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## ACT test prepthat actually works ### BRAIN-COMPATIBLE ACT PREP New ACT Math Test Content Part 1: The Factorial Combinatorics has been added to the ACT math test—and 36U is here to get you ready! Don’t worry. Combinatorics may sound intimidating, but it’s really about fancy, advanced counting techniques that can be a lot of fun and save lots of time. Let’s get right to it. Factorials First, let’s work on a possibly new concept with some new notation—the factorial. 6 factorial, written 6!, means 6 • 5 • 4 • 3 • 2 • 1. And, again: 5! = 5 • 4 • 3 • 2 • 1 To simplify, 5! = 5 • 4 • 3 • 2 • 1 = 120. Let’s put the factorial to work… Example 1: In how many different ways can you arrange the letters of the word OVERCAST? ____ ____ ____ ____ ____ ____ ____ ____ You have 8 options for the first letter, 7 options for the 2nd, 6 options for the 3rd, and so on. Mathematically, that looks like this: _8 options _7 options _6 options _5 options _4 options _3 options _2 options _1 option or 8! –> 40,320 There are 40,320 different ways the letters of OVERCAST can be arranged. Easy so far. Let’s move on… Example 2: In how many different ways can you arrange the letters of the word CANYON? This is a slightly trickier item because CANYON has 2 N’s! Fortunately, if you understood our first example, this isn’t much more difficult. Step 1: There are six different spots to place the letters in the word CANYON: _6 options _5 options _4 options _3 options _2 options _1 option = 6! There are 6! or 720 different ways of arranging C-A-N-Y-O-N, but…You have to account for repeated options because there are 2 Ns. Here’s how: Take your total number of possible arrangements (6!) and divide by 2! to account for N appearing twice. 6!/2! –> (6 • 5 • 4 • 3 • 2 • 1)/(2 • 1) –> 360 There are 360 different ways the letters of the word CANYON can be arranged! That’s your introduction to new counting techniques (combinatorics) that are being tested on the ACT. Take time to brush up on your combination and permutations, too. For more instruction and practice on these topics, check out our online program. -Dr. Kendal Shipley, 36U 10/13/17
IB Vectors • Feb 9th 2008, 12:55 PM overduex IB Vectors Find the position vectors that join the origin to the points with coordinates A(2, -1) and B(-3, 2). Express your answers as column vectors. Hence find line AB. All help is welcome :) • Feb 9th 2008, 01:13 PM earboth Quote: Originally Posted by overduex Find the position vectors that join the origin to the points with coordinates A(2, -1) and B(-3, 2). Express your answers as column vectors. Hence find line AB. Vector $\vec a$ points at the point A and $\vec b$ points at the point B: $\vec a=\left(\begin{array}{c}2\\-1\end{array}\right)$ $\vec b=\left(\begin{array}{c}-3\\2\end{array}\right)$ The direction of the line AB is determined by $\vec a - \vec b = \left(\begin{array}{c}5\\-3\end{array}\right)$ The line AB passes either through A or B. Therefore the equation of the line is: $\vec x = \left(\begin{array}{c}x\\y\end{array}\right) = \left(\begin{array}{c}2\\-1\end{array}\right) + r \cdot \left(\begin{array}{c}5\\-3\end{array}\right)~,~r \in \mathbb{R}$ • Feb 9th 2008, 01:23 PM mr fantastic Quote: Originally Posted by overduex Find the position vectors that join the origin to the points with coordinates A(2, -1) and B(-3, 2). Express your answers as column vectors. Hence find line AB. All help is welcome :) $\vec{OA} = \left( \begin{array}{c} 2 \\ -1 \end{array} \right) $ and $\vec{OB} = \left( \begin{array}{c} -3 \\ 2 \end{array} \right) $ . $\vec{AB} = \vec{AO} + \vec{OB} = -\vec{OA} + \vec{OB} = - \left( \begin{array}{c} 2 \\ -1 \end{array} \right) + \left( \begin{array}{c} -3 \\ 2 \end{array} \right) = \left( \begin{array}{c} -2 \\ 1 \end{array} \right) + \left( \begin{array}{c} -3 \\ 2 \end{array} \right)$ $= \left( \begin{array}{c} -5 \\ 3 \end{array} \right)$ . Earboth is obviously a quicker hand at latex than me. Note that his/her vector and mine are both in the direction of the line .... just in opposite directions. It doesn't matter which one you use.
It looks like you're using Internet Explorer 11 or older. This website works best with modern browsers such as the latest versions of Chrome, Firefox, Safari, and Edge. If you continue with this browser, you may see unexpected results. # Calculus ## Basic Trigonometric Integrals The trigonometric integrals of the six functions are \begin{align} \int\sin u \,du&=-\cos u+C\\ \int\cos u \,du&=\sin u+C \\ \int\sec^2 u \,du&=\tan u+C \\ \int\csc^2 u \,du&=-\cot u+C \\ \int\sec u\tan u \,du&=\sec u+C \\ \int\csc u\cot u \,du&=-\csc u+C\end{align} Example 1: Integrate $\int 0.3 \sec^2 3\theta \,d\theta$ Solution: From the antiderivatives above, we know the integral of $$\sec^2 u$$ turns into $$\tan u$$. Let $$u=3\theta$$, then $$du=3\,d\theta$$ \begin{align} &0.3\int \sec^2 u \left(\frac{du}{3}\right) \\ =& \frac{0.3}{3} \tan u+C \\ =& \frac{1}{10} \tan 3\theta+C \end{align} Example 2: Integrate $\int \left(\cos^2 4x-\sin^2 4x\right)dx$ Solution: We want to simplify the expression inside the integral by using the double angle identity $$\cos 2u = \cos^2 u-\sin^2 u$$ \begin{align} \int \left(\cos^2 4x-\sin^2 4x\right)dx &= \int \cos 2(4x) dx \\ &= \int \cos 8x dx \\ &= \frac{\sin 8x}{8} +C \end{align} ## Integrals of Tangent and Reciprocal Trigonometric Functions We know that $\tan u = \frac{\sin u}{\cos u}$ Thus, $\int \tan u \,du = \int \frac{\sin u}{\cos u}du$ We can integrate using the logarithmic form by letting $$x=\cos u$$. Thus, $$dx=-\sin u \,du$$ \begin{align} &\int \frac{\sin u}{\cos u}du \\ =& -\int \frac{1}{x}dx \\ =& -\ln\|x\|+C \\ =& -\ln\|\cos u\|+C \end{align} By a similar method we can show that $\int \cot u \,du=\ln\|\sin u\|+C$ For the integral of $$\sec u\,du$$, we want to multiply the top and bottom of the function by $$\sec u + \tan u$$. Not that the derivative of $$\sec u + \tan u$$ is $$\left(\sec u\tan u + \sec^2 u\right)$$ Thus, \begin{align} \int \sec u\,du &= \int \frac{\sec u(\sec u + \tan u)du}{\sec u + \tan u} \\&= \int \frac{\sec^2 u + \sec u \tan u}{\sec u + \tan u}du \\ &= \ln\|\sec u + \tan u\|+C \end{align} By multiplying the top and bottom by $$\csc u - \cot u$$, we can prove that $\csc u \,du=\ln\|\csc u - \cot u\|+C$ Example 1: Solve $\int_{0.5}^{1} x^2\cot x^3\,dx$ Solution: Let $$u=x^3$$, then $$du=3x^2\,dx$$ \begin{align} \int_{0.5}^{1} x^2\cot x^3\,dx &= \int_{0.5}^{1} \cot u\left(\frac{du}{3}\right) \\ &= \left(\ln\|\sin u\|\right)_{0.5}^{1} \\ \left(\ln\|\sin x^3\|\right)_{0.5}^{1} \\ &= \ln\|\sin (1)^3 - \ln\|\sin (0.5)^3 = 2.079\end{align} Example 2: Find the volume generated by revolving the region bounded by $$y=\sec x$$, $$x=0$$, $$x=\frac{\pi}{3}$$, and $$y=0$$ about the x-axis. Solution: The volume is given by the formula $V=\int_{a}^{b} \pi[f(x)]^2dx$ \begin{align} V&=\int_{0}^{\frac{\pi}{3}} \pi[\sec x]^2dx \\ &=\pi\left(\ln\|\sec x+\tan x\|\right)_{0}^{\frac{\pi}{3}} \\&=4.137 \end{align}
### Similar presentations Factors Factors and Multiples If one number divides another number without remainder then it is a factor of the number. Example 1: The factors of 10 are: 1, 2, 5, and 10 since: 10  1 = 10 10  2 = 510  5 = 210  10 = 1 Example 2: The factors of 15 are: 1, 3, 5, and 15 since: 15  1 = 15 15  3 = 515  5 = 315  15 = 1 17  1 = 17 17  17 = 1 Example 3: The factors of 17 are: 1, and 17 since: A prime number has only 2 factors, 1 and itself Factors and Multiples Find the factors of the following: 8121518 24303157 50568199 abcd efgh ijkl 1,2,4,81,2,3,4,6,121,3,5,151,2,3,6,9,18 1,2,3,4,6,8,12,241,2,3,5,6,10,15,30 1,31 1,2,5,10,25,501,2,4,7,8,14,28 1,3,9,27,811,3,9,11,33,99 1,3,19,57 If one number divides another number without remainder then it is a factor of the number. Factors and Multiples Some of the numbers in the pentagons are factors of the number in the central decagon. Which ones? 52 13 4 9 7 8 26 20 2 6 12 Factors and Multiples Some of the numbers in the pentagons are factors of the number in the central decagon. Which ones? 36 13 15 18 7 3 9 20 14 6 16 Factors and Multiples 64 32 10 18 8 4 9 24 14 6 16 Some of the numbers in the pentagons are factors of the number in the central decagon. Which ones? Factors and Multiples 84 32 12 15 8 4 3 28 14 6 16 Some of the numbers in the pentagons are factors of the number in the central decagon. Which ones? Factors and Multiples 144 36 12 74 8 2 4 28 14 72 18 Some of the numbers in the pentagons are factors of the number in the central decagon. Which ones? Prime Factors Factors and Multiples A prime number which divides into another number without leaving a remainder is called a prime factor of the number. Example 1: The prime factors of 12 are 2 and 3 since both of these are prime numbers that divide 12. Example 2: The prime factors of 15 are 3 and 5 since both of these are prime numbers that divide 15. Example 3: The prime factors of 42 are 2,3, and 7since these are prime numbers that divide 42. Example 4: The prime factors of 55 are 5 an 11 since these are prime numbers that divide 55. Factors and Multiples Find the prime factors of the following: 16202425 28493623 3042110210 abcd efgh ijkl 22 and 52 and 35 2 and 772 and 323 2, 3, and 52, 3 and 72, 5, and 112,3,5 and 7 A prime number which divides into another number without leaving a remainder is called a prime factor of the number. Factors and Multiples Find prime factors of the number in the middle. 18 13 4 9 7 8 1 12 2 6 3 Factors and Multiples Find prime factors of the number in the middle. 30 15 4 5 7 10 1 30 2 6 3 Factors and Multiples Find prime factors of the number in the middle. 42 3 4 5 7 1 21 6 2 9 Factors and Multiples Find prime factors of the number in the middle. 77 10 4 5 7 42 11 77 6 2 3 Factors and Multiples Find prime factors of the number in the middle. 105 3 4 5 7 15 11 21 6 2 9 HCF Factors and Multiples The highest common factor (HCF) of two or more numbers is the largest factor that is common to them all. Example 1: The HCF of 12 and 18 is 6 since this is largest number that will divide both. Example 2: The HCF of 15 and 30 is 15 since this is largest number that will divide both. Example 3: The HCF of 23 and 18 is 1 since this is largest number that will divide both (23 is prime). Example 4: The HCF of 8, 12, 28 is 4 since this is largest number that will divide all three numbers. The highest common factor (HCF) of two or more numbers is the largest factor that is common to them all. Find the HCF of the sets of numbers shown. HCF 9 and 15 HCF 12 and 30HCF 18 and 4 HCF 20 and 35 HCF 42 and 28HCF 36 and 27 HCF 9, 54, 18 HCF 24, 56, 16 HCF 60, 36, 24 3 62 5 14 9 9 8 12 a bc d e f g h i Factors and Multiples Find the HCF of the two numbers shown in the middle. 10 12 6 9 7 8 5 1 2 4 10 12 Factors and Multiples Find the HCF of the two numbers shown in the middle. 12 6 9 7 8 5 1 2 4 10 20 Factors and Multiples Find the HCF of the two numbers shown in the middle. 15 25 7 9 35 3 5 15 2 4 10 35 Factors and Multiples Find the HCF of the two numbers shown in the middle. 20 18 12 36 9 5 15 2 4 10 36 Factors and Multiples Find the HCF of the two numbers shown in the middle. 72 24 18 12 36 9 8 15 2 4 10 48 Factors and Multiples Find the HCF of the three numbers shown in the middle. 18 24 18 12 28 2 8 15 6 4 9 28 48 Factors and Multiples Find the HCF of the three numbers shown in the middle. 18 24 18 12 36 2 8 15 6 4 9 36 48 Factors and Multiples Find the HCF of the three numbers shown in the middle. 60 24 4 12 30 20 9 15 6 5 10 90 105 Multiples Factors and Multiples If a number is multiplied by another whole number then the new number formed is a multiple of the original. Example 1: Some multiples of 3 are: 3, 6, 9, and 15 since: 1 x 3 = 3 2 x 3 = 63 x 3 = 95 x 3 = 15 3 x 8 = 245 x 8 = 408 x 8 = 6410 x8 = 80 Example 2: Some multiples of 8 are: 24, 40, 64, and 80 since: Example 3: The first 4 multiples of 9 are 9, 18, 27, 36 since: 1 x 9 = 92 x 9 = 183 x 9 = 274 x 9 = 36 Factors and Multiples Find the first four multiples of the following: 2345 6879 10111213 abcd efgh ijkl 2, 4, 6, 83, 6, 9, 124, 8, 12, 165, 10, 15, 20 6,12,18,24 8,16,24,32 7,14,21,28 10,20,30,4011,22,33,4412,24,36,4813,26,39,52 If a number is multiplied by another whole number then the new number formed is a multiple of the original. 9,18,27,36 Factors and Multiples Some of the numbers in the pentagons are multiples of the number in the central decagon. Which ones? 4 13 4 9 7 8 36 20 2 6 12 Factors and Multiples Some of the numbers in the pentagons are multiples of the number in the central decagon. Which ones? 9 13 4 9 54 18 36 20 2 6 12 Factors and Multiples Some of the numbers in the pentagons are multiples of the number in the central decagon. Which ones? 8 4 8 9 56 18 32 25 2 16 12 Factors and Multiples Some of the numbers in the pentagons are multiples of the number in the central decagon. Which ones? 6 8 8 9 54 19 3 36 2 16 14 LCM Factors and Multiples The lowest common multiple (LCM) of two or more numbers is the smallest number that they will all divide into evenly without leaving a remainder. Example 1: The LCM of 6 and 12 is 12 since 12 is the smallest number that both 6 and 12 divide into without remainder. Example 2: The LCM of 8 and 20 is 40 since 40 is the smallest number that both 8 and 20 divide into without remainder. Example 3: The LCM of 9 and 5 is 45 since 45 is the smallest number that both 9 and 5 divide into without remainder. Example 4: The LCM of 2, 3 and 8 is 24 since 24 is the smallest number that 2, 3 and 8 divide into without remainder. Find the LCM of the sets of numbers shown. LCM 2 and 3 LCM 4 and 10LCM 6 and 18 LCM 9 and 36 LCM 7 and 9LCM 36 and 12 LCM 9, 54, 18 LCM 14, 28, 7 LCM 5, 6, 7 6 2018 36 63 36 54 28 210 The lowest common multiple (LCM) of two or more numbers is the smallest number that they will all divide into evenly without leaving a remainder. a bc d e f g h i Factors and Multiples Find the LCM of the numbers shown in the centre. 4 12 4 9 7 8 36 24 2 6 12 Factors and Multiples Find the LCM of the numbers shown in the centre. 6 12 48 4 24 8 48 18 2 6 9 8 Factors and Multiples Find the LCM of the numbers shown in the centre. 3 20 48 4 24 30 60 15 5 6 3 4 5 Factors and Multiples Find the LCM of the numbers shown in the centre. 15 90 4 24 30 20 60 5 6 3 12 4 Find the factors of the following: 8121518 24303157 505681100 abcd efgh ijkl If one number divides another number without remainder then it is a factor of the number. Worksheet 1 Find the prime factors of the following: 16202425 28493623 3042110210 abcd efgh ijkl A prime number which divides into another number without leaving a remainder is called a prime factor of the number. Worksheet 2 The highest common factor (HCF) of two or more numbers is the largest factor that is common to them all. Find the HCF of the sets of numbers shown. HCF 9 and 15 HCF 12 and 30HCF 18 and 4 HCF 20 and 35 HCF 42 and 28HCF 36 and 27 HCF 9, 54, 18 HCF 24, 56, 16 HCF 60, 36, 24 a bc d e f g h i Worksheet 3 Find the first four multiples of the following: 2345 6879 10111213 abcd efgh ijkl If a number is multiplied by another whole number then the new number formed is a multiple of the original. Worksheet 4 Find the LCM of the sets of numbers shown. LCM 2 and 3 LCM 4 and 10LCM 6 and 18 LCM 9 and 36 LCM 7 and 9LCM 36 and 12 LCM 9, 54, 18 LCM 14, 28, 7 LCM 5, 6, 7 The lowest common multiple (LCM) of two or more numbers is the smallest number that they will all divide into evenly without leaving a remainder. a bc d e f g h i Worksheet 5
# Product of 4 Consecutive Integers is One Less than Square ## Theorem Let $a$, $b$, $c$ and $d$ be consecutive integers. Then: $\exists n \in \Z: a b c d = n^2 - 1$ That is, the product of $a$, $b$, $c$ and $d$ is one less than a square. ## Proof 1 ### Lemma Let $a$, $b$, $c$ and $d$ be consecutive integers. Let us wish to prove that the product of $a$, $b$, $c$ and $d$ is one less than a square. Then it is sufficient to consider $a$, $b$, $c$ and $d$ all strictly positive. $\Box$ As $a$, $b$, $c$ and $d$ are all consecutive, we can express them as: $a$, $a + 1$, $a + 2$ and $a + 3$ where $a \ge 1$. Hence: $\ds a \paren {a + 1} \paren {a + 2} \paren {a + 3} + 1$ $=$ $\ds a^4 + 6 a^3 + 11 a^2 + 6 a + 1$ $\ds$ $=$ $\ds \paren {a^2 + 3 a + 1}^2$ by inspection Hence the result. ## Proof 2 ### Lemma Let $a$, $b$, $c$ and $d$ be consecutive integers. Let us wish to prove that the product of $a$, $b$, $c$ and $d$ is one less than a square. Then it is sufficient to consider $a$, $b$, $c$ and $d$ all strictly positive. $\Box$ As $a$, $b$, $c$ and $d$ are all consecutive, we can express them as: $a$, $a + 1$, $a + 2$ and $a + 3$ where $a \ge 1$. Then: $\ds a \paren {a + 3}$ $=$ $\ds a^2 + 3 a$ $\ds \paren {a + 1} \paren {a + 2}$ $=$ $\ds a^2 + 3 a + 2$ $\ds \leadsto \ \$ $\ds a \paren {a + 1} \paren {a + 2} \paren {a + 3}$ $=$ $\ds \paren {a^2 + 3 a} \paren {a^2 + 3 a + 2}$ $\ds$ $=$ $\ds \paren {n - 1} \paren {n + 1}$ where $n = a^2 + 3 a + 1$ $\ds$ $=$ $\ds n^2 - 1$ Difference of Two Squares $\blacksquare$
Education.com Try Brainzy Try Plus # The Six Hyperbolic Functions Help (page 2) (not rated) By — McGraw-Hill Professional Updated on Oct 3, 2011 ## The Six ### Two To Start Let x be a real number. The hyperbolic sine and the hyperbolic cosine can be defined in terms of powers of e , like this: sinh x = ( e xe x )/2 cosh x = ( e x + e x )/2 If these look intimidating, just remember that using them involves nothing more than entering numbers into a calculator and hitting certain keys in the correct sequence. In a theoretical course, you will find other ways of expressing the hyperbolic sine and cosine functions, but for our purposes, the above two formulas are sufficient. ### The Other Four The remaining four hyperbolic functions follow from the hyperbolic sine and the hyperbolic cosine, like this: tanh x = sinh x /cosh x csch x = 1/sinh x sech x = 1/cosh x coth x = cosh x /sinh x In terms of exponential functions, they are expressed this way: tanh x = ( e xe x )/( e x + e x ) csch x = 2/( e xe x ) sech x = 2/( e x + e x ) coth x = ( e x + e x )/( e xe x ) Now let’s look at the graphs of the six hyperbolic functions. As is the case with the inverses of the circular functions, the domain and/or range of the inverse of a hyperbolic function may have to be restricted to ensure that there is never more than one ordinate ( y value) for a given abscissa ( x value). ## Hyperbolic Sine Figure 4-1 is a graph of the function y = sinh x . Its domain and range both extend over the entire set of real numbers. Fig. 4-1. Graph of the hyperbolic sine function. ## Hyperbolic Cosine Figure 4-2 is a graph of the function y = cosh x . Its domain extends over the whole set of real numbers, and its range is the set of real numbers y greater than or equal to 1. Fig. 4-2 . Graph of the hyperbolic cosine function. ## Hyperbolic Tangent Figure 4-3 is a graph of the function y = tanh x . Its domain encompasses the entire set of real numbers. The range of the hyperbolic tangent function is limited to the set of real numbers y between, but not including, –1 and 1; that is, –1 < y < 1. Fig. 4-3. Graph of the hyperbolic tangent function. ## Hyperbolic Cosecant Figure 4-4 is a graph of the function y = csch x . Its domain encompasses the set of real numbers x such that x ≠ 0. The range of the hyperbolic cotangent function encompasses the set of real numbers y such that y ≠ 0. Fig. 4-4. Graph of the hyperbolic cosecant function. ## Hyperbolic Secant Figure 4-5 is a graph of the function y = sech x. Its domain encompasses the entire set of real numbers. Its range is limited to the set of real numbers y greater than 0 but less than or equal to 1; that is, 0 < y ≤ 1. Fig. 4-5. Graph of the hyperbolic secant function. ## Hyperbolic Cotangent Figure 4-6 is an approximate graph of the function y = coth x . Its domain encompasses the entire set of real numbers x such that x ≠ 0. The range of the hyperbolic cotangent function encompasses the set of real numbers y less than –1 or greater than 1; that is, y < –1 or y > 1. Fig. 4-6. Graph of the hyperbolic cotangent function. 150 Characters allowed ### Related Questions #### Q: See More Questions Top Worksheet Slideshows
USING OUR SERVICES YOU AGREE TO OUR USE OF COOKIES # Is 32 A Prime Number? • No the number 32 is not a prime number. • Thirty-two is a composite number. Because 32 has more divisors than 1 and itself. ## Prime Factorization Of 32 • Prime factors of number 32 are: 2 • Equcation of 32 is: 2 * 2 * 2 * 2 * 2 • The smallest common factor of 32 is number 2 • Highest or greatest common factor GCF of 32 is number 2 ## How To Calculate Prime Number Factors • How do you calculate natural number factors? To get the number that you are factoring just multiply whatever number in the set of whole numbers with another in the same set. For example 7 has two factors 1 and 7. Number 6 has four factors 1, 2, 3 and 6 itself. • It is simple to factor numbers in a natural numbers set. Because all numbers have a minimum of two factors(one and itself). For finding other factors you will start to divide the number starting from 2 and keep on going with dividers increasing until reaching the number that was divided by 2 in the beginning. All numbers without remainders are factors including the divider itself. • Let's create an example for factorization with the number nine. It's not dividable by 2 evenly that's why we skip it(Remembe 4,5 so you know when to stop later). Nine can be divided by 3, now add 3 to your factors. Work your way up until you arrive to 5 (9 divided by 2, rounded up). In the end you have 1, 3 and 9 as a list of factors. ## Mathematical Information About Numbers 3 2 • About Number 3. Three is the first odd prime number and the second smallest right after number two. At the same time it is the first Mersenne prime (2 ^ 2-1), the first Fermat prime (2 ^ {2 ^ 0} +1), the second Sophie Germain prime and the second Mersenne prime exponent. It is the fourth number of the Fibonacci sequence and the second one that is unique. The triangle is the simplest geometric figure in the plane. With the calculation of its sizes deals trigonometry. Rule of three: If the sum of the digits of a number is a multiple of three, the underlying number is divisible by three. • About Number 2. Two is the smallest and the only even prime number. Also it's the only prime which is followed by another prime number three. All even numbers are divisible by 2. Two is the third number of the Fibonacci sequence. Gottfried Wilhelm Leibniz discovered the dual system (binary or binary system) that uses only two digits to represent numbers. It witnessed the development of digital technology for a proliferation. Because of this, it is the best known and most important number system in addition to the commonly used decimal system. ## What is a prime number? Prime numbers or primes are natural numbers greater than 1 that are only divisible by 1 and with itself. The number of primes is infinite. Natural numbers bigger than 1 that are not prime numbers are called composite numbers. Primes can thus be considered the basic building blocks of the natural numbers. There are infinitely many primes, as demonstrated by Euclid around 300 BC. The property of being prime (or not) is called primality. In number theory, the prime number theorem describes the asymptotic distribution of the prime numbers among the positive integers. It formalizes the intuitive idea that primes become less common as they become larger. Primes are used in several routines in information technology, such as public-key cryptography, which makes use of properties such as the difficulty of factoring large numbers into their prime factors. © Mathspage.com \| Privacy \| Contact \| info [at] Mathspage [dot] com
[next] [prev] [prev-tail] [tail] [up] 1.8Exact equations Note: 1–2 lectures, §1.6 in [EP], §2.6 in [BD] Another type of equation that comes up quite often in physics and engineering is an exact equation. Suppose is a function of two variables, which we call the potential function. The naming should suggest potential energy, or electric potential. Exact equations and potential functions appear when there is a conservation law at play, such as conservation of energy. Let us make up a simple example. Let We are interested in the lines of constant energy, that is lines where the energy is conserved; we want curves where , for some constant . In our example, the curves are circles. See Figure 1.15. We take the total derivative of : For convenience, we will make use of the notation of and . In our example, We apply the total derivative to , to ﬁnd the diﬀerential equation . The diﬀerential equation we obtain in such a way has the form An equation of this form is called exact if it was obtained as for some potential function . In our toy example, we obtain the equation Since we obtained this equation by diﬀerentiating , the equation is exact. We often wish to solve for in terms of . In our example, An interpretation of the setup is that at each point is a vector in the plane, that is, a direction and a magnitude. As and are functions of , we have a vector ﬁeld. The particular vector ﬁeld that comes from an exact equation is a so-called conservative vector ﬁeld, that is, a vector ﬁeld that comes with a potential function , such that Let be a path in the plane starting at and ending at . If we think of as force, then the work required to move along is That is, the work done only depends on endpoints, that is where we start and where we end. For example, suppose is gravitational potential. The derivative of given by is the gravitational force. What we are saying is that the work required to move a heavy box from the ground ﬂoor to the roof, only depends on the change in potential energy. That is, the work done is the same no matter what path we took; if we took the stairs or the elevator. Although if we took the elevator, the elevator is doing the work for us. The curves are those where no work need be done, such as the heavy box sliding along without accelerating or breaking on a perfectly ﬂat roof, on a cart with incredibly well oiled wheels. An exact equation is a conservative vector ﬁeld, and the implicit solution of this equation is the potential function. 1.8.1Solving exact equations Now you, the reader, should ask: Where did we solve a diﬀerential equation? Well, in applications we generally know and , but we do not know . That is, we may have just started with , or perhaps even It is up to us to ﬁnd some potential that works. Many diﬀerent will work; adding a constant to does not change the equation. Once we have a potential function , the equation gives an implicit solution of the ODE. Example 1.8.1: Let us ﬁnd the general solution to . Forget we knew what was. If we know that this is an exact equation, we start looking for a potential function . We have and . If exists, it must be such that . Integrate in the variable to ﬁnd (1.5) for some function . The function is the “constant of integration”, though it is only constant as far as is concerned, and may still depend on . Now diﬀerentiate (1.5) in and set it equal to , which is what is supposed to be: Integrating, we ﬁnd . We could add a constant of integration if we wanted to, but there is no need. We found . Next for a constant , we solve for in terms of . In this case, we obtain as we did before. Exercise 1.8.1: Why did we not need to add a constant of integration when integrating ? Add a constant of integration, say , and see what you get. What is the diﬀerence from what we got above, and why does it not matter? The procedure, once we know that the equation is exact, is: (i) Integrate in resulting in . (ii) Diﬀerentiate this in , and set that equal to , so that we may ﬁnd by integration. The procedure can also be done by ﬁrst integrating in and then diﬀerentiating in . Pretty easy huh? Let’s try this again. Example 1.8.2: Consider now . OK, so and . We try to proceed as before. Suppose exists. Then . We integrate: for some function . Diﬀerentiate in and set equal to : But there is no way to satisfy this requirement! The function cannot be written as plus a function of . The equation is not exact; no potential function exists. Is there an easier way to check for the existence of , other than failing in trying to ﬁnd it? Turns out there is. Suppose and . Then as long as the second derivatives are continuous, Let us state it as a theorem. Usually this is called the Poincarè Lemma5. Theorem 1.8.1 (Poincarè). If and are continuously diﬀerentiable functions of , and . Then near any point there is a function such that and . The theorem doesn’t give us a global deﬁned everywhere. In general, we can only ﬁnd the potential locally, near some initial point. By this time, we have come to expect this from diﬀerential equations. Let us return to the example above where and . Notice and , which are clearly not equal. The equation is not exact. Example 1.8.3: Solve We write the equation as so and . Then The equation is exact. Integrating in , we ﬁnd Diﬀerentiating in and setting to , we ﬁnd So , and will work. Take . We wish to solve . First let us ﬁnd . As then . Therefore , so . Now we solve for to get Example 1.8.4: Solve We leave to the reader to check that . This vector ﬁeld is not conservative if considered as a vector ﬁeld of the entire plane minus the origin. The problem is that if a curve was a circle around the origin, say starting at and ending at going counterclockwise, then if existed we would expect That is nonsense! We leave the computation of the path integral to the interested reader, or you can consult your multivariable calculus textbook. So there is no potential function deﬁned everywhere outside the origin . If we think back to the theorem, it did not guarantee such a function anyway. It only guaranteed a potential function locally. As we start at the point . Considering and integrating in or in we ﬁnd So the implicit solution is . Solving, . That is, the solution is a straight line. Solving gives us that , and so is the desired solution. See Figure 1.16, and note that the solution only exists for . Example 1.8.5: Solve The reader should check that this equation is exact. Let and . We follow the procdure for exact equations and Therefore or and . We try to solve . We easily solve for and then just take the square root: When , the term in front of vanishes. You can also see that our solution is not valid in that case. However, one could in that case try to solve for in terms of starting from the implicit solution . In this case the solution is somewhat messy and we leave it as implicit. 1.8.2Integrating factors Sometimes an equation is not exact, but it can be made exact by multiplying with a function . That is, perhaps for some nonzero function , is exact. Any solution to this new equation is also a solution to . In fact, a linear equation is always such an equation. Let be the integrating factor for a linear equation. Multiply the equation by and write it in the form of . Then , so , while , so . In other words, we have an exact equation. So integrating factors for linear functions are just a special case of integrating factors for exact equations. But how do we ﬁnd the integrating factor ? Well, given an equation should be a function such that Therefore, At ﬁrst it may seem we replaced one diﬀerential equation by another one. True, but hope is not lost. A strategy that often works is to look for a that is a function of alone, or a function of alone. If is a function of alone, that is , then we write instead of , and is just zero. Then In particular, ought to be a function of alone (not depend on ). If so, then we have a linear equation Letting , we solve using the standard integrating factor method, to ﬁnd . The constant in the solution is not relevant, we need any nonzero solution, we take . So is the integrating factor. Similarly we could try a function of the form . Then In particular ought to be a function of alone. If so, then we have a linear equation Letting we ﬁnd , and we can take . So is the integrating factor. Example 1.8.6: Solve Let and . Compute As this is not zero, the equation is not exact. We notice is a function of alone. We compute the integrating factor We multiply our given equation by to obtain which is an exact equation that we solved in Example 1.8.5. The solution was Example 1.8.7: Solve First compute As this is not zero, the equation is not exact. We observe is a function of alone. We compute the integrating factor Therefore we look at the exact equation The reader should double check that this equation is exact. We follow the procdure for exact equations and (1.6) Consequently or . Thus . It is not possible to solve for in terms of elementary functions, so let us be content with the implicit solution: We are looking for the general solution and we divided by above. We should check what happens when , as the equation itself makes perfect sense in that case. We plug in to ﬁnd the equation is satisﬁed. So is also a solution. 1.8.3Exercises Exercise 1.8.2: Solve the following exact equations, implicit general solutions will suﬃce: a) b) c) d) Exercise 1.8.3: Find the integrating factor for the following equations making them into exact equations: a) b) c) d) Exercise 1.8.4: Suppose you have an equation of the form: . a) Show it is exact. b) Find the form of the potential function in terms of and . Exercise 1.8.5: Suppose that we have the equation . a) Is this equation exact? b) Find the general solution using a deﬁnite integral. Exercise 1.8.6: Find the potential function of the exact equation in two diﬀerent ways. a) Integrate in terms of and then diﬀerentiate in and set to . b) Integrate in terms of and then diﬀerentiate in and set to . Exercise 1.8.7: A function is said to be harmonic function if . a) Show that is an exact equation. Therefore there exists (at least locally) the so-called harmonic conjugate function such that and . Verify that the following are harmonic and ﬁnd the corresponding harmonic conjugates : b) c) d) Exercise 1.8.101: Solve the following exact equations, implicit general solutions will suﬃce: a) b) c) d) Exercise 1.8.102: Find the integrating factor for the following equations making them into exact equations: a) b) c) d) Exercise 1.8.103: a) Show that every separable equation can be written as an exact equation, and verify that it is indeed exact. b) Using this rewrite as an exact equation, solve it and verify that the solution is the same as it was in Example 1.3.1. 5Named for the French polymath Jules Henri Poincarè (1854–1912).
# 4 Chapter 4: Two Dimensional Motion Chapter 4 of the textbook # Section 4.1: Motion in Two and Three Dimensions We began in the last chapter with the most simple kind of motion – motion in a straight line. Now we’re going to generalize this motion from a single dimension (up/down, left/right, or forward/back) to combinations of all of these dimensions. We’ll use the same three equations to describe this motion, but now instead of restricting ourselves to one direction at a time, we’ll use a combination of straight-line motion in two or three dimensions at once. The important thing to realize here is that it’s the same motion we saw in the last chapter – we’re still moving at a constant acceleration, but now we might have different motion in the direction than we did in the direction. It’s vital to remember that changing your velocity, position, or acceleration does not change anything about your position, velocity, or acceleration in the and directions. The three directions are to each other – changing one does not change the others. Because we will now be talking about motion in the , , and sometimes direction, we’re going to introduce a new way of writing position that doesn’t limit us to constantly referring to the direction. The new position vector is written as: The three equations above are all equivalent ways of writing the position vector . As a result, the , the straight-line displacement between any two locations in three dimensions, is given by: Now that we have a position vector, we can re-define our velocity vector in terms of the position vector. The three equations above are equivalent statements of the velocity vector. Finally, we can do the same with the acceleration vector . Textbook 4.2: Acceleration Vector # Section 4.2: Projectile Motion An object subject to a constant acceleration at an angle relative to its motion will move in a two-dimensional path known as its . Most of the objects we encounter will be subject to the constant acceleration due to the gravitational force. Any object subject to gravity is known as a . Two-dimensional kinematics works the same way as one-dimensional kinematics, but now acting in two (perpendicular) directions. The important points to remember here are (1) that the and directions are orthogonal, which means they are perpendicular and that one does not rely on the other (a change in does not mean you have a change in , and (2) that we will have a constant acceleration acting on our object, in either the direction, the direction, or in some combination of the and directions. Our derivation of the kinematics equations depended on a constant, non-changing acceleration. We again use the one-dimensional kinematics equations, but now we have one set in the direction and one set in the direction. The only variable they have in common is the time measured in seconds. As we did with the one-dimensional kinematics problems, begin by drawing a picture of the situation and labeling all the variables you know as well as identifying those you are looking for. Remember you may need to break your velocity, given distance, and/or acceleration into two dimensions to make sure all of your vectors are along the or axes. The overall problem-solving strategies are the same. Textbook 4.3: Projectile Motion Khan Academy: Projectile Motion # Section 4.3: Kinematics Problem-Solving We’ll start out using a set of steps to solve a kinematics problem. Often the hardest part in solving these problems is simply getting started, so use these steps as a road map. Start by reading the problem. Then, draw a picture. This doesn’t have to be super-accurate, but it should include everything involved in the problem. You might want to include labels for things and arrows indicating where things are going. Include the path you think objects will take. Pick a coordinate system. Right now, a normal coordinate system is a good place to start. It doesn’t matter where you put the origin ( = 0, = 0), you’ll get the same answer no matter where you put it, but some choices can make the problem a bit easier to solve, as you’ll discover. Draw the coordinate system on your picture. Label things you know that appear in your equations – positions, velocities, accelerations. Write down what you know, and then, nearby, what you’re looking for. Write down the kinematics equations. Figure out how to get what you want out of what you have, now that you’ve got it all written out in front of you. Let’s take a look at an example in one dimension. Suppose you’re standing in the window of your dorm room, and a friend is standing outside on the ground. You toss a ball straight up in the air at 8 m/s. 2.85 seconds later, your friend on the ground catches the ball. How far below you is your friend? Here’s my picture. I’m throwing the ball straight up. I know it goes up for a while, slows to a stop, then starts heading back down until my friend catches it. Next, we’ll pick a coordinate system. To make the problem a bit simpler, I’m going to choose the height at which my friend catches the ball to be the = 0 m position, and the direction goes up from there. Then I’m going to label everything I know. Now let’s write down a list of everything we know, what we’re looking for, and include our equations. Since the ball is only going straight up and down, not side-to-side, we only have to worry about one dimension here. However, there are three different positions we know the ball passes through – starting at our hand, reaching it’s highest point, and reaching it’s lowest point at my friend’s hand. So I’m going to label these positions , , and . At those positions, the ball has velocities , , and . The ball arrives at those positions at s, , and . We will often deal with problems where we need to consider multiple time frames. If we see that ahead of time by drawing a picture and labeling it appropriately, it can make the problem easier to solve. Here, we want to know the distance from me to my friend. That’s the same thing as the displacement of the ball from the beginning to end of it’s path. Alternatively, you could also solve this problem by determining how far up the ball goes until it reaches it’s , then subtract that from the distance it falls from there. Either way you decide to solve this problem, you need to keep track of which position, velocity, and time you’re talking about. The kinematics equations apply to any one part of the problem, or the entire problem as a whole, but you need to make sure you’re not, say, trying to find the distance from (from your hand to the apex) by using the time between the apex and your friend’s hand (. That’s not going to work out correctly! Whichever way you choose to solve this problem, make sure the positions and velocities and accelerations and times all correspond to the same interval! The easiest way to solve this problem is to note we have , , and a constant acceleration that applies at all times. Therefore, we can use for and solve for in one step. Doing this, we find m. Why is it negative? Because what we’ve found is , which in this case is , so we’re subtracting a lower position from a higher one. That’s where the negative sign comes from – we need to go down from our initial height to reach our final height. So our friend is 17.0 m below us. Let’s look at the alternative way of solving this problem, where we use the intermediate step of finding the height of the ball at it’s apex. The problem is solved in the figure below. Note the use of positions 1, 2, and 3 and the notation. Also, note we got the same answer in the end! You’ll often find there are multiple ways of solving a problem in physics to get to the same answer. One might take less time/steps than another, but it’s just as correct in the end. # In Class Group Problem 4.1 You kick a soccer ball. Its resulting trajectory is shown below. The ball moved from left to right. Three specific positions the ball passed through are labeled , , and . is the position of the ball as it leaves the ground, and is the position of the ball just before it hits the ground. At which labeled position(s) has the ball reached its apex? Draw the range of the maximum value of on the graph. Draw the range of the maximum value of on the graph. Where does the ball have the largest velocity? Why? Where does the ball have the smallest velocity? Why? Where does the ball feel the largest acceleration? Why? The time it takes the ball to go from position to position is . How long does it take the ball to go from position to position ? How do you know? Under which circumstances would your answer to the previous question be incorrect? # In Class Group Work Problem 4.2 You have two identical balls. You drop one from a height of 2.0 m, and throw the other parallel to the ground at = 3.5 m/s from a height of 2.0 m. How long does it take the first ball to hit the ground? How long does it take the second ball to hit the ground? Compare your answer to your answer for the first ball. # In Class Group Problem 4.3 A soccer player kicks the ball with an initial velocity of 17 m/s in the horizontal direction and 13 m/s in the vertical direction. How fast is the ball moving when it hits the ground? Assume it hits the ground at the same level at which it left the ground. Note: the answer is not zero! is always the velocity just before you hit the ground. How long was the ball in the air? How high did the ball get above the ground? How far did the ball travel in the direction? Why can’t you assume that is equal to zero? # In Class Group Problem 4.4 A golf ball is hit at an angle of 25 and an initial velocity of 45 m/s. 2.5 seconds later, the ball hits the side of a building. What were the initial velocities of the ball in the and directions, and ? Did the ball hit the building before or after it reached its peak in the direction? Show your work when answering this question. How far had the ball traveled in the and directions when it hit the building? # In Class Group Problem 4.5 Arya is trying to jump her horse over a canyon. If the horse is running at 11 m/s and runs off the cliff horizontally, as least how far down does the next cliff 6.5 m away need to be in order for her to safely make the jump? Draw a picture of this situation and label everything you know. Check your picture with others at your table before moving on to make sure you all agree. How far down does the other side of the cliff need to be for her to just barely make the jump? # In Class Group Problem 4.6 At the Bristol 4th of July carnival, you can win a stuffed panda bear if you manage to toss a quarter into a cup on a shelf. The shelf is 2.5 m away from you. If you toss the quarter with a velocity of 7.2 m/s at an angle of 48, the quarter lands in the cup. How high above the level where you threw the quarter does the cup sit? # In Class Group Problem 4.7 Because of your physics background, you have been hired as a consultant for a new movie about Galileo. In one scene, he climbs up to the top of a tower and, in frustration over the people who ridicule his theories, throws a rock at a group of them standing on the ground. The rock leaves his hand at 30 below the horizontal. The script calls for the rock to land 15 m from the base of the tower near a group of his detractors. It is important for the script that the rock take precisely 3.0 seconds to hit the ground so that there is time for a good expressive close-up. The set coordinator is concerned that the rock will hit the ground with too much speed causing cement chips from the plaza to injure one of the high priced actors. You are told to calculate the speed of the rock hitting the ground. # In Class Group Problem 4.8 Because of your knowledge of physics, you have been hired as a consultant for a new James Bond movie, “Oldfinger”. In one scene, Bond jumps horizontally off the top of a cliff to escape a villain. To make the stunt more dramatic, the cliff has a horizontal ledge a distance beneath the top of the cliff which extends a distance from the vertical face of the cliff. The stunt coordinator wants you to determine the minimum horizontal speed, in terms of and , with which Bond must jump so that he misses the edge. # In Class Group Problem 4.9 Your friend has decided to make some money at the next Bristol Carnival by inventing a game of skill. In the game she has developed so far, the customer shoots a rifle at a 5.0 cm diameter target falling straight down. Anyone who hits the target in the center wins a stuffed animal. Each shot costs 50 cents. The rifle would be mounted on a pivot 1.0 meter above the ground so that it can point in any direction at any angle. When shooting, the customer stands 100 m from where the target would hit the ground if the bullet misses. At the instant the bullet leaves the rifle (with a muzzle velocity of 1200 ft/sec according to the manual) the target is released from its holder 7.0 meters above the ground. Your friend asks you to try out the game which she has set up on a farm outside of town. Before you fire the gun, calculate where you should aim. # In Class Group Problem 4.10 A dare-devil is planning to jump a narrow section of the Grand Canyon on his rocket-propelled motorcycle. He fires up the rockets (which accelerate the motorcycle in the horizontal direction) and drives towards the cliff, leaving the edge of the cliff traveling horizontal to the ground at 45 m/s. The rockets maintain a constant acceleration all the way across the canyon, and he lands 30 m horizontally and 2 m vertically below where he left the other edge of the canyon. What minimum acceleration did the rockets have to provide for him to make a successful landing on the other side? # Practice Exam Question 4.1 These questions are provided for practice purposes. There is no guarantee a problem similar to this one will be on your exam. Wile E. Coyote is standing on a cliff 13 vertical meters above a canyon in which the Road Runner is napping. If the Road Runner is napping 62 horizontal meters from the bottom of the vertical cliff, how fast does Coyote need to run off the edge of the horizontal cliff to land directly on top of the Road Runner? # Practice Exam Question 4.2 These questions are provided for practice purposes. There is no guarantee a problem similar to this one will be on your exam. Your friend challenges you to a go-kart race. She is driving at her constant maximum speed of 9.2 m/s when she passes you, and you are driving at 1.8 m/s as she passes. You hit the accelerator as she passes and accelerate at 0.20 m/s until you pass her. How long does it take you to catch up to her? # Practice Exam Question 4.3 These questions are provided for practice purposes. There is no guarantee a problem similar to this one will be on your exam. During a tournament, a golfer hits her ball off the tee at a velocity of 55 m/s at an angle of 22 up from the horizontal. The ball lands on the fairway 13 m below her position. How far did the ball travel horizontally (what is )? # Practice Exam Question 4.4 These questions are provided for practice purposes. There is no guarantee a problem similar to this one will be on your exam. You are watching people practicing archery when you wonder how fast an arrow is shot from a bow. With a flash of insight you remember your physics and see how you can easily determine what you want to know by a simple measurement. You ask one of the archers to pull back her bowstring as far as possible and shoot an arrow horizontally. The arrow strikes the ground at an angle of 86 degrees from the vertical at 100 feet from the archer. What was the initial velocity (magnitude and direction) of the arrow? # Practice Exam Question 4.5 These questions are provided for practice purposes. There is no guarantee a problem similar to this one will be on your exam. A cliff diver positions herself on a cliff that angles downwards towards the edge. The length of the top of the cliff () is 44.0 m and the angle of the cliff is = 19.0 below the horizontal. The cliff diver runs towards the edge of the cliff with a constant speed, and reaches the edge of the cliff in a time of 5.80 s. After running straight off the edge of the cliff (without jumping up), the diver falls = 26.0 m before hitting the water. Assume there is no air friction. How long does it take her to hit the water after leaving the edge of the cliff? How far does she travel horizontally () after leaving the edge of the cliff before hitting the water? # Practice Exam Question 4.6 These questions are provided for practice purposes. There is no guarantee a problem similar to this one will be on your exam. Starbuck and Apollo have landed their spaceship, a Viper, on a strangely shaped moon while looking for water. This particular moon has a sideways gravitational field with a magnitude of 7.38 m/s in a direction pointing 42 down from the axis. While standing on a ledge 8.00 m above the level ground, Starbuck kicks a rock at Apollo, who is standing on the ground. The rock hits the ground at his feet. If she kicked the rock up into the air at 12.5 m/s and at an angle of 17 above the horizontal, how long does it take for the rock to hit the ground at Apollo’s feet? What is the displacement between Starbuck and Apollo?
Courses Courses for Kids Free study material Offline Centres More Store # A man in a balloon rising vertically with an acceleration of $4.9\,m/{s^2}$ releases a ball 2 sec after the balloon is let go from the ground. The greatest height above the ground reaches by the ball is $(g = 9.8\,m/{s^2})$A) $14.7\,m$B) $19.6\,m$C) $9.8\,m$D) $24.5\,m$ Last updated date: 20th Sep 2024 Total views: 81.3k Views today: 0.81k Verified 81.3k+ views Hint: In this solution, we will use the second and the third equation of kinematics. We will first find the height of the balloon at 2 seconds and then the maximum possible height that can be achieved by the ball. Formula used: In this solution, we will use the following formulae: 1- First equation of kinematics: $v = u + at$ where $v$ is the final velocity of an object with initial velocity $u$ under acceleration $a$ in time $t$ 2- Second equation of kinematics: $d = ut + \dfrac{1}{2}a{t^2}$where $d$ is the distance travelled by the object with initial velocity $u$ under acceleration $a$ in time $t$ 3-Third equation of kinematics: ${v^2} = {u^2} + 2ad$ where $v$ is the final velocity of the object, $d$ is the distance the object travels We know that the balloon starts rising with an acceleration of $4.9\,m/{s^2}$ . Initially, the velocity of the balloon is zero. Then in two seconds, it will achieve a height of $d = 0(2) + \dfrac{1}{2}(4.9){(2)^2}$ $\Rightarrow d = 9.8\,m$ At this point, the man releases the ball with an initial velocity equal to the velocity of the balloon. The velocity of the balloon at 2 seconds will be $v = 0 + (4.9)(2)$ $\Rightarrow v = 9.8\,m/s$ This will also be the velocity of the ball when it is released. Then the extra height achieved by the ball will be determined from the third equation of kinematics as ${0^2} = {2^2} - 2(9.8)(h)$ Which give us the height $h = \dfrac{4}{{2 \times 9.8}}$ $\Rightarrow h = 4.9\,m$ Hence the maximum height of the balloon above the ground will be $h' = h + d$ $\Rightarrow h' = 4.9 + 9.8$ Which gives us $h' = 14.7\,m$ hence the correct choice will be option (A). Note: When the ball is released from the balloon, the only force that will act on it is gravitational force and hence it will experience gravitational acceleration. The gravitational acceleration will be downwards, which is why we use a negative sign in the third equation of kinematics. After reaching the topmost point, the ball will start falling back to the ground.
# Plus One Math's Solution Ex 3.1 Chapter 3 Trigonometric Functions The solutions for the Trigonometric Functions are not readily available. While many schools may have a ready-made solution for this set in their school textbook, some might not. This is where our solution will be useful. In this article, you will find detailed solutions provided by us for the above set.Trigonometric Functions (Key Concept Reference) describes some basic and advanced uses of trigonometric functions, including identities, graph transformations, inverse functions, solutions of triangles, and polar coordinates. Ncert Plus one Maths chapter-wise textbook solution for chapter 3 Trigonometric Functions Exercise 3.1. It contains detailed solutions for each question which have prepared by expert teachers to make each answer easily understand the students. they are well arranged solutions so that students would be able to understand easily. Board SCERT, Kerala Text Book NCERT Based Class Plus One Subject Math's Textbook Solution Chapter Chapter 3 Exercise Ex 3.1 Chapter Name Trigonometric Functions Category Plus One Kerala ## Kerala Syllabus Plus One Math's Textbook Solution Chapter 3 Trigonometric Functions Exercises 3.1 ### Chapter 3 Trigonometric Functions Textbook Solution Kerala plus One maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state. ### Chapter 3 Trigonometric Functions Exercise 3.1 Find the radian measures corresponding to the following degree measures: (i) 25° (ii) – 47° 30' (iii) 240° (iv) 520° (i) 25° We know that 180° = π radian (ii) –47° 30' –47° 30' = degree [1° = 60'] degree (iii) 240° We know that 180° = π radian (iv) 520° We know that 180° = π radian Find the degree measures corresponding to the following radian measures . (i) (ii) – 4 (iii) (iv) (i) We know that π radian = 180° (ii) – 4 We know that π radian = 180° (iii) We know that π radian = 180° (iv) We know that π radian = 180° A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? Number of revolutions made by the wheel in 1 minute = 360 ∴Number of revolutions made by the wheel in 1 second = In one complete revolution, the wheel turns an angle of 2π radian. Hence, in 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e., Thus, in one second, the wheel turns an angle of 12π radian. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm. We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then Therefore, forr = 100 cm, l = 22 cm, we have Thus, the required angle is 12°36′. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord. Diameter of the circle = 40 cm ∴Radius (r) of the circle = Let AB be a chord (length = 20 cm) of the circle. In ΔOAB, OA = OB = Radius of circle = 20 cm Also, AB = 20 cm Thus, ΔOAB is an equilateral triangle. ∴θ = 60° = We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then. Thus, the length of the minor arc of the chord is. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii. Let the radii of the two circles be and. Let an arc of length l subtend an angle of 60° at the centre of the circle of radius r1, while let an arc of length subtend an angle of 75° at the centre of the circle of radius r2. Now, 60° =and 75° = We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then. Thus, the ratio of the radii is 5:4. Find the angle in radian though which a pendulum swings if its length is 75 cm and the tip describes an arc of length (i) 10 cm (ii) 15 cm (iii) 21 cm We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then. It is given that r = 75 cm (i) Here, l = 10 cm (ii) Here, = 15 cm (iii) Here, = 21 cm Find the values of other five trigonometric functions if x lies in third quadrant. Since lies in the 3rd quadrant, the value of sec x will be negative. Find the values of other five trigonometric functions if x lies in fourth quadrant. Since x lies in the 4th quadrant, the value of sin x will be negative. Find the values of other five trigonometric functions if x lies in second quadrant. Since x lies in the 2nd quadrant, the value of sec x will be negative. ∴sec x = Find the value of the trigonometric function sin 765° It is known that the values of sin repeat after an interval of 2π or 360°. #### Chapter 3 Trigonometric Functions EX 3.1 Solution- Preview Feel free to comment and share this article if you found it useful. Give your valuable suggestions in the comment session or contact us for any details regarding HSE Kerala Plus one syllabus, Previous year question papers, and other study materials.
# COMPARISON MEASURES OF CENTRAL TENDENCY & VARIABILITY EXERCISE 8/5/2013. MEASURE OF CENTRAL TENDENCY: MODE (Mo) MEASURE OF CENTRAL TENDENCY: MODE (Mo) Save this PDF as: Size: px Start display at page: Download "COMPARISON MEASURES OF CENTRAL TENDENCY & VARIABILITY EXERCISE 8/5/2013. MEASURE OF CENTRAL TENDENCY: MODE (Mo) MEASURE OF CENTRAL TENDENCY: MODE (Mo)" ## Transcription 1 COMPARISON MEASURES OF CENTRAL TENDENCY & VARIABILITY Prepared by: Jess Roel Q. Pesole CENTRAL TENDENCY -what is average or typical in a distribution Commonly Measures: 1. Mode. Median 3. Mean quantified characteristics of a distribution VARIABILITY -extent to which scores are spread out in a distribution Measures: 1. Range. Standard 3. Variance (Mo) The most frequent score in the distribution For ungroupeddata, it is the score or category that occurs most often in a distribution Example (for ungrouped data): 1,, 3, 4, 4, 5, 5, 6, 6, 6, 7, 8, 9, 10 Mo = 6 (Mo) For grouped data, it is the midpoint of the interval containing the largest number of cases Example: Mo = (44 +40) / = 4 Therefore, Mo = 4. Class Interval f (Mo) MAJOR MINOR Find the mode of the following series of numbers. 1. 4, 1, 0, 19, 18, 17, 16, 15, 14, 1., 0, 19, 18, 18, 18, 18, 16, 16, , 7, 7, 6, 6, 6, 6, 4, 4, 1 2 Determine the mode for the following frequency distribution. Class Interval f cf c% N = 50 MEDIAN (Mdn) Scale value below which 50 percent of scores fall Similar to P 50 For ungroupeddata, it is the centermost score if the number of scores are odd If it is even, it is the average of the two centermost scores Position of the median = (N + 1) / MEDIAN (Mdn) Example (for ungrouped scores): 11, 1, 13, 16, 17, 0, 5 N = 7 (odd) Therefore, the median is 16 11, 1, 13, 16, 17, 0, 5, 6 N = 8 Mdn= () = 16.5 Therefore, the median is 16.5 Position of Median = (N+ 1) / = (7 + 1) / = 8 / = 4 Position of Median = (N+ 1) / = (8 + 1) / = 9 / = 4.5 MEDIAN (Mdn) For grouped data, you can use the same formula as P 50 Example: Mdn= + = (150)(.) 63 = (0.1667)(75 63) = (0.1667)(1) = = Therefore, the median is Class Interval f cf C% N =150 Find the median of the following series of numbers. 1. 4, 1, 0, 19, 18, 17, 16, 15, 14, 1., 0, 19, 18, 18, 18, 18, 16, 16, , 7, 7, 6, 6, 6, 6, 4, 4, Determine the median for the following frequency distribution. Class Interval f cf c% N = 50 3 Sum of the scores divided by the number of scores = = Where: = mean of a sample = mean of a = summation = raw scores N= number of scores Example: The following are scores from a sample of 10 math scores. Solve for the mean of the scores. 11, 1, 13, 14, 16, 17, 17, 0, 5, 6 = = = =17.1 CONCLUSION: The mean from the sample of math scores is If you are given grouped data: = = Where: = mean of a sample = mean of a = summation = frequency of the interval m= midpoint of the interval N= number of scores OVERALL The weighted mean, the mean of means The sum of the mean of each group times the number of scores in the group, divided by the sum of the number of scores in each group = Where: = number of cases in a particular group = mean of a particular group = total number of cases in all groups combined Example: The following are the means of final grades from three blocks in psychology, as well as the number of students in each block. Determine the overall mean of the final grades for the three blocks. Section 1: 1 = 85; N 1 = 95 Section : = 7; N = 5 Section 3: 3 = 79; N 3 = 18 = = (18)(79) = = 11, = =81.86 CONCLUSION: The mean final grade for the three blocks is The following are productivity scores for four departments in a company, as well as its corresponding number of workers. Compute for the overall mean productivity for all four departments: Section 1: N 1 = 0; 1 = 10 Section : N = 15; = 14 Section 3:N 3 = 18; 3 = 15 Section 4:N 4 = ; 4 = 8 3 4 1 = 4 11 Deviation distance and direction of any raw score from the mean Deviation = X - Example: 9, 8, 6, 5, Mean = 6 X DEVIATION MEASURES OF CENTRAL TENDENCY AND SYMMETRY WHICH TO USE? -Used forany measurement scale, especiallythe nominal scale -Used when haste is necessary MEDIAN -Usedfor ordinalor interval data -Used when data is skewed, since it is not as sensitive to extreme scores compared to the mean - Used for interval or ratio data -Used when one has a symmetrical or normal distribution -Can also be useful in skewed distributions, since it is more flexible to advanced statistical analysis PROPERTIES OF THE 1. The mean is sensitive to the exact value of all the scores in the distribution.. The sum of the s about the mean equals zero. = (X i -) = 0 3. The mean is very sensitive to extreme scores. 4. The sum of the squared s of all the scores about their mean is a minimum. 5. Under most circumstances, of the measures used for central tendency, the mean is least subject to sampling variation. PROPERTIES OF THE MEDIAN 1. The median is less sensitive than the mean to extreme scores.. Under usual circumstances, the median is more subject to sampling variability than the mean but less subject to sampling variability than the mode. 4 5 RANGE Range-Difference between the highest and lowest scores in a distribution Range = Highest score lowest score Example: Range = 17 1 = 16 Measure of variability that reflects the typical from the mean = = = 1 = 1 SS= sum of squared s = s= sample ) The equation (Deviation = X -) was only limited to only two values Adding all s in a distribution would only lead to ZERO. In the past, the mean was used to determine the variability of a distribution. In this process, absolute s were used: = These days, the mean is no longer used since it is hard to use in advanced statistical analysis. To overcome the problems faced by a mean s, we use squared sinstead of absolute s From this we can get this formula, which is the equation for deviance: = However, the variance has the weakness of expressing variability in squared units (EXAMPLE: If you were asked to determine the variability of exam scores in a class, the variance will express it in terms of squared exam score.) As such, we get the square root of the deviance in order to reflect variability in the appropriate units; this becomes our sum of squares (SS) Raw Score Formula for Standard Deviation: = = 1 = s= sample ) X = sum of the squared raw scores N= total number of scores = mean square 5 6 Example: On a measure of authoritarianism (higher scores reflect greater tendency toward prejudice, ethnocentrism and submission to authority), seven students scored as follows: X X X = 37 X = 47 N = 7 Mean = Mean Square = PROPERTIES OF THE STANDARD DEVIATION 1. It gives a measure of dispersion relative to the mean.. It is sensitive to each score in the distribution. 3. It is stable with regards to sampling fluctuations. VARIANCE Square of the = = 1 SS= sum of squared s = s= sample ) VARIANCE Raw Score Formula for Variance: = = 1 = s= sample ) X = sum of the squared raw scores N= total number of scores = mean square RAWSCORES UNGROUPED GROUPED SUMMARY Scorewith the greatest frequency Scorewith the greatest frequency Midpointof the interval with the greatest frequency MEDIAN Middlemostscore, as determined by the position of the median Middlemost score, as determined by the position of the median Mdn (50 TH percentile)= + RAW SCORES SUMMARY UNGROUPED GROUPED = = = 6 7 SUMMARY PRACTICE RAW SCORES UNGROUPED GROUPED (sor ) VARIANCE (s or ) The scores of attitudes toward older people for 30 students were arranged in the following simple frequency distribution (higher scores indicate more favorable attitudes towards older people): Attitude Score Value f N = 30 Find the (a) mode, (b) median, and (c)mean. PRACTICE The scores on a religiosity scale (higher scores indicate greater commitment to religious expression) were obtained for 46 adults. For the following simple frequency distribution, calculate the three measures of central tendency. Score Value f cf N =46 7 ### CHAPTER 3 CENTRAL TENDENCY ANALYSES CHAPTER 3 CENTRAL TENDENCY ANALYSES The next concept in the sequential statistical steps approach is calculating measures of central tendency. Measures of central tendency represent some of the most simple ### 10-3 Measures of Central Tendency and Variation 10-3 Measures of Central Tendency and Variation So far, we have discussed some graphical methods of data description. Now, we will investigate how statements of central tendency and variation can be used. ### LearnStat MEASURES OF CENTRAL TENDENCY, Learning Statistics the Easy Way. Session on BUREAU OF LABOR AND EMPLOYMENT STATISTICS LearnStat t Learning Statistics the Easy Way Session on MEASURES OF CENTRAL TENDENCY, DISPERSION AND SKEWNESS MEASURES OF CENTRAL TENDENCY, DISPERSION AND SKEWNESS OBJECTIVES At the end of the session, ### DESCRIPTIVE STATISTICS. The purpose of statistics is to condense raw data to make it easier to answer specific questions; test hypotheses. DESCRIPTIVE STATISTICS The purpose of statistics is to condense raw data to make it easier to answer specific questions; test hypotheses. DESCRIPTIVE VS. INFERENTIAL STATISTICS Descriptive To organize, ### Homework 3. Part 1. Name: Score: / null Name: Score: / Homework 3 Part 1 null 1 For the following sample of scores, the standard deviation is. Scores: 7, 2, 4, 6, 4, 7, 3, 7 Answer Key: 2 2 For any set of data, the sum of the deviation scores ### MCQ S OF MEASURES OF CENTRAL TENDENCY MCQ S OF MEASURES OF CENTRAL TENDENCY MCQ No 3.1 Any measure indicating the centre of a set of data, arranged in an increasing or decreasing order of magnitude, is called a measure of: (a) Skewness (b) ### We will use the following data sets to illustrate measures of center. DATA SET 1 The following are test scores from a class of 20 students: MODE The mode of the sample is the value of the variable having the greatest frequency. Example: Obtain the mode for Data Set 1 77 For a grouped frequency distribution, the modal class is the class having ### Chapter 3: Central Tendency Chapter 3: Central Tendency Central Tendency In general terms, central tendency is a statistical measure that determines a single value that accurately describes the center of the distribution and represents ### Measures of Central Tendency and Variability: Summarizing your Data for Others Measures of Central Tendency and Variability: Summarizing your Data for Others 1 I. Measures of Central Tendency: -Allow us to summarize an entire data set with a single value (the midpoint). 1. Mode : ### A frequency distribution is a table used to describe a data set. A frequency table lists intervals or ranges of data values called data classes A frequency distribution is a table used to describe a data set. A frequency table lists intervals or ranges of data values called data classes together with the number of data values from the set that ### CHINHOYI UNIVERSITY OF TECHNOLOGY CHINHOYI UNIVERSITY OF TECHNOLOGY SCHOOL OF NATURAL SCIENCES AND MATHEMATICS DEPARTMENT OF MATHEMATICS MEASURES OF CENTRAL TENDENCY AND DISPERSION INTRODUCTION From the previous unit, the Graphical displays ### Each exam covers lectures from since the previous exam and up to the exam date. Sociology 301 Exam Review Liying Luo 03.22 Exam Review: Logistics Exams must be taken at the scheduled date and time unless 1. You provide verifiable documents of unforeseen illness or family emergency, ### 1) Write the following as an algebraic expression using x as the variable: Triple a number subtracted from the number 1) Write the following as an algebraic expression using x as the variable: Triple a number subtracted from the number A. 3(x - x) B. x 3 x C. 3x - x D. x - 3x 2) Write the following as an algebraic expression ### Chapter 15 Multiple Choice Questions (The answers are provided after the last question.) Chapter 15 Multiple Choice Questions (The answers are provided after the last question.) 1. What is the median of the following set of scores? 18, 6, 12, 10, 14? a. 10 b. 14 c. 18 d. 12 2. Approximately ### MEASURES OF VARIATION NORMAL DISTRIBTIONS MEASURES OF VARIATION In statistics, it is important to measure the spread of data. A simple way to measure spread is to find the range. 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Data Understanding and Preparation 2.1 Data Understanding Spring 2010 Data Mining Part 2. and Preparation 2.1 Spring 2010 Instructor: Dr. Masoud Yaghini Introduction Outline Introduction Measuring the Central Tendency Measuring the Dispersion of Data Graphic Displays References ### Module 3: Correlation and Covariance Using Statistical Data to Make Decisions Module 3: Correlation and Covariance Tom Ilvento Dr. Mugdim Pašiƒ University of Delaware Sarajevo Graduate School of Business O ften our interest in data analysis ### Measures of Central Tendency Measures of Central Tendency TABLE OF CONTENTS Measures of Central Tendency... 1 What are MEASURES OF CENTRAL TENDENCY?... 1 Measures of Central Tendency... 1 Measures of Central Tendency... 1 Mode... ### Content DESCRIPTIVE STATISTICS. Data & Statistic. Statistics. Example: DATA VS. STATISTIC VS. STATISTICS Content DESCRIPTIVE STATISTICS Dr Najib Majdi bin Yaacob MD, MPH, DrPH (Epidemiology) USM Unit of Biostatistics & Research Methodology School of Medical Sciences Universiti Sains Malaysia. Introduction ### Research Methods 1 Handouts, Graham Hole,COGS - version 1.0, September 2000: Page 1: Research Methods 1 Handouts, Graham Hole,COGS - version 1.0, September 000: Page 1: DESCRIPTIVE STATISTICS - FREQUENCY DISTRIBUTIONS AND AVERAGES: Inferential and Descriptive Statistics: There are four ### MATH BOOK OF PROBLEMS SERIES. New from Pearson Custom Publishing! MATH BOOK OF PROBLEMS SERIES New from Pearson Custom Publishing! The Math Book of Problems Series is a database of math problems for the following courses: Pre-algebra Algebra Pre-calculus Calculus Statistics ### Frequency Distributions Descriptive Statistics Dr. Tom Pierce Department of Psychology Radford University Descriptive statistics comprise a collection of techniques for better understanding what the people in a group look like ### Variables and Data A variable contains data about anything we measure. For example; age or gender of the participants or their score on a test. The Analysis of Research Data The design of any project will determine what sort of statistical tests you should perform on your data and how successful the data analysis will be. For example if you decide ### Descriptive Statistics. Purpose of descriptive statistics Frequency distributions Measures of central tendency Measures of dispersion Descriptive Statistics Purpose of descriptive statistics Frequency distributions Measures of central tendency Measures of dispersion Statistics as a Tool for LIS Research Importance of statistics in research ### Describing Data: Measures of Central Tendency and Dispersion 100 Part 2 / Basic Tools of Research: Sampling, Measurement, Distributions, and Descriptive Statistics Chapter 8 Describing Data: Measures of Central Tendency and Dispersion In the previous chapter we ### Summarizing Data: Measures of Variation Summarizing Data: Measures of Variation One aspect of most sets of data is that the values are not all alike; indeed, the extent to which they are unalike, or vary among themselves, is of basic importance ### CHAPTER 2 DESCRIPTIVE MEASURES CHAPTER 2 DESCRIPTIVE MEASURES Decide whether the following statements are true (T) or false (F). The arithmetic mean always describes the distribution of data well. The median and the mode can always
Subject Area Lessons More Lessons Like This... ## #2569. Estimation Using Goldfish Crackers Mathematics, level: Elementary Posted Fri May 3 13:39:10 PDT 2002 by RaeJean A. Brown (raejean@paulbunyan.net). Student, Bemidji, USA Activity Time: 30-35 Mins. Concepts Taught: Estimation and Rounding Please Note: If you use my lesson plan, please give me the credit, I did this all by myself. ฉ RaeJean Brown 2002 Grade Level: 2nd grade Area 4A Number Sense. A student shall use number relationships to represent information and solve problems by: (1)... (2).... (3) solving problems and justifying thinking by selecting appropriate numbers and representations; using operations, patterns, and estimation; generating multiple solutions; organizing data using pictures and charts; and using concrete objects, diagrams, or maps to solve simple problems involving counting, arrangements, or routes. Concept: Estimation and Rounding Objective(s): Students will be able to solve problems by estimating and rounding numbers. Students will be able to pick numbers within reason to represent certain objects. Materials: Small goldfish bowl Goldfish crackers Copy of the estimation song Individual snack bags with 10 goldfish Review Worksheets Procedure: First I will explain that we are doing a unit on estimation and rounding. I will hold up the fish bowl and ask them if they can guess how many goldfish crackers are in there. I will call on students and hear their guesses. I will hand out the copy of the estimation song (the snack bag will be stapled with it). I will say, seeing that there are 10 goldfish in the bag, how many bags of goldfish will fill up the fish bowl. I will introduce the estimation song and have us sing it all the way through. Then I will tell them how many gold fish are actually in the bowl. We will also discuss how estimation can be used in other ways other than just guessing how many objects there are, we can use it in math for addition and subtraction as well as other operations. This is where rounding comes in. I will show some examples. Example: 26 + 12 = ? (30 + 10) If the last digit in the number we are working with is 5 or greater, we are going to round up. So 26 would become 30 (the tenths place will also have to change to a greater number so it does not become 20). If the last digit in the number we are working with is below 5, we will round down. 12 will become 10. Now that we are doing the math, we will say 30 + 10 = 40. That will be our estimate answer by rounding. Another example is: 41- 23 = ? (40-20) By applying what we just learned about rounding, 41 would be rounded down to 40 and 23 will also be rounded down, but to 20. Then we get 40-20 and we know this is 20. This isn't the "correct" answer. But it is right in terms of estimating. Evaluation: I will see if the students are grasping the concept of rounding and estimating by giving them a worksheet that has math problems that they can estimate. I will also ask if they know anything else that can be used in estimation (such as time, measurement, area). And we can focus on that later in the unit. The Song: Estimate* 1. we estimate (we estimate) so we can know (so we can know) how many things (how many things) there are to show (there are to show) we estimate so we can know, how many things there are to show 2. we can round up (we can round up) we can round down (we can round down) just do the math (just do the math) what have we found (what have we found) we can round up, we can round down, just do the math, what have we found 3. it's fun to guess (it's fun to guess) so we can know (so we can know) how many fish (how many fish) are in that bowl (are in that bowl) it's fun to guess, so we can know, how many fish are in that bowl 4. now don't you fret (now don't you fret) and don't you frown (and don't you frown) is never wrong (is never wrong) now don't you fret and don't you frown, the answer you get, is never wrong 5. if all of math (if all of math) was just this fun (was just this fun) well boy oh boy (well boy oh boy) what we'd get done! (what we'd get done!) if all of math was just this fun, well boy oh boy what we'd get done! 6. that's all there is (that's all there is) there is no more (there is no more) until we work (until we work) on unit four (on unit four) that's all there is, there is no more, until we work, on unit four. *"Estimate" is to the tune of the song, "The Bear" which can be found on pp. 38-39 in the Wee Sing Fun 'N' Folk book by Pamela Conn Beall and Susan Hagen Nipp.1989.
# Assign stalls to K cows to maximize the minimum distance between them Given a sorted array arr[] consisting of N integers which denote the position of a stall. You are also given an integer K which denotes the number of aggressive cows. You are given the task of assigning stalls to K cows such that the minimum distance between any two of them is the maximum possible. Examples: Input: N = 5, K = 3, arr[] = {1, 2, 4, 8, 9} Output: 3 Explanation: We can place cow 1 at position 1, cow 2 at position 4 and cow 3 at position 9. So, the maximum possible minimum distance between two cows is 3. Input: N = 6, K = 4, arr[] = {6, 7, 9, 11, 13, 15} Output: 2 Explanation: We can place cow 1 at position 6, cow 2 at position 9 and cow 4 at position 15. So, the maximum possible minimum distance between two cows is 2. Recommended Practice Naive Approach: The basic way to solve the problem is as follows: The maximum distance possible between two cows can be the difference between the maximum and minimum element (Which is less than the maximum element of the array) of the array and the minimum possible distance can be 1. Follow the steps to solve the problem: • We will start iterating from 1 to the maximum element of the array and for each distance, • We will check whether it is possible to place all the cows in the barn by maintaining this minimum distance. • If it is possible then we will store it as a possible answer and continue to search for a better answer. • Else, break the loop, and return the answer. Illustration: Consider: N = 5, K = 3, arr = {1, 2, 4, 8, 9}. • Maximum element = 9. So we will iterate between 1 to 9. • At i = 1, we can place the cows at positions 1, 2, 4 maintaining the minimum distance of 1. • At i = 2, we can place the cows at positions 1, 4, 8 maintaining the minimum distance of 2. • At i = 3, we can place the cows at 1, 4, and 8 maintaining the minimum distance of 3. • At i = 4, we can not place the cows by maintaining the minimum distance of 4. • So, the largest possible distance is 3. Below is the implementation of the above approach. ## C++ `// C++ code to implement the approach` `#include ``using` `namespace` `std;` `// Function to check whether a distance is``// keeping all the cows in the barn``bool` `ok(vector<``int``>& v, ``int` `x, ``int` `c)``{`` ``int` `n = v.size();`` ``int` `count = 1;`` ``int` `d = v[0];`` ``for` `(``int` `i = 1; i < n; i++) {`` ``if` `(v[i] - d >= x) {`` ``d = v[i];`` ``count++;`` ``}`` ``else` `{`` ``continue``;`` ``}`` ``}`` ``if` `(count >= c) {`` ``return` `true``;`` ``}`` ``return` `false``;``}` `// Function to find the maximum possible``// minimum distance between two cows``int` `aggressive_cows(vector<``int``>& v, ``int` `n, ``int` `k)``{`` ``long` `long` `ans = -1;`` ``int` `maxi = 0;`` ``for` `(``int` `i = 0; i < n; i++) {`` ``maxi = max(maxi, v[i]);`` ``}` ` ``// Loop from 1 to max limit of the`` ``// barn length (here = 10^9)`` ``for` `(``long` `long` `i = 1; i <= maxi; i++) {`` ``if` `(ok(v, i, k)) {`` ``ans = i;`` ``}`` ``else` `{`` ``break``;`` ``}`` ``}` ` ``return` `ans;``}` `// Driver code``int` `main()``{`` ``int` `K = 3;`` ``vector<``int``> arr = { 1, 2, 4, 8, 9 };`` ``int` `N = arr.size();` ` ``// Function call`` ``int` `ans = aggressive_cows(arr, N, K);`` ``cout << ans << endl;` ` ``return` `0;``}` ## Java `// Java code to implement the approach``import` `java.util.;` `public` `class` `Main {`` ``// Driver code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `K = ``3``;`` ``int``[] arr = { ``1``, ``2``, ``4``, ``8``, ``9` `};`` ``int` `N = arr.length;` ` ``// Function call`` ``System.out.println(aggressive_cows(arr, N, K));`` ``}` ` ``// Function to check whether a distance is`` ``// keeping all the cows in the barn`` ``public` `static` `boolean` `ok(``int``[] v, ``int` `x, ``int` `c)`` ``{`` ``int` `n = v.length;`` ``int` `count = ``1``;`` ``int` `d = v[``0``];`` ``for` `(``int` `i = ``1``; i < n; i++) {`` ``if` `(v[i] - d >= x) {`` ``d = v[i];`` ``count++;`` ``}`` ``else` `{`` ``continue``;`` ``}`` ``}`` ``if` `(count >= c) {`` ``return` `true``;`` ``}`` ``return` `false``;`` ``}` ` ``// Function to find the maximum possible`` ``// minimum distance between two cows`` ``public` `static` `int` `aggressive_cows(``int``[] v, ``int` `n, ``int` `k)`` ``{`` ``long` `ans = -``1``;`` ``int` `maxi = ``0``;`` ``for` `(``int` `i = ``0``; i < n; i++) {`` ``maxi = Math.max(maxi, v[i]);`` ``}` ` ``// Loop from 1 to max limit of the`` ``// barn length (here = 10^9)`` ``for` `(``long` `i = ``1``; i <= maxi; i++) {`` ``if` `(ok(v, (``int``)i, k)) {`` ``ans = i;`` ``}`` ``else`` ``break``;`` ``}`` ``return` `(``int``)ans;`` ``}``}` `// This code is contributed by Tapesh(tapeshdua420)` ## Python3 `# Python code to implement the approach` `# Function to check whether a distance is``# keeping all the cows in the barn` `def` `ok(v,x,c):`` ``n``=``len``(v)`` ``count``=``1`` ``d``=``v[``0``]`` ``for` `i ``in` `range``(``1``,n):`` ``if``(v[i]``-``d>``=``x):`` ``d``=``v[i]`` ``count``=``count``+``1`` ``else``:`` ``continue`` ``if``(count>``=``c):`` ``return` `True`` ` ` ``return` `False`` ` `# Function to find the maximum possible``# minimum distance between two cows``def` `aggressive_cows(v,n,k):`` ``ans``=``-``1`` ``maxi``=``0`` ``for` `i ``in` `range``(n):`` ``maxi``=``max``(maxi,v[i])`` ` ` ``# Loop from 1 to max limit of the`` ``# barn length (here = 10^9)`` ` ` ``for` `i ``in` `range``(``1``,maxi``+``1``):`` ``if``(ok(v,i,k)):`` ``ans``=``i`` ``else``:`` ``break`` ` ` ``return` `ans`` ` `# Driver code``K``=``3``arr``=``[``1``,``2``,``4``,``8``,``9``]``N``=``len``(arr)` `# Function call``ans``=``aggressive_cows(arr,N,K)``print``(ans)` `# This code is contributed by Pushpesh Raj.` ## C# `// C# code to implement the approach``using` `System;` `class` `Program {`` ``// Driver code`` ``static` `void` `Main(``string``[] args)`` ``{`` ``int` `K = 3;`` ``int``[] arr = { 1, 2, 4, 8, 9 };`` ``int` `N = arr.Length;` ` ``// Function call`` ``Console.WriteLine(aggressive_cows(arr, N, K));`` ``}` ` ``// Function to check whether a distance is`` ``// keeping all the cows in the barn`` ``public` `static` `bool` `ok(``int``[] v, ``int` `x, ``int` `c)`` ``{`` ``int` `n = v.Length;`` ``int` `count = 1;`` ``int` `d = v[0];` ` ``for` `(``int` `i = 1; i < n; i++)` ` ``if` `(v[i] - d >= x) {`` ``d = v[i];`` ``count++;`` ``}`` ``else`` ``continue``;` ` ``if` `(count >= c)`` ``return` `true``;` ` ``return` `false``;`` ``}` ` ``// Function to find the maximum possible`` ``// minimum distance between two cows`` ``public` `static` `int` `aggressive_cows(``int``[] v, ``int` `n, ``int` `k)`` ``{`` ``long` `ans = -1;`` ``int` `maxi = 0;` ` ``for` `(``int` `i = 0; i < n; i++)`` ``maxi = Math.Max(maxi, v[i]);` ` ``// Loop from 1 to max limit of the`` ``// barn length (here = 10^9)`` ``for` `(``long` `i = 1; i <= maxi; i++)`` ``if` `(ok(v, (``int``)i, k))`` ``ans = i;`` ``else`` ``break``;` ` ``return` `(``int``)ans;`` ``}``}` `// This code is contributed by Tapesh(tapeshdua420)` ## Javascript ` ``// JavaScript code for the above approach` ` ``// Function to check whether a distance is`` ``// keeping all the cows in the barn`` ``function` `ok(v, x, c) {`` ``let n = v.length;`` ``let count = 1;`` ``let d = v[0];`` ``for` `(let i = 1; i < n; i++) {`` ``if` `(v[i] - d >= x) {`` ``d = v[i];`` ``count++;`` ``}`` ``else` `{`` ``continue``;`` ``}`` ``}`` ``if` `(count >= c) {`` ``return` `true``;`` ``}`` ``return` `false``;`` ``}` ` ``// Function to find the maximum possible`` ``// minimum distance between two cows`` ``function` `aggressive_cows(v, n, k) {`` ``let ans = -1;`` ``let maxi = 0;`` ``for` `(let i = 0; i < n; i++) {`` ``maxi = Math.max(maxi, v[i]);`` ``}` ` ``// Loop from 1 to max limit of the`` ``// barn length (here = 10^9)`` ``for` `(let i = 1; i <= maxi; i++) {`` ``if` `(ok(v, i, k)) {`` ``ans = i;`` ``}`` ``else` `{`` ``break``;`` ``}`` ``}` ` ``return` `ans;`` ``}` ` ``// Driver code` ` ``let K = 3;`` ``let arr = [1, 2, 4, 8, 9];`` ``let N = arr.length;` ` ``// Function call`` ``let ans = aggressive_cows(arr, N, K);`` ``console.log(ans);` `// This code is contributed by Potta Lokesh` Output `3` Time Complexity: O(N(max element in the input array)) Auxiliary Space: O(1). Efficient Approach: In the above approach, the search for the answer can be optimized using Binary search. As we have seen in the case of brute force solution we are iterating from 1 to the maximum element of the array and at each distance, we check whether it can be our possible answer or not. Instead of doing this, we can run a binary search from 1 to the maximum element of the array. Follow the steps to solve the problem: • Apply a binary search between 1 and the maximum element of the array. • Each time find the middle element of the search space. • If that middle element can be a possible minimum distance store it as a possibility until we find a better answer, then move in the right direction in the search space. • Else, we will move left in our search space. • When the binary search is complete, return the answer. Illustration: Consider: N = 5, K =3, arr = {1, 2, 4, 8, 9}. • Maximum element = 9. So our search space will be 1 to 9. • First, L = 1 and R = 9, mid = 5; we can not place the cows by maintaining the minimum distance of 5. So, we will move left in our search space. • Now, L = 1 and R = 4, mid = 2; we can place the cows at positions 1, 4, and 8 maintaining the minimum distance of 2. So, we will store 2 as a possible answers and move right into our search space. • Now, L=3 and R=4, mid = 3; we can place the cows at positions 1, 4, and 8 maintaining the minimum distance of 3. So, we will store 3 as a possible answers and move right into our search space. • Now, L = 4 and R = 4, mid = 4; we can not place the cows by maintaining the minimum distance of 5. So, we will move left in our search space. • Now, L = 4 and R = 3, Since, L > R, our binary search is complete, and the largest possible answer is 3. Below is the implementation of the above approach. ## C++ `// C++ code to implement the approach` `#include ``using` `namespace` `std;` `// Function to check if it is possible``// to position the cow x distance apart``bool` `ok(vector<``int``>& v, ``int` `x, ``int` `c)``{`` ``int` `n = v.size();` ` ``// We already place it at the first`` ``// available slot i.e v[0](Greedy)`` ``int` `count = 1;`` ``int` `d = v[0];`` ``for` `(``int` `i = 1; i < n; i++) {`` ``if` `(v[i] - d >= x) {`` ``d = v[i];`` ``count++;`` ``}`` ``else` `{`` ``continue``;`` ``}`` ``}`` ``if` `(count >= c) {`` ``return` `true``;`` ``}`` ``return` `false``;``}` `// Function to find the maximum possible``// minimum distance between the cows``int` `aggressive_cows(vector<``int``>& v, ``int` `n, ``int` `k)``{`` ``// Just take l = 1 and high = 1000000`` ``// or any large number`` ``int` `l = 1;`` ``int` `r = 1e9;`` ``int` `ans = -1;` ` ``// Applying Binary search`` ``while` `(r >= l) {`` ``int` `mid = l + (r - l) / 2;`` ``if` `(ok(v, mid, k)) {`` ``ans = mid;`` ``l = mid + 1;`` ``}`` ``else` `{`` ``r = mid - 1;`` ``}`` ``}` ` ``return` `ans;``}` `// Driver code``int` `main()``{`` ``int` `K = 3;`` ``vector<``int``> arr = { 1, 2, 8, 4, 9 };`` ``int` `N = arr.size();` ` ``// Function call`` ``int` `ans = aggressive_cows(arr, N, K);`` ``cout << ans << endl;` ` ``return` `0;``}` ## Java `// Java code to implement the approach``import` `java.util.;` `class` `GFG {` ` ``// Function to check if it is possible`` ``// to position the cow x distance apart`` ``static` `boolean` `ok(``int``[] v, ``int` `x, ``int` `c)`` ``{`` ``int` `n = v.length;` ` ``// We already place it at the first`` ``// available slot i.e v[0](Greedy)`` ``int` `count = ``1``;`` ``int` `d = v[``0``];`` ``for` `(``int` `i = ``1``; i < n; i++) {`` ``if` `(v[i] - d >= x) {`` ``d = v[i];`` ``count++;`` ``}`` ``else` `{`` ``continue``;`` ``}`` ``}`` ``if` `(count >= c) {`` ``return` `true``;`` ``}`` ``return` `false``;`` ``}`` ``// Function to find the maximum possible`` ``// minimum distance between the cows`` ``static` `int` `aggressive_cows(``int``[] v, ``int` `n, ``int` `k)`` ``{`` ``// Just take l = 1 and high = 1000000`` ``// or any large number`` ``int` `l = ``1``;`` ``int` `r = ``1000000000``;`` ``int` `ans = -``1``;` ` ``// Applying Binary search`` ``while` `(r >= l) {`` ``int` `mid = l + (r - l) / ``2``;`` ``if` `(ok(v, mid, k)) {`` ``ans = mid;`` ``l = mid + ``1``;`` ``}`` ``else` `{`` ``r = mid - ``1``;`` ``}`` ``}` ` ``return` `ans;`` ``}` ` ``// Driver code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `K = ``3``;`` ``int``[] arr = ``new` `int``[] { ``1``, ``2``, ``8``, ``4``, ``9` `};`` ``int` `N = arr.length;` ` ``// Function call`` ``System.out.println(aggressive_cows(arr, N, K));`` ``}``}` `// This code is contributed by Tapesh(tapeshdua420)` ## Python3 `# Python code to implement the approach` `# Function to check if it is possible``# to position the cow x distance apart` `def` `ok(v, x, c):`` ``n ``=` `len``(v)` ` ``# We already place it at the first`` ``# available slot i.e v[0](Greedy)`` ``count ``=` `1`` ``d ``=` `v[``0``]`` ``for` `i ``in` `range``(``1``, n):`` ``if` `(v[i] ``-` `d >``=` `x):`` ``d ``=` `v[i]`` ``count ``+``=` `1`` ``else``:`` ``continue` ` ``if` `(count >``=` `c):`` ``return` `True`` ``return` `False` `# Function to find the maximum possible``# minimum distance between the cows` `def` `aggressive_cows(v, n, k):`` ``# Just take l = 1 and high = 1000000`` ``# or any large number`` ``l ``=` `1`` ``r ``=` `1000000000`` ``ans ``=` `-``1` ` ``# Applying Binary search`` ``while` `(r >``=` `l):`` ``mid ``=` `l ``+` `(r ``-` `l)``/``/``2`` ``if` `(ok(v, mid, k)):`` ``ans ``=` `mid`` ``l ``=` `mid ``+` `1`` ``else``:`` ``r ``=` `mid ``-` `1` ` ``return` `ans` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:`` ``K ``=` `3`` ``arr ``=` `[``1``, ``2``, ``8``, ``4``, ``9``]`` ``N ``=` `len``(arr)` ` ``# Function call`` ``print``(aggressive_cows(arr, N, K))` `# This code is contributed by Tapesh(tapeshdua420)` ## C# `// C# code to implement the approach``using` `System;` `class` `Program {` ` ``// Function to check if it is possible`` ``// to position the cow x distance apart`` ``static` `bool` `ok(``int``[] v, ``int` `x, ``int` `c)`` ``{`` ``int` `n = v.Length;` ` ``// We already place it at the first`` ``// available slot i.e v[0](Greedy)`` ``int` `count = 1;`` ``int` `d = v[0];`` ``for` `(``int` `i = 1; i < n; i++) {`` ``if` `(v[i] - d >= x) {`` ``d = v[i];`` ``count++;`` ``}` ` ``else`` ``continue``;`` ``}` ` ``if` `(count >= c)`` ``return` `true``;` ` ``return` `false``;`` ``}` ` ``// Function to find the maximum possible`` ``// minimum distance between the cows`` ``static` `int` `aggressive_cows(``int``[] v, ``int` `n, ``int` `k)`` ``{`` ``// Just take l = 1 and high = 1000000`` ``// or any large number`` ``int` `l = 1;`` ``int` `r = 1000000000;`` ``int` `ans = -1;` ` ``// Applying Binary search`` ``while` `(r >= l) {`` ``int` `mid = l + (r - l) / 2;` ` ``if` `(ok(v, mid, k)) {`` ``ans = mid;`` ``l = mid + 1;`` ``}` ` ``else`` ``r = mid - 1;`` ``}`` ``return` `ans;`` ``}` ` ``// Driver code`` ``public` `static` `void` `Main()`` ``{`` ``int` `K = 3;`` ``int``[] arr = ``new` `int``[] { 1, 2, 8, 4, 9 };`` ``int` `N = arr.Length;` ` ``// Function call`` ``Console.WriteLine(aggressive_cows(arr, N, K));`` ``}``}` `// This code is contributed by Tapesh(tapeshdua420)` ## Javascript `// JavaScript code to implement the approach` `// Function to check if it is possible``// to position the cow x distance apart``function` `ok(v, x, c) {`` ``var` `n = v.length;` ` ``// We already place it at the first`` ``// available slot i.e v[0](Greedy)`` ``var` `count = 1;`` ``var` `d = v[0];`` ``for` `(``var` `i = 1; i < n; i++) {`` ``if` `(v[i] - d >= x) {`` ``d = v[i];`` ``count++;`` ``} ``else` `{`` ``continue``;`` ``}`` ``}` ` ``if` `(count >= c) {`` ``return` `true``;`` ``}`` ``return` `false``;``}` `// Function to find the maximum possible``// minimum distance between the cows``function` `aggressive_cows(v, n, k) {`` ``// Just take l = 1 and high = 1000000`` ``// or any large number`` ``var` `l = 1;`` ``var` `r = 1e9;` ` ``var` `ans = -1;` ` ``// Applying Binary search`` ``while` `(r >= l) {`` ``var` `mid = (l + (r - l) / 2) \| 0;`` ``if` `(ok(v, mid, k)) {`` ``ans = mid;`` ``l = mid + 1;`` ``} `` ``else` `{`` ``r = mid - 1;`` ``}`` ``}`` ``return` `ans;``}` `// Driver code``var` `K = 3;``var` `arr = [1, 2, 8, 4, 9];``var` `N = arr.length;` `// Function call``var` `ans = aggressive_cows(arr, N, K);``console.log(ans);` `// This code is contributed by Tapesh(tapeshdua420)` Output `1` Time Complexity: O(N log(maximum element of the array )) Auxiliary Space: O(1) ### More Efficient Approach: As we have seen in the case of brute force solution we are iterating from 0 to the maximum element of the array and at each distance, we check whether it can be our possible answer or not. In Efficient Process we can run a binary search from 0 to the maximum element in the array. But more than all we can run binary search from 0 to Maximum difference of the array. In sorted array is A[N-1] – A[0] (last – first element in the array) is the maximum difference possible. It works perfectly fine, as our search space is that only, we can never get a difference that is greater than this. This saves a lot of iterations in many cases, one such example is • Arr = [70, 70, 70, 70] N=4, K=3 • Here Efficient method runs binary search from 0 to 70. • More Efficient method (this method) runs binary search from 0 to 0. (A[N-1] – A[0] = 0) ## C++ `// C++ code to implement the approach` `#include ``using` `namespace` `std;` `// Function to check if it is possible``// to position the cow x distance apart``bool` `ok(vector<``int``>& v, ``int` `x, ``int` `c)``{`` ``int` `n = v.size();` ` ``// We already place it at the first`` ``// available slot i.e v[0](Greedy)`` ``int` `count = 1;`` ``int` `last = 0;`` ``for` `(``int` `i = 0; i < n; i++) {`` ``if` `(v[i] - v[last] >= x) {`` ``last = i; ``//cow placed`` ``count++;`` ``}`` ``if` `(count >= c) {`` ``return` `true``;`` ``}`` ``}`` ``return` `false``;``}` `// Function to find the maximum possible``// minimum distance between the cows``int` `aggressive_cows(vector<``int``>& v, ``int` `n, ``int` `k)``{`` ``// Just take l = 1 and high = M (MAX DIFF POSSIBLE)`` ``// M = last - first element in sorted array`` ``int` `l = 0;`` ``int` `r = v[n-1] - v[0];`` ``int` `ans = -1;` ` ``// Applying Binary search`` ``while` `(l<=r) {`` ``int` `mid = l + (r - l) / 2;`` ``if` `(ok(v, mid, k)) {`` ``ans = mid;`` ``l = mid + 1;`` ``}`` ``else` `{`` ``r = mid - 1;`` ``}`` ``}` ` ``return` `ans;``}` `// Driver code``int` `main()``{`` ``int` `K = 3;`` ``vector<``int``> arr = { 1, 2, 4, 8, 9 };`` ``int` `N = arr.size();` ` ``// Function call`` ``int` `ans = aggressive_cows(arr, N, K);`` ``cout << ans << endl;` ` ``return` `0;``}` `//Code contributed by Balakrishnan R (rbkraj000)` ## Java `//Java code for the above approach``import` `java.util.Vector;` `class` `GFG {` ` ``// Function to check if it is possible`` ``// to position the cow x distance apart`` ``static` `boolean` `ok(Vector v, ``int` `x, ``int` `c) {`` ``int` `n = v.size();` ` ``// We already place it at the first`` ``// available slot i.e v[0](Greedy)`` ``int` `count = ``1``;`` ``int` `last = ``0``;`` ``for` `(``int` `i = ``0``; i < n; i++) {`` ``if` `(v.get(i) - v.get(last) >= x) {`` ``last = i; ``//cow placed`` ``count++;`` ``}`` ``if` `(count >= c) {`` ``return` `true``;`` ``}`` ``}`` ``return` `false``;`` ``}` ` ``// Function to find the maximum possible`` ``// minimum distance between the cows`` ``static` `int` `aggressive_cows(Vector v, ``int` `n, ``int` `k) {`` ``// Just take l = 1 and high = M (MAX DIFF POSSIBLE)`` ``// M = last - first element in sorted array`` ``int` `l = ``0``;`` ``int` `r = v.get(n-``1``) - v.get(``0``);`` ``int` `ans = -``1``;` ` ``// Applying Binary search`` ``while` `(l <= r) {`` ``int` `mid = l + (r - l) / ``2``;`` ``if` `(ok(v, mid, k)) {`` ``ans = mid;`` ``l = mid + ``1``;`` ``} ``else` `{`` ``r = mid - ``1``;`` ``}`` ``}`` ``return` `ans;`` ``}``//Driver code`` ``public` `static` `void` `main(String[] args) {`` ``int` `K = ``3``;`` ``Vector arr = ``new` `Vector();`` ``arr.add(``1``);`` ``arr.add(``2``);`` ``arr.add(``4``);`` ``arr.add(``8``);`` ``arr.add(``9``);`` ``int` `N = arr.size();` ` ``// Function call`` ``int` `ans = aggressive_cows(arr, N, K);`` ``System.out.println(ans);`` ``}``}` ## Python3 `# Python code to implement the approach``import` `math` `# Function to check if it is possible``# to position the cow x distance apart``def` `ok(v, x, c):`` ``n ``=` `len``(v);` ` ``# We already place it at the first`` ``# available slot i.e v[0](Greedy)`` ``count ``=` `1``;`` ``last ``=` `0``;`` ``for` `i ``in` `range``(``0``,n):`` ``if` `(v[i] ``-` `v[last] >``=` `x) :`` ``last ``=` `i; ``#cow placed`` ``count``+``=``1``;`` ` ` ``if` `(count >``=` `c) :`` ``return` `True``;`` ` ` ``return` `False``;` `# Function to find the maximum possible``# minimum distance between the cows``def` `aggressive_cows(v, n, k):`` ``# Just take l = 1 and high = M (MAX DIFF POSSIBLE)`` ``# M = last - first element in sorted array`` ``l ``=` `0``;`` ``r ``=` `v[n``-``1``] ``-` `v[``0``];`` ``ans ``=` `-``1``;` ` ``# Applying Binary search`` ``while` `(l<``=``r):`` ``mid ``=` `l ``+` `math.floor((r ``-` `l) ``/` `2``);`` ``if` `(ok(v, mid, k)):`` ``ans ``=` `mid;`` ``l ``=` `mid ``+` `1``;`` ` ` ``else` `:`` ``r ``=` `mid ``-` `1``;`` ` ` ``return` `ans;` `# Driver code``K ``=` `3``;``arr ``=` `[ ``1``, ``2``, ``4``, ``8``, ``9` `];``N ``=` `len``(arr);` `# Function call``ans ``=` `aggressive_cows(arr, N, K);``print``(ans);` `# This code is contributed by ritaagarwal.` ## Javascript `// Javascript code to implement the approach` `// Function to check if it is possible``// to position the cow x distance apart``function` `ok(v, x, c)``{`` ``let n = v.length;` ` ``// We already place it at the first`` ``// available slot i.e v[0](Greedy)`` ``let count = 1;`` ``let last = 0;`` ``for` `(let i = 0; i < n; i++) {`` ``if` `(v[i] - v[last] >= x) {`` ``last = i; ``//cow placed`` ``count++;`` ``}`` ``if` `(count >= c) {`` ``return` `true``;`` ``}`` ``}`` ``return` `false``;``}` `// Function to find the maximum possible``// minimum distance between the cows``function` `aggressive_cows(v, n, k)``{`` ``// Just take l = 1 and high = M (MAX DIFF POSSIBLE)`` ``// M = last - first element in sorted array`` ``let l = 0;`` ``let r = v[n-1] - v[0];`` ``let ans = -1;` ` ``// Applying Binary search`` ``while` `(l<=r) {`` ``let mid = l + Math.floor((r - l) / 2);`` ``if` `(ok(v, mid, k)) {`` ``ans = mid;`` ``l = mid + 1;`` ``}`` ``else` `{`` ``r = mid - 1;`` ``}`` ``}` ` ``return` `ans;``}` `// Driver code``let K = 3;``let arr = [ 1, 2, 4, 8, 9 ];``let N = arr.length;` `// Function call``let ans = aggressive_cows(arr, N, K);``console.log(ans);`` ` ` ``// This code is contributed by poojaagarwal2.` ## C# `// C# implementation of the above approach` `using` `System;``using` `System.Collections.Generic;` `class` `GFG {`` ` ` ``// Function to check if it is possible`` ``// to position the cow x distance apart`` ``static` `bool` `ok(List<``int``> v, ``int` `x, ``int` `c)`` ``{`` ``int` `n = v.Count;`` ` ` ``// We already place it at the first`` ``// available slot i.e v[0](Greedy)`` ``int` `count = 1;`` ``int` `last = 0;`` ``for` `(``int` `i = 0; i < n; i++) {`` ``if` `(v[i] - v[last] >= x) {`` ``last = i; ``//cow placed`` ``count++;`` ``}`` ``if` `(count >= c) {`` ``return` `true``;`` ``}`` ``}`` ``return` `false``;`` ``}`` ` ` ``// Function to find the maximum possible`` ``// minimum distance between the cows`` ``static` `int` `aggressive_cows(List<``int``> v, ``int` `n, ``int` `k)`` ``{`` ``// Just take l = 1 and high = M (MAX DIFF POSSIBLE)`` ``// M = last - first element in sorted array`` ``int` `l = 0;`` ``int` `r = v[n-1] - v[0];`` ``int` `ans = -1;`` ` ` ``// Applying Binary search`` ``while` `(l<=r) {`` ``int` `mid = l + (r - l) / 2;`` ``if` `(ok(v, mid, k)) {`` ``ans = mid;`` ``l = mid + 1;`` ``}`` ``else` `{`` ``r = mid - 1;`` ``}`` ``}`` ` ` ``return` `ans;`` ``}`` ` ` ``// Driver code`` ``static` `public` `void` `Main()`` ``{`` ``int` `K = 3;`` ``List<``int``> arr = ``new` `List<``int``>{ 1, 2, 4, 8, 9 };`` ``int` `N = arr.Count;`` ` ` ``// Function call`` ``int` `ans = aggressive_cows(arr, N, K);`` ``Console.Write(ans);`` ` ` ``}``}` Output `3` Time Complexity: O(N * log(M)) • N = Size of the array. • M = A[N-1] – A[0] (last – first element) Maximum possible difference in the array. Auxiliary Space: O(1) The above More Efficient Method Idea, Algorithm, and Code are contributed by Balakrishnan R (rbkraj000 – GFG ID). Previous Next Share your thoughts in the comments Similar Reads
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # Tests of Single Variance ## A test of a single variance may be right-tailed, left-tailed, or two-tailed. One example is the chi-squared test. Estimated10 minsto complete % Progress Practice Tests of Single Variance MEMORY METER This indicates how strong in your memory this concept is Progress Estimated10 minsto complete % Tests of Single Variance ### Tests of Single Variance We have learned how the chi-square test can help us assess the relationships between two variables. In addition to assessing these relationships, the chi-square test can also help us test hypotheses surrounding variance, which is the measure of the variation, or scattering, of scores in a distribution. There are several different tests that we can use to assess the variance of a sample. The most common tests used to assess variance are the chi-square test for one variance, the F\begin{align}F\end{align}-test, and the Analysis of Variance (ANOVA). Both the chi-square test and the F\begin{align}F\end{align}-test are extremely sensitive to non-normality (or when the populations do not have a normal distribution), so the ANOVA test is used most often for this analysis. However, in this Concept, we will examine in greater detail the testing of a single variance using the chi-square test. #### Testing a Single Variance Hypothesis Using the Chi-Square Test Suppose that we want to test two samples to determine if they belong to the same population. The test of variance between samples is used quite frequently in the manufacturing of food, parts, and medications, since it is necessary for individual products of each of these types to be very similar in size and chemical make-up. This test is called the test for one variance. To perform the test for one variance using the chi-square distribution, we need several pieces of information. First, as mentioned, we should check to make sure that the population has a normal distribution. Next, we need to determine the number of observations in the sample. The remaining pieces of information that we need are the standard deviation and the hypothetical population variance. For the purposes of this exercise, we will assume that we will be provided with the standard deviation and the population variance. Using these key pieces of information, we use the following formula to calculate the chi-square value to test a hypothesis surrounding single variance: χ2=df(s2)σ2\begin{align}\chi^2=\frac{df(s^2)}{\sigma^2}\end{align} where: χ2\begin{align}\chi^2\end{align} is the chi-square statistical value. df=n1\begin{align}df=n-1\end{align}, where n\begin{align}n\end{align} is the size of the sample. s2\begin{align}s^2\end{align} is the sample variance. σ2\begin{align}\sigma^2\end{align} is the population variance. We want to test the hypothesis that the sample comes from a population with a variance greater than the observed variance. Let’s take a look at an example to help clarify. #### Testing the Randomness of a Sample with Respect to Variance Suppose we have a sample of 41 female gymnasts from Mission High School. We want to know if their heights are truly a random sample of the general high school population with respect to variance. We know from a previous study that the standard deviation of the heights of high school women is 2.2. To test this question, we first need to generate null and alternative hypotheses. Our null hypothesis states that the sample comes from a population that has a variance of less than or equal to 4.84 (σ2\begin{align}\sigma^2\end{align} is the square of the standard deviation). Null Hypothesis H0:σ24.84\begin{align}H_0:\sigma^2 \le 4.84\end{align} (The variance of the female gymnasts is less than or equal to that of the general female high school population.) Alternative Hypothesis Ha:σ2>4.84\begin{align}H_a:\sigma^2 > 4.84\end{align} (The variance of the female gymnasts is greater than that of the general female high school population.) Using the sample of the 41 gymnasts, we compute the standard deviation and find it to be s=1.2\begin{align}s=1.2\end{align}. Using the information from above, we calculate our chi-square value and find the following: χ2=(40)(1.22)4.84=11.9\begin{align}\chi^2=\frac{(40)(1.2^2)}{4.84}=11.9\end{align} Therefore, since 11.9 is less than 55.758 (the value from the chi-square table given an alpha level of 0.05 and 40 degrees of freedom), we fail to reject the null hypothesis and, therefore, cannot conclude that the female gymnasts have a significantly higher variance in height than the general female high school population. #### Calculating a Confidence Interval for a Population Variance Once we know how to test a hypothesis about a single variance, calculating a confidence interval for a population variance is relatively easy. Again, it is important to remember that this test is dependent on the normality of the population. For non-normal populations, it is best to use the ANOVA test, which we will cover in greater detail in another lesson. To construct a confidence interval for the population variance, we need three pieces of information: the number of observations in the sample, the variance of the sample, and the desired confidence interval. With the desired confidence interval, α\begin{align}\alpha\end{align} (most often this is set at 0.10 to reflect a 90% confidence interval or at 0.05 to reflect a 95% confidence interval), we can construct the upper and lower limits around the significance level. #### Constructing a Confidence Interval 1. We randomly select 30 containers of Coca Cola and measure the amount of sugar in each container. Using the formula that we learned earlier, we calculate the variance of the sample to be 5.20. Find a 90% confidence interval for the true variance. In other words, assuming that the sample comes from a normal population, what is the range of the population variance? To construct this 90% confidence interval, we first need to determine our upper and lower limits. The formula to construct this confidence interval and calculate the population variance, σ2\begin{align}\sigma^2\end{align}, is as follows: dfs2χ20.05σ2dfs2χ20.95\begin{align}\frac{dfs^2}{\chi^2_{0.05}} & \le \sigma^2 \le \frac{dfs^2}{\chi^2_{0.95}}\end{align} Using our standard chi-square distribution table, we can look up the critical χ2\begin{align}\chi^2\end{align} values for 0.05 and 0.95 at 29 degrees of freedom. According to the χ2\begin{align}\chi^2\end{align} distribution table, we find that χ20.05=42.557\begin{align}\chi^2_{0.05}=42.557\end{align} and that χ20.95=17.708\begin{align}\chi^2_{0.95}=17.708\end{align}. Since we know the number of observations and the standard deviation for this sample, we can then solve for σ2\begin{align}\sigma^2\end{align} as shown below: dfs242.557150.8042.5573.54σ2dfs217.708σ2150.8017.708σ28.52\begin{align}\frac{dfs^2}{42.557} & \le \sigma^2 \le \frac{dfs^2}{17.708}\\ \frac{150.80}{42.557} & \le \sigma^2 \le \frac{150.80}{17.708}\\ 3.54 & \le \sigma^2 \le 8.52\end{align} In other words, we are 90% confident that the variance of the population from which this sample was taken is between 3.54 and 8.52. 2. Assume the following data is from a normal population. Construct a 95% confidence interval for the standard deviation. 68.7 27.4 26 60.5 34.6 61.1 68.6 48.4 43.6 39.5 85.3 26.3 43.4 83.7 68.9 First, we need to find the sample standard deviation. This is easy enough to do by entering the data into a list and then use one variable statistics commands on a graphing calculator to determine the sample standard deviation, which you will find to be equal to 20.1. Note that there are 15 data points so the chi-square will have 14 degrees of freedom. Next we have to find the formula for a confidence interval of the standard deviation, based off of the one for the variance: dfs2χ20.05dfs2χ20.05sdfχ20.0520.11426.11914.72σ2dfs2χ20.95σ2dfs2χ20.95σsdfχ20.95σ20.1145.79σ31.42\begin{align}\frac{dfs^2}{\chi^2_{0.05}} & \le \sigma^2 \le \frac{dfs^2}{\chi^2_{0.95}}\\ \sqrt{\frac{dfs^2}{\chi^2_{0.05}}} & \le \sqrt{\sigma^2} \le \sqrt{\frac{dfs^2}{\chi^2_{0.95}}}\\ s\sqrt{\frac{df}{\chi^2_{0.05}}} & \le \sigma \le s\sqrt{\frac{df}{\chi^2_{0.95}}}\\ 20.1\sqrt{\frac{14}{26.119}} & \le \sigma \le 20.1\sqrt{\frac{14}{5.79}}\\ 14.72 & \le \sigma \le 31.42 \end{align} We believe that the population standard deviation is between 14.72 and 31.42. ### Example #### Example 1 Suppose a random sample of twenty boxes of crackers has a mean weight of 7.45 grams and a standard deviation of 4.1 grams. Assume the population is normally distributed. Find a 95% confidence interval for the standard deviation of the population. Use the formula we derived in example C to find the confidence interval for the standard deviation: sdfχ20.054.11932.853.12σ5.99σsdfχ20.95σ4.1198.91\begin{align} s\sqrt{\frac{df}{\chi^2_{0.05}}} & \le \sigma \le s\sqrt{\frac{df}{\chi^2_{0.95}}}\\ 4.1\sqrt{\frac{19}{32.85}} & \le \sigma \le 4.1\sqrt{\frac{19}{8.91}}\\ 3.12\leq \sigma \leq 5.99 \end{align} We are 95% confident that the population standard deviation is between 3.12 and 5.99. ### Review 1. We use the chi-square distribution for the: 1. goodness-of-fit test 2. test for independence 3. testing of a hypothesis of single variance 4. all of the above 2. True or False: We can test a hypothesis about a single variance using the chi-square distribution for a non-normal population. 3. In testing variance around the population mean, our null hypothesis states that the two population means that we are testing are: 1. equal with respect to variance 2. not equal 3. none of the above 4. In the formula for calculating the chi-square statistic for single variance, σ2\begin{align}\sigma^2\end{align} is: 1. standard deviation 2. number of observations 3. hypothesized population variance 4. chi-square statistic 5. If we knew the number of observations in a sample, the standard deviation of the sample, and the hypothesized variance of the population, what additional information would we need to solve for the chi-square statistic? 1. the chi-square distribution table 2. the population size 3. the standard deviation of the population 4. no additional information is needed 6. We want to test a hypothesis about a single variance using the chi-square distribution. We weighed 30 bars of Dial soap, and this sample had a standard deviation of 1.1. We want to test if this sample comes from the general factory, which we know from a previous study to have an overall variance of 3.22. What is our null hypothesis? 7. Compute χ2\begin{align}\chi^2\end{align} for Question 6. 8. Given the information in Questions 6 and 7, would you reject or fail to reject the null hypothesis? 9. Let’s assume that our population variance for this problem is unknown. We want to construct a 90% confidence interval around the population variance, σ2\begin{align}\sigma^2\end{align}. If our critical values at a 90% confidence interval are 17.71 and 42.56, what is the range for σ2\begin{align}\sigma^2\end{align}? 10. What statement would you give surrounding this confidence interval? 1. Consider the population mean and variance to be unknown. A random sample of size 8 is taken from a normal distribution. The sample has a variance of 0.64. A statistician is interested in testing the hypothesis: σ2=0.36\begin{align}\sigma^2=0.36\end{align} at the α=0.05\begin{align}\alpha=0.05\end{align} level. What is the result of this test? Explain. 2. What is the p-value for the test in problem 11? 3. Find a 90% confidence interval for the situation in problem 11. 4. For a confidence interval for the population variance find the upper and lower critical values for a 95% confidence interval with ten degrees of freedom. 5. A random sample of the population of 17 US state capitals has a mean of 330,731 and a standard deviation of 371,691. Assume that the population is normally distributed. Find a 90% confidence interval for the standard deviation of all US state capitals. 6. For a random sample of size 25 and a sample variance of 12.2 what is the probability of observing a value this low or lower if, in fact, the true population variance is 15.4? 7. Based on past experience a researcher believes the standard deviation of a population to be 12. A pilot study of n = 15 indicates a sample standard deviation of 9.25. At the 0.10 level of significance, what are the critical boundaries for rejecting H0:σ=12\begin{align}H_0: \sigma=12\end{align}? 8. A cereal company claims that their boxes of cereal weigh 15 ounces with a standard deviation of at most .5 ounces. Customers were complaining that this was not the case. The company needed to determine if the filling machine needed to be recalibrated. They took a sample of 84 boxes and determined that the standard deviation of their sample was 0.54. Does the machine need to be recalibrated? 9. Does the cost of a graphing calculator vary from store to store? To answer this question a survey was taken of 43 stores. This survey yielded a sample mean of $84 and a sample standard deviation of$13. Test the claim that the standard deviation is greater than \$15. 10. Given the following survey data. Number of births Frequency 0 5 1 30 2 10 3 5 Does this data indicate that the population standard deviation is greater than 0.75? 1. In a statistical hypothesis test, s2\begin{align}s^2\end{align} is the sample variance and σ2\begin{align}\sigma^2\end{align} is the population variance and n is the sample size. What is the formula for the chi-square test for a single variance? 2. Suppose a company claims that on average the cell phone batteries they produce last 60 minutes with a standard deviation of 4 minutes. The company randomly selects 7 batteries. The standard deviation of these batteries is 6 minutes. 1. What is the chi-square statistic for this test? 2. What are the null and alternative hypotheses for this test? 3. What is the decision? Explain. 3. Suppose the cell phone battery company takes a new random sample of 7 batteries. What is the probability that the standard deviation in the new test would be greater than 6 minutes? 4. A doctor’s records show that height of a random sample 25 infants at age 12 months to be 29.530 inches with a standard deviation of 1.0953 inches. Construct a 95 percent confidence interval for population variance. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English TermDefinition ANOVA The Analysis Of Variance, popularly known as the ANOVA, can be used in cases where there are more than two groups. chi-squared distribution The distribution of the chi-square statistic is called the chi-square distribution. chi-squared statistic The chi-squared statistic (X^2) is used to evaluate how well a set of observed data fits a corresponding expected set. confidence intervals A confidence interval is the interval within which you expect to capture a specific value. The confidence interval width is dependent on the confidence level. variance A measure of the spread of the data set equal to the mean of the squared variations of each data value from the mean of the data set.
# WRITING RATIOS WORKSHEET Writing ratios worksheet : Worksheet on ratios is much useful to the students who would like to practice problems on ratios. ## Writing ratios worksheet The number of dogs compared to the number of cats in an apartment complex is represented by the model shown. 1. Write a ratio that compares the number of dogs to the number of cats. 2. Complete the statement: In the apartment complex, there are __________ cats per dog. 3. If there are 15 cats in the apartment complex, how many dogs are there ? 4. How many cats are there if there are 5 dogs in the apartment complex ? The contents of Dana’s box of muffins are shown. Write each ratio in three different ways. 5. Banana nut muffins to corn muffins. 6. Corn muffins to total muffins Kaine's videos collection is shown. 7. Write the ratio of comedies to dramas in three different ways. 8. Write the ratio of dramas to total videos in three different ways. 8 cups made by party mix is shown. Write each ratio in three different ways. 9. Bagel chips to peanuts 10. Cheese crackers to peanuts ## Writing ratios worksheet - Solution Question 1 : Write a ratio that compares the number of dogs to the number of cats. 1 : 5 Question 2 : Complete the statement: In the apartment complex, there are __________ cats per dog. There are 5 cats per dog Question 3 : If there are 15 cats in the apartment complex, how many dogs are there ? Let "a" be the number of dogs for 15 cats. There is 1 dog for 5 cats and "a" number of dogs for 15 cats. So, we have 1 : 5 = a : 15 5 x a = 1 x 15 -----> a = 3 If there are 15 cats in the apartment complex, then 3 dogs will be there. Question 4 : How many cats are there if there are 5 dogs in the apartment complex ? Let "a" be the number of cats for 5 dogs. There are 5 cats for 1 dog and "a" number of cats for 5 dogs. So, we have 5 : 1 = a : 5 1 x a = 5 x 5 -----> a = 25 If there are 5 dogs in the apartment complex, then 25 cats will be there. Question 5 : Banana nut muffins to corn muffins There are 2 banana nut muffins and 5 corn muffins So, we have 2 banana nut muffins to 5 corn muffins, 2 : 5, 2/5 Question 6 : Corn muffins to total muffins There are 5 corn muffins and 12 muffins in total So, we have 5 corn muffins to 2 total muffins, 5 : 2, 5/2 Question 7 : Write the ratio of comedies to dramas in three different ways. There are 8 comedies and 3 dramas. So, we have 8 comedies to 3 dramas, 8 : 3, 8/3 Question 8 : Write the ratio of dramas to total videos in three different ways. There are 3 dramas and 14 total videos. So, we have 3 dramas to 14 total videos, 3 : 14, 3/14 Question 9 : Bagel chips to peanuts There are 3 bagel chips and 1 peanuts. So, we have 3 bagel chips to 1 peanuts, 3 : 1, 3/1 Question 10 : Cheese crackers to peanuts There are 1 cheese crackers and 1 peanuts. So, we have 1 cheese crackers to 1 peanuts, 1 : 1, 1 : 1 After having gone through the stuff given above, we hope that the students would have understood "Writing ratios worksheet". Apart from "Writing ratios worksheet", if you need any other stuff in math, please use our google custom search here. WORD PROBLEMS HCF and LCM word problems Word problems on simple equations Word problems on linear equations Algebra word problems Word problems on trains Area and perimeter word problems Word problems on direct variation and inverse variation Word problems on unit price Word problems on unit rate Word problems on comparing rates Converting customary units word problems Converting metric units word problems Word problems on simple interest Word problems on compound interest Word problems on types of angles Complementary and supplementary angles word problems Double facts word problems Trigonometry word problems Percentage word problems Profit and loss word problems Markup and markdown word problems Decimal word problems Word problems on fractions Word problems on mixed fractrions One step equation word problems Linear inequalities word problems Ratio and proportion word problems Time and work word problems Word problems on sets and venn diagrams Word problems on ages Pythagorean theorem word problems Percent of a number word problems Word problems on constant speed Word problems on average speed Word problems on sum of the angles of a triangle is 180 degree OTHER TOPICS Profit and loss shortcuts Percentage shortcuts Times table shortcuts Time, speed and distance shortcuts Ratio and proportion shortcuts Domain and range of rational functions Domain and range of rational functions with holes Graphing rational functions Graphing rational functions with holes Converting repeating decimals in to fractions Decimal representation of rational numbers Finding square root using long division L.C.M method to solve time and work problems Translating the word problems in to algebraic expressions Remainder when 2 power 256 is divided by 17 Remainder when 17 power 23 is divided by 16 Sum of all three digit numbers divisible by 6 Sum of all three digit numbers divisible by 7 Sum of all three digit numbers divisible by 8 Sum of all three digit numbers formed using 1, 3, 4 Sum of all three four digit numbers formed with non zero digits Sum of all three four digit numbers formed using 0, 1, 2, 3 Sum of all three four digit numbers formed using 1, 2, 5, 6
# Derivative and the Tangent Line Problem ## Presentation on theme: "Derivative and the Tangent Line Problem"— Presentation transcript: Derivative and the Tangent Line Problem The beginnings of Calculus Tangent Line Problem Definition of Tangent to a Curve Now to develop the equation of a line we must first find slope Definition of Tangent Line with Slope m Slope of Secant Line If f is defined on an open interval containing c, and if the limit exists, then the line passing through (c, f(c)) with slope m is the tangent line to the graph of f at the point (c, f(c)). Definition of Tangent Line The limit is telling us that the distance between the two points is getting smaller and smaller. If we let the limit approach zero, we are actually approaching being on one point and not two. That is how we can say the line is tangent and no longer a secant line. Definition of Tangent Line Find some slopes using this definition. Definition of the Derivative of a Function Demonstration Symbols for Derivative Sir Isaac Newton: f’(x) or y’ Gottfried Leibniz: This is read as “the derivative of y with respect to x.” Finding the Derivative by the Limit Process Examples: Quadratic Equation Square Root Function Rational Function Finding the Derivative Summation A very good summation of this information Differentiation Definition Alternative form of derivative: (This is when given one point) When is a Function not Differentiable When the graph has a sharp point. (This is because the derivative (slope) from the right and the derivative from the left are different values. When is a Function not Differentiable When the graph has a vertical line tangent. Remember the slope of a vertical line is undefined. When is a Function not Differentiable Where the function is not continuous at that point. Differentiability Implies Continuity If f is differentiable at x = c, then f is continuous at x = c. The converse is not true. Just because a function is continuous does not make it differentiable everywhere. (See the prior slides)
# Math Blog ### Worksheet on Word problem on Multiplication \| Multiplication Facts Practice the worksheet on word problem on multiplication. 1. One complete set of class IV costs $264. How much money did a class of 42 children pay to the bookshop owner, if all of them bought their books from him? 2. One pair of football shoes costs$ 628. Find the cost of Continue reading "Worksheet on Word problem on Multiplication \| Multiplication Facts" ### Word Problems on Multiplication \|3rd Grade Math\|Multiplication Problem Solved examples on word problems on multiplication.For a school trip 6 buses were hired. Each bus carried 42 children. How many children went on the trip? Solution: 2. The product of two numbers is 96. If one number is 8, find the other. Multiplicand × Multiplier = Product ###### Jul 25, 2018 Practice the worksheet on word problem on addition and subtraction. 1. In a village, there are 4,318 men, 3,624 women and 5,176 children. What is the total population of the village? 2. In a school, there are 860 children in the pre-primary section, 1,200 children in th ### Making the Numbers From Given Digits \| Write Smallest/Greatest Number We can make numbers from the given digits. Let us see the rules. Rule I. To get the smallest number, arrange the digits in ascending order from left to right. Rule II. To get the greatest number, arrange the digits in descending order from left to right. Example: Write the Continue reading "Making the Numbers From Given Digits \| Write Smallest/Greatest Number" ### Word Problems on Addition and Subtraction \| Mixed Add & Subtract Solved examples on Word problems on addition and subtraction . 1. In a school there are 2,392 boys and 2,184 girls. Find the total number of students in the school. Solution: Number of boys in the school = 2392 Number of girls in the school = + 2184 Total students in the ### Worksheet on Facts about Division \| Division with Small Numbers Practice the worksheet on facts about division. We know, dividend is always equal to the product of the divisor and the quotient added to the remainder. This will help us to solve the given questions. 1. Fill in the blanks: (i) Division is __ subtraction. Continue reading "Worksheet on Facts about Division \| Division with Small Numbers" ###### Jul 13, 2018 Practice the worksheet on facts about multiplication. We know in multiplication, the number being multiplied is called the multiplicand and the number by which it is being multiplied is called the multiplier. This will help us to solve the given questions. ### Worksheet on Facts about Subtraction \| Subtraction with Small Numbers Practice the worksheet on facts about subtraction. Subtraction with small numbers can be worked out horizontally and subtraction with large numbers is worked out vertically. 1. Fill in the missing numbers. (i) Take away 14 from 80 is ______ (ii) 150 decreased by 80 is ____ Continue reading "Worksheet on Facts about Subtraction \| Subtraction with Small Numbers" ###### Jul 12, 2018 Practice the worksheet on facts about addition. Addition of small numbers can be done horizontally and large numbers are added in vertical columns. 1. Fill in the missing number/word. (i) 4315 + 101 = 101 + ______ = ______ (ii) 1795 + 241 = 241 + ______ ### Facts about Multiplication \| Multiplication Operation \| Multiplicand We have learnt multiplication of numbers with 2digit multiplier. Now, we will learn more. Let us know some facts about multiplication. 1. In multiplication, the number being multiplied is called the multiplicand and the number by which it is being multiplied is called the ### Facts about Subtraction \| Subtraction of Small Numbers\|Solved Examples The operation to finding the difference between two numbers is called subtraction. Let us know some facts about subtraction which will help us to learn subtraction of large numbers. 1. Subtraction with small numbers can be worked out horizontally. Example: 8 – 5 = 3 24 – 4 = ###### Jul 11, 2018 The operation to find the total of different values is called addition. Let us know some facts about addition which will help us to learn to add 4-digit and 5-digit numbers. 1. Addition of small numbers can be done horizontally. Example: 6 + 2 + 3 = 11 ### Facts about Division \| Basic Division Facts \| Learn Long Division We have already learned division by repeated subtraction, equal sharing/distribution and by short division method. Now, we will read some facts about division to learn long division. 1. If the dividend is ‘zero’ then any number as a divisor will give the quotient as ‘zero’. Continue reading "Facts about Division \| Basic Division Facts \| Learn Long Division" ### Expanded Form and Short Form of a Number \| Numbers in Expanded Form When we write a number as a sum of place value of its digits, the number is said to be in expended form and when we write a number using digits, the number is said to be in short form. There are 3 ways to write the expanded form. There are 3 ways to write the expanded form Continue reading "Expanded Form and Short Form of a Number \| Numbers in Expanded Form" ### Long Division \| Division by One-Digit Divisor and Two-Digit Divisors As we know that the division is to distribute a given value or quantity into groups having equal values. In long division, values at the individual place (Thousands, Hundreds, Tens, Ones) are dividend one at a time starting with the highest place. Continue reading "Long Division \| Division by One-Digit Divisor and Two-Digit Divisors" ### Multiplication of Matrices \| How to Multiply Matrices? \|Rules\|Examples Two matrices A and B are said to be conformable for the product AB if the number of columns of A be equal to the number of rows of B. If A be an m × n matrix and B an n × p matrix then their product AB is defined to be the m × p matrix whose (ij)th element is obtained by Continue reading "Multiplication of Matrices \| How to Multiply Matrices? \|Rules\|Examples" ### Worksheet on Addition of Matrices \| Find the Sum of Two Matrices \| Ans Practice the problems given in the worksheet on addition of matrices. If M and N are the two matrices of the same order, then the matrices are said conformable for addition, and their sum is obtained by adding the corresponding elements of M and N. 1. Find the sum of A and B Continue reading "Worksheet on Addition of Matrices \| Find the Sum of Two Matrices \| Ans" ### Properties of Scalar Multiplication of a Matrix \|Scalar Multiplication We will discuss about the properties of scalar multiplication of a matrix. If X and Y are two m × n matrices (matrices of the same order) and k, c and 1 are the numbers (scalars). Then the following results are obvious. I. k(A + B) = kA + kB II. (k + c)A = kA + cA III. k(cA) Continue reading "Properties of Scalar Multiplication of a Matrix \|Scalar Multiplication" ### Scalar Multiplication of a Matrix \| Examples on Scalar Multiplication The operation of multiplying variables by a constant scalar factor may properly be called scalar multiplication and the rule of multiplication of matrix by a scalar is that the product of an m × n matrix A = [aij] by a scalar quantity c is the m × n matrix [bij] where bij Continue reading "Scalar Multiplication of a Matrix \| Examples on Scalar Multiplication" ### Subtraction of Matrices \| Examples on Difference of Two Matrices We proceed to develop the algebra of subtraction of matrices. Two matrices A and B are said to be conformable for subtraction if they have the same order (i.e. same number of rows and columns) and their difference A - B is defined to be the addition of A and (-B). Continue reading "Subtraction of Matrices \| Examples on Difference of Two Matrices" ### Properties of Addition of Matrices \| Commutative Law \| Associative Law We will discuss about the properties of addition of matrices. 1. Commutative law of addition of matrix: Matrix multiplication is commutative. This says that, if A and B are matrices of the same order such that A + B is defined then A + B = B + A. Proof: Let A = [aij]m × n Continue reading "Properties of Addition of Matrices \| Commutative Law \| Associative Law" ### Addition of Matrices \| Example on Sum of Two Matrices We proceed to develop the algebra of matrices. Two matrices A and B are said to be conformable for addition if they have the same order (same number of rows and columns). If A = (aij)m, n and B = (bij)m,n then their sum A + B is the matrix C = (cij)m,n where cij = aij + bij Continue reading "Addition of Matrices \| Example on Sum of Two Matrices" ### Triangular Matrix \| Upper Triangular Matrix \| Lower Triangular Matrix There are two types of triangular matrices. 1. Upper Triangular Matrix: A square matrix (aij) is said to be an upper triangular matrix if all the elements below the principal diagonal are zero (0). That is, [aij]m × n is an upper triangular matrix if (i) m = n and (ii) aij Continue reading "Triangular Matrix \| Upper Triangular Matrix \| Lower Triangular Matrix" ### Height and Distance with Two Angles of Elevation \| Solved Problems We will solve different types of problems on height and distance with two angles of elevation. Another type of case arises for two angles of elevations. In the given figure, let PQ be the height of pole of ‘y’ units. QR be the one of the distance between the foot of the pole Continue reading "Height and Distance with Two Angles of Elevation \| Solved Problems" ### Angle of Elevation \| How to Find out the Angle of Elevation We have already learnt about trigonometry in previous units in detail. Trigonometry has its own applications in mathematics and in physics. One such application of trigonometry in mathematics is “height and distances”. To know about height and distances, we have to start Continue reading "Angle of Elevation \| How to Find out the Angle of Elevation" ### Identity Matrix \| Unit Matrix \|If [d] is a scalar matrix then [d] = dI A scalar matrix whose diagonal elements are all equal to 1, the identity element of the ground field F, is said to be an identity (or unit) matrix. The identity matrix of order n is denoted by In. A scalar matrix is said to be a unit matrix, if diagonal elements are unity. Continue reading "Identity Matrix \| Unit Matrix \|If [d] is a scalar matrix then [d] = dI" ### Definition of Equal Matrices \| Examples of Equal Matrices Equality of two matrix: Two matrices [aij] and [bij] are said to be equal when they have the same number of rows and columns and aij = bij for all admissible values of i and j. Definition of Equal Matrices: Two matrices A and B are said to be equal if A and B have the same Continue reading "Definition of Equal Matrices \| Examples of Equal Matrices" ### Null Matrix \| Null or Zero Matrix\|Zero Matrix\|Problems on Null Matrix If each element of an m × n matrix be 0, the null element of F, the matrix is said to be the null matrix or the zero matrix of order m × n and it is denoted by Om,n. It is also denoted by O, when no confusion regarding its order arises. Null or zero Matrix: Whether A is a Continue reading "Null Matrix \| Null or Zero Matrix\|Zero Matrix\|Problems on Null Matrix" ### Column Matrix \| Definition of Column Matrix \|Examples of Column Matrix Here we will discuss about the column matrix with examples. In an m × n matrix, if n = 1, the matrix is said to be a column matrix. Definition of Column Matrix: If a matrix have only one column then it is called column matrix. Examples of column matrix: Continue reading "Column Matrix \| Definition of Column Matrix \|Examples of Column Matrix" ### Row Matrix \| Definition of Row Matrix \| Examples of Row Matrix In an m × n matrix, if m = 1, the matrix is said to be a row matrix. Definition of Row Matrix: If a matrix have only one row then it is called row matrix. Here we will discuss about the row matrix with examples. Examples of row matrix: Continue reading "Row Matrix \| Definition of Row Matrix \| Examples of Row Matrix" ### Square Matrix \| Definition of Square Matrix \|Diagonal of Square Matrix If square matrixes have n rows or columns then the matrix is called the square matrix of order n or an n-square matrix. Definition of Square Matrix: An n × n matrix is said to be a square matrix of order n. In other words when the number of rows and the number of columns in Continue reading "Square Matrix \| Definition of Square Matrix \|Diagonal of Square Matrix" ### Matrix \| Definition of a Matrix \| Examples of a Matrix \| Elements A rectangular array of mn elements aij into m rows and n columns, where the elements aij belongs to field F, is said to be a matrix of order m × n (or an m × n matrix) over the field F. Definition of a Matrix: A matrix is a rectangular arrangement or array of numbers Continue reading "Matrix \| Definition of a Matrix \| Examples of a Matrix \| Elements" ### Joint Variation \| Solving Joint Variation Problems and Application 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.A ∝ BCD or A = kBCD (k = constant ), then A varies jointly Continue reading "Joint Variation \| Solving Joint Variation Problems and Application" ### Indirect Variation \| Inverse Variation \| Inverse or Indirect Variation When two variables change in inverse proportion it is called as indirect variation. In indirect variation one variable is constant times inverse of other. If one variable increases other will decrease, if one decrease other will also increase. This means that the variables Continue reading "Indirect Variation \| Inverse Variation \| Inverse or Indirect Variation" ### Direct Variation \| Solving Direct Variation Word Problem When two variables change in proportion it is called as direct variation. In direct variation one variable is constant times of other. If one variable increases other will increase, if one decrease other will also decease. This means that the variables change in a same ratio Continue reading "Direct Variation \| Solving Direct Variation Word Problem" ### Methods of Solving Simultaneous Linear Equations \| Solved Examples There are different methods for solving simultaneous linear Equations: 1. Elimination of a variable 2. Substitution 3. Cross-multiplication 4. Evaluation of proportional value of variables This topic is purely based upon numerical examples. So, let us solve some examples Continue reading "Methods of Solving Simultaneous Linear Equations \| Solved Examples" ### Method of Cross Multiplication\|Solve by Method of Cross Multiplication The next method of solving linear equations in two variables that we are going to learn about is method of cross multiplication. Let us see the steps followed while soling the linear equation by method of cross multiplication: Assume two linear equation be A1 x + B1y + C1= Continue reading "Method of Cross Multiplication\|Solve by Method of Cross Multiplication" ### Properties of Angles of a Triangle \|Sum of Three Angles of a Triangle We will discuss about some of the properties of angles of a triangle. 1. The three angles of a triangle are together equal to two right angles. ABC is a triangle. Then ∠ZXY + ∠XYZ + ∠YZX = 180° Using this property, let us solve some of the examples. Solved examples Continue reading "Properties of Angles of a Triangle \|Sum of Three Angles of a Triangle " ### Geometrical Property of Altitudes\|Altitudes of Triangle are Concurrent The three altitudes of triangle are concurrent. The point at which they intersect is known as the orthocentre of the triangle. In the adjoining figure, the three altitudes XP, YQ and ZR intersect at the orthocentre O. Continue reading "Geometrical Property of Altitudes\|Altitudes of Triangle are Concurrent" ### Medians and Altitudes of a Triangle \|Three Altitudes and Three Medians Here we will discuss about Medians and Altitudes of a Triangle. Median: The straight line joining a vertex of a triangle to the midpoint of the opposite side is called a median. A triangle has three medians. Here XL, YM and ZN are medians. A geometrical property of medians Continue reading "Medians and Altitudes of a Triangle \|Three Altitudes and Three Medians" ### Classification of Triangles on the Basis of Their Sides and Angles Here we will discuss about classification of triangles on the basis of their sides and angles Equilateral triangle: An equilateral triangle is a triangle whose all three sides are equal. Here, XYZ is an equilateral triangle as XY = YZ = ZX. Isosceles triangle: An isosceles Continue reading "Classification of Triangles on the Basis of Their Sides and Angles" ### Triangle \| Exterior Opposite Angles\|Interior Opposite Angles\|Perimeter A triangle is a plane figure bounded by three straight lines. A triangle has three sides and three angles, and each one of them is called an element of the triangle. Here, PQR is a triangle, its three sides are line segments PQ, QR and RP; ; ∠PQR, ∠QRP and ∠RPQ are its Continue reading "Triangle \| Exterior Opposite Angles\|Interior Opposite Angles\|Perimeter" ### Problem on Change the Subject of a Formula \| Changing the Subject We will solve different types of problems on change the subject of a formula. The subject of a formula is a variable whose relation with other variables of the context is sought and the formula is written in such a way that subject is expressed in terms of the other Continue reading "Problem on Change the Subject of a Formula \| Changing the Subject" ### Worksheet on Change of Subject \| Change the Subject as Indicated Practice the questions given in the worksheet on change of subject When a formula involving certain variables is known, we can change the subject of the formula. What is the subject in each of the following questions? Change the subject as indicated. Continue reading "Worksheet on Change of Subject \| Change the Subject as Indicated" ### Worksheet on Framing a Formula \| Framing Formulas \| Frame an Equation Practice the questions given in the worksheet on framing a formula. I. Frame a formula for each of the following statements: 1. The side ‛s’ of a square is equal to the square root of its area A. Continue reading "Worksheet on Framing a Formula \| Framing Formulas \| Frame an Equation" ### Establishing an Equation \| Framing a Formula \| Framing Linear Equation We will discuss here about establishing an equation. In a given context, the relation between variables expressed by equality (or inequality) is called a formula. When a formula is expressed by an equality, the algebraic expression is called an equation. Continue reading "Establishing an Equation \| Framing a Formula \| Framing Linear Equation" ### Rectangular Cartesian Co-ordinates \| Abscissa \| Ordinate \| Oblique Co-ordinate What is Rectangular Cartesian Co-ordinates? Let O be a fixed point on the plane of this page; draw mutually perpendicular straight line XOX’ and YOY’ through O. Clearly, these lines divide the plane Continue reading "Rectangular Cartesian Co-ordinates \| Abscissa \| Ordinate \| Oblique Co-ordinate " ### What is Co-ordinate Geometry? \| Analytical Geometry\| Cartesian Co-ordinate What is co-ordinate geometry? The subject co-ordinate geometry is that particular branch of mathematics in which geometry is studied with the help of algebra. This branch of mathematics was fir Continue reading "What is Co-ordinate Geometry? \| Analytical Geometry\| Cartesian Co-ordinate" ### Table of Tangents and Cotangents \| Natural Tangents and Natural Cotangents We will discuss here the method of using the table of tangents and cotangents. This table shown below is also known as the table of natural tangents and natural cotangents. Using the table we can Continue reading "Table of Tangents and Cotangents \| Natural Tangents and Natural Cotangents "
Transcript 1.5 Analyzing Graphs of Functions 1.5 Analyzing Graphs of Functions (2,4) Find: a. the domain [-1,4) b. the range [-5,4] c. f(-1) = -5 (4,0) (-1,-5) d. f(2) = 4 Vertical Line Test for Functions Do the graphs represent y as a function of x? no yes yes Increasing and Decreasing Functions -2 -1 1 2 3 4 5 1. The function is decreasing on the interval (-2,0). 2. The function is constant on the interval (0,3). 3. The function is increasing on the interval (3,5). A function f is increasing on an interval if, for any x1 and x2 in the interval, x1 < x 2 implies f(x1) < f(x2) A function f is decreasing on an interval if, for any x1 and x2 in the interval, x1 < x 2 implies f(x1) > f(x2) A function f is constant on an interval if, for any x1 and x2 in the interval, f(x1) = f(x2) go to page 57 Tests for Even and Odd Functions A function is y = f(x) is even if, for each x in the domain of f, f(-x) = f(x) An even function is symmetric about the y-axis. A function is y = f(x) is odd if, for each x in the domain of f, f(-x) = -f(x) An odd function is symmetric about the origin. Ex. g(x) = x3 - x g(-x) = (-x)3 – (-x) = -x3 + x = -(x3 – x) Therefore, g(x) is odd because f(-x) = -f(x) Ex. h(x) = x2 + 1 h(-x) = (-x)2 + 1 = x2 + 1 h(x) is even because f(-x) = f(x) Summary of Graphs of Common Functions f(x) = c y=x y x y x y=x3 y = x2 The Greatest Integer Function y  [[ x]] x y 0 .2 .5 .8 .99 0 0 0 0 0 -4 -3 x y 1.1 1.4 1.7 1.8 1.99 1 1 1 1 1 -2 -1 0 1 2 1 3 2 4 5 3 Average Rate of Change of a Function Find the average rates of change of f(x) = x3 - 3x from x1 = -2 to x2 = 0. The average rate of change of f from x1 to x2 y f (x 2 )  f (x1)   x x 2  x1 f (x 2 )  f (x1 ) f (0)  f (2) 0  (2)   1 x 2   x1 0  2 2  Finding Average Speed The average speed of s(t) from t1 to t2 is s s(t 2 )  s(t1 )   t t 2  t1 Ex. The distance (in feet) a moving car is from a stoplight is given by the function s(t) = 20t3/2, where t is thetime (in seconds). Find the average speed of the car from t1 = 0 to t2 = 4 seconds. s(t 2 )  s(t1 ) s(4)  s(0) 160 0     40 ft /sec t 2  t1 40 4 What’s the average speed of the car from  540160  76ft /sec  5 4 to 9 seconds?
# Math Insight ### Introduction to triple integrals Remember how double integrals can be written as iterated integrals. Triple integrals are essentially the same thing as double integrals. (We just add a third dimension.) We will turn triple integrals into (triple) iterated integrals. Just as with double integrals, the only trick is determining the limits on the iterated integrals. (Unfortunately, it's harder to draw in three dimensions.) Before discussing how to set up the iterated integrals, we first address how to define triple integrals in the same way we define most of our integrals: with a Riemann sum. #### Defined by Riemann sums Let $f(x,y,z)$ be the density of a three-dimensional solid $\dlv$ at the point $(x,y,z)$ inside the solid. We want to define the triple integral of $f$ over $\dlv$ to be the total mass of $\dlv$. As with double integrals, we define the integral with Riemann sums. We chop up the solid $\dlv$ into small boxes, say with dimensions $\Delta x$, $\Delta y$, $\Delta z$. If $\dlv$ happened to be a cube, this chopping might look something like this. The volume of each small box is \begin{align} \Delta V = \Delta x \Delta y \Delta z. \end{align} Think of the boxes as being arranged in layers, with each layer arranged into rows and columns. We can then index the boxes so that box $ijk$ refers to the box in the $i$th row, the $j$th column, and the $k$th layer. For each box, we pick a point in the box to represent that box. For box $ijk$, we call that point $(x_{ijk}, y_{ijk}, z_{ijk})$. Pretend that the density of box $ijk$ is constant, i.e., that the density is $f(x_{ijk}, y_{ijk}, z_{ijk})$ everywhere in that box. The mass of box $ijk$ is its density times it volume: \begin{align} f(x_{ijk}, y_{ijk}, z_{ijk}) \Delta V. \end{align} We sum up these approximate masses to estimate the total mass of the solid $\dlv$. We obtain the Riemann sum \begin{align} \sum_{ijk} f(x_{ijk}, y_{ijk}, z_{ijk})\Delta V, \end{align} where the sum is over all small boxes. Let $\Delta x \to 0$, $\Delta y \to 0$, and $\Delta z \to 0$ (and let the number of small boxes go to infinity). The Riemann sum approaches the triple integral over the solid $\dlv$, \begin{align} \iiint_\dlv f\, dV = \lim_{\Delta x, \Delta y, \Delta z \to 0} \sum_{ijk} f(x_{ijk}, y_{ijk}, z_{ijk}) \Delta V, \end{align} assuming $f$ is continuous. The triple integral is the actual mass of $\dlv$. #### Triple iterated integrals If the solid $\dlv$ is a cube defined by $a \le x \le b$, $c \le y \le d$, and $p \le z \le q$, then we can easily write the triple integral as an iterated integral. We could first integrate $x$ from $a$ to $b$, then integrate $y$ from $c$ to $d$, and finally integrate $z$ from $p$ to $q$, \begin{align} \iiint_\dlv f\, dV = \int_p^q \left(\int_c^d \left(\int_a^b f(x,y,z) dx \right ) dy \right) dz. \end{align} This order of integration corresponds a certain way of ordering the terms in the Riemann sum: first, we sum over rows $i$, then we sum over columns $j$, and finally we sum over layers $k$. Just as with double integrals, other orders of integration are possible. We could, for example, first integrate with respect to $z$, then integrate with respect to $x$, and lastly integrate with respect to $y$, \begin{align} \iiint_\dlv f\, dV = \int_c^d \left(\int_a^b \left(\int_p^q f(x,y,z) dz \right ) dx \right) dy \end{align} The integration always proceeds from the inside to the outside. The integration order coincides with the order that the differentials (e.g. $dx$) appear, so we can specify the integration order by a list of the differentials. The previous integral is in the order $dz\,dx\,dy$. With this understanding, the parentheses become optional, and we usually omit them, writing the previous integral as \begin{align} \iiint_\dlv f\, dV = \int_c^d \int_a^b \int_p^q f(x,y,z) dz\, dx \, dy. \end{align} The iterated integral is simple when the solid $\dlv$ is a rectangular solid (like a cube, but where all edges aren't necessarily the same length). For more complicated shapes, finding the limits of integration can be tricky. As a first step, just remember these rules, which are analogous to the rules we had for limits on double iterated integrals. 1. The outer limits have to be constant. They cannot depend on any of the variables. 2. The middle limits can depend on the variable from the outer integral only. They cannot depend on the variable from the inner integral. 3. The inner limits can depend on the variable from the outer integral and the variable from the middle integral. For example, the following integral makes sense \begin{align} \iiint_\dlv f\, dV = \int_2^3 \int_{1-z}^0 \int_{-y^2-z^2}^{y^2+z^2} f(x,y,z) dx\, dy \, dz. \end{align} It describes the integral of $f$ over the region $\dlv$ defined by \begin{gather} 2 \le z \le 3,\\ 1-z \le y \le 0,\\ -y^2-z^2 \le x \le y^2+z^2 \end{gather} (not that you should be able to visualize what $\dlv$ looks like). $\cancel{}$ The following integral does not make sense \begin{align} \iiint_\dlv f\, dV = \color{red}{\xcancel{\color{black}{\int_y^x \int_{1}^{2x} \int_0^1 f(x,y,z) dx \, dy \, dz.}}} \end{align} Can you see why? The outer integral limits depend on both $x$ and $y$ (but $y$ isn't defined until you go inside the middle integral, and $x$ isn't defined until you go inside the inner integral). Also, the middle integral limits depend on $x$. A tricky part of triple integrals is determining the limits of integration (or bounds). Two methods for determining bounds are the shadow method and the cross section method. Nothing beats practice in learning to compute triple integrals, and you can try your hand on some of these triple integral examples.
LCM that 15 and 25 is the the smallest number amongst all typical multiples that 15 and 25. The first couple of multiples that 15 and 25 room (15, 30, 45, 60, 75, . . . ) and also (25, 50, 75, 100, 125, 150, 175, . . . ) respectively. There space 3 frequently used methods to uncover LCM the 15 and also 25 - by listing multiples, by department method, and also by element factorization. You are watching: What is the lcm of 15 and 25 1 LCM that 15 and 25 2 List of Methods 3 Solved Examples 4 FAQs Answer: LCM that 15 and also 25 is 75. Explanation: The LCM of 2 non-zero integers, x(15) and y(25), is the smallest positive integer m(75) the is divisible by both x(15) and y(25) without any remainder. The methods to discover the LCM that 15 and 25 are described below. By element Factorization MethodBy Listing MultiplesBy department Method ### LCM the 15 and 25 by prime Factorization Prime factorization of 15 and also 25 is (3 × 5) = 31 × 51 and (5 × 5) = 52 respectively. LCM that 15 and also 25 can be obtained by multiply prime factors raised to your respective highest possible power, i.e. 31 × 52 = 75.Hence, the LCM of 15 and 25 by element factorization is 75. ### LCM the 15 and 25 through Listing Multiples To calculate the LCM of 15 and 25 through listing the end the typical multiples, we deserve to follow the given listed below steps: Step 1: list a couple of multiples that 15 (15, 30, 45, 60, 75, . . . ) and 25 (25, 50, 75, 100, 125, 150, 175, . . . . )Step 2: The typical multiples native the multiples that 15 and 25 are 75, 150, . . .Step 3: The smallest common multiple the 15 and 25 is 75. ∴ The least usual multiple that 15 and also 25 = 75. ### LCM the 15 and also 25 by department Method To calculation the LCM that 15 and 25 by the division method, we will certainly divide the numbers(15, 25) by your prime components (preferably common). The product of this divisors offers the LCM that 15 and also 25. Step 3: continue the actions until just 1s are left in the critical row. See more: 270 Park Avenue New York Ny 10017, 270 Park Avenue, New York, Ny 10017 The LCM of 15 and also 25 is the product of every prime number on the left, i.e. LCM(15, 25) by department method = 3 × 5 × 5 = 75.
# Quadratic Functions. The graph of any quadratic function is called a parabola. Parabolas are shaped like cups, as shown in the graph below. If the coefficient. ## Presentation on theme: "Quadratic Functions. The graph of any quadratic function is called a parabola. Parabolas are shaped like cups, as shown in the graph below. If the coefficient."— Presentation transcript: The graph of any quadratic function is called a parabola. Parabolas are shaped like cups, as shown in the graph below. If the coefficient of x 2 is positive, the parabola opens upward; otherwise, the parabola opens downward. The vertex (or turning point) is the minimum or maximum point. Graphs of Quadratic Functions The Standard Form of a Quadratic Function The quadratic function f (x)  a(x  h) 2  k, a  0 is in standard form. The graph of f is a parabola whose vertex is the point (h, k). The parabola is symmetric to the line x  h. If a  0, the parabola opens upward; if a  0, the parabola opens downward. Graphing Parabolas With Equations in Standard Form To graph f (x)  a(x  h) 2  k: 1.Determine whether the parabola opens upward or downward. If a  0, it opens upward. If a  0, it opens downward. 2.Determine the vertex of the parabola. The vertex is (h, k). 3.Find any x-intercepts by replacing f (x) with 0. Solve the resulting quadratic equation for x. 4.Find the y-intercept by replacing x with zero. 5.Plot the intercepts and vertex. Connect these points with a smooth curve that is shaped like a cup. Step 1 Determine how the parabola opens. Note that a, the coefficient of x 2, is -2. Thus, a  0; this negative value tells us that the parabola opens downward. Standard form f (x)  a(x  h) 2  k a  -2 h  3 k  8 Given equation f (x)   2(x  3) 2  8 Solution We can graph this function by following the steps in the preceding box. We begin by identifying values for a, h, and k. Text Example Graph the quadratic function f (x)   2(x  3) 2  8.   2(x  3) 2  8 Find x-intercepts, setting f (x) equal to zero. Step 3 Find the x-intercepts. Replace f (x) with 0 in f (x)   2(x  3) 2  8. Step 2 Find the vertex. The vertex of the parabola is at (h, k). Because h  3 and k  8, the parabola has its vertex at (3, 8). (x  3) 2  4 Divide both sides by 2. (x  3)   2 Apply the square root method. x  3  2 or x  3  2 Express as two separate equations. x  1 or x  5 Add 3 to both sides in each equation. The x-intercepts are 1 and 5. The parabola passes through (1, 0) and (5, 0). 2(x  3) 2  8 Solve for x. Add 2(x  3) 2 to both sides of the equation. Text Example cont. Step 5 Graph the parabola. With a vertex at (3, 8), x-intercepts at 1 and 5, and a y-intercept at –10, the axis of symmetry is the vertical line whose equation is x  3.  f   2(0  3) 2  8   2(  3) 2  8   2(9)  8   10 Step 4 Find the y-intercept. Replace x with 0 in f (x)   2(x  3) 2  8. The y-intercept is –10. The parabola passes through (0,  10). Text Example cont. The Vertex of a Parabola Whose Equation Is f (x)  ax 2  bx  c Consider the parabola defined by the quadratic function f (x)  ax 2  bx  c. The parabola's vertex is at Example Graph the quadratic function f (x)   x 2  6x  Solution: Step 1 Determine how the parabola opens. Note that a, the coefficient of x 2, is -1. Thus, a  0; this negative value tells us that the parabola opens downward. Step 2 Find the vertex. We know the x-coordinate of the vertex is x = -b/(2a). We identify a, b, and c to substitute the values into the equation for the x-coordinate: x = -b/(2a) = -6/2(-1)=3. The x-coordinate of the vertex is 3. We substitute 3 for x in the equation of the function to find the y-coordinate: y=f(3) = -(3)^2+6(3)-2=-9+18-2=7, the parabola has its vertex at (3,7). Example Graph the quadratic function f (x)   x 2  6x  Step 3 Find the x-intercepts. Replace f (x) with 0 in f (x)   x 2  6x  2. 0 =  x 2  6x  2 Example Graph the quadratic function f (x)   x 2  6x  Step 4 Find the y-intercept. Replace x with 0 in f (x)   x 2  6x  2. f   0 2  6 0  2    The y-intercept is –2. The parabola passes through (0,  2).  Step 5 Graph the parabola. Minimum and Maximum: Quadratic Functions Consider f(x) = ax 2 + bx +c. 1.If a > 0, then f has a minimum that occurs at x = -b/(2a). This minimum value is f(-b/(2a)). 2.If a < 0, the f has a maximum that occurs at x = -b/(2a). This maximum value is f(-b/(2a)).
New Zealand Level 6 - NCEA Level 1 # Units of Volume and Capacity Lesson As we have been learning, everything in maths that relates to the ‘real world’ has units. If an amount represents a real life quantity, it has units attached to it. The units that we use depend on what we are measuring. There are some units we need to be familiar with, and know how to convert between. LENGTH/DISTANCE mm, cm, m, km mm2, cm2, m2, km2 mm3, cm3, m3, km3 mL, L, kL, ML mg, g, kg, metric tonne sec, mins, hrs, days, weeks, months, years ## Units of capacity Capacity is the amount of liquid that a container can hold. Capacity is usually measured using one of the following units: • millilitres (mL • litres (L • kilolitres (kL • megalitres (ML) You would be used to most of these through previous experiences in the size of your milk container, measuring ingredients when cooking, measuring medicines or even in science experiments you may have done at school. $1$1 = $1000$1000 mL $1$1 kL = $1000$1000 L = $1000000$1000000 mL ($1000\times1000$1000×1000) $1$1 ML = $1000$1000 kL = $1000000$1000000 L = $1000000000$1000000000 mL Thats about 200 million full teaspoons! To move from larger capacity units to smaller capacity units we multiply at each step. To move from smaller capacity units to larger capacity units we divide at each step. Do you see some patterns when looking at changing units of capacity? For example, notice that there is a factor of 1000 between each step. ## The prefixes You see here that we also use the prefixes of milli and kilo again. Remember that the prefix milli means $\frac{1}{1000}$11000 th of something. So we can see here that a millilitre is $\frac{1}{1000}$11000th of a litre, which means that there are $1000$1000 mL in $1$1 litre. Also the prefix kilomeans $1000$1000 lots of something, so a kilolitre is $1000$1000 litres. #### Example ##### Question 1 Question: Change $6.732$6.732kL into mL. Think: Think about the steps needed to move from kL to mL, (kL -> L -> mL) and identify the multiplication amounts for each step. I suggest moving through one step at a time. Do: First convert to litres: $6.732\times1000=6732$6.732×1000=6732 L Then convert to millilitres: $6732\times1000=6732000$6732×1000=6732000 mL It really doesn't matter if you think about it like I did, or if you do it differently. What is important is to keep track of your steps. See how my units changed at each calculation. Here is another: #### Example ##### Question 2 Question: Convert $468296$468296 mL into litres Think: Think about the steps needed to move from ml to litres, (mL -> L ) and identify the division amounts for each step. Do: $468296\div1000$468296÷1000 $=$= $468.296$468.296 L ## Units of Volume It is probably worthwhile to remind ourselves of the units that are often used for calculations involving volume. Units of Volume cubic millimetres = mm3 (picture a cube with side lengths of 1 mm each - pretty small this one!) cubic centimetres = cm3 (picture a cube with side lengths of 1 cm each - about the size of a dice) cubic metres = m3 (picture a cube with side lengths of 1 m each - what could be this big?) AND to convert to capacity - 1cm3 = 1mL ### Outcomes #### GM6-2 Apply the relationships between units in the metric system, including the units for measuring different attributes and derived measures #### 91030 Apply measurement in solving problems
# Into Math Grade 7 Module 5 Lesson 1 Answer Key Understand Multiplication and Division of Rational Numbers We included HMH Into Math Grade 7 Answer Key PDF Module 5 Lesson 1 Understand Multiplication and Division of Rational Numbers to make students experts in learning maths. ## HMH Into Math Grade 7 Module 5 Lesson 1 Answer Key Understand Multiplication and Division of Rational Numbers I Can evaluate expressions and solve real-world problems involving rational number multiplication and division. Arnot wins a $50 gift card for a virtual reality arcade. If he does not use the card for a whole year, the balance on the card will be reduced by$5 each month that it continues to go unused. What will be the change in the value of the card if Arnot doesn’t use it for 18 months? Arnot wins $50 gift card If he does not use the card for a whole year the balanced is reduced by$5 each month 1 year = 12 months = 12 Ã— -5 = -$60 The change in value if Arnot doesn’t use it for 18 months is =$18 Ã— -5 = -$90 Turn and Talk Did you use the same method? If not, explain your reasoning to make sure both methods are correct. Build Understanding 1. Jordan is scuba diving and stops each time she descends 15 feet to take a photo. Starting at the surface, she does this 4 times until she reaches her final depth. What is her overall change in elevation? Answer: Jordan descends 15 feet to take a photo = -15 Ã— 4 = -60 feet A. What number represents each change in elevation? _______________________ Answer: – 15 feet. B. Use the number line to model Jordanâ€™s entire descent. _______________________ C. Write an addition expression to represent your model. _______________________ Answer: 15 + 15 + 15 + 15 + 15 D. Write a multiplication expression to represent your model. The number line shows arrows moving ___ units ___ times. That means the multiplication expression would be ___ (___). Answer: The number line shows arrows moving 15 units 4 times. That means the multiplication expression would be – 60. E. How should the value of the addition expression in Step C compare to the value of the multiplication expression in Step D? Explain your reasoning. __________________________ F. Make a conjecture about the result of multiplying a negative number by a positive number. Explain your reasoning. __________________________ __________________________ Answer: Multiplying a negative number by a positive number we get negative number. (-) Ã— (+) = (-) Example: -10 Ã— +5 = -50 G. Use your conjecture in Step F to complete the last column in the table indicating the sign of the product pq. Answer: H. Complete the rule for multiplication of rational numbers with different signs. + The product of two rational numbers with different signs is a ___ number. I. What is Jordanâ€™s overall change in elevation? Answer: Jordanâ€™s overall change in elevation is -60 feet. Turn and Talk Steps A- E support the first rule in the Products of Rational Numbers table. Which mathematical property supports the second rule? Explain your reasoning. Answer: Commutative property. Explanation: The mathematical property that supports the second rule is called the Commutative property. a + b = b + a for addition a Ã— b = b Ã— a for mutiplication 2. In order for operations like multiplication to work with negative numbers the same way they work with positive numbers, the operations have to satisfy the properties of operations. A. Use the rule you wrote in the first task to complete the following. 3(6 + (-4)) = 3(6) + 3(___) 3(2) = 18 + (____) 6 = ___ Answer: 3(6 + (-4)) = 3(6) + 3(___) 3(6 + (-4)) = 3(6) + 3(-4) 3(2) = 18 – 12 6 = 6 B. What property is being applied in the first equation in Step A? Does applying the rule for multiplying numbers with different signs give the correct result? Why or why not? __________________________ __________________________ Answer: The property is applied in the first equation in Step A is called Associative property. Applying this rule for multiplying numbers with different signs gives the correct result. C. Complete the following to make the statements true. Explain how you figured out the answer. -3(6 + (-4)) = (-3)(6) + (-3)(-4) -3(2) = -18 + ___ -6 = -6 This means that -3(-4) = ___. __________________________ __________________________ Answer: -3(6 + (-4)) = (-3)(6) + (-3)(-4) -3(2) = -18 + 12 -6 = -6 This means that -3(-4) = 12. By solving the given expression and by using the BODMAS rule. Solve the numbers which are braces (6 + (-4)) =(6 – 4) = 2 – Ã— – = + so (-3)(-4) = 12 D. Complete the table. Answer: E. Write a rule for multiplication of rational numbers with the same sign. The product of two rational numbers with the same sign is a ___ number. (+) Ã— (+) = (+) The product of two rational numbers with the same sign is a positive number. Turn and Talk Write another example like the example in Step C to show why the product (-5)(-4) must be equal to 20. Answer: The product of two rational numbers with the same sign is a positive number. the product (-5)(-4) must be equal to 20. (-) Ã— (-) = (+) (-5) Ã— (-4) = 20 3. You can use what you know about multiplying signed numbers to figure out the rules for dividing signed numbers. In multiplication, if one factor is 0, the product will be 0. In division, the divisor cannot be 0. Division by 0 is undefined. A. Use the fact that division and multiplication are inverse operations to complete the number statements in the table. Answer: B. Use your results from Step A to complete the table. C. Complete the rules for division of rational numbers. • The quotient of two rational numbers with different signs is a ________ number. The quotient of two rational numbers with different signs is a negative number. • The quotient of of two rational numbers with the same sign is a ____ number. The quotient of two rational numbers with the same sign is a positive number. Answer: Check Understanding Question 1. Kaleb’s football team lost 2 yards on each of 3 consecutive plays. Complete the statement to represent the change in position relative to the line of scrimmage after these three plays. How does this example show that the product of a positive number and a negative number is negative? 3(___) = -2 + (-2) + (-2) = ___ Answer: They lost total of 6 yards. Explanation: They lost 2 yards on every 3 consecutive plays. That means the change is 6 yards. Question 2. How can you use the equation in Problem 1 to illustrate that the quotient of two negative numbers is positive? Answer: The quotient of two negative numbers is a positive rational number. Explanation: Let us take -3 and -2 as an example If we multiply – Ã— – we get positive -3 Ã— -2 = 6 Question 3. Compare the rules for finding the signs of products and quotients. Answer: The sign rules are the same for the multiplication and division of numbers. And multiplication can also be expressed as division. Example: a Ã— b and a Ã· b On Your Own Model with Mathematics Model the situation with a multiplication and an addition expression involving negative numbers. Then evaluate. Question 4. Every day that Annabelle takes the train to work, she uses an app to charge the parking fee shown to her credit card. What will be her card balance for commuter parking if she parks and takes the train to work 10 times? Answer:$50 Explanation: All a day parking is $5. So 10 times bill is 10 Ã—$5 = $50 Question 5. Alejandro makes donations in the amount of$100 every December to 5 of his favorite charities. He pays for the donations by check. What is the change in his checking account balance after making these donations? $500 Explanation: The donation will reduce his checking account balance by$500. Use Repeated Reasoning Complete the statements. Explain how you figured out the answer. Question 6. 8(-5 + 7) = 8(-5) + 8(7) 8(2) = ___ + 56 16 = 16 This means that 8(-5) = ___ -40 Explanation: Given, 8(-5 + 7) = 8(-5) + 8(7) 8(-5) = -40 8(2) = -40 + 56 16 = 16 (+) + (-) = (-) This means that 8(-5) = -40 Question 7. -2(-5 + 9) = -2(-5) + (-2(9)) -2(4) = ___ + (-18) -8 = -8 This means that -2(-5) = ___ 10 Explanation: Given, 2(-5 + 9) = -2(-5) + (-2(9)) -2(-5) = 10 -2(4) = _10__ + (-18) -8 = -8 This means that -2(-5) = 10 (-) Ã— (-) = (+) Identify the sign of the product or quotient. Do not evaluate. Question 8. 253 Ã— (-185) ________ The sign of the product is – Question 9. -59 Ã— 819 ________ The sign of the product is (-). Question 10. -1,200 Ã— (-490) ________ The sign of the product is (+). Question 11. 1,242 Ã· (-18) ________ The sign of the Quotient is (-). Explanation: Given, 1,242 Ã· (-18) The product of (+) Ã— (-) = (-). Question 12. -1,890 Ã· (-15) ________ The sign of the product is (+). Explanation: Given, -1,890 Ã· (-15) The product of (-) Ã— (-) = (+). Question 13. -18,175 Ã· 125 ________ The sign of the product is (-). Explanation: Given, -18,175 Ã· 125 The product of (-) Ã— (+) = (-). Explanation: Given, -18,175 Ã· 125 The product of (-) Ã— (+) = (-). Question 14. Complete each step to show that (-1)(-1) = 1. (-1)(0) = ____ (- 1)(- 1 + 1) = 0 Addition Property of opposites (- 1)(- 1) + (- 1)(1) = 0 (- 1)(- 1) + ___ = 0 Identity Property of Multiplication (- 1)(- 1) + (- 1) + ___ = 0 + ___ (- 1)(- 1) + (- 1 + 1) = 0 + 1 (- 1)(- 1) + ___ = 0 + 1 Addition Property of Opposites (- 1)(- 1) = ___ Identity Property of Addition (-1)(0) = 0 (- 1)(- 1 + 1) = 0 Addition Property of opposites (- 1)(- 1) + (- 1)(1) = 0 (- 1)(- 1) + 0 = 0 Identity Property of Multiplication (- 1)(- 1) + (- 1) + 0 = 0 + 0 (- 1)(- 1) + (- 1 + 1) = 0 + 1 (- 1)(- 1) + 0 = 0 + 1 Addition Property of Opposites (- 1)(- 1) = 1 Identity Property of Addition Question 15. The value of a collectible baseball card decreased as shown. A. Use Tools To find the average monthly change in the value of the card, use the number line to find -12 Ã· 6. -12 Ã· 6 = ___ _______________________ -2 Explanation: Given, -12 Ã· 6 By dividing the given expression the answer is -2. B. Construct Arguments What is the sign (positive or negative) of the quotient? Explain why this makes sense in the context of the problem. _______________________ _______________________ The sign of the quotient is (-). From the given question after dividing the expression the resultant will be negative. Question 16. Open Ended Write a different division problem involving negative numbers that can be solved using the number line given in Problem 15. let us assume the negative numbers is -14 Ã· -7 = 2 -10 Ã· -2 = 5 I’m in a Learning Mindset! How can I use my understanding of rational number multiplication to help support my understanding of rational number division? Lesson 5.1 More Practice/Homework Model with Mathematics For Problems 1 and 2, model the situation with a multiplication and an addition expression involving negative numbers. Then evaluate. Question 1. After a gymnastics competition, the coaches reviewed all the gymnastsâ€™ scores to identify areas for improvement. A missed landing on an aerial cartwheel deducts 2 points from the score. The coaches found that aerial cartwheels were missed 5 times across the competition. How did these missed landings affect the scores overall? 10 points will be deducted. Explanation: A missed landing on an aerial cartwheel deducts 2 points from the score. The coaches found that aerial cartwheels were missed 5 times across the competition Hence 5 Ã— 2 = 10 points would be deducted. Question 2. The temperature fell by 3Â°F every hour during a 6-hour period. What was the overall change in temperature during the 6-hour period? _______________________ 18 degrees Explanation: The temperature fell by 3Â°F the time period is 6 hour The overall change in temperature during the 6-hour period is 3 Ã— 6 = 18 degrees. Question 3. Use Repeated Reasoning Complete the statements. Explain. -4(5 + (-2)) = (___) (5) + (___) (-2) -4(___) = -20 + ___ This means that (-4)(-2) = ___ -12 = -12 _______________________ _______________________ (-4)(-2) = 8 Explanation: Given, 4(5 + (-2)) = (_4_) (5) + (_-4_) (-2) -4(_3_) = -20 + _8_ This means that (-4)(-2) = _8_ -12 = -12 Identify the sign of the product (positive or negative). Do not evaluate. Question 4. -819 Ã— (-324) The sign of the product is (+). Explanation: The multiplication of (-) Ã— (-) = (+) Question 5. -1,201 Ã— 54 The sign of the product is (-). Explanation: The multiplication of (-) Ã— (+) = (-) Question 6. 10,005 Ã— (-84) The sign of the product is (-). Explanation: The multiplication of (+) Ã— (-) = (-) Question 7. Construct Arguments Dario said that if the dividend and divisor have the same sign, then the quotient also has that same sign. Do you agree or disagree? Explain. I disagree. Explanation: If the dividend and divisor are positive numbers a quotient is a positive number. Example: 2 Ã· 2 = 1 If the dividend and divisor are negative numbers the quotient is a positive number. Example: -5 Ã· -5 = 1 Question 8. Use Structure Explain how you can use the fact that multiplication and division are inverse operations to help you determine whether the quotient -10 Ã· (-5) is positive or negative. -10 Ã· (-5) = 2 is positve. Explanation: Multiplying and dividing by the same number does not change the original number. For example (11 Ã— 6) Ã· 6 = 11. And the given quotient -10 Ã· (-5) = 2 is positive. Test Prep Question 9. Which quotients are negative? Select all that apply. A. $$\frac{-52}{13}$$ B. 14 Ã· (-2) C. -36 Ã· (-9) D. $$\frac{-20}{-5}$$ E. -27 Ã· 3 F. $$\frac{7}{-1}$$ A. $$\frac{-52}{13}$$ B. 14 Ã· (-2) E. -27 Ã· 3 F. $$\frac{7}{-1}$$ The options which are selected quotients are negative. Question 10. Marley’s bank charges a $3 service fee each time money is withdrawn from another bank’s ATM. Marley is traveling and must withdraw money from another bank’s ATM 4 times. Which expressions model the change in the balance of her account due to the service fees? Select all that apply. A. -3 + (-3) + (-3) + (-3) B. -4 + (-4) + (-4) + (-4) C. 4 Ã— (-3) D. -4 Ã— (-3) E. 3 Ã— 4 Answer: C. 4 Ã— (-3) Explanation: Marley’s bank charges a$3 service fee He must withdraw money ATM 4 times. 4 Ã— (-3) Question 11. An equation is shown. a âˆ™ b = c Which statement is true? A. If a > 0 and b > 0, then c < 0. B. If a > 0 and b < 0, then c < 0. C. If a < 0 and b > 0, then c > 0. D. If a < 0 and b < 0, then c < 0. C. If a < 0 and b > 0, then c > 0. Explanation: The statement that is true is option C. C. If a < 0 and b > 0, then c > 0. Spiral Review Question 12. Miguel takes $50 to the mall. He buys a flannel shirt for$18.99 and a hat for $12.49. On his way home, he stops at the bank and withdraws$25. How much money does Miguel have now? $43.52 Explanation: Given, He buys a flannel shirt =$18.99 And a hat = $12.49$18.99 + $12.49 =$31.48 $50 –$31.48 = $18.52$25 + $18.52 =$43.52 Question 13. A store marks up its inventory 45%. The wholesale price of an item was $130. What is the retail price of the item? Answer: The retail price of the item is 188.5. Explanation: Form the given question, find 130 Ã— (1 + 45%) Calculate the expression 130 Ã— 1.45 188.5 Question 14. Stephanie has a -$40 balance on her credit card. She makes two additional charges to pay $20 for gas and$25 for parking. Then she makes a credit card payment of $30. What is her remaining balance? Answer:$25 left Explanation: She has a total of $40 She makes two additional charges to pay$20 for gas and $25 for parking. Credit card payment of$30 = $30 –$25 = $5 On the card, she has a balance of$25. Scroll to Top Scroll to Top
# 1971 AHSME Problems/Problem 26 (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) ## Problem $[asy] size(2.5inch); pair A, B, C, E, F, G; A = (0,3); B = (-1,0); C = (3,0); E = (0,0); F = (1,2); G = intersectionpoint(B--F,A--E); draw(A--B--C--cycle); draw(A--E); draw(B--F); label("A",A,N); label("B",B,W); label("C",C,dir(0)); label("E",E,S); label("F",F,NE); label("G",G,SE); //Credit to chezbgone2 for the diagram[/asy]$ In $\triangle ABC$, point $F$ divides side $AC$ in the ratio $1:2$. Let $E$ be the point of intersection of side $BC$ and $AG$ where $G$ is the midpoints of $BF$. The point $E$ divides side $BC$ in the ratio $\textbf{(A) }1:4\qquad \textbf{(B) }1:3\qquad \textbf{(C) }2:5\qquad \textbf{(D) }4:11\qquad \textbf{(E) }3:8$ ## Solution We will use mass points to solve this problem. $AC$ is in the ratio $1:2,$ so we will assign a mass of $2$ to point $A,$ a mass of $1$ to point $C,$ and a mass of $3$ to point $F.$ We also know that $G$ is the midpoint of $BF,$ so $BG:GF=1:1.$ $F$ has a mass of $3,$ so $B$ also has a mass of $3.$ In line $BC,$ $B$ has a mass of $3$ and $C$ has a mass of $1.$ Therefore, $BE:EC = 1:3.$ The answer is $\textbf{(B)}.$ -edited by coolmath34
# 3.3 Rates of change and behavior of graphs Page 1 / 15 In this section, you will: • Find the average rate of change of a function. • Use a graph to determine where a function is increasing, decreasing, or constant. • Use a graph to locate local maxima and local minima. • Use a graph to locate the absolute maximum and absolute minimum. Gasoline costs have experienced some wild fluctuations over the last several decades. [link] http://www.eia.gov/totalenergy/data/annual/showtext.cfm?t=ptb0524. Accessed 3/5/2014. lists the average cost, in dollars, of a gallon of gasoline for the years 2005–2012. The cost of gasoline can be considered as a function of year. $y$ 2005 2006 2007 2008 2009 2010 2011 2012 $C\left(y\right)$ 2.31 2.62 2.84 3.3 2.41 2.84 3.58 3.68 If we were interested only in how the gasoline prices changed between 2005 and 2012, we could compute that the cost per gallon had increased from $2.31 to$3.68, an increase of $1.37. While this is interesting, it might be more useful to look at how much the price changed per year . In this section, we will investigate changes such as these. ## Finding the average rate of change of a function The price change per year is a rate of change because it describes how an output quantity changes relative to the change in the input quantity. We can see that the price of gasoline in [link] did not change by the same amount each year, so the rate of change was not constant. If we use only the beginning and ending data, we would be finding the average rate of change over the specified period of time. To find the average rate of change, we divide the change in the output value by the change in the input value. The Greek letter $\text{Δ}\text{\hspace{0.17em}}$ (delta) signifies the change in a quantity; we read the ratio as “delta- y over delta- x ” or “the change in $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ divided by the change in $\text{\hspace{0.17em}}x.$ ” Occasionally we write $\text{\hspace{0.17em}}\text{Δ}f\text{\hspace{0.17em}}$ instead of $\text{\hspace{0.17em}}\text{Δ}y,\text{\hspace{0.17em}}$ which still represents the change in the function’s output value resulting from a change to its input value. It does not mean we are changing the function into some other function. In our example, the gasoline price increased by$1.37 from 2005 to 2012. Over 7 years, the average rate of change was On average, the price of gas increased by about 19.6¢ each year. Other examples of rates of change include: • A population of rats increasing by 40 rats per week • A car traveling 68 miles per hour (distance traveled changes by 68 miles each hour as time passes) • A car driving 27 miles per gallon (distance traveled changes by 27 miles for each gallon) • The current through an electrical circuit increasing by 0.125 amperes for every volt of increased voltage • The amount of money in a college account decreasing by \$4,000 per quarter ## Rate of change A rate of change describes how an output quantity changes relative to the change in the input quantity. The units on a rate of change are “output units per input units.” The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values. $\frac{\Delta y}{\Delta x}=\frac{f\left({x}_{2}\right)-f\left({x}_{1}\right)}{{x}_{2}-{x}_{1}}$ what is the answer to dividing negative index In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c. give me the waec 2019 questions the polar co-ordinate of the point (-1, -1) prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x tanh`(x-iy) =A+iB, find A and B B=Ai-itan(hx-hiy) Rukmini what is the addition of 101011 with 101010 If those numbers are binary, it's 1010101. If they are base 10, it's 202021. Jack extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve 1+cos²A/cos²A=2cosec²A-1 test for convergence the series 1+x/2+2!/9x3 a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he? 100 meters Kuldeep Find that number sum and product of all the divisors of 360 Ajith exponential series Naveen yeah Morosi prime number? Morosi what is subgroup Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
# Chapter 3 - The Derivative - 3.1 Limits - 3.1 Exercises - Page 137: 28 100 #### Work Step by Step see: Rules for Limits Let $a, A$, and $B$ be real numbers, and let $f$ and $g$ be functions such that $\displaystyle \lim_{x\rightarrow a}f(x)=A$ and $\displaystyle \lim_{x\rightarrow a}g(x)=B$. 1. If $k$ is a constant, then $\displaystyle \lim_{x\rightarrow a}k=k$ and $\displaystyle \lim_{x\rightarrow a}[k\cdot f(x)]=k\cdot\lim_{x\rightarrow a}f(x)=k\cdot A$. 2. $\displaystyle \lim_{x\rightarrow a}[f(x)\pm g(x)]=\lim_{x\rightarrow a}f(x)\pm\lim_{x\rightarrow a}g(x)=A\pm B$ 6. For any real number $k,$ $\displaystyle \lim_{x\rightarrow a}[f(x)]^{k}=[\lim_{x\rightarrow a}f(x)]^{k}=A^{k}$, provided this limit exists. --------------- $\displaystyle \lim_{x\rightarrow 4}[1+f(x)]^{2}=\qquad$... use rule 6 =$[\displaystyle \lim_{x\rightarrow 4}(1+f(x))]^{2}\qquad$... use rule $2$ $=[\displaystyle \lim_{x\rightarrow 4}1+\lim_{x\rightarrow 4}f(x)]^{2}\qquad$... use rule $1$ $=(1+9)^{2}=10^{2}$ $=100$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
Lesson Explainer: Newton’s Third Law of Motion \| Nagwa Lesson Explainer: Newton’s Third Law of Motion \| Nagwa # Lesson Explainer: Newton’s Third Law of Motion Mathematics In this explainer, we will learn how to solve problems on Newton’s third law. An understanding of Newton’s first and second laws of motion is important for acquiring an understanding of Newton’s third law of motion. Let us define Newton’s first law of motion. ### Definition: Newton’s First Law of Motion A body in uniform motion will continue in that uniform motion unless a nonzero net force acts on it. Newton’s first law tells us that the momentum of a body will not change if the forces acting on the body are in equilibrium. Let us define Newton’s second law of motion. ### Definition: Newton’s Second Law of Motion When a net force acts on a rigid body, the body changes its velocity in the direction of the force. For a body of constant mass, the magnitude of the acceleration depends on the magnitude of the force and on the mass of the body, according to the formula where is the mass of the body and is the acceleration of the body. The rate of change of momentum of a body is equal to the net force acting on the body. For a constant-mass body, the rate of change of momentum of the body is directly proportional to the rate of change of velocity of the body. Let us now define Newton’s third law of motion. ### Definition: Newton’s Third Law of Motion Any force that acts on a body is one of a pair of forces that meet the following conditions: 1. The forces act on different interacting bodies. 2. The forces act in opposite directions. 3. The forces act along the same line. 4. The forces have the same magnitude. 5. The forces act for the same time interval. Bodies must exert forces on each other to interact. Consider the pair of forces and that act on bodies and and the pair of forces and that act on bodies and , as shown in the following figure. Each of the pairs of forces are exerted during collisions between the bodies that the forces act on. The forces and might seem to meet the conditions for a pair of forces defined by Newton’s third law of motion, as might the forces and . Body does not act on body to produce , however, and body does not act on body to produce . Only the pair of forces and and the pair of forces and are examples of pairs of forces that meet the conditions for a pair of forces defined by Newton’s third law of motion, as only these forces act on different interacting bodies. Newton’s third law is equivalent to the principle of the conservation of momentum. A force accelerating a body changes the momentum of a body. If when a force acts, an equal-magnitude force acts in the opposite direction for the same time interval on another body, then the sum of the changes of momentum of the bodies due to these forces is zero. Consider two bodies of equal mass that move at equal speeds toward each other along a smooth horizontal surface, approaching a position at which they will collide, as shown in the following figure. The sum of the momenta of the bodies is given by After the collision, suppose that the velocities of both of the bodies change sign but retain the same magnitude, as shown in the following figure. The net momentum after the collision is also zero. The change of the momentum of one of the bodies is equal to the change in the momentum of the other body; hence, the forces exerted on the bodies are equal. Newton’s third law applies to collisions between moving bodies but can apply to bodies that are at rest. A common example of pairs of forces that satisfy Newton’s third law of motion is the forces that act when a body is in contact with a surface on Earth. A body that is in contact with a surface on Earth is acted on by the weight of the body. The weight of the body is a force that depends on the mass of the body and the acceleration of the body. On Earth, the value of a due to the gravitational force produced by the gravitational field of Earth is 9.8 m/s2 and is represented by . A body in contact with a surface is also acted on by a reaction force normal to the surface. For a body in contact with a surface on Earth, two pairs of forces that satisfy Newton’s third law of motion exist: 1. the gravitational force that Earth exerts on the body and the gravitational force that the body exerts on Earth, 2. the reaction force on the body and the reaction force on Earth due to their mutual contact. The forces acting on the body and on Earth are shown in the following figure, where the forces are shown acting at the centers of mass of the body and of Earth. The pair of forces labeled that act to attract Earth and the body are gravitational forces, while the pair of forces labeled that act to repel Earth and the body are reaction forces. Each pair of forces that act on both the body and Earth satisfies Newton’s third law of motion. The magnitudes of the forces in either pair equal the magnitudes of the forces in the other pair. Questions involving the equilibrium of forces on a body at rest on the surface of Earth, where the equilibrium of the forces is the result of Newton’s third law, do not necessarily consider forces that act on Earth. Such questions may only show the forces that act on the body in contact with Earth, as is the case in the following figure of a body at rest on a surface on Earth. The force in the diagram is the weight, which is the force exerted on the body due to the masses of Earth and the body. The force is the reaction force exerted on the body by the surface of Earth. The forces are of equal magnitude. The gravitational and reaction forces that act on the body meet all the conditions for a pair of forces that satisfy Newton’s third law of motion except for the condition of acting on different bodies. Considering only the forces acting on a body can give a misleading impression that Newton’s third law can sometimes not apply. Consider a body that is not in contact with Earth that falls toward Earth. Before the falling body reaches the surface of Earth, there is no reaction force on the body and it seems as if the weight of the body acts to increase the momentum of the body without there being a corresponding change in the momentum of another body. According to Newton’s third law, however, as the body falls and so accelerates toward Earth, Earth accelerates toward the body, as the gravitational forces on the Earth and on the body have equal magnitude. The gravitational force on Earth increases the momentum of Earth in the opposite direction to the momentum of the body, conserving the momentum of the system consisting of the body and Earth. The change in the velocity of Earth due to its change in the momentum is negligible as the mass of Earth is so much greater than the mass of the body. Earth is treated as effectively being at a fixed position when considering bodies that fall toward Earth, as the acceleration of Earth due to the force acting on it due to the falling body is negligible. The examples in this explainer deal with bodies that are in contact with surfaces, where the surfaces may accelerate either in or opposite to the direction of the gravitational force on the bodies. To help us understand how to approach such examples, we can use free-body diagrams. A body at rest on a surface that is at rest is shown in the following figure. The net force on the body is zero. Suppose that the surface on which the body is at rest accelerates uniformly with an acceleration , where vertically upward is positive. The net force on the body must be given by The net force on the body is not the only force acting vertically upward on the body. A vertically upward force with a magnitude equal to the weight of the body acts on the body even when the body is at rest. When the body accelerates vertically upward, the upward force on the body is given by If the surface that the body rests on has acceleration vertically downward, then and the vertically upward force on the body to give it this acceleration is given by The vertically upward force on the body is the reaction force on the body. Let us look at an example involving forces acting on an object in contact with a moving surface. ### Example 1: Finding the Reaction Force between a Man and an Elevator Moving at a Constant Speed An elevator is moving vertically upward at a constant speed. A man of mass 90 kg is standing inside. Determine the reaction force of the floor on the man. Take . The elevator is moving at a constant speed. The man in the elevator is at rest relative to the elevator and so is also moving at a constant speed. A body moving at a constant speed is not accelerating, and the net force acting on a body with no acceleration must be zero. As the man is in a region where the force of Earth’s gravitational field accelerates unsupported bodies vertically downward at 9.8 m/s2, there is a vertically downward force acting on the man. This force is the weight of the man. If the vertically downward direction is taken as negative, the force of the man’s weight is given by The net force on the man is zero, so the floor of the elevator must be exerting a vertically upward force on the man and this force must have the same magnitude as the weight of the man, so the reaction force from the elevator floor has a magnitude of 882 N. The constant-velocity motion of a body does not involve the action of a net force, as Newton’s first law of motion states. Motion of a body where there is a change in velocity, however, does involve the action of a net force. Let us now look at an example involving the forces acting on a body that has a constant acceleration. ### Example 2: Finding the Reaction Force between a Man and an Elevator Accelerating Upward An elevator is accelerating vertically upward at 3.3 m/s2. A man of mass 80 kg is standing inside. Determine the reaction force of the floor on the man. The following figure shows the forces acting on the man. The elevator is accelerating vertically upward at 3.3 m/s2. The man in the elevator is at rest relative to the elevator, so he is also accelerating vertically upward at 3.3 m/s2. The magnitude of the net force acting on the man to provide this acceleration, taking vertically upward motion as positive, is given by The force of the man’s weight, , acts vertically downward. and it’s given by taking the magnitude of as 9.8 m/s2. The reaction force on the man from the elevator floor must provide a vertically upward force of this magnitude to keep the man at rest relative to the elevator floor. If the elevator was not accelerating, the reaction force vertically upward would have a magnitude of 784 N. The elevator is accelerating vertically upward at 3.3 m/s2, however, and because the man is at rest relative to the elevator, the force required to accelerate the man at this rate is the net force on the man, 264 N. The reaction force must therefore provide a vertically upward force on the man to both keep him at rest relative to a point in the gravitational field of Earth and to move that point away from Earth at 3.3 m/s2. The reaction force, , has a magnitude given by This result can also be obtained using the formula Let us consider how the reaction force would be affected by constant acceleration in the direction of the weight of a man rather than in the opposite direction to that weight. ### Example 3: Finding the Reaction Force between a Man and an Elevator Accelerating Downward An elevator was accelerating vertically downward at 1.7 m/s2. Given that the acceleration due to gravity is , find the reaction force of the floor to a passenger of mass 103 kg. The following figure shows the forces acting on the passenger. If the passenger was falling freely, the only force acting on the passenger would be the force of their weight and they would accelerate vertically downward at 9.8 m/s2. As the passenger is accelerating vertically downward at only 1.7 m/s2, a vertically upward force must be acting. The vertically upward acting force is the reaction force from the elevator floor. The reaction force on the passenger must be less than the force of their weight, as the acceleration of the passenger is downward. The force of the passenger’s weight, taking vertically downward motion as negative, is given by The net force acting on the passenger to provide their acceleration is given by The reaction force, , is therefore given by This result can also be obtained using the formula The reaction force on a person is what provides a person with the sensation of their body’s weight. The subjective experience of weight is the sensation of a resistance from the ground preventing motion through the ground, which is distinct from mere contact with the ground. Suppose that a person stands on a platform that is in an aircraft flying high above the ground. While the person is standing on the platform, the aircraft floor below the platform is removed. The platform and the person both accelerate vertically downward at 9.8 m/s2 provided that two assumptions are made. The first assumption is that resistive forces on the person and on the platform due to the motion of the air are negligible. The second assumption is that the gravitational forces exerted by the person and the platform on each other are negligible. If both assumptions are made, then the net force on the platform is equal to the weight of the platform and the net force on the person is equal to the weight of the person. The feet of the person remain in contact with the platform, but for both platform and person, the only forces that act on them are their respective weights. Therefore, no reaction forces are exerted on either the person or the platform. The person would experience no sensation of the platform pushing their feet upward and so would not feel as if their body weighed anything and would experience apparent weightlessness. The apparent weight of a body is, therefore, equal to the reaction force on the body. We have seen that the reaction force on a body is given by where is the acceleration of the surface that produces the reaction force. The following figures show how the apparent weight depends on the mass of a body and the acceleration of the surface that the body rests on. We can define the apparent weight of a body as follows. ### Definition: The Apparent Weight of a Body A body with a mass that is supported by a surface that has a uniform acceleration , within a uniform gravitational field of strength , exerts a reaction force, , on the body given by The reaction force on the body is known as the apparent weight of the body. Let us look at an example involving the apparent weight of a body. ### Example 4: Finding the Reading on a Spring Balance in an Accelerating Elevator A body of mass 45 kg was placed in a spring balance fixed to the ceiling of an elevator. If the elevator was accelerating downward at 105 cm/s2, what would the apparent weight of the body be? Consider the acceleration due to gravity to be 9.8 m/s2 and round your answer to two decimal places. A spring balance does not have a surface that supports a body, but rather suspends a body from a spring. The tension of the spring produced by the weight of the body acts equivalently to the reaction force that a surface supporting the body would produce. The only difference between the tension force produced by a spring and the reaction force produced by a surface is the point at which the forces act on the body; a surface reaction force acts at a point below the body and a spring tension acts at a point above the body. This difference is of no importance for the purposes of this question as the body that the force acts on has no stated features that distinguish it from a particle; it simply has a mass. The following figure shows the forces acting on the body. The magnitudes of the accelerations are not given in the same units, so one of the accelerations must be converted to have the unit of the other. As is a standard value and is given in metres per second squared, both of the accelerations will be expressed in metres per second squared, m/s2. The question states that the elevator accelerates downward, so the apparent weight of the body as measured by the spring balance has a magnitude given by Let us look at another example involving apparent weight. ### Example 5: Studying the Motion of an Elevator Using the Weight of a Body in Different Conditions A man was standing on a set of scales in an elevator, recording the readings as the elevator moved. His first reading was when the elevator was accelerating upward at a rate of . He made another reading when the elevator was accelerating downward at a rate of . Given that the ratio between the two readings was , determine where is the acceleration due to gravity. The variable used in representing the acceleration of the elevator should not be confused with the variable used in the expression for the apparent weight The value of in this expression is the acceleration of a body, not of an arbitrary variable. The acceleration of a body can be written in the form but is not equal to the acceleration of the body unless equals 1. The upward acceleration rate is stated to be . This acceleration is upward, hence in the opposite direction to . The equation is in this case equivalent to Hence, The downward acceleration rate is stated to be . The acceleration direction is the same as the direction of , so We are told that We see, therefore, that The factor of can be eliminated to give The bracketed terms can be expanded to give which can be rearranged to give The ratios of the magnitudes of and , therefore, is . ### Key Points • Newton’s third law of motion states that any force that acts is one of a pair of equal-magnitude forces acting for the same time interval along the same line in opposite directions and on different bodies. • The weight of a body is due to the gravitational force on the body, while the apparent weight of a body is due to the reaction force on the body from a surface with which the body is in contact. • If a body and its support move uniformly, the apparent weight of the body is equal to the weight of the body. • If a body and its support have an acceleration in the opposite direction to gravitational acceleration, the apparent weight of the body is greater than the weight of the body. • If a body and its support have an acceleration in the same direction as gravitational acceleration, the apparent weight of the body is less than the weight of the body. • If a body and its support have an acceleration in the same direction as gravitational acceleration equal to or greater than the magnitude of the gravitational acceleration, the apparent weight of the body is zero.
# Winding Firstly, you need to draw one big loop through all the given pieces, by connecting their ends together. The connections all have to be unit length. This means that any given end can be connected to at most 7 other ends. Secondly, the numbers indicate how often the loop winds around that point. Intuitively, the winding number of a point with respect to a loop is the number of times the path winds around it. Think of it as putting a nail into the paper, and letting the loop fall loose: the number of times it is still hanging around the nail is the winding number. If it falls down completely, the winding number is 0. For mathematicians: the given number is the absolute value of the winding number. The loop is allowed to cross itself, but it may not have any sharp angles. The puzzle-applet allows only one way to connect two given ends (with distance 1). ## Example Next to the puzzle you'll see a figure which indicates which connections are allowed from the top-right end of the bottom-left piece. Note that every end can reach at most 7 other ends. First look at the top ends of the top two pieces. They can only reach each other, as all other ends are too far away. So we can draw that line already. The puzzle applet allows only one way to connect these two ends without sharp angle. The same holds for the bottom ends of the bottom two pieces. Now look at the "1" to the right of the piece on the right. The only manner this number can come inside the loop is when we connect the right two ends of that piece. The remaining end of the piece in the top right can now only reach one end, and the same holds for the remaining end of the bottom-right piece. This means we can complete the right half of the puzzle. On the left, the "0" cannot come in the loop, so the leftmost ends of the left piece cannot be connected. The bottom two ends of the left piece also cannot connect, because then the remaining end in the bottom left piece cannot connect to any end. This means that the bottom left end of the left piece connects to the remaining end of the bottom left piece. In the same way, the top right end of the piece on the left connects with the top left piece. Now only two ends remain, which must be connected to each other. Verify for yourself that all numbers in the solution have the winding number as indicated. Puzzles in this genre. Info \| Statistics
# Angle Aod Has What Measurement According to the Protractor EstateName.com – Angle Aod Has What Measurement According to the Protractor Construction of angles is one of the essential part of geometry. An angle is a shape formed by two rays (called arms of angle) that shares a common point (called vertex). We can use protractor to construct various types of angles. Also, there are methods by which we can construct some specific angles such as 60°, 30°, 120°, 90°, 45°, etc., without using protractor. Hence, these angles can be constructed using a compass and ruler. Construction is an important concept where we learn to construct angles, lines and different shapes, in geometry. In this article, we will learn to construct the angles using protractor, compass, ruler and pencil. ## Types of Angles for Construction Before talking about construction of angles, let us quickly recall the different classifications of angles in Mathematics. Depending on the inclination between the two arms, the six different types of angles are: • Acute angle (less than 90 degrees) • Obtuse angle (more than 90 degrees) • Right angle (exactly 90 degrees) • Straight angle (equal to 180 degrees) • Reflex angle (more than 180 degrees) • Full rotation (equal to 360 degrees) ## Construction of Angles Using Protractor In our primary classes, we are taught to construct angles using protractors. It is easy to mark an angle of any measure using a protractor and construct it. We just need here a protractor, a ruler and a pencil. Let us see the steps of constructing angles using a protractor here. • Step 1: Draw a line segment AB • Step 2: Now place the center of the protractor on point A, such that the line segment AB is aligned with the line of the protractor • Step 3: Starting from 0 (in the protractor) mark the point C in the paper as per the required angle. • Step 4: Join points A and C.∠BAC is the required angle See the figure below to understand the steps. ## Construction of Angles Using Compass and Ruler Another method of constructing angles is by using a compass and a ruler. But we can use this method to construct some particular angles only such as 60°, 30°, 90°, 45°, etc. We can construct angles line 23°, 44°, 57°, etc., with accuracy using a compass and ruler. Let us construct few angles here using a compass. Read: Which Story Premise Most Clearly Lacks Conflict ### Construction of angle 60 degrees (60°) 60 degrees is one of the most basic constructions, which facilitates constructing angles of several other measures. The steps are: Step 1: Draw a line segment. Mark the left end as point O and the right end as point B. Step 2: Take the compass and open it up to a convenient radius. Place its pointer at O and with the pencil-head make an arc which meets the line OB at say, P. Step 3: Place the compass pointer at P and mark an arc that passes through O and intersects the previous arc at a point, say A. Step 4: Draw a line from O through A. We get the required angle i.e. ∠AOB = 60-degree angle. ### Construction of 30 degrees Angle (30°) A 30-degree angle is half of the 60-degree angle. • Follow the same steps as we have used for the construction of 60 degrees angles • Bisect the angle into two equal parts • Each part will be a 30-degree angle Hence, a 30-degree angle is constructed. ### Construction of angle 120 degrees (120°) A 120-degree angle is twice the angle of 60-degree. The steps for construction of 120 degrees angle is given below: Step 1: Draw a line segment. Mark the left end as point O and the right end as point B. Step 2: Take the compass and open it up to a convenient radius. Place its pointer at O and with the pencil-head make an arc which meets the line OB at say, P. Step 3: Without disturbing the radius, place the pointer at P and make an arc that cuts the previous arc at a point, say Q. Step 4: Similarly, with the same radius on the compass, place the pointer at point Q. Mark another arc on the first arc. Mark the point where they intersect as A. Step 5: Draw a line from O through A. Hence, ∠AOB is the required 120 degrees angle. ### Construction of angle 90-degree (90°) A 90-degree angle lies exactly between a 120-degree angle and a 60-degree angle. If we know the construction of 60 degrees and 120 degrees angles, then we can easily construct 90 degrees angle. Hence, follow the below steps: • Draw a line segment OA • Taking O as center and using a compass draw an arc of some radius, that cuts OA at B • Taking B as center and with the same radius draw another arc, that cuts the first arc at C • Taking C as center and with same radius draw an arc, that cuts the first arc at D • Now taking C and D as centers and radius greater than the arc CD, draw two arcs, such that they intersect at E. • Join OE such that∠AOE is a 90-degree angle Read: Which Polynomial is Represented by the Algebra Tiles Hence, ∠AOE is the required 90-degree angle. ## Solved Examples Let us learn to construct more angles using a compass and a ruler. ### Construction of angle 45-degree (45°) A 45-degree angle is the half of 90° angle. Hence, follow the below steps to get the construct 45 degrees angle. • Construct a 90-degree angle • Construct an angle bisector that bisects the 90-degree angle into two equal parts. • Each of the obtained angles is 45 degrees angle Hence, a 45-degree angle is constructed ### Construction of angle 75-degree (75°) Angle 75 degrees can also be constructed using a compass and ruler. Let us see the steps. • Draw a line segment OA • Taking O as center and using a compass draw an arc of some radius, that cuts OA at C • Taking C as center and with the same radius draw another arc, that cuts the first arc at M • Taking M as center and with same radius draw an arc, that cuts the first arc at L • Now taking L and M as centers and radius greater than the arc LM, draw two arcs, such that they intersect at B. • Join OB such that∠AOB is a 90-degree angle • Now taking N and M as centers, draw two arcs cutting at point P • Join OP Hence,∠AOP is the required 75 degrees angle. ### Construction of angle 150 degrees 150 degrees is equal to the sum of angles 30 degrees and 120 degrees. 30° + 120° = 150° Thus, steps to construct the 150 degrees angle are: • Construct angle∠AOC = 120 • Extend AO to OB • ∠BOC is equal to 60-degree angle (Since AOB = ∠AOC + ∠BOC = 180 degrees) • Bisect the angle∠BOC to form ∠COB = 30 Read: Which Phrase Best Defines the Term Figurative Language Hence,∠AOD is the 150-degree angle. ## Practice Question on Construction of Angles 1. Construct an angle of 135 degrees using a compass. 2. Construct a 105-degree angle using a compass. 3. Construct a 210-degree angle using a protractor. 4. Construct a 245-degree angle using a compass and a ruler From the above discussion, one would be able to understand the importance of special angles in the field of geometry. To learn more about constructing angles of different measures, download BYJU’S- The Learning App. ## Frequently Asked Questions – FAQs ### What is construction of angle? Construction of angle explains the construction of different angles (such as 30°,45°, 60°,90°) in geometry. These angles can be drawn using protractor or a compass and a ruler. ### How to construct angle? We can use protractor to construct an angle. For specific angles such as 30°,45°, 60°,90°, 120°, 150°, etc., we can use a compass and a rule to construct the angles. ### How to construct a right angle? A right angle is equal to 90°. Draw a line segment OA Taking O as center and using a compass draw an arc of some radius, that cuts OA at C Taking C as center and with the same radius draw another arc, that cuts the first arc at M Taking M as center and with the same radius draw an arc, that cuts the first arc at L Now taking L and M as centers and radius greater than the arc LM, draw two arcs, such that they intersect at B. Join OB such that ∠AOB is a 90-degree angle How to measure an angle using protractor? ### How to measure an angle using protractor? Keep the protractor above the vertex of the angle, such that the base arm of angle coincides with the line on the protractor. The measured angle will be a line on the protractor that is coinciding with the other arm of the angle. ### Angle Aod Has What Measurement According to the Protractor Sumber: https://byjus.com/maths/construction-of-angle/ ## 8g 10 35 3g 8g 10 35 3g (52% Off) Kaufen Günstig Pinuslongaeva 3g 5g 8g 10g 15g 20g …
Contents Result: 0 ### How to Use the Adding Fractions Calculator 1. Choose the Number of Fractions: Use the dropdown menu to select how many fractions you want to add or subtract. The default setting is 4 fractions, but you can choose between 2 and 10. 2. Enter Numerators and Denominators: For each fraction, enter the numerator (top number) and denominator (bottom number). The input fields are clearly labeled and aligned, making it easy to enter your values. 3. Select the Operator: By default, the calculator is set to add fractions. However, you can choose to subtract any of the fractions by selecting the “−” option from the dropdown next to the relevant fraction. 4. Calculate: Click the “Calculate” button to perform the addition or subtraction of the fractions. The result will be displayed at the bottom of the calculator. 5. View the Explanation: Below the result, the calculator will show a step-by-step explanation of how the calculation was performed, including how the fractions were adjusted to have the same denominator, the sum of the numerators, and how the result was simplified. ### Example Calculations Let’s add the fractions $\frac{1}{4} + \frac{2}{3}$ • Step 1: The LCD of 4 and 3 is 12. • Step 2: Adjust the fractions to have the same denominator: $\frac{1}{4}×\frac{3}{3}=\frac{3}{12},\phantom{\rule{1em}{0ex}}\frac{2}{3}×\frac{4}{4}=\frac{8}{12}$ • Step 3: Add the numerators: $\frac{3}{12}+\frac{8}{12}=\frac{11}{12}$ • Final Result: $\frac{11}{12}$ Example 2: Subtracting Fractions Now, let’s subtract the fractions $\frac{5}{6} – \frac{1}{2}$ • Step 1: The LCD of 6 and 2 is 6. • Step 2: Adjust the fractions to have the same denominator: $\frac{5}{6}×\frac{1}{1}=\frac{5}{6},\phantom{\rule{1em}{0ex}}\frac{1}{2}×\frac{3}{3}=\frac{3}{6}$ • Step 3: Subtract the numerators: $\frac{5}{6}-\frac{3}{6}=\frac{2}{6}$ • Step 4: Simplify the result: $\frac{2}{6}=\frac{1}{3}$ • Final Result: $\frac{1}{3}$ #### Quick Tips • Denominators Matter: Make sure the denominator you enter is not zero, as dividing by zero is undefined. • Check Your Inputs: Ensure all inputs are correct before calculating to avoid errors. • Simplify: The calculator automatically simplifies the result, but understanding how it’s done can help reinforce your knowledge of fractions.
# Developmental Maths Part 2 #### Learn intermediate maths using examples from the Monterey Institute. Created by Khan Academy.Recommended that you have a good foundation in Developmental Maths Part 1 before taking this course. 12 hours Intermediate ##### Pick a lesson 3: Reading Pie Graphs (Circle Graphs) 5: u08_l1_t2_we3 Stem-and-leaf Plots 6: Mean Median and Mode 7: Range and Mid-range 8: Measures of Center 10: Box-and-Whisker Plots 12: Simple Probability 13: Probability of More Complex Outcome 14: Coin Flipping Example 15: Variables and Expressions 1 16: Evaluating Expressions with Two Variables 17: Locate integers on a number line 18: Absolute Value of Integers 19: Opposite of a given number 20: u09 l1 t2 we4 Adding Real Numbers Application 21: Points on a number line 22: Number Sets 23: Comparing Rational Numbers 24: Identifying Rational Numbers 26: Adding integers with different signs 27: Adding real numbers with different signs 28: Adding fractions with different signs 29: Identity property of 0 30: Application of addition of real numbers 31: Subtracting Real Numbers 32: Adding and subtracting real numbers 33: Adding and subtracting real numbers application 34: Multiplying negative real numbers 35: Dividing real numbers with different signs 36: Identity property of 1 (second example) 37: Multiplying real number application 39: Associative property for multiplication 40: Distributive property 3 41: Order of Operations 1 42: Order of Operations 2 43: Solving One-Step Equations 44: Solving One-Step Equations 2 45: Multi-step equations 1 46: One-Step Equation Involving Fractions 47: Multi-step equations 2 48: More Involved Multi-Step Equation 49: Equation Special Cases 50: Application problems with equation in one variable 51: Evaluate a formula using substitution 52: Rearrange formulas to isolate specific variables 53: Inequalities on a number line 54: One-Step Inequalities 55: One-Step Inequalities 2 56: Multi-Step Inequalities 57: Multi-Step Inequalities 2 58: Compund Inequalities 59: Compund Inequalities 2 60: Compound Inequalities 3 61: Compound Inequalities 4 62: Absolute Value Equations Example 1 63: Absolute Value Inequalities Example 2 64: Absolute Value Equations 2 65: Evaluating exponential expressions 66: Evaluating exponential expressions 2 67: Negative and Positive Exponents 68: Exponent Properties 1 69: Exponent Properties 2 70: Exponent Properties 3 71: Exponent Properties 4 72: Exponent Properties 5 73: Exponent Properties 6 74: Exponent Properties 7 75: Scientific Notation I 76: Scientific Notation Example 2 77: Scientific Notation 3 78: Multiplying in Scientific Notation 79: Terms coefficients and exponents in a polynomial 80: Evaluating a polynomial at a given value 81: Simply a polynomial 83: Opposite of a Polynomial 84: Subtracting Polynomials 85: Multiplying Monomials 86: Multiplying Monomials by Polynomials 87: Multiplying Binomials 88: Multiplying Polynomials 89: Special Polynomials Products 1 90: Square a Binomial 91: Polynomial divided by monomial 92: Dividing Monomials 93: Dividing polynomials 1 94: Dividing polynomials with remainders 95: Evaluating an expression with multiple variables 96: Simplifying an expression with multiple variables 97: Simplifying Multivariable Polynomial 98: Subtracting polynomials with multiple variables 99: More multiplying polynomials 100: Dividing multivariable polynomial with monomial 101: Monomial Greatest Common Factor 102: GCF to Factor a Polynomial 103: Factor expressions by grouping 104: Factoring trinomials with a leading 1 coefficient 105: Factoring trinomials with a common factor 106: Factoring trinomials with a non-1 leading coefficient by grouping 107: Factoring perfect square trinomials 108: Factoring difference of squares 109: Factoring Sum of Cubes 110: Difference of Cubes Factoring 111: Solving factored equations 114: Plotting (x,y) relationships 116: Graphing using X and Y intercepts 117: Ordered Pair Solutions of Equations 118: Application problem with graph 119: Graphical Slope of a Line 120: Slope of a Line 2 121: Slope of a Line 3 122: Hairier Slope of Line 123: Equation of a line 1 124: Equation of a line 2 125: Equation of a line 3 126: Perpendicular Line Slope 127: Equation of a Line hairier example 128: Parallel Line Equation 129: Graphing Inequalities 1 130: Graphing Inequalities 2 131: Plot ordered pairs
# Fraction calculator This calculator adds two fractions. First, all fractions are converted to a common denominator when fractions have different denominators. Find the Least Common Denominator (LCD) or multiply all denominators to find a common denominator. When all denominators are the same, subtract the numerators and place the result over the common denominator. Then, simplify the result to the lowest terms or a mixed number. ## The result: ### 1/3 + 5/6 = 7/6 = 1 1/6 ≅ 1.1666667 Spelled result in words is seven sixths (or one and one sixth). ### How do we solve fractions step by step? 1. Add: 1/3 + 5/6 = 1 · 2/3 · 2 + 5/6 = 2/6 + 5/6 = 2 + 5/6 = 7/6 It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(3, 6) = 6. It is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 3 × 6 = 18. In the following intermediate step, it cannot further simplify the fraction result by canceling. In other words - one third plus five sixths is seven sixths. #### Rules for expressions with fractions: Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts. Mixed numerals (mixed numbers or fractions) keep one space between the integer and fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2. Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3. Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45. ### Math Symbols SymbolSymbol nameSymbol MeaningExample -minus signsubtraction 1 1/2 - 2/3 asteriskmultiplication 2/3 3/4 ×times signmultiplication 2/3 × 5/6 :division signdivision 1/2 : 3 /division slashdivision 1/3 / 5 :coloncomplex fraction 1/2 : 1/3 ^caretexponentiation / power 1/4^3 ()parenthesescalculate expression inside first-3/5 - (-1/4) The calculator follows well-known rules for the order of operations. The most common mnemonics for remembering this order of operations are: PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction. GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction. MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule. Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right.
# Chapter 5 - Trigonometric Functions - Section 5.2 Trigonometric Functions: Unit Circle Approach - 5.2 Assess Your Understanding - Page 402: 17 $\sin{t} = \dfrac{\sqrt{2}}{2}$ $\cos{t} = -\dfrac{\sqrt{2}}{2}$ $\tan{t} = -1$ $\csc{t} =\sqrt{2}$ $\sec{t} = -\sqrt{2}$ $\cot{t} =-1$ #### Work Step by Step With $P= \left(-\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2} \right) = (x,y)$, then $x = -\dfrac{\sqrt{2}}{2} \text{ and } \hspace{15pt} y =\dfrac{\sqrt{2}}{2}$. Thus, $\sin{t} = y$ $\sin{t} = \dfrac{\sqrt{2}}{2}$ $\cos{t} = x$ $\cos{t} = -\dfrac{\sqrt{2}}{2}$ $\tan{t} = \dfrac{y}{x}$ $\tan{t} = \dfrac{\dfrac{\sqrt{2}}{2}}{ -\dfrac{\sqrt{2}}{2} } = -1$ $\csc{t} = \dfrac{1}{y}$ $\csc{t} = \dfrac{1}{\dfrac{\sqrt{2}}{2}} = \sqrt{2}$ $\sec{t} = \dfrac{1}{x}$ $\sec{t} = \dfrac{1}{-\dfrac{\sqrt{2}}{2}}= -\sqrt{2}$ $\cot{t} = \dfrac{x}{y}$ $\cot{t} = \dfrac{-\dfrac{\sqrt{2}}{2} }{\dfrac{\sqrt{2}}{2}} = -1$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
### Factoring 2s^2+4s-15 Solution The variable we want to find is s We will solve for s using quadratic formula -b +/- sqrt(b^2-4ac)/(2a), graphical method and completion of squares. ${x}_{}=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$ Where a= 2, b=4, and c=-15 Applying values to the variables of quadratic equation -b, a and c we have ${s}_{}=\frac{-4±\sqrt{{4}^{2}-4x\mathrm{2x}\mathrm{-15}}}{2x2}$ This gives ${s}_{}=\frac{-4±\sqrt{{4}^{2}-\mathrm{-120}}}{4}$ ${s}_{}=\frac{-4±\sqrt{136}}{4}$ ${s}_{}=\frac{-4±11.6619037897}{4}$ ${\mathrm{s1}}_{}=\frac{-4+11.6619037897}{4}$ ${\mathrm{s2}}_{}=\frac{-4-11.6619037897}{4}$ ${s}_{1}=\frac{7.66190378969}{4}$ ${s}_{1}=\frac{-15.6619037897}{4}$ The s values are s1 = 1.91547594742 and s2 = -3.91547594742 ### Factoring Quadratic equation 2s^2+4s-15 using Completion of Squares 2s^2+4s-15 =0 Step1: Divide all terms by the coefficient of s2 which is 2. ${s}^{2}+\frac{4}{2}x-\frac{15}{2}=0$ Step 2: Keep all terms containing x on one side. Move the constant to the right. ${s}^{2}+\frac{4}{2}s=\frac{15}{2}$ Step 3: Take half of the x-term coefficient and square it. Add this value to both sides. ${s}^{2}+\frac{4}{2}s+{\left(\frac{4}{4}\right)}^{2}=\frac{15}{2}+{\left(\frac{4}{4}\right)}^{2}$ Step 4: Simplify right hand sides of expression. ${s}^{2}+\frac{4}{2}s+{\left(\frac{4}{4}\right)}^{2}=\frac{136}{16}$ Step 2: Write the perfect square on the left. ${\left(s+\frac{4}{4}\right)}^{2}=\frac{136}{16}$ Step 2: Take the square root on both sides of the equation. $s+\frac{4}{4}=±\sqrt{\frac{136}{16}}$ Step 2: solve for root s1. ${s}_{1}=-\frac{4}{4}+\frac{11.6619037897}{4}=\frac{7.66190378969}{4}$ ${s}_{1}=1.91547594742$ Step 2: solve for root s2. ${s}_{2}=-\frac{4}{4}-\frac{11.6619037897}{4}=\frac{-15.6619037897}{4}$ ${s}_{2}=-3.91547594742$ ### Solving equation 2s^2+4s-15 using Quadratic graph s2 + s + = 0 Solutions how to factor polynomials? Polynomials can be factored using this factoring calculator how to factor trinomials Trinomials can be solved using our quadratic solver Can this be used for factoring receivables, business, accounting, invoice, Finance etc No this cannot be used for that If you spot an error on this site, we would be grateful if you could report it to us by using the contact email provided. send email to contact on our site. Other Variants of 2s^2+4s-15 are below
# Help! I don’t understand my child’s maths homework! Here’s how many of us were taught division: “How many 25s are there in 3, non, carry and move on. ” “How many 25s in 36? 1. Move on and take the remainder with you. ” and so on. How you dealt with the remainder depended on your teacher and whether the book wanted a decimal answer or not. Is it straight forward? No not really, but the method is easy to figure out. For example, from a child’s point of view, there’s one 25 in 36, and four 25’s in 115, therefore there’s fourteen 25’s in 365… eh? It’s very hard to conceptualise for a child. One 250 in 360, with 110 left over, and four 25’s in 115 equate to fourteen 25’s in 365 isn’t much better either. Either way, that method isn’t very common anymore. Where it is used, it looks more like this these days: This is supposedly clearer. The other more common method is known as ‘chunking’, or the ‘partial quotient method’ (most likely known as chunking!). This is essentially a method whereby we break off pieces of the number and divided them by 25 separately. In the example above, we ‘know’ (we assume we know) that there are ten 25’s in 250, which is a big chunk of 365. So our answer has to be bigger than 10. We put 10 to the side (on the right in the picture above), and subtract 250 from our original number, 365. Now we have 115 left over. We ‘know’ that four 25’s are 100, so that chunk is dealt with too, and we subtract it from 115, and store the ‘4’ on the right hand side. So we have 10 + 4 groups of 25, and we’re left at this point with 15 left over from the original number 365. Well, 15 is less than our divisor 25, so it must be our remainder. And you wonder why kids find division difficult to do, let alone comprehend! ## 4 thoughts on “Help! I don’t understand my child’s maths homework!” 1. God knows why anybody needs to learn how to use division algorithms for BIG numbers these days. “Hey, if we start down this road then adding fractions will be next.”
# What is a Prime number? ## Definition A natural number that is divisible only by both one and itself is called a prime number. ### Introduction There are infinite natural numbers and every natural number is divisible by either one or more natural numbers but some natural numbers, which are primarily divisible only by one and itself. So, those natural numbers are called the prime numbers. Now, let’s learn clearly from some examples that what a prime number is. $(1).\,\,$ The natural number $1$ is not a prime number. $\implies$ $1 \div 1$ $\,=\,$ $1$ $(2).\,\,$ The natural number $2$ is a prime number because it is completely divisible only by $1$ and $2$. $\implies$ $2 \div 1$ $\,=\,$ $2$ and $2 \div 2$ $\,=\,$ $1$ $(3).\,\,$ The natural number $3$ is a prime number because it is completely divisible only by $1$ and $3$. $\implies$ $3 \div 1$ $\,=\,$ $3$ and $3 \div 3$ $\,=\,$ $1$ $(4).\,\,$ The natural number $4$ is not a prime number because it is completely divisible only by $1$ and $4$ but it is also divisible by $2$. $\implies$ $4 \div 1$ $\,=\,$ $4,$ $4 \div 2$ $\,=\,$ $2$ and $4 \div 4$ $\,=\,$ $1$ $(5).\,\,$ The natural number $5$ is a prime number because it is completely divisible only by $1$ and $5$. $\implies$ $5 \div 1$ $\,=\,$ $5$ and $5 \div 5$ $\,=\,$ $1$ The above examples help you to understand the fundamental definition of a prime number. In this way, the prime numbers can be identified from the infinite natural numbers in mathematics. ###### Math Questions The math problems with solutions to learn how to solve a problem. Learn solutions Practice now ###### Math Videos The math videos tutorials with visual graphics to learn every concept. Watch now ###### Subscribe us Get the latest math updates from the Math Doubts by subscribing us.
# DAV Class 8 Maths Chapter 16 Worksheet 2 Solutions The DAV Class 8 Maths Solution and DAV Class 8 Maths Chapter 16 Worksheet 2 Solutions of Rotational Symmetry offer comprehensive answers to textbook questions. ## DAV Class 8 Maths Ch 16 WS 2 Solutions Question 1. Name 2 figures having both line symmetry and rotational symmetry. Solution: Cricle and equilateral triangles have their rotational symmetry as well as line symmetry. Question 2. Make cutouts of the shapes given in Column I and complete the given table: Solution: ### DAV Class 8 Maths Chapter 16 Value Based Questions Question 1. On the ocassion of Independence Day, a school organised a Rajigoli Competition. Mehak made a beautiful rangoli design and won the first prize in the competition. (i) Does rotational symmetry exist in the given design? If yes, then find the angle and order of rotation. (ii) What is the importance of celebrating Independence Day? Solution: (i) Yes, the rotational symmetry exists in the given design. If the given design is rotated about its centre, it repeats itself six times in a full turn. So, it has a rotational symmetry of order 6. Angle of rotation of the design = $$\frac{360^{\circ}}{6}$$ = 60°. (ii) Celebrating Independence Day develops the spirit of unity among the citizens. It is also a way of tribute to martyrs. Question 2. A school celebrated ‘Road Safety Week’ and the children made circular posters of various traffic signs. The teacher asked Rohan and Sohan to display all the posters on the bulletin board. (i) Which of these signs exhibit rotational symmetry? Also find their angle and order of rotation. Solution: (i) Signs (b), (c) and (d) exhibit rotational symmetry. Angle of rotation of sign (b) = 180° Order of rotation of sign (b) = 2 Angle of rotation of sign (c) = 90° Order of rotation of sign (c) = 4 Angle of rotation of sign (d) = 120° Order of rotation of sign (d) = 3 (ii) Write the meaning of these safety signs. Solution: (a) Compulsory right turn (b) No parking (c) No stopping
2.4 Exponents (Page 2/2) Page 2 / 2 ${\left(8x\right)}^{3}$ means $\left(8x\right)\left(8x\right)\left(8x\right)$ since the parentheses indicate that the exponent 3 is directly connected to the factor $8x$ . Remember that the grouping symbols indicate that the quantities inside are to be considered as one single number. $34{\left(a+1\right)}^{2}$ means $34\cdot \left(a+1\right)\left(a+1\right)$ since the exponent 2 applies only to the factor $\left(a+1\right)$ . Practice set b Write each of the following without exponents. $4{a}^{3}$ $4aaa$ ${\left(4a\right)}^{3}$ $\left(4a\right)\left(4a\right)\left(4a\right)$ Sample set c Select a number to show that ${\left(2x\right)}^{2}$ is not always equal to $2{x}^{2}$ . Suppose we choose $x$ to be 5. Consider both ${\left(2x\right)}^{2}$ and $2{x}^{2}$ . $\begin{array}{lll}{\left(2x\right)}^{2}\hfill & \hfill & 2{x}^{2}\hfill \\ {\left(2\cdot 5\right)}^{2}\hfill & \hfill & 2\cdot {5}^{2}\hfill \\ {\left(10\right)}^{2}\hfill & \hfill & 2\cdot 25\hfill \\ 100\hfill & \ne \hfill & 50\hfill \end{array}$ Notice that ${\left(2x\right)}^{2}=2{x}^{2}$ only when $x=0$ . Practice set c Select a number to show that ${\left(5x\right)}^{2}$ is not always equal to $5{x}^{2}$ . Select $x=3$ . Then ${\left(5\cdot 3\right)}^{2}={\left(15\right)}^{2}=225$ , but $5\cdot {3}^{2}=5\cdot 9=45$ . $225\ne 45$ . In ${x}^{n}$ , Base $x$ is the base Exponent $n$ is the exponent Power The number represented by ${x}^{n}$ is called a power . $x$ To the $n$ Th power The term ${x}^{n}$ is read as " $x$ to the $n$ th power," or more simply as " $x$ to the $n$ th." $x$ Squared and $x$ Cubed The symbol ${x}^{2}$ is often read as " $x$ squared," and ${x}^{3}$ is often read as " $x$ cubed." A natural question is "Why are geometric terms appearing in the exponent expression?" The answer for ${x}^{3}$ is this: ${x}^{3}$ means $x\cdot x\cdot x$ . In geometry, the volume of a rectangular box is found by multiplying the length by the width by the depth. A cube has the same length on each side. If we represent this length by the letter $x$ then the volume of the cube is $x\cdot x\cdot x$ , which, of course, is described by ${x}^{3}$ . (Can you think of why ${x}^{2}$ is read as $x$ squared?) Cube with length $=x$ width $=x$ depth $=x$ Volume $=xxx={x}^{3}$ The order of operations In Section [link] we were introduced to the order of operations. It was noted that we would insert another operation before multiplication and division. We can do that now. The order of operations 1. Perform all operations inside grouping symbols beginning with the innermost set. 2. Perform all exponential operations as you come to them, moving left-to-right. 3. Perform all multiplications and divisions as you come to them, moving left-to-right. 4. Perform all additions and subtractions as you come to them, moving left-to-right. Sample set d Use the order of operations to simplify each of the following. ${2}^{2}+5=4+5=9$ ${5}^{2}+{3}^{2}+10=25+9+10=44$ $\begin{array}{ll}{2}^{2}+\left(5\right)\left(8\right)-1\hfill & =4+\left(5\right)\left(8\right)-1\hfill \\ \hfill & =4+40-1\hfill \\ \hfill & =43\hfill \end{array}$ $\begin{array}{ll}7\cdot 6-{4}^{2}+{1}^{5}\hfill & =7\cdot 6-16+1\hfill \\ \hfill & =42-16+1\hfill \\ \hfill & =27\hfill \end{array}$ $\begin{array}{ll}{\left(2+3\right)}^{3}+{7}^{2}-3{\left(4+1\right)}^{2}\hfill & ={\left(5\right)}^{3}+{7}^{2}-3{\left(5\right)}^{2}\hfill \\ \hfill & =125+49-3\left(25\right)\hfill \\ \hfill & =125+49-75\hfill \\ \hfill & =99\hfill \end{array}$ $\begin{array}{ll}{\left[4{\left(6+2\right)}^{3}\right]}^{2}\hfill & ={\left[4{\left(8\right)}^{3}\right]}^{2}\hfill \\ \hfill & ={\left[4\left(512\right)\right]}^{2}\hfill \\ \hfill & ={\left[2048\right]}^{2}\hfill \\ \hfill & =4,194,304\hfill \end{array}$ $\begin{array}{ll}6\left({3}^{2}+{2}^{2}\right)+{4}^{2}\hfill & =6\left(9+4\right)+{4}^{2}\hfill \\ \hfill & =6\left(13\right)+{4}^{2}\hfill \\ \hfill & =6\left(13\right)+16\hfill \\ \hfill & =78+16\hfill \\ \hfill & =94\hfill \end{array}$ $\begin{array}{ll}\frac{{6}^{2}+{2}^{2}}{{4}^{2}+6\cdot {2}^{2}}+\frac{{1}^{3}+{8}^{2}}{{10}^{2}-\left(19\right)\left(5\right)}\hfill & =\frac{36+4}{16+6\cdot 4}+\frac{1+64}{100-95}\hfill \\ \hfill & =\frac{36+4}{16+24}+\frac{1+64}{100-95}\hfill \\ \hfill & =\frac{40}{40}+\frac{65}{5}\hfill \\ \hfill & =1+13\hfill \\ \hfill & =14\hfill \end{array}$ Practice set d Use the order of operations to simplify the following. ${3}^{2}+4\cdot 5$ 29 ${2}^{3}+{3}^{3}-8\cdot 4$ 3 ${1}^{4}+{\left({2}^{2}+4\right)}^{2}÷{2}^{3}$ 9 ${\left[6\left(10-{2}^{3}\right)\right]}^{2}-{10}^{2}-{6}^{2}$ 8 $\frac{{5}^{2}+{6}^{2}-10}{1+{4}^{2}}+\frac{{0}^{4}-{0}^{5}}{{7}^{2}-6\cdot {2}^{3}}$ 3 Exercises For the following problems, write each of the quantities using exponential notation. $b$ to the fourth ${b}^{4}$ $a$ squared $x$ to the eighth ${x}^{8}$ $\left(-3\right)$ cubed 5 times $s$ squared $5{s}^{2}$ 3 squared times $y$ to the fifth $a$ cubed minus $\left(b+7\right)$ squared ${a}^{3}-{\left(b+7\right)}^{2}$ $\left(21-x\right)$ cubed plus $\left(x+5\right)$ to the seventh $xxxxx$ ${x}^{5}$ $\left(8\right)\left(8\right)xxxx$ $2\cdot 3\cdot 3\cdot 3\cdot 3xxyyyyy$ $2\left({3}^{4}\right){x}^{2}{y}^{5}$ $2\cdot 2\cdot 5\cdot 6\cdot 6\cdot 6xyyzzzwwww$ $7xx\left(a+8\right)\left(a+8\right)$ $7{x}^{2}{\left(a+8\right)}^{2}$ $10xyy\left(c+5\right)\left(c+5\right)\left(c+5\right)$ $4x4x4x4x4x$ ${\left(4x\right)}^{5}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}{4}^{5}{x}^{5}$ $\left(9a\right)\left(9a\right)\left(9a\right)\left(9a\right)$ $\left(-7\right)\left(-7\right)\left(-7\right)aabbba\left(-7\right)baab$ ${\left(-7\right)}^{4}{a}^{5}{b}^{5}$ $\left(a-10\right)\left(a-10\right)\left(a+10\right)$ $\left(z+w\right)\left(z+w\right)\left(z+w\right)\left(z-w\right)\left(z-w\right)$ ${\left(z+w\right)}^{3}{\left(z-w\right)}^{2}$ $\left(2y\right)\left(2y\right)2y2y$ $3xyxxy-\left(x+1\right)\left(x+1\right)\left(x+1\right)$ $3{x}^{3}{y}^{2}-{\left(x+1\right)}^{3}$ For the following problems, expand the quantities so that no exponents appear. ${4}^{3}$ ${6}^{2}$ $6\text{\hspace{0.17em}}·\text{\hspace{0.17em}}6$ ${7}^{3}{y}^{2}$ $8{x}^{3}{y}^{2}$ $8\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}y\text{\hspace{0.17em}}·\text{\hspace{0.17em}}y$ ${\left(18{x}^{2}{y}^{4}\right)}^{2}$ ${\left(9{a}^{3}{b}^{2}\right)}^{3}$ $\left(9aaabb\right)\left(9aaabb\right)\left(9aaabb\right)\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9aaaaaaaaabbbbbb$ $5{x}^{2}{\left(2{y}^{3}\right)}^{3}$ $10{a}^{3}{b}^{2}{\left(3c\right)}^{2}$ $10aaabb\left(3c\right)\left(3c\right)\text{\hspace{0.17em}or}\text{\hspace{0.17em}}10\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3aaabbcc$ ${\left(a+10\right)}^{2}{\left({a}^{2}+10\right)}^{2}$ $\left({x}^{2}-{y}^{2}\right)\left({x}^{2}+{y}^{2}\right)$ $\left(xx-yy\right)\left(xx+yy\right)$ For the following problems, select a number (or numbers) to show that ${\left(5x\right)}^{2}$ is not generally equal to $5{x}^{2}$ . ${\left(7x\right)}^{2}$ is not generally equal to $7{x}^{2}$ . Select $x=2.$ Then, $196\ne 28.$ ${\left(a+b\right)}^{2}$ is not generally equal to ${a}^{2}+{b}^{2}$ . For what real number is ${\left(6a\right)}^{2}$ equal to $6{a}^{2}$ ? zero For what real numbers, $a$ and $b$ , is ${\left(a+b\right)}^{2}$ equal to ${a}^{2}+{b}^{2}$ ? Use the order of operations to simplify the quantities for the following problems. ${3}^{2}+7$ 16 ${4}^{3}-18$ ${5}^{2}+2\left(40\right)$ 105 ${8}^{2}+3+5\left(2+7\right)$ ${2}^{5}+3\left(8+1\right)$ 59 ${3}^{4}+{2}^{4}{\left(1+5\right)}^{3}$ $\left({6}^{2}-{4}^{2}\right)÷5$ 4 ${2}^{2}\left(10-{2}^{3}\right)$ $\left({3}^{4}-{4}^{3}\right)÷17$ 1 ${\left(4+3\right)}^{2}+1÷\left(2\cdot 5\right)$ ${\left({2}^{4}+{2}^{5}-{2}^{3}\cdot 5\right)}^{2}÷{4}^{2}$ 4 ${1}^{6}+{0}^{8}+{5}^{2}{\left(2+8\right)}^{3}$ $\left(7\right)\left(16\right)-{9}^{2}+4\left({1}^{1}+{3}^{2}\right)$ 71 $\frac{{2}^{3}-7}{{5}^{2}}$ $\frac{{\left(1+6\right)}^{2}+2}{19}$ $\frac{51}{19}$ $\frac{{6}^{2}-1}{5}+\frac{{4}^{3}+\left(2\right)\left(3\right)}{10}$ $\frac{5\left[{8}^{2}-9\left(6\right)\right]}{{2}^{5}-7}+\frac{{7}^{2}-{4}^{2}}{{2}^{4}-5}$ 5 $\frac{{\left(2+1\right)}^{3}+{2}^{3}+{1}^{3}}{{6}^{2}}-\frac{{15}^{2}-{\left[2\left(5\right)\right]}^{2}}{5\cdot {5}^{2}}$ $\frac{{6}^{3}-2\cdot {10}^{2}}{{2}^{2}}+\frac{18\left({2}^{3}+{7}^{2}\right)}{2\left(19\right)-{3}^{3}}$ $\frac{1070}{11}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}97.\overline{27}$ Exercises for review ( [link] ) Use algebraic notation to write the statement "a number divided by eight, plus five, is equal to ten." ( [link] ) Draw a number line that extends from $-5$ to 5 and place points at all real numbers that are strictly greater than $-3$ but less than or equal to 2. ( [link] ) Is every integer a whole number? ( [link] ) Use the commutative property of multiplication to write a number equal to the number $yx$ . $xy$ ( [link] ) Use the distributive property to expand $3\left(x+6\right)$ . what is the stm is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.? Rafiq industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong Damian How we are making nano material? what is a peer What is meant by 'nano scale'? What is STMs full form? LITNING scanning tunneling microscope Sahil how nano science is used for hydrophobicity Santosh Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq Rafiq what is differents between GO and RGO? Mahi what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq Rafiq what is Nano technology ? write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? why we need to study biomolecules, molecular biology in nanotechnology? ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. why? what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe 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 How can I make nanorobot? Lily 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 how can I make nanorobot? Lily 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 how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good Please keep in mind that it's not allowed to promote any social groups (whatsapp, facebook, etc...), exchange phone numbers, email addresses or ask for personal information on QuizOver's platform.
# How do you write 2(x+4)^2 + 3(y-1)^2 = 24 in standard form? May 3, 2016 ${\left(x - \left(- 4\right)\right)}^{2} / {\left(\sqrt{12}\right)}^{2} + {\left(y - 1\right)}^{2} / {\left(\sqrt{8}\right)}^{2} = 1$ #### Explanation: The standard form for an ellipse is $\textcolor{w h i t e}{\text{XXX}} {\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$ Given $\textcolor{w h i t e}{\text{XXX}} 2 {\left(x + 4\right)}^{2} + 3 {\left(y - 1\right)}^{2} = 24$ $\Rightarrow$ $\textcolor{w h i t e}{\text{XXX}} {\left(x + 4\right)}^{2} / 12 + {\left(y - 1\right)}^{2} / 8 = 1$ $\textcolor{w h i t e}{\text{XXX}} {\left(x - \left(- 4\right)\right)}^{2} / {\left(\sqrt{12}\right)}^{2} + {\left(y - 1\right)}^{2} / {\left(\sqrt{8}\right)}^{2} = 1$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ If you like you could replace ${\left(\sqrt{12}\right)}^{2}$ with ${\left(2 \sqrt{3}\right)}^{2}$ and ${\left(\sqrt{8}\right)}^{2}$ with ${\left(2 \sqrt{2}\right)}^{2}$ ...but this really doesn't make the form any simpler.
14-4 Arcs of Circles 1 / 11 # 14-4 Arcs of Circles - PowerPoint PPT Presentation 14-4 Arcs of Circles. Learn vocabulary; Central Angle, Minor arc, Major arc, Semi circle. Learn definition of degree measure. A central angle of a circle is an angle whose vertex is the center of the circle. A. P. B. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. ## PowerPoint Slideshow about ' 14-4 Arcs of Circles' - aqua 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 ### 14-4 Arcs of Circles Learn vocabulary; Central Angle, Minor arc, Major arc, Semi circle. Learn definition of degree measure. Let C be a circle with center P, and let A and B be points which lie on C but are not the end points of the same diameter. • Then the minor arc AB is the union A, B, and all points of C that lie in the interior of <APB. A C P B Let C be a circle with center P, and let A and B be points which lie on C but are not the end points of the same diameter. • Then the Major arc AXB is the union A, B, and all points of C that lie in the exterior of <APB. A C P B x What is the intersection of major and minor arc AB and AXB? What is the Union of the major and minor arc AB and AXB • A semicircle AXB is the union of A, B, and the points of C that lie in a given half-plane with AB as edge. C x A B P (1) The degree measure of a minor arc is the measure of the corresponding central angle. A X r B mAXB = r. (3) the degree measure of a major arc is equal to 360 minus the measure of the corresponding minor arc. A r X B mAXB = 360-r Theorem 14-15 The Arc Addition Theorem • If B is a point of AC • then mABC = mAB + mBC. A B C
Question Video: Dividing by 3 Using Models \| Nagwa Question Video: Dividing by 3 Using Models \| Nagwa # Question Video: Dividing by 3 Using Models Mathematics Find 18 ÷ 3 using the cubes shown. 02:39 ### Video Transcript Find 18 divided by three using the cubes shown. In this question, we need to practice our skills at dividing by three. And we know that we need to use a model to help us because we’re given a picture of some cubes. And we’re told that we need to find the answer using the cubes shown. Now when you see the calculation 18 divided by three, what do you think of? We could think of it as 18 split into groups of three. And we try to find how many groups. Or we could think of it as 18 shared into three equal groups. And then we’d be trying to find out how many in each group. In the picture, we’re given 18 cubes. And you know, it doesn’t really matter which way we think of 18 divided by three. We can find the answer in both ways. And both ways involve drawing the lines on our model. Let’s start with the first idea. We could think of this as 18 split into groups of three. And we’re going to count how many groups we can make. Perhaps this is the way you’d think of the division. Well by looking at our cubes, can you see how to split them into groups of three? If we look carefully, we can see some groups of three already. We can see one, two, three, four, five, six groups of three. So if we think of our division like this, we can split 18 into groups of three and see that we can make six groups. But perhaps you thought of the model another way. Perhaps you thought of it as 18 shared into three equal groups. Let’s show the model again. And we’re going to draw our lines in a different way this time. Can you see how you draw the lines to share 18 into three equal groups? We could do it like this: one group, two groups, three groups. And they’re all the same size. We can see that they all contain one, two, three, four, five, six. So if we start with 18 cubes and we share them into three equal groups, there’ll be six cubes in each group. So it doesn’t really matter how we think of this division. Whether we want to find the number of groups or the number in each group, the answer is still the same. We’ve used a model in two different ways to find the answer to 18 divided by three. And that answer is six. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions
$$\require{cancel}$$ # 8.6: The general Solution to the Equation [ "article:topic", "authorname:ucd7" ] The general Solution to the Equation A more general way to write this equation that emphasizes its applicability to all cases of SHM is as follows: We will arbitrarily use the symbol “y”, but we just as well have used $$\theta$$, or x, or any other symbol. $\dfrac{d^2}{dt^2} y(t) = \left(\dfrac{- 2 \pi} { T} \right)^2 y(t)$ The general solution in standard form is, $y(t) = A \sin \left( \dfrac{2 \pi t}{T} + \phi \right)$ What is the meaning of the constants? The sine function goes through a complete cycle every $$2\pi$$ radians. This occurs each time increases by an amount T . Thus T is the period of the oscillations, as was mentioned before. The reciprocal of the period is the frequency of oscillations, $$\mathcal{f}$$: $\mathcal{f} = \dfrac{1}{T}$ T is the time required to complete the cycle, while $$\mathcal{f}$$ is the number of cycles per second. T is measured in seconds;$$\mathcal{f}$$, in reciprocal seconds (1/s), which are called hertz and abbreviated Hz. The maximum value of the sine function is +1 and the minimum value is -1. Thus, A is the amplitude of the oscillations. That is, the maximum value of y is +A and the minimum value is -A . The angle $$\phi$$ is determined by the value of y at the particular time t = 0. If y has its positive maximum value at t = 0, then $$\phi$$ has to be $$90^\circ$$ or $$\pi/2$$ radians. We say that $$\phi$$ depends on the “initial conditions”. The angle $$\phi$$ is often n referred to as the phase angle. By including the phase angle, we can make the sine function fit any particular physical situation. Without the phase angle, we would always have to start timing the oscillation when the position had the value zero. By including the phase angle, we have a perfectly general solution. The solutions we have written down describe the position as a function of time for any object vibrating in simple harmonic motion. They give the specific time dependence of the position of whatever it is that is vibrating. The three constants depend on the particular situation. Let's explore our solution to SHM further. The equillibrium value of y is the value y has when no oscillation is occurring. For the way we have written the solutions, this value is zero. Thus, the amplitude is the change in y that occurs in going from the equilibrium value to the maximum value of y. The picture(Figure 8.6.1) is for the angle $$\phi = 0$$. Changing the value of the phase angle $$\phi$$ shifts the curve sideways. Compare this plot to the plot of the sine solution with $$\phi = \pi/4$$ radians, in the following graph. You can see that they are just the same except for a sideways shift. Figure 8.6.1 To summarize, then, the solutions we have written down describe the position as a function of time for any object vibrating in simple harmonic motion. They give the specific time dependence of the position of whatever it is that is vibrating. The three constants, A, T and $$\phi$$ characterize the motion and depend on the particular situation. Figure 8.6.2 The period, T, or frequency, $$\mathcal{f}$$, of an oscillating system is determined by the constants appearing as the coefficient of the linear term of the differential equation when written in standard form: $\dfrac{d^2}{dt^2} y(t) = \left(\dfrac{- 2 \pi} { T} \right)^2 y(t)$ For a mass on a spring, we found that, $(\dfrac{2\pi}{T})^2 = \dfrac{k}{m} \rightarrow T = 2\pi \sqrt{\dfrac{m}{k}}$ For the pendulum, $(\dfrac{2\pi}{T})^2 = \dfrac{g}{l} \rightarrow T = 2\pi \sqrt{\dfrac{g}{l}}$ Note that you do not need to write out the solution of the differential equation to get the frequency or period. It comes directly from applying Newton’s 2nd law and “reading off” the constants. ### Energy for a Mass Hanging from a Spring As a mass hanging from a spring oscillates, very little mechanical energy is converted to thermal energy or sound, so we expect the mechanical energy to remain essentially constant for many periods. The potential energy is defined in terms of the work required to move the spring, and has the value, $PE = \dfrac{1}{2} ky^2$ assuming that the origin is placed at the equilibrium value of y and that the potential energy is defined as zero there. As usual kinetic energy is, $KE = \dfrac{1}{2} mv^2$ Let's substitute the general solution for $$y(t)$$ into these expressions: $PE = \dfrac{1}{2}ky^2 = \dfrac{1}{2}kA^2 \sin \left( \dfrac{2 \pi t}{T} + \phi \right)$ To calculate the kinetic energy, we differentiate the expression for y(t) to get v(t): $v(t) = \dfrac{dy}{dt} = \dfrac{2\pi}{T}A \cos \left( \dfrac{2 \pi t}{T} + \phi \right)$ so that $KE = \dfrac{1}{2}m \left(\dfrac{ 2\pi}{T} \right)^2A^2 \cos^2 \left( \dfrac{2 \pi t}{T} + \phi \right)$ But for a mass spring system, (2 $$\pi$$/T)2 = k/m. We can substitute this relation into the equation 8.6.11, finally getting the expression for the kinetic energy: $KE = \dfrac{1}{2}k A^2 \cos^2 \left( \dfrac{2 \pi t}{T} + \phi \right)$ We note that the maximum values of the PE and KE are the same. Also, the time average values of the KE and the PE are the same: 1/2 of the maximum values, since the average of $$sin^2$$ or $$cos^2$$ is just 1/2. In Chapter 3 we could merely say this was plausible in our discussion of equipartition of energy. Now we see why we were justified in saying the average energy in the KE mode is the same as the average in the PE mode. When we add the kinetic energy to the potential energy to get the total energy, we notice that we have the sum $$(sin^2 + cos^2)$$ which always has value unity. Hence the total energy is, $E_{tot}= 1/2A^2$ 1. the total energy is a constant just as we expected, and 2. the total energy is proportional to the square of the amplitude. This is a characteristic of all kinds of oscillating systems and extends even to wave motions, as will see in the next chapter in Part 3 of this text. ### Energy Graphs The Potential Energy Function, $PE = \frac{1}{2}y^2$ is a parabola when plotted as a function of y, while the total energy, being constant, is a horizontal line (Figure 8.4.3). Figure 8.6.3 The kinetic energy is the difference, $$KE=E_{tot} - PE$$. But the kinetic energy cannot be negative, since the mass is never negative and $$v^2$$ is never negative regardless of the sign of $$v$$ itself. This means that the oscillation is limited, and can go from $$y_{max}$$ to $$y_{min}$$. Of course this just reinforces what we already know, since $$y_{max} = A$$ and $$y_{min}=-A$$. But for any object that oscillates about an equilibrium position, even when the force law is not as simple as F=-ky, this graphical analysis provides an accurate description for small oscillations. As an example, look at the plot below(Figure 8.6.4): Figure 8.6.4 This is the shape of the potential energy between two bound atoms that we encountered in chapter 3. Even though the pair-wise potential energy function is not a parabola over all distances, near the minimum it is approximately so, and therefore the oscillations of this system will be simple harmonic if they are at sufficiently small amplitude about the minimum in the potential energy curve. Thus, we can make a very strong statement: essentially every system that vibrates, does so in SHM for small amplitudes of vibration. All molecules, including those in our bodies, and all atoms in solids and liquids move essentially in simple harmonic motion. ### Applying our Results to the Universe! Let’s now consider our model for matter. For liquids and solids, we picture molecules that are bound to each other as if they have little springs attaching them together. They bounce around, and in liquids tumble and change positions. What is the nature of the oscillations these molecules undergo? And what happens when we exert external forces on the matter and bend it or compress it. How then does the matter respond on a macroscopic scale? What we saw in the discussion above on SHM, is that if the restoring force is proportional to the displacement, but in the opposite direction of the displacement, then SHM results. portional to the displacement, but in the opposite direction of the displacement, then SHM results. This will always be the case if the displacement from equilibrium is sufficiently small. The result is that everything, on both the atomic scale and macroscopic scale, tends to vibrate in SHM. This is very nice, because we know the solution! The period of oscillation depends on factors like the mass of the particles or object being considered and the strength of the restoring forces. If we can identify the forces acting and write down Newton’s 2nd law, then for small oscillations, we have the problem solved! Think about what we have accomplished. For any kind of matter, we know how to go about finding its vibration frequencies when subjected to external forces as well as how its internal parts oscillate. We have a very general and powerful approach that works for almost all vibrating phenomena on any scale!
## Unfamiliar Mathematical Properties of 15 Familiar Numbers Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reason to believe that it is a mystery into which the human mind will never penetrate. Leonhard Euler Numbers play a significant role in our daily lives, from the time we wake up and check the clock to the time we go to bed and count sheep. However, despite their prevalence, we often take them for granted and overlook their fascinating mathematical properties. In this blog post, we will explore some of the intriguing properties of familiar numbers that may have gone unnoticed. • 2 is the only prime number without an e in its name. • The only number equal to its factorial since 2! = 2. • 2 is the only natural number satisfying the property 2 + 2 = 2 × 2. • For any convex polyhedron, the number of faces plus the number of vertices, minus the number of edges, is 2: VE + F = 2. • The sum of the reciprocals of the factors of any perfect number is equal to 2. • 2 is the base of the binary number system. • 2 related items are often called a pair, and words like dual, duel, couple, twin, and double emphasize the significance of the number two. • We can express π as an infinite product containing only 2 and its reciprocal ½. • 73 is the 21st prime number. Its mirror, 37, is the 12th prime number (which is the mirror of 21). The sum of the digits in both numbers is 10, and their product is 21. • If you add 100 to both 37 and 73, you get 137 and 173, respectively, which are both prime numbers. • The number of days in an ordinary year is 365, the product of the two primes 5 and 73. • The alphabetic value of the word NUMBER (= 2 + 5 + 13 + 14 + 18 + 21) is 73. • The number 73 is the only EMIRP that is one less than the double of its reversal: 73 = 2 × 37 ⎼ 1. • If you strip off the six triangular points from the 73-circle hexagram, you are left with a hexagon of 37 circles. • All three-digit repunits are divisible by 37. • If the multiples of 37 be mirrored and then separated by a zero, they will be another multiple of 37. For example, 518 and 80105. • Any multiple of 37 that has a three-digit repunit inserted will generate another multiple of 37. Examples of such numbers include 30007 and 78884. • From a three-digit number abc, form two more numbers cab and bca, in that order. Add all three numbers together. The resulting number is always divisible by 37. From 456, create 645 and 564. The sum of them is divisible by 37. • The chances of a wedding for the Sultan’s Dowry Problem are about 37%! • 18 is the only two-digit number that is twice the sum of its digits. • 18 is the only number where the sum of its digits 1 + 8 = 9 is equal to half of itself (18/2 = 9). • The number 18 = 31 − 13 = 97 − 79 is the smallest positive difference between an EMIRP and its reverse. • The cube of 18 and the fourth power of 18 together use each digit exactly once: 183 = 5832 and 184 = 104976. • The 18-point problem is to place a sequence of 18 points on an interval so that the first two points are in separate halves of the interval, the first three points are in separate thirds of the interval, the first four points are in separate fourths of the interval, and so on. • The sum of the digits of 81 (= 8 + 1 = 9) when multiplied by its reversed self, yields the original number (9 × 9 = 81). • 81 is the smallest square such that the sum of its divisors, 1 + 3 + 9 + 27 + 81 = 121 = 112, is also a square. • It is the only two-digit number that is the square of the sum of its digits: 81 = (8 + 1)2. • In the decimal expansion of 1/81 = 0.012345679012345679 . . . , each digit appears (in order) except 8. • There are 81 rectangles on a Shogi board (Japanese chess). • 69 days = 6 days 9 weeks • The sum of the divisors of 69 is equal to its reversal: 96 = 1 + 3 + 23 + 69. • The numbers 692 = 4761 and 693 = 328509 together contain each of the digits exactly once. • The alphabetical value of the letters in 69’s Roman numeral LXIX is 12 + 24 + 9 + 24 = 69. • 69 is equal to 105 octal, while 105 is equal to 69 hexadecimal. • The numeral 69 looks the same when rotated by 180º. • 96 is the second smallest number with six prime factors: 96 = 2 × 2 × 2 × 2 × 2 × 3. • A number with one-eighth the number of divisors of the original number. • There are 96 spokes on a daisy wheel printer. • The Courier game is an old variant of chess played on a board with 96 squares. • 96 is an untouchable number. • 96 appears the same when turned upside down. • Skilling’s figure, a degenerate uniform polyhedron, has a Euler characteristic χ = ‒ 96. • 96 is 88 in base 11. • 23 is the smallest prime number with consecutive digits. • The smallest prime whose reversal is a power: 32 = 25. • Using congruence, we can write 23 ≡ 2 (mod 3) and 23 ≡ 3 (mod 2). • 23 differs from its reversal 32 by 32. • There are 23 definitions in Book I of Euclid’s Elements. • W is the 23rd letter of the modern English alphabet. Have you ever noticed that it has 2 points down and 3 points up? • In a room of just 23 people, a greater than 50% chance exists that two of the people will share a common birthday. • Representation using its digits: 23 = 1! + (2! + 2!) + (3! + 3! + 3!). • The sum of squares of its digits 22 + 32 = 4 + 9 = 13 is also a prime number. • If we add the square of the digits of the number 23 successively, we eventually reach 1: 22 + 32 = 13; 12 + 32 = 10; 12 + 02 = 1. • The number 32 is (possibly) the highest power of 2 with all prime digits. • As a sum of products: 32 = 1 × 2 + 1 × 2 × 3 + 1 × 2 × 3 × 4. • As a sum of powers: 32 = 11 + 22 + 33 • There are 32 fluid ounces to the quart, 32 gills to the gallon, 32 dry quarts to the bushel, and 32 ells to the bolt. • The freezing point of pure water at sea level is 32˚F. • The acceleration of gravity on the surface of the Earth is about 32 feet per second per second. • The number 63 is the smallest number n whose Roman numeral has alphabetic value n. Indeed, the value of LXIII is 12 + 24 + 9 + 9 + 9 = 63. • It is the sum of the powers of 2 from 0 to 5: 63 = 20 + 21 + 22 + 23 + 24 + 25 • There are 63 posets with 5 unlabelled elements. • A barrel that holds 63 gallons is called a hogshead. • 36 is the number of degrees in the interior angle of each tip of a regular pentagram. • The 36 officers’ problem (proposed by Euler) is a mathematical puzzle with no solution. • It is the number of possible outcomes in the roll of two distinct dice. • 36 (numerals 0–9 and the letters A‒Z) is the number of alpha-numeric keys. • The truncated cube and the truncated octahedron are Archimedean solids with 36 edges. • 36 is the number of inches in a yard (3 feet). • It is 3 dozen, or a quarter of a gross. • 13 is the smallest EMIRP. • A baker’s dozen is a group of 13. • A reflectional type property: 132 = 169 and its reversal 312 = 961. • There are 13 Archimedean solids. • Three planes can cut a donut into a maximum of 13 parts. • The sum of primes up to 13 is equal to the 13th prime. • Consider the primes up to 13. While dividing 13 by each of these primes individually, the sum of the remainders is 13. • The fear of the number 13 is called triskaidekaphobia. • 13 is the number of cards in one suit out of a deck of fifty two cards. • The dice game Yahtzee consists of 13 rounds. • 13 is ELEVEN + TWO = TWELVE + ONE. Can you solve this anagrammatically? • 31 is the number of days in seven months. • It is the sum of the first two primes raised to themselves: 22 + 33 = 31. • The reverse of 31 is 13 which is also a prime. So 31 is an EMIRP. • 31 is the smallest prime that is the sum of the divisors of two different numbers, 16 and 25. • 31 is the number of possible Steiner topologies for Steiner trees with 4 terminals. • The cube root of 31 is the value of pi correct to four significant figures (∛31 = 3.1413…). • 17 is the (1 × 7)th prime. • It can be written as the powers of first two primes: 17 = 23 + 32. • 17 is an EMIRP. • It is the smallest prime that is the sum of consecutive primes: 17 = 2 + 3 + 5 + 7. • 17 is the smallest number of givens (prefilled squares) that a Sudoku puzzle can have. • It is equal to the sum of digits of its cube: 173 = 4913 and 4 + 9 + 1 + 3 = 17. • 17 is the smallest prime sandwiched between two non-square free numbers (42 < 17 < 2 × 32).
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 4.2: The Mean Value Theorem The Mean Value Theorem is one of the most important theorems in calculus. We look at some of its implications at the end of this section. First, let’s start with a special case of the Mean Value Theorem, called Rolle’s theorem. ## Rolle’s Theorem Informally, Rolle’s theorem states that if the outputs of a differentiable function $$f$$ are equal at the endpoints of an interval, then there must be an interior point $$c$$ where $$f'(c)=0$$. Figure illustrates this theorem. Figure 1: If a differentiable function f satisfies f(a)=f(b), then its derivative must be zero at some point(s) between a and b. Rolle’s Theorem Let $$f$$ be a continuous function over the closed interval $$[a,b]$$ and differentiable over the open interval $$(a,b)$$ such that $$f(a)=f(b)$$. There then exists at least one $$c∈(a,b)$$ such that $$f'(c)=0.$$ Proof Let $$k=f(a)=f(b).$$ We consider three cases: 1. $$f(x)=k$$ for all $$x∈(a,b).$$ 2. There exists $$x∈(a,b)$$ such that $$f(x)>k.$$ 3. There exists $$x∈(a,b)$$ such that $$f(x)<k.$$ Case 1: If $$f(x)=0$$ for all $$x∈(a,b)$$, then $$f'(x)=0$$ for all $$x∈(a,b).$$ Case 2: Since $$f$$ is a continuous function over the closed, bounded interval $$[a,b]$$, by the extreme value theorem, it has an absolute maximum. Also, since there is a point $$x∈(a,b)$$ such that $$f(x)>k$$, the absolute maximum is greater than $$k$$. Therefore, the absolute maximum does not occur at either endpoint. As a result, the absolute maximum must occur at an interior point $$c∈(a,b)$$. Because $$f$$ has a maximum at an interior point $$c$$, and $$f$$ is differentiable at (c\), by Fermat’s theorem, $$f'(c)=0.$$ Case 3: The case when there exists a point $$x∈(a,b)$$ such that $$f(x)<k$$ is analogous to case 2, with maximum replaced by minimum. An important point about Rolle’s theorem is that the differentiability of the function $$f$$ is critical. If f is not differentiable, even at a single point, the result may not hold. For example, the function f(x)=\|x\|−1 is continuous over $$[−1,1]$$ and $$f(−1)=0=f(1)$$, but $$f'(c)≠0$$ for any $$c∈(−1,1)$$ as shown in the following figure. Figure 2: Since $$f(x)=\|x\|−1$$ is not differentiable at $$x=0$$, the conditions of Rolle’s theorem are not satisfied. In fact, the conclusion does not hold here; there is no $$c∈(−1,1)$$ such that $$f'(c)=0.$$ Let’s now consider functions that satisfy the conditions of Rolle’s theorem and calculate explicitly the points c where $$f'(c)=0.$$ Example $$\PageIndex{1}$$: Using Rolle’s Theorem For each of the following functions, verify that the function satisfies the criteria stated in Rolle’s theorem and find all values $$c$$ in the given interval where $$f'(c)=0.$$ 1. $$f(x)=x^2+2x$$ over $$[−2,0]$$ 2. $$f(x)=x^3−4x$$ over $$[−2,2]$$ Solution Since $$f$$ is a polynomial, it is continuous and differentiable everywhere. In addition, $$f(−2)=0=f(0).$$ Therefore, $$f$$ satisfies the criteria of Rolle’s theorem. We conclude that there exists at least one value $$c∈(−2,0)$$ such that $$f'(c)=0$$. Since $$f'(x)=2x+2=2(x+1),$$ we see that $$f'(c)=2(c+1)=0$$ implies $$c=−1$$ as shown in the following graph. Figure 3: This function is continuous and differentiable over [−2,0], f'(c)=0 when c=−1. b. As in part a. $$f$$ is a polynomial and therefore is continuous and differentiable everywhere. Also, $$f(−2)=0=f(2).$$ That said, $$f$$ satisfies the criteria of Rolle’s theorem. Differentiating, we find that $$f'(x)=3x^2−4.$$ Therefore, $$f'(c)=0$$ when $$x=±\frac{2}{\sqrt{3}}$$. Both points are in the interval $$[−2,2]$$, and, therefore, both points satisfy the conclusion of Rolle’s theorem as shown in the following graph. Figure 4: For this polynomial over $$[−2,2], f'(c)=0$$ at $$x=±2/\sqrt{3}$$. Exercise $$\PageIndex{1}$$ Verify that the function $$f(x)=2x^2−8x+6$$ defined over the interval $$[1,3]$$ satisfies the conditions of Rolle’s theorem. Find all points $$c$$ guaranteed by Rolle’s theorem. Hint: Find all values $$c$$, where $$f'(c)=0$$. Solution: $$c=2$$ ## The Mean Value Theorem and Its Meaning Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions $$f$$ that are zero at the endpoints. The Mean Value Theorem generalizes Rolle’s theorem by considering functions that are not necessarily zero at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’s theorem (Figure). The Mean Value Theorem states that if $$f$$ is continuous over the closed interval $$[a,b]$$ and differentiable over the open interval $$(a,b)$$, then there exists a point $$c∈(a,b)$$ such that the tangent line to the graph of $$f$$ at $$c$$ is parallel to the secant line connecting $$(a,f(a))$$ and $$(b,f(b)).$$ Figure 5: The Mean Value Theorem says that for a function that meets its conditions, at some point the tangent line has the same slope as the secant line between the ends. For this function, there are two values $$c_1$$ and $$c_2$$ such that the tangent line to $$f$$ at $$c_1$$ and $$c_2$$ has the same slope as the secant line. Mean Value Theorem Let $$f$$ be continuous over the closed interval $$[a,b]$$ and differentiable over the open interval $$(a,b)$$. Then, there exists at least one point $$c∈(a,b)$$ such that $f'(c)=\frac{f(b)−f(a)}{b−a}$ Proof The proof follows from Rolle’s theorem by introducing an appropriate function that satisfies the criteria of Rolle’s theorem. Consider the line connecting $$(a,f(a))$$ and $$(b,f(b)).$$ Since the slope of that line is $\frac{f(b)−f(a)}{b−a}$ and the line passes through the point $$(a,f(a)),$$ the equation of that line can be written as $y=\frac{f(b)−f(a)}{b−a}(x−a)+f(a).$ Let $$g(x)$$ denote the vertical difference between the point $$(x,f(x))$$ and the point $$(x,y)$$ on that line. Therefore, $g(x)=f(x)−[\frac{f(b)−f(a)}{b−a}(x−a)+f(a)].$ Figure 6: The value g(x) is the vertical difference between the point (x,f(x)) and the point (x,y) on the secant line connecting (a,f(a)) and (b,f(b)). Since the graph of $$f$$ intersects the secant line when $$x=a$$ and $$x=b$$, we see that $$g(a)=0=g(b)$$. Since $$f$$ is a differentiable function over $$(a,b)$$, $$g$$ is also a differentiable function over $$(a,b)$$. Furthermore, since $$f$$ is continuous over $$[a,b], g$$ is also continuous over $$[a,b]$$. Therefore, $$g$$ satisfies the criteria of Rolle’s theorem. Consequently, there exists a point $$c∈(a,b)$$ such that $$g'(c)=0.$$ Since $g'(x)=f'(x)−\frac{f(b)−f(a)}{b−a},$ we see that $g'(c)=f'(c)−\frac{f(b)−f(a)}{b−a}.$ Since $$g'(c)=0,$$ we conclude that $f'(c)=\frac{f(b)−f(a)}{b−a}.$ In the next example, we show how the Mean Value Theorem can be applied to the function $$f(x)=\sqrt{x}$$ over the interval $$[0,9]$$. The method is the same for other functions, although sometimes with more interesting consequences. Example $$\PageIndex{2}$$: Verifying that the Mean Value Theorem Applies For $$f(x)=\sqrt{x}$$ over the interval $$[0,9]$$, show that $$f$$ satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value $$c∈(0,9)$$ such that $$f′(c)$$ is equal to the slope of the line connecting $$(0,f(0))$$ and $$(9,f(9))$$. Find these values $$c$$ guaranteed by the Mean Value Theorem. Solution We know that $$f(x)=\sqrt{x}$$ is continuous over $$[0,9]$$ and differentiable over $$(0,9).$$ Therefore, $$f$$ satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value $$c∈(0,9)$$ such that $$f′(c)$$ is equal to the slope of the line connecting $$(0,f(0))$$ and $$(9,f(9))$$ (Figure). To determine which value(s) of $$c$$ are guaranteed, first calculate the derivative of $$f$$. The derivative $$f′(x)=\frac{1}{(2\sqrt{x})}$$. The slope of the line connecting $$(0,f(0))$$ and $$(9,f(9))$$ is given by $\frac{f(9)−f(0)}{9−0}=\frac{\sqrt{9}−\sqrt{0}}{9−0}=\frac{3}{9}=\frac{1}{3}.$ We want to find $$c$$ such that $$f′(c)=\frac{1}{3}$$. That is, we want to find $$c$$ such that $\frac{1}{2\sqrt{c}}=\frac{1}{3}.$ Solving this equation for $$c$$, we obtain $$c=\frac{9}{4}$$. At this point, the slope of the tangent line equals the slope of the line joining the endpoints. Figure 7: The slope of the tangent line at c=9/4 is the same as the slope of the line segment connecting (0,0) and (9,3). One application that helps illustrate the Mean Value Theorem involves velocity. For example, suppose we drive a car for 1 h down a straight road with an average velocity of 45 mph. Let $$s(t)$$ and $$v(t)$$ denote the position and velocity of the car, respectively, for $$0≤t≤1$$ h. Assuming that the position function $$s(t)$$ is differentiable, we can apply the Mean Value Theorem to conclude that, at some time $$c∈(0,1)$$, the speed of the car was exactly $v(c)=s′(c)=\frac{s(1)−s(0)}{1−0}=45mph.$ Example $$\PageIndex{1}$$: Mean Value Theorem and Velocity If a rock is dropped from a height of 100 ft, its position $$t$$ seconds after it is dropped until it hits the ground is given by the function $$s(t)=−16t^2+100.$$ 1. Determine how long it takes before the rock hits the ground. 2. Find the average velocity $$v_{avg}$$ of the rock for when the rock is released and the rock hits the ground. 3. Find the time $$t$$ guaranteed by the Mean Value Theorem when the instantaneous velocity of the rock is $$v_{avg}.$$ Solution a. When the rock hits the ground, its position is $$s(t)=0.$$ Solving the equation $$−16t^2+100=0$$ for $$t$$, we find that $$t=±\frac{5}{2}sec$$. Since we are only considering $$t≥0$$, the ball will hit the ground $$\frac{5}{2}$$ sec after it is dropped. b. The average velocity is given by $$v_{avg}=\frac{s(5/2)−s(0)}{5/2−0}=\frac{1−100}{5/2}=−40ft/sec.$$ c. The instantaneous velocity is given by the derivative of the position function. Therefore, we need to find a time $$t$$ such that $$v(t)=s′(t)=vavg=−40ft/sec.$$ Since s(t) is continuous over the interval $$[0,5/2]$$ and differentiable over the interval $$(0,5/2),$$ by the Mean Value Theorem, there is guaranteed to be a point $$c∈(0,5/2)$$ such that $$s′(c)=\frac{s(5/2)−s(0)}{5/2−0}=−40.$$ Taking the derivative of the position function $$s(t)$$, we find that $$s′(t)=−32t.$$ Therefore, the equation reduces to $$s′(c)=−32c=−40.$$ Solving this equation for $$c$$, we have $$c=\frac{5}{4}$$. Therefore, $$\frac{5}{4}$$ sec after the rock is dropped, the instantaneous velocity equals the average velocity of the rock during its free fall: $$−40$$ ft/sec. Figure 8: At time $$t=5/4$$ sec, the velocity of the rock is equal to its average velocity from the time it is dropped until it hits the ground. Exercise $$\PageIndex{2}$$: Suppose a ball is dropped from a height of 200 ft. Its position at time $$t$$ is $$s(t)=−16t^2+200.$$ Find the time $$t$$ when the instantaneous velocity of the ball equals its average velocity. Hint: First, determine how long it takes for the ball to hit the ground. Then, find the average velocity of the ball from the time it is dropped until it hits the ground. Solution $$\frac{5}{2\sqrt{2}}$$ sec ## Corollaries of the Mean Value Theorem Let’s now look at three corollaries of the Mean Value Theorem. These results have important consequences, which we use in upcoming sections. At this point, we know the derivative of any constant function is zero. The Mean Value Theorem allows us to conclude that the converse is also true. In particular, if $$f′(x)=0$$ for all $$x$$ in some interval $$I$$, then $$f(x)$$ is constant over that interval. This result may seem intuitively obvious, but it has important implications that are not obvious, and we discuss them shortly. Corollary 1: Functions with a Derivative of Zero Let $$f$$ be differentiable over an interval $$I$$. If $$f′(x)=0$$ for all $$x∈I$$, then $$f(x)=$$ constant for all $$x∈I.$$ Proof Since $$f$$ is differentiable over $$I$$, $$f$$ must be continuous over $$I$$. Suppose $$f(x)$$ is not constant for all $$x$$ in $$I$$. Then there exist $$a,b∈I$$, where $$a≠b$$ and $$f(a)≠f(b).$$ Choose the notation so that $$a<b.$$ Therefore, $$\frac{f(b)−f(a)}{b−a}≠0.$$ Since $$f$$ s a differentiable function, by the Mean Value Theorem, there exists $$c∈(a,b)$$ such that $$f′(c)=\frac{f(b)−f(a)}{b−a}$$. Therefore, there exists $$c∈I$$ such that $$f′(c)≠0$$, which contradicts the assumption that $$f′(x)=0$$ for all $$x∈I$$. From Note, it follows that if two functions have the same derivative, they differ by, at most, a constant. Corollary 2: Constant Difference Theorem If $$f$$ and $$g$$ are differentiable over an interval $$I$$ and $$f′(x)=g′(x)$$ for all $$x∈I$$, then $$f(x)=g(x)+C$$ for some constant $$C$$. Proof Let $$h(x)=f(x)−g(x).$$ Then, $$h′(x)=f′(x)−g′(x)=0$$ for all $$x∈I.$$ By Corollary 1, there is a constant $$C$$ such that $$h(x)=C$$ for all $$x∈I$$. Therefore, $$f(x)=g(x)+C$$ for all $$x∈I.$$ The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recall that a function $$f$$ is increasing over $$I$$ if $$f(x_1)<f(x_2)$$ whenever $$x_1<x_2$$, whereas $$f$$ is decreasing over $$I$$ if $$f(x)_1>f(x_2)$$ whenever $$x_1<x_2$$. Using the Mean Value Theorem, we can show that if the derivative of a function is positive, then the function is increasing; if the derivative is negative, then the function is decreasing (Figure). We make use of this fact in the next section, where we show how to use the derivative of a function to locate local maximum and minimum values of the function, and how to determine the shape of the graph. This fact is important because it means that for a given function $$f$$, if there exists a function $$F$$ such that $$F′(x)=f(x)$$; then, the only other functions that have a derivative equal to $$f$$ are $$F(x)+C$$ for some constant $$C$$. We discuss this result in more detail later in the chapter. Figure 7: If a function has a positive derivative over some interval $$I$$, then the function increases over that interval $$I$$; if the derivative is negative over some interval $$I$$, then the function decreases over that interval $$I$$. Corollary 3: Increasing and Decreasing Functions Let $$f$$ be continuous over the closed interval $$[a,b]$$ and differentiable over the open interval $$(a,b)$$. 1. If $$f′(x)>0$$ for all $$x∈(a,b)$$, then $$f$$ is an increasing function over $$[a,b].$$ 2. If $$f′(x)<0$$ for all $$x∈(a,b)$$, then $$f$$ is a decreasing function over $$[a,b].$$ Proof We will prove i.; the proof of ii. is similar. Suppose $$f$$ is not an increasing function on $$I$$. Then there exist $$a$$ and $$b$$ in $$I$$ such that $$a<b$$, but $$f(a)≥f(b)$$. Since $$f$$ is a differentiable function over $$I$$, by the Mean Value Theorem there exists $$c∈(a,b)$$ such that $f′(c)=\frac{f(b)−f(a)}{b−a}.$ Since $$f(a)≥f(b)$$, we know that $$f(b)−f(a)≤0$$. Also, $$a<b$$ tells us that $$b−a>0.$$ We conclude that $f′(c)=\frac{f(b)−f(a)}{b−a}≤0.$ However, $$f′(x)>0$$ for all $$x∈I$$. This is a contradiction, and therefore $$f$$ must be an increasing function over $$I$$. ## Key Concepts • If $$f$$ is continuous over $$[a,b]$$ and differentiable over $$(a,b)$$ and $$f(a)=0=f(b)$$, then there exists a point $$c∈(a,b)$$ such that $$f′(c)=0.$$ This is Rolle’s theorem. • If $$f$$ is continuous over $$[a,b]$$ and differentiable over $$(a,b)$$, then there exists a point $$c∈(a,b)$$ such that $$f'(c)=\frac{f(b)−f(a)}{b−a}.$$ This is the Mean Value Theorem. • If $$f'(x)=0$$ over an interval $$I$$, then $$f$$ is constant over $$I$$. • If two differentiable functions $$f$$ and $$g$$ satisfy $$f′(x)=g′(x)$$ over $$I$$, then $$f(x)=g(x)+C$$ for some constant $$C$$. • If $$f′(x)>0$$ over an interval $$I$$, then $$f$$ is increasing over $$I$$. If $$f′(x)<0$$ over $$I$$, then $$f$$ is decreasing over $$I$$. ## Glossary mean value theorem if $$f$$ is continuous over $$[a,b]$$ and differentiable over $$(a,b)$$, then there exists $$c∈(a,b)$$ such that $$f′(c)=\frac{f(b)−f(a)}{b−a}$$ rolle’s theorem if $$f$$ is continuous over $$[a,b]$$ and differentiable over $$(a,b)$$, and if $$f(a)=f(b)$$, then there exists $$c∈(a,b)$$ such that $$f′(c)=0$$
# 2 solve x 2 1 6 2 5 2 x solution we begin by finding • Notes • 13 This preview shows page 3 - 6 out of 13 pages. 2. Solve x 0 = 2 1 6 0 2 5 0 0 2 x . Solution. We begin by finding the eigenvalues. Because our matrix is an upper triangular matrix, the eigenvalues are the diagonal entries, hence λ = 2 mult. 3. We now find the eigenvectors for λ = 2: 2 - 2 1 6 0 0 2 - 2 5 0 0 0 2 - 2 0 0 1 6 0 0 0 5 0 0 0 0 0 The second row tells us that x 3 = 0, and the first row tells us that x 2 = - 6 x 3 . But this implies that x 2 = 0. x 1 is our free variable, so we can choose it to be 1, and we get the eigenvector v 1 = 1 0 0 We only found one eigenvector, which is less than the multiplicity of our eigenvalue. This means two things: we must find more eigenvectors, and we must have a t 2 e 2 t and te 2 t in our solution (i.e. we must “bump up” our solution). We find the next eigenvector by solving the equation ( A - ) v 2 = v 1 : 0 1 6 1 0 0 5 0 0 0 0 0 The second row tells us x 3 = 0, and the first row tells us that x 2 = - 6 x 3 +1, which means that x 2 = 1. x 1 is free, so if we choose it to be 0 then the second eigenvector is v 2 = 0 1 0 We need to find another eigenvector, so we solve the equation ( A - ) v 3 = v 2 : 0 1 6 0 0 0 5 1 0 0 0 0 The second row tells us that x 3 = 1 5 , the first row tells us that x 2 = - 6 x 3 = - 6 5 . x 1 is free, so if we choose it to be 0 then the third eigenvector is v 3 = 0 - 6 5 1 5 . Therefore the solution is x = c 1 1 0 0 e 2 t + c 2 1 0 0 te 2 t + 0 1 0 e 2 t + c 3 1 0 0 t 2 2 e 2 t + 0 1 0 te 2 t + 0 - 6 5 1 5 e 2 t . 3 Subscribe to view the full document. 3. Solve x 0 = 4 0 1 0 6 0 - 4 0 4 x . Solution. We begin by finding the eigenvalues: 4 - λ 0 1 0 6 - λ 0 - 4 0 4 - λ = (4 - λ ) 6 - λ 0 0 4 - λ - 0 + 1 0 6 - λ - 4 0 = (4 - λ )(6 - λ )(4 - λ ) + 4(6 - λ ) = (6 - λ )[(4 - λ )(4 - λ ) + 4] = (6 - λ )( λ 2 - 8 λ + 20) We use the quadratic formula to find the eigenvalues of λ 2 - 8 λ + 20: λ = 8 ± p 64 - 4(1)(20) 2 = 8 ± - 16 2 = 8 ± 4 i 2 = 4 ± 2 i Therefore the eigenvalues are λ 1 = 6 , λ 2 = 4 + 2 i, λ 3 = 4 - 2 i . When λ 1 = 6: 4 - 6 0 1 0 0 6 - 6 0 0 - 4 0 4 - 6 0 - 2 0 1 0 0 0 0 0 - 4 0 - 2 0 - 2 0 1 0 0 0 0 0 0 0 - 4 0 The last row tells us that x 3 = 0, the first row tells us that 2 x 1 = x 3 which implies that x 1 = 0. x 2 is free, so if we choose it to be 1 then we have the eigenvector v 1 = 0 1 0 When λ 2 = 4 + 2 i : 4 - (4 + 2 i ) 0 1 0 0 6 - (4 + 2 i ) 0 0 - 4 0 4 - (4 + 2 i ) 0 - 2 i 0 1 0 0 2 - 2 i 0 0 - 4 0 - 2 i 0 The second row tells us that (2 - 2 i ) x 2 = 0, or that x 2 = 0. The first row tells us that 2 ix 1 = x 3 . If we choose x 1 = 1, then we have the eigenvector v 2 = 1 0 2 i . This implies that the third eigenvector is v 3 = 1 0 - 2 i . Therefore the solution to the system is x = c 1 0 1 0 e 6 t + c 2 1 0 2 i e (4+2 i ) t + c 3 1 0 - 2 i e (4 - 2 i ) t , 4 however we want to write our solution as a real solution. Therefore we must expand v 2 e λ 2 t (you can also do the same for v 3 e λ 3 t , but it is enough to just do one expansion): 1 0 2 i e 4 t e i 2 t = 1 0 2 i e 4 t (cos 2 t + i sin 2 t ) = e 4 t cos 2 t + i sin 2 t 0 2 i cos 2 t - 2 sin 2 t = e 4 t cos 2 t 0 - 2 sin 2 t Subscribe to view the full document. {[ snackBarMessage ]} ###### "Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found." — Christopher R., University of Rhode Island '15, Course Hero Intern ### What students are saying • As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students. Kiran Temple University Fox School of Business ‘17, Course Hero Intern • I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero. Dana University of Pennsylvania ‘17, Course Hero Intern • The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time. Jill Tulane University ‘16, Course Hero Intern
How Do You Go About Dividing a Polynomial by a Monomial? # How Do You Go About Dividing a Polynomial by a Monomial? To divide a polynomial by a monomial, first identify which one is which. Then, separate the division into parts. Finally, once each part is processed, put them back together. 1. Figure out which is the polynomial and which is the monomial An algebraic expression with just one term is called a monomial. That could be a variable, number, or a number attached with a variable. For example x and y are both variables, 11 and 12 are numbers, and 3x and -8xy are numbers attached to variables. A polynomial is an expression with more than one term, all joined by either a plus or minus sign. For example, 6x+6y and 7x+3xy+z+5 are both polynomials 2. Split the operation into parts. Process each term of the polynomial by the single term of the monomial. For example, if dividing 3xy+9x+12 by 3, the first operation would be 3xy divided by 3. The answer is xy. Then, 9x should be divided by 3, getting 3x. Finally, 12 would be divided by 3, getting 4. 3. Put the parts back together Once the parts have been divided, it's time to put them back again. Simply connect the divisions from the previous step by the original operation that connected them, either a plus or minus sign. To continue the example from the previous step, the answer would be xy+3x+4. Similar Articles
# Maths The Year 2 Learner Working mathematically By the end of Year 2, children will solve problems with one or a small number of simple steps. Children will discuss their understanding and begin to explain their thinking using appropriate mathematical vocabulary, hands-on resources and different ways of recording. They will ask simple questions relevant to the problem and begin to suggest ways of solving them. Number Counting and understanding numbers Children will develop their understanding of place value of numbers to at least 100 and apply this when ordering, comparing, estimating and rounding. Children begin to understand zero as a place holder as this is the foundation for manipulating larger numbers in subsequent years. Children will count fluently forwards and backwards up to and beyond 100 in multiples of 2, 3, 5 and 10 from any number. They will use hands-on resources to help them understand and apply their knowledge of place value in two digit numbers, representing the numbers in a variety of different ways. Calculating Children learn that addition and multiplication number sentences can be re-ordered and the answer remains the same (commutativity) such as 9+5+1= 5+1+9. They learn that this is not the case with subtraction and division. They solve a variety of problems using mental and written calculations for +, -, x, ÷ in practical contexts. These methods will include partitioning which is where the number is broken up into more manageable parts (e.g. 64 = 60 + 4 or 50 + 14), re-ordering (e.g. moving the larger number to the beginning of the number sentence when adding several small numbers) and using a number line. Children will know the 2, 5 and 10 times tables, as well as the matching division facts (4 x 5 = 20, 20 ÷ 5 = 4) and can recall them quickly and accurately. They apply their knowledge of addition and subtraction facts to 20 and can use these to work out facts up to 100. Fractions including decimals Throughout year 2, children will develop their understanding of fractions and the link to division. They explore this concept using pictures, images and hands-on resources. They will solve problems involving fractions (e.g. find 1/3 of the hexagon or ¼ of the marbles) and record what they have done. They will count regularly and fluently in fractions such as ½ and ¼ forwards and backwards and, through positioning them on a number line, understand that some have the same value (equivalent) e.g. ½ = ¼. Measurement Children will estimate, choose, use and compare a variety of measurements for length, mass, temperature, capacity, time and money. By the end of year 2, they will use measuring apparatus such as rulers accurately. They will use their knowledge of measurement to solve problems (e.g. how many ways to make 50p). They extend their understanding of time to tell and write it on an analogue clock to 5 minute intervals, including quarter past / to the hour. They will know key time related facts (minutes in an hour, hours in a day) and relate this to their everyday life. Geometry Children will identify, describe, compare and sort common 2-D and 3-D shapes according to their properties (sides, vertices, edges, faces) and apply this knowledge to solve simple problems. They develop their understanding by finding examples of 3-D shapes in the real world and exploring the 2-D shapes that can be found on them (e.g. a circle is one of the faces on a cylinder). Children begin to describe position, direction and movement in a range of different situations, including understanding rotation (turning through right angles clockwise and anti-clockwise). They use their knowledge of shape in patterns and sequences. Statistics Children sort and compare information, communicating findings by asking and answering questions. They will draw simple pictograms, tally charts and tables.
## q3: Shikha is 30 years older than her son. I year ago she was four times as old as her son. Find their present a Question q3: Shikha is 30 years older than her son. I year ago she was four times as old as her son. Find their present ages. in progress 0 1 month 2021-08-14T05:48:28+00:00 2 Answers 0 views 0 1. Let , the age of son be ” x “ First Condition : Shikha is 30 years older than her son Thus , Shikha’s age = ” x + 30 “ Second Condition : 1 year ago she was four times as old as her son Thus , (x + 30 – 1) = 4(x – 1) x + 29 = 4x – 4 33 = 3x x = 33/3 x = 11 The present ages of shikha and her son are 11 and 41 Step by step explanation: Given: • Shikha is 30 years older than her son. • 1 year ago she was four times as old as her son. To Find: • Their present ages. Let us take the present age of son be x . Then the present age of Shikha will be (x+30) . Now , • Age of Shikha 1 year ago = (x+30)-1=x+29. • Age if son one year ago = (x-1). Atq, Hence • Present age of Shikha = x+30 = 11+30=41 years. • Present age of son = x = 11 years.
# 3.3: Conditional Probability and Independent Events skills to develop • To learn the concept of a conditional probability and how to compute it. • To learn the concept of independence of events, and how to apply it. Suppose a fair die has been rolled and you are asked to give the probability that it was a five. There are six equally likely outcomes, so your answer is $$1/6$$. But suppose that before you give your answer you are given the extra information that the number rolled was odd. Since there are only three odd numbers that are possible, one of which is five, you would certai: nly revise your estimate of the likelihood that a five was rolled from $$1/6$$ to $$1/3$$. In general, the revised probability that an event A has occurred, taking into account the additional information that another event $$B$$ has definitely occurred on this trial of the experiment, is called the conditional probability of $$A$$ given $$B$$ and is denoted by $$P(A\mid B)$$. The reasoning employed in this example can be generalized to yield the computational formula in the following definition. Definition: conditional probability The conditional probability of $$A$$ given $$B$$, denoted $$P(A\mid B)$$, is the probability that event $$A$$ has occurred in a trial of a random experiment for which it is known that event $$B$$ has definitely occurred. It may be computed by means of the following formula: $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)} \label{CondProb}$ Example $$\PageIndex{1}$$: Rolling a Die A fair (unbiased) die is rolled. 1. Find the probability that the number rolled is a five, given that it is odd. 2. Find the probability that the number rolled is odd, given that it is a five. Solution: The sample space for this experiment is the set $$S={1,2,3,4,5,6}$$ consisting of six equally likely outcomes. Let $$F$$ denote the event “a five is rolled” and let $$O$$ denote the event “an odd number is rolled,” so that $F={5}\; \; \text{and}\; \; O={1,3,5}$ 1. This is the introductory example, so we already know that the answer is $$1/3$$. To use Equation \ref{CondProb} to confirm this we must replace $$A$$ in the formula (the event whose likelihood we seek to estimate) by $$F$$ and replace $$B$$ (the event we know for certain has occurred) by $$O$$: $P(F\mid O)=\dfrac{P(F\cap O)}{P(O)}$Since $F\cap O={5}\cap {1,3,5}={5},\; P(F\cap O)=1/6$Since $O={1,3,5}, \; P(O)=3/6.$Thus $P(F\mid O)=\dfrac{P(F\cap O)}{P(O)}=\dfrac{1/6}{3/6}=\dfrac{1}{3}$ 2. This is the same problem, but with the roles of $$F$$ and $$O$$ reversed. Since we are given that the number that was rolled is five, which is odd, the probability in question must be $$1$$. To apply Equation \ref{CondProb} to this case we must now replace $$A$$ (the event whose likelihood we seek to estimate) by $$O$$ and $$B$$ (the event we know for certain has occurred) by $$F$$:$P(O\mid F)=\dfrac{P(O\cap F)}{P(F)}$Obviously $$P(F)=1/6$$. In part (a) we found that $$P(F\mid O)=1/6$$. Thus$P(O\mid F)=\dfrac{P(O\cap F)}{P(F)}=\dfrac{1/6}{1/6}=1$ Just as we did not need the computational formula in this example, we do not need it when the information is presented in a two-way classification table, as in the next example. Example $$\PageIndex{2}$$: Marriage and Gender In a sample of $$902$$ individuals under $$40$$ who were or had previously been married, each person was classified according to gender and age at first marriage. The results are summarized in the following two-way classification table, where the meaning of the labels is: • $$M$$: male • $$F$$: female • $$E$$: a teenager when first married • $$W$$: in one’s twenties when first married • $$H$$: in one’s thirties when first married $$E$$ $$W$$ $$H$$ Total $$M$$ 43 293 114 450 $$F$$ 82 299 71 452 Total 125 592 185 902 The numbers in the first row mean that $$43$$ people in the sample were men who were first married in their teens, $$293$$ were men who were first married in their twenties, $$114$$ men who were first married in their thirties, and a total of $$450$$ people in the sample were men. Similarly for the numbers in the second row. The numbers in the last row mean that, irrespective of gender, $$125$$ people in the sample were married in their teens, $$592$$ in their twenties, $$185$$ in their thirties, and that there were $$902$$ people in the sample in all. Suppose that the proportions in the sample accurately reflect those in the population of all individuals in the population who are under $$40$$ and who are or have previously been married. Suppose such a person is selected at random. 1. Find the probability that the individual selected was a teenager at first marriage. 2. Find the probability that the individual selected was a teenager at first marriage, given that the person is male. Solution: It is natural to let $$E$$ also denote the event that the person selected was a teenager at first marriage and to let $$M$$ denote the event that the person selected is male. 1. According to the table, the proportion of individuals in the sample who were in their teens at their first marriage is $$125/902$$. This is the relative frequency of such people in the population, hence $$P(E)=125/902\approx 0.139$$ or about $$14\%$$. 2. Since it is known that the person selected is male, all the females may be removed from consideration, so that only the row in the table corresponding to men in the sample applies: $$E$$ $$W$$ $$H$$ Total $$M$$ 43 293 114 450 The proportion of males in the sample who were in their teens at their first marriage is $$43/450$$. This is the relative frequency of such people in the population of males, hence $$P(E/M)=43/450\approx 0.096$$ or about $$10\%$$. In the next example, the computational formula in the definition must be used. Example $$\PageIndex{3}$$: Body Weigth and hypertension Suppose that in an adult population the proportion of people who are both overweight and suffer hypertension is $$0.09$$; the proportion of people who are not overweight but suffer hypertension is $$0.11$$; the proportion of people who are overweight but do not suffer hypertension is $$0.02$$; and the proportion of people who are neither overweight nor suffer hypertension is $$0.78$$. An adult is randomly selected from this population. 1. Find the probability that the person selected suffers hypertension given that he is overweight. 2. Find the probability that the selected person suffers hypertension given that he is not overweight. 3. Compare the two probabilities just found to give an answer to the question as to whether overweight people tend to suffer from hypertension. Solution: Let $$H$$ denote the event “the person selected suffers hypertension.” Let $$O$$ denote the event “the person selected is overweight.” The probability information given in the problem may be organized into the following contingency table: $$O$$ $$O^c$$ $$H$$ 0.09 0.11 $$H^c$$ 0.02 0.78 1. Using the formula in the definition of conditional probability (Equation \ref{CondProb}), $P(H\|O)=\dfrac{P(H\cap O)}{P(O)}=\dfrac{0.09}{0.09+0.02}=0.8182$ 2. Using the formula in the definition of conditional probability (Equation \ref{CondProb}), $P(H\|O)=\dfrac{P(H\cap O^c)}{P(O^c)}=\dfrac{0.11}{0.11+0.78}=0.1236$ 3. $$P(H\|O)=0.8182$$ is over six times as large as $$P(H\|O^c)=0.1236$$, which indicates a much higher rate of hypertension among people who are overweight than among people who are not overweight. It might be interesting to note that a direct comparison of $$P(H\cap O)=0.09$$ and $$P(H\cap O^c)=0.11$$ does not answer the same question. ### Independent Events Although typically we expect the conditional probability $$P(A\mid B)$$ to be different from the probability $$P(A)$$ of $$A$$, it does not have to be different from $$P(A)$$. When $$P(A\mid B)=P(A)$$, the occurrence of $$B$$ has no effect on the likelihood of $$A$$. Whether or not the event $$A$$ has occurred is independent of the event $$B$$. Using algebra it can be shown that the equality $$P(A\mid B)=P(A)$$ holds if and only if the equality $$P(A\cap B)=P(A)\cdot P(B)$$ holds, which in turn is true if and only if $$P(B\mid A)=P(B)$$. This is the basis for the following definition. Definition: Independent and Dependent Events Events $$A$$ and $$B$$ are independent (i.e., events whose probability of occurring together is the product of their individual probabilities). if $P(A\cap B)=P(A)\cdot P(B)$ If $$A$$ and $$B$$ are not independent then they are dependent. The formula in the definition has two practical but exactly opposite uses: • In a situation in which we can compute all three probabilities $$P(A), P(B)\; \text{and}\; P(A\cap B)$$, it is used to check whether or not the events $$A$$ and $$B$$ are independent: • If $$P(A\cap B)=P(A)\cdot P(B)$$, then $$A$$ and $$B$$ are independent. • If $$P(A\cap B)\neq P(A)\cdot P(B)$$, then $$A$$ and $$B$$ are not independent. • In a situation in which each of $$P(A)$$ $$P(B)$$can be computed and it is known that $$A$$ and $$B$$ are independent, then we can compute $$P(A\cap B)$$ by multiplying together $$P(A) \; \text{and}\; P(B)$$$$P(A\cap B)=P(A)\cdot P(B)$$. Example $$\PageIndex{4}$$: Rolling a Die again A single fair die is rolled. Let $$A=\{3\}$$ $$B=\{1,3,5\}$$Are $$A$$ and $$B$$ independent? Solution: In this example we can compute all three probabilities $$P(A)=1/6$$$$P(B)=1/2$$$$P(A\cap B)=P(\{3\})=1/6$$. Since the product $$P(A)\cdot P(B)=(1/6)(1/2)=1/12$$ is not the same number as $$P(A\cap B)=1/6$$, the events $$A$$ and $$B$$ are not independent. Example $$\PageIndex{5}$$ The two-way classification of married or previously married adults under $$40$$ according to gender and age at first marriage produced the table E W H Total M 43 293 114 450 F 82 299 71 452 Total 125 592 185 902 Determine whether or not the events $$F$$: “female” and $$E$$: “was a teenager at first marriage” are independent. Solution: The table shows that in the sample of $$902$$ such adults, $$452$$ were female, $$125$$ were teenagers at their first marriage, and $$82$$ were females who were teenagers at their first marriage, so that $P(F)=\dfrac{452}{902},\; \; P(E)=\dfrac{125}{902},\; \; P(F\cap E)=\dfrac{82}{902}$ Since $P(F)\cdot P(E)=\dfrac{452}{902}\cdot \dfrac{125}{902}=0.069$ is not the same as $P(F\cap E)=\dfrac{82}{902}=0.091$ we conclude that the two events are not independent. Example $$\PageIndex{6}$$ Many diagnostic tests for detecting diseases do not test for the disease directly but for a chemical or biological product of the disease, hence are not perfectly reliable. The sensitivity of a test is the probability that the test will be positive when administered to a person who has the disease. The higher the sensitivity, the greater the detection rate and the lower the false negative rate. Suppose the sensitivity of a diagnostic procedure to test whether a person has a particular disease is $$92\%$$. A person who actually has the disease is tested for it using this procedure by two independent laboratories. 1. What is the probability that both test results will be positive? 2. What is the probability that at least one of the two test results will be positive? Solution: 1. Let $$A_1$$ denote the event “the test by the first laboratory is positive” and let $$A_2$$ denote the event “the test by the second laboratory is positive.” Since $$A_1$$ and $$A_2$$ are independent, $P(A_1\cap A_2)=P(A_1)\cdot P(A_2)=0.92\times 0.92=0.8464$ 2. Using the Additive Rule for Probability and the probability just computed, $P(A_1\cup A_2)=P(A_1)+P(A_2)-P(A_1\cap A_2)=0.92+0.92-0.8464=0.9936$ Example $$\PageIndex{7}$$: specificity of a diagnostic test The specificity of a diagnostic test for a disease is the probability that the test will be negative when administered to a person who does not have the disease. The higher the specificity, the lower the false positive rate. Suppose the specificity of a diagnostic procedure to test whether a person has a particular disease is $$89\%$$. 1. A person who does not have the disease is tested for it using this procedure. What is the probability that the test result will be positive? 2. A person who does not have the disease is tested for it by two independent laboratories using this procedure. What is the probability that both test results will be positive? Solution: 1. Let $$B$$ denote the event “the test result is positive.” The complement of $$B$$ is that the test result is negative, and has probability the specificity of the test, $$0.89$$. Thus $P(B)=1-P(B^c)=1-0.89=0.11$ 2. Let $$B_1$$ denote the event “the test by the first laboratory is positive” and let $$B_2$$ denote the event “the test by the second laboratory is positive.” Since $$B_1$$ and $$B_2$$ are independent, by part (a) of the example $P(B_1\cap B_2)=P(B_1)\cdot P(B_2)=0.11\times 0.11=0.0121$ The concept of independence applies to any number of events. For example, three events $$A,\; B,\; \text{and}\; C$$ are independent if $$P(A\cap B\cap C)=P(A)\cdot P(B)\cdot P(C)$$. Note carefully that, as is the case with just two events, this is not a formula that is always valid, but holds precisely when the events in question are independent. Example $$\PageIndex{8}$$: redundancy The reliability of a system can be enhanced by redundancy, which means building two or more independent devices to do the same job, such as two independent braking systems in an automobile. Suppose a particular species of trained dogs has a $$90\%$$ chance of detecting contraband in airline luggage. If the luggage is checked three times by three different dogs independently of one another, what is the probability that contraband will be detected? Solution: Let $$D_1$$ denote the event that the contraband is detected by the first dog, $$D_2$$ the event that it is detected by the second dog, and $$D_3$$ the event that it is detected by the third. Since each dog has a $$90\%$$ of detecting the contraband, by the Probability Rule for Complements it has a $$10\%$$ chance of failing. In symbols, $P(D_{1}^{c})=0.10,\; \; P(D_{2}^{c})=0.10,\; \; P(D_{3}^{c})=0.10$ Let $$D$$ denote the event that the contraband is detected. We seek . It is easier to find because although there are several ways for the contraband to be detected, there is only one way for it to go undetected: all three dogs must fail. Thus $$D^c=D_{1}^{c}\cap D_{2}^{c}\cap D_{3}^{c}$$ and $P(D)=1-P(D^c)=1-P(D_{1}^{c}\cap D_{2}^{c}\cap D_{3}^{c})$But the events $$D_1$$, $$D_2$$, and $$D_3$$ are independent, which implies that their complements are independent, so $P(D_{1}^{c}\cap D_{2}^{c}\cap D_{3}^{c})=P(D_{1}^{c})\cdot P(D_{2}^{c})\cdot P(D_{3}^{c})=0.10\times 0.10\times 0.10=0.001$ Using this number in the previous display we obtain $P(D)=1-0.001=0.999$ That is, although any one dog has only a $$90\%$$ chance of detecting the contraband, three dogs working independently have a $$99.9\%$$ chance of detecting it. ### Probabilities on Tree Diagrams Some probability problems are made much simpler when approached using a tree diagram. The next example illustrates how to place probabilities on a tree diagram and use it to solve a problem. Example $$\PageIndex{9}$$: A jar of Marbles A jar contains $$10$$ marbles, $$7$$ black and $$3$$ white. Two marbles are drawn without replacement, which means that the first one is not put back before the second one is drawn. 1. What is the probability that both marbles are black? 2. What is the probability that exactly one marble is black? 3. What is the probability that at least one marble is black? Solution: A tree diagram for the situation of drawing one marble after the other without replacement is shown in Figure $$\PageIndex{1}$$. The circle and rectangle will be explained later, and should be ignored for now. Figure $$\PageIndex{1}$$:Tree Diagram for Drawing Two Marbles The numbers on the two leftmost branches are the probabilities of getting either a black marble, $$7$$ out of $$10$$, or a white marble, $$3$$ out of $$10$$, on the first draw. The number on each remaining branch is the probability of the event corresponding to the node on the right end of the branch occurring, given that the event corresponding to the node on the left end of the branch has occurred. Thus for the top branch, connecting the two Bs, it is $$P(B_2\mid B_1)$$, where $$B_1$$ denotes the event “the first marble drawn is black” and $$B_2$$ denotes the event “the second marble drawn is black.” Since after drawing a black marble out there are $$9$$ marbles left, of which $$6$$ are black, this probability is $$6/9$$. The number to the right of each final node is computed as shown, using the principle that if the formula in the Conditional Rule for Probability is multiplied by then the result is $P(B\cap A)=P(B)\cdot P(A\mid B)$ 1. The event “both marbles are black” is $$B_1\cap B_2$$ and corresponds to the top right node in the tree, which has been circled. Thus as indicated there, it is $$0.47$$. 2. The event “exactly one marble is black” corresponds to the two nodes of the tree enclosed by the rectangle. The events that correspond to these two nodes are mutually exclusive: black followed by white is incompatible with white followed by black. Thus in accordance with the Additive Rule for Probability we merely add the two probabilities next to these nodes, since what would be subtracted from the sum is zero. Thus the probability of drawing exactly one black marble in two tries is $$0.23+0.23=0.46$$. 3. The event “at least one marble is black” corresponds to the three nodes of the tree enclosed by either the circle or the rectangle. The events that correspond to these nodes are mutually exclusive, so as in part (b) we merely add the probabilities next to these nodes. Thus the probability of drawing at least one black marble in two tries is $$0.47+0.23+0.23=0.93$$. Of course, this answer could have been found more easily using the Probability Law for Complements, simply subtracting the probability of the complementary event, “two white marbles are drawn,” from 1 to obtain $$1-0.07=0.93$$. As this example shows, finding the probability for each branch is fairly straightforward, since we compute it knowing everything that has happened in the sequence of steps so far. Two principles that are true in general emerge from this example: Probabilities on Tree Diagrams • The probability of the event corresponding to any node on a tree is the product of the numbers on the unique path of branches that leads to that node from the start. • If an event corresponds to several final nodes, then its probability is obtained by adding the numbers next to those nodes. ### Key Takeaway • A conditional probability is the probability that an event has occurred, taking into account additional information about the result of the experiment. • A conditional probability can always be computed using the formula in the definition. Sometimes it can be computed by discarding part of the sample space. • Two events $$A$$ and $$B$$ are independent if the probability $$P(A\cap B)$$ of their intersection $$A\cap B$$ is equal to the product $$P(A)\cdot P(B)$$ of their individual probabilities.
# Many graphs, one relation ## April 04, 2021 \| 8 minutes, 30 seconds This post will be fairly brief because the processes involved will be fairly handwavey in this post. Admittedly, a lot of it requires experimentation, and thanks to graphing tools I've mentioned in the past, it's fairly easy to do so. I'll admit, "many graphs, one relation" is also a bit clickbait-y. But it's not entirely inaccurate - there's far more room here to combine more than just two equations, and potentially any number of them, but it's not nearly as clear as our previous method. ## Conceptualisation Consider two relations in $$x$$ and $$y$$. For example $$x = f(y)$$ or $$y = g(x)$$, or whatever combination of relations you want to play with. From the last post detailing piecewise maximum/minimum functions, we established properties of the maximum and minimum functions that apply to sets. Let's see if we can move it to functions. So rather than manipulating sets of real values, we'll manipulate sets of real-valued functions that combine to form relations themselves. Let's consider $$\operatorname{max}\{a,b\} = 0$$. Why this? Well, if $$a \geq b$$ then $$a = 0$$ but $$b$$ can be any value below. Similarly, if $$a \leq b$$, then $$b = 0$$ but $$a$$ can be less than or equal to 0. In a set of real values, this doesn't really matter much. But with relations? It can be pretty cool. For our example here, we'll use $$y = f(x)$$ and $$y = g(x)$$ for our inputs to the maximum function. There are several ways to combine these, and you'll see why they matter: 1. Pair 1: $$y - f(x) = 0$$ and $$y - g(x) = 0$$ gives us $$\operatorname{max}\{y-f(x),y-g(x)\}=0$$. 2. Pair 2: $$y - f(x) = 0$$ and $$-y + g(x) = 0$$ gives us $$\operatorname{max}\{y-f(x),-y+g(x)\}=0$$. 3. Pair 3: $$-y+f(x) = 0$$ and $$y - g(x) = 0$$ gives us $$\operatorname{max}\{-y+f(x),y-g(x)\}=0$$. 4. Pair 4: $$-y+f(x) = 0$$ and $$-y+g(x) = 0$$ gives us $$\operatorname{max}\{-y+f(x),-y+g(x)\}=0$$. Essentially - they all graph the same pairs of functions, but the places where the functions are graphed vary based on how you write them. These are the only ways to write them - given that when you divide or multiply a negative number in an inequality, you 'flip' the sign, we only have two possibilities that really matter for that variable: $$-1$$ and $$1$$. Any other value is simply cancelled out. Also notice that the maximum of Pair 4 is the minimum of Pair 1 and vice versa. Likewise the maximum of Pair 3 is the minimum of Pair 2. To see where each graph is plotted, you can apply the inequality to each pair (and I encourage anyone reading to derive this by hand using the piecewise notation for the maximum function - it's far more compact and quite fun to see for yourself!): • Pair 1: • $$y=f(x)$$ is plotted when $$g(x)\geq f(x)$$. • $$y = g(x)$$ is plotted when $$f(x)\geq g(x)$$. • Pair 2: • $$y=f(x)$$ is plotted when $$f(x)\geq g(x)$$. • $$y=g(x)$$ is plotteed when $$f(x)\geq g(x)$$ (Yes! This is entirely possible, having two graphs in the same domain!) • Pair 3: • $$y=f(x)$$ is plotted when $$g(x)\geq f(x)$$. • $$y=g(x)$$ is plotted when $$g(x)\geq f(x)$$. • Pair 4: • $$y=f(x)$$ is plotted when $$f(x)\geq g(x)$$. • $$y=g(x)$$ is plotted when $$g(x)\geq f(x)$$. Try plotting $$\operatorname{max}\{\alpha \left(y-x\right),\beta \left(y-x^2\right)\} = 0$$ where $$\alpha, \beta \in \{-1, 1\}$$ in something like Desmos. It becomes notably more complicated if we combine functions in $$x$$ and $$y$$. But as you can see, simply changing the way we represent our relations changes where we can mix our graphs, which is a really useful tool for creating our own relations. ## Application So why did I say this post was a bit handwavey? Well, I want to focus on the general application of it rather than how we can achieve what we do. For example, taking the maximum of 3 relations to achieve our own specific relation could be rigourously derived - but it is so much more effort than we probably want to go to (at least the way I know how, if you can find a technique please do share!! I would be super excited if you could!). Instead, we can tinker, with the help of graphing tools. Let's start off with 3 equations. Let's create a simple triangle with little overhead calculations (as an additional challenge, construct a triangle that uses $$x^2+y^2 = 1$$ to determine vertices, using the technique outlined here). Picking 3 vertices; $$\{(0,1), (-1,-1), (1,-1)\}$$ we can construct the lines $$y=-1$$, $$y=-2x+1$$ and $$y=2x+1$$. From here, to combine, we have the following relation: $$\operatorname{max}\{\alpha(y+1),\beta(y+2x-1),\gamma(y-2x-1)\} = 0$$. If you plot this with all positive values for $$\alpha, \beta, \gamma$$, you'll notice the angled lines that we want aren't plotted where we want. In fact, both of them aren't. Given that the equation $$y=-1$$ intersects both of them, we can reason that perhaps $$\alpha$$ must be negative, and perhaps $$\beta, \gamma$$ are positive. Turns out, yes, this is the case, and we're left with: $\operatorname{max}\{-y-1,y+2x-1,y-2x-1\} = 0$ Note: This line of reasoning is perhaps a decent way of determining whether certain equations should be represented in other ways, but I've yet to test it out yet further. Explicitly points out that if you want one equation changed, you change another that interacts with it in a certain way. Cool. If you want, you can apply this similar process say using Desmos, and try yourself! We can simplify using the rules we established in our earlier post: $$\operatorname{max}\{-y,y+\|2x\|\}=1$$, or, alternatively, in a more compact form, $$2\operatorname{max}\{y+\left\|x\right\|,0\}-y = 1$$ (this form's coefficients also show you what you can do to transform the triangle we made, definitely worth trying out). Another example I wanted before was to combine several lines: $$y=x$$, $$y=-\frac{1}{2}x$$ and $$y=-1-x$$, at their intersection points to create a weird line that almost looks like the absolute value function. It was achievable, but the form of it was relatively ugly and also took a bit of experimentation. $$y = \operatorname{max}\left\{\left\|x+\frac{1}{2}\right\| -\frac{1}{2},-\frac{1}{2}x\right\}$$ when simplified. Also shows that equations we can combine like this can also be simplified down to only be in terms of absolute value functions - but would look rather messy... cool, nonetheless, but messy. Honestly, there are plenty of examples. The beautiful thing is you can try this one all by yourself with no complicated algebra, but a lot of experimentation. You can create closed shapes! ### The diamond In one of our posts, we derived the equation for a diamond through two methods. But actually, we can derive it using a third as well, using this outlined technique. Take our lines $$y=1-x$$, $$y=1+x$$, $$y=x-1$$ and $$y=-x-1$$ and combine them using our new technique: $$\operatorname{max}\left\{\alpha(y+x-1), \beta(y-x-1), \gamma(y-x+1), \delta(y+x+1)\right\}=0$$. If we plot this, we see that our lines lie below where we want. Given the lines intersect $$y=x-1$$ and $$y=-x-1$$, let's suggest that, for our values, we have $$\alpha=1, \beta=1, \gamma=-1, \delta=-1$$. We can additionally motivate this decision by the sign of the $$1$$ in each of our elements in the input for the maximum function. Turns out this works, so we have (and simplifying): \begin{align} \operatorname{max}\left\{y+x-1,y-x-1,-y+x-1,-y-x-1\right\}&=0 \\ \operatorname{max}\left\{y+x,y-x,-y+x,-y-x\right\}&=1 \\ \operatorname{max}\left\{y+\operatorname{max}\left\{x,-x\right\},-y+\operatorname{max}\left\{x,-x\right\}\right\}&=1 \\ \operatorname{max}\left\{y+\left\|x\right\|,-y+\left\|x\right\|\right\}&=1 \\ \left\|x\right\|+\operatorname{max}\left\{y,-y\right\}&=1 \\ \left\|x\right\|+\left\|y\right\|&=1 \end{align} An alternative form we might consider: \begin{align} \operatorname{max}\left\{y+x,y-x,-y+x,-y-x\right\}&=1 \\ \operatorname{max}\left\{\operatorname{max}\left\{y+x,-y-x\right\},\operatorname{max}\left\{y-x,-y+x\right\}\right\}&=1 \\ \operatorname{max}\left\{\left\|x+y\right\|,\left\|x-y\right\|\right\}&=1 \end{align} And we're done! ### The square In fact, one of the things I'm building up to is a fun closed shape - the square. As a challenge, if we define a square by the lines on the $$x$$ and $$y$$ axes at $$\frac{1}{2}$$ and $$-\frac{1}{2}$$ each, see if you can combine them to form a single relation describing a square. Added bonus; simplify it. When I get around to writing my posts about squares, we can compare forms!
# Multiply Strings \| LeetCode \| Observation Based Approach, Explained LeetCode 43. Multiply Strings. I explain the Observation Based Approach using simple maths in detail along with the implementation in Python3. Problem Link. ## 0. Multiply Strings: Problem Discussion Multiply two numbers given as strings, and return their output as a string. ## 1. Multiply Strings: Observations & Logic ### 1.0. Intuitions • How do we multiply two numbers? Take for example 243 and 19. • The first thing we do is take the first digit of num2 and multiply all the digits in num1 with it. • For each of the xy operations, we also have to take care of the carry-overs. We can do that by `temp = carry + x + y`. • The green part is now the remainder on division with 10, and the yellow part is the divisor. • Finally, when we come to the second digit in num2, we add one more zero at the rightmost position (to account that 1 is actually 10, as in, 19 = 10 + 9) ## 2. Multiply Strings: Implementation Once we understand the method, much of remaining solution is implementation. Here is my naming scheme: • `final_ans` stores the total of all the intermediate green rows • `ans` is the answer for the intermediate row • `multipler` is what takes care of how many zeros have to be added in the rightmost position • `carry` keeps a track of the current carry-over (yellow) • `rem` is the remainder (green single digits, constituting the green intermediate row) • `position` specifies what position in the `ans` the current `rem` is – whatever power of 10 we have to raise it ### 2.0. Code ```class Solution: def multiply(self, num1: str, num2: str) -> str: multiplier = 0 final_ans = 0 for y in reversed(num2): carry = 0 ans = 0 position = multiplier for x in reversed(num1): carry, rem = divmod(carry+int(x)int(y), 10) ans += 10*position rem position += 1 if carry: ans += 10*position carry multiplier += 1 final_ans += ans return str(final_ans)``` ### 2.1. Complexity Analysis • Time: O(MN). We iterate over all num1 digits and num2 digits (nested) • Space: O(1). We only ever store variables: multiplier, position, carry, rem, ans, final_ans. ##### Tanishq Chaudhary Producing high-quality intuitive explanations of interview problems. Currently covering LeetCode and InterviewBit. ### Recommended Posts ##### Gas Station \| InterviewBit \| Solution Explained June 13, 2022 ##### Disjoint Intervals \| InterviewBit \| Solution Explained June 12, 2022 ##### Reverse integer \| InterviewBit \| LeetCode \| Solution Explained January 24, 2022 This site uses Akismet to reduce spam. Learn how your comment data is processed.
Velocity and acceleration Category : Straight Line Motion The motion of objects in one dimension is described in a problem-solution based approach. Velocity and acceleration Displacement and Distance Velocity and acceleration Most of objects and phenomena in physics have motion. Therefore, to have a better understanding of how nature works, we should have an exact and clear definition of motion and ability of its computation. Motion is defined as the change of an object's location with time which can be taken place in linear, circular or projectile paths. In a linear motion, whole parts of an object move in one direction or simply they have equal displacement and direction. Circular motion is the change of object's angular position. This is a movement along a circular path or circumference of a circle. Once an object moves or fired with an angle through space, we could call this type of motion as projectile one. Previously, we defined displacement of an object and mentioned it is the building block of the other kinematical concepts which describe the motion of a particle along an arbitrary path. Here, we study the concepts of velocity and acceleration in a problem-solution based approach. We believe that in this way the AP or college students can succeed more. Average velocity and speed: We need usually to know at what rate does the particle moves? To answer this question, we define an important quantity as follows: $Average\ speed=\frac{distance\ traveled}{elapsed\ time}=\frac{D}{\Delta t}$ $Average\ velocity=\frac{displacement}{elapsed\ time}$ The average speed defines in terms of the distance traveled ($D$) so it is a scalar quantity. But the average velocity is a vector that points in the same direction as the displacement vector and defines as the change in position of a body divided by the time interval in which that changes occur. Average velocity is a more useful quantity than the other since it describes both how fast and in what direction the objects move. Therefore, in one direction, say $x$,the average velocity is defined as $\vec{v}_{ave-x}=\frac{\Delta \vec{x}}{\Delta t}$ The SI unit of velocity is meters per second ($\rm m/s$). Other common units include Kilometers per second ($\rm Km/h$), feet per second ($\rm ft/s$) and miles per hour ($\rm mi/h$). Note that, in general, the magnitude of the average velocity and average speed are not equal in a given time interval. For example, if the above car moves from $A$ to $B$ in $4\,{\rm s}$ we have $Average\ speed = \frac{10\,{\rm m}}{4\,{\rm s}}=2.5\ {\rm m/s}$ whereas $Average\ velocity= \frac{\Delta \vec{x}}{\Delta t}=+\frac{4\,{\rm m}}{4\,{\rm s}}=+1\ {\rm m/s}$ The positive indicates that the overall motion is in the positive $x$ direction. Example $1$: A bird is flying $100\,{\rm m}$ due east at $10\,{\rm m/s}$ and then it turns around and flying west in $15\,{\rm s}$ at $20\,{\rm m/s}$. Find the average velocity and average speed during the overall time interval. Solution: First we must find the total time. The first and second parts are done in $\Delta t_1=\frac{\Delta x}{v}=\frac{100}{10}=10\,{\rm s}$ and $\Delta t_2=15\, {\rm s}$. Thus the overall time of flying of the bird is $\Delta t_{tot}=10+15=25\,{\rm s}$. The distance traveled of the bird due west is $d_w=v\Delta t_2=20 \times 15=300\,{\rm m}$. Therefore, by definition of the average speed, we have $Average\ speed =\frac{(100+300)\,{\rm m}}{25\,{\rm s}}=16\ {\rm m/s}$ As we can see from the figure, the displacement vector is $\Delta x=x_f-x_i=-200-0=-200\,{\rm m}$ so the average velocity is found as $\vec{v}_{ave}=\frac{\Delta \vec{x}}{\Delta t}=\frac{-200\,{\rm m}}{25\,{\rm s}}=-8\ {\rm m/s}$ The negative indicates that the average velocity is towards the $–x$ direction. To show the motion of a particle usually a position–versus-time ($x-t$) graph (or simply position graph) is used. In such plots, every point in the plane represents the coordinate and its corresponding time of a moving object. You should not be mistaken the path of motion with the curves in the $x-t$ graph. As you can see later, straight line movement of a car does not yield a straight line!. Further, time is specified on the horizontal axis. Using these graphs one can extract (or simplify) more information about the type of motion. Let's consider the velocity of a moving particle is constant during any successive equal-time intervals, then the $x-t$ graph is straight-line. This type of motion is called uniform motion. In this case, the average velocity is the slope of the position-versus-time graph. In one dimensional motion, this is simply $v_{ave-x}=\frac{\Delta x}{\Delta t}={\rm slope\ of\ the\ x-t\ graph}$ Therefore, from the mathematical viewpoint, in the uniform motion, equal displacements occur during any successive equal-time interval. If the velocity is not constant, the motion's graph is a curve as shown in the figure below (the actual path of the motion could be a straight line, this is a reason why the position graph is not a picture of the path!). In such a case, the object's average velocity during each particular time interval is the slope of the $x-t$ graph that connects the initial and final points in that interval $\Delta t$. It is worth noting that in mathematics the slope of line connecting two points $A$ and $B$, in a plane, defines as $\tan$ of smaller angle with the $x$ axis. Therefore, we can relate the average speed or velocity (in which those two points get closer together) to the $\tan$ of smaller angle as $\text{average speed}=\text{slope of line AB}=\tan \alpha$ Instantaneous velocity The average velocity is not a useful quantity for analyzing the non-uniform motions. Very often it is necessary to know the object’s velocity at a single instant of time. In this case, the instantaneous velocity can be defined. Instantaneous velocity is the limit of average velocity $\vec{v}_{ave}=\frac{\Delta \vec{x}}{\Delta t}$ as $\Delta t$ approaches zero. In the language of calculus, this limit is called the derivative of $x$ with respect to $t$ and is written $\frac{dx}{dt}$. Thus in the mathematical form, we can state $\vec{v}_x=\lim_{\Delta t \to 0}\frac{\Delta \vec{x}}{\Delta t}=\frac{dx}{dt}$ Therefore, the instantaneous velocity at time $t$ is the slope of the line that is tangent to the position-versus-time graph at the instance $t$. The time interval $\Delta t$ is always positive, so $v_x$ has the same sign as $\Delta x$. In many textbooks, the velocity refers to the instantaneous velocity, so from now on, we adopt this convension. Example ($2$): The position of a particle varies with time according to the equation $x(t) = 40-5t-5t^2$ where $x$ and $t$ are in $\rm m$ and $\rm s$, respectively. Find ($a$) The particle's average velocity between $t=1\,{\rm s}$ and $t=2\,{\rm s}$. ($b$) The velocity at $t=2\,{\rm s}$. Solution: (a) By definition of the average velocity, we must first find the displacement during that interval. \begin{eqnarray} x(t=2\,{\rm s})=40-5(2)-5(2)^2=10\ {\rm m}\\ x(t=1\,{\rm s})=40-5(1)-5(1)^2=30\ {\rm m}\\ \Delta x=x(t=2\,{\rm s})-x(t=1\,{\rm s})=10-30=-20\ {\rm m} \end{eqnarray} Therefore, the average velocity becomes $v_{ave-x}=\frac{\Delta x}{\Delta t}=\frac{-20\,{\rm m}}{(2-1)\,{\rm s}}=-20\ {\rm m/s}$ (b) The velocity is defined $v_x=\frac{dx}{dt}$ $v_x=\frac{dx}{dt}=\frac{d}{dt} \left(40-5t-5t^2\right)=(-5-10\,t)\ {\rm m/s}$ Evaluating the velocity at $t=2\,{\rm s}$ gives $v_x (t=2\,{\rm s})=-5-10(2)=-25\ {\rm m/s}$ In both cases, the negatives indicate the particle moves in the $-x$ direction. Example $3$: A student to arrive at her school, first walks $900\,{\rm m}$ to the north then $300\,{\rm m}$ toward the east. Finally walks $500\,{\rm m}$ along the south. The total time of walking is $5$ minutes. ($a$) Find the distance traveled by her? ($b$) Find the average velocity. Solution: ($a$) To find the distance traveled, one should add lengths of each part of the path with each other. So \begin{eqnarray} \text{distance} &=& AB+BC+CD \\ &=& 900+300+500=1700\,{\rm m} \end{eqnarray} ($b$) As mentioned, average velocity is a vector quantity which depends on the displacement of an object between initial and final points. Thus, draw a vector which connects those points to each other as figure below (this is a simple geometrical problem) \begin{eqnarray} \end{eqnarray} In the first line, the definition of displacement as vector addition is represented and using Pythagoras theorem its magnitude is computed. Now, by definition, the magnitude of average velocity $\|\vec{v}_{av}\|$ read as \begin{eqnarray} \|\vec{v}_{av}\| &=& \frac{\text{displacement}}{\text{total time}}\\ &=& \frac{500\,{\rm m}}{5 \times 60\,{\rm s}}\\ &=& \frac {5}{3}\,{\rm \frac m s} \end{eqnarray} We can go further and find its direction. Recall that average velocity is a vector in the same direction as displacement vector. Therefore, \begin{eqnarray} \theta &=& \tan^{-1} \left(\frac{\text{y-component of $\overrightarrow{AD}$}}{\text{x-component of $\overrightarrow{AD}$}} \right)\\ &=& \tan^{-1} \left(\frac{400}{300}\right)\\ &\cong& 53^\circ \end{eqnarray} Note: in everyday language, we usually define velocity as the distance traveled divided by the elapsed time, say here $v=\frac{1700}{5\times 60}=\frac{17}{3}\,{\rm m/s}$. But in physics, there is distinction between these concepts. Practice $1$: A car moves along the $x$ axis as following figure. Its position at instant $t_1=2\,{\rm s}$ is at $A$. It is positioned at $B$ at time $t_2=5\,{\rm s}$ then returns and passes through the point $C$ at $t_3=10\,{\rm s}$. (a) What is the average speed and velocity of the car between the times $t_1$ and $t_2$? b) Now find the above quantities in the time interval $t_1$ and $t_3$. Solution is here. Acceleration Acceleration means the rate of change of velocity. That is, acceleration measures how quickly or slowly an object's velocity changes. This change may be the result of the change in the magnitude or direction of the velocity or both of them. Thus if the magnitude of the velocity remains constant and only its direction varies, we can still have an accelerating motion (for example in the circular motions). The average acceleration during a particular time interval $\Delta t$ is defined as $\vec{a}_{ave}=\frac{\Delta \vec{v}}{\Delta t}$ The $\text{SI}$ unit of acceleration is (meters per second) per second, abbreviated $\rm m/s^2$. The average acceleration like other kinematic parameters position $r$ and velocity $v$ is a vector and points in the same direction as velocity change $\Delta \vec{v}$. If the velocity of a car changes from $0$ to $50\,{\rm km/h}$ in $10\,{\rm s}$, its average acceleration is found as $\vec{a}_ave=\frac{50\,(\rm Km/h)}{10\,{\rm s}}$ This means that the velocity of the car increases by $5\, {\rm km/h}$ every $1\,{\rm s}$. Instantaneous acceleration is the limit of the ratio $\frac{\Delta \vec{v}}{\Delta t}$ as $\Delta t$ approaches zero or in the mathematical language is the derivative of the velocity with respect to time. $\vec{a}=\lim_{\Delta t \to 0}\frac{\Delta \vec{v}}{\Delta t}=\frac{d\vec{v}}{dt}$ The algebraic sign of the acceleration does not tell us about the object is speeding up or slowing down. To determine this, we must take into account the signs of both of acceleration and velocity. In one dimensional motion, say $x$ direction, if $\vec{v}_x$ and $\vec{a}_x$ are both positive, the speed of the object is increasing and it is speeding up. When $\vec{v}_x$ and $\vec{a}_x$ have opposite signs, the speed is decreasing. In summary, when $\vec{v}_x$ and $\vec{a}_x$ have the same sign, the speed is increasing and when they have opposite signs, the speed is decreasing which usually called decelerating motion. To find the type of motion, speeding up or slowing down, the exact way is that plot the $v-t$ graph and from it characterize the sign of acceleration and velocity as follows (see Examples below). In general, in constant(uniform) acceleration, the position and velocity graph are parabola and straight-line, respectively. But in case of non-uniform one, the $v-t$ graph is a smooth curve. Example ($4$): A particle at time $t=3\, {\rm s}$ and at position $x=7\, {\rm m}$ has velocity $v=4\, {\rm m/s}$. At $t=7\, {\rm s}$ it is located at $x=-5\, {\rm m}$ with velocity $v=-2\, {\rm m/s}$. Find: (a) The average velocity of the particle. (b) The average acceleration of the particle. Solution (a) By definition of the average velocity, we must first determine the displacement vector.so $\Delta \vec{x}=\vec{x}_f-\vec{x}_i=(-5\,{\rm m})-(7\, {\rm m})=-12\ {\rm m}$ $average\ velocity : \vec{v}_{ave}=\frac{\Delta \vec{x}}{\Delta t}=\frac{-12\, {\rm m}}{(7-3)\, {\rm s}}=\frac{-12}{4}=-3\ {\rm m/s}$ The minus sign indicates that the direction of the motion of the particle points to the $–x$ axis. (b) To find the average acceleration, first determine the change in the velocity of the particle during that interval time. $\Delta \vec{v}=\vec{v}_f-\vec{v}_i=-2-4=-6\ {\rm m/s}$ By definition we have $\vec{a}_{ave}=\frac{\Delta \vec{v}}{\Delta t}=\frac{-6\, {\rm m/s}}{(7-3)\, {\rm s}}=\frac{-6}{4}=-1.5\ {\rm m/s^2}$ Example ($5$): A bus accelerates uniformly from an initial velocity of $15\,{\rm m/s}$ to a final velocity of $7\,{\rm m/s}$ in $4$ seconds. Calculate the acceleration of the bus. Let the direction of motion of the bus be the positive direction . Solution: Let's consider the positive direction of $+x$ axis as the direction of velocity of the car. We know that the subtracting the initial velocity from the final value divided by the time taken gets the average acceleration as \begin{eqnarray} \vec{a}_{ave} &=&\frac{\Delta \vec{v}}{\Delta t}\\ &=& \frac{\vec{v}_f-\vec{v}_i}{t_f-t_i}=\frac{7-15}{4}\\ &=& -2\,{\rm m/s} \end{eqnarray} As you can see, the direction of the car does not change (always toward the $+x$ direction) but the obtained acceleration is to the opposite side (the minus sign in front of it indicated), thus we here encounter a decelerating motion since the product of acceleration and velocity is negative i.e. $a_x v_x <0$. Practice $2$: The position-versus-time graph of a moving object along a straight line is as follows. Discuss the type of motion, speeding up or slowing down, in the time interval $1\,{\rm s}$ to $3\,{\rm s}$? Solution: To practice and see this solution go here. Example ($6$): The equation of motion of a car in a straight line (in SI units) is as follows: $x(t)=t^3 - 6t^2 +9t$. (a) What is the magnitude of the average acceleration in the interval $t_1=1\,{\rm s}$ and $t_2=2\,{\rm s}$. (b) At what points, the car has changed its direction? (c) Determine the type of motion along the entire path? Solution: (a) The average acceleration is defined as the change of initial and final instantaneous velocity in a specific time interval $\Delta t$. The variation of position with time is given as above. To obtain the desired quantity, we should take two steps, one calculate the velocity of the car at the given instants, next divide their difference by the time interval. \begin{eqnarray} v(t)=\frac{dx}{dt}=\left(3\,t^{2}-12\,t+9\right) \end{eqnarray} In above, we omitted the vector sign over the velocity $v$, since we are in straight line and the sign of velocity or acceleration imply the direction of the motion. Given the equation of velocity, $v(t)$, one can construct their initial and final values \begin{eqnarray} v(t_1=1) &=& 3(1)^2-12(1)+9=0\,{\rm m/s}\\ v(t_2=2)&=&3(2)^2-12(2)+9=-3\,{\rm m/s}\\ \Delta v=v(t_2)-v(t_1)=-3-0=-3\,{\rm m/s} \end{eqnarray} As noted above, the minus sign reveals the direction of motion toward the $-x$ axis. Therefore, the magnitude of average acceleration read \begin{eqnarray} a_{ave}=\frac{\Delta v}{\Delta t}=\frac{-3}{2-1}=-3\,{\rm m/s^2 } \end{eqnarray} Which as previously, the negative is direction of the acceleration. Because of $a_x v_x>0$, the motion is accelerating. (b) A moving object to change direction, firstly should be stopped. To do this, simply find the roots of equation of velocity with time or roots of derivative of equation of motion. \begin{eqnarray} v(t)=\frac{dx}{dt} &=&\frac{d}{dt}(t^3 )-\frac{d}{dt}(6t^2 )+\frac{d}{dt}(9t)=0\\ &=& 3t^2 -12t+9=0 \end{eqnarray} This is a quadratic equation, $a x^2 +b x+c=0$ with the following well-know solutions \begin{eqnarray} x_{1,2}=\frac{-b\pm \sqrt{b^2 - 4ac}}{2a} \end{eqnarray} Therefore, \begin{eqnarray} t_{1,2} &=&\frac{-(-12)\pm\sqrt{(-12)^2 - 4(3)(9)}}{2(3)}\\ &=& t_{1,2}=\frac{12\pm\sqrt{144-108}}{6}\\ &\Longrightarrow& t_1=\frac{12+6}{6}=3\,{\rm s}\\ & & t_2=\frac{12-6}{6}=1\,{\rm s} \end{eqnarray} For better understanding, we plotted $x-t$ graph of this motion. It is worth noting that since this plot is not a parabola so the motion has a non-uniform acceleration. Recall that the slope of $x-t$ graph is the velocity(speed). In a graph, extrema are where the first derivative of a function is zero. As one can see, in this plot, there are two points where the instantaneous velocity (first derivative of position with time) is zero. These points denotes when the car changes its direction or called turning points (c) As mentioned, first we should plot velocity graph. In part (a), velocity obtained by taking derivative of position equation $x(t)$. Here, we plotted the profile of $v(t)$ by a third-party drawing software like FX Draw or Google. The car starts out moving to the right ($v>0$) with an initial velocity $v(t=0)=9\,{\rm m/s}$. The negative acceleration (the slope of $v-t$ graph indicates the acceleration) causes the speed to decrease (since the acceleration ($\vec{a}<0$) and velocity ($\vec{v}>0$) are in opposite directions) until the car reaches its first turning point (where the instantaneous velocity is zero, $v=0$) just before $t=1\,{\rm s}$. Now the car speeding up to the left ($v<0$ and $a<0$ so $a v>0$) until reaching maximum speed $v=3\,{\rm m/s}$ at $t=2\,{\rm s}$. In the time interval $t=2\,{\rm s}$ until $t=3\,{\rm s}$ when the car stops and changes its direction (the second turning point), the direction of motion is still to the left with an positive acceleration (slowing down). After the final turning point at $t=3\,{\rm s}$, the car moves to the right ($v> 0$) and the positive acceleration makes the velocity ever more positive (that is speeding up motion). In summary, the motion in any time interval behaves as $\left\{ \begin{array}{lll} a_x<0 & v_x>0 & a_x v_x<0\, \text{(slowing down)} & 0\leq t\leq 1 \\ a_x<0 & v_x<0 & a_x v_x>0\, \text{(speeding up)} & 1\leq t\leq 2 \\ a_x>0 & v_x<0 & a_x v_x<0\, \text{(slowing down)} & 2\leq t\leq 3 \\ a_x>0 & v_x>0 & a_x v_x>0\, \text{(speeding up)} & t\geq 3 \\ \end{array} \right.$ Displacement, Average Velocity, Average acceleration, Speed, Instantaneous velocity, Instantaneous acceleration, kinematic
# 4 Examples Of Rational Numbers In the context of mathematics, a rational number is a number that can be expressed as the ratio of two integers. A rational number is a number that is equal to the quotient of two integers p and q. In other words, a rational number can be expressed as some fraction where the numerator and denominator are integers. Some examples of rational numbers include: 1. The number 8 is rational because it can be expressed as the fraction 8/1 (or the fraction 16/2) 2. the fraction 5/7 is a rational number because it is the quotient of two integers 5 and 7 3. the decimal number 1.5 is rational because it can be expressed as the fraction 3/2 4. the repeating decimal 0.333… is equivalent to the rational number 1/3 Traditionally, the set of all rational numbers is denoted by a bold-faced Q. Rational numbers are distinguished from the natural number, integers, and real numbers, being a superset of the former 2 and a subset of the latter. There also exist irrational numbers; numbers that cannot be expressed as a ratio of two integers. An example of an irrational number is √2. √2 cannot be written as the quotient of two integers. Let’s take a step back and talk about the different kinds of numbers. ## Kinds Of Numbers It may come as a surprise to some that there exist different classes of numbers. After all, a number is a number, so how can some numbers be fundamentally different than other numbers? In a nutshell, numbers can be differentiated by how they behave when being added, subtracted, multiplied, or divided. ### Natural Numbers Credit: Good Free Photos CC0 1.0 Let’s start with the most basic group of numbers, the natural numbers. The set of natural numbers (denoted with N) consists of the set of all ordinary whole numbers {1, 2, 3, 4,…} The natural numbers are also sometimes called the counting numbers because they are the numbers we use to count discrete quantities of things. A key feature of natural numbers is that they can be represented without some fractional or decimal component. There are an infinite amount of natural numbers stretching from 1 to infinity. The natural numbers are considered the most basic kind of number because all other kinds of numbers can be defined as extensions of the natural numbers. Traditionally, the natural numbers do not contain the number zero (0), though some mathematicians consider 0 to be a natural number. The natural numbers are closed under addition and multiplication. This means that if you add or multiply any two natural numbers, your answer will be another natural number. Adding 4 and 4 gives equals the natural number 8 and multiplying 5 by 1,000,000 equals the natural number 5,000,000. Adding or multiplying two natural numbers will always give you another natural number, no exceptions. What about subtraction though? The natural numbers are not closed under subtraction. This means that if you subtract two natural numbers, your answer may not always be a natural number, which leads us to… ### Integers Next up are the integers. The integers (denoted with Z) consists of all natural numbers and all negative whole numbers (…-4, -3, -2, -1) The set of integers is constructed by adding the additive inverse of every natural number, so it contains all positive and negative whole numbers {…-4, -3, -2, -1, 0, 1, 2, 3, 4,…}. As a consequence, all natural numbers are also integers. Like the naturals, there are an infinite amount of integers spanning from negative infinity to positive infinity. Credit: WikiCommons CC0 1.0 Like the natural numbers, the integers are closed under addition and subtraction. Adding or multiplying any two integers will always give you another integer. Introducing negatives into our number systems makes it so that the integers are also closed under subtraction. Subtracting any two integers will always give you another integer. 6−3 = -3 and 12−40 = -28. Now we have a set of numbers that is closed under addition, multiplication, and subtraction. What about division though? A moment’s thinking should tell you that no, the integers are not closed under division. Dividing two integers may not always result in another integer. This realization leads us to the next set of numbers… ### Rational Numbers Credit: Pixabay CC0 1.0 Enter the rational numbers. The addition of rational numbers (denoted Q) allows us to express numbers as the quotient of two integers. Every rational number can be uniquely represented by some irreducible fraction. The number 3/2 is a rational number because it is expressed as a fraction in simplest form. Consequently, the rational number 6/4 is also equal to 3/2, because 6/4 can be simplified to 3/2. All integers (and so all natural numbers) can be expressed as an irreducible fraction (8 = 8/1 and -5 = -5/1), so all integers and natural numbers are also rational numbers. Rational numbers are added to the number system to allow that numbers also be closed under division (with the lone exception of division by 0). The quotient of any two rational numbers can always be expressed as another rational number. This insight can be seen in the general rule for dividing fractions (i.e. rational numbers). a/b ÷ c/d = ad/bc when d and b ≠ 0 a/b and c/d are rational numbers, meaning that by definition a, b, c, and d are all integers. Since the integers are closed under multiplication, ad and bc are also integers. ad/bc is represented as a ratio of two integers, which is the exact definition of a rational number. Therefore, the rational numbers are closed under division. Rational numbers can also be expressed as decimals. Converting from fraction to decimal notation is easy: all you have to do is set up a long division problem and divide the numerator by the denominator. Dividing out an irreducible fraction will give you one of two results: either (i) long division will terminate in some finite decimal sequence or (ii) long division will produce an infinitely repeating sequence of decimals (e.g. 1/3 = 0.333… and 6/11 = 0.5454…). Converting from a decimal to a fraction is likewise easy. All you have to do is multiply the decimal by some power of 10 to get rid of the decimal point and simplify the resulting fraction. eg) 0.25 × 100/100 = 25/100 = 1/4 The rational numbers are the simplest set of numbers that is closed under the 4 cardinal arithmetic operations, addition, subtraction, multiplication, and division. This property makes them extremely useful to work with in everyday life. Rational numbers are not the end of the story though, as there is a very important class of numbers that cannot be expressed as a ratio of two integers. ### Irrational Numbers An irrational number is a number that cannot be expressed as a ratio of two integers. Irrational numbers cannot be represented as a fraction in lowest form. The two sets of rational and irrational numbers are mutually exclusive; no rational number is irrational and no irrational number is rational. Common examples of irrational numbers include π, Euler’s number e, and the golden ratio φ. None of these three numbers can be expressed as the quotient of two integers. How do we even know irrational number exist? Here is a simple proof by contradiction which shows that √2 is an irrational number: Assume √2 is a rational number. If √2 is a rational number, then that means it can be expressed as an irreducible fraction of two integers. Let’s call those two integers p and q. (1.) √2 = p/q Since p/q is an irreducible fraction (per the definition of a rational number) they do not have any factors in common. Squaring both sides to get rid of the left hand radical gives us: (2.) 2 = p2/q2 which we can rearrange into: (3.) 2q2 = p2 This result implies that p2 is an even number because 2 is one of its factors. The only way p2 could be even is if p itself is even. If p is even, then there is some number k such that p = 2k. Substituting 2k for p in equation (3.) gives us: (4.) 2q2 = (2k)2 (5.) 2q2 = 4k2 Equation (5.) can be rewritten as: (6.) q2 = 2k2 By similar reasoning, q2 and must be even. Therefore, both p and q are even numbers. However, this contradicts our requirement from (1.) that p and q do not share any factors. Since we derived a contradiction, our initial assumption (that √2 is rational) must be false. Therefore, √2 is an irrational number and cannot be expressed as the quotient of two integers. Irrational numbers rear their head all over the place. As it turns out, the square roots of most natural numbers are irrational. Many commonly seen numbers in mathematics are irrational. For example, the number π which is the ratio of the diameter of a circle to its circumference is irrational Additionally, Euler’s number e, the unique number whose natural logarithm is 1, is also irrational. For some time, it was thought that all numbers were rational numbers. The preoccupation with rational numbers stems back to ancient Greece with the teaching of the Pythagoreans. The Pythagoreans were a quasi-religious sect who believed that numbers are the basic constituents of the universe. Central to their beliefs was the idea that all quantities could be expressed as rational numbers. The legend goes that the Pythagorean Hippasus first discovered the existence of irrational numbers when trying to solve for the hypotenuse of a right triangle with sides of equal length. Hippasus discovered that the length of the hypotenuse could not be understood as proportional to the lengths of its sides, and in doing so discovered irrational numbers. Reportedly, his discovery so greatly distressed the other Pythagoreans that they had Hippasus drowned as punishment for sacrilege. Nowadays, we understand that not only do irrational numbers exist but that the vast majority of numbers are actually irrational. Comparatively, the set of rational numbers (which includes the integers and natural numbers) is incomprehensibly dwarfed by the size of the set of irrational numbers. To sum up, rational numbers are numbers that can be expressed as the quotient of two integers. Rational numbers form an important class of numbers and are the simplest set of numbers that is closed under the 4 cardinal arithmetic operations of addition, subtraction, multiplication, and division. Rational numbers are distinguished from irrational numbers; numbers that cannot be written as some fraction. ## About Alex Bolano PRO INVESTOR When Alex isn't nerdily stalking the internet for science news, he enjoys tabletop RPGs and making really obscure TV references. Alex has a Masters's degree from the University of Missouri-St. Louis. ## Questions & Answers (0) Have a question? Our panel of experts willanswer your queries.Post your question
# 35094 in words 35094 in words is written as Thirty Five Thousand and Ninety Four. In 35094, 3 has a place value of ten thousand, 5 is in the place value of thousand, 9 is in the place value of ten and 4 is in the place value of one. The article on Place Value gives more information. The number 35094 is used in expressions that relate to money, distance, social media views, and many more. For example, “The postal code in Alabama, USA, is Thirty Five Thousand and Ninety Four.” 35094 in words Thirty Five Thousand and Ninety Four Thirty Five Thousand and Ninety Four in Numbers 35094 ## How to Write 35094 in Words? We can convert 35094 to words using a place value chart. The number 35094 has 5 digits, so let’s make a chart that shows the place value up to 5 digits. Ten thousand Thousands Hundreds Tens Ones 3 5 0 9 4 Thus, we can write the expanded form as: 3 × Ten thousand + 5 × Thousand + 0 × Hundred + 9 × Ten + 4 × One = 3 × 10000 + 5 × 1000 + 0 × 100 + 9 × 10 + 4 × 1 = 35094. = Thirty Five Thousand and Ninety Four. 35094 is the natural number that is succeeded by 35093 and preceded by 35095. 35094 in words – Thirty Five Thousand and Ninety Four. Is 35094 an odd number? – No. Is 35094 an even number? – Yes. Is 35094 a perfect square number? – No. Is 35094 a perfect cube number? – No. Is 35094 a prime number? – No. Is 35094 a composite number? – Yes. ## Solved Example 1. Write the number 35094 in expanded form Solution: 3 x 10000 + 5 x 1000 + 0 x 100 + 9 x 10 + 4 x 1 We can write 35094 = 30000 + 5000 + 000 + 90 + 4 = 3 x 10000 + 5 x 1000 + 0 x 100 + 9 x 10 + 4 x 1. ## Frequently Asked Questions on 35094 in words Q1 ### How to write the number 35094 in words? 35094 in words is written as Thirty Five Thousand and Ninety Four. Q2 ### Is 35094 a perfect square number? No. 35094 is not a perfect square number. Q3 ### Is 35094 a prime number? No. 35094 is not a prime number.
• ### Unit 7: Work and Energy Energy describes the capacity of a physical system to perform work. It plays an essential role in everyday events and scientific phenomena. You can probably name many forms of energy: from the energy our food provides us, to the energy that runs our cars, to the sunlight that warms us on the beach. Not only does energy have many interesting forms, but it is involved in almost all phenomena and is one of the most important concepts of physics. Energy can change forms, but it cannot appear from nothing or disappear without a trace. Thus, energy is one of a handful of physical quantities that we say is conserved. Completing this unit should take you approximately 5 hours. • ### 7.1: Calculating Work and Force Work is done on a system when a constant applied force causes the system to be displaced or moved in the direction of the applied force. We can describe work using the equation $W = Fd\cos\theta$, where $F$ is force, $d$ is displacement, and $\theta$ (the Greek letter theta) is the angle between $F$ and $d$. From the equation for work, we can see that the unit for work must be the Newton-meter: the unit for force is the Newton and the unit for displacement (distance) is the meter. We define the Newton-meter as the unit joule. Consequently, we use joules as the unit for work and energy. • ### 7.2: Work, Potential Energy, and Linear Kinetic Energy We define kinetic energy as the energy associated with motion. We calculate kinetic energy as ${KE} = \frac{1}{2}mv^2$. When work is done on a system, energy is transferred to the system. We define net work as the total of all work done on a system by all external forces. We can think of the sum of all the external forces acting on a system as a net force, or $F_{\mathrm{net}}$. We can write the equation for net work in a similar way to how we wrote the equation for work earlier: $W_{\mathrm{net}} = F_{\mathrm{net}}d\cos\theta$, where $W_{\mathrm{net}}$ is net work, $F_{\mathrm{net}}$ is net force, $d$ is displacement, and $\theta$ is the angle between force and displacement. • ### 7.3: Conservative Forces and Potential Energy A non-conservative force is a force that depends on the path an object takes. In other words, a non-conservative force depends on how an object got from its initial state to its final state. Non-conservative forces change the amount of mechanical energy in a system. This differs from conservative forces, which do not depend on the path taken from initial to final state, and do not change the amount of mechanical energy in a system. An important example of a non-conservative force is friction. We know that friction is the force between two surfaces. We see friction when rolling a ball on a carpet versus a hardwood floor. The ball rolls farther on the hardwood floor than it does on a carpet. This is because the fuzzy carpet has more friction than the smooth hardwood. Friction converts some of the kinetic energy of the ball to thermal energy, or heat. As kinetic energy is converted to thermal energy, the balls slows to a stop. On the other hand, a conservative force is a force which does work that only depends on the beginning point and the end point of the system. The work done by a conservative force does not depend on the path the system takes to get from beginning to end. Conservative forces exist in ideal systems with no friction. An idealized spring that does not experience friction would be an example of conservative forces. • ### 7.4: Conservation of Energy The Law of Conservation of Energy states that the total energy in any process is constant. Energy can be transformed between different forms, and energy can be transferred between objects. However, energy cannot be created or destroyed. This is a broader law than the conservation of mechanical energy because this applies to all energy, not just energy when only conservative forces are applied. We can write the Law of Conservation of Energy as $KE + PE + OE = \mathrm{Constant}$ or as $KE_{i} + PE_{i} + OE_{i} = KE_{f} + PE_{f} + OE_{f}$ In the second equation, the $KE_{i}$, $PE_{i}$, and $OE_{i}$ are initial conditions and $KE_{f}$, $PE_{f}$, and $OE_{f}$ are final conditions. The new term, $OE$, is other energy. This is a collected term for all forms of energy that are not kinetic energy or potential energy. Other forms of energy include: thermal energy (heat), nuclear energy (used in nuclear power plants), electrical energy (used to power electronics), radiant energy (light), and chemical energy (energy from chemical reactions). When solving Conservation of Energy problems, it is important to identify the system of interest, and all forms of energy that can occur in the system. To do this, we need to first identify all forces acting on the system. Then, we can plug equations for different types of energy into the Law of Conservation of Energy equation to solve for the unknown in the problem. • ### 7.5: Rotational Kinetic Energy Why do tornadoes spin so rapidly? The answer is that the air masses that produce tornadoes are themselves rotating, and when the radii of the air masses decrease, their rate of rotation increases. An ice skater increases their spin in an exactly analogous way. The skater starts their rotation with outstretched limbs and increases their spin by pulling them in toward their body. The same physics describes the spin of a skater and the wrenching force of a tornado. Clearly, force, energy, and power are associated with rotational motion. We cover these and other aspects of rotational motion in this unit. We will see that important aspects of rotational motion have already been defined for linear motion or have exact analogs in linear motion. We can write an equation for the rotational kinetic energy (the energy of rotational motion) as: $KE_{\mathrm{rot}}=\frac{1}{2}I \omega^{2}$ • ### 7.6: Power We define power as the rate at which work is done. We can write this as $P = \frac{W}{\Delta t}$, where $w$ is work and $\Delta t$ is the duration of the work being done. The unit for power is the watt, W. One watt equals one joule per second. Higher power means more work is done in a shorter time. This also means that more energy is given off in a shorter time. For example, a 60 W light bulb uses 60 J of work in a second, and also gives off 60 J of radiant and heat energy every second.
# 9.2: Pressure [ "article:topic", "ChemPrime", "pressure", "weight", "atmosphere", "pascal", "newton", "atmospheric pressure", "standard atmosphere", "authorname:chemprime", "showtoc:no" ] You are probably familiar with the general idea of pressure from experiences in pumping tires or squeezing balloons. A gas exerts force on any surface that it contacts. The force per unit surface area is called the pressure and is represented by P. The symbols F and A represent force and area, respectively. On the image below, a force is pushing down on the circular area of a barometer. The pressure is then the amount of force pushing on a unit area of the circle of the barometer. $\text{Pressure}=\frac{\text{force}}{\text{area}}\text{ }P=\frac{F}{A}$ As a simple example of pressure, consider a rectangular block of lead which measures 20.0 cm by 50.0 cm by 100.0 cm (Fig. $$\PageIndex{1}$$ ). The volume V of the block is 1.00 × 105 cm3, and since the density ρ of Pb is 11.35 g cm–3, the mass m is $m=V\rho =\text{1}\text{.00 }\times \text{ 10}^{\text{5}}\text{ cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{11}\text{.35 g}}{\text{1 cm}^{\text{3}}}\text{ }\label{1} =\text{1}\text{.135 }\times \text{ 10}^{\text{6}}\text{ g}=\text{1}\text{.135 }\times \text{ 10}^{\text{3}}\text{ kg}$ Figure $$\PageIndex{1}$$ When the block stands upright, the weight of the block (11.1 kN) is distributed over an area of 0.1 m3. Laid flat, this same force is now exerted over an area 5 times larger (0.5 m3), exerting a pressure 5 times smaller than before, even though the weight of the block remains the same. According to the second law of motion, discovered by British physicist Isaac Newton, the force on an object is the product of the mass of the object and its acceleration a: $F = ma\label{4}$ At the surface of the earth, the acceleration of gravity is 9.81 m s–2. Substituting the mass of the lead block into Eq. $$\ref{4}$$, we have $F = 1.135 \times 10^{3} \text{ kg} \times \text{ m }\text{s}^{\text{-2}} = 11.13 \times 10 \text{ kg} \text{ m}\text{ s}^\text{-2}$ The units kilogram meter per square second are given the name newton in the International System and abbreviated N. Thus the force which gravity exerts on the lead block (the weight of the block) is 11.13 × 103 N. A block that is resting on the floor will always exert a downward force of 11.13 kN. The pressure exerted on the floor depends on the area over which this force is exerted. If the block rests on the 20.0 cm by 50.0 cm side (Fig. 9.2a), its weight is distributed over an area of 20.0 cm × 50.0 cm = 1000 cm3. Thus: $P=\frac{F}{A}=\frac{\text{11}\text{.13 kN}}{\text{1000 cm}^{\text{2}}}=\frac{\text{11}\text{.13 kN}}{\text{1000 cm}^{\text{2}}}\text{ }\times \text{ }\left( \frac{\text{100 cm}}{\text{1 m}} \right)^{\text{2}} =\frac{\text{11}\text{.13 kN}}{\text{10}^{\text{3}}\text{ cm}^{\text{2}}}=\frac{\text{10}^{\text{4}}\text{ cm}^{\text{2}}}{\text{1 m}^{\text{2}}}=\text{111}\text{.3 }\frac{\text{kN}}{\text{m}^{\text{2}}} =\text{111}\text{.3 }\times \text{ 10}^{\text{3}}\text{ N m}^{-\text{2}}$ Thus we see that pressure can be measured in units of newtons (force) per square meter (area). The units newton per square meter are used in the International System to measure pressure, and they are given the name pascal (abbreviated Pa). Like the newton, the pascal honors a famous scientist, in this case Blaise Pascal (1623 to 1662), one of the earliest investigators of the pressure of liquids and gases. If the lead block is laid on its side (Fig. 1b), the pressure is altered. The area of contact with the floor is now 50.0 cm × 100.0 cm = 5000 cm2, and so $P=\frac{F}{A}=\frac{\text{11}\text{.13 }\times \text{ 10}^{\text{3}}\text{ N}}{\text{5000 cm}^{\text{2}}}=\frac{\text{11}\text{.13 }\times \text{ 10}^{\text{3}}\text{ N}}{\text{0}\text{.500 m}^{\text{2}}} =\text{22}\text{.26 }\times \text{ 10}^{\text{3}}\text{ N m}^{-\text{2}}=\text{22}\text{.26 kPa}$ When the block is lying flat, its pressure on the floor (22.26 kPa) is only one-fifth as great as the pressure (111.3 kPa) when it stands on end. This is because the area of contact is 5 times larger. The air surrounding the earth is pulled toward the surface by gravity in the same way as the lead block we have been discussing. Consequently the air also exerts a pressure on the surface. This is called atmospheric pressure. The following video shows the "power" of atmospheric pressure. A metal can full of water is heated until the water inside boils, creating a high internal pressure. The can is the put upside down into a bowl of cold ice water, causing the formerly hot water vapor to cool and decrease in volume. This cooling causes a decrease in the internal pressure of the can. The lower pressure exerts less force on the can and can no longer counter the atmospheric pressure coming from the outside of the can, which pushes inward, crushing the can. Because winds may add more air or take some away from the vertical column above a given area on the surface, atmospheric pressure will vary above and below the result obtained in Example 9.1. Pressure also decreases as one moves to higher altitudes. The tops of the Himalayas, the highest mountains in the world at about 8000 m (almost 5 miles), are above more than half the atmosphere. The lower pressure at such heights makes breathing very difficult—even the slightest exertion leaves one panting and weak. For this reason jet aircraft, which routinely fly at altitudes of 8 to 10 km, have equipment to maintain air pressure in their cabins artificially. It is often convenient to express pressure using a unit which is about the same as the average atmospheric pressure at sea level. As we saw in Example 1, atmospheric pressure is about 101 kPa, and the standard atmosphere (abbreviated atm) is defined as exactly 101.325 kPa. Since this unit is often used, it is useful to remember that $1 \text{ atm} = 101.325 \text{ kPa}$ Example $$\PageIndex{1}$$: Atmospheric Pressure The total mass of air directly above a 30 cm by 140 cm section of the Atlantic Ocean was 4.34 × 103 kg on July 27, 1977. Calculate the pressure exerted on the surface of the water by the atmosphere. Solution First calculate the force of gravitational attraction on the air: $F = ma = 4.34 \times 10^{3} \text{ kg} \times 9.81 \text{ m }\text{s}^{\text{-2}} = 4.26 \times 10^{4} \text{ kg m s}^\text{-2} = 4.26 \times 10^{4} \text{ N}$ The area is $A=\text{30 cm }\times \text{ 140 cm}=\text{4200 cm}^{\text{2}}\text{ }\times \text{ }\left( \frac{\text{1 m}}{\text{100 cm}} \right)^{\text{2}}\text{ }=\text{0}\text{.42 m}^{\text{2}}$ Thus the pressure is $P=\frac{F}{A}=\frac{\text{4}\text{.26 }\times \text{ 10}^{\text{4}}\text{ N}}{\text{0}\text{.42 m}^{\text{2}}}=\text{1}\text{.01 }\times \text{ 10}^{\text{5}}\text{ Pa}=\text{101 kPa}$
# 1.2: Square Root Property, Complete the Square, and The Quadratic Formula $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ $$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$ $$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$ $$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vectorC}[1]{\textbf{#1}}$$ $$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$ $$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$ $$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$ $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left\|#1\right\|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$ ## Solve Quadratic Equations by Using the Square Root Property A quadratic equation in standard form is $$a x ^ { 2 } + b x + c = 0$$ where $$a, b$$, and $$c$$ are real numbers and $$a ≠ 0$$. Quadratic equations can have two real solutions, one real solution, or no real solution—in which case there will be two complex solutions. Reviewing a few definitions pertaining to complex numbers: • A complex number is a number in the form $$a+bi$$ where $$a$$ and $$b$$ are real numbers. The real part is $$a$$ and the imaginary part is $$bi$$. • The imaginary unit is denoted as $$i$$ and is defined as $$i=\sqrt{-1}$$ and so $$i^2=-1$$. • The square root of a negative number should always be written in its complex number form: $$\sqrt{-b} = i \sqrt{b}$$ When it is factorable, a quadratic equation can be solved by factoring. In this section, an approach to solving the special case $$a x ^ { 2 } + c = 0$$ where there is no linear term (i.e. $$b=0$$), is examined. As an example, $$4x^{2} − 9 = 0$$ can be solved by factoring as follows: \begin{aligned} 4 x ^ { 2 } - 9 & = 0 \\ ( 2 x + 3 ) ( 2 x - 3 ) & = 0 \end{aligned} $$\begin{array} { c } { 2 x + 3 = 0 \quad \text { or }\quad 2 x - 3 = 0 } \\ { 2 x = - 3 \quad \quad\:\:\:2 x = 3 } \\ { x = - \frac { 3 } { 2 } \quad\quad\:\:\:\: x = \frac { 3 } { 2 } } \end{array}$$ The solution set is $$\Big\{ ± \dfrac{3}{2} \Big\}$$. Here we use $$±$$ to write the two solutions in a more compact form. An alternative method that can be used to more easily solve this equation comes from first isolating $$x^{2}$$ and then taking the square root of both sides of the equation. Besides its simplicity, this method allows us to solve equations that do not factor. Continuing the above example to illustrate this alternative approach: \begin{aligned} 4 x ^ { 2 } - 9 & = 0 \\ 4 x ^ { 2 } & = 9 && \text{First, isolate the squared term}\\ x ^ { 2 } & = \dfrac { 9 } { 4 } \\ \sqrt { x ^ { 2 } } & = \pm \sqrt { \dfrac { 9 } { 4 } } && \text{Take the square root of both sides of the equal sign. Always remember the } \pm \text{ !!! } \\ x & = \pm \frac { 3 } { 2 } && \text{Simplify the radical} \end{aligned} In summary, when there is no linear term in a quadratic equation, one method to solve it is to use the square root property. In this approach, the $$x^2$$ term (or more generally the squared term) is isolated first, and then the square root of both sides of the equal sign is taken. The Square Root Property Given an algebraic expression $$u$$, and a nonzero real number $$k$$ , then the equation $$u^2=k$$ has exactly two solutions. If $$u^2 = k$$, then $$u = \pm \sqrt{k}$$ which is equivalent to $$u=\sqrt{k}$$ and $$u=-\sqrt{k}$$ How to: Use the Square Root Property to Solve an Equation. This method can only be used on equations that have a squared expression and a constant term, but no linear term. 1. Isolate the squared expression $$u^2$$ on one side of the equal sign. Here $$u$$ could simply be a single variable like $$x$$, or it could be an expression involving $$x$$. 2. Take the square root of both sides of the equation, putting a $$±$$ sign before the expression on the side opposite the squared expression. 3. Simplify the numbers on the side with the $$±$$ sign. Example $$\PageIndex{1}$$: Solve: $$9 x ^ { 2 } - 8 = 0$$. Solution \begin{aligned} 9 x ^ { 2 } - 8 & = 0 && \text {The Square Root Property can be used} \\ 9 x ^ { 2 } & = 8 && \text{Isolate the squared term} \\ x ^ { 2 } & = \frac { 8 } { 9 } \\ x & = \pm \sqrt { \frac { 8 } { 9 } } && \text{Square root both sides. Remember the } \pm \text{ !!!}\\ & = \pm \frac { 2 \sqrt { 2 } } { 3 } && \text{Simplify. } \quad \textbf{Solution Set: } \Big\{ ± \frac { 2 \sqrt { 2 } } { 3 } \Big\} \end{aligned} For completeness, these two solutions can be checked. Check $$x = - \frac { 2 \sqrt { 2 } } { 3 }$$ Check $$x = \frac { 2 \sqrt { 2 } } { 3 }$$ \begin{aligned} 9 x ^ { 2 } - 8 & = 0 \\ 9 \left(\color{Cerulean}{ - \frac { 2 \sqrt { 2 } } { 3 }} \right) ^ { \color{black}{2} } - 8 & = 0 \\ 9 \left( \frac { 4(2) } { 9 } \right) - 8 & = 0 \\ 8 - 8 & = 0 \\ 0 & = 0\:\:\color{Cerulean}{✓} \end{aligned} \begin{aligned} 9 x ^ { 2 } - 8 & = 0 \\ 9 \left(\color{Cerulean}{ \frac { 2 \sqrt { 2 } } { 3 }} \right) ^ { \color{black}{2} } - 8 & = 0 \\ 9 \left( \frac { 4(2) } { 9 } \right) - 8 & = 0 \\ 8 - 8 & = 0 \\ 0 & = 0\:\:\color{Cerulean}{✓} \end{aligned} Sometimes quadratic equations have no real solution. In this case, the solutions will be complex numbers. Example $$\PageIndex{2}$$: Solve: $$x^{2}+31=6$$. Solution \begin{aligned} x ^ { 2 } + 31 & = 6 && \text {The Square Root Property can be used}\\ x ^ { 2 } & = -25 && \text{Isolate the squared term}\\ x & = \pm \sqrt { - 25 } && \text{Square root both sides. Remember the } \pm \text{ !!!}\\ & = \pm \sqrt {- 1} \cdot \sqrt{25} &&\text{Simplify. } \\ & = \pm i \cdot 5 \\ & = \pm 5 i &&\textbf{Solution Set: } \{ ± 5i \}\end{aligned} The Square Root Property can also be used on equations that have a squared expression rather than just a simple squared variable. Example $$\PageIndex{3}$$: Solve $$( x + 5 ) ^ { 2 } = 9$$. Solution \begin{aligned} ( x + 5 ) ^ { 2 } & = 9 && \text {The Square Root Property can be used. The squared term is already isolated.}\\ x + 5 & = \pm \sqrt { 9 } && \text{Square root both sides. Remember the } \pm \text{ !!!}\\ x + 5 & = \pm 3 &&\text{Simplify. }\\ x & = - 5 \pm 3 \end{aligned} At this point, separate the “plus or minus” into two equations and solve each individually. $$\begin{array} { l } { x = - 5 + 3 \quad\text { or } \quad x = - 5 - 3 } \\ { x = - 2 \quad\quad\quad \quad\quad x = - 8 } \\ {\textbf{Solution Set: { -2, -8 } } } \end{array}$$ In addition to fewer steps, this method allows us to solve equations that do not factor. Example $$\PageIndex{4}$$: Solve: $$2 ( x - 2 ) ^ { 2 } - 1 = 4$$. Solution \begin{aligned} 2 ( x - 2 ) ^ { 2 } - 1 & = 4 && \text {The Square Root Property can be used}\\ 2 ( x - 2 ) ^ { 2 } & = 5 && \text{Isolate the squared term}\\ ( x - 2 ) ^ { 2 } & = \frac { 5 } { 2 } \\ x - 2 & = \pm \sqrt { \frac { 5 } { 2 } } && \text{Square root both sides. Remember the } \pm \text{ !!!}\\ x & = 2 \pm \frac { \sqrt { 5 } } { \sqrt { 2 } } \cdot \color{Cerulean}{ \frac { \sqrt { 2 } } { \sqrt { 2 } } } &&\text{Rationalize the denominator.} \\ x & = 2 \pm \frac { \sqrt { 10 } } { 2 } &&\text{Simplify. }\\ x & = \frac { 4 \pm \sqrt { 10 } } { 2 } &&\quad \textbf{Solution Set: } \Bigg\{ \frac { 4 - \sqrt { 10 } } { 2 }, \frac { 4 + \sqrt { 10 } } { 2 }\Bigg\} \end{aligned} Example $$\PageIndex{5}$$: Solving a Quadratic Equation Using the Square Root Property Solve $$25x^4-9=7$$. Solution First, isolate the $$x^4$$ term. Then take the square root of both sides. \begin{align} 25x^4-9&= 7 && \text {The Square Root Property can be used}\\ 25x^4&= 16 && \text{Isolate the squared term}\\ (x^2)^2&= \frac{16}{25} \\ x^2 &= \pm \dfrac{16}{25} && \text{Square root both sides. Remember the } \pm \text{ !!!}\\ x^2 &= \pm \dfrac{4}{5} &&\text{Simplify. } \end{align} $\begin{array} {cccl} x^2=\dfrac{4}{5} & \text{ and } & x^2=-\dfrac{4}{5} & \text{Isolate the squared term} \\ x=\pm \sqrt{\dfrac{4}{5}} && x=\pm \sqrt{\dfrac{-4}{5}} & \text{Square root both sides. Remember the } \pm \text{ !!!}\\ x=\pm \dfrac {\sqrt{4}} {\sqrt{5}} && x=\pm \dfrac {\sqrt{-4}} {\sqrt{5}} & \text{Simplify. } \\ x=\pm \dfrac {2} {\sqrt{5}} \dfrac{\sqrt{5}}{\sqrt{5}} && x=\pm \dfrac {2i} {\sqrt{5}} \dfrac{\sqrt{5}}{\sqrt{5}} & \\ x=\pm \dfrac {2\sqrt{5}}{5} && x=\pm \dfrac {2i\sqrt{5}}{5} & \quad \textbf{Solution Set: } \Big\{ \dfrac {2\sqrt{5}}{5}, -\dfrac {2\sqrt{5}}{5}, \dfrac {2i\sqrt{5}}{5}, -\dfrac {2i\sqrt{5}}{5} \Big\} \\ \end{array}$ Try It $$\PageIndex{6}$$ Solve. a) $$2x^{2}+3=0$$ b) $$3{(x−4)}^2=15$$ c) $$2 ( 3 x - 1 ) ^ { 2 } + 9 = 0$$ a) $$\pm \dfrac { \sqrt { 6 } } { 2 } i$$ b) $$x=4±\sqrt{5}$$ c) $$\dfrac { 1 }{ 3 } \pm \dfrac { \sqrt{2}}{2} i$$ ## Completing the Square Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, $$a$$, must equal $$1$$. If it is not, then divide the entire equation by $$a$$ before beginning the complete the square process. Then, we can use the following procedures to solve a quadratic equation by completing the square. We will use the example $$x^2+6x+1=0$$ to illustrate each step. This is a quadratic equation that cannot be factored and with $$a=1$$. \begin{align} x^2+6x+1&= 0 && \text{Put the constant on the right of the equal sign; the } x \text{ and } x^2 \text{ terms on the other side}\\ x^2+6x&= -1 && \text{Isolate the } x \text{ and } x^2 \text{ terms.}\\ x^2+6x + \large\Box &= -1 + \large\Box && \text{Add a constant } {\Box} \text{ to create a trinomial that is also a binomial square: } (x +\bigcirc )^2\\ (x +\bigcirc )^2 &= -1 + \large\Box && \text{To accomplish this, }(x +{\color{Cerulean}{\bigcirc }})^2 = x^2+{\color{Cerulean}{2\bigcirc x}} + \bigcirc ^2 = x^2 + 6x + \Box \\ & && \text{The binomial square constant, } \bigcirc \text{, is always HALF the }x \text{ coefficient}\\ (x+3)^2 &= -1+ 9 && \text{The number added to both sides of the equal sign, } \Box \text{, is the square of that constant: } \square = \bigcirc ^2 \\ (x+3)^2 &=8 && \text{Now use the square root property to solve the resulting equation.}\\ \sqrt{{(x+3)}^2}&= \pm \sqrt{8} \\ x+3&= \pm 2 \sqrt{2}\\ x&= -3 \pm 2\sqrt{2} && \textbf{Solution Set: } \{ −3+2\sqrt{2}, −3−2\sqrt{2} \} \end{align} How to: Use Completing the Square to Solve an Equation. 1. The coefficient of the $$x^2$$ term MUST be 1. If it is not 1, divide both sides of the equal sign by the coefficient of the $$x^2$$ term to make it 1. 2. Isolate the variable ( $$x$$ and $$x^2$$ ) terms on one side of the equal sign. 3. Add a constant to both sides of the equal sign that combined with the ($$x$$ and $$x^2$$ ) terms will create a trinomial that is a perfect square binomial. 1. Add the square of half the coefficient of $$x$$ to both sides of the equation. 2. Factor the resulting trinomial into a binomial square. 4. Use the Square Root Property to solve for $$x$$. Remember the $$±$$ sign !! This technique can be used to solve ANY quadratic equation, whereas factoring works only some of the time. Example $$\PageIndex{7}$$: Solve by completing the square: $$x ^ { 2 } - 8 x - 2 = 0$$. Solution \begin{aligned} x ^ { 2 } - 8 x - 2 &= 0 && \text{The leading coefficient is 1}\\ x ^ { 2 } - 8 x &= 2 && \text{Isolate the variable terms} \\ x ^ { 2 } - 8 x +4^2 &= 2 + 4^2 && \text{Add a constant - the square of half the coefficient of } x \\ x ^ { 2 } - 8 x +16 &= 2 + 16 \\ (x-4)^2 &= 18 && \text{Factor. } \\ \sqrt{(x-4)^2} &= \pm \sqrt{18} && \text{Solve using the square root property } \\ x-4 &= \pm 3\sqrt{2} \\ x &= 4 \pm 3\sqrt{2} && \textbf{Solution Set: } \{ 4+3\sqrt{2}, 4−3\sqrt{2} \}\\ \end{aligned} Example $$\PageIndex{8}$$: Solve by completing the square: $$x ^ { 2 } - 10 x + 26 = 0$$. Solution \begin{aligned} x ^ { 2 } - 10 x + 26 & = 0 && \text{The leading coefficient is 1} \\ x ^ { 2 } - 10 x & = - 26 && \text{Isolate the variable terms} \\ x ^ { 2 } - 10 x + \color{Cerulean}{25} & \color{black}{=} - 26 + \color{Cerulean}{25} && \text{Add a constant - the square of half the coefficient of } x, ( - 5 ) ^ { 2 } = \color{Cerulean}{25} \\ (x-5)^2 & =-1 && \text{Factor into a binomial square}\\ \sqrt{(x-5)^2} & = \pm \sqrt{-1} && \text{Solve using the square root property } \\ x - 5 & = \pm i \\ x & = 5 \pm i && \textbf{Solution Set: } \{ 5+i, 5-i \} \end{aligned} Example $$\PageIndex{9}$$: Solve by completing the square: $$x ^ { 2 } + 2 x - 48 = 0$$. Solution \begin{aligned} x ^ { 2 } + 2 x - 48 &= 0 && \text{The leading coefficient is 1} \\ x ^ { 2 } + 2 x &= 48 && \text{Isolate the variable terms} \\ x ^ { 2 } + 2 x + {\color{Cerulean}{1}} &= 48 +{\color{Cerulean}{1}} && \text{Add a constant - the square of half the coefficient of } x, ( 1 ) ^ { 2 } = \color{Cerulean}{1}\\ ( x + 1 ) ( x + 1 ) &= 49 && \text{Factor into a binomial square}\\ ( x + 1 ) ^ { 2 } &= 49 && \text{Solve using the square root property } \\ x + 1 &= \pm \sqrt { 49 } \\ x + 1 &= \pm 7 \\ x &= - 1 \pm 7 && x = -1+7 \text{ or } x=-1-7 \quad \textbf{Solution Set: } \{ 6, -8 \} \end{aligned} When the coefficient of $$x$$ is not divisible by $$2$$, the constant added in the complete the square process will be a fraction. Example $$\PageIndex{10}$$: Solve by completing the square: $$x^{2} + 3x+4=0$$. Solution \begin{aligned} x ^ { 2 } + 3 x + 4 &= 0 && \text{The leading coefficient is 1} \\ x ^ { 2 } + 3 x &= - 4 && \text{Isolate the variable terms} \\ x ^ { 2 } + 3 x +\color{Cerulean}{\frac { 9 } { 4 }} & \color{black}{=} - 4 + \color{Cerulean}{\frac { 9 } { 4 }} && \text{Add a constant - the square of half the coefficient of } x, \left( \frac { 3 } { 2 } \right) ^ { 2 } = \color{Cerulean}{\frac { 9 } { 4 }}\\ \left( x + \frac { 3 } { 2 } \right) \left( x + \frac { 3 } { 2 } \right) & = \frac { - 16 } { 4 } + \frac { 9 } { 4 } && \text{Factor into a binomial square}\\ \left( x + \frac { 3 } { 2 } \right) ^ { 2 } & = \frac { - 7 } { 4 } && \text{Solve using the square root property } \\ x + \frac { 3 } { 2 } & = \pm \sqrt { \frac { - 1 \cdot 7 } { 4 } } \\ x + \frac { 3 } { 2 } & = \pm \frac { i \sqrt { 7 } } { 2 } \\ x & = - \frac { 3 } { 2 } \pm \frac { \sqrt { 7 } } { 2 } i &&\quad \textbf{Solution Set: } \Bigg\{ - \frac { 3 } { 2 } \pm \frac { \sqrt { 7 } } { 2 } i \Bigg\} \end{aligned} So far, all of the examples have had a leading coefficient of $$1$$. If this is not the case, remove it. This can be done by dividing both sides of the equal sign by the leading coefficient before completing the square. Example $$\PageIndex{11}$$: Solve by completing the square: $$3 x ^ 2 -12 x +17 =0$$. Solution \begin{aligned} 3 x ^ 2 -12 x +17 &= 0 && \text{The leading coefficient is NOT 1} \\ \frac { 3 x ^ 2 -12 x +17 } {3 } &= \frac { 0 } { 3} && \text{Divide by the coefficient to remove it from the } x^2 \text{ term} \\ \frac { 3 x ^ { 2 } } { 3 } - \frac { 12 x } { 3} + \frac { 17 } { 3 } &= 0 \\ x ^ { 2 } - 4x &= \frac { -17 } { 3 } && \text{Isolate the variable terms} \\ x ^ { 2 } - 4 x + {\color{Cerulean}{4}} &= \frac { -17 } { 3 } + {\color{Cerulean}{4}} && \text{Add a constant - the square of half the coefficient of } x, \left( \frac { 4} { 2 } \right) ^ { 2 } = \color{Cerulean}{4} \\ (x-2)(x-2) &= \frac { -17 } { 3} + \frac { 12 } { 3 } && \text{Factor into a binomial square}\\ (x-2)^2 & = \frac { -5 } { 3 } \\ x -2 & = \pm \sqrt { \frac { -5 } { 3 } } && \text{Solve using the square root property } \\ x & = 2 \pm \frac { i \sqrt { 5 } } { \sqrt{3} } \cdot \frac{\sqrt{3}}{\sqrt{3}}\\ x & = \frac { 6 } { 3 } \pm \frac { i\sqrt { 15 } } { 3 } &&\quad \textbf{Solution Set: } \Bigg\{ \frac { 6 \pm i\sqrt { 15 } } { 3 } \Bigg\}\\ \end{aligned} Example $$\PageIndex{12}$$: Solve by completing the square: $$2 x ^ { 2 } + 5 x - 1 = 0$$. Solution \begin{aligned} 2 x ^ { 2 } + 5 x - 1 &= 0 && \text{The leading coefficient is NOT 1} \\ \frac { 2 x ^ { 2 } + 5 x - 1 } { 2 } &= \frac { 0 } { 2} && \text{Divide by the coefficient to remove it from the } x^2 \text{ term} \\ \frac { 2 x ^ { 2 } } { 2 } + \frac { 5 x } { 2 } - \frac { 1 } { 2 } &= 0 \\ x ^ { 2 } + \frac { 5 } { 2 } x &= \frac { 1 } { 2 } && \text{Isolate the variable terms} \\ x ^ { 2 } + \frac { 5 } { 2 } x + {\color{Cerulean}{\frac { 25 } { 16 }}} &= \frac { 1 } { 2 } + {\color{Cerulean}{\frac { 25 } { 16 }}} && \text{Add a constant - the square of half the coefficient of } x, \left( \frac { 5 } { 4 } \right) ^ { 2 } = \color{Cerulean}{\frac { 25 } { 16 }} \\ \left( x + \frac { 5 } { 4 } \right) \left( x + \frac { 5 } { 4 } \right) &= \frac { 8 } { 16 } + \frac { 25 } { 16 } && \text{Factor into a binomial square}\\ \left( x + \frac { 5 } { 4 } \right) ^ { 2 } & = \frac { 33 } { 16 } \\ x + \frac { 5 } { 4 } & = \pm \sqrt { \frac { 33 } { 16 } } && \text{Solve using the square root property } \\ x + \frac { 5 } { 4 } & = \pm \frac { \sqrt { 33 } } { 4 } \\ x & = - \frac { 5 } { 4 } \pm \frac { \sqrt { 33 } } { 4 } &&\quad \textbf{Solution Set: } \Bigg\{ \frac { - 5 \pm \sqrt { 33 } } { 4 } \Bigg\}\\ \end{aligned} Try It $$\PageIndex{13}$$ Solve by completing the square. a) $$x^2−6x=13$$ b) $$x ^ { 2 } - 2 x - 17 = 0$$ c) $$3 x ^ { 2 } - 2 x + 1 = 0$$ a) $$x=3±\sqrt{22}$$ b) $$x = 1 \pm 3 \sqrt { 2 }$$ c) $$x = \frac { 1 } { 3 } \pm \frac { \sqrt { 2 } } { 3 } i$$ The fourth method of solving a quadratic equation is by using the quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number. We can derive the quadratic formula by completing the square. Given $$ax^2+bx+c=0, a \ne 0$$, we will complete the square as follows: \begin{aligned} ax^2+bx+c &= 0 && \text{The leading coefficient is NOT 1} \\ x^2+\dfrac{b}{a}x + \dfrac{c}{a} &= 0 && \text{Divide by the coefficient to remove it from the } x^2 \text{ term} \\ x^2 + \dfrac{b}{a}x &= −\dfrac{c}{a} && \text{Isolate the variable terms} \\ x^2 + \dfrac{b}{a}x + {\color{Cerulean}{\frac {b^2} {4a^2}}} &= −\dfrac{c}{a} \cdot \frac{4a}{4a} + {\color{Cerulean}{\frac {b^2} {4a^2}}} && \text{Add a constant - the square of half the coefficient of } x, \left( \frac { b } { 2a } \right) ^ { 2 } = \color{Cerulean}{\frac {b^2} {4a^2}} \\ { \left( x+\dfrac{b}{2a} \right) }^2 &= \dfrac{b^2-4ac}{4a^2} && \text{Factor into a binomial square}\\ x+\dfrac{b}{2a} & = ±\sqrt{\dfrac{b^2-4ac}{4a^2}} && \text{Solve using the square root property } \\ x & = -\dfrac{b}{2a} ±\dfrac{\sqrt{b^2-4ac}}{\sqrt{4a^2}} &&\quad \textbf{Solution Set: } \Bigg\{ \dfrac{-b±\sqrt{b^2-4ac}}{2a} \Bigg\}\\ \end{aligned} Written in standard form, $$ax^2+bx+c=0$$ where $$a$$, $$b$$, and $$c$$ are real numbers and $$a≠0$$, any quadratic equation can be solved using the quadratic formula: $x=\dfrac{-b±\sqrt{b^2-4ac}}{2a} \nonumber$ How to: Use the Quadratic Formula to Solve an Equation. 1. Make sure the equation is in standard form: $$ax^2+bx+c=0$$. 2. Make note of the values of the coefficients and constant term, $$a$$, $$b$$, and $$c$$. 3. Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula. 4. Calculate and solve. Example $$\PageIndex{14}$$: Solve using the quadratic formula: $$2 x^{2}-7 x-15=0$$ Solution The coefficients are: $$a=2,b=-7,c=-15$$. Substitute these values into the quadratic formula and simplify. \begin{aligned} x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ &=\frac{-(\color{OliveGreen}{-7}\color{black}{)} \pm \sqrt{(\color{OliveGreen}{-7}\color{black}{)}^{2}-4(\color{OliveGreen}{2}\color{black}{)}(\color{OliveGreen}{-15}\color{black}{)}}}{2(\color{OliveGreen}{2}\color{black}{)}} \\ &=\frac{7 \pm \sqrt{49+120}}{4} \\ &=\frac{7 \pm \sqrt{169}}{4} \\ &=\frac{7 \pm 13}{4} \end{aligned} $$\textbf{Solution Set: } \Big\{ -\dfrac{3}{2}, 5 \Big\} \text{ because } x = \dfrac{7-13}{4}=\dfrac{-6}{4}=-\dfrac{3}{2} \text{ and } x =\dfrac{7+13}{4} = \dfrac{20}{4} =5$$ Example $$\PageIndex{15}$$: Solve using the quadratic formula: $$3 x^{2}+6 x-2=0$$. Solution The coefficients are: $$a=3 \quad b=6 \quad c=-2$$ \begin{aligned} x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ &=\frac{-(\color{OliveGreen}{6}\color{black}{)} \pm \sqrt{(\color{OliveGreen}{6}\color{black}{)}^{2}-4(\color{OliveGreen}{3}\color{black}{)}(\color{OliveGreen}{-2}\color{black}{)}}}{2(\color{OliveGreen}{3}\color{black}{)}} \\ &=\frac{-6 \pm \sqrt{36+24}}{6} \\ &=\frac{-6 \pm \sqrt{60}}{6} \\ &= \frac{-6 \pm 2 \sqrt{15}}{6} = \frac{\cancel{2}(-3 \pm \sqrt{15})}{\cancel{6}} = \frac{-3 \pm \sqrt{15}}{3} \\ &\qquad\qquad\qquad \text{ or } =\frac{-6}{6} \pm \frac{2\sqrt{15}}{6} =-1 \pm \frac{\sqrt{15}}{3} \end{aligned} Two ways the solution set can be written are: $$\Big\{ \dfrac{-3 \pm \sqrt{15}}{3} \Big\}$$ or $$\Big\{ -1 \pm \dfrac{\sqrt{15}}{3} \Big\}$$ Sometimes terms are missing. When this is the case, use $$0$$ as the coefficient. Also make sure the terms are written in descending powers of $$x$$ so the correct values of the parameters $$a, b,$$ and $$c$$ are used in the formula. Example $$\PageIndex{16}$$: Solve using the quadratic formula: $$45-x^{2}=0$$ Solution This equation is equivalent to $$-1 x^{2} + 0 x + 45=0$$ so $$a=-1 \quad b=0 \quad c=45$$ \begin{aligned} x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ &=\frac{-(\color{OliveGreen}{0}\color{black}{)} \pm \sqrt{(\color{OliveGreen}{0}\color{black}{)}^{2}-4(\color{OliveGreen}{-1}\color{black}{)}(\color{OliveGreen}{45}\color{black}{)}}}{2 (\color{OliveGreen}{-1}\color{black}{)}} \\ &=\frac{0 \pm \sqrt{0+180}}{-2} \\ &=\frac{\pm \sqrt{36(5)}}{-2} \\ &=\frac{\pm 6 \sqrt{5}}{2} \\ &=\pm 3 \sqrt{5} \quad \text{ Solution Set: } \{\pm 3 \sqrt{5} \}\end{aligned} Often solutions to quadratic equations are not real. Example $$\PageIndex{17}$$: Solve using the quadratic formula: $$x^{2}-4 x+29=0$$. Solution The coefficients are: $$a=1 \quad b=-4 \quad c=29$$ \begin{aligned} x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ &=\frac{-(\color{OliveGreen}{-4}\color{black}{)} \pm \sqrt{(\color{OliveGreen}{-4}\color{black}{)}^{2}-4(\color{OliveGreen}{1}\color{black}{)}(\color{OliveGreen}{29}\color{black}{)}}}{2 (\color{OliveGreen}{1}\color{black}{)}} \\ &=\frac{4 \pm \sqrt{16-116}}{2} \\ &=\frac{4 \pm \sqrt{-100}}{2} \\ &=\frac{4 \pm 10 i}{2} \\ &=\frac{4}{2} \pm \frac{10 i}{2} \\ &=2 \pm 5 i \qquad\qquad \text{Solution Set: } \{ 2 \pm 5 i \} \end{aligned} Example $$\PageIndex{18}$$: Use the quadratic formula to solve $$x^2+x+2=0$$. Solution The coefficients are: $$a=1, \; b=1, \;$$ and $$c=2$$. Substitute these values into the quadratic formula. \begin{align} x&= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\ &= \dfrac{-(1) \pm \sqrt{(1)^2-4(1)(2)}}{2(1)}\\ &= \dfrac{-1 \pm \sqrt{1-8}}{2}\\ &= \dfrac{-1 \pm \sqrt{-7}}{2}\\ &= \dfrac{-1 \pm i\sqrt{7}}{2} \qquad\qquad \text{Solution Set: } \Big\{ \ \dfrac{-1 \pm i\sqrt{7}}{2} \Big\} \end{align} Try It $$\PageIndex{20}$$ Solve the quadratic equation using the quadratic formula: $$9x^2+3x−2=0$$. $$x=-\dfrac{2}{3},x=\dfrac{1}{3}$$ ## Solving Cubic Equations If multiple roots and complex roots are counted, then the fundamental theorem of algebra implies that every polynomial with one variable will have as many roots as its degree. For example, we expect $$f (x) = x^{3} − 8$$ to have three roots. In other words, the equation $$x^{3}-8=0$$ should have three solutions. To find them one might first think of trying to extract the cube roots just as we did with square roots, \begin{aligned} x^{3}-8 &=0 \\ x^{3} &=8 \\ x &=\sqrt[3]{8} \\ x &=2 \end{aligned} As you can see, this leads to one solution, the real cube root. There should be two others; let’s try to find them. Example $$\PageIndex{19}$$: Find the set of all roots: $$f(x)=x^{3}-8$$. Solution Notice that the expression $$x^{3}-8$$ is a difference of cubes and recall that $$a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$$. Here $$a=x$$ and $$b=2$$ and we can write \begin{aligned} x^{3}-8 &=0 \\(x-2)\left(x^{2}+2 x+4\right) &=0 \end{aligned} Next apply the zero-product property and set each factor equal to zero. After setting the factors equal to zero we can then solve the resulting equation using the appropriate methods. \begin {array}{l c l} { x-2=0} &\text{or} & x^{2}+2x+4=0 \\ {x=2}& & { \begin{aligned} x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ &=\frac{-(2) \pm \sqrt{(2)^{2}-4 (1)(4)}}{2(1)} \\ &=\frac{-2 \pm \sqrt{-12}}{2} \\ &=\frac{-2 \pm 2 i \sqrt{3}}{2} \\ &=-1 \pm i \sqrt{3} & \text{Solution Set: } \{2,-1 \pm i \sqrt{3}\} \end{aligned} } \end{array} Using this method, we were able to obtain the set of all three roots $$\{2,-1 \pm i \sqrt{3}\}$$, one real and two complex. Sometimes factoring will produce a quadratic factor that needs to be solved using the quadratic formula or complete the square. Example $$\PageIndex{21}$$: Solve $$2x^4+2000x=0$$ Solution: Factor first. $$2x^4+2000x= 2x(x^3+1000) = 2x(x+10)(x^2-10x+100)$$ Use the zero factor property and complete the square on: $$\quad 2x(x+10)(x^2-10x+100) = 0$$ $$\begin{array} {lll} 2x=0 & (x+10) = 0 & (x^2-10x+100) = 0 \\ x=0& x=-10 & x^2-10x +\square = -100+\square \\ && (x-5)^2 = -100+25 \\ && x-5 = \pm \sqrt {-75} \\ && x = 5 \pm 5i\sqrt{3} \\ \end{array}$$ The solution set is $$\{ 0, -10, 5 \pm 5i\sqrt{3} \}$$ When an expression is both a difference of squares and a sum or difference of cubes, if factoring as a difference of squares is done first, a more complete factorization is obtained. For example, if given the equation $$64x^6-1=0$$ to solve, when it is first factored as a difference of squares as $$(8x^3-1) (8x^3+1)$$, and then as a difference of cubes, the solutions obtained are $$\pm \frac{1}{2}$$ and $$\pm \frac{1}{4}({1 \pm i\sqrt{3} })$$. In contrast, if it is first factored as a difference of cubes as $$(4x^2-1) (16x^4 +4x^2+1)$$, the solutions eventually obtained are $$\pm \frac{1}{2}$$ and $$\pm \frac{1}{4}\sqrt {-2 \pm 2i\sqrt{3} }$$. ## The Discriminant The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant, or the expression under the radical, $$b^2−4ac$$. The discriminant tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. The table below relates the value of the discriminant to the solutions of a quadratic equation. Value of Discriminant Results $$b^2−4ac=0$$ $$b^2−4ac>0$$, and is a perfect square $$b^2−4eac>0$$, and is not a perfect square $$b^2−4ac<0$$ One rational solution (double solution) Two rational solutions Two irrational solutions Two complex solutions The Discriminant For $$ax^2+bx+c=0$$, where $$a$$, $$b$$, and $$c$$ are real numbers, the discriminant is the expression under the radical in the quadratic formula: $$b^2−4ac$$. It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect. Example $$\PageIndex{22}$$: Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation Use the discriminant to find the nature of the solutions to the following quadratic equations: 1. $$x^2+4x+4=0$$ 2. $$8x^2+14x+3=0$$ 3. $$3x^2−5x−4=0$$ 4. $$3x^2−10x+15=0$$ Solution. Calculate the discriminant $$b^2−4ac$$ for each equation and state the expected type of solutions. a. $$\quad x^2+4x+4=0$$ $$\qquad b^2-4ac={(4)}^2-4(1)(4)=0$$ There will be one rational double solution. b. $$\quad 8x^2+14x+3=0$$ $$\qquad b^2-4ac={(14)}^2-4(8)(3)=100$$ As $$100$$ is a perfect square, there will be two rational solutions. c. $$\quad 3x^2−5x−4=0$$ $$\qquad b^2-4ac={(-5)}^2-4(3)(-4)=73$$ As $$73$$ is positive but not a perfect square, there will be two irrational solutions. d. $$\quad 3x^2−10x+15=0$$ $$\qquad b^2-4ac={(-10)}^2-4(3)(15)=-80$$ There will be two complex solutions. 1.2: Square Root Property, Complete the Square, and The Quadratic Formula is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
### Consecutive Numbers An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore. ### Roll These Dice Roll two red dice and a green dice. Add the two numbers on the red dice and take away the number on the green. What are all the different possible answers? ### Domino Square Use the 'double-3 down' dominoes to make a square so that each side has eight dots. # Magic Vs ##### Age 7 to 11Challenge Level Somin, Kim and Georgia from the American International School Chennai in India sent in the following: To do this problem, we started with the lowest circle on the V, because this would affect both arms. Then, we took the highest of the remaining numbers, and the lowest of the remaining numbers and added them together. We then compared the sum of those to the sum of the two middle/in-between remaining numbers. If the sums were the same, we entered the addins used to make those two sums into the 2 spots on each arm of the V. We know that we found every solution because when trying to mix up the pairs of numbers to make other sums, none of the sums were equal. The only way you could make another V is if you mirrored one we already made, or switched the places of the top and middle circles of each arm. A pattern we noticed during this project is that every even number in the sequence (the 2nd and 4th numbers, not 2 and 4, for example) did not work as a center number. The other numbers could not be paired to create equal sums. We are including a picture of a small picture we made representing our work. From Sudbury Primary School in Suffolk we had solutions sent in from Maisy, Jessica, Luke, Amelia, Molli & Amelia, Elena, Lili, Ben, Tashi, Megan C, Charlotte and Belle. Here is Maisy and Jessica's submission: We first found our first three Vs then realised we could find 8 solutions, by just changing one V. This also happened with the other two Vs. We multiplied 8 by 3 because our three vs had 8 solutions. Our answer was 24 solutions. Then we changed round the sides to get 14523. Afterwards, we just swapped one side then another till we couldn't swap it anymore without getting a solution we already had. Here is Mollie and Amelia's results: To solve all of the solutions,first we decided to put 1 as the bottom number and worked systematically to get all 8 solutions. Then, we picked then number 3 for the bottom number as we knew it couldn't be even. After finding 8 answers, we got 5 for the number at the bottom. We found all the answers for that number. We stopped here since we couldn't go above 5.There are 24 solutions. Matty from The British School Manila in the Philippines sent in the following exhaustive solution: If we have the numbers 1, 2, 3, 4 and 5, and our goal is to arrange all 5 numbers in a way such that the 2 'arms' have the same total. If we get the total of all the give numbers, we get 15. In the middle, we must have an odd number, because if we have an even, say 2, 15-2=13 and 13 is not divisible among the other 2 circles of arms. If we pick 1 as the middle, then each side is required to have a total of 7 on the other 2 circles. For 3, we need a total of 6 on the 2 circles, and for 5, we need a total of 5 on the other 2 circles. Thanks for the question, Nrich. (which was published at http://bit.ly/2pCieMH) 5,2,1,3,4 5,2,1,4,3 2,5,1,3,4 2,5,1,4,3 3,4,1,5,2 3,4,1,2,5 4,3,1,5,2 4,3,1,2,5 1,5,3,2,4 1,5,3,4,2 5,1,3,2,4 5,1,3,4,2 2,4,3,1,5 2,4,3,5,1 4,2,3,1,5 4,2,3,5,1 1,4,5,2,3 1,4,5,3,2 4,1,5,2,3 4,1,5,3,2 2,3,5,1,4 2,3,5,4,1 3,2,5,1,4 3,2,5,4,1 We have 24 different permutations. Here is the formula: Slot 1: 4 possibilities Slot 2: 1 possibility Slot 3: 2 possibilities Slot 4: 1 possibility Middle number: 3 possibilities (3 odd numbers between 1 to 5, inclusive) So 3(4121) = 24. We also had some very interesting solutions from North London Collegiate School that are certainly worth viewing here. magic Vs Y4.pdf magic Vs 3S group.pdf magic Vs Ronis group.pdf Thank you all so much for these solutions you certainly went about it in good ways to find how many there were.
# What is the second derivative of f(x) = ln x/x^2 ? Dec 19, 2015 $\frac{6 \ln \left(x\right) - 5}{{x}^{4}}$ #### Explanation: By the quotient rule, the first derivative is $f ' \left(x\right) = \frac{{x}^{2} \cdot \frac{1}{x} - \ln \left(x\right) \cdot 2 x}{{x}^{4}} = \frac{x - 2 x \ln \left(x\right)}{{x}^{4}} = \frac{1 - 2 \ln \left(x\right)}{{x}^{3}}$. Use the quotient rule again to find the second derivative: $f ' ' \left(x\right) = \frac{{x}^{3} \cdot - \frac{2}{x} - \left(1 - 2 \ln \left(x\right)\right) \cdot 3 {x}^{2}}{{x}^{6}}$ $= \frac{- 2 {x}^{2} - 3 {x}^{2} + 6 {x}^{2} \ln \left(x\right)}{{x}^{6}} = \frac{6 \ln \left(x\right) - 5}{{x}^{4}}$ For extra interest, the fact that $f ' \left(x\right) = \frac{1 - 2 \ln \left(x\right)}{{x}^{3}}$ implies that $f$ is increasing for $0 < x < {e}^{\frac{1}{2}} \approx 1.65$ and decreasing for $x > {e}^{\frac{1}{2}}$, with a local maximum value of $f \left({e}^{\frac{1}{2}}\right) = \frac{\frac{1}{2}}{{\left({e}^{\frac{1}{2}}\right)}^{2}} = \frac{1}{2 e} \approx 0.18$ at $x = {e}^{\frac{1}{2}} \approx 1.65$. (Make sure you check all this!) The fact that $f ' ' \left(x\right) = \frac{6 \ln \left(x\right) - 5}{{x}^{4}}$ implies that $f$ is concave down for $0 < x < {e}^{\frac{5}{6}} \approx 2.30$ and concave up for $x > {e}^{\frac{5}{6}}$, with an inflection point at $\left(x , y\right) = \left({e}^{\frac{5}{6}} , f \left({e}^{\frac{5}{6}}\right)\right) = \left({e}^{\frac{5}{6}} , \frac{\frac{5}{6}}{{\left({e}^{\frac{5}{6}}\right)}^{2}}\right) = \left({e}^{\frac{5}{6}} , \frac{5}{6 {e}^{\frac{5}{3}}}\right) \approx \left(2.30 , 0.16\right)$. (Make sure you check all this!) Here's the graph to allow you to see these features. graph{ln(x)/(x^2) [-0.562, 4.438, -0.77, 1.73]}
# 2014 AMC 8 Problems/Problem 21 ## Problem The $7$-digit numbers $\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}$ and $\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}$ are each multiples of $3$. Which of the following could be the value of $C$? $\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8$ ## Solution The sum of a number's digits $\mod{3}$ is congruent to the number $\pmod{3}$. $74A52B1 \mod{3}$ must be congruent to 0, since it is divisible by 3. Therefore, $7+4+A+5+2+B+1 \mod{3}$ is also congruent to 0. $7+4+5+2+1 \equiv 1 \pmod{3}$, so $A+B\equiv 2 \pmod{3}$. As we know, $326AB4C\equiv 0 \pmod{3}$, so $3+2+6+A+B+4+C =15+A+B+C\equiv 0 \pmod{3}$, and therefore $A+B+C\equiv 0 \pmod{3}$. We can substitute 2 for $A+B$, so $2+C\equiv 0 \pmod{3}$, and therefore $C\equiv 1\pmod{3}$. This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is $\boxed{\textbf{(A) }1}$.
# How do you use the remainder theorem to find P(c) p(x)=x^4+3x^3-4x^2+6x-31; 2? Jul 16, 2015 Divide $P \left(x\right)$ by $x - 2$, the remainder is $P \left(2\right)$ #### Explanation: $P \left(x\right) = {x}^{4} + 3 {x}^{3} - 4 {x}^{2} + 6 x - 31$. The Remainder Theorem tells us that if we divide $P \left(x\right)$, by $x - c$, then the remainder is $P \left(c\right)$. So we'll divide $P \left(x\right)$ by $x - 2$. You could use long division, but synthetic division requires less writing and we can use it for any linear divisor with leading coefficient $1$, so: $\text{2} \| \|$ $\text{1 " " " "3 " " " "-4 " " " "6" " " } - 31$ $+$ $\text{ " " " " "" "2" " " " 10" " " "12" " " " " "36}$ $\text{ " }$$- - - - - - - - -$ $\text{ " " " 1" " " " 5" " "" " "6" " " "18" " " "\|\|" " 5}$ The remainder is $5$, so $P \left(2\right) = 5$ If you use long division, it will look something like: $\text{ " " " " " }$ ${x}^{3}$ $+ 5 {x}^{2}$$+ 6 x$ $+ 18$ $\text{ " " }$$- - - - - - - - - - -$ x-2 ) ${x}^{4}$ $+ 3 {x}^{3}$ $- 4 {x}^{2}$ $+ 6 x$ $- 31$ $\text{ " " }$ ${x}^{4}$ $- 2 {x}^{3}$ $\text{ " " }$$- - - - - - - - - - -$ $\text{ " " " " "" }$ $5 {x}^{3}$$\text{ }$ $- 4 {x}^{2}$ $+ 6 x$ $- 31$ $\text{ " " " " "" }$ $5 {x}^{3}$$\text{ }$ $- 10 {x}^{2}$ $\text{ " " }$$- - - - - - - - - - -$ $\text{ " " " " " " " " "" " " " " }$ $6 {x}^{2}$ $+ 6 x$ $- 31$ $\text{ " " " " " " " " "" " " " " }$ $6 {x}^{2}$ $- 12 x$ $\text{ " " }$$- - - - - - - - - - -$ $\text{ " " " " " " " " "" " " " " " " " " }$ $18 x$ $- 31$ $\text{ " " " " " " " " "" " " " " " " " " }$ $18 x$ $- 36$ $\text{ " " }$$- - - - - - - - - - -$ $\text{ " " " " " " " " "" " " " " " " " " " " " " " " }$ $5$ Again, the remainder is $5$, so $P \left(2\right) = 5$
Edit Article # wikiHow to Graph a Quadratic Equation When graphed, quadratic equations of the form ax2 + bx + c or a(x - h)2 + k give a smooth U-shaped or a reverse U-shaped curve called a parabola. Graphing a quadratic equation is a matter of finding its vertex, direction, and, often, its x and y intercepts. In the cases of relatively simple quadratic equations, it may also be enough to plug in a range of x values and plot a curve based on the resulting points. See Step 1 below to get started. ## Steps 1. 1 Determine which form of quadratic equation you have. The quadratic equation can be written in three different forms: the standard form, vertex form, and the quadratic form. You can use either form to graph a quadratic equation; the process for graphing each is slightly different. If you're doing a homework problem, you'll usually receive the problem in one of these two forms - in other words, you won't be able to choose, so it's best to understand both. The two forms of quadratic equation are: • Standard form. In this form, the quadratic equation is written as: f(x) = ax2 + bx + c where a, b, and c are real numbers and a is not equal to zero. • For example, two standard form quadratic equations are f(x) = x2 + 2x + 1 and f(x) = 9x2 + 10x -8. • Vertex form. In this form, the quadratic equation is written as: f(x) = a(x - h)2 + k where a, h, and k are real numbers and a does not equal zero. Vertex form is so named because h and k directly give you the vertex (central point) of your parabola at the point (h,k). • Two vertex form equations are f(x) = 9(x - 4)2 + 18 and -3(x - 5)2 + 1 • To graph either of these types of equations, we need to first find the vertex of the parabola, which is the central point (h,k) at the "tip" of the curve. The coordinates of the vertex in standard form are given by: h = -b/2a and k = f(h), while in vertex form, h and k are specified in the equation. 2. 2 Define your variables. To be able to solve a quadratic problem, the variables a, b, and c (or a, h, and k) usually need to be defined. An average algebra problem will give you a quadratic equation with the variables filled in, usually in standard form, but sometimes in vertex form. • For example, for the standard form equation f(x) = 2x2 +16x + 39, we have a = 2, b = 16, and c = 39. • For the vertex form equation f(x) = 4(x - 5)2 + 12, we have a = 4, h = 5, and k = 12. 3. 3 Calculate h. In vertex form equations, your value for h is already given, but in standard form equations, it must be calculated. Remember that, for standard form equations, h = -b/2a. • In our standard form example (f(x) = 2x2 +16x + 39), h = -b/2a = -16/2(2). Solving, we find that h = -4. • In our vertex form example (f(x) = 4(x - 5)2 + 12), we know h = 5 without doing any math. 4. 4 Calculate k. As with h, k is already known in vertex form equations. For standard form equations, remember that k = f(h). In other words, you can find k by replacing every instance of x in your equation with the value you just found for h. • We have determined in our standard form example that h = -4. To find k, we solve our equation with our value for h replacing x: • k = 2(-4)2 + 16(-4) + 39. • k = 2(16) - 64 + 39. • k = 32 - 64 + 39 = 7 • In our vertex form example, again, we know the value of k (which is 12) without having to do any math. 5. 5 Plot your vertex. The vertex of your parabola will be the point (h, k) - h specifies the x coordinate, while k specifies the y coordinate. The vertex is the central point in your parabola - either the very bottom of a "U" or the very top of an upside-down "U." Knowing the vertex is an essential part of graphing an accurate parabola - often, in schoolwork, specifying the vertex will be a required part of a question. • In our standard form example, our vertex will be at (-4,7). So, our parabola will peak 4 spaces to the left of 0 and 7 spaces above (0,0). We should plot this point on our graph, being sure to label coordinates. • In our vertex form example, our vertex is at (5,12). We should plot a point 5 spaces to the right and 12 spaces above (0,0). 6. 6 Draw the parabola's axis (optional). A parabola's axis of symmetry is the line running through its middle which divides it perfectly in half. Across this axis, the left side of the parabola will mirror the right side. For quadratics of the form ax2 + bx + c or a(x - h)2 + k, the axis is a line parallel to the y-axis (in other words, perfectly vertical) and passing through the vertex. • In the case of our standard form example, the axis is a line parallel to the y-axis and passing through the point (-4, 7). Though it's not part of the parabola itself, lightly marking this line on your graph can eventually help you see how the parabola curves symmetrically. 7. 7 Find the direction of opening. After having figured out the vertex and axis of the parabola, we next need to know whether the parabola opens upwards or downwards. Luckily, this is easy. If "a" is positive, the parabola will open upwards, while if "a" is negative, the parabola will open downwards (i.e., it will be turned upside-down.) • For our standard form example (f(x) = 2x2 +16x + 39), we know we have a parabola opening upwards because, in our equation, a = 2 (positive). • For our vertex form example (f(x) = 4(x - 5)2 + 12), we know we have also have a parabola opening upwards because a = 4 (positive). 8. 8 If necessary, find and plot x intercepts. Often, on schoolwork, you'll be asked to find a parabola's x-intercepts (which are either one or two points where the parabola meets the x axis). Even if you're not to find them, these two points can be invaluable for drawing an accurate parabola. However, not all parabolas have x-intercepts. If your parabola has a vertex opens upward and has a vertex above the x axis or if it opens downward and has a vertex below the x axis, it won't have any x intercepts. Otherwise, solve for your x intercepts with one of the following methods: • Simply set f(x) = 0 and solve the equation. This method may work for simple quadratic equations, especially in vertex form, but will prove exceedingly difficult for more complicated ones. See below for an example • f(x) = 4(x - 12)2 - 4 • 0 = 4(x - 12)2 - 4 • 4 = 4(x - 12)2 • 1 = (x - 12)2 • SqRt(1) = (x - 12) • +/- 1 = x -12. x = 11 and 13 are the parabola's x-intercepts. • Factor your equation. Some equations in the ax2 + bx + c form can be easily factored into the form (dx + e)(fx +g), where dx × fx = ax2, (dx × g + fx × e) = bx, and e × g = c. In this case, your x intercepts are the values for x which make either term in parentheses = 0. For example: • x2 + 2x + 1 • = (x + 1)(x + 1) • In this case, your only x intercept is -1 because setting x equal to -1 will make either of the factored terms in parentheses equal 0. • Use the quadratic formula. If you can't easily solve for your x intercepts or factor your equation, use a special equation called the quadratic formula designed for this very purpose. If it isn't already, get your equation into the form ax2 + bx + c, then plug a, b, and c into the formula x = (-b +/- SqRt(b2 - 4ac))/2a. Note that this often gives you two answers for x, which is OK - this just means your parabola has two x intercepts. See below for an example: • -5x2 + 1x + 10 gets plugged into the quadratic formula as follows: • x = (-1 +/- SqRt(12 - 4(-5)(10)))/2(-5) • x = (-1 +/- SqRt(1 + 200))/-10 • x = (-1 +/- SqRt(201))/-10 • x = (-1 +/- 14.18)/-10 • x = (13.18/-10) and (-15.18/-10). The parabola's x intercepts are at approximately x = -1.318 and 1.518 • Our previous standard form example, 2x2 + 16x + 39 gets plugged into the quadratic formula as follows: • x = (-16 +/- SqRt(162 - 4(2)(39)))/2(2) • x = (-16 +/- SqRt(256 - 312))/4 • x = (-16 +/- SqRt(-56)/-10 • Because finding the square root of a negative number is impossible, we know that no x intercepts exist for this particular parabola. 9. 9 If necessary, find and plot the y intercept. Though it's often not necessary to find an equation's y intercept (the point at which the parabola passes through the y axis), you may eventually be required to, especially if you're in school. This process is fairly easy - just set x = 0, then solve your equation for f(x) or y, which gives you the y value at which your parabola passes through the y axis. Unlike x intercepts, standard parabolas can only have one y intercept. Note - for standard form equations, the y intercept is at y = c. • For example, we know our quadratic equation 2x2 + 16x + 39 has a y intercept at y = 39, but it can also be found as follows: • f(x) = 2x2 + 16x + 39 • f(x) = 2(0)2 + 16(0) + 39 • f(x) = 39. The parabola's y intercept is at y = 39. As noted above, the y intercept is at y = c. • Our vertex form equation 4(x - 5)2 + 12 has a y intercept that can be found as follows: • f(x) = 4(x - 5)2 + 12 • f(x) = 4(0 - 5)2 + 12 • f(x) = 4(-5)2 + 12 • f(x) = 4(25) + 12 • f(x) = 112. The parabola's y intercept is at y = 112. 10. 10 If necessary, plot additional points, then graph. You should now have a vertex, direction, x intercept(s), and, possibly, a y intercept for your equation. At this point, you can either attempt to draw your parabola using the points you have as a guideline, or you can find more points to "fill out" your parabola so that the curve you draw is more accurate. The easiest way to do this is simply to plug in a few x values on either side of your vertex, then plot these points using the y values you obtain. Often, teachers will require you to obtain a certain number of points before you draw your parabola. • Let's revisit the equation x2 + 2x + 1. We already know its only x intercept is at x = -1. Because it only touches the x intercept at one point, we can infer that its vertex is its x intercept, which means its vertex is (-1,0). We effectively only have one point for this parabola - not nearly enough to draw a good parabola. Let's find a few more to ensure we draw an accurate graph. • Let's find the y values for the following x values: 0, 1, -2, and -3. • For 0: f(x) = (0)2 + 2(0) + 1 = 1. Our point is (0,1). • For 1: f(x) = (1)2 + 2(1) + 1 = 4. Our point is (1,4). • For -2: f(x) = (-2)2 + 2(-2) + 1 = 1. Our point is (-2,1). • For -3: f(x) = (-3)2 + 2(-3) + 1 = 4. Our point is (-3,4). • Plot these points to the graph and draw your U-shaped curve. Note that the parabola is perfectly symmetrical - when your points on one side of the parabola lie on whole numbers, you can usually save yourself some work by simply reflecting a given point across the parabola's axis of symmetry to find the corresponding point on the other side of the parabola. ## Community Q&A Search • On my homework, I got x^2-x-2 and I graphed it, so how do I find the axis of symmetry? wikiHow Contributor Did you find the vertex when graphing it? If so, the axis of symmetry is the vertical line running through the vertex. To check your answer, it should have the equation x = 1/2. • How do I solve a system of equations algebraically? See the wikiHow article Solve Systems of Equations. • Does it matter what points you plot after the vertex? You will want to plot several points near (and on either side of) the vertex, and then not so many as you get farther from the vertex. 200 characters left ## Tips • Note that in f(x) = ax2 + bx + c, if b or c equal zero, those numbers disappear. For example, 12x2 + 0x + 6 becomes 12x2 + 6 because 0x is 0.
# Case Study Question 03 ## Chapter 11: Constructions ### Class 10 Linking San Francisco with Marin County, the Golden Gate Bridge is a 1.7 mile-long suspension bridge that can be crossed by car, on bicycles or on foot. The Golden Gate Bridge, completed after more than four years of construction, is a landmark recognized almost universally. The Golden Gate Bridge’s 4,200-foot-long main suspension span was a world record that stood for 27 years. The bridge’s two towers rise 746 feet making them 191 feet taller than the Washington Monument. The five-lane bridge crosses Golden Gate Strait. Question.1. To divide a line segment AB in the ratio m:n (m, n are positive integers), we draw a ray AX so that \angleBAX is an acute angle and then mark points on ray AX at equal distances. What will be the minimum number of these points? As we want to divide the given line segment in the ratio m:n, we mark (m + n) points at equal distances on the ray AX drawn as given in the question. This is because we will use Basic proportionality theorem in dividing the line segment in the ratio m : n by joining A_{m+n} to B and then drawing a line through A_{m} parallel to A_{m+n}B to intersect AB at C. Then AC : CB = m:n (By BPT). Hence the minimum number of points will be m+n. Question.2. Suppose we want to divide the main suspension span of the bridge, which is 4200 feet long, in the ratio 3:4. What will be the minimum number of points which should be marked on ray AX. As we want to divide the given line segment in the ratio 3:4, we mark 7 (3 + 4) points at equal distances on the ray AX drawn as given in the question. This is because we will use Basic proportionality theorem in dividing the line segment in the ratio 3:4 by joining A to B and then drawing a line through A_{3} parallel to AB to intersect AB at C Then AC : CB = 3:4 (By BPT). Hence the minimum number of points to be marked are 7. Question.3. In the above question, points A_{1}, A_{2}, A_{3}, … are located at equal distances on the ray AX drawn such that \angleBAX is an acute angle. To which point should point B joined to? We will use Basic proportionality theorem in dividing the line segment in the ratio 3:4 by joining A_{7} to B and then drawing a line through A_{3} parallel to A_{3}B to intersect AB at C. Then AC : CB = 3:4 (By BPT). Hence, the point joined should be A_{7}. Question.4. What is the principal theorem or rule used in the steps to divide the given line segment in any ratio? Basic Proportionality Theorem. Question.5. In the part (B), suppose a ray AX is drawn such that \angleBAX is an acute angle. Then a ray BY is drawn parallel to AX and the points A_{1}, A_{2}, A_{3}, … and B_{1}, B_{2}, B_{3}, … are located at equal distances on ray AX and BY, respectively. Then which two points will be joined together? A_{3} and B_{4} will be joined together. error: Content is protected !!
# Power Show/Hide Sub-topics (Work, Energy & Power \| O Level) Power is defined as the rate of work done or energy converted with respect to time. $P = \frac{W}{t}$ OR $P = \frac{E}{t}$ , where W = work, t = time, E = Energy SI Unit for power is watt (W), scalar quantity • One watt (W) is defined as the rate of work done or energy conversion of one joule per second. • $\text{One watt} = \frac{\text{one joule}}{\text{one second}}$ $\rightarrow$ $1 \, \text{W} = 1 \, \text{J s}^{-1}$ Power tells us how fast work is being done or how fast energy is being converted from one form to another. Another useful equation for power: $$P = Fv$$ , where F = force, v = velocity (Simple derivation below) Simple derivation of P = Fv In some questions, the formulation of power in terms of force and velocity will be useful in the problem-solving. The formula $P = Fv$ can be simply derived as seen below: \begin{aligned} P &= \frac{W}{t} \\ &= \frac{F \times d}{t} \\ &= Fv \text{ where } v = \frac{d}{t} \end{aligned} ## Efficiency From the Principle of Conservation of Energy, we know that the total energy output of a machine must be equal to its energy input. However, it is found that the energy output of a machine is always less than the energy input. This phenomena can be attributed to the work done against frictional forces, which is considered as wasted energy output. Hence, we have: Energy input = useful energy output + wasted energy output Efficiency of a system is given by $$\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100 \%$$ ## Self-Test Question 1: Fat Man & Thin Man ### A fat man and a thin man (with half of the mass of the fat man) ran to the top of a hill in the same time. Whose power output is higher? Why? Recall that power is given by $P = \frac{W}{t}$. Notice that the time, t, is the same for both fat man and thin man. Hence, we will have to look at the work done by both men. The work done by the fat man will be larger than the work done by the thin man. Hence, the power output by the fat man is higher. ## Worked Example 1: Power of engine ### An engine does 60 000 J of work in ten minutes. What is the power of the engine? \begin{aligned} P &= \frac{W}{t} \\ &= \frac{60 000 \, \text{J}}{10 \times 60 \, \text{s}} \\ &= 100 \, \text{W} \end{aligned} ## Worked Example 2: Efficiency of Electric Motor ### An electric motor is rated at 1.0 kW. If 60% of the input energy is lost as heat and sound, find the amount of useful energy produced in half hour. Hence or otherwise, find the efficiency of the electric motor. Let’s begin by calculating the total energy output of the electric motor: \begin{aligned} P &= \frac{W}{t} \\ W &= P \times t \\ &= 1000 \, \text{W} \times 30 \times 60 \, \text{s} \\ &= 1.8 \times 10^{6} \, \text{J} \end{aligned} Since 60% of the energy from the electric motor is lost as heat and sound, only 40% is useful energy output. Hence, \begin{aligned} \text{Useful Energy Output} &= \frac{40}{100} \times 1.8 \times 10^{6} \\ &= 7.2 \times 10^{5} \, \text{J} \end{aligned} As given in the question, the efficiency of the electric motor is 40%. ## Worked Example 3: Electric Motor ### An electric motor is used to lift a 10 N load through 5 m. The total amount of electrical energy input is 65 J. Calculate the amount of energy wasted by the motor. Hence or otherwise, calculate the efficiency of the motor. Since efficiency is given by: $\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100 \%$, we have to find the useful energy output. In this case, the useful energy output is the lifting of the 10 N load through 5 m. Hence, useful energy output: \begin{aligned} W_{\text{useful}} &= F \times d \\ &= 10 \times 5 \\ &= 50 \text{ J} \end{aligned} Since the total amount of electrical energy input is given to be 65 J, the energy wasted will be: \begin{aligned} E_{\text{wasted}} &= 65-50 \\ &= 15 \text{ J} \end{aligned} Hence, the efficiency of the electric motor will be: \begin{aligned} \text{Efficiency} &= \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100 \% \\ &= \frac{50}{65} \times 100 \% \\ &= 76.9 \% \end{aligned} ### 11 thoughts on “Power” 1. A machine is 12% efficient if useful work done by the machine is 6 J then calculate the total energy consume by the machine? 2. A pendulum Bob swings from one end to the other. At which point (s) will the gravitational potential energy of the pendulum Bob be maximum or minimum? 3. A softballer throws a ball into the air and catches it on the way down.state the energy changes that take place ? 4. hello! I have a question the input power to a lamp is 6.0W. The lamp wasted 2.7J of energy in 3.0s. What is the efficiency of the lamp? • Convert Power into Total input Energy using formulae. Subtract your answer from Wasted Energy , you will get you Useful Energy.Then use Efficiency formulae to solve your answer by dividing useful energy by total energy multiply by 100. Solution:- Wasted Energy=2.7 J Total Energy input=Power x Time =6 x 3 =18 J Useful Energy= Total Energy – Wasted Energy =18 – 2.7 =15.3 J Efficiency=Useful/Total x 100 =15.3/18 x 100
## Tangent and Normal Question7 In this page tangent and normal question7 we are going to see solution of some practice questions from the worksheet. (7) Let P be a point on the curve y = x³ and suppose that the tangent line at P intersects the curve again at Q. Prove that the slope at Q is four times the slope at P. Solution: The tangent drawn at the point "P" on the curve touches the curve again at the point "Q" Let P (a,a³) be a point on the curve y = x³ --- (1) dy/dx = 3 x² slope at the point "P" (dy/dx) = 3 (a)² = 3 a² Equation of the tangent at the point "P" (y - y₁) = m (x - x₁) (y - a³) = 3 a² (x - a) y - a³ = 3 a² x - 3 now we are going to apply the value of y in this equation x³ - a³ = 3 a² x - 3 x³ - a³ - 3 a² x + 3 a³ = 0 x³ - 3 a² x + 2 a³ = 0 (x - a)² (x + 2a) = 0 (x - a)² = 0 (x + 2a) = 0 (x - a) = 0 x = - 2 a x = a Here a is the x-coordinate value of the point "P". So we have to take -2a is the x-coordinate value of the point "Q" Slope at the point Q dy/dx = 3 x² = 3 (-2a)² = 3 (4a²) = 4 (3a²) Slope at the point Q = 4 (Slope at the point P) (8) Prove that the curve 2x²+4y²=1 and 6x²-12y²=1 cut each other at right angles. Solution: Let (x₁,y₁) be the common point on the curve 2 x₁² + 4 y₁² = 1 ----- (1) 6 x₁² - 12 y₁² = 1 ----- (2) to find the point of intersection we have to solve the given equations (1) x 3 => 6 x₁² + 12 y₁² = 3 6 x₁² - 12 y₁² = 1 ------------------ 12 x₁² = 4 x₁² = 4/12 x₁² = 1/3 now we are going to apply the value of x₁² in the first equation 2 (1/3) + 4 y₁² = 1 (2/3) + 4 y₁² = 1 4 y₁² = 1 - 2/3 4 y₁² = 1/3 y₁² = 1/12 Therefore the common point on both curves is (1/3 , 1/12) differentiate the first equation with respect to x 2 x² + 4 y² = 1 4 x + 8 y(dy/dx) = 0 8 y(dy/dx) = - 4 x (dy/dx) = - 4 x/8y = - x/2y 6 x₁² - 12 y₁² = 1 12 x - 24 y (dy/dx) = 0 - 24 y (dy/dx) = -12 x dy/dx = -12 x/-24y = x/2y If two curves are intersecting orthogonally then m₁ x m₂ = -1 (- x/2y) (x/2y) = -1 - x²/4y² = -1 -(1/3)/4(1/12) = -1 - (1/3)/(1/3) = - 1 - 1 = -1 tangent and normal question7 tangent and normal question7
# Quarter Three: Week #27th February 26th-March 2nd Lesson Plans (Math/Science/S. Studies Homeroom Only) Teacher Michael Lee Math/Science/S. Studies 4th Grade Week #27th February 26th-March 2nd Lesson Plans Angle Measurement /Motion of Objects/Florida Statehood Standard(s) Taught Math: 4.MD.3.5, 4.MD.3.6, 4.G.1.1 (not parallel, perpendicular, or 2-D figures) 1. Recognize angles as geometric shapes that are formed wherever two rays share a common endpoint, and understand concepts of angle measurement. a. An angle is measured with reference to a circle with its center at the common endpoint of the rays, by considering the fraction of the circular arc between the points where the two rays intersect the circle. An angle that turns through 1/360 of a circle is called a “one-degree angle,” and can be used to measure angles. b. An angle that turns through n one-degree angles is said to have an angle measure of n degrees. 2. Measure angles in whole-number degrees using a protractor. Sketch angles of specified measure. 3. Use a protractor to sketch an angle given a specific measurement. Science: SC.4.P.12.1, SC.4.P.12.2 1. Recognize that an object in motion always changes its position and may change its direction. 2. Investigate and describe that the speed of an object is determined by the distance it travels in a unit of time and that objects can move at different speeds. Learning Targets and Learning Criteria Math: 1. Recognize an angle as a geometric shape that is formed when two rays share a common endpoint. 1. Explain the relationship between a circle and the number of degrees in an angle (i.e., an angle is measured in reference to a circle—its center is the endpoint for each of the rays that make up the angle). 2. Explain an angle as a series of “one-degree turns” and the total number of “one-degree turns” is the measure of the angle in degrees. E.g., A water sprinkler rotates one-degree at each interval. If the sprinkler rotates a total of 100 one-degree turns, what is the measure of the sprinkler’s rotation in degrees? 1. Explain that since it takes 360 “one-degree turns” to rotate through a circle, 1/360 of a circle is a “one-degree angle”. 1. Measure an angle to the nearest whole number degrees using a protractor. 1. Use a protractor to sketch an angle given a specific measurement. Science: 1. Describe an object’s position and motion in space. 2. Explain that motion is a change of an object’s position. 3. Demonstrate that moving objects always change position. 4. Demonstrate that moving objects may change direction. 5. Explain that the speed of an object is determined by the distance it travels within a unit of time. 6. Investigate and compare the speeds of different objects by measuring the distance each object travels during a set amount of time using tools and technology. 7. Investigate and compare the speeds of different objects by measuring the amount of time it takes each object to travel a set amount of distance using tools and technology. 8. Display obtained speeds in chart, table and graph format. Classroom Activities Math Centers (Math Games, App “Media” Resources, Teacher Time, Hands-On Activities) Multiplication Facts timed test Splash Math My Math Vol. 2 Chapter 14 (p.905-924) Study Jams: Force and Motion Study Jams: Geometry Science Fusion Book p. 341-342 Assignments Due Math IXL Codes this week: (Z.3, J.4, O.8) Math Weekly Homework Q3: Week 6 (Comparing Decimals) Science Fusion Unit 8: Lesson #1-2 Quiz Science Fusion Unit 8: Review and Summative Assessment Science SMT2 Assessment
451 views ### Can you explain the formula method for solving quadratic equations? We can use the formula method (for solving quadratic equations) to find 'roots' or values of x that satisfy or 'work out' for a given quadratic equation of an unknown variable (say x.) The formula is: x=-b(+or- sqrt[b2-4ac])/2a Note that the 'plus or minus' can give us 2 possible values or 'roots' for the unknown 'x'. These may be 2 positive roots, 2 negative roots, or a negative and a positive root. These roots are the coordinates where a curve/line intersects with the x axis (we know that y=0 on the x-axis already.) We may compare our quadratic equation to the general format (ax2+bx+c) to obtain the values for a, b, and c, which are coefficients of x (c is the coefficient of x0 which equals 1.) Our 2 values may then be substituted back into our original equation to show that the 2 sides 'match' and thus the equation is valid. We let the quadratic equation equal zero to display that the 2 sides are balanced or 'homogeneous'. Example: Solve the quadratic equation 3x2+9x+3 via the formula method. Firstly, we must compare the above quadratic equation with the general format (ax2+bx+c) to obtain values for the coefficients of x. We can see that a=3, b=9, and c=3. Our general formula: x=-b(+or- sqrt[b2-4ac])/2a is thus x=-9(+or- sqrt[(9)2-4(3)(3)])/2(3) So that by solving for x, x=-0.381 (3 d.p.) and x=-2.618 (3 d.p.). We obtained these answers by adding and subtracting the square root terms (respectively) and performing the arithmetic. We can check that these are correct by equating the quadratic to zero and substituting in our x values: 3(-0.381)2+9(-0.381)+3=0.0064833 3(-2.618)2+9(-2.618)+3=-0.000228 Thus our roots are correct! The equations do not equal zero exactly as we have rounded our roots to 3 decimal places. 10 months ago Answered by Daniel , a GCSE Maths tutor with MyTutor ## Still stuck? Get one-to-one help from a personally interviewed subject specialist #### 534 SUBJECT SPECIALISTS £36 /hr Steven A. Degree: Bioscience (Masters) - Durham University Subjects offered:Maths, Science+ 5 more Maths Science Human Biology English Chemistry -Personal Statements- “Hey! I'm Steven, 18 years old and I'm studying Biochemistry at Durham University. I love tutoring people, to put it simply. At my college i was a tutor to a student in the year below me for chemistry and growing up with a younger sist...” £18 /hr Giorgos A. Degree: Mechanical Engineering with Renewable Energy (Masters) - Edinburgh University Subjects offered:Maths, Physics Maths Physics “Feel rewarded helping younger students. Nobody is born knowing everything, life is a learning process and my aim is to help you achieve your goals.” MyTutor guarantee £22 /hr Ayusha A. Degree: BEng electrical and electronics engineering (Bachelors) - Newcastle University Subjects offered:Maths, Physics+ 1 more Maths Physics Further Mathematics “About me: I am a final year Electrical and Electronic Engineering student at Newcastle University. I took Mathematics, Further Mathematics, Chemistry and Physics as my A-level subjects. I did peer mentoring in university and also have...” Daniel M. Currently unavailable: for new students Degree: Physics (Bachelors) - Bath University Subjects offered:Maths, Science+ 4 more Maths Science Physics Chemistry Biology -Personal Statements- “Hi! I'm Daniel! As a patient, understanding, and industrious tutor, I aim to nourish your child's academic curiosity in a caring, understandable and encouraging manner so that they may be the best that they can be. I'm currently readi...” ### You may also like... #### Other GCSE Maths questions Expand (2x-3)(4x+4) using the FOIL method. How do you find the equation of a straight line on a graph?
# How do you solve the system x+4y-6z=8, 2x-y+3z=-10, and 3x-2y+3z=-18? Nov 30, 2016 The solution is $\left(x , y , z\right) = \left(- 4 , \text{ "4," } \frac{2}{3}\right)$. #### Explanation: Write the coefficients in an augmented matrix: $\left[\begin{matrix}1 & 4 & - 6 & \| & 8 \\ 2 & - 1 & 3 & \| & - 10 \\ 3 & - 2 & 3 & \| & - 18\end{matrix}\right]$ Our goal is to change this into a matrix that is upper-right triangular; that is, where everything below the diagonal is a zero. We do this by adding linear combinations of the rows together to create new rows (that replace the current ones) and by multiplying rows by constants. Remember: this does not change the information implied by the augmented matrix, it just presents it in a new (and hopefully more beneficial) way. (Brief notation recap: ${R}_{1}$, ${R}_{2}$, and ${R}_{3}$ mean "row 1", "row 2", and "row 3" respectively.) Let's start with the first cell in row 2. Its value is $2$, which is $\frac{2}{1} = 2$ times as much as the $1$ above it. If we want the $2$ to become a $0$, we need to subtract $2 {R}_{1} - {R}_{2}$: $2 \times \left[\left(1 , 4 , \text{-"6,\|,8" }\right)\right]$ ul(" "-[(2,"-"1,3,\|,"-"10)]) "= "[(0,9,"-"15,\|,26)] This new row represents a new valid equation (that is, $0 x + 9 y - 15 z = 26$) which can replace one of the two rows used in forming it. Choosing to keep row 1 the same, we replace row 2: $\left[\begin{matrix}1 & 4 & - 6 & \| & 8 \\ \textcolor{red}{0} & \textcolor{red}{9} & \textcolor{red}{- 15} & \| & \textcolor{red}{26} \\ 3 & - 2 & 3 & \| & - 18\end{matrix}\right]$ A similar linear combination is then used to replace row 3—that combination is $3 {R}_{1} - {R}_{3}$, and the following shorthand is used to indicate what row operation has been done to form a new row: $\implies \textcolor{w h i t e}{\begin{matrix}\null \\ \null \\ \textcolor{b l a c k}{3 {R}_{1} - {R}_{3}}\end{matrix}} \left[\begin{matrix}1 & 4 & - 6 & \| & 8 \\ 0 & 9 & - 15 & \| & 26 \\ \textcolor{red}{0} & \textcolor{red}{14} & \textcolor{red}{- 21} & \| & \textcolor{red}{42}\end{matrix}\right]$ All the coefficients in ${R}_{3}$ are multiples of 7, so let's reduce: $\implies \textcolor{w h i t e}{\begin{matrix}\null \\ \null \\ \textcolor{b l a c k}{\frac{1}{7} \cdot {R}_{3}}\end{matrix}} \left[\begin{matrix}1 & 4 & - 6 & \| & 8 \\ 0 & 9 & - 15 & \| & 26 \\ 0 & 2 & - 3 & \| & 6\end{matrix}\right]$ $\implies \textcolor{w h i t e}{\begin{matrix}\null \\ \null \\ \textcolor{b l a c k}{9 {R}_{3} - 2 {R}_{2}}\end{matrix}} \left[\begin{matrix}1 & 4 & - 6 & \| & 8 \\ 0 & 9 & - 15 & \| & 26 \\ 0 & 0 & 3 & \| & 2\end{matrix}\right]$ Our newest ${R}_{3}$ says $3 z = 2$, which means that $z = \frac{2}{3}$. And since the equation directly above ${R}_{3}$ only involves one additional unknown (that is, $y$), we can plug this solved value for $z$ into it to solve for $y$: $9 y - 15 z = 26$ $\implies 9 y - 15 \left(\frac{2}{3}\right) = 26$ $\implies 9 y - 10 = 26$ $\implies 9 y = 36$ $\implies y = 4$ With the matrix in the upper-triangular form, solving for variables is like falling dominoes. Knowing $z$ gives us $y$, and knowing $z$ & $y$ gives us $x$: $x + 4 y - 6 z = 8$ $\implies x + 4 \left(4\right) - 6 \left(\frac{2}{3}\right) = 8$ $\implies x + 16 - 4 = 8$ $\implies x = \text{-} 4$ Our final solution is $\left(x , y , z\right) = \left(- 4 , \text{ "4," } \frac{2}{3}\right)$. (It is best to check your solution to make sure it works; I will save that as an exercise.) ## Note: Many teachers will encourage you to reduce each new row you create so that the leading coefficients are $1$. This is perfectly fine, and you'll get the same solutions. It just means your matrix will likely have fractions in it. I chose not to reduce my rows for the sake of making the matrices easier to read. Nov 30, 2016 Here is an alternate approach that turns the coefficient matrix into the identity matrix, thus making the augmented column the solution. #### Explanation: $\text{ } \left[\begin{matrix}1 & 4 & - 6 & \| & 8 \\ 2 & - 1 & 3 & \| & - 10 \\ 3 & - 2 & 3 & \| & - 18\end{matrix}\right]$ $\implies \textcolor{w h i t e}{\begin{matrix}\null \\ \textcolor{b l a c k}{2 {R}_{1} - {R}_{2}} \\ \null\end{matrix}} \left[\begin{matrix}1 & 4 & - 6 & \| & 8 \\ 0 & 9 & - 15 & \| & 26 \\ 3 & - 2 & 3 & \| & - 18\end{matrix}\right]$ $\implies \text{ } \textcolor{w h i t e}{\begin{matrix}\null \\ \textcolor{b l a c k}{\frac{1}{9} \cdot {R}_{2}} \\ \null\end{matrix}} \left[\begin{matrix}1 & 4 & - 6 & \| & 8 \\ 0 & 1 & - \frac{5}{3} & \| & \frac{26}{9} \\ 3 & - 2 & 3 & \| & - 18\end{matrix}\right]$ $\implies \textcolor{w h i t e}{\begin{matrix}\null \\ \null \\ \textcolor{b l a c k}{3 {R}_{1} - {R}_{3}}\end{matrix}} \left[\begin{matrix}1 & 4 & - 6 & \| & 8 \\ 0 & 1 & - \frac{5}{3} & \| & \frac{26}{9} \\ 0 & 14 & - 21 & \| & 42\end{matrix}\right]$ $\implies \text{ } \textcolor{w h i t e}{\begin{matrix}\null \\ \null \\ \textcolor{b l a c k}{\frac{1}{7} \cdot {R}_{3}}\end{matrix}} \left[\begin{matrix}1 & 4 & - 6 & \| & 8 \\ 0 & 1 & - \frac{5}{3} & \| & \frac{26}{9} \\ 0 & 2 & - 3 & \| & 6\end{matrix}\right]$ $\implies \textcolor{w h i t e}{\begin{matrix}\null \\ \null \\ \textcolor{b l a c k}{{R}_{3} - 2 {R}_{2}}\end{matrix}} \left[\begin{matrix}1 & 4 & - 6 & \| & 8 \\ 0 & 1 & - \frac{5}{3} & \| & \frac{26}{9} \\ 0 & 0 & \frac{1}{3} & \| & \frac{2}{9}\end{matrix}\right]$ $\implies \text{ } \textcolor{w h i t e}{\begin{matrix}\null \\ \null \\ \textcolor{b l a c k}{3 \cdot {R}_{3}}\end{matrix}} \left[\begin{matrix}1 & 4 & - 6 & \| & 8 \\ 0 & 1 & - \frac{5}{3} & \| & \frac{26}{9} \\ 0 & 0 & 1 & \| & \frac{2}{3}\end{matrix}\right]$ $\implies \textcolor{w h i t e}{\begin{matrix}\null \\ \textcolor{b l a c k}{{R}_{2} + \frac{5}{3} {R}_{1}} \\ 3 \cdot {R}_{3}\end{matrix}} \left[\begin{matrix}1 & 4 & - 6 & \| & 8 \\ 0 & 1 & 0 & \| & 4 \\ 0 & 0 & 1 & \| & \frac{2}{3}\end{matrix}\right]$ $\implies \textcolor{w h i t e}{\begin{matrix}\textcolor{b l a c k}{{R}_{1} - 4 {R}_{2}} \\ \null \\ 3 \cdot {R}_{3}\end{matrix}} \left[\begin{matrix}1 & 0 & - 6 & \| & - 8 \\ 0 & 1 & 0 & \| & 4 \\ 0 & 0 & 1 & \| & \frac{2}{3}\end{matrix}\right]$ $\implies \textcolor{w h i t e}{\begin{matrix}\textcolor{b l a c k}{{R}_{1} + 6 {R}_{3}} \\ \null \\ 3 \cdot {R}_{3}\end{matrix}} \left[\begin{matrix}1 & 0 & 0 & \| & - 4 \\ 0 & 1 & 0 & \| & 4 \\ 0 & 0 & 1 & \| & \frac{2}{3}\end{matrix}\right]$ These equations now say $x \text{ "="-} 4$ $\text{ "y" } = 4$ $\text{ } z = 2 / 3$ So our answer is as before.
Keyword Search Quick Order Lesson ID: 183 # PEMDAS: Order of Operations Mathematics, Algebra and Function, Operations (+, -, x, /, etc.), Problem Solving ## Objective Students learn about the order of operations. ## Directions In algebra, there is a specific order in which operations are performed when evaluating an expression or solving equations. This is the algebraic order of operations: 1. Do any work within parentheses ( ) or other grouping symbols [ ] first. 2. Do any work with exponents (powers) or roots. 3. Do any multiplication or division in order from left to right. 4. Do any addition and subtraction in order from left to right. The acronym for this order of operations is PEMDAS. Parentheses Exponents Multiplication Division Addition Subtraction A popular expression for remembering this is: Please Excuse My Dear Aunt Sally. Sample A 42 + (9 - 3) = ? Do the work in the parentheses first. 42 + 6 = ? Do the exponents next. 42 = 4 x 4 = 16, so16 + 6 = ? Finally, add the numbers together. 16 + 6 = 22 42 + (9 - 3) = 22 Sample B 52 - (4 x 3) + 9 / 3 x 4 = ? Do work in the parentheses first. 52 - 12 + 9 / 3 x 4 = ? Do exponents next. 52 = 5 x 5 = 25, so25 - 12 + 9 / 3 x 4 = ? Multiply and divide in order from left to right. 25 - 12 + 3 x 4 = ? 9 divided by 3 is 3, and 3 times 4 equals 12. 25 - 12 + 12 = ? Add and subtract in order from left to right. 13 + 12 = ? 25 minus 12 is 13 and 13 plus 12 equals 25. 13 + 12 = 25
# Investigation: Constructions with a compass Lesson Every profession has particular tools that they use all the time, a chef has their knives, the sewer a machine and the mathematician, well they have their construction tools. No, not hammers and hard hats - Mathematicians use a compass, pencil and a straight edge. From just these three things we can create (nearly) all of what we require in geometry. Notice how I used the word straight edge and not ruler. Well that's because all we really need for the constructions is a straight edge, we often don't even need numbers! A great Greek mathematician named Euclid, who is credited to have written the first mathematics textbook over 2000 years ago, went to great lengths to detail many of the mathematical constructions we will look at today. Geometrical constructions were so important to mathematics at the time because most problems were solved graphically, not arithmetically. ### Copy a segment To construct congruent line segments with a compass and straightedge, follow the instructions below. Play and pause the video at each step to help you. #### Steps for construction 1. Start with the line segment $\overline{AB}$AB on the page that you want to copy. 2. Mark a point on the page where you want the start of the copied line segment. Call this point $C$C. 3. Set the compass width to the length of $\overline{AB}$AB. 4. Without changing the compass width, move the compass to $C$C 5. Draw an arc. The endpoint of the new line segment can be anywhere on the arc. 6. Choose a point ($D$D) on the arc and draw $\overline{CD}$CD with the straightedge. ### Copy an angle As we have seen, congruent means same, so congruent angles are 2 (or more) angles that are exactly the same size. They can be facing in any direction, but if they are the same size then the angles are congruent. You can think of congruent angles as copies of each other. To construct congruent angles you need a compass and a straight edge. #### Steps for construction 1. Start with the angle $BAC$BAC ($\angle BAC$BAC) on the page that you want to copy. 2. Draw a ray, $\overrightarrow{PQ}$PQ, on the page. 3. Position the compass on point $A$A, adjust to any width. 4. Draw an arc that crosses both legs of $\angle BAC$BAC, name the points where the arc intersects the rays $F$F and $G$G 5. Position the compass on point on point $P$P, and draw another arc. Where it crosses $\overrightarrow{PQ}$PQ call it $M$M. 6. Measure the distance $FG$FG with the compass. 7. Place the compass at point $M$M, and cross the arc. Call this intersection $N$N. 8. Draw in a ray from $P$P, through $N$N. Call it $\overrightarrow{PR}$PR 9. Now you have created $\angle RPQ$RPQ. 10. $\angle RPQ\equiv\angle BAC$RPQBAC ### Outcomes #### 7.G.2 Draw (freehand, with ruler and protractor, and with technology) two-dimensional geometric shapes with given conditions. Focus on constructing triangles from three measures of angles or sides, noticing when the conditions determine a unique triangle, more than one triangle, or no triangle.
# Into Math Grade 8 Module 4 Lesson 2 Answer Key Investigate Angle-Angle Similarity We included HMH Into Math Grade 8 Answer Key PDF Module 4 Lesson 2 Investigate Angle-Angle Similarity to make students experts in learning maths. ## HMH Into Math Grade 8 Module 4 Lesson 2 Answer Key Investigate Angle-Angle Similarity I Can use Angle-Angle similarity to test triangles for similarity and find unknown angle measures. Asa is comparing the architect’s model of a barn with the finished building. He looks at the triangle that forms the front of the roof in the model and in the completed barn. Are the triangles similar? Why or why not? Turn and Talk Which is the easier way of deciding if the triangles are similar, comparing the angles or comparing the side lengths? Why? Build Understanding Two triangles are similar if all three pairs of corresponding angles are congruent. What if only two out of three pairs of corresponding angles are congruent? 1. Using the given drawing by the set designer, Shawna is making a flag for a play. She has measured Angles ABC and C with a protractor, and they match Angles E and F, respectively, on the set designer’s notes. Can she be sure that her flag is similar to the drawing without measuring the third angle? A. What do you know about the sum of the measures of the angles in a triangle? ___________________________ ___________________________ B. How does knowing the sum of the measures of the angles in a triangle help you solve this problem? ___________________________ C. Write and evaluate an expression to find the measure of the third angle of Triangle ABC, using the angle measures given. ___________________________ D. Write and evaluate an expression to find the measure of the third angle of Triangle DEF using the angle measures given. ___________________________ E. Based on your calculation of the measure of the third angle, can you now state with confidence whether Shawna’s flag is similar to the drawing without measuring the third angle? Explain. ___________________________ The Angle-Angle Similarity Postulate states that two triangles are similar if they have two pairs of corresponding angles that are congruent. Turn and Talk When two triangles have two angle measures in common, will the third angle measure always be the same for both triangles? Explain. Step It Out 2. In the two triangles shown, mâˆ A = mâˆ D and mâˆ C = mâˆ F. A. Do you know enough to say whether the two triangles are similar? Explain. ___________________________ ___________________________ B. What does your answer to Part A imply about Angles B and E? Explain how you know. ___________________________ C. What equation can you write relating the measures of Angles B and El ___________________________ D. Solve your equation from Part C for x. What are the measures of Angles B and E? ________ ___________________________ E. How can you find the measure of Angle F? ______________ ___________________________ ___________________________ F. How can you find the measure of Angle C? ___________________________ G. Fill in the measures of the six angles: mâˆ A = ___ mâˆ B = ____ mâˆ C = ____ mâˆ D = ___ mâˆ E = ____ mâˆ F = ____ Turn and Talk If you were given the measures of Angles A, C, D, and E from Part G, how could you determine whether the triangles are similar? 3. The illustration shows a wheelchair ramp from the side. The support at $$\overline{B C}$$ binds the ramp to the floor. Are Triangles ABC and ADE similar? ______________________ B. Do Triangles ABC and ADE both contain the same angle? If so, name it. ______________________ C. Does that mean both triangles have two pairs of corresponding congruent angles? If so, name the corresponding congruent angles. ______________________ D. Are Triangles ABC and ADE similar? Explain. ______________________ Check Understanding Question 1. Which two triangles are similar? âˆ†ABC and âˆ†XYZ are similar as the angles of both the triangles are the same. 58 + 42 + x = 180 100 + x = 180 x = 180 – 100 x = 80 Question 2. Explain why âˆ†XYZ is similar to âˆ†LMZ. Answer: If any two angles of the triangles are the same then the triangles are said to be similar. The angles of the given triangles âˆ†XYZ and âˆ†LMZ are the same. Thus âˆ†XYZ is similar to âˆ†LMZ. Question 3. Reason A graphic designer wants to reproduce a logo of a mountain-climbing club for some club stationery. The logo is a triangle with the angles shown. A. The designer wants the base to be 2 inches long. How should the triangle be drawnâ€™ B. How can the Angle-Angle Similarity Postulate help the graphic designer make sure the triangle is reproduced correctly? Question 4. Angles A, D, and G are congruent, and Angles C, F, and J are congruent. A. Write an expression to find the measures of Angles B, F, and G. ______________________ B. What is the measure of Angle E? Since the angles A, D, and G are congruent then the sum of given angles is equal to 180 degrees. x + 20 + 2x + x = 180 4x + 20 = 180 4x = 180 – 20 4x = 160 x = 160/4 x = 40 Thus Angles B, E, and H are congruent. E = x + 20 40 + 20 = 60 So Angle E = 60 Use Triangles QRS and TUV to solve Problems 5â€”6. Question 5. Are Triangles QRS and TUV similar? How do you know? Answer: Yes Triangles QRS and TUV similar are similar as the shape of both the triangles are the same. Question 6. What is the measure of Angle R in terms of x? x + (x + 15) + 90 = 180 2x + 15 + 90 = 180 2x + 105 = 180 2x = 180 – 105 2x = 75 x = 75/2 x = 37.5 degrees Question 7. Reason Are the triangles shown similar? Why or why not? Answer: No, the angles of both the right triangles are not the same. So, the given triangles are not similar. Question 8. Reason Are the triangles shown similar? How do you know? Answer: Yes both the triangles are similar as the sides of both the triangles are the same. Question 9. Reason Are the triangles similar? Explain. Answer: No the triangles are not similar as the angles of both the triangles are not the same. 25 + 90 + x = 180 115 + x = 180 x = 180 – 115 x = 65 65 â‰ 75 Question 10. Angle C is congruent to Angle F. What is the measure of Angle E? Question 11. Does the diagram show similar triangles? Explain. Answer: No, the triangles are not similar as the angles are not the same. I’m in a Learning Mindset! What can I apply from previous work with triangles to better understand whether two triangles are similar? _______________________ Lesson 4.2 More Practice/Homework Question 1. One triangle shown is formed by the tree, its shadow, and a line of sight from the ground to the top of the tree. The other is formed by Manny, his shadow, and a line of sight from the ground to the top of his head. A. Are the two triangles similar? How do you know? Answer: Yes the two triangles are similar as the shape of the triangles are the same. B. Manny is 5$$\frac{1}{2}$$ feet tall, and his shadow is 4$$\frac{1}{4}$$ feet long. The shadow of the tree is 17 feet long. Can you determine the height of the tree? If so, how? Given, Manny is 5$$\frac{1}{2}$$ feet tall, and his shadow is 4$$\frac{1}{4}$$ feet long. The shadow of the tree is 17 feet long. 4.25 Ã· 5.5 = 17 Ã· x 0.77 = 17/x x = 17/0.77 x = 22 feet Thus the height of the tree is 22 feet. C. What is the height of the tree? The height of the tree is 22 feet. Question 2. Reason Are the triangles similar? Explain. Yes both the triangles are similar as the angles are the same. Question 3. Consider the diagram for Parts A and B. A. The value of x is ___. 90 + 2x + x = 180 90 + 3x = 180 3x = 180 – 90 3x = 90 x = 90/3 x = 30 Thus the value of x is 30. B. The measure of Angle A is ____ âˆ BAC = 2x Substitute the value of x. 2(30) = 60 The measure of Angle A is 60Â° Test Prep Question 4. Angle L is congruent to Angle P, and Angle N is congruent to Angle R. What is the measure of Angle 0? Angle L is congruent to Angle P, and Angle N is congruent to Angle R. âˆ L = 45 (Î”LMN âˆ¼ Î”PQR) âˆ R = 20 (Î”LMN âˆ¼ Î”PQR) 45 + 20 + âˆ Q = 180 mâˆ Q = 180 – 65 mâˆ Q = 115 Question 5. Name the triangles that are similar. triangle A: 95 + 27 + x = 180 122 + x = 180 x = 180 – 122 x = 58 triangle B: 27 + 58 + y = 180 85 + y = 180 y = 180 – 85 y = 95 triangle C: 95 + 58 + z = 180 153 + z = 180 z = 180 – 153 z = 27 Triangle A, B, C are similar. Question 6. The measures of two pairs of corresponding angles of two triangles are 24Â° and 55Â°. Explain why the two triangles are similar. The measures of two pairs of corresponding angles of two triangles are 24Â° and 55Â°. The corresponding angles of the triangle are the same. Hence the two triangles are similar. Spiral Review Question 7. Mrs. Kato has 6 bags of dried beans and a 2-pound bag of rice in one shopping bag. In another shopping bag she has a 5-pound bag of flour. The two shopping bags weigh the same amount. What is the weight of each bag of dried beans? Given, Mrs. Kato has 6 bags of dried beans and a 2-pound bag of rice in one shopping bag. In another shopping bag she has a 5-pound bag of flour. 6x + 2 = 5 6x = 5 – 2 6x = 3 x = 3/6 x = 1/2 or 0.5 Thus the weight of each bag of dried beans is 1/2 pound. Question 8. A triangle is dilated with scale factor 4. Write a true statement about the image and preimage of the triangle.
# Trying to figure out a card game • Jun 12th 2009, 04:23 PM ceasar_19134 Trying to figure out a card game My friend and I play a Vietnamese card game called Thirteen (or Tien len). I'd like to know, what are the chances of having at least two pairs after drawing thirteen cards? • Jun 12th 2009, 05:05 PM HallsofIvy The simplest way to do this is to ask the opposite questions: 1) What is the probability of drawing 13 cards with NO pair? The first card can be anything. The second card can be any of the 48 cards that do not pair the first card: probability 48/51. The third card can be any of the 44 cards that do not match either of first two cards: probability 44/50. Do you see the pattern? For each card we subtract 4 from the numerator and 1 from the denominator. Multiplying all the fractions together, the denominator will be 51(50)(49)(48)...(39)= 51!/38! and the numerator is 48(44)(40)...(4)= 412(12)(11)(10)...(1)=41212!. The probability of getting 13 cards with no pair is $\displaystyle \frac{4^12 12! 38!}{51!}$. 2) What is the probability of drawing 13 cards with exactly one pair? First calculate the probability that the first two cards pair and the other 11 do not. The first card can be anything. The second must match that- probability 3/51. The third can be anything that does not match those: probability 48/50. The fourth can be anything other than the first two or that: 44/49 and so on: (3/51)(48/50)(44/49)... (8/39). The probability of drawing 13 cards with at most one pair is sum of those two and the probability of drawing 13 cards with at least two pair is 1 minus that probability. • Jun 12th 2009, 05:24 PM ceasar_19134 My friend and I thought similar to you when we first looked at the problem. However, our numbers were a bit skewed when we calculated things out. In response to you first point, you neglect to count the hands where players have three of a kind and no pair or four of a kind and no pair. In response to the second point, your method suggest only one order of getting those cards. You could also not get a pair on the second draw, thus increasing your chances of getting a pair as now the next card can match any of the two drawn before it. And in addition to all of that, you could draw a pair and then a three of kind or a four of a kind, thus increasing your odds even more. I just don't know how to manipulate probabilities in order to account for all of that. • Jun 12th 2009, 06:15 PM Soroban Hello, ceasar_19134! A variation of HallsofIvy's solution . . . Quote: My friend and I play a Vietnamese card game called Thirteen (or Tien len). What is the probability of having at least two pairs after drawing thirteen cards? There are: .$\displaystyle _{52}C_{13}$ possible hands. The opposite of "at least two pairs" is "no pairs" or "one pair." No Pairs We must draw one of each of the 13 values. There are: .$\displaystyle _4C_1 = 4$ ways to draw each value. Hence, there are: .$\displaystyle 4^{13}$ hands with No Pairs. One Pair There are 13 choices for the value of the Pair. There are: .$\displaystyle _4C_2 = 6$ ways to get the Pair. The other 11 cards must not match the Pair or each other. There are: .$\displaystyle _{12}C_{11} = 12$ choices for their values. And: .$\displaystyle _4C_1 = 4$ ways to draw each value. So there are are: .$\displaystyle 12\cdot4^{12}$ choices for the other 11 cards. Hence, there are: .$\displaystyle 13\!\cdot\!6\!\cdot\!12\!\cdot\!4^{12}$ ways to get One Pair. Then there are: .$\displaystyle 4^{13} + 936\!\cdot\!4^{12}\:=\:4^{13}(235)$ ways to get No Pairs or One Pair. Hence: .$\displaystyle P(\text{No Pair or 1 Pair}) \;=\;\frac{4^{13}(235)} {_{52}C_{13}}$ Therefore: .$\displaystyle P(\text{at least 2 Pairs}) \;=\;1 - \frac{4^{13}(235)}{_{52}C_{13}}$
# How do you find the vertex of f(x)=-5/2x^2+10x+1/3? Aug 7, 2015 The vertex is at $\left(2 , \frac{31}{3}\right)$ #### Explanation: The general vertex form for a quadratic is $\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = m {\left(x - a\right)}^{2} + b$ $\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$with its vertex at $\left(a , b\right)$ Given $f \left(x\right) = - \frac{5}{2} {x}^{2} + 10 x + \frac{1}{3}$ Extract the $m = - \frac{5}{2}$ $\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = \left(- \frac{5}{2}\right) \left({x}^{2} - 4 x\right) + \frac{1}{3}$ Complete the square $\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = \left(- \frac{5}{2}\right) \left({x}^{2} - 4 x + 4\right) + \frac{1}{3} - \left(- \frac{5}{2}\right) \cdot 4$ Write as a squared binomial and simplfy $\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = \left(- \frac{5}{2}\right) {\left(x - 2\right)}^{2} + \frac{31}{3}$ Note that this is in explicit vertex form. The graph looks like: graph{-5/2x^2+10x+1/3 [-12.22, 16.25, -2.78, 11.46]}
Find the area of the region bounded by the curves $y^2 = 9x, y = 3x.$ It is mentioned in the question that, $y\textsuperscript{2} = 9x$ belongs to a parabola and $y = 3x$ belong to a straight line which passes through origin. Starting with a rough figure showing those equations below, From the parabola equations it is seen that x cannot be negative, so the graph would be on the right of the X-axis. The parabola would thus be opening to the right. In order to find the points of the two equations mentioned, you have to solve the two equations simultaneously. $\begin{array}{l} \text { Put } y=3 x \text { in } y^{2}=9 x \\ \Rightarrow(3 x)^{2}=9 x \\ \Rightarrow 9 x^{2}=9 x \\ \Rightarrow x^{2}-x=0 \\ \Rightarrow x(x-1)=0 \\ \Rightarrow x=0 \text { and } x=1 \end{array}$ We got the coordinates of x, for finding the coordinates of y. Put x = 1 and x = 0 in the equations of y = 3x. It is found that y coordinated are y = 3 and y = 0 respectively. Therefore, the point of interaction of parabola and straight line is (1, 3) and (0, 0) Now, calculate the area enclosed between the parabola and the straight line. For calculating the area we have to minus the area under the straight line which extends from x = 0 to x = 1 from the area under the parabola. It can be written as, Area between parabola and straight line = area under parabola – area under straight line …. (1) On calculating the area under parabola, $y\textsuperscript{2 = }9x \\$ $\item y = 3 \sqrt x\\$ On integrating the above equation from 0 to 1 $\\ \Rightarrow \int_{0}^{1} y d x=3 \int_{0}^{1} \sqrt{x} d x \\ \Rightarrow \int_{0}^{1} y d x=3 \int_{0}^{1} x^{\frac{1}{2}} d x \\ \Rightarrow \int_{0}^{1} y d x=3\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} y d x=3\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1} \\$ $\\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=3 \frac{2}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=2\left[1^{\frac{3}{2}}-0\right] \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=2$ Now, for calculating the area under line y = 3x i.e. area of triangle OAB $\item y = 3x \\$ On integrating the above equation from 0 to 1 $\\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\int_{0}^{1} 3 \mathrm{xdx} \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=3\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=3\left(\frac{1^{2}}{2}-0\right) \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\frac{3}{2}$ Using equation mentioned above (1) area between parabola and straight line $= 2 - \frac{3}{2} = 1/2\ unit\textsuperscript{2 }\\$ Therefore, It is found that the area was 1/2 unit2 between the curves $y^2 = 9x\: \: \: and\: \: \: y = 3x.$
Right Fashion Style # Multiplication Sums for Class 1 Multiplication Sums for Class 1- the process of calculating the product of two or more numbers in mathematics. It is a basic arithmetic operation that is frequently used in everyday life. When we need to combine groups of equal size, we use multiplication. ## Multiplication Sums for Class 1 Multiplication is a mathematical operation that represents the fundamental concept of repeated addition of the same number. The numbers that are multiplied are known as the factors, and the result of multiplying two or more numbers is known as the product of those numbers. Multiplication is used to simplify the task of repeating the same number’s addition. Different symbols are used in mathematics. One of the most commonly used math symbols is the multiplication symbol. The multiplication of two numbers 6 and 9 was demonstrated in the preceding example. When we look at the multiplication expression (6 9 = 54), we can see that the symbol () connects the two numbers and completes the equation. Multiplication is represented by the mid-line dot operator () and the asterisk sign (*) in addition to the cross symbol (). ## Multiplication Sums Examples The order of the numbers in multiplication is irrelevant. So go with the order that feels most natural to you. When using the multiplication tables, students may remember 4 9 more easily than 9 4. When multiplying three numbers, choose the two that are the easiest to multiply. Multiplying 5 17 2 is difficult, for example, if we try to multiply 5 17 first. Multiplying 5 and 2 instead yields 10, which can be easily multiplied by 17 to get 170. When multiplying a two-digit number by a one-digit number, it can be useful to divide the two-digit number by the place values. Then multiply and add each part. For example, 37 4 can be mentally solved by dividing 37 by 30 + 7. So 30 4 = 120 and 7 4 = 28. As a result, the final answer is 120 + 28 = 148. While this may appear to be more difficult to solve when written down, it is much easier to solve mentally. Multiplication Sums for Class 1- Even if you don’t remember the multiplication fact, you can easily figure it out mentally. For example, the number 17 9 is difficult to remember. However, this can be mentally restructured as 17 (10 – 1). As a result, the answer is 170 – 17 = 153.
# Chapter 10 ## Section 10.1 (Area of parallelograms) Formula for the Area of a Parallelogram: A=bh (Pic Below+Practice Question) Area=base times height +Pic ## How to find the area of a parrallelogram How to Find the Area of a Parallelogram ## Section 10.1 (Area of Trapezoids) Area of a Trapezoid: A=1/2(b1+b2)h (See pic Below) +Pic ## Vocab for section 1 Base of a parallelogram- The length of any one of its sides. Height of a parallelogram- The perpendicular distance between the base and the opposite side. Bases of trapezoids- Its two parallel sides. Height of a trapezoid- The perpendicular distance between the bases. ## Section 10.2 (Areas of Circles) Formula for the area of a circle: A=Pier^2 (Below) ## Practice question A=Pier^2 Formula A= 3.14(5)^2 Substitute 3.14 for pie and 5 for r Pic Below Math Antics - Circles, Circumference And Area ## Vocab for section 2 Area- The number of square units covered by a figure. Circle- All the set points in a plane that are the same distance from a fixed point. Radius- The distance from the center point to any point. Diameter- The distance across the circle through the center, or double the radius. Circumference- The distance around a circle. ## Section 10.4 (Surface Area of a Prisms) Surface Area of a Prism: S= 2B+Ph B= Area of the base P= Base's perimeter ## Practice question S=2B+Ph S=2(1/21012) + (13+13+10) 15 S=660 ## Video for the surface area of prisms and cylinders Algebra Man - Surface Area of Prisms and Cylinders ## Section 10.4 (Surface Area of a Cylinder) Surface Area of a Cylinder: S= 2Pier^2+2Pierh +Pic ## Vocab for section 4 Net- A two-dimensional pattern that forms a solid when it is folded. Surface Area- The sum of the areas of its faces. ## Section 10.5 (Surface Area of Pyramids) Surface area of a pyramid: S=B+1/2PSlant height P= Base's perimeter +Pic ## Video to find the surface area of a pyramid How to Find the Surface Area of a Pyramid ## Section 10.5 (Surface area of a cone) Surface area of a cone: S=Pier^2+Pierslant height +Pic ## Vocab for section 5 Slant height- The height of a lateral face, that is, any face that is not the base. ## Section 10.6 (Volume of prisms) Volume of a prism: V=Bh B=Area of the base +Pic ## Real life tie in If you wanted to know how much space the 8 kleenex boxes were taking up on your counter top, all you have to do is find the volume of one prism and then multiply it by 8, which is how many kleenex boxes you have. ## Section 10.6 (Volume of Cylinders) Volume of a cylinder: V=Pier^2h +Pic ## Real life tie in If you wanted to know how much space your disinfecting wipes take up in classroom, all you have to do is find the volume of a cylinder, V=Pier^2h, and then you have your answer. ## Vocab for section 6 Volume- A measurement of the amount of space it occupies. ## Section 10.7 (Volume of a pyramid) Volume of a pyramid: V=1/3Bh B= area of the base +Pic ## Section 10.7 (Volume of a cone) Volume of a cone: V=1/3Pier^2h +Pic ## Real life tie in If we wanted to know how much space a Native Americans teepee took up all we would have to do is find the volume using the formula, V=1/3Pier^2h, and we would get are answer. ## Vocab for section 7 Pyramid- A polyhedron, have one base, and the other faces are triangles. Cone- A solid with one circular base. Volume- A measurement of the amount of space it occupies. ## Formulas for every section 10.1: Area of parallelograms: A=bh Area of trapezoids: A=1/2(b1+b2)h 10.2: Area of circles: A=Pier^2 10.4: Surface area of prisms: S=2B+Ph Surface area of cylinders: S=2Pier^2+2Pierh 10.5: Surface area of pyramids: S=B+1/2Pslant height Surface area of cones: S=Pier^2+Pierslant height 10.6: Volume of prisms: V=Bh Volume of cylinders: V=Pier^2h 10.7: Volume of prisms: V=1/3Bh Volume of cones: V=1/3Pier^2h Volume of spheres: V=4/3Pie*r^3 ## Vocab for every section 10.1: Base of a parallelogram- The length of any one of its sides. Height of a parallelogram- The perpendicular distance between the base and the opposite side. Bases of trapezoids- Its two parallel sides. Height of a trapezoid- The perpendicular distance between the bases. 10.2: Area- The number of square units covered by a figure. Circle- All the set points in a plane that are the same distance from a fixed point. Radius- The distance from the center point to any point. Diameter- The distance across the circle through the center, or double the radius. Circumference- The distance around a circle. 10.4: Net- A two-dimensional pattern that forms a solid when it is folded. Surface Area- The sum of the areas of its faces. 10.5: Slant height- The height of a lateral face, that is, any face that is not the base. 10.6: Volume- A measurement of the amount of space it occupies. 10.7: Pyramid- A polyhedron, have one base, and the other faces are triangles. Cone- A solid with one circular base. Volume- A measurement of the amount of space it occupies.
# Finding the summation of $c_i$ for $c_i=\begin{cases}i &\quad\text{if$i-1$is exact power of$2$}\\1&\quad\text{otherwise.}\\ \end{cases}$ While reading the text Introduction to Algorithms by Cormen et. al. I came across a few mathematical step which I felt like proving in a more detailed manner, as I could not get steps of the mathematics which they did in short. Below is the excerpt from the text. $$c_i = \begin{cases} i &\quad\text{if i-1 is an exact power of 2 }\\ 1&\quad\text{otherwise.}\\ \end{cases}$$ So, $$\sum_{i=1}^{n}c_i\leq n+\sum_{j=0}^{\lfloor lg(n) \rfloor}2^j\tag 1$$ $$ The following is my attempt to understand the step $$(1)$$ $$\sum_{i=1}^{n}c_i=\sum_{\text{i-1 is a power of 2}}c_i +\sum_{\text{i-1 is not a power of 2}}c_i$$ $$=\sum_{\text{j is a power of 2}}(j+1) +\sum_{\text{j is not a power of 2}}(1) ,\quad\quad\text{where j=i-1}$$ $$=\sum_{\text{j is a power of 2}}(j) +\sum_{\forall j}(1) = \left (\sum_{\text{j is a power of 2}}j\right )+n \tag 2$$ $$\text{where 0\leq j \leq n-1}$$ for the situation in which $$j$$ is power of $$2$$ let $$2^k$$ be the maximum possible value of $$j$$. So, $$2^k=n-1 \implies k=\lfloor \log_2(n-1) \rfloor$$ Now we know, $$n-1 Let $$j=2^t$$ , $$t=0$$ to $$k$$ So from $$(2)$$ and $$(3)$$ we have, $$\sum_{i=1}^{n}c_i\leq n+\sum_{t=0}^{\lfloor lg(n) \rfloor}2^t \tag 4$$ The step which the authors achieved directly in $$(1)$$ took me so many steps to understand or derive in $$(4)$$. Is there a shorter method available or some intuition which the authors used to achieve the result directly? • I did not get you Commented Jul 9, 2020 at 21:00 • I jumped the step, substituted $c_i$ with $i$ or $1$ and then changed the variables to $j$ Commented Jul 9, 2020 at 21:06 If $$i-1=2^j$$, where $$i\le n$$ then $$j=\lg(i-1)<\lg n$$. Moreover, $$j$$ is an integer, so $$j\le\lfloor\lg n\rfloor$$. Thus, each term of $$\sum_{i=1}^nc_i$$ is either $$1$$ or $$2^j+1$$ for some $$j$$ such that $$0\le j\le\lfloor\lg n\rfloor$$. Thus, we automatically get a contribution of $$1$$ from each of the $$n$$ terms, for a total of $$n$$. We get another $$2^j$$ for the terms with $$0\le j\le\lfloor\lg n\rfloor$$, which contribute $$\sum_{j=0}^{\lfloor\lg n\rfloor}2^j=2^{\lfloor\lg n\rfloor+1}-1<2\cdot2^{\lg n}=2n\;.$$ Thus, $$\sum_{i=1}^nc_i\le n+\sum_{j=0}^{\lfloor\lg n\rfloor}2^j • $\text{lg}=\log_2$? Commented Jul 9, 2020 at 21:10 • @UmbQbify-Key20-: Yes; it’s a standard abbreviation. (And you can get it with \lg.) Commented Jul 9, 2020 at 21:11 • Specifically for base 2 logarithm? Thanks! Commented Jul 9, 2020 at 21:14 • @UmbQbify-Key20-: Yes, specifically for the binary log. Commented Jul 9, 2020 at 21:15 • @AbhishekGhosh: Oops! You’re absolutely right, and I need to change the answer slightly. Thanks! (Because I can’t see the question when I’m typing an answer, I sometimes lose track of details.) Commented Jul 10, 2020 at 16:01
# NCERT Solutions for Class 12 Maths Find 100% accurate solutions for NCERT Class XII Maths. All questions have been explained and solved step-by-step, to help you understand thoroughly. Free Download option available! 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. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84. 85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 99. 100. 101. 102. 103. 104. Comparing coefficients x, of 2A = 6 ⇒ A = 3 Comparing constants, -9A + B = 7 On solving, we get A = 3, B = 34 Putting the values of A and B in eq. (ii), 6x + 7 = 3(2x – 9) + 34 Putting this value of 6x + 7 in eq. (i), 105. Comparing coefficients of x, –2A = 1 ⇒ A = Comparing constants, 4A + B = 2 On solving, we get A = , B = 4 Putting the values of A and B in eq. (ii), 106. Comparing coefficients of x, 2A = 1 ⇒ A = Comparing constants, 2A + B = 2 On solving, we get A = , B = 1 Putting the values of A and B in eq. (ii), 107. Comparing coefficients of x, 2A = 1 ⇒ A = Comparing constants, -2A + B = 3 On solving, we get ⇒ A = , B = 4 Putting the values of A and B in eq.(ii), x + 3 = (2x – 2) + 4 Putting this value of x + 3 in eq.(i), 108. 109. 110. 111. Comparing coefficients of x on both sides A + B = 1 …….(ii) Comparing constants 2A + B = 0 …….(iii) Solving eq. (ii) and (iii), we get A = -1 and B = 2 Putting these values of A and B in eq. (i), 112. 113. Comparing coefficients of x2 : A + B + C = 0 ….(ii) Comparing coefficients of x : -5A – 4B – 3C = 3 ⇒ 5A + 4B + 3C = -3 …(iii) Comparing constants: 6A + 3B + 2C = –1 ……….(iv) On solving eq. (i), (ii) and (iii), we get A = 1, B = –5, C = 4 Putting the values of A, B and C in eq. (i), 114. Comparing coefficients of x2 : A + B + C = 0 ….(ii) Comparing coefficients of x : -5A – 4B – 3C = 1 ⇒ 5A + 4B + 3C = -1 ….(iii) Comparing constants: 6A + 3B + 2C = 0 ……….(iv) On solving eq. (i), (ii) and (iii), we get Putting the values of A, B and C in eq. (i), 115. Comparing coefficients of x on both sides A + B = 2 ……….(ii) Comparing constants 2A + B = 0 …….(iii) Solving eq. (ii) and (iii), we get A = -2 and B = 4 Putting these values of A and B in eq. (i), 116. Comparing coefficients of on both sides Comparing constants A = 1 ……….(iv) Solving eq. (ii) and (iii), we get Putting these values of A and B in eq. (ii), 117. Comparing coefficients of A + C = 0 ……….(ii) Comparing coefficients of , –A + B = 1 ……….(iii) Comparing constant terms, –B + C = 0 ……….(iv) Solving eq. (ii), (iii) and (iv), we get Putting the values of A, B and C in eq.(i), 118. Comparing coefficients of x2 : A + C = 0 ……….(ii) Comparing coefficients of x : A + B – 2C= 1 ……….(iii) Comparing constants: –2A + 2B + C = 0 ……….(iv) On solving eq. (i), (ii) and (iii), we get Putting the values of A, B and C in eq. (i), 119. Comparing coefficients of x2 : A + C = 0 ……….(ii) Comparing coefficients of x : B – 2C= 3 ……….(iii) Comparing constants: –2A + B + C = 5 ……….(iv) On solving eq. (i), (ii) and (iii), we get Putting the values of A, B and C in eq. (i), 120. Comparing coefficients of x2 : 2A + 2B + C = 0 ……….(ii) Comparing coefficients of x : 5A + B = 2 ……….(iii) Comparing constants: 3A – 3B – C = –3 ……….(iv) On solving eq. (i), (ii) and (iii), we get Putting the values of A, B and C in eq. (i), 121. Comparing coefficients of x2 : A + +B + C = 0 ……….(ii) Comparing coefficients of x : –B + 3C = 5 ……….(iii) Comparing constants: –4A – 2B + 2C = 0 ……….(iv) On solving eq. (i), (ii) and (iii), we get Putting the values of A, B and C in eq. (i), 122. 123. Comparing the coefficients of x2 A – B = 0 ……….(ii) Comparing the coefficients of x B – C = 0 ……….(iii) Comparing constants A + C = 2 ……….(iv) On solving eq. (ii), (iii) and (iv), we get A = 1, B = 1, C = 1 Putting these values of A, B and C in eq. (i), 124. 125. Comparing the coefficients of y A + B = 0 ………(ii) Comparing constants A – B = 1 ……….(iii) On solving the eq. (ii) and (iii), we get Putting the values of A, B and in eq. (i), 126. 127. 128. Comparing coefficients of y A + B = –4 ……….(v) Comparing constants 4A + 3B = –10 ……….(vi) On solving eq. (v) and (vi), we get A = 2, B = –6 Putting the values of A, B and y in eq. (iii), 129. 130. 131. 132. Comparing coefficients of x A + B = 1 ……….(ii) Comparing constants –2A – B = 0 ……….(iii) On solving eq. (ii) and (iii), we get A = –1, B = 2 Putting these values of A and B in eq. (i), Therefore, option (B) is correct. 133. 134. 135. 136. 137. 138. 139. 140. 141. 142. 143. 144. 145. 146. 147. 148. 149. 150. 151. 152. 153. 154. 155. 156. 157. 158. 159. 160. 161. 162. 163. 164. 165. 166. 167. 168. 169. 170. 171. 172. 173. 174. 175. 176. 177. 178. 179. 180. 181. 182. 183. 184. 185. 186. 187. 188. 189. 190. 191. 192. 193. 194. 195. 196. 197. To change the limits of integration from x to t when x = 0, t = x2 + 1 = 0 + 1 = 1 when x = 1, t = x2 + 1 = 1 + 1 = 2 ∴ From eq. (i), 198. 199. 200. 201. 202. 203. 204. 205. 206. 207. 208. 209. 210. 211. 212. 213. 214. 215. 216. 217. 218. 219. 220. 221. 222. 223. 224. 225. 226. 227. 228. Comparing the coefficients of x2 –A + B – C = 0 ……….(ii) Comparing the coefficients of x B + C = 0 ……….(iii) Comparing constants A = 1 ……….(iv) On solving eq. (ii), (iii) and (iv), we get Putting these values in eq. (i), 229. 230. 231. 232. 233. Comparing coefficients of x2 A + B = 0 ……….(iii) Comparing coefficients of x B + C = 5 ……….(iv) Comparing constants 9A + C = 0 ……….(v) On solving eq. (iii), (iv) and (v), we get Putting these values of A, B and C in eq. (ii), 234. 235. 236. 237. 238. 239. 240. 241. 242. 243. 244. 245. 246. 247. 248. 249. Comparing coefficients of x2 A + C = 1 ……….(iii) Comparing coefficients of x 3A + B + 2C = 1 ……….(iv) Comparing constants 2A + 2B + C = 1 ……….(v) On solving eq. (iii), (iv) and (v), we get A = –2, B = 1, C = 3 Putting these values of A, B and C in eq. (ii), 250. 251. 252. 253. 254. 255. Putting sin x – cos x = t ⇒ (cos x + sin x) dx = dt Again (sin x – cos x)2 = t2 ⇒ sin2 x + cos2 x – 2 sin x cos x = t2 ⇒ sin 2x = 1 – t2 256. 257. Putting sin x – cos x = t ⇒ (cos x + sin x) dx = dt Again (sin x – cos x)2 = t2 ⇒ sin2 x + cos2 x – 2 sin x cos x = t2 ⇒ sin 2x = 1 – t2 Limits of integration when x = 0, t = 0 – 1 = -1 and 258. 259. 260. 261. Taking RHS Comparing coefficients of x A + C + 0 ……….(iii) Comparing coefficients of x A + B = 0 ……….(iv) Comparing constants B = 1 On solving eq. (iii), (iv) and (v), we get A = -1, B = 1, C = 1 Putting these values of A, B and C in eq. (ii), 262. 263. 264. 265. 266. 267. 268. 269. 270. 271. MySchoolPage connects you with exceptional, certified maths tutors who help you stay focused, understand concepts better and score well in exams!
# How do you explain scale in math? Scale is the ratio that defines the relation between the actual figure and its model. It is used in maps to represent the actual figures in smaller units. For example, a scale of 1:5 means 1 on the map represents the size of 5 in the real world. ## What is scale in mathematics? What is scale maths? Scale maths is a way of enlarging an object. If we have two shapes that are similar, one will be a scale diagram of the other. We can calculate the scale factors for length, area and volume. ## How do you describe a scale? Scale is defined as the ratio of the length of any object on a model (blueprint) to the actual length of the same object in the real world. When we draw a real-world object on a piece of paper, we use a scale to describe our measurements accurately. ## How do you explain scale to a child? A scale factor is defined as the ratio between the scale of a given original object and a new object, which is its representation but of a different size (bigger or smaller). For example, if we have a rectangle of sides 2 cm and 4 cm, we can enlarge it by multiplying each side by a number, say 2. ## What does a scale 1 3 mean? If the scale factor is one-third that means that the new object or figure is one-third the size of the original. ## Is a scale of 1 to 5 or 1 to 10 better? There's more variance in larger scales, like the 1-10 scale rating. This is usually undesirable; making the 1-5 Likert scale the most common survey scale. ## What does 1 to 5 mean in scale? A 1 to 5 rating scale is a simple and effective way to rate the severity or magnitude of something. It typically goes from 1, the lowest rating, to 5, the highest rating. The 1 to 5 scale allows respondents to answer quickly and can be applied to a variety of things, such as pain, temperature, and brightness. ## How do you teach a scale? Usually a scale is played just by going up and down the staircase: 1,2,3,4 up to 8 and down again. You could try playing around with the notes in different patterns for example: 1,3,2,4,3,5,4,6 etc. or 1,2,3,2,1,3,4,3,2,4,5 etc. ## How do you teach students scales? Ten other ways to learn scales 1. Chanting fingering. 2. Focussing on fingering patterns. 3. Playing notes in clusters by depressing each group of fingers at the same time. 4. Playing with different rhythms and articulations – staccato, tongued, slurred and stressing different beats. ## What is a scale very short answer? A scale is a set of levels or numbers which are used in a particular system of measuring things or are used when comparing things. ... an earthquake measuring five-point-five on the Richter scale. The patient rates the therapies on a scale of zero to ten. ## What are the three ways to describe scale? There are three types of scales commonly used on maps: written or verbal scale, a graphic scale, or a fractional scale. A written or verbal scale uses words to describe the relationship between the map and the landscape it depicts such as one inch represents one mile. ## How do you solve scales? As long as you know that the two shapes are similar, you can use one dimension on both figures to calculate the scale factor. For example, if you know the width of the shape, divide one width by the other to find the scale factor. ## How do I learn scale theory? Even if you already know several scales, start with some easy ones to get used to this process. Start with C, super easy: it's just the alphabet. Then go around the Circle of Fifths and add a sharp (or a flat), then add another, and another, etc. ## What do children learn from scales? Incorporating a set of scales into the learning environment encourages children to explore and experiment the mathematical concepts of weight differentiation and balance. Depending on what is being weighed, children can also create a link between quantity and weight. ## What is the easiest scale to learn? As mentioned above, the C major scale doesn't have any sharps or flats, so it's easy to remember. Starting from a low C note, there are 7 total notes in the C major scale – 8 if you count the final note, which is the same as the 1st (or root) note. ## What is a 1 to 500 scale? A scale of 1:500 means that the actual real-life measurements are 500 times greater than those on the plan or map. This means that it does not matter whether you take the measurements on the plan in millimetres (mm), centimetres (cm) or metres (m) – the measurements will be 500 times as much in real life. ## How to do a scale drawing? So when you're working with scale drawings: 1. Find out what the scale on the drawing is. 2. Measure the distance on the drawing using a ruler (or count the number of squares, if that's an option). 3. Multiply the distance you measure by the scale to give the distance in real life. ## What is the formula for scale drawing? A scale is a ratio of a length in the drawing to the corresponding length in the actual object. To find the scale used in a drawing or model, divide the drawing length by the actual length. ## What scale is yes or no? In research activities a YES/NO scale is nominal. It has no order and there is no distance between YES and NO. ## How do you calculate scale 1 50? For example, the scale of 1:50 means that 1 mm on the drawing represents 50 mm on the object. This means that the object is 50 times larger than the drawing of it. An object 450 mm long would be represented by a line 9 mm long (450 mm/50). Figure 7 shows one of the three sides of a metric scale. ## What is a 5-point rating scale examples? For example, a 5-point scale will score “Strongly Agree” 5 points and “Strongly Disagree” 1 point. You can switch the scores if you want. Use these numbers to aggregate the results. Likert scale results are usually best presented using a statistical mode. ## How do you manipulate a scale? One way is to place a weight on the other side of the scale that is heavier than the object you are trying to weigh. This will cause the scale to read lower than the actual weight. Another way is to put something light on one side of the scale and something heavy on the other side.
# DAV Class 4 Maths Chapter 9 Worksheet 7 Solutions The DAV Class 4 Maths Solutions and DAV Class 4 Maths Chapter 9 Worksheet 7 Solutions of Fractions offer comprehensive answers to textbook questions. ## DAV Class 4 Maths Ch 9 WS 7 Solutions Question 1. Change into mixed number. (a) $$\frac{4}{3}$$ $$\frac{4}{3}$$ = 4 ÷ 3 = 1$$\frac{1}{3}$$ (b) $$\frac{41}{10}$$ $$\frac{41}{10}$$ = 41 ÷ 10 = 4$$\frac{1}{10}$$ (c) $$\frac{11}{4}$$ $$\frac{11}{4}$$ = 11 ÷ 4 = 2$$\frac{3}{4}$$ (d) $$\frac{29}{6}$$ $$\frac{29}{6}$$ = 29 ÷ 6 = 4$$\frac{5}{6}$$ (e) $$\frac{7}{5}$$ $$\frac{7}{5}$$ = 7 ÷ 5 = 1$$\frac{2}{5}$$ (f) $$\frac{50}{9}$$ $$\frac{50}{9}$$ = 50 ÷ 9 = 5$$\frac{5}{9}$$ (g) $$\frac{68}{11}$$ $$\frac{68}{11}$$ = 68 ÷ 11 = 6$$\frac{2}{11}$$ (h) $$\frac{74}{15}$$ $$\frac{74}{15}$$ = 74 ÷ 15 = 4$$\frac{14}{15}$$ (i) $$\frac{54}{7}$$ $$\frac{54}{7}$$ = 54 ÷ 7 = 7 $$\frac{5}{7}$$ Question 2. Change into improper fractions. (a) 2$$\frac{1}{3}$$ = (2 × 3) + 1 = 6 + 1 = 7 $$\frac{7}{3}$$ (b) 6$$\frac{1}{4}$$ = (4 × 6) + 1 = 24 + 1 = 25 = $$\frac{25}{4}$$ (c) 1$$\frac{4}{9}$$ = (1 × 9) + 4 = 9 + 4 = 13 = $$\frac{13}{9}$$ (d) 33$$\frac{1}{3}$$ = (33 × 3) + 1 = 99 + 1 = 100 = $$\frac{100}{3}$$ (e) 10$$\frac{1}{10}$$ = (10 × 10) + 1 = 100 + 1 = 101 = $$\frac{101}{10}$$ (f) 7$$\frac{5}{6}$$ = (7 × 6) + 5 = 42 + 5 = 47 = $$\frac{47}{6}$$ (g) 12$$\frac{3}{7}$$ = (12 × 7) + 3 = 84 + 3 = 87 = $$\frac{87}{7}$$ (h) 9$$\frac{1}{9}$$ = (9 × 9) + 1 = 81 + 1 = 82 = $$\frac{82}{9}$$ (i) 4$$\frac{7}{8}$$ = (4 × 8) = 32 + 7 = 39 = $$\frac{39}{8}$$ ### DAV Class 4 Maths Chapter 9 Worksheet 7 Notes Changing Fractions 1. Imprper fractrion into mixed number. E.g. Change $$\frac{17}{2}$$ into mixed number. $$\frac{17}{2}$$ can be written as 17 ÷ 2 so $$\frac{17}{2}$$ = 8$$\frac{1}{2}$$ E.g. Change 2$$\frac{1}{2}$$ into fraction. 2$$\frac{1}{3}$$ = 2 × 3 = 6 2$$\frac{1}{3}$$ = 6 + 1 = $$\frac{7}{3}$$
## BEATCALC: Beat the Calculator! Back to Calculation Tips & Tricks From B. Lee Clay Squaring Numbers Multiplying Numbers Dividing Numbers Subtracting Numbers Finding Percents Calculation Practice Exercises Full List ### Multiplying a 2-digit number by 11 1/2 1. Select a 2-digit number. 2. Divide by 2. 3. To this, add 11 times the original number. #### Example: 1. The 2-digit number chosen to multiply by 11 1/2 is 21. 2. Divide by 2: 21/2 = 10 1/2. 3. To this, add 11 times original number: 10 1/2 + 11(21) = 10 1/2 + 231 = 241 1/2. 4. So 11 1/2 × 21 = 241 1/2. Recall that to multiply a number by 11 (for example, 11 × 21), the pattern is: 1. Last digit is second digit of number (continuing the example above, 1) 2. Middle digit is sum of the numbers' two digits (2 + 1 = 3) 3. First digit is first digit of number (2) 4. So 11 × 21 is 231. See the pattern? 1. The 2-digit number chosen to multiply by 11 1/2 is 85. 2. Divide by 2: 85/2 = 42 1/2. 3. To this, add 11 times original number: 42 1/2 + 11(85) = 42 1/2 + 935 = 975 + 2 1/2 = 977 1/2. 4. So 85 × 11 1/2 = 977 1/2. Practice multiplying by 11. Remember that when sum of the two digits is 10 or more, add one to the first digit for the first digit of the product.
THE ANGLES OF A POLYGON - Quadrilaterals and Other Polygons - PLANE GEOMETRY - SAT SUBJECT TEST MATH LEVEL 1 ## CHAPTER 10Quadrilaterals and Other Polygons • The Angles of a Polygon • Perimeter and Area of Quadrilaterals • Exercises A polygon is a closed geometric figure made up of line segments. The line segments are called sides, and the endpoints of the line segments are called vertices (each one is called a vertex). Line segments inside the polygon drawn from one vertex to another are called diagonals. Three-sided polygons, called triangles, were discussed in Chapter 9. Although in this section our main focus will be on four-sided polygons, which are called quadrilaterals, we will discuss other polygons as well. There are special names for many polygons with more than four sides. The ones you need to know for the Math 1 test are given in the following chart. A regular polygon is a polygon in which all the sides have the same length and all the angles have the same measure. A regular three-sided polygon is an equilateral triangle, and, as we shall see, a regular quadrilateral is a square. Pictured below are a regular pentagon, regular hexagon, and regular octagon. pentagon hexagon octagon Remember In a regular polygon, all the angles are congruent and all the sides are congruent. ### THE ANGLES OF A POLYGON A diagonal of a quadrilateral divides it into two triangles. Since the sum of the measures of the three angles in each of the triangles is 180°, the sum of the measures of the angles in the quadrilateral is 360°. a + b + c = 180 and d + e + f = 180 a + (b + e ) + (c + d ) + f = 360 Key Fact I1 In any quadrilateral, the sum of the measures of the four angles is 360°. Similarly, any polygon can be divided into triangles by drawing in all of the diagonals emanating from one vertex. pentagon hexagon octagon Notice that a five-sided polygon can be divided into three triangles, and a six-sided polygon can be divided into four triangles. In general, an n-sided polygon can be divided into (n – 2) triangles, which leads to KEY FACT I2. Key Fact I2 The sum of the measures of the n angles in a polygon with n sides is (n2) 180°. EXAMPLE 1: To find the measure of each angle of a regular octagon, first use KEY FACT I2 to get that the sum of all eight angles is (8 – 2) 180° = 6 180° = 1,080°. Then since in a regular octagon all eight angles have the same measure, the measure of each one is 1,080° 8 = 135°. An exterior angle of a polygon is formed by extending a side. Surprisingly, in all polygons, the sum of the measures of the exterior angles is the same. Key Fact I3 In any polygon, the sum of the measures of the exterior angles, taking one at each vertex, is 360°. x + y + z = 360 a + b + c + d + e = 360 EXAMPLE 2: KEY FACT I3 gives us an alternative method of calculating the measure of each angle in a regular polygon. In Example 1 we used KEY FACT I2 to find the measure of each angle in a regular octagon. By KEY FACT I3, the sum of the measures of the eight exterior angles of any octagon is 360°. As a result, in a regular octagon, the measure of each exterior angle is 360° 8 = 45°. Therefore, the measure of each interior angle is 180° – 45° = 135°.
# Lesson 6 Find That Factor These materials, when encountered before Algebra 1, Unit 5, Lesson 6 support success in that lesson. ## 6.1: Multiplication and Division (10 minutes) ### Warm-up In this warm-up, students recall what they know about related multiplication and division equations. ### Launch Display this image for all to see, and ask students how they would express the relationship between the quantities pictured: Language students might use is “2 groups of 3” or “6”. Next to the image, write two corresponding equations: $$2 \boldcdot 3 = 6$$ and $$6 \div 2 = 3$$. Before students begin working, tell them that it’s not necessary to draw diagrams unless they find the diagrams helpful. The task is to come up with the missing equation, so that each relationship is expressed using both multiplication and division. ### Student Facing Here are some multiplication and division equations. Write the missing pieces. The first one is completed, as an example. 1. $$6 \div 2 = 3$$ and $$2 \boldcdot 3 = 6$$ 2. $$20 \div 4 = 5$$ and $$\underline{\hspace{.5in}}$$ 3. $$\underline{\hspace{.5in}}$$ and $$1.5 \boldcdot 12 = 18$$ 4. $$9 \div \frac14 = 36$$ and $$\underline{\hspace{.5in}}$$ 5. $$12 \div 15 = \underline{\hspace{.5in}}$$ and $$\underline{\hspace{.5in}}$$ 6. $$a \div b = c$$ and $$\underline{\hspace{.5in}}$$ ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis Either display the corresponding equations, or ask students to share their responses. It is possible to write equations that are correct but look different. For example, a response of $$5 \boldcdot 4 = 20$$ or $$18 \div 12 = 1.5$$. All of these equivalent forms should be validated. The important point is that a relationship between three numbers involving division can also be expressed as a relationship involving multiplication. ## 6.2: Scaling Segments (20 minutes) ### Activity This representation creates a bridge between multiplication-as-scaling and the distance of a point from the $$x$$-axis given its $$y$$-coordinate. In the associated Algebra 1 lesson, students will need to interpret a graph representing an exponential relationship and determine the growth factor. This activity pares that down to focusing on only two points on the graph, and provides students a tool (multiplication and division equations) to extract relevant information. ### Launch It’s recommended that students try this task without using a calculator, but provide access to calculators if the calculations present too great a barrier. ### Student Facing For each question, the length of the second segment (on the right) is some fraction of the length of the first segment (on the left). Complete the division and multiplication equations that relate the lengths of the segments. $$7 \div 14 = \frac12$$ $$14 \boldcdot \frac12 = 7$$ $$\boxed{\phantom{33}} \div \boxed{\phantom{33}} = 3$$ $$\boxed{\phantom{33}} \boldcdot \boxed{\phantom{33}} = 12$$ $$8 \div 12 = \boxed{\phantom{33}}$$ $$12 \boldcdot \boxed{\phantom{33}}=8$$ $$\boxed{\phantom{33}} \div \boxed{\phantom{33}}=\boxed{\phantom{33}}$$ $$\boxed{\phantom{33}} \boldcdot \boxed{\phantom{33}}=\boxed{\phantom{33}}$$ $$\boxed{\phantom{33}} \div \boxed{\phantom{33}}=\boxed{\phantom{33}}$$ $$\boxed{\phantom{33}} \boldcdot \boxed{\phantom{33}}=\boxed{\phantom{33}}$$ $$\boxed{\phantom{33}} \div \boxed{\phantom{33}}=\boxed{\phantom{33}}$$ $$\boxed{\phantom{33}} \boldcdot \boxed{\phantom{33}}=\boxed{\phantom{33}}$$ ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis Ask students to take a few minutes to look at the equations associated with segments that grow in length, and those associated with segments that shrink in length. Ask, “How can you tell from an equation if a second segment will be longer or shorter?” If students struggle to answer this question, use the examples to point out that some factors are greater than 1 and some are less than 1. ## 6.3: Medicine Wears Off (15 minutes) ### Activity In this practice activity, students can continue to write corresponding division and multiplication equations (as in the previous activity) in order to determine a decay factor. ### Launch Provide access to calculators so that students can focus on looking for a common decay factor rather than on doing computations. ### Student Facing Some different medications were given to patients in a clinical trial, and the amount of medication remaining in the patient’s bloodstream was measured every hour for the first three hours after the medicine was given. Here are graphs representing these measurements. 1. For one of these medicines, the relationship between medicine remaining and time is not exponential. Which one? Explain how you know. 2. For the other four medicines: 1. How much was given to the patient? 2. By what factor does the medicine remaining change with each passing hour? 3. How much medicine will remain at 4 hours? 3. Which medicine leaves the bloodstream the quickest? The slowest? Explain how you know. ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis Invite students to share their responses and their reasoning process. If students write division and multiplication equations to make sense of the given information and extract the growth factor, write these equations next to the graph. In order to find the amount of medicine at 4 hours, highlight the approach of multiplying the amount at 3 hours by the decay factor that was found earlier in the previous question.
# Expanding Brackets and Simplifying Expanding brackets is a process that helps us simplify expressions by multiplying each term inside the brackets with the term outside. It’s based on the distributive property, which states that when we have a number multiplied by a sum, like , we can multiply the number by each term in the sum separately and then add the results: . • Therefore, Also: Remember that . ## FOIL Method The FOIL method is a technique used to expand brackets containing two binomials, which are expressions with two terms. The acronym “FOIL” stands for “First, Outer, Inner, Last,” representing the order in which the terms in the binomials are multiplied. This systematic approach ensures that each term in one binomial is multiplied by each term in the other binomial. Here’s how the FOIL method works: 1. First: Multiply the first terms of both binomials. 2. Outer: Multiply the outer terms of both binomials. 3. Inner: Multiply the inner terms of both binomials. 4. Last: Multiply the last terms of both binomials. After completing these steps, combine any like terms to simplify the resulting expression. Example: Consider the following expression: (A + B)(C + D) Applying the FOIL method, we perform the following multiplications: 1. First: A × C = AC 2. Outer: A × D = AD 3. Inner: B × C = BC 4. Last: B × D = BD Now, combine the results: AC + AD + BC + BD So, the expanded form of (A + B)(C + D) using the FOIL method is AC + AD + BC + BD. Let’s look at some examples. ## Examples Example 1: Expand and simplify the following: i) ii) iii) iv) v) i) ii) iii) iv) v) Example 2: Expand and simplify the following: i) ii) iii) iv) v) i) ii) iii) iv) v)
You are on page 1of 4 July 3, 2019 Wednesday I. OBJECTIVES At the end of the lesson, the students must be able to: 1. write quadratic equation in standard forms; 2. identify the values of a, b and c of quadratic equation; 4. perform the indicated operations with speed and accuracy. II. SUBJECT MATTER References: Learner’s Module for Mathematics Grade 9, pp. 47 - 55 Teacher’s Guide for Mathematics Grade 9, pp. 33 - 38 Materials: cartolina and marker, activity sheets III. PROCEDURES A. Priming/Activating Prior Knowledge Module 1, Lesson 2D, p. 48 Directions: Write the following quadratic equations in standard form, ax 2 + bx + c = 0. Then, identify the values of a, b and c. 1. -2x2 = 2 - 7x 3. x(5 - 2x) + 15 = 0 5. 3(x - 5)2 + 10 = 0 2. 10 + 7x - 3x2 = 0 4. (x - 6)(x - 9) = 0 B. Activity Activity 4: Lead Me to the Formula Module 1, Lesson 2D, p. 49 Directions: Find the solutions of the following quadratic equation by completing the square. 2x2 + 9x + 10 = 0 C. Analysis 1. How did you use completing the square in solving the given equation? Show the complete 2. What are the solutions of the given equation? 3. How would you describe the solutions obtained? 4. Compare your work with those of other groups. Did you obtain the same solution? If not, explain. 5. In the equation 2x2 + 9x + 10 = 0, what would be the resulting equation if 2, 9, and 10 were replaced by a, b, and c, respectively? 6. Using the resulting equation in item 5, how are you going to find the value of x if you follow the same procedure in finding the solutions of 2x 2 + 9x + 10 = 0? 7. What equation or formula would give the value of x? 8. Do you think the equation or formula that would give the value of x can be used in solving D. Abstraction To solve any quadratic equation ax2 + bx + c = 0 using the quadratic formula, determine the values of a, b, and c, then substitute these in the equation  b  b 2  4ac x 2a Simplify the result if possible then check the solutions obtained against the original equation. Example: 2x2 + 3x = 27 Solution: 2x2 + 3x - 27 = 0 a = 2, b = 3, c = 27  (3)  (3) 2  4( 2)(27) x 2( 2)  3  15  3  15 x 3 9  216 x 4 x 4 4  18 12 x  3  225 x 4 x 4 9 4 x3 x  3  15 2 x 4 Checking: 2x2 + 3x = 27 x=3 x = -9/2 2(3)2 + 3(3) = 27 2(-9/2)2 + 3(-9/2) = 27 2(9) + 9 = 27 2(81/4) -27/2 = 27 18 + 9 = 27 81/2 - 27/2 = 27 27 = 27 54/2 = 27 27 = 27 E. Application Find the solutions of the following quadratic equations using the quadratic formula. Tell whether the solutions are real numbers or not real numbers. Explain your answer. 1. x2 + 2x + 9 = 0 3. (2x - 5)2 - 4 = 0 2. 2x2 + 4x + 7 = 0 4. (x + 2)2 = 3x + 10 IV. EVALUATION Activity 5: Is the Formula Effective? Module 1, Lesson 2D, p. 52 Directions: Find the solutions of the following quadratic equations by using quadratic formula. 1. x2 + 10x + 9 = 0 3. x2 + 5x - 14 = 0 5. 2x2 + 4x = 3 2. x2 - 12x + 35 = 0 4. 2x2 + 7x + 9 = 0 6. Do you think the quadratic formula is more appropriate to use in solving quadratic equations? V. CLOSURE Activity 8: Show me the Best Floor Plan!!! Module 1, Lesson 2D, p. 55 Directions. Use the situation below to answer the questions that follow. Mr. Luna would like to construct a new house with a floor area of 72 m2. He asked an architect to prepare a floor plan that shows the following: a. 2 bedrooms d. Comfort room b. Living room e. Kitchen c. Dining room f. Laundry area 1. Suppose you were the architect asked by Mr. Luna to prepare a floor plan. How will you do it? Draw the floor plan. 2. Formulate as many quadratic equations using the floor plan that you prepared. Solve the equations using the quadratic formula. Index of Mastery No. of Learners Within Mastery Level Needing Reinforcement Needing Remediation NAME: _______________________ GRADE 9 - _____________ DATE: ____________ Activity No. _____ Directions. Use quadratic formula to find the roots of the following quadratic expressions. Write the correct words of the roots for you to answer the trivia. What is BLENNOPHOBIA? 1. 2x2 - 8x + 3 = 0 4. 3x2 + 2 = 5x 2. x2 + 14x - 7 = 0 5. 3x2 + 5x - 3 = 5 + 3x - 4x2 3. x2 +4(x + 3) = x  7  2 14 4  10 2 x x  1, 2 3 SLIME FEAR WATER  1  57 4  10 x  3,1 x x 7 6 OF HEIGHTS THE 3 1 5 2
# Converting Fahrenheit to Celsius Fahrenheit and Celsius are two distinct but widely used temperature scales. The unit of measure for Fahrenheit is written using the degree Fahrenheit symbol (°F), whereas the unit for Celsius is denoted by the degree Celsius symbol (°C). Fahrenheit is the most commonly used temperature measurement in the United States. However, if you travel to other countries, you will notice that Celsius is more commonly used. Therefore, learning how to convert temperatures between the two scales can help you understand what kind of temperature is being discussed no matter where you are in the world. ## How to Convert from Fahrenheit to Celsius When converting from Fahrenheit to Celsius, we use the following standard formula: If we take a closer look at the formula, below are the steps to convert a given Fahrenheit temperature to Celsius. 1. Subtract ${32}$ from the given Fahrenheit temperature (simplify operations that are inside the parenthesis first) 2. Multiply the difference by $\Large{5 \over 9}$ 3. Express or write the answer with the degrees Celsius symbol (°C) ## Alternate Method (°F to °C) Another formula that we can use as an alternative to convert temperatures from the Fahrenheit scale to the Celsius scale is dividing the quantity $\left( {^\circ F – 32} \right)$ by $\textbf{{1.8}}$. But you might ask, how did we get this alternate method? Well, multiplying the quantity $\left( {^\circ F – 32} \right)$ by $\Large{5 \over 9}$ ${\Large{5 \over 9}}\left( {^\circ F – 32} \right)$ is equivalent to dividing the quantity $\left( {^\circ F – 32} \right)$ by $\Large{9 \over 5}$. Notice that the numerator and denominator have switched. $\Large{{\left( {^\circ F – 32} \right)} \over {\large{{9 \over 5}}}}$ Since ${\Large{9 \over 5}} = 1.8$, we have the alternate formula: Now, let’s convert 86° Fahrenheit to degrees Celsius using both formulas to see if we get the same answer. And we did! Using both formulas, we get the same answer of 30° C. Therefore, we can say that 86 degrees Fahrenheit is 30 degrees Celsius. ### Converting Temperatures from Fahrenheit to Celsius Examples There is no such thing as too much practice. For a better understanding of the formulas and steps involved in Fahrenheit to Celsius conversion, let’s go through a few examples. Example 1: Convert 140° F to °C. For this first example, we’ll use the standard formula for converting temperature to the Celsius scale. Answer: 140 degrees Fahrenheit is 60 degrees Celsius OR 140° F = 60° C. Example 2: What is 96.8 degrees Fahrenheit in Celsius? The Fahrenheit temperature in this example contains a decimal. Will it also have a decimal once converted to the Celsius scale? Let’s find out. Using the alternative formula, we have Our answer came up as a whole number. Therefore, just because a Fahrenheit temperature has a decimal, it does not mean that the corresponding Celsius temperature will also have a decimal. Example 3: Convert -13° F to °C. Sometimes, you’ll also be given temperatures in the negatives. This is common especially if you’re talking about very cold or below freezing temperatures. However, as long as you know the formula, converting them from Fahrenheit to Celsius is easy. Answer: -13 degrees Fahrenheit is -25 degrees Celsius OR -13° F = -25° C. Example 4: The normal body temperature of a cheetah is at 101° F. After sprinting during a hunt, its body temperature increased 1.2° F. What was its increased body temperature in Celsius? Before we do any conversion, we first need to find out the cheetah’s increased body temperature in Fahrenheit. We’ll do so by adding 1.2° F to it’s normal body temperature. • Cheetah’s Normal Body Temperature: 101° F • Cheetah’s Body Temperature After Sprinting: 101° + 1.2° = 102.2° F Now that we know it’s increased body temperature in Fahrenheit, we can proceed with converting 102.2° F to Celsius. We’ll use the alternative formula for this one. Answer: The cheetah’s increased body temperature was 39° C. Example 5: When I woke up on Christmas morning last year, the temperature outside was 25° F. By dinner time, the outside temperature has decreased by 6° F. What was the decreased temperature in Celsius? First, we need to get the decreased temperature in Fahrenheit. • Morning Outside Temperature: 25° F • Dinner Time Outside Temperature: 25° – 6° = 19° F Next, let’s find out what 19° F is in Celsius using the Fahrenheit to Celsius conversion formula. Answer: It was about -7.2° C outside by dinner time on Christmas day last year. You may also be interested in these related math lessons or tutorials: Converting Fahrenheit to Celsius Practice Problems with Answers Converting Celsius to Fahrenheit
, 29.10.2019dedrick31 # Which shows the expressions rewritten with the least common denominator? 7x-2/4x^2 and x-1/8x The last one. Explanation: When comparing two fractions with variables like this, it's important to get to the same denominator in order to compare apples with apples and then be able to do not only comparisons but also perform additions/subtractions. Question is which denominator to use and how to reach it. In this case, the question and the answer choices do the work for you. The question asks which one is the LEAST common denominator, and the answers show denominators 8x², 16x² and 16x³. The smallest of these is 8x². So, how do we transform each fraction to get to an 8x² denominator? We multiply them by 1, expressed in a different way. Since we're multiplying by one, we're not affecting the value, just the way it looks. Let's start with the first one, to get the denominator to go from 4x² to 8x², we need to multiply it by 2... so we'll multiply by 2/2 (which is 1, neutral for the multiplication). We do a similar process with the second fraction, but this one to go from 8x to 8x², has to be multiplied by x. And now you have both fractions on the same denominator, without having changed their value, just their looks The last one. Explanation: When comparing two fractions with variables like this, it's important to get to the same denominator in order to compare apples with apples and then be able to do not only comparisons but also perform additions/subtractions. Question is which denominator to use and how to reach it. In this case, the question and the answer choices do the work for you. The question asks which one is the LEAST common denominator, and the answers show denominators 8x², 16x² and 16x³. The smallest of these is 8x². So, how do we transform each fraction to get to an 8x² denominator? We multiply them by 1, expressed in a different way. Since we're multiplying by one, we're not affecting the value, just the way it looks. Let's start with the first one, to get the denominator to go from 4x² to 8x², we need to multiply it by 2... so we'll multiply by 2/2 (which is 1, neutral for the multiplication). We do a similar process with the second fraction, but this one to go from 8x to 8x², has to be multiplied by x. And now you have both fractions on the same denominator, without having changed their value, just their looks Step-by-step explanation: Least common denominator is the smallest number or expression that can be a common denominator for a set of fractions. Comparing the two expressions, and , the common denominator = 24, and the highest common factor is 2p. Therefore, the least common denominator = = 12p Thus, = and = The expression that shows the least common denominator when rewritten is . and . Step-by-step explanation: The given expression is: and . The least common denominator is: . We collect LCM for the denominator to obtain; and . We multiply out the parenthesis to obtain; and . Therefore the correct answer is B or the second option. ### Other questions on the subject: Mathematics The correct option is 3.Step-by-step explanation:It is given that a vertical line passing through the point (−5, −1).If a vertical line passes through a point (a,b), then the equat...Read More Mathematics, 21.06.2019, paaoolav By the polynomial remainder theorem, if is divisible by , then the remainder is 0 andNow,...Read More -a rotation 90° counterclockwise about the origin followed by a reflection across the y-axis...Read More
# Divide and Conquer Algorithms: Divide and Conquer on Trees - Cartesian Tree ## Introduction Divide and conquer algorithms are a powerful technique used by programmers to efficiently solve complex problems. In this tutorial, we will explore the concept of "Divide and Conquer on Trees" and specifically delve into the idea of the Cartesian Tree. By the end of this tutorial, you will have a solid understanding of how to apply the divide and conquer paradigm to tree structures. ## What is a Cartesian Tree? A Cartesian Tree is a binary tree derived from a sequence of elements, where each node represents an element and satisfies the following properties: 1. The binary tree is in a heap-ordered arrangement, meaning the value of each node is greater than the values of its children. 2. The "Cartesian Property" is also satisfied, which means that the inorder traversal of the tree will give us the original sequence. ## The Structure of a Cartesian Tree To understand the structure of a Cartesian Tree, let's consider an example: Suppose we have the following input sequence: ``````[3, 2, 6, 1, 9, 4] `````` The Cartesian Tree derived from this input sequence would look like this: `````` 9 / \ 6 4 / \ / 2 1 3 `````` Observing this tree, we can notice that the maximum value in the input sequence is at the root, and all the elements less than this maximum value create the left subtree. Similarly, all the elements greater than this maximum value create the right subtree. This property holds recursively for each subtree. ## Building a Cartesian Tree: The Divide and Conquer Approach Now that we understand the structure of a Cartesian Tree, let's dive into the process of building one using the divide and conquer approach. The steps to build a Cartesian Tree are as follows: 1. Find the index of the maximum element in the input sequence. 2. Create a node with this maximum element as the root of the tree. 3. Recursively build the left subtree using the elements to the left of the maximum element's index. 4. Recursively build the right subtree using the elements to the right of the maximum element's index. 5. Connect the left and right subtrees to the root node. Here's a code snippet that implements the divide and conquer approach to build a Cartesian Tree in Python: ``````class TreeNode: def __init__(self, val): self.val = val self.left = None self.right = None def build_cartesian_tree(arr): if not arr: return None max_val = max(arr) max_idx = arr.index(max_val) root = TreeNode(max_val) root.left = build_cartesian_tree(arr[:max_idx]) root.right = build_cartesian_tree(arr[max_idx + 1:]) return root `````` ## Traversing a Cartesian Tree Once we have built a Cartesian Tree, we can perform various operations on it. One common operation is traversing the tree. Let's consider the following code snippet to traverse a Cartesian Tree using an inorder traversal: ``````def inorder_traversal(root): if root: inorder_traversal(root.left) print(root.val) inorder_traversal(root.right) `````` ## Conclusion In this tutorial, we explored the concept of Divide and Conquer on Trees, focusing on the Cartesian Tree. We learned how to build a Cartesian Tree by employing the divide and conquer paradigm, as well as how to traverse it using an inorder traversal. The Cartesian Tree is a powerful tool that can be applied to efficiently solve problems on tree structures. By understanding and utilizing this algorithmic paradigm, programmers can enhance their problem-solving abilities in various domains. Try experimenting with different input sequences and explore more applications of the Cartesian Tree algorithm. Happy coding! (Note: This blog post is written in Markdown format and can be converted to HTML using any Markdown-to-HTML converter.)
Lesson Explainer: Linear Inequalities \| Nagwa Lesson Explainer: Linear Inequalities \| Nagwa # Lesson Explainer: Linear Inequalities Mathematics In this explainer, we will learn how to solve linear inequalities and two linear inequalities combined using βandβ or βorβ in one variable. We can solve linear inequalities using very similar techniques to those of solving linear equations. Recall that we can solve a linear equation by isolating the variable (typically ) on one side of the equation. For instance, consider the equation By subtracting 7 from both sides, we get Then, by dividing both sides by 2, we have Thus, we have shown that the solution is . Letβs suppose instead that we had an inequality sign (e.g., ββ) in place of the equal sign: Once again, we can subtract 7 from both sides: Then, we can divide both sides by 2: We can see that the steps for solving the inequality are exactly the same. We just rearrange the inequality so that we can isolate the variable . The main difference is that the solution to a linear inequality is itself in the form of an inequality rather than a single value. Thus, the solution is actually a set of values for which the inequality is valid. The difference can be appreciated by considering a number line. Any value of greater than or equal to satisfies the inequality ; however, only the value of is a solution to the equation . Letβs work through a similar example of solving a straightforward linear inequality. ### Example 1: Solving an Inequality Find the set of values of that satisfy the inequality . To solve an inequality in , we want to isolate on one side of the inequality by applying the same operations to both sides of the inequality. We can start by subtracting 7 from both sides of the inequality to obtain We now want to divide through by 3; note that since this value is positive, we do not need to switch the direction of the inequality. We have Hence, the set of values of that satisfy the inequality is . It is very important to consider the direction of the inequality after applying operations to both sides of the inequality transformations. For example, imagine we want to solve . We can do this by asking which values are greater than 1 after being multiplied by . We can note that , and so satisfies the inequality. In fact, any value below will satisfy the inequality. We can justify why this happens by considering a geometric interpretation. Multiplying a number by on a number line switches its sign; it is the same as reflecting the number line through 0. This tells us that multiplying both sides of an inequality by a negative number will switch the direction of the inequality since it involves a reflection. We have the following properties that we can use to solve inequalities. ### Property: Properties of Inequalities • We can add any value to or subtract any value from both sides of an inequality to obtain an equivalent inequality. This can also be a variable. • We can multiply or divide both sides of an inequality by a positive constant to obtain an equivalent inequality. • We can multiply or divide both sides of an inequality by a negative constant to obtain an equivalent inequality by reversing the direction of the inequality. We can use these properties to solve linear inequalities by isolating the variable on one side of the inequality. In our next example, we will solve an inequality with -terms on either side. ### Example 2: Solving an Inequality with Multiple π₯-Terms Find the set of values of that satisfy the inequality . To solve an inequality in , we want to isolate on one side of the inequality by applying the same operations to both sides of the inequality. There are a few different ways we could do this. The first method involves adding to both sides of the inequality to obtain We can then subtract 1 from both sides of the inequality to get We can then divide the inequality through by 2; we note that this is positive, so the direction of the inequality is maintained. We have We usually write inequalities with the variable on the left, and we can do this by switching the direction of the inequality and swapping the sides to get The second method involves adding to both sides of the inequality to obtain We then subtract 5 from both sides of the inequality to get We now need to divide both sides of the inequality by to isolate . We recall that since this value is negative, we also need to switch the direction of the inequality. We get Hence, the set of values of that satisfy the inequality is . In our next example, we will solve an inequality by expanding the brackets and simplifying. ### Example 3: Solving an Inequality by Expansion Find the set of values of that satisfy the inequality . To solve an inequality in , we want to isolate on one side of the inequality by applying the same operations to both sides of the inequality. In order to do this, we first need to expand the brackets on both sides of the inequality. We do this by distributing each factor: We can then solve the inequality by isolating the variable. We add to both sides of the inequality to obtain We now add 4 to both sides of the inequality to get We now divide both sides of the inequality by 8 and note that this is a positive value, so we do not need to reverse the inequality. We obtain Hence, the set of values of that satisfy the inequality is . In our next example, we will solve an inequality by expanding the brackets and simplifying. ### Example 4: Solving an Inequality by Expansion Find the set of values of that satisfy the inequality . To solve an inequality in , we want to isolate on one side of the inequality by applying the same operations to both sides of the inequality. In order to do this, we first need to expand the brackets on both sides of the inequality: We now want to combine the like terms by rewriting the inequality. We can add to both sides of the inequality to get We can now add to both sides of the inequality. This gives us We now want to isolate , so we subtract 3 from both sides of the inequality: and then we divide both sides of the inequality by 9; we note that this is a positive value, so we do not need to reverse the inequality. We obtain We usually write the variable on the left-hand side of the inequality. We can do this by switching the direction of the inequality and swapping the sides to get Hence, the set of values of that satisfy the inequality is . Sometimes, we will want to solve multiple linear inequalities. We can do this by solving each linear inequality and then combining the solution sets at the end. It is also a good idea to sketch the solution set to each inequality on a number line to get a good idea of exactly what is happening. For example, letβs say we want to find the values of that solve both and . We start by solving each inequality by using the properties of the inequality. We subtract 1 from both sides of the first inequality to get We then divide both sides of the second inequality by 2 to obtain We want the values that satisfy both inequalities; we can find this by sketching both solution sets on a number line. We want the values that are in both solution sets, so they must be overlapping on the number line. We see that this is all the values between 1 and 3, including 3. We can write this as the compound inequality , which is really two separate inequalities: and . We can write this in set notation in two different ways. We can use the compound inequality to get . However, we can also use the individual inequalities and the fact that the intersection of sets means all the values in both sets to write this as . It is also worth noting that if there is no overlap between the two inequalities, then there are no values that satisfy both inequalities. In a similar way, we could also find the solutions to either of two inequalities. Letβs say we want to represent the solutions to either or . We can sketch these sets on a number line. We see that there is no overlap, so we must leave these as separate inequalities. We can also write this in set notation by recalling that the union of two sets gives us all the values in either set. Therefore, the solutions to either inequality are given by the set . In our next example, we will find and represent the solution to either of two given inequalities by using set notation. ### Example 5: Finding the Solutions to Either of Two Inequalities Find the set of values of that satisfy either the inequality or the inequality . Give your answer in set notation. We first note that we are looking for the values of that solve either of the two given inequalities. Therefore, we can start by finding the solutions of each inequality separately and then combining the solution sets afterward. Letβs start by solving . We expand the brackets on both sides of the inequality to obtain We now want to isolate on one side of the inequality; we can do this by subtracting from both sides of the inequality to get We then add 15 to both sides of the inequality. This gives us We usually write the variable on the left-hand side of the inequality. We can do this by switching the direction of the inequality and swapping the sides to get We can follow a similar process to find the solution set of the other inequality. We start by expanding the brackets to obtain We then add to both sides of the inequality to get We now subtract 20 from both sides of the inequality. This gives us We now divide both sides of the inequality by 11; we note that this is a positive value, so we do not need to reverse the inequality. We obtain We have found that the values of that satisfy one of the two inequalities are the values of that satisfy or . In order to help us visualize this set, we can represent both of the inequalities on a number line. We recall that we use a solid dot to show that the endpoint is included and an arrow to show the direction of the inequality. We get the following. There is no overlap in the two sets, so we have to leave these inequalities separate. We are told to represent the solution set by using set notation. We can represent each individual inequality in set notation. We have Finally, we want to say that we can take any value of in either of these sets. We recall that we do this by taking their union: In our final example, we will find and represent the solution to two given inequalities by using set notation. ### Example 6: Finding the Solution Set to Two Inequalities Find the set of values of that satisfy the inequalities and . Give your answer in set notation. We first note that we are looking for the values of that solve both of the two given inequalities. In order to do this, we first need to solve each inequality separately and then find the values inside both solution sets. Letβs start by solving . We expand the brackets on both sides of the inequality to obtain We now want to isolate on one side of the inequality; we can do this by subtracting from both sides of the inequality to get We then add 3 to both sides of the inequality. This gives us We can follow a similar process to find the solution set of the other inequality. We start by expanding the brackets to obtain We then add to both sides of the inequality to get We now subtract 15 from both sides of the inequality. This gives us We now divide both sides of the inequality by 11; we note that this is a positive value, so we do not need to reverse the inequality. We obtain We have found that the values of that satisfy one of the two inequalities are the values of that satisfy and . In order to help us visualize this set, we can represent both inequalities on a number line. We recall that we use a solid dot to show that the endpoint is included, a hollow dot to show that the endpoint is not included, and an arrow to show the direction of the inequality. We get the following. We want the values that are in both solution sets. In other words, they are overlapping. We can see from the diagram that this is all the values between and 1, and is not included. We can write this as an inequality: . However, we are asked to write this in set notation. Thus, the solution set is Letβs finish by recapping some of the important points from this explainer. ### Key Points • Adding or subtracting a constant or algebraic expression from either side of an inequality gives an equivalent inequality. • Multiplying or dividing both sides of an inequality by a positive number gives an equivalent inequality. • Multiplying or dividing both sides of an inequality by a negative number reverses the direction of the inequality. • We can represent solutions to inequalities on a number line by using a solid dot to represent an endpoint being included in the solution set, a hollow dot to represent an endpoint not being included in the solution set, and an arrow to show the direction of the solution set if it is unbounded. • We can also represent solutions to multiple inequalities by using set notation. We can use the union of two solution sets to show that a value from either set will satisfy one or more inequalities, and we can use the intersection of two solution sets to show that a value must be in both sets to satisfy one or more inequalities.
# How do you write an equation of a line given (-8,8) and (0,1)? Jul 28, 2017 See a solution process below: #### Explanation: First, we need to determine the slope of the line. The slope can be found by using the formula: $m = \frac{\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}}{\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}}$ Where $m$ is the slope and ($\textcolor{b l u e}{{x}_{1} , {y}_{1}}$) and ($\textcolor{red}{{x}_{2} , {y}_{2}}$) are the two points on the line. Substituting the values from the points in the problem gives: $m = \frac{\textcolor{red}{1} - \textcolor{b l u e}{8}}{\textcolor{red}{0} - \textcolor{b l u e}{- 8}} = \frac{\textcolor{red}{1} - \textcolor{b l u e}{8}}{\textcolor{red}{0} + \textcolor{b l u e}{8}} = - \frac{7}{8}$ We can now use the point-slope formula to write and equation for the line running between the two points in the problem. The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$ Where $\textcolor{b l u e}{m}$ is the slope and $\left(\textcolor{red}{{x}_{1} , {y}_{1}}\right)$ is a point the line passes through. Substituting the slope we calculated and the values from the first point in the problem gives: $\left(y - \textcolor{red}{8}\right) = \textcolor{b l u e}{- \frac{7}{8}} \left(x - \textcolor{red}{- 8}\right)$ $\left(y - \textcolor{red}{8}\right) = \textcolor{b l u e}{- \frac{7}{8}} \left(x + \textcolor{red}{8}\right)$ We can also substitute the slope we calculated and the values from the second point in the problem giving: $\left(y - \textcolor{red}{1}\right) = \textcolor{b l u e}{- \frac{7}{8}} \left(x - \textcolor{red}{0}\right)$ $\left(y - \textcolor{red}{1}\right) = \textcolor{b l u e}{- \frac{7}{8}} x$ We can also solve this equation for $y$ to put it into slope-intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$ Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value. $y - \textcolor{red}{1} = - \frac{7}{8} x$ $y - \textcolor{red}{1} + 1 = - \frac{7}{8} x + 1$ $y - 0 = - \frac{7}{8} x + 1$ $y = \textcolor{red}{- \frac{7}{8}} x + \textcolor{b l u e}{1}$
Informative line # Addition of Integers (Same Signs) • A negative number means loss and when we add two losses then the sum represents more loss. • A positive number means gain and when we add two gains then the sum represents more gain. • To add two integers having same signs: • Ignore the signs. • Find the sum of the given numbers. • The answer will have the same sign of the numbers. For example: We want to add two negative numbers, $$-5$$ and $$-15$$. $$-5+(-15)=-20$$ The answer is $$-20$$. Note: We can add the integers just like the whole numbers and keep the sign same. #### Find: $$(-75)+(-24)$$ A $$-99$$ B $$99$$ C $$100$$ D $$75+24$$ × When we add two integers with the same sign, the sign stays same for the sum as well. Ignore the signs and add directly. $$\begin{array}\ &7&5\\ +&2&4\\ \hline &9&9\\ \hline \end{array}$$ Thus, $$(-75)+(-24)=-99$$ Hence, option (A) is correct. ### Find: $$(-75)+(-24)$$ A $$-99$$ . B $$99$$ C $$100$$ D $$75+24$$ Option A is Correct # Addition of Integers (Different Signs) • We can add the integers with different signs. Steps to add integers having different signs: 1. Ignore the signs. 2. Find the difference of the given numbers. 3. The answer will have the sign of the greater number. • For example: Add $$-9$$ and $$3$$. $$-9+3=\Box$$ • First, ignore the sign and find the difference of $$9$$ and $$3$$. $$9-3=6$$ • Now, $$6$$ will have the sign of the greater addend. $$9$$ is greater than $$3$$. So, the sum will have the sign same as $$9,$$ i.e. '–' Thus, $$-9+3=-6$$ • Example: $$-15+3=-12$$ $$12+(-7)=5$$ $$-5+2=-3$$ #### What is the sum of $$35$$ and $$-25$$ ? A $$10$$ B $$-10$$ C $$-60$$ D $$60$$ × To find the sum of $$35$$ and $$-25$$, first, ignore the signs and find the difference of $$35$$ and $$25$$. $$35-25=10$$ Now $$10$$ will have the sign of the greater addend. $$35$$ is greater than $$25$$ with a positive sign. So, $$10$$ will also have the positive sign. Thus, $$35+(-25)=10$$ Hence, option (A) is correct. ### What is the sum of $$35$$ and $$-25$$ ? A $$10$$ . B $$-10$$ C $$-60$$ D $$60$$ Option A is Correct # Subtraction of Integers (Same Signs) • We can subtract the integers having same signs. • Follow the given steps to subtract the integers: Step 1: Look at the integer being subtracted. Step 2: Take the opposite of that integer (which is being subtracted). Step 3: Add the opposite to the first integer. For example: Subtract $$-9$$ from $$-15$$. $$-15-(-9)=\Box$$ • The integer being subtracted is $$-9$$, i.e. negative. Opposite of $$-9$$ is $$9$$. • Now, add $$-15$$ and $$9$$. Thus, $$-15+9=-6$$ Examples: 1. $$-8-(-4)=-4$$ 2. $$7-12=-5$$ 3. $$-20-(-10)=-10$$ #### Find: $$-28-(-30)$$ A $$-58$$ B $$2$$ C $$-2$$ D $$58$$ × The integer being subtracted is $$-30$$, i.e. negative. Opposite of $$-30$$ is $$30$$. Now, add $$-28$$ and $$30$$. $$-28+30=2$$ Hence, option (B) is correct. ### Find: $$-28-(-30)$$ A $$-58$$ . B $$2$$ C $$-2$$ D $$58$$ Option B is Correct # Subtraction of Integers (Different Signs) • We can subtract integers with different signs. • Follow the given steps to subtract the integers: Step 1: Look at the integer being subtracted. Step 2: Take the opposite of that integer (which is subtracted). Step 3: Add the opposite to the first integer. • For example: Subtract $$2$$ from $$-5$$. $$-5\,-2=\Box$$ • The integer being subtracted is $$2$$, i.e. positive. • Opposite of $$2$$ is $$-2$$. Now, add $$-2$$ to $$-5$$. $$-5+(-2)=-7$$ • Examples: 1. $$70-(-30)=100$$ 2. $$-15\,-45=-60$$ 3. $$55-(-25)=80$$ #### Find: $$-456-44$$ A $$-500$$ B $$-412$$ C $$412$$ D $$500$$ × The integer being subtracted is $$44$$, i.e. positive. Opposite of $$44$$ is $$-44$$. Now, add $$-44$$ to $$-456$$, $$-456+(-44)=-500$$ Hence, option (A) is correct. ### Find: $$-456-44$$ A $$-500$$ . B $$-412$$ C $$412$$ D $$500$$ Option A is Correct
# How do you divide (8x^3+5x^2-12x+10)/(x^2-3)? May 13, 2016 $\frac{8 {x}^{3} + 5 {x}^{2} - 12 x + 10}{{x}^{2} - 3} = 8 x + 5 + \frac{12 x + 25}{{x}^{2} - 3}$ #### Explanation: I like to long divide just the coefficients, not forgetting to include $0$'s for any missing powers of $x$. In our current example that means the missing term in $x$ in the divisor... The process is similar to long division of numbers. Write the dividend under the bar and the divisor to the left. Write the first term $\textcolor{b l u e}{8}$ of the quotient above the bar, choosing it so that when multiplied by the divisor $1 , 0 , - 3$ the product matches the first term of the dividend. Write the product $8 , 0 , - 24$ of $8$ and the divisor under the dividend and subtract it to give a remainder $5 , 12$. Bring down the next term from the dividend alongside it, then choose the next term $\textcolor{b l u e}{5}$ of the quotient to match the leading term of our running remainder. Write the product $5 , 0 , - 15$ under the running remainder and subtract it to give the final remainder $12 , 25$. There are no more terms to bring down from the dividend and the remainder is now shorter than the divisor, so this is where we stop. Our quotient is $8 , 5$, meaning $8 x + 5$ and our final remainder is $12 , 25$ meaning $12 x + 25$ So: $\frac{8 {x}^{3} + 5 {x}^{2} - 12 x + 10}{{x}^{2} - 3} = 8 x + 5 + \frac{12 x + 25}{{x}^{2} - 3}$
R " COOL" STUFF ABOUT R RR POLAR EQUATIONS RR (The Journey from Polar to Rectangular) by Dixie Williford Polar equations are composed of the polar coordinateswhere r is the directed distance of a point P on the polar curve to the origin; and is the angle measure, in radians, of the point P from the x-axis. Therefore, the coordinates determine the location of the point P. It is possible to rewrite an equation that is in polar coordinates as rectangular coordinates using the following coversion identities: R R R R The following is an exploration into three specific polar functions: R R R FIRST, Let's take a look at the graph of . The graph appears to be an ellipse. But, how can we know for sure? If it is an ellipse, how can we determine, algebraically, information about the center and the major and minor axes? Merely, by observation, we can estimate the following information : R Major Axis: 5 R Minor Axis: 4 R Center: (3,0) We could, therefore, guess that the equation for this figure, in rectangular coordinates is: .This information, however, would be more difficult to observe for various figures. Click here to see if we have indeed confirmed that the figure is the ellipse described above. SECONDLY,let us consider the polar equation: As you will notice, this equation is very similar to the previous one. The only difference between the two is the angular measure of which the cosine is being taken for each . What is the difference between verses ? Let's take a look at the graphs of these two polar equations together : in blue; in green. In rectangular coordinates, cos(x-2) is the translation of the function cos(x) 2 units to the right. Similarly, in polar coordinates, is the rotation of about the orgin by 45 degrees in the positive direction. Let's take this knowledge about the cosine function as we look at the polar equation of real interest: and how it compares with: . Let's look at the graphs of these two functions togethter : in red; in blue. Again, it seems as though the blue graph is a rotation of the red about the origin. Does it hold that the rotation angle is by 45 degrees as we saw in the graphs of cosine? The following table shows values of both as well as as increases . for for (0,8) (0,5.5581) (,5.5581) (,8) (,3.2) (,5.5581) (, 5.5581) (,3.2) By looking at this table of values, you can see that indeed is the a or 45 degrees rotation of . Click here for a Graphing Calculator 2.2 animation demonstrating how this fact. THIRDLY, let's investigate the function . The following is a graph of the function: Why is it that we get an assymptote at x=2 and y=-2? It is much easier to answer these questions by looking at the equation in its rectangular coordinates. It is easy to see that has a vertical assymptote at x=(-2) because this value of x gives a zero in the denominator of the function. The horizontal assymptote can be found by taking the limit, L, of as x approaches infinity. Since the highest power of x in the numerator is equivalent to the highest power of x in the denominator, L = (-2/1) = -2. Therefore, a horizontal assymptote occurs at y=-2. Therfore, because the vertical assymptote is determined by the "2" in the original polar equation, we see that whatever constant, c, appears in the numerator x= -c is the vertical assymptote. To test this out, let's try graphing As you can see, the vertical assymptote is at x= -5, as predicted. Notice, however, that by altering the numerator, you also alter the location of the horizontal asssymptote as well. By reverting to our rectangular coordinates, we will find that the new horizontal assymptote lies as y = - 5. IN SUMMARY, we find that it is fairly easy (dispite slightly messy Algebra) to converty from polar equations to rectangular equations. Through this fairly simple conversion we learn some things about our graphs that are not so obvious from the polar form of the equation. It is very useful to be able to do such conversions in order to show students the relationships between the polar and rectangular coordinate sysytem. This conversion process can help students understand that one graph can be represented in numerous ways, one of which is its polar equation. RETURN
Education.com # Proportion Word Problems Practice Questions (not rated) To review these concepts, go to Proportion Word Problems Study Guide. ## Proportion Word Problems Practice Questions ### Practice 1 #### Problems Circle true or false for each statement that follows. 1 The proportion is a true proportion. True False 2 The proportion is a true proportion. True False 3 In the proportion , x = 3 makes a true proportion. True False 4 In the proportion , x = 5 makes a true proportion. True False #### Solutions 1. True; each cross product is equal to 90. 2. False; the cross products are 39 and 48, and they are not equal. 3. False; use x = 4 to make it a true proportion. 4. True. When x = 5, each of the cross products is 120. ### Practice 2 #### Problems 1. If the scale on a map is 1 inch = 100 miles, what is the actual distance between two locations 3 inches apart on the map? 2. The scale on a map is inch = 15 miles. How far apart on a map are two cities that are actually located 150 miles apart? 3. The height of a window on a scale drawing is 1.5 inches. If the scale is 1 inch = 4 feet, what is the actual height of the window? #### Solutions 1. The proportion could be set up as: 2. Cross multiply the means and the extremes, and solve the equation. 1x = 3 × 100 x = 300 miles The actual distance between the two locations is 300 miles. 3. The proportion could be set up as: 4. Cross multiply the means and the extremes, and solve the equation. 15x = 75 miles Divide each side of the equation by 15. x = 5 inches The distance on the map between the two locations is 5 inches. 5. The proportion could be set up as: 6. Cross multiply the means and the extremes, and solve the equation. 1x = 1.5 × 4 x = 6 feet The actual height of the window is 6 feet. ### Practice 3 #### Problems 1. Justine ran the first 6 miles of a 10-mile race in 48 minutes. At this pace, how long will it take her to run the entire 10 miles? 2. Harold makes \$204 when he works a shift of 5 hours. How long does he need to work in order to make \$326.40? 3. The scale on a drawing is 1 inch = 12 inches. If the length of a wall on the drawing is 9 inches, what is the actual length of the wall? 4. The ratio of flour to salt in a recipe for molding dough is 4:1. If Sam is using 2 cups of salt in his mixture, how many cups of flour should he use? ### Ask a Question 150 Characters allowed ### Related Questions #### Q: See More Questions ### Today on Education.com #### WORKBOOKS May Workbooks are Here! #### ACTIVITIES Get Outside! 10 Playful Activities #### PARENTING 7 Parenting Tips to Take the Pressure Off Welcome!
Question # The ages of A and B are in the ratio 3 : 5; eight years later their ages will be in the ratio 5 : 7. Find their current ages. Hint: Assume the ages of A and B as variables and apply the two conditions to get the system equations with two variables. Solve these equations and you will get the current ages of A and B. To solve the given problem we will assume the current age of A as â€˜xâ€™ and the current age of B as â€˜yâ€™. By using the notations we will write the given data as follows, The current ages of A and B are in the ratio 3 : 5 Therefore, x : y = 3 : 5 The above equation can also be written as, $\dfrac{x}{y}=\dfrac{3}{5}$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1) Also, After eight years the ratio of ages of A and B will become 5 : 7 Therefore, x + 8 : y + 8 = 5 : 7 The above equation can also be written as, $\dfrac{x+8}{y+8}=\dfrac{5}{7}$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (2) As we have written the given data therefore we will write the equation (1) and equation (2) one by one, Therefore equation (1) will become, $\dfrac{x}{y}=\dfrac{3}{5}$ By cross multiplication in the above equation we will get, $\Rightarrow 5\times x=3\times y$ If we shift 5 on the right hand side of the equation we will get, $\Rightarrow x=\dfrac{3\times y}{5}$ $\Rightarrow x=\dfrac{3}{5}y$ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (3) Also equation (2) will become, $\dfrac{x+8}{y+8}=\dfrac{5}{7}$ By cross multiplication in the above equation we will get, $\Rightarrow 7\times \left( x+8 \right)=5\times \left( y+8 \right)$ If we multiply the constants inside the bracket we will get, $\Rightarrow 7\times x+7\times 8=5\times y+5\times 8$ $\Rightarrow 7x+56=5y+40$ $\Rightarrow 56-40=5y-7x$ $\Rightarrow 16=5y-7x$ By rearranging the above equation we will get, $\Rightarrow 5y-7x=16$ Now we will put the value of equation (3) in the above equation therefore we will get, $\Rightarrow 5y-7\times \left( \dfrac{3}{5}y \right)=16$ $\Rightarrow 5y-\dfrac{21}{5}y=16$ If we multiply by 5 on both sides of the equation we will get, $\Rightarrow 5\times 5y-5\times \dfrac{21}{5}y=5\times 16$ $\Rightarrow 25y-21y=80$ $\Rightarrow 4y=80$ $\Rightarrow y=\dfrac{80}{4}$ Therefore, y = 20 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (4) Therefore the current age of B is 20 years. Now we will put the value of equation (4) in equation (3), therefore we will get, $\Rightarrow x=\dfrac{3}{5}\times 20$ $\Rightarrow x=3\times 4$ Therefore, x = 12 Therefore the current age of A is 12 years. Therefore the ages of A and B are 12 years and 20 years respectively. Note: You can use A and B as the ages directly in place of using the variables so that you can get the direct answers in terms of A and B.
## Section12.1Kepler's Laws of Planetary Motion ### Subsection12.1.1Math of Ellipse Ellipse plays central role in planetary motion. Here, we review some of the basics of ellipse. Ellipse is an oval-shaped planar curve with two focal points $F_1$ and $F_2$ as shown in Figure 12.1.1. The sum of the distances $d_1$ and $d_2$ from the focal points to the points of the ellipse is constant, \begin{equation} d_1 + d_2 = \text{constant}. \end{equation} Let $c$ denote the distance of a focus from the center, $a$ the distance from the center to the end of the wider side, and $b$ the distance from the center to the end of the narrower side. The quantities $a$ and $b$ are also called semi-major and semi-minor axes, and quantity $c$ is called the focal distance. They are related by $$c^2 = a^2 - b^2,\label{eq-ellipse-csq-asq-bsq}\tag{12.1.1}$$ The ratio $c/a$ is called the eccentricity and denoted by $e\text{.}$ $$\text{Eccentricity, } e = \dfrac{c}{a}.\tag{12.1.2}$$ Ellipse in Cartesian Form Suppose we place an ellipse with wider side along $x$ axis and the narrower side along $y$ axis. This will place the center of the ellipse at the origin. The foci will be symmetrically located on $x$ axis at $(-c,0)$ and $(c,0)\text{.}$ The definition of ellipse says that any point $(x,y)$ on ellipse must obey $$\sqrt{ (x+c)^2 + y^2 } + \sqrt{ (x-c)^2 + y^2 } = \text{constant}\tag{12.1.3}$$ The points $(a,0)$ and $(-a,0)$ will be the end points of the ellipse on the $x$ axis. Then, $$\sqrt{ (x+c)^2 + y^2 } + \sqrt{ (x-c)^2 + y^2 } = 2a.\label{eq-ellipse-defining-equation}\tag{12.1.4}$$ By appropriately rearranging terms in Eq. (12.1.4) and squaring to remove radicals, and then using Eq. (12.1.1) we can obtain the standard form of equation of ellipse. $$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1.\label{eq-ellipse-defining-equation-standard}\tag{12.1.5}$$ Ellipse in Polar Form With Origin at Focus Sometimes, it is better to work with polar form of an ellipse that has origin at the focus and not at the center. Let $r$ denote the radial coordinate and $\theta$ counterclockwise from the $x$ axis that is along the major axis. Then, it is possible to show that if origin is at the focal point on the right then $$r = \dfrac{a\left( 1 - e^2\right)}{1 + e\,\cos\theta},\tag{12.1.6}$$ and if origin is at the focal point on the left then $$r = \dfrac{a\left( 1 - e^2\right)}{1 - e\,\cos\theta}.\tag{12.1.7}$$ If the origin was not at one of the foci, but at the center of the ellipse, we would get \begin{equation} r = \dfrac{b}{ \sqrt{1-e^2\,\cos^2\theta} }. \end{equation} Finally, we will find the formula of the area of an ellipse useful. $$\text{Area of an ellipse, } A = \pi\,a\,b.\tag{12.1.8}$$ ### Subsection12.1.2Kepler's Laws Johannes Kepler (1571-1630) analyzed the voluminous data of planetary positions recorded by Tycho Brahe (1546-1601) and deduced the following three laws of planetary motion. A good place to get planetary data now a days is at the NASA website. First Law or the Law of Elliptical Orbits Planets travel in elliptical orbits about the Sun with the Sun at one focus. Second Law or the Law of Equal Areas The line joining the Sun and a planet covers equal area in equal time. Thus when a planet is nearer to the Sun, it has a higher speed than when it is further out. Figure 12.1.3 illustrates this law. Third Law or the Law of Harmonies The ratio of square of the period of revolution about the Sun to the cube of the semi major axis of the elliptical orbit of two planets are equal to each other. Thus if $T_1$ and $T_2$ are time for planets 1 and 2 to go around the Sun, and $a_1$ and $a_2$ the semi-major axes of their elliptical orbits, we have the following equality (see Figure 12.1.4). $$\frac{T_1^2}{a_1^3} = \frac{T_2^2}{a_2^3},\ \ \text{or,}\ \ \left( \frac{T_1}{T_2} \right)^2 = \left( \frac{a_1}{a_2} \right)^3.\tag{12.1.9}$$ The closest and farthest distance of the Earth from the Sun are 0.98 AU and 1.02 AU, where, the Astronomical Unit, 1 AU = 149,598,000 km. Find (a) the semi-major axis, (b) the eccentricity of Earth's orbit, and (c) the semi-minor axis. Hint $2a =$ sum of the two distances given. (a) $1.0\text{ AU}\text{,}$ (b) $0.02\text{,}$ (c) $b=0.9998\text{ AU}\approx 1.00\text{ AU}\text{.}$ Solution (a) We will refer to the following figure. From the figure, we see that \begin{equation} a = \dfrac{0.98+1.02}{2} = 1.0\text{ AU}. \end{equation} (b) Therefore, the focal distance is \begin{equation} c = a - r_\text{min} = 1.0 - 0.98 = 0.02\text{ AU}. \end{equation} From $a$ and $c$ we get the eccentricity of the orbit. \begin{equation} e = \dfrac{c}{a} = \dfrac{0.02}{1.0} = 0.02\text{ (no units)} \end{equation} (c) \begin{equation} b^2 = a^2 - c^2 = 1.0^2 - 0.02^2 = 0.9996. \end{equation} Therefore, $b=0.9998\text{ AU}\approx 1.00\text{ AU}$ = a. Since $a\approx b\text{,}$ the orbit of Earth is very close to circular. The closest and farthest distance of Mars from the Sun are $1.38\text{ AU}$ and $1.67\text{ AU}\text{,}$ where, the Astronomical Unit, $1\text{ AU} = 149,598,000\text{ km}\text{.}$ Data: Semimajor axis of Earth's orbit = $1.0\text{ AU}$ and orbital period of Earth = $365.2\text{ days}\text{.}$ Find (a) the semi-major axis, (b) the eccentricity , and (c) the period of Mars's orbit. The semi-major axis of Earth is $1.0\text{ AU}$ and the orbital period is $365.2\text{ days}$ Hint $2a =$ sum of the two distances given. (a) $1.0\text{ AU}\text{,}$ (b) $0.02\text{,}$ (c) $687.8\text{ days} \text{.}$ Solution We will refer to the following figure. (a) From the figure, we see that \begin{equation} a = \dfrac{1.38+1.67}{2} = 1.525\text{ AU}. \end{equation} (b) Therefore, the focal distance is \begin{equation} c = a - r_\text{min} = 1.525 - 1.38 = 0.145 \text{ AU}. \end{equation} From $a$ and $c$ we get the eccentricity of the orbit. \begin{equation} e = \dfrac{c}{a} = \dfrac{0.145}{1.525} = 0.095\text{ (no units)} \end{equation} (c) From the semi-major axis of Mars and the semi-major axis and orbital period of Earth, we can find the orbital period of Mars by using Kepler's third law. \begin{equation} T_\text{Mars} = \left( \dfrac{1.525\text{ AU}}{1.0\text{ AU}} \right)^{3/2}\times 365.2\text{ days} = 687.8\text{ days}. \end{equation} The orbit of Halley's comet is approximately elliptical with $e=0.967\text{.}$ Halley's comet comes around every 76 years. Find (a) the distance of the closest approach to the Sun and (b) the farthest distance the comet goes from the Sun. Data: The semi-major axis of Earth is $1.0\text{ AU}$ and the orbital period is $365.2\text{ days}\text{,}$ where $1\text{ AU} = 149,598,000\text{ km}\text{.}$ Hint (a) and (b) : Use third law to find $a$ of Haley's comet first. (a) $0.59\text{ AU}\text{,}$ (b) $35.29\text{ AU}\text{.}$ Solution First we use Kepler's third law on orbits of Haley's comet and Earth to obtain the semi-major axis of Heley's comet. \begin{align} a_H \amp = \left( \dfrac{T_H}{T_E} \right)^{2/3}\, a_E, \\ \amp = \left( \dfrac{76\text{ yr}}{1\text{ yr}} \right)^{2/3}\, 1\text{ AU} = 17.94\text{ AU}. \end{align} Now, we use the definition of eccentricity to find the focal distance. \begin{equation} c = e\,a = 0.967\times 17.94\text{ AU} = 17.35\text{ AU}. \end{equation} Now, refer to the figure to obtain the min and max distances. \begin{align} \amp \text{(a) } r_\text{min} = a - c = 17.94-17.35= 0.59\text{ AU}.\\ \amp \text{(b) } r_\text{max} = a + c = 17.94+17.35= 35.29\text{ AU}. \end{align} The perihelion and aphelion distances of Mercury are $46.0$ and $69.8$ in units of $10^{6}\text{ km}\text{,}$ and that of Earth are $147.1$ and $152.1$ in the same units. What is the orbital period of Mars in Earth days if the orbital period of Earth is $365.2\text{ days}\text{?}$ Compare your answer to the listed value of $88.0\text{ days}\text{.}$ Hint Use Kepler's third law. $87.9 \text{ days}\text{.}$ Solution We can use Kepler's third law. \begin{equation} \left( \dfrac{T_1}{T_2} \right)^2 = \left( \dfrac{a_1}{a_2} \right)^3. \end{equation} Here, the semimajor axis $a$ is the average of the perihelion and aphelion distances. We don't even have to average them, we can just add them, since the dividing factor 2 will be canceled in the ratio. \begin{align} T_\text{mer} \amp = \left( \dfrac{a_\text{mer}} {a_\text{earth}} \right)^{3/2}\ T_\text{earth}, \\ \amp = \left( \dfrac{ 46.0 + 69.8}{ 147.1 + 152.1} \right)^{3/2}\times 365.2 = 87.9 \text{ days}, \end{align} This is impressively close to $88.0\text{ days}\text{.}$ The distances of various planets from Sun and their orbital periods are given in NASA website as follows. (Unit of distance is $10^6\text{ km}$ and the unit of time is Earth $\text{day}\text{.}$) The list order is [Mercury, Venus, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto]. The orbital periods are [88.0, 224.7, 365.2, 687.0, 4331, 10747, 30589, 59800, 90560] and the distances from the Sun are [57.9, 108.2, 149.6, 227.9, 778.6, 1433.5, 2872.5, 4495.1, 5906.4]. You suspect that $T \propto r^n\text{,}$ where $T$ is the period and $r$ the distance. But you don't know $n$ yet. We pretend we don't know - since this has already been found by Kepler, but here we will find it our own way from the data. Find $n$ from the given data. Hint Plot $\ln T$ versus $\ln r\text{.}$ $n = 1.50\text{.}$ Solution Since we suspect $T\propto r^n \text{,}$ we make it an equality first and set \begin{equation} T = c r^n, \end{equation} where $c$ is an unknown constant - at the moment of no interest. By taking natural log of both sides you get \begin{equation} \ln T = \ln c + n \ln r. \end{equation} Thus if we plot $\ln T$ on the $y$-axis and $\ln r$ on the $x$-axis, and fit the data to a straight line, the slope of this line will give us $n\text{.}$ The following figure shows the calculations and result done in Microsoft Excel. ### Subsection12.1.3Kepler's Impact on the Development of Physics Kepler's three laws played critical role in the discovery of the universal law of gravitation, and had major impact on the acceptance of science by the general public. By 1666 Newton had early versions of his three laws of motion, but yet did not have the law of gravitation. In that year he had an insight that Earth's gravity extended also to the Moon and was counterbalanced by the centrifugal force. Newton used the balance of the two forces to find that the gravitational force must decrease as inverse square of the distance. The calculation was, however, only for a circular motion. Newton did not work on the planetary motion problem any further till 1679 when another physicist, Robert Hooke, which you have met in the Hooke's law of the spring force, went to see him about the elliptical orbit problem of the planets. Hooke had conjectured that a planet moving in an ellipse must be acted on by a central force by the Sun. Hooke had also come to the conclusion that this force must vary as the square of the inverse of the distance of the planet from the Sun, but he could not prove his conjectures mathematically. After Hooke's visit, Newton went back to work on the planetary motion problem. First he showed mathematically that, if a body obeys Kepler's second law, then the force on the body must be central. Isaac Newton also showed that the angular momentum of a body is conserved if the body is acted upon by a central force. This finding demonstrated the physical basis of Kepler's second law. Next, Newton showed that if a body is in an elliptical path, then the force must be pointed towards one of the foci and must vary as the square of the inverse of the distance from the focus. However, Newton did not publish any of these results until after the great Astronomer, Edmund Halley (1656-1742), asked him in 1684 if he could prove Hooke's conjecture. To Haley's surprise, Newton immediately replied that he had proved it five years earlier. Upon Halley's insistence Newton wrote up his treatise on mechanics and its applications to the celestial mechanics. This work is called the Philosophiae naturalis principia mathematica, or Principia and was published in 1687. Halley used Newtonian calculations to ascertain that comets appearing in the sky in 1531, 1607 and 1682 were the same object and it was regularly appearing every 76 years. Sure enough the comet, now called Halley's comet, arrived on Christmas day in 1758 as predicted by Edmund Halley. The last time Halley's comet observable on Earth was in 1986 shown in Figure 12.1.10.
# Law of floatation and Archimedes principle difference One of the key points for the difference between the law of floatation and Archimedes’ principle is that the law of floatation is related to the weight of the liquid displaced by a floating body while Archimedes’ principle is related to the buoyant force exerted by the liquid on a partially or fully submerged body. Let’s understand them before looking at the difference. ## Law of floatation: According to the law of floatation, a floating object displaces the amount of liquid that weighs equal to the weight of the object. ∴ Weight of liquid displaced by floating object = Weight of the object Example:- The above figure shows the object of weight ‘W’ floating in the liquid. In this case the weight of the liquid displaced by the floating object is given by, \mathbf{W_{\text{object}}} = \mathbf{W_{\text{liquid displaced}}} ## Archimedes’ principle: Archimedes’ principle says that the upward force or buoyant force exerted by the liquid on a partially or fully submerged object is equal to the weight of the liquid displaced by the object. Thus according to Archimedes’ law, \mathbf{F_{\text{Buoyant}}} = Weight of liquid displaced Example:- The below figure shows a cube of side ‘a’ floating in liquid up to the depth of \frac{a}{2}. The liquid displaced by an object is equal to the volume of the portion of the object immersed in liquid. V_{\text{displaced}} = V_{\text{Immersed}} Weight of liquid displaced is given by, W_{\text{liquid displaced}} = V_{\text{displaced}} x \rho_{\text{liquid}} According to Archimedes’ principle, the buoyant force acting on an object is given by, F_{\text{buoyant}} = W_{\text{liquid displaced}} F_{\text{buoyant}} = V_{\text{displaced}} x \rho_{\text{liquid}} For the object shown in the above figure, the buoyant force is given by, F_{\text{buoyant}} = V_{\text{displaced}} x \rho_{\text{liquid}} F_{\text{buoyant}} = V_{\text{Immersed}} x \rho_{\text{liquid}} F_{\text{buoyant}} = (a x a x \frac{a}{2}) x \rho_{\text{liquid}} F_{\text{buoyant}} = \frac{a^{3} \times \rho_{\text{liquid}}}{2} ## Law of floatation and Archimedes principle difference: Pratik is a Graduated Mechanical engineer. He enjoys sharing the engineering knowledge learned by him with people.
# Base-Ten Numeral – Definition with Examples Welcome to Brighterly, where the world of mathematics is at your fingertips! We’re here to illuminate the fascinating journey through numbers, starting with one of the most fundamental concepts: the base-ten numeral system. A system so ingrained in our daily lives that its importance often goes unnoticed. But rest assured, a solid understanding of the base-ten numeral system will light the way to a brighter mathematical future. Base-ten numerals, also known as decimal numbers, are the backbone of our everyday number system. They form the foundation upon which the towering edifices of mathematics are built. In this blog post, we will delve into the intriguing world of base-ten numerals, unravel their intricacies, and illustrate their ubiquitous application. We’ll go step by step, introducing the base-ten place value chart, explaining how to find the place value of a digit, and much more. Along the way, we’ll be providing examples and practice questions to enhance your understanding and strengthen your numerical abilities. ## What Are Base-Ten Numerals? The world of mathematics is abundant with complex concepts, but at the core of them all, there’s the simple, yet crucial concept of base-ten numerals. But what are base-ten numerals? Quite simply, base-ten numerals, also known as decimal numbers, are the standard numbers we use every day: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Each number in this system is a combination of these ten digits, placed in different positions to signify varying values. The base-ten system is a part of the broader field known as positional notation, where a digit’s value depends on its position in a number. The position of a digit, going left to right, signifies an increasing power of ten, which forms the basis of this system. ## Base-Ten Place Value Chart Understanding a base-ten place value chart is pivotal to grasping the base-ten system. The chart shows how each digit’s position in a number determines its value. The digit farthest to the right in a number is in the ones place, then tens, hundreds, thousands, and so on. Each place value is ten times the place value to its immediate right. For example, the number 2435 is broken down as 2000 (2 thousands), 400 (4 hundreds), 30 (3 tens), and 5 (5 ones). Learning how to read a base-ten place value chart enables one to comprehend and manipulate numbers effectively. ## How to Find the Place Value of a Digit? Finding the place value of a digit is simple when you know the base-ten system. Look at the digit and its position from the right. If it’s the first digit from the right, its place value is ones (10^0), the second is tens (10^1), the third is hundreds (10^2), and so forth. For example, in the number 5312, the place value of 3 is hundreds. You can find more detailed guidance on the official education website. ## Base-10 (Decimal Number System) The base-10, or decimal number system, is the foundation of our daily numerical language. The term “decimal” comes from the Latin ‘decimus’, meaning tenth. As the name suggests, this system is based on powers of ten. The decimal system is universally used due to its simplicity and convenience. Its wide use in everyday life can be attributed to humans historically counting on their ten fingers. The decimal number system is also widely used in scientific notation, financial calculations, computing, and more. ## Numbers in Decimal Number System When we refer to numbers in the decimal number system, we are talking about numbers composed of ten unique digits ranging from 0 to 9. Every natural number we use daily belongs to this system. For instance, when we say the number 6537, we mean six thousands (10^3), five hundreds (10^2), three tens (10^1), and seven ones (10^0). ## Base of a Number System The base of a number system defines how many unique digits can be used in that system. It’s the number of different digits or combination of digits and letters that a system of counting uses to represent numbers. For the base-ten numeral system, the base is 10, meaning it uses ten unique digits (0-9). There are other systems as well, like binary (base-2), octal (base-8), and hexadecimal (base-16), which are frequently used in computer science. ## How to Show the Base-Ten? To denote a number as base-ten, we often use subscript notation. However, because the decimal system is so commonly used, numbers without a specified base are typically assumed to be in base-ten. For example, we write 6537_(10) to represent the number 6537 in base-ten. ## Expanded Form The expanded form of a base-ten number displays the value of each digit according to its place value. For instance, 4653 in expanded form would be 4000 + 600 + 50 + 3. This representation allows children to understand the importance of each digit’s position in a number. ## Place Values in Decimals Just like in whole numbers, place values in decimals follow the base-ten system, but they extend to the right of the decimal point. The first place to the right of the decimal point is tenths (1/10 or 10^-1), the second is hundredths (1/100 or 10^-2), and so forth. ## Solved Examples on Base-Ten Understanding base-ten numerals can be greatly assisted through practical examples. Let’s explore this concept further with some detailed examples: Example 1: Consider the number 2356.78. When expanded according to the base-ten system, we break this number down into each place’s value. Hence, it translates to: • 2000 (2 in the thousands place or 2 * 10^3), • 300 (3 in the hundreds place or 3 * 10^2), • 50 (5 in the tens place or 5 * 10^1), • 6 (6 in the ones place or 6 * 10^0), • 0.7 (7 in the tenths place or 7 * 10^-1), • 0.08 (8 in the hundredths place or 8 * 10^-2). So, 2356.78 can be written in expanded form as 2000 + 300 + 50 + 6 + 0.7 + 0.08. Example 2: Let’s take another number, say 4096. The expanded form would be 4000 + 0 + 90 + 6. Example 3: For a number with more decimal places, like 45.1203, the expanded form would be 40 + 5 + 0.1 + 0.02 + 0.003. These examples should help students understand how each digit contributes to a number’s total value in the base-ten system. ## Practice Questions on Base-Ten Applying what you’ve learned to practice questions is a vital part of understanding the base-ten numeral system. Let’s go through some exercises of varying complexity: Exercise 1: Write the number 628 in expanded form. Exercise 2: What is the place value of the digit 3 in the number 3456? Exercise 3: Write the number 9876.543 in expanded form. Exercise 4: What is the place value of the digit 8 in the number 0.789? Challenge 1: Convert the binary number 1011 (base-2) to base-ten. Challenge 2: Convert the hexadecimal number F4 (base-16) to base-ten. Exercise 1: 600 + 20 + 8 Exercise 2: 3 is in the hundreds place, so its value is 300. Exercise 3: 9000 + 800 + 70 + 6 + 0.5 + 0.04 + 0.003 Exercise 4: 8 is in the hundredths place, so its value is 0.08. Challenge 1: Binary 1011 = 1(2^3) + 0(2^2) + 1(2^1) + 1(2^0) = 8 + 0 + 2 + 1 = 11 in base-ten. Challenge 2: Hexadecimal F4 = 15(16^1) + 4(16^0) = 240 + 4 = 244 in base-ten. ## Conclusion We’ve embarked on a fascinating journey through the realm of the base-ten numeral system, exploring its structure, working through examples, and reinforcing our knowledge with practice questions. As we’ve seen, the base-ten numeral system is an integral part of our daily lives, an invisible framework underlying every numeric transaction we engage in, from checking the time to calculating the cost of groceries. At Brighterly, we believe that understanding this fundamental concept will illuminate your path to further mathematical exploration. Just as the base-ten system underpins our number system, so does a solid foundation in mathematics underpin a brighter future. Therefore, we encourage you to revisit this material, apply it, and use it as a stepping stone to delve deeper into the intriguing world of mathematics. Remember, every math expert started with understanding the basics, just like the base-ten system. ## Frequently Asked Questions on Base-Ten ### Why do we use base-ten? We use the base-ten numeral system, also known as the decimal system, because it is simple and efficient. This system is believed to have originated from humans counting on their ten fingers. Its widespread use makes it universally understandable and it forms the foundation of many mathematical operations. ### What is the significance of base-ten in real life? The base-ten system is crucial in our everyday life. It’s the system we use for counting, measuring, rating, and virtually any other numerical representation. It’s also the fundamental basis for arithmetic operations: addition, subtraction, multiplication, and division. ### How is base-ten related to place value? Place value is a positional system of notation in which the position of a number with respect to the point determines its value. In the base-ten system, each place represents a power of ten. For example, in the number 345, 3 is in the hundreds place (10^2), 4 is in the tens place (10^1), and 5 is in the ones place (10^0). ### What’s the difference between base-ten and other number systems? Other number systems have different bases. For instance, the binary system uses base-2, the octal system uses base-8, and the hexadecimal system uses base-16. This means that they use a different number of symbols to represent numbers. The base-ten system uses ten symbols (0-9), the binary system uses two symbols (0,1), and so on. Each system has its unique applications, like binary in digital computing and hexadecimal in computing and digital electronics. Information Sources: After-School Math Program • Boost Math Skills After School! After-School Math Program Boost Your Child's Math Abilities! Ideal for 1st-8th Graders, Perfectly Synced with School Curriculum!
Basic Rules for Finding Indefinite Integrals 1. Power Rule: • Formula:${x}^{n}$ dx = $\frac{{x}^{\left(n+1\right)}}{\left(n+1\right)}$ + C (except when n = -1) • Explanation: Integrating a power function involves increasing the exponent by 1 and dividing by the new exponent. • Examples: • ${x}^{4}$ dx = $\frac{{x}^{4}}{4}$ + C • ${x}^{\left(-2\right)}$ dx = -${x}^{\left(-1\right)}$ + C 2. Constant Rule: • Formula: ∫ k dx = kx + C (where k is a constant) • Explanation: Integrating a constant simply involves multiplying the constant by the variable of integration. • Examples: • ∫ 5 dx = 5x + C • ∫ (-2) dx = -2x + C 3. Sum/Difference Rule: • Formula: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx • Explanation: Integrating a sum or difference of functions can be split into individual integrals. • Example: • ∫ (3${x}^{2}$ + 4x) dx = ∫ 3${x}^{2}$ dx + ∫ 4x dx = ${x}^{3}$ + 2${x}^{2}$ + C 4. Linearity of Integration: • Property: ∫[af(x) + bg(x)] dx = a ∫ f(x) dx + b ∫ g(x) dx • Explanation: Constants can be factored out of the integral, and the integral distributes over addition and subtraction. • Example: • ∫ (2x + 3) dx = 2 ∫ x dx + 3 ∫ 1 dx = ${x}^{2}$+ 3x + C 5. Constant of Integration: • Property: ∫ f(x) dx + C • Explanation: Every indefinite integral has a constant of integration (C), representing a family of functions. • Example: • ∫ 2x dx = ${x}^{2}$ + C 6. Exponential and Logarithmic Rules: • Exponential Function:${e}^{x}$ dx = ${e}^{x}$ + C • Natural Logarithm:$\frac{1}{x}$ dx = ln\|x\| + C • Explanation: Integrals of exponential and logarithmic functions have simple forms and are essential in various calculations. 7. Trigonometric Integrals: • Sine Function: ∫ sin(x) dx = -cos(x) + C • Cosine Function: ∫ cos(x) dx = sin(x) + C • Tangent Function: ∫ tan(x) dx = -ln\|cos(x)\| + C • Explanation: Integrals of basic trigonometric functions have specific forms, often requiring knowledge of trigonometric identities.

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