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## Find the missing number in math puzzles:
### Question 1: Find the next number in the series:
This series is logical. Its logic is simple. The series follows as: 14, 28, 20, 40, 32, 64, ….
We need to find the next number in the series.
The series starts with 14, next number in the series is twice the first number i.e. 28, third number in this series is previous number - 8
i.e. 28 – 8 = 20
We repeat as: 20 + 20 = 40
40 – 8 = 32
Repeat again:
32 + 32 = 64
Now we need to find the missing number which is previous number – 8.
Hence,
64 – 8 = 56
Hence 56 is the missing number.
### Question 2: Find the missing number:
Find unknown number in math puzzles
This question is pretty straight forward because it can be solved with logic. You need to find a perfect square of every number in the series. Hence we start from left, 1 is encircled & it is a perfect square of 1. Proceeding in clockwise direction, in the triangular pattern, we find 4 which is a perfect square of 2. Further proceeding in clockwise direction will take us to 9 which is a perfect square of 3. So it shows that we need to find a perfect square of number series starting from 1. We can guess the next number in the series would be 16 & it is a perfect square is 4.
So our answer is 16. Now we are going to prove whether we are right or wrong by following the same pattern.
Solve math triangle puzzle
This question shows us a path that we should follow. When we follow the arrow then we find our next number is 25 which is a perfect square of 5. Following number is 36, square of 6. 49 comes after which is a square of 7. Then we see 64, which is a square of 8. Final number in the number series is 81 which is a square of 9 & then we reach 1 again. Hence this question shows that we need to find square of every number in clockwise direction, in a triangular pattern. |
# How do you tell whether the graph opens up or down, find the vertex, and find the axis of symmetry given y=-x^2-1?
Jun 16, 2018
Graph opens down.
Vertex at $\left(0 , - 1\right)$;Axis of Symmetry: $x = 0$
#### Explanation:
$y = - {x}^{2} - 1$
$y$ is a quadratic function of the from: $a {x}^{2} + b x + c$
Where: $a = - 1 , b = 0 \mathmr{and} c = - 1$
Since $y$ is a quadratic its graph will be a parabola.
Since $a < 0$ $y$ will have a maximum value at its vertex.
Now, since $y$ is a parabola with a maximum value it must open downwards.
The vertex of $y$ will lie on its axis of symmetry where $x = \frac{- b}{2 a}$ i.e. where $x = 0$
Since, $y \left(0\right) = - 1 \to {y}_{\text{vertex}} = \left(0 , - 1\right)$
The axis of symmetry of $y$ is the vertical line $x = 0$ which is the $y -$axis.
We can see these results from the graph of $y$ below.
graph{ -x^2-1 [-13.28, 12.03, -7.17, 5.49]} |
# Often asked: What Is Function In Gen Math?
## What is function in general mathematics with example?
A function is a mapping from a set of inputs (the domain) to a set of possible outputs (the codomain). The definition of a function is based on a set of ordered pairs, where the first element in each pair is from the domain and the second is from the codomain.
## WHAT IS function and its types?
1. Injective (One-to-One) Functions: A function in which one element of Domain Set is connected to one element of Co-Domain Set. 2. Surjective (Onto) Functions: A function in which every element of Co-Domain Set has one pre-image.
## What is a function in algebra?
A function is an equation that has only one answer for y for every x. A function assigns exactly one output to each input of a specified type. It is common to name a function either f(x) or g(x) instead of y. f(2) means that we should find the value of our function when x equals 2.
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## WHAT IS function and relation in general mathematics?
A relation is a set of inputs and outputs, and a function is a relation with one output for each input.
## What is a real life example of a function?
A weekly salary is a function of the hourly pay rate and the number of hours worked. Compound interest is a function of initial investment, interest rate, and time. Supply and demand: As price goes up, demand goes down.
## How do you describe a function?
A function relates an input to an output. It is like a machine that has an input and an output. And the output is related somehow to the input. “f(x) = ” is the classic way of writing a function.
## What are the 4 types of functions?
The various types of functions are as follows:
• Many to one function.
• One to one function.
• Onto function.
• One and onto function.
• Constant function.
• Identity function.
• Polynomial function.
## What is a function easy definition?
A technical definition of a function is: a relation from a set of inputs to a set of possible outputs where each input is related to exactly one output. We can write the statement that f is a function from X to Y using the function notation f:X→Y.
## What are the two main types of functions?
What are the two main types of functions? Explanation: Built-in functions and user defined ones. The built-in functions are part of the Python language.
## What is not a function?
The y value of a point where a vertical line intersects a graph represents an output for that input x value. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because that x value has more than one output.
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## How do you tell if a graph is a function?
Use the vertical line test to determine whether or not a graph represents a function. If a vertical line is moved across the graph and, at any time, touches the graph at only one point, then the graph is a function. If the vertical line touches the graph at more than one point, then the graph is not a function.
## How do you know if a function is not a function?
Determining whether a relation is a function on a graph is relatively easy by using the vertical line test. If a vertical line crosses the relation on the graph only once in all locations, the relation is a function. However, if a vertical line crosses the relation more than once, the relation is not a function.
## What is difference between relation and function?
Relation – In maths, the relation is defined as the collection of ordered pairs, which contains an object from one set to the other set. Functions – The relation that defines the set of inputs to the set of outputs is called the functions. In function, each input in the set X has exactly one output in the set Y.
## What is the example of function and relation?
In mathematics, a function can be defined as a rule that relates every element in one set, called the domain, to exactly one element in another set, called the range. For example, y = x + 3 and y = x2 – 1 are functions because every x- value produces a different y- value. A relation is any set of ordered-pair numbers.
## What is a relation example?
What is the Relation? In other words, the relation between the two sets is defined as the collection of the ordered pair, in which the ordered pair is formed by the object from each set. Example: {(-2, 1), (4, 3), (7, -3)}, usually written in set notation form with curly brackets. |
# Distance Between Line And Plane Pdf
By Denise D.
In and pdf
17.05.2021 at 15:02
File Name: distance between line and plane .zip
Size: 29263Kb
Published: 17.05.2021
In mathematics , a plane is a flat, two- dimensional surface that extends infinitely far.
By now, we are familiar with writing equations that describe a line in two dimensions. To write an equation for a line, we must know two points on the line, or we must know the direction of the line and at least one point through which the line passes. In two dimensions, we use the concept of slope to describe the orientation, or direction, of a line. In three dimensions, we describe the direction of a line using a vector parallel to the line. In this section, we examine how to use equations to describe lines and planes in space.
Lines and planes are perhaps the simplest of curves and surfaces in three dimensional space. They also will prove important as we seek to understand more complicated curves and surfaces. A plane does not have an obvious "direction'' as does a line. It is possible to associate a plane with a direction in a very useful way, however: there are exactly two directions perpendicular to a plane. Any vector with one of these two directions is called normal to the plane. So while there are many normal vectors to a given plane, they are all parallel or anti-parallel to each other. Example
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up. Hint: The line and the plane as you have noted are parallel. The distance from the plane to the line is therefore the distance from the plane to any point on the line. So just pick any point on the line and use "the formula" to find the distance from this point to the plane. Sign up to join this community.
## Math Insight
In this lesson, we will see how to compute the distance between a point and a plane. We will also see how to compute the distance between two parallel lines and planes. In other words, we can say that the shortest distance between a point and a plane is the length of the perpendicular line from the point to a plane. Let suppose there is a plane, , and point P as shown below:. The distance between and P will be a perpendicular line drawn from point P to the plane.
Его массивная шея зажала ей рот, и Росио чуть не задохнулась. Боже, поскорей бы все это закончилось, взмолилась она про. - Si. Si! - вскрикивала она в интервалах между его рывками и впивалась ногтями ему в спину, стараясь ускорить его движения. Все смешалось в ее голове - лица бесчисленных мужчин, склонявшиеся над ней, потолки гостиничных номеров, в которые она смотрела, мечты о том, что когда-нибудь все это кончится и она заведет детей… Внезапно, без всякого предупреждения, тело немца выгнулось, замерло и тут же рухнуло на. Это .
Ключ… - Ее передернуло. - Коммандер Стратмор отправил кого-то в Испанию с заданием найти ключ. - И что? - воскликнул Джабба. - Человек Стратмора его нашел. Сьюзан, больше не в силах сдержать слезы, разрыдалась. - Да, - еле слышно сказала. - Полагаю, что .
## Distance Between a Point and a Plane
Беккер изобразил улыбку. - Я должен идти. Он извинился перед немцем за вторжение, в ответ на что тот скромно улыбнулся. - Keine Ursache. Беккер вышел в коридор.
- Слово разница многозначно.
Нет. По-испански говорила очень плохо. - Она не испанка? - спросил Беккер. - Нет. Думаю, англичанка. И с какими-то дикими волосами - красно-бело-синими. Беккер усмехнулся, представив это зрелище.
Я хочу знать. Бринкерхофф уже пожалел, что не дал ей спокойно уйти домой. Телефонный разговор со Стратмором взбесил. После истории с Попрыгунчиком всякий раз, когда Мидж казалось, что происходит что-то подозрительное, она сразу же превращалась из кокетки в дьявола, и, пока не выясняла все досконально, ничто не могло ее остановить. - Мидж, скорее всего это наши данные неточны, - решительно заявил Бринкерхофф. - Ты только подумай: ТРАНСТЕКСТ бьется над одним-единственным файлом целых восемнадцать часов.
Набирая скорость на последнем отрезке Матеус-Гаго, он увидел впереди горой вздымающийся готический собор XI века. Рядом с собором на сто двадцать метров вверх, прямо в занимающуюся зарю, поднималась башня Гиральда. Это и был Санта-Крус, квартал, в котором находится второй по величине собор в мире, а также живут самые старинные и благочестивые католические семьи Севильи. Беккер пересек мощенную камнем площадь. Единственный выстрел, к счастью, прозвучал слишком поздно. Беккер на своем мотоцикле скрылся в узком проходе Каллита-де-ля-Вирген. ГЛАВА 88 Фара веспы отбрасывала контрастные тени на стены по обе стороны от узкой дорожки.
Он бросил быстрый взгляд на Сьюзан, которая по-прежнему сидела на стуле, обхватив голову руками и целиком уйдя в. Фонтейн не мог понять, в чем дело, но, какими бы ни были причины ее состояния, выяснять это сейчас не было времени.
Пора. Она должна немедленно поговорить со Стратмором. Сьюзан осторожно приоткрыла дверь и посмотрела на глянцевую, почти зеркальную стену шифровалки.
В такой одежде ты тут ничего не добьешься. Беккер нахмурился. - Я вовсе не хочу с ней переспать.
И тут же забилась, задыхаясь от удушья.
Простите, что я на вас накричала. Я так испугалась, увидев. - Не стоит, - удивился Беккер - Я зашел куда не следовало.
Да хватит тебе, Эдди! - Но, посмотрев в зеркало, он убедился, что это вовсе не его закадычный дружок. Лицо в шрамах и следах оспы. Два безжизненных глаза неподвижно смотрят из-за очков в тонкой металлической оправе.
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ii) A plane with normal. −→. N and containing a point Q. −→. N. Ingredients: i) A point P,. −−→. PQ. −−→. PQ. The distance from P to the plane is d.
James D. |
# How do solve the following linear system?: -2x+3y=-1 , -x-7y=14 ?
Aug 29, 2016
First, solve equation #2 for x.
#### Explanation:
This gives you $x = - 7 y - 14$ .
Now substitute this value of x into the 1st equation, like this:
$- 2 \left(- 7 y - 14\right) + 3 y = - 1$
Multiply the negative 2 into the parentheses:
$14 y + 28 + 3 y = - 1$
Combine like terms:
$17 y + 28 = - 1$
Move the 28 to the other side of the equation:
$17 y = - 29$
Divide both sides by 17 to isolate the y-term:
$17 \frac{y}{17} = - \frac{29}{17}$
which gives you:
$y = - \frac{29}{17}$.
Since 17 and 29 are both prime numbers, the answer cannot be reduced.
Now you have the value of y. Plug that into the either eqn:
$- 2 x + 3 \left(- \frac{29}{17}\right) = - 1$
and solve for x.
$- 2 x - \frac{87}{17} = - 1$
$- 2 x = - 1 + \frac{87}{17} = \frac{70}{17}$
Divide both sides by negative 2:
$x = \frac{\frac{70}{17}}{-} 2 = \frac{70}{17} \cdot \left(- \frac{1}{2}\right) = - \frac{70}{34}$
Reduce:
$x = - \frac{35}{17}$
Now plug $x = - \frac{35}{17}$ and $y = - \frac{29}{17}$ into either eqn to check the answers:
Does $- 2 \left(- \frac{35}{17}\right) + 3 \left(- \frac{29}{17}\right) = - 1$ ?
$\frac{70}{17} - \frac{87}{17} = - 1$ ?
$- \frac{17}{17} = - 1$
The fractions look odd, but it checks out. |
In mathematics, an equation is a mathematical formula that expresses the equality of two expressions, by connecting them with the equals sign =.[2][3] The word equation and its cognates in other languages may have subtly different meanings; for example, in French an équation is defined as containing one or more variables, while in English, any well-formed formula consisting of two expressions related with an equals sign is an equation.[4]
Solving an equation containing variables consists of determining which values of the variables make the equality true. The variables for which the equation has to be solved are also called unknowns, and the values of the unknowns that satisfy the equality are called solutions of the equation. There are two kinds of equations: identities and conditional equations. An identity is true for all values of the variables. A conditional equation is only true for particular values of the variables.[5][6]
The "=" symbol, which appears in every equation, was invented in 1557 by Robert Recorde, who considered that nothing could be more equal than parallel straight lines with the same length.[1]
Description
An equation is written as two expressions, connected by an equals sign ("=").[2] The expressions on the two sides of the equals sign are called the "left-hand side" and "right-hand side" of the equation. Very often the right-hand side of an equation is assumed to be zero. This does not reduce the generality, as this can be realized by subtracting the right-hand side from both sides.
The most common type of equation is a polynomial equation (commonly called also an algebraic equation) in which the two sides are polynomials. The sides of a polynomial equation contain one or more terms. For example, the equation
${\displaystyle Ax^{2}+Bx+C-y=0}$
has left-hand side ${\displaystyle Ax^{2}+Bx+C-y}$, which has four terms, and right-hand side ${\displaystyle 0}$, consisting of just one term. The names of the variables suggest that x and y are unknowns, and that A, B, and C are parameters, but this is normally fixed by the context (in some contexts, y may be a parameter, or A, B, and C may be ordinary variables).
An equation is analogous to a scale into which weights are placed. When equal weights of something (e.g., grain) are placed into the two pans, the two weights cause the scale to be in balance and are said to be equal. If a quantity of grain is removed from one pan of the balance, an equal amount of grain must be removed from the other pan to keep the scale in balance. More generally, an equation remains in balance if the same operation is performed on its both sides.
Properties
Two equations or two systems of equations are equivalent, if they have the same set of solutions. The following operations transform an equation or a system of equations into an equivalent one – provided that the operations are meaningful for the expressions they are applied to:
• Adding or subtracting the same quantity to both sides of an equation. This shows that every equation is equivalent to an equation in which the right-hand side is zero.
• Multiplying or dividing both sides of an equation by a non-zero quantity.
• Applying an identity to transform one side of the equation. For example, expanding a product or factoring a sum.
• For a system: adding to both sides of an equation the corresponding side of another equation, multiplied by the same quantity.
If some function is applied to both sides of an equation, the resulting equation has the solutions of the initial equation among its solutions, but may have further solutions called extraneous solutions. For example, the equation ${\displaystyle x=1}$ has the solution ${\displaystyle x=1.}$ Raising both sides to the exponent of 2 (which means applying the function ${\displaystyle f(s)=s^{2))$ to both sides of the equation) changes the equation to ${\displaystyle x^{2}=1}$, which not only has the previous solution but also introduces the extraneous solution, ${\displaystyle x=-1.}$ Moreover, if the function is not defined at some values (such as 1/x, which is not defined for x = 0), solutions existing at those values may be lost. Thus, caution must be exercised when applying such a transformation to an equation.
The above transformations are the basis of most elementary methods for equation solving, as well as some less elementary ones, like Gaussian elimination.
Examples
Analogous illustration
An equation is analogous to a weighing scale, balance, or seesaw.
Each side of the equation corresponds to one side of the balance. Different quantities can be placed on each side: if the weights on the two sides are equal, the scale balances, and in analogy, the equality that represents the balance is also balanced (if not, then the lack of balance corresponds to an inequality represented by an inequation).
In the illustration, x, y and z are all different quantities (in this case real numbers) represented as circular weights, and each of x, y, and z has a different weight. Addition corresponds to adding weight, while subtraction corresponds to removing weight from what is already there. When equality holds, the total weight on each side is the same.
Parameters and unknowns
Equations often contain terms other than the unknowns. These other terms, which are assumed to be known, are usually called constants, coefficients or parameters.
An example of an equation involving x and y as unknowns and the parameter R is
${\displaystyle x^{2}+y^{2}=R^{2}.}$
When R is chosen to have the value of 2 (R = 2), this equation would be recognized in Cartesian coordinates as the equation for the circle of radius of 2 around the origin. Hence, the equation with R unspecified is the general equation for the circle.
Usually, the unknowns are denoted by letters at the end of the alphabet, x, y, z, w, ..., while coefficients (parameters) are denoted by letters at the beginning, a, b, c, d, ... . For example, the general quadratic equation is usually written ax2 + bx + c = 0.
The process of finding the solutions, or, in case of parameters, expressing the unknowns in terms of the parameters, is called solving the equation. Such expressions of the solutions in terms of the parameters are also called solutions.
A system of equations is a set of simultaneous equations, usually in several unknowns for which the common solutions are sought. Thus, a solution to the system is a set of values for each of the unknowns, which together form a solution to each equation in the system. For example, the system
{\displaystyle {\begin{aligned}3x+5y&=2\\5x+8y&=3\end{aligned))}
has the unique solution x = −1, y = 1.
Identities
An identity is an equation that is true for all possible values of the variable(s) it contains. Many identities are known in algebra and calculus. In the process of solving an equation, an identity is often used to simplify an equation, making it more easily solvable.
In algebra, an example of an identity is the difference of two squares:
${\displaystyle x^{2}-y^{2}=(x+y)(x-y)}$
which is true for all x and y.
Trigonometry is an area where many identities exist; these are useful in manipulating or solving trigonometric equations. Two of many that involve the sine and cosine functions are:
${\displaystyle \sin ^{2}(\theta )+\cos ^{2}(\theta )=1}$
and
${\displaystyle \sin(2\theta )=2\sin(\theta )\cos(\theta )}$
which are both true for all values of θ.
For example, to solve for the value of θ that satisfies the equation:
${\displaystyle 3\sin(\theta )\cos(\theta )=1\,,}$
where θ is limited to between 0 and 45 degrees, one may use the above identity for the product to give:
${\displaystyle {\frac {3}{2))\sin(2\theta )=1\,,}$
yielding the following solution for θ:
${\displaystyle \theta ={\frac {1}{2))\arcsin \left({\frac {2}{3))\right)\approx 20.9^{\circ }.}$
Since the sine function is a periodic function, there are infinitely many solutions if there are no restrictions on θ. In this example, restricting θ to be between 0 and 45 degrees would restrict the solution to only one number.
Algebra
Algebra studies two main families of equations: polynomial equations and, among them, the special case of linear equations. When there is only one variable, polynomial equations have the form P(x) = 0, where P is a polynomial, and linear equations have the form ax + b = 0, where a and b are parameters. To solve equations from either family, one uses algorithmic or geometric techniques that originate from linear algebra or mathematical analysis. Algebra also studies Diophantine equations where the coefficients and solutions are integers. The techniques used are different and come from number theory. These equations are difficult in general; one often searches just to find the existence or absence of a solution, and, if they exist, to count the number of solutions.
Polynomial equations
In general, an algebraic equation or polynomial equation is an equation of the form
${\displaystyle P=0}$, or
${\displaystyle P=Q}$ [a]
where P and Q are polynomials with coefficients in some field (e.g., rational numbers, real numbers, complex numbers). An algebraic equation is univariate if it involves only one variable. On the other hand, a polynomial equation may involve several variables, in which case it is called multivariate (multiple variables, x, y, z, etc.).
For example,
${\displaystyle x^{5}-3x+1=0}$
is a univariate algebraic (polynomial) equation with integer coefficients and
${\displaystyle y^{4}+{\frac {xy}{2))={\frac {x^{3)){3))-xy^{2}+y^{2}-{\frac {1}{7))}$
is a multivariate polynomial equation over the rational numbers.
Some polynomial equations with rational coefficients have a solution that is an algebraic expression, with a finite number of operations involving just those coefficients (i.e., can be solved algebraically). This can be done for all such equations of degree one, two, three, or four; but equations of degree five or more cannot always be solved in this way, as the Abel–Ruffini theorem demonstrates.
A large amount of research has been devoted to compute efficiently accurate approximations of the real or complex solutions of a univariate algebraic equation (see Root finding of polynomials) and of the common solutions of several multivariate polynomial equations (see System of polynomial equations).
Systems of linear equations
A system of linear equations (or linear system) is a collection of linear equations involving one or more variables.[b] For example,
{\displaystyle {\begin{alignedat}{7}3x&&\;+\;&&2y&&\;-\;&&z&&\;=\;&&1&\\2x&&\;-\;&&2y&&\;+\;&&4z&&\;=\;&&-2&\\-x&&\;+\;&&{\tfrac {1}{2))y&&\;-\;&&z&&\;=\;&&0&\end{alignedat))}
is a system of three equations in the three variables x, y, z. A solution to a linear system is an assignment of numbers to the variables such that all the equations are simultaneously satisfied. A solution to the system above is given by
{\displaystyle {\begin{alignedat}{2}x&\,=\,&1\\y&\,=\,&-2\\z&\,=\,&-2\end{alignedat))}
since it makes all three equations valid. The word "system" indicates that the equations are to be considered collectively, rather than individually.
In mathematics, the theory of linear systems is a fundamental part of linear algebra, a subject which is used in many parts of modern mathematics. Computational algorithms for finding the solutions are an important part of numerical linear algebra, and play a prominent role in physics, engineering, chemistry, computer science, and economics. A system of non-linear equations can often be approximated by a linear system (see linearization), a helpful technique when making a mathematical model or computer simulation of a relatively complex system.
Geometry
Analytic geometry
In Euclidean geometry, it is possible to associate a set of coordinates to each point in space, for example by an orthogonal grid. This method allows one to characterize geometric figures by equations. A plane in three-dimensional space can be expressed as the solution set of an equation of the form ${\displaystyle ax+by+cz+d=0}$, where ${\displaystyle a,b,c}$ and ${\displaystyle d}$ are real numbers and ${\displaystyle x,y,z}$ are the unknowns that correspond to the coordinates of a point in the system given by the orthogonal grid. The values ${\displaystyle a,b,c}$ are the coordinates of a vector perpendicular to the plane defined by the equation. A line is expressed as the intersection of two planes, that is as the solution set of a single linear equation with values in ${\displaystyle \mathbb {R} ^{2))$ or as the solution set of two linear equations with values in ${\displaystyle \mathbb {R} ^{3}.}$
A conic section is the intersection of a cone with equation ${\displaystyle x^{2}+y^{2}=z^{2))$ and a plane. In other words, in space, all conics are defined as the solution set of an equation of a plane and of the equation of a cone just given. This formalism allows one to determine the positions and the properties of the focuses of a conic.
The use of equations allows one to call on a large area of mathematics to solve geometric questions. The Cartesian coordinate system transforms a geometric problem into an analysis problem, once the figures are transformed into equations; thus the name analytic geometry. This point of view, outlined by Descartes, enriches and modifies the type of geometry conceived of by the ancient Greek mathematicians.
Currently, analytic geometry designates an active branch of mathematics. Although it still uses equations to characterize figures, it also uses other sophisticated techniques such as functional analysis and linear algebra.
Cartesian equations
In Cartesian geometry, equations are used to describe geometric figures. As the equations that are considered, such as implicit equations or parametric equations, have infinitely many solutions, the objective is now different: instead of giving the solutions explicitly or counting them, which is impossible, one uses equations for studying properties of figures. This is the starting idea of algebraic geometry, an important area of mathematics.
One can use the same principle to specify the position of any point in three-dimensional space by the use of three Cartesian coordinates, which are the signed distances to three mutually perpendicular planes (or, equivalently, by its perpendicular projection onto three mutually perpendicular lines).
The invention of Cartesian coordinates in the 17th century by René Descartes revolutionized mathematics by providing the first systematic link between Euclidean geometry and algebra. Using the Cartesian coordinate system, geometric shapes (such as curves) can be described by Cartesian equations: algebraic equations involving the coordinates of the points lying on the shape. For example, a circle of radius 2 in a plane, centered on a particular point called the origin, may be described as the set of all points whose coordinates x and y satisfy the equation x2 + y2 = 4.
Parametric equations
A parametric equation for a curve expresses the coordinates of the points of the curve as functions of a variable, called a parameter.[7][8] For example,
{\displaystyle {\begin{aligned}x&=\cos t\\y&=\sin t\end{aligned))}
are parametric equations for the unit circle, where t is the parameter. Together, these equations are called a parametric representation of the curve.
The notion of parametric equation has been generalized to surfaces, manifolds and algebraic varieties of higher dimension, with the number of parameters being equal to the dimension of the manifold or variety, and the number of equations being equal to the dimension of the space in which the manifold or variety is considered (for curves the dimension is one and one parameter is used, for surfaces dimension two and two parameters, etc.).
Number theory
Diophantine equations
A Diophantine equation is a polynomial equation in two or more unknowns for which only the integer solutions are sought (an integer solution is a solution such that all the unknowns take integer values). A linear Diophantine equation is an equation between two sums of monomials of degree zero or one. An example of linear Diophantine equation is ax + by = c where a, b, and c are constants. An exponential Diophantine equation is one for which exponents of the terms of the equation can be unknowns.
Diophantine problems have fewer equations than unknown variables and involve finding integers that work correctly for all equations. In more technical language, they define an algebraic curve, algebraic surface, or more general object, and ask about the lattice points on it.
The word Diophantine refers to the Hellenistic mathematician of the 3rd century, Diophantus of Alexandria, who made a study of such equations and was one of the first mathematicians to introduce symbolism into algebra. The mathematical study of Diophantine problems that Diophantus initiated is now called Diophantine analysis.
Algebraic and transcendental numbers
An algebraic number is a number that is a solution of a non-zero polynomial equation in one variable with rational coefficients (or equivalently — by clearing denominators — with integer coefficients). Numbers such as π that are not algebraic are said to be transcendental. Almost all real and complex numbers are transcendental.
Algebraic geometry
Algebraic geometry is a branch of mathematics, classically studying solutions of polynomial equations. Modern algebraic geometry is based on more abstract techniques of abstract algebra, especially commutative algebra, with the language and the problems of geometry.
The fundamental objects of study in algebraic geometry are algebraic varieties, which are geometric manifestations of solutions of systems of polynomial equations. Examples of the most studied classes of algebraic varieties are: plane algebraic curves, which include lines, circles, parabolas, ellipses, hyperbolas, cubic curves like elliptic curves and quartic curves like lemniscates, and Cassini ovals. A point of the plane belongs to an algebraic curve if its coordinates satisfy a given polynomial equation. Basic questions involve the study of the points of special interest like the singular points, the inflection points and the points at infinity. More advanced questions involve the topology of the curve and relations between the curves given by different equations.
Differential equations
A differential equation is a mathematical equation that relates some function with its derivatives. In applications, the functions usually represent physical quantities, the derivatives represent their rates of change, and the equation defines a relationship between the two. They are solved by finding an expression for the function that does not involve derivatives. Differential equations are used to model processes that involve the rates of change of the variable, and are used in areas such as physics, chemistry, biology, and economics.
In pure mathematics, differential equations are studied from several different perspectives, mostly concerned with their solutions — the set of functions that satisfy the equation. Only the simplest differential equations are solvable by explicit formulas; however, some properties of solutions of a given differential equation may be determined without finding their exact form.
If a self-contained formula for the solution is not available, the solution may be numerically approximated using computers. The theory of dynamical systems puts emphasis on qualitative analysis of systems described by differential equations, while many numerical methods have been developed to determine solutions with a given degree of accuracy.
Ordinary differential equations
An ordinary differential equation or ODE is an equation containing a function of one independent variable and its derivatives. The term "ordinary" is used in contrast with the term partial differential equation, which may be with respect to more than one independent variable.
Linear differential equations, which have solutions that can be added and multiplied by coefficients, are well-defined and understood, and exact closed-form solutions are obtained. By contrast, ODEs that lack additive solutions are nonlinear, and solving them is far more intricate, as one can rarely represent them by elementary functions in closed form: Instead, exact and analytic solutions of ODEs are in series or integral form. Graphical and numerical methods, applied by hand or by computer, may approximate solutions of ODEs and perhaps yield useful information, often sufficing in the absence of exact, analytic solutions.
Partial differential equations
A partial differential equation (PDE) is a differential equation that contains unknown multivariable functions and their partial derivatives. (This is in contrast to ordinary differential equations, which deal with functions of a single variable and their derivatives.) PDEs are used to formulate problems involving functions of several variables, and are either solved by hand, or used to create a relevant computer model.
PDEs can be used to describe a wide variety of phenomena such as sound, heat, electrostatics, electrodynamics, fluid flow, elasticity, or quantum mechanics. These seemingly distinct physical phenomena can be formalised similarly in terms of PDEs. Just as ordinary differential equations often model one-dimensional dynamical systems, partial differential equations often model multidimensional systems. PDEs find their generalisation in stochastic partial differential equations.
Types of equations
Equations can be classified according to the types of operations and quantities involved. Important types include:
Notes
1. ^ As such an equation can be rewritten PQ = 0, many authors do not consider this case explicitly. |
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Algebraic Equations Study Guide: GED Math (page 2)
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Updated on Mar 23, 2011
Solving Multiple-Step Algebraic Equations
The algebraic equations you've solved so far in this lesson have mostly required only one inverse operation. But some algebraic equations require more than one inverse operation to isolate the variable and then solve the equation.
When solving multiple-step algebraic equations, you should first add or subtract. Then, multiply or divide. In other words, you follow the order of operations in inverse order. Another way to look at it is to solve for the number attached to the variable last. Let's look at some examples.
Examples
1. Solve the following algebraic expression: 2m + 5 = 13.
2. First, look for numbers that are being added or subtracted to the term with the variable. In this equation, 5 is added to the 2m. To simplify, subtract 5 from both sides of the equation.
Perform the inverse operation for any multiplication or division. The variable in this equation is multiplied by 2, so you must divide each side of the equation by 2 in order to isolate the variable.
The answer is m = 4.
3. Solve the following algebraic equation: 5p + 24 = 3p – 4.
4. This equation has a variable on each side of the equal sign. So, you first need to get the variable terms on one side of the equation. You can do this by subtracting the smaller of the two variables from each side, because they have like terms.
Look for numbers that are being added to or subtracted from the term with the variable. In this equation, 24 is added to the 2p. So you subtract 24 from both sides of the equation.
Perform the inverse operation for any multiplication or division. The variable in this equation is multiplied by 2, so you divide each side of the equation by 2 to isolate the variable.
The answer is p = –14.
5. Solve the following algebraic equation: 4(b + 1) = 20.
6. This equation has parentheses, so you need to remove the parentheses by distributing the 4. 4(b + 1) = 4b + 4
So, your equation becomes 4b + 4 = 20.
Look for numbers that are being added to or subtracted from the term with the variable. Subtract 4 from both sides of the equation.
Perform the inverse operation for any multiplication or division. Divide each side of the equation by 4.
The answer is b = 4.
Practice problems for these concepts can be found at:
Algebra and Functions Practice Problems: GED Math |
# What is the Square Root of 25?
The square root of a number is a value that, when multiplied by itself, gives the original number. In the case of 25, let’s find out what its square root is.
To determine the square root of 25, we need to find the number that, when multiplied by itself, equals 25. Let’s go through the process step by step:
1. Start by considering some possible numbers. Since 5 multiplied by 5 equals 25, it’s a good candidate to test as the square root of 25.
2. Calculate: 5 multiplied by 5 equals 25.
Thus, the square root of 25 is 5.
In mathematical notation, we can represent the square root of 25 as √25 = 5.
It’s important to note that every positive number has two square roots: a positive and a negative root. In this case, the positive square root of 25 is 5, and the negative square root is -5, as -5 multiplied by -5 also equals 25.
## FAQs about the Square Root of 25
### FAQ 1: Can the square root of a number be negative?
The square root of a positive number is usually expressed as both a positive and negative value. This is because multiplying a negative number by itself also results in a positive value. In the case of the square root of 25, we have both the positive square root (5) and the negative square root (-5).
### FAQ 2: What is the difference between the square root and the exponentiation of a number?
The square root of a number (√x) represents the value that, when multiplied by itself, equals the original number (x). Exponentiation, on the other hand, involves raising a number to a power. For example, squaring a number (x^2) means multiplying it by itself. The square root is the inverse operation of squaring.
### FAQ 3: Can the square root of a negative number be determined?
No, the square root of a negative number cannot be determined using real numbers. The square root operation is defined for non-negative real numbers only. However, in the realm of complex numbers, the concept of imaginary numbers allows us to define the square root of negative numbers.
### FAQ 4: Can the square root of 25 be expressed as a fraction?
No, the square root of 25 is an irrational number, which means it cannot be expressed as a simple fraction or a terminating or repeating decimal. The decimal representation of the square root of 25 is approximately 5.
### FAQ 5: What is the relationship between the square root and the area of a square?
The square root is closely related to the concept of area in geometry. When we take the square root of a number, we are essentially finding the length of one side of a square with an area equal to that number. For example, the square root of 25 gives us the length of one side of a square with an area of 25 square units.
## Conclusion
In conclusion, the square root of 25 is 5. The square root represents the value that, when multiplied by itself, gives the original number. It’s important to note that every positive number has both a positive and a negative square root. The square root of 25 can be expressed as √25 = 5 or -5. The concept of square roots is significant in various mathematical applications, including geometry, algebra, and calculus. |
# How do you use the Squeeze Theorem to find lim xcos(50pi/x) as x approaches zero?
Oct 15, 2015
See the explanation.
#### Explanation:
From trigonometry $- 1 \le \cos \theta \le 1$ for all real numbers $\theta$.
So,
$- 1 \le \cos \left(\frac{50 \pi}{x}\right) \le 1$ for all $x \ne 0$
From the right
For $x > 0$, we can multiply without changing the directions of the inequalities, so we get:
$- x \le \cos \left(\frac{50 \pi}{x}\right) \le x$ for $x > 0$.
Observe that , ${\lim}_{x \rightarrow {0}^{+}} \left(- x\right) = {\lim}_{x \rightarrow {0}^{+}} \left(x\right) = 0$,
so, ${\lim}_{x \rightarrow {0}^{+}} \cos \left(\frac{50 \pi}{x}\right) = 0$
From the left
For $x < 0$, when we multiply we must change the directions of the inequalities, so we get:
$- x \ge \cos \left(\frac{50 \pi}{x}\right) \ge x$ for $x < 0$.
Observe that , ${\lim}_{x \rightarrow {0}^{-}} \left(- x\right) = {\lim}_{x \rightarrow {0}^{-}} \left(x\right) = 0$,
so, ${\lim}_{x \rightarrow {0}^{-}} \cos \left(\frac{50 \pi}{x}\right) = 0$
Two-sided Limit
Because both the left and rights limits are $0$, we conclude that:
${\lim}_{x \rightarrow 0} \cos \left(\frac{50 \pi}{x}\right) = 0$ |
# How do you convert to vertex form
## Standard Form to Vertex Form
In this mini-lesson, we will explore the process of converting standard form to vertex form and vice-versa. The standard form of a parabola is y = ax2 + bx + c and the vertex form of a parabola is y = a (x - h)2 + k. Here, the vertex form has a square in it. So to convert the standard to vertex form we need to complete the square.
1 Standard Form and Vertex Form of a Parabola 2 How to Convert Standard Form to Vertex Form? 3 How to Convert Vertex Form to Standard Form? 4 FAQs on Standard Form to Vertex Form
### Standard Form and Vertex Form of a Parabola
The equation of a parabola can be represented in multiple ways like: standard form, vertex form, and intercept form. One of these forms can always be converted into the other two forms depending on the requirement. In this article, we are going to learn how to convert
• standard form to vertex form and
• vertex form to standard form
Let us first explore what each of these forms means.
#### Standard Form
The standard form of a parabola is:
• y = ax2 + bx + c
Here, a, b, and c are real numbers (constants) where a ≠ 0. x and y are variables where (x, y) represents a point on the parabola.
#### Vertex Form
The vertex form of a parabola is:
• y = a (x - h)2 + k
Here, a, h, and k are real numbers, where a ≠ 0. x and y are variables where (x, y) represents a point on the parabola.
### How to Convert Standard Form to Vertex Form?
In the vertex form, y = a (x - h)2 + k, there is a "whole square". So to convert the standard form to vertex form, we just need to complete the square. But apart from this, we have a formula method also for doing this. Let us look into both methods.
#### By Completing the Square
Let us take an example of a parabola in standard form: y = -3x2 - 6x - 9 and convert it into the vertex form by completing the square. First, we should make sure that the coefficient of x2 is 1. If the coefficient of x2 is NOT 1, we will place the number outside as a common factor. We will get:
y = −3x2 − 6x − 9 = −3 (x2 + 2x + 3)
Now, the coefficient of x2 is 1. Here are the steps to convert the above expression into the vertex form.
Step 1: Identify the coefficient of x.
Step 2: Make it half and square the resultant number.
Step 3: Add and subtract the above number after the x term in the expression.
Step 4: Factorize the perfect square trinomial formed by the first 3 terms using the suitable identity.
Here, we can use x2 + 2xy + y2 = (x + y)2.
In this case, x2 + 2x + 1= (x + 1)2
The above expression from Step 3 becomes:
Step 5: Simplify the last two numbers and distribute the outside number.
Here, -1 + 3 = 2. Thus, the above expression becomes:
This is of the form y = a (x - h)2 + k, which is in the vertex form. Here, the vertex is, (h, k)=(-1,-6).
#### By Using the Formula
In the above method, ultimately we could find the values of h and k which are helpful in converting standard form to vertex form. But the values of h and k can be easily found by using the following steps:
• Find h using h = -b/2a.
• Since (h, k) lies on the given parabola, k = ah2 + bh + c. Just use this to find k by substituting the value of 'h' from the above step.
Let us convert the same example y = -3x2 - 6x - 9 into standard form using this formula method. Comparing this equation with y = ax2 + bx + c, we get a = -3, b = -6, and c = -9. Then
(i) h = -b/2a = -(-6) / (2 × -3) = -1
(ii) k = -3(-1)2 - 6(-1) - 9 = -3 + 6 - 9 = -6
Substitute these two values (along with a = -3) in the vertex form y = a (x - h)2 + k, we get y = -3 (x + 1)2 - 6. Note that we have got the same answer as in the other method.
Which method is easier? Decide and go ahead.
Tips and Tricks:
If the above processes seem difficult, then use the following steps:
• Compare the given equation with the standard form (y = ax2 + bx + c) and get the values of a,b, and c.
• Apply the following formulas to find the values the values of h and k and substitute it in the vertex form (y = a(x - h)2 + k):
h = -b/2a
k = -D/4a
Here, D is the discriminant where D = b2 - 4ac.
### How to Convert Vertex Form to Standard Form?
To convert vertex form into standard form, we just need to simplify a (x - h)2 + k algebraically to get into the form ax2 + bx + c. Technically, we need to follow the steps below to convert the vertex form into the standard form.
• Expand the square, (x − h)2.
• Distribute 'a'.
• Combine the like terms.
Example: Let us convert the equation y = -3 (x + 1)2 - 6 from vertex to standard form using the above steps:
y = -3 (x + 1)2 - 6
y = -3 (x + 1)(x + 1) - 6
y = -3 (x2 + 2x + 1) - 6
y = -3x2 - 6x - 3 - 6
y = -3x2 - 6x - 9
Important Notes on Standard Form to Vertex Form:
• In the vertex form, (h, k) represents the vertex of the parabola where the parabola has either maximum/minimum value.
• If a > 0, the parabola has the minimum value at (h, k) and
if a < 0, the parabola has the maximum value at (h, k).
Related Topics:
• Vertex Calculator
### Standard Form to Vertex Form Examples
1. Example 1: Find the vertex of the parabola y = 2x2 + 7x + 6 by completing the square.
Solution:
The given equation of parabola is y = 2x2 + 7x + 6. To find its vertex, we will convert it into vertex form.
To complete the square, first, we will make the coefficient of x2 as 1.
We will take the coefficient of x2 (which is 2 in this case) as the common factor.
2x2 + 7x + 6 = 2 (x2 + 7/2 x + 3)
The coefficient of x is 7/2, half it is 7/4, and its square is 49/16. Adding and subtracting it from the quadratic polynomial that is inside the brackets of the above step,
2x2 + 7x + 6 = 2 (x2 + 7/2 x + 49/16 - 49/16 + 3)
Factorizing the quadratic polynomial x2 + 7/2 x + 49/16, we get (x + 7/4)2. Then
2x2 + 7x + 6 = 2 ((x + 7/4)2 - 49/16 + 3)
= 2 ((x + 7/4)2 - 1/16)
= 2 (x + 7/4)2 - 1/8
By comparing this with a (x - h)2 + k, we will get (h, k) = (-7/4, -1/8).
Answer: The vertex of the given parabola is (-7/4, -1/8).
2. Example 2: Find the vertex of the same parabola as in Example 1 without converting into vertex form.
Solution:
The given parabola is y = 2x2 + 7x + 6. So a = 2, b = 7, and c = 6.
The x-coordinate of the vertex is, h = -b/2a = -7/[2(2)] = -7/4.
The y-coordinate of the vertex is, k = 2(-7/4)2 + 7(-7/4) + 6 = -1/8
Answer: We have got the same answer as in Example 1 which is (h, k) = (-7/4, -1/8).
3. Example 3: Find the equation of the following parabola in standard form.
Solution:
We can see that the parabola has the maximum value at the point (2, 2).
So the vertex of the parabola is, (h, k) = (2, 2).
So the vertex form of the above parabola is, y = a (x - 2)2 + 2 . .. (1).
To find 'a' here, we have to substitute any known point of the parabola in this equation.
The graph clearly passes through the point (1, 0) = (x, y).
Substitute it in (1):
0 = a (1 - 2)2 + 2
0 = a + 2
a = -2
Substitute it back into (1) and expand the square to convert it into the standard form:
y = -2 (x - 2)2 + 2
y = -2 (x2 - 4x + 4) + 2
y = -2x2 + 8x - 8 + 2
y = -2x2 + 8x - 6
Answer: Thus, the standard form of the given parabola is: y = -2x2 + 8x - 6.
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### Practice Questions on Standard Form to Vertex Form
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### FAQs on Standard Form to Vertex Form
#### How Do You Convert Standard Form to Vertex Form?
To convert standard form to vertex form,
• Convert y = ax2 + bx + c into the form y = a (x - h)2 + k by completing the square.
• Then y = a (x - h)2 + k is the vertex form.
#### How Do You Convert Vertex Form to Standard Form?
Converting vertex form into standard form is so easy. Just expand the square in y = a (x - h)2 + k, then expand the brackets, and finally simplify.
#### How to Convert Standard Form to Vertex Form Using Completing the Square?
To convert standard form to vertex form by using completing the square method,
• Take the coefficient of x2 as the common factor if it is other than 1.
• Make the coefficient of x half and square it.
• Add and subtract this number to the quadratic expression of the first step.
• Then apply algebraic identities to write it in the vertex form.
#### How to Find the Vertex of a Parabola in Standard Form?
Vertex can't be directly identified from standard form. Convert standard form into vertex form y = a (x - h)2 + k, then (h, k) would give the vertex of the parabola.
#### How to Convert Standard Form to Vertex Form Without Completing the Square?
To convert y = ax2 + bx + c into y = a (x - h)2 + k without completing the square, just find 'h' and 'k' using the following formulas
• h = -b / 2a
• k = - (b2 - 4ac) / 4a
#### What is the Use of Converting Standard Form into Vertex Form?
Vertex form is more helpful in graphing quadratic functions where we can easily identify the vertex, and by finding one/two points on either side of the vertex would give the perfect shape of a parabola.
## Vertex Form Calculator
Created by Wojciech Sas, PhD
Reviewed by Steven Wooding
Last updated: Oct 16, 2022
• How to find the vertex of a parabola? Vertex equation
• What is the vertex form of a quadratic equation?
• How to convert from the standard form to the vertex form?
• Vertex form to standard form converter
• How to use the vertex form calculator?
This is the vertex form calculator (also known as vertex calculator or even find the vertex calculator). If you want to know how to find the vertex of a parabola, this is the right place to begin. Moreover, our tool teaches you what the vertex form of a quadratic equation is and how to derive the equation of the vertex form or the vertex equation itself.
And this is not the end! This calculator also helps you convert from the standard to the vertex form of a parabola, or even the other way round, in a blink of an eye!
### How to find the vertex of a parabola? Vertex equation
The vertex of a parabola is a point that represents the extremal value of a quadratic curve. The quadratic part stands because the most significant power of our variable (`x`) is two. The vertex can be either a minimum (for a parabola opening up) or a maximum (for a parabola opening down).
Alternatively, we can say that the vertex is the intersection of the parabola and its symmetry axis.
Typically, we denote the vertex as a point `P(h,k)`, where `h` stands for the x-coordinate, and `k` indicates the y-coordinate.
That's enough on the definitions. But how to find the vertex of a quadratic function? It may be a surprise, but we don't need to evaluate any square root to do so!
Whenever we face a standard form of a parabola `y = a·x² + b·x + c`, we can use the equations of the vertex coordinates:
`h = -b/(2a)`,
`k = c - b²/(4a)`.
Knowing how to find these ratios, we can move one step further and ask: What is the vertex form of a parabola?
### What is the vertex form of a quadratic equation?
Intuitively, the vertex form of a parabola is the one that includes the vertex’s details inside. We can write the vertex form equation as:
`y = a·(x-h)² + k`.
As you can see, we need to know three parameters to write a quadratic vertex form. One of them is `a`, the same as in the standard form. It tells us whether the parabola is opening up (`a > 0`) or down (`a < 0`). The parameter `a` can never equal zero for a vertex form of a parabola (or any other form, strictly speaking).
The remaining parameters, `h` and `k`, are the components of the vertex. That's where the vertex form equation gets its name.
Additionally, it's worth mentioning that it's possible to draw a quadratic function graph having only the parameter `a` and the vertex.
### How to convert from the standard form to the vertex form?
We can try to convert a quadratic equation from the standard form to the vertex form using completing the square method (you can read more about this method in our completing the square calculator):
1. Write the parabola equation in the standard form: `y = a·x² + b·x + c`;
2. Extract `a` from the first two terms: `y = a · (x² + b/a · x) + c`;
3. Complete the square for the expressions with `x`. The missing fraction is `(b/(2a))²`. Add and subtract this term in the parabola equation: `y = a · [x² + b/a · x + (b/(2a))² - (b/(2a))²] + c`;
4. We can compress the three leading terms into a shortcut version of multiplication: `y = a · [(x + b/(2a))² - (b/(2a))²] + c`;
5. Remove the square bracket by multiplying the terms by `a`: `y = a·(x + b/(2a))² - b²/(4a) + c`;
6. Compare the outcome with the vertex form of a quadratic equation: `y = a·(x-h)² + k`;
7. As a result of the comparison, we know how to find the vertex of a parabola: `h = -b/(2a)`, and `k = c - b²/(4a)`.
That is one way of how to convert to vertex form from a standard one. The second (and quicker) one is to use our vertex form calculator - the way we strongly recommend! It only requires typing the parameters `a`, `b`, and `c`. Then, the result appears immediately at the bottom of the calculator space.
### Vertex form to standard form converter
Our find the vertex calculator can also work the other way around by finding the standard form of a parabola. In case you want to know how to do it by hand using the vertex form equation, this is the recipe:
1. Write the parabola equation in the vertex form: `y = a·(x-h)² + k`;
2. Expand the expression in the bracket: `y = a·(x² - 2·h·x + h²) + k`;
3. Multiply the terms in the parenthesis by `a`: `y = a·x² - 2·a·h·x + a·h² + k`;
4. Compare the outcome with the standard form of a parabola: `y = a·x² + b·x + c`;
5. Estimate the values of parameters: `b = -2·a·h`, `c = a·h² + k`.
### How to use the vertex form calculator?
There are two approaches you can take to use our vertex form calculator:
We've already described the last one in one of the previous sections. Let's see what happens for the first one:
• Type the values of parameter `a`, and the coordinates of the vertex, `h` and `k`. Let them be `a = 0.25`, `h = -17`, `k = -54`;
• That's all! As a result, you can see a graph of your quadratic function, together with the points indicating the vertex, y-intercept, and zeros.
Below the chart, you can find the detailed descriptions:
• Both the vertex and standard form of the parabola: `y = 0.25(x + 17)² - 54` and `y = 0.25x² + 8.5x + 18.25` respectively;
• The vertex: `P = (-17, -54)`;
• The y-intercept: `Y = (0, 18.25)`;
• The values of the zeros: `X₁ = (-31.6969 , 0)`, `X₂ = (-2.3031, 0)`. In case you're curious, we round the outcome to five significant figures here.
Wojciech Sas, PhD
What do you want to do?
Vertex form: y = a(x-h)² + k
Results
Vertex form equation:
y = x²
Standard form equation:
y = x²
Characteristic points:
Vertex P(0, 0)
Y-intercept Y(0, 0)
Zero X₁(0, 0)
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## Eco Man without poisons and nitrates
Food as medicine
"The biggest threat to our planet is the belief that
that someone else will save it. "
Robert Swan, modern Antarctic explorer .
Ecology is a science that cares about the conservation of the natural resources of the Earth's biosphere, in order to make human life better and longer.
Did you know that most foods, including baked goods, contain trace amounts of agrochemicals that are harmful to humans?
Nearly all fruits and vegetables from supermarkets have been treated with about 15 chemicals. Once in your body, they will not have the best effect on your health. For example, strawberries contain an average of almost 8 pesticides per sample.
Hormonal fungicides have been found in almost 90% of non-organic citrus fruits such as tangerines, oranges and grapefruits... And the neurotoxic insecticide (permethrin) found in spinach will be a ticking time bomb even in vegetarians.
The use of garden greens, fruits and vegetables is the basis of all healthy diets. Most followers of a healthy diet do not think about the fact that these products are dangerous to health and can do more harm than good.
If you can cut down on unhealthy vegetables and fruits, what about bread? In the memory of the Russian people there are many proverbs and sayings about bread, about the attitude towards it. Bread is the basis of life and well-being, a guarantee of future success, a symbol of human joy and happiness. "There will be bread, so everything will be." But not everything is so simple with bread in our time!
Accent
According to foreign scientists , the average residual content of pesticides in non-organic bread is 61.49%, which is much higher than the total pesticide residue in all products combined, which is approximately 40%. It is alarming that the pesticide content in bread has increased from 28.24% in 2001 to 63.43% today.
“God, forgive them, for they know what they are doing.
Karl Kraus, Austrian writer
### Agriculture - contrary to common sense?
According to Professor Pimmenthal (Purdue University), we have ignored the natural processes in the soil - microbiology and physics, especially, we have increased the amount of fertilizer and oil-based pesticides applied too much. If at 19In the 1940s, 10-50 kg of water-soluble fertilizers were enough to apply to the fields, but now this amount has increased by almost 10 times.
### Pesticide use has increased by 3000% in the last 50 years!
At first, we used fallow land, crop rotation with green manure to grow foods rich in vitamins and minerals, and in the past received the same or even higher yields than we get today with extensive use of fertilizers and pesticides. Modern science better understands the biological processes that take place in the soil, and also understands how to grow foods rich in minerals. We can reverse the downward trend in minerals and vitamins in our food. Every year of using the right biological approach to growing food resonates well with the soil and increases its productivity.
So the question is... How can we make sure that foods including bread, vegetables and fruits have been grown without using any toxic chemicals that could put us in danger? How to protect your loved ones? There are not many options.
For example, buy bread at an organic food store, buy fruits and vegetables at a rural market. But the most reliable is to grow it yourself "Eco-garden without poisons and nitrates"!
### Obtaining a fully mineralized soil
All diseases originate in the soil - a bold statement, but true. The quality of what we eat—vitamins, minerals, and enzymes—determines our health. The body will not be strong and the immune system strong without the essential minerals that are the building blocks of our food. This building material comes from the soil and is made available to plants and animals through the activity of soil organisms.
So, in order to provide plants with plenty of nutrients, rich taste, and no pesticides, it is necessary to make sure that the soil has a balanced amount of minerals, including trace elements and a vibrant diverse community of beneficial microbes.
The most effective way to get well-mineralized soil is to focus on abundant calcium and microbes in the soil. Most soil tests show percent of the main saturation with positively charged elements - calcium - 70%, magnesium - 15% , etc. To achieve these indicators, you need to put a lot of effort and knowledge.
### One of the simplest and most affordable solutions is PRK "White Pearl Drip BioCalcium + BioMagnesium":
extract of the vegetative mass of oceanic bioflora on an organo-mineral basis. A unique phytocorrector for deficiency of Ca + Mg elements, directed for emergency action. "Canned" chlorophyll in a natural set of 72 bioelements accumulated by cells marine photosynthesis products. A gift from nature in perfect proportion.
Principle of action: "Plant to plant"
Unlike mineral fertilizers, all elements are in an accessible natural form, bioactive, balanced (no antagonism), do not require processing by the plant. Ca + Mg freely penetrate into the metabolic system, transported by xylem and phloem currents to all parts of plants. Organic complexes of nutrients are quickly absorbed by leaf tissues, through stomata, through ion tunnels, through the cuticle.
Fast, affordable source of bionutrients. 10 times less required!..
— Manufactured to pharmaceutical guidelines, ultrafine particle size.
— No limiting factors, i.e. no sulfates/chlorides found in most calcium-containing foods.
- High solubility and mobility.
- Low salt index (5). For comparison, calcium nitrate has a salt index of 52.5.
- Compatible with many blends.
- Stable in all types of soil.
In contrast to traditional forms of fertilizers, small-cluster calcium of White Pearl Drip Ca+Mg PRK quickly moves both in xylem and phloem (with an ascending and descending flow of substances), while being evenly distributed between plant organs.
Efficiency:
- prevention of plant leaf chlorosis and fruit sunburn;
- reducing the risk of bud and fruit shedding under adverse conditions;
- prevention of the development of root rot, fruit top rot, bitter pitting and other physiological diseases;
- reduction of pesticide load due to an increase in the immune status of plants and resistance to diseases and pests;
- increase in marketability of products (caliber and size of fruits), transportability and keeping quality during storage.
“Battery density?! Is this about agriculture in general?”
Phyllis Tichinin, Hawkes Bay
Battery density is a term that has become popular in the last decade. The term describes foods and meats that are high in minerals, vitamins, and secondary metabolites. Thus, cabbages that receive a full fertilizer program are tastier, weigh more than cabbages of the same variety and size, and are likely to have higher nutrient densities.
This is the most common way for growers to check if the grapes are ready for harvest. The refractometer values (numbers) are the percentage of dissolved acids - sugars, vitamins and minerals in the juice. And since this value can vary from 3% to 15-20%, of course, for your money you will want to get 5 times more minerals and choose cabbage with a value of at least 15%.
)
Control Experience Weight 365 g Weight 430 g (+65 g or 17. 8% to control) Brix 20.4% N:K ratio = 1:8.5 N:K ratio = 1:13.9
### , mineral composition of products and the amount of secondary metabolites. The Royal Society of Chemistry of Britain (RSC) conducted a study of food. As it turned out, there was an average decrease in vitamins and minerals in our food
by 60% . We simply do not get the same concentration of nutrients that is necessary for our health with the products.
### Food as medicine
Nutrient density determines the taste, medicinal qualities and shelf life of fresh foods. Plant nutrition in agriculture involves the management of minerals, microbes and humus to ensure maximum nutrient density.
If you are looking for real food grown in balanced soils and completely free of chemical pollution. If you're looking for a meal with a forgotten taste and an extended shelf life in the refrigerator that will support, not harm, your health and well-being. Read on!
We adhere to the concept of "food as medicine", and food products using the "Eco-Garden without poisons and nitrates" technology are produced with the aim of achieving this result.
Humus and Nutrient Density
So, how do you grow high mineral, low chemical foods in a biological program?
• We know how to enlist the support of microbes in the formation of synergistic systems for the production of agricultural products.
• We can use humic substances to reduce the use of fertilizers.
• We use micronutrients in foliar applications to shift the plant's metabolism for better yield, flavor, keeping quality and resistance.
In short, we focus on growing humus in the soil.
Humus is a highly complex, stable by-product of the digestion of soil organic matter by microbes. It is the most complex natural substance on earth and is very difficult to test. A practical indicator of the presence of humus in the soil is the darkening of the soil to the lower layers, the roots of plants covered with thick hairs, velvety soil and rising Brix levels of plant sap.
Yes, Brix again. The more sugar and minerals in the plant sap, the more sugar it will pump into the soil to feed the microbes in the root space. Normally, a healthy plant gives soil microbes an average of 20-50% of all sugars produced. Beneficial microbes in the soil increase the amount and type of minerals, antibiotics and enzymes that enter the plant through the roots.
More minerals increase plant productivity, root growth and humus levels, since most humus is formed by microbes from dead plant roots.
The main component of humus is glomalin, which is produced by soil fungi. Soil fungi are very sensitive to pure chemical fertilizers and pesticides and are not fast growing soil bacteria. Every time we apply pesticides or unbuffered fertilizers, we are reducing the soil's ability to create humus and nutrient rich food.
Biological agriculture is a concept and technology for gradually increasing the amount of humus in the soil to produce more mineral rich food and improve environmental regeneration.
Our path to profitable farming with environmental benefits lies in the direction of fruits and vegetables rich in minerals and taste. The same processes and benefits apply to growing animal feed crops. To increase the nutritional value, it is necessary to grow humus in the soil. Over time, in such biological systems, crop yields will increase, proving the relationship between nutrient density and crop productivity.
PRK "Black Pearl Humus" - is a granular soil elixir for improving soil fertility with a complex of available elements.
Contributing to the conservation of soil moisture, Black Pearl Humus creates a nutrient medium for the development of microorganisms and bacteria. There is an accumulation of humus, soil fertility increases.
PRK "CHZH Humus" has a positive effect on the soil ecosystem, protects the fertile layer from caking, cracking. It inhibits the development of pathogens, has a beneficial effect on the development of beneficial soil bacteria and microorganisms.
The mechanism of action of PRK "Black Pearl Humus":
- activates the process of soil maturation in early spring;
- starts microbiological processes;
- improves the structure of the soil, it becomes loose and crumbly, but does not crack when dried;
- optimizes the pH level of the soil, increases the availability of soil nutrients for plants;
- increases the efficiency of fertilizers of the mineral group by 2-3 times;
- increases soil moisture retention during the dry period, increases drought and salt resistance of plants;
- contributes to the development of a powerful root system of plants, increase the number and improve the caliber of root crops, increases the yield of marketable products;
- increases the immune status of plants.
Marketable potatoes using Black Pearl Humus PRK
Interesting
A way to increase the paramagnetism and productive potential of the soil
Paramagnetism - guide to the productive potential of your soil. This phenomenon was first described by an excellent American scientist, Professor Phil Callahan. He pointed out that the productive fertility of volcanic soils is directly related to their paramagnetic quality. In fact, the higher the reading of the PCSM meter, the less problems with the soil and the higher the profit from it.
Phil Callahan and PCSM Meter
Here's how it works. Volcanic soils serve as antennas or receivers that attract and store atmospheric energy - extra long frequency radio waves. This energy originally comes from lightning bolts, but in the atmosphere, their explosive energy passes into a more subtle and stable form. Volcanic soils not only attract and store this energy, they can convert it into tiny particles of light called biophotons . The release of these tiny particles of light into the soil effectively gives light to the roots of the plant and the armies of organisms that surround them. Light energy enhances root growth and nodule formation in legumes and stimulates positive microbes.
Available paramagnetism
PRK "Black Pearl Humus", a line of micronized suspensions of PRK "White Pearl Drip BioCalcium + BioMagnesium" contains micronized volcanic rocks.
Nutrient Density and Mineral Fertilizers
Agrochemistry has come to this to some extent, since even with a weak transport system, all highly soluble substances such as calcium nitrate and complex NPKs are simply transported with water. Although water dilutes the juice, it flows very easily due to its low density. That is why foods grown on chemical additives have a rough, watery cell structure, as well as low nutritional value and poor keeping quality.
The Many Faces of Calcium
Calcium is the most important and often overlooked mineral in the vegetable garden or home garden. The first thing to understand is the relationship between soil pH and crop nutrient density. Acidic soil consumes far fewer nutrients than soil with an ideal pH of 6.4. Most minerals are most available at pH 6.4. If you neglect calcium, you are, in fact, neglecting your health.
Calcium is king!
The sooner you get an optimal level of calcium, the sooner the minerals will become more available to plants, and the microbial community will feel great ... Calcium is always the first mineral to be corrected in the soil as it has a very strong effect on other minerals. We often refer to calcium as the "carrier of all minerals" because it directly stimulates the intake of seven other minerals. It also indirectly affects the consumption of all minerals, being the gatekeeper of cell membranes through which all minerals enter the cell itself.
Note that nitrogen goes where there is calcium. It is the basis of amino acid formation, protein chemistry, and DNA copying. As soon as nitrogen appears in the field of view of any proteins, the production of enzymes and hormones begins, and various complex systems are launched, which include such trace elements as iron, zinc, copper, manganese, cobalt, molybdenum, etc.
In the soil calcium is the element that opens up the soil itself. This allows easy penetration into the soil oxygen and easy to leave the soil CO 2 for photosynthesis (gas exchange). Simply put, calcium allows the soil to breathe efficiently.
In plants, calcium is responsible for the vigor of cells and the resulting plant resistance to external influences. It also promotes cell division, plant growth and improved crop quality. With a lack of this mineral, we observe, for example, the top rot of tomatoes and capsicum.
Calcium plays an important role in determining whether your crop is harvested by devastating microbes or sap-sucking insects. A weak cell wall is the hallmark of all pests.
Calcium affects the incidence of bacterial diseases in several ways. First, calcium plays an important role in the formation of healthy, stable cell walls. Adequate calcium levels reduce the formation of enzymes produced by fungi and bacteria that dissolve the middle layer of the lamella, allowing pests to enter and infect plants. Calcium deficiency triggers the accumulation of sugars and amino acids in the apoplast, which reduces disease resistance. Fruit tissues with low calcium content are more susceptible to bacterial diseases and physiological disorders that lead to rot during storage.
To the agronomist's library
Calcium plays an important role in nitrogen fixation and amino acid chemistry, is responsible for the balance of charges in proteins, and is especially important in cell division, which occurs in fruits or seeds immediately after pollination. Without calcium, there will be no fruits or seeds.
For example, for corn, calcium targets in foliar analysis are between 0.3 and 1.0%, and they rise as the corn approaches the heading phase, and should be even higher during kernel formation. If calcium does not reach the required level, the grains at the end of the ear do not fill up. A soybean-type crop needs double or triple the amount of calcium compared to corn to fully set the pods and avoid pod cracking, which is a common soybean problem.
Do you want all the cobs to be filled so that a pod comes out of each soybean flower? This is possible only with the interaction of boron, silicon and calcium.
• Magnesium ( Mg ) is the king of chlorophyll Since photosynthesis requires magnesium, it is the fifth element in the biochemical sequence, in first place among all trace elements. Of course, photosynthesis is not just the production of energy by chlorophyll. Energy must be converted to produce sugars from carbon dioxide and water, which requires phosphorus to convert energy. Otherwise, the chlorophyll burns out, and the leaves turn wine-red.
Magnesium is the central part of chlorophyll , the green pigment found inside the sugar factory (chloroplast) that produces glucose, the building block of life. Magnesium deficiency leads to a decrease in photosynthesis, which will cost us a tidy sum. Magnesium also stimulates phosphorus uptake. Magnesium deficiency reduces the yield and resistance of plants to diseases.
The calcium/magnesium ratio is one of the most important mineral interactions in the soil, allowing the soil to breathe. And also affects the optimal availability for plants of both of these minerals. Excess of any of them seriously affects the consumption of the other. In fact, all the main cations are highly interconnected and an excess of one of them will affect the intake of all the others. That is why concept cation balance is critical for soil.
Calcium opens the soil, while excess magnesium has the opposite effect. Soils with a lot of magnesium cannot breathe normally. On such soils, you will get bogged down in dense, sticky mud. On such soils, mud clods appear during tillage, their poor gas exchange (oxygen in and CO 2 out) reduces photosynthesis and favors pathogens that do not need oxygen to live. Soils with more magnesium need more nitrogen due to its fixation, reuse and availability of nitrogen being impaired. In order to “make money on such soils”, you first need to fix calcium/magnesium ratio .
PRK "White Pearl Drip BioCalcium + BioMagnesium" can be effectively fertigated to correct magnesium deficiency in the root zone .
We invite partners and specialists from research institutes interested in new technologies for plant nutrition, restoration of soil fertility, increase in yield and quality of agricultural crops, to maximize the genetic potential of modern varieties, even in risky farming areas.
### For more information, contact the specialists of AgroPlus Group of Companies LLC:
AgroPlus Group of Companies LLC
350072, Krasnodar Territory, Krasnodar
st. Shosseynaya (Poplar residential area ter.), No. 2/2.
tel.: 8 (861) 252-33-32, 8 (918) 436-36-49, 8 (918) 076-21-05.
e-mail: [email protected]
www.agroplus-group.ru
The Organic Farming Union is an independent social movement. Growth in the production and consumption of healthy, organic products, training, consumer education, scientific research, the introduction of eco-agrotechnologies in the agro-industrial complex.
Health of soils, ecosystems and people.
Enter the Union of Organic Agriculture
Subscribe to us on social networks:
https://vk.com/union_of_organic_agriculture
## ponimayka1 - p. 36
Kemetal in Op. Box of the Reverse operation and in the attributes of the polyNormal node, the Normal Mode parameter. If you change it from Reverse and Extract to just Reverse, the vertices will not be duplicated when the normal is reversed. However, MAYA will still draw the boundary edges, signaling a problem.
Select another face and flip the normal with the Reverse option.
In this case you get really bad geometry. Bad in terms of mathematics. Two faces protrude from one edge, the normals of which are directed in different directions. Important note. The most important detector of problems and bad geometry is the conversion of a polygonal surface into subdivisions. For a quick surface quality check, use the standard Alt-` hotkey in combination with Undo. In case of an error, read the Script Editor, there will be a diagnosis and even a prescription for treatment.
If you try to convert a plane with a bad edge to a subdiv (Modify=>Convert =>Polygons to Subdiv), MAYA will "swear", indicating that some edges are nonmanifold (nonmanifold), that is, they do not connect faces smoothly:
//Error: line 2: polyToSubdivl (Poly To Subdiv Node): One or more edges is nonmanifold.
In this case, the cure is obvious: you just need to reverse the normal with the same Reverse option. However, to find all such non-manifold edges on a surface, it is easier to use the universal cleaner, the Cleanup operation.
Select the surface itself and open the Option Box of the Cleanup operation.
Check the Nonmanifold Geometry checkbox. By default, the Normals and
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Geometry treatment method is selected, using which MAYA first tries to unfold the normals to get rid of bad edges, and if this does not help, then duplicates these edges, making them boundary and cutting adjacent faces. If the Geometry Only option is selected, it does not touch the normals, but immediately duplicates the bad edges.
Experiment with both options by clicking Apply and Undo. See how MAYA then converts the surface into a subdiv.
The concept of "non-manifold geometry" can be translated as follows: a mesh that cannot be unfolded into one flat non-overlapping patch.
I will give additional examples of non-manifold geometry.
If you killed the faces in a checkerboard pattern, the remaining faces are joined by vertices, not edges. Such geometry also cannot be converted to a subdiv, and any anti-aliasing does not work on it, although it will be rendered and animated without problems. In this case, the Cleanup operation simply duplicates the corner vertices.
Another example of a non-manifold mesh is the result of an Extrude Edge operation on internal edges when more than two faces protrude from the same edge. Such a design cannot
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be unfolded into a flat flap and converted into subdivisions.
At the same time, it can be smoothed out and shouldn't cause problems when rendering.
The Cleanup operation in this case duplicates the problematic edge so that each face sticking out of it has its own individual edge.
The main sources of bad geometry are unwashed user hands, chronic sleep deprivation, Extrude Edge, Normals=>Reverse, Delete Face, Collapse, Reduce operations.
Let me give you another example, often encountered for the first two reasons.
Create a plane.
Copy it, as if by accident, without noticing the copy. Select both planes and do Combine.
Like good boys, do Merge Vertices after this.
You now have a visually decent surface that has double edges. And
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is quite difficult to detect. By dragging the vertices, you are synchronously changing these double faces, which have all edges in common. Such faces are called lamina faces. By selecting any face and dragging it, you will simultaneously drag all the edges, and, as a result, the second face. And just by selecting a face not by clicking, but by circling its center with a box, you can notice (which is unlikely) in the title bar of the MAYA window that two faces are selected.
Tip. Turn on the display of polygonal statistics on the screen; Disptay=>Heads Up Display. This will allow you to control how many and which components are selected at any given time.
The treatment of such a "laminar" virus is quite intricate.
Try to convert the resulting plane into a subdiv. You will get the standard error message (stating that the geometry is non-manifold):
//Error: line 2: polyToSubdivl (Poly To Subdiv Node): One or more edges is nonmanifold.
//Nonmanifold geometry cannot be converted to a subdivision surface.
//To clean up nonmanifold edges, use Polygons->Cleanup with the nonmanifold option. / /
Here, of course, you will call "antivirus", that is, in the Option Box of the Cleanup operation, check the Nonmanifold Geometry checkbox.
However, this is not the option to use in this case. However, you may not know this, and therefore, by selecting the default Normal and Geometry option and pressing the Apply button, you will get a rather strange set of selected faces. If you're lucky enough to be able to show normals, you'll see that the surface has normals on both sides!
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If you don't guess, you'll see: MAYA simply cut the surface into separate faces. If you are not completely discouraged by the result, you will probably move vertices or edges to find out what is wrong, and finally find that the surface is double!
Once you see the double layered surface, press Undo as many times as needed to return to the "uncooked" surface.
Open again the Option Box of the Cleanup operation, uncheck the Nonmanifold Geometry checkbox and check the Lamina Faces checkbox at the very bottom.
The treatment will now be more effective.
The Remove Geometry section also contains options that allow you to remove faces that are less than a certain area threshold, or edges that are less than a given value. Since this process will be completely automatic, I would advise you to use it with great care, as such a removal may cause another cleaning of the surface.
The Tessetate Geometry section contains parameters that are not related to cleaning itself
and surface treatment. They allow you to break the mesh into triangles, and specify the criteria for determining which faces to break: quadrilaterals, polygons, concave faces, faces with holes, or non-planar faces. This can be useful when importing into a game engine.
In principle, if you are not going to convert the model into subdivisions, the question of what is considered bad geometry becomes quite subjective. Especially if you do not intend to intensively bend the geometry in the future.
Of course, double vertices or faces are in any case a defect. But other types of non-manifold geometry can only cause partial problems when smoothing the surface. Therefore, if these problems do not concern you, you can not waste time solving them.
However, the popularity of Subdivision Surfaces at the moment is such that it would be rather frivolous to dismiss them (I'm speaking mostly in the organic modeling camp). Draw your own conclusions.
Note. Faces with holes are not "officially" considered a defect, and some operations support them. However, smoothing operations do not work correctly on them, and when converted to subdivs, holes will simply be ignored. But the most important thing is that the Split Polygon Tool does not work on such faces. Therefore, edges with holes are prohibited from use in most Polizen schools.
Very good geometry. About the benefits of rectangles and the dangers of triangles. Frank* Polizen Masters
I'll warn you right away: anyone involved in architecture, design or toys can skip this section. It will focus on the quality of models intended for character animation. I do not pretend to be objective, therefore, perhaps something may seem unobvious or controversial to you.
After speaking with the Polizen masters, I discovered that absolutely all of them mystically avoid triangles. Not satisfied with the answer that triangles have a bad effect on karma, and the number "four" is a symbol of stability, I did some research of my own, which I summarized in the following material.
There are three technical reasons why craftsmen avoid triangles and try to work only with quadrilaterals. And also - a few esoteric reasons, about which a little later. The first reason is based on the popularity of subdivs for organic modeling. It's the subdivs that really dislike triangular faces. Do the following exercise. Take the polycube. Select one of the vertices and cut it with the Chamfer Vertex operation.
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Delete history. Duplicate the cube and move the copy to the right.
Select the cut face and cut it into three quadrilaterals using the Subdivide operation. Then take the Split Polygon Tool and cut the three faces adjacent to the cut one in half, diagonally from the middle of the cut edge. Delete history.
Now select both cubes and convert them to subdivs by pressing 'Alt-'. Play with the smoothness of the display by pressing "0,1,2,3".
Press "0" and note that the cube, which had more faces, has turned into a "lighter" subdiv. This happened precisely because it was correctly divided into quadrangles.
In addition, grids containing triangles, pentagons and other polygons are difficult to parameterize, and even generate “special” points (“asterisks”) on subdivisions, where three, five or more “isoparms” converge. These points are a potential source of problems, since all sorts of "buckling" can appear in their vicinity when deformations occur.
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Quadrilaterals tend to be well parametrized from the very beginning because they are pieces of a rectangular grid, so subdivisions “like” quadrilaterals very much.
To finally verify this, do the final experiment. Select both subdivs and convert them to NURBS surfaces: Modify"Convert=>Subdiv
to NURBS.
Now count the number of spline patches on each of the former cubes. There will be 27 of them on the first cube, and only 9 on the sawn one.
The second technical reason is that Polyzen masters often use Renderman to render characters. And he, oh, how he does not like triangles! And not only Renderman: after all, until recently, Mental Ray, built into MAYA, could not render Subdivision Surfaces, which did not consist of only quadrilaterals, and any triangular face in the base mesh caused him allergies. Of course, now there are effective methods for rendering triangles in subdivs (such as Loop Scheme), but the sediment, as you understand, remains, and triangles still have a reputation as a hindrance to rendering subdivs.
The third reason is not related to subdivs, but to character animation. Very often, non-quad faces located at surface folds cause unpredictable deformations. Also, in the same places, it is undesirable to have a vanishing point of several "patches", that is, a vertex from which more than four edges emerge. All this leads to unpleasant consequences in the form of "disturbances", screeds, creases, etc. Of course, if you cannot do without triangles at all, you can leave them in some places, using the following rule: the main thing is that a given non-quadrangular (triangle, pentagon, etc.) .d.) did not cause perturbations on the surface after smoothing or when translated into subdiv. And behaved decently during the animation. Practice shows that such non-quadrangular polygons cannot be left in those places that are actively stretched, compressed during animation. Especially if the character's skin is not covered with fur. Roughly reformulate this rule as follows: if you want to effectively bend or somehow deform the surface, keep its edges quadrangular. Let it need an extra chain of edges.
There are also some existential reasons for avoiding non-quadrangles. If you look around, you can find a lot of examples that in nature, quadrilateral patches are often preferred, when talking about surfaces, not about volumes. Take, for example, fabrics: their fibrous structure is a rectangular grid. In confirmation of this, no doubt, controversial hypothesis, I would like to cite a story that one of the Polidzen masters Sergey Lutsenko told me. The story is very instructive
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and at the same time incomprehensible to those who have not yet felt what “the form does not live” means. Read and try to feel it.
“I made my personal discovery about how the edges and faces of the model should be distributed when I once had to model the cap of an American policeman. I got stuck on modeling the tulle (the very top of this headdress |. An attempt to just formally lift up the front of the tulle did not lead to anything good. The cap was not a cap anyway. The wife who was modeling clothes came to the rescue. She, looking from the side ", immediately pointed out the mistake and suggested paying attention exactly to how the fabric is located on a real object, what and how many pieces of fabric the cap consists of? How and where it is sewn. And when I reproduced all these fragments, exactly according to the original, the cap "came to life ". The design is working!"
Some orthodox Polyzen masters of the old Indian school still make the initial blank of the model with spline patches, which are then converted into polygons, getting flawless edges and perfect texture mapping to boot.
At this point I would like to finish the ode to quadrilaterals and briefly highlight three operations for splitting faces into the required number of vertices.
To spoil the model and turn all its faces (or not all, but selected ones) into triangles, there is a special operation: Triangulate. (This is a joke, and often this operation is needed to export the model to the game engine or feed it to the input of a program that, apart from triangles, does not understand anything - after all, there are others).
When I started learning MAYA, I expected that the Quadrangulate operation does the same, not the opposite, that is, it finds all non-quad faces and cuts them into four vertices. But no, this operation is the opposite of Triangulate: it only searches for triangular faces, and if the angle between adjacent triangles is less than the specified one, these two triangles are turned into a quadrilateral by deleting a common edge. The process, as you know, is not very predictable.
But if you want to ensure that the selected faces turn into quadrilaterals, you should use the Subdivide operation. She breaks the edges in the right way. In addition, it also works for selected edges.
You can select only triangular or only quadrilateral faces using the Edit Polygons=>Selection=>Selection Constraints window. Also keep in mind that its content depends on the active components.
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Also, don't forget: one of the added bonuses of the Smooth operation is that it only creates quad faces during the smoothing process (for the Exponent method).
Let's talk a little about vertex normals. Let me remind you that these are normals sticking out of vertices, the number of which in each vertex is equal to the number of faces adjoining it. These normals have a decorative function and affect how soft or hard transitions will be in the edges when lit.
Basically, if you are going to convert the surface to subdivs, vertex normals shouldn't bother you at all: in that case they are ignored. However, they often disfigure the appearance of the surface on the screen so much that it is useful to find some kind of control on them. Very often the model "arrives" in the scene with bad vertex normals when imported from other packages. In this case, it is useful to do the following procedure to treat corrupted vertex normals.
The easiest way is to select the entire surface and perform the Edit Polygons=>Normals=>Set Vertex Normal operation, and in the Option Box, check the Unlock Normals checkbox.
This checkbox with an illogical and incomprehensible name says that you need to not only unlock vertex normals, but also recalculate them to default values (these values are determined by the softness / hardness of the edges set earlier).
After that you can set the hardness/softness of the edges with the Soften/Hard en operation. The mechanism of blocking vertex normals is not obvious, since it is NOT explicitly displayed anywhere on the screen whether vertex normals are blocked or not.
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New SAT Math Workbook
# 12 6 a b c d e 5 6 43 8 62 273 274 chapter 16 4 in
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Unformatted text preview: s are all multiples of 12, but no other elements. (B) The shaded region to the left of the y-axis accounts for all values of x that are less than or equal to –3 . In other words, this region is the graph of x ≤ –3. The shaded region to the right of the y-axis accounts for all values of x that are greater than or equal to 3 . In other words, this region is the graph of x ≥ 3. (C) Note that (–2)5 = –32. So, the answer to 1 3. 4. the problem must involve the number 5. However, the 2 in the number 2 is in the denominator, and you must move it to the numerator. Since a negative number reciprocates its base, − 1 5. (E) f ( 2) x +1 x +1 −5 = −32 . Substitute x + 1 for x, then simplify: 1 x +1 1 () 1= x +1 = 1= +1 1 x +1 1 + = 1+ ( x +1) x +1 1 x +1 = x +1 1 + ( x + 1) x + 2 www.petersons.com 272 Chapter 15 6. (C) According to the function, if x = 0, then y = 1. (The function’s range includes the number 1.) If you square any real number x other than 0, the result is a number greater than 0. Accordingly, for any non-zero value of x, 1 – x2 < 1. The range of the function includes 1 and all numbers less than 1. (C) The graph of f is a straight line, one point on which is (–6,–2). In the general equation y = mx + b, m = –2. To find the value of b, substitute the (x,y) value pair (–6,–2) for x and y, then solve for b: y = −2 x + b (−2) = −2(−6) + b −2 = 12 + b −14 = b 9. (A) The graph of any quadratic equation of the incomplete form x = ay2 (or y = ax2 ) is a parabola with vertex at the origin (0,0). Isolating x in the equation 3x = 2y2 shows that the equation is of that form: x= 2 y2 3 7. To confirm that the vertex of the graph of 2 y2 x= lies at (0,0), substitute some simple 3 values for y and solve for x in each case. For example, substituting 0, 1, and –1 for y gives us the three (x,y) pairs (0,0), ( 2 ,1), and ( 2 ,–1). 3 3 Plotting these three points on the xy-plane, then connecting them with a curved line, suffices to show a parabola with vertex (0,...
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## This note was uploaded on 08/15/2010 for the course MATH a4d4 taught by Professor Colon during the Spring '10 term at Embry-Riddle FL/AZ.
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# What is the formula of differentiation?
## What is the formula of differentiation?
Some of the general differentiation formulas are; Power Rule: (d/dx) (xn ) = nx. Derivative of a constant, a: (d/dx) (a) = 0. Derivative of a constant multiplied with function f: (d/dx) (a.
## Which graph is a one to one function?
Horizontal Line test: A graph passes the Horizontal line test if each horizontal line cuts the graph at most once. Using the graph to determine if f is one-to-one A function f is one-to-one if and only if the graph y = f(x) passes the Horizontal Line Test.
## How do you use differentiation in the classroom?
Teachers who practice differentiation in the classroom may:
1. Design lessons based on students’ learning styles.
2. Group students by shared interest, topic, or ability for assignments.
3. Assess students’ learning using formative assessment.
4. Manage the classroom to create a safe and supportive environment.
## Where is differentiation used in real life?
Application of Derivatives in Real Life. To calculate the profit and loss in business using graphs. To check the temperature variation. To determine the speed or distance covered such as miles per hour, kilometre per hour etc.
## Are parabolas one to one functions?
The function f(x)=x2 is not one-to-one because f(2) = f(-2). Its graph is a parabola, and many horizontal lines cut the parabola twice. The function f(x)=x 3, on the other hand, IS one-to-one. If two real numbers have the same cube, they are equal.
## What are some real life examples of functions?
You might draw from the following examples:
• A soda, snack, or stamp machine. The user puts in money, punches a specific button, and a specific item drops into the output slot.
• Measurement: Â Thermometer.
• Miles per gallon.
• Basic economics and money math:
• Geometric Patterns.
## What are examples of differentiation?
Examples of differentiating content at the elementary level include the following:
• Putting text materials on tape;
• Using spelling or vocabulary lists at readiness levels of students;
• Presenting ideas through both auditory and visual means;
## How are inverse functions used in real life?
An inverse lets you take information that tells you how to get y from x, and in return it tells you how to get to x from y. The formula for finding the Celsius temperature given the Fahrenheit temperature is the inverse of the function for finding the Fahrenheit temperature given the Celsius temperature.
## What are the three things that help you in representing real life situations?
Answer. Answer: Equations representing direct, inverse, and joint variation are examples of rational formulas that can model many real-life situations.
## What is the purpose of differentiation?
Differentiation allows us to find rates of change. For example, it allows us to find the rate of change of velocity with respect to time (which is acceleration). It also allows us to find the rate of change of x with respect to y, which on a graph of y against x is the gradient of the curve.
## What is a one to one function example?
A one-to-one function is a function in which the answers never repeat. For example, the function f(x) = x^2 is not a one-to-one function because it produces 4 as the answer when you input both a 2 and a -2, but the function f(x) = x – 3 is a one-to-one function because it produces a different answer for every input.
## What are three types of product differentiation?
There are several different factors that can differentiate a product, however, there are three main categories of product differentiation. Those include horizontal differentiation, vertical differentiation, and mixed differentiation.
## What is the physical meaning of differentiation?
Physical meaning of differentiation is that it represent rate of change of parameter. For example differential of velocity graph will indicate accelaration. Integration inficated area uder a curve. For example integral of a graph of force against distance will indicate work done. 1.
## How do you find an inverse?
Finding the Inverse of a Function
1. First, replace f(x) with y .
2. Replace every x with a y and replace every y with an x .
3. Solve the equation from Step 2 for y .
4. Replace y with f−1(x) f − 1 ( x ) .
5. Verify your work by checking that (f∘f−1)(x)=x ( f ∘ f − 1 ) ( x ) = x and (f−1∘f)(x)=x ( f − 1 ∘ f ) ( x ) = x are both true.
## Why is it called differentiation?
The etymological root of “differentiation” is “difference”, based on the idea that dx and dy are infinitesimal differences. If I recall correctly, this usage goes back to Leibniz; Newton used the term “fluxion” instead.
## What differentiation means?
Differentiation is a process of finding a function that outputs the rate of change of one variable with respect to another variable. Informally, we may suppose that we’re tracking the position of a car on a two-lane road with no passing lanes.
## What are one and onto functions?
A function f from A (the domain) to B (the range) is BOTH one-to-one and onto when no element of B is the image of more than one element in A, AND all elements in B are used. Functions that are both one-to-one and onto are referred to as bijective. Bijections are functions that are both injective and surjective.
## What is differentiation in real life?
Differentiation and integration can help us solve many types of real-world problems. We use the derivative to determine the maximum and minimum values of particular functions (e.g. cost, strength, amount of material used in a building, profit, loss, etc.).
## What are the three things that help you representing real life?
Any function of the form , where and , and is a rational function….Three things that may help you representing real-life situation to rational function are the following;
• The average velocity of a vehicle:
• The universal gravitation:
• Work rate problems:
## How do you write a one to one function?
If the graph of a function f is known, it is easy to determine if the function is 1 -to- 1 . Use the Horizontal Line Test. If no horizontal line intersects the graph of the function f in more than one point, then the function is 1 -to- 1 .
## What is differentiation in simple words?
Differentiation means finding the derivative of a function f(x) with respect to x. Differentiation is used to measure the change in one variable (dependent) with respect to per unit change in another variable (independent).
## What is product differentiation?
Product differentiation is a marketing strategy designed to distinguish a company’s products or services from the competition. If successful, product differentiation can create a competitive advantage for the product’s seller and ultimately build brand awareness.
## What is the first principle of differentiation?
In this section, we will differentiate a function from “first principles”. This means we will start from scratch and use algebra to find a general expression for the slope of a curve, at any value x. First principles is also known as “delta method”, since many texts use Δx (for “change in x) and Δy (for “change in y”). |
## Closure Property and Imaginary Numbers
If an operation is performed on any two numbers in a set and the result is still in that set, we say that the set is closed under that operation. For example, we can say that the set of positive integers is closed under addition because if we add any two positive integers, the result is still a positive integer. On the other hand, the set of positive integers is NOT closed under subtraction because in many cases, the result is a negative integer (e.g 9 – 12 = -3). The video below is one of the best videos I’ve seen so far explaining closure property This video is just one of a series of videos about Imaginary Numbers.
Did you enjoy the video? If yes, kindly share it to your friends.
## Properties of Addition and Multiplication of Real Numbers
We are all familiar with some properties of real numbers. Real numbers are commutative, associative, and closed under addition and multiplication. We have also discussed that multiplication of real numbers is distributive over addition.
In this post, we formalize our knowledge of these properties of real numbers and add two more to the list: the identity and inverse properties. Notice also that almost all properties under addition have their corresponding equivalents under multiplication. » Read more
## Military drills and the closure property of real numbers
In military drills, we are familiar with the commands that let the soldiers face or turn to certain directions. Left face (sometimes called left turn), right face, and about face are probably the most common commands used. If we are facing north, a left face would mean turning left 90 degrees, which means facing west. In the following discussion, we will agree that our starting position is facing north. We will call this position, the standard position.
Let us represent the turns with letters:A for about face, R for right face, and L for left face. Notice that whatever combination of turns we do, LR or A, the result is confined to the four directions. An R followed by an A is equivalent to an L (facing west) with respect to the standard position. Likewise an L followed by another L is equivalent to A.We will to denote our starting position P as reference; it is not turn command, so if the soldier is not facing north (see second figure), a P will just mean that the soldiers remain in their current position. » Read more |
# Data Science Probability Interview Questions
##### Categories
Top companies require top data scientists. The ones that are good at solving probability problems, which is what we’ll do here.
Probability is a part of statistics. But at the job interviews, this is often a completely separate topic. That’s why within the non-coding questions we have a category dedicated solely to the probability interview questions.
There are questions of different levels of complexity, but they all boil down to one basic concept: the probability definition.
## What is a Probability?
A probability is defined as a ratio between the number of favorable and all possible outcomes or cases.
Mathematically, we can show it this way:
$Probability\ = \ \frac{Number\ of\ Favorable\ Cases}{Number\ of\ All\ Possible\ Cases}$
Let’s have a look at how you can apply this to something practical. Namely, to solve the probability interview questions.
To follow what comes easier, you might want to have a look at our guide explaining probability and other statistics concepts you need as a data scientist.
### Data Science Probability Interview Question #1: Two Cards Same Suit
We’ll solve this probability interview question with the assumption that the cards are not replaced.
#### Intuitive Solution
Having in mind the probability definition, you can solve this data science probability interview question intuitively. There are four suits in the card deck. In every suit, there are 13 cards. That means there are 52 cards in total.
If you draw one card and have a look at its suit, the second card you draw has to be the same suit. Since you already drew one card, that means there are still 51 possible cases. There are also 12 favorable cases.
If you translate this to a formula, you get the following:
$\frac{Favorable\ Cases}{All\ Possible\ Cases}\ =\ \frac{12}{51}\ =\ \frac{4}{17}$
#### Mathematical Solution
The number of all possible cases expressed mathematically is:
\begin{aligned} All\ Possible\ Cases\ &= \ {}^{52}P_{2} \\ &=\ 52\ \times\ 51 \\ &=\ 2,652 \end{aligned}
This is because any two cards can be drawn from a 52-card deck.
To calculate the favorable outcome, we have to choose which of the four suits we want to double up. Once we have determined the suit, we can draw two cards from the same suit. In other words:
\begin{aligned} Favorable\ Cases\ &= \ 4\ \times\ {}^{13}P_{2} \\ &=\ 4\ \times\ 13\ \times\ 12 \\ &=\ 624 \end{aligned}
Put this into the probability formula and you’ll get:
\begin{aligned} Probability\ =\ \frac{624}{2,652} \\ =\ \frac{4}{17} \end{aligned}
### Data Science Probability Interview Question #2: Where Are the Birthday People?
This one’s a little bit more difficult question. It’s asked by Yammer:
Here we’re going to assume that there are 365 days in a year and k ≤ 365.
Unlike the previous question, where repetition was not possible, here the birthdays might fall on the same date.
For each person in the room, there are 365 possible scenarios. Therefore:
\begin{aligned} All\ Possible\ Cases\ &=\ 365\ \times\ 365\ \times\ ...\ k\ times \\ &=\ 365^k \end{aligned}
You would expect we’re going to find the favorable cases as a next step. We will, but not just yet. Since here it’s harder to find favorable cases than unfavorable ones, we’ll first find the unfavorable cases. From there on, it’ll be easy to find favorable cases.
Unfavorable cases mean none of the k number of people has the birthday on the same date. Let’s assume there are 365 seats in the room, each numbered with a calendar date:
1st Jan 2nd Jan 3rd Jan 29th Dec 30th Dec 31st Dec
Each person has to sit on a seat which will be numbered with their date of birth. The first person has 365 options. The second person can’t sit on the same seat as the first person, so there remain 364 options, and so on.
Mathematically this equals to:
\begin{aligned} All\ Possible\ Cases\ &=\ 365\ \times\ 364\ \times\ ...\ k\ times \\ &=\ {}^{365}P_k \end{aligned}
From it follows:
\begin{aligned} Favorable\ Cases\ &=\ All\ Possible\ Cases\ -\ Unfavorable\ Cases \\ &=\ 365^k\ -\ {}^{365}P_k \end{aligned}
Probability is, therefore:
$Probability\ =\ \frac{365^k\ -\ {}^{365}P_k}{365^k}$
### Data Science Probability Interview Question #3: Two Out of Three Tails
The last probability interview question we’ll talk about is asked by Jane Street:
This question too can be answered intuitively and mathematically.
#### Intuitive Solution
You can quite easily visualize the scenarios. For example, if you flip four coins with two sides, the number of possible outcomes is:
$All\ Possible\ Cases\ =\ 2^4\ =\ 16$
The distribution can be shown in the following way:
Scenario # Cases 4 Heads 1: HHHH 3 Heads, 1 Tail 4: HHHT, HHTH, HTHH, THHH 2 Heads, 2 Tails 6: HHTT, HTHT, HTTH, THHT, THTH, TTHH 1 Heads, 3 Tails 4: TTTH, TTHT, THTT, HTTT 4 Tails 1: TTTT
As you can see, there are really 16 possible outcomes. However, the question states there are at least two tails. That means that the first two scenarios are not possible. This leaves us with:
$All\ Possible\ Cases\ =\ 11$
Of the remaining cases, the only favorable ones are where we get one head and three tails. Therefore:
$Favorable\ Cases\ =\ 4$
Finally, the probability is:
$Probability\ =\ \frac{4}{11}$
#### Conclusion
The three questions we’ve covered give you a picture of what you can expect at the data science interviews when it comes to the probability interview questions. The solutions we provided are also an outline of how you should approach solving the data science probability interview questions.
This should be just a start for you to practice more of these questions. Solving many probability interview questions is the only way to increase, ahem, the probability of becoming a data scientist at the top company. These 30 probability and statistics interview questions that’ll give a good kick in that direction. You can always have some more in the non-coding questions section on our platform.
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# Video: Converting Improper Fractions to Mixed Numbers
Tim Burnham
The height of a building is 65/7 meters. Write that height as a mixed number.
02:31
### Video Transcript
The height of a building is sixty-five over seven meters. Write that height as a mixed number.
So we’ve been given our measurement, our height, in meters. And that, in this case, is a top heavy or an improper fraction. And the question is asking us to convert that into a height as a mixed number. Now a mixed number is a number that’s got a whole part and then to that we add a fractional part, which is less than one. So we need to analyse sixty-five over seven and split it down into a whole number plus a fractional number.
So we can visualize this as some whole meters plus a fraction of a meter. Now each of these whole meters are gonna be split into sevenths, because our denominator on our improper fraction is seven. So each of those whole meters could be split into seven. And then we’ll have some of the sevenths of a meter left over which are going to make up our fractional part. Now at this stage, I don’t know how many whole meters we’re gonna have, but each whole meter is seven seven seven meter. So we’ve got seven seventh of a meter, plus another seven seven seventh of a meter, plus another seven seven seventh of a meter and so on, and finally another seven seven seventh of a meter. And then we’re gonna have some number of other sevenths left over. So the first thing we need to address is how many of these seven seven sevenths of a meter are we gonna get out of our sixty-five sevenths of a meter.
So for the whole part, how many times does seven go into sixty-five? Well seven times nine is sixty-three, so I could have nine lots of seven. Seven times ten is seventy, so that would be too many. Seventy is bigger than sixty-five. So the largest whole number that I can multiply seven by and get an answer of sixty-five or less is nine, so nine is gonna be the whole part of our answer. So the whole number accounts for sixty-three sevenths, but I want sixty-five sevenths. So to work out the fractional part, I need to find the difference between these two. So that’s sixty-five minus sixty-three sevenths, which is two-sevenths. So to get our mixed number, all I have to do is add the nine for the whole number to the two-seventh we’ve got here. And that gives us our answer of nine and two-sevenths meters. |
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# What are key features of a rational function?
## What are key features of a rational function?
Two important features of any rational function r(x)=p(x)q(x) r ( x ) = p ( x ) q ( x ) are any zeros and vertical asymptotes the function may have. These aspects of a rational function are closely connected to where the numerator and denominator, respectively, are zero.
What are the 5 examples of rational function?
Rational Functions
• f(x)=x+2x.
• g(x)=x−1x−2.
• h(x)=x(x−1)(x+5)
• k(x)=x2−1×2−9.
• l(x)=x2−1×2+1.
### What are the 3 types of rational functions?
Rational functions can have 3 types of asymptotes: Horizontal Asymptotes. Vertical Asymptotes.
What defines a rational function?
Definition of rational function : a function that is the quotient of two polynomials also : polynomial.
## What are the distinct features of rational inequalities?
A rational inequality is an inequality that contains a rational expression, where a rational expression is a ratio of two polynomials. That is, a rational expression is of the form R(x) / Q(x), where R(x) and Q(x) are polynomials and Q(x) is not zero.
What is the meaning of rational function?
### What is the distinct feature of rational equation?
A rational equation is an equation containing at least one fraction whose numerator and denominator are polynomials, \frac{P(x)}{Q(x)}. Q(x)P(x). These fractions may be on one or both sides of the equation.
What are the two things importantly used in graphing rational function?
Two important properties of rational functions are: The zeros of the function are the zeros of the numerator. These values cannot be the zeros of the denominator. The vertical asymptotes of the graph are determined by calculating the zeros of the denominator.
## What is rational function in general mathematics?
A rational function is one that can be written as a polynomial divided by a polynomial. Since polynomials are defined everywhere, the domain of a rational function is the set of all numbers except the zeros of the denominator. Example 1. f(x) = x / (x – 3). The denominator has only one zero, x = 3.
Why is it called rational function?
A function that is the ratio of two polynomials. It is “Rational” because one is divided by the other, like a ratio.
### What are distinct features of rational inequality?
Why are rational functions important?
Rational equations can be used to solve a variety of problems that involve rates, times and work. Using rational expressions and equations can help you answer questions about how to combine workers or machines to complete a job on schedule.
## Why rational function is important?
What is the test point method?
The ‘test point method’ involves identifying important intervals, and then ‘testing’ a number from each interval—so the name is appropriate. There are two slightly different ‘flavors’ of the test point method, but only one is discussed here.
### What have you learned in rational functions?
Lesson Summary Rational functions are functions where you have a polynomial in both the numerator and denominator. We can write these functions if we are given the vertical and horizontal asymptotes.
How are rational functions used in everyday?
There are several applications of rational functions in everyday life. We can form rational equations and formulas to calculate speeds or distances, calculate the work rate of people or machines, and we can solve mixing problems. |
# 2019 AIME I Problems/Problem 2
## Problem 2
Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$. Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$. The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
## Solution
Realize that by symmetry, the desired probability is equal to the probability that $J - B$ is at most $-2$, which is $\frac{1-P}{2}$ where $P$ is the probability that $B$ and $J$ differ by 1 (no zero, because the two numbers are distinct). There are $20 * 19 = 380$ total possible combinations of $B$ and $J$, and $1 + 18 * 2 + 1 = 38$ ones that form $P$, so $P = \frac{38}{380} = \frac{1}{10}$. Therefore the answer is $\frac{9}{20} \rightarrow \boxed{029}$.
## Solution 2
This problem is basically asking how many ways there are to choose 2 distinct elements from a 20 element set such that no 2 elements are adjacent. Using the well-known formula $\dbinom{n-k+1}{k}$, there are $\dbinom{20-2+1}{2} = \dbinom{19}{2} = 171$ ways. Dividing 171 by 380, our desired probability is $\frac{171}{380} = \frac{9}{20}$. Thus, our answer is $9+20=\boxed{029}$. -Fidgetboss_4000
## Solution 3
Simply create a grid using graph paper, with 20 columns for the values of J from 1 to 20 and 20 rows for the values of B from 1 to 20. Since we know that $B$ cannot equal $C$, we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since $B - J$ must be at least $2$, we can mark the line where $B - J = 2$. Now we sum the number of squares that are on this line and below it. We get $171$. Then we find the number of total squares, which is $400 - 20 = 380$. Finally, we take the ratio $\frac{171}{380}$, which simplifies to $\frac{9}{20}$. Our answer is $9+20=\boxed{029}$. |
Loops
Math Lair Home > Topics > Loops
"Mathematicians have their own version of stunt flying. They use numbers instead of airplanes. (Wouldn't you know it!) They get their thrills from numerical loopings and from discovering patterns while they're doing their tricks. One person's pattern is another's loop-the-loop."
Math for Smarty Pants
Here are some interesting looping techniques (involving starting with some number and repeating some process on it over and over again) that you might want to try out:
Start with any number you like and follow the following rules:
1. If your number is even, divide it by two. Otherwise, multiply it by 3, then add 1.
2. Loop back to step 1. Keep on repeating step one again and again and again...
For example, Start with 10. It's even, so divide by two. That gives you 5. It's odd, so multiply it by 3 and add 1: ( 5 × 3) + 1 = 16. 16 is even, so take half and get 8. Half again gives you 4. Half again gets you to 2, and half again gives you 1. Since 1 is odd, multiply by 3 and add 1 to get 4. Half of 4 gives you 2, and half of that gets you back to 1. You're in a loop now and will be forever if you keep at it.
Here's another example, using 33. You may want to try it yourself to verify that you get the following: 33, 100, 50, 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, ...
A calculator might come in handy here.
Have some fun trying some more, but note that some numbers can take a while to loop back. For example, if you were to start with 27, it takes 109 steps! Other time-consuming yet small numbers are 31, 41, 47, 55, 62, 63, 71, and 73. Of course, any of these numbers multiplied by two, four, or any other power of two will also give you a long sequence.
You might wonder whether every integer will eventually loop into this 4-2-1 pattern at the end. I have personally tested this for every number up to 1,000,000 (with a computer program, of course), and every number falls into that loop. Others have tested this conjecture to much higher numbers. Naturally, checking every number doesn't satisfy mathematicians, as it will never answer the question of whether every number will eventually loop into the pattern. This is known as the Collatz conjecture, the idea that every number does. While it seems likely to be true, proving it to be either true or false is quite difficult. No-one has found a number that doesn't work, but no-one has been able to prove that every number works.
Here's another looping procedure somewhat similar to the previous one. This one was developed by Clifford Pickover, who calls it the Juggler Sequence. The difference between this sequence and the loop described above is that, to get the next number, you take the square root of the current number if it is even, and multiply by the square root of the current number if it is odd, and then round down to the next-lowest integer. Most sequences end . . ., 6, 2, 1. Some take a while, though. You might want to try a few numbers. Actually proving that all sequences end in 1 would probably be quite difficult. To explain why, notice that the the next number in each sequence can be found by taking the number to the power of 1.5 if the number is odd and taking it to the power of 0.5 if it is even. If you replaced 1.5 with 1.49 and 0.5 with 0.51, some very different sequences would be produced. You may want to experiment with this.
Another looping procedure is known as aliquot chains. For this procedure, start with any natural number. Find the sum of its proper divisors (also known as aliquot parts). Now take the sum, find the sum of its aliquot parts, and so on and so on.
This loop behaves in a variety of ways. For some numbers, such as 6 and 28, the sum of their aliquot parts is equal to themselves; these are known as perfect numbers. Other numbers loop back to themselves after two (for amicable numbers) or more (for sociable numbers) iterations. Many numbers reach 1 eventually (if you start with a prime number to begin with, you'll get there really quickly).
Another well-known loop is as follows:
1. Select a four-digit number whose digits are not all the same
2. Rearrange the digits of the number so that they form the largest number possible.
3. Rearrange the digits of the number so that they form the smallest number possible
4. Subtract the smaller number from the larger number
5. Take the result of the subtraction and repeat, starting at step 2.
If you repeat the process long enough, you'll get 6,174. It isn't known why. This loop was discovered by the Indian mathematician Dattathreya Ramachandra Kaprekar in 1946. You might want to see what happens if you try numbers of different lengths.
For the next looping procedure, start with any two numbers from 0 to 9 and follow this rule: Add the two numbers and write down just the digit that is in the ones place. Here's an example: Suppose you start with 8 and 9. Adding them gives you 17. Keep just the 7, which is in the ones place. Add the last two numbers, the 9 and the 7. That gives 16; keep the 6, then you have 8-9-7-6. Keep going, adding the last two numbers in the series each time, keeping only the digit in the ones place. Do this until you get 8 and 9 again. Then the loop starts all over. The 8-9 pattern has twelve numbers in the loop before it repeats. The pattern is: 8-9-7-6-3-9-2-1-3-4-7-1-8-9.
If going around in a numerical circle appeals to you, you may have the makings of a terrific mathematician. Hang in there. But beware. If you start with the same two numbers, but in the opposite order, and follow the same rule: 9-8-7-5, and so on, it will take 60 numbers before it starts to repeat! Don't tackle that one unless you're sure you have the time. For a quickie, try 2 and 6.
Here's some questions you might want to think about:
1. How many different possible pairs of numbers are there to start with? (It's okay to start with two numbers that are the same).
2. Do all pairs of numbers eventually return to the starting point?
3. What's the shortest loop you can find?
4. Is there a pattern of odds and evens in the loop?
Here is a numerical example using words. Start with any number (in the example below, 39). Write it as a word: thirty-nine. Then continue as shown.
Start with any number 39 Write it as a word thirty-nine Count the letters 10 Write that as a word ten Count the letters 3 Write that as a word three Count the letters 5 Write that as a word five Count the letters 4 Write that as a word four Count the letters 4
You'll get 4 forever and ever. In fact, you will always end up with 4, no matter what number you start with originally. Try a different number and see. Convince yourself with some examples, then see if you can figure out why you'll always get to four. A good starting point would be to look at how to write out big numbers. You may also want to determine what the result would be in other languages.
Loops appear in many different areas of mathematics. If you're interested in looping, you may want to check out the pages on palindromes, sociable numbers, chaos and iterative systems, fractals, and Armstrong numbers.
Sources used (see bibliography page for titles corresponding to numbers): 32, 57, . |
# 5 Times Table – Multiplication Table
## Multiplication Table of 5
Learning multiplication tables can be a crucial milestone in a child’s mathematical development. The 5 times table is one of the most interesting and easy-to-learn tables for kids. It provides a foundation that aids in understanding patterns and numerical relationships.
The 5 times table is particularly appealing because every product in the series ends in either 0 or 5. Children often learn this table with the help of rhymes or by counting in fives on their fingers. You can see a pattern emerging as you go through the sequence: 5, 10, 15, 20, 25, and so on. Brighterly’s 5 Times Table Chart offers a colorful, interactive way to get acquainted with this sequence.
## 5 Times Table
Understanding the 5 times table goes beyond mere memorization; it encourages pattern recognition and develops a foundational understanding of multiplication. Multiplying by 5 can be visualized as adding five identical numbers or adding 5 five times:
5 1 5 5 2 10 5 3 15 5 4 20 5 5 25 5 6 30 5 7 35 5 8 40 5 9 45 5 10 50
This pattern continues, and you can create fun games or quizzes to master this table. The 5 times table also introduces children to the concept of multiples and helps them recognize these in everyday contexts.
## Tips for 5 Times Table
The 5 times table is often considered one of the most approachable multiplication tables. Here are some engaging ways to learn and practice:
1. Use Your Hands: Counting on fingers by fives is a tactile and visual way to understand the pattern.
2. Rhymes and Songs: Create or find catchy rhymes or songs that integrate the 5 times table.
3. Double then Add: You can use known 2 times table facts and then double the result, adding another set of the number to get the 5 times result.
## 5 Times Table Examples
The 5 times table extends into higher numbers, and the pattern continues to remain consistent. Here are some examples:
• 5 x 10 = 50
• 5 x 20 = 100
• 5 x 30 = 150
• 5 x 100 = 500
These examples show how the 5 times table acts as a bridge to understanding larger numbers and prepares children for more advanced multiplication tasks.
## More Multiplication Tables:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
## Practice Problems 5 Times Table
Practicing the 5 times table can be both fun and rewarding. Here are some practice problems:
1. 5 x 6 = ?
2. 5 x 12 = ?
3. 5 x 7 = ?
4. 5 x 15 = ?
## Conclusion
The 5 times table is a beautiful introduction to the world of multiplication. Its patterns are accessible and engaging, offering children a fun and intuitive way to grasp essential mathematical concepts. With resources like rhymes, visual aids, and online tools from Brighterly, the journey of learning this table can be an enriching experience that opens doors to further mathematical exploration.
## FAQs on 5 Times Table
### Is the 5 times table useful for older students?
Yes, it serves as a foundation for understanding multiplication with larger numbers.
### Can learning the 5 times table help in daily life?
Absolutely, it assists in quick mental calculations in various everyday situations like shopping.
### What are some creative ways to teach the 5 times table?
Songs, games, and visual aids can make learning this table a joyful experience.
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# College Algebra : Rational Exponents
## Example Questions
← Previous 1
### Example Question #1 : Rational Exponents
Simplify:
Explanation:
An option to solve this is to split up the fraction. Rewrite the fractional exponent as follows:
A value to its half power is the square root of that value.
Substitute this value back into .
### Example Question #2 : Rational Exponents
Which of the following is equivalent to ?
Explanation:
Which of the following is equivalent to ?
When dealing with fractional exponents, keep the following in mind: The numerator is making the base bigger, so treat it like a regular exponent. The denominator is making the base smaller, so it must be the root you are taking.
This means that is equal to the fifth root of b to the fourth. Perhaps a bit confusing, but it means that we will keep , but put the whole thing under.
So if we put it together we get:
### Example Question #3 : Rational Exponents
Evaluate
Explanation:
When dealing with fractional exponents, remember this form:
is the index of the radical which is also the denominator of the fraction, represents the base of the exponent, and is the power the base is raised to. That value is the numerator of the fraction.
With a negative exponent, we need to remember this form:
represents the base of the exponent, and is the power in a positive value.
### Example Question #4 : Rational Exponents
Explanation:
First, distribute the exponent to both the numerator and denominator of the fraction.
The numerator of a fractional exponent is the power you take the number to and the denominator is the root that you take the number to.
You can take the cubed root and square the numbers in either order but if you can do the root first that is often easier.
This is the answer. Alternatively, you could have squared the numbers first before taking the cubed root.
### Example Question #5 : Rational Exponents
Evaluate:
Explanation:
In order to evaluate fractional exponents, we can express them using the following relationship:
In this formula, represents the index of the radical from the denominator of the fraction and is the exponent that raises the base: . When exponents are negative, we can express them using the following relationship:
We can then rewrite and solve the expression in the following way.
Dividing by a fraction is the same as multiplying by its reciprocal.
### Example Question #6 : Rational Exponents
Evaluate.
Explanation:
Exponents raised to a power of <1 can be written as the root of the denominator.
So:
Recall that a square root can give two answers, one positive and one negative.
### Example Question #7 : Rational Exponents
Evaluate
Explanation:
The denominator of the exponent "N" is the same as the "N" root of that number.
So
### Example Question #8 : Rational Exponents
Evaluate
Explanation:
This appears more complicated than it is.
is really just where
and
Re-written this equation looks like
### Example Question #9 : Rational Exponents
Evaluate
Explanation:
can be seen as , in a scientific calculator use the button where .
### Example Question #10 : Rational Exponents
Evaluate the given rational exponent: |
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# Probability
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### Probability
1. 1. Probability: Study Time Situation: The final provincial math exam is coming up and Christine and 4 of her friends have decided to create a study group. They meet up at John's house everyday. Since Christine is the one that lives the farthest from John, she decides that she will pick up her other three friends on the way.
2. 2. This is a layout of how to get to everyone's houses. Christine's home Lucas Store Kelly Shelly John
3. 3. First. • If Christine could only travel South and East, how many ways can she get to Lucas? Christine's home Lucas Store Kelly Shelly John
4. 4. Answer.. There is SIX different ways to get to Lucas.
5. 5. How I got this answer. Well the first thing you have to know is where both of the points are. Point A Point A would be Christine's house Point B would be Lucas' house. Point B The first thing you have to find out is 'how many ways can you get to the first corner'. A 1 Well as you can see in this diagram, the outer corner of the street is labelled as 1. This is because there is only one way for you to to go in each direction, 1 South or East. A 1 To get the resultant for the next point, what you would have to do is add the + two numbers together to get the answer for the bottom corner of the box. 1 2 This technique was discovered by Pascal's Triangle.
6. 6. Pascal's Triangle Let me tell you a little more of what it is. Pascal's triangle is a arrangement of numbers (binomial numbers) that is formed into a triangle. *Hence the name* There are many patterns that you can see on Pascal's triangle. These are just some examples of them. 0 Example 1 2 All the numbers appear in 2 1 Example 3 order, diagonally. Fibonacci numbers. This 22 is the one that we will be using. Each term is the sum of the previous terms. 1,1,2,3,5,8,13,21 ... Example 2 All the sums of the rows in the triangle are equal to the power of 2
7. 7. This is how it would look when you apply Pascal's Triangle into the problem. ( The Fibonacci numbers ) Christine's home 1 1 1 1 1 1 2 3 4 5 6 1 3 6 10 15 21 Lucas 1 4 10 20 35 56 15 1 5 Store 35 70 126 Kelly 1 6 21 56 126 252 1 7 28 84 462 210 1 1 1 Shelly 1 2 3 4 From here, you can see the maximum amount of ways to get to each 1 3 6 10 1 1 person's home. 2 3 1 John
8. 8. Christine's home 1 1 1 1 1 Now how many 1 4 5 6 possible ways can you get from Christine's 2 3 1 3 Lucas 6 10 15 21 house the John's 1 4 10 20 35 56 house? 15 1 5 Store 35 70 126 Kelly 1 6 21 56 126 252 7 28 84 462 1 210 1 1 1 Shelly Using the Fibonacci numbers you can 1 2 3 4 determine how many ways you can get there now. 1 3 6 10 1 1 All you have you have to do is take the sum 2 3 1 of every corner and you multiply them to get John to John's house.
9. 9. So you would do 462 x 10 x 3 = 12,780 ways to get to John's house. For Christine to go to John's house, she would get 12,780 different possible way's that she can get there. |
# Neperian logarithm calculator
Calculating a neperian logarithm o logaritmo natural es muy sencillo gracias a nuestra calculadora online. Recuerda que el logaritmo neperiano es aquel cuya base es el número 'e', por lo tanto, podemos decir que el logaritmo neperiano de un número x es el exponente al que debe ser elevado el número 'e' para obtener dicho número, o lo que es lo mismo, the exponential function.
## How to clear the neperian logarithm of an equation
For clear the neperian logarithm of an equationyou have to apply the exponential function to both sides of the equality.
For example, imagine that you start from the following equation and you want to clear ln(x):
y = ln(x)
Applying the exponential function on both sides, we have:
ey = eln(x)
Like the exponential function and the neperian logarithm can be simplified with each otherthen we are left with:
x = ey
If the equation is more complicated, obviously it will be more difficult to clear the unknown in which the neperian logarithm is located, but in short, this would be the procedure to be performed to clear it.
## Neperian logarithm of 1
By definition, the neperian logarithm of x (ln x) is equal to the potencia a la que debemos elevar el número 'e' para obtener x.
Therefore, if we want to calculate the neperian logarithm of 1, es decir, ln(1), tenemos que encontrar una potencia que al elevar el número 'e', nos de como resultado 1. Esto nos condiciona el resultado a una única opción:
e0 = 1
And as you may already know, any number raised to zero is equal to 1.
Therefore, we have that ln (1) = 0. The neperian logarithm of 1 is 0.
## Neperian logarithm of 1437
We have received many e-mails from you asking us what is the neperian logarithm of 1437. Using our calculator we can see that:
ln (1437) = 7.27
Let's do the demonstration seen in the first point to check that the result obtained is the right one:
7,27 = ln (1437)
e7,27 = eln(1437)
1437 = e7,27
Indeed, the equality is fulfilled and we have demonstrated that the ln of 1437 is 7.27.
## What is the natural logarithm of e?
The natural logarithm of e is equal to 1. how did we calculate it? Let's see it step by step as in the previous examples:
1 = ln (e)
e1 = eln(e)
e = e1
## How to calculate the Neperian logarithm in Excel
If you wish, you can use Excel for calculating the neperian logarithm of a number. To do this, you must use the LN function that will give you the result you are looking for.
For solving neperian logarithms with ExcelIf you have a cell, you simply take a cell and type the following function:
=LN(A1)
Remember that A1 is the cell in which you have written the number for which you want to calculate its neperian logarithm or if you prefer, you can substitute the cell coordinates and write the quantity directly between the parentheses.
If you have doubts about how to solve neperian logarithms in Excel, you can take a look at the video we have recorded and in which we perform several examples to see how to use the LN function in Microsoft's spreadsheet program.
If you want to calculate the logarithm of a number in any baseyou can do it with our other logarithm calculator which will allow you to select the specific value of the base, be it ten or any other. |
Lesson 3.3: Solving Equations: Using the Addition & Multiplication Properties of Equality Together
# Lesson 3.3: Solving Equations: Using the Addition & Multiplication Properties of Equality Together
## Lesson 3.3: Solving Equations: Using the Addition & Multiplication Properties of Equality Together
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##### Presentation Transcript
1. Lesson 3.3: Solving Equations: Using the Addition & Multiplication Properties of Equality Together
2. REVIEW x + 4 = 7 – 4 – 4 To undo addition you should To undo subtraction you should To undo multiplication you should To undo division you should subtract Addition Property of Equality x – 5 = 3 +5 +5 add 3x = 12 ÷3 ÷3 divide * Multiplication Property of Equality x/6 = 5 *6 *6 multiply *EXCEPTION: When multiplying by a fraction, don’t divide. Instead multiply by the reciprocal. 3/4x = 5 *4/3 *4/3
3. MULTISTEP PROBLEMS Most problems require more than one step to solve. For example: 3x + 4 = 19 coefficient in front of variable something added 3x + 4 = 19 – 4 – 4 3x = 15 Use the Addition Property first to isolate the term with the variable. ÷3 ÷3 x = 5 Then use the Multiplication Property to make the coefficient before the variable be a ‘1’. √ 3*5 + 4 = 19 Always check your answer (plug ‘n chug)! TRY THESE: 1. 5x + 6 = 16 2. –7x + 6 = 13 3. 5 = 4y – 12 4. –4c + 3 = 12
4. MULTISTEP PROBLEMS: SIMPLIFICATION Sometimes problems need to be simplified before solving. Simplification includes: 1) Collecting like terms on the same side of the = sign 2) Distributing through parentheses AFTER these steps are done, you can use the Addition & Multiplication Properties to solve For example: 3x + 7x = 30 Combine them and then solve 3x + 7x = 30 10x = 30 ÷10 ÷10 x = 3 like terms on same side of = sign TRY THESE: 1. 6x + 2x = 16 2. 7x – 3x = 6 3. -2a + 3 + 6a = 14
5. MULTISTEP PROBLEMS: SIMPLIFICATION Sometimes problems need to be simplified before solving. Simplification includes: 1) Collecting like terms on the same side of the = sign 2) Distributing through parentheses AFTER these steps are done, you can use the Addition & Multiplication Properties to solve For example: 2(2y + 3) = 14 Distribute and then solve 2(2y + 3) = 14 4y + 6 = 14 -6 -6 4x = 8 ÷4 ÷4 x = 2 parentheses TRY THESE: 1. 8(3x + 2) = 30 2. 5X +5(4X – 1) = 20 |
# Lesson 6
The Pythagorean Identity (Part 2)
Let’s use the Pythagorean Identity.
### 6.1: Math Talk: Which Quadrant?
For an angle $$\theta$$ in the quadrant indicated, use mental estimation to identify the values of $$\cos(\theta)$$, $$\sin(\theta)$$, and $$\tan(\theta)$$ as either positive or negative.
Quadrant 1
Quadrant 2
Quadrant 3
Quadrant 4
### 6.2: Andre's Calculations
Suppose that the angle $$\theta$$, in radians, is in quadrant 4 of the unit circle. If $$\cos(\theta)=0.28$$, what are the values of $$\sin(\theta)$$ and $$\tan(\theta)$$?
Andre uses the Pythagorean Identity and determines that the value of $$\sin(\theta)$$ is -0.96. Using the values of sine and cosine, he then calculates the value of tangent:
\begin{align} \tan(\theta)&=\dfrac{\sin(\theta)}{\cos(\theta)} \\ &=\frac{\text-0.96}{0.28} \\ &\approx \text-3.43 \end{align}
Do you agree with Andre? Explain or show your reasoning.
### 6.3: Card Sort: Where's the Point?
Your teacher will give you a set of cards that should be arranged face up with cards showing values for sine, cosine, and tangent on one side and cards showing quadrants on the other.
1. Take turns with your partner matching pairs of cards. Identify 2 pairs that are possible on the unit circle and 2 pairs that are not possible, in any order.
1. For each pair, explain to your partner how you know if the pair is or is not possible on the unit circle. Once a pair is identified, place the cards in front of you to use later.
2. For each pair that your partner draws, listen carefully to their explanation. If you disagree, discuss your thinking and work to reach an agreement.
2. Once you and your partner have identified 4 pairs each, pick 1 of your possible matches and then calculate the values of the two missing trigonometric ratios for that match. When you finish, trade calculations with your partner and check each other's work. If you disagree, discuss your thinking and work to reach an agreement.
Suppose $$\tan(\theta)=2$$ and the angle $$\theta$$ is in quadrant 1. Find the values of $$\cos(\theta)$$ and $$\sin(\theta)$$.
### Summary
Say we know that $$\sin(\theta)=\text-\frac{\sqrt{2}}{2}$$ and that $$\theta$$ is an angle in quadrant 3. What can we say about the values of cosine and tangent at $$\theta$$?
Since we can think of $$\sin(\theta)$$ as the $$y$$-coordinate of a point $$P$$ in quadrant 3, let’s start with a sketch of the unit circle showing point $$P$$.
The sketch helps us see that the $$x$$-coordinate, which is $$\cos(\theta)$$, is also negative. Using the Pythagorean Identity, we can calculate the value of $$\cos(\theta)$$:
\begin{align} \cos^2(\theta)+\sin^2(\theta)&=1^2 \\ \cos^2(\theta)+\left(\text-\frac{\sqrt{2}}{2}\right)^2 &= 1 \\ \cos^2(\theta)&=1-\left(\text-\frac{\sqrt{2}}{2}\right)^2 \\ \cos^2(\theta)&=1-\frac{2}{4} \\ \cos(\theta)&=\text-\sqrt{1-\frac{1}{2}} \\ \cos(\theta)&=\text-\sqrt{\frac{1}{2}} \end{align}
Now that we know the value of cosine, we can calculate the value of tangent with some division:
$$\tan(\theta)=\dfrac{\text-\frac{\sqrt{2}}{2}}{\text-\sqrt{\frac{1}{2}}}=\dfrac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=1$$
We can use one piece of information and the structure of the unit circle to figure out a whole bunch more, similar to how we used the value of one side length, the hypotenuse, and the structure of right triangles in the past to figure out the other side length of the right triangle.
### Glossary Entries
• Pythagorean identity
The identity $$\sin^2(x) + \cos^2(x) = 1$$ relating the sine and cosine of a number. It is called the Pythagorean identity because it follows from the Pythagorean theorem.
• period
The length of an interval at which a periodic function repeats. A function $$f$$ has a period, $$p$$, if $$f(x+p) = f(x)$$ for all inputs $$x$$.
• periodic function
A function whose values repeat at regular intervals. If $$f$$ is a periodic function then there is a number $$p$$, called the period, so that $$f(x + p) = f(x)$$ for all inputs $$x$$.
• unit circle
The circle in the coordinate plane with radius 1 and center the origin. |
# Effective Rate of Interest
By now, you have a clear understanding of simple and compound interest. However, when interest is compounded, for more than one year, the actual interest rate per annum is lesser than the effective rate of interest. In this article, we will look at the definition, formula, and some examples of calculating the effective rate of interest.
## Effective Rate of Interest
Source: Pixabay
### Definition
The effective rate of interest is the equivalent annual rate of interest which is compounded annually. Further, the compounding must happen more than once every year. Let’s look at an example for better clarity:
Example 1:Â Peter invests Rs. 10,000 for one year at the rate of 6% per annum. The interest is compounded semi-annually. Let’s calculate the interest earned in the first six months (I1).
Solution: I1 = 10,000 x $$\frac {6}{100}$$ x $$\frac {6}{12}$$ = Rs. 300. Since the interest is compounded, the principal for the next 6 months = 10,000 + 300 = Rs. 10,300. Therefore, the interest earned in the next six months (I2) is,
I2 = 10,300 x $$\frac {6}{100}$$ x $$\frac {6}{12}$$ = Rs. 309.
Hence, the total interest earned during the year I = I1 + I2 = 300 + 309 = Rs. 609. We know the formula for interest is I = PNR … where ‘I’ is the interest, ‘P’ is the principal amount, ‘N’ is the time period, and ‘R’ is the rate of interest. In the case of this example, R = E or the effective rate of interest. Therefore, we have,
E = $$\frac {I}{PN}$$ = $$\frac {609}{10, 000 × 1}$$ = 0.0609 or 6.09%.
### Formula for Calculation of Effective Rate of Interest
You can use the following formula to calculate the effective rate of interest:
E = (1 + i)n – 1 … (1)
Where ‘E’ is the effective rate of interest, ‘i’ is the actual rate of interest in decimal, and ‘n’ is the number of conversion periods.
Example 2:Â John invests Rs. 5,000 in a term deposit scheme. The scheme offers an interest rate of 6% per annum, compounded quarterly. How much interest will John earn after one year? Also, what is the effective rate of interest?
Solution: We know that,
• Principal amount = P = Rs. 5,000
• Actual rate of interest = i = 6% p.a. = 0.06 p.a. = 0.015 per quarter
• Number of conversion periods = n = 4 (since we are calculating for one year and compounding happens every quarter)
Therefore, the compound interest (I) is,
I = P x [(1 + i)n – 1] = 5000 x [(1 + 0.015)4 – 1] = 5000 x 0.06136355 = 306.82
Hence, after one year, John earns a total interest (I) of Rs. 306.82. Further, the effective rate of interest (E) is,
E =Â (1 + i)n – 1 =Â (1 + 0.015)4 – 1 =Â 0.0613 or 6.13%.
## Solved Examples on Effective Rate of Interest
Example 3: In a bank, an amount of Rs. 20,000 is deposited for one year. The rate of interest is 8% per annum and is compounded semi-annually. What is the effective rate of interest?
1. 8 percent
2. 8.08 percent
3. 8.16 percent
4. 8.22 percent
Solution: To calculate the effective rate of interest, we do not need the amount. As per equation (1) above,
E = (1 + i)n – 1 … where ‘E’ is the effective rate of interest, ‘i’ is the actual rate of interest in decimal, and ‘n’ is the number of conversion periods.
In this problem, we know that,
• The actual rate of interest = i = 8% p.a. = 0.08 p.a. = 0.04 per semi-year (6 months).
• Number of conversion periods = n = 2 (since we are calculating for one year and compounding happens once every six months)
Therefore, the effective rate of interest is,
E =Â (1 + i)n – 1 =Â (1 + 0.04)2 – 1 = 1.0816 – 1 = 0.0816 or 8.16%.
Hence, the correct answer is option c – 8.16 percent.
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# 2011 AIME I Problems/Problem 12
## Problem
Six men and some number of women stand in a line in random order. Let $p$ be the probability that a group of at least four men stand together in the line, given that every man stands next to at least one other man. Find the least number of women in the line such that $p$ does not exceed 1 percent.
## Solution
Let $n$ be the number of women present, and let _ be some positive number of women between groups of men. Since the problem states that every man stands next to another man, there cannot be isolated men. Thus, there are five cases to consider, where $(k)$ refers to a consecutive group of $k$ men:
_(2)_(2)_(2)_
_(3)_(3)_
_(2)_(4)_
_(4)_(2)_
_(6)_
For the first case, we can place the three groups of men in between women. We can think of the groups of men as dividers splitting up the $n$ women. Since there are $n+1$ possible places to insert the dividers, and we need to choose any three of these locations, we have $\dbinom{n+1}{3}$ ways.
The second, third, and fourth cases are like the first, only we need to insert two dividers among the $n+1$ possible locations. Each gives us $\dbinom{n+1}{2}$ ways, for a total of $3\dbinom{n+1}{2}$.
The last case gives us $\dbinom{n+1}{1}=n+1$ ways.
Therefore, the total number of possible ways where there are no isolated men is
$$\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1).$$
The total number of ways where there is a group of at least four men together is the sum of the third, fourth, and fifth case, or
$$2\dbinom{n+1}{2}+(n+1).$$
Thus, we want to find the minimum possible value of $n$ where $n$ is a positive integer such that
$$\dfrac{2\dbinom{n+1}{2}+(n+1)}{\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1)}\le\dfrac{1}{100}.$$
The numerator is equal to
$$2\cdot\dfrac{(n+1)!}{2!(n-1)!}+(n+1)=2\cdot\dfrac{(n+1)(n)}{2}+(n+1)=n(n+1)+1(n+1)=(n+1)^2.$$
For the denominator, we get
\begin{align*}\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1)&=\dfrac{(n+1)!}{3!(n-2)!}+3\dfrac{(n+1)!}{2!(n-1)!}+(n+1)\\ &=\dfrac{(n+1)(n)(n-1)}{6}+3\dfrac{(n+1)(n)}{2}+(n+1)\\ &=(n+1)\left[\dfrac{n^2-n}{6}+\dfrac{3n}{2}+1\right]\\ &=\dfrac{1}{6}(n+1)(n^2-n+9n+6)\\ &=\dfrac{1}{6}(n+1)(n^2+8n+6). \end{align*}
So, we get
$$\dfrac{(n+1)^2}{\frac{1}{6}(n+1)(n^2+8n+6)}=\dfrac{6(n+1)}{n^2+8n+6}\le\dfrac{1}{100}.$$
$100(n^2+8n+6)>0$ since $n$ must be positive. So, when multiplying both sides of the inequality by that expression, it will not change the inequality sign. After multiplying by it, we get
\begin{align*}100\cdot6(n+1)&\le n^2+8n+6\\ 600n+600&\le n^2+8n+6\\ 0&\le n^2-592n-594. \end{align*}
Thus we seek the smallest positive integer value of $n$ such that $n^2-592n-594\ge0$. Since the quadratic function's discriminant, or $\sqrt{592^2-4(-594)}=\sqrt{592^2+4\cdot594}$, is positive, the polynomial has two distinct real roots.
Also, since the polynomial has a positive leading coefficient, the graph of the polynomial is concave up, and the value of $n$ we want must be either slightly larger than the positive root (if the other, smaller root is negative) or equal to $1$ (if the smaller root is positive). We see that $n=1$ does not satisfy the inequality, so there must be a positive and negative root.
The solution to the polynomial is
(Error compiling LaTeX. ! Missing \$ inserted.)\dfrac{
Since $\dfrac{n+1}{n+2}$ is slightly less than 1 when $n$ is large, $\dfrac{6}{n+6}$ will be close to $\dfrac{1}{100}$. They equal each other when $n = 594$.
If we let $n= 595$ or $593$, we will notice that the answer is $\boxed{594}$
2011 AIME I (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. |
# Subtract with Pictures Worksheet || Grade 1 Math
This post is collection of subtraction questions where you have to do calculation with the help of images. All the questions are provided with options and you have to select the right answer.
## Subtraction with Cubes
In this section you have to count the number of cubes and do the subtraction when some of the cubes are taken away.
(01) Below are Six Cubes. Take away One cube and find the number of cubes left.
(a) 6 – 1 = 5
(b) 4 – 2 = 2
(c) 6 – 2 = 4
(d) 7 – 1 = 6
Solution
Number of Cubes present = 6
Number of Cubes removed = 1
Cubes Left ⟹ 6 – 1 = 5
Option (a) is the right answer
(02) Below are four cubes. Take away two cubes and find the number of cubes left.
(a) 4 – 1 = 3
(b) 4 – 2 = 2
(c) 4 – 3 = 1
(d) 4 – 4 = 0
Solution
Number of Cubes present = 4
Number of Cubes removed = 2
Cubes Left ⟹ 4 – 2 = 2
Option (b) is the right answer
(03) Below are seven cubes. Take away three cubes and find the number of cubes left.
(a) 8 – 4 = 4
(b) 8 – 3 = 5
(c) 7 – 2 = 5
(d) 7 – 3 = 4
Solution
Number of Cubes present = 7
Number of Cubes removed = 3
Cubes Left ⟹ 7 – 3 = 4
Option (d) is the right answer
(04) Below are ten cubes. Take away five cubes and find the number of cubes left.
(a) 9 – 3 = 6
(b) 9 – 4 = 5
(c) 10 – 5 = 5
(d) 10 – 6 = 4
Solution
Number of Cubes present = 10
Number of Cubes removed = 5
Cubes Left ⟹ 10 – 5 = 5
Option (c) is the right answer
(05) Below are three cubes. Take away three cubes and find the number of cubes left.
(a) 4 – 3 = 1
(b) 3 – 3 = 0
(c) 3 – 1 = 2
(d) 3 – 2 = 1
Solution
Number of Cubes present = 3
Number of Cubes removed = 3
Cubes Left ⟹ 3 – 3 = 0
Option (b) is the right answer
(06) Below are nine cubes. Take away five cubes and find the number of cubes left.
(a) 9 – 3 = 6
(b) 9 – 4 = 5
(c) 9 – 2 = 7
(d) 9 – 5 = 4
Solution
Number of Cubes present = 9
Number of Cubes removed = 5
Cubes Left ⟹ 9 – 5 = 4
Option (d) is the right answer
(07) Below are two cubes. Take away one cube and find the number of cubes left
(a) 5 – 3 = 2
(b) 4 – 4 = 0
(c) 2 – 1 = 1
(d) 3 – 1 = 2
Solution
Number of Cubes present = 2
Number of Cubes removed = 1
Cubes Left ⟹ 2 – 1 = 1
Option (c) is the right answer
(08) Below are eight cubes. Take away seven cube and find the number of cubes left
(a) 8 – 7 = 1
(b) 8 – 6 = 2
(c) 7 – 6 = 1
(d) 7 – 7 = 0
Solution
Number of Cubes present = 8
Number of Cubes removed = 7
Cubes Left ⟹ 8 – 7 = 1
Option (a) is the right answer
(09) Below are ten cubes. Take away six cubes and find the number of cubes left
(a) 10 – 4 = 6
(b) 7 – 6 = 1
(c) 9 – 6 = 3
(d) 10 – 6 = 4
Solution
Number of Cubes present = 10
Number of Cubes removed = 6
Cubes Left ⟹ 10 – 6 = 4
Option (d) is the right answer
(10) Below are seven cubes. Take away four cubes and find the number of cubes left
(a) 7 – 7 = 0
(b) 7 – 4 = 3
(c) 7 – 6 = 1
(d) 7 – 5 = 2
Solution
Number of Cubes present = 7
Number of Cubes removed = 4
Cubes Left ⟹ 7 – 4 = 3
Option (b) is the right answer
(11) Below are nine cubes. Take away one cube and find the number of cubes left
(a) 10 – 3 = 7
(b) 9 – 3 = 6
(c) 9 – 1 = 8
(d) 10 – 1 = 9
Solution
Number of Cubes present = 9
Number of Cubes removed = 1
Cubes Left ⟹ 9 – 1 = 8
Option (c) is the right answer
(12) Below are five cubes. Take away four cubes and find the number of cubes left
(a) 5 – 4 = 1
(b) 5 – 3 = 2
(c) 4 – 3 = 1
(d) 4 – 2 = 2
Solution
Number of Cubes present = 5
Number of Cubes removed = 4
Cubes Left ⟹ 5 – 4 = 1
Option (a) is the right answer
(13) Below are three cubes. Take away one cube and find the number of cubes left
(a) 4 – 3 = 1
(b) 3 – 3 = 0
(c) 3 – 1 = 2
(d) 3 – 2 = 1
Solution
Number of Cubes present = 3
Number of Cubes removed = 1
Cubes Left ⟹ 3 – 1 = 2
Option (d) is the right answer
(14) Below are Eight cubes. Take away three cubes and find the number of cubes left
(a) 8 – 3 = 5
(b) 8 – 4 = 4
(c) 8 – 2 = 6
(d) 9 – 4 = 5
Solution
Number of Cubes present = 8
Number of Cubes removed = 3
Cubes Left ⟹ 8 – 3 = 5
Option (a) is the right answer
(15) Below are Ten cubes. Take away two cubes and find the number of cubes left
(a) 10 – 3 = 7
(b) 10 – 2 = 8
(c) 9 – 2 = 7
(d) 9 – 4 = 5
Solution
Number of Cubes present = 10
Number of Cubes removed = 2
Cubes Left ⟹ 10 – 2 = 8
Option (b) is the right answer
## Subtract with Pictures
In this question set, a illustration is provided to you which will show some collection of objects. By looking at the illustration you have to do subtraction and select the right answer.
(01) Do the subtraction of below illustration and select the right answer
(a) 2
(b) 3
(c) 4
(d) 5
Solution
Total number of scooters = 5
Number of scooters cancelled = 2
Scooter Left = 5 – 2 = 3
Option (b) is the right answer
(02) Observe the below illustration and do the required subtraction
(a) 4
(b) 5
(c) 6
(d) 7
Solution
Total number of owls = 7
Number of owls removed = 3
Owls Left = 7 – 3 = 4
Option (a) is the right answer
(03) Calculate the subtraction for the following picture
(a) 3
(b) 2
(c) 1
(d) 0
Solution
Total number of cars = 3
Number of cars removed = 3
Car Left = 3 – 3 = 0
Option (d) is the right answer
(04) Do the subtraction for below image and select the right answer
(a) 3
(b) 2
(c) 1
(d) 0
Solution
Total number of airplanes = 4
Number of airplanes removed = 3
Airplanes Left = 4 – 3 = 1
Option (c) is the right answer
(05) Do the subtraction and select the right answer
(a) 5
(b) 8
(c) 9
(d) 6
Solution
Total number of cups = 8
Number of cups removed = 2
Cups Left = 8 – 2 = 6
Option (d) is the right answer
(06) Subtract and select the right answer
(a) 8
(b) 4
(c) 7
(d) 6
Solution
Total number of keys = 5
Number of keys removed = 1
Keys Left = 5 – 1 = 4
Option (b) is the right answer
(07) Subtract and select the right answer
(a) 5
(b) 7
(c) 9
(d) 11
Solution
Total number of flowers = 10
Number of flowers removed = 5
Flowers Left = 10 – 5 = 5
Option (a) is the right answer
(08) Observe the following image and do subtraction accordingly
(a) 5
(b) 2
(c) 3
(d) 4
Solution
Total number of hammers = 3
Number of hammer removed = 1
Hammers Left = 3 – 1 = 2
Option (b) is the right answer
(09) Observe the following image and do subtraction accordingly
(a) 6
(b) 5
(c) 2
(d) 1
Solution
Total number of oranges = 5
Number of oranges removed = 4
Oranges Left = 5 – 4 = 1
Option (d) is the right answer
(10) Count the number of mangoes and do subtraction as given in below images
(a) 10
(b) 4
(c) 3
(d) 6
Solution
Total number of mangoes = 8
Number of mangoes removed = 5
Mangoes Left = 8 – 5 = 3
Option (c) is the right answer
(11) Count the number of kites and do the required subtraction
(a) 3
(b) 0
(c) 1
(d) 2
Solution
Total number of kites = 2
Number of kites removed = 2
Kites Left = 2 – 2 = 0
Option (b) is the right answer
(12) Do the subtraction for following images
(a) 3
(b) 0
(c) 1
(d) 2
Solution
Total number of kites = 2
Number of kites removed = 2
Kites Left = 2 – 2 = 0
Option (b) is the right answer
(13) Count the number of parrots and do the required subtraction
(a) 7
(b) 8
(c) 9
(d) 6
Solution
Total number of parrots = 10
Number of parrots removed = 3
Parrots Left = 10 – 3 = 7
Option (a) is the right answer
(14) Count the number of stairs and do the required subtraction
(a) 3
(b) 0
(c) 1
(d) 2
Solution
Total number of stairs = 3
Number of stairs removed = 3
Stairs Left = 3 – 3 = 0
Option (b) is the right answer
(15) Count the number of lanterns and do the required subtraction
(a) 1
(b) 3
(c) 4
(d) 2 |
# As a math tutor, you teach and review this method constantly.
Back in May, I began a series of posts about factoring polynomials. To refresh the topic, you can check here, here, and here.
Factoring polynomials is a make-or-break skill for high school students taking academic math. It encompasses about five techniques, of which easy trinomial factoring is probably the best known. Let’s have a quick look:
Example 1: Factor x2 -3x -28
Solution: Since the coefficient of x2 is 1 (which we know because there is no number written in front of it), we can use the easy trinomial method.
Step 1: Write (x )(x )
Step 2: After the x’s, write the numbers that will multiply to give -28, but add to give -3.
You have to do some mental math: 7×4=28, but one of the numbers has to be negative to give -28. The numbers must be -7 and +4, since -7+4=-3.
(x -7)(x +4)
The answer is (x – 7)(x + 4). You can verify using the foil method:
First: x*x=x2
Outer: x*4=4x
Inner: -7*x=-7x
Last: -7*4=-28 (remember: negative times positive gives negative)
Now, line up the four terms we just obtained:
x2 +4x -7x -28
We can combine the like terms: 4x – 7x = -3x
Finally we get
x2 -3x -28.
If you foil out your answer and get back the original trinomial, you know it’s right.
Example 2: Factor x2 + 5x + 4
Solution: The numbers that multiply to give 4 but add to give 5 are 1 and 4: 1*4=4, 1+4=5.
Therefore, the answer is (x + 4)(x + 1)
Example 3: Factor x2 -10x + 16
Solution: The numbers that multiply to give 16 but add to give -10 are -8 and -2 (recall that negative times negative gives positive): -8*-2=16, -8+-2=-10
The answer is (x – 8)(x – 2)
Example 4: Factor x2 +5x – 14
The answer is (x + 7)(x – 2)
Good luck with this method. Most people like it once they get used to it:)
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC. |
# 90 Degree Angle Triangle
In 90 degree angle triangle, two of the sides are sides of the right angle. These are called the legs of the right triangle. The third side is opposite the right angle. It is called the hypotenuse of the right triangle. Call the lengths of the legs a and b, call the length of the hypotenuse c. the Pythagorean Theorem says the a2+b2=c2. This fact only applies to right triangles.
Please express your views of this topic Right Angle Math by commenting on blog.
## The Pythagorean Theorem 90 degree angle triangles:
This is due to finding right triangle in abundance in the physical structures of the world around us and the importance of finding perpendicular distance. The very important tools in 90 degree angle triangle problem are the Pythagorean Theorem
The Pythagorean Theorem:
More unlike proofs have been written for the Pythagorean Theorem than for any other theorem in geometry. Pythagoras is given credit for developing the follow proof based on similar triangles.
Given right ∆ABC with the right angle at C, prove a2+b2=c2.
Altitude `barCbarD` form two 90 degree angle triangles that are similar to each other and to the original triangle. Therefore,
`c/b`=`b/m` or `c/a`=`a/m`
So, b2=_(anyvalue)_ and a2=_(anyvalue)_, adding, a2+b2=c2
## Example for 90 degree angle triangles:
Problem 1:
Derive a formula for the diagonal d of a square in terms of any side s using Pythagorean Theorem.
Solution:
c2=a2+a2
c2=2a2
Problem 2:
Derive a formula for the altitude h of any equilateral triangle in terms of any side s using Pythagorean Theorem.
Solution:
h="s-(`s/2`)
h="s-`s/4`=`(3s)/(4)`
Two basic properties of similar triangles using Pythagorean Theorem:
Rule 1:
If two 90 degree angle triangles are similar using Pythagorean theorem
a) Their corresponding angles are congruent.
Thus, if ∆I~∆I’ in figure
Then
`angle`C’`~=``angle`C
`angle`A’`~=``angle`A
`angle`B’`~=``angle`B
And m `angle`C=90o
m`angle`A=40o
m `angle`B=50o
Between, if you have problem on these topics Trig Special Angles ,please browse expert math related websites for more help on cbse previous year papers class 12.
b) The ratios of their corresponding sides are equal.
Thus, if ∆I ~∆I’ in figure
Then c="15" since `c/15`=`9/3`
And b="12" since `b/4`=`9/3`.
Rule 2:
Ratios of corresponding sides of 90 degree angle triangle are equal using Pythagorean Theorem
If ∆II’~∆II in figure, find x and y by using the data indicated.
Solution:
Since ∆II’ ~ ∆II,
`(x)/(32)`=`(15)/(20)` `(y)/(26)`=`(15)/(20)`
x="`15/20`32" y="`15/20`26
x="24" y="19`1/2` |
# Given the function y=(2x-3)/(x-1)*(x-2). What is the antiderivative of the function [y-1/(x-2)]^2010?
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
The function y = (2x-3)/(x-1)*(x-2)
We have to find the antiderivative of [y-1/(x-2)]^2010
[y-1/(x-2)]^2010
=> [(2x-3)/(x-1)*(x-2) - 1/(x - 2)]^2010
=> [(2x-3)/(x-1)*(x-2) - (x - 1)/(x - 1)(x - 2)]^2010
=> [((2x-3) - (x - 1))/(x-1)*(x-2)]^2010
=> [(2x - 3 - x + 1)/(x-1)*(x-2)]^2010
=> [(x - 2)/(x-1)*(x-2)]^2010
=> [1/(x-1)]^2010
Int[ 1/(x - 1)^2010 dx]
let x - 1 = y, dy = dx
=> Int[1/y^2010 dy]
=> Int[y^-2010 dy]
=> y^-2009/(-2009)
substitute y = x - 1
=> -1/2009*(x - 1)^2009 + C
The required antiderivative is -1/2009*(x - 1)^2009 + C
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
First, we'll try to decompose the function y into partial fractions.
(2x-3)/(x-1)*(x-2) = A/(x-1) + B/(x-2)
2x - 3 = x(A+B) - 2A - B
Comparing both sides, we'll get:
A+B = 2 => A = 2 - B (1)
-2A-B = -3 => -2(2 - B) - B = -3
We'll remove the brackets:
-4 + 2B - B = -3
B = 4 - 3 => B = 1 => A = 2-1 = 1
y = (2x-3)/(x-1)*(x-2) = 1/(x-1) + 1/(x-2)
Now, we'll calculate the difference:
y - 1/(x-2) = 1/(x-1) + 1/(x-2) - 1/(x-2)
We'll eliminate like terms:
y - 1/(x-2) = 1/(x-1)
We'll raise both sides to 2010 power;
[y - 1/(x-2)]^2010 = 1/(x-1)^2010
We'll calculate the antiderivative of the function [y - 1/(x-2)]^2010, integrating both sides:
Int [y - 1/(x-2)]^2010 dx = Int dx/(x-1)^2010
We'll replace x - 1 by t:
x-1 = t
We'll differentiate both sides:
dx = dt
Int dx/(x-1)^2010 = Int dt/t^2010
We'll apply negative power rule:
Int dt/t^2010 =Int t^(-2010)dt
Int t^(-2010)dt = t^(-2010+1)/(-2010+1)
Int t^(-2010)dt = -1/2009*t^2009 + C
The requested antiderivative of the function [y - 1/(x-2)]^2010 is F(x) = -1/2009*(x-1)^2009 + C. |
### Similar presentations
9.3 Surface Area, Volume, and Capacity
3 Surface Area
4 Suppose you wish to paint a box whose edges are each 3 ft, and you need to know how much paint to buy. To determine this you need to find the sum of the areas of all the faces—this is called the surface area.
5 Example 1 – Find the surface area of a box Find the amount of paint needed for a box with edges 3 ft. Solution: A box (cube) has 6 faces of equal area. 6 9 ft 2 = 54 ft 2 You need enough paint to cover 54 ft 2. Number of faces of cube Area of each face
6 Surface Area In the next example, we consider a box without a top and a can with a bottom but not a top.
7 Example 2 – Find the outside surface area Find the outside surface area.
8 Example 2 – Solution a. We find the sum of the areas of all the faces: Front: 30 50 = 1,500 Back: 1,500 Side: 80 30 = 2,400 Side: 2,400 Bottom: 80 50 = 4,000 Total: 11,800 cm 2 Same as front Sides are the same size.
9 Example 2 – Solution b. To find the surface area, find the area of a circle (the bottom) and think of the sides of the can as being “rolled out.” The length of the resulting rectangle is the circumference of the can and the width of the rectangle is the height of the can. cont’d
10 Example 2 – Solution Side: A = /w = ( d)w = (6)(6) = 36 Bottom: A = r 2 = (3) 2 = 9 Surface area: 36 + 9 = 45 141.37167 The surface area is about 141 cm 2. cont’d
11 Volume
12 Volume To measure area, we covered a region with square units and then found the area by using a mathematical formula. A similar procedure is used to find the amount of space inside a solid object, which is called its volume. We can imagine filling the space with cubes.
13 Volume A cubic inch and a cubic centimeter are shown in Figure 9.12. If the solid is not a cube but is a box (called a rectangular parallelepiped) with edges of different lengths, the volume can be found similarly. Figure 9.12 Common units of measuring volume a. 1 cubic inch (1 cu in. or 1 in. 3 ) b. 1 cubic centimeter (1 cu cm, cc, or 1 cm 3 )
14 Volume
15 Example 4 – Find the volume of a box Find the volume of a box that measures 4 ft by 6 ft by 4 ft.
16 Example 4 – Solution There are 24 cubic feet on the bottom layer of cubes. Do you see how many layers of cubes will fill the solid? Since there are four layers with 24 cubes in each, the total is 4 24 = 96 The volume is 96 ft 3.
17 Volume This example leads us to the following formula for the volume of a box.
18 Example 5 – Find the volume of a solid Find the volume of each solid.
19 Example 5 – Solution a. V = s 3 = (10 cm) 3 = (10 10 10) cm 3 = 1,000 cm 3 b. V = /wh = (25 cm)(10 cm)(4 cm) = (25 10 4) cm 3 = 1,000 cm 3
20 Example 5 – Solution c. V = /wh = (11 in.)(7 in.)(3 in.) = (11 7 3) in. 3 Sometimes the dimensions for the volume we are finding are not all given in the same units. cont’d
21 Example 5 – Solution In such cases, you must convert all units to a common unit. The common conversions are as follows: 1 ft = 12 in. To convert feet to inches, multiply by 12. To convert inches to feet, divide by 12. 1 yd = 3 ft To convert yards to feet, multiply by 3. To convert feet to yards, divide by 3. 1 yd = 36 in. To convert yards to inches, multiply by 36. To convert inches to yards, divide by 36. cont’d
22 Capacity
23 Capacity One of the most common applications of volume involves measuring the amount of liquid a container holds, which we refer to as its capacity. For example, if a container is 2 ft by 2 ft by 12 ft, it is fairly easy to calculate the volume: 2 2 12 = 48 ft 3
24 Capacity
25 Capacity Some capacity statements from purchased products are listed in Table 9.1. Table 9.1 Capacities of Common Grocery Items, as Shown on Labels
26 Example 7 – Measure the quantity of a liquid Measure the amount of liquid in the measuring cup in Figure 9.14, both in the U.S. system and in the metric system. Figure 9.14 Standard measuring cup with both metric and U.S. measurements
27 Example 7 – Solution Metric: 240 mL U.S.: About 1 c or 8 oz
28 Capacity
29 Example 8 – Capacity of a container How much water would each of the following containers hold?
30 Example 8 – Solution a. V = 90 cm 80 cm 40 cm = 288,000 cm 3 Since each 1,000 cm 3 is 1 liter, This container would hold 288 liters.
31 Example 8 – Solution b. V = 7 in. 22 in. 6 in. = 924 in. 3 Since each 231 in. 3 is 1 gallon, This container would hold 4 gallons. cont’d
32 Example 8 – Solution Note that there is no need to estimate part a since the arithmetic is fairly easy. However, for part b, the estimate might be cont’d |
# Lesson 1.2 - PowerPoint PPT Presentation
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Lesson 1.2. Intro to Geometry. L earning T arget. I can understand basic geometric terms and postulates. 1. Points. Points do not have actual size. How to Sketch: Using dots How to label: Use capital letters. A. B. 2. Postulates. An accepted statement of fact.
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Lesson 1.2
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## Lesson 1.2
Intro to Geometry
Learning Target
I can understand basic geometric terms and postulates.
### 1. Points
• Points do not have actual size.
• How to Sketch:
Using dots
• How to label:
Use capital letters
A
B
### 2. Postulates
• An accepted statement of fact.
### 3. Line POSTULATE – Through any two points there is exactly ONE line
• Lines extend indefinitely and have no thickness or width.
• How to sketch : using arrows at both ends.
• How to name: 2 ways
(1) small script letter – line n
(2) any 2points on the line -
• Never name a line using three points -
n
A
B
C
### 4. Segment
A part of a line that consists of two endpoints and all the points in between them.
How to name: by 2 endpoints
A
B
C
### 5. Ray
• A part of a line that consists of one endpoint and all the points to one side of that endpoint.
A
B
C
How to name: 2 points with ENDPOINT FIRST
Not like this…
### 6. Collinear Points
• Collinear points are points that lie on the same line.
C
A
A
B
C
B
Non collinear
Collinear
### 7. Planes POSTULATE – Through any 3 non-collinear points there is exactly ONE plane
• A plane is a flat surface that extends indefinitely in all directions.
• How to sketch: Use a parallelogram (four sided figure)
• How to name: 2 ways
(1) Capital script letter – Plane M
(2) Any 3 non collinear points in the plane - Plane: ABCor BCA or BAC, etc.
A
M
B
C
A
B
Plane BCG
Plane ABF
Plane CDH
Etc.
D
C
E
F
H
G
### Other planes in the same figure:
Any three non collinear points determine a plane!
Plane ACG
Plane ACH
Etc.
### 8. CoplanarObjects
Coplanar objects (points, lines, etc.) are objects that lie on the same plane.
Are the following points coplanar?
A, B, C ?
Yes
A, B, C, F ?
No
H, G, F, E ?
Yes
E, H, C, B ?
Yes
A, G, F ?
Yes
C, B, F, H ?
No
### 9. Intersection of Two Lines POSTULATE
The intersection of two lines is a point.
m
Line m and line n intersect at point P.
P
n
### 10. Intersection of Two Planes POSTULATE
The intersection of two planes is exactly ONE line.
B
P
A
R
Plane P and Plane R intersect at the line |
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Lesson 1: Decomposing fractions
# Decomposing a mixed number
Sal uses fraction models to decompose 2 1/4. Created by Sal Khan.
## Want to join the conversation?
• 1/1 = 8/8 because 1/1 is 1 out of 1 which is 1, and 8 out of 8 which is 8. the fraction sign is also the division symbol, so 1 divided by 1 =1, and 8 divided by 8= 1.
• They help you with cooking. You will need fractions if you are trying to figure out how much sugar you need for a cake. (‾◡◝)
• bro stop the cap
(1 vote)
• i can't do anything with fraction is just too hard.
• So start with a 20x20 times table and learning equivalent fractions. Pick any two numbers (like 2 and 5) and start going down both lines 2/5 = 4/10 = 6/15 = 8/20 ... This is where I like to start talking about fractions. So if you have 25/65, you can see that they are both on the 5 line, so going backwards, you get 5/13. This helps to find and reduce fractions.
Next, learn to multiply fractions (3/2)(5/4) multiply top and bottom 15/8. Since there is no line (except the one line) that has both 15 and 8, it cannot be reduced. Move to multiplications that can reduce, (4/5)(10/2) so when you multiply, you get 40/10. Since both 40 and 10 are on the 10 line, you can go up to 4/1, and you do not need to divide by 1, so you get 4.
Once you get good at these, then you can start to learn to add and subtract fractions, divide fractions, and then start working with mixed numbers.
• Can a fraction be cubed?
• Sure can! Take 4/5 for example. 4 cubed is 64, and 5 cubed is 125. So 4/5 cubed is 64/125.
• 8/8 is longer form for 1/1 I think
• I do not think that longer form is language we use in math, we would say that 8/8 and 1/1 are equivalent fractions (meaning they are equal to each other).
• We should find the l.c.m while addition and subtraction or we can use the same method for multiplication and division?
• When multiplying or dividing (which is basically multiplying the first fraction by the reciprocal of the second fraction, "flipped") fractions, you don't have to find the LCM for the denominator because you're multiplying the denominators. It's a good idea to simplify the fractions before, though, so the numbers you get at the end aren't too large.
• Can a fraction be squared (with an exponent)?
• Yes a fraction can be squared 4/5 squared = 16/25 because 4 squared is 16 and 5 squared is 25 but it is not the same because you don't multiply the numerator( the 4) and the denominator( the 5) by the same thing |
1. ## Solving Graphic Equations
How do you solve this question? Please help me. The weights of two beluga whales and 3 orca whales add up to 36,000 pounds and one of the orcas add up to 13,000 lbs. Using this info only... How much do the beluga's weigh? How much so the orcas weigh?
2. Are you sure you don't mean: the 2 orca whales add up to 13,000 lbs?
Because I'm getting an odd answer...
Let $b$ be the weight of a beluga whale
Let $o$ be the weight of an orca whale
You know: "The weights of two beluga whales and 3 orca whales add up to 36,000 pounds"
So: $2b + 3o = 36,000$
And: "...one of the orcas add up to 13,000 lbs."
So $o = 13,000$
We have 2 orca whales, so their weight is two times the weight of one.
So its weight is: $2(13,000) = 26,000\,lbs$.
Use the value of $o$ to get the weight of one beluga whale.
$2b + 3(13,000) = 36,000$
$2b + 39,000 = 36,000$
$2b = -3,000$
$b = -1,500$
A negative weight? I don't see how... ?
3. Originally Posted by Jonboy
Are you sure you don't mean: the 2 orca whales add up to 13,000 lbs?
Because I'm getting an odd answer...
Let $b$ be the weight of a beluga whale
Let $o$ be the weight of an orca whale
You know: "The weights of two beluga whales and 3 orca whales add up to 36,000 pounds"
So: $2b + 3o = 36,000$
And: "...one of the orcas add up to 13,000 lbs."
So $o = 13,000$
We have 2 orca whales, so their weight is two times the weight of one.
So its weight is: $2(13,000) = 26,000\,lbs$.
Use the value of $o$ to get the weight of one beluga whale.
$2b + 3(13,000) = 36,000$
$2b + 39,000 = 36,000$
$2b = -3,000$
$b = -1,500$
A negative weight? I don't see how... ?
Wouldn't it be
2b + 3o = 36,000
o = 13,000
2o (13,000)=26,000
2b + 26,000=36,000
Does this make any sense to you? at least up to here?
4. Originally Posted by noelly_01
Wouldn't it be
2b + 3o = 36,000
o = 13,000
2o (13,000)=26,000
2b + 26,000=36,000
Does this make any sense to you? at least up to here?
2(5000) + 26,000 = 36,000
Is this OK or am I more confused?
5. Originally Posted by noelly_01
How do you solve this question? Please help me. The weights of two beluga whales and 3 orca whales add up to 36,000 pounds and one of the orcas add up to 13,000 lbs. Using this info only... How much do the beluga's weigh? How much so the orcas weigh?
Please do NOt ask questions in LiveChat. That's what the Math Forum is for. |
# Value of Cos 120
When we study the relationship between angles and sides of a triangle, it is called Trigonometry. Trigonometry is used in almost all fields. These applications include engineering, phonetics or game development, and so on. In all these areas, we use trigonometry functions for various purposes. The values of these functions like the value of Cos 120 or Cos 0 is important in these fields.
Value of Cos 120 is -½.
It has many other applications too. In some cases, it is used indirectly. For example, in creating computer music as sound travels in the form of waves. These waves follow a pattern of sine or cosine functions to develop computer music.
## How to find the value of Cos 1200
As mentioned in the solution given below, 120° can be represented in terms of two angles i.e. either 90° or 180°.
We can show that 120 degrees can be represented in two angles, whose value can be taken from trigonometry table.
90 degree and 180 degree
180° – 60° = 120° ———– (1)
90° + 30° = 120° ———— (2)
Let’s use these now.
Cos 120° = cos(180° – 60°) = – cos 60° = -½ (since cos(180° – x) = – cos x)
Cos 120° = cos(90° + 30°) = – sin 30° = -½ (we know that cos (90° + x) = -sin x)
## Cos 120
Other trigonometric ratios for different angles are:
Trigonometry Ratio Table Angles (In Degrees) 0 30 45 60 90 180 270 360 Angles (In Radians) 0 π/6 π/4 π/3 π/2 π 3π/2 2π sin 0 1/2 1/√2 √3/2 1 0 −1 0 cos 1 √3/2 1/√2 1/2 0 −1 0 1 tan 0 1/√3 1 √3 Not Defined 0 Not Defined 0 cot Not Defined √3 1 1/√3 0 Not Defined 0 Not Defined cosec Not Defined 2 √2 2/√3 1 Not Defined −1 Not Defined sec 1 2/√3 √2 2 Not Defined −1 Not Defined 1
Visit BYJU’S to learn other trigonometric formulas and concepts. |
# Multiplying and Dividing Integers
Multiply and dividing with integers is much like multiplying and dividing with whole numbers. You just have to be careful of the signs.
Examples:2 x 3 = 6
Multiplying is really just showing repeated adding. So, we need to add 2 three times. 2 + 2 + 2 = 6
How would this look on a number line?
Now, let's take a look at how this works with negatives.
Examples:-2 x 3 = -6
This one is asking us to add -2 three times. That means that we would have -2 + -2 + -2 = -6.
We can take a look at this on the number line as well.
What if there were two negatives?
Examples:-2 x -3 = 6
This example says that we should add -2 negative 3 times. How do you add something negative times?
Well, we can also think of the negative symbol as meaning "the opposite". So if we are going to add - 2 the opposite of 3 times, we will move in the opposite direction on the number line. We had moved to the left, so now we move to the right.
Let's take what we have seen in the examples and write some simpler rules.
Rule #1:If the signs are the same, the answer is positive.
Examples:
Rule #2:If the signs are different, the answer is negative.
Examples:
The great news is that the rules for dividing integers are the same as the rules for multiplying integers.
Remember that dividing is the opposite of multiplying. So we can use the same rules to solve.
Rule #1:If the signs are the same, the answer is positive.
Examples:
Rule #2:If the signs are different, the answer is negative.
Examples:
When trying to remember the rule, you can say to yourself, "Are my signs the same?"
If you answer oooooo, then the answer is egative.
If you answer Yes, then the answer is Positive. Otherwise, the multiplying and dividing is the same as with whole numbers.
Related Links:
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Fractions
Factors |
# Thread: solving initial value problem!
1. ## solving initial value problem!
I have
xy' - 3y = x^4(e^x + cos x) - 2x^2 ; y(pi) = e^pi + (2/pi).
Could someone solve this out for me?
Thank you.
2. ## Re: solving initial value problem!
I'm not going to post a full solution, but I will be glad to help you get one. Can you think of a way to write the ODE as a linear equation?
3. ## Re: solving initial value problem!
By linear, he means an equation of the form $y'+p(x)y = q(x)$. But that's easy:
$y' - \frac{3}{x}y = x^3(e^x + \cos {x}) - 2x$
Now, how do you solve a first-order linear equation?
- Hollywood
4. ## Re: solving initial value problem!
So i set p(x)= -3/x
integrate p(x)dx = - integrate 3/x dx = - ln x^3
then I did e^(-ln x^3) = 1/x^(3) which is the integrating factor... what do I do next??
5. ## Re: solving initial value problem!
Multiply the equation in standard linear form by this integrating factor. You should then find the left-hand side can be expressed as the derivative of a product.
The equation in standard linear form:
$\frac{dy}{dx}-\frac{3}{x}y=x^3(e^x+\cos(x))-2x$
Now, multiply through by $\mu(x)=\frac{1}{x^3}$
$\frac{1}{x^3}\cdot\frac{dy}{dx}-\frac{3}{x^4}y=e^x+\cos(x)-\frac{2}{x^2}$
Rewrite the left side as the derivative of a product:
$\frac{d}{dx}\left(\frac{1}{x^3}\cdot y \right)=e^x+\cos(x)-\frac{2}{x^2}$
Now, integrate with respect to $x$.
6. ## Re: solving initial value problem!
Hello, I have a question how did 1/x^3 x dy/dx - 3y/x^4 became d/dx(1/x^3 x y) ??
7. ## Re: solving initial value problem!
If the integrating factor is calculated correctly, you should always get the left side to be the derivative of the product of the integrating factor and the dependent variable.
Observe that:
$\frac{d}{dx}\left(\frac{1}{x^3}\cdot y \right)=\frac{1}{x^3}\cdot\frac{dy}{dx}-\frac{3}{x^4}y$
Your textbook should give you a development of this method which will demonstrate why this always works.
On a side note, I highly recommend that you do not use the character "x" to denote multiplication, as it can be confused with the variable $x$.
8. ## Re: solving initial value problem!
I integrated the right side respect to x then i got
y/x^3 = e^x + cos x - 2/x + c
Then I multiplied x^3 to both side getting
y= x^3 e^x + x^3 sin x + 2/x^2 + c/x^3
Am I right so far? because I have this equation y(pi) = e^pi + (2/pi)
But I tried and it does not seem to work
9. ## Re: solving initial value problem!
You have integrated the trigonometric term on the right incorrectly (wait, I see you changed it in the second line), so not counting you first line, what you have done incorrectly is you have integrated the rational term incorrectly and divided the parameter (constant of integration) by x^3 rather than multiplied. So, what you should have is the general solution:
$y(x)=x^3\left(e^x+\sin(x)+\frac{2}{x}+c \right)$
Now, use the given initial condition to determine $c$.
10. ## Re: solving initial value problem!
Deos the constant C becomes Cx^3? after multiplying?
11. ## Re: solving initial value problem!
Yes, if you choose to distribute, then everything withing the parentheses gets a factor of x^3.
12. ## Re: solving initial value problem!
Hello so I got this equation
e^pi + 2/pi = pi^3 e^pi + 2pi^2 +Cpi^3 from y(pi) = e^pi + 2/pi
I tried moving everything to left leaving Cpi^3 on the right. But I find that the equation gets very messy? Is it supposed to be like that?
13. ## Re: solving initial value problem!
Yes, the value I found for C is not a simple number. Post what you find for C and I will tell you if our results agree.
14. ## Re: solving initial value problem!
I have got (pi e^pi - 2 - pi^4 e^pi - 2pi^3)/pi^4
15. ## Re: solving initial value problem!
That is similar to what I have, not not the same. Could you post your derivation?
Page 1 of 2 12 Last |
A Program of The Actuarial Foundation. Aligned with Common Core State and NCTM Standards.
What is an actuary? An actuary is an expert in statistics who works with businesses, governments, and organizations to help them plan for the future. Actuarial science is the discipline that applies math and statistical methods to assess risk.
6-8
DURATION
1 null
# Lesson 3: Converting Percentages
In this lesson, students will understand key ideas about percentages and how to convert them to equivalent decimals and fractions.
STANDARDS (CCSS & NCTM)
• Grade 6: Unit Rate (CCSS 6.RP.A.2)
• Grade 7: Multi-Step Real-Life Problems With Fractions, Decimals, and Whole Numbers (CCSS 7.EE.B.3)
• Grade 6–8: Making Sense of Problems, Reasoning, Constructing an Argument, Modeling, Using Appropriate Tools, Attending to Precision (CCSS MP1–6); NCTM Number and Operations (CCSS MP3–6); NCTM Number and Operations
OBJECTIVE
Students will be able to:
• Convert percentages to equivalent decimals and fractions; and
• Students will be able to set up and solve proportions.
MATERIALS
DIRECTIONS
Time required: 20 minutes, plus additional time for worksheet(s)
1. Ask students where they have seen percentages in their everyday lives. Answers might include merchandise on sale, stock market share price increases and decreases, test scores, etc.
2. Give an example of a test score, say 93 correct answers out of 100 questions. Ask the class what the percentage of correct answers on the test is (93%). Remind students that percent literally means "out of 100." So 93%, for example, is 93 out of 100.
3. Tell the class that there are situations that call for converting a percentage to a fraction. For example, if you have a problem with a mix of fractions and percentages and need to add them, it might be easier to convert the percentages to fractions. Convert a percentage to a fraction by making the percentage into the numerator of a fraction with 100 as the denominator, i.e., 93% is 93/100. In another example, 80% could be converted to a fraction by creating a fraction with 80 in the numerator and 100 in the denominator. The next step is to express the fraction in simplest terms, in this case 4/5.
4. Tell students that it is also possible to convert a percentage to a decimal by dividing the percentage by 100. This always results in the decimal point moving two places to the left. For example, 20% = 20 ÷ 100 = 0.2.
5. Tell the students that percentages aren’t always nice, neat whole numbers. For example, suppose you want to convert 37.5% to a fraction and a decimal. To convert to a decimal, drop the % sign and divide by 100, i.e., move the decimal point two places to the left, resulting in a decimal of 0.375. Say the name of the decimal, i.e., three hundred seventy-five thousandths, to make it easier to convert to a fraction of 375/1,000.
6. Percentages can also be used to express part of a whole, just like fractions and decimals. For example, if you read in an article that 35% of 400 people surveyed said they like rock music, how many people like rock music? To determine the answer, set up the proportion 35/100 = x/400. Solving for x, the number of people who like rock music is 140. A simpler way of completing the calculation is to convert the percentage to a decimal (35% converts to 0.35) and multiplying the decimal by the number of people surveyed, i.e., 0.35 x 400 = 140.
7. Tell students that they can also use proportions to calculate percentages. For example, if 450 students out of 1,000 own a fitness tracker, what percentage of students own one? To calculate a percentage, the goal is to arrive at a ratio with 100 as the denominator. Set up a proportion equation where 450/1,000 = x/100. x = 45 or 45%.
8. Guided Practice: Group students into pairs and ask them to convert 30%, 95%, and 12.5% to fractions and decimals [Answers: 30% = 30/100 (3/10 in simplest terms) = 0.30, 95% = 95/100 (19/20 in simplest terms) = 0.95, 12.5% = 125/1000 (1/8 in simplest terms) = 0.125.]
9. Independent Practice: Distribute Worksheet 3: “Show Me the Money With Percentages” (PDF)for classwork or homework.
10. Check for Understanding: Review worksheet answers with the class using the Worksheet Answer Key (PDF).
11. Optional: For additional reinforcement or practice, distribute Bonus Worksheet 3: "Time and Money: It’s a Matter of Math" (PDF).Review worksheet answers with the class using the Worksheet Answer Key (PDF).
REAL-WORLD MATH EXTENSIONS
One or both extensions could be used in conjunction with any of the three lessons in Conversions Rock, as the teacher sees fit.
• Ask students if they can think of professions that involve math. Discuss with students what an actuary is. Actuaries use statistics in their job to calculate risks for many different industries, and they look at data in terms of fractions, decimals, and percentages. Actuaries also use ratios and proportions in predicting the likelihood of events. For example, by analyzing past experience, an insurance company determines that 1 in every 20 drivers will have an accident in a given year. If they insure 10,000 drivers this year, the insurance company can plan ahead and put aside money to pay for 500 accidents (based on the proportion 1/20 = 500/10,000).
• The Series of Unfortunate Events books contain types of events for which actuaries may estimate the likelihood of occurring. For example, they may find that 1/3 of all skiers have accidents. Or that 40% of all skydivers injure their feet. Or 0.20 of all residents in a Kansas town have experienced tornado damage. Can you think of other events actuaries might analyze?
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### Final Project2
1. 1. Probability By TN All of the calculations base on TI-84 Plus
2. 2. What’s the probability if randomly selected 1 of the numbers which have 5 digit numbers, greater than 65000 , and the number cannot repeat ? ضطش
3. 3. FIRST STEP …
4. 4. Because we calculate from 65000 to 69000. So that it must has 1 choice. 6 5 0 0 0 1 At first, we need to know how many number from 65000 to 65999 and the numbers cannot be repeat.
5. 5. Because we calculate from 65000 to 69000. So that it must has 1 choice. We have 4 number greater than 5, except 6 (1 st number).So that it must be 5-1=4 6 5 0 0 0 1 4 At first, we need to know how many number from 65000 to 65999 and the numbers cannot be repeat.
6. 6. Because we calculate from 65000 to 69000. So that it must has 1 choice. We have 4 number greater than 5, except 6 (1 st number).So that it must be 5-1=4 6 5 0 0 0 1 4 8 We have 8 numbers to chose (except 2 letters before) At first, we need to know how many number from 65000 to 65999 and the numbers cannot be repeat.
7. 7. Because we calculate from 65000 to 69000. So that it must has 1 choice. We have 4 number greater than 5, except 6 (1 st number).So that it must be 5-1=4 6 5 0 0 0 1 4 8 We have 8 numbers to chose (except 2 letters before) 7 We have 7 numbers to chose (except 3 letters before) At first, we need to know how many number from 65000 to 65999 and the numbers cannot be repeat.
8. 8. Because we calculate from 65000 to 69000. So that it must has 1 choice. We have 4 number greater than 5, except 6 (1 st number).So that it must be 5-1=4 6 5 0 0 0 1 4 8 We have 8 numbers to chose (except 2 letters before) 7 We have 7 numbers to chose (except 3 letters before) 6 At first, we need to know how many number from 65000 to 65999 and the numbers cannot be repeat. We have 7 numbers to chose (except 3 letters before)
9. 9. Because we calculate from 65000 to 69000. So that it must has 1 choice. We have 4 number greater than 5, except 6 (1 st number).So that it must be 5-1=4 6 5 0 0 0 1 4 8 We have 8 numbers to chose (except 2 letters before) 7 We have 7 numbers to chose (except 3 letters before) 6 x x x x Multiply it, we will have the total of number greater than 65000 and smaller than 69999 At first, we need to know how many number from 65000 to 65999 and the numbers cannot be repeat. We have 7 numbers to chose (except 3 letters before)
10. 10. Because we calculate from 65000 to 69000. So that it must has 1 choice. We have 4 number greater than 5, except 6 (1 st number).So that it must be 5-1=4 6 5 0 0 0 1 4 8 We have 8 numbers to chose (except 2 letters before) 7 We have 7 numbers to chose (except 3 letters before) 6 x x x x Multiply it, we will have the total of number greater than 65000 and smaller than 69999 At first, we need to know how many number from 65000 to 65999 and the numbers cannot be repeat. 1 x 4 x 8 x 7 x 6 =1344 numbers We have 7 numbers to chose (except 3 letters before)
11. 11. SECOND STEP
12. 12. 7,8,9 at 1 st position can make the numbers greater than 7000 (the numbers after cannot be repeated => exclude 70000) 0 0 3 0 0 7 Then, we need to know how many number greater than 65999, from 70000 to 99999 and the numbers cannot be repeat.
13. 13. 7,8,9 at 1 st position can make the numbers greater than 7000 (the numbers after cannot be repeated => exclude 70000) 0 0 3 0 0 7 Then, we need to know how many number greater than 65999, from 70000 to 99999 and the numbers cannot be repeat. 9 We have 9 numbers to chose (except 1 letters before)
14. 14. 7,8,9 at 1 st position can make the numbers greater than 7000 (the numbers after cannot be repeated => exclude 70000) 0 0 3 0 0 7 Then, we need to know how many number greater than 65999, from 70000 to 99999 and the numbers cannot be repeat. 9 We have 9 numbers to chose (except 1 letters before) 8 We have 8 numbers to chose (except 2 letters before)
15. 15. 7,8,9 at 1 st position can make the numbers greater than 7000 (the numbers after cannot be repeated => exclude 70000) 0 0 3 0 0 7 Then, we need to know how many number greater than 65999, from 70000 to 99999 and the numbers cannot be repeat. 9 We have 9 numbers to chose (except 1 letters before) 8 We have 8 numbers to chose (except 2 letters before) 7 We have7 numbers to chose (except 3 letters before)
16. 16. 7,8,9 at 1 st position can make the numbers greater than 7000 (the numbers after cannot be repeated => exclude 70000) 0 0 3 0 0 7 Then, we need to know how many number greater than 65999, from 70000 to 99999 and the numbers cannot be repeat. 9 We have 9 numbers to chose (except 1 letters before) 8 We have 8 numbers to chose (except 2 letters before) 7 We have7 numbers to chose (except 3 letters before) 6 We have 6 numbers to chose (except 4 letters before)
17. 17. 7,8,9 at 1 st position can make the numbers greater than 7000 (the numbers after cannot be repeated => exclude 70000) 0 0 3 0 0 7 Then, we need to know how many number greater than 65999, from 70000 to 99999 and the numbers cannot be repeat. 9 We have 9 numbers to chose (except 1 letters before) 8 We have 8 numbers to chose (except 2 letters before) 7 We have7 numbers to chose (except 3 letters before) 6 We have 6 numbers to chose (except 4 letters before) x x x x
18. 18. 7,8,9 at 1 st position can make the numbers greater than 7000 (the numbers after cannot be repeated => exclude 70000) 0 0 3 0 0 7 Then, we need to know how many number greater than 65999, from 70000 to 99999 and the numbers cannot be repeat. 9 We have 9 numbers to chose (except 1 letters before) 8 We have 8 numbers to chose (except 2 letters before) 7 We have7 numbers to chose (except 3 letters before) 6 We have 6 numbers to chose (except 4 letters before) x x x x Multiply it, we will have the total of number greater than 65999 and smaller than 99999 3 x 9 x 8 x 7 x 6=9075 numbers
19. 19. And now, we add it together to get the 5 digit number greater than 65000. 1344+9072=10416 numbers
20. 20. To calculate the probability, we need to know how many number have 5 digit number
21. 21. To calculate the probability, we need to know how many number have 5 digit number Exclude 0 because if 0 lay at 1 st position, it will make the number with 4 digit. 9
22. 22. To calculate the probability, we need to know how many number have 5 digit number 9 Exclude 0 because if 0 lay at 1 st position, it will make the number with 4 digit. 10 10 10 10 The number can be 0 and repeated
23. 23. To calculate the probability, we need to know how many number have 5 digit number x x x x 9 Exclude 0 because if 0 lay at 1 st position, it will make the number with 4 digit. 10 10 10 10 The number can be 0 and repeated = 90000 Multiply it, we will have the total number have 5 digit numbers
24. 24. PROBABILITY IS 10416/90000 217/1875 (11.57%) |
# Proportion
We will discuss about the proportion of four quantities.
Four quantities w, x, y, and z are in proportion if w : x :: y : z or, $$\frac{w}{x}$$ = $$\frac{y}{z}$$.
Definition: If four quantities w, x, y and z are such that the ratio w : x is equal to the ratio y : z then we say w, x, y and z are in proportion or, w, x, y and z are proportional. We express it by writing w : x :: y : z.
w : x :: y : z if and only if $$\frac{w}{x}$$ = $$\frac{y}{z}$$ i.e., wz = xy.
In w : x : : y : z, w, x, y and z are the first, second, third and fourth terms respectively.
Also, w and z are called the extreme terms while x and y are called the middle terms or mean terms.
That is in a proportion the first and fourth terms are called extremes, while the second and third terms are called means.
Product of extremes = product of means
or, wz = xy.
If w, x, y and z are in proportion then the last term z is also called the fourth proportional.
For example, 2 : 7 :: 8 : 28 because $$\frac{2}{7}$$ = $$\frac{8}{28}$$.
Solved examples on Proportion:
Check whether the following numbers form a proportion or not.
(i) 2.5, 4.5, 5.5, 9.9
(ii) 1$$\frac{1}{4}$$, 1$$\frac{3}{4}$$, 1.5, 1.4
Solution:
(i) 2.5 : 4.5 = $$\frac{2.5}{4.5}$$ = $$\frac{25}{45}$$ = $$\frac{5}{9}$$
5.5 : 9.9 = $$\frac{5.5}{9.9}$$ = $$\frac{55}{99}$$ = $$\frac{5}{9}$$
Therefore, $$\frac{2.5}{4.5}$$ = $$\frac{5.5}{9.9}$$
Hence, 2.5, 4.5, 5.5 and 9.9 are in proportion.
(ii) 1$$\frac{1}{4}$$ : 1$$\frac{3}{4}$$ = $$\frac{5}{4}$$ : $$\frac{7}{4}$$ = $$\frac{5}{4}$$ × 4: $$\frac{7}{4}$$ × 4 = 5 : 7 = $$\frac{5}{7}$$
1.5 : 1.4 = $$\frac{1.5}{1.4}$$ = $$\frac{15}{14}$$
Since, $$\frac{5}{7}$$ ≠ $$\frac{15}{14}$$
Hence, 1$$\frac{1}{4}$$, 1$$\frac{3}{4}$$, 1.5 and 1.4 are not in proportion.
● Ratio and proportion
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# Decompose Rectangular Prisms
Videos to help Grade 5 students learn how to compose and decompose right rectangular prisms using layers.
New York State Common Core Math Module 5, Grade 5, Lesson 3
Common Core Standards: 5.MD.3, 5.MD.4
Related Topics:
Lesson Plans and Worksheets for Grade 5
Lesson Plans and Worksheets for all Grades
Lesson 3 Concept Development
Lesson 3 Problem Set
1. Use the prisms to find the volume.
• Build the rectangular prism pictured below to the left with your cubes, if necessary.
• Decompose it into layers in three different ways, and show your thinking on the blank prisms.
• Complete the missing information in the table.
2. Josh and Jonah were finding the volume of the prism to the right. The boys agree that 4 layers can be added together to find the volume. Josh says that he can see on the end of the prism that each layer will have 16 cubes in it. Jonah says that each layer has 24 cubes in it. Who is right? Explain how you know using words, numbers, and/or pictures.
Lesson 3 Homework
This video shows how to find the volume of rectangular prisms using a method of decomposing in layers.
1. Use the prisms to find the volume.
• The rectangular prisms pictured below were constructed with 1-cm cubes
• Decompose each prism into layers in three different ways, and show your thinking on the blank prisms.
• Complete each table
2. Stephen and Chelsea want to increase the volume of this prism by 72 cubic centimeters. Chelsea wants to add eight layers and Stephen says they only need to add four layers.
Their teacher tells them they are both correct. Explain how this is possible.
Lesson 3 Homework
4. Imagine the rectangular prism below is 4 meters long, 3 meters tall, and 2 meters wide. Draw horizontal lines to show how the prism could be decomposed into layers that are 1 meter in height.
It has _____ layers from left to right.
Each layer contains ______ cubic units.
The volume of this prism is __________.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
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# Rational Expressions Involving Addition, Subtraction and Multiplication
We will learn how to simplify rational expressions involving addition, subtraction and multiplication.
Solved examples on rational expressions:
1. Simplify: (-12/7 × 35/16) - (9/5 × 4/3)
Solution:
(-12/7 × 35/16) - (9/5 × 4/3)
= [(-12) × 35/7 × 16] - [9 × 4/5 × 3]
= -420/112 - 36/15
= -15/4 - 12/5
= -15 × 5/4 × 5 - 12 × 4/5 × 4
= -75/20 - 48/20
= (-75 - 48)/20
= -123/20
2. Simplify: (-3/2 × 4/5) + (9/5 × -10/3) - (1 × 3/2 × 4)
Solution:
(-3/2 × 4/5) + (9/5 × -10/3) - (1 × 3/2 × 4)
= (-3 × 4/2 × 5) + (9 × -10/5 × 3) - 3/8
= -12/10 + -90/15 - 3/8
= -6/5 + -6/1 – 3/8
= -6 × 8/5 × 8 +(-6) × 40/1 × 40 – 3 × 5/8 × 5
= -48/40 + (-240)/40 – 15/40
= -48 + (-240) - 15/40
= -303/40
3. Simplify: (-7/18 × 15/(-7)) – (1 × (1/4)) + (1/2 × 1/4)
Solution:
[-7/18 × 15/(-7)] – (1 × (1/4)) + (1/2 × 1/4)
= [(-7) × 15/18 × (-7)] – ¼ + (1 × 1)/(2 × 4)
= (-105)/(-126) – ¼ + 1/8
= 5/6 – ¼ + 1/8
= 5 × 4/6 × 4 – 1 × 6/4 × 6 + 1 × 3/8 × 3
= 20/24 – 6/24 + 3/24
= (20 - 6 + 3)/24
= 17/24
The above examples help us how to solve different types of rational expressions involving addition, subtraction and multiplication.
Rational Numbers
What is Rational Numbers?
Is Every Rational Number a Natural Number?
Is Zero a Rational Number?
Is Every Rational Number an Integer?
Is Every Rational Number a Fraction?
Positive Rational Number
Negative Rational Number
Equivalent Rational Numbers
Equivalent form of Rational Numbers
Rational Number in Different Forms
Properties of Rational Numbers
Lowest form of a Rational Number
Standard form of a Rational Number
Equality of Rational Numbers using Standard Form
Equality of Rational Numbers with Common Denominator
Equality of Rational Numbers using Cross Multiplication
Comparison of Rational Numbers
Rational Numbers in Ascending Order
Representation of Rational Numbers on the Number Line
Addition of Rational Number with Same Denominator
Addition of Rational Number with Different Denominator
Addition of Rational Numbers
Properties of Addition of Rational Numbers
Subtraction of Rational Number with Different Denominator
Subtraction of Rational Numbers
Properties of Subtraction of Rational Numbers
Simplify Rational Expressions Involving the Sum or Difference
Multiplication of Rational Numbers
Properties of Multiplication of Rational Numbers
Rational Expressions Involving Addition, Subtraction and Multiplication
Division of Rational Numbers
Rational Expressions Involving Division
Properties of Division of Rational Numbers
To Find Rational Numbers
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# Continuous functions
We define a continuous function $$f(x)$$ if it satisfies that for any point of its domain, the function is continuous at this point. In a more formal way: $$\displaystyle \mbox{ For all } x=a \mbox{ belonging to } Dom(f) , \lim_{x \to x^{\pm}} f(x)=f(a)$$$We observe that in a continuous function we have continuity and side continuity in each and every point. Attention: this definition is not equivalent to that of "a function is continuous if it is possible to draw it without raising the pencil from the paper". Let's observe the following example to show the difference: Let's take the function $$\displaystyle f(x)=\frac{1}{x}$$. If we represent it graphically we have the graph: and we can observe that we have two different branches that are not connected, from that we can think that the function is not continuous. This function satisfies the definition of continuity that we have given, if at any point the definition is satisfied, but one can think that at $$x=0$$ there could be some problem: The side limits do not coincide: $$\displaystyle \begin{array} {l} \lim_{x \to 0^+}f(x)=\lim_{x \to 0^+}\frac{1}{x}=+ \infty \\ \lim_{x \to 0^-}f(x) = \lim_{x \to 0^-}\frac{1}{x}= -\infty \end{array}$$$ but the fact is that the function is not defined at point zero, for we cannot compare with the value of $$f(0)$$.
Note that the definition of continuity is about all the points that are in the domain of the function, and not about the points that are outside the domain.
Consequently, the function in the example has two separated branches, both of them continuous. |
# Simple Interest
In this lesson, you will learn to compute simple interest. Interest is the amount of money charged or paid for borrowing or using money. When you deposit money into a savings account, you are paid interest. When you borrow money, you pay interest. Following is the formula used to compute interest.
The interest is the money earned. The principal is the amount of money deposited or borrowed. The rate is the percent that is used and should always be changed to a decimal. The time is calculated in years.
Multiply the principal by the rate by the time to get the interest. Look at a few examples and see how simple it is to use the formula.
### Example 1
Find the interest for the following conditions:
The principal is \$225, the rate is 3%, and the time is 2 years.
First, write the formula.
I = P * r * t Next, substitute the values given in the conditions.
I = \$225 * .03 * 2 Now, just multiply, using your calculator.
I = \$13.50 The interest is \$13.50.
### Example 2
Find the rate for the following conditions:
The interest is \$300, the Principal is \$1,000, and the time is 5 years.
First, write the formula.
I = P*r*t Next, substitute the values given in the conditions.
300 = 1000 * 5 * r Now, multiply the right side to get an equation.
300 = 5000r As you can see, you only have to divide both sides by 5000 to find r.
The rate is 6%.
Now, let us put the conditions into a word problem.
### Example 3
Erica deposits \$7000 in an account that earns 7% simple interest. About how long will it take for her account balance to reach \$8,000?
Write the formula: I = P * r * t
List the information
Principal = 7,000. Rate =0.07 Balance (including interest will be 8,000). This means she will have earned \$1000 interest! Now substitute and solve.
1000 = 7000 * 0.07 * t
1000 = 490t
or it will take just a bit over two years for her to get \$8000 in her account with these conditions.
Did you notice that as long as there are three known values, the other value can be found using the formula for simple interest?
Complete this learning activity before moving on to the assignments.
## Simple Interest Practice
Principal Rate Time Interest Total Amt
\$225 5% 3 (1) (1)
\$4250 7% 1.5 (2) (2)
\$397 5% 1 (3) (3)
\$700 6.25% 2 (4) (4)
\$775 8% 1 (5) (5)
\$650 4.5% 2 (6) (6)
\$2975 6% 1 (7) (7)
\$500 9% 3 (8) (8)
\$1422 3% 5 (9) (9)
\$1500 3.85% 6 (10) (10)
(source) |
# 9. Determinants
### Definition
In many applications we will use the notion of determinant of a matrix. The determinant of a matrix makes sense for square matrices only and is defined recursively:
•
where is matrix with -th row and -th column crossed out. So (the determinant is denoted by or by using absolute value style brackets around a matrix):
therefore:
And so on. E.g.:
### Laplace expansion
The above definition is only a special case of a more general fact called Laplace expansion. Instead of using the first row we can use any row or column (choose always the one with most zeros). So:
for any row . Analogical fact is true for any column.
E.g. for the below matrix it is easiest to use the third column:
### Determinant and operations on a matrix
Notice first that from the Laplace expansion we easily get that if a matrix has a row of zeros (or column) its determinant equals zero.
Consider now different operations on rows of a matrix, which we use to calculate a ,,stair-like” form of a matrix. Using Laplace expansion we can prove that swapping two rows multiplies the determinant by — indeed calculating the determinant using the first column we see that the signs in the sum may change, but also the rows in the minor matrices get swapped.
Immediately we can notice that multiplying a row by a number multiplies also the determinant by this number — you can see it easily calculating Laplace expansion using this row.
Therefore multiplying whole matrix by a number multiplies the determinant by this number many times, precisely:
where is a matrix of size .
Notice also, that the determinant of a matrix with two identical rows equals zero, because swapping those rows does not change the matrix but multiplies the determinant by , so , therefore . So because of the row multiplication rule, if two rows in a matrix are linearly dependent, then its determinant equals .
Also the Laplace expansion implies that if matrices , , differ only by -th row in the way that this row in matrix is a sum of -th rows in matrices and , then the determinant of is the sum of determinants of and , e.g.:
But it can be easily seen that in general !
Finally, consider the most important operation of adding to a row another row multiplied by a number. Then we actually deal with the situation described above. The resulting matrix is matrix , which differs from and only by the row we sum to. Matrix is the original matrix and matrix is matrix , in which we substitute the row we sum to with the row we are summing multiplied by a number. Therefore , but has two linearly dependent rows, so and . Therefore the operation of adding a row multiplied by a number to another row does not change the determinant of a matrix.
Finally, the matrix multiplication. All the above operations can be written as multiplication by a special matrix. E.g. swapping of -nd and -rd rows in a matrix of size , is actually the following:
multiplication of the -rd row by scalar , is:
adding to the third row the second multiplied by is:
It is relatively easy to see that even in the general case the matrix, which changes rows has determinant , the matrix multiplying a row by scalar , has determinant , and matrix of adding to a row another row multiplied by a scalar has determinant . Therefore multiplying by those matrices (called the elementary matrices) multiplies the determinant of a matrix by their determinants. Moreover, every matrix can be created by multiplying elementary matrix, which gives the following important conclusion:
### Calculating the determinant via triangular form of matrix
If you look closely enough you will see that the Laplace expansion also implies that the determinant of a matrix in an echelon form (usually called triangular for square matrices) equals the product of elements on the diagonal of the matrix, so e.g.:
Because we know how the elementary operations change, to calculate the determinant of a matrix we can calculate a triangular form, calculate its determinant and recreate the determinant of the original matrix. This method is especially useful for large matrices, e.g.:
Therefore, the determinant of the last matrix is . On our way we have swapped rows once and we have multiplied one row by , therefore the determinant of the first matrix equals .
The above fact also implies how to calculate the determinant of a matrix which is in the block form: with left bottom block of zeros. The determinant of such a matrix equals , e.g.: |
# Background Information on Exponentials and Logarithms
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1 Background Information on Eponentials and Logarithms Since the treatment of the decay of radioactive nuclei is inetricably linked to the mathematics of eponentials and logarithms, it is important that you have some epertise in using them. This is intended as a brief outline of how to use them. Note that you calculator will have these operations built into it, allowing you to get the result of the operation with a single key stroke. We might note in passing that the mathematics below is not all limited to the topic in the class, that is radioactive decay. It is not even limited to physics or the other sciences. For eample if you establish an account which pays compound interest at a fied rate, and reinvest the interest into the account, then the value of the account increases eponentially with time, following all of the same mathematical formulae below. Step 1 - Integral Powers of 10 Let us start with something with which (I hope) you are already familiar, that is powers of 10. The notation 10 means 10 multiplied by itself times, C 10² = 10 * 10 = 100 C 10³ = 10 * 10 * 10 = C 10 = 10 * 10 * 10 * 10 * 10 * 10 = 1,000,000 C and so on Step 2 - NonIntegral Powers of 10 Let us take the previous idea on step further, suppose that the power is not a whole number, say Then what is 10 now? You will need your calculator to solve this one. Depending on your calculator the eact keystrokes will be different, but there is a good chance that there is a key labeled 10. Try entering followed by 10, and you should get the result 10 = Step 3 - Base 10 Logarithms The previous step solved the equation y = 10, if you are given then you can calculate y. If = then y = However, suppose you are given y and are required to calculate. For eample what is the value of if y = 69. As an equation solve for the variable if 69 = 10. The key to solving this equation is the logarithm (abbreviated to just log). It is defined as the opposite operation to the power of 10: If y = 10 then = log y Again your calculator probably has a key labeled log which performs this operation for you. Type in 69 followed by log, and you should get the result Just a check try using the method above and you
3 Again these work even if the eponent is not an integer 8.81 C 2 = (Your calculator should have a y key to perform this operation. Type 2, then y, then 8.81, then the = key) 9.45 C 16 = (Type 16, then y, then 9.45, then the = key) C 3 = (Type 3, then y, then 0.45, then the change sign key, then the = key) Step 6 - Non integral bases So far we have considered the case when the power is not an integer. Can we do the same if the base is not an integer? The answer is yes. C 4.234³ = 4.234*4.234*4.234 = (Type 4.234, then y, then 3, then the = key) 5 C (-0.023) = (-0.023)*(-0.023)*(-0.023)*(-0.023)*(-0.023) = C = (Type 4.123, then y, then 7.76, then the = key) Step 7 - e One of these non-integer numbers is so special in mathematics it is given its own symbol, e which stands for (The reasons why it is so special have to do with calculus and need not concern us here.) However, it in all respects it behaves just like all other bases C You can raise e to any power. In this case the power is also referred to as an eponent. Your calculator should have a key to do this for you. It is likely labeled e or ep, depending on brand. C e² = C e = C e = C e = C You can take logarithms using e as the base. Note, when taking logarithms above we were using base 10, and the logarithm should more precisely be written as log 10 since we could use any base we like. For eample in base 4 we would write log, and in base 16 we would write log In base e we should write log e, but since e is special this is given its own designation, ln. Your calculator will likely also have a key labeled ln also. C if 5 = e then = ln(5) = (Type 5 and then the ln key. You should not have to hit the = key to get the answer.) C if = e then = ln(6778.3) = C if = e then = ln( ) = C if = e then = ln( ) = C if = e then = ln( ) = Step 8 - Putting powers and logarithms together A simple rule is that, for the same base, raising the base to a power and finding the logarithm are opposite actions. One followed by the other has no overall effect
4 C log y(y ) = 4 C log 10(10 ) = C log 4(4 ) = C log e(e ) = ln(e ) = C log e(e ) = ln(e ) = log () C y y = log (12) C 4 4 = 12 log(3.456) C 10 = ln(2) C e = 2 ln( ) C e = Note though that this rule does not work if the bases are different. For eample ln(10³) is not three. The power operation used base 10, but the logarithm used base e. Step 9 - A radioactive half life problem -kt The equation which describes the radioactive decay of an nucleus is N = N o e where N o is the initial number of nuclei, N is the number remaining after a time t, and k is a constant related to the lifetime (k = ln2/t = /T). With a little bit of rearranging this can be written as y = e if y stands for the fraction N/N o and stands for -kt. Using the rules developed above we can solve these radioactive decay problems. C Eample I An isotope has a half life of 25 minutes. What fraction remains after one hour? C First of all we need to find k. Its value is ln2/t = ln2/25 = /25 = C Putting in t = 60 minutes, we have = -kt = *60 = C Then y = e = e = = 18.9 % C Eample II For this same isotope how much would remain the net day C Now put t = 24 hours, which is the same as 24*60 = 1440 minutes C Then = -kt = *1440 = C The fraction remaining is now y = e = e = , which is a very small fraction. This sample has now almost completely disappeared. 14 C Eample III Suppose that you have a sample to be dated using C. By measuring the activity of the sample you discover that it has only one third the activity it must have had when it was new. How old is it? 14 C First of all calculate k. The half life of C is 5730 years, and so k = ln2/t = / = C In this case we know y = a = , and we need to solve for knowing that = e. C If y = e then = ln y = ln = C Since stands for -kt, to calculate t we have t = -/k = -( )/ = 9082 years. As mentioned above this mathematical treatment relates to similar problems in many walks of life. Let s go
5 through the following eamples: C Financial Eample 1 You put \$1000 into an account which bears an annual interest rate of 6% paid monthly, and reinvest the interest into the account. How much is the account worth after 5 years? (Assuming that there are no subsequent deposits or withdrawals.) C The monthly interest rate is 6%/12 = ½% = After one month the account will have a value of \$1000*( ) = \$1000*(1.005), after 2 months \$1000*(1.005)*(1.005) = \$1000*(1.005)², after 3 months \$1000*(1.005)²*(1.005) = \$1000*(1.005)³, and so on. Generalizing this, after months the value of the account is \$1000*(1.005). We have then same mathematics as above (y = b ) if we use a base of for the calculations. 60 C After 5 years (60 months) the value of the account will be \$1000*(1.005) = \$1000* = \$ C Financial Eample II How long will it take for the value of the account to rise to \$1,000,000? n C We have to solve the equation 1,000,000 = 1000* After canceling a factor of 1000 n from both sides of the equation this becomes 1000 = C The solution to this equation (see above) is n = log If your calculator could do calculations in base you could solve this directly, but this is unlikely. Instead we can make use of a useful relationship for logarithms, that log = ln/lnb. b C For our eample n = log 1000 = ln1000/ln1.005 = / = months or a little over 115 years.
### The numerical values that you find are called the solutions of the equation.
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1. Bellwork Copy and Answer • 5 – 7 • 6 + -15 • -14 - -10 • 6 x -7 • -5 ÷ -5 • -10 x -10
2. Tuesday 10/18
3. 2-5 • Solving Equations Containing Integers • pg. 100
4. Essential Question • How can you solve equations that contain integers?
5. Who REMEMBERS? • Monday we talked about 3 steps to solve an equation. • What are the 3 steps? • I___________ • S___________ • C___________ 5
6. Steps to Solving Equations: • Use the opposite operation to ISOLATE the variable to one side of the equation • SOLVE for the Variable • CHECK the answer to the equation by substituting the number in for the variable
7. Equation Mat • Today we will use an Equation Mat and Integer Chips to model solving an equation. • LAB 7
8. Equation Mat Directions… • Yellow=positive 1 • Red= negative 1 • Use the chips for our numbers. • If the equation is x+3=5…on your mat you should have • x + = • Then, use your red chips to solve the equation. • The red and yellow chips cancel out. • We are left with 2 yellow chips, so x=2.
9. Equation Mat Equations • x + 3=5 • x + 4=9 • x + 5=8 • x + 6=6 • x - 4=3 • x - 2=8 • x - 5=-5 • x - 7=0 X=2 X=5 X=3 X=0 X=7 X=10 X=0 X=7
10. Wednesday 10/19
11. 2-5 • Solving Equations Containing Integers • pg. 100
12. Essential Question • How can you solve equations that contain integers?
13. Algebra Geek Patrol The Pocket Protectors of Algebra
14. My name is Buck Einstein, but you can call me Buck. Hello! Pick my nose to continue
15. It’s that we on the Algebra Geek Patrol are… well…different. I have a secret to tell you. What is it? You might even say that we are the opposite of everyone else. Pick my nose to continue
16. When you worry about what you wear, We worry about having our Pocket Protectors. Pick my nose to continue
17. When you worry that you won’t meet your friends, We worry that we won’t have our calculators. Pick my nose to continue
18. We are just a little bit opposite. That may be why we are so good at Algebra, because that is all Algebra is… …doing the opposite. Pick my nose to continue
19. Think of some words that are opposites. For example, what do you think of as the opposite of word below. Old = ______ New Pick my nose to continue
20. Now try some other ones. Make sure to write your answers on your paper, and when you are finished click on me. Up = __________ Out = __________ Small = __________ Stinky = __________ Pick my nose to continue
21. Some possible answers are… Down Up = __________ In Out = __________ Big Small = __________ Fragrant Stinky = __________ (or non-stinky) Pick my nose to continue
22. Now another set. Make sure to write your answers on your paper, and when you are finished click on me. Add = __________ Subtract = __________ Multiply = __________ Divide = __________ Pick my nose to continue
23. Your answers should have been… Subtract Add = __________ Add Subtract = __________ Divide Multiply = __________ Multiply Divide = __________ Pick my nose to continue
24. One last set. Make sure to write your answers on your paper, and when you are finished click on me. + = __________ - = __________ x = __________ ÷ = __________ Pick my nose to continue
25. Your answers should have been… - + = __________ + - = __________ ÷ x = __________ x ÷ = __________ Pick my nose to continue
26. Congratulations! You just did some ALGEBRA Scary, isn’t it! Pick my nose to continue
27. One Step Equations Algebra Geek Patrol Now you are going to work on solving some Algebra problems. Don’t be afraid of the A-word (Algebra). Remember, it is all about being opposite. Pick my nose to continue
28. One Step Equations Algebra Geek Patrol You will complete 5 sections. After each section you will be awarded a Geek Pen to put in your official AGP Pocket Protector. Geek Luck! Pick my nose to continue
29. One Step Equations Algebra Geek Patrol X + 3 = 17 Click on the answer below that is the opposite of add 3. Add 3 Subtract 3 Multiply 3 Algebra Geek Patrol Divide 3
30. One Step Equations Algebra Geek Patrol Close, but that’s not quite it. Remember, we are looking for the opposite of add 3. Click on Buck and give it another try. Algebra Geek Patrol
31. One Step Equations Algebra Geek Patrol Good Job! The correct answer is subtract 3. Write the answer on your paper, next to the original problem, and then click on Buck. Algebra Geek Patrol
32. One Step Equations Algebra Geek Patrol 4x = 20 Click on the answer below that is the opposite of multiply by 4. Add 4 Subtract 4 Multiply by 4 Algebra Geek Patrol Divide by 4
33. One Step Equations Algebra Geek Patrol No, not really. Remember, we are looking for the opposite of multiply by 4. Click on Buck and give it another try. Algebra Geek Patrol
34. One Step Equations Algebra Geek Patrol Nice! You’re halfway to your first Geek Pen. The correct answer is divide by 4. Write the answer on your paper next to the original problem, and then click on Buck. Algebra Geek Patrol
35. One Step Equations Algebra Geek Patrol x 5 = 10 Click on the answer below that is the opposite of divide by 5. Add 5 Subtract 5 Multiply by 5 Algebra Geek Patrol Divide by 5
36. One Step Equations Algebra Geek Patrol Uhh, in a word, no. Remember, we are looking for the opposite of divide by 5. Click on Buck and give it another try. Algebra Geek Patrol
37. One Step Equations Algebra Geek Patrol Good!Are you sure you’re not a Geek? The correct answer is multiply by 5. Write the answer on your paper next to the original problem, and then click on Buck. Algebra Geek Patrol
38. One Step Equations Algebra Geek Patrol X – 7 = 8 Click on the answer below that is the opposite of - 7. + 7 - 7 x 7 Algebra Geek Patrol ÷ 7
39. One Step Equations Algebra Geek Patrol Almost Remember, we are looking for the opposite of - 7. Click on Buck and give it another try. Algebra Geek Patrol
40. One Step Equations Algebra Geek Patrol Congratulations! And here is your first Geek Pen The correct answer is + 7. Write the answer on your paper next to the original problem, and then click on Buck. Algebra Geek Patrol
41. One Step Equations Algebra Geek Patrol Alright, now you are going to actually solve some problems. This is nothing to be worried about, just remember one word… Opposite! Geek Luck! Pick my nose to continue
42. One Step Equations X + 5 = 13 1 Click on the answer below that is the opposite of + 5. + 5 - 5 x 5 Algebra Geek Patrol ÷ 5
43. One Step Equations C’mon, You know this. 1 Remember, we are looking for the opposite of + 5. Click on Buck and give it another try. Algebra Geek Patrol
44. One Step Equations Click on +5 or -5 to cancel them. Good! X + 5 = 13 1 - 5 - 5 _________ X = 8 That’s right! Look at the right side. All we have left is 13 and – 5. So,what is 13 – 5? Now to the hard part. That’s right, zero. And zero is nothing, so let’s cancelout + 5 and -5. That just leaves X on the left side. So, let’s put – 5 on each side of the equation. Let’s look at the left side. What will you get when you combine + 5 and -5 ? Copy the entire problem and all the steps on your paper, then click on Buck. Just teasing! Click on the pen when you are ready. Algebra Geek Patrol
45. One Step Equations Algebra Geek Patrol Congratulations,You just completed your first Algebra problem. See it wasn’t that bad. Now let’s try another. Pick my nose to continue
46. Who REMEMBERS? • What are the 3 steps to solve a one step equation? • I___________ • S___________ • C___________ 46
47. Steps to Solving Equations: • ISOLATE the variable by doing the opposite, Inverse, Operation. • SOLVE the equation. • CHECK your solution.
48. One Step Equations X + 9 = 15 2 Click on the answer below that is the opposite of + 9. + 9 - 9 x 9 Algebra Geek Patrol ÷ 9
49. One Step Equations Umm, I don’t think so. 2 Remember, we are looking for the opposite of + 9. Click on Buck and give it another try. Algebra Geek Patrol
50. One Step Equations Nice Job! X + 9 = 15 2 _________ - 9 - 9 This time I am going to give you the steps, which you will need to write down, and then you will work the problem on your paper. X = 6 Second, cancel out one side. Third, combine the other side. Make sure you have copied all the steps on your paper and then click on Buck. First, put the opposite (- 9) on both sides. Write this on your paper, and work this part of the problem. Algebra Geek Patrol Click the pen to continue Pick my nose to continue |
# Natural Number Multiplication is Associative/Proof 2
## Theorem
The operation of multiplication on the set of natural numbers $\N$ is associative:
$\forall x, y, z \in \N: \paren {x \times y} \times z = x \times \paren {y \times z}$
## Proof
We are to show that:
$\left({x \times y}\right) \times n = x \times \left({y \times n}\right)$
for all $x, y, n \in \N$.
From the definition of natural number multiplication, we have that:
$\displaystyle \forall m, n \in \N: \ \$ $\displaystyle m \times 0$ $=$ $\displaystyle 0$ $\displaystyle m \times \left({n + 1}\right)$ $=$ $\displaystyle \left({m \times n}\right) + m$
Let $x, y \in \N$ be arbitrary.
For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
$\left({x \times y}\right) \times n = x \times \left({y \times n}\right)$
### Basis for the Induction
$P \left({0}\right)$ is the case:
$\displaystyle \left({x \times y}\right) \times 0$ $=$ $\displaystyle 0$ $\displaystyle$ $=$ $\displaystyle x \times 0$ $\displaystyle$ $=$ $\displaystyle x \times \left({y \times 0}\right)$
and so $P \left({0}\right)$ holds.
This is our basis for the induction.
### Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true.
So this is our induction hypothesis:
$\left({x \times y}\right) \times k = x \times \left({y \times k}\right)$
Then we need to show:
$\left({x \times y}\right) \times \left({k + 1 }\right) = x \times \left({y \times \left({k + 1}\right)}\right)$
### Induction Step
This is our induction step:
$\displaystyle \left({x \times y}\right) \times \left({k + 1 }\right)$ $=$ $\displaystyle \left({\left({x \times y}\right) \times k}\right) + \left({x \times y}\right)$ Definition of Natural Number Multiplication $\displaystyle$ $=$ $\displaystyle \left({x \times \left({y \times k}\right)}\right) + \left({x \times y}\right)$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \left({x \times y}\right) + \left({x \times \left({y \times k}\right)}\right)$ Natural Number Addition is Commutative $\displaystyle$ $=$ $\displaystyle x \times \left({y + \left({y \times k}\right)}\right)$ Natural Number Multiplication Distributes over Addition $\displaystyle$ $=$ $\displaystyle x \times \left({\left({y \times k}\right) + y}\right)$ Natural Number Addition is Commutative $\displaystyle$ $=$ $\displaystyle x \times \left({y \times \left({k + 1 }\right)}\right)$ Definition of Natural Number Multiplication
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
$\blacksquare$ |
# Eureka Math Algebra 1 Module 1 Lesson 27 Answer Key
## Engage NY Eureka Math Algebra 1 Module 1 Lesson 27 Answer Key
### Eureka Math Algebra 1 Module 1 Lesson 27 Example Answer Key
Example
Review Exercise 3 from the previous lesson: Using a generic initial value, a0, and the recurrence relation, ai+1=2ai+5, for i≥0, find a formula for a1, a2, a3, a4 in terms of a0.
a1=2a0+5,
a2=2a1+5=2(2a0+5)+5=22∙a0+15,
a3=2a2+5=2(2∙2a0+15)+5=23∙a0+35,
a4=2a3+5=2(23∙a0+35)+5=24∙a0+75.
### Eureka Math Algebra 1 Module 1 Lesson 27 Exercise Answer Key
Mathematical Modeling Exercise/Exercise 1
Using one of the four formulas from Example 1, write an inequality that, if solved for a0, will lead to finding the smallest starting whole number for the double and add 5 game that produces a result of 1,000 or greater in 3 rounds or fewer.
This exercise is loaded with phrases that students will need to interpret correctly in order to formulate an equation (do not expect this to be easy for them). Start with simple questions and build up:
→ What does a2 mean in terms of rounds?
→ The result of round two
→ Write what the statement, “produce a result of 1,000 or greater in two rounds,” means using a term of the sequence.
→ The result of round two, a2, must be greater than or equal to 1,000. Ask students to write the equation, a2≥1000, for that statement.
→ After replacing a2 in the inequality, a2≥1000, with the expression in terms of a0, what do the numbers a0 that satisfy the inequality, 4a0+15≥1000, mean?
→ The numbers a0 that satisfy the inequality are the starting numbers for the double and add 5 game that produce a result of 1,000 or greater in two rounds or fewer. The “or fewer” in the previous sentence is important and can be understood by thinking about the question, “Do we need two rounds to reach 1,000, starting with number 999? 800? 500?”
Let students solve for a0 in 4a0+15≥1000, and let them find the smallest whole number a0 for exactly two rounds (Answer: 247). Then continue with your questioning:
→ What inequality in terms of a0 would you write down to find the smallest starting number for the double and add 5 game that produces a result of 1,000 or greater in three rounds or fewer?
→ 8a0+35≥1000
Exercise 2.
Solve the inequality derived in Exercise 1. Interpret your answer, and validate that it is the solution to the problem. That is, show that the whole number you found results in 1,000 or greater in three rounds, but the previous whole number takes four rounds to reach 1,000.
8a0+35≥1000
8a0+35-35≥1000-35
8a0≥965
$$\frac{1}{8}$$ (8a0 )≥$$\frac{1}{8}$$ (965)
a0≥$$\frac{965}{8}$$
Students should write or say something similar to the following response: I interpret a0≥$$\frac{965}{8}$$ or a0≥120.625 as the set of all starting numbers that reach 1,000 or greater in three rounds or fewer. Therefore, the smallest starting whole number is 121. To validate, I checked that starting with 121 results in 1,003 after three rounds, whereas 120 results in 995 after three rounds.
Exercise 3.
Find the smallest starting whole number for the double and add 5 game that produces a result of 1,000,000 or greater in four rounds or fewer.
16∙a0+75≥1 000 000
16a0+75-75≥1 000 000-75
16a0≥999 925
$$\frac{1}{16}$$ (16a0 )≥$$\frac{1}{16}$$ (999 925)
a0≥$$\frac{999 925}{16}$$
Students should write or say something similar to the following response: I interpreted a0≥$$\frac{999 925}{16}$$ or a0≥62495.3125 as the set of all starting numbers that reach 1,000,000 or greater in four rounds or fewer. Therefore, the smallest starting whole number is 62,496. To validate, I checked that starting with 62,496 results in 1,000,011 after four rounds, whereas 62,495 results in 999,995 after four rounds.
### Eureka Math Algebra 1 Module 1 Lesson 27 Problem Set Answer Key
Question 1.
Your older sibling came home from college for the weekend and showed you the following sequences (from her homework) that she claimed were generated from initial values and recurrence relations. For each sequence, find an initial value and recurrence relation that describes the sequence. (Your sister showed you an answer to the first problem.)
a. (0,2,4,6,8,10,12,14,16,…)
a1=0 and ai+1=ai+2 for i≥1
b. (1,3,5,7,9,11,13,15,17,…)
a1=1 and ai+1=ai+2 for i≥1
c. (14,16,18,20,22,24,26,…)
a1=14 and ai+1=ai+2 for i≥1
d. (14,21,28,35,42,49,…)
a1=14 and ai+1=ai+7 for i≥1
e. (14,7,0,-7,-14,-21,-28,-35,…)
a1=14 and ai+1=ai-7 for i≥1
f. (2,4,8,16,32,64,128,…)
a1=2 and ai+1=2ai for i≥1
g. (3,6,12,24,48,96,…)
a1=3 and ai+1=2ai for i≥1
h. (1,3,9,27,81,243,…)
a1=1 and ai+1=3ai for i≥1
i. (9,27,81,243,…)
a1=9 and ai+1=3ai for i≥1
Question 2.
Answer the following questions about the recursive sequence generated by initial value, a1=4, and recurrence relation, ai+1=4ai for i≥1.
a. Find a formula for a1, a2, a3, a4, a5 in terms of powers of 4.
a1=41
a2=42
a3=43
a4=44
a5=45
b. Your friend, Carl, says that he can describe the nth term of the sequence using the formula, a_n=4n. Is Carl correct? Write one or two sentences using the recurrence relation to explain why or why not.
Yes. The recurrence relation, ai+1=4ai for i≥0, means that the next term in the sequence is always 4 times larger than the current term, i.e., one more power of 4. Therefore, the nth term will be n powers of 4, or 4n.
Question 3.
The expression, 2n (a0+5)-5, describes the nth term of the double and add 5 game in terms of the starting number a0 and n. Verify that it does describe the nth term by filling out the tables below for parts (b) through (e). (The first table is done for you.)
a. Table for a0=1
b. Table for a0=8
c. Table for a0=9
d. Table for a0=120
e. Table for a0=121
Question 4.
Bilbo Baggins stated to Samwise Gamgee, “Today, Sam, I will give you $1. Every day thereafter for the next 14 days, I will take the previous day’s amount, double it and add$5, and give that new amount to you for that day.”
a. How much did Bilbo give Sam on day 15? (Hint: You don’t have to compute each term.)
a15=215 (1+5)-5=196,603. Bilbo gave Sam $196,603 on day 15. b. Did Bilbo give Sam more than$350,000 altogether?
Yes. He gave $98,299 on day 14,$49,147, on day 13, \$24,571 on day 12, and so on.
Question 5.
The formula, an=2n-1 (a0+5)-5, describes the nth term of the double and add 5 game in terms of the starting number a0 and n. Use this formula to find the smallest starting whole number for the double and add 5 game that produces a result of 10,000,000 or greater in 15 rounds or fewer.
Solving 214 (a0+5)-5≥10 000 000 for a0 results in a0≥300.1759….
Hence, 301 is the smallest starting whole number that will reach 10,000,000 in 15 rounds or fewer.
### Eureka Math Algebra 1 Module 1 Lesson 27 Exit Ticket Answer Key
Write a brief report about the answers you found to the double and add 5 game problems. Include justifications for why your starting numbers are correct.
Results for finding the smallest starting number in the double and add 5 game:
1. Reaching 100 in three rounds or fewer: The starting number 9 results in 107 in round three. The starting number 8 results in 99 in round three, requiring another round to reach 100. Numbers 1–8 take more than three rounds to reach 100.
2. Reaching 1,000 in three rounds or fewer: The starting number 121 results in 1,003 in round three. The starting number 120 results in 995 in round three, requiring another round to reach 1,000. All other whole numbers less than 120 take more than three rounds to reach 1,000.
3. Reaching 1,000,000 in four rounds or fewer: The starting number 62,496 results in 1,000,011 in round four. The starting number 62,495 results in 999,995 in round four, requiring another round to reach 1,000,000. All other whole numbers less than 62,495 take more than four rounds to reach 1,000,000.
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# Determine which functions are ODD (symmetry with origin): a. f(x) = 2x^3 + x + 3 b. f(x) = 2x^3 + X c. f(x) = 2x^3 + x^2 + x + 3 d. f(x) = 4x e. f(x) = x^3 + the absolute value of x f....
Determine which functions are ODD (symmetry with origin):
a. f(x) = 2x^3 + x + 3
b. f(x) = 2x^3 + X
c. f(x) = 2x^3 + x^2 + x + 3
d. f(x) = 4x
e. f(x) = x^3 + the absolute value of x
f. f(x) = 2x + 1/x
jeew-m | Certified Educator
A real valued function is said to be odd when the following condition satisfies.
`-f(x) = f(-x)`
Let us consider our functions.
`a. f(x) = 2x^3 + x + 3`
`f(-x) = -2x^3-x+3 != -f(x) `
`b. f(x) = 2x^3 + x`
`f(-x) = -2x^3-x = -(2x^3+x) = -f(x)`
`c. f(x) = 2x^3 + x^2 + x + 3`
`f(-x) = -2x^3+x^2-x+3 != -f(x)`
`d. f(x) = 4x`
`f(-x) = -4x = -f(x)`
`e. f(x) = x^3 +|x|`
`f(-x) = -x^3+|-x| = -x^3+|x| != -f(x)`
`f. f(x) = 2x + 1/x`
`f(-x) = -2x-1/x = -(2x+1/x) = -f(x)`
So the odd functions are the functions at b,d and f.
Note:
If you have a constant term in your function that will never be a odd function (as a and c)
If you have a even degree polynomial you will never get a odd function. |
## How do I find the exterior angle?
Exterior Angles of a Regular Polygon with n sides: Exterior angle = 360°/n. We will use the formula of the sum of interior angles and exterior angles to answer this question. Explanation: The sum of interior angles is given by 180 (n – 2), where n is the number of sides.
## What is exterior angle answer?
Exterior angles are defined as the angles formed between the side of the polygon and the extended adjacent side of the polygon.
## What is each exterior angle?
The Exterior Angle is the angle between any side of a shape, and a line extended from the next side. Another example: When we add up the Interior Angle and Exterior Angle we get a straight line 180°. They are “Supplementary Angles”.
## What is Interior Angle with example?
Mathwords: Interior Angle. An angle on the interior of a plane figure. Examples: The angles labeled 1, 2, 3, 4, and 5 in the pentagon below are all interior angles. Angles 3, 4, 5, and 6 in the second example below are all interior angles as well (parallel lines cut by a transversal).
## What is outer angle?
The Exterior Angle is the angle between any side of a shape, and a line extended from the next side. Another example: When we add up the Interior Angle and Exterior Angle we get a straight line 180°. They are “Supplementary Angles”.
## What is exterior angle property class 7?
An exterior angle of a triangle is equal to the sum of the opposite interior angles. In the above figure, ∠ACD is the exterior angle of the Δ ABC.
## What is the meaning of exterior in maths?
Definition of exterior angle
1 : the angle between a side of a polygon and an extended adjacent side. 2 : an angle formed by a transversal as it cuts one of two lines and situated on the outside of the line.
## How do you find the exterior angle of a triangle?
An exterior angle of a triangle is equal to the sum of the two opposite interior angles. Example: Find the values of x and y in the following triangle. y + 92° = 180° (interior angle + adjacent exterior angle = 180°.)
## Which are alternate exterior angles?
Alternate exterior angles are formed by a transversal intersecting two parallel lines . They are located “outside” the two parallel lines but on opposite sides of the transversal, creating two pairs (four total angles) of alternate exterior angles. |
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### Course: Class 8 (Foundation)>Unit 11
Lesson 2: Transversal
# Missing angles with a transversal
When a third line called a transversal crosses two parallel lines, we can find the measures of angles using properties like corresponding angles, vertical angles, and supplementary angles. If we know just one of the angle measurements, these properties help us find all the missing angle measurements. Created by Sal Khan.
## Want to join the conversation?
• can transversal line be perpendicular to parallel lines?
• Yes it can, it crosses through two lines at a certain point, so I would consider it transversal. It would create perpendicular lines.
• all of you are super smart
do your dreams i know you can do it.
• What is a congruent angle?
• Congruent angles are, by definition, angles that have the same degree measure.
• How do you remember the difference between supplementary and complementary?
• Remember:
Complementary angles add up to 90°
- example: 20° & 70°
(added together, they form a right angle)
-and-
Supplementary angles add up to 180°
- example: 60° & 120°
(added together, they form a straight line)
Two facts:
(1) 90° comes before 180° on the number line
(2) "C" comes before "S" in the alphabet
90° goes with "C" for complementary
so complementary angles add up to 90°
180° goes with "S" for supplementary
so supplementary angles add up to 180°
Hope this helps!
• How would I know if the lines are parallel? Therefore how would I know it actually is a transversal
• Bvanplane,
Either it has to be given that the lines are parallel,
Or you have to be given two angles that allow you to determine that like angles are equal,
Or you could be given an equation for the lines like
2x+3y=2 and
2x+3y = 8 which lines would have no solution, therefore they do not cross, therefore they are parallel.
• Why was "x" invented for math?!
(2 year later edit: Hi now that im in 6th i understand x and different kind of variables i was just too young to understand)
• "x" was invented as a variable to replace unknown numbers. So if I had 10 cookies in a jar and ate 3, (I know the answer is obvious) but say I didn't know how many are left. I could use x instead of a question mark. So 10 - 3 = x
Variables are also helpful because if I had two unknown numbers that were different in value, I could use x and y, instead of using the same ? for different values.
(x) is very helpful in life and I first learned about it in Pre-Algebra.
I hope I helped you.
"You only need to know one thing, you can learn anything"
• How did he find out what the pink angle was?
• The line where the transversal intercepts one of the parallel lines create 180 degrees due to the rule of supplementary angles. Supplementary is when two or more angles add up to 180 degrees. So Mr. Khan knew that the one measurement was 110 degrees. Using the rule of Supplementary angles, you know that the other side of the line must be 70 degrees, since 110 + 70 = 180. I hope you find this helpful!
• If complementary angles add up to 90 degrees and supplementary angles add up to 180, are there terms for angles that add up to 270 degrees and 260 degrees?
• Two angles that sum to a complete circle (1 turn, 360°) are called explementary angles or conjugate angles.
• how did he figure out that the angles are 70 degrees |
# Difference between revisions of "1998 AJHSME Problems/Problem 6"
## Problem
Dots are spaced one unit apart, horizontally and vertically. The number of square units enclosed by the polygon is
$[asy] for(int a=0; a<4; ++a) { for(int b=0; b<4; ++b) { dot((a,b)); } } draw((0,0)--(0,2)--(1,2)--(2,3)--(2,2)--(3,2)--(3,0)--(2,0)--(2,1)--(1,0)--cycle); [/asy]$
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$
## Solutions
### Solution 1
We could count the area contributed by each square on the $3 \times 3$ grid:
Top-left: $0$
Top: Triangle with area $\frac{1}{2}$
Top-right: $0$
Left: Square with area $1$
Center: Square with area $1$
Right: Square with area $1$
Bottom-left: Square with area $1$
Bottom: Triangle with area $\frac{1}{2}$
Bottom-right: Square with area $1$
Adding all of these together, we get $6$ which is the same as $\boxed{B}$
### Solution 2
By Pick's Theorem, we get the formula, $A=I+\frac{b}{2}-1$ where $I$ is the number of lattice points in the interior and $b$ being the number of lattice points on the boundary. In this problem, we can see that $I=1$ and $B=12$. Substituting gives us $A=1+\frac{12}{2}-1=6$ Thus, the answer is $\boxed{\text{(B) 6}}$
### Solution 3
Notice that the extra triangle on the top with area $1$ can be placed (like a jigsaw puzzle) at the bottom of the grid where there is a triangular hole, also with area $1$. This creates a $2*3$ rectangle, with a area of $6$. The answer is $\boxed{\text{(B) 6}}$ ~sakshamsethi |
# Find the derivative of -6/((5x-1)^(1/3))
## \frac{d}{dx}\left(\frac{-6}{\left(5x-1\right)^{\frac{1}{3}}}\right)
Go!
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◻/◻
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ln
log
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sin
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csc
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atan
acot
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sinh
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acsch
$\frac{10}{\sqrt[3]{\left(5x-1\right)^{4}}}$
## Step by step solution
Problem
$\frac{d}{dx}\left(\frac{-6}{\left(5x-1\right)^{\frac{1}{3}}}\right)$
1
Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$
$\frac{6\frac{d}{dx}\left(\sqrt[3]{5x-1}\right)+\sqrt[3]{5x-1}\cdot\frac{d}{dx}\left(-6\right)}{\left(\sqrt[3]{5x-1}\right)^2}$
2
The derivative of the constant function is equal to zero
$\frac{6\frac{d}{dx}\left(\sqrt[3]{5x-1}\right)+0\sqrt[3]{5x-1}}{\left(\sqrt[3]{5x-1}\right)^2}$
3
Any expression multiplied by $0$ is equal to $0$
$\frac{6\frac{d}{dx}\left(\sqrt[3]{5x-1}\right)+0}{\left(\sqrt[3]{5x-1}\right)^2}$
4
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{6\cdot \frac{1}{3}\left(5x-1\right)^{-\frac{2}{3}}\cdot\frac{d}{dx}\left(5x-1\right)+0}{\left(\sqrt[3]{5x-1}\right)^2}$
5
The derivative of a sum of two functions is the sum of the derivatives of each function
$\frac{6\cdot \frac{1}{3}\left(5x-1\right)^{-\frac{2}{3}}\left(\frac{d}{dx}\left(-1\right)+\frac{d}{dx}\left(5x\right)\right)+0}{\left(\sqrt[3]{5x-1}\right)^2}$
6
The derivative of the constant function is equal to zero
$\frac{6\cdot \frac{1}{3}\left(5x-1\right)^{-\frac{2}{3}}\left(0+\frac{d}{dx}\left(5x\right)\right)+0}{\left(\sqrt[3]{5x-1}\right)^2}$
7
The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function
$\frac{6\cdot \frac{1}{3}\left(5x-1\right)^{-\frac{2}{3}}\left(0+5\frac{d}{dx}\left(x\right)\right)+0}{\left(\sqrt[3]{5x-1}\right)^2}$
8
The derivative of the linear function is equal to $1$
$\frac{6\cdot \left(0+1\cdot 5\right)\cdot \frac{1}{3}\left(5x-1\right)^{-\frac{2}{3}}+0}{\left(\sqrt[3]{5x-1}\right)^2}$
9
Multiply $5$ times $1$
$\frac{\left(0+5\right)\cdot 2\left(5x-1\right)^{-\frac{2}{3}}+0}{\left(\sqrt[3]{5x-1}\right)^2}$
10
Add the values $5$ and $0$
$\frac{5\cdot 2\left(5x-1\right)^{-\frac{2}{3}}+0}{\left(\sqrt[3]{5x-1}\right)^2}$
11
Multiply $2$ times $5$
$\frac{10\left(5x-1\right)^{-\frac{2}{3}}+0}{\left(\sqrt[3]{5x-1}\right)^2}$
12
$x+0=x$, where $x$ is any expression
$\frac{10\left(5x-1\right)^{-\frac{2}{3}}}{\left(\sqrt[3]{5x-1}\right)^2}$
13
Applying the power of a power property
$\frac{10\left(5x-1\right)^{-\frac{2}{3}}}{\sqrt[3]{\left(5x-1\right)^{2}}}$
14
Simplifying the fraction by $5x-1$
$10\left(5x-1\right)^{\left(-\frac{2}{3}-\frac{2}{3}\right)}$
15
Subtract the values $-\frac{2}{3}$ and $-\frac{2}{3}$
$10\left(5x-1\right)^{-\frac{4}{3}}$
16
Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number
$10\frac{1}{\sqrt[3]{\left(5x-1\right)^{4}}}$
17
Apply the formula: $a\frac{1}{x}$$=\frac{a}{x}$, where $a=10$ and $x=\sqrt[3]{\left(5x-1\right)^{4}}$
$\frac{10}{\sqrt[3]{\left(5x-1\right)^{4}}}$
$\frac{10}{\sqrt[3]{\left(5x-1\right)^{4}}}$
### Main topic:
Differential calculus
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Q)
# The management committee of a residential colony decided to award some of its members (say x) for honesty,some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean . The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added in two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards.
This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.
Comment
A)
$\textbf{Step 1}$:
Given that $x$ = # of members awarded for honesty, $y$ = # of members awarded for for helping others and $z$ = members awarded for supervising the workers to keep the colony neat and clean .
By the given condition; the equations are:
$x+y+z = 12\; (1)$
$2x+3y+3z=33\;(2)$
$x+z=2y \;(3)$
We can express this in the form: $\begin{bmatrix} 1 & 1 & 1\\ 2 & 3 & 3 \\ 1 & -2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end {bmatrix}=\begin{bmatrix} 12 \\ 33 \\ 0 \end{bmatrix}$
This is of the form $Ax=B$
Evaluating $A = 1 (3 \times 1 - (-2) \times 3) - 1 (2 \times 1 - 1 \times 3) + 1 (2 \times -2 - 1 \times 3) = =9+1-7=3$
Therefore $A^{-1}$ exists.
$\textbf{Step 2}$:
$A_{11}=3+6=9$
$A_{12}=-2(2-3)=+1$
$A_{13}=-4-3=-7$
$A_{21}=-(1+2)=-3$
$A_{22}=(1-1)=0$
$A_{23}=-(-2-1)=3$
$A_{31}=(3-3)=0$
$A_{32}=-(3-2)=-1$
$A_{33}=(3-2)=1$
Therefore Adjoint of A=$\begin{bmatrix} 9 & -3 & 0\\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{bmatrix}$
Therefore $A^{-1}=\large \frac{1}{3}$$\begin{bmatrix} 9 & -3 & 0\\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{bmatrix}$
$\textbf{Step 3}$:
$AX=B\qquad => X=A^{-1}B$
$\begin{bmatrix} x \\ y \\ z \end {bmatrix} = \frac{1}{3} \begin{bmatrix} 9 & -3 & 0\\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{bmatrix}\begin{bmatrix} 12 \\ 33 \\ 0 \\ \end {bmatrix}=\begin{bmatrix} \frac{9}{3} \\ \frac{12}{3} \\ \frac{15}{3} \\ \end {bmatrix}$
Hence $\begin{bmatrix} x \\ y \\ z \end {bmatrix}=\begin{bmatrix} 3 \\ 4 \\ 5 \end {bmatrix}$
Therefore $x=3,y=4,z=5$
Apart from honesty, cooperaton and supervision, another value which can be added is punctuality among the workers |
# Common Core Standard 2.G.3 Lesson Plans
In 2.G.3, 2nd grade geometry item 3, the Common Core State Standards for Mathematics reads:
2.G.3 Partition circles and rectangles into two, three, or four equal shares, describe the shares using the words halves, thirds, half of, a third of, etc., and describe the whole as two halves, three thirds, four fourths. Recognize that equal shares of identical wholes need not have the same shape.
# Common Core Standard 6.RP.A.3 Lesson Plans
In (6.RP.A.3), the Common Core State Standards for Mathematics reads ‘Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.’
• CCSS.MATH.CONTENT.6.RP.A.3.A
Make tables of equivalent ratios relating quantities with whole-number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
• CCSS.MATH.CONTENT.6.RP.A.3.B
Solve unit rate problems including those involving unit pricing and constant speed.For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
• CCSS.MATH.CONTENT.6.RP.A.3.C
Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
• CCSS.MATH.CONTENT.6.RP.A.3.D
Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
# Common Core Standard 7.G.B.4 Lesson Plans
In (7.G.B.4), the Common Core State Standards for Mathematics reads ‘Know the formulas for the area and circumference of a circle and use them to solve problems; give an informal derivation of the relationship between the circumference and area of a circle.’
# Common Core Standard 1.G.3. Lesson Plans
1.G.3 (1st grade geometry item 3) in the Common Core Standards for Mathematics states:
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares.
# Common Core Standard 3.MD.4 Lesson Plans
In (3.MD.10), the Common Core State Standards for Mathematics reads ‘Draw a scaled picture graph and a scaled bar graph to represent a data set with several categories. Solve one- and two-step “how many more” and “how many less” problems using information presented in scaled bar graphs. For example, draw a bar graph in which each square in the bar graph might represent 5 pets.’
# Common Core Standard 2.MD.10 Lesson Plans
(2.MD.10), the Common Core State Standards for Mathematics reads ‘Draw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories. Solve simple put- together, take-apart, and compare problems using information presented in a bar graph.’ |
# 2014 AMC 12B Problems/Problem 11
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
A list of $11$ positive integers has a mean of $10$, a median of $9$, and a unique mode of $8$. What is the largest possible value of an integer in the list?
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 31\qquad\textbf{(D)}\ 33\qquad\textbf{(E)}\ 35$
## Solution
We start off with the fact that the median is $9$, so we must have $a, b, c, d, e, 9, f, g, h, i, j$, listed in ascending order. Note that the integers do not have to be distinct.
Since the mode is $8$, we have to have at least $2$ occurrences of $8$ in the list. If there are $2$ occurrences of $8$ in the list, we will have $a, b, c, 8, 8, 9, f, g, h, i, j$. In this case, since $8$ is the unique mode, the rest of the integers have to be distinct. So we minimize $a,b,c,f,g,h,i$ in order to maximize $j$. If we let the list be $1,2,3,8,8,9,10,11,12,13,j$, then $j = 11 \times 10 - (1+2+3+8+8+9+10+11+12+13) = 33$.
Next, consider the case where there are $3$ occurrences of $8$ in the list. Now, we can have two occurrences of another integer in the list. We try $1,1,8,8,8,9,9,10,10,11,j$. Following the same process as above, we get $j = 11 \times 10 - (1+1+8+8+8+9+9+10+10+11) = 35$. As this is the highest choice in the list, we know this is our answer. Therefore, the answer is $\boxed{\textbf{(E) }35}$ |
Thread: Determine the equation of the median from vertex
1. Determine the equation of the median from vertex
ABC has the following co-ordinates: A(3,7)B(-1,-6) and C(-5,3). Determine the equation of the median from vertex C.
This question is giving me a bit of trouble. If someone could please help me out i would greatly appreciate it.
2. Originally Posted by Scott9909
ABC has the following co-ordinates: A(3,7)B(-1,-6) and C(-5,3). Determine the equation of the median from vertex C.
This question is giving me a bit of trouble. If someone could please help me out i would greatly appreciate it.
Part the First, find the coordinates of midpoint: By definition, the median from vertex $\displaystyle C=(-5,3)$ is a line which joins the midpoint of side $\displaystyle AB$. But by the midpoint formula the midpoint of $\displaystyle AB$ is $\displaystyle (\frac{3-1}{2},\frac{7-6}{2})=(1,1/2)$. Thus, the median passes through points $\displaystyle C=(-5,3)$ and $\displaystyle (1,1/2)$.
Part the Second, find the equation of median: Using the slope-point formula which states that the equation of a line passing through point $\displaystyle (x_0,y_0)$ having slope $\displaystyle m$ is $\displaystyle y-y_0=m(x-x_0)$. Thus, the slope of $\displaystyle (1,1/2),(-5,3)$ is $\displaystyle m=-5/12$. Thus, the equation of line is (use any point for $\displaystyle (x_0,y_0)$)
$\displaystyle y+5=-5/12(x-3)$ Open and simplify,
$\displaystyle y=-\frac{5}{12}x-\frac{15}{4}$
Q.E.D.
3. Im a bit confused on part 2.
Do you have to find the slope of the line? Im not furmilur with the formula you put up. Ive been taught to do it Y=X2-X1/Y2-y1
and i dont seem to be getting the same slope.
4. That is exactly what I did $\displaystyle (y_2-y_1)/(x_2-x_1)$. You mean the formula for the equation of the line?
5. Im not exactly sure. I really dont understand math that well.
are you supposed to do y2-y/x2-x1 with your midpoint and C(-5,3)?
And also if it is what is considered y2 and x2? C or the midpoint.
Sorry if these questions are stupid.
6. Originally Posted by Scott9909
Im not exactly sure. I really dont understand math that well.
are you supposed to do y2-y/x2-x1 with your midpoint and C(-5,3)?
And also if it is what is considered y2 and x2? C or the midpoint.
Hello,
1. If you have 2 points $\displaystyle P_1,\ P_2$ with the coordinates $\displaystyle P_1(x_1,\ y_1),\ P_2(x_2,\ y_2)$ then you'll get the midpoint $\displaystyle M \left( \frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$
2. A line through 2 points is described completely by the following equation: $\displaystyle \frac{y-y_1}{x-x_1} = \frac{y_2 - y_1}{x_2 - x_2}$
Solve this equation for y and you'll get: $\displaystyle y = \frac{y_2 - y_1}{x_2 - x_2}\cdot (x-x_1) + y_1$ where $\displaystyle \frac{y_2 - y_1}{x_2 - x_2}$ is the slope of the line.
I hope that these additional remarks helped a little bit.
Greetings
EB
,
,
,
,
,
,
,
,
,
,
,
formula of medians when vertices are given
Click on a term to search for related topics. |
# Question: How Could You Use 25% Of A Number To Find The Number?
## How do you find 15% of a number?
15% is 10% + 5% (or 0.15 = 0.1 + 0.05, dividing each percent by 100).
Thinking about it this way is useful for two reasons.
First, it’s easy to multiply any number by 0.1; just move the decimal point left one digit.
For example, 75.00 x 0.1 = 7.50, or 346.43 x 0.1 = 34.64 (close enough)..
## How do I find the percentage of two numbers without a calculator?
If you need to find a percentage of a number, here’s what you do – for example, to find 35% of 240: Divide the number by 10 to find 10%. In this case, 10% is 24. Multiply this number by how many tens are in the percentage you’re looking for – in this case, that’s 3, so you work out 30% to be 24 x 3 = 72.
## How do you find what number a number is a percentage of?
Learning how to calculate the percentage of one number vs. another number is easy. If you want to know what percent A is of B, you simple divide A by B, then take that number and move the decimal place two spaces to the right. That’s your percentage!
## What number is 15% of 75?
11.25Percentage Calculator: What is 15 percent of 75? = 11.25.
## How do you find 30% of a number?
Once you have the decimal figure, multiply it by the number for which you seek to calculate the percentage; i.e., if you need to know 30 percent of 100, you convert 30 percent to a decimal (0.30) and multiply it by 100 (0.30 x 100, which equals 30).
## How can you find 75% of any number?
Calculate the percent value:75 ÷ 100 =0.75 =0.75 × 100/100 =75/100 =
## How do I calculate a percentage between two numbers?
First: work out the difference (increase) between the two numbers you are comparing. Then: divide the increase by the original number and multiply the answer by 100.
## What is a percentage of a number?
In mathematics, a percentage is a number or ratio that represents a fraction of 100. It is often denoted by the symbol “%” or simply as “percent” or “pct.” For example, 35% is equivalent to the decimal 0.35, or the fraction. 35.
## How do I calculate mean?
The mean is the average of the numbers. It is easy to calculate: add up all the numbers, then divide by how many numbers there are. In other words it is the sum divided by the count.
## How do you calculate percentages quickly?
To figure out the decimal form of a percent, simply move the decimal two places to the left. For example, the decimal form of 10 percent is 0.1. Then, to calculate what 10 percent of is, say, 250 students, simply multiply the number of students by 0.1.
## What is the formula for calculating percentage?
To calculate the percentage, multiply this fraction by 100 and add a percent sign. 100 * numerator / denominator = percentage . In our example it’s 100 * 2/5 = 100 * 0.4 = 40 . Forty percent of the group are girls.
## What number is 15% of 100?
0.1515 percent of 100? = 0.15.
## How do you find 25 percent of a number on a calculator?
3. How to find X if P percent of it is Y. Use the percentage formula Y/P% = XConvert the problem to an equation using the percentage formula: Y/P% = X.Y is 25, P% is 20, so the equation is 25/20% = X.Convert the percentage to a decimal by dividing by 100.Converting 20% to a decimal: 20/100 = 0.20.More items…
## What number of 15% is 80?
1212 is 15% of 80 . |
# Theoretical Yield Worked Problem
Amount of Reactant Needed to Produce a Product
This example problem demonstrates how to calculate the amount of reactant needed to produce a product.
## Problem
Aspirin is prepared from the reaction of salicylic acid (C7H6O3) and acetic anhydride (C4H6O3) to produce aspirin (C9H8O4) and acetic acid (HC2H3O2). The formula for this reaction is
C7H6O3 + C4H6O3 → C9H8O4 + HC2H3O2
How many grams of salicylic acid are needed to make 1,000 1-gram tablets of aspirin? (Assume 100 percent yield.)
## Solution
Step 1: Find the molar mass of aspirin and salicylic acid.
From the periodic table:
Molar Mass of C = 12 grams
Molar Mass of H = 1 grams
Molar Mass of O = 16 grams
MMaspirin = (9 x 12 grams) + (8 x 1 grams) + (4 x 16 grams)
MMaspirin = 108 grams + 8 grams + 64 grams
MMaspirin = 180 grams
MMsal = (7 x 12 grams) + (6 x 1 grams) + (3 x 16 grams)
MMsal = 84 grams + 6 grams + 48 grams
MMsal = 138 grams
Step 2: Find the mole ratio between aspirin and salicylic acid.
For every mole of aspirin produced, 1 mole of salicylic acid was needed. Therefore the mole ratio between the two is one.
Step 3: Find the grams of salicylic acid needed.
The path to solving this problem starts with the number of tablets. Combining this with the number of grams per tablet will give the number of grams of aspirin. Using the molar mass of aspirin, you get the number of moles of aspirin produced. Use this number and the mole ratio to find the number of moles of salicylic acid needed. Use the molar mass of salicylic acid to find the grams needed.
Putting all this together:
grams salicylic acid = 1,000 tablets x 1 g aspirin/1 tablet x 1 mol aspirin/180 g of aspirin x 1 mol sal/1 mol aspirin x 138 g of sal/1 mol sal
grams salicylic acid = 766.67 |
# 11.7 Archimedes’ principle (Page 4/9)
Page 4 / 9
## Specific gravity
Specific gravity is the ratio of the density of an object to a fluid (usually water).
## Calculating average density: floating woman
Suppose a 60.0-kg woman floats in freshwater with $97.0%$ of her volume submerged when her lungs are full of air. What is her average density?
Strategy
We can find the woman’s density by solving the equation
$\text{fraction submerged}=\frac{{\overline{\rho }}_{\text{obj}}}{{\rho }_{\text{fl}}}$
for the density of the object. This yields
${\overline{\rho }}_{\text{obj}}={\overline{\rho }}_{\text{person}}=\left(\text{fraction submerged}\right)\cdot {\rho }_{\text{fl}}.$
We know both the fraction submerged and the density of water, and so we can calculate the woman’s density.
Solution
Entering the known values into the expression for her density, we obtain
${\overline{\rho }}_{\text{person}}=0\text{.}\text{970}\cdot \left({\text{10}}^{3}\frac{\text{kg}}{{\text{m}}^{3}}\right)=\text{970}\frac{\text{kg}}{{\text{m}}^{3}}.$
Discussion
Her density is less than the fluid density. We expect this because she floats. Body density is one indicator of a person’s percent body fat, of interest in medical diagnostics and athletic training. (See [link] .)
There are many obvious examples of lower-density objects or substances floating in higher-density fluids—oil on water, a hot-air balloon, a bit of cork in wine, an iceberg, and hot wax in a “lava lamp,” to name a few. Less obvious examples include lava rising in a volcano and mountain ranges floating on the higher-density crust and mantle beneath them. Even seemingly solid Earth has fluid characteristics.
## More density measurements
One of the most common techniques for determining density is shown in [link] .
An object, here a coin, is weighed in air and then weighed again while submerged in a liquid. The density of the coin, an indication of its authenticity, can be calculated if the fluid density is known. This same technique can also be used to determine the density of the fluid if the density of the coin is known. All of these calculations are based on Archimedes’ principle.
Archimedes’ principle states that the buoyant force on the object equals the weight of the fluid displaced. This, in turn, means that the object appears to weigh less when submerged; we call this measurement the object’s apparent weight . The object suffers an apparent weight loss equal to the weight of the fluid displaced. Alternatively, on balances that measure mass, the object suffers an apparent mass loss equal to the mass of fluid displaced. That is
what is Andromeda
what is velocity
displacement per unit time
Murlidhar
the ratec of displacement over time
Jamie
the rate of displacement over time
Jamie
the rate of displacement over time
Jamie
did you need it right now
Pathani
up to tomorrow
Santosh
i need a description and derivation of kinetic theory of gas
Santosh
pls the sum of change in kinetic and potential energy is always what ?
Faith
i need a description and derivation of kinetic theory of gas
Santosh
did you need it right now
Pathani
A few grains of table salt were put in a cup of cold water kept at constant temperature and left undisturbed. eventually all the water tasted salty. this is due to?
Aunt Faith,please i am thinking the dissolution here from the word "solution" exposed the grains of salt to be dissolved in the water.Thankyou
Junior
Junior
Aunt Faith,please i am thinking the dissolution here from the word "solution" exposed the grains of salt to be dissolved in the water.Thankyou
Junior
it is either diffusion or osmosis. just confused
Faith
due to solvation....
Pathani
what is solvation pls
Faith
water molecule surround the salt molecules . solute solute attraction break in the same manner solvent solvent interaction also break. as a result solute and solvent attraction took place.
Pathani
okay thanks
Faith
my pleasure
Pathani
what is solvation pls
Faith
water act as a solvent and salt act as solute
Pathani
okay thanks
Faith
its ok
Pathani
due to solvation....
Pathani
water molecule surround the salt molecules . solute solute attraction break in the same manner solvent solvent interaction also break. as a result solute and solvent attraction took place.
Pathani
what is magnetism
physical phenomena arising from force caused by magnets
is the phenomenon of attracting magnetic substance like iron, cobalt etc.
Faith
what is heat
Heat is a form of energy where molecules move
saran
topic-- question
Salman
I know this is unrelated to physics, but how do I get the MCQs and essay to work. they arent clickable.
20cm3 of 1mol/dm3 solution of a monobasic acid HA and 20cm3 of 1mol/dm3 solution of NaOH are mixed in a calorimeter and a temperature rise of 274K is observed. If the heat capacity of the calorimeter is 160J/K, calculate the enthalpy of neutralization of the acid.(SHCw=4.2J/g/K) Formula. (ms*cs+C)*T
why is a body moving at a constant speed able to accelerate
20cm3 of 1mol/dm3 solution of a monobasic acid HA and 20cm3 of 1mol/dm3 solution of NaOH are mixed in a calorimeter and a temperature rise of 274K is observed. If the heat capacity of the calorimeter is 160J/K, calculate the enthalpy of neutralization of the acid.(SHCw=4.2J/g/K) Formula. (ms*cs+C)*T
Lilian
because it changes only direction and the speed is kept constant
Justice
Why is the sky blue...?
It's filtered light from the 2 forms of radiation emitted from the sun. It's mainly filtered UV rays. There's a theory titled Scatter Theory that covers this topic
Mike
A heating coil of resistance 30π is connected to a 240v supply for 5min to boil a quantity of water in a vessel of heat capacity 200jk. If the initial temperature of water is 20°c and it specific heat capacity is 4200jkgk calculate the mass of water in a vessel
A thin equi convex lens is placed on a horizontal plane mirror and a pin held 20 cm vertically above the lens concise in position with its own image the space between the undersurface of d lens and the mirror is filled with water (refractive index =1•33)and then to concise with d image d pin has to
Be raised until its distance from d lens is 27cm find d radius of curvature
Azummiri
what happens when a nuclear bomb and atom bomb bomb explode add the same time near each other
A monkey throws a coconut straight upwards from a coconut tree with a velocity of 10 ms-1. The coconut tree is 30 m high. Calculate the maximum height of the coconut from the top of the coconut tree? Can someone answer my question
v2 =u2 - 2gh 02 =10x10 - 2x9.8xh h = 100 ÷ 19.6 answer = 30 - h.
Ramonyai
why is the north side is always referring to n side of magnetic |
# McGraw Hill My Math Grade 1 Chapter 1 Lesson 7 Answer Key Ways to Make 4 and 5
All the solutions provided in McGraw Hill My Math Grade 1 Answer Key PDF Chapter 1 Lesson 7 Ways to Make 4 and 5 will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 1 Answer Key Chapter 1 Lesson 7 Ways to Make 4 and 5
Explore and Explain
Teacher Directions: Use to model. Mia’s dad put 1 piece of wood on the fire. Then he put 3 more pieces on it. How many pieces of wood are on the fire in all? Trace the counters you used. Write the addition number sentence.
Explanation:
The sum of 1 and 3 is 4
1 + 3 = 4
See and Show
There are many ways to make a sum of 4 and 5.
Use Work Mat 3 and to show different ways to make a sum of 4. Color the Write the numbers.
Ways to Make 4
Question 1.
Explanation:
The sum of 2 and 2 is 4
2 + 2 = 4
Question 2.
Explanation:
The sum of 3 and 1 is 4
3 + 1 = 4
Question 3.
Explanation:
The sum of 0 and 1 is 1
0 + 1 = 1
Talk Math What is another way to make 4?
On My Own
Use Work Mat 3 and to show different ways to make a sum of 5. Color the Write the numbers.
Question 4.
Explanation:
The sum of 1 and 3 is 4
1 + 3 = 4
Question 5.
Explanation:
The sum of 4 and 0 is 4
4 + 0 = 4
Question 6.
Explanation:
The sum of 0 and 4 is 4
0 + 4 = 4
Question 7.
Explanation:
The sum of 2 and 2 is 4
2 + 2 = 4
Question 8.
1 + 4 = ____
5
Explanation:
The sum of 1 and 4 is 5
1 + 4 = 5
Question 9.
3 + 1 = ____
4
Explanation:
The sum of 3 and 1 is 4
3 + 1 = 4
Question 10.
3 + 2 = ____
5
Explanation:
The sum of 3 and 2 is 5
3 + 2 = 5
Question 11.
2 + 2 = ____
4
Explanation:
The sum of 2 and 2 is 4
2 + 2 = 4
Question 12.
Explanation:
The sum of 1 and 4 is 5
1 + 4 = 5
Question 13.
Explanation:
The sum of 4 and 0 is 4
4 + 0 = 4
if we add 0 to a number the number remains the answer
Question 14.
Explanation:
The sum of 0 and 5 is 5
0 + 5 = 5
if we add 0 to a number the number remains the answer
Problem Solving
Question 15.
Kylie has 3 maps. Taye has 2 maps. How many maps do they have in all?
_____ + _____ = _____ maps
3 + 2 = maps
Explanation:
Kylie has 3 maps.
Taye has 2 maps.
3 + 2 = 5
5 maps that they have in all
Question 16.
Kate saw 1 turkey. Malik saw 3 other turkeys. How many turkeys did they see in all?
_____ + ____ = _____ turkeys
1 + 3 = 4 turkeys
Explanation:
Kate saw 1 turkey.
Malik saw 3 other turkeys.
1 + 3 = 4
4 turkeys that they see in all
Write Math Is there more than one way to make 5? Explain.
_______________________________
_______________________________
_______________________________
Explanation:
yes, there are more possible ways to make 5
1 + 4 = 5
2 + 3 = 5
3 + 2 = 5
4 + 1 = 5
0 + 5 = 5
5 + 0 = 5
### McGraw Hill My Math Grade 1 Chapter 1 Lesson 7 My Homework Answer Key
Practice
Write different ways to make 4.
Question 1.
_____ + _____ = 4
3 + 1 = 4
Explanation:
The sum of 3 and 1 is 4
3 + 1 = 4
Question 2.
_____ + _____ = 4
2 + 2 = 4
Explanation:
The sum of 2 and 2 is 4
2 + 2 = 4
Question 3.
_____ + _____ = 4
1 + 3 = 4
Explanation:
The sum of 1 and 3 is 4
1 + 3 = 4
Question 4.
_____ + _____ = 4
4 + 0 = 4
Explanation:
The sum of 4 and 0 is 4
4 + 0 = 4
if we add 0 to a number the number remains the answer
Write different ways to make 5.
Question 5.
_____ + _____ = 5
4 + 1 = 5
Explanation:
The sum of 4 and 1 is 5
4 + 1 = 5
Question 6.
_____ + _____ = 5
3 + 2 = 5
Explanation:
The sum of 3 and 2 is 5
3 + 2 = 5
Question 7.
_____ + _____ = 5
2 + 3 = 5
Explanation:
The sum of 2 and 3 is 5
2 + 3 = 5
Question 8.
_____ + _____ = 5
1 + 4 = 5
Explanation:
The sum of 1 and 4 is 5
1 + 4 = 5
Question 9.
_____ + _____ = 5
0 + 5 = 5
Explanation:
The sum of 0 and 5 is 5
if we add 0 to a number the number remains the answer
Question 10.
Jose saw 3 green frogs and I red frog. How many frogs did he see in all?
______ frogs
4 frogs
Explanation:
Jose saw 3 green frogs and I red frog.
3 + 1 = 4
4 frogs that Jose see in all.
Test Practice
Question 11.
How many rainbows are there in all?
(A) 3
(B) 4
(C) 5
(D) 6 |
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Select Page
A decimal is a numerical representation that includes a whole number part and a fractional part separated by a decimal point. The fractional part consists of digits representing less than one, such as tenths, hundredths, and thousandths.
Decimals are used to express quantities that are not whole numbers, showing precise measurements and quantities between whole number values.
Personalized Online Tutoring
## Adding Decimals with Different Decimal Places
Like decimals have an equal number of digits after the decimal.
For Example, 0.15 and 2.25 are like decimals, whereas,
Unlike decimals, they have different numbers of digits after the decimal.
For Example, 7.31 and 5.825 are unlike decimals.
Here’s a step-by-step guide on how to add like decimals:
Step 1: Align the Decimal Points
Place the numbers vertically, aligning the decimal points.
4.37
+ 2.19
——-
Step 2: Add the Digits in the Ones Place
Start by adding the digits in the rightmost column (the ones place). 7 + 9 = 16. Write down 6 and carry over 1.
4.37
+ 2.19
——-
6
Step 3: Move Left and Add the Tenths Place
Move one column to the left (the tenths place) and add the digits in that column, including the carryover from the previous step. 1 (carryover) + 3 + 1 = 5.
4.37
+ 2.19
——-
56
Step 4: Move Left and Add the Hundredths Place
Move one more column to the left (the hundredths place) and add the digits. 4 + 2 = 7. Make sure to include the decimal point in the correct position based on the original numbers.
4.37
+ 2.19
——-
6.56
Step 1: Align the Decimal Points
Place the numbers you want to add vertically, aligning their decimal points. Convert unlike decimals to like decimals by adding the required number of zeros.
Example:
12.345
+ 3.560
———
Start by adding the digits in the rightmost column (the ones place). Write down the sum below the line.
Example:
12.345
+ 3.560
———
5
Step 3: Move to the Next Column
Move one column to the left (the tenths place) and add the digits in that column, including the carry from the previous step if there are any.
Example:
12.345
+ 3.560
———
05
Repeat this process for each successive column, adding the digits and any carryover from the previous step and placing the decimal point accordingly.
Example:
12.345
+ 3.560
———
15.905
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## Adding Decimals with Whole Numbers
To add a decimal with a whole number, convert the whole number into a decimal number. A whole number can be changed to a decimal by adding a decimal point after the whole number and writing the required number of zeros after the decimal point to make both addends of the same length.
Example: Add 3 + 4.93 + 5
Step 1: Convert whole numbers to decimals and make all decimals like decimals.
The highest decimal place is 2, so let’s rewrite all numbers as decimal numbers with 2 decimal places to make them like decimals.
3 = 3.00
4.93
5 = 5.00
Step 2: Now, place the numbers to add vertically, aligning their decimal points.
3.00
4.93
+5.00
———
12.93
## FAQs
#### What is the first step in adding decimals?
The first step is to align the decimal points of the numbers you want to add. It ensures that you’re adding digits that have the same place value.
#### How do I align decimals when adding?
To add decimals, align the numbers vertically by placing the decimal points directly above each other. If one number has fewer decimal places, add zeros to the end of that number to make the decimal places equal.
#### Can I add decimals and whole numbers together?
Yes, you can add decimals and whole numbers together. Treat the whole number as a decimal by adding a decimal point at the end of it, and then follow the same steps for adding decimals.
#### How do I write the final answer with the correct decimal point?
The final answer should have a decimal point placed in the same position as the original numbers. Make sure to count the same number of decimal places from the right as the original numbers.
Gloria Mathew writes on math topics for K-12. A trained writer and communicator, she makes math accessible and understandable to students at all levels. Her ability to explain complex math concepts with easy to understand examples helps students master math. LinkedIn
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# LCM of 9 and 16
LCM of 9 and 16 is 144. The smallest number among all common multiples of 9 and 16 is 144. We can determine the LCM of 9 and 16 with the help of prime factorisation, division method and listing multiples. (9, 18, 27, 36, 45, 54, 63, ….) and (16, 32, 48, 64, 80, 96….) are the first few multiples of 9 and 16. Students can make use of the article HCF and LCM to learn tricks of finding LCM and HCF more effectively. Let us grasp the technique of finding the least common multiple of 9 and 16 with examples and FAQs in a comprehensive manner in this article.
## What is LCM of 9 and 16?
The Least Common Multiple of 9 and 16 is 144.
## How to Find LCM of 9 and 16?
The following methods help us to find the LCM of 9 and 16:
• Prime Factorisation
• Division method
• Listing the multiples
### LCM of 9 and 16 Using Prime Factorisation Method
By prime factorisation method, we can write 9 and 16 as the product of prime numbers, such that;
9 = 3 × 3
16 = 2 × 2 × 2 × 2
LCM (9, 16) = 2 × 2 × 2 × 2 × 3 × 3 = 144
### LCM of 9 and 16 Using Division Method
Here, we divide the numbers 9 and 16 by their prime factors to find their LCM. The product of these divisors represents the least common multiple of 9 and 16.
2 9 16 2 9 8 2 9 4 2 9 2 3 9 1 3 3 1 x 1 1
No more further division can be done.
Therefore, LCM (9, 16) = 2 × 2 × 2 × 2 × 3 × 3 = 144
### LCM of 9 and 16 Using Listing the Multiples
To determine the least common multiple of 9 and 16 using listing multiples, we list out the multiples of 9 and 16 as shown below.
Multiples of 9: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, ……., 126, 135, 144, ……..
Multiples of 16: 16, 32, 48, 64, 80, 96, 112, 128, 144, ……..
LCM (9, 16) = 144
## Related Articles
Least Common Multiple (LCM)
LCM of Two Numbers
Prime Factorization and Division Method for LCM and HCF
LCM Formula
LCM with Examples
LCM Calculator
## Solved Examples
Q. 1: Find the LCM of 9 and 16 using listing multiples?
Solution: First, list the multiples of 9 and 16
Multiples of 9 = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, 117, 126, 135, 144, 153, …..
Multiples of 16 = 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, ……..
Here, the LCM is 144
Hence the LCM of 9 and 16 is 144.
Q. 2: The product of two numbers is 144. If their GCD is 1, calculate their LCM.
Solution: Given
GCD = 1
Product of numbers = 144
We know that,
LCM × GCD = Product of numbers
LCM = Product of numbers/GCD
LCM = 144/1
LCM = 144
Thus the LCM is 144.
## Frequently Asked Questions on LCM of 9 and 16
### What is the LCM of 9 and 16?
The LCM of 9 and 16 is 144.
### How do you find the LCM of 9 and 16 by Prime Factorisation?
9 = 3 × 3
16 = 2 × 2 × 2 × 2
LCM (9, 16) = 2 × 2 × 2 × 2 × 3 × 3 = 144
Hence the LCM of 9 and 16 is 144.
### Name the methods used to determine the LCM of 9 and 16.
The following methods are used to determine the LCM of 9 and 16
Prime Factorisation Method
Division Method
Listing the Multiples |
# PROBLEMS ON DIAGONALOF RHOMUS
Diagonals in rhombus :
Diagonals will bisect each other at right angles.
Area of rhombus with diagonals :
(1/2) x d1 x d2
Note :
The diagonals will not be equal.
Problem 1 :
Find the side length of the rhombus ABCD.
Solution :
Let E be the point of intersection of two diagonals. Considering the triangle DEC,
DE = 20 ft, EC = 30 ft
DC2 = DE2 + EC2
DC2 = 202 + 302
DC2 = 400 + 900
DC2 = 1300
DC = 10√13
Problem 2 :
The area of a rhombus is 150 cm2. The length of one of its diagonal is 10 cm. The length of their other diagonal is .
(a) 25 cm (b) 30 cm (c) 35 cm (d) 40 cm
Solution :
Area of rhombus = (1/2) x d1 x d2
(1/2) x 10 x d2 = 150
d2 = 150 / 5
d2 = 30 cm
Problem 3 :
One of the diagonals of a rhombus is double the other diagonal. Its area is 25 sq.cm. The sum of the diagonal is
(a) 10 cm (b) 12 cm (c) 15 cm (d) 16 cm
Solution :
d1 = 2d2
(1/2) x d1 x d2 = 25
(1/2) x 2d2 x d2 = 25
d22 = 25
d2 = 5
then d1 = 2(5) ==> 10 cm
Sum of the diagonals = 10 + 5 ==> 15 cm
Problem 4 :
If the diagonals of a rhombus are 24 cm and 10 cm, find the area and perimeter of rhombus.
Solution :
Area of rhombus = (1/2) x d1 x d2
d1 = 24 cm and d2 = 10 cm
= (1/2) x 24 x 10
= 120 square cm
To find perimeter of rhombus, we should find the side length of rhombus.
side2 = 122 + 52
side2 = 144 + 25
side2 = 169
√169
= 13
Perimeter of rhombus = 4(13)
= 52 cm
Problem 5 :
Each side of a rhombus is 26 cm and one of its diagonal is 48 cm long. The area of the rhombus is.
(a) 240 cm2 (b) 300 cm2 (c) 360 cm2 (d) 480 cm2
Solution :
The diagonals will bisect each other at right angle.
Half length of given diagonal = 24
half length of another diagonal = x
262 = 242 + x2
676 = 576 + x2
x2 = 676 - 576
x2 = 100
x = 10
Length of another diagonal = 20
Area of rhombus = (1/2) x 48 x 20
= 480 square cm
Problem 6 :
The length one diagonal of a rhombus is 80% of the other diagonal. The area of the rhombus is how many times the square of length of the other diagonal ?
(a) 4/5 (b) 2/5 (c) 3/4 (d) 1/4
Solution :
Let d1 and d2 be length of diagonals.
d1 = 80% of d==> 0.80d2
Area of rhombus = (1/2) x 0.80dx d2
= (1/2) x 0.80(d2)2
= 0.40(d2)2
= (40/100)(d2)2
= 2/5(d2)2
## Recent Articles
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2. ### Positive and Negative Numbers Connecting in Real Life Worksheet
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Positive and Negative Numbers Connecting in Real Life Worksheet |
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#### Resources tagged with Making and proving conjectures similar to Walkabout:
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Broad Topics > Using, Applying and Reasoning about Mathematics > Making and proving conjectures
### Close to Triangular
##### Stage: 4 Challenge Level:
Drawing a triangle is not always as easy as you might think!
### Janine's Conjecture
##### Stage: 4 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
##### Stage: 4 Challenge Level:
The points P, Q, R and S are the midpoints of the edges of a non-convex quadrilateral.What do you notice about the quadrilateral PQRS and its area?
### To Prove or Not to Prove
##### Stage: 4 and 5
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
##### Stage: 4 Challenge Level:
Explore the relationship between quadratic functions and their graphs.
##### Stage: 4 Challenge Level:
Points D, E and F are on the the sides of triangle ABC. Circumcircles are drawn to the triangles ADE, BEF and CFD respectively. What do you notice about these three circumcircles?
### DOTS Division
##### Stage: 4 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### Dice, Routes and Pathways
##### Stage: 1, 2 and 3
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### Tri-split
##### Stage: 4 Challenge Level:
A point P is selected anywhere inside an equilateral triangle. What can you say about the sum of the perpendicular distances from P to the sides of the triangle? Can you prove your conjecture?
### Curvy Areas
##### Stage: 4 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
### Multiplication Arithmagons
##### Stage: 4 Challenge Level:
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
### Pericut
##### Stage: 4 and 5 Challenge Level:
Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?
### How Old Am I?
##### Stage: 4 Challenge Level:
In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
### On the Importance of Pedantry
##### Stage: 3, 4 and 5
A introduction to how patterns can be deceiving, and what is and is not a proof.
### Center Path
##### Stage: 3 and 4 Challenge Level:
Four rods of equal length are hinged at their endpoints to form a rhombus. The diagonals meet at X. One edge is fixed, the opposite edge is allowed to move in the plane. Describe the locus of. . . .
### What's Possible?
##### Stage: 4 Challenge Level:
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
### Always a Multiple?
##### Stage: 3 Challenge Level:
Think of a two digit number, reverse the digits, and add the numbers together. Something special happens...
### Alison's Mapping
##### Stage: 4 Challenge Level:
Alison has created two mappings. Can you figure out what they do? What questions do they prompt you to ask?
### Few and Far Between?
##### Stage: 4 and 5 Challenge Level:
Can you find some Pythagorean Triples where the two smaller numbers differ by 1?
### Charlie's Mapping
##### Stage: 3 Challenge Level:
Charlie has created a mapping. Can you figure out what it does? What questions does it prompt you to ask?
### A Little Light Thinking
##### Stage: 4 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
### Triangles Within Squares
##### Stage: 4 Challenge Level:
Can you find a rule which relates triangular numbers to square numbers?
### Polycircles
##### Stage: 4 Challenge Level:
Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?
### Pentagon
##### Stage: 4 Challenge Level:
Find the vertices of a pentagon given the midpoints of its sides.
### Rotating Triangle
##### Stage: 3 and 4 Challenge Level:
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
##### Stage: 4 Challenge Level:
The points P, Q, R and S are the midpoints of the edges of a convex quadrilateral. What do you notice about the quadrilateral PQRS as the convex quadrilateral changes?
### Helen's Conjecture
##### Stage: 3 Challenge Level:
Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Happy Numbers
##### Stage: 3 Challenge Level:
Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general.
### Loopy
##### Stage: 4 Challenge Level:
Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?
### Multiplication Square
##### Stage: 3 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Triangles Within Pentagons
##### Stage: 4 Challenge Level:
Show that all pentagonal numbers are one third of a triangular number.
### Triangles Within Triangles
##### Stage: 4 Challenge Level:
Can you find a rule which connects consecutive triangular numbers?
### Consecutive Negative Numbers
##### Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Exploring Simple Mappings
##### Stage: 3 Challenge Level:
Explore the relationship between simple linear functions and their graphs.
### Epidemic Modelling
##### Stage: 4 and 5 Challenge Level:
Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths. |
# 6th Grade Math: 1-4 HW.
## Presentation on theme: "6th Grade Math: 1-4 HW."— Presentation transcript:
6th Grade Math: 1-4 HW
Course 1 1-5 Mental Math 6th Grade Math: 1-5 HW Page 26 #13-24 all
1-5 Mental Math Warm Up Problem of the Day Lesson Presentation
Course 1 Warm Up Problem of the Day Lesson Presentation
1-5 Mental Math Warm Up Find each sum or product. 1. 17 + 15
Course 1 1-5 Mental Math Warm Up Find each sum or product. 3. 8(24) 4. 7(12) 5. 3(91) 6. 6(15) 32 68 192 84 273 90
1-5 Mental Math Problem of the Day
Course 1 1-5 Mental Math Problem of the Day Determine the secret number from the following clues: The number is a multiple of 5. It is divisible by 3. It is less than 200. Its tens digit equals the sum of its other two digits. 165
Learn to use number properties to compute mentally.
Course 1 1-5 Mental Math Today’s Learning Goal Learn to use number properties to compute mentally.
Insert Lesson Title Here
Course 1 1-5 Mental Math Insert Lesson Title Here Vocabulary Commutative Property Associative Property Distributive Property
Course 1 1-5 Mental Math Mental math means “doing math in your head.”
COMMUTATIVE PROPERTY (Ordering)
Course 1 1-5 Mental Math COMMUTATIVE PROPERTY (Ordering) Words Numbers You can add or multiply numbers in any order. = 15 2 = 2 15
ASSOCIATIVE PROPERTY (Grouping)
Course 1 1-5 Mental Math ASSOCIATIVE PROPERTY (Grouping) Words Numbers When you are only adding or only multiplying, you can group any of the numbers together. (17 + 2) + 9 = 17 + (2 + 9) (12 2) 4 = 12 (2 4)
Course 1 1-5 Mental Math Additional Example 1A: Using Properties to Add and Multiply Whole Numbers A. Evaluate Look for sums that are multiples of 10 Use the Commutative Property. Use the Associative Property to make groups of compatible numbers. (17 + 3) + (5 + 15) Use mental math to add. 40
Course 1 1-5 Mental Math Additional Example 1B: Using Properties to Add and Multiply Whole Numbers B. Evaluate 4 13 5. Look for products that are multiples of 10 4 13 5 13 4 5 Use the Commutative Property. Use the Associative Property to group compatible numbers. 13 (4 5) 13 20 Use mental math to multiply. 260
1-5 Mental Math Try This: Example 1A A. Evaluate 12 + 5 + 8 + 5.
Course 1 1-5 Mental Math Try This: Example 1A A. Evaluate Look for sums that are multiples of 10 Use the Commutative Property. Use the Associative Property to make groups of compatible numbers. (12 + 8) + (5 + 5) Use mental math to add. 30
1-5 Mental Math Try This: Example 1B B. Evaluate 8 3 5.
Course 1 1-5 Mental Math Try This: Example 1B B. Evaluate 8 3 5. Look for products that are multiples of 10 8 3 5 3 8 5 Use the Commutative Property. Use the Associative Property to group compatible numbers. 3 (8 5) 3 40 Use mental math to multiply. 120
DISTRIBUTIVE PROPERTY
Course 1 1-5 Mental Math DISTRIBUTIVE PROPERTY Words Numbers When you multiply a number times a sum, you can find the sum first and then multiply, or multiply by each number in the sum and then add. 6 (10 + 4) = 6 14 = 84 6 (10 + 4) = (6 10) + (6 4) = =
Course 1 1-5 Mental Math When you multiply two numbers, you can “break apart” one of the numbers into a sum and then use the Distributive Property. Break the greater factor into a sum that contains a multiple of 10 and a one-digit number. You can add and multiply these numbers mentally. Helpful Hint
Additional Example 2A: Using the Distributive Property to Multiply
Course 1 1-5 Mental Math Additional Example 2A: Using the Distributive Property to Multiply Use the Distributive Property to find the product. A. Evaluate 6 35. “Break apart” 35 into 6 35 = 6 (30 + 5) Use the Distributive Property. = (6 30) + (6 5) Use mental math to multiply. = = Use mental math to add.
Additional Example 2B: Using the Distributive Property to Multiply
Course 1 1-5 Mental Math Additional Example 2B: Using the Distributive Property to Multiply Use the Distributive Property to find the product. B. Evaluate 9 x 87. “Break apart” 87 into 9 87 = 9 (80 + 7) Use the Distributive Property. = (9 80) + (9 7) Use mental math to multiply. = = Use mental math to add.
1-5 Mental Math Try This: Example 2A
Course 1 1-5 Mental Math Try This: Example 2A Use the Distributive Property to find the product. A. Evaluate 4 x 27. “Break apart” 27 into 4 27 = 4 (20 + 7) Use the Distributive Property. = (4 20) + (4 7) Use mental math to multiply. = = Use mental math to add.
1-5 Mental Math Try This: Example 2B
Course 1 1-5 Mental Math Try This: Example 2B Use the Distributive Property to find the product. B. Evaluate 6 43. “Break apart” 43 into 6 43 = 6 (40 + 3) Use the Distributive Property. = (6 40) + (6 3) Use mental math to multiply. = = Use mental math to add.
Insert Lesson Title Here
Course 1 1-5 Mental Math Insert Lesson Title Here Lesson Quiz Evaluate. 5 3 Use the Distributive Property to find each product. 4. 8 15 6. Angie wants to buy 3 new video games. How much will she need to save if each game costs \$27? 50 150 70 96 90 \$81
Weekly Review! Thank you for all of your hard work this week! I am so proud to be your teacher! Mrs. Burgess |
# My number has 2 hundreds and 1 thousand. It has 4 ones and 1 more ten block's than ones blocks
Answer: 1254 i believe because 1000 is 1 thousand, 2 hundreds, 1 more tens block than ones block is 5 because there are 4 ones blocks
The question involves identifying a number based on given information about its digits and blocks. To solve this problem, we use place value to determine the number.
### Explanation:
The subject of this question is Mathematics. The question involves identifying a number based on given information about its digits and blocks. To solve this problem, we can break down the information and use place value to determine the number.
The number has 2 hundreds, which means it has a value of 200. It also has 1 thousand, which is equal to 1000. Adding those two values, we get 200 + 1000 = 1200.
Next, the number has 4 ones and 1 more ten block than ones blocks. Since there are 4 ones, this value is simply 4. For the ten blocks, we add 1 to 4 to get 5.
Combining the values, we can write the number as 1200 + 50 = 1250.
So, the number described in this question is 1250.
brainly.com/question/35447631
#SPJ2
## Related Questions
Given: PKHE - inscribed, m∠E=120°, m∠EPK=51°, PF ∥ EH .
Find: m∠KPF, m∠H.
C is the answer they are equal.
Step-by-step explanation:
the order pairs(1,1),(2,8),(3,27),(4,64),(5,125) represent a function what is the rule that represent
if the first number is x and the second number is y
1^3 = 1
2^3 =8
3^3=27
4^3 = 64
y=x^3
f(x)=x^3
Step-by-step explanation:
Jill earns \$15 per hour babysitting plus a transportation fee of \$5 per job. Write a formula for E, Jill's earnings per babysitting job, in terms of h, the number of hours for a job. Then solve your formula for h.
Jill earns \$15 per hour babysitting plus a transportation fee of \$5 per job. Write a formula for E, Jill's earnings per babysitting job, in terms of h, the number of hours for a job
E is the Jill's earnings
h is the number of hours for a job
\$15 is the constant rate per hour so we use linear equation
y= mx+b, Instead of x and y we use h and E
E = mh + b
m is the slope (constant rate \$15)
b is the transportation fee \$5
So equation becomes E = 15h + 5
Now we solve for h
E = 15h + 5 (subract 5 on both sides)
E - 5 = 15h (divide by 15 )
We solved for h ,
since you know the amount of what she earns you would use the number 15 and you dont know how long she worked you would put "x" next to the 15. so you would have 15x and then put a plus 5 to show her transportation fee. so you would have the equation of 15x+5:)
How to simplify 28/168
Clueless
28/168
14/84
7/42
1/6
i usually divide them by 2 till i get a odd number
7 goes into 42 6 times so its 1/6
Write a word phrase for each algebraic expression 25-0.6x
The word phrase for each algebraic expression 25-0.6x is twenty five minus zero point six. multiply x ( the unknown numerical value).
### What is word phrase?
A word phrase is a collection of two or more words that together communicate one thought, but do not constitute a whole sentence.
The word phrase of "25" is twenty five.
The word phrase of "-"is minus.
The word phrase of "0.6" is zero point six.
And x is any unknown numerical number.
0.6x means the two numbers are multiplied.
So, the word phrase of "25-0.6x" is twenty five minus zero point six multiply x ( the unknown numerical value).
So, the word phrase of "0.6x" is zero point six multiply x ( the unknown numerical value).
Therefore, the word phrase for each algebraic expression 25-0.6x is twenty five minus zero point six multiply x ( the unknown numerical value). |
# Thread: Continuity and Differential Problem
1. ## Continuity and Differential Problem
I'm very confused on how im suppose to prove that any K works for the graph to be continous in part A.
Also , how do i even approach part B? I'm just kinda lost in how i should approach the whole problem!
Here's the problem : https://imgur.com/a/POws4
2. ## Re: Continuity and Differential Problem
a) Start with the definition of continuity of a function at a point. This should enable you to write down an equation that you can solve for $k$.
b) Again, start with the definition of the derivative at a point. This again leads to an equation that you can solve for $k$.
If you work out how to get a), you will see the method for b).
3. ## Re: Continuity and Differential Problem
Well, it helps to know what "continuous" means! A function, f(x), is said to be "continuous at x= a" if and only if $\lim_{x\to a} f(x)= f(a)$. Taking the limit of f(x) as x approaches 0 from below, $lim_{x\to 0} 4x= 0$. Taking the limit of f(x) as x approaches 0 from above, $\lim_{x\to 0} kx+ 2kx^2= 0$. So the limits is 0, for any k. Further, f(0)= 0. This function is continuous at 0 for all k. And, each part is a polynomial so it is necessarily continuous for x positive or negative.
This function is continuous for all values of k.
For the second part, you need to know what "differentiable" means! The "derivative of f at x= a" is $\lim_{h\to 0}\frac{f(a+ h)- f(a)}{h}$. A function is "differentiable" if and only if that derivative exists. Again, every polynomial is differentiable for all x so f is clearly differentiable for all x other than 0. To find the derivative at x= 0, take the "limit from below" $\lim_{h\to 0}\frac{4(x+ h)- 4x}{h}= \lim_{h\to 0} \frac{4x+ 4h- 4x}{h}= \lim_{h\to 0} \frac{4h}{h}= 4$ and then take the "limit from above" $\lim_{h\to 0} \frac{k(x+h)+ 2k(x+ h)^2- kx- 2kx^2}{h}= \lim_{h\to 0}\frac{kx+ kh+ 2kx^2+ 4khx+ kh^2- kx- 2x^2}{h}= \lim_{h\to 0}\frac{kh+ 4khx+ kh^2}{h}= \lim_{h\to 0} k+ 4kx+ kh= k$.
In order that f be differentiable at x= 0, those two limits must be the same: k= 4.
4. ## Re: Continuity and Differential Problem
Originally Posted by HallsofIvy
Well, it helps to know what "continuous" means! A function, f(x), is said to be "continuous at x= a" if and only if $\lim_{x\to a} f(x)= f(a)$. Taking the limit of f(x) as x approaches 0 from below, $lim_{x\to 0} 4x= 0$. Taking the limit of f(x) as x approaches 0 from above, $\lim_{x\to 0} kx+ 2kx^2= 0$. So the limits is 0, for any k. Further, f(0)= 0. This function is continuous at 0 for all k. And, each part is a polynomial so it is necessarily continuous for x positive or negative.
This function is continuous for all values of k.
For the second part, you need to know what "differentiable" means! The "derivative of f at x= a" is $\lim_{h\to 0}\frac{f(a+ h)- f(a)}{h}$. A function is "differentiable" if and only if that derivative exists. Again, every polynomial is differentiable for all x so f is clearly differentiable for all x other than 0. To find the derivative at x= 0, take the "limit from below" $\lim_{h\to 0}\frac{4(x+ h)- 4x}{h}= \lim_{h\to 0} \frac{4x+ 4h- 4x}{h}= \lim_{h\to 0} \frac{4h}{h}= 4$ and then take the "limit from above" $\lim_{h\to 0} \frac{k(x+h)+ 2k(x+ h)^2- kx- 2kx^2}{h}= \lim_{h\to 0}\frac{kx+ kh+ 2kx^2+ 4khx+ kh^2- kx- 2x^2}{h}= \lim_{h\to 0}\frac{kh+ 4khx+ kh^2}{h}= \lim_{h\to 0} k+ 4kx+ kh= k$.
In order that f be differentiable at x= 0, those two limits must be the same: k= 4.
I have one question about the second part. Why doesn't continuity mean that it is differentiable too? How the the last part equal K ? I get k +4kx as the limit after plugging in h for 0.
5. ## Re: Continuity and Differential Problem
Originally Posted by lc99
I have one question about the second part. Why doesn't continuity mean that it is differentiable too? How the the last part equal K ? I get k +4kx as the limit after plugging in h for 0.
Consider the function $f(x) = |x| = \begin{cases}-x & x \le 0 \\ x & x \ge 0\end{cases}$. It is easy to show that $f(x)$ is continuous. For all values less than zero, the derivative is negative 1. For all values greater than zero, the derivative is positive 1. What is the derivative at zero? It does not exist because the left-handed limit does not equal the right handed limit. In other words, continuity does not imply differentiability.
6. ## Re: Continuity and Differential Problem
"Differentiability" is a stronger condition that "continuity". If a function is differentiable at point then it must be continuous there. But, as SlipEternal showed, a function may be continuous at a point without being differentiable.
7. ## Re: Continuity and Differential Problem
In otherword, the slope at x=0 should be the same for the function at x <=0 and x>0 to be differentiable?
8. ## Re: Continuity and Differential Problem
Originally Posted by lc99
In otherword, the slope at x=0 should be the same for the function at x <=0 and x>0 to be differentiable?
Consider this example:
$f(x)=\begin{cases}x^3-2 &: x\ne 2 \\ 8 &: x= 2\end{cases}$
Is it clear to you that $\displaystyle{\lim _{x \to {2^ + }}}f(x) = {\lim _{x \to {2^ - }}}f(x) = 6~?$
But the function $f(x)$ does not have a derivative at $x=2$ Do you see why? |
Approximating the Poisson Distribution(Edexcel International A Level Maths: Statistics 2)
Revision Note
Author
Dan
Expertise
Maths
Calculating probabilities using a binomial or Poisson distribution can take a while. Under certain conditions we can use a normal distribution to approximate these probabilities. As we are going from a discrete distribution (binomial or Poisson) to a continuous distribution (normal) we need to apply continuity corrections.
Continuity Corrections
What are continuity corrections?
• The binomial and Poisson distribution are discrete and the normal distribution is continuous
• A continuity correction takes this into account when using a normal approximation
• The probability being found will need to be changed from a discrete variable, X to a continuous variable, XN
• For example, X = 4 for Poisson can be thought of as for normal as every number within this interval rounds to 4
• Remember that for a normal distribution the probability of a single value is zero so
How do I apply continuity corrections?
• Think about what is largest/smallest integer that can be included in the inequality for the discrete distribution and then find its upper/lower bound
• You add 0.5 as you want to include in the inequality
• You subtract 0.5 as you don't want to include in the inequality
• You subtract 0.5 as you want to include in the inequality
• You add 0.5 as you don't want to include in the inequality
• For a closed inequality such as
• Think about each inequality separately and use above
• Combine to give
Normal Approximation of Poisson
When can I use a normal distribution to approximate a Poisson distribution?
• A Poisson distribution can be approximated by a normal distribution provided
• is large
• Remember that the mean and variance of a Poisson distribution are approximately equal, therefore the parameters of the approximating distribution will be:
• The greater the value of λ in a Poisson distribution, the more symmetrical the distribution becomes and the closer it resembles the bell-shaped curve of a normal distribution
Why do we use approximations?
• If there are a large number of values for a Poisson distribution there could be a lot of calculations involved and it is inefficient to work with the Poisson distribution
• These days calculators can find Poisson probabilities so approximations are no longer necessary
• However it can still be easier to work with a normal distribution
• You can calculate the probability of a range of values quickly
• You can use the inverse normal distribution function (most calculators don't have an inverse Poisson distribution function)
How do I approximate a probability?
• STEP 1: Find the mean and variance of the approximating distribution
• STEP 2: Apply continuity corrections to the inequality
• STEP 3: Find the probability of the new corrected inequality
• Find the standard normal probability and use the table of the normal distribution
• The probability will not be exact as it is an approximation but provided λ is large enough then it will be a close approximation
Worked example
The number of hits on a revision web page per hour can be modelled by the Poisson distribution with a mean of 40. Use a normal approximation to find the probability that there are more than 50 hits on the webpage in a given hour.
Exam Tip
• The question will make it clear if an approximation is to be used, λ will be bigger than the values in the formula booklet.
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# How do you use Heron's formula to determine the area of a triangle with sides of that are 15, 16, and 22 units in length?
Jan 13, 2016
A ≈ 120
#### Explanation:
Heron's formula is a two step process :
step 1 : Calculate half of the Perimeter (s )
If the lengths of the sides are a , b and c , then
$s = \frac{a + b + c}{2}$
In this question let a = 15 , b = 16 and c = 22
$\Rightarrow s = \frac{15 + 16 + 22}{2} = \frac{53}{2} = 26.5$
step 2 : Calculate Area (A ) using :
$A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$
substitute in values :
A = sqrt(26.5(26.5 - 15 )(26.5 -16 )(26.5 - 22 )
rArr A = sqrt(26.5 xx 11.5 xx 10.5 xx 4.5 ) = sqrt14399.4375 ≈ 120 |
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# 9.5: Collisions in Multiple Dimensions
[ "article:topic", "authorname:openstax", "collisions", "license:ccby" ]
Skills to Develop
• Express momentum as a two-dimensional vector
• Write equations for momentum conservation in component form
• Calculate momentum in two dimensions, as a vector quantity
It is far more common for collisions to occur in two dimensions; that is, the angle between the initial velocity vectors is neither zero nor 180°. Let’s see what complications arise from this.
The first idea we need is that momentum is a vector; like all vectors, it can be expressed as a sum of perpendicular components (usually, though not always, an x-component and a y-component, and a z-component if necessary). Thus, when we write down the statement of conservation of momentum for a problem, our momentum vectors can be, and usually will be, expressed in component form.
The second idea we need comes from the fact that momentum is related to force:
$$\vec{F} = \frac{d \vec{p}}{dt} \ldotp$$
Expressing both the force and the momentum in component form,
$$F_{x} = \frac{dp_{x}}{dt}, F_{y} = \frac{dp_{y}}{dt}, F_{z} = \frac{dp_{z}}{dt} \ldotp$$
Remember, these equations are simply Newton’s second law, in vector form and in component form. We know that Newton’s second law is true in each direction, independently of the others. It follows therefore (via Newton’s third law) that conservation of momentum is also true in each direction independently.
These two ideas motivate the solution to two-dimensional problems: We write down the expression for conservation of momentum twice: once in the x-direction and once in the y-direction.
$$p_{f,x} = p_{1,i,x} + p_{2,i,x} \tag{9.18}$$
$$p_{f,y} = p_{1,i,y} + p_{2,i,y}$$
This procedure is shown graphically in Figure 9.22.
Figure $$\PageIndex{1}$$: (a) For two-dimensional momentum problems, break the initial momentum vectors into their x- and y-components. (b) Add the x- and y-components together separately. This gives you the x- and y-components of the final momentum, which are shown as red dashed vectors. (c) Adding these components together gives the final momentum.
We solve each of these two component equations independently to obtain the x- and y-components of the desired velocity vector:
$$v_{f,x} = \frac{m_{1} v_{1,i,x} + m_{2} v_{2,i,x}}{m}$$
$$v_{f,y} = \frac{m_{1} v_{1,i,y} + m_{2} v_{2,i,y}}{m}$$
(Here, m represents the total mass of the system.) Finally, combine these components using the Pythagorean theorem,
$$v_{f} = |\vec{v}_{f}| = \sqrt{v_{f,x}^{2} + v_{f,y}^{2}} \ldotp$$
Problem-Solving Strategy: Conservation of Momentum in Two Dimensions
The method for solving a two-dimensional (or even three-dimensional) conservation of momentum problem is generally the same as the method for solving a one-dimensional problem, except that you have to conserve momentum in both (or all three) dimensions simultaneously:
1. Identify a closed system.
2. Write down the equation that represents conservation of momentum in the x-direction, and solve it for the desired quantity. If you are calculating a vector quantity (velocity, usually), this will give you the x-component of the vector.
3. Write down the equation that represents conservation of momentum in the y-direction, and solve. This will give you the y-component of your vector quantity.
4. Assuming you are calculating a vector quantity, use the Pythagorean theorem to calculate its magnitude, using the results of steps 3 and 4.
Example 9.14
##### Traffic Collision
A small car of mass 1200 kg traveling east at 60 km/hr collides at an intersection with a truck of mass 3000 kg that is traveling due north at 40 km/hr (Figure 9.23). The two vehicles are locked together. What is the velocity of the combined wreckage?
Figure $$\PageIndex{2}$$: A large truck moving north is about to collide with a small car moving east. The final momentum vector has both x- and y-components.
##### Strategy
First off, we need a closed system. The natural system to choose is the (car + truck), but this system is not closed; friction from the road acts on both vehicles. We avoid this problem by restricting the question to finding the velocity at the instant just after the collision, so that friction has not yet had any effect on the system. With that restriction, momentum is conserved for this system.
Since there are two directions involved, we do conservation of momentum twice: once in the x-direction and once in the y-direction.
##### Solution
Before the collision the total momentum is
$$\vec{p} = m_{c} \vec{v}_{c} + m_{T} \vec{v}_{T} \ldotp$$
After the collision, the wreckage has momentum
$$\vec{p} = (m_{c} + m_{T}) \vec{v}_{w} \ldotp$$
Since the system is closed, momentum must be conserved, so we have
$$m_{c} \vec{v}_{c} + m_{T} \vec{v}_{T} = (m_{c} + m_{T}) \vec{v}_{w} \ldotp$$
We have to be careful; the two initial momenta are not parallel. We must add vectorially (Figure 9.24).
Figure $$\PageIndex{3}$$: Graphical addition of momentum vectors. Notice that, although the car’s velocity is larger than the truck’s, its momentum is smaller.
If we define the +x-direction to point east and the +y-direction to point north, as in the figure, then (conveniently),
$$\vec{p}_{c} = p_{c}\; \hat{i} = m_{c} v_{c}\; \hat{i}$$
$$\vec{p}_{T} = p_{T}\; \hat{j} = m_{T} v_{T}\; \hat{j} \ldotp$$
Therefore, in the x-direction:
$$m_{c} v_{c} = (m_{c} + m_{T}) v_{w,x}$$
$$v_{w,x} = \left(\dfrac{m_{c}}{m_{c} + m_{T}}\right) v_{c}$$
and in the y-direction:
$$m_{T} v_{T} = (m_{c} + m_{T}) v_{w,y}$$
$$v_{w,y} = \left(\dfrac{m_{T}}{m_{c} + m_{T}}\right) v_{T} \ldotp$$
Applying the Pythagorean theorem gives
$$\begin{split} |\vec{v}_{w}| & = \sqrt{\Big[ \left(\dfrac{m_{c}}{m_{c} + m_{T}}\right) v_{c} \Big]^{2} + \Big[ \left(\dfrac{m_{T}}{m_{c} + m_{T}} \right) v_{T} \Big]^{2}} \\ & = \sqrt{\Big[ \left(\dfrac{1200\; kg}{4200\; kg}\right) (16.67\; m/s) \Big]^{2} + \Big[ \left(\dfrac{3000\; kg}{4200\; kg}\right) (11.1\; m/s) \Big]^{2}} \\ & = \sqrt{(4.76\; m/s)^{2} + (7.93\; m/s)^{2}} \\ & = 9.25\; m/s \approx 33.3\; km/hr \ldotp \end{split}$$
As for its direction, using the angle shown in the figure,
$$\theta = \tan^{-1} \left(\dfrac{v_{w,x}}{v_{w,y}}\right) = \tan^{-1} \left(\dfrac{7.93\; m/s}{4.76\; m/s}\right) = 59^{o} \ldotp$$
This angle is east of north, or 31° counterclockwise from the +x-direction.
##### Significance
As a practical matter, accident investigators usually work in the “opposite direction”; they measure the distance of skid marks on the road (which gives the stopping distance) and use the work-energy theorem along with conservation of momentum to determine the speeds and directions of the cars prior to the collision. We saw that analysis in an earlier section.
Suppose the initial velocities were not at right angles to each other. How would this change both the physical result and the mathematical analysis of the collision?
Example 9.15
##### Exploding Scuba Tank
A common scuba tank is an aluminum cylinder that weighs 31.7 pounds empty (Figure 9.25). When full of compressed air, the internal pressure is between 2500 and 3000 psi (pounds per square inch). Suppose such a tank, which had been sitting motionless, suddenly explodes into three pieces. The first piece, weighing 10 pounds, shoots off horizontally at 235 miles per hour; the second piece (7 pounds) shoots off at 172 miles per hour, also in the horizontal plane, but at a 19° angle to the first piece. What is the mass and initial velocity of the third piece? (Do all work, and express your final answer, in SI units.)
Figure $$\PageIndex{4}$$: A scuba tank explodes into three pieces.
##### Strategy
To use conservation of momentum, we need a closed system. If we define the system to be the scuba tank, this is not a closed system, since gravity is an external force. However, the problem asks for the just the initial velocity of the third piece, so we can neglect the effect of gravity and consider the tank by itself as a closed system. Notice that, for this system, the initial momentum vector is zero.
We choose a coordinate system where all the motion happens in the xy-plane. We then write down the equations for conservation of momentum in each direction, thus obtaining the x- and y-components of the momentum of the third piece, from which we obtain its magnitude (via the Pythagorean theorem) and its direction. Finally, dividing this momentum by the mass of the third piece gives us the velocity.
##### Solution
First, let’s get all the conversions to SI units out of the way:
$$31.7\; lb \times \frac{1\; kg}{2.2\; lb} \rightarrow 14.4\; kg$$
$$10\; lb \rightarrow 4.5\; kg$$
$$235\; \frac{miles}{hour} \times \frac{1\; hour}{3600\; s} \times \frac{1609\; m}{mile} = 105\; m/s$$
$$7\; lb \rightarrow 3.2\; kg$$
$$172 \frac{mile}{hour} = 77\; m/s$$
$$m_{3} = 14.4\; kg - (4.5\; kg + 3.2\; kg) = 6.7\; kg \ldotp$$
Now apply conservation of momentum in each direction.
x-direction:
$$\begin{split} p_{f,x} & = p_{0,x} \\ p_{1,x} + p_{2,x} + p_{3,x} & = 0 \\ m_{1} v_{1,x} + m_{2} v_{2,x} + p_{3,x} & = 0 \\ p_{3,x} & = -m_{1} v_{1,x} - m_{2} v_{2,x} \end{split}$$
y-direction:
$$\begin{split} p_{f,y} & = p_{0,y} \\ p_{1,y} + p_{2,y} + p_{3,y} & = 0 \\ m_{1} v_{1,y} + m_{2} v_{2,y} + p_{3,y} & = 0 \\ p_{3,y} & = -m_{1} v_{1,y} - m_{2} v_{2,y} \end{split}$$
From our chosen coordinate system, we write the x-components as
$$\begin{split} p_{3,x} & = - m_{1} v_{1} - m_{2} v_{2} \cos \theta \\ & = - (14.5\; kg)(105\; m/s) - (4.5\; kg)(77\; m/s) \cos (19^{o}) \\ & = -1850\; kg\; \cdotp m/s \ldotp \end{split}$$
For the y-direction, we have
$$\begin{split} p_{3,y} & = 0 - m_{2} v_{2} \sin \theta \\ & = - (4.5\; kg)(77\; m/s) \sin (19^{o}) \\ & = -113\; kg\; \cdotp m/s \ldotp \end{split}$$
This gives the magnitude of p3:
$$\begin{split} p_{3} & = \sqrt{p_{3,x}^{2} + p_{3,y}^{2}} \\ & = \sqrt{(-1850\; kg\; \cdotp m/s)^{2} + (-113\; kg\; \cdotp m/s)} \\ & = 1854\; kg\; \cdotp m/s \ldotp \end{split}$$
The velocity of the third piece is therefore
$$v_{3} = \frac{p_{3}}{m_{3}} = \frac{1854\; kg\; \cdotp m/s}{6.7\; kg} = 277\; m/s \ldotp$$
The direction of its velocity vector is the same as the direction of its momentum vector:
$$\phi = \tan^{-1} \left(\dfrac{p_{3,y}}{p_{3,x}}\right) = \tan^{-1} \left(\dfrac{113\; kg\; \cdotp m/s}{1850\; kg\; \cdotp m/s}\right) = 3.5^{o} \ldotp$$
Because $$\phi$$ is below the −x -axis, the actual angle is 183.5° from the +x-direction.
##### Significance
The enormous velocities here are typical; an exploding tank of any compressed gas can easily punch through the wall of a house and cause significant injury, or death. Fortunately, such explosions are extremely rare, on a percentage basis.
Notice that the mass of the air in the tank was neglected in the analysis and solution. How would the solution method changed if the air was included? How large a difference do you think it would make in the final answer?
### Contributors
• Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0). |
## Thursday, 4 December 2014
### Maths: Snowflakes
I have been thinking of ways to incorporate maths into fun Christmas time activities recently and this one went down well with my 11 and 9 year old. This a great for a winter project and a way for kids to practise two skills:
• They will learn to recognize and describe symmetries of designs by finding centres of rotation and axis of symmetry.
• They will learn to make their own design with one or more specified symmetries.
Read on if you want to know how we made our snowflake and how I used it to teach my girls some maths...
You will need:
• Isometric paper - I used isometric paper with the girls to make it easier for them.
• A sharp pencil
• A ruler
STEP ONE
Draw a line down the centre of the page. We measured it out to be an even number. 12-16cm is best.
STEP TWO
Draw the two lines through the centre. To do this you will have to find the centre of the first line so you know where the other two lines should go through.
This is where the isometric paper helped as a protractor wasn't needed. If you do use a plain bit of paper the lines should all be at a 60 degree angle to have 6 equal parts.
STEP THREE
Draw your snowflake pattern. The rule we stuck to when drawing ours was to copy exactly what we did on one of the 6 branches on each of the other five branches. They must be precise, they must be exactly the same on the other 5 branches...
Involving more maths...
This is where you can talk about reflection. We used a mirror and placed it in between the branches to demonstrate why it is a reflection. The pattern we had copied on to each branch should, if it is done correctly, look exactly the same as the reflection in the mirror.
For my seven year old we used the mirror as we went along. She didn't get it right but she had fun and it introduced her to why, with symmetry, it is important to get it right.
With my older girls we talked more about symmetry; we talked about how when we place the mirror in certain places on our snowflake sometimes the mirror reflected the same drawing we had made on the paper...
It has reflection symmetry! The snow flake looked the same with the mirror as without it. At this point you have found the axis of symmetry: this is a line through a shape so that each side is a mirror image.
And sometimes it didn't...
Then I encouraged the girls to turn the paper and see how many rotations they could make where the shape still looked the same. This is called rotational symmetry. If a shape will only fit itself in one way, if you rotate it it will always look different, that means there is no rotational symmetry in it.
When the drawing of a snowflake has been conquered, you might like to follow this project on further and think more about reflections and rotations. You might also like to think about translations and turning your snowflakes into a tessellation
You might like to predict what shape your snowflake will be before unfolding it. Can you get it right?
Can you make your pattern using Hama beads? |
# Ways to multiply n elements with an associative operation
Given a number n, find the number of ways to multiply n elements with an associative operation.
Examples :
```Input : 2
Output : 2
For a and b there are two ways to multiply them.
1. (a * b)
2. (b * a)
Input : 3
Output : 12```
Explanation(Example 2) :
```For a, b and c there are 12 ways to multiply them.
1. ((a * b) * c) 2. (a * (b * c))
3. ((a * c) * b) 4. (a * (c * b))
5. ((b * a) * c) 6. (b * (a * c))
7. ((b * c) * a) 8. (b * (c * a))
9. ((c * a) * b) 10. (c * (a * b))
11. ((c * b) * a) 12. (c * (b * a))```
Approach: First, we try to find out the recurrence relation. From above examples, we can see h(1) = 1, h(2) = 2, h(3) = 12 . Now, for n elements there will be n – 1 multiplications and n – 1 parentheses. And, (a1, a2, …, an ) can be obtained from (a1, a2, …, a(n – 1)) in exactly one of the two ways :
1. Take a multiplication (a1, a2, …, a(n – 1))(which has n – 2 multiplications and n – 2 parentheses) and insert the nth element ‘an’ on either side of either factor in one of the n – 2 multiplications. Thus, for each scheme for n – 1 numbers gives 2 * 2 * (n – 2) = 4 * (n – 2) schemes for n numbers in this way.
2. Take a multiplication scheme for (a1, a2, .., a(n-1)) and multiply on left or right by (‘an’). Thus, for each scheme for n – 1 numbers gives two schemes for n numbers in this way.
So after adding above two, we get, h(n) = (4 * n – 8 + 2) * h(n – 1), h(n) = (4 * n – 6) * h(n – 1). This recurrence relation with same initial value is satisfied by the pseudo-Catalan number. Hence, h(n) = (2 * n – 2)! / (n – 1)!
## C++
`// C++ code to find number of ways to multiply n ` `// elements with an associative operation` `# include ` `using` `namespace` `std;` `// Function to find the required factorial` `int` `fact(``int` `n)` `{` ` ``if` `(n == 0 || n == 1) ` ` ``return` `1 ;` ` ``int` `ans = 1; ` ` ``for` `(``int` `i = 1 ; i <= n; i++) ` ` ``ans = ans * i ; ` ` ``return` `ans ;` `}` `// Function to find nCr` `int` `nCr(``int` `n, ``int` `r)` `{` ` ``int` `Nr = n , Dr = 1 , ans = 1;` ` ``for` `(``int` `i = 1 ; i <= r ; i++ ) {` ` ``ans = ( ans * Nr ) / ( Dr ) ;` ` ``Nr-- ;` ` ``Dr++ ;` ` ``}` ` ``return` `ans ;` `}` `// function to find the number of ways` `int` `solve ( ``int` `n )` `{` ` ``int` `N = 2*n - 2 ;` ` ``int` `R = n - 1 ; ` ` ``return` `nCr (N, R) * fact(n - 1) ;` `}` `// Driver code` `int` `main()` `{` ` ``int` `n = 6 ;` ` ``cout << solve (n) ; ` ` ``return` `0 ;` `}`
## Java
`// Java code to find number of ` `// ways to multiply n elements ` `// with an associative operation` `import` `java.io.*;` `class` `GFG ` `{` `// Function to find the` `// required factorial` `static` `int` `fact(``int` `n)` `{` ` ``if` `(n == ``0` `|| n == ``1``) ` ` ``return` `1` `;` ` ``int` `ans = ``1``; ` ` ``for` `(``int` `i = ``1` `; i <= n; i++) ` ` ``ans = ans * i ; ` ` ``return` `ans ;` `}` `// Function to find nCr` `static` `int` `nCr(``int` `n, ``int` `r)` `{` ` ``int` `Nr = n , Dr = ``1` `, ans = ``1``;` ` ``for` `(``int` `i = ``1` `; i <= r ; i++ ) ` ` ``{` ` ``ans = ( ans * Nr ) / ( Dr ) ;` ` ``Nr-- ;` ` ``Dr++ ;` ` ``}` ` ``return` `ans ;` `}` `// function to find` `// the number of ways` `static` `int` `solve ( ``int` `n )` `{` ` ``int` `N = ``2` `* n - ``2` `;` ` ``int` `R = n - ``1` `; ` ` ``return` `nCr (N, R) * fact(n - ``1``) ;` `}` `// Driver Code` `public` `static` `void` `main (String[] args) ` `{` `int` `n = ``6` `;` `System.out.println( solve (n)) ; ` `}` `}` `// This code is contributed by anuj_67.`
## Python3
`# Python3 code to find number` `# of ways to multiply n ` `# elements with an` `# associative operation` `# Function to find the` `# required factorial` `def` `fact(n):` ` ``if` `(n ``=``=` `0` `or` `n ``=``=` `1``): ` ` ``return` `1``;` ` ``ans ``=` `1``; ` ` ``for` `i ``in` `range``(``1``, n ``+` `1``): ` ` ``ans ``=` `ans ``*` `i; ` ` ``return` `ans;` `# Function to find nCr` `def` `nCr(n, r):` ` ``Nr ``=` `n ; Dr ``=` `1` `; ans ``=` `1``;` ` ``for` `i ``in` `range``(``1``, r ``+` `1``):` ` ``ans ``=` `int``((ans ``*` `Nr) ``/` `(Dr));` ` ``Nr ``=` `Nr ``-` `1``;` ` ``Dr ``=` `Dr ``+` `1``;` ` ``return` `ans;` `# function to find ` `# the number of ways` `def` `solve ( n ):` ` ``N ``=` `2``*` `n ``-` `2``;` ` ``R ``=` `n ``-` `1` `; ` ` ``return` `(nCr (N, R) ``*` ` ``fact(n ``-` `1``));` `# Driver code` `n ``=` `6` `;` `print``(solve (n) ); ` `# This code is contributed` `# by mits`
## C#
`// C# code to find number of ` `// ways to multiply n elements ` `// with an associative operation` `using` `System;` `class` `GFG {` ` ` ` ``// Function to find the` ` ``// required factorial` ` ``static` `int` `fact(``int` `n)` ` ``{` ` ``if` `(n == 0 || n == 1) ` ` ``return` `1 ;` ` ` ` ``int` `ans = 1; ` ` ``for` `(``int` `i = 1 ; i <= n; i++) ` ` ``ans = ans * i ; ` ` ` ` ``return` `ans ;` ` ``}` ` ` ` ``// Function to find nCr` ` ``static` `int` `nCr(``int` `n, ``int` `r)` ` ``{` ` ``int` `Nr = n , Dr = 1 , ans = 1;` ` ``for` `(``int` `i = 1 ; i <= r ; i++ ) ` ` ``{` ` ``ans = ( ans * Nr ) / ( Dr ) ;` ` ``Nr-- ;` ` ``Dr++ ;` ` ``}` ` ``return` `ans ;` ` ``}` ` ` ` ``// function to find` ` ``// the number of ways` ` ``static` `int` `solve ( ``int` `n )` ` ``{` ` ``int` `N = 2 * n - 2 ;` ` ``int` `R = n - 1 ; ` ` ``return` `nCr (N, R) * fact(n - 1) ;` ` ``}` ` ` ` ``// Driver Code` ` ``public` `static` `void` `Main () ` ` ``{` ` ``int` `n = 6 ;` ` ``Console.WriteLine( solve (n)) ; ` ` ``}` `}` `// This code is contributed by anuj_67.`
## PHP
``
## Javascript
``
Output :
`30240`
Time Complexity: O(n).
Auxiliary Space: O(1). |
# 5.5 – Solving Polynomial Equations
Key Terms
• Complex Conjugate Theorem – If a + bi is a root, then a – bi is also a root.
• Conjugate Radical Theorem – If $a+ \sqrt{b}$ is a root, then $a- \sqrt{b}$ is also a root.
• Fundamental Theorem of Algebra – A polynomial of degree n has exactly n complex roots.
• Lower Bound – A value that is less than the smallest root of a polynomial.
• Multiplicity – A count of the number of times a specific value occurs as a root of a polynomial function.
• Upper Bound – A value that is larger than the largest root of a polynomial.
Review
• If the leading coefficient has no sign, it is a positive number.
• Complex Numbers
• Square roots of negative numbers produce complex numbers, which are written in the form a + bi (the combination of a real and an imaginary number).
Notes
• Roots
• If a solution set has imaginary or irrational roots, those numbers will not be listed when you use the rational root theorem: $\frac{p}{q}$.
• Fundamental Theorem of Algebra
• The degree of a polynomial function tells you how many roots, or zeros, it has!
• Multiplicity
• When you factor a polynomial like $x^2-10x+25$, you get two identical factors: (x − 5) and (x − 5); so, there are two identical roots: 5 and 5.
• This root touches the x-axis once, at (5, 0); and, it is called a double root since there are two of the same.
• Multiplicities must be included when you say that a polynomial of degree n has n roots.
• Complex Roots and Imaginary Numbers
• Complex Conjugate Theorem
• Complex roots and imaginary roots always come in pairs.
• If a + bi is a root, then a − bi is also a root.
• When graphed, imaginary roots do not cross the x-axis.
• Imaginary (complex) numbers cannot be graphed.
• Conjugate Radicals – the two roots: $a+ \sqrt{b}$ and $a- \sqrt{b}$
• Possible Roots
• If a polynomial has a degree of three, it can have 1 real root AND 2 complex roots or it may have 3 real roots.
• Descartes’ Rule of Signs
• The number of positive roots (or zeros) is less than or equal to the number of sign changes for the terms of a polynomial function, f(x).
• The number of negative roots (or zeros) is less than or equal to the number of sign changes for the terms of a polynomial function, f(−x).
• In each case, if the number of roots is less than the number of sign changes, then the number of roots differs from the number of sign changes by a multiple of 2.
• Example: Count the number of times the term changes from positive to negative or vice-versa
• $f(x)=x^3+x^2-x+2$ changes from positive to negative, then back to positive
• Two sign changes: + to – and then – to +
• There will either be 2 or 0 positive roots (we know this because we tested a positive x in the function: f(x))
• $f(-x)=(-x)^3+(-x)^2-(-x)+2$ simplifies to: $f(x)=-x^3+x^2+x+2$, which changes from negative to positive
• Just one sign change: – to +
• There will be 1 negative root (we know this because we tested a negative x in the function: f(-x))
• Bounds
• Upper bound: A number greater than or equal to all real roots.
• Upper Bound Theorem: for a positive number c, if f(x) is divided by (x − c) and the resulting quotient polynomial and remainder have no changes in sign (all positive terms or all negative terms), then f(x) has no real roots greater than c.
• Lower bound: A number less than or equal to all real roots.
• Lower Bound Theorem: for a negative number c, if f(x) is divided by (x − c) and the resulting quotient and remainder have alternating signs, then f(x) has no real roots less than c.
• Each type of bound has a specific pattern in synthetic division.
• If synthetic division by a possible root gives NO changes in sign on the bottom row, then this number is the upper bound; and, any numbers larger than this number can be deleted from the list of possible roots.
• Steps for Solving a Polynomial Equation
1. Use the fundamental theorem of algebra to identify the total number of roots in a polynomial.
2. Use the rational root theorem to create a list of possible rational roots.
3. Use Descartes’ rule of signs to determine the number of positive and negative roots. Remember that multiplicity applies.
4. Use synthetic division to try to find roots.
5. While using synthetic division, pay attention to the sign changes of the results. Use the upper bound and lower bound
6. theorems to eliminate higher or lower values from the list of possible roots.
7. Factor the polynomial using synthetic division, factoring rules, or the quadratic formula. Remember that the conjugate radical theorem and complex conjugate theorem apply.
8. Identify the roots that are solutions.
Examples
• Ex 1. What are the roots of the polynomial below:
• $f(x)=x^3-x^2-9x+9$
• Step 1: list the possible roots: p = 9, q = 1
• $\frac{p}{q}$: Since q = 1, you won’t have a denominator: 9, -9, 3, -3
• Step 2: try each root using synthetic division
• If the remainder is zero, you have a root!
• Answer: The roots are 1, 3, and -3.
• Notice that when the root is 1, the resulting polynomial quotient is: $x^2-9$.
• This is factored into $(x-3)(x+3)$, so you know that 3 and -3 are roots without having to do all of this synthetic division!
• Ex 2. What are all the possible rational roots of the polynomial below:
• $f(x)=2x^3+5x^2-8x-10=0$
• List the possible roots: p = 10, q = 2
• $\frac{p}{q}$: $\frac{\pm 1, \pm 2, \pm 5, \pm 10}{\pm 1, \pm 2}$
• Simplified: $\pm 1, \pm \frac{1}{2}, \pm 2, \pm 5, \pm \frac{5}{2}, \pm 10$
• Ex 3. Which of the following expresses the possible number of positive real solutions for the polynomial equation shown below?
• $5x^3+x^2+7x-28=0$
• Ex 4. Which of the following expresses the possible number of positive real solutions for the polynomial equation shown below?
• $5x^3+x^2-7x-28=0$
• Ex 5. A polynomial has one root that equals 4 + 17i. What is the other root of this polynomial?
• Ex 6. How many solutions over the complex number system does this polynomial have?
• $3x^6-x^3-4x^2+3x+52=0$
• Ex 7. Express the polynomial as a product of linear factors.
• $f(x)=x^3-5x^2-18x+72$
• Find all possible roots, then try them using sythetic division. Once you find one, factor the remaining quotient to find the rest!
• Answer: $(x-3)(x+4)(x-6)$
• Ex 8. What is the sum of the roots of the polynomial shown below?
• $f(x)=3x^3+12x^2+3x-18$
• Find all of the roots, then combine them.
• Answer: $-3-2+1=4$ |
# Volume of Cone
Volume of Cone : From the above diagram we can see that there is right circular cylinder and a right circular cone of the same base radius and same height.
When fill up cone up to the brim is emptied into the cylinder 3 times then the cylinder will be completely filled up to the brim. So from that we can conclude that cone volume is 1/3 rd that of the volume of the cylinder.
The formula is :
volume of - cone = 1/3 π r2 h l2 = r2 + h2
Some solved examples :
1) Find the volume of a cone the radius of whose base is 21 cm and height is 28 cm.
Solution : r = 21 cm and h = 28 cm
Volume of cone = 1/3 π r
2 h
V = 1/3 ( 3.14 x 21 x 21 x 28)
V = 1/3 x 38772.72
∴ Volume of a cone = 12924.24 cm
3
_________________________________________________________________
2) If the height of a cone is 15 cm and its volume is 770 cu.cm; find the radius of its base.
Solution : h = 15 cm and V = 770 cu.cm
volume of cone = 1/3 π r
2 h
⇒ 770 = 1/3 x 3.14 x r
2 x 15
⇒ 770 = 3.14 x r
2 x 5
⇒ 770 = 15.7 x r
2
⇒ r
2 = 770 / 15.7 = 49
2 = 49
∴ r = 7 cm.
_________________________________________________________________
3) A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Solution : As the triangle revolved about the side 12 cm.
∴ radius = r = 5m and height = h = 12 cm
Volume = 1/ 3 π r
2 h
V = 1/3 x 3.14 x 5 x 5 x 12
V = 314 cm
3
Volume :
Volume Formulas
Volume of Irregular Shape
Volume of a Cube
Volume of a Rectangular Prism(Cuboid)
Volume of a Cylinder
Volume of Cone
Volume of a Sphere
Volume of a Hemisphere
Volume of a Prism
Volume of a Pyramid
To Mensuration |
# Fundamental Hardware Elements of Computers: Simplifying boolean equations
← Boolean algebra Simplifying boolean equations Boolean identities →
A common question is to give you a complex boolean equation, which you will then have to work out a simpler exact equivalent. This is useful when you are designing circuits and want to minimise the number of gates you are using or make circuits that only use particular types of gates. To simplify boolean equations you must be familiar with two methods. You can normally use either, but try to master both:
• Truth tables
• Boolean algebra - identities and De Morgans Law
Example: Simplifying boolean equations with Truth Tables
Draw the truth table for the following:
$(A . B) + A$
We are going to solve this using a truth table and we need to break the problem down into its component parts:
As the equation uses A and B list the different values they can take (4 in total)
$A$ $B$
0 0
0 1
1 0
1 1
Next we are going to work out the brackets first: $A.B$ and add this to our truth table
$A$ $B$ $A.B$
0 0 0
0 1 0
1 0 0
1 1 1
Finally we will OR this result ($A.B$) with A to find $(A.B)+A$, the final column of the truth table
$A$ $B$ $A.B$ $(A.B)+A$
0 0 0 0
0 1 0 0
1 0 0 1
1 1 1 1
Now that our truth table is complete, look at the final column, is there a simpler way of writing this? Why aye! The final column is true when, and only when, A is true, it doesn't require B's input at all. So we can simplify to A telling us that:
$(A.B)+A = A$
Example: Simplifying boolean equations with Truth Tables
Let's look at another example
$\overline{A . \overline{B}}$
We first of all need to break down the equation into its component parts. Starting off with A and B we the work out $\overline{B}$, then $A.\overline{B}$ and finally $\overline{A . \overline{B}}$.
$A$ $B$ $\overline{B}$ $A.\overline{B}$ $\overline{A . \overline{B}}$
0 0
1
0
1
0 1
0
0
1
1 0
1
1
0
1 1
0
0
1
This can be simplified to $\overline{A} + B$ telling us that: $\overline{A . \overline{B}} = \overline{A} + B$. How did we jump to this conclusion? Let's take a look at all the places where the result is true:
$A$ $B$ $\overline{A . \overline{B}}$
0 0
1
0 1
1
1 0
0
1 1
1
We need to get a combination of A, B that gives the result shown above. We can see that whenever A is false ($\overline{A}$)the answer is true:
$A$ $B$ $\overline{A . \overline{B}}$
0 0
1
0 1
1
1 0
0
1 1
1
We can also see that whenever the B value is true then the answer is also true:
$A$ $B$ $\overline{A . \overline{B}}$
0 0
1
0 1
1
1 0
0
1 1
1
So we know that we need to combine $\overline{A} with B$ to get an equation solving all cases.
If we AND them $\overline{A} . B$ this only gives us one of the scenarios, so that's not the answer.
If we OR them $\overline{A} + B$ then this gives us three answers, matching all the responses above. This is our solution.
Exercise: Simplifying boolean equations with Truth Tables
Give a simplified boolean description, ?, for the following truth tables:
$A$ $B$ $?$
0 0 0
0 1 1
1 0 0
1 1 1
$B$
$A$ $B$ $?$
0 0 0
0 1 0
1 0 0
1 1 1
$A.B$
$A$ $B$ $?$
0 0 1
0 1 1
1 0 1
1 1 1
$1$
$A$ $B$ $?$
0 0 1
0 1 0
1 0 1
1 1 1
$A+\overline{B}$
Simplify the following boolean equations using truth tables:
$(\overline{A.B}).A$
$A$ $B$ ${A.B}$ $\overline{A.B}$ $\overline{A.B}.A$
0 0
0
1
0
0 1
0
1
0
1 0
0
1
1
1 1
1
0
0
This can be simplified to: $A.\overline{B}$
$(\overline{A.B})+A$
$A$ $B$ ${A.B}$ $\overline{A.B}$ $\overline{A.B}+A$
0 0
0
1
1
0 1
0
1
1
1 0
0
1
1
1 1
1
0
1
The answer in all cases is a plain $1$
$(A.\overline{B})+B$
$A$ $B$ $\overline{B}$ $(A.\overline{B})$ $(A.\overline{B})+B$
0 0
1
0
0
0 1
0
0
1
1 0
1
1
1
1 1
0
0
1
This can be simplified to: $\overline{\overline{A}.\overline{B}}$ OR $A+B$ (because of De Morgan's Laws)
$\overline{B}+A.B$
Remember that we deal with the AND before the OR, meaning we can read the equation as: $\overline{B}+(A.B)$
$A$ $B$ $(A.B)$ $\overline{B}$ $\overline{B}+(A.B)$
0 0
0
1
1
0 1
0
0
0
1 0
0
1
1
1 1
1
0
1
This can be simplified to: $A+\overline{B}$
$(\overline{A}+\overline{B}).B$
$A$ $B$ $\overline{A}$ $\overline{B}$ $(\overline{A}+\overline{B})$ $(\overline{A}+\overline{B}).B$
0 0
1
1
1
0
0 1
1
0
1
1
1 0
0
1
1
0
1 1
0
0
0
0
This can be simplified to: $\overline{A}.B$ |
# Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 62: 60
(a) There are 2 chances to jump on the bus; first at 3.0 seconds and then at 6 seconds. (b) The person can wait a maximum time of 0.25 seconds.
#### Work Step by Step
(a) The distance covered when running must be 9 meters more than the bus moves; $v~t=\frac{1}{2}at^2+9$ $(4.5~m/s)~t=\frac{1}{2}(1.0~m/s^2)~t^2+9.0~m$ $(1.0~m/s^2)~t^2-(9.0~m/s)~t+18.0~m=0$ $(t-3.0)(t-6.0)=0$ $t = 3.0~s, 6.0~s$ There are 2 chances to jump on the bus; first at 3.0 seconds and then at 6 seconds. (b) If we wait until the last possible moment, the bus's speed when we reach the door will be 4.5 m/s. We can find the time $t_b$ that the bus is moving. $(1.0~m/s^2)~t_b = 4.5~m/s$ $t_b = 4.5 ~s$ We can find the distance $x$ the bus moves in this time. $x = \frac{1}{2}at_b^2$ $x = \frac{1}{2}(1.0~m/s^2)(4.5~s)^2$ $x = 10.125~m$ The person needs to run a total distance of 19.125 meters. We can find the time $t_p$ it takes the person to run this distance. $t_p = \frac{19.125~m}{4.5~m/s}$ $t_p = 4.25~s$ The person can wait a maximum time of 4.5 s - 4.25 s which is 0.25 seconds.
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. |
# What is 159/184 as a decimal?
## Solution and how to convert 159 / 184 into a decimal
159 / 184 = 0.864
159/184 or 0.864 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Fractions and decimals represent parts of a whole, sometimes representing numbers less than 1. Depending on the situation, decimals can be more clear. We don't say 1 and 1/2 dollar. We use the decimal version of \$1.50. Same goes for fractions. We will say 'the student got 2 of 3 questions correct'. If we need to convert a fraction quickly, let's find out how and when we should.
## 159/184 is 159 divided by 184
The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 159 is being divided into 184. Think of this as our directions and now we just need to be able to assemble the project! The numerator is the top number in a fraction. The denominator is the bottom number. This is our equation! We must divide 159 into 184 to find out how many whole parts it will have plus representing the remainder in decimal form. This is our equation:
### Numerator: 159
• Numerators are the portion of total parts, showed at the top of the fraction. 159 is one of the largest two-digit numbers you'll have to convert. The bad news is that it's an odd number which makes it harder to covert in your head. Large numerators make converting fractions more complex. Now let's explore the denominator of the fraction.
### Denominator: 184
• Denominators represent the total parts, located at the bottom of the fraction. 184 is one of the largest two-digit numbers to deal with. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Have no fear, large two-digit denominators are all bark no bite. Next, let's go over how to convert a 159/184 to 0.864.
## Converting 159/184 to 0.864
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 184 \enclose{longdiv}{ 159 }$$
Use long division to solve step one. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 184 \enclose{longdiv}{ 159.0 }$$
Uh oh. 184 cannot be divided into 159. So that means we must add a decimal point and extend our equation with a zero. This doesn't add any issues to our denominator but now we can divide 184 into 1590.
### Step 3: Solve for how many whole groups you can divide 184 into 1590
$$\require{enclose} 00.8 \\ 184 \enclose{longdiv}{ 159.0 }$$
How many whole groups of 184 can you pull from 1590? 1472 Multiply by the left of our equation (184) to get the first number in our solution.
### Step 4: Subtract the remainder
$$\require{enclose} 00.8 \\ 184 \enclose{longdiv}{ 159.0 } \\ \underline{ 1472 \phantom{00} } \\ 118 \phantom{0}$$
If there is no remainder, you’re done! If you still have numbers left over, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. And the same is true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But they all represent how numbers show us value in the real world. Here are just a few ways we use 159/184, 0.864 or 86% in our daily world:
### When you should convert 159/184 into a decimal
Dining - We don't give a tip of 159/184 of the bill (technically we do, but that sounds weird doesn't it?). We give a 86% tip or 0.864 of the entire bill.
### When to convert 0.864 to 159/184 as a fraction
Carpentry - To build a table, you must have the right measurements. When you stretch the tape measure across the piece of wood, you won't see 10.6 inches. You'll see a tick mark at 10 and 3/5 inches.
### Practice Decimal Conversion with your Classroom
• If 159/184 = 0.864 what would it be as a percentage?
• What is 1 + 159/184 in decimal form?
• What is 1 - 159/184 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.864 + 1/2? |
# Question 8e21a
Dec 7, 2016
$43.10 m$, rounded to two decimal places.
#### Explanation:
Constant speed of Car A $= 100 k m / h = 100 \times \frac{1000}{3600} = 27. \overline{7} m {s}^{-} 1$
Let us consider Car B first.
It enters the acceleration lane at a speed $= 25 k m / h r = 6.9 \overline{4} m {s}^{-} 1$.
It accelerates uniformly and enters the main traffic lane after traveling $70 m \text{ in } 5 s$.
Using the following kinematic equation to obtain acceleration $a$
$s = u t + \frac{1}{2} a {t}^{2}$
$70 = 6.9 \overline{4} \times 5 + \frac{1}{2} a \times {5}^{2}$
$\implies \frac{1}{2} a \times {5}^{2} = 70 - 6.9 \overline{4} \times 5$
Dividing both sides by $5$ we get
$\frac{5}{2} a = 14 - 6.9 \overline{4}$
$\implies a = 2.8 \overline{2} m {s}^{-} 2$
To calculate the final speed at the time of entering the main traffic we use the following kinematic equation
$v = u + a t$
$v = 6.9 \overline{4} + 2.8 \overline{2} \times 5$
$\implies v = 21.0 \overline{5} m {s}^{-} 1$
It then continues to accelerate at the same rate until it reaches a speed of $27. \overline{7} m {s}^{-} 1$, which it then maintains.
Time taken to reach top speed is calculated from the kinematic equation
$v = u + a t$
$27. \overline{7} = 6.9 \overline{4} \times 2.8 \overline{2} t$
=>t=(27.bar7 – 6.9bar4) / {2.8bar2 } = 7.382 s, rounded to three decimal places.
During this period of acceleration average speed
$= \frac{v + u}{2} = \frac{6.9 \overline{4} + 27. \overline{7}}{2} = 17.36 \overline{1} m {s}^{-} 1$
Distance traveled in the main traffic lane$= 17.36 \overline{1} \times 7.382 = 128.16 m$, rounded to 2 decimal places
Now Car A
Distance traveled during this time period $= 27. \overline{7} \times 7.382 = 205.0 \overline{5} m$
As the car A was $120 m$ behinf car B when car B entered the main traffic lane. Therefore, distance of car A from car B
=120 + 128.16– 205.0bar5 m = 43.10 m#, rounded to two decimal places |
### Moving toward simpler ways of solving linear systems
In this section we'll revisit systems of linear equations. You've see them before: Two linear equations with two unknowns, say x and y, where you're trying to find the point of intersection (if there is one). Or you've seen three equations and three unknowns problems (x, y and z, or x1, x2, x3), and tried to find the point or line of intersection of three planes in space, and so on ...
Each equation is a piece of information, and we always need n unique pieces of information to solve for n variables.
It's actually not uncommon to need to solve very large systems of equations, even with hundreds of variables. Using matrices, we'll work toward much simpler – and much more expandable – methods to do that.
n unique pieces of information (equations) are needed to solve for n variables.
Let's solve this system "the old-fashioned way":
One way to do this (and it's not the only way) is to solve for one variable in one of the equations, we'll try z in the first equation here,
... then replace z with the result (-2x + y + 3) in the second and third equations. We've thus eliminated z from the second two equations and reduced the 3-equations / 3 unknowns problem to a two-equations / 2 unknowns problem:
These reduce to:
Conveniently, the value of x falls right out of the second equation, and from that we can easily solve for y using -5x + 5y = 5, then z using any of the original three equations:
### Solving the same system with an augmented matrix
Now let's see how matrices can make solving systems a little easier. We'll take that same system of equations,
and write it in the form of an augmented matrix, which looks like this:
The augmented matrix is just a block of numbers that are the coefficients of x, y and z from the left side of the = sign, and a column of numbers from the right side of the = sign that we'll call the result vector. Now how to solve the system using this weird matrix?
Our goal will to be to make the 3×3 matrix on the left side of the line upper triangular. Recall that an upper-triangular matrix has only zeros below the diagonal. You'll see how it works if you keep going. We'll accomplish that goal by applying elementary row operations.
#### Elementary row operations
Elementary row operations include adding a multiple of one row of a matrix or an augmented matrix to another or multiplying an entire row by a constant.
### Same system, new method:
Here is our augmented matrix again (below), and right next to it, a note about what the first elementary row operation will be. "R3+R2" means "add Row 2 to Row 3 and replace Row 3 with the result."
There's nothing special about the "R3+R2" notation. It's how I remember what I'm doing, but you might come up with your own notation. The aim of this operation is to get rid of that -1 in the lower-left corner of the matrix. Here's the result:
We're on our way to an upper-triangular 3×3. That targeted matrix element is now zero, so we'll do the next operation, 2R2-R1, which means "multiply row 2 by 2, subtract row 1 from it, and place the result back in row 2." The aim this time is to turn the leading 1 in the second row into a zero. Here's the result of that operation:
The next operation, 5R3-3R2, is designed to zero out the second element (3) of the third row. It means "multiply row 3 by 5, subtract from it 3 times row 2, and place the result back in row 3." Here's the result:
Now this system is essentially solved because it's in upper-triangular form. The solutions are simple; we just start with z, plug it into the next equation to get y, and so on:
The process of making the matrix upper-triangular through a series of elementary row operations is called Gaussian elimination (named for mathematician Carl Friedrich Gauss).
### Going a little further: Gauss-Jordan elimination
In that last example, we used elementary row operations (Gaussian elimination) to reduce the matrix to upper-triangular form, and it was basically solved, except for a couple of algebra moves. We can go further, though, to make an identity matrix out of the left side, in which case the remaining result vector on the right of the augmented matrix will be the solution (x, y, z) with no extra algebra needed.
That technique is called Gauss-Jordan elimination, and here's how that might work. We'll pick up from our upper-triangular matrix:
The operation "R2+R3" is designed to zero-out the last 5 in the second row of the 3 × 3 matrix. The result is below, and to that we'll add an operation to zero out the 1 in the 3rd position of the first row.
Next we'll zero out the remaining non-diagonal element:
The result is a diagonal matrix, and we only have to divide each row by its diagonal element to finish:
Here's the result, and notice that the vector on the right contains our expected result, x = 1, y = 2 and z = 3.
While Gauss-Jordan elimination isn't strictly needed to solve a system of equations, it's certainly a good goal because the solution just falls right in your lap. In other sections we'll find even better ways of solving systems that are much larger than 3-dimensional.
### The way to learn this is to practice it.
Hopefully you can see that solving some linear systems, especially ones with more equations and more unknowns, can be made easier with Gaussian or Gauss-Jordan elimination using augmented matrices. There are practice problems below, and you can look at some video problem solving examples (magenta bar below).
In sections ahead, you'll learn two other important methods of solving linear systems with matrices: The inverse matrix method, and Cramer's rule.
### Practice problems
Solve each of these linear systems using Gaussian or Gauss-Jordan elimination with an augmented matrix.
1 2 3 4
5 6 7 8
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S k i l l
i n
A R I T H M E T I C
Lesson 5
# ADDING WHOLE NUMBERS AND DECIMALS
In this Lesson, we will answer the following:
1. How do we add whole numbers in writing?
2. How do we add decimals?
1. How do we add whole numbers in writing? 4,674 8,422 5,533 3,840 Write the numbers in a column and align the same units; that is, align the ones, the tens, etc. Then, starting with the ones on the right, add each column. When the sum of a column is 9 or less, write that sum. But when that sum is more than 9, write the ones of that sum and regroup the tens onto the next column. For, we may compose 10 of a lower unit into 1 of the next higher unit.
4 thousands + 6 hundreds + 7 tens + 4 ones 8 thousands + 4 hundreds + 2 tens + 2 ones 5 thousands + 5 hundreds + 3 tens + 3 ones 3 thousands + 8 hundreds + 4 tens + 0 ones 22 thousands + 4 hundreds + 6 tens + 9 ones
On adding the ones column, we get 9. But on adding the tens column, we get 16. Now,
16 tens = 10 tens + 6 tens
(Lesson 3.) Write 6, and regroup 10 tens onto the hundreds column -- because 10 tens are equal to 1 hundred. Therefore when we add the hundreds, we have
1 + 6 + 4 + 5 + 8 = 24
Again,
24 hundreds = 20 hundreds + 4 hundreds
Write 4, and regroup 20 hundreds onto the thousands column -- because 20 hundreds are equal to 2 thousands. (Lesson 3, Example 3c.)
When we add the thousands, then, we get
2 + 4 + 8 + 5 + 3 = 22.
The sum of those numbers is 22,469.
Example. 9545 + 8982 18527
"5 + 2 is 7." Write 7.
"4 + 8 is 12." Write 2 -- carry 1 mentally.
"5 + 9 is 14, plus 1 is 15." Write 5 -- carry 1 mentally.
"9 + 8 is 17, plus 1 is 18." Write 18.
2. How do we add decimals? Add decimals in the same way as whole numbers, taking care to align the same units. Specifically, align the ones, because every number has ones.
Example 1. Write in a column and add: 4,785 + 9 + 2.307
Solution.. The ones are shown:
4,785 + 9 + 2.307
(Lesson 2, Question 5.) Therefore, align as follows:
4,785 9 + 2 .307 4,796 .307
Example 2. Write in a column and add: .58 + 5.8 + 58
Solution. Here are the ones:
_.58 + 5.8 + 58
As for .58, the ones are at the first place to the left of the decimal point.
Align as follows:
.58 5 .8 + 58 64 .38
When there are decimal points, align them. The decimal point in the answer will fall in the same place. (But this is true only in addition and subtraction, not in multiplication. Lesson 9, Question 3.)
As for a whole number such as 58, to help with alignment we may imagine a decimal point after the 8.
58 = 58.
Whole numbers, however, are normally written without a decimal point because the decimal point means "and" -- here come the fractions!
Example 3. .5 + .5 + .5
a) 15
b) .15
c) 1.5
d) .015
For, if we aligned them and wrote .5 as 0.5 --
1 0.5 0.5 + 0.5 1.5
-- we would see that the 1 (of 15 tenths) carries over into the next column.
The point is:
will have the same number of decimal places as the numbers themselves.
Example 4. .007 + .003 + .004
The numbers being added have three decimal places. Therefore the answer also will have three decimal places.
Example 5. Perimeters. The perimeter of a plane (flat) figure is its boundary.
This figure is a rectangle, which is a four-sided figure in which all the angles are right angles. In a rectangle, the opposite sides are equal. Therefore the perimeter of that rectangle is:
12 + 12 + 6.5 + 6.5 = 24 + 13 = 37 in
2 0.83 7 0.49 6 0.26 + 8 0.58
Technique. Do not break this up into separate pieces. Add each entire column, starting on the right.
2 0.83 7 0.49 "12" 6 0.26 "18" + 8 0.58 "26" 6
Write 6, carry 2. To add the middle column, say
2 2 0.83 "10" 7 0.49 "14" 6 0.26 "16" + 8 0.58 "21" 0.16
Write 1, carry 2. To add the last column, say
2 2 "4" 2 0.83 "11" 7 0.49 "17" 6 0.26 "25" 8 0.58 25 0.16
Write 25.
The decimal point in the answer is aligned with the decimal points above. (This is true only in addition and subtraction, not in multiplication. Lesson 9.) We may write the decimal point in the answer when we come to it; that is, upon adding the middle column.
Please "turn" the page and do some Problems.
or
www.proyectosalonhogar.com |
# Golden Ratio in Nature
This nautilus shell demonstrates a perfect golden ratio spiral.
Are there patterns in nature? There are likely many, as successful patterns tend to be repeated. However, there is one pattern that is very recognizable and is found in a variety of organisms and even inorganic things like funnel clouds and galaxies. It is called the golden ratio and has a mathematical base.
The golden ratio is simply 1.6180339887498948482...(an infinite number) or its inverse, 0.618.... and is represented by the Greek letter, phi ɸ. The ratio is this: with the line, A____________C_________B, if AB/AC=AC/CB (the ratio of line AB to AC is the same as the ratio of line AC to CB) it is a golden ratio.
How it is achieved and how it applies take a bit of explaining. Start with the Fibonacci sequence. This is an infinite series of numbers derived from adding a sequence of numbers this way: 1, 0+1=1, 1+1=2, 1+2=3, 2+3=5, 3+5=8, 5+8=13, 8+13=21, 13+21=34, 21+34=55 and so on, where each number in the sequence is the sum of the two numbers that precede it. Each summed number is a Fibonacci number (1, 2, 3, 5, 8, 13, 21, 34, 55 and so on) and the ratio between the numbers is the golden ratio.
One application of this sequence creates spirals in nature that are precise and functional. To best illustrate this, start with a single square, 1x1 inch. Next to it, add to it a second 1x1 box forming a 1x2 rectangle. On top of this, add a 2x2 inch square, forming a rectangle 3 inches tall by 2 inches wide. On the side of this, add a 3x3 inch square, forming a new rectangle, 3x5 inches. Now add another square, this one 5x5 inches, forming a rectangle 5x8 inches. Continue as long as you want. It should look something like this:
When you run an arc through opposite corners of each square, a specific spiral occurs. That is cool, and it turns out that nature uses this spiral in many designs. The most commonly used example is the shell of the sea creature called a nautilus. While some mathematicians argue (they rarely agree on anything) that the distance measurement between the curves is logarithmic and not Fibonacci, the curve itself perfectly aligns with the shell.
Other well-known examples of a Fibonacci sequence and the golden ratio include the seeds in the head of a sunflower, the scales on a pinecone and the rings on the outside of a pineapple. Many seashells also show this curving design. Not only do they have the spiral pattern, the numbers of curves are almost always (very little is ever absolute in nature) Fibonacci numbers. For instance, the cone of a ponderosa pine in my collection has 13 spirals one direction and 8 in the other direction.
Flowers often have Fibonacci-numbered petals and leaves. And even the leaf arrangement is derived from what is considered the golden angle, 137.5 degrees. When leaves are oriented in this fashion, all the leaves receive maximum exposure to sunlight without being shaded excessively by neighboring leaves. It is not that the plants have an internal ruler, but growth inhibitors do keep the leaves from growing too close to each other and over time plants have found this perfect angle.
For eons, humans have either inadvertently or purposely built using the golden ratio. Many ancient places, including some of the pre-Columbian structures at Mesa Verde National Park, are built upon multiples of the golden ratio.
When you observe carefully, you will find the golden ratio and the Fibonacci sequence in many aspects of nature. Of course, it isn’t the only design out there, but it is a fascinating one and lends credence to the concept of an intelligent design. I can’t begin to do justice to this topic in the space allotted, so, for more on the subject, I refer you to the clever and entertaining three-part series by Vi Hart: https://www.youtube.com/watch?v=ahXIMUkSXX0, and, It’s Okay to be Smart’s presentation on the same subject: https://www.youtube.com/watch?v=1Jj-sJ78O6M.
Help Idaho Wildlife
When we traveled across the state in October 2017, most of the vehicles we saw using the wildlife management areas did not have wildlife plates. Buying wildlife plates is a great way for non-hunters and hunters alike to support wildlife-based recreation like birding.
C'mon folks, let's help Idaho's wildlife by proudly buying and displaying a wildlife license plate on each of our vehicles!
See below for information on Idaho plates. Most states have wildlife plates so if you live outside Idaho, check with your state's wildlife department or vehicle licensing division for availability of state wildlife plates where you live.
And tell them that you heard about it from Nature-track.com!
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# Absolute Value Functions and Graphs – Real World Applications
There are many ways to WOW your students with math. This is a lesson that can be extremely fun for both you and them. Absolute Value Functions and Graphs are something that can be related to one of the things they understand most in the world because they have no choice but to use it everyday… Distance and Travel!
## Absolute Value Functions Real World Applications
As you know Absolute Value is a way to eliminate negative numbers in situations in which they are not called for. For example, if your student walks to school and it is a 2 mile walk we would say he or she walked 2 miles. However, when the student walks home from school that night we would not say it’s a (Negative) -2 mile walk. For instance, when dealing with direction we can illustrate many of these situations so that the students can have visual aids to prove everything you teach them. Here is an example…
## Graphing Absolute Functions
Next, have them graph their travel. The time in minutes will be the X axis and the distance from their home will be on the Y axis. Their graph should look something like the one above…
## Absolute Value Functions Project
A more advanced lesson (or even a Project) could involve them using the GPS on their phone or tablet to plan a class trip. Say they fly from the nearest airport to Hawaii and then back. Have them chart the time and distance of the trip using the GPS system (Google Maps, Maps for iPhone, ect.). If you really want to blow their minds have them each plan different trips and then graph their travel using the same method above. They will all start to notice that all of their graphs are the same shape. Better yet, they will all notice when someone’s graph is different and/or has a mistake. They will notice that some graphs are tall and skinny and some are low and stretched out across the X axis. This will lead you into your lesson on Transforming Absolute Value Functions and Graphs.
# Here is your Free Content for this Lesson!
## Absolute Value Functions and Graphs – Word Docs & PowerPoints
To gain access to our editable content Join the Algebra 2 Teacher Community! Here you will find hundreds of lessons, a community of teachers for support, and materials that are always up to date with the latest standards.
## Absolute Value Functions and Graphs Worksheet – PDFs
2-5 Assignment – Absolute Value Functions and Graphs
2-5 Bellwork – Absolute Value Functions and Graphs
2-5 Exit Quiz – Absolute Value Functions and Graphs
2-5 Guided Notes SE – Absolute Value Functions and Graphs
2-5 Guided Notes TE – Absolute Value Functions and Graphs
2-5 Lesson Plan – Absolute Value Functions and Graphs
2-5 Online Activity – Absolute Value Functions and Graphs
2-5 Slide Show – Absolute Value Functions and Graphs |
Difference between revisions of "2010 AIME II Problems/Problem 14"
Problem
Triangle $ABC$ with right angle at $C$, $\angle BAC < 45^\circ$ and $AB = 4$. Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$. The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$, where $p$, $q$, $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$.
Solution
Let $O$ be the circumcenter of $ABC$ and let the intersection of $CP$ with the circumcircle be $D$. It now follows that $\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}$. Hence $ODP$ is isosceles and $OD = DP = 2$.
Denote $E$ the projection of $O$ onto $CD$. Now $CD = CP + DP = 3$. By the pythagorean theorem, $OE = \sqrt {2^2 - \frac {3^2}{2^2}} = \sqrt {\frac {7}{4}}$. Now note that $EP = \frac {1}{2}$. By the pythagorean theorem, $OP = \sqrt {\frac {7}{4} + \frac {1^2}{2^2}} = \sqrt {2}$. Hence it now follows that,
$$\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}$$
This gives that the answer is $\boxed{007}$.
An alternate finish for this problem would be to use Power of a Point on $BA$ and $CD$. By Power of a Point Theorem, $CP\cdot PD=1\cdot 2=BP\cdot PA$. Since $BP+PA=4$, we can solve for $BP$ and $PA$, giving the same values and answers as above.
$[asy] /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(250); real lsf = 0.5; /* changes label-to-point distance */ pen xdxdff = rgb(0.49,0.49,1); pen qqwuqq = rgb(0,0.39,0); pen fftttt = rgb(1,0.2,0.2); /* segments and figures */ draw((0.2,0.81)--(0.33,0.78)--(0.36,0.9)--(0.23,0.94)--cycle,qqwuqq); draw((0.81,-0.59)--(0.93,-0.54)--(0.89,-0.42)--(0.76,-0.47)--cycle,qqwuqq); draw(circle((2,0),2)); draw((0,0)--(0.23,0.94),linewidth(1.6pt)); draw((0.23,0.94)--(4,0),linewidth(1.6pt)); draw((0,0)--(4,0),linewidth(1.6pt)); draw((0.23,(+0.55-0.94*0.23)/0.35)--(4.67,(+0.55-0.94*4.67)/0.35)); /* points and labels */ label("1", (0.26,0.42), SE*lsf); draw((1.29,-1.87)--(2,0)); label("2", (2.91,-0.11), SE*lsf); label("2", (1.78,-0.82), SE*lsf); pair parametricplot0_cus(real t){ return (0.28*cos(t)+0.23,0.28*sin(t)+0.94); } draw(graph(parametricplot0_cus,-1.209429202888189,-0.24334747753738661)--(0.23,0.94)--cycle,fftttt); pair parametricplot1_cus(real t){ return (0.28*cos(t)+0.59,0.28*sin(t)+0); } draw(graph(parametricplot1_cus,0.0,1.9321634507016043)--(0.59,0)--cycle,fftttt); label("\theta", (0.42,0.77), SE*lsf); label("2\theta", (0.88,0.38), SE*lsf); draw((2,0)--(0.76,-0.47)); pair parametricplot2_cus(real t){ return (0.28*cos(t)+2,0.28*sin(t)+0); } draw(graph(parametricplot2_cus,-1.9321634507016048,0.0)--(2,0)--cycle,fftttt); label("2\theta", (2.18,-0.3), SE*lsf); dot((0,0)); label("B", (-0.21,-0.2),NE*lsf); dot((4,0)); label("A", (4.03,0.06),NE*lsf); dot((2,0)); label("O", (2.04,0.06),NE*lsf); dot((0.59,0)); label("P", (0.28,-0.27),NE*lsf); dot((0.23,0.94)); label("C", (0.07,1.02),NE*lsf); dot((1.29,-1.87)); label("D", (1.03,-2.12),NE*lsf); dot((0.76,-0.47)); label("E", (0.56,-0.79),NE*lsf); clip((-0.92,-2.46)--(-0.92,2.26)--(4.67,2.26)--(4.67,-2.46)--cycle); [/asy]$
Solution 2
Let $AC=b$, $BC=a$ by convention. Also, Let $AP=x$ and $BP=y$. Finally, let $\angle ACP=\theta$ and $\angle APC=2\theta$.
We are then looking for $\frac{AP}{BP}=\frac{x}{y}$
Now, by arc interceptions and angle chasing we find that $\triangle BPD \sim \triangle CPA$, and that therefore $BD=yb.$ Then, since $\angle ABD=\theta$ (it intercepts the same arc as $\angle ACD$) and $ADB$ is right,
$\cos\theta=\frac{DB}{AB}=\frac{by}{4}$.
Using law of sines on $APC$, we additionally find that $\frac{b}{\sin 2\theta}=\frac{x}{\sin\theta}.$ Simplification by the double angle formula $\sin 2\theta=2\sin \theta\cos\theta$ yields
$\cos \theta=\frac{b}{2x}$.
We equate these expressions for $\cos\theta$ to find that $xy=2$. Since $x+y=AB=4$, we have enough information to solve for $x$ and $y$. We obtain $x,y=2 \pm \sqrt{2}$
Since we know $x>y$, we use $\frac{x}{y}=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}$
Solution 3
Let $\angle{ACP}$ be equal to $x$. Then by Law of Sines, $PB = -\frac{\cos{x}}{\cos{3x}}$ and $AP = \frac{\sin{x}}{\sin{3x}}$. We then obtain $\cos{3x} = 4\cos^3{x} - 3\cos{x}$ and $\sin{3x} = 3\sin{x} - 4\sin^3{x}$. Solving, we determine that $\sin^2{x} = \frac{4 \pm \sqrt{2}}{8}$. Plugging this in gives that $\frac{AP}{PB} = \frac{\sqrt{2}+1}{\sqrt{2}-1} = 3 + 2\sqrt{2}$. The answer is $\boxed{007}$.
Solution 4 (The quickest and most elegant)
Let $\alpha=\angle{ACP}$, and$\beta=\angle{ABC}$ . By Law of Sines,
$\frac{1}{sin(\beta)}=\frac{BP}{sin(90-\alpha)}\implies sin(\beta)=\frac{cos(\alpha)}{BP}$ (1), and
$\frac{AP}{sin(\alpha)}=\frac{4sin(\beta)}{sin(2\alpha)} \implies 4-BP=\frac{2sin(\beta)}{cos(\alpha)}$. (2)
Then, substituting (1) into (2), we get
$4-x=\frac{2}{BP} \implies BP^2-4BP+2=0 \implies BP=2-\sqrt{2} \implies \frac{BP}{AP}=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}$
The answer is $\boxed{007}$. |
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4.13
Precalculus
Skip to 0 minutes and 11 seconds Hello and welcome back to a step in practice. We are dealing here with equations involving a radical. In exercise 1, we want to solve the equation 3 plus square of x minus 1 equals x.
Skip to 0 minutes and 34 seconds In every equation, we have first of all to study the domain. The domain here is the set of numbers, such that x minus 1 is greater or equal than 0. Otherwise, the square root of x minus 1 does not make sense. So it’s the interval 1 plus infinity. So we write here that we need that x be greater than 1.
Skip to 1 minute and 5 seconds We must remember this condition.
Skip to 1 minute and 10 seconds Now, we put the square on the left-hand side, and the linear terms on the right-hand side, and we write that the equation is equivalent when x is in D. So this was the first step, this is the second step. For x in D, the equation is equivalent to square root of x minus 1, equals x minus 3. And now recall what Francis told us studying this kind of equations. This is equivalent to two conditions. x minus 3 greater or equal than 0, and the squared the equation, x minus 1, which is square root of x minus 1 to the square equals x minus 3 to the square. So the first condition gives us x greater or equal than 3.
Skip to 2 minutes and 12 seconds So another constraint, and we have to take care and remember it. Now we studied the squared equation. So x minus 1 equals x minus 3 to the square is equivalent to x minus 1 equals x squared minus 6x plus 9, and this turns out to be equivalent to x squared minus 7x plus 10 equal to 0. Now, the discriminant of the equation is 49 minus 40. It is 9, 3 to the square. And thus, there are two roots, 1 is equal equal– x equals 7 minus 3 over 2, or x equals 7 plus 3 over 2. This gives 2 or x equals 5. Now we look at the compatibility with the other conditions that we found.
Skip to 3 minutes and 32 seconds We must have x greater than 3 and x greater than 1, that is, x greater than 3, so we exclude 2 and it remains x equals 5. So the solution is 5. Let us check that it works. Square root of 5 minus 1 is square root of 4, that is 2, and 2 plus 3 gives 5. And it is true that when x is 5, so the equality is satisfied.
Skip to 4 minutes and 7 seconds In exercise 2, we are asked to solve the equation 2x plus square root of 2 minus x equals 1. Again, we study the natural domain of the equation, which is the set D equals to the x satisfying 2 minus x greater or equal than 0 because we want the square root of 2 minus x to be defined. So this means the interval minus infinity 2. So take care. And what follows, we must be sure that, at the end, x will be less or equal than 2. Let us write it here, apart.
Skip to 4 minutes and 58 seconds Now, for x in D, the equation is equivalent to square root of 2 minus x equals to 1 minus 2x, and we know as Francis showed us, that this is equivalent to a couple of facts. One, 1 minus 2x is greater or equal than 0, and 2 minus x, we square the equation 2 minus x equals 1 minus 2x to the square. Now, take care. This implies that x should be less or equal than 1/2. So we write here x less or equal than 1/2.
Skip to 5 minutes and 48 seconds And now we studied the squared equation.
Skip to 5 minutes and 52 seconds So it is equivalent to 2 minus x equals 1 plus 4x to the square minus 4x, which is 4 x square minus 3x minus 1 equals to 0. Now, you see there is one evident root, x equal to 1, 4 minus 3 minus 1 gives 0.
Skip to 6 minutes and 23 seconds And the other root, it’s minus 1 over 4.
Skip to 6 minutes and 32 seconds We can reach the result also by studying the discriminant, by means of the formulas. Now, we study the compatibility of the solutions with our conditions. We know that x must be less or equal than 1/2. So of course, we must exclude x equal to 1. It remains x equal to minus 1/4, which is actually less than 2 and less than 1/2. So the solution is minus 1/4.
Skip to 7 minutes and 9 seconds And this ends our step.
Skip to 7 minutes and 14 seconds See you in the next step.
Equations involving a radical in practice
The following exercises are solved in this step.
We invite you to try to solve them before watching the video.
In any case, you will find below a PDF file with the solutions.
Exercise 1.
Solve the equation $$3+\sqrt{x-1}=x$$.
Exercise 2.
Solve the equation $$2x+\sqrt{2-x}=1$$. |
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# Evaluate the following: $\sin 30^\circ - \cos 30^\circ$ .
Last updated date: 22nd Jul 2024
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Hint: We can see that in the above we have been a trigonometric expression. We will try to remember the values of the given trigonometric ratios for the given corresponding angles and then we substitute the values in the equation. After that we will simplify the equation to get the answer.
Here we have $\sin 30^\circ - \cos 30^\circ$ .
We should know the values of the following trigonometric ratios:
$\theta$ $0^\circ$ $30^\circ$ $45^\circ$ $60^\circ$ $90^\circ$ $\sin$ $0$ $\dfrac{1}{2}$ $\dfrac{1}{{\sqrt 2 }}$ $\dfrac{{\sqrt 3 }}{2}$ $1$ $\cos$ $1$ $\dfrac{{\sqrt 3 }}{2}$ $\dfrac{1}{{\sqrt 2 }}$ $\dfrac{1}{2}$ $0$
From the above table we get that the value of
$\sin 30^\circ = \dfrac{1}{2}$
And the value of
$\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}$
Now we will substitute the values in the given equation:
$\dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}$
On simplifying we have
$\dfrac{{1 - \sqrt 3 }}{2}$
Hence the required value is $\dfrac{1}{2}\left( {1 - \sqrt 3 } \right)$ .
Note:
We should always remember the values of all other trigonometric ratios with their corresponding angles too. The values of the angles that lie in the first quadrant are positive for all.
ratios, while in the second quadrant the value of sine and cosec is positive.
Similarly in the third quadrant, the value of tangent $(\tan )$ and cotangent $(\cot )$ is positive. And in the fourth quadrant the value of cosine $(\cos )$ and secant i.e. $(\sec )$ is positive. |
Kirchhoff's Current and Voltage Laws
(Difference between revisions)
Kirchhoff's Current Law and Nodal Analysis
Kirchhoff's Current Law (KCL) says that the current going into a junction or node is equal to the current going out of a node. In other words, the sum of the currents entering the node must be zero (if we consider currents leaving the node to be a negative current entering the node). Consider the following diagram:
For the node A in the center, i1 and i2 are entering the node, and i3 and i4 are leaving the node. We would write:
$i1+i2+(-i3)+(-i4)=0\,$ which can also be written as $i1+i2=i3+i4\,$
Note that i7 is equal to i2; we can prove this by analyzing node B. We can also treat everything between node C and D as one big node, and conclude that i5 is equal to i6 without having to know the value of any of the currents within.
When solving for the currents in a real problem, we can choose arbitrarily in which direction the arrows point. If the current flows opposite to the direction of our arrow, the value we obtain after solving for the current will be negative. If you draw i4 as leaving node A as in the diagram above, then i4 must be entering node D.
If we had drawn the currents for node A as following:
Then our node equation looks like:
$i1+i2+i3+i4=0\,$
Unless all the currents are zero, one or more of the currents must turn out to be negative. The negative currents flow in the direction opposite from that which the arrow is pointing.
Kirchhoff's Voltage Law and Loop Analysis
Kirchhoff's Voltage Law (KVL) states that the sum of the voltages around any closed loop is equal to zero. Also, the voltage between any two nodes is the same no matter which path is taken. In the following diagram, it doesn't matter which path you choose(or which direction you go in) to add the voltages of the components from point A back to point A, they all must add up to zero.
Also, the voltage between any two points does not depend on which path you take. In the following diagram, the voltage between point A and point B is the same no matter which path is chosen.
To keep all the positive and negative signs lined up when solving a circuit, we use the following sign convention for the voltage and current for the components:
With the sign convention above, iv > 0 means that the element is storing power (for a capacitor or inductor) or dissipating power (for a resistor). If iv < 0, the element is sourcing, or providing, power. Note that resistors cannot store power (and hence cannot supply power); the energy absorbed by a resistor is lost as heat energy.
The circuit below shows a battery, resistor, inductor, and capacitor connected in series.
In this circuit, we have chosen to draw the current flowing clockwise and assign the voltages for our components according to our sign convention. We then add up the voltages of the components, following the loop. It doesn't matter if you go clockwise or counter-clockwise. If you go from a "+" sign to a "-" sign when going around the loop, subtract the voltage across that element. Otherwise, add it. Once you've completed the loop, set the summation equal to zero.
In the example above, if we choose to start in the bottom left corner and move in the clockwise direction, our equation looks like:
$V_{V1}-V_{R1}-V_{L1}-V_{C1}=0\,$
or if we go in the counter-clockwise direction, we get:
$V_{C1}+V_{L1}+V_{R1}-V_{V1}=0\,$
which is the same thing.
If we had drawn the current going in a counter-clockwise direction, our circuit diagram would look like this:
$V_{V1}+V_{R1}+V_{L1}+V_{C1}=0\,$
Solving Problems Using Kirchhoff's Laws
To solve for the currents and voltages in a circuit, simply write all KCL and KVL equations and constitutive laws for each element and solve simultaneously. Note that some KCL and KVL equations are redundant and can be eliminated.
1. Write all KCL equations for each node
2. Write all KVL equations for each loop
3. Write all constitutive equations
4. Solve simultaneous equations
Example and Practice Problems
Example 1
Find the voltage and current of each resistor at the steady state of the system.
Answer: At steady state, the capacitor becomes charged and acts like an open circuit, and the inductor acts like a short circuit. Therefore, at steady state, our system is equivalent to:
Notice in this circuit, we have two nodes (one at the top and one at the bottom) and three loops (the loop on the left consisting of Vs,R1, and R3; the loop on the right consisting of R2 and R3; and the loop around the entire perimeter, consisting of V2,R1, and R2.
We solve problems like this by writing out the loop equations, node equations, and constitutive laws. Then, we solve the equations simultaneously.
In this example, we draw our current arrows at the nodes as follows:
The directions for the currents were chosen arbitrarily. Keep in mind that there is no "wrong" direction when choosing the direction of the arrow, as long as we are consistent with each current's direction.
We can now write our equations:
KCL equations:
Node A: i1 = i2 + i3
Node B: i2 + i3 = i1
As we can see, the two equations are exactly the same, so we won't be needing both of them.
KVL equations:
Loop 1 (left "window pane"): 10 − 200i1 − 300i3 = 0
Loop 2 (right "window pane"): 300i3 − 30i2 = 0
Loop 3 (entire "window frame"): 10 − 200i1 − 30i2 = 0
Notice that the equation for Loop 3 is equal to the sum of the equations for Loop 1 and Loop 2. This means that only 2 of the are linearly independent; the third one is redundant. We can choose any two of the three equations to use when solving the system of equations.
Now we take our node equation and two of our loop equations to make a system of equations (3 equations and 3 unknowns):
i1 = i2 + i3
10 − 200i1 − 300i3 = 0
300i3 − 30i2 = 0
Solving the system reveals the currents through each resistor: i1 = 0.044A,i2 = 0.040A, and i3 = 0.004A We then find the voltages across each resistor using the constitutive law for resistors (Ohm's law): V = IR.
V1 = (0.044)(200) = 8.8V
V2 = (0.040)(30) = 1.2V
V3 = (0.004)(300) = 1.2V
Alternative Method
First, simplify the network of resistors. The combined resistance of the three resistors is equal to:
$R_{total}=\frac{(30)(300)}{30+300}+200=227.273\Omega$
The voltage across the resistors is 10V, and using ohm's law, we find that the current going through the circuit is
i1 = 10 / 227.273 = 0.044A
This is the current flowing through the voltage source R1. Kirchhoff's voltage law tells us that the voltage or R2 and R3 are the same, and the sum of R1 and R2 is equal to 10V. We can find the voltage of R1 by using Ohm's law again:
V = (0.044A)(200Ω) = 8.8V
Then R2 and R3 must have
10 − 8.8 = 1.2V
across them, and from that we can calculate that R2 's current is
i1 = 1.2 / 30 = 0.04A
and R3 's current is
i3 = 1.2 / 300 = 0.004A
Note that the sum of the currents for R2 and R3 is equal to the current in R1.
Example 2
Find the voltage and current of each element in the system.
We start by assigning currents to the nodes:
The we write our node equations, loop equations and constitutive laws.
KCL equations:
Node A: i1i2i3 = 0
Node B: i1 + i2 + i3 = 0
Again, these equations are equivalent, so we only need one of them.
KVL equations:
Left-side loop: 1000i1 − 1000i3 − 10 = 0
Right-side loop: 10 + 1000i3 − 2000i2 − 20 = 0
Beware the sign convention that we use: the voltage of a resistor is high at the end the current is entering into.
These loops already cover the entire circuit. The equation for the "outside frame" will only be the superposition of these two equations, so we omit it.
The constitutive law for the resistor is V = iR.
Solving the loop and node equations yields:
i1 = 0.008A
i2 = − 0.006A
i3 = − 0.002A
Using V=iR, we can find the voltages across each resistor.
VR1 = 8V
VR2 = − 12V
VR3 = − 2V
The negative value simply means that the voltage is actually higher on at the (-) terminal drawn in the diagram we used to derive our equations.
The voltage across the voltage source is constant, so:
VV1 = 10V
VV2 = 20V
Alternative Method
Kirchhoff's current law says that a current flowing into a node also has to flow out of it. If we trace the current's path in a loop, we can also solve for this current. In the figure below, we have two "current loops," and solving for i1 and i2 will give us the currents flowing through each element. The current through V1 and R3 will be the superposition of i1 and i2.
We can solve this by writing two loop equations using KVL--one for each loop. Note that R3 and V1 have both i1 and i2 going through them; therefore, the total currents going through R3 and V1 will be the superposition of i1 and i2.
For the loop on the left (in the direction of the current):
0 = VV1VR3VR1
0 = V1 − (i1i2)R3 − VR1
0 = 10 − 1000(i1i2) − 1000(i1),
and for the loop on the right:
0 = VV2VR2VR3V1
0 = VV2i2R2 − (i2i1)R3VV1
0 = 20 − 2000i2 − 1000(i2i1) − 10
Now we have a system of two equations and two unknowns, which is easily solved:
i1 = 8mA and i2 = 6mA
Now we use ohm's law and calculate our voltages:
VV1 = 10V
VV2 = 20V
VR1 = 8V
VR2 = 12V
VR3 = 2V
And our currents:
IV1 = 6mA
IV2 = 2mA
IR1 = 8mA
IR2 = 6mA
IR3 = 2mA
Note that we could also have drawn our loops like this:
In this case, our equations would be:
for the small loop:
0 = VV1VR3VR1
0 = 10 − 1000(i1) − 1000(i1 + i2)
and for the big loop:
0 = VV2VR2VR3
0 = 20 − 2000i2 − 1000(i2 + i1)
Solving for i1 and i2, we find that:
i1 = 2mA
and
i2 = 6mA
The current going through R1 is equal to i1 + i2 = 8mA, which matches our previous answer.
Practice Problem 1
Solve for the voltage across and current through each of the components:
Practice Problem 2
Solve for the voltage across and current through each of the resistors at time $t=0\,$ and $t=\infty$. Assume that both the inductor and the capacitor are uncharged in their initial states. |
# CE 201 - Statics Lecture 2. Contents Vector Operations – Multiplication and Division of Vectors – Addition of Vectors – Subtraction of vectors – Resolution.
## Presentation on theme: "CE 201 - Statics Lecture 2. Contents Vector Operations – Multiplication and Division of Vectors – Addition of Vectors – Subtraction of vectors – Resolution."— Presentation transcript:
CE 201 - Statics Lecture 2
Contents Vector Operations – Multiplication and Division of Vectors – Addition of Vectors – Subtraction of vectors – Resolution of a Vector Vector Addition of Forces Analysis of Problems
Vector Operations Multiplication and Division of a Vector by a Scalar Vector A Scalar a A a = aA Magnitude of aA Direction of A if a is positive (+) Direction of –A (opposite) if a is negative (-) A -A 1.5 A
Vector Addition Vectors are added according to the parallelogram law The resultant R is the diagonal of the parallelogram If two vectors are co-linear (both have the same line of action), they are added algebraically A B A B R = A + B A B B A A B
Vector Subtraction The resultant is the difference between vectors A and B A B -B A R R A
Resolution of a Vector If lines of action are known, the resultant R can be resolved into two components acting along those lines (i.e. a and b). a b A B R
Vector addition of Forces Force! Is it vector OR scalar? Why? The two common problems encountered in STATICS are: 1. Finding the RESULTANT (by knowing the COMPONENTS). OR 2. Resolving a FORCE into its COMPONENTS (by applying the parallelogram law). If more than two forces are to be added!! Apply the same law more than once depending on the number of forces.
3 Forces F1 F2 F3 R1=F1+F2 R2=R1+F3
Four Forces F1 F2 F3 R1=F1+F2 R2=R1+F3 F4 R3=R2+F4
Analysis of Problems Two procedures to be followed: Parallelogram law Trigonometry sine and/or cosine laws may be used
Sine Law A/sin (a) = B/sin (b) = C/sin (c) a A B b C c
Cosine Law C = A 2 +B 2 -2ABcos (c) a A B b C c
Examples 2.1 2.2 2.3 2.4 Problem 2-8 Problem 2-25
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Solve for x: 3x + 5 < 5
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Updated 328 days ago|9/25/2017 2:49:01 AM
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User: Solve for x: 3x + 5 < 5
Weegy: Solution: 3x+6?15 First, let's work on the left hand side of your inequality, the 3x+6 This means, for instance, [ to see if it can be simplified at all. Multiply x and 3 Multiply x and 1 The x just gets copied along. The answer is x x 3*x evaluates to 3x 3*x+6 evaluates to 3x+6 So, all-in-all, the left hand side of your inequality can be written as: 3x+6 Now, let's work on the right hand side of your inequality, the 15 The right hand side of your inequality can be written as: 15 So with these (any) simplifications, the inequality we'll set out to solve is: 3x+6 ? 15 Move the 6 to the right hand side by subtracting 6 from both sides, like this: From the left hand side: 6 - 6 = 0 The answer is 3x From the right hand side: 15 - 6 = 9 The answer is 9 Now, the inequality reads: 3x ? 9 To isolate the x, we have to divide both sides of the inequality by the other "stuff" (variables or coefficients) around the x on the left side of the inequality. The last step is to divide both sides of the inequality by 3 like this: To divide x by 1 The x just gets copied along in the numerator. The answer is x 3x ? 3 = x 9 ? 3 = 3 The solution to your inequality is: x ? 3 So, your solution is: x must be less than 3 ]
bongche|Points 2230|
User: Solve for x: 3x + 5 < 5
Weegy: Solution: 3x+6?15 First, let's work on the left hand side of your inequality, the 3x+6 This means, for instance, [ to see if it can be simplified at all. Multiply x and 3 Multiply x and 1 The x just gets copied along. The answer is x x 3*x evaluates to 3x 3*x+6 evaluates to 3x+6 So, all-in-all, the left hand side of your inequality can be written as: 3x+6 Now, let's work on the right hand side of your inequality, the 15 The right hand side of your inequality can be written as: 15 So with these (any) simplifications, the inequality we'll set out to solve is: 3x+6 ? 15 Move the 6 to the right hand side by subtracting 6 from both sides, like this: From the left hand side: 6 - 6 = 0 The answer is 3x From the right hand side: 15 - 6 = 9 The answer is 9 Now, the inequality reads: 3x ? 9 To isolate the x, we have to divide both sides of the inequality by the other "stuff" (variables or coefficients) around the x on the left side of the inequality. The last step is to divide both sides of the inequality by 3 like this: To divide x by 1 The x just gets copied along in the numerator. The answer is x 3x ? 3 = x 9 ? 3 = 3 The solution to your inequality is: x ? 3 So, your solution is: x must be less than 3 ]
bongche|Points 2230|
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3x + 5 < 5;
3x < 5 - 5;
3x < 0;
x < 0/3;
x < 0
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#### NCERT Class 12 Chapter – 8: Application of Integrals
Area between two curves: Based on formula given in Application of Integrals
CASE – 1:
For finding the Area bounded by the curve y = f(x), x-axis and the lines x = a, x = b let us consider a very thin vertical strip of length y and width dx. Therefore, Area of the strip (dA) = y dx [Since, y = f(x)]
Hence the total Area enclosed by the curve y = f(x), x- axis and the lines x = a, x = b:
$$\boldsymbol{A =\int_{a}^{b} dA=\int_{a}^{b}y\;dx}$$
Therefore, A = $$\int_{a}^{b}f(x)\;dx$$
CASE – 2:
Similarly the Area bounded by the curve x = g(y), y-axis and the lines y = c, y = d is given by:
$$A=\int_{c}^{d} dA=\int_{c}^{d} x\;dy$$
Therefore, A = $$\int_{c}^{d}g(y)\;dy$$
CASE – 3:
If the curve lies below x-axis, then the Area bounded by the curve y = f(x), x-axis and the lines x = a, x = b will come negative.
Since, the area cannot be negative therefore we will neglect the negative sign and considering its absolute value only.
$$A = \left |\int_{a}^{b}f(x)\;dx \right|$$
CASE – 4:
As shown above, some portion of the curve lies above the x-axis and some portion of the curve lies below the x-axis.
Here in this case, Area-1 > 0 and Area-2 < 0
Therefore, total Area bounded by the curve y= f(x), x-axis and the lines x=a, x=b is given by: A = |A|+B
NOTE:
$$\boldsymbol{\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}}$$
Example 1: Find the area enclosed by equation: x2 + y2 = 9
Sol:Based on formula given in Application of Integrals
Equation x2 + y2 = 32 represents the circle with centre (0,0) and radius 3 units.
Therefore, y = $$\sqrt{3^{2}-x^{2}}$$
From the figure, the Area enclosed by the circle = 4 × (Area enclosed by the curve ABOA)
Now, Area enclosed by the curve ABOA:
$$\boldsymbol{\Rightarrow }$$ $$\int_{0}^{3}y\;dx = \int_{0}^{3} \sqrt{9-x^{2}}\;dx$$
Since, $$\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{3}\sqrt{3^{2}-x^{2}}\;dx=\left | \frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin^{-1}\frac{x}{3} \right | _{0}^{3}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{3}{2} \sqrt{9-9}+\frac{9}{2} \sin^{-1}(1) – \frac{0}{2} \sqrt{9-0}-\frac{0}{2} \sin^{-1}\frac{0}{3}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{9}{2}\times \frac{\pi }{2}=\frac{9\pi }{4}}$$ unit2
Therefore, the Area enclosed by the circle = 4 × (area enclosed by curve ABOA)
$$\boldsymbol{\Rightarrow }$$ $$4\times \frac{9\pi }{4}=9\pi$$
Hence the Area enclosed by the circle = 9π unit2
Example 2: Find the area enclosed by the curve $$\frac{x^{2}}{9}+\frac{y^{2}}{4} = 1$$
Sol:Based on formula given in Application of Integrals
Equation $$\frac{x^{2}}{9}+\frac{y^{2}}{4} = 1$$ represents an ellipse with major axis = 3 units and minor axis = 2 units
Since, $$\frac{x^{2}}{9}+\frac{y^{2}}{4} = 1$$
$$\boldsymbol{\Rightarrow }$$ 4x2 + 9y2 = 36
$$\boldsymbol{\Rightarrow }$$ 9y2 = 62 – (2x)2
Therefore, $$y=\sqrt{\frac{6^{2}-(2x)^{2}}{9}}$$
Therefore, the Area of region enclosed by the ellipse = 4 × (Area enclosed by the curve ABOA)
Now, Area enclosed by the curve ABOA = $$\int_{0}^{3} y\;dx$$
Since, $$\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{3}\times \int_{0}^{3}\sqrt{6^{2}-(2x)^{2}}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{3}\int_{0}^{3}\;\frac{2x}{2}\sqrt{6^{2}-(2x)^{2}}+\frac{36}{2}\sin^{-1}\frac{2x}{6}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{3}\left | \frac{2x}{2}\sqrt{6^{2}-(2x)^{2}}+\frac{36}{2}\sin^{-1}\frac{2x}{6} \right |_{0}^{3}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{3}[\frac{6}{2}\sqrt{36-36}+18\sin^{-1}(1)]=3\pi }$$
Therefore, the Area enclosed by the curve ABOA = 3π unit2
Hence, the total Area enclosed by an ellipse ABA’B’ = 4×3π = 12π unit2.
Area of region bounded by the curve and line:
Example 3: Find the area of the region enclosed by the curve y = x2 and the line y = 3.
Sol: Based on formula given in Application of Integrals
y = x2 represents a parabola, symmetrical about y-axis as shown in the above figure. By substituting y =9 in equation of parabola y = x2 we will get coordinates of point M.
i.e. x2 = 9
Therefore, x = +3 or -3
Hence the coordinates of point M = (3, 9)
The area of the region bounded by the curve y = x2 and the line y = 9 is the area enclosed by curve POMP.
Now, Area of region POMP = 2(area of region AOMA)
Since, x2 = y
Therefore, x = $$\sqrt{y}$$
Thus, the Area of region bounded by the curve AOMA:
$$\boldsymbol{\Rightarrow }$$ $$\int_{0}^{9}x\;dy$$ = $$\int_{0}^{9}\sqrt{y}\;dy$$
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{{\left | \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{0}^{9}=\frac{2}{3}\times (9^{\frac{3}{2}})}}\\$$
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{2}{3}\times 27 = 18}$$unit2
Therefore, the Area of region bounded by the curve AOMA = 18 unit2
Hence, the Area of region bounded by the curve POMP = 2 × (area of region bounded by curve AOMA) = 36 unit2
Example 4: Find the area enclosed by the curve x2 + y2 = 50, x-axis and the line y = x in the 1st quadrant.
Sol: Based on formula given in Application of Integrals
Equation x2 + y2 = 50 represents a circle with radius $$\sqrt{50}$$ units.
Since y = x
Therefore x2 + x2 = 50 [for points of intersection of both the curves]
Hence, x = +5 and -5
Similarly, y = +5 and -5
Thus the coordinates of point B are (5, 5)
Form the figure, the area of region bounded by the curve x2 + y2 = 50, x = y and x – axis in the 1st quadrant is the Area enclosed by the curve OMABO.
i.e. Area of triangle OMB + Area under the curve MBAM.
Now, the Area of triangle = $$\frac{1}{2}\times Base \times Altitude$$
$$\boldsymbol{\Rightarrow }$$ $$\frac{1}{2}\times OM \times BM\;\;=\;\frac{1}{2}\times 5\times 5 =\frac{25}{2}$$ unit2.
Now, the Area under the curve MBAM = $$\int_{5}^{5\sqrt{2}}y\;dx$$
Since, x2 +y2 = 50. Therefore, y2 = 50 – x2
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{y=\sqrt{50-x^{2}}}$$
Therefore, the Area under curve MBAM [x2 + y2 = 50]:
Since, $$\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{5}^{5\sqrt{2}}\sqrt{50-x^{2}}\;dx=\left | \frac{x}{2}\sqrt{50-x^{2}}+\frac{50}{2}\;\sin^{-1}\frac{x}{5\sqrt{2}} \right |_{5}^{5\sqrt{2}}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{[\frac{5\sqrt{2}}{2}\times\ \sqrt{50-50}]+[\frac{50}{2}\;\sin^{-1}(1)]-[\frac{5}{2}\times \sqrt{50-25}]-[\frac{50}{2}\times \sin^{-1}\frac{1}{\sqrt{2}}]}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{0+\frac{25\pi }{2}-\frac{25}{2}-\frac{25\pi }{4}=\frac{25}{4}(\pi -2)}$$unit2
Therefore, the Area under the curve MBAM = $$\boldsymbol{\frac{25}{4}(\pi -2)}$$ unit2
Now, the total Area under the shaded region = Area of triangle OMB + Area under the curve MBAM
$$\boldsymbol{\Rightarrow }$$ $$\frac{25}{2}+\frac{25}{4}(\pi -2)=\frac{25}{2}+\frac{25\pi }{4}-\frac{25}{2}\boldsymbol{=\frac{25\pi }{4}}$$unit2
Hence, the Area of shaded region OMABO = $$\boldsymbol{\frac{25\pi }{4}}$$unit2
#### Exercise 8.1
Q.1: Find the area enclosed by the curve y = x2 and the lines y = 2, y = 4 and the y-axis.
Sol: Based on formula given in Application of Integrals
Equation y = x2 represents a parabola symmetrical about y-axis.
The Area of the region bounded by the curve y = x2, y = 2, and y = 4, is the Area enclosed by the curve AA’B’BA.
Now, the Area of region AA’B’BA = 2 (Area of region ABNMA)
Since, x2 = y
Therefore, x = $$\sqrt{y}$$
Thus, the Area of region bounded by the curve ABNMA [y = x2]:
$$\boldsymbol{\Rightarrow }$$ $$\int_{2}^{4}x\;dy$$ = $$\int_{2}^{4}\sqrt{y}\;dy$$
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{{\left | \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{2}^{4}=\frac{2}{3}\times [(4^{\frac{3}{2}})-(2)^{\frac{3}{2}}}]}$$
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{2}{3}\times (8-2\sqrt{2}) = \frac{4}{3}(4-\sqrt{2})}$$ unit2
Therefore, the Area of region bounded by the curve ABNMA = $$\boldsymbol{\frac{4}{3}(4-\sqrt{2})}$$unit2
Hence, the Area of region bounded by the curve AA’B’BA = 2(Area of region bounded by the curve ABNMA)= $$\boldsymbol{\frac{8}{3}(4-\sqrt{2})}$$unit2
Q.2: Find the area enclosed by the curve y2 = 4x and lines x = 1, x = 3 and the x- axis in the first quadrant.
Sol : Based on formula given in Application of Integrals
Equation y2 = 4x represents a parabola, symmetrical about the x-axis as shown in the above figure.
The area of the region bounded by the curve y2 = 4x, x = 1, x = 3 and the x-axis is the area enclosed by the curve ABCDA.
Now, the Area of region bounded by the curve ABCDA:
Since, y2 = 4x
Therefore, y = $$2\sqrt{x}$$
Hence, the area of region bounded by the curve ABCDA [y2 = 4x]:
$$\boldsymbol{\Rightarrow }$$ $$\int_{1}^{3}y\;dx$$ = $$\int_{1}^{3}2\sqrt{x}\;dx$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{{\left | 2\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{1}^{3}=\frac{4}{3}\times [(3^{\frac{3}{2}})-(1)}]}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{4}{3}\times (3\sqrt{3}-1) = \frac{-4\;+\;12\sqrt{3}}{3}}$$ unit2
Therefore, the area of region bounded by the curve ABCDA $$\boldsymbol{= \frac{-4\;+\;12\sqrt{3}}{3}}$$unit2
Q.3: Find the value of k if the line x = k divides the area enclosed by the curve y2 = 9x and the line x = 4 in to two equal parts.
Sol: Based on formula given in Application of Integrals
Equation y2 = 9x represents a parabola, symmetrical about the x-axis as shown in the above figure.
Since, the line x = k divides the Area OCBB’O in to two equal halves and the curve is symmetrical to x-axis. Therefore, Area of the region OADO = Area of the region ABCDA.
The Area of the region bounded by the curve y2 = 9x and the line x = k is the Area of region enclosed by the curve OADO.
Since, y2 = 9x
Therefore, y = $$3\sqrt{x}$$
Hence, the Area of region bounded by the curve OADO:
$$\boldsymbol{\Rightarrow }$$ $$\int_{0}^{k}y\;dx$$ = $$\int_{0}^{k}3\sqrt{x}\;dx$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{{\left | 3\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{1}^{3}=2\times [(k^{\frac{3}{2}})-(0)}]}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{2k^{\frac{3}{2}}}$$unit2
Therefore, the Area of region bounded by the curve OADO = $$\boldsymbol{2k^{\frac{3}{2}}}$$unit2
The Area of the region bounded by the curve y2 = 9x and the lines x = k and x = 4 is the Area under the curve ABCDA
Now, the Area of region bounded by the curve ABCDA [y2 = 9x]:
$$\boldsymbol{\Rightarrow }$$ $$\int_{k}^{4}y\;dx$$ = $$\int_{k}^{4}3\sqrt{x}\;dx$$
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{{\left | 3\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{k}^{4}=2\times [(4^{\frac{3}{2}})-(k)^{\frac{3}{2}}}]}$$
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{(16-2k^{\frac{3}{2}})}$$ unit2
Therefore, the Area of region bounded by the curve ABCDA = $$\boldsymbol{\Rightarrow (16-2k^{\frac{3}{2}})}$$ unit2
Since, the Area of region OADO = Area of region ABCDA [GIVEN]
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{(16-2k^{\frac{3}{2}})=2k^{\frac{3}{2}}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{k^{\frac{3}{2}}=4}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{k = 4^{\frac{2}{3}}}$$
Therefore, the value of k = $$\boldsymbol{4^{\frac{2}{3}}}$$
Q.4: Find the area enclosed by the curve y2 = 16x, y-axis and the line y = 2.
Sol: Based on formula given in Application of Integrals
Equation y2 = 16x represents a parabola, symmetrical about the x-axis as shown in the above figure.
The Area of the region bounded by the curve y2 = 16x, y = 2 and the y-axis is the Area of region enclosed by the curve OABO.
Now, the Area of region enclosed by the curve OABO:
Since, y2 = 16x
Therefore, x = $$\frac{y^{2}}{16}$$
Thus, the Area of region bounded by the curve OABO:
$$\\\boldsymbol{\Rightarrow }$$ $$\int_{0}^{2}x\;dy$$ = $$\int_{0}^{2}\frac{y^{2}}{16}\;dy$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{{\left | \frac{1}{16}\times \frac{y^{3}}{3}\right |_{0}^{2}=\frac{1}{16}\times [\frac{8}{3}-0}]}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{6}}$$ unit2
Therefore, Area of region bounded by the curve OABO $$\boldsymbol{=\frac{1}{6}}$$unit2
Q.5: Find the area enclosed by the curve y2 = 25x and the line x = 3
Sol: Based on formula given in Application of Integrals
Equation y2 = 25x represents a parabola, symmetrical about x-axis as shown in the above figure.
The area of the region bounded by the curve y2 = 25x and x = 3 is the Area enclosed by the curve BOCAB.
Now, the Area of region BOCAB = 2(Area of region OABO)
Since, y2 = 25x
Therefore, y = $$5\sqrt{x}$$
Hence, the Area of region bounded by the curve OABO:
$$\\\boldsymbol{\Rightarrow }$$ $$\int_{0}^{3}y\;dx$$ = $$\int_{0}^{3}5\sqrt{x}\;dx$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{{\left | 5\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{0}^{3}=\frac{10}{3}\times [(3^{\frac{3}{2}})-(0)}]}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{10}{3}\times (3\sqrt{3})=10\sqrt{3}}$$unit2
Therefore, the Area of region bounded by the curve OABO$$\boldsymbol{=10\sqrt{3}}$$ unit2
Now, the Area of region BOCAB = 2 × (Area of region OABO) $$\boldsymbol{=20\sqrt{3}}$$ unit2
Q.6: Find the area enclosed by the curve x2 = 8y, y = 1, y = 9 and the y-axis in the first quadrant.
Sol: Based on formula given in Application of Integrals
Equation x2 = 8y represents a parabola, symmetrical about the y-axis as shown in the above figure.
The area of the region bounded by the curve x2 = 8y, y = 1, y = 9 and the first quadrant is the area enclosed by the curve ABDCA.
Now, the Area of the region ABDCA:
Since, x2 = 8y
Therefore, x = $$2\sqrt{2y}$$
Hence, the Area of region bounded by the curve ABDCA:
$$\\\boldsymbol{\Rightarrow }$$ $$\int_{1}^{9}x\;dy$$ = $$\int_{1}^{9}2\sqrt{2y}\;dy\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{{\left | 2\sqrt{2}\times \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{1}^{9}=\frac{4\sqrt{2}}{3}\times [(9^{\frac{3}{2}})-(1)}]}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{4\sqrt{2}}{3}\times (27-1)=104\times \frac{\sqrt{2}}{3}}$$ unit2
Therefore, Area of region bounded by the curve ABDCA $$\boldsymbol{=104\times \frac{\sqrt{2}}{3}}$$unit2
Q.7: Find the area bounded by the curve whose equation is x2 = 9y and the line x = 6y – 3.
Sol: Based on formula given in Application of Integrals
Equation x2 = 9y represents a parabola, symmetrical about the y-axis as shown in the above figure.
The Area of the region bounded by parabola x2 = 9y and the line x =6y – 3 is the Area enclosed under the curve ABC0A.
Since, the parabola x2 = 9y and the line x = 6y – 3 intersect each other at points A and C, hence the coordinates of points A and C are given by:
(6y-3)2 = 9y
$$\boldsymbol{\Rightarrow }$$ 36y2 – 45y + 9 = 0
By Hit and Trial method solutions of this quadratic equation are:
y = 1 and y = $$\frac{1}{4}$$
Hence, the co-ordinates of point A and point C are (3,1) and $$(\frac{-3}{2},\frac{1}{4})$$ respectively
Since, x2 = 9y
Therefore, x = $$3\sqrt{y}$$
The Area of region bounded by the curve ABCOA = [Area of region OBCbOArea of region OCbO] + [Area of region ABOaAArea of region OAaO]
The Area enclosed by the curve OBCbO:
$$\\\boldsymbol{\Rightarrow }$$ $$\int_{0}^{3} y\;dx = \boldsymbol{\int_{0}^{3}\frac{x+3}{6}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{6}\left | \frac{x^{2}}{2}+ 3x \right |_{0}^{3}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{6}(\frac{9}{2}+9)=\frac{9}{4}}$$ unit2
The Area enclosed by the curve ABOaA:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{\frac{-3}{2}}^{0}\;y\;dx} = \boldsymbol{\int_{\frac{-3}{2}}^{0}\frac{x+3}{6}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{6}\left | \frac{x^{2}}{2}+ 3x \right |_{\frac{-3}{2}}^{0}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{6}(0-\frac{9}{8}+\frac{9}{2})=\frac{9}{16}}$$ unit2
The Area enclosed by the curve OAaO:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{\frac{-3}{2}}^{0}\;y\;dx\; =\;\int_{\frac{-3}{2}}^{0} \frac{x^{2}}{9}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{9}\left | \frac{x^{3}}{3}\right |_{\frac{-3}{2}}^{0}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{27}(0-\frac{-27}{8})=\frac{1}{8}}$$unit2
The Area enclosed by the curve OCbO:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{3} y\;dx\; = \;\int_{0}^{3}\frac{x^{2}}{9}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{9}\left | \frac{x^{3}}{3}\right |_{0}^{3}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{9}(\frac{27}{3}-0)=1}$$ unit2
Since, the Area of region bounded by the curve ABCOA = [Area of region OBCbO – Area of region OCbO] + [Area of region ABOaA – Area of region OAaO]
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{[\frac{9}{4}-1]+[\frac{9}{16}-\frac{1}{8}]=\frac{27}{16}}$$
Therefore, the Area of region bounded by the curve ABCOA $$=\frac{27}{16}$$ unit2
Q.8: Find the area enclosed by the curve 4y = x2 and y = |x|
Sol:Based on formula given in Application of Integrals
Equation x2 = 4y represents a parabola, symmetrical about the y-axis as shown in the above figure.
The area of the region bounded by the curve x2 = 4y and y = |x| is 2(OAEO) i.e. (area OCFO+ area OAEO)
Now, Area of region OAEO = OABO – OEABO
Since, x2 = 4y
$$\boldsymbol{\Rightarrow }$$ $$y=\frac{x^{2}}{4}$$
Now, the Area of region bounded by the curve OEABO:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{4}y\;dx\;=\;\int_{0}^{4}\frac{x^{2}}{4}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{4}\left | \frac{x^{3}}{3} \right |_{0}^{4}=\frac{16}{3}}$$ unit2
Therefore, the area of region bounded by the curve OEABO = $$=\frac{16}{3}$$ unit2
Now, the Area of region bounded by the curve OABO:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{4}y\;dx\;\Rightarrow \;\int_{0}^{4}x\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left | \frac{x^{2}}{2} \right |_{0}^{4}=8}$$ unit2
Therefore, the Area of the region bounded by the curve OABO = 8 unit2
Now, Area of region OAEO = Area of region (OABO – OEABO)
$$\boldsymbol{\Rightarrow }$$ $$8-\frac{16}{3}$$ = $$\frac{8}{3}$$ unit2
Therefore, the total Area of shaded region = 2×$$\frac{8}{3}$$ = $$\frac{16}{3}$$unit2
Q.9: Find the area enclosed by the curve $$\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1$$
Sol:Based on formula given in Application of Integrals
Equation $$\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1$$ represents an ellipse.
Since, $$\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\frac{y^{2}}{9}=1-\frac{x^{2}}{4}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$y=\frac{3}{2}\sqrt{4-x^{2}}$$
Therefore, the Area of region enclosed by the ellipse = 4 × (Area enclosed by the curve ABOA)
Now, the Area enclosed by the curve ABOA:
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{2} y\;dx\;=\; \frac{3}{2}\times \int_{0}^{2}\sqrt{2^{2}-(x)^{2}}\;dx}$$
Since, $$\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}$$
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{3}{2}\;\left [ \frac{x}{2}\sqrt{2^{2}-(x)^{2}}+\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]_{0}^{2}}\\$$
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{3}{2}[\frac{2}{2}\sqrt{4-4}+2\sin^{-1}(1)-0]=\frac{3\pi }{2}}$$ unit2
Therefore, the Area enclosed by the curve ABOA $$\boldsymbol{=\frac{3\pi}{2}}$$ unit2
Hence, the total Area enclosed by the ellipse ABA’B’ = 4 × $$\boldsymbol{\frac{3\pi}{2}}$$ unit2
= 6π unit2
Q.10: Find the area enclosed by the curve x2 + y2 = 9, line x = $$2\sqrt{2}y$$ and the first quadrant.
Sol:
Equation x2 + y2 = 9 represents a circle with radius equal to 3units.
Since, x = $$2\sqrt{2}y$$
Therefore ($$2\sqrt{2}y$$)2 + y2 = 9 (for points of intersection of both the curves)
Hence, the coordinates of point B = ($$2\sqrt{2}y$$,1)
Now, the area of region bounded by the curve x2 + y2 = 9, x = $$2\sqrt{2}y$$, and the first quadrant is the area enclosed by the curve OMABO.
i.e. Area of triangle OMB + Area under the curve MBAM.
Now, the Area of triangle = $$\frac{1}{2}\times Base \times Altitude$$
$$\boldsymbol{\Rightarrow }$$ $$\frac{1}{2}\times OM \times BM\;=\;\frac{1}{2}\times2\sqrt{2}y\times 1 =\sqrt{2}\;y\;$$unit2
Now, the Area under the curve MBAM:
$$\boldsymbol{\Rightarrow }$$ $$\int_{2\sqrt{2}}^{3}\;y\;dx$$
Since, x2 +y2 =9
Therefore, y2 = 9 – x2
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{y=\sqrt{9-x^{2}}}$$
Since, $$\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{2\sqrt{2}}^{3}\sqrt{3^{2}-x^{2}}\;dx=\left | \frac{x}{2}\sqrt{9-x^{2}}+\frac{9}{2}\;\sin^{-1}\frac{x}{3} \right |_{2\sqrt{2}}^{3}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\;\boldsymbol{[\frac{3}{2}\times\ \sqrt{9-9}]+[\frac{9}{2}\;\sin^{-1}(1)]-[\frac{2\sqrt{2}}{2}\times \sqrt{9-8}]-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{9\pi }{4}-\sqrt{2}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\\$$ unit2
Therefore, the Area under the curve MBAM:
= $$\boldsymbol{\frac{9\pi }{4}-\sqrt{2}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}$$unit2
Now, total Area under the shaded region = Area under the curve MBAM + Area of the triangle OMB
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{9\pi }{4}-\sqrt{2}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]+\sqrt{2}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{9\pi }{4}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}$$ unit2
Hence, the required Area is given by the region OMABO:
$$\boldsymbol{=\frac{9\pi }{4}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}$$unit2
Q.11: Find the area of larger part of circle x2 + y2 = 16 cut off by the line x = $$\frac{4}{\sqrt{2}}$$ in the first quadrant.
Sol:Based on formula given in Application of Integrals
Equation x2 + y2 = 16 represents a circle with radius = 4 units.
For coordinates of point B:
$$\boldsymbol{\Rightarrow }$$ $$(\frac{4}{\sqrt{2}})^{2}+y^{2}=16$$
$$\boldsymbol{\Rightarrow }$$ 8 + y2 = 16
$$\boldsymbol{\Rightarrow }$$ y = $$2\sqrt{2}$$
Hence, the coordinates of point B are: $$\boldsymbol{(\frac{4}{\sqrt{2}},2\sqrt{2})}$$
The required Area is given by the curve OMBCO:
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{2\sqrt{2}}y\;dx\;=\;\int_{0}^{2\sqrt{2}}\sqrt{16-x^{2}}\;dx}$$
Since, $$\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left | \frac{x}{2}\sqrt{16-x^{2}}+\frac{16}{2}\sin^{-1} \frac{x}{4} \right |_{0}^{2\sqrt{2}}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{[\frac{2\sqrt{2}}{2}\sqrt{16-8}+8\sin^{-1}\frac{1}{\sqrt{2}}]-0}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\sqrt{2}\times 2\sqrt{2}+8\times \frac{\pi }{4}=[4+2\pi] }$$ unit2
Therefore, the Area of shaded region OMBCO = [4 + 2π]unit2
Let us assume two curves represented by the equation y = f(x) and y = g(x) in [a, b] as shown in the above figure. In this case the height of an elementary strip will be [f(x) – g(x)] and its width will be dx.
Now, Area of the elementary strip (dA) = [f(x) – g(x)] dx
Hence, the total Area of shaded region (A) = $$\int_{a}^{b} [f(x)-g(x)]\;dx$$
Example – 1: Find the area enclosed between two curves whose equations are: x2 = 4y and y2 = 4x.
Sol:Based on formula given in Application of Integrals
The Equation x2 = 4y represents a parabola symmetrical about y-axis and the equation y2 = 4x represents a parabola symmetrical about x-axis.
On solving both the equations:
$$\boldsymbol{\Rightarrow }$$ $$\left ( \frac{y^{2}}{4} \right )^{2}=4y$$
$$\boldsymbol{\Rightarrow }$$ y3 = 64 i.e. y = 4
Which gives x = 4. Hence, the coordinates of point N are (4, 4).
Now, The Area of region enclosed by the curve NAOBN = Area of region enclosed by the curve OANMO – Area of region enclosed by the curve OBNMO.
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{4}2\sqrt{x}\;dx-\int_{0}^{4}\frac{x^{2}}{4}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ 2\times \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{4}-\left [ \frac{x^{3}}{12} \right ]_{0}^{4}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{4}{3}\times (4)^{\frac{3}{2}}-0 \right ]-\left [ \frac{64}{12}-0 \right ]= \frac{16}{3}}$$ unit2
Therefore, the Area of shaded region = $$\frac{16}{3}$$ unit2
EXAMPLE – 2: Find the area enclosed by the sides of a triangle whose vertices have coordinates (1, 0) (3, 5) and (5, 4).
Sol:Based on formula given in Application of Integrals
Form the above figure:
Let, A (1, 0), B (3, 5) and C (5, 4) be the vertices of triangle ABC.
Now, the equation of line AB:
Since, $$(y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]$$
$$\boldsymbol{\Rightarrow }$$ $$(y-0)=(x-1)\times \left[\frac{5\;-\;0}{3\;-\;1}\right]$$
$$\boldsymbol{\Rightarrow }$$ 2y = 5x – 5
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{y=\frac{5x\;-\;5}{2}}$$
The Equation of line BC:
Since, $$(y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]$$
$$\boldsymbol{\Rightarrow }$$ $$(y-5)=(x-3)\times \left[\frac{4-5}{5-3}\right]$$
$$\boldsymbol{\Rightarrow }$$ 2y – 10 = 3 – x
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{y=\frac{13\;-\;x}{2}}$$
The Equation of line AC:
Since, $$(y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right]$$
$$\boldsymbol{\Rightarrow }$$ $$(y-0)=(x-1)\times \left[\frac{4-0}{5-1}\right]$$
$$\boldsymbol{\Rightarrow }$$ 4y = 4x – 4
$$\boldsymbol{\Rightarrow }$$ y = x – 1
Now, the Area of triangle ABC = Area under the curve ABMA + Area under the curve MBCN – Area under the curve ACNA.
The Area under the curve ABMA [2y = 5x – 5]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{1}^{3}y\;dx\;=\;\int_{1}^{3}\frac{5x-5}{2}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{5x^{2}}{4}-\frac{5x}{2} \right ]_{1}^{3}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{45}{4}-\frac{15}{2}-\frac{5}{4}+\frac{5}{2}=5}$$ unit2
Therefore, Area under the curve ABMA = 5 unit2
The Area under the curve MBCN [2y = 13 – x]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{3}^{5}y\;dx\;=\;\int_{3}^{5}\frac{13-x}{2}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{13x}{2}-\frac{x^{2}}{4} \right ]_{3}^{5}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{65}{2}-\frac{25}{4}-\frac{39}{2}+\frac{9}{4}=9}$$ unit2
Therefore, Area under curve MBCN = 9 unit2
The Area under the curve ACNA [y = x – 1]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{1}^{5}y\;dx\;=\;\int_{1}^{5}(x-1)\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x^{2}}{2}-x \right ]_{1}^{5}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{25}{2}-5-\frac{1}{2}+1=8}$$ unit2
Therefore, Area under the curve ACNA = 8 unit2
Now, Area of triangle ABC = Area under curve ABMA + Area under curve MBCNArea under curve ACNA.
Therefore, the Area of triangle ABC = 5 + 9 – 8 = 6 unit2
#### Exercise – 8.2
Q.1: Find the area lying above the x-axis enclosed between two curves whose equations are given as: x2 + y2 = 6x and y2 = 3x.
Sol:Based on formula given in Application of Integrals
The Equation y2 = 3x represents a parabola symmetrical about x-axis.
The Equation x2 + y2 = 6x i.e. (x – 3)2 + y2 = 9 represents a circle with centre (3, 0) and radius 3 units.
Substituting the equation of parabola in equation of circle:
(x – 3)2 + 3x = 9 $$\boldsymbol{\Rightarrow}$$ x2 + 9 – 6x + 3x = 9
$$\boldsymbol{\Rightarrow}$$ x2 – 3x = 0 i.e. x = 0 or x = 3 which gives y = 0 and y = ± 3
Therefore, the coordinates of point A above the x-axis are (3, 3)
Now, the Area of region bounded by the curve OQABO = Area of region bounded by the curve OQAMO + Area of region bounded by the curve ABMA
The Area of region bounded by the curve OQAMO [y2 = 3x]:
$$\\\boldsymbol{\Rightarrow}$$ $$\boldsymbol{\int_{0}^{3}y\;dx=\int_{0}^{3}\sqrt{3x}\;dx}$$
$$\\\boldsymbol{\Rightarrow}$$ $$\boldsymbol{\sqrt{3}\times\left [ \frac{2x^{\frac{3}{2}}}{3} \right ]_{0}^{3}=\sqrt{3}\times \frac{2}{3}\times 3^{\frac{3}{2}}}\\$$
$$\\\boldsymbol{\Rightarrow}$$ $$\boldsymbol{\sqrt{3}\times \frac{2}{3}\times 3\sqrt{3}=6}$$ unit2
Therefore, the Area of region bounded by the curve OQAMO = 6 unit2
Now, the Area of region bounded by the curve ABMA [(x – 3)2 + y2 = 9)]:
$$\boldsymbol{\Rightarrow}$$ $$\boldsymbol{\int_{3}^{6}y\;dx=\int_{3}^{6}\sqrt{9-(x-3)^{2}}\;dx}\\$$
Since, $$\int \sqrt{a^{2}-(x-b)^{2}}\;dx = \frac{x-b}{2}\sqrt{a^{2}-(x-b)^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x-b}{a}\\$$
$$\boldsymbol{\Rightarrow}$$ $$\boldsymbol{\left [ \frac{x-3}{2}\sqrt{\left (3 \right )^{2}-\left ( x-3\right )^{2}}+\frac{9}{2}\sin^{-1}\frac{x-3}{3} \right ]_{3}^{6}}\\$$
$$\boldsymbol{\Rightarrow \;\;\;\left [\frac {6-3}{2}\sqrt{{9}-\left (6-3\right )^{2}}+\frac{9}{2}\sin^{-1}\frac{6-3}{3}\right]-\left [ \frac{x-3}{2}\;\sqrt{9-\left (3-3 \right )^{2}}\;+\frac{9}{2}\;\sin^{-1}\frac{3-3}{3}\;\right]}\\$$
$$\\\boldsymbol{\Rightarrow}$$ $$\boldsymbol{\left [0 +\frac{9}{2}\times \sin^{-1}1-0\right ]=\frac{9}{2}\times \frac{\pi }{2}}\\$$
$$\\\boldsymbol{\Rightarrow}$$ $$\boldsymbol{\frac{9\pi }{4}}$$ unit2
Therefore, the Area of region bounded by the curve ABMA $$=\frac{9\pi }{4}$$ unit2
The Area of region bounded by the curve OQABO = Area of region bounded by the curve OQAMO + Area of region bounded by the curve ABMA
$$\boldsymbol{\Rightarrow}$$ $$6+\frac{9\pi }{4}=\frac{24+9\pi }{4}$$ unit2
Therefore, the Area of shaded region (OQABO)$$=\frac{24+9\pi }{4}$$ unit2
Q.2: Find the area enclosed between two curves: 9x2 + 9y2 = 4 and (x – $$\frac{2}{3}$$)2 + y2 = $$\frac{4}{9}$$
Sol:Based on formula given in Application of Integrals
Equation 9x2 + 9y2 = 4 . . . . . . . (1), represents a circle with centre (0, 0) and radius 3 units.
Equation (x – $$\frac{2}{3}$$ )2 + y2 = $$\frac{4}{9}$$ . . . . . . . . . . .(2), represents a circle with centre (3, 0) and radius 3 units.
On solving equation (1) and equation (2), we will get:
$$\\\boldsymbol{\Rightarrow }$$ $$(x-\frac{2}{3})^{2}+\frac{4-9x^{2}}{9}=\frac{4}{9}$$
$$\\\boldsymbol{\Rightarrow }$$ $$x^{2}+\frac{4}{9}-\frac{4x}{3}+\frac{4}{9}-x^{2}=\frac{4}{9}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\frac{4}{9}=\frac{4}{3}\;x$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{x=\frac{1}{3}}$$ which gives $$\boldsymbol{y=\pm \frac{1}{\sqrt{3}}}$$
Therefore, the coordinates of points M and N are: $$(x=\frac{1}{3})\;\;(y=\pm \frac{1}{\sqrt{3}})$$
Now, the Area enclosed by region BMONB = 2 × [Area enclosed by the curve BMOB].
And the Area enclosed by the curve BMOB = Area of region enclosed by the curve (MOPM+MPBM).
Now, the Area of region enclosed by the curve MPBM [9x2 + 9y2 = 4]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{\frac{1}{3}}^{\frac{2}{3}}y\;dx=\int_{\frac{1}{3}}^{\frac{2}{3}}\sqrt{\frac{4}{9}-x^{2}}\;dx}$$
Since, $$\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x}{2}\sqrt{\left ( \frac{2}{3} \right )^{2}-x^{2}}+\frac{4}{2\times 9}\sin^{-1}\frac{3x}{2} \right ]_{\frac{1}{3}}^{\frac{2}{3}}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [\frac{1}{3}\times \sqrt{\frac{4}{9}-\frac{4}{9}}+\frac{2}{9}\sin^{-1}(1) \right ] -\left [ \frac{1}{6}\times \sqrt{\frac{4}{9}-\frac{1}{9}}+\frac{2}{9}\sin^{-1}\frac{1}{2}\right ] }\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ 0+\frac{2}{9}\times \frac{\pi }{2} \right ]-\left [ \frac{1}{6}\times \frac{1}{\sqrt{3}}+\frac{2}{9}\times \frac{\pi }{6} \right ]=\left [\frac{2\pi }{27}-\frac{ 1}{6\sqrt{3}} \right ]}$$ unit2
Therefore, area of region enclosed by the curve MPBM $$=\left [\frac{2\pi }{27}-\frac{ 1}{6\sqrt{3}} \right ]$$ unit2
Now, the Area of region enclosed by the curve MOPM $$\left [ \left ( x-\frac{2}{3} \right )^{2}+y^{2}=\frac{4}{9} \right ]$$ :
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\frac{1}{3}}y\;dx=\int_{0}^{\frac{1}{3}}\sqrt{\left ( \frac{2}{3} \right )^{2}-\left ( x-\frac{2}{3} \right )^{2}}\;dx}\\$$
Since, $$\\\int \sqrt{a^{2}-(x-b)^{2}}\;dx = \frac{x-b}{2}\sqrt{a^{2}-(x-b)^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x-b}{a}$$
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x-\frac{2}{3}}{2}\sqrt{\left ( \frac{2}{3} \right )^{2}-\left ( x-\frac{2}{3} \right )^{2}}+\frac{2}{9}\sin^{-1}\frac{x-\frac{2}{3}}{\frac{2}{3}} \right ]_{0}^{\frac{1}{3}}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{\frac{1}{3}-\frac{2}{3}}{2}\sqrt{\frac{4}{9}-\left ( \frac{1}{3}-\frac{2}{3} \right )^{2}}+\frac{2}{9}\sin^{-1}\frac{\frac{1}{3}-\frac{2}{3}}{\frac{2}{3}} \right]- \left [ \frac{0-\frac{2}{3}}{2}\;\sqrt{\frac{4}{9}-\left ( 0-\frac{2}{3} \right )^{2}}\;+\frac{2}{9}\;\sin^{-1}\frac{0-\frac{2}{3}}{\frac{2}{3}}\; \right ]}\\$$
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{-1}{6}\times \sqrt{\frac{4}{9}-\frac{1}{9}}+\frac{2}{9}\times \sin^{-1}\frac{-1}{2} \right ]- \left [ \frac{-1}{3}\times \sqrt{\frac{4}{9}-\frac{4}{9}}+\frac{2}{9}\times \sin^{-1}(-1) \right ]}\\$$
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{-1}{6\sqrt{3}}+\frac{2}{9}\times \frac{-\pi }{6} \right ]-\left [ \frac{2}{9}\times \frac{-\pi }{2} \right ]= \left [ \frac{2\pi }{27}-\frac{1}{6\sqrt{3}} \right ]}$$ unit2
Therefore, the Area of region enclosed by the curve MOPM $$= \left [ \frac{2\pi }{27}-\frac{1}{6\sqrt{3}} \right ]$$ unit2
Now, the Area enclosed by the curve BMOB = Area of region enclosed by the curve (MOPM+MPBM)
$$\boldsymbol{\Rightarrow }$$ $$\left [ \frac{2\pi }{27}-\frac{1}{6\sqrt{3}} \right ] +\left [\frac{2\pi }{27}-\frac{ 1}{6\sqrt{3}} \right ]\\$$
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{4\pi }{27}-\frac{1}{3\sqrt{3}} \right ]}$$ unit2
Now, the Area enclosed by the curve BMONB = 2 [Area enclosed by the curve BMOB] $$=2\times \left [ \frac{4\pi }{27}-\frac{1}{3\sqrt{3}} \right ]$$ unit2
Therefore, the Area of shaded region = $$\left [ \frac{8\pi }{27}-\frac{2}{3\sqrt{3}} \right ]$$ unit2
Q.3: Find the area lying above the x-axis enclosed between two curves whose equations are given as: 4x2 + 4y2 = 9 and x2 = 4y.
Sol:Based on formula given in Application of Integrals
The Equation x2 = 4y represents a parabola symmetrical about y-axis.
The Equation 4x2 + 4y2 = 9 i.e. x2 + y2 = $$\frac{3}{2}$$ represents a circle with centre (0, 0) and radius $$\frac{3}{2}$$ units.
Now, on substituting the equation of parabola in the equation of circle we will get:
4(4y) + 4y2 =9 i.e. 4y2 + 16y – 9 = 0
From the above quadratic equation: a = 4, b = 16 and c = -9
Substituting the values of a, b and c in quadratic formula:
$$\\\boldsymbol{\Rightarrow }$$ $$\\y=\frac{-16+\sqrt{(16)^{2}-4(4\times-9)}}{2\times 4}\;and\;y=\frac{-16-\sqrt{(16)^{2}-4(4\times -9)}}{2\times 4}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\\ y=\frac{-16+\sqrt{256+144}}{8}\;and\;y=\frac{-16-\sqrt{256+144}}{8}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\\ y=\frac{-16+\sqrt{400}}{8}\;and\;y= \frac{-16-\sqrt{400}}{8}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\\ y=\frac{-16+20}{8}\;and\;y=\frac{-16-20}{8}$$
$$\\\boldsymbol{\Rightarrow }$$ $$y = \frac{1}{2}\;\;and\;\;y = \frac{-9}{2}$$
Which gives x = $$\pm \;\sqrt{2}$$ [Neglecting y = $$\frac{-9}{2}$$ as it gives absurd results]
Therefore, the coordinates of point B are $$\left (\sqrt{2},\;\frac{1}{2} \right )$$
Now, Area of region bounded by the curve ODCBO = 2 × (Area of region bounded by the curve OBCO)
Now, Area of region bounded by curve OBCO = (Area of region bounded by the curve OCBMO + Area of region bounded by the curve OBMO)
Area of region bounded by the curve OCBMO [4x 2 + 4y2 = 9]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\sqrt{2}}y\;dx=\int_{0}^{\sqrt{2}}\sqrt{\left ( \frac{3}{2} \right )^{2}-x^{2}}\;\;dx}$$
Since, $$\\\int \sqrt{a^{2}-b^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x}{2}\sqrt{\frac{9}{4}- x^{2}}+\frac{9}{2\times 4}\sin^{-1}\frac{2\times x}{3} \right ]_{0}^{\sqrt{2}}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [\frac {\sqrt{2}}{4}\times \sqrt{{\frac{9}{4}-2}}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right]-0}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [\sqrt{2}\times \frac{1}{4}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right ]\\}$$ unit2
Therefore, Area of region bounded by the curve ABMA = $$\left [\sqrt{2}\times \frac{1}{4}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right]$$ unit2
Area of region bounded by the curve OBMO [x2 = 4y]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\sqrt{2}}y\;dx=\int_{0}^{\sqrt{2}}\frac{x^{2}}{4}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x^3}{12} \right ]_{0}^{\sqrt{2}}=\frac{2\sqrt{2}}{12}-0}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{3\sqrt{2}}}$$ unit2
Therefore, the area of region bounded by the curve OBMO = $$\frac{1}{3\sqrt{2}}$$ unit2
Now, the Area of region bounded by curve OBCO = [Area of region bounded by the curve OCBMO – Area of region bounded by the curve OBMO]
$$\\\boldsymbol{\Rightarrow }$$ $$\left [\sqrt{2}\times \frac{1}{4}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right]-\frac{1}{3\sqrt{2}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\left [ \frac{\sqrt{2}}{12}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]$$ unit2
Therefore, Area of region bounded by the curve OBCO = $$\left [ \frac{\sqrt{2}}{12}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]$$ unit2
Now, the Area of region bounded by the curve ODCBO = 2 × [Area of region bounded by the curve OBCO]
$$\\\boldsymbol{\Rightarrow }$$ $$2\times \left [\frac{\sqrt{2}}{12}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]$$
Therefore, the Area of shaded region = $$\left [ \frac{1}{2\sqrt{2}}+\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]$$ unit2
Q.4: Find the area enclosed by the sides of a triangle whose vertices have coordinates (-2, 0) (3, 4) and (5, 2).
Sol:Based on formula given in Application of Integrals
Form the above figure:
Let, A (1, 0), B (3, 5) and C (5, 4) be the vertices of triangle ABC
Now, the equation of line AB:
Since, $$(y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]$$
$$\\\boldsymbol{\Rightarrow }$$ $$(y-0)\;=\;(x+2)\times \left[\frac{4\;-\;0}{3\;+\;2}\right]$$
$$\\\boldsymbol{\Rightarrow }$$ 5y = 4x + 8
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{y=\frac{4x\;+\;8}{5}}$$
The Equation of line BC:
Since, $$(y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]$$
$$\\\boldsymbol{\Rightarrow }$$ $$(y-4)=(x-3)\times \left[\frac{2\;-\;4}{5\;-\;3}\right]$$
$$\\\boldsymbol{\Rightarrow }$$ 2y – 8 = 6 – 2x
$$\\\boldsymbol{\Rightarrow }$$ y = 7 – x
The Equation of line AC:
Since, $$(y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]$$
$$\\\boldsymbol{\Rightarrow }$$ $$(y-0)=(x+2)\times \left[\frac{2\;-\;0}{5\;+\;2}\right]$$
$$\\\boldsymbol{\Rightarrow }$$ 7y = 2x + 4
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{ y= \frac{2\;x+4}{7}}$$
Now, the Area of triangle ABC = Area under the curve ABMA + Area under the curve MBCN – Area under the curve ACNA.
The Area under the curve ABMA:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{-2}^{3}y\;dx\;=\;\int_{-2}^{3}\frac{4x\;+\;8}{5}\;dx}\\$$ = $$\boldsymbol{\left [ \frac{4x^{2}}{10}+\frac{8x}{5} \right ]_{-2}^{3}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{36}{10}+\frac{24}{5}-\frac{16}{10}-\frac{-16}{5}=10}$$ unit2
Therefore, the Area under the curve ABMA = 10 unit2
The Area under the curve MBCN:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{3}^{5}y\;dx\;=\;\int_{3}^{5}=(7-x)\;dx}\\$$ = $$\boldsymbol{\left [ 7x-\frac{x^{2}}{2} \right ]_{3}^{5}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{35-\frac{25}{2}- 21+\frac{9}{2}=6}$$ unit2
Therefore, the Area under the curve MBCN = 6 unit2
The Area under the curve ACNA:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{-2}^{5}y\;dx\;=\;\int_{-2}^{5}\frac{2x\;+\;4}{7}\;dx}\\$$ = $$\boldsymbol{\left [ \frac{2x^{2}}{14}+\frac{4x}{7}\right ]_{-2}^{5}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{25}{7}+\frac{20}{7}-\frac{4}{7}-\frac{-8}{7}=7}$$ unit2
Therefore, the Area under the curve ACNA = 7 unit2
Now, Area of triangle ABC = Area under curve ABMA + Area under curve MBCN – Area under curve ACNA.
Therefore, the Area of triangle ABC = 10 + 6 – 7 = 9 unit2
Q.5: Find the area enclosed by the sides of a triangle whose equations are: y = 4x + 2, y = 3x + 2 and x = 5.
Sol:Based on formula given in Application of Integrals
From the above figure,
The Equation of side AC: y = 4x + 2 . . . . . . (1)
The Equation of side BC: y =3x + 2 . . . . . . (2)
And, x = 5 . . . . . . . . . . (3)
From equation (1) and equation (3):
y = 4(5) + 2 = 22 [Since, x = 5]
Therefore, the coordinates of point A are (5, 22).
From equation (2) and equation (3):
y = 3(5) + 2 = 17 [Since, x = 5]
Therefore, the coordinates of point B are (5, 17).
Now, substituting equation (1) in equation (2):
3x + 2 = 4x + 2 i.e. x = 0 and y = 2
Therefore, the coordinates of point C are (0, 2).
Now, the Area of triangle ABC = Area enclosed by the curve ACOMA – Area enclosed by the curve COMBC.
The Area under the curve ACOMA [y = 4x + 2]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{5}y\;dx\;=\;\int_{0}^{5}(4x+2)\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{4x^{2}}{2}+2x \right ]_{0}^{5}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{50+10-0=60}$$ unit2
Therefore, the Area under the curve ABMA = 60 unit2.
The Area under the curve MBCN [y = 3x + 2]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{5}y\;dx\;=\;\int_{0}^{5}(3x+2)\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{3x^{2}}{2}+2x \right ]_{0}^{5}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{75}{2}+10-0=47.5}$$ unit2
Therefore, the Area under the curve MBCN = 47.5 unit2.
Now, the Area of triangle ABC = Area enclosed by the curve ACOMA – Area enclosed by the curve COMBC
$$\\\boldsymbol{\Rightarrow }$$ 60 – 47.5 = 12.5 unit2
Therefore, the Area of triangle ABC = 12.5 unit2
Q.6: Find the area enclosed by the curves y = x2 + 3, y = 2x, x = 2 and x =0.
Sol:Based on formula given in Application of Integrals
The Equation y = x2 + 3 represents a parabola symmetrical about y-axis.
On substituting equation of line x = 2 in the equation of parabola we will get the coordinates of point C i.e. (2, 7).
On substituting x = 2 in the equation of line y = 2x we will get the coordinates of point B i.e. (2, 4).
From the above figure,
The Area of region enclosed by the curve ODCBO = Area of region enclosed by the curve ODCAOArea of region enclosed by the curve OBAO
Now, the Area of region enclosed by the curve ODCAO [y = x2 + 3]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{2}y\;dx = \int_{0}^{2}(x^{2}+3)\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x^{3}}{3}+3x \right ]_{0}^{2}=\frac{8}{3}+6=\frac{26}{3}}$$unit2
Therefore, the Area of region enclosed by the curve ODCAO = $$\frac{26}{3}$$ unit2
Now, the Area of region enclosed by the curve OBAO [y = 2x]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{2}y\;dx = \int_{0}^{2}(2x)\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [x^{2} \right ]_{0}^{2}=4}$$ unit2
Therefore, the Area of region enclosed by the curve OBAO = 4 unit2
Now, the Area of region enclosed by the curve ODCBO = Area of region enclosed by the curve ODCAO – Area of region enclosed by the curve OBAO
$$\Rightarrow \frac{26}{3}-4=\frac{2}{3}$$ unit2
Therefore, The Area of shaded region (ODCBO) = $$\frac{2}{3}$$unit2
Q.7: Find the area enclosed between the curve y2 = 6x and line y = 3x.
Sol:Based on formula given in Application of Integrals
Equation y2 = 6x represents a parabola, symmetrical about x-axis.
Now, substituting the Equation of line y = 3x in the equation of parabola:
(3x)2 = 6x $$\Rightarrow x = \frac{2}{3}$$ which gives y = 2
Hence the coordinates of point A are $$(\frac{2}{3},2)$$
The area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO
Now, the Area enclosed by the curve OMABO [y2 = 6x]:
Since, y2 = 6x
Therefore, y = $$\sqrt{6x}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\frac{2}{3}} y\;dx = \int_{0}^{\frac{2}{3}}\sqrt{6x}\;dx}$$
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\sqrt{6}\left [ \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{\frac{2}{3}}=\sqrt{6}\times \frac{2}{3}\times \left ( \frac{2}{3} \right )^{\frac{3}{2}}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\sqrt{2} \times \sqrt{3}\times \frac{2}{3}\times 2\sqrt{2}\times \frac{1}{3\sqrt{3}}=\frac{8}{9}}\\$$ unit2
Therefore, the Area enclosed by the curve OMABO = $$\frac{8}{9}$$ unit2
Now, the Area enclosed by the curve OAMO [y = 3x]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\frac{2}{3}} y\;dx\Rightarrow \int_{0}^{\frac{2}{3}}3x\;dx}$$
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{3\left [\frac{x^{2}}{2}\right ]_{0}^{\frac{2}{3}}=\frac{3}{2}\times \left ( \frac{2}{3} \right )^2}$$ = $$\boldsymbol{\frac{2}{3}}$$ unit2
Therefore, the Area enclosed by the curve OAMO = $$\frac{2}{3}$$ unit2
Now, the area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO
$$\\\boldsymbol{\Rightarrow }$$ $$\frac{8}{9}-\frac{2}{3} = \frac{2}{9}$$ unit2
Therefore, the Area enclosed by the curve OABO = $$\frac{2}{9}$$ unit2
#### Miscellaneous Exercise
Q.1: Find the area enclosed by the curve whose equations are: y = 2x2, x = 2, x = 3 and x-axis.
Sol:
The equation y = 2x2 represents a parabola symmetrical about y-axis.
Now, the Area of region enclosed by the curve ABCDA [y = 2x2]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{2}^{3}y\;dx=\int_{2}^{3}2x^{2}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{2x^{3}}{3} \right ]_{2}^{3}=\left [ 18-\frac{16}{3} \right ]=\frac{38}{3}}$$ unit2
Therefore, the Area of shaded region $$=\frac{38}{3}$$ unit2
Q.2: Find the area enclosed by the curve whose equations are: y = 5x4, x = 3, x = 7 and x-axis.
Sol:Based on formula given in Application of Integrals
The equation y = 2x2 represents a quartic parabola symmetrical about y-axis.
Now, the Area of region enclosed by the curve ABCDA [y = 5x4]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{3}^{7}y\;dx=\int_{3}^{7}5x^{4}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{5x^{5}}{5} \right ]_{3}^{7}=\left [ 16807-243 \right ]=16564}$$ unit2
Therefore, the Area of shaded region = 16807 unit2
Q.3: Find the area enclosed between the curve y2 = 3x and line y = 6x.
Sol:Based on formula given in Application of Integrals
Equation y2 = 3x represents a parabola, symmetrical about x-axis.
Now, substituting the Equation of line y = 6x in the equation of parabola:
(6x)2 = 3x $$\Rightarrow x = \frac{1}{12}$$ which gives y = $$\frac{1}{2}$$
Hence the coordinates of point A are $$(\frac{1}{12},\frac{1}{2})$$
The Area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO
Now, the Area enclosed by the curve, OMABO [y2 = 3x]:
Since, y2 = 3x
Therefore, y = $$\sqrt{3x}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\frac{1}{12}} y\;dx = \int_{0}^{\frac{1}{12}}\sqrt{3x}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\sqrt{3}\left [ \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{\frac{1}{12}}=\sqrt{3}\times \frac{2}{3}\times \left ( \frac{1}{12} \right )^{\frac{3}{2}}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\sqrt{3} \times \frac{2}{3}\times \frac{1}{24\sqrt{3}}=\frac{1}{36}}$$ unit2
Therefore, the Area enclosed by the curve OMABO = $$\frac{1}{36}$$unit2
Now, the Area enclosed by the curve OAMO [y = 6x]:
$$\\\boldsymbol{\Rightarrow }$$ $$Area\;of \;\Delta OAM = \frac{1}{2}\times Base\times Altitude$$
$$\\\boldsymbol{\Rightarrow }$$ $$\frac{1}{2}\times |OM|\times |AM|=\boldsymbol{\frac{1}{2}\times \frac{1}{12}\times \frac{1}{2}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{48}}$$unit2
Therefore, the Area enclosed by the curve OAMO = $$\frac{1}{48}$$unit2
Now, the area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO
$$\\\boldsymbol{\Rightarrow }$$ $$\frac{1}{36}-\frac{1}{48} = \frac{1}{144}$$unit2
Therefore, the Area enclosed by the curve OABO = $$\frac{1}{144}$$unit2
Q.4 Find the area enclosed by the curve y = 2x2 and the lines y = 1, y = 3 and the y-axis.
Sol:Based on formula given in Application of Integrals
Equation y = 2x2 represents a parabola symmetrical about y-axis.
The Area of the region bounded by the curve y = 2x2, y = 1, and y = 3, is the Area enclosed by the curve AA’B’BA.
Now, the Area of region AA’B’BA = 2 (Area of region ABNMA)
Since, 2x2 = y
Therefore, x = $$\sqrt{\frac{y}{2}}$$
Thus, the Area of region bounded by the curve ABNMA [y = 2x2]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{1}^{3}x\;dy=\int_{1}^{3}\sqrt{\frac{y}{2}}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{\sqrt{2}}\times {\left | \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{1}^{3}=\frac{\sqrt{2}}{3}\times [(3^{\frac{3}{2}})-(1)^{\frac{3}{2}}}}]\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{\sqrt{2}}{3}\times (3\sqrt{3}-1) = \frac{\sqrt{2}}{3}(3\sqrt{3}-1)}$$unit2
Therefore, the Area of region bounded by the curve ABNMA = $$\frac{\sqrt{2}}{3}(3\sqrt{3}-1)$$unit2
Hence, the Area of region bounded by the curve AA’B’BA = 2(Area of region bounded by the curve ABNMA)$$= \frac{2\sqrt{2}}{3}(3\sqrt{3}-1)$$unit2
Q.5: Find the area enclosed between the curve y2 = 9ax and line y = mx.
Sol:Based on formula given in Application of Integrals
Equation y2 = 9ax represents a parabola, symmetrical about x-axis.
Now, substituting the Equation of line y = mx in the equation of parabola:
9ax = (mx)2 i.e x = $$\frac{9a}{m^{2}}$$ which gives y = $$\frac{9a}{m}$$
Hence the co-ordinates of point A are $$\left ( \frac{9a}{m^{2}},\frac{9a}{m}\right )$$
The Area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO
Now, the Area enclosed by the curve OMABO [y2 = 9ax]:
Since, y2 = 9ax
Therefore, y = $$3\sqrt{ax}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\frac{9a}{m^{2}}} y\;dx = \int_{0}^{\frac{9a}{m^{2}}}3\sqrt{ax}\;dx}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{3\sqrt{a}\left [ \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{\frac{9\;a}{m^{2}}}=3\sqrt{a}\times \frac{2}{3}\times \left ( \frac{9\;a}{m^{2}} \right )^{\frac{3}{2}}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{2\sqrt{a} \times 27\times a\sqrt{a}\times \frac{1}{m^{3}}=\frac{54\;a^{2}}{m^{3}}}$$ unit2
Therefore, the Area enclosed by the curve OMABO $$=\frac{54\;a^{2}}{m^{3}}$$ unit2
Now, the Area enclosed by the curve OAMO [y = mx]:
$$\\\boldsymbol{\Rightarrow }$$ $$Area\;of \;\Delta OAM = \frac{1}{2}\times Base\times Altitude$$
$$\\\boldsymbol{\Rightarrow }$$ $$\frac{1}{2}\times |OM|\times |AM|=\boldsymbol{\frac{1}{2}\times \frac{9\;a}{m^{2}}\times \frac{9\;a}{m}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{81\;a^{2}}{2\;m^{3}}}$$ unit2
Therefore, the Area enclosed by the curve OAMO =$$\frac{81\;a^{2}}{2\;m^{3}}$$ unit2
Now, the Area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO
i.e. $$\frac{54\;a^{2}}{m^{3}}-\frac{81\;a^{2}}{2\;m^{3}} = \frac{27\;a^{2}}{2\;m^{3}}$$ unit2
Therefore, the Area enclosed by the curve OABO = $$\frac{27\;a^{2}}{2\;m^{3}}$$unit2
Q.6: Find the area bounded by the curve whose equation is 3x2 = 4y and the line 2y – 12 = 3x.
Sol:Based on formula given in Application of Integrals
Equation 3x2 = 4y represents a parabola, symmetrical about the y-axis as shown in the above figure.
The Area of the region bounded by parabola 3x2 = 4y and the line 2y -12 = 3x is the Area enclosed under the curve ABC0A.
Since, the parabola 3x2 = 4y and the line 3x = 2y – 12 intersect each other at points A and C, hence the coordinates of points A and C are given by:
Since, $$\;x=\frac{2y-12}{3}$$
$$\\\boldsymbol{\Rightarrow }$$ $$3\left ( \frac{2y-12}{3} \right )^{2}=4y\;\;\;i.e. \;\;\;(2y-12)^{2}=12y$$
$$\\\boldsymbol{\Rightarrow }$$ 4y2 +144 – 48y – 12 y = 0
$$\\\boldsymbol{\Rightarrow }$$ y2 – 15y + 36 = 0
By splitting the middle term Method solutions of this quadratic equation are:
y2 – (12+3)y + 36 = 0 $$\Rightarrow$$ y(y – 12) –3(y – 12) = 0
$$\Rightarrow$$ (y – 3) (y – 12) = 0
Therefore, y = 12 and y = 3 which gives x = 4 and x = -2 respectively.
Hence, the co-ordinates of point A and point C are (-2, 3) and (4, 12) respectively.
Since, 3x2 = 4y
Therefore, y = $$\frac{3x^{2}}{4}$$
The Area of region bounded by the curve ABCOA = [Area of region aACba] – [Area of region OCbO + Area of region OAaO ]
The Area enclosed by the curve aACBa [2y – 12 = 3x]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{-2}^{4} y\;dx \Rightarrow \int_{-2}^{4}\frac{3x+12}{2}\;dx}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{2}\left | \frac{3x^{2}}{2}+ 12x \right |_{-2}^{4}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{2}\left [ 24+48-6-(-24) \right ]=45}$$ unit2
The Area enclosed by the curve OAaO [3x2 = 4y]:
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{-2}^{0}\;y\;dx\;\Rightarrow \int_{-2}^{0} \frac{3x^{2}}{4}\;dx}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{3}{4}\left | \frac{x^{3}}{3}\right |_{-2}^{0}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left | \frac{3}{4}(0-\frac{-8}{3}) \right |=2}$$ unit2
The Area enclosed by the curve OCbO [3x2 = 4y]:
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{4}\;y\;dx\;\Rightarrow \int_{0}^{4} \frac{3x^{2}}{4}\;dx}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{3}{4}\left | \frac{x^{3}}{3}\right |_{0}^{4}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left | \frac{3}{4}(\frac{64}{3}-0) \right |=16}$$ unit2
Now, the Area of region bounded by the curve ABCOA = [ Area of region aACba ] – [ Area of region OCbO + Area of region OAaO ]
$$\\\boldsymbol{\Rightarrow }$$ 45 – [2 + 16] = 27 unit2
Therefore, the Area of shaded region ABCOA = 27 unit2
Q.7: Find the area enclosed by the curves {(x , y) : 6y x2 and y = |x|}
Sol:Based on formula given in Application of Integrals
Equation x2 = 6y represents a parabola, symmetrical about the y-axis as shown in the above figure.
The Area of the region bounded by the curve x2 = 6y and y = |x| is 2(OAEO) i.e. (area OCFO+ area OAEO)
Now, Area of region OAEO = OABO – OEABO
Since, x2 = 6y
Therefore, $$y=\frac{x^{2}}{6}$$
Now, the Area of region bounded by the curve OEABO:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{6}y\;dx\;\Rightarrow \;\int_{0}^{6}\frac{x^{2}}{6}\;dx}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{6}\left | \frac{x^{3}}{3} \right |_{0}^{6}=12}$$ unit2
Therefore, the Area of region bounded by the curve OEABO = 12 unit2
Now, the Area of region bounded by the curve OABO:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{6}y\;dx\;\Rightarrow \;\int_{0}^{6}x\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left | \frac{x^{2}}{2} \right |_{0}^{6}=18}$$ unit2
Therefore, the Area of the region bounded by the curve OABO = 18 unit2
Now, Area of region OAEO = Area of region (OABO – OEABO)
$$\\\boldsymbol{\Rightarrow }$$ 18 – 12 = 6 unit2
Therefore, the total Area of shaded region = 2 × 6 = 12 unit2
Q.8: Find the area enclosed by the sides of a triangle whose vertices have coordinates (3, 0) (5, 8) and (7, 5).
Sol:Based on formula given in Application of Integrals
Form the above figure:
Let, A (3, 0), B (5, 8) and C (7, 5) be the vertices of triangle ABC.
Now, the equation of line AB:
Since, $$(y-y_{1})=(x-x_{1})\times \left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)$$
$$\boldsymbol{\Rightarrow }$$ $$(y-0)=(x-3)\times \left(\frac{8-0}{5-3}\right)$$
$$\boldsymbol{\Rightarrow }$$ 2y = 8x – 24
$$\boldsymbol{\Rightarrow }$$ y=4x-12
The Equation of line BC:
Since, $$(y-y_{1})=(x-x_{1})\times \left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)$$
$$\boldsymbol{\Rightarrow }$$ $$(y-8)=(x-5)\times \left(\frac{5-8}{7-5}\right)$$
$$\boldsymbol{\Rightarrow }$$ 2y – 16 = -3x + 15
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\Rightarrow y=\frac{31-3x}{2}}$$
The Equation of line AC:
Since, $$(y-y_{1})=(x-x_{1})\times \left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)$$
$$\boldsymbol{\Rightarrow }$$ $$(y-0)=(x-3)\times \left(\frac{5-0}{7-3}\right)$$
$$\boldsymbol{\Rightarrow }$$ 4y=5x-15
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\Rightarrow y=\frac{5x-15}{4}}$$
Now, the Area of triangle ABC = Area under the curve ABMA + Area under the curve MBCN – Area under the curve ACNA.
The Area under the curve ABMA [y = 4x – 12]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{3}^{5}y\;dx\;=\;\int_{3}^{5}(4x-12)\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{4x^{2}}{2}-12x\right ]_{3}^{5}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{[50-60]-[18-36]=8}$$unit2
Therefore, Area under the curve ABMA = 8 unit2
The Area under the curve MBCN [2y + 3x = 31]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{5}^{7}y\;dx\;=\;\int_{5}^{7}\frac{31-3x}{2}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{2}\times \left [ 31x-\frac{3x^{2}}{2}\right ]_{5}^{7}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{[\frac{217}{2}-\frac{147}{4}]-[\frac{155}{2}-\frac{75}{4}]=13}$$ unit2
Therefore, Area under the curve MBCN = 13 unit2
The Area under the curve ACNA [4y = 5x – 15]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{3}^{7}y\;dx\;=\;\int_{3}^{7}\left ( \frac{5x-15}{4} \right )dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{5x^{2}}{8}-\frac{15x}{4} \right ]_{3}^{7}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{245}{8}-\frac{105}{4} \right ]-\left [ \frac{45}{8}-\frac{45}{4} \right ]=10}$$ unit2
Therefore, Area under the curve ACNA = 10 unit2
Now, Area of triangle ABC = Area under curve ABMA + Area under curve MBCN – Area under curve ACNA.
Therefore, the Area of triangle ABC = 8 + 13 – 10 = 11 unit2
Q.9: Find the area enclosed by the sides of a triangle whose equations are: 2x – 4 = y, – 2y = -3x + 6 and x – 3y = -5.
Sol:Based on formula given in Application of Integrals
From the above figure:
The Equation of line AB: 3y = x + 5 . . . . . . (1)
The Equation of line BC: y = 4 – 2x . . . . . . (2)
The Equation of line AC: 2y = 3x – 6 . . . . . . (3)
From equation (1) and equation (2):
3(4 – 2x) = x + 5 i.e. x = 1, which gives y = 2
Therefore, the coordinates of point B are (1, 2)
From equation (2) and equation (3):
2(4 – 2x) = 3x – 6 i.e. x = 2 which gives y = 0
Therefore, the coordinates of point C are (2, 0).
From equation (1) and equation (3):
2y = 3(3y – 5) – 6 i.e. y = 3 which gives x = 4
Therefore, the coordinates of point A are (4, 3).
Now, the Area of triangle ABC = Area enclosed by the curve ABMNA Area enclosed by the curve BMCB – Area enclosed by the curve ACNA
The Area under the curve ABMNA [3y = x + 5]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{1}^{4}y\;dx\;=\;\int_{1}^{4}\frac{x+5}{3}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x^{2}}{6}+\frac{5x}{3} \right ]_{1}^{4}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{16}{6}+\frac{20}{3}-\frac{1}{6}-\frac{5}{3}=\frac{15}{2}}$$ unit2
Therefore, the Area under the curve ABMNA $$=\frac{15}{2}$$ unit2
The Area under the curve BMCB [y = 4 – 2x]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{1}^{2}y\;dx\;=\;\int_{1}^{2}(4-2x)\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ 4x-x^{2} \right ]_{1}^{2}}$$
$$\\\boldsymbol{\Rightarrow }$$ [8 – 4] – [4 – 1] =1 unit2
Therefore, the Area under the curve MBCN = 1 unit2
The Area under the curve ACNA [2y = 3x – 6]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{2}^{4}y\;dx\;=\;\int_{2}^{4}\frac{3x-6}{2}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{3x^{2}}{4}-\frac{6x}{2} \right ]_{2}^{4}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{3\times 16}{4}-\frac{6\times 4}{2} \right ]- \left [ \frac{3\times 4}{4}-\frac{6\times 2}{2} \right ]=3}$$ unit2
Therefore, the Area under the curve ACNA = 3 unit2
The Area of triangle ABC = Area enclosed by the curve ABMNA Area enclosed by the curve BMCB – Area enclosed by the curve ACNA
$$\\\boldsymbol{\Rightarrow }$$ $$\frac{15}{2}-1-3 = 3.5$$ unit2
Therefore, the Area of triangle ABC = 3.5 unit2
Q.10: Find the area enclosed by the curve 2x2 = y and the line y = 2x + 12 and x – axis.
Sol:Based on formula given in Application of Integrals
Equation 2x2 = y represents a parabola, symmetrical about the y-axis as shown in the above figure.
The Area of the region bounded by parabola 2x2 = y and the line y = 2x + 12 and x-axis is the Area enclosed under the curve ABCOA.
Since, the parabola 2x2 = y and the line y = 2x + 12 intersect each other at points A and C, hence the coordinates of points A and C are given by:
Since, y = 2x + 12
$$\\\boldsymbol{\Rightarrow }$$ 2x2 = (2x+12)
$$\\\boldsymbol{\Rightarrow }$$ x2 – x – 6 = 0
By splitting the middle term Method solutions of this quadratic equation are:
x2 – (3 – 2)x – 6 = 0 $$\Rightarrow$$ x(x – 3) +2(x – 3) = 0
$$\Rightarrow$$ (x – 3) (x + 2) = 0
Therefore, x = 3 and x = -2 which gives y = 18 and y = 8 respectively.
Hence, the co-ordinates of point E and point A are (3, 18) and (-2, 8) respectively.
Since, 2x2 = y
Therefore, y = 2x2
The Area of region bounded by the curve ABCOA = [Area of region ACOA] + [Area of region ABC]
The Area enclosed by the curve ACOA [ 2x2 = y ]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{-2}^{0}\;y\;dx\;\Rightarrow \int_{-2}^{0} 2x^{2}\;dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left | \frac{2\;x^{3}}{3}\right |_{-2}^{0}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left |(0-\frac{-16}{3}) \right |=\frac{16}{3}}$$ unit2
The Area enclosed by the curve ABC [ y = 2x + 12 ] :
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{Area\;of\;\Delta ABC=\frac{1}{2}\times Base\times Altitude}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{2}\times |BC|\times |AC|=\frac{1}{2}\times \left | 4 \right |\times |8|=16}$$ unit2
The Area of region bounded by the curve ABCOA = [Area of region ACOA] + [Area of region ABC]
$$\Rightarrow \frac{16}{3}+16=\frac{64}{3}$$ unit2
Therefore, the Area of shaded region ABCOA $$=\frac{64}{3}$$unit2
Q.11: Plot the curve y = |x + 4| and hence evaluate $$\int_{-9}^{0}|x+4|\;dx$$
Sol:Based on formula given in Application of Integrals
From the given equation the corresponding values of x and y are given in the following table.
X -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 Y 5 4 3 2 1 0 1 2 3 4 5 6 7
Now, on using these values of x and y, we will plot the graph of y = |x + 4|
From the above graph, the required Area = the Area enclosed by the curve ABCA + the Area enclosed by the curve CDOC
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{-9}^{0}|x+4|\;dx+\int_{-9}^{-4}(x+4)\;dx+\int_{-4}^{0}(x+4)\;dx}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \left | \frac{x^{2}}{2}+4x \right | \right ]_{-9}^{-4}+\left [ \left | \frac{x^{2}}{2}+4x \right | \right ]_{-4}^{0}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left | 8-16-\frac{81}{2}+36 \right |+\left | 0-(8-16) \right |=\left | \frac{-25}{2} \right |+8}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{41}{2}}$$ unit2
Therefore, the area of shaded region = $$\frac{41}{2}$$ unit2
Q.12: Find the area enclosed by the curve y = sin x between 0 x ≤ 2π
Sol:Based on formula given in Application of Integrals
From the above figure, the required Area is represented by the curve OABCD.
Now, the Area bounded by the curve OABCD = the Area bounded by the curve OABO + the Area bounded by the curve BCDB
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\pi } sin(x)\;dx+\left | \int_{\pi }^{2\pi } sin(x)\;dx\right |}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ -cos(x)\right ]_{0}^{\pi }+\left | \left [ -cos (x) \right ] _{\pi }^{2\pi }\right |}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{[-cos(\pi )+cos(0)]+\left | [-cos(2\pi )+cos(\pi )] \right |}\\$$
$$\boldsymbol{\Rightarrow }$$ [1+1] + [ |– 1 – 1| ] unit2
Therefore, the area of shaded region = 4 unit2
Q.13: Find the area of smaller region enclosed by the curve $$\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1$$ and the line $$\frac{x}{2}+\frac{y}{3}=1$$
Sol:Based on formula given in Application of Integrals
The Equation $$\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1$$ represents an ellipse.
The Equation $$\frac{x}{2}+\frac{y}{3}=1$$ represents a line with x and y intercepts as 2 and 3 respectively.
Since, $$\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1$$
$$\\\boldsymbol{\Rightarrow }$$ $$\frac{y^{2}}{9}=1-\frac{x^{2}}{4}$$
$$\\\boldsymbol{\Rightarrow }$$ $$y=\frac{3}{2}\sqrt{4-x^{2}}$$
Therefore, the Area of smaller region enclosed by the Ellipse $$\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1$$ and the line $$\frac{x}{2}+\frac{y}{3}=1$$ is represented by curve ACBA
Now, the Area enclosed by the curve ACBA = Area enclosed by the curve ACBOA – Area enclosed by the curve ABOA
Now, the Area enclosed by the curve ACBOA:
$$\boldsymbol{\int_{0}^{2} y\;dx\Rightarrow\frac{3}{2}\times \int_{0}^{2}\sqrt{2^{2}-(x)^{2}}\;dx}\\$$
Since, $$\ \int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{3}{2}\;\left [ \frac{x}{2}\sqrt{2^{2}-(x)^{2}}+\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]_{0}^{2}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{3}{2}[\frac{2}{2}\sqrt{4-4}+2\sin^{-1}(1)-0]=\frac{3\pi }{2}}$$ unit2
Therefore, the Area enclosed by the curve ACBOA = $$\boldsymbol{\frac{3\pi}{2}}$$unit2
Now, the Area enclosed by the curve ABOA:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{Area \;of\;\Delta ABO = \frac{1}{2}\times AO\times BO}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{2}\times 2\times 3=3}$$ unit2
Therefore, the Area enclosed by the curve ABOA = 3 unit2
Since, the Area enclosed by the curve ACBA = Area enclosed by the curve ACBOA – the Area enclosed by the curve ABOA
$$\\\boldsymbol{\Rightarrow }$$ $$\frac{3\pi }{2}-3=\frac{3}{2}(\pi -2)$$unit2
Therefore, the Area of shaded region $$=\frac{3}{2}(\pi -2)$$ unit2
Q.14: Find the area enclosed by the curve |x| + |y| = 2, by using the method of integration.
Sol: Based on formula given in Application of Integrals
Equation |x| + |y| = 2 represent a region bounded by the lines:
x + y = 2 . . . . . . (1)
x – y = 2 . . . . . . (2)
-x + y = 2 . . . . . . (3)
-x – y = 2 . . . . . . (4)
From equations (1), (2), (3) and (4) we conclude that the curve intersects x-axis and y-axis axis at points A (0, 2), B (2, 0), C (0, -2) and D (-2, 0) respectively.
From the above figure:
Since, the curve is symmetrical to x-axis and y-axis. Therefore, the Area of region bounded by the curve ABCDA = 4 × Area of region bounded by the curve ABOA
Now, the Area of region bounded by the curve ABOA:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{2}y\;dx=4\int_{0}^{2}(2-x)dx}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ 2x-\frac{x^{2}}{2} \right ]_{0}^{2}=(4-2)=2}\\$$ unit2
Therefore, the Area of region bounded by the curve ABOA = 2unit2
Since, the Area of region bounded by the curve ABCDA = 4 × Area of region bounded by the curve ABOA
Therefore, the Area of region bounded by the curve ABCDA = (2 × 4) = 8 unit2
Therefore, the Area of shaded region = 8 unit2
Q.15: Find the area which is exterior to curve x2 = 2y and interior to curve x2 + y2 = 15.
Sol: Based on formula given in Application of Integrals
The Equation x2 = 2y represents a parabola symmetrical about y-axis.
The Equation x2 + y2 = 15 represents a circle with centre (0, 0) and radius units.
Now, on substituting the equation of parabola in the equation of circle we will get:
(2y) + y2 = 15 i.e. y2 + 2y – 15 = 0
Now, by splitting the middle term method solutions of this quadratic equation are:
y2 + (5 – 3)y – 15 = 0 $$\Rightarrow$$ y(y + 5) – 3(y +5) = 0
$$\Rightarrow$$ (y – 3) (y + 5) = 0
Neglecting y = -5 [gives absurd values of x]
Therefore, y = 3 which gives x = $$\pm \sqrt{6}$$
Hence, the coordinates of point B and point D are ($$\sqrt{6}$$, 3) ($$-\sqrt{6}$$, 3) respectively.
Now, the Area of region bounded by the curve BAC’A’DOB = 2 × (Area of region bounded by the curve OBMO + Area of region bounded by the curve BAMB+ Area of region bounded by the curve OAC’O)
Area of region bounded by the curve BAMB [ x 2 + y2 = 15 ]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{\sqrt{6}}^{\sqrt{15}}y\;dx=\int_{\sqrt{6}}^{\sqrt{15}}\sqrt{\left ( \sqrt{15} \right )^{2}-x^{2}}\;\;dx}\\$$
Since, $$\\\int \sqrt{a^{2}-b^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x}{2}\sqrt{15- x^{2}}+\frac{15}{2}\sin^{-1}\frac{x}{\sqrt{15}} \right ]_{\sqrt{6}}^{\sqrt{15}}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac {\sqrt{15}}{2}\times \sqrt{{15-15}}+\frac{15}{2}\sin^{-1}\frac{\sqrt{15}}{\sqrt{15}} \right ]-}\\$$ $$\\\boldsymbol{\left [\frac {\sqrt{6}}{2}\times \sqrt{{15-6}}+\frac{15}{2}\sin^{-1}\frac{\sqrt{6}}{\sqrt{15}}\right]}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ 0+\frac{15\pi }{4} \right ]-\left [\frac{3\sqrt{6}}{2}+\frac{15}{2}\sin^{-1}\frac{\sqrt{10}}{5}\right]}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{15\pi }{4}=\frac{3\sqrt{6}}{2}-\frac{15}{2}\sin^{-1}\frac{\sqrt{10}}{5} \right ]}$$unit2
Area of region bounded by the curve OBMO [ x2 = 2y ]:
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\sqrt{6}}y\;dx=\int_{0}^{\sqrt{6}}\frac{x^{2}}{2}\;dx}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x^3}{6} \right ]_{0}^{\sqrt{6}}=\frac{6\sqrt{6}}{6}-0}\\$$ = $$\boldsymbol{\sqrt{6}}$$ unit2
Area of region bounded by the curve OAC’O:
$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{Area\;of\;circle}{4}=\frac{\pi \times \left ( \sqrt{15} \right )^{2} }{4}}\\$$ = $$\boldsymbol{\frac{15\pi }{4}}$$unit2
Now, the Area of region bounded by the curve BAC’A’DOB = 2 × (Area of region bounded by the curve OBMO + Area of region bounded by the curve BAMB + Area of region bounded by the curve OAC’O)
Therefore, the Area of region bounded by the curve BAC’A’DOB:
$$\boldsymbol{\Rightarrow }$$ $$2\left [ \frac{15\pi }{4}-\frac{3\sqrt{6}}{2}-\frac{15}{2}\sin^{-1}\frac{\sqrt{10}}{5}+\sqrt{6} +\frac{15\pi }{4}\right ]\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ 15\pi -\sqrt{6}-15\left ( \sin^{-1}\frac{\sqrt{10}}{5} \right ) \right ]}\\$$unit2
Therefore, the Area of shaded region: $$\boldsymbol{ \left [ 15\pi -\sqrt{6}-15\left ( \sin^{-1}\frac{\sqrt{10}}{5} \right ) \right ]}$$unit2
Q.16: Find the area enclosed by the curve y = x3, x-axis and the lines x = -2 and x = 2.
Sol. Based on formula given in Application of Integrals
The Equation y = x3 represents a cubic parabola which intersects the line x = 2 and x = -2 at points A and D respectively
From the above figure, the required Area = Area enclosed by the curve ABOA + Area enclosed by the curve CODC
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{2}x^{3}\;dx+\left | \int_{-2}^{0}x^{3}\;dx \right |}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x^{4}}{4} \right ]_{0}^{2}+\left | \left [ \frac{x^{4}}{4} \right ]_{-2}^{0} \right |}$$
$$\\\boldsymbol{\Rightarrow }$$ 4 – 0 + 0 + 4 =16 unit2
Therefore, the Area of shaded region = 16 unit2
Q.17: Find the area enclosed by the curve y = x|x|, y – axis and the lines y = -1 and y = 3.
Sol: Based on formula given in Application of Integrals
Now, y = x|x| is equal to [y = x2] when x > o
And y = x|x| is equal to [y = -x2] when x < o
From the above figure, the required Area = Area enclosed by the curve ABOA + Area enclosed by the curve CODC
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{3}x^{2}\;dx+\left | \int_{-1}^{0}-x^{2}\;dx \right |}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x^{3}}{3}\right ]_{0}^{3}+\left | \left [ \frac{x^{3}}{3} \right ]_{-1}^{0} \right |}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{9+\frac{1}{3}=\frac{28}{3}}$$unit2
Therefore, the area of shaded region $$=\frac{28}{3}$$unit2
Q.18: Find the area bounded by the curve y = Cos (x) and y = Sin (x) and y – axis, when [0 ≤ x ≤ $$\frac{\pi }{2}$$].
Sol: Based on formula given in Application of Integrals
y = Cos(x) . . . . . . . . (1)
y = Sin(x) . . . . . . . . . (2)
Now, from equation (1) and equation (2):
Cos (x) = Sin (x) $$\Rightarrow$$ Cos (x) = Cos $$\left [ \frac{\pi }{2}-x \right ]$$
$$\Rightarrow$$ x = $$\left [ \frac{\pi }{2}-x \right ]$$ $$\Rightarrow$$ x = $$\frac{\pi }{4}$$
Therefore, the coordinates of point A are: $$\left ( \frac{\pi }{4},\frac{1}{\sqrt{2}} \right )$$
Now, from the above figure:
The required Area = Area enclosed by the curve ADMA + Area enclosed by the curve AMOA
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\frac{1}{\sqrt{2}}}sin^{-1}y\;dy+ \int_{\frac{1}{\sqrt{2}}}^{1}cos^{-1}y\;dy}\\$$
Since, $$\int \sin^{-1}y\;dy= y\sin^{-1}y+\sqrt{1-y^{2}}\\$$
And, $$\\\int \cos^{-1}y\;dy= y \cos^{-1}y-\sqrt{1-y^{2}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\frac{1}{\sqrt{2}}}sin^{-1}y\;dy+ \int_{\frac{1}{\sqrt{2}}}^{1}cos^{-1}y\;dy}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ y\sin^{-1}y+\sqrt{1-y^{2}} \right ]_{0}^{\frac{1}{\sqrt{2}}}+\left [ y\cos^{-1}y-\sqrt{1-y^{2}} \right ]_{\frac{1}{\sqrt{2}}}^{1}}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{1}{\sqrt{2}}\times \frac{\pi }{4}+\frac{1}{\sqrt{2}}-(0+1) \right ]+\left [ 0-\left ( \frac{1}{\sqrt{2}}\times \frac{\pi }{4}-\frac{1}{\sqrt{2}} \right ) \right ]}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [\sqrt{2}-1 \right ]}$$unit2
Therefore, the Area of shaded region = $$\left [\sqrt{2}-1 \right ]$$unit2
Q.20: Find the area which is exterior to curve y2 = 6x and interior to curve x2 + y2 = 16.
Sol: Based on formula given in Application of Integrals
The Equation y2 = 6x represents a parabola symmetrical about x – axis.
The Equation x2 + y2 = 16 represents a circle with centre (0, 0) and radius 4 units.
Now, on substituting the equation of parabola in the equation of circle we will get:
x2 + (6x) = 16 i.e. x2 + 6x – 16 = 0
Now, by splitting of middle term method solutions of this quadratic equation are:
x2 + (8 – 2)x – 16 = 0 $$\Rightarrow$$ x(x + 8) – 2(x + 8) = 0
$$\Rightarrow$$ (x – 2) (x + 8) = 0
Neglecting x = -8 [gives absurd values of y]
Therefore, x = 2 which gives y = $$\pm 2\sqrt{3}$$
Hence, the coordinates of point B and point D are (2, $$2\sqrt{3}$$) (2, $$-2\sqrt{3}$$) respectively.
Now, the Area of region bounded by the curve PBA’B’NOP = 2 × (Area of region bounded by the curve BPMOB – Area of region bounded by the curve POMP + Area of region bounded by the curve BOA’B)
Area of region bounded by the curve BPMOB [x 2 + y2 = 16]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{2}y\;dx=\int_{0}^{2}\sqrt{\left ( 4\right )^{2}-x^{2}}\;\;dx}\\$$
Since, $$\\\int \sqrt{a^{2}-b^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x}{2}\sqrt{16- x^{2}}+\frac{16}{2}\sin^{-1}\frac{x}{4} \right ]_{0}^{2}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{2}{2}\times \sqrt{{16-4}}+8\times \sin^{-1}\frac{1}{2} \right ]-0}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ 2\sqrt{3}+\frac{8\times \pi }{6} \right ]=\left [ 2\sqrt{3}+\frac{4\pi }{3} \right ]}\\$$unit2
Therefore, Area of region bounded by the curve BPMOB = $$\left [ 2\sqrt{3}+\frac{4\pi }{3} \right ]$$unit2
Area of region bounded by the curve POMP [y2 = 6x]:
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{2}y\;dx=\int_{0}^{2}\sqrt{6x}\;dx}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\left [ \frac{\sqrt{6}\times 2}{3}\times x^\frac{3}{2} \right ]_{0}^{2}=\frac{2\sqrt{6}}{3}\times 2\sqrt{2}-0\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\frac{8\sqrt{3}}{3}$$unit2
Therefore, Area of region bounded by the curve POMP : $$\frac{8\sqrt{3}}{3}$$unit2
Area of region bounded by the curve BOA’B:
$$\\\boldsymbol{\Rightarrow \frac{Area\;of\;circle}{4}=\frac{\pi \times \left ( 4 \right )^{2} }{4}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ 4π unit2
Therefore, Area of region bounded by the curve BOA’B = 4π unit2
Now, the Area of region bounded by the curve PBA’B’NOP = 2 × (Area of region bounded by the curve BPMOB – Area of region bounded by the curve POMP + Area of region bounded by the curve BOA’B)
Therefore, the Area of region bounded by the curve PBA’B’NOP:
$$\\\boldsymbol{\Rightarrow }$$ $$2\left [2\sqrt{3}+\frac{4\pi }{3}-\frac{8\sqrt{3}}{3}+4\pi \right ]\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{32\pi }{3}-\frac{4\sqrt{3}}{3} \right ]=\frac{4}{3}\times \left [ 8\pi -\sqrt{3} \right ]}\\$$unit2
Therefore, the Area of shaded region = $$\frac{4}{3}\times \left [ 8\pi -\sqrt{3} \right ]$$unit2
Q.21: Find the area bounded by the curve y = Cos (x) and y = Sin (x) and x – axis, when [0 ≤ x ≤ ].
Sol:Based on formula given in Application of Integrals
y = Cos(x) . . . . . . . . (1)
y = Sin(x) . . . . . . . . . (2)
Now, from equation (1) and equation (2):
Cos (x) = Sin (x) $$\Rightarrow$$ Cos (x) = Cos $$\left [ \frac{\pi }{2}-x \right ]\\$$
$$\\\Rightarrow$$ x = $$\left [ \frac{\pi }{2}-x \right ]\\$$ $$\\\Rightarrow$$ x = $$\frac{\pi }{4}$$
Therefore, the coordinates of point A are: $$\left ( \frac{\pi }{4},\frac{1}{\sqrt{2}} \right )$$
Now, from the above figure:
The required Area = Area enclosed by the curve AONA + Area enclosed by the curve ANBA
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ -cos(x) \right ]_{0}^{\frac{\pi }{4}}+[sin (x)]_{\frac{\pi }{4}}^{\frac{\pi }{2}}}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ -cos\frac{\pi }{4}+cos(0)+sin\frac{\pi }{2}-sin\frac{\pi }{4}\right ]}\\$$
$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{-1}{\sqrt{2}}+1+1-\frac{1}{\sqrt{2}} \right ]=[2-\sqrt{2}\;]}$$unit2
Therefore, the Area of shaded region $$= [2-\sqrt{2}\;]$$unit2 |
# How do you solve r^2+25=0 using the quadratic formula?
Dec 3, 2016
$r = - 5 i \mathmr{and} r = + 5 i$
#### Explanation:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{g r e e n}{c}}}{2 \textcolor{red}{a}}$
Converting the given: ${r}^{2} + 25 = 0$ into standard form:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{1} {r}^{2} + \textcolor{b l u e}{0} r + \textcolor{g r e e n}{25} = 0$
$\textcolor{w h i t e}{\text{XXX}} r = \frac{\textcolor{b l u e}{0} \pm \sqrt{{\textcolor{b l u e}{0}}^{2} - 4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{25}}}{2 \cdot \textcolor{red}{1}}$
$\textcolor{w h i t e}{\text{XXX}} = \pm \frac{2 \sqrt{- 25}}{2} = \pm \sqrt{- 25} = \pm 5 i$ |
# Fibonacci Roulette System
Leonardo de Pisa was a mathematician born in 1170 AD who introduced a number sequence now known as the Fibonacci. Although his work was known to Indian mathematicians as early as the 6th century, his book Liber Abaci, posed and solved a problem involving the growth of a population of rabbits by adding the parents (1+1) and then continuing with a calculation that included the sum of the previous two numbers in a sequence. His work, while noteworthy at the time, is now most often linked to the game of roulette after game aficionados applied the sequence to wagers and had some moderate success.
Like the d’Alembert roulette system, the Fibonacci calls for a sequence of wagers to be played for maximum success. The Fibonacci sequence is: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233. When followed, a winning number will return the two previous losses.
## How to Play the Fibonacci System
The Fibonacci is usually played on even-money roulette bets (and European rules that offer En Prison) such as Red/Black and Odd/Even where the player makes a single unit wager and awaits the outcome. The unit may be any specific amount of their choosing, such as €5. If they win, they continue with a single unit until a loss. After a loss, the next wager is the sum of the previous two losses, so the next bet is also 1, but a third loss would call for a wager of 2, then 3, then 5 etc. When a new win takes place, the two previous wagers are dropped and the betting starts up with the number three back from the high point.
Unlike the Martingale, the Fibonacci is not designed to recoup all losses with a single win. Instead, the allure is that the risk is lower than the Martingale and the player can still get ahead with less wins than losses. An example is a 6-loss streak where the player has wagered 1+1+2+3+5+8+13 and finally wins with the wager of 13. This wipes out the 8 and 5 bets, leaving a loss of just 7 units and the next wager is 3 units.
However, some players tweak the betting just a tad, and instead of reversing back three spaces (13, 8, 5) and wagering 3 units, they only go back two spaces and wager 5 units. This adds more risk, but allows the player to make a profit after just a few wins! |
The 30-60-90 Triangle
The 30-60-90 Triangle Video Lecture | CSAT Preparation - UPSC
CSAT Preparation
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FAQs on The 30-60-90 Triangle Video Lecture - CSAT Preparation - UPSC
1. What is a 30-60-90 triangle?
Ans. A 30-60-90 triangle is a special type of right triangle where the angles measure 30 degrees, 60 degrees, and 90 degrees. The ratio of the side lengths in this triangle is 1:√3:2, where the hypotenuse is twice the length of the shorter leg.
2. How do you find the side lengths of a 30-60-90 triangle?
Ans. To find the side lengths of a 30-60-90 triangle, you can start with a known side length, such as the shorter leg, and use the ratios 1:√3:2 to determine the lengths of the other sides. For example, if the shorter leg is 5 units, the longer leg would be 5√3 units, and the hypotenuse would be 10 units.
3. What are the properties of a 30-60-90 triangle?
Ans. A 30-60-90 triangle has several properties: - The lengths of the sides are in the ratio 1:√3:2. - The shorter leg is opposite the 30-degree angle. - The longer leg is opposite the 60-degree angle. - The hypotenuse is opposite the 90-degree angle. - The sum of the angles in a 30-60-90 triangle is always 180 degrees.
4. How can a 30-60-90 triangle be used in real-world applications?
Ans. 30-60-90 triangles have various applications in fields like engineering, architecture, and physics. For example, they can be used to calculate the height of a building or the length of a ramp. These triangles are also helpful in solving trigonometric problems involving angles of 30 and 60 degrees.
5. Can a 30-60-90 triangle be a right triangle with sides of any length?
Ans. No, a 30-60-90 triangle is a specific type of right triangle with angles measuring 30, 60, and 90 degrees. The side lengths of a 30-60-90 triangle are determined by a fixed ratio of 1:√3:2. Therefore, a right triangle with different side lengths would not be considered a 30-60-90 triangle.
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# Bisector: what is it, how to find it, theorem
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bisector is the inner ray of an angle drawn from its vertex, dividing it in two angles congruent. The angle bisectors of a triangle meet at a point known as the incenter, which is the center of the circle inscribed in that polygon.
From the bisector, two important theorems were elaborated: the internal angle and the external angle, developed in triangles that use proportion to relate the sides of that polygon. In the Cartesian plane, it is possible to trace the bisector in odd and even quadrants.
Read too: Notable points of a triangle
## bisector summary
• A bisector is a ray that divides an angle into two congruent angles.
• We can plot the bisectors of interior angles of triangles.
• The interior angle theorem was developed from the bisector of an angle of the triangle.
• There are two bisectors in the Cartesian plane, even quadrants and odd quadrants.
## What is bisector?
Given an angle AOB, we call the ray OC bisector, which starts at the point O and divides the angle AOB into two congruent angles.
In the image, ray OC bisects angle AOB.
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## How to find the bisector?
To find the bisector, a ruler and a compass are used as instruments and the following steps are followed:
• 1st step: The dry point of the compass is placed under the vertex O and an arc is made over the rays OA and OB.
• 2nd step: The dry point of the compass is placed at the point of intersection of the arc with the ray OA and an arc is made with the compass facing the inner part of the angle.
• 3rd step: At the point of intersection of the arc with the ray OB, place the dry point of the compass and repeat the previous process.
• 4th step: Finally, by drawing a ray from the vertex of the angle that passes through the points of intersection between the arcs, the angle bisector is found.
Read too: Barycenter — one of the notable points of a triangle
## Bisector of a triangle
When bisectors of the interior angles of a triangle are traced, we can find its remarkable point, known as incenter, which is the meeting pointThe of bisectors and also the center of circumference inscribed in the polygon.
## Internal Bisector Theorem
segments are formed proportional adjacent sides of a triangle when we bisector one of its interior angles.
Example:
Given the following triangle, find the length of side AC.
Resolution:
Applying the internal bisector theorem, we calculate:
## External Bisector Theorem
When the bisector of one of the exterior angles of a triangle is drawn, the prolongation of the side opposite the exterior angle forms proportional segments to adjacent sides.
Example:
Find the value of x.
Applying the outer bisector theorem, we have:
## Bisector of quadrants of the Cartesian plane
It is possible to plot the bisector in the Cartesian plane. There are two possibilities: the bisector that passes through the even quadrants and the one that passes through the odd quadrants.
THE bisector of quadrants odd numbers pass through the 1st and 3rd quadrants. When the bisector cuts the odd quadrants, The your equation is y = x. Therefore, the points belonging to the bisector of the even quadrants have the same abscissa and ordinate.
The second case concerns when the bisector passes through the even quadrants, that is, by the 2nd and 4th quadrants. When this occurs, the equation of the line will be y = – x. Therefore, the points have abscissa and ordinate as symmetric numbers.
Read too: Fundamental similarity theorem — the relationship between a parallel line and the side of a triangle
## Solved exercises on bisector
question 1
In the following image, knowing that OC is the bisector of angle AOB, we can say that the measure of angle AOB is equal to
A) 15th
B) 30°
C) 35°
D) 60°
E) 70º
Resolution:
Alternative E
Since OC is a bisector, we have the following:
3x – 10 = 2x + 5
3x – 2x = 10 + 5
x = 15°
It is known that x = 15 and that the value of half the angle AOB is equal to 2x + 5. Substituting x by 15, we get:
2 · 15 + 5
30 + 5
35°
Half of the angle AOB is 35°. Therefore, the angle AOB is equal to twice 35°, that is,
AOC = 35 · 2 = 70°.
question 2
In a triangle, its three internal bisectors were drawn. After tracing them, it was possible to notice that they meet at a point. The point where the angle bisectors of a triangle meet is known as
A) centroid.
B) incenter.
C) circumcenter.
D) orthocenter.
Resolution:
Alternative B
When the internal bisectors of a triangle are drawn, their meeting point is known as the incenter.
By Raul Rodrigues de Oliveira
Maths teacher
Would you like to reference this text in a school or academic work? Look:
OLIVEIRA, Raul Rodrigues de. "Bisetrix"; Brazil School. Available in: https://brasilescola.uol.com.br/matematica/bissetriz.htm. Accessed January 20, 2022.
Teachs.ru
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# How do you solve 6-7x=3x²?
Jun 22, 2017
$x = \frac{2}{3} , x = - 3$
#### Explanation:
Put all terms onto one side of the equation so that everything is equal to $0$.
$6 - 7 x = 3 {x}^{2}$
$6 = 3 {x}^{2} + 7 x$
$0 = 3 {x}^{2} + 7 x - 6$
Factor.
$3 {x}^{2} + 7 x - 6 = 0$
$\left(3 x - 2\right) \left(x + 3\right) = 0$
Set each factor equal to $0$. Make sure to do this for both.
$3 x - 2 = 0$
$3 x = 2$
$x = \frac{2}{3}$
$x + 3 = 0$
$x = - 3$
Answers: $x = \frac{2}{3} , x = - 3$
Jun 22, 2017
$\frac{2}{3} \mathmr{and} - 3$
#### Explanation:
$y = 3 {x}^{2} + 7 x - 6 = 0$
Use the new transforming method (Socratic Search)
Transformed equation:
$y ' = {x}^{2} + 7 x - 18 = 0 \text{ } \rightarrow \left(a \times c = 3 \left(- 6\right) = - 18\right)$
Proceed. Find 2 real roots of y', then, divide them by $a = 3.$
Find $2$ numbers knowing the sum (-b = -7))
and the product $\left(a \times c = - 18\right)$.
They are: $2 , \mathmr{and} \left(- 9\right) .$
Consequently, the $2$ real roots of $y$ are:
${x}_{1} = \frac{2}{a} = \frac{2}{3}$, and ${x}_{2} = - \frac{9}{3} = - 3$
Note. This new method avoids the lengthy factoring by grouping and solving the 2 binomials. |
# The Weierstrass Substitution
A standard way to calculate $\int{\frac{dx}{1+\text{sin}x}}$ is via a substitution $u=\text{tan}(x/2)$. As I'll show in a moment, this substitution leads to
\begin{align} \text{sin}x&=\frac{2u}{1+u^2} \\ \text{cos}x&=\frac{1-u^2}{1+u^2} \\ \text{tan}x&=\frac{2u}{1-u^2} \\ dx&=\frac{2du}{1+u^2} \end{align}
Using these,
\begin{align} \int{\frac{dx}{1+\text{sin}x}}&=\int{\frac{1}{1+2u/(1+u^{2})}\frac{2}{1+u^2}du} \\ &=\int{\frac{2du}{1+2u+u^2}} \\ &=\int{\frac{2du}{(1+u)^2}} \\ &=-\frac{2}{1+u}+C \\ &=-\frac{2}{1+\text{tan}(x/2)}+C. \end{align}
The method is known as the Weierstrass substitution. It applies to trigonometric integrals that include a mixture of constants and trigonometric function. H. Anton, though, warns the student that the substitution can lead to cumbersome partial fractions decompositions and consequently should be used only in the absence of finding a simpler method.
For another example,
\begin{align} \int{\frac{dx}{\text{sin}x+\text{tan}x}}&=\int{\frac{1}{\frac{2u}{1+u^2}+\frac{2u}{1-u^2}}\frac{2}{1+u^2}du} \\ &=\int{\frac{2(1-u^{2})}{2u}du} \\ &=\int{(\frac{1}{u}-u)du} \\ &=\text{ln}|u|-\frac{u^2}{2} + C \\ &=\text{ln}|\text{tan}(x/2)|-\frac{\text{tan}^2(x/2)}{2} + C. \end{align}
Now, let's return to the substitution formulas. It's not difficult to derive them using trigonometric identities. But here is a proof without words due to Sidney Kung:
The diagram is suggestive of
$\text{sin}\theta=\frac{AC}{AB}=\frac{2u}{1+u^2}$ and $\text{cos}\theta=\frac{BC}{AB}=\frac{1-u^2}{1+u^2}$.
Finally, it must be clear that, since $\text{tan}x$ is undefined for $\frac{\pi}{2}+k\pi$, $k$ any integer, the substitution is only meaningful when restricted to intervals that do not contain those values, e.g., for $-\pi\lt x\lt\pi$.
## References
1. H. Anton, Calculus, John Wiley & Sons, 1995, pp 470-471 |
# The angle of elevation of the top of a tower from a point 40 m away from its foot is 60∘. Find the height of the tower.
Video Solution
Text Solution
Generated By DoubtnutGPT
## To find the height of the tower using the given information, we can follow these steps:Step 1: Understand the problemWe have a tower and a point from which the angle of elevation to the top of the tower is given. We need to find the height of the tower.Step 2: Draw a diagramLet's denote:- The height of the tower as h.- The distance from the point to the foot of the tower as d=40 m.- The angle of elevation as θ=60∘.Step 3: Identify the right triangleFrom the point at a distance of 40 m from the foot of the tower, we can form a right triangle:- The height of the tower h is the opposite side.- The distance from the point to the foot of the tower (40 m) is the adjacent side.- The angle of elevation θ is at the point.Step 4: Use the tangent functionIn a right triangle, the tangent of an angle is the ratio of the opposite side to the adjacent side. Therefore, we can write:tan(θ)=OppositeAdjacent=hdSubstituting the known values:tan(60∘)=h40Step 5: Calculate tan(60∘)We know that:tan(60∘)=√3So we can substitute this into our equation:√3=h40Step 6: Solve for hTo find the height h, we can rearrange the equation:h=40⋅√3Step 7: Calculate the numerical valueUsing the approximate value of √3≈1.732:h≈40⋅1.732≈69.28 mFinal AnswerThe height of the tower is approximately 69.28 meters.---
|
Updated on:7/8/2024
### Knowledge Check
• Question 1 - Select One
## The angle of elevation of the top of a tower from a point 20 m away from its base is 45∘.What is the height of the tower?
A10 m
B20m
C30m
D40m
• Question 2 - Select One
## The angle of elevation of the top of a tower form a point 20 m away from its base is 45∘. What is the height of the tower?
A10 m
B20m
C30m
D40m
• Question 3 - Select One
## The angle of elevation of the top of a tower from a point 20 metres away from its base is 45∘. The height of the tower is
A10 m
B20m
C30m
D40m
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# What are all the possible rational zeros for y=30x^3-x^2-6x+1 and how do you find all zeros?
Sep 10, 2016
Use the Rational Zero Theorem, synthetic division, and factoring.
#### Explanation:
According to the Rational Zero Theorem, the list of all possible rational zeros is obtained by dividing all the factors of the constant term 1 by all the factors of the leading coefficient term 30.
The possible factors of 1 are $\pm 1$
The possible factors of 30 are $\pm 1 , 2 , 3 , 5 , 6 , 10 , 15 , 30$
The possible zeros are:
$\pm \frac{1}{1} , \pm \frac{1}{2} , \pm \frac{1}{3} , \pm \frac{1}{5} , \pm \frac{1}{6} , \pm \frac{1}{10} , \pm \frac{1}{15} , \pm \frac{1}{30}$
To find all the zeros, use synthetic division. Pick one of the possible zeros as a divisor. If the remainder is zero, the divisor is a zero. If the remainder is not zero, choose another possible zero and try again. I chose 1/3 because I "cheated" and first checked the zeros using a graphing utility.
$\frac{1}{3} \rceiling 30 \textcolor{w h i t e}{a a} - 1 \textcolor{w h i t e}{a a a} - 6 \textcolor{w h i t e}{a a a a} 1$
$\textcolor{w h i t e}{a a a a a a a a a} 10 \textcolor{w h i t e}{a a a a a} 3 \textcolor{w h i t e}{1 a} - 1$
$\textcolor{w h i t e}{a a a}$_________
$\textcolor{w h i t e}{a a a} 30 \textcolor{w h i t e}{a a a a a} 9 \textcolor{w h i t e}{a a a} - 3 \textcolor{w h i t e}{a a a a} 0$
$\frac{1}{3}$ is a zero because the remainder is zero.
Write the quotient using the coefficients found in synthetic division and set it equal to zero.
$30 {x}^{2} + 9 x - 3 = 0$
Factor and solve to find the remaining zeros:
$3 \left(10 {x}^{2} + 3 x - 1\right) = 0$
$3 \left(5 x - 1\right) \left(2 x + 1\right) = 0$
$x = \frac{1}{5}$ and $x = - \frac{1}{2}$
The three zeros are $x = - \frac{1}{2} , x = \frac{1}{5} , x = \frac{1}{3}$ |
Friday, June 28, 2024
HomeEducationWhat are the concepts of Log and Antilog? How to calculate?
# What are the concepts of Log and Antilog? How to calculate?
## What is the Concept Logarithm
The power to which a number must be raised to attain certain other values is defined as the logarithm. The logarithms are the best approach for long numbers to express conveniently in the form of powers or exponents. A logarithm contains many key characteristics that indicate that logarithms can also be divided and multiplied just as the addition and subtraction of logarithm.Â
We can define the logarithms as, the logarithm of a positive real number a with regard to base b is the exponent by which b must be raised to obtain a, whereas the b is a positive real number which should be not equal to 1[nb 1]. The logarithm in this case i.e by= a, is read as the logarithm of a to the base of b.
We can use various bases for a logarithm, but the basis of the common logarithm and the natural logarithm are generally the bases frequently used in logs. The common logarithm has base up to 10 and is shown as log(x). The natural logarithm has a well-known irrational number base e, which is represented by ln(x).
## Common Logs
The logs up to the base of 10 are often known as common logarithms. “log10” bases are commonly written as “log” or “lg.” Using a scientific calculator, common logarithms can easily be calculated.
According to the logarithm definition, the common logs can be expressed mathematically as
log Y = X ↔ Y = 10X
Although calculating the logarithm is not a complex process but people like to follow an online log calculator for solving logarithm problems online.
## Natural Logs
In addition to base 10 of logarithms, e is also an important base of logs. Base e Logarithms are known as natural logarithms. The abbreviation used to denote “loge” commonly is given as “ln”. Just as common logs the use of a scientific calculator is necessary to calculate natural logarithms.
ln Y = X ↔ Y = eX
The expression mentioned above is used to denote natural logarithms. The common, as well as natural logarithms, can be used to solve ax = b problems, particularly when b cannot be represented as an.
## What is Antilog
Antilog is an abbreviation used for the term anti logarithms, which are the exponents in other words. When we determine a number’s logarithm, a specific procedure is followed, the reverse technique of that process is used to get a number’s anti-log.Â
For instance, a number’s log (where the number is) b with base x, is a. We can then say b is the antilog of a. Moreover, the M is called antilog of x if log M =x and is represented as M = antilog x.
## How to Calculate Antilog
For step by step calculation of antilog of a certain number let’s suppose the number value 3. 7645.
1. Separate Component and Mantissa
The first step implies the separation of the characteristic and the mantissa part from the number. The characteristic part as per our example is 3 whereas the mantissa part is 7645.
1. Find Mantissa Value
The Antilog Table is used to obtain an appropriate value of the mantissa part. Then we have to find the corresponding value with the antilog table. As a first component locate the row number beginning with the .76 column and for the column number you have to look for the number 4. Thus the correspondent value for .76 in columns is found to be 5.808.
1. Find Mean Difference
Next the value of the mean difference of columns is found. Use the same number of row again i.e. .76 and obtain the column 3 value.
1. Sum up Both Values
Obtained values from the step 2 or step 3 are then summed up.
1. Place Decimal Point
Lastly, put the decimal point in the summed-up value. However, before placing a decimal point, we have to add 1 to the value of characteristic. Now the 3 becomes 4 in our example and it implies that after 4 digits we have to place our decimal point. The antilog for our example 3. 7645 is therefore calculated as 5814.3343.
Except for following all these steps one by one, We can also calculate antilog of logarithmic function by using an antilog calculator with steps that provide us complete detailed solution.
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