picture of a piece of a mosaic tile. If the radius of each smaller circle is inch, find the area of red section, the white section, and the blue section (round to the nearest hundredth). Red = . in2, White = . in2, Blue = . in2 Closing (2 minutes) Present the questions to the class, and have a discussion, or have students answer individually in writing. Use this as a method of informal assessment.  Explain how to find the area of a sector of a circle if you know the measure of the arc in degrees.   Find the fraction of the circumference by dividing the measure of the arc in degrees by 360, and then multiply by the circumference 2. Explain how to find the arc length of an arc if you know the central angle.  Find the fraction of the area by dividing the measure of the central angle in degrees by 360, then multiply by the circumference 2 . Exit Ticket (5 minutes) Lesson 10: Date: Unknown Length and Area Problems 2/6/16 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 124 Lesson 10 NYS COMMON CORE MATHEMATICS CURRICULUM M5 GEOMETRY Name Date Lesson 10: Unknown Length and Area Problems Exit Ticket 1. Given circle , find the following (round to the nearest hundredth): ̂ in degrees. a. The b. 2. ̂. The area of sector Find the shaded area (round to the nearest hundredth). 62⁰ Lesson 10: Date: Unknown Length and Area Problems 2/6/16 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 125 Lesson 10 NYS COMMON CORE MATHEMATICS CURRICULUM M5 GEOMETRY Exit Ticket Sample Solutions 1. Given circle , find the following (round to the nearest hundredth): a. ̂ in degrees. The . ° b. ̂. The area of sector 2. Find the shaded area. . 62⁰ Problem Set Sample Solutions Students should continue the work they began in class for homework. 1. Find the area of the shaded region if the diameter is inches (round to the nearest hundredth). . in2 2. Find the area of the entire circle given the area of the sector. in2 Lesson 10: Date: Unknown Length and Area Problems 2/6/16 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 126 NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 10 M5 GEOMETRY 3. ̂ are arcs of concentric circles with ̅̅̅̅ ̂ and ̅̅̅̅ lying on the radii of the larger circle. Find the area of the region (round to the nearest hundredth). . cm2 4. Find the radius of the circle, , , and (round to the nearest hundredth). Radius = . cm, = . °, = . °, = cm 5. In the figure, the radii of two concentric circles are cm and cm. ̂ = ⁰. If a chord ̅̅̅̅ of the larger circle intersects the smaller circle only at , find the area of the shaded region in terms of . Lesson 10: Date: Unknown Length and Area Problems 2/6/16 © 2014 Common Core, Inc. Some rights reserved. commoncore.org 127 ``` 12 cards 20 cards 14 cards 23 cards 20 cards

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