text
stringlengths 22
1.01M
|
|---|
# Monic polynomial
On this post you will find what a monic polynomial is and examples of monic polynomials. You will also see the properties of this type of polynomial and how a polynomial is transformed into a monic polynomial.
## What is a monic polynomial?
The definition of a monic polynomial is as follows:
In mathematics, a monic polynomial is a univariate polynomial (polynomial with only one variable) whose leading coefficient is equal to 1.
For example, the following polynomial of degree 2 is monic because it is a single-variable polynomial and its leading coefficient is 1:
Remember that the leading coefficient of a polynomial is the coefficient of its highest degree term.
## Examples of monic polynomials
Once we have seen what a monic polynomial means, let’s see several examples of this type of polynomial:
Example of monic polynomial of degree 2:
Example of monic polynomial of degree 3:
Example of monic polynomial of degree 4:
## How to transform any polynomial into a monic polynomial
Now that we know the meaning of a monic polynomial, let’s see how to convert a polynomial into a monic polynomial.
So, we are going to solve a problem step by step to see how to do it:
To transform the polynomial into a monic one, we have to divide all the terms of the polynomial by the coefficient of the highest degree term. In this case, the coefficient of the highest degree term is 4, therefore:
Now we simplify the fractions of the polynomial:
And we have already converted the polynomial of the problem into a monic polynomial.
## Properties of monic polynomials
Monic polynomials meet the following characteristics:
• The product of a monic polynomial and another monic polynomial always results in a monic polynomial.
This is due to the properties of the polynomial multiplication. To know why this always happens, see the properties of the product of polynomials.
• If a monic polynomial is composed only of integer coefficients, the roots of that monic polynomial will also be integers.
The roots (or zeros) of a polynomial are numbers that define a polynomial, therefore, it is a very important concept. If you do not know what they are or how to calculate them, you should see:
• Even if the leading coefficient of a multivariate polynomial is one, this type of polynomial is never considered a monic polynomial because it has more than one variable.
### 3 thoughts on “Monic polynomial”
1. Thank you for the clear explanation. It helped me to understand the concept.
|
# How do you find the distance between (2/3,4) and (5/3,7)?
Feb 27, 2017
The distance, d, between two points, $\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$ is:
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
#### Explanation:
Using the distance formula with the given points, $\left(\frac{2}{3} , 4\right) \mathmr{and} \left(\frac{5}{3} , 7\right)$:
$d = \sqrt{{\left(\frac{5}{3} - \frac{2}{3}\right)}^{2} + {\left(7 - 4\right)}^{2}}$
$d = \sqrt{{\left(\frac{3}{3}\right)}^{2} + {\left(3\right)}^{2}}$
$d = \sqrt{{\left(1\right)}^{2} + {\left(3\right)}^{2}}$
$d = \sqrt{1 + 9}$
$d = \sqrt{10}$
|
# How do you find 3.4% of 340 to the nearest tenth?
Dec 10, 2016
11.6 is 3.4% of 340 rounded to the nearest tenth.
#### Explanation:
"Percent" or "%" means "out of 100" or "per 100", Therefore 3.4% can be written as $\frac{3.4}{100}$.
When dealing with percents the word "of" means "times" or "to multiply".
Finally, lets call the number we are looking for "n".
Putting this altogether we can write this equation and solve for $n$ while keeping the equation balanced:
$n = \frac{3.4}{100} \times 340$
$n = \frac{1156}{100}$
$n = 11.56$ rounded to the nearest tenth is $11.6$
|
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# Tara is $66$ inches tall and her son, Tom is $59\dfrac{7}{{12}}$ inches tall. How much taller is tara?
Last updated date: 09th Aug 2024
Total views: 355.2k
Views today: 4.55k
Verified
355.2k+ views
Hint: According to the question, we shall see the difference in height of a mother with her son tom. We can see the difference by subtracting the height of Tom from the height of Tara. By this we can find out who is taller.
Complete step-by-step solution:
By the given statistics the mother is $66$ inches tall and her son Tom is $59\dfrac{7}{{12}}$ inches tall.
As per the analytics the equation can be written as,
$66 - 59\dfrac{7}{{12}}$ inches tall
In the above equation, the mother looks taller by the analytics. So, that is the reason why we are putting her height first.
Now, by solving the equation written above, we get:
$= 66 - 59\dfrac{7}{{12}}$
Now, when we solve the mixed fraction, we get:
$= 66 - 59.5833$
$= 6.4166$
Therefore, this is the difference between the height of Tara and Tom. So, we can say that Tara is $6.4166$inches taller than Tom.
Note: To find out the difference between their height, we subtracted the height of Tom from the height of Tara, because Tara’s height is much more than Tom’s height. If we had done the opposite by subtracting the height of Tara from the height of Tom, then the answer would have come in a negative value, which would be incorrect.
|
# Question Video: Differentiating Rational Functions Using the Quotient Rule at a Point Mathematics • Higher Education
Find dπ¦/dπ₯ at π₯ = 2 when π¦ = π₯Β³/(2π₯ β 3)β΄.
03:51
### Video Transcript
Find dπ¦ by dπ₯ at π₯ is equal to two when π¦ is equal to π₯ cubed divided by two π₯ minus three all raised to the fourth power.
Weβre given π¦ as a function in π₯, and weβre asked to determine dπ¦ by dπ₯ when π₯ is equal to two. So, the first thing weβre going to need to do is differentiate π¦ with respect to π₯. And we can see that π¦ is the quotient of two functions. So, weβll do this by using the quotient rule. Letβs start by recalling the quotient rule.
The quotient rule tells us for differentiable functions π’ of π₯ and π£ of π₯, the derivative of π’ of π₯ over π£ of π₯ with respect to π₯ is equal to π’ prime of π₯ times π£ of π₯ minus π£ prime of π₯ times π’ of π₯ all divided by π£ of π₯ all squared. And of course, we know this wonβt be valid when π£ of π₯ is equal to zero. So, weβll set π’ of π₯ to be the function in our numerator, thatβs π₯ cubed, and π£ of π₯ to be the function in our denominator. Thatβs two π₯ minus three all raised to the fourth power.
And we can see that both π’ of π₯ and π£ of π₯ are polynomials, so we can differentiate this by using the quotient rule. To use the quotient rule, weβre going to need to find the expressions for π’ prime of π₯ and π£ prime of π₯. Letβs start with π’ prime of π₯. Thatβs the derivative of π₯ cubed with respect to π₯. Well, we can do this by using the power rule for differentiation. We want to multiply by our exponent of π₯ and reduce this exponent by one. This gives us π’ prime of π₯ is equal to three π₯ squared.
Next, weβre going to want to find an expression for π£ prime of π₯. However, we can see this is the composition of two functions. Weβre raising a linear function to the fourth power. So, we have a few options to do this. We could use the chain rule or we could distribute this by using the binomial expansion formula. Weβll find π£ prime of π₯ by using the general power rule, which is just another example of the chain rule. We recall this tells us for differentiable function π of π₯ and constant π, the derivative of π of π₯ all raised to the πth power with respect to π₯ is equal to π times π prime of π₯ multiplied by π of π₯ raised to the power of π minus one.
To use this on our function π£ of π₯, weβll set π of π₯ to be our inner function, thatβs two π₯ minus three, and π to be our exponent, which is four. Then, to use the general power rule, we need to find an expression for π prime of π₯. Thatβs the derivative of two π₯ minus three with respect to π₯, which is equal to two. So, by using the general power rule with π equal to two, π prime of π₯ equal to two, and π of π₯ equal to two π₯ minus three, we get π£ prime of π₯ is equal to four times two multiplied by two π₯ minus three all cubed, which, of course, simplifies to give us eight times two π₯ minus three all cubed.
And now that we found the expressions for π’ prime of π₯ and π£ prime of π₯, we can apply the quotient rule to our original function. Substituting our expressions for π’ of π₯, π£ of π₯, π’ prime of π₯, and π£ prime of π₯ into our quotient rule formula, we get dπ¦ by dπ₯ is equal to three π₯ squared times two π₯ minus three all raised to the fourth power minus eight times two π₯ minus three all cubed times π₯ cubed all divided by two π₯ minus three all raised to the fourth power all squared. And weβll simplify this equation slightly. First, we have two π₯ minus three all raised to the fourth power all squared. By using our laws of exponents, we can just write this as two π₯ minus three all raised to the eighth power. And we could simplify this even further.
However, remember, we only need to find the derivative when π₯ is equal to two. So weβre going to substitute π₯ is equal to two into this expression. Substituting π₯ is equal to two into our expression for dπ¦ by dπ₯ gives us the following expression. And all we need to do now is evaluate this expression. We could do this by using a calculator. However, this is not necessary.
First, we need to notice that two times two minus three is equal to one. So, all three of these factors just simplify to give us one. And, of course, one to the fourth power, one cubed, and one to the eighth power are all equal to one. So, this entire expression simplifies to give us three times two squared minus eight times two cubed. And we can just calculate this; itβs equal to negative 52.
Therefore, by using the quotient rule, we were able to show if π¦ is equal to π₯ cubed over two π₯ minus three all raised to the fourth power, then dπ¦ by dπ₯ when π₯ is equal to two is given by negative 52.
|
# Difference between revisions of "2014 AMC 10B Problems/Problem 20"
## Problem
For how many integers $x$ is the number $x^4-51x^2+50$ negative?
$\textbf {(A) } 8 \qquad \textbf {(B) } 10 \qquad \textbf {(C) } 12 \qquad \textbf {(D) } 14 \qquad \textbf {(E) } 16$
## Solution 1
First, note that $50+1=51$, which motivates us to factor the polynomial as $(x^2-50)(x^2-1)$. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so $x^2-50<0. Solving this inequality, we find $1. There are exactly 12 integers $x$ that satisfy this inequality, $\pm 2,3,4,5,6,7$.
Thus our answer is $\boxed{\textbf {(C) } 12}$
## Solution 2
Since the $x^4-51x^2$ part of $x^4-51x^2+50$ has to be less than $-50$ (because we want $x^4-51x^2+50$ to be negative), we have the inequality $x^4-51x^2<-50$ --> $x^2(x^2-51) <-50$. $x^2$ has to be positive, so $(x^2-51)$ is negative. Then we have $x^2<51$. We know that if we find a positive number that works, it's parallel negative will work. Therefore, we just have to find how many positive numbers work, then multiply that by 2. If we try $1$, we get $1^4-51(1)^4+50 = -50+50 = 0$, and 0 therefore doesn't work. Test two on your own, and then proceed. Since two works, all numbers above $2$ that satisfy $x^2<51$ work, that is the set${2,3,4,5,6,7}$. That equates to $6$ numbers. Since each numbers' parallel counterparts work, $6\cdot2=\boxed{\textbf{(C) }12}$.
|
Area of Parallelograms
Areas of Parallelograms and Triangles
Serial order wise
### Transcript
Question 4 In the given figure, P is a point in the interior of a parallelogram ABCD. Show that ar (APB) + ar (PCD) = 1/2 ar (ABCD) Given: A parallelogram ABCD To prove: ar (APB) + ar(PCD) = 1/2 ar (ABCD) Proof: Since ABCD is a parallelogram AB ∥ CD & AD ∥ BC We draw line EF passing through P, parallel to AB & DC i.e. EF ∥ AB ∥ CD Here, AE ∥ BF & AB ∥ EF ∴ EFBA is a parallelogram Similarly, EFCD is a parallelogram Now, ΔAPB and parallelogram EFBA are on the same base AB and between the same parallel lines AB and EF, ∴ Area (ΔAPB) = 1/2 Area (EFBA) Similarly, ΔPCD and parallelogram EFCD are on the same base CD and between the same parallel lines CD and EF, ∴ Area (ΔPCD) = 1/2 Area (EFCD) Adding (1) & (2), we obtain Area (∆APB) + Area (∆PCD) = 1/2 Area (EFAB) + 1/2 Area (EFCD) Area (ΔAPB) + Area (ΔPCD) = 1/2 Area (ABCD) Question 4 In the given figure, P is a point in the interior of a parallelogram ABCD. Show that (ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD) Given: A parallelogram ABCD To prove: ar (APD) + ar(PBC) =ar (APB) + ar (PCD) Proof: Since ABCD is a parallelogram AB ∥ CD & AD ∥ BC We draw line GH passing through P, parallel to AD & BC i.e. GH ∥ AD ∥ BC Here, GH ∥ AD & AG ∥ DH ∴ AGHD is a parallelogram Similarly, GBCH is a parallelogram Now, ΔAPD and parallelogram AGHD are on the same base AD and between the same parallel lines AD and GH, ∴ Area (ΔAPD) = 1/2 Area (AGHD) Similarly, ΔPBC and parallelogram GBCH are on the same base BC and between the same parallel lines BC and GH, ∴ Area (ΔPCB) = 1/2 Area (GBCH) Adding (1) & (2), we obtain Area (∆APD) + Area (∆PBC) = 1/2 Area (AGHD) + 1/2 Area (GBCH) Area (∆APD) + Area (∆PBC) = 1/2 Area (ABCD) Area (ΔAPD) + Area (ΔPBC) = Area (ΔAPB) + Area (ΔPCD) Hence proved
|
# Concise Class-9 Solids Surface Area and Volume of 3D ICSE Maths
Concise Class-9 Solids Surface Area and Volume of 3D ICSE ICSE Maths Solutions Chapter-21. We provide step by step Solutions of Exercise / lesson-21 Solids Surface Area and Volume of 3D for ICSE Class-9th Concise Selina Mathematics by R K Bansal.
Our Solutions contain all type Questions with Exe-21 A, Exe-21 B, and Exe-21 C, to develop skill and confidence. Visit official Website for detail information about ICSE Board Class-9th Mathematics .
## Concise Class-9 Solids Surface Area and Volume of 3D ICSE ICSE Maths Solutions Chapter-21
–: Select Topic :–
Exercise 21 A,
Exercise-21 B,
Exercise-21 C,
### Exercise 21 A, Concise Class-9 Solids Surface Area and Volume of 3D ICSE ICSE Maths Solutions
Question 1
The length breadth and height of a rectangular solid are in the ratio of 5 : 4 : 2 if the total surface area of 1216 cm². Find the length breadth and height of the solid
Therefore, the length, breadth and height of rectangular solid are
Question 2
The volume of a cube is 729 cm3. Find its total surface area.
Question 3
The dimensions of a Cinema Hall are 100 m, 60 m and 15 m. How many persons can sit in the hall, if each requires 150 m3 of air?
Question 4
75 persons can sleep in a room 25 m by 9.6 m. If each persons requires 16 m3 of air; find the height of the room.
Question 5
The edges of three cubes of metal are 3 cm, 4 cm and 5 cm. They are melted and formed into a single cube. Find the edge of the new cube.
Question 6
Three cubes, whose edges are x cm, 8 cm and 10 cm respectively, are melted and recasted into a single cube of edge 12 cm. Find ‘x’.
Question 7
Three equal cubes are placed adjacently in a row. Find the ratio of the total surfaced area of the resulting cuboid to that of the sum of the total surface areas of the three cubes.
Question 8
The cost of papering the four walls of a room at 75 paisa per square meter Rs. 240. The height of the room is 5 metres. Find the length and the breadth of the room, if they are in the ratio 5 : 3.
Question 9
The area of a playground is 3650 m2. Find the cost of covering it with gravel 1.2 cm deep, if the gravel costs Rs. 6.40 per cubic metre.
The area of the playground is 3650 m2 and the gravels are 1.2 cm deep. Therefore the total volume to be covered will be:
3650 x 0.012 =43.8 m3.
Since the cost of per cubic meter is Rs. 6.40, therefore the total cost will be:
43.8 x Rs.6.40 = Rs.280.32
Question 10
A square plate of side ‘x’ cm is 8 mm thick. If its volume is 2880 cm3; find the value of x.
Question 11
The external dimensions of a closed wooden box are 27 cm, 19 cm and 11 cm. If the thickness of the wood in the box is 1.5 cm; find:
(i) Volume of the wood in the box;
(ii) The cost of the box, if wood costs Rs. 1.20 per cm3;
(iii) Number of 4 cm cubes that could be placed into the box.
Question 12
A tank 20 m long, 12 m wide and 8 m deep is to be made of iron sheet. If it is open at the top. Determine the cost of iron-sheet, at the rate of Rs. 12.50 per metre, if the sheet is 2.5 m wide.
Question 13
A closed rectangular box is made of wood of 1.5 cm thickness. The exterior length and breadth are respectively 78 cm and 19 cm, and the capacity of the box is 15 cubic decimetres. Calculate the exterior height of the box.
Let exterior height is h cm. Then interior dimensions are 78-3=75, 19-3=16 and h-3 (subtract two thicknesses of wood). Interior volume = 75 x 16 x (h-3) which must = 15 cu dm
Question 14
The square on the diagonal of a cube has an area of 1875 sq. cm. Calculate:
(i) The side of the cube.
(ii) The total surface area of the cube.
Question 15
A hollow square-shaped tube open at both ends is made of iron. The internal square is of 5 cm side and the length of the tube is 8 cm. There are 192 cm3 of iron in this tube. Find its thickness.
Question 16
Four identical cubes are joined end to end to form a cuboid. If the total surface area of the resulting cuboid as 648 m2; find the length of edge of each cube.
Also, find the ratio between the surface area of resulting cuboid and the surface area of a cube.
### Concise Class-9 Solids Surface Area and Volume of 3D ICSE ICSE Maths Solutions Exercise-21(B)
Question 1
The following figure shows a solid of uniform cross-section. Find the volume of the solid. All measurements are in centimetres.
Assume that all angles in the figures are right angles.
The given figure can be divided into two cuboids of dimensions 6 cm, 4 cm, 3 cm, and 9 cm respectively. Hence, volume of solid
Question 2
A swimming pool is 40 m long and 15 m wide. Its shallow and deep ends are 1.5 m and 3 m deep respectively. If the bottom of the pool slopes uniformly, find the amount of water in litres required to fill the pool.
Question 3
The cross-section of a tunnel perpendicular to its length is a trapezium ABCD as shown in the following figure; also given that:
AM = BN; AB = 7 m; CD = 5 m. The height of the tunnel is 2.4 m. The tunnel is 40 m long. Calculate:
(i) The cost of painting the internal surface of the tunnel (excluding the floor) at the rate of Rs. 5 per m2 (sq. metre).
(ii) The cost of paving the floor at the rate of Rs. 18 per m2.
The cross section of a tunnel is of the trapezium shaped ABCD in which AB = 7m, CD = 5m and AM = BN. The height is 2.4 m and its length is 40m.
Question 4
Water is discharged from a pipe of cross-section area 3.2 cm2 at the speed of 5m/s. Calculate the volume of water discharged:
(i) In cm3 per sec.
(ii) In litres per minute.
Question 5
A hose-pipe of cross-section area 2 cm2 delivers 1500 litres of water in 5 minutes. What is the speed of water in m/s through the pipe?
Question 6
The cross-section of a piece of metal 4 m in length is shown below. Calculate:
(i) The area of the cross-section;
(ii) The volume of the piece of metal in cubic centimetres.
If 1 cubic centimetre of the metal weighs 6.6 g, calculate the weight of the piece of metal to the nearest kg.
Question 7
A rectangular water-tank measuring 80 cm 60 cm 60 cm is filled form a pipe of cross-sectional area 1.5 cm2, the water emerging at 3.2 m/s. How long does it take to fill the tank?
Question 8
A rectangular card-board sheet has length 32 cm and breadth 26 cm.Squares each of side 3 cm, are cut from the corners of the sheet and the sides are folded to make a rectangular container. Find the capacity of the container formed.
Question 9
A swimming pool is 18 m long and 8 m wide. Its deep and shallow ends are 2 m and 1.2 m respectively. Find the capacity of the pool, assuming that the bottom of the pool slopes uniformly.
Question 10
The following figure shows a closed victory-stand whose dimensions are given in cm.
Find the volume and the surface are of the victory stand.
### Concise Class-9 Solids Surface Area and Volume of 3D Exercise 21-(C ) ICSE Maths Solutions Chapter-21
Question 1
Each face of a cube has perimeter equal to 32 cm. Find its surface area and its volume.
Question 2
A school auditorium is 40 m long, 30 m broad and 12 m high. If each student requires 1.2 m2 of the floor area; find the maximum number of students that can be accommodated in this auditorium. Also, find the volume of air available in the auditorium, for each student.
Question 3
The internal dimensions of a rectangular box are 12 cm x cm 9 cm. If the length of the longest rod that can be placed in this box is 17 cm; find x.
Question 4
The internal length, breadth and height of a box are 30 cm, 24 cm, and 15 cm. Find the largest number of cubes which can be placed inside this box if the edge of each cube is
(i) 3 cm (ii) 4 cm (iii) 5 cm
Question 5
A rectangular field is 112 m long and 62 m broad. A cubical tank of edge 6 m is dug at each of the four corners of the field and the earth so removed is evenly spread on the remaining field. Find the rise in level.
Question 6
When length of each side of a cube is increased by 3 cm, its volume is increased by 2457 cm3. Find its side. How much will its volume decrease, if length of each side of it is reduced by 20%?
Question 7
A rectangular tank 30 cm × 20 cm × 12 cm contains water to a depth of 6 cm. A metal cube of side 10 cm is placed in the tank with its one face resting on the bottom of the tank. Find the volume of water, in litres, that must be poured in the tank so that the metal cube is just submerged in the water.
Question 8
The dimensions of a solid metallic cuboid are 72 cm × 30 cm × 75 cm. It is melted and recast into identical solid metal cubes with each of edge 6 cm. Find the number of cubes formed.
Also, find the cost of polishing the surfaces of all the cubes formed at the rate Rs. 150 per sq. m.
Question 9
The dimensions of a car petrol tank are 50 cm × 32 cm × 24 cm, which is full of petrol. If car’s average consumption is 15 km per litre, find the maximum distance that can be covered by the car.
Question 10
The dimensions of a rectangular box are in the ratio 4 : 2 : 3. The difference between cost of covering it with paper at Rs. 12 per m2 and with paper at the rate of 13.50 per m2 is Rs. 1,248. Find the dimensions of the box.
|
### How to simplify fractions easy
Here's your refresher on two methods to reduce fractions. For example, this is how you would list the factors of the numerator and denominator of the fraction, 24/ Find the greatest common factor (GCF) of the numerator and denominator. Understand the trick of simplifying fractions. Solve within seconds. See examples & explanations. Get access to FREE math worksheets on. Kids learn about Simplifying and Reducing Fractions including the steps needed such as finding the greatest common factor.
## how to simplify fractions step by step worksheet
To simplify a fraction, divide the top and bottom by the highest number that Simplifying (or reducing) fractions means to make the fraction as simple as possible. Reducing fractions to their lowest terms involves division. But because you Here's an easier way to reduce fractions after you get comfortable with the concept. Okay, how do you actually simplify fractions? The easiest way is to start by factoring the numerator and denominator of the fraction you're trying to simplify.
Fractions may have numerators and denominators that are composite numbers ( numbers that has more factors than 1 and itself). How to simplify a fraction. Simplifying fractions is often required when your answer is not in the form required by the assignment. As a matter of fact, most math instructors will demand that. Essentially, you want to know an easy way to find the gcf of 2 large numbers. My favorite is the euclidean algorithm. Start by placing the.
## fraction calculator
Here is probably the way you were taught to reduce fractions: Divisibility by 11 is easy to see in two digits (22, 33, 44, etc.) Another trick is to. Simplify fractions the fun way by using the gap between the numerator How to Find the LCM (aka LCD) in Two Easy StepsIn Mental Math. This fraction worksheet is great for testing children in their reducing of fractions. The problems may be selected from easy, medium or hard level of difficulty. Most fractions can be reduced through simple division, and there is a real-world situation where reducing fractions comes in handy (hint: it involves pizza). Simplifying Fractions. Nine-tenths is an easier way to think about the information. It is also easier to convert to a percent or work with in other ways. However. We need to simplify these fractions in order to compare them more easily. The numerator and denominator of a fraction are called its terms. If we simplify a. You can simplify fractions by dividing the numerator and denominator by the same number. Find out more in this Bitesize Primary KS2 Maths guide. Method 2: Simplify Fractions using the Greatest Common Factor . an easier way to simplify fractions is to divide both the numerator and denominator by its. We reduce a fraction to lowest terms by finding an equivalent fraction in which the numerator and denominator are as small as possible. This means that there is. This article will show you how to simplify fractions and improper fractions.I am sure you are already familiar with the concept of fractions but let's have a brief look.
|
### Pre-Algebra
We ended the last quiz with a dot puzzle, where the number of dots per square was a variable that we could adjust to change the total number of dots in the image.
In this quiz we continue working with dot puzzles, to develop our intuition for solving equations with variables. We will even go so far as to solve problems with more than one variable, without needing to use any formal algebraic methods!
First, we'll warm up with another problem where you can find the solution by experimenting with a slider.
# Dot Puzzles
Three squares each contain an identical number of dots.
How many dots should go in each square so that the total number of dots $($including the $4$ outside of the squares$)$ is $13?$
# Dot Puzzles
In the previous problem, the number of dots per square was a variable serving as an unknown quantity. If we represent the number of dots per square with the letter $D,$ then the problem was asking when
$4 + D + D + D = 13$
or, equivalently,
$4 + 3 \times D = 13$
is true.
To solve, we were able to simplify the problem into an arithmetic calculation: first, subtracting $4$ from $13,$ and then dividing the result by $3.$
The dot puzzles that follow don't include a slider; try to work out how you might solve the problem with arithmetic, rather than experimenting with dots.
# Dot Puzzles
Three squares each contain an identical number of dots.
How many dots should go in each square so that the total number of dots (including those outside of the squares) is $107?$
# Dot Puzzles
Both squares contain the same number of dots. The number of dots in a square and the number of dots in a triangle may or may not be the same.
Two squares and one triangle together have $10$ dots. How many dots must be in the triangle?
# Dot Puzzles
Using $T$ and $S$ to stand for the number of dots in triangles and squares, respectively, the last problem could be represented by
$T + S + S = 10$
As there are multiple values of $T$ that work, there is no set "solution." This does not mean the algebra is meaningless! A variable is not just a box waiting for a number; it can be used to indicate relationships between numbers, without having to fix what those numbers are. If I state
"My brother is five years older than me."
both "my brother's age" and "my age" are variables with a set relationship, and this might be the only information given.
This can have practical value; for instance, in a scientific study involving siblings who are five years apart in age, the study might want to make a statement about this relationship that applies no matter what the specific ages are.
# Dot Puzzles
If given a relationship without a clear solution as in the previous question, adding more information may be enough to find one.
Two squares and one triangle together have $10$ dots:
Additionally, just one square and one triangle together have $7$ dots:
With all rules as before, where each square contains the same number of dots and each triangle contains the same number of dots, how many dots are in each square?
# Dot Puzzles
The rules here are the same as before (all squares have the same number of dots, as do all triangles).
Including the dots that are not in a triangle or square, there are $11$ dots in the following figure:
Two squares and one triangle together have $10$ dots:
How many dots are in each square?
# Dot Puzzles
We have now seen how to solve dot puzzles without using the slider to guess and check our answers, even after complicating the puzzles with an additional variable. And while we connected some puzzles with the practice of using letters to represent variables, we didn't need to use those letters to solve the problems, demonstrating that algebra is not simply arithmetic with letters.
As problems get more abstract, we will see that representing variables with letters can streamline the problem-solving process. However, this requires us to first be comfortable with the meanings behind some of the formal rules of algebra. In the following chapters, we will discover those meanings using what we already know about geometry and arithmetic!
# Dot Puzzles
×
|
# What is 10 out of 33 as a percentage?
Now we can see that our fraction is 30.30303030303/100, which means that 10/33 as a percentage is 30.303%.
## How do you find the percentage of 10 subjects?
To calculate how to compute a student’s percentage of marks in an exam, a student’s total marks should be divided by the maximum marks, and then multiplied by 100. For instance, if a student receives 95 out of 100 in English, 85 out of 100 in Hindi, 75 out of 100 in history.
## How do you find 10 percent of a number?
As finding 10% of a number means to divide by 10, it is common to think that to find 20% of a number you should divide by 20 etc. Remember, to find 10% of a number means dividing by 10 because 10 goes into 100 ten times. Therefore, to find 20% of a number, divide by 5 because 20 goes into 100 five times.
## What is 3 30 as a percent?
Convert 3/30 to Percentage by Changing Denominator Our percent fraction is 10/100, which means that 330 as a percentage is 10%.
## What is 6 out of 30 as a percentage?
What is this? Now we can see that our fraction is 20/100, which means that 6/30 as a percentage is 20%.
## What is the percentage of 23 33?
Now we can see that our fraction is 69.69696969697/100, which means that 23/33 as a percentage is 69.697%.
## What is 12 out of 33 as a percentage?
Now we can see that our fraction is 36.363636363636/100, which means that 12/33 as a percentage is 36.3636%.
## What is the percent of 88 and 96?
Now we can see that our fraction is 91.666666666667/100, which means that 88/96 as a percentage is 91.6667%.
## How do you find the percentage of all subjects?
To find the percentage of the marks, divide the marks obtained in the examination with the maximum marks and multiply the result by 100.
## How do you find 30 percent?
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).
## What is 11 out of 25 as a percentage?
Solution and how to convert 11 / 25 into a percentage 0.44 times 100 = 44. That’s all there is to it!
## What is 7 out of 30 as a percentage?
Now we can see that our fraction is 23.333333333333/100, which means that 7/30 as a percentage is 23.3333%.
## What is 19 out of 30 as a percentage?
Now we can see that our fraction is 63.333333333333/100, which means that 19/30 as a percentage is 63.3333%.
## What is a 65 out of 100 grade?
What is this? Now we can see that our fraction is 65/100, which means that 65/100 as a percentage is 65%.
## What is 14 out of 17 as a percentage?
Solution and how to convert 14 / 17 into a percentage 0.82 times 100 = 82.35. That’s all there is to it!
## What is 21 out of 33 as a percentage?
Now we can see that our fraction is 63.636363636364/100, which means that 21/33 as a percentage is 63.6364%.
## What is 11 out of 33 as a percentage?
Now we can see that our fraction is 33.333333333333/100, which means that 11/33 as a percentage is 33.3333%. Once we have the answer to that division, we can multiply the answer by 100 to make it a percentage: What is this?
## What is 60 out of 80 as a percentage?
60/80 x 100 = 6/8 = 3/4 =75% of 100! that’s how you work out the percentage. you have 20 left out of 80 when you take 60.
|
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 9.2: Summation Notation
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
In the previous section, we introduced sequences and now we shall present notation and theorems concerning the sum of terms of a sequence. We begin with a definition, which, while intimidating, is meant to make our lives easier.
Summation Notation
Given a sequence $$\left\{ a_{n} \right\}_{n=k}^{\infty}$$ and numbers $$m$$ and $$p$$ satisfying $$k \leq m \leq p$$, the summation from $$m$$ to $$p$$ of the sequence $$\left\{a_{n}\right\}$$ is written
$\sum_{n=m}^{p} a_{n} = a_{m} + a_{m + 1} + \ldots + a_{p} \label{sigmanotation}$
The variable $$n$$ is called the index of summation. The number $$m$$ is called the \index{summation notation ! lower limit of summation} \textbf{lower limit of summation} while the number $$p$$ is called the upper limit of summation.
In English, Definition \ref{sigmanotation} is simply defining a short-hand notation for adding up the terms of the sequence $$\left\{ a_{n} \right\}_{n=k}^{\infty}$$ from $$a_{m}$$ through $$a_{p}$$. The symbol $$\Sigma$$ is the capital Greek letter sigma and is shorthand for sum'. The lower and upper limits of the summation tells us which term to start with and which term to end with, respectively. For example, using the sequence $$a_{n} = 2n-1$$ for $$n \geq 1$$, we can write the sum $$a_{\mbox{\tiny$$3\)}} +a_{\mbox{\tiny\)4\)}} + a_{\mbox{\tiny\)5\)}} + a_{\mbox{\tiny\)6\)}}\) as
$\begin{array}{rcl} \displaystyle{\sum_{n=3}^{6}(2n-1) } & = & (2(3)-1) + (2(4)-1) + (2(5)-1) + (2(6)-1) \\ & = & 5 + 7 + 9 + 11 \\ & = & 32 \\ \end{array}$
The index variable is considered a dummy variable' in the sense that it may be changed to any letter without affecting the value of the summation. For instance,
$\displaystyle{\sum_{n=3}^{6}(2n-1)} = \displaystyle{\sum_{k=3}^{6}(2k-1)} = \displaystyle{\sum_{j=3}^{6}(2j-1)}$
One place you may encounter summation notation is in mathematical definitions. For example, summation notation allows us to define polynomials as functions of the form
$f(x) = \displaystyle{\sum_{k=0}^{n} a_{k} x^{k}}$
for real numbers $$a_{k}$$, $$k = 0, 1, \ldots n$$. The reader is invited to compare this with what is given in Definition \ref{polynomialfunction}. Summation notation is particularly useful when talking about matrix operations. For example, we can write the product of the $$i$$th row $$R_{i}$$ of a matrix $$A = [a_{ij}]_{m \times n}$$ and the $$j^{\mbox{\scriptsize th}}$$ column $$C_{j}$$ of a matrix $$B = [b_{ij}]_{n \times r}$$ as
$Ri \cdot Cj = \displaystyle{\sum_{k=1}^{n} a_{ik}b_{kj}}$
Again, the reader is encouraged to write out the sum and compare it to Definition \ref{rowcolumnproduct}. Our next example gives us practice with this new notation.
Example $$\PageIndex{1}$$:
1. Find the following sums.
• $$\displaystyle{\sum_{k=1}^{4} \dfrac{13}{100^k} }$$
• $$\displaystyle{\sum_{n=0}^{4} \dfrac{n!}{2}}$$
• $$\displaystyle{\sum_{n=1}^{5} \dfrac{(-1)^{n+1}}{n} (x-1)^n}$$
1. \item Write the following sums using summation notation.
• $$1 + 3 + 5 + \ldots + 117$$
• $$1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + - \ldots + \dfrac{1}{117}$$
• $$0.9 + 0.09 + 0.009 + \ldots 0. \! \! \! \! \underbrace{0 \cdots 0}_{\text{$$n-1\) zeros}} \! \! \! \! 9\)
Solution
1. We substitute $$k=1$$ into the formula $$\frac{13}{100^k}$$ and add successive terms until we reach $$k=4.$$ $\begin{array}{rcl} \displaystyle{\sum_{k=1}^{4} \dfrac{13}{100^k} } & = & \dfrac{13}{100^1} + \dfrac{13}{100^2} + \dfrac{13}{100^3} + \dfrac{13}{100^4} \\ & = & 0.13 + 0.0013 + 0.000013 + 0.00000013 \\ & = & 0.13131313 \\ \end{array}$
2. Proceeding as in (a), we replace every occurrence of $$n$$ with the values $$0$$ through $$4$$. We recall the factorials, $$n!$$ as defined in number Example \ref{seqex1}, number \ref{factorialintroex} and get: $\begin{array}{rcl} \displaystyle{\displaystyle{\sum_{n=0}^{4} \dfrac{n!}{2}}} & = & \dfrac{0!}{2} + \dfrac{1!}{2} + \dfrac{2!}{2} + \dfrac{3!}{2} = \dfrac{4!}{2} \\ & = & \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{2 \cdot 1}{2} + \dfrac{3 \cdot 2 \cdot 1}{2} + \dfrac{4 \cdot 3 \cdot 2 \cdot 1 }{2} \\ & = & \dfrac{1}{2} + \dfrac{1}{2} + 1 + 3 + 12 \\ & = & 17 \\ \end{array}$
3. We proceed as before, replacing the index $$n$$, but \emph{not} the variable $$x$$, with the values $$1$$ through $$5$$ and adding the resulting terms. $\begin{array}{rcl} \displaystyle{\sum_{n=1}^{5} \dfrac{(-1)^{n+1}}{n} (x-1)^n} & = & \dfrac{(-1)^{1+1}}{1} (x-1)^1 + \dfrac{(-1)^{2+1}}{2} (x-1)^2 + \dfrac{(-1)^{3+1}}{3} (x-1)^3 \\ && + \dfrac{(-1)^{1+4}}{4} (x-1)^4 + \dfrac{(-1)^{1+5}}{5} (x-1)^5 \\ [10pt] & = & (x-1) - \dfrac{(x-1)^2}{2} + \dfrac{(x-1)^3}{3} - \dfrac{(x-1)^4}{4} + \dfrac{(x-1)^5}{5} \\ \end{array}$
4. The key to writing these sums with summation notation is to find the pattern of the terms. To that end, we make good use of the techniques presented in Section \ref{Sequences}.
\begin{enumerate}
\item The terms of the sum $$1$$, $$3$$, $$5$$, etc., form an arithmetic sequence with first term $$a = 1$$ and common difference $$d = 2$$. We get a formula for the $$n$$th term of the sequence using Equation \ref{arithgeoformula} to get $$a_{n} = 1 + (n-1)2 = 2n-1$$, $$n \geq 1$$. At this stage, we have the formula for the terms, namely $$2n-1$$, and the lower limit of the summation, $$n=1$$. To finish the problem, we need to determine the upper limit of the summation. In other words, we need to determine which value of $$n$$ produces the term $$117$$. Setting $$a_{n} = 117$$, we get $$2n-1=117$$ or $$n = 59$$. Our final answer is
$\begin{array}{rcl} 1 + 3 + 5 + \ldots + 117 & = & \displaystyle{\sum_{n=1}^{59} (2n-1)} \end{array}$
\item We rewrite all of the terms as fractions, the subtraction as addition, and associate the negatives \)-\)' with the numerators to get
$\dfrac{1}{1} + \dfrac{-1}{2} + \dfrac{1}{3} + \dfrac{-1}{4} + \ldots + \dfrac{1}{117}$
The numerators, $$1$$, $$-1$$, etc. can be described by the geometric sequence\footnote{This is indeed a geometric sequence with first term $$a = 1$$ and common ratio $$r=-1$$.} $$c_{n} = (-1)^{n-1}$$ for $$n \geq 1$$, while the denominators are given by the arithmetic sequence\footnote{It is an arithmetic sequence with first term $$a=1$$ and common difference $$d=1$$.} $$d_{n} = n$$ for $$n \geq 1$$. Hence, we get the formula $$a_{n} = \frac{(-1)^{n-1}}{n}$$ for our terms, and we find the lower and upper limits of summation to be $$n=1$$ and $$n = 117$$, respectively. Thus
$\begin{array}{rcl} 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + - \ldots + \dfrac{1}{117} & = & \displaystyle{\sum_{n=1}^{117} \dfrac{(-1)^{n-1}}{n}} \end{array}$
\item Thanks to Example \ref{seqex2}, we know that one formula for the $$n^{\mbox{\scriptsize th}}$$ term is $$a_{n} = \frac{9}{10^{n}}$$ for $$n \geq 1$$. This gives us a formula for the summation as well as a lower limit of summation. To determine the upper limit of summation, we note that to produce the $$n-1$$ zeros to the right of the decimal point before the $$9$$, we need a denominator of $$10^{n}$$. Hence, $$n$$ is the upper limit of summation. Since $$n$$ is used in the limits of the summation, we need to choose a different letter for the index of summation.\footnote{To see why, try writing the summation using \)n\)' as the index.} We choose $$k$$ and get
$\begin{array}{rcl} 0.9 + 0.09 + 0.009 + \ldots 0.\! \! \! \! \underbrace{0 \cdots 0}_{\text{\)n-1\) zeros}} \! \! \! \! 9 & = & \displaystyle{\sum_{k=1}^{n} \dfrac{9}{10^{k}}} \end{array}$
The following theorem presents some general properties of summation notation. While we shall not have much need of these properties in Algebra, they do play a great role in Calculus. Moreover, there is much to be learned by thinking about why the properties hold. We invite the reader to prove these results. To get started, remember, When in doubt, write it out!''
Properties of Summation Notation
Suppose $$\left\{a_{n}\right\}$$ and $$\left\{b_{n}\right\}$$ are sequences so that the following sums are defined. \index{summation notation ! properties of}
• $$\displaystyle{ \sum_{n=m}^{p} \left(a_{n} \pm b_{n} \right) = \sum_{n=m}^{p} a_{n} \pm \sum_{n=m}^{p} b_{n} }$$
• $$\displaystyle{\sum_{n=m}^{p} c \, a_{n} = c \sum_{n=m}^{p} a_{n}}$$, for any real number $$c$$.
• $$\displaystyle{\sum_{n=m}^{p} a_{n} = \sum_{n=m}^{j} a_{n} + \sum_{n=j+1}^{p} a_{n}}$$, for any natural number $$m \leq j < j+1 \leq p$$.
• $$\displaystyle{\sum_{n=m}^{p} a_{n} = \sum_{n=m+r}^{p+r} a_{n-r}}$$, for any whole number $$r$$.
We now turn our attention to the sums involving arithmetic and geometric sequences. Given an arithmetic sequence $$a_{k} = a + (k-1) d$$ for $$k \geq 1$$, we let $$S$$ denote the sum of the first $$n$$ terms. To derive a formula for $$S$$, we write it out in two different ways
$\begin{array}{ccccccccccc} S & = & a & + & (a + d) & + & \ldots & + & (a + (n-2)d) & + & (a + (n-1)d) \\ S & = & (a + (n-1)d) & + & (a + (n-2)d) & + & \ldots & + & (a + d) & + & a \\ \end{array}$
If we add these two equations and combine the terms which are aligned vertically, we get
$2S = (2a + (n-1)d) + (2a + (n-1)d) + \ldots + (2a + (n-1)d) + (2a + (n-1)d)$
The right hand side of this equation contains $$n$$ terms, all of which are equal to $$(2a + (n-1)d)$$ so we get $$2S = n(2a + (n-1)d)$$. Dividing both sides of this equation by $$2$$, we obtain the formula
$S = \dfrac{n}{2} (2a + (n-1)d)$
If we rewrite the quantity $$2a + (n-1)d$$ as $$a + (a + (n-1)d) = a_{\mbox{\tiny$$1\)}} + a_{n}\), we get the formula
$S = n \left(\dfrac{a_{\mbox{\tiny\)1\)}} + a_{n}}{2}\right)$
A helpful way to remember this last formula is to recognize that we have expressed the sum as the product of the number of terms $$n$$ and the \textit{average} of the first and $$n^{\mbox{\scriptsize th}}$$ terms.
To derive the formula for the geometric sum, we start with a geometric sequence $$a_{k} = ar^{k-1}$$, $$k \geq 1$$, and let $$S$$ once again denote the sum of the first $$n$$ terms. Comparing $$S$$ and $$rS$$, we get
$\begin{array}{ccccccccccccccc} S & = & a & + & ar & + & ar^2 & + & \ldots & + & ar^{n-2} & + & ar^{n-1} & & \\ r S & = & & & ar & + & ar^2 & + & \ldots & + & ar^{n-2} & + & ar^{n-1} & + & ar^{n} \\ \end{array}$
Subtracting the second equation from the first forces all of the terms except $$a$$ and $$ar^{n}$$ to cancel out and we get $$S - rS = a - ar^{n}$$. Factoring, we get $$S(1-r) = a \left(1-r^{n}\right)$$. Assuming $$r \neq 1$$, we can divide both sides by the quantity $$(1-r)$$ to obtain
$S = a \left( \dfrac{1-r^n}{1-r}\right)$
If we distribute $$a$$ through the numerator, we get $$a - ar^{n} = a_{\mbox{\tiny$$1\)}} - a_{n\mbox{\tiny\) + 1\)}}\) which yields the formula
$S = \dfrac{a_{\mbox{\tiny\)1\)}}-a_{n\mbox{\tiny\) + 1\)}}}{1-r}$
In the case when $$r=1$$, we get the formula
$S = \underbrace{a + a + \ldots +a }_{\text{\)n\) times}} = n \, a$
Our results are summarized below.
Sums of Arithmetic and Geometric Sequences
\item The sum $$S$$ of the first $$n$$ terms of an arithmetic sequence $$a_{k}= a + (k-1)d$$ for $$k \geq 1$$ is
$S = \displaystyle{\sum_{k=1}^{n} a_{k}} = n \left(\dfrac{a_{\mbox{\tiny\)1\)}} + a_{n}}{2}\right) = \dfrac{n}{2} (2a + (n-1)d) \label{arithgeosum}$
\item The sum $$S$$ of the first $$n$$ terms of a geometric sequence $$a_{k}= ar^{k-1}$$ for $$k \geq 1$$ is
1. \item $$S = \displaystyle{\sum_{k=1}^{n} a_{k}} = \dfrac{a_{\mbox{\tiny$$1\)}} - a_{n\mbox{\tiny\) + 1\)}}}{1-r} =a \left( \dfrac{1-r^n}{1-r}\right)\), if $$r \neq 1$$. \index{sequence ! arithmetic ! sum of first $$n$$ terms}
2. \item $$S = \displaystyle{\sum_{k=1}^{n} a_{k} = \sum_{k=1}^{n} a =n a}$$, if $$r =1$$. \index{sequence ! geometric ! sum of first $$n$$ terms}
While we have made an honest effort to derive the formulas in Equation \ref{arithgeosum}, formal proofs require the machinery in Section \ref{Induction}. An application of the arithmetic sum formula which proves useful in Calculus results in formula for the sum of the first $$n$$ natural numbers. The natural numbers themselves are a sequence\footnote{This is the identity function on the natural numbers!} $$1$$, $$2$$, $$3$$, \ldots which is arithmetic with $$a = d = 1$$. Applying Equation \ref{arithgeosum},
$\begin{array}{rcl} 1 + 2 + 3 + \ldots + n & = & \dfrac{n(n+1)}{2} \end{array}$
So, for example, the sum of the first $$100$$ natural numbers\footnote{There is an interesting anecdote which says that the famous mathematician \href{http://en.wikipedia.org/wiki/Carl_Fr...underline{Carl Friedrich Gauss}} was given this problem in primary school and devised a very clever solution.} is $$\frac{100(101)}{2} = 5050$$.
An important application of the geometric sum formula is the investment plan called an \index{annuity ! ordinary ! definition of} \textbf{annuity}. Annuities differ from the kind of investments we studied in Section \ref{ExpLogApplications} in that payments are deposited into the account on an on-going basis, and this complicates the mathematics a little.\footnote{The reader may wish to re-read the discussion on compound interest in Section \ref{ExpLogApplications} before proceeding.} Suppose you have an account with annual interest rate $$r$$ which is compounded $$n$$ times per year. We let $$i = \frac{r}{n}$$ denote the interest rate per period. Suppose we wish to make ongoing deposits of $$P$$ dollars at the \textit{end} of each compounding period. Let $$A_{k}$$ denote the amount in the account after $$k$$ compounding periods. Then $$A_{\mbox{\tiny$$1\)}} = P\), because we have made our first deposit at the \textit{end} of the first compounding period and no interest has been earned. During the second compounding period, we earn interest on $$A_{\mbox{\tiny$$1\)}}\) so that our initial investment has grown to $$A_{\mbox{\tiny$$1\)}}(1+i) = P(1+i)\) in accordance with Equation \ref{simpleinterest}. When we add our second payment at the end of the second period, we get
$A_{\mbox{\tiny\)2\)}} = A_{\mbox{\tiny\)1\)}}(1+i) + P = P(1+i) + P = P(1+i)\left(1 + \dfrac{1}{1+i}\right)$
The reason for factoring out the $$P(1+i)$$ will become apparent in short order. During the third compounding period, we earn interest on $$A_{\mbox{\tiny$$2\)}}\) which then grows to $$A_{\mbox{\tiny$$2\)}}(1+i)\). We add our third payment at the end of the third compounding period to obtain
$A_{\mbox{\tiny\)3\)}} = A_{\mbox{\tiny\)2\)}}(1+i) + P = P(1+i)\left(1 + \dfrac{1}{1+i}\right)(1+i) + P = P(1+i)^2\left(1 + \dfrac{1}{1+i} + \dfrac{1}{(1+i)^2}\right)$
During the fourth compounding period, $$A_{\mbox{\tiny$$3\)}}\) grows to $$A_{\mbox{\tiny$$3\)}}(1+i)\), and when we add the fourth payment, we factor out $$P(1+i)^3$$ to get
$A_{\mbox{\tiny\)4\)}} = P(1+i)^3 \left(1 + \dfrac{1}{1+i} + \dfrac{1}{(1+i)^2} + \dfrac{1}{(1+i)^3}\right)$
This pattern continues so that at the end of the $$k$$th compounding, we get
$A_{k} = P(1+i)^{k-1} \left(1 + \dfrac{1}{1+i} + \dfrac{1}{(1+i)^2} + \ldots + \dfrac{1}{(1+i)^{k-1}}\right)$
The sum in the parentheses above is the sum of the first $$k$$ terms of a geometric sequence with $$a = 1$$ and $$r = \frac{1}{1+i}$$. Using Equation \ref{arithgeosum}, we get
$1 + \dfrac{1}{1+i} + \dfrac{1}{(1+i)^2} + \ldots + \dfrac{1}{(1+i)^{k-1}} = 1 \left(\dfrac{1 - \dfrac{1}{(1+i)^k}}{1 - \dfrac{1}{1+i}}\right) = \ \dfrac{(1+i)\left(1 - (1+i)^{-k}\right)}{i}$
Hence, we get
$A_{k} = P(1+i)^{k-1} \left(\dfrac{(1+i)\left(1 - (1+i)^{-k}\right)}{i}\right) = \dfrac{P\left((1+i)^k - 1\right)}{i}$
If we let $$t$$ be the number of years this investment strategy is followed, then $$k = nt$$, and we get the formula for the future value of an \index{annuity ! ordinary ! future value} \textbf{ordinary annuity}.
Future Value of an Ordinary Annuity
Suppose an annuity offers an annual interest rate $$r$$ compounded $$n$$ times per year. Let $$i = \frac{r}{n}$$ be the interest rate per compounding period. If a deposit $$P$$ is made at the end of each compounding period, the amount $$A$$ in the account after $$t$$ years is given by
$A = \dfrac{P\left((1+i)^{nt} - 1\right)}{i} \label{fvannuity}$
The reader is encouraged to substitute $$i = \frac{r}{n}$$ into Equation \ref{fvannuity} and simplify. Some familiar equations arise which are cause for pause and meditation. One last note: if the deposit $$P$$ is made a the \textit{beginning} of the compounding period instead of at the end, the annuity is called an \index{annuity ! annuity-due} \textbf{annuity-due}. We leave the derivation of the formula for the future value of an annuity-due as an exercise for the reader.
Example $$\PageIndex{1}$$
An ordinary annuity offers a $$6 \%$$ annual interest rate, compounded monthly.
1. If monthly payments of $$\$$50\) are made, find the value of the annuity in $$30$$ years.
2. How many years will it take for the annuity to grow to $$\$$100,\! 000\)?
Solution
1. We have $$r = 0.06$$ and $$n = 12$$ so that $$i = \frac{r}{n} = \frac{0.06}{12} = 0.005$$. With $$P=50$$ and $$t=30$$, $A = \dfrac{50\left((1+0.005)^{(12)(30)} - 1\right)}{0.005} \approx 50225.75$ Our final answer is $$\$$50,\!225.75\).
2. To find how long it will take for the annuity to grow to $$\$$100,\!000\), we set $$A = 100000$$ and solve for $$t$$. We isolate the exponential and take natural logs of both sides of the equation.
$\begin{array}{rcl} 100000 & = & \dfrac{50\left((1+0.005)^{12t} - 1\right)}{0.005} \\ 10 & = & (1.005)^{12t} - 1 \\ (1.005)^{12t} & = & 11 \\ \ln\left((1.005)^{12t}\right) & = & \ln(11) \\ 12t \ln(1.005) & = & \ln(11) \\ t & = & \frac{\ln(11)}{12 \ln(1.005)} \approx 40.06 \\ \end{array}$
This means that it takes just over $$40$$ years for the investment to grow to $$\$$100,\!000\). Comparing this with our answer to part 1, we see that in just $$10$$ additional years, the value of the annuity nearly doubles. This is a lesson worth remembering.
We close this section with a peek into Calculus by considering \textit{infinite} sums, called \index{series} \textbf{series}. Consider the number $$0.\overline{9}$$. We can write this number as
$0.\overline{9} = 0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + \ldots$
From Example \ref{seriesex1}, we know we can write the sum of the first $$n$$ of these terms as
$0.\underbrace{9 \cdots 9}_{\text{\)n\) nines}} = .9 + 0.09 + 0.009 + \ldots 0.\! \! \! \! \underbrace{0 \cdots 0}_{\text{\)n-1\) zeros}} \! \! \! \! 9 = \displaystyle{\sum_{k=1}^{n} \dfrac{9}{10^{k}}}$
Using Equation \ref{arithgeosum}, we have
$\displaystyle{\sum_{k=1}^{n} \dfrac{9}{10^{k}}} = \dfrac{9}{10} \left( \dfrac{1 - \dfrac{1}{10^{n+1}}}{1 - \dfrac{1}{10}} \right) = 1 - \dfrac{1}{10^{n+1}}$
It stands to reason that $$0.\overline{9}$$ is the same value of $$1 - \frac{1}{10^{n+1}}$$ as $$n \rightarrow \infty$$. Our knowledge of exponential expressions from Section \ref{IntroExpLogs} tells us that $$\frac{1}{10^{n+1}} \rightarrow 0$$ as $$n \rightarrow \infty$$, so $$1 - \frac{1}{10^{n+1}} \rightarrow 1$$. We have just argued that $$0.\overline{9} = 1$$, which may cause some distress for some readers.(To make this more palatable, it is usually accepted that $$0.\overline{3} = \frac{1}{3}$$ so that $$0.\overline{9} = 3\left(0.\overline{3}\right) = 3\left(\frac{1}{3} \right) = 1$$). Any non-terminating decimal can be thought of as an infinite sum whose denominators are the powers of $$10$$, so the phenomenon of adding up infinitely many terms and arriving at a finite number is not as foreign of a concept as it may appear. We end this section with a theorem concerning geometric series.
Definition: Geometric Series
Given the sequence $$a_{k} = ar^{k-1}$$ for $$k \geq 1$$, where $$|r| < 1$$,
$a + ar + ar^2 + \ldots = \displaystyle{\sum_{k=1}^{\infty} ar^{k-1}} = \dfrac{a}{1-r} \label{geoseries}$
If $$|r| \geq 1$$, the sum $$a + ar + ar^2 + \ldots$$ is not defined.
The justification of the result in Theorem \ref{geoseries} comes from taking the formula in Equation \ref{arithgeosum} for the sum of the first $$n$$ terms of a geometric sequence and examining the formula as $$n \rightarrow \infty$$. Assuming $$|r|<1$$ means $$-1 < r < 1$$, so $$r^{n} \rightarrow 0$$ as $$n \rightarrow \infty$$. Hence as $$n \rightarrow \infty$$,
$\displaystyle{\sum_{k=1}^{n} a r^{k-1}} = a \left( \dfrac{1-r^n}{1-r}\right) \rightarrow \dfrac{a}{1-r}$
As to what goes wrong when $$|r| \geq 1$$, we leave that to Calculus as well, but will explore some cases in the exercises.
## Contributors
• Carl Stitz, Ph.D. (Lakeland Community College) and Jeff Zeager, Ph.D. (Lorain County Community College)
|
Basic math operations
Basic math operations include four basic operations:
Addition (+)
Subtraction (-)
Multiplication (* or x) and
Division ( : or /)
These operations are commonly called arithmetic operations. Arithmetic is the oldest and most elementary branch of mathematics. In this and other related lessons we will briefly explain basic math operations. Keep in mind that, even though the operations and the examples shown here are pretty simple, they provide the basis for even the most complex operations used in mathematics.
Addition
Addition is a mathematical operation that explains the total amount of objects when they are put together in a collection. For example, let’s say that Jimmy has 2 apples and Laura has 3 apples, and that we want to find out how many apples they have together. By adding them together, we see that both of them combined have 5 apples (2 Jimmy’s apple + 3 Laura’s apples = 5 apples in total). As you can see, the addition is signified by the “plus sign (+)”.
Addition can also be used to perform operations with negative numbers, fractions, decimal numbers, functions etc. There are several properties that are typical for addition:
1. Commutativity
2. Associativity
3. Identity element
Subtraction
Subtraction is the arithmetic operation that is the opposite of addition. Subtraction is used when you want to know how many objects are left in the group after you take away a certain amount of objects from that group. For example, Maggie has 5 apples and she gives 2 apples to her friend Paul. How many apples does she have? She has 3 apples (5 apples that she had – 2 apples that she gave to Paul = 3 apples that are left to her).
As you can see, subtraction is determined by the “minus (-) symbol”. Subtraction can also be used to perform operations with negative numbers, fractions, decimal numbers, functions, etc.
Multiplication
Multiplication of two numbers is equivalent to the addition of a number to itself as many times as the value of the other one number is. Think of it like this: you have 5 groups of apples and each group has 3 apples. One of the ways you can find out how many apples you have is this one:
3 apples + 3 apples + 3 apples + 3 apples + 3 apples = 15 apples in total
You can see that it is way too much work (especially if you have larger numbers), so you can use multiplication to solve this problem:
5 group of apples x 3 apples in every group = 15 apples in total
This could be even easier by using the table of multiplication:
Multiplication is signified with multiplication sign “x”, and it is often read as “times” or “multiplied by”. So if you had an expression like “3 x 4”, you could read it as “3 times 4” or “3 multiplied by 4”. In the other words, the expression of multiplication signifies the number of times one number is multiplied by another number.
$\ 3 * 4= 12$ -> The number 3 is multiplied in this equation 4 times, and when you multiply 3 by 4 you get the number 12 as a result.
Division
Division is the fourth basic math operation. Basically, you can say that dividing means splitting objects into equal parts or groups. For example, you have 12 apples that need to be shared equally between 4 people. So, how many apples will each person get? Each person will get 3 apples (12 apples / 4 people = 3 apples per person). The division is the opposite of multiplication:
$\ 3 * 4 = 12$
$\ 4 * 3= 12$
$\frac{12}{4} = 3$
$\frac{12}{3}=4$
For easier understanding of the division of one number by another, use the table of division:
In mathematics when you perform computational actions you must have in mind that there is a sequence that need to be respected in order to do calculation properly.
Addition and subtraction are first degree mathematical operations, and multiplication and division are second degree mathematical operations. Which means:
● if same degree operations, we resolve them by their order (from left to right):
For example,
$\ 18 – 2 + 4 = 16 + 4 = 20$ -> this only applies if there are no brackets in the equation. If there are brackets, we firstly resolve numbers in the brackets.
$\ 18 – ( 2 + 4 ) = 18 – 6 = 12$ -> notice the difference in results, even with same numbers
● if there are different degree operations, we resolve it by the degree order – multiplication and division first and addition and subtraction after.
For example:
$\ 2 + 3 * 4 = 2 + 12 = 14$
Numbers in brackets are need to be resolve firstly in any case!
|
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Triangle Sum Theorem
## Interior angles add to 180 degrees
Estimated6 minsto complete
%
Progress
Practice Triangle Sum Theorem
MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated6 minsto complete
%
Understanding the Angle Measures of Triangles
A roof truss is a pre-fabricated, triangular wooden structure used to support the roof of a home or building. The triangular shape is used because it is incredibly strong. A particular roof truss is an isosceles triangle such that the base angle has a measure of 50°. How can the worker figure out the measure of the other angles of the triangular roof truss to ensure that all trusses for this roof are identical?
In this concept, you will learn to understand the angle measures of triangles.
### Equiangular Triangles
The sum of the interior angles of a triangle is 180°. This fact is true for all triangles regardless of the type of triangle. An equiangular triangle has three equal angles which each measure 60°. The following diagram is an equiangular triangle.
Let’s look at another triangle to see how this information can be applied.
The above triangle has two acute angles that each measure 25° and one obtuse angle. The measure of the obtuse angle is not known.
The sum of the three angles of the triangle equals 180°. Write an equation to represent this statement.
\begin{align*}m \angle A + m \angle B + m \angle C = 180^{\circ}\end{align*}
Next, substitute the known information into the equation.
\begin{align*}m \angle A + 25^{\circ} + 25^{\circ} = 180^{\circ}\end{align*}
Next, simplify the left side of the equation.
\begin{align*}\begin{array}{rcl} m \angle A + 25 ^{\circ} + 25^{\circ} &=& 180^{\circ} \\ m \angle A + 50^{\circ} &=& 180^{\circ} \end{array}\end{align*}
Then, subtract 50° from both sides of the equation to solve for \begin{align*}m \angle A\end{align*}.
\begin{align*}\begin{array}{rcl} m \angle A + 50^{\circ} &=& 180^{\circ} \\ m \angle A + 50^{\circ} - 50^{\circ} &=& 180^{\circ} - 50^{\circ} \\ m \angle A &=& 130^{\circ} \end{array}\end{align*}
The measure of the obtuse angle is 130°.
### Examples
#### Example 1
Earlier, you were given a problem about the worker with the roof truss. He needs to figure out the measure of the interior angles in the roof truss. How can he do this?
He can use the facts that the triangle is isosceles and the sum of the interior angles of the triangle equals 180°.
First, draw and label an isosceles triangle to model the roof truss.
The triangular roof truss is an isosceles triangle. The angles opposite the equal sides are equal in measure.
\begin{align*}\begin{array}{rcl} m \angle A &=& 50^{\circ} \\ m \angle A &=& m \angle B \\ m \angle B &=& 50^{\circ} \end{array} \end{align*}
Next, write an equation to represent the sum of the interior angles of the triangle.
\begin{align*}\angle A + \angle B + \angle C = 180^{\circ}\end{align*}
Next, fill into the equation, the measures of the equal angles.
\begin{align*}\begin{array}{rcl} \angle A + \angle B + \angle C &=& 180^{\circ} \\ 50^{\circ} + 50^{\circ} + \angle C &=& 180^{\circ} \end{array}\end{align*}
Next, simplify the left side of the equation.
\begin{align*}\begin{array}{rcl} 50^{\circ} + 50^{\circ} + \angle C &=& 180^{\circ} \\ 100^{\circ} + \angle C &=& 180^{\circ} \end{array}\end{align*}
Next, subtract 100° from both sides of the equation to determine the measure of \begin{align*}\angle C\end{align*}.
\begin{align*}\begin{array}{rcl} 100^{\circ} + \angle C &=& 180^{\circ} \\ 100^{\circ} - 100^{\circ} + \angle C &=& 180^{\circ} - 100^{\circ} \\ \angle C &=& 80^{\circ} \end{array}\end{align*}
The measures of the other two interior angles of the roof truss are 50° and 80°.
#### Example 2
If the measure of the vertex angle of an isosceles triangle is 50°, what is the measure of the base angles of the triangle?
First, draw and label a triangle to represent the given information.
Next, write down what you know from the problem.
\begin{align*}\begin{array}{rcl} m \angle 1 + m \angle 2 + m \angle 3 &=& 180^{\circ} \quad \text{Sum of the angles of a triangle} \\ m \angle 3 &=& 50^{\circ} \quad \ \ \text{Measure of the vertex angle} \\ m \angle 1 &=& m \angle 2 \quad \text{Equal angles of an isosceles triangle} \end{array}\end{align*}
Next, write down an equation to represent the information.
\begin{align*}m \angle 1 + m \angle 2 + 50^{\circ} = 180^{\circ}\end{align*}
Next, let ‘\begin{align*}x\end{align*}’ represent each of the equal angles of the isosceles triangle.
\begin{align*}x + x + 50^{\circ} = 180^{\circ}\end{align*}
Next, simplify the left side of the equation.
\begin{align*}2x + 50^{\circ} = 180^{\circ}\end{align*}
Next, subtract 50° from both sides of the equation and simplify to isolate the variable.
\begin{align*}\begin{array}{rcl} 2x + 50^{\circ} &=& 180{^\circ} \\ 2x + 50^{\circ} - 50^{\circ} &=& 180^{\circ} - 50^{\circ} \\ 2x &=& 130^{\circ} \end{array}\end{align*}
Then, divide both sides of the equation by ‘2’ to solve for ‘\begin{align*}x\end{align*}.’
\begin{align*}\begin{array}{rcl} 2x &=& 130^{\circ} \\ \frac{\overset{1}{\cancel{2}}x}{\cancel{2}} &=& \frac{130^{\circ}}{2} \\ x &=& 65^{\circ} \end{array}\end{align*}
\begin{align*}m \angle 1 = 65^{\circ}\end{align*} and \begin{align*}m \angle 2 = 65^{\circ}\end{align*}.
#### Example 3
For the following diagram, determine the measure of angles ‘\begin{align*}x\end{align*}’ and ‘\begin{align*}y\end{align*}.’
First, write down what you know from the diagram.
\begin{align*}\begin{array}{rcl} m \angle x + 140^{\circ} &=& 180^{\circ} \quad \text{Straight angle formed by adjacent angles} \\ m \angle y + 125^{\circ} &=& 180^{\circ} \quad \text{Straight angle formed by adjacent angles} \\ m \angle x + m \angle y + 85^{\circ} &=& 180^{\circ} \quad \text{Sum of the angles of a triangle} \end{array}\end{align*}
Next, use what you have written down to determine the measure of \begin{align*}\angle x\end{align*}.
\begin{align*}m \angle x + 140^{\circ} = 180^{\circ}\end{align*}
Next, subtract 140° from both sides of the equation.
\begin{align*}\begin{array}{rcl} m \angle x + 140^{\circ} &=& 180^{\circ} \\ m \angle x + 140^{\circ} - 140^{\circ} &=& 180^{\circ} - 140^{\circ} \end{array}\end{align*}
Next, simplify both sides of the equation.
\begin{align*}\begin{array}{rcl} m \angle x + 140^{\circ} - 140^{\circ} &=& 180^{\circ} - 140^{\circ} \\ m \angle x &=& 40^{\circ} \end{array}\end{align*}
The measure of \begin{align*}\angle x\end{align*} is 40°.
Now, use what you know about the measures of the interior angles of the triangle to solve for the measure of \begin{align*}\angle y\end{align*}.
\begin{align*}m \angle x + m \angle y + 85^{\circ} = 180^{\circ}\end{align*}
Next, substitute the measure of \begin{align*}\angle x\end{align*} into the equation.
\begin{align*}\begin{array}{rcl} m \angle x + m \angle y + 85^{\circ} &=& 180^{\circ} \\ 40^{\circ} + m \angle y + 85^{\circ} &=& 180^{\circ} \end{array}\end{align*}
Next, simplify the left side of the equation.
\begin{align*}\begin{array}{rcl} 40^{\circ} + m \angle y + 85^{\circ} &=& 180^{\circ} \\ 125^{\circ} + m \angle y &=& 180^{\circ} \end{array}\end{align*}
Next, subtract 125° from both sides of the equation.
\begin{align*}\begin{array}{rcl} 125^{\circ} + m \angle y &=& 180^{\circ} \\ 125^{\circ} - 125^{\circ} + m \angle y &=& 180^{\circ} - 125^{\circ} \end{array}\end{align*}
Next, simplify both sides of the equation.
\begin{align*}\begin{array}{rcl} 125^{\circ} - 125^{\circ} + m \angle y &=& 180^{\circ} - 125^{\circ} \\ m \angle y &=& 55^{\circ} \end{array}\end{align*}
The measure of \begin{align*}\angle y\end{align*} is 55°.
#### Example 4
Given \begin{align*} \triangle ABC\end{align*}, an obtuse scalene triangle with \begin{align*}\angle A = 37^{\circ}\end{align*} and \begin{align*}\angle b = 28^{\circ}\end{align*}, what is the measure of the obtuse angle?
First, draw and label a triangle to model the problem.
Next, write an equation to represent the sum of the interior angles of the triangle.
\begin{align*}\angle A + \angle B + \angle C = 180^{\circ}\end{align*}
Next, fill into the equation, the measures of the angles given in the diagram.
\begin{align*}\begin{array}{rcl} \angle A + \angle B + \angle C &=& 180^{\circ} \\ 37^{\circ} + 28^{\circ} + \angle C &=& 180^{\circ} \end{array}\end{align*}
Next, simplify the left side of the equation.
\begin{align*}\begin{array}{rcl} 37^{\circ} + 28^{\circ} + \angle C &=& 180^{\circ} \\ 65^{\circ} + \angle C &=& 180^{\circ} \end{array}\end{align*}
Then, subtract 65° from both sides of the equation to solve for the measure of \begin{align*}\angle C\end{align*}.
\begin{align*}\begin{array}{rcl} 65^{\circ} + \angle C &=& 180^{\circ} \\ 65^{\circ} - 65^{\circ} + \angle C &=& 180^{\circ} - 65^{\circ} \\ \angle C &=& 115^{\circ} \end{array}\end{align*}
The measure of the obtuse angle is 115°.
#### Example 5
Given \begin{align*}\triangle DEF\end{align*}, such that \begin{align*}\angle D = 42^{\circ}\end{align*} and \begin{align*}\angle E = 123^{\circ}\end{align*}, what is the measure of \begin{align*}\angle F\end{align*}?
First, draw and label a triangle to model the problem.
Next, write an equation to represent the sum of the measures of the interior angles of the triangle.
\begin{align*}\angle D + \angle E + \angle F = 180^{\circ}\end{align*}
Next, fill into the equation, the measures of the angles given in the diagram.
\begin{align*}\begin{array}{rcl} \angle D + \angle E + \angle F &=& 180^{\circ} \\ 42^{\circ} + 123^{\circ} + \angle F &=& 180^{\circ} \end{array}\end{align*}
Next, simplify the left side of the equation.
\begin{align*}\begin{array}{rcl} 42^{\circ} + 123^{\circ} + \angle F &=& 180^{\circ} \\ 165^{\circ} + \angle F &=& 180^{\circ} \end{array}\end{align*}
Then, subtract 165° from both sides of the equation to solve for the measure of \begin{align*}\angle C\end{align*}.
\begin{align*}\begin{array}{rcl} 165^{\circ} + \angle F &=& 180^{\circ} \\ 165^{\circ} - 165^{\circ} + \angle F &=& 180^{\circ} - 165^{\circ} \\ \angle F &=& 15^{\circ} \end{array}\end{align*}
The measure of \begin{align*}\angle F = 15^{\circ}\end{align*}.
### Review
Using what you have learned about the interior angles of a triangle, determine the missing angle in each triangle.
1. 45°, 45°, ?
2. 60°, 60°, ?
3. 90°, 50°, ?
4. 100°, 40°, ?
5. 110°, 30°, ?
6. 50°, 10°, ?
7. 145°, 15°, ?
8. 55°, 45°, ?
9. 70°, 35°, ?
10. 50°, 50°, ?
11. 63°, 42°, ?
12. 18°, 75°, ?
Identify three triangles in the room around you.
13.
14.
15.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Acute Triangle
An acute triangle has three angles that each measure less than 90 degrees.
Congruent
Congruent figures are identical in size, shape and measure.
Equilateral Triangle
An equilateral triangle is a triangle in which all three sides are the same length.
Isosceles Triangle
An isosceles triangle is a triangle in which exactly two sides are the same length.
Obtuse Triangle
An obtuse triangle is a triangle with one angle that is greater than 90 degrees.
Right Triangle
A right triangle is a triangle with one 90 degree angle.
Scalene Triangle
A scalene triangle is a triangle in which all three sides are different lengths.
Triangle Sum Theorem
The Triangle Sum Theorem states that the three interior angles of any triangle add up to 180 degrees.
1. [1]^ License: CC BY-NC 3.0
2. [2]^ License: CC BY-NC 3.0
3. [3]^ License: CC BY-NC 3.0
4. [4]^ License: CC BY-NC 3.0
5. [5]^ License: CC BY-NC 3.0
6. [6]^ License: CC BY-NC 3.0
7. [7]^ License: CC BY-NC 3.0
8. [8]^ License: CC BY-NC 3.0
### Explore More
Sign in to explore more, including practice questions and solutions for Triangle Sum Theorem.
|
# Search by Topic
#### Resources tagged with Positive-negative numbers similar to Women in Maths:
Filter by: Content type:
Stage:
Challenge level:
### There are 19 results
Broad Topics > Numbers and the Number System > Positive-negative numbers
### The History of Negative Numbers
##### Stage: 3, 4 and 5
This article -useful for teachers and learners - gives a short account of the history of negative numbers.
### Negative Numbers
##### Stage: 3
A brief history of negative numbers throughout the ages
### Strange Bank Account (part 2)
##### Stage: 3 Challenge Level:
Investigate different ways of making £5 at Charlie's bank.
### Vector Racer
##### Stage: 2, 3 and 4 Challenge Level:
The classic vector racing game brought to a screen near you.
### Missing Multipliers
##### Stage: 3 Challenge Level:
What is the smallest number of answers you need to reveal in order to work out the missing headers?
### Strange Bank Account
##### Stage: 3 Challenge Level:
Imagine a very strange bank account where you are only allowed to do two things...
### Making Sense of Positives and Negatives
##### Stage: 3
This article suggests some ways of making sense of calculations involving positive and negative numbers.
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Playing Connect Three
##### Stage: 3 Challenge Level:
In this game the winner is the first to complete a row of three. Are some squares easier to land on than others?
### Up, Down, Flying Around
##### Stage: 3 Challenge Level:
Play this game to learn about adding and subtracting positive and negative numbers
### Minus One Two Three
##### Stage: 4 Challenge Level:
Substitute -1, -2 or -3, into an algebraic expression and you'll get three results. Is it possible to tell in advance which of those three will be the largest ?
### Pair Sums
##### Stage: 3 Challenge Level:
Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers?
### First Connect Three
##### Stage: 2 and 3 Challenge Level:
The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for?
### Adding and Subtracting Positive and Negative Numbers
##### Stage: 3
How can we help students make sense of addition and subtraction of negative numbers?
### Consecutive Numbers
##### Stage: 2 and 3 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Weights
##### Stage: 3 Challenge Level:
Different combinations of the weights available allow you to make different totals. Which totals can you make?
### Consecutive Negative Numbers
##### Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Negatively Triangular
##### Stage: 4 Challenge Level:
How many intersections do you expect from four straight lines ? Which three lines enclose a triangle with negative co-ordinates for every point ?
### Connect Three
##### Stage: 3 and 4 Challenge Level:
Can you be the first to complete a row of three?
|
# Difference between revisions of "2017 AMC 10B Problems/Problem 14"
## Problem
An integer $N$ is selected at random in the range $1\leq N \leq 2020$ . What is the probability that the remainder when $N^{16}$ is divided by $5$ is $1$?
$\textbf{(A)}\ \frac{1}{5}\qquad\textbf{(B)}\ \frac{2}{5}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{4}{5}\qquad\textbf{(E)}\ 1$
## Solution 1
By Fermat's Little Theorem, $N^{16} = (N^4)^4 \equiv 1 \text{ (mod 5)}$ when N is relatively prime to 5. However, this happens with probability $\boxed{\textbf{(D) } \frac 45}$.
## Solution 2
Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits $0-9$ . The pattern for $0$ is $0$, no matter what power, so $0$ doesn't work. Likewise, the pattern for $5$ is always $5$. Doing the same for the rest of the digits, we find that the units digits of $1^{16}$, $2^{16}$ ,$3^{16}$, $4^{16}$ ,$6^{16}$, $7^{16}$ ,$8^{16}$ and $9^{16}$ all have the remainder of $1$ when divided by $5$, so $\boxed{\textbf{(D) } \frac 45}$.
2017 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
|
# Plot Points and Draw Lines on the Plane
Related Topics:
Lesson Plans and Worksheets for Grade 5
Lesson Plans and Worksheets for all Grades
Videos, solutions and examples to help grade 5 students learn how to plot points, using them to draw lines in the plane, and describe patterns within the coordinate pairs.
### New York State Common Core Math Module 6, Grade 5, Lesson 7
Lesson 7 Concept Development
Problem 1: Describe patterns in coordinate pairs and name the rule.
Lesson 7 Problem Set
1. Complete the chart. Then, plot the points on the coordinate plane below.
a. Use a straightedge to draw a line connecting these points.
b. Write a rule showing the relationship between the x- and y-coordinates of points on the line.
c. Name 2 other points that are on this line.
2. Complete the chart. Then, plot the points on the coordinate plane below.
a. Use a straightedge to draw a line connecting these points.
b. Write a rule showing the relationship between the x- and y-coordinates.
c. Name 2 other points that are on this line.
3. Use the coordinate plane below to answer the following questions.
a. Give the coordinates for 3 points that are on line a.
b. Write a rule that describes the relationship between the x- and y-coordinates for the points on line a.
c. What do you notice about the -coordinates of every point on line b?
d. Fill in the missing coordinates for points on line d.
e. For any point on line c, the x-coordinate is _______.
f. Each of the points lies on at least 1 of the lines shown in the plane above. Identify a line that contains each of the following points. Lesson 7 Homework
1. Complete the chart. Then, plot the points on the coordinate plane.
a. Use a straightedge to draw a line connecting these points.
b. Write a rule showing the relationship between the x- and y- coordinates of points on this line.
c. Name two other points that are also on this line.
2. Complete the chart. Then, plot the points on the coordinate plane.
a. Use a straightedge to draw a line connecting these points.
b. Write a rule showing the relationship between the x- and y- coordinates for points on the line.
c. Name two other points that are also on this line.
3. Use the coordinate plane to answer the following questions.
a. For any point on line m, the x- coordinate is ______.
b. Give the coordinates for 3 points that are on line n.
c. Write a rule that describes the relationship between the x- and y-coordinates on line q.
d. Give the coordinates for 3 points that are on line q.
e. Write a rule that describes the relationship between the x- and y-coordinates on line q.
f. For each point, identify a line on which each of these points lie.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
|
# Dot Product
A vector has magnitude (how long it is) and direction:
Here are two vectors:
They can be multiplied using the "Dot Product" (also see Cross Product).
## Calculating
The Dot Product is written using a central dot:
a · b
This means the Dot Product of a and b
We can calculate the Dot Product of two vectors this way:
a · b = |a| × |b| × cos(θ)
Where:
|a| is the magnitude (length) of vector a
|b| is the magnitude (length) of vector b
θ is the angle between a and b
So we multiply the length of a times the length of b, then multiply by the cosine of the angle between a and b
OR we can calculate it this way:
a · b = ax × bx + ay × by
So we multiply the x's, multiply the y's, then add.
Both methods work!
And the result is a number (called a "scalar" to show it is not a vector).
### Example: Calculate the dot product of vectors a and b:
a · b = |a| × |b| × cos(θ)
a · b = 10 × 13 × cos(59.5°)
a · b = 10 × 13 × 0.5075...
a · b = 65.98... = 66 (rounded)
OR we can calculate it this way:
a · b = ax × bx + ay × by
a · b = -6 × 5 + 8 × 12
a · b = -30 + 96
a · b = 66
Both methods came up with the same result (after rounding)
Also note that we used minus 6 for ax (it is heading in the negative x-direction)
## Why cos(θ) ?
OK, to multiply two vectors it makes sense to multiply their lengths together but only when they point in the same direction.
So we make one "point in the same direction" as the other by multiplying by cos(θ):
We take the component of a
that lies alongside b
Like shining a light to see
THEN we multiply !
It works exactly the same if we "projected" b alongside a then multiplied.
Because it doesn't matter which order we do the multiplication:
|a| × |b| × cos(θ) = |a| × cos(θ) × |b|
In effect, the dot product multiplies the aligned lengths.
### Remembering Cos
To remember to multiply by cos(θ) think "dot cos".
## Right Angles
When two vectors are at right angles to each other the dot product is zero.
### Example: calculate the dot product for:
a · b = |a| × |b| × cos(θ)
a · b = |a| × |b| × cos(90°)
a · b = |a| × |b| × 0
a · b = 0
or we can calculate it this way:
a · b = ax × bx + ay × by
a · b = -12 × 12 + 16 × 9
a · b = -144 + 144
a · b = 0
This can be a handy way to find out if two vectors are at right angles.
## Same Direction
The dot product of two vectors that point in the same direction is the simple product of their lengths, because the angle is 0 degrees which has a cosine of 1
a · b = |a| × |b| × cos(0°)
a · b = a × b × 1
a · b = ab
## Right-Angled Triangle
Let's use the dot product on a right-angled triangle!
Vector c is the sum of a and b:
c = a + b
Do a dot product on both sides:
c · c = (a + b) · (a + b)
Expand:
c · c = a · (a + b) + b · (a + b)
Expand Again:
c · c = a · a + a · b + b · a + b · b
a · b = b · a = 0 (right angles):
c · c = a · a + b · b
c · c = c2 etc:
c2 = a2 + b2
We just proved the Pythagorean Theorem!
Note: when we allow angles other than 90 degrees we can create the Law of Cosines. Have a go yourself, but be careful how you define the angle!
## Physics
The Dot Product is used a lot in Physics
### Example: Work
In Physics Work is force times distance, but only the aligned part.
So work is the dot product of force and distance.
Alex pushes a box 3 m straight forward using 200 N of force. But his push is a little upwards by 20°.
Work =
Force · Distance
=
Force × Distance × cos θ
=
200 N × 3 m × cos 20°
=
200 N × 3 m × 0.9397...
=
564 J (to nearest Joule, where J = N m)
(Without cos θ, the wrong value would be 600 J)
## Three or More Dimensions
This all works fine in 3 (or more) dimensions, too.
And can actually be very useful!
### Example: Sam has measured the end-points of two poles, and wants to know the angle between them:
We have 3 dimensions, so don't forget the z-components:
a · b = ax × bx + ay × by + az × bz
a · b = 9 × 4 + 2 × 8 + 7 × 10
a · b = 36 + 16 + 70
a · b = 122
Now for the other formula:
a · b = |a| × |b| × cos(θ)
But what is |a| ? It is the magnitude, or length, of the vector a. We can use Pythagoras:
• |a| = √(42 + 82 + 102)
• |a| = √(16 + 64 + 100)
• |a| = √180
Likewise for |b|:
• |b| = √(92 + 22 + 72)
• |b| = √(81 + 4 + 49)
• |b| = √134
And we know from the calculation above that a · b = 122, so:
a · b = |a| × |b| × cos(θ)
122 = √180 × √134 × cos(θ)
cos(θ) = 122 / (√180 × √134)
cos(θ) = 0.7855...
θ = cos-1(0.7855...) = 38.2...°
Done!
I tried a calculation like that once, but worked all in angles and distances ... it was very hard, involved lots of trigonometry, and my brain hurt. The method above is much easier.
## Cross Product
The Dot Product gives a scalar (ordinary number) answer, and is sometimes called the scalar product.
But there is also the Cross Product which gives a vector as an answer, and is sometimes called the vector product.
3036, 3037, 3030, 3031, 3032, 3033, 3034, 3035, 3903, 3904
|
# Subtracting Mixed Numbers
## Calculatrice associée: Calculateur de fractions
Subtracting mixed numbers is quite easy.
We know that mixed number consists of integer part and fractional part.
To subtract mixed numbers three steps are needed:
1. Convert each mixed number to improper fraction.
2. Subtract improper fractions (using subtraction of fractions with unlike denominators)
3. Convert improper fraction to mixed number if needed (and if possible).
Example 1. Find ${1}\frac{{3}}{{5}}-{2}\frac{{4}}{{9}}$.
Convert each mixed number into improper fraction: ${1}\frac{{3}}{{5}}=\frac{{8}}{{5}}$ and ${2}\frac{{4}}{{9}}=\frac{{22}}{{9}}$.
Now subtract fractions: $\frac{{8}}{{5}}-\frac{{22}}{{9}}=\frac{{{8}\cdot{9}}}{{{5}\cdot{9}}}-\frac{{{22}\cdot{5}}}{{{9}\cdot{5}}}=\frac{{72}}{{45}}-\frac{{110}}{{45}}=-\frac{{38}}{{45}}$.
Convert fraction to mixed number: can't convert because fraction is proper.
Answer: $-\frac{{38}}{{45}}$.
Next example.
Example 2. Find $-{2}\frac{{1}}{{6}}-{3}\frac{{1}}{{2}}$.
Convert each mixed number into improper fraction: $-{2}\frac{{1}}{{6}}=-\frac{{13}}{{6}}$ and ${3}\frac{{1}}{{2}}=\frac{{7}}{{2}}$.
Now subtract fractions: $-\frac{{13}}{{6}}-\frac{{7}}{{2}}=-\frac{{13}}{{6}}-\frac{{{7}\cdot{3}}}{{{2}\cdot{3}}}=-\frac{{13}}{{6}}-\frac{{21}}{{6}}=-\frac{{34}}{{6}}$.
Reduce fraction: $-\frac{{34}}{{6}}=-\frac{{17}}{{3}}$.
Convert fraction to mixed number: $-\frac{{17}}{{3}}=-{5}\frac{{2}}{{3}}$.
Answer: $-\frac{{17}}{{3}}=-{5}\frac{{2}}{{3}}$.
Next example.
Example 3. Find $-{2}-{5}\frac{{8}}{{11}}$.
Since first number has no fractional part, we can easier subtract numbers.
Subtract integer parts: $-{2}-{5}=-{7}$ and fractional part leave the same.
Warning. This works only when we subtract either negative and positive numbers (this case) or positive and negative numbers (like ${2}-{\left(-{5}\frac{{8}}{{11}}\right)}={7}\frac{{8}}{{11}}$).
Answer: $-{7}\frac{{8}}{{11}}=-\frac{{85}}{{11}}$.
Now, take pen and paper and do following exercises.
Exercise 1. Find ${5}\frac{{1}}{{6}}-{3}\frac{{7}}{{8}}$.
Answer: $\frac{{31}}{{24}}={1}\frac{{7}}{{24}}$.
Next exercise.
Exercise 2. Find ${7}\frac{{4}}{{9}}-{\left(-{2}\frac{{5}}{{6}}\right)}$.
Answer: $\frac{{185}}{{18}}={10}\frac{{5}}{{18}}$.
Next exercise.
Exercise 3. Find $-{5}\frac{{1}}{{6}}-{\left(-{3}\frac{{7}}{{8}}\right)}$.
Answer: $-\frac{{31}}{{24}}=-{1}\frac{{7}}{{24}}$.
Next exercise.
Exercise 4. Find ${3}\frac{{5}}{{6}}-\frac{{5}}{{7}}$.
${3}\frac{{5}}{{6}}-\frac{{5}}{{7}}=\frac{{23}}{{6}}-\frac{{5}}{{7}}=\frac{{161}}{{42}}-\frac{{30}}{{42}}=\frac{{131}}{{42}}={3}\frac{{5}}{{42}}$.
Answer: $\frac{{131}}{{42}}={3}\frac{{5}}{{42}}$.
Next exercise.
Exercise 5. Find ${5}\frac{{1}}{{6}}-{7}$.
Here we just can't add fractional parts to obtain $-{2}\frac{{1}}{{6}}$. This is not correct, because both numbers are positive.
We do it as always.
${5}\frac{{1}}{{6}}-{7}=\frac{{31}}{{6}}-\frac{{42}}{{6}}=-\frac{{11}}{{6}}=-{1}\frac{{5}}{{6}}$.
Answer: $-\frac{{11}}{{6}}=-{1}\frac{{5}}{{6}}$.
If you are not sure whether it is possible to subtract integer parts, use the three-step method. It guarantees correct answer.
|
# 4.4. Conditional Distributions¶
To understand the relation between two variables we must examine the conditional behavior of each of them given the value of the other. Towards this goal, we will start by examining the example of the previous section and then develop the general theory.
In our example, the joint distribution of $$X$$ and $$Y$$ is given by joint_table. Here we also display the marginal distribution of $$X$$.
joint_table.marginal('X')
X=0 X=1 X=2
Y=5 0.00000 0.0000 0.03125
Y=4 0.00000 0.0625 0.09375
Y=3 0.03125 0.1875 0.09375
Y=2 0.09375 0.1875 0.03125
Y=1 0.09375 0.0625 0.00000
Y=0 0.03125 0.0000 0.00000
Sum: Marginal of X 0.25000 0.5000 0.25000
Now suppose we know that $$Y = 3$$. Then the outcome space is reduced to just the cells in the row labeled Y=3.
def indicator_Y_equals_3(i, j):
return j == 3
joint_table.event(indicator_Y_equals_3, 'X', 'Y')
P(Event) = 0.3125
X=0 X=1 X=2
Y=5
Y=4
Y=3 0.03125 0.1875 0.09375
Y=2
Y=1
Y=0
Of course, the probabilities along this row don’t sum to 1. Their sum is $$P(Y = 3) = 0.3125$$.
By the division rule, for each $$x = 0, 1, 2$$ we have
$P(X = x \mid Y = 3) ~ = ~ \frac{P(X = x, Y = 3)}{P(Y = 3)}$
By normalizing all the probabilities in the row by their sum, we get the conditional distribution of $$X$$ given $$Y=3$$.
\begin{split} \begin{align*} P(X = 0 \mid Y = 3) ~ &= ~ \frac{0.03125}{0.3125} = 0.1 \\ \\ P(X = 1 \mid Y = 3) ~ &= ~ \frac{0.1875}{0.3125} = 0.6 \\ \\ P(X = 2 \mid Y = 3) ~ &= ~ \frac{0.09375}{0.3125} = 0.3 \end{align*} \end{split}
Compare this conditional distribution to the unconditional distribution of $$X$$:
$P(X = 0) ~ = ~ 0.25, ~~~~~ P(X = 1) ~ = ~ 0.5, ~~~~~ P(X = 2) ~ = ~ 0.25$
The two distributions are different. Given $$Y = 3$$, the chance that $$X$$ is large is higher than it is if we don’t have that condition.
This shows that $$X$$ and $$Y$$ are dependent. We will define dependence and independence formally in the next section.
Quick Check
Find the following (without using Python) based on the joint distribution in the example above.
(a) $$P(X=0 \mid Y=1)$$
(b) the conditional distribution of $$X$$ given $$Y=1$$
(c) the conditional distribution of $$X$$ given $$Y=5$$
## 4.4.1. Conditional Distribution of $$X$$ given $$Y = y$$¶
The conditional_dist method operates on a joint distribution object and displays conditional distributions, as follows.
# conditional distribution of X given each different value of Y
joint_table.conditional_dist('X', 'Y')
X=0 X=1 X=2 Sum
Dist. of X | Y=5 0.00 0.0 1.00 1.0
Dist. of X | Y=4 0.00 0.4 0.60 1.0
Dist. of X | Y=3 0.10 0.6 0.30 1.0
Dist. of X | Y=2 0.30 0.6 0.10 1.0
Dist. of X | Y=1 0.60 0.4 0.00 1.0
Dist. of X | Y=0 1.00 0.0 0.00 1.0
Marginal of X 0.25 0.5 0.25 1.0
To understand this table, start with the row labeled Y=3. The entries are the probabilities in the conditional distribution of $$X$$ given $$Y=3$$.
In the row labeled Y=1, the entries are the probabilities in the conditional distribution of $$X$$ given $$Y=1$$. Notice that if $$Y=1$$ then $$X$$ can’t be 2. You can go back and confirm that in the joint distribution table, $$P(X = 2, Y = 1) = 0$$.
All the other rows can be understood in the same way. In row $$y$$, the given condition is $$Y=y$$, and the entries are the probabilities in the conditional distribution of $$X$$ given $$Y=y$$.
It is easy to see why each row in the table of conditional distributions sums to 1. The value in each cell in the row is obtained from the joint distribution table by taking the corresponding cell in that table and dividing its entry by the sum of the entries in the row.
## 4.4.2. The Theory¶
We can now generalize the calculations we did in the example above.
Let $$X$$ and $$Y$$ be two random variables defined on the same space. If $$x$$ is a possible value of $$X$$, and $$y$$ and possible value of $$Y$$, then
$P(X = x \mid Y = y) = \frac{P(X = x, Y = y)}{P(Y = y)}$
The conditional probability $$P(X = x \mid Y = y)$$ is displayed in the $$(x, y)$$ cell of the table of conditional distributions above.
For a fixed value $$y^*$$ of $$Y$$, the conditional distribution of $$X$$ given $$Y = y^*$$ is the collection of probabilities
$P(X = x \mid Y = y^*) = \frac{P(X = x, Y = y^*)}{P(Y = y^*)}$
where $$x$$ ranges over all the values of $$X$$. Keep in mind that $$x$$ represents the values of the variable here. The value $$y^*$$ is the particular value of $$Y$$ that was observed, so it is a constant.
## 4.4.3. The Probabilities in a Conditional Distribution Sum to 1¶
In a distribution, the probabilities have to sum to 1. To see that this is true for the conditional distribution defined above, start by using the fundamental rule.
Find $$P(Y = y^*)$$ by partitioning the event $$\{ Y = y^* \}$$ according to the values of $$X$$:
$P(Y = y^*) = \sum_{\text{all }x} P(X = x, Y = y^*)$
Now sum the probabilities in the conditional distribution of $$X$$ given $$Y = y^*$$:
\begin{split} \begin{align*} \sum_{\text{all }x} P(X = x \mid Y = y^*) ~ &= ~ \sum_{\text{all }x} \frac{P(X = x, Y = y^*)}{P(Y = y^*)} \\ \\ &= ~ \frac{1}{P(Y = y^*)} \sum_{\text{all }x} P(X = x, Y = y^*) \\ &= ~ \frac{1}{P(Y = y^*)} \cdot P(Y = y^*) \\ \\ &= ~ 1 \end{align*} \end{split}
Thus the conditional distribution is just an ordinary probability distribution: a set of values with probabilities that sum to 1.
|
Name: Rilla Collins
Topic: Jack and the Beanstalk
Objectives (P.A.S.S.): Standard 5:
Data Analysis & Probability -- The student will demonstrate an understanding of data collection, display & interpretation.
1. Data Analysis
a. Pose questions, collect, record, & interpret data to help answer questions (e.g., Which was the most popular booth at our carnival?).
b. Read graphs & charts; identify the main idea, draw conclusions, make predictions based on the data (e.g., predict how many children will bring their lunch based on a menu).
c. Construct a bar graph or pictograph with labels & a title from a set of data.
Introduction: This lesson is a great beginning lesson for gathering information from one scource to form graphs. This lesson guides the students into a mind frame of how to gather information from a book and graph the information in a bar, line, or pictograph.
Instructional process: I first introduce how to make graphs on the board. I also show examples of different kinds of graphs, how we use graphs and why graphs are important. We read different kinds of graphs and evaluate the information that each graph contains. We then pick out 10 words that we think will appear most often - the students are involved with this process. We divided into groups and each group took time to skim the book looking for these 10 words.
Closure: In closing, the groups gathered the information, I then took the information and made a chart to save time. The kids could have made the chart, but we then took the chart and proceeded to make 3 different kinds of graphs. This assignment takes 3-4 days to create the graphs for the first time.
Assessment: Assessment can be done on the group work, the skimming process to see how accurately they found the words, and how accuarely they formed the graphs to show the information obtained.
Modifications/Accommodations: For students with special needs I paired them with higher achieving students so they could aid them with the work that needed to be done. Also special needs kids could only be given 3-5 word to graph instead of 10 words.
Reflection: This project worked well in my class. The kids enjoyed making the graphs and most of them did a good job. We will continue to work on this skill because some really struggled changing kinds of graphs. They could do the pictograph (pictogram) easily, but the line and bar graphs confused them.
This activity teaches many skills that are needed in the real world of life and graphs and charts are a major part of the PASS Skills.
Word / Tally mark / Total of times
hide/ hiding / 4
mother / 18
down / 11
came / 4
giant / 27
climbed / 21
again / 3
up / 5
money / 9
Jack / 57
Word / Tally mark / Total of times
hide/ hiding / 4
mother / 18
down / 11
came / 4
giant / 27
climbed / 21
again / 3
up / 5
money / 9
Jack / 57
Word / Tally mark / Total of times
hide/ hiding / 4
mother / 18
down / 11
came / 4
giant / 27
climbed / 21
again / 3
up / 5
money / 9
Jack / 57
|
### Showcase: Ocean Drifters— Investigating Ocean
```NATIONAL INSTITUTE FOR LITERACY
LITERACY INFORMATION AND COMMUNICATION SYSTEM (LINCS)
Science & Numeracy Special Collection
Showcase: Ocean Drifters—
Investigating Ocean Currents
ACTIVITY 2: THE RACE IS ON
(TEACHER VERSION)
A.Speed
Speed is a measure of how fast an object (including people) moves across a given distance.
To calculate speed we take the distance between two points and divide by the time it takes to
cross that distance:
Speed (m/s) = Distance (m) ∏ Time (s)
Many different units can be used to describe speed. For instance, if you are calculating the
speed of a train, you may want to use mph (miles per hour) or kph (kilometers per hour).
However, if you are tracking the path of a snail, you may want to use cm/h (centimeters per
hour) or in/h (inches per hour).
Practice calculating speed using different units for the examples given below.
* nautical miles
When making charts to show currents, scientists often use arrows to represent speed. The
size of the arrow is proportional to the speed. If 1 cm = 1 mph, then the airplane’s speed
could be represented by an arrow that is 350 cm long. Yikes! That is a long arrow! In this
case it would be better to use 1 cm = 35 mph. Then our arrow would be only 10 cm long.
If we use 1 cm = 1mph, then the snail’s arrow would only be 0.8 cm long. That arrow is a
short arrow. Using 1 cm = 10 mph, on the next page, draw the arrows which would represent
the speeds of the whale and cheetah.
B. Direction of movement:
Speed describes how fast or slowly an object moves, but it does
not tell us anything about the direction of movement. If we combine
the speed and direction of an object, it is called velocity. Velocity
can be shown using a vector (an arrow which shows direction and
speed).
On maps and charts there is usually a compass rose. The compass
rose shows north-south-east-west directions on a map. To the right
is an example.
N = North, S = South, W = West, E = East
If a direction is between 2 of the major directions (N,S, E and W), then the directions are
combined. An arrow pointing to a direction between North and East, would be pointing
Northeast (NE). What direction do these arrows point?
Example:
East
a.
North
b.
Southeast
For each of the arrows below, measure its length. Using a scale of 1 cm = 10 mph, determine
what speed it represents and name the direction it is pointing (this is its velocity).
a Example: West (20 mph)
b.
Southwest (10 mph)
c.
South (23 mph)
Draw arrows which represent the speed and direction for the following velocities.
a. 20 mph, south
b. 30 mph, southwest
These activities were adapted from “Track a NOPP Drifter” written by Anna C. Switzer for the
NOPP-Consortium of Oceanographic Activities for Students and Teachers (COAST). To find
|
## Exploding Dots
### 3.3 (Optional) The Traditional Algorithm
Lesson materials located below the video overview.
How does this dots-and-boxes approach to addition compare to the standard algorithm most people know?
Let’s go back to the example $$358+287$$. Most people are surprised (maybe even perturbed) by the straightforward left-to-right answer $$5|13|15$$.
This is because the traditional algorithm has us work from right to left, looking at $$8+7$$ first.
But in the algorithm we don’t write down the answer $$15$$. Instead, we explode ten dots right away and write on paper a $$5$$ in the answer line together with a small $$1$$ tacked on to the middle column. People call this carrying the one and it – correctly – corresponds to adding an extra dot in the tens position.
Now we attend to the middle boxes. Adding gives $$14$$ dots in the tens box ($$5+8$$ gives thirteen dots, plus the extra dot from the previous explosion).
And we perform another explosion.
On paper, one writes “$$4$$” in the tens position of the answer line, with another little “$$1$$” placed in the next column over. This matches the idea of the dots-and-boxes picture precisely.
And now we finish the problem by adding the dots in the hundreds position.
So the traditional algorithm works right to left and does explosions (“carries”) as one goes along. On paper it is swift and compact and this might be why it has been the favored way of doing long addition for centuries.
The Exploding Dots approach works left to right, just as we are taught to read in English, and leaves all the explosions to the end. It is easy to understand and kind of fun.
Both approaches, of course, are good and correct. It is just a matter of taste and personal style which one you choose to do. (And feel free to come up with your own new, and correct, approach too!)
Please join the conversation on Facebook and Twitter and kindly share this page using the buttons below.
Facebook
Twitter
## Books
Take your understanding to the next level with easy to understand books by James Tanton.
BROWSE BOOKS
## Guides & Solutions
Dive deeper into key topics through detailed, easy to follow guides and solution sets.
BROWSE GUIDES
## Donations
Consider supporting G'Day Math! with a donation, of any amount.
Your support is so much appreciated and enables the continued creation of great course content. Thanks!
## Ready to Help?
Donations can be made via PayPal and major credit cards. A PayPal account is not required. Many thanks!
DONATE
|
## Introduction
In our daily life, there are many situations where we need to compare two quantities.
• When you want to know which is more
• When you want to know which is better
For e.g., if you want to buy a television for ₹ 20,000, you compare the features of all the televisions under ₹ 20,000 and purchase the best.
• When you want to find an increase in quantity.
For eg., to calculate the increase in the population of the country, we compare the population of the current year to the previous year.
To understand the growth of the company, we need to compare the growth from the previous years to the current year.
Comparison is used in every field. We compare quantities to understand:
• Which is more and which is less
• The right thing that is suited for a particular purpose.
Percentages are widely used to compare different quantities. Watch the video to understand the need for percentages.
Percentage is a way to express any given number as a fraction of 100, i.e., the value in each 100.
32% means 32 out of 100. We use the symbol % to represent a percentage.
Anu scored 80 marks out of 100 in the maths exam. How do we write Anu's marks? Anu’s marks can be written as 80 100.
The percentage of Anu’s marks = 80%
#### Concepts
The chapter ‘Comparing Quantities’ covers the following concepts:
#### Percentage
Two friends Reeta and Anu are discussing the marks they scored in their term exams.
Who do you think did better in the exam, Reeta or Anu? Since Reeta scored 450 marks out of 600 and Anu scored 425 marks out of 500, can we say Reeta did better? No, we cannot decide who did better just by comparing their marks. What do we do in this case? We should convert the marks into a percentage.
Per cent means per hundred.
So, we should convert the marks of Reeta and Anu per 100.
• If Reeta scored 450 out of 600, what is her score out of 100?
• If Anu scored 425 out of 500, what is her score out of 100?
By converting marks per hundred we can compare the marks and decide who did better.
Per cent is derived from the Latin word ‘per centum’ meaning ‘per hundred.’ Per cent is represented using the symbol %.
For example, 2% marks mean 2 marks out of total 100 marks. Which can be written as 2% = 2 100
In other words, percentages are numerators of fractions with denominator 100. That is, 20 100 = 20 per cent.
There are 100 children in a library. Out of the 100 children, 34 are from class 5, 26 from class 6 and 40 from class 7.
Let us find the fractions of class 5, class 6, and class 7 children.
The fraction of class 5 children = 34 100
The fraction of class 6 children = 26 100
The fraction of class 7 children = 40 100
Let us find the percentages of class 5, class 6, and class 7 children.
The percentage of class 5 children = 34%
The percentage of class 6 children = 26%
The percentage of class 7 children = 40%
Anu, Mahesh and Anish are playing in a park. Anu’s brother saw them playing. He comes and asks them a question.
All 3 of them calculate the percentage of girls in the park. Mahesh calculates as shown,
Anu calculates as shown:
Anish calculates as shown,
All of them tell Akshay that the percentage of girls is 48%.
Did you notice, all 3 of them have found the percentage in different methods? Yes! Hence, we can calculate the percentage in different methods and still arrive at the same result.
Out of the 20 balls in a basket, 15 are red in colour. Let us find the percentage of red balls in the basket. What is the fraction of red balls?
The fraction of red balls = 15 20
We need to find the percentage of red balls, that is, we need to find, out of the 100 balls how many will be red balls. Out of the 20 balls, 15 are red balls, then out of 100 balls how many will be red balls?
The percentage of red balls = 15 20 × 100
= 15 × 5
= 75%
To express a given quantity as a percentage of a whole follow this strategy.
• Express the given quantity as a fraction of the whole quantity.
• Multiply the fraction by 100.
• Put the per cent sign '%'.
#### Converting Fraction/ Decimal to Percentage
To compare fractional numbers, we need a common denominator. What do we do in such cases? We can convert the fractions into percentage. That is, to compare fractional numbers, we need a common denominator and we have seen that it is more convenient to compare if our denominator is 100. The maximum marks are different in both tests, in such cases if we convert the fraction into percentage, it becomes easier to compare.
Converting to percentage means: Finding how many marks out of 100.
The fraction 1625 means 16 out of 25, so let us find how many marks out of 100.
By cross multiplying, we get 16 25 × 100
= 16 × 4
= 64
That is 64 out of 100 = 64 100 = 64%
In which test did Kavita do better?
Let us convert test 2 marks into a percentage, since we already converted test 1 marks as 64%:
32 40 = ( 32 40 × 100)% = 320 4% = 80%
Hence, Kavita did better in the second test.
For e.g., convert 0.8 into a percentage:
0.8 = 8 10 = 8 10 × 100% = 80%
#### Converting Percentage to Fraction/ Decimal
We know how to convert fraction or decimal into a percentage. Now, let us understand how to convert a percentage into fraction or decimal?
Hence, 34% is equal to the fraction 17 50
Let us understand how to convert a percentage into decimal.
Hence, 22% is equal to the decimal 0.22.
To convert the percentage into a fraction:
• Divide the percentage by 100.
• Reduce the fraction to the simplest form and remove the sign %.
To convert percentage into decimal:
• Convert the percentage to a fraction by dividing the percentage by 100.
• Write the fraction in the decimal form.
#### Application of Percentages
Akshay goes to a shop to buy a toy. The cost of that toy is ₹600 but he has only ₹500.
Luckily, there is a sale going on in that shop. There is a discount of 25% for every purchase. Now Akshay wants to calculate the cost of the toy that he has selected, after the discount.
Do you think he has enough money to buy the toy after the discount?
Let us help Akshay find 25% of 600.
• First, write the percentage as a fraction out of 100
That is, 25% = 25 100
As you know ‘of’ means multiplication.
• Hence, multiply the fraction and the given number
25 100× 600 = 150
There will be a discount of ₹150 for the toy. So, the cost of the toy after discount is ₹600 – ₹150 = ₹450.
Hence, Akshay has enough money to buy that toy and he must pay ₹450 after applying the discount.
To find the percentage of a given quantity:
• Write the percentage as a fraction out of 100 or convert the percentage into a fraction
• Multiply the fraction by the given quantity.
You can also find the whole quantity if the percentage of the quantity is given. Let us take an example:
15% of 200 = 15 100 × 200 = 30
15% of 200 is 30 which means if 15% of the number is 30, then the number is 200. Now how do we find a number, if 15% of it is 30? Let the required number be ‘p.’ We have 15% of ‘p’ as 30.
We write 15% as 15 100
Hence, 15 100 × p = 30
p = 30 × 100 15 = 200
Hence, 15% of 200 is 30.
All the children in class 7B are eagerly waiting for their test marks. The teacher comes to the class and announces that she is going to give the marks in the form of a ratio. Neha and Ajay get their marks.
• The ratio of Neha’s mark to the total marks is 3:4.
• The ratio of Ajay’s marks to the total marks is 5:8.
By looking at the ratios can you decided who got more marks? Is it Ajay or Neha? It is not possible to say who got more marks by just looking at the ratios. To compare the ratios, we need to make the second parts of the ratios equal by taking the LCM of the second parts of both ratios. Is there any easy way to compare the marks?
Yes! If the marks are in the form of percentage, it becomes easy to compare, so here we need to convert the ratio into a percentage for easier comparison.
For e.g., let us convert the ratio 4:5 into a percentage:
4:5 = 4 5 = 4 5 × 100 = 4 × 20 = 80%
Anu and Vishal are supposed to pack 200 sweet boxes. The sweet boxes are divided between Anu and Vishal in the ratio 3:1. Now how do we find what percentage of the sweet box should be packed by Anu and Vishal?
Here total parts = 3 + 1 = 4
• Percentage of the sweet box that Anu must pack:
3 5 × 100 = 3 × 25 = 75%
• Percentage of the sweet box that Vishal must pack:
1 4 × 100 = 25%
To find the percentage of each part of the ratio:
• Find the fraction of each part to the total parts
• Multiply each fraction by 100 and put '%' sign
Anu had 2 pencils and her mother gave her 2 more pencils. Raghu had 4 pencils and his mother gave him 2 more pencils.
Do you think both have the same increase in the number of pencils? Yes, both got 2 pencils from their moms, hence, they had an equal increase in the number of pencils. If you see, the number of pencils that Anu had has doubled. Do you think the number of pencils that Raghu had also doubled? No! Then how can we say who received a better increase? We can compare the increase by converting it into a percentage
Can you say they both had an equal percentage of increase in the number of pencils?
Let us understand how to convert the increase or decrease in a quantity into a percentage of the initial amount.
To convert the increase or decrease into a percentage:
Percentage increase/ decrease = (amount of change / original amount × 100)%
Anu: Number of pencils increased from 2 to 4:
Here the increase in the number of pencils
= 4 - 2 = 2
Percentage increase = 2 2 × 100 = 100%
Raghu: Number of pencils increased from 4 to 6:
Here the increase in the number of pencils
= 6 - 4 = 2
Percentage increase = 24 × 100 = 50%
Hence, Anu had a better increase in the number of pencils.
• If the buying price, that is the Cost Price (CP) is less than the Selling Price (SP), then you make a profit.
• If the buying price, that is the Cost Price (CP) is more than the Selling Price (SP), then you have made a loss.
• If the buying price, that is the Cost Price (CP) is the same as the Selling Price (SP), then you have made no profit and no loss.
• To find the profit or gain:
Selling Price – Cost Price
• To find the loss:
Cost Price – selling Price
Anu and Akshay are discussing the loss they made.
We can see that both Anu and Akshay made a loss of ₹4. Can we say both suffered the same loss? To understand who suffered more loss, we need to convert the loss into a percentage.
In general, profit and loss are expressed as a percentage. Profit and loss percentage are calculated on the cost price.
To convert loss into a percentage:
Loss % = CP-SP CP × 100
We know that: CP – SP = Loss
Hence, loss % = Loss CP × 100
In a similar way, we find the profit per cent using the formula:
Profit % = SP- CP CP × 100
We know that SP – CP = Profit
Hence, profit % = Profit CP × 100
To understand who suffered more loss, we need to convert the loss into percentage.
• Anu bought a crayon for ₹20 and made a loss of ₹4 by selling it at ₹16.
Hence, Loss % = CP-SP CP × 100
= 4 20 × 100
= 4 × 5 = 20%
Hence, Anu made a loss of 20%
• Akshay bought a book for ₹200 and made a loss of ₹4 by selling it at ₹196.
Hence, Loss % = CP-SP CP × 100
= 4 200 × 100
= 2%
Hence, Akshay made a loss of 2%.
We know how to calculate the profit% or loss% when the cost price and selling price are known. But how do we find the selling price when cost price and profit/loss% are given? Let us take an example:
The cost price is ₹250 and loss% is 12%. Find the selling price.
Given:
• The cost price = ₹250
• The loss% = 12%
Find: We need to find the selling price
We can see that the situation is loss.
Hence, Selling price = Cost price - Loss
• Let us find the loss
We know that the loss is always calculated on the cost price.
Loss = 12% of ₹250
= ₹(12 100 × 250)
= ₹30
• Let us find the selling price
Selling price = Cost price - Loss
= ₹250 - ₹30 = ₹220
Hence, the selling price is ₹220.
To find the selling price when the cost price and the profit% are given:
• Calculate the amount of profit on the cost price.
• Calculate the selling price by using the formula.
Selling price = Cost price + Profit
We know how to calculate the selling price when the profit% or loss% are known. But how do we find the cost price when the selling price and loss% are given?
To find the cost price when the selling price and loss% are given. Let us take an example:
The selling price is ₹308 and loss% is 12%, find the cost price.
• The selling price = ₹308
• Loss% = 12%
We need to find the cost price.
Since the situation is loss, the selling price should be less than the cost price. We know that loss is 12% of the cost price.
• Let us take the cost price = ₹100
• Then the loss
= 12% of ₹100 = ₹12
• The selling price = Cost price – loss
= ₹100 – ₹12 = ₹88
If the cost price is ₹100, then the selling price is ₹88. Then how much is the cost price when the selling price is ₹308?
By cross multiplying we get,
The cost price
= ₹( 308 × 100 88 )
= ₹ 30800 88
= ₹350
How to find the cost price when the selling price and the loss% are known?
To find the cost price when the selling price and loss% are given:
• Calculate the loss on the cost price
• Calculate the Cost price using the formula
Cost price = Selling price + Loss
To find the cost price when the selling price and profit% are given:
• Take the cost price as ₹100
• Find the selling price when the cost price is ₹100
Selling price = Cost price + Profit
• Perform cross multiplication to find the cost price for the given selling price.
#### Simple Interest
• The money borrowed from a bank is called the principal or the sum.
• To the borrowed money at the end, we need to pay an extra amount that is called interest.
• The total money paid by the borrower at the end is called the amount.
Amount = Principal + Interest
• Interest is calculated as the percentage for a period of one year that is per annum. Interest on ₹100 for 1 year is known as the rate per cent per annum.
Now, what is simple interest? The interest-based only on the principal amount is called simple interest. In short, we write simple interest as SI. For e.g., if you have taken a loan of ₹12,000 from a bank, to keep this money for two years, you must pay ₹1,000 for the first year and ₹1,000 for the second year. ₹2,000 is called simple interest.
If the money borrowed is ₹2,000 and the interest is 10% per annum:
Then the interest that must be paid for one year is 10% of ₹2,000.
• Let us calculate 10% of ₹2,000
10% of ₹2000 = ₹( 10 100 ×2000)
= ₹(10 × 20) = ₹200
Hence, ₹200 is the simple interest to be paid for the principal ₹2,000 for one year.
To find the simple interest/amount to be paid on a given sum of money for more than one year:
We calculate the simple interest using the formula,
SI = P × R × T 100 100
Where P is principal, R is the rate of interest, T is time period.
To calculate the amount, we use the following formula,
How do we calculate if the principal, the time, and the rate of interest are given?
How do we calculate if the rate of interest, when the time period, SI and principal are given?
How do we calculate if the time period, when the principal, SI and the rate of interest are given?
#### Common Errors
The following are topics in which students make common mistakes when dealing with comparing quantities:
• 1. Percentage means out of 100
• 2. Profit and Loss
• 3. Profit or loss percentage is always calculated on the cost price
• 4. Increase or decrease percentage is always calculated on the original price
• 5. How to calculate simple interest when the time period is in the form of months?
#### Percentage Means Out Of 100
To find what per cent of 500 is 20
• Correct method: 20 500 × 100
• Incorrect method: 20 100 × 500
To find 20% of 500:
• Correct method: 20 100 × 500
• Incorrect method: 20 500 × 100
#### Profit and Loss
• When the selling price is more than the cost price the situation is profit.
• When the selling price is less than the cost price the situation is a loss.
#### Profit Or Loss Percentage Is Always Calculated On The Cost Price
We calculate the profit or loss percentage on the cost price of an item, not the selling price. For e.g., The shopkeeper buys a shirt for ₹200 and sell it for ₹250. Here the situation is profit.
Profit = Cost price – Selling price = ₹250 - ₹200 = ₹50. Now,
#### Increase Or Decrease Percentage Is Always Calculated On The Original Price
We calculate the increase or decrease percentage on the original price of an item, not the increased /decreased price.
#### How To Calculate Simple Interest When The Time Period Is In The Form Of Months?
To calculate simple interest when the time period is in the form of months, first, we need to convert the months to years and then use the formula to find the simple interest. For e.g., Principal P = ₹2,000, Rate of interest R = 8% and Time period T = 9 months We need to convert 9 months to years.
9 months = 9 12 years
Then,
Simple Interest = P × T × R 100
= 2000 × 9 × 8 100 × 12
#### Conclusion
We compare quantities almost daily in our lives. Something as simple as ordering food involves comparing quantities, as you compare different restaurants for food options, customer reviews, prices, etc. and then make your choice. Can you solve the riddle below based on comparing quantities?
What percentage of the figure is coloured?
|
# Lesson 12Units in Scale Drawings
Let's use different scales to describe the same drawing.
### Learning Targets:
• I can tell whether two scales are equivalent.
• I can write scales with units as scales without units.
## 12.1Centimeters in a Mile
There are 2.54 cm in an inch, 12 inches in a foot, and 5,280 feet in a mile. Which expression gives the number of centimeters in a mile? Explain your reasoning.
## 12.2Scales Card Sort
Your teacher will give you some cards with a scale on each card.
1. Sort the cards into sets of equivalent scales. Be prepared to explain how you know that the scales in each set are equivalent. Each set should have at least two cards.
2. Trade places with another group and check each other’s work. If you disagree about how the scales should be sorted, work to reach an agreement.
Pause here so your teacher can review your work.
1. Next, record one of the sets with three equivalent scales and explain why they are equivalent.
## 12.3The World’s Largest Flag
As of 2016, Tunisia holds the world record for the largest version of a national flag. It was almost as long as four soccer fields. The flag has a circle in the center, a crescent moon inside the circle, and a star inside the crescent moon.
1. Complete the table. Explain or show your reasoning.
flag length flag height height of
crescent moon
actual 396 m 99 m
at 1 to 2,000 scale 13.2 cm
2. Complete each scale with the value that makes it equivalent to the scale of 1 to 2,000. Explain or show your reasoning.
1. 1 cm to ____________ cm
2. 1 cm to ____________ m
3. 1 cm to ____________ km
4. 2 m to _____________ m
1. 5 cm to ___________ m
2. ____________ cm to 1,000 m
3. ____________ mm to 20 m
1. What is the area of the large flag?
2. What is the area of the smaller flag?
3. The area of the large flag is how many times the area of the smaller flag?
## 12.4Pondering Pools
Your teacher will give you a floor plan of a recreation center.
1. What is the scale of the floor plan if the actual side length of the square pool is 14 m? Express your answer both as a scale with units and without units.
2. Find the actual area of the large rectangular pool. Show your reasoning.
3. The kidney-shaped pool has an area of 3.2 cm2 on the drawing. What is its actual area? Explain or show your reasoning.
### Are you ready for more?
1. Square A is a scaled copy of Square B with scale factor 2. If the area of Square A is 10 units2, what is the area of Square B?
2. Cube A is a scaled copy of Cube B with scale factor 2. If the volume of Cube A is 10 units3, what is the volume of Cube B?
3. The four-dimensional Hypercube A is a scaled copy of Hypercube B with scale factor 2. If the “volume” of Hypercube A is 10 units4, what do you think the “volume” of Hypercube B is?
## Lesson 12 Summary
Sometimes scales come with units, and sometimes they don’t. For example, a map of Nebraska may have a scale of 1 mm to 1 km. This means that each millimeter of distance on the map represents 1 kilometer of distance in Nebraska. The same scale without units is 1 to 1,000,000, which means that each unit of distance on the map represents 1,000,000 units of distance in Nebraska. This is true for any choice of unit.
To see that these two scales are equivalent, notice that there are 1,000 millimeters in 1 meter and 1,000 meters in 1 kilometer. This means there are or 1,000,000 millimeters in 1 kilometer. So the actual distances in Nebraska are 1,000,000 times as far as the distances on the map.
A scale tells us how a length on a drawing corresponds to an actual length, and it also tells us how an area on a drawing corresponds to an actual area.
For example, if 1 centimeter on a scale drawing represents 2 meters in actual distance, what does 1 square centimeter on the drawing represent in actual area? The square on the left shows a square with side lengths 1 cm, so its area is 1 square cm.
The square on the right shows the actual dimensions represented by the square on the left. Because each side length in the actual square is 2 m, the actual square has an area of or 4 square meters.
We can use this relationship to find the actual area of any region represented on this drawing. If a room has an area of 18 cm2 on the drawing, we know that it has an actual area of or 72 m2.
In general, if 1 unit on the drawing represents actual units, then one square unit on the drawing represents actual square units.
## Lesson 12 Practice Problems
1. The Empire State Building in New York City is about 1,450 feet high (including the antenna at the top) and 400 feet wide. Andre wants to make a scale drawing of the front view of the Empire State Building on an -inch-by--inch piece of paper. Select a scale that you think is the most appropriate for the scale drawing. Explain your reasoning.
1. 1 inch to 1 foot
2. 1 inch to 100 feet
3. 1 inch to 1 mile
4. 1 centimeter to 1 meter
5. 1 centimeter to 50 meters
6. 1 centimeter to 1 kilometer
2. Elena finds that the area of a house on a scale drawing is 25 square inches. The actual area of the house is 2,025 square feet. What is the scale of the drawing?
3. Which of these scales are equivalent to 3 cm to 4 km? Select all that apply. Recall that 1 inch is 2.54 centimeters.
1. 0.75 cm to 1 km
2. 1 cm to 12 km
3. 6 mm to 2 km
4. 0.3 mm to 40 m
5. 1 inch to 7.62 km
4. These two triangles are scaled copies of one another. The area of the smaller triangle is 9 square units. What is the area of the larger triangle? Explain or show how you know.
5. Water costs \$1.25 per bottle. At this rate, what is the cost of:
1. 10 bottles?
1. 20 bottles?
1. 50 bottles?
6. The first row of the table shows the amount of dish detergent and water needed to make a soap solution.
number of batches cups of water cups of detergent
1 6 1
2
3
4
1. Complete the table for 2, 3, and 4 batches.
2. How much water and detergent is needed for 8 batches? Explain your reasoning.
|
# Basic Statistical Vocabulary and Displaying Distributions with Graphs
Date 18.10.2016 Size 73.16 Kb.
Basic Statistical Vocabulary and Displaying Distributions with Graphs
AP Statistics – Section 1.1
We'll begin our study of statistics by looking at some basic vocabulary and some graphical displays of data.
I. What Exactly is Statistics?
The topic of statistics can be divided as follows:
Statistics
the science of collecting, organizing, and analyzing numerical data.
Descriptive Statistics
collecting and presenting data
Inferential Statistics
drawing conclusions from data that has already been collected
Data Analysis
presenting data in the form of charts and graphs; summarizing data numerically
Data Production
studying how data is collected
II. Variables
Definitions
1. An individual is an object described by a set of data.
Example____________________________________________________
1. A variable is a characteristic of an individual.
Example____________________________________________________
Example 1
Consider the set of students in this class.
The individuals in this set are_____________________________________
Some examples of variables that might describe them are
Variables used in statistics break down into the following categories
Qualitative or Categorical Variables Quantitative Variables
examples: examples:
Now we consider another distinction:
Discrete Variables Continuous Variables
definition: definition:
examples: examples:
Example 2 (from IPS)
Popular magazines often rank cities in terms of how desirable it is to live and work in each city. List five variables that you would measure for each city if you were designing a study. Give reasons for your choices.
Example 3 (from Iman)
Indicate whether the following variables are quantitative or qualitative.
1. Marital status _____________________________________________
1. Sex _____________________________________________________
1. Occupation _______________________________________________
1. Social Security # ___________________________________________
1. Number of children at home____________________________________
1. Annual income______________________________________________
1. Number of telephones in your home _______________________________
1. Whether you own or rent a home _________________________________
1. Type of credit card you use _____________________________________
1. Street number _____________________________________________
We will be examining the different values that variables take on and how often these variables assume these values. Most of the information is summarized in what is called a distribution.
Definition
The distribution of a variable indicates what values a variable takes on and the frequency (i.e., how often) at which it takes on these values. We are often interested in graphing the distribution of a variable. There are a few types of graphs we could use, depending on what type of variable we are examining and what type of information we’d like to display.
Here is a list of graphs we will examine now. Boxplots and scatterplots will be covered later.
Qualitative or Categorical Variable Graphs Quantitative Variable Graphs
• bar chart
• pie chart
III. Categorical Graphs
We'll concentrate on categorical variables and their graphs for now.
Bar Charts (Qualitative / Categorical Data)
• Horizontal axis = categories
• Vertical axis = values
Note: As the categories are distinctly unique and one category does not flow into the next, like real numbers on the x-axis, there should be spaces between the bars. (That's not to say everyone gets this right.)
Example: Rainfall (Brase/Brase, Understanding Statistics)
The information listed below gives the average monthly rainfall throughout the year in Honolulu, Hawaii:
Month Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec. Rainfall (in.) 4.4 2.46 3.18 1.36 0.96 0.32 0.60 0.76 0.67 1.51 2.99 3.64
1. Make a bar graph of this information with month on the horizontal axis and rainfall on the vertical.
1. There is the rainy season and the dry season. From the graph, which 6 months would you say make up the rainy season?
2. Without rain insurance, which winter month (Nov., Dec., Jan., or Feb.) would be the best time to plan a trip?
Pie Charts or Circle Graphs (Qualitative / Categorical Data)
• Circle divided up into sectors to represent categories.
• Angles for sectors are proportional to percentage weights of categories.
Example: Causes for Lateness (Brase/Brase, Understanding Statistics)
Suppose you want to arrive at your college 15 minutes before your first class so that you can feel relaxed when you walk into class. An early arrival time allows you room for unexpected delays. However, you always find yourself arrive "just in time" or slightly late. What causes you to be late? One student made a list of possible causes and then kept a checklist for 2 months. On some days, more than one item was checked because several events occurred that caused the student to be late. Construct a pie chart using this data.
Causes for Lateness
Cause Frequency
Snoozing after alarm goes off 15
Car trouble 5
Too long over breakfast 13
Last-minute studying 20
Finding something to wear 8
Talking too long with roommate 9
Other 3
Excel:
Note that Excel's chart wizard is a great way to make graphs. I'll post a how-to on the Resources page on the course site. Other spreadsheet programs have similar utilities.
We now turn to several examples of graphs for quantitative data.
IV. Dotplots
Example 6
Let X = the number of letters in the first name of a student in this class. Construct a dotplot of the data obtained from this class.
Definition (BPS)
An outlier in any graph is an individual observation that falls outside the overall pattern of the graph.
Question: Does the dotplot above appear to have any outliers?
Note: We will study more precise ways of determining if an observation is an outlier of a data set in the weeks to come.
V. Histograms
In many more cases, we find a histogram to be a more useful visualization.
We begin with a definition:
Definition: Range
The range of a set of numbers is the difference between the largest and smallest numbers in the set, i.e.
range = (largest value) - (smallest value)
Before we can construct a histogram, we need to build a slightly different sort of frequency table. Here we divide the range into equally-sized classes and look at how many observations fall into each class.
We need to formalize a few more definitions:
Definitions and Properties for Grouped Data:
• We divide the range into equally-sized intervals we call classes or class intervals.
• The smallest number in a class is called the lower class limit.
• The largest number in a class is called the upper class limit.
• The classes must be designed so that each number in the set falls into exactly one class.
• The midpoint of a class lies halfway between the limits, i.e.
midpoint =
• The class width is the difference between consecutive lower class limits.
• When we divide data into classes, we produce what we call grouped data.
We examine histograms is in the context of an example:
Example 7 (Understanding Statistics): Nurses
Nurses on the eighth floor of Community Hospital believe they need extra staffing at night. To estimate the night workload, a random sample of 35 nights was used. For each night the total number of room calls to the nurses' station on the eighth floor was record as follows:
68 60 69 70 83 58 90 86 71 71
92 95 70 74 46 18 84 82 75 63
101 77 102 80 86 85 73 86 62 100
90 37 88 70 87
Build a histogram by following these steps:
1. Enter this data in a list on the calculator and sort the list.
2. Compute the range of this data set.
3. We'll use a total of five classes for our histogram. How wide should these classes be? What are they?
4. Find the midpoints of each of the classes.
1. Tally the frequencies for each class.
Classes Frequency
1. Draw the histogram. Put the frequency of each class on the y-axis and plot the point at the midpoint on the x-axis.
Notes of importance:
1. Once we turn data into grouped data, we can never go back and recover the original data.
2. Histograms can be created using the TI graphing calculators. (It's under "Stat Plot," accessed by hitting 2nd then Y=. You need to turn on the plot, choose the histogram, and choose the list. You'll also want to go to ZOOM and select 9:ZoomStat.)
3. When constructing a histogram, it is important to remember that all classes should have the same size (width). It is recommended that a histogram have somewhere between 5 and 20 classes (probably closer to 5). To find an appropriate number of classes, it is helpful to use the range, as we did above. Note that the intervals can go beyond the range a bit, within reason. In other words, it is usually more sensible to have integer class widths and go a little too high or a little to low than to use a class width like 14.2.
4. Bars of a histogram, in general, should be connected (unlike the bars of a bar graph, which are not).
5. A relative frequency histogram has the same shape as a histogram with the exception that the vertical axis measures relative frequencies (percents) instead of frequencies.
Example 8
Construct a relative frequency histogram using the data in example #7.
We want to be able to describe data by interpreting its histogram. The key features of a histogram worth noticing are:
1. The center of the histogram (more than one way to measure this)
2. The spread of the histogram (more than one way to measure this too)
3. The shape of the histogram (usually only one way to describe this)
There are three basic shapes you need to know:
1. Symmetric
2. Skewed right
3. Skewed left
Be careful with skewed left and skewed right!! It's very easy to confuse them!
One way you could remember the distinction is to think of starting with a symmetric graph and having someone step on the left, pushing more observations to the right. This is the skewed left situation. For skewed right, imagine someone stepping on the right side of the graph and pushing more observations to the left.
VI. Stemplots or Stem and Leaf Displays
Example 9 (Understanding Statistics): How old are rich people?
Forbes Richest People gives the profile of the world’s wealthiest men and women. Do you have to be old to be worth at least \$2 billion? You can answer this question yourself by studying the following data- ages of men and women worth at least \$2 billion:
40 66 43 82 52 58 77 52 50 48 47
68 66 73 76 53 67 88 40 79 73 66
65 70 72 77 48 75 82 54 76 41 93
65 60 57 74 70 83 67 68 77 66 34
66 59 48 56 71 40 53 63 52 57 83
52 60 56 71 64 61 53 53 73 70
Make a stem and leaf diagram for this data (feel free to use your graphing calculator to sort the numbers in increasing order).
Notes of importance:
1. To construct a stem and leaf plot, simply remove the last digit of each number in the data (these will be used as leaves) and use the remaining digits as a row label (called a stem label). An entire row of a stem plot is called a stem.
2. Leaves should be arranged in increasing order along each stem.
3. A stemplot preserves the original data in a data set whereas a histogram does not.
4. Stems may also be split (which is often done when there are a large number of leaves in one row. For example, suppose we use the ages in the last example for our discussion. Instead of listing all ages that were in the 50s, one could list all the ages between 50 and 54 on one stem and 55-59 on another stem. This part of the stemplot would look like this:
5 0 2 2 2 2 3 3 3 3 4
5 * 6 6 7 7 8 9
Example 10 (Understanding Statistics)
Tel-a-Message is experimenting with computer-delivered telephone advertisements. Of primary concern is how much of the 4-minute advertisement is heard. A study was done to see how long the advertisement ran before the listeners hung up. A random sample of 30 calls gave the information listed below. Construct a stemplot using this data.
1.3 0.7 2.1 0.5 0.2 0.9 1.1 3.2 4.0 3.8
1.4 3.1 2.5 0.6 0.5 2.1 4.0 4.0 0.3 1.2
1.0 1.5 0.4 4.0 2.3 2.7 4.0 0.7 0.5 4.0
Back-to-back Stemplots
Example 11 (BPS)
Here are the number of home runs that Babe Ruth hit in his 15 years with the New York Yankees, 1920-1934:
54 59 35 41 46 25 47 60
54 46 49 46 41 34 22
Babe Ruth’s home run record for a single season was broken by another Yankee, Roger Maris, who hit 61 home runs in 1961. Here are Maris’ home run totals for his 10 years in the American League:
13 23 26 16 33 61 28 39 14 8
Draw a back-to-back stemplot illustrating this data. Comment on the shapes of the two distributions.
Example 12 (SDA): National League Stadiums
Consider the seating capacities of baseball stadiums in the National League, as shown in the table below:
Atlanta Braves 52,003
Chicago Cubs 38,170
Cincinnati Reds 52,952
Florida Marlins 47,226
Houston Astros 54,816
L.A.Dodgers 56,000
Montreal Expos 43,739
New York Mets 55,601
Pittsburgh Pirates 58,727
St. Louis Cardinals 56,227
San Francisco Giants 58,000
a. Draw a stem and leaf plot, using only the first two digits of each number.
b. Draw a histogram illustrating this data.
VII. Time Plots
Time plots are used to illustrate bivariate quantitative data where the independent variable represents time.
Example 13 (text)
The Virginia Department of Motor Vehicles publishes data each year on the number of road fatalities, pedestrian fatalities and alcohol-related fatalities in the state. This information is used as a stimulus for public safety awareness programs, legislation on speed limits and the use of seat belts, highway engineering projects and similar purposes. Here are the results for an eleven-year period:
Year Total Pedestrian Alcohol-related
1986 1118 141 492
1987 1022 117 418
1988 1069 131 522
1989 999 141 480
1990 1071 116 535
1991 938 112 429
1992 839 93 379
1993 875 112 397
1994 925 101 376
1995 900 93 360
1996 869 114 346
1. Make a time plot for the total number of road fatalities. Does there appear to be a trend? If so, describe it. Can you give some possible reasons for what has happened?
1. Make a time plot for the number of alcohol-related fatalities. Answer the same question as in (a).
VIII. Cumulative Relative Frequency and Ogives
Definitions:
1. A cumulative frequency is the number of observations less than or equal to a given number.
2. A cumulative relative frequency is the cumulative frequency divided by the total number of observations. (So what kind of numbers do we get?)
3. The graph of a cumulative relative frequency is called a cumulative relative frequency graph (mmmm…. creative) or an ogive.
Example 14 (Iman)
The following is a list of 20 typing test scores (net words per minute):
68 72 91 47 52 75 53 55 65 35
84 45 58 61 69 22 46 55 66 71
Let's first write down the cumulative frequency and cumulative relative frequency for this data:
Scores Cumulative Frequency Cumulative Relative Frequency 22 35 45 46 47 52 55 (two of them) 58 61 63 65 66 68 69 71 72 75 84 91
Definition (Iman): An empirical distribution function (e.d.f.) is a graph of the cumulative relative frequency vs. the raw data in the sample. It is a form of a step function.
On the back, draw the empirical distribution function for the above data.
Homework: #1.1-1.11, 1.12a-d, 1.14, 1.17, 1.20-1.25, 1.28
Page of
|
# Difference between revisions of "2011 AIME II Problems/Problem 6"
## Problem 6
Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a, and $a+d>b+c$. How many interesting ordered quadruples are there?
## Solution 1
Rearranging the inequality we get $d-c > b-a$. Let $e = 11$, then $(a, b-a, c-b, d-c, e-d)$ is a partition of 11 into 5 positive integers or equivalently: $(a-1, b-a-1, c-b-1, d-c-1, e-d-1)$ is a partition of 10 into 5 non-negative integer parts. Via a standard balls and urns argument, the number of ways to partition 10 into 5 non-negative parts is $\binom{6+4}4 = \binom{10}4 = 210$. The interesting quadruples correspond to partitions where the second number is less than the fourth. By symmetry there as many partitions where the fourth is less than the second. So, if $N$ is the number of partitions where the second element is equal to the fourth, our answer is $(210-N)/2$.
We find $N$ as a sum of 4 cases:
• two parts equal to zero, $\binom82 = 28$ ways,
• two parts equal to one, $\binom62 = 15$ ways,
• two parts equal to two, $\binom42 = 6$ ways,
• two parts equal to three, $\binom22 = 1$ way.
Therefore, $N = 28 + 15 + 6 + 1 = 50$ and our answer is $(210 - 50)/2 = \fbox{80.}$
## Solution 2
Let us consider our quadruple (a,b,c,d) as the following image xaxbcxxdxx. The location of the letter a,b,c,d represents its value and x is a place holder. Clearly the quadruple is interesting if there are more place holders between c and d than there are between a and b. 0 holders between a and b means we consider a and b as one unit ab and c as cx yielding $\binom83 = 56$ ways, 1 holder between a and b means we consider a and b as one unit axb and c as cxx yielding $\binom 63 = 20$ ways, 2 holders between a and b means we consider a and b as one unit axxb and c as cxxx yielding $\binom43 = 4$ ways and there cannot be 3 holders between a and b so our total is 56+20+4=$\fbox{80.}$.
## See also
2011 AIME II (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 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.
|
science physics
# What is the constant of variation in math?
Last Updated: 19th April, 2020
25
The constant of variation is the number that relates two variables that are directly proportional or inversely proportional to one another.
Moreover, how do you find the constant variation?
Since k is constant (the same for every point), we can find k when given any point by dividing the y-coordinate by the x-coordinate. For example, if y varies directly as x, and y = 6 when x = 2, the constant of variation is k = = 3. Thus, the equation describing this direct variation is y = 3x.
Also Know, what is the meaning of variation in math? Variation. Variation problems involve fairly simple relationships or formulas, involving one variable being equal to one term. When one variable increases the other decreases in proportion so that the product is unchanged.
Just so, what is a constant in math?
A fixed value. In Algebra, a constant is a number on its own, or sometimes a letter such as a, b or c to stand for a fixed number. Example: in "x + 5 = 9", 5 and 9 are constants. See: Variable. Algebra - Definitions.
How do you find the variation in math?
To calculate the variance follow these steps:
1. Work out the Mean (the simple average of the numbers)
2. Then for each number: subtract the Mean and square the result (the squared difference).
3. Then work out the average of those squared differences. (Why Square?)
Professional
## What is variation with example?
Major causes of variation include mutations, gene flow, and sexual reproduction. DNA mutation causes genetic variation by altering the genes of individuals in a population. Examples of genetic variation include eye color, blood type, camouflage in animals, and leaf modification in plants.
Professional
## What is the constant of variation example?
The constant of variation in a direct variation is the constant (unchanged) ratio of two variable quantities. where k is the constant of variation . Example 1: If y varies directly as x and y=15 when x=24 , find x when y=25 .
Professional
## How do you find the inverse variation?
Inverse Variation
1. An inverse variation can be represented by the equation xy=k or y=kx .
2. That is, y varies inversely as x if there is some nonzero constant k such that, xy=k or y=kx where x≠0,y≠0 .
3. Suppose y varies inversely as x such that xy=3 or y=3x . That graph of this equation shown.
Pundit
## What is a constant in graphing?
A constant function is a linear function for which the range does not change no matter which member of the domain is used. With a constant function, for any two points in the interval, a change in x results in a zero change in f(x) . Example: Graph the function f(x)=3 .
Pundit
## What is the constant of variation for the quadratic variation?
Algebra Examples
Since the equation can be written in the form y=kx2 y = k x 2 , y varies directly with x2 and k . The constant of variation, k , is −7 .
Pundit
## Is 0 a constant number?
Since c occurs in a term that does not involve x, it is called the constant term of the polynomial and can be thought of as the coefficient of x0; any polynomial term or expression of degree zero is a constant.
Pundit
## Is 0 a positive constant?
The most common usage in English is that zero is neither positive nor negative. That is "positive" is normally understood to be "strictly positive". In the same way, "greater than" is normally understood to mean "strictly greater than", as in k>j (not k≥j).
Pundit
## What are coefficients?
In mathematics, a coefficient is a multiplicative factor in some term of a polynomial, a series, or any expression; it is usually a number, but may be any expression. For example, if y is considered as a parameter in the above expression, the coefficient of x is −3y, and the constant coefficient is 1.5 + y.
Teacher
## What are different types of constants?
Types of C constant:
• Integer constants.
• Real or Floating point constants.
• Character constants.
• String constants.
• Backslash character constants.
Teacher
## Is Pi a constant?
The number π (/pa?/) is a mathematical constant. Originally defined as the ratio of a circle's circumference to its diameter, it now has various equivalent definitions and appears in many formulas in all areas of mathematics and physics. It is approximately equal to 3.14159.
Teacher
## Is a decimal a constant?
A constant (also called a literal) specifies a value. Constants are classified as string constants or numeric constants. A decimal constant specifies a decimal number as a signed or unsigned number that consists of no more than 63 digits and either includes a decimal point or is not within the range of binary integers.
Teacher
## What is the constant of an equation?
Parts of an Equation
A Variable is a symbol for a number we don't know yet. It is usually a letter like x or y. A number on its own is called a Constant.
Reviewer
## What is the variation in math?
Variation is defined by any change in some quantity due to change in another. We often come across with different types of variation problems in mathematics. Variation problems involve fairly simple relationships or formulas, involving one variable being equal to one term.
Reviewer
## What is mean variation?
English Language Learners Definition of variation
: a change in the form, position, condition, or amount of something. : something that is similar to something else but different in some way. music : a repeated version of a short piece of music with changes in its rhythm, tune, or harmony.
Reviewer
## What are the 4 types of variation?
Examples of types of variation include direct, inverse, joint, and combined variation.
Reviewer
## What is a direct variation in math definition?
Direct Variation. Directly Proportional. A relationship between two variables in which one is a constant multiple of the other. In particular, when one variable changes the other changes in proportion to the first. If b is directly proportional to a, the equation is of the form b = ka (where k is a constant).
Supporter
## What is variation function?
A variation function is a function in which the variables are related by how they change in relation to each other. For instance, in this function, if x increases or decreases then D does the same. There are two types of variation functions, direct and inverse variation functions.
Supporter
## What are the types of variation?
Variation in a species is not uncommon, but there are actually two major categories of variation in a species: continuous variation and discontinuous variation. Continuous variation is where the different types of variations are distributed on a continuum.
Supporter
## How do you interpret variance?
Subtract the mean from each data value and square each of these differences (the squared differences). 3. Find the average of the squared differences (add them and divide by the count of the data values). This will be the variance.
Co-Authored By:
6
19th April, 2020
43
|
# What are the two opposite rays?
## What are the two opposite rays?
Opposite rays are two rays that both start from a common point and go off in exactly opposite directions. Because of this the two rays (QA and QB in the figure above) form a single straight line through the common endpoint Q. When the two rays are opposite, the points A,Q and B are collinear
## Are rays and opposite rays the same?
A ray is part of a line, has one fixed endpoint, and extends infinitely along the line from the endpoint. Opposite math rays are rays with a common endpoint, extending in opposite directions and forming a line.
## How do you find opposite rays?
Naming of rays By two points. In the figure at the top of the page, the ray would be called AB because starts at point A and passes through B on its way to infinity. Recall that points are usually labelled with single upper-case (capital) letters.
## How do you find two opposite rays?
The two rays are called the sides of the angle. The common endpoint is called the vertex. 6. The figure formed by opposite rays is often referred to as a straight angle.
## Which rays are the opposite rays?
Examples of Opposite Rays Ray CD and ray CG are the opposite rays. You can tell because points G, C, and D lie on one straight line. If you picked any other pair of rays, the points that are on those rays do not lie in one straight line.
## How do you find pairs of opposite rays?
A pair of opposite rays are two rays that have the same endpoint and extend in opposite directions. Rays are always named with two points and the first point in the name must be the endpoint. So, when you name opposite rays, the first letter in the name of both rays must be the same.
## How do you know if two rays are opposite?
A pair of opposite rays are two rays that have the same endpoint and extend in opposite directions. Rays are always named with two points and the first point in the name must be the endpoint. So, when you name opposite rays, the first letter in the name of both rays must be the same.
## How do you find opposite rays on a line?
Opposite rays are two rays that both start from a common point and go off in exactly opposite directions. Because of this the two rays (QA and QB in the figure above) form a single straight line through the common endpoint Q. When the two rays are opposite, the points A,Q and B are collinear
## What are opposite rays?
Two rays are opposite rays, by definition, if. (1) they have the same endpoint, and. (2) their union is a line. The first letter in the name of a ray always refers to the endpoint of the ray. Therefore, has its endpoint at and has its endpoint at .
## What two rays are opposite rays?
Opposite rays are two rays that both start from a common point and go off in exactly opposite directions. Because of this the two rays (QA and QB in the figure above) form a single straight line through the common endpoint Q. When the two rays are opposite, the points A,Q and B are collinear
## How do you find two rays?
The two sides of an angle are the two rays that compose it. Each of these rays begins at the vertex and proceeds out from there. In naming a ray, we always begin with the letter of the endpoint (where the ray starts) followed by another point on the ray in the direction it travels.
## What is the measure of opposite rays?
Two geometric rays that share the same endpoint form an angle with each other. If these angles measure 180 degrees, they point in opposite directions. These are named opposite rays.
## How do you identify opposite rays?
Opposite rays are two rays that both start from a common point and go off in exactly opposite directions. Because of this the two rays (QA and QB in the figure above) form a single straight line through the common endpoint Q. When the two rays are opposite, the points A,Q and B are collinear.
## How do you find a ray on a line?
Two rays are opposite rays, by definition, if. (1) they have the same endpoint, and. (2) their union is a line. The first letter in the name of a ray always refers to the endpoint of the ray. Therefore, has its endpoint at and has its endpoint at .
## What is an opposite Ray example?
Examples of Opposite Rays Ray CD and ray CG are the opposite rays. You can tell because points G, C, and D lie on one straight line. If you picked any other pair of rays, the points that are on those rays do not lie in one straight line. The point that lies on a straight line with L and M is point P.
## What is opposite rays in math?
Opposite math rays are rays with a common endpoint, extending in opposite directions and forming a line. ray line endpoint opposite rays. If we had a line so it extends infinitely.
## What rays are opposite rays?
Opposite rays are two rays that both start from a common point and go off in exactly opposite directions. Because of this the two rays (QA and QB in the figure above) form a single straight line through the common endpoint Q. When the two rays are opposite, the points A,Q and B are collinear.
## What are 2 rays called?
Two rays are opposite rays, by definition, if. (1) they have the same endpoint, and. (2) their union is a line. The first letter in the name of a ray always refers to the endpoint of the ray. Therefore, has its endpoint at and has its endpoint at .
## How do you find rays?
Two rays are opposite rays, by definition, if. (1) they have the same endpoint, and. (2) their union is a line. The first letter in the name of a ray always refers to the endpoint of the ray.
## What does 2 rays mean?
The two rays are called the sides of the angle. The common endpoint is called the vertex. 6. The figure formed by opposite rays is often referred to as a straight angle.
## What is the measure between two opposite rays?
Answer: THE ANGLE BETWEEN TWO OPPOSITE RAYS IS EQUAL TO INTERIOR ANGLE.
## What is the opposite of rays?
Opposite Rays – Definition A pair of opposite rays are two rays that have the ‘same endpoint and extend in opposite directions. So, together a pair of opposite rays always forms a straight line.
## Are opposite rays equal in measure?
Two angles whose sides are opposite rays of each other are called vertical angles. Vertical angles have the same measure. Notice that angles x26lt;1 and x26lt;3. in the above diagram are formed by opposite rays (are formed by the same two lines, but are across from each other) .
|
Resource Lesson Significant Figures and Scientific Notation
Whereas when counting objects, everyone would count the same number, and have an equally exact answer; when making measurements, there is no such thing as an EXACT measurement. Instead objects are measured to an accepted level of precision based on the limitations of the measuring apparatus. A digit is said to be a significant figure if it is either known with certainty or if it is the first estimated digit in a measurement.
For example, suppose that there are three slips of paper on a desk. No matter which student counts them, they will all tell you the same answer, "There are three pieces of paper." However, if each student then measures the length of each slip of paper, there will most likely be a difference in their answers. One student might report the length as 8.20 cm, another as 8.19 cm and yet another as 8.22 cm. In each case, each answer has three significant figures. All three student agree that the slip of paper is greater than 8 cm long. All three would round off their answers to 8.2 cm. But they have each estimated a final digit. All three answers, within the limitations of their rulers, should be considered accept.
Let's look at some examples.
Refer to the following information for the next six questions.
If each of the numbers (1, 2, 3, 4, 5) represent centimeters, then what is the reading for each of the specified locations? Note that in each of these measurements there should be two (2) certain digits and a third estimated digit.
A
B
C
D
E
F
Suppose you were now asked to state the TOTAL length of the ruler diagrammed? A reasonable value, to 3 significant figures, might be 5.28 cm. This measurement could also be stated as 52800 µm, 52.8 mm, 0.528 dm, 0.0528 m, 0.0000528 km (lesson on metric conversions). Because the value cannot become more accurate by converting it to other units, each of these new representations must also have only three significant figures. The question arises, when is zero a significant figure?
A zero is said to be significant if:
(1) it is between two non-zero digits 3001 m, 30.001 m 4 SD, 5 SD (2) it is at the end of a decimal expression 0.00310 km 3 SD (3) it is required when expressing the number in scientific notation 3.10 x 106 m 3 SD
Otherwise a zero is considered to be only a placeholder.
150,000 m all four zeros are placeholders 1.5 x 105 m 0.0015 km all three zeros are placeholders 1.5 x 10-3 km 150. Gm no zeros are placeholders(note the deliberate inclusion of the decimal) 1.50 x 1011 m
When multiplying or dividing two measurements, your answer should be rounded off so that it only has accurate as many significant digits as your least accurate original value.
When adding or subtracting two measurements, first convert them to the same unit of measurement, then line up the decimals. Your final answer should be rounded off so that it only has as many decimal places as your least accurate original value.
Numerical constants (π, e, ½) do not have significant digits.
For example, the volume of sphere is calculated with the formula V = 4/3 πr3.
Using this formula, a sphere with a measured diameter of 24 cm would have a volume equal to
4/3 π(12)3 = 4/3 π(1728) = 2304π cm3
Since 12 only had two significant digits, your final value for the sphere's volume should only have 2 SD.
This means that a calculated value of V = 7238.229 cm3 should be expressed in final form as
7200 cm3 = 7.2 x 103 cm3
Scientific Notation. Express your value so that it has one digit to the left of the decimal and all other significant digits to the right of the decimal. It should then be multiplied by an appropriate power of 10.
(1) When the absolute value of the original number is greater than one, then moving the decimal point will require the resulting number to be multiplied by 10 raised to a positive exponent.
(2) When the absolute value of the original number is less than one, then moving the decimal point will require the resulting number to be multiplied by 10 raised to a negative exponent.
where the decimal was moved .... 0.066 g 6.6 x 10-2 g |0.066|<1 two decimal places to the right 0's are placeholders 200.0 g 2.000 x 102 g |200.0|>1 two decimal places to the left all three zeros are significant 0.543 g 5.43 x 10-1 g |0.543|<1 one decimal place to the right 0 is a placeholder 1600 g 1.6 x 103 g |1600|>1 three decimal places to the left both zeros are placeholders 75.2 g 7.52 x 101 g |75.2|>1 one decimal place to the left - - - - -
Refer to the following information for the next five questions.
State each of the following numbers in scientific notation with the correct number of significant digits.
126 sec
126000 m/sec
0.0126 m/sec2
12060 m
0.001260 m/sec
|
Math Calculators, Lessons and Formulas
It is time to solve your math problem
mathportal.org
# Quadratic equation solver
This calculator solves quadratic equations and provides an easy-to-understand step-by-step explanation. The solver uses three methods - completing the square, quadratic formula and factoring method.
show help ↓↓ examples ↓↓ tutorial ↓↓
$2x^2 + x - 3 = 0$
$\frac{1}{2}x^2+\frac{3}{4}x - 11 = 0$
$x^2+\sqrt{2}x - 4 = 0$
Solve using factoring Solve by completing the square Solve by using quadratic formula
Allow the calculator to select the best method
working...
EXAMPLES
example 1:ex 1:
Solve for $x^2 + 3x - 4 = 0$ by factoring.
example 2:ex 2:
Solve $4x^2 - x - 3 = 0$ by completing the square.
example 3:ex 3:
Solve $-2x^2 - 0.5x + 0.75 = 0$ using the quadratic formula.
example 4:ex 4:
Solve $\frac{2}{3} x^2 - \frac{1}{3} x - 5 = 0$.
Search our database of more than 200 calculators
TUTORIAL
## Solving quadratic equations
The most commonly used methods for solving quadratic equations are:
1. Factoring method
2. Completing the square
3. Using quadratic formula
In the following sections, we'll go over these methods.
### Method 1A : Factoring method
If a quadratic trinomial can be factored, this is the best solving method.
We often use this method when the leading coefficient is equal to 1 or -1. If this is not the case, then it is better to use some other method.
Example 01: Solve $x^2 \color{red}{-8}x \color{blue}{+ 15} = 0$ by factoring.
Here we see that the leading coefficient is 1, so the factoring method is our first choice.
To factor this equation, we must find two numbers ( $a$ and $b$ ) with a sum is $a + b = \color{red}{8}$ and a product of $a \cdot b = \color{blue}{15}$.
After some trials and errors, we see that $a = 3$ and $b = 5$.
Now we use formula $x^2 - 8x + 15 = (x - a)(x - b)$ to get factored form:
$$x^2 - 8x + 15 = (x - 3)(x - 5)$$
Divide the factored form into two linear equations to get solutions.
\begin{aligned} x^2 - 8x + 15 &= 0 \\ (x - 3)(x - 5) &= 0 \\ x -3 &= 0 ~~ \text{or} ~~ x - 5= 0 \\ x &= 3 ~~ \text{or} ~~ x = 5 \end{aligned}
### Method 1B : Factoring - special cases
Example 02: Solve $x^2 -8x = 0$ by factoring.
In this case, (when the coefficient c = 0 ) we can factor out $\color{blue}{x}$ out of $x^2 - 8x$.
\begin{aligned} x^2 - 8x &= 0 \\ \color{blue}{x} \cdot ( x - 8 ) &= 0 \\ x &= 0 ~~ \text{or} ~~ x - 8 = 0 \\ x &= 0 ~~ \text{or} ~~ x = 8 \end{aligned}
Example 03: Solve $x^2 - 16 = 0$ by factoring.
In this case, ( when the middle term is equal 0) we can use the difference of squares formula.
\begin{aligned} x^2 - 16 &= 0 \\ x^2 - 4^2 &= 0 \text{ use } a^2 - b^2 = (a-b)(a+b) \\ (x - 4)(x+4) &= 0 \\ x - 4 &= 0 ~~ \text{or} ~~ x + 4 = 0 \\ x &= 4 ~~ \text{or} ~~ x = -4 \end{aligned}
### Method 3 : Solve using quadratic formula
This method solves all types of quadratic equations. It works best when solutions contain some radicals or complex numbers.
Example 05: Solve equation $2x^2 + 3x - 2 = 0$ by using quadratic formula.
Step 1: Read the values of $a$, $b$, and $c$ from the quadratic equation. ( $a$ is the number in front of $x^2$ , $b$ is the number in front of $x$ and $c$ is the number at the end)
$$a = 2 ~~ b = 3 ~~ \text{and} ~~ c = -2$$
Step 2:Plug the values for a, b, and c into the quadratic formula and simplify.
\begin{aligned} x_1, x_2 &= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ x_1, x_2 &= \frac{-3 \pm \sqrt{3^2-4 \cdot 2 \cdot (-2) }}{2 \cdot 2} \\ x_1, x_2 &= \frac{-3 \pm \sqrt{9+ 16 }}{4} \\ x_1, x_2 &= \frac{-3 \pm \sqrt{25}}{4} \\ x_1, x_2 &= \frac{-3 \pm 5}{4} \end{aligned}
Step 3: Solve for $x_1$ and $x_2$
\begin{aligned} x_1 = & \frac{-3 \color{blue}{+} 5}{4} = \frac{2}{4} = \frac{1}{2} \\ x_2 = & \frac{-3 \color{blue}{-} 5}{4} = \frac{-8}{4} = -2 \end{aligned}
### Method 2 : Completing the square
This method can be used to solve all types of quadratic equations, although it can be complicated for some types of equations. The method involves seven steps.
Example 04: Solve equation $2x^2 + 8x - 10= 0$ by completing the square.
Step 1: Divide the equation by the number in front of the square term.
\begin{aligned} 2x^2 + 8x - 10 & = 0 ~~ / ~ \color{orangered}{:2} \\ \frac{2x^2}{2} + \frac{8x}{2} - \frac{10}{2} & = \frac{0}{2} \\ x^2 + 4x - 5 & = 0 \end{aligned}
Step 2: move $-5$ to the right:
$$x^2 + 4x = 5$$
Step 3: Take half of the x-term coefficient $\color{blue}{\dfrac{4}{2}}$, square it $\color{blue}{\left(\dfrac{4}{2} \right)^2}$ and add this value to both sides.
$$x^2 + 4x + \color{blue}{\left(\frac{4}{2} \right)^2} = 5 + \color{blue}{\left(\frac{4}{2} \right)^2}$$
Step 4: Simplify left and right side.
$$x^2 + 4x + 2^2 = 9$$
Step 5: Write the perfect square on the left.
$$\left( x + 2 \right)^2 = 9$$
Step 6: Take the square root of both sides.
\begin{aligned} x + 2 &= \pm \sqrt{9} \\\\ x + 2 &= \pm 3 \end{aligned}
Step 7: Solve for $x_1$ and $x_2$ .
\begin{aligned} x_1 & = +3 - 2 = 1 \\ x_2 & = -3 - 2 = - 5 \end{aligned}
|
# Math in Focus Grade 4 Chapter 5 Answer Key Data and Probability
Go through the Math in Focus Grade 4 Workbook Answer Key Chapter 5 Data and Probability to finish your assignments.
## Math in Focus Grade 4 Chapter 5 Answer Key Data and Probability
Math Journal
Write the steps to solve the problem.
Neil bought 5 books. The average price of 2 of the books is $5. The average price of the rest of the books is$4. Find the total amount of money Neil paid for the 5 books.
Then, following your steps above, solve the problem.
The total amount of money Neil paid for the 5 books is $16. Explanation: Given that Neil bought 5 books and the average price of 2 of the books is$5 and the average price of the rest of the books is $4. So the total amount of money Neil paid for the 5 books is as the price of 2 books is$5 and the price of 1 book is $4. So 5-2 is 3, and the price of the 4 books will be 4×3 which is$12. So the cost for 5 books it will be $12+$4 which is \$16.
Challenging Practice
Question 1.
Michelle got an average score of 80 on two tests. What score must she get on the third test so that her average score for the three tests is the same as the average score for the first two tests?
The score she got on the third test is 80.
Explanation:
Given that Michelle got an average score of 80 on two tests and the sum of score in 2 tests will be 80×2 which is 160. Let the score for third test be x,
so the new average will be $$\frac{160+x}{3}$$,
and the average score is 80, so $$\frac{160+x}{3}$$ = 80
160+x = 80×3
160+x = 240
x = 240 – 160
= 80.
Question 2.
The line plot shows the shoe sizes of students in Ms. George’s class.
a. How many students are in the class?
25 students.
Explanation:
The total number of students are in the class is 25 students.
b. What is the mode of the set of data?
3$$\frac{1}{2}$$.
Explanation:
The mode of the set of data is 3$$\frac{1}{2}$$ as the number that appears most often.
c. How many students in the class wear a size 3$$\frac{1}{2}$$ shoe?
10 students.
Explanation:
The number of students in the class wear a size 3$$\frac{1}{2}$$ shoe is 10 students.
d. Suppose you looked at 100 pairs of shoes for the grade, which includes 3 other classes. How many pairs of size 3$$\frac{1}{2}$$ would there be? Explain your answer.
Problem Solving
Question 1.
The average height of Andy, Chen, and Chelsea is 145 centimeters. Andy and Chen are of the same height and Chelsea is 15 centimeters taller than Andy. Find Andy’s height and Chelsea’s height.
The Andy’s height and Chelsea’s height is 140 cm.
Explanation:
Given that the average height of Andy, Chen, and Chelsea is 145 centimeters and Andy and Chen are of the same height and Chelsea is 15 centimeters taller than Andy, so let the height of Andy and Chen be x and the height of Chelsea is 15 centimeters taller than Andy which is x+15. So Andy’s height and Chelsea’s height will be
$$\frac{x+x+x+15}{3}$$ = 145
$$\frac{3x+15}{3}$$ = 145
3x+15 = 145×3
3x+15 = 435
3x = 435-15
3x = 420
x = 420÷3
= 140.
So the Andy’s height and Chelsea’s height is 140 cm.
Question 2.
Eduardo has 3 times as many stamps as Sally. The average number of stamps they have is 450. How many more stamps does Eduardo have than Sally?
$$\frac{1}{2}$$ of total number of stamps extra.
Explanation:
Given that Eduardo has 3 times as many stamps as Sally and the average number of stamps they have is 450. Here, if Sally has 1 stamp then Eduardo has 3 stamps. So total stamps will be 4, Eduardo has 2 extra and Eduardo has $$\frac{2}{4}$$ which is $$\frac{1}{2}$$ of total number of stamps extra.
Question 3.
Bag A and Bag B each contain 2 marbles — 1 white and 1 red. Troy picks 1 marble from Bag A and 1 from Bag B. What is the probability that the following are picked?
a. 2 white marbles
$$\frac{2}{4}$$
The probability of picking up 2 white marbles is $$\frac{2}{4}$$.
|
Publicité
1 sur 19
Publicité
### CLASS 9 LINEAR EQUATIONS IN TWO VARIABLES PPT
1. Linear Equations In Two Variables Class-09 FA-3 (2014-2015)
2. Linear Equations In Two Variables “ The principal use of the analytic art is to bring mathematical problem to equations and to exhibit those equations in the most simple terms that can be .”
3. Contents • Introduction • Linear equations • Points for solving a linear equation • Solution of a linear equation • Graph of a linear equation in two variables • Equations of lines parallel to x-axis and y-axis • Examples and solutions • Summary
4. Introduction • An excellent characteristic of equations in two variables is their adaptability to graphical analysis. The rectangular coordinate system is used in analyzing equations graphically. This system of horizontal and vertical lines, meeting each other at right angles and thus forming a rectangular grid, is called the Cartesian coordinate system. Cartesian Plane
5. Introduction • A simple linear equation is an equality between two algebraic expressions involving an unknown value called the variable. In a linear equation the exponent of the variable is always equal to 1. The two sides of an equation are called Right Hand Side (RHS) and Left-Hand Side (LHS). They are written on either side of equal sign. Equation LHS RHS 4x + 3 = 5 4x + 3 5 2x + 5y = 0 2x + 5y 0 -2x + 3y = 6 -2x + 3y 6
6. Cont… • A linear equation in two variables can be written in the form of ax + by = c, where a, b, c are real numbers, and a, b are not equal to zero. Equation a b c 2x+3y=9 2 3 -9 X+y/4-4=0 1 1/4 -4 5=2x 2 0 5 Y-2=0 0 1 -2 2+x/3=0 1/3 0 2
7. Linear equation :- • A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. Linear equations can have one or more variables. Linear equations occur with great regularity in applied mathematics. While they arise quite naturally when modeling many phenomena, they are particularly useful since many non-linear equations may be reduced to linear equations by assuming that quantities of interest vary to only a small extent from some "background" state -3 -2 -1 0 1 2 3 X + 2 = 0 X = -2
8. Solution of a linear equation Every linear equation has a unique solution as there is a single variable in the equation to be solved but in a linear equation involving two variables in the equation, a solution means a pair of values, one for x and one for y which satisfy the given equation Example- p (x)=2x+3y (1) y in terms of x If x=3 2x + 3y = (2x3) + (3xy) = 12 6 + 3y = 12 y = 2 therefore the solution is (3,2) (2)If x = 2 2x + 3y = (2x2) + (3xy) = 12 4 + 3y = 12 y= 8/3 therefore the solution is (2,8/3) Similarly many another solutions can be taken out from this single equation. That is ,a linear equation in two variables has infinitely many solutions.
9. Graph of a linear equation is representation of the linear equation . Observations on a graph Every point whose coordinates satisfy the equation lies on the line. Every point on the line gives a solution of the equation. Any point, which does not lie on the line is not a solution of equation. X+2Y=6 Graph of a linear equation in two variables
10. A linear equation in one variable represents a point on a number line and a straight line parallel to any of the axes in a coordinate plane.
11. • Example 1: Represent the equation 2y + 3 = 0 graphically on the number line in the Cartesian plane • Solution: (i) The equation 2y + 3 = 0 has a unique solution . Now, the geometrical representation of 2y + 3 = 0 i.e., y = 1.5 on a number line is as follows. Equations of lines parallel to x-axis
12. • (ii) The equation 2y + 3 = 0 can be written as . Thus, represents a straight line in the Cartesian plane parallel to x-axis and at a distance of i.e., −1.5 from x-axis. The graph of this equation has been shown in the following figure
13. Equations of lines parallel to y-axis • Equations of lines parallel to y-axis The graph of x=a is a straight line parallel to the y- axis • In two variables, 2x + 9 = 0 represents a straight line passing through point (−4.5, 0) and parallel to y-axis. It is a collection of all points of the plane, having their x- coordinate as 4.5.
14. Summary • An equation of the form ax +by + c =0,wherea,b and c are real numbers, such that a and b are not both zero, is called a linear equation in two variables. • A linear equation in two variables has infinitely many solutions. • The graph of every linear equation in two variables is a straight line. • X=0 is the equation of the y-axis and y=0 is the equation of the x-axis • The graph of x=a is a straight line parallel to the y-axis. • The graph of y=a is a straight line parallel to the x-axis. • An equation of the type y=mx represents a line passing through the origin. • Every point on the graph of a linear equation in two variables is a solution of the linear equation. • Every solution of the linear equation is a point on the graph of the linear equation.
15. Question Bank • 1) Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case: i) x = 3y 1x − 3y + 0 = 0 Comparing this equation with ax + by + c = 0, a = 1, b = −3, c = 0 ii) 2x = −5y 2x + 5y + 0 = 0 Comparing this equation with ax + by + c = 0, a = 2, b = 5, c = 0 iii) 3x + 2 = 0 3x + 0.y + 2 = 0 Comparing this equation with ax + by + c = 0, a = 3, b = 0, c = 2
16. • 2) Write 4solutions for the following equation : 2x + y = 7 y in terms of x For x = 0, 2(0) + y = 7 y = 7 Therefore, (0, 7) is a solution of this equation. For x = 1, 2(1) + y = 7 y = 5 Therefore, (1, 5) is a solution of this equation. For x = −1, 2(−1) + y = 7 y = 9 Therefore, (−1, 9) is a solution of this equation. For x = 2, 2(2) + y = 7 y = 3 Therefore, (2, 3) is a solution of this equation.
17. 3) Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k. Putting x = 2 and y = 1 in the given equation, 2x + 3y = k 2(2) + 3(1) = k 4 + 3 = k k = 7 Therefore, the value of k is 7. •
18. 4) Draw the graph of the following linear equations in two variables: i) x + y = 4 y in terms of x y = 4 – x If x = 0 If x = 4 y = 4 y = 0 (0,4) is a solution. (4,0) is a solution. ii) y = 3x y in terms of x y = 3x If x = -1 If x = 1 y = -3 y = 3 (-1,-3)is a solution. (1,3) is a solution. x 0 4 y 4 0 x -1 1 y -3 3
19. Thank You Done by: IX-A Boys S. Siva nithish N.M. Kamalanathan K.K. Kanishkar V. Rahul R. Hariharasudhan
Publicité
|
# What is a secant line of a circle
Service Unavailable in EU region
A secant is an extension of a chord in a circle which is a straight line segment of which the endpoints lie on the circle. If the same chord passes through the centre of the circle, then it is a diameter. So an extended Diameter is a secant. The line is a secant because it intersects the circle twice.
A tangent line is a line that intersects a circle wha one point. Such a line is said to be tangent to that circle. The point at lnie the circle and the line intersect is the cidcle of tangency. Point T is the point of tangency. When a radius of a circle secang drawn to a point of tangency from the center, of the circle, of coursethat radius is perpendicular to the tangent line containing that point of tangency. This means that for how to hide a chest in minecraft tangent line, there exists a perpendicular radius.
A tangent segment is a segment with one endpoint at the point of tangency and its other endpoint somewhere on the tangent line. A tangent segment is also perpendicular to the radius of the circle whose endpoint is the point of tangency. A secant line is a line that intersects a circle at two points. Every secant line, therefore, contains a chord of the circle it intersects. Summary Tangent Lines and Secant Lines. Tangent Lines A tangent line is a line that intersects a circle at one point.
Previous section Problems Next section Problems. Take a Study Break.
The #1 FREE study site for students
A secant line is the equivalent of the average rate of change or the slope between two points. Once you have calculated the slope of a line we can find the equation of the line through the two points. The two points of a secant line are denoted by: (x 1, y 1) and (x 2, y 2). The blue line in the figure above is called the "secant to the circle c". As you move one of the points P,Q, the secant will change accordingly. If the two points coincide at the same point, the secant becomes a tangent, since it now touches the circle at just one point. The line segment inside the circle between P and Q is called a chord. In geometry, a secant of a curve is a line that intersects the curve in at least two (distinct) points. In the case of a circle, a secant will intersect the circle in exactly two points and a chord is the line segment determined by these two points, that is the interval on a secant whose endpoints are these points. Click to .
Click to see full answer Similarly, you may ask, how do you find the secant of a circle? Two Secants Intersecting. Beside above, can a secant be a diameter?
A secant line, or just secant , is the infinite line extension of a chord. More generally, a chord is a line segment joining two points on any curve, for instance, an ellipse. A chord that passes through a circle's center point is the circle's diameter. Every diameter is a chord, but not every chord is a diameter. Chords, Secants , and Tangents. A line will meet a circle at no more than two points.
We call a line a secant if it intersects twice, and a tangent if it intersects once just touching at a single point. Chords are segments connecting two points on a circle , so chords become secants when extended. In geometry, the tangent line or simply tangent to a plane curve at a given point is the straight line that "just touches" the curve at that point. Leibniz defined it as the line through a pair of infinitely close points on the curve. The word " tangent " comes from the Latin tangere, "to touch".
A line that intersects a circle in exactly one point is called a tangent and the point where the intersection occurs is called the point of tangency. A secant is a line that intersects a circle in exactly two points. Secant Lines. A secant line is a line that intersects a circle at two points. Every secant line , therefore, contains a chord of the circle it intersects.
A line which touches a circle or ellipse at just one point. Below, the blue line is a tangent to the circle c. Note the radius to the point of tangency is always perpendicular to the tangent line. A special property can be noted when the two circles are tangent to each other : As this common tangent line is perpendicular to both radii of each circle at that point, it follows that the two radii and the tangent point lie along the same line.
Also, the two circles can be tangent in a different way. Two Tangents Theorem : If two tangent segments are drawn from the same external point, then the segments are equal.
Secant line. The word secant comes from the Latin word secare, meaning to cut. In the case of a circle, a secant will intersect the circle in exactly two points and a chord is the line segment determined by these two points, that is the interval on a secant whose endpoints are these points. The secant sec? It is the ratio of the hypotenuse to the side adjacent to a given angle in a right triangle. A line intersecting a circle in two places is referred to as a secant.
The portion of the secant contained within the circle is called a chord. If a line intersects a circle at only a single point, it is called a tangent. A tangent is always perpendicular to the radius of the circle at the point of tangency. Secant sec - Trigonometry function In a right triangle, the secant of an angle is the length of the hypotenuse divided by the length of the adjacent side.
Of the six possible trigonometric functions, secant , cotangent, and cosecant , are rarely used. To construct a secant of a circle, follow these steps: i Draw a circle of the given radius. Thus, your diagram is done.
If you are taking only one point ,which is external to the circle then two tangent and infinite secant can be drawn. If you are taking the whole circle , then infinite tangent and infinite secant can be drawn on it. Chord is a line segment joining two points on the circle, while secant is a line 'l' intersecting the circle in two distinct points. A minor arc left figure is an arc of a circle having measure less than or equal to radians. A line segment that has the endpoints on the circle and passes through the midpoint is called the diameter.
The diameter is twice the size of the radius. A line segment that has its endpoints on the circular border but does not pass through the midpoint is called a chord. A chord of a circle divides the circle into two regions, which are called the segments of the circle. The minor segment is the region bounded by the chord and the minor arc intercepted by the chord. The major segment is the region bounded by the chord and the major arc intercepted by the chord. What is the secant of a circle?
Category: science space and astronomy. In geometry, a secant of a curve is a line that intersects the curve in at least two distinct points. In the case of a circle , a secant will intersect the circle in exactly two points and a chord is the line segment determined by these two points, that is the interval on a secant whose endpoints are these points.
Can a secant be a tangent? What is the secant line of a circle? What is a tangent in math? Can we draw two tangents perpendicular to each other on a circle? What is the two tangent theorem? Why is it called Secant? What is Secant equal to? What is Secant the inverse of? Is a secant always a chord? What is a secant in trigonometry?
How do you construct a secant? How many tangents can a circle have? How many Secant can a circle have? Is a chord a secant? What is a minor arc of a circle? What is a line segment diameter? What is segment of a circle? Similar Asks.
When constructing an inscribed regular hexagon and you are given a point on the circle How many arcs will be drawn on the circle? Popular Asks.
## 1 thoughts on “What is a secant line of a circle”
1. Bagul:
The Kimi Raikkonen treatment is great.
|
# What are the first and second derivatives of f(x)=ln(x^(2x+1) ) ?
Nov 27, 2015
First derivative: $\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \ln \left(x\right) - \frac{1}{x} + 2$
2nd derivative: $\frac{2}{x} + \frac{1}{{x}^{2}}$ Did not have time to do the $\frac{{d}^{2}}{\mathrm{dx}}$ properly so just gave the answer!
#### Explanation:
Given:$y = \ln \left({x}^{2 x + 1}\right)$
Write as : $y = \left(2 x - 1\right) \ln \left(x\right)$
Using standard form $\frac{\mathrm{dy}}{\mathrm{dx}} = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let $u = 2 x - 1 \to \frac{\mathrm{du}}{\mathrm{dx}} = 2$
Let $v = \ln \left(x\right) \to \textcolor{w h i t e}{\ldots} \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{x}$
Then$\frac{\mathrm{dy}}{\mathrm{dx}} = \ln \left(x\right) \left(2\right) + \left(2 x - 1\right) \left(\frac{1}{x}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \ln \left(x\right) + \frac{2 x}{x} - \frac{1}{x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \ln \left(x\right) - \frac{1}{x} + 2$
|
# Construction of a right triangle
It's a high school level question which we can't seem to solve. Here it is:
Given $2$ lines, one of the length of the hypotenuse and the other with the length of the sum of the $2$ legs, construct with straightedge and compass the corresponding right triangle
I didn't make much progress. It seems that there's a theorem or a few basic facts about right triangles that I'm missing. What path do you suggest I take?
• Does this mean that you are given the length of the hypotenuse, and the sum of the lengths of the two legs, and you are supposed to find a triangle that satisfies those constraints? When it says "construct", does it mean "construct with straightedge and compass", or what? – MJD Oct 8 '12 at 17:06
• yes, I've edited my question - I hope it's more clear now. – Amihai Zivan Oct 8 '12 at 17:12
Let $ABC$ be the triangle you want to construct, with $\angle A =90^\circ$.
Then you are given $AB+AC$ and $BC$.
Extend $BA$ past $A$ to $BB'$ by a length equal to $AC$, that is $AB'=AC$. Then, the triangle $ACB'$ is a right isosceles triangle.
This means that in the triangle $BB'C$ you know $BB'=AB+AC$, $BC$ and the angle $B'=45^\circ$.
This suggests how you can construct it: construct the triangle $BB'C$, and then construct the height from $C$. The leg of the height will be $A$.
Since you are constructing $BB'C$ by $SSA$, there should be two solutions for $C$.
Here is the actual construction:
Start by drawing an angle of $45^\circ$. Denote the vertex of the angle $B'$.
On one side pick a point $B$ so that $BB'=AC+AB$.
Next draw a circle of centre $B$ and radius $BC$. This will intersect the other ray of the angle in two points $C_1, C_2$. Pick the one which makes the angle $CBB'$ acute.
P.S. I always solve the problems the way i.m.s. did, I like more the algebraic approach, but since you mention that this is high school level you are probably looking for the geometric approach.
You know the hypotenuse length $c$, and the sum of the lengths of the legs $u=a+b$. Thus you also know $v=u^2-c^2=a^2+b^2+2ab-a^2-b^2=2ab$. Then $b=u-a=u-v/2b$ so b satisfies the quadratic equation $b^2-ub+v/2=0$ which can be constructed using ruler and compass.
The algebra can guide the geometry. We want to construct a triangle with given hypotenuse $c$ and legs $a$ and $b$, where we are given $a+b$.
Note that $$(a-b)^2=2(a^2+b^2)-(a+b)^2=2c^2-(a+b)^2.$$ It is easy to construct a line of length $\sqrt{2} c$. Then construct the right triangle with hypotenuse $\sqrt{2}c$ and one leg equal to $a+b$. Easy, with center the middle of the hypotenuse, draw a circle that passes through the endpoints. Then through one endpoint draw a circle of radius $a+b$.
The other leg of our just constructed triangle has length $|a-b|$. Subtract this from $a+b$. The result is twice a leg of our target triangle.
|
Home Explore the BBC
Life | The Universe | Everything | Advanced Search
A Beginner's Guide to Mean, Median and Mode
An average summarises a group of numbers. There are three main types of averages: mean, median and mode. Each of these will be looked at in turn.
Mean
This is the most commonly used average. The mean is calculated by adding up the numbers in a sample and dividing that answer by the sample size. This is the only type of average that takes into account all the numbers in the sample.
Example - Here is a sample of numbers:
1,2,2,2,3,3,4,6
The first thing to do is add the numbers up. In this case, the result is 23.
All that is left to do is divide this number by the sample size. In this case, we are dividing 23 by 8 (because there are eight numbers altogether) and we get the result 2.875 .
Potential Problems - Mean averages have two linked problems:
• If you have a large number of small values with a few very large values in your sample, mean averages get skewed: the mean is nearer to the bigger values even though the small values there are more smaller numbers. If you have a few small values and a few large values, the mean average can get skewed this way too.
• If you have one, or more, outlying values that do not follow the general trend of the numbers in a sample, the mean average can be affected more dramatically than intended.
Example of the Effect of Outliers - In this case, the number 100 has been added to the sample above:
1,2,2,2,3,3,4,6,100
The sum of these numbers is 123. If we divide this by the sample size - 9, we get 13.6666 (recurring) which does not represent the earlier numbers.
Median
This type of average is the middle number in a sample and requires the numbers to be in order.
Example - Here is a sample of numbers to illustrate median:
4,2,2,6,3,3,1,2
To use the median, these numbers must be placed in order, like this:
1,2,2,2,3,3,4,6
Here the median is 2.5 - it comes out as this because there is a even sample size here. Therefore there is no middle number. To work out the median, you need to take the 2 and the 3 which are the middle numbers and get the mean of them - which is 2.5.
With an odd sample it is much easier: you just take the middle number as the median.
Potential Problems - One problem with using median is that it requires the numbers to be put in order first. For a large set of numbers, this task can be extremely labourious.
Mode
This type of average is the number that occurs the most times in the sample. Where the mean has problems with 'representativeness', mode focuses on the most common numbers and gives less or no attention to less frequently-occuring numbers.
Example - another list of numbers:
1,2,2,2,3,3,4,6
The mode here is 2 as it appears three times. Note: if there are two numbers which are equally common in the sample, then you take both as the mode.
Potential Problems - Mode is less useful when you have a lot of values that are close together but have not been rounded to the nearest whole number. This means an inaccurate mode of the numbers will be taken. It would be better in this example to round the numbers first before using mode.
Real-world Examples
A class of 15 students took a test to be marked out of 10. Seven students got 8 marks, 4 got 7 marks, 2 got 6 marks, 1 got 5 marks, and 1 got 4 marks.
Mean - The total number of marks the students got is 105. Divide that by the number of students, 15, and you get 7. That is the mean number of the students' marks.
Median - Below, the marks the students received are shown, in order from the highest to the lowest:
4,5,6,6,7,7,7,7,8,8,8,8,8,8,8
The median number is the middle number, so the median of the students' marks is 7.
Mode - The marks the students received were:
4,5,6,6,7,7,7,7,8,8,8,8,8,8,8
The mode number is the number that occurs the most times, so the mode number of the students marks is 8.
The average mark the students got depends on which average is used.
Another Real-world Application
From telephony, there are some interesting statistics that may help to understand the differences between mean, median and mode more easily.
• The mean telephone call - normally around two minutes 30 seconds.
• The median telephone call - normally around 40 seconds.
• The modal telephone call - normally less than five seconds (about two seconds is usual).
The interesting thing is why we get these figures for telephone calls. Phone calls split into about three different types:
• Very short - A failed call. You phoned and got somebody's voicemail and don't leave a message; a fax failed (you put in the person's voice number by mistake). With a relatively high number of failed faxes and conversations being just a few seconds long (ie within a small range of possible values), this causes a modal around two seconds, maybe three.
• Fax - A single page takes between 35 and 50 seconds. This causes a huge block of calls that nearly always falls in the middle of an ordered list of call lengths (but still is a smaller figure than the connected but failed calls, and furthermore, in a greater range of possible values).
• Conversation - Well... people can talk for quite some while. Because some people can talk for a very long time it skews the mean much higher. The calls are long, but infrequent, however, and so they don't affect the median or modal at all.
A graph of number of calls against time shows a huge peak between one and ten seconds and a slightly smaller peak between 35 and 50 seconds. After that, it tails off.
not so mean (Last Posting: Mar 5, 2003) Wishful thinking (Last Posting: Mar 4, 2003)
Add your Opinion!There are tens of thousands of h2g2 Guide Entries, written by our Researchers. If you want to be able to add your own opinions to the Guide, simply become a member as an h2g2 Researcher. Tell me More!
Entry Data
Entry ID: A956289 (Edited)
Edited by:
SchrEck Inc.
Date: 03 March 2003
Referenced Guide Entries Numbers Students Mathematics Telephones A History of Numbers
Related BBC Pages SOS teacher
Most of the content on this site is created by h2g2's Researchers, who are members of the public. The views expressed are theirs and unless specifically stated are not those of the BBC. The BBC is not responsible for the content of any external sites referenced. In the event that you consider anything on this page to be in breach of the site's House Rules, please click here to alert our Moderation Team. For any other comments, please start a Conversation below.
Most of the content on h2g2 is created by h2g2's Researchers, who are members of the public. The views expressed are theirs and unless specifically stated are not those of the BBC. The BBC is not responsible for the content of any external sites referenced. In the event that you consider anything on this page to be in breach of the site's House Rules, please click here. For any other comments, please start a Conversation above.
|
# Odd/Even Permutations
How do you classify a permutation as odd or even (composition of an odd or even number of transpositions)? I somewhat understand the textbook definition of it but I'm having hard time conceptualizing and determining how it is actually determined if it's odd or even.
Every permutation can be expressed as the product of one and only one of the following:
• an odd number of transpositions $\iff$ odd permutation
• an even number of transpositions $\iff$ even permutation
There are many ways to write a permutation as the product of transpositions, and they can vary in length, but those products will have either an odd or an even number of factors, never both.
If you know cycle notation, knowing the parity (oddness/evenness) can be found fairly easily.
One can always resort to following the pattern:
$$(a_1, a_2, a_3, a_4, a_5) = (a_1, a_5)(a_1, a_4)(a_1, a_3)(a_1, a_2)$$ which is even because there are four transpositions.
Alternatively:
$$(a_1, a_2, a_3, a_4, a_5) = (a_1, a_2)(a_2, a_3)(a_3, a_4)(a_4, a_5)$$
Again, an even number of transpositions $\iff$ the permutation is even.
You'll see that the number of transpositions in a product corresponding to a permutation that is a cycle of length $n$ can be expressed as the product of $n - 1$ transpositions. So a cycle with a length that is even (has an even number of elements) is ODD, and a cycle with a length that is odd (has an odd number of elements) is EVEN.
If you have a permutation that is the product of disjoint cycles: say three cycles, corresponding to lengths $n_1, n_2, n_3$, then the number of transpositions representing this permutation can be computed by the parity of $(n_1 - 1)+(n_2 - 1) + (n_3 - 1)$ or simply the parity (oddness/evenness) of $n_1+n_2+n_3 - 1$
One final note: the identity permutation (i.e., the "do nothing" permutation): the permutation which can be represented as the product of one-cycles sending $1 \mapsto 1,\;2\mapsto 2,\; \cdots , n\mapsto n\;$ is always considered to be an EVEN permutation. Why? Well, note that we can represent the identity permutation by the product, say, of $(12)(12) = (12)(12)(3)\cdots(n) = (1)(2)(3)\cdots (n)$, so it is indeed an even permutation.
• Such a nice descriptive answer, hope you get feedback! +1 Commented Apr 15, 2013 at 4:22
• I didn't get how >$(12)(12) = (12)(12)(3)\cdots(n) = (1)(2)(3)\cdots (n)$ shows that identity is an even permutation. Commented Dec 1, 2015 at 6:28
• How do you write so clearly? Namaste! Commented Mar 11, 2019 at 7:00
• Thank you so much, apparently it was hard for my teacher to write the first 3 sentences of your response.
– AMRO
Commented Jan 1, 2020 at 12:18
• @sequence That's because permutation $(1 2)(1 2)(3)...(n)$ has even number of transpositions ($(1 2)$ and $(1 2)$) Commented May 11, 2020 at 14:23
Simply put, it's like adding odd and even numbers. If you add two even numbers, you'll only ever get another even number. If you add two odd numbers, you'll get an even number. If you add an odd and an even number, you'll get another odd number.
It works the same for permutations. "Odd" and "Even" are defined in terms of how they interact. Define the identity permutation (that is, the one that doesn't move any elements) as an even permutation, since applying it twice will produce itself.
Now, consider the smallest possible permutations - the ones that interchange two elements, but otherwise leave everything as-is. For instance, looking at {1,2,3,4,5}, you might have {1,3,2,4,5} as an example. These are defined to be "odd" permutations, and are referred to as "transpositions".
The use of this concept is that, when expressing a permutation in terms of other permutations, oddness and evenness is preserved, no matter how you might express it. Like how, if you have the number 37, then it doesn't matter how many ways you express it as a sum of integers, there will necessarily be an odd number of odd numbers in the sum - 36+1 has one odd, 32+3+1+1 has three odds, and so on.
The permutation {3,2,1} could be expressed as "switch one and three", a single transposition. That makes it an odd permutation. It could also be expressed as "switch positions 1 and 2 ({2,1,3}), then switch positions 2 and 3 ({2,3,1}), then switch positions 1 and 2 ({3,2,1})" - three transpositions, again odd. No matter how you express it, it will always require an odd number of odd permutations.
To demonstrate it in action, consider the function $$f(a,b,c) = \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$ Now, if you apply an odd permutation over the three variables, you'll get a different result. But if you apply an even permutation, it will produce the same result.
One can also think of oddness in terms of "cycles". Every permutation can be expressed in terms of a set of mutually exclusive cycles. For instance, {3,7,4,1,6,2,5} can be expressed as (431)(6572), where the notation means "move 4 to 3, move 3 to 1, and move 1 to 4" and "move 6 to 5, move 5 to 7, move 7 to 2, and move 2 to 6". How does this help? Quite simply, if you count the number of elements within a cycle, then if the result is even, it's an odd permutation, and vice versa. So here, we have 3 elements and 4 elements, making an even and an odd permutation, respectively.
Why does 3 elements make an even permutation? Because you can express it as two transpositions: (431) becomes (43)(31). Similarly, a 4 element cycle is odd - (6572) becomes (65)(57)(72).
• You have given the example of going from {1,2,3} to {3,2,1} and said that "No matter how you express it, it will always require an odd number of odd permutations." Can you please prove that in general, if something requires an odd number of permutations to go from one state x to another state y, then there is no way that we can go from state x to state y using an even number of permutations ? Commented Aug 29, 2022 at 8:54
• @HemantAgarwal - this is not the appropriate forum for that level of proof. The question was about how they're classified. I would have suggested that you ask it as a new question on math.SE, but it's already been asked, here math.stackexchange.com/questions/46403/… - or here, math.stackexchange.com/questions/4768/… Commented Aug 29, 2022 at 10:31
• @HemantAgarwal - if, instead, you're asking for a general proof in the specific case of [3,2,1], then it's easy enough to obtain by noting that, if you label [1,2,3], [2,3,1], and [3,1,2] as "even" and [1,3,2], [2,1,3], and [3,2,1] as "odd", then any single transposition (you can check by exhaustion) will switch you from one to the other. Commented Aug 29, 2022 at 10:34
Making it even more simple:
Every permutation can be reduced to a sequence of "two-element swaps": for example, the permutation that changes 123 into 312 can be written as (13)(12): first swap 1 and 3: 123-> 321, then swap 1 and 2: 321->312.
Of course, there are many different ways to do that. Any one permutation will consist of either an even number of swaps or an odd number no matter how that is done.
An even permutation is one that requires and even number of "swaps", an odd permutation is one that requires an odd number of "swaps".
Any permutation may be written as a product of transpositions. If the number of transpositions is even then it is an even permutation, otherwise it is an odd permutation. For example $(132)$ is an even permutation as $(132)=(13)(12)$ can be written as a product of 2 transpositions.
To determine whether $(a_1a_2\cdots a_n)(b_1b_2\cdots b_m)\cdots$ is an even permutation break each cycle down into transpositions : $(a_1a_2\cdots a_m)=(a_1a_2)(a_1a_3)\cdots(a_1a_n)$. The total number of transpositions for all cycles should be an even number for an even permutation.
There is another definition of even/odd permutation I found in one of the books, which I found to be easier to understand. It is:
Let P is a permutation function on a set S. For a pair (i,j) of elements in S such that i < j , if P(i) > P(j), then the permutation is said to invert the order of (i,j). The number of such pairs is known as the parity of the permutation. If permutation inverts even number of such pairs it is an even permutation else it is an odd permutation.
• Please state the book's name. Thanks in anticipation. Commented Aug 29, 2018 at 13:01
• Hi, I answered this 3 years back. Let me search for the book. can't promise that I will be able to find it. Ill surely try. Commented Aug 29, 2018 at 13:19
• The book "Aspects of Symmetry", by Robert Howlett uses a similar definition (see p.44 of online PDF from author's website Commented Apr 20, 2022 at 3:16
One other definition of the sign of a permutation comes from linear algebra, a permutation is a map of sets $$\sigma:X\rightarrow X$$, so we can view it as a linear map on a vector space with a basis indexed by our set X. Then the sign of sigma is simply the determinant of this linear map, interpreted in our field.
This perspective can be helpful for computing the signs of permutations arising from other actions too, such as the symmetric group acting on $$k$$ element subsets of $$X$$. It also suggests to look at other coefficients of the characteristic polynomial, for example, the trace counts the number of fixed points of $$\sigma$$ on $$X$$.
The results follow naturally upon considering the action of $$S_n$$ on arrangements :
A list $$[k_1, \ldots, k_n]$$ made by taking $$1, 2, \ldots, n$$ in some order is called an arrangement.
For $$\sigma \in S_n$$ and arrangement $$[k_1, \ldots, k_n]$$, we can define
$$\sigma * [k_1, \ldots, k_n] := ( [k_1, \ldots, k_n] \text{ after putting each } k_i \text{ into slot } \sigma(i) )$$
That is, $$[k_1, \ldots, k_n] \rightsquigarrow \sigma * [k_1, \ldots, k_n]$$ amounts to putting whatever is in slot $$i$$ into slot $$\sigma(i)$$.
Note $$\sigma * [k_1, \ldots, k_n] = [k_{\sigma^{-1} (1)}, \ldots, k_{\sigma^{-1} (n)}]$$
[Because in $$[k_1, \ldots, k_n] \rightsquigarrow \sigma * [k_1, \ldots, k_n]$$ the $$k_t$$ which gets sent to slot $$j$$ satisfies $$\sigma(t) = j$$ i.e. $$t = \sigma ^{-1} (j)$$.]
Also $$\sigma * ( \tau * [k_1, \ldots, k_n] ) = (\sigma \tau) * [k_1, \ldots, k_n]$$
[Because in $$[k_1, \ldots, k_n] \rightsquigarrow \sigma * (\tau * [k_1, \ldots, k_n])$$, $$k_i$$ is first sent to slot $$\tau(i)$$ and then to slot $$\sigma(\tau(i))$$. And $$[k_1, \ldots, k_n] \rightsquigarrow (\sigma \tau)*[k_1, \ldots, k_n]$$ has the same effect.]
We'll write $$\, [k_1, \ldots, k_n] \stackrel{\sigma}{\rightsquigarrow} [l_1, \ldots, l_n]"$$ to mean $$ \, [l_1, \ldots, l_n] = \sigma * [k_1, \ldots, k_n]"$$.
Example 1. Cycle $$(1 \, \, 2 \, \, 3 \, \, 4) = (1 \, \, 4) (1 \, \, 3) (1 \, \, 2)$$. It's easy to check this is true, but here is one way we can come up with the decomposition :
So $$(1 \, \, 2 \, \, 3 \, \, 4)*[1,2,3,4] = (1 \, \, 4)*\bigg( (1 \, \, 3) * ( (1 \, \, 2) * [1,2,3,4]) \bigg)$$
i.e.
$$(1 \, \, 2 \, \, 3 \, \, 4)*[1,2,3,4] = (1 \, \, 4) (1 \, \, 3) (1 \, \, 2) * [1,2,3,4]$$
i.e.
$$(1 \, \, 2 \, \, 3 \, \, 4) = (1 \, \, 4) (1 \, \, 3) (1 \, \, 2)$$
This suggests in general $$(a_1 \, \, a_2 \, \, \ldots \, \, a_k) = (a_1 \, \, a_k) (a_1 \, \, a_{k-1}) \ldots (a_1 \, \, a_2)$$, which is readily verified to be true.
Also any $$\sigma \in S_n$$ is a product of disjoint cycles, and each cycle decomposes by $$(a_1 \, \, a_2 \, \, \ldots \, \, a_k) = (a_1 \, \, a_k) (a_1 \, \, a_{k-1}) \ldots (a_1 \, \, a_2)$$. So each $$\sigma \in S_n$$ is a product of transpositions.
Example 2.
So
$$(1 \, \, 4) = (1 \, \, 2) (2 \, \, 3) (3 \, \, 4) (2 \, \, 3) (1 \, \, 2)$$
Similarly in general any transposition is a product of an odd number of "elementary transpositions" [i.e. transpositions of the form $$(j \, \, j+1)$$]
For $$\sigma \in S_n$$, a pair $$i < j$$ such that $$\sigma(i) > \sigma(j)$$ is called an inversion in $$\sigma$$. Also $$\text{inv}(\sigma)$$ denotes the number of inversions in $$\sigma$$.
To keep track of parity of $$\text{inv}(\sigma)$$, we can look at sign $$\text{sgn}(\sigma) := (-1)^{\text{inv}(\sigma)}$$. Permutations with sign $$1$$ are called even, and those with sign $$(-1)$$ odd.
By an "inversion in arrangement $$[k_1, \ldots, k_n]$$", we'll mean an inversion in $$\sigma = \begin{pmatrix} 1 &2 &\ldots &n \\ k_1 &k_2 &\ldots &k_n \end{pmatrix}$$ [that is, a pair $$k_i, k_j$$ in $$[k_1, \ldots, k_n]$$ where the larger of the two is to the left of the smaller]. Similarly sign of an arrangement is also meaningful.
Notice for any $$(j \, \, j+1) \in S_n$$, $$[k_1, \ldots, k_n] \rightsquigarrow (j \, \, j+1)*[k_1, \ldots, k_n]$$ changes number of inversions by $$\pm 1$$ (and therefore flips sign).
Hence for any transposition $$\tau \in S_n$$, writing it as a product of odd number of elementary transpositions (example 2) gives that $$[k_1, \ldots, k_n] \rightsquigarrow \tau * [k_1, \ldots, k_n]$$ reverses sign.
Let $$\tau_1, \ldots, \tau_k \in S_n$$ be transpositions. Looking at their product $$\sigma = \tau_1 \ldots \tau_k$$,
$$(\tau_1 \ldots \tau_k)*[1, \ldots, n] = \tau_1 * \big( \ldots * (\tau_k * [1, \ldots, n]) \ldots \big)$$
As RHS has sign $$(-1)^k$$,
$$\text{sgn}(\tau_1 \ldots \tau_k * [1, \ldots, n]) = (-1)^k$$
i.e.
$$\text{sgn}([\sigma^{-1}(1), \ldots, \sigma^{-1}(n)]) = (-1)^k$$
i.e. $$\text{sgn}(\sigma ^{-1}) = (-1)^k$$, that is $$\text{sgn}(\tau_k \ldots \tau_1) = (-1)^k$$.
So product of any $$k$$ transpositions has sign $$(-1)^k$$.
Let $$\sigma, \pi \in S_n$$. Writing them as product of transpositions (example 1)
$$\sigma = \tau_1 \ldots \tau_k \\ \pi = \tau'_1 \ldots \tau'_l$$
we have $$\text{sgn}(\sigma \pi) = \text{sgn}(\tau_1 \ldots \tau_k \tau'_1 \ldots \tau'_l)$$ $$= (-1)^{k+l} = (-1)^k (-1)^l$$ $$= \text{sgn}(\tau_1 \ldots \tau_k) \text{sgn}(\tau'_1 \ldots \tau'_l)$$ $$= \text{sgn}(\sigma) \text{sgn}(\pi)$$.
So finally for any $$\sigma, \pi \in S_n$$, $$\fbox{ \text{sgn}(\sigma \pi) = \text{sgn}(\sigma) \text{sgn} (\pi) }$$
Example 3. As $$(a_1 \, \, a_2 \, \, \ldots \, \, a_k) = (a_1 \, \, a_k) (a_1 \, \, a_{k-1}) \ldots (a_1 \, \, a_2)$$, $$\text{sgn}(a_1 \, \, a_2 \, \, \ldots \, \, a_k) = (-1)^{k-1}$$.
Ref: A similar discussion can be found in E.B.Vinberg's "Course in Algebra".
• In this example if you do the transpositions from right to left using $\sigma(1)=3$, $\sigma(2)=4$, $\sigma(3)=5$, $\sigma(4)=2$, $\sigma(5)=1$ you get the same result if you do the transpositions from left to right without making any $\sigma$ associations. Is these always possible?
– koy
Commented Mar 30 at 20:53
[This is one approach, mentioned in Jacobson's Basic Algebra Vol 1]
Fix $$n \in \mathbb{Z} _{\gt 0}.$$ For $$\sigma \in S _n,$$ let $$N(\sigma)$$ be the number of cycles in the cycle decomposition of $$\sigma.$$
Th: Let $$\sigma \in S _n,$$ and let $$\tau \in S _n$$ be a transposition. Then $$N(\tau \sigma) = N(\sigma) \pm 1$$ (especially $$\sigma \rightsquigarrow \tau \sigma$$ flips the parity of number of cycles).
Pf: Let $$\tau = (\alpha \text{ } \beta),$$ and let $$\sigma = C _1 \ldots C _{N(\sigma)}$$ be the cycle decomposition of $$\sigma.$$
$$\underline{\textbf{Case-1}}$$ (Both $$\alpha, \beta$$ appear in the same cycle of $$\sigma$$)
WLOG $$C _1$$ has both $$\alpha, \beta$$ in it. Now the cycle decomposition of $$(\alpha \text{ } \beta) C _1$$ is: $${ ({\color{purple}{\alpha}} \text{ } {\color{green}{\beta}}) ({\color{purple}{\alpha}} \text{ } x _1 \ldots \text{ } x _k \text{ } {\color{green}{\beta}} \text{ } x _{k+1} \text{ } \ldots \text{ } x _l) }$$ $${ = ({\color{purple}{\alpha}} \text{ } x _1 \text{ } \ldots \text{ } x _k) ({\color{green}{\beta}} \text{ } x _{k+1} \text{ } \ldots \text{ } x _l ) }.$$ So $${ N(\tau \sigma) = N(\sigma) + 1}$$ in this case.
$$\underline{\textbf{Case-2}}$$ ($$\alpha, \beta$$ appear in different cycles of $$\sigma$$)
WLOG $$C _1$$ has $$\alpha$$ and $$C _2$$ has $$\beta$$ in it. Now the cycle decomposition of $${ (\alpha \text{ } \beta) C _1 C _2 }$$ is: $${ ({\color{purple}{\alpha}} \text{ } {\color{green}{\beta}} ) ({\color{purple}{\alpha}} \text{ } x _1 \text{ } \ldots \text{ } x _k) ( {\color{green}{\beta}} \text{ } y _1 \text{ } \ldots \text{ } y _l) }$$ $${ = (x _1 \text{ } \ldots \text{ } x _k \text{ } {\color{green}{\beta}} \text{ } y _1 \text{ } \ldots \text{ } y _l \text{ } {\color{purple}{\alpha}} ). }$$ So $$N(\tau \sigma) = N(\sigma) -1$$ in this case.
Cor: If two products of transpositions $$\tau _1 \ldots \tau _k$$ and $$\tau' _1 \ldots \tau' _l$$ are equal in $$S _n,$$ then $$k,l$$ have the same parity.
Every $$\sigma \in S _n$$ is a product of transpositions.
$$\sigma$$ is a product of disjoint cycles, and each cycle $$(a _1 \text{ } \ldots \text{ } a _k)$$ is $$\underbrace{(a _1 \text{ } a _k) \ldots (a _1 \text{ } a _2)} _{k-1 \text{ transpositions}} .$$
But from the corollary, no matter how we express $$\sigma$$ as a product of transpositions in $$S _n$$ the quantity $$(-1)$$ raised to the number of transpositions remains the same. We call this invariant $$\text{sgn}(\sigma).$$
Now we have $$\text{sgn}(\sigma _1 \sigma _2) = \text{sgn}(\sigma _1) \text{sgn}(\sigma _2)$$ and $$\text{sgn}((a _1 \text{ } \ldots \text{ } a _k)) = (-1) ^{k-1}.$$
|
# Which Value Must Be Added To The Expression X2 + X To Make It A Perfect-square Trinomial?
0
44
In this article, we will discuss how to find the exact value of a perfect-square trinomial or trinomial equation. The trick is to use the factoring method.
## Identifying the term with an odd number of occurrences of X
When determining the term with an even number of occurrences of X, there are two main factors to take into account.
The first is the term’s magnitude or size, and the second is how many times the expression has appeared in your equation.
When having a large expression such as (X + 5)2 + 10 = 35 + 25 + 15 = 55, having the addition appear twice gives you a term with five times the magnitude of X, so you would look for a term with an even number of occurrences of X.
When finding a perfect-square Trinomial, remember that 3 multiplied by 3 equals 6, so in order for there to be an addition appearing twice, there needed to be an equation involving 6.
## Determining the difference between the term with an odd number of occurrences of X and the square root of 2
The term with an odd number of occurrences of X is called an odd term and the square root of 2 is called a positive term.
When preparing a Trinomial Expression, the first step is to determine if there are even or even terms or if the term has an odd number of occurrences of X.
If there are even terms, then the middle term must be added to create a perfect-square Trinomial. The total amount of squares required can be found by adding up all the angles in the parentheses.
If there are even terms but one has an odd number of occurrences of X, then the middle term must be added again to create a perfect-square Trinomial with an Even Number of Angles.
This process is repeated until all angles in the expression have been accounted for.
## Calculating all terms that have an odd number of occurrences of X
The trick to finding all the odd terms of an expression is to add together the exponents of each term. For example, the expression 2 + 2•3 + 4 + 5•6 + 7 + 8 adds up to 12, so we have to add the exponents.
The same rule applies to Trinomials, except that we have to subtract off the top term. The rule for Trinominales is that they must have only even terms, so we must subtract off 1 from each term.
The trick to finding all the perfect-square Trinominals is to divide by 2.
## Calculating all terms that do not have an odd number of occurrences of X
This is one of the most common mistakes made when trying to solve Trinomial equations. The easiest way to calculate all of the non-X terms is to multiply by X itself!
This can be done easily by adding the two sides of the equation, but in order for this to work, there must be an even number of terms. In order for this to be true, there must be an odd number of terms because if there were an even amount of them, then 1 + 1 + 1 + 1 + 1 would equal 2, which doesn’t exist.
So, in order for this formula to work, there must be an addition on each side! Luckily for us, computers have a way to do this for us.
## Finding the difference between each term with an odd number of occurrences of X and its corresponding squared root value
The trick to finding the square root of a Trinomial is to find the difference between each term with an even number of occurrences of X and its corresponding term with an odd number of occurrences.
The square root of an Trinomial can be found by adding up all the terms with an even number of occurrences and finding their square. The term with the higher value will be the square root, and its corresponding term will be decreased in size.
This is done by using the 2 × 2 matrix introduced earlier in this article. In this matrix, side A represents the term with an even number of occurrences, side B represents its correspondingtermwithanoddnumberofoccurrencesthatnumberofweeks, and side C represents increased size.
When finding the difference between two values, it is important to know whether one value is larger or smaller than the other.
## Coming up with a final value to add to each term with an even number or occurrence to make it a perfect square trinomial
This is a difficult math problem to solve, so most students will not bother. However, the solution can be found in the bullet point below.
The trinomial equation can be rewritten as a special case of the perfect-square-trinomial equation. The new equation has three terms instead of two, but the way to add them up is still the same.
When solving trinomial equations, many times what students do is take one term of each type and add them together, and then take the other term of each type and add together. This gives them an answer to simplify the third term on the equation.
## Understanding how to simplify perfect square trinomial expressions
When a perfect square trinomial expression has more than one variable, it is easy to simplify by adding additional variables.
For example, the variable y can be replaced by the value of x in the equation. The same happens with z, the variable for t, the quantity of x.
So far, we have only used these variables in place of t and x, but there is a potential solution that does not need any new variables. This potential solution can be found by doing a search on web sites that help you solveperfect squaretrinomials.
## Practicing by trying some examples
Trying some examples is the best way to learn new concepts. Even though this section is called Try Some Examples, these examples can be done on a computer or mobile device.
This section of the article is for those who do not have a computer or mobile device. You can still try these examples!
There are several problems you can try these examples on. For example, you can try trying trying trying trying trying trying tries on: 3, 5, 7, and 11 times the expression x2 + x + 1 = 0. Or you can try trying tries like adding 1 to each term and finding the solution that does not change unless x + 1 is removed.
Harry Potter, the famed wizard from Hogwarts, manages Premier Children's Work - a blog that is run with the help of children. Harry, who is passionate about children's education, strives to make a difference in their lives through this platform. He involves children in the management of this blog, teaching them valuable skills like writing, editing, and social media management, and provides support for their studies in return. Through this blog, Harry hopes to inspire others to promote education and make a positive impact on children's lives. For advertising queries, contact: [email protected]
|
# What Every Student Needs to Know for Multiplication (Part 6)
***This is part 6 of a series. Click the links to go back and read part 1, part 2, part 3, part 4, and part 5***
In the latest series of blog posts by the Coordinating Units team, we explored our recent work in the classroom developing multiplicative reasoning with students. One area that we focused on was creating different contexts to present the task. If students can generalize the mathematics and transfer it to different contexts, then they can use it in problem solving instead of memorized procedures.
To begin the school year we used cubes from which the students built towers.
In October, we changed the context of the task, sending them out “trick or treating,” by having them place a predetermined number of candies (small blocks) in bags.
In November, we changed the context yet again, having the students retrieve turkey cut-outs, for their Thanksgiving dinner, with a specified number of feathers on each turkey.
In all the variations, the essence of the task was preserved as the students were asked to make equal groups of items and then asked a series of questions. For example, they were asked to make towers with three cubes, bags of candies with three candies, and turkeys with three feathers, and then asked to determine the number of groups (towers, bags, turkeys), the number of items in each group (three), and the total number of items (cubes, candies, feathers).
Many teachers have experienced students struggling when the context of a concept is changed.
One example of this is:
14 x 10 = 140
So, when multiplying by 10 just add a zero to the end of the number.
Except, an example like this:
1.4 x 10 ≠ 1.40
doesn’t follow this rule even though the problems look the same.
One of the reasons that students struggle is that they are taught to focus in on keywords or formats and then to use a set of prescribed procedures that go along with them. Problems arise when the keywords are not found or the problem is formatted differently. The students see the situation as different and a new set of procedures is needed to solve it. The result is the impression by some students that mathematics is complex with lots of rules and procedures that must be memorized in order to be successful. Teachers want their students to generalize concepts and extrapolate them to situations that have a common underlying mathematical structure even when they are not familiar with the problem. Changing the context gives the student the opportunity to solve problems by understanding the underlying concept instead of memorizing a set of procedures.
In our last teaching session before the Thanksgiving break, one of the students asked us how the task would be changed when we returned. This prompted the other students to start making suggestions such as placing ornaments on trees, presents under trees, bells on a string, and many others. What remained the same in the ideas was that they all could be made into equal groups with a constant number of items in each group. The students have generalized the tasks and realize that the situations contained a common underlying structure of distributing a set of items over a set of groups. They can now produce their own situations using this structure, knowing that the context could change while the math remains constant.
Click here to continue on to the final installment of this series
|
# Factorial Calculator
Factorial Calculator
In Mathematics, factorial means multiplying the given number by every natural number which comes below it. For example, if we say that we want to calculate the factorial of n, we are going to multiply n by every natural number below it.
## Factorial Calculator
Formula –
n! = n × (n-1) × (n-2) × ….. × 3 × 2 × 1
You can use our Factorial calculator to calculate the factorial of the given number.
### How to calculate factorial?
Well, to calculate the factorial of some number, you just need to find a product of all integers from 1 to that particular number. For example, if we are calculating the factorial of 5, then we are going to calculate the product of all the integers from 1 to 5. This will go like – 5×4×3×2×1, which comes out to be 120. So, the factorial of 5 comes out to be 120. We can write it like this –
5! = 120
You can try solving several examples, and also use our calculator to calculate the factorial of some numbers.
1. What is 0!?
Ans: Well, the answer to 0! is 1.
2. How do you calculate factorial?
Ans: If we want to calculate the factorial of n, we can simply multiply that n by every natural number below it. In other words, we can use this formula – n! = n × (n-1) × (n-2) × ….. × 3 × 2 × 1
3. What is the use of calculating factorial?
Ans: Factorial can be required in doing many calculations, like when we are dealing with permutations and combinations. For example, if we want to arrange 3 different toys, in how many ways can we do it? This goes a little bit tricky, but if you try to solve it, it comes out to be 6, which is 3!
Scroll to Top
|
# Find the value of $x$, if:(i) $4x = (52)^2 – (48)^2$(ii) $14x = (47)^2 – (33)^2$(iii) $5x = (50)^2 – (40)^2$
Given:
(i) $4x = (52)^2 – (48)^2$
(ii) $14x = (47)^2 – (33)^2$
(iii) $5x = (50)^2 – (40)^2$
To do:
We have to find the value of $x$ in each case.
Solution:
Here, we have to find the value of $x$ in each expression. The given expressions are the difference of two squares. So, to find the value of $x$ we can simplify the RHS in each case using the identity:
$(a – b) (a + b) = a^2 – b^2$.
Therefore,
(i) $4x = (52)^2 – (48)^2$
This implies,
$4x=(52+48)\times(52-48)$
$4x=100\times4$
$4x=400$
$x=\frac{400}{4}$
$x=100$
Hence, the value of $x$ is $100$.
(ii) $14x = (47)^2 – (33)^2$
This implies,
$14x=(47+33)\times(47-33)$
$14x=80\times14$
$x=\frac{80\times14}{14}$
$x=80$
Hence, the value of $x$ is $80$.
(iii) $5x = (50)^2 – (40)^2$
This implies,
$5x=(50+40)\times(50-40)$
$5x=90\times10$
$x=\frac{90\times10}{5}$
$x=90\times2$
$x=180$
Hence, the value of $x$ is $180$.
|
💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!
# Use the table to evaluate each expression.(a) $f (g(1))$(b) $g (f(1))$(c) $f (f(1))$(d) $g (g(1))$(e) $(g \circ f) (3)$(f) $(f \circ g) (6)$
## a) $f(g(1))=5$b) $g(f(1))=2$c) $f(f(1))=4$d) $g(g(1))=3$e) $(g \circ f)(3)=1$f) $(f \circ g)(6)=4$
### Discussion
You must be signed in to discuss.
##### Kristen K.
University of Michigan - Ann Arbor
##### Samuel H.
University of Nottingham
##### Michael J.
Idaho State University
Lectures
Join Bootcamp
### Video Transcript
Let's use the table to find f of G of one. So first we're going to find G of one. So we used the G row and the one column, and we find that g of oneness. Six. So let's replace GF one with six, and now we're looking for F of six. So let's use the six column and the F row and F of six is five. So the answer is that Now let's find G of f of one. So first, let's find f of one. So we use the one column and the F ro and we get f of one is three. So now we're looking for G of three. So we used the three column and the G Row and G of three is, too. So your answer is to now for part C. Let's find af f of one. So we start by finding f of one. So we use the one row for the one column and the F row and 1/2 of one is three. So we substitute that in, and now we're finding f of three. So to find f of three, we used the F ro and the three column and we get F of three is four. So the answer is for now, let's find G of G of one. So starting with the inside g of one, we use the one column and the G row, and we get six g of 1 to 6. So we substitute that in, and now we're finding G of six. So we use the six column and the G Row and G of sixes three. So our answer is three. Next for party. Let's find G of F of three and another way to write. That would be G of f of three just with parentheses like that. So let's start by finding f of three. So we used the three column and the F Row F of three is four. So we substitute that in, and now we're finding Geo for So use the four column and the G Row G affords one. So our answer is one. Now let's do part F, and we could write Rewrite this as f of G of six with parentheses like so? So we'll start on the inside and find G of six. So we use the six column and the G Row G of six is three, So we'll substitute that in and we have f of three. So now we're going to use the three column and the F Row F of three is four. So our answer is for
Oregon State University
##### Kristen K.
University of Michigan - Ann Arbor
##### Samuel H.
University of Nottingham
##### Michael J.
Idaho State University
Lectures
Join Bootcamp
|
Question Video: Determining How Many Circles Pass Through Two Points | Nagwa Question Video: Determining How Many Circles Pass Through Two Points | Nagwa
# Question Video: Determining How Many Circles Pass Through Two Points Mathematics • Third Year of Preparatory School
## Join Nagwa Classes
Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!
How many circles can pass through two points?
03:38
### Video Transcript
How many circles can pass through two points?
We can define a circle as a set of points in a plane that are a constant distance from a point in the center. Letβs imagine that these two points are here, and we can define them as π sub one and π sub two. These two points, π sub one and π sub two, will both lie on the same circle if they are the same distance from the center of the circle. For example, if we took this point π, could this be the center of a circle which passes through both the points π sub one and π sub two? And the answer would be no, because although the circle with center π passes through π sub one, itβs much too small to pass through π sub two as well.
And so, to find a center of a circle which passes through π sub one and π sub two, we really need to consider this question: How do we find a point or a set of points, which are equidistant from two other points? Remember that this word βequidistantβ simply means the same distance away. To do this, weβre going to construct the perpendicular bisector of the line segment between π sub one and π sub two. To do this accurately, we need this tool, which will be called a pair of compasses or a compass, depending on where you live. It can be helpful to draw in the line segment between the two points, and we start by putting the sharp pointed end of the compass into one of the points.
So, letβs start with π sub one. We then stretch-open the compass so that the pencil tip will lie more than halfway along the length of the line segment. We then use the pencil in the compass to create an arc above the line segment and below it. We then repeat the process, this time putting the pointed end of the compass onto π sub two. We now have a pair of arcs above the line segment and below it. We observed that in each pair of arcs there is a point of intersection. We join these two points of intersection with a straight line. This is the perpendicular bisector of the line segment between the two points. This line has divided the line segment between π sub one and π sub two into two congruent pieces and at 90 degrees.
However, in this problem, weβre not really interested in the line segment between the two points. But letβs consider this perpendicular bisector and what it actually represents. It will represent all the points that are equidistant from π sub one and π sub two. For example, here is a point on the line. Letβs call this point π΄. And because itβs equidistant from π sub one and π sub two, then we could create a circle of center π΄ and the circle with center π΄ will pass through both π sub one and π sub two. We could repeat this as many times as we wanted. For example, hereβs a smaller circle and a larger circle. In fact, why even limit ourselves to circles which can fit on the screen? We know that this line will extend infinitely in both directions, and that means that we could draw an infinite number of circles.
And therefore for any two points, whether those points are really close together or thousands of kilometers apart, we know that we can draw an infinite number of circles that pass through those two points.
## Join Nagwa Classes
Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!
• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions
Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy
|
# What is a Probability Distribution?
A probability distribution is a table or an equation that links each outcome of a statistical experiment with its probability of occurrence.
## Probability Distribution Prerequisites
To understand probability distributions, it is important to understand variables. random variables, and some notation.
• A variable is a symbol (A, B, x, y, etc.) that can take on any of a specified set of values.
• When the value of a variable is the outcome of a statistical experiment, that variable is a random variable.
Generally, statisticians use a capital letter to represent a random variable and a lower-case letter, to represent one of its values. For example,
• X represents the random variable X.
• P(X) represents the probability of X.
• P(X = x) refers to the probability that the random variable X is equal to a particular value, denoted by x. As an example, P(X = 1) refers to the probability that the random variable X is equal to 1.
## Probability Distributions
An example will make clear the relationship between random variables and probability distributions. Suppose you flip a coin two times. This simple statistical experiment can have four possible outcomes: HH, HT, TH, and TT. Now, let the variable X represent the number of Heads that result from this experiment. The variable X can take on the values 0, 1, or 2. In this example, X is a random variable; because its value is determined by the outcome of a statistical experiment.
A probability distribution is a table or an equation that links each outcome of a statistical experiment with its probability of occurrence. Consider the coin flip experiment described above. The table below, which associates each outcome with its probability, is an example of a probability distribution.
Number of heads Probability
0 0.25
1 0.50
2 0.25
The above table represents the probability distribution of the random variable X.
## Cumulative Probability Distributions
A cumulative probability refers to the probability that the value of a random variable falls within a specified range.
Let us return to the coin flip experiment. If we flip a coin two times, we might ask: What is the probability that the coin flips would result in one or fewer heads? The answer would be a cumulative probability. It would be the probability that the coin flip experiment results in zero heads plus the probability that the experiment results in one head.
P(X < 1) = P(X = 0) + P(X = 1) = 0.25 + 0.50 = 0.75
Like a probability distribution, a cumulative probability distribution can be represented by a table or an equation. In the table below, the cumulative probability refers to the probability than the random variable X is less than or equal to x.
Number of heads:
x
Probability:
P(X = x)
Cumulative Probability:
P(X < x)
0 0.25 0.25
1 0.50 0.75
2 0.25 1.00
## Uniform Probability Distribution
The simplest probability distribution occurs when all of the values of a random variable occur with equal probability. This probability distribution is called the uniform distribution.
Uniform Distribution. Suppose the random variable X can assume k different values. Suppose also that the P(X = xk) is constant. Then,
P(X = xk) = 1/k
Example 1
Suppose a die is tossed. What is the probability that the die will land on 5?
Solution: When a die is tossed, there are 6 possible outcomes represented by: S = { 1, 2, 3, 4, 5, 6 }. Each possible outcome is a random variable (X), and each outcome is equally likely to occur. Thus, we have a uniform distribution. Therefore, the P(X = 5) = 1/6.
Example 2
Suppose we repeat the dice tossing experiment described in Example 1. This time, we ask what is the probability that the die will land on a number that is smaller than 5?
Solution: When a die is tossed, there are 6 possible outcomes represented by: S = { 1, 2, 3, 4, 5, 6 }. Each possible outcome is equally likely to occur. Thus, we have a uniform distribution.
This problem involves a cumulative probability. The probability that the die will land on a number smaller than 5 is equal to:
P( X < 5 ) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
P( X < 5 ) = 1/6 + 1/6 + 1/6 + 1/6 = 2/3
|
## By MathPlanetVideos
Solve the given equation using the quadratic formula x2+2x−8=0
Sal introduces the quadratic formula and how it should be used in order to solve quadratic equations.
Sal introduces the quadratic formula and how it should be used in order to solve quadratic equations.
Sal introduces the quadratic formula and how it should be used in order to solve quadratic equations.
# Solving Quadratic Equations using the Quadratic Formula - Example 2, Complex Solutions
## By PatrickJMT
YouTube presents Solving Quadratic Equations using the Quadratic Formula - Example 2, Complex Solutions an educational video resources on math.
# Solving Quadratic Equations using the Quadratic Formula - Example 2, Complex Solutions
## By PatrickJMT
YouTube presents Solving Quadratic Equations using the Quadratic Formula - Example 2, Complex Solutions an educational video resources on math.
Sal solves the equation -7q^2+2q+9=0 by using the quadratic formula.
Sal solves the equation -x^2+8x=1 by first bringing it to standard form and the using the quadratic formula.
Sal solves the equation -x^2+8x=1 by first bringing it to standard form and the using the quadratic formula.
Sal solves the equation -x^2+8x=1 by first bringing it to standard form and the using the quadratic formula.
Sal solves the equation -7q^2+2q+9=0 by using the quadratic formula.
Sal solves the equation -7q^2+2q+9=0 by using the quadratic formula.
Watch this video to learn how the quadratic formula is proven from completing the square for the general quadratic equation Ax2 + Bx + C = 0.
Sal rewrites the equation 6x^2+3=2x-6 in standard form and identifies a, b, and c, that can later be used within the quadratic formula.
Sal rewrites the equation 6x^2+3=2x-6 in standard form and identifies a, b, and c, that can later be used within the quadratic formula.
Sal rewrites the equation 6x^2+3=2x-6 in standard form and identifies a, b, and c, that can later be used within the quadratic formula.
|
Upcoming SlideShare
×
# Section 4.1
316 views
Published on
Published in: Education
0 Likes
Statistics
Notes
• Full Name
Comment goes here.
Are you sure you want to Yes No
• Be the first to comment
• Be the first to like this
Views
Total views
316
On SlideShare
0
From Embeds
0
Number of Embeds
15
Actions
Shares
0
2
0
Likes
0
Embeds 0
No embeds
No notes for slide
### Section 4.1
1. 1. Learning Target:Students will learn how to calculate decimal numbers and whole numbers by multiplying. Students will learn the communicative and associative properties of multiplication. <br />
2. 2. Rule:Multiply like normal. Then count the number of place values behind (to the right) the decimal point in the problem.Start at the right of the product and move the decimal the number of placed to the left that you counted from in the step above. <br />
3. 3. Find the product. Then write the product in words.2.36 64 0.024<br />
4. 4. Answers:2.36 6 =14.16Fourteen and sixteen hundredths4 0.024 =0.096Ninety-six thousandths<br />
5. 5. Find the product. Use estimation to check.12.706 31.505 8<br />
6. 6. Answers: 12.706 338.118est. 13 3 = 391.505 8 12.04est. 2 8 = 16; 1 8 = 8<br />
7. 7. Properties of Multiplication:Commutative Property – You can multiply numbers in any order2 6.5 = 6.5 2a b = b aAssociative Property – Changing the grouping of factors will not change the product.(2 6.5) 4 = 2 (6.5 4)(a b) c = a (b c)<br />
8. 8. Tell which property of multiplication is shown. Explain your choice. (3.25 5) 2 = 3.25 (5 2)0.4 32 5 = 32 0.4 5<br />
9. 9. (3.25 5) 2 = 3.25 (5 2)Associative property of multiplication(changing the grouping) 0.4 32 5 = 32 0.4 5Commutative property of multiplication(changing the order)<br />
10. 10. You complete the first leg of a bike race in about 12 hours over two days. Your cyclometer shows that your average speed for this leg was 15.95 miles per hour. About how long was this leg of the race?<br />
|
Roots & Radicals - GMAT Math Study Guide
Definitions
• Nth Root of X - the number that, when multiplied together N times, yields X.
For example, the N = 2 root (also called the square root) of 9 is the number, 3, that when multiplied together 2 times yields 9.
• Square Root of X - the number that, when multiplied together two times, yields X.
For example, the square root of 16 is 4 because four multiplied together two times (i.e., 4*4) is 16.
• Cube Root of X - the number that, when multiplied together three times, yields X.
For example, the cube root of 64 is 4 because four multiplied together three times (i.e., 4*4*4) is 64.
• Radical - the sign used to denote the square or nth root of a number.
For example, the value of "radical 4" is 2 and the value of "radical 9" is 3.
• Radicand - the number that is beneath the radical sign and from which the square root (or nth root) is derived.
For example, when saying, "2 is the square root of 4," the number 4 is the radicand.
Relationship to Exponents
The relationship between roots and exponents is effectively an inverse relationship.
If MN = Z, then the Nth root of Z is M
As will be shown shortly in the formulas section of this study guide, roots relate to exponents through this extremely important formulaic relationship.
Examples of the Rules of Roots & Radicals
In order to improve your ability to work mathematics questions successfully and quickly, it is extremely helpful to memorize a few commonly used exponents and roots:
There are four different cases that encompass the world of radicals.
Case 1: Positive Radicand and Even Root
In Case 1, it is common mathematical convention that there is only one solution and this solution is positive. For example, the square root of 4 is only positive 2, even though (technically speaking) -2 is a root since (-2)(-2)=+4. Consequently, when asked, "what is the square root of 25?" the correct answer is "only positive 5." Similarly, if asked, "what is the 4th root of 16?" the correct answer is "only positive 2" even though (-2)4 = 16 too.
Case 2: Positive Radicand and Odd Root
In Case 2, there is only one possible answer. By comparison and for clarification's sake, in Case 1, there were two possible answers, although mathematical convention agreed that only one of those answers (i.e., the positive one) was correct. Due to the nature of how negative numbers behave when multiplied together an odd number of times (i.e., they retain their negative sign), it is impossible to have a negative answer in Case 2.
Case 3: Negative Radicand and Odd Root
In Case 3, there is only one possible answer. Since the only way to have a negative product is by multiplying a negative number an odd number of times, the answer in Case 3 is always negative. In order to more clearly see this, notice that multiplying together a negative number an even number of times produces a positive number. Similarly, multiplying a positive number by itself an odd number of times produces a positive number.
Case 4: Negative Radicand and Even Root
In Case 4, there is no answer to this type of question within the domain of what are called real numbers. Technically, this Case can be answered using what are known as imaginary numbers--but this is well beyond the scope of what is tested. If you encounter a question such as "what is the square root of -4" the correct answer is "there is no real solution."
|
Posted on by Kalkicode
Code Recursion
# Reverse a number using recursion
In this article, we will explore the concept of reversing a number using recursion. Reversing a number means changing the order of its digits from left to right. For example, reversing the number 12345 would result in 54321. We will discuss the problem statement, provide an explanation with suitable examples, present the pseudocode and algorithm for the recursive solution, and finally, provide an explanation of the resultant output with the time complexity of the code.
## Problem Statement
The problem is to write a recursive function that takes an integer as input and returns its reverse as output. If the input number is positive, the function should reverse its digits and return the new number. If the input number is negative, the function should reverse its absolute value and then return the new number with a negative sign.
## Explanation with Suitable Example
Let's take the number 12345 as an example. To reverse this number using recursion, we follow these steps:
1. Extract the last digit of the number: 12345 % 10 = 5
2. Recur with the remaining digits: reverse_num(1234, 5)
3. Extract the last digit of the remaining number: 1234 % 10 = 4
4. Recur with the remaining digits: reverse_num(123, 54)
5. Repeat this process until no digits are left: reverse_num(1, 5432)
Now, the base condition is reached since 1 is less than or equal to 0. So, we return the result, which is 54321.
## Pseudocode and Algorithm
The recursive function `reverse_num(number, result)` takes two parameters: `number` (the remaining digits) and `result` (the reversed number obtained so far).
The algorithm can be represented using standard pseudocode as follows:
``````function reverse_num(number, result):
if number > 0:
return reverse_num(number / 10, result * 10 + (number % 10))
return result
function reverse(number):
if number < 0:
return -reverse_num(-number, 0)
else:
return reverse_num(number, 0)
``````
## Explanation of the Algorithm
1. The `reverse_num` function takes two arguments: `number` and `result`. If the `number` is greater than 0, it means there are still digits left to process.
2. In each recursive call, the last digit of the `number` is extracted using `(number % 10)` and added to the reversed result `result * 10`.
3. The function is called recursively with the remaining digits obtained by dividing the number by 10 (`number / 10`).
4. This process continues until there are no digits left (`number <= 0`), at which point the result is returned.
5. The `reverse` function first checks if the input number is negative. If so, it calls `reverse_num` with the absolute value of the number and then returns the result with a negative sign.
6. If the input number is positive, the `reverse` function directly calls `reverse_num` with the input number.
## Resultant Output Explanation
Let's consider the provided test cases:
1. For the number 12345:
• Before reversing, the number is displayed as [12345].
• After reversing, the output is [54321].
2. For the number 78942:
• Before reversing, the number is displayed as [78942].
• After reversing, the output is [24987].
3. For the number 1020:
• Before reversing, the number is displayed as [1020].
• After reversing, the output is [201].
4. For the number -28:
• Before reversing, the number is displayed as [-28].
• After reversing, the output is [-82].
## Time Complexity of the Code
The time complexity of the code is O(log N), where N is the input number. In each recursion, the input number is divided by 10 until it becomes 0. The number of recursive calls depends on the number of digits in the input number, which is logarithmic with respect to the input number. Hence, the time complexity is logarithmic.
## Comment
Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.
Categories
Relative Post
|
# 180 Days of Math for Second Grade Day 126 Answers Key
By accessing our 180 Days of Math for Second Grade Answers Key Day 126 regularly, students can get better problem-solving skills.
## 180 Days of Math for Second Grade Answers Key Day 126
Directions: Solve each problem.
Question 1.
66 = Tens Ones
66 = 6 tens 6 ones
Explanation:
Six tens is 60 and then the digit in the ones place is 6.
Question 2.
3 + 49 = ___________
By adding 3 and 49 we get 52.
Question 3.
What is 50 less than 76?
Explanation:
76 – 50 = 26
By subtracting 50 from 76 we get 26.
Question 4.
15 – = 7
Explanation:
Let the missing number be x.
15 – x = 7
15 – 7 = x
x = 8
Question 5.
Count the angles.
The name of the above shape is the circle. A circle has a total of 360 degrees.
Question 6.
Write the month that comes after April.
Answer: The month that comes after April is May.
Question 7.
Library Books Checked Out
Week 1 Week 2 Week 3 Jody 4 5 5 Emily 4 4 4 Brenda 5 7 6 Alison 6 3 6
What was the total number of library books that Jody checked out?
Number of books checked out in week 1 = 4
Number of books checked out in week 2 = 5
Number of books checked out in week 3 = 5
4 + 5 + 5 = 14
Thus the total number of library books that Jody checked out is 14.
Question 8.
Write the number sentence.
Seventy-three plus seventeen equals ninety.
Seventy-three plus seventeen equals ninety means we have to add the two-digit numbers.
73 +17 = 90
Scroll to Top
Scroll to Top
|
UPSKILL MATH PLUS
Learn Mathematics through our AI based learning portal with the support of our Academic Experts!
Rules to convert decimals to fractions:
Step 1: Write the decimal number in the numerator and remove the decimal point.
Step 2: In the denominator, followed by $$1$$ add the number of zeroes that were equal to the number digits present after the decimal point.
Step 3: Write the lowest fraction by dividing numerator and denominator by the common factor.
Example:
1. Convert $$54.63$$ to fraction.
Step 1: Write $$54.63$$ as $$5463$$ in the numerator.
Step 2: There are two digits after the decimal point. So add two zeros in the denominator followed by $$1$$.
Therefore, the fraction is $\frac{5463}{100}$.
2. Write $$3.2$$ as the lowest fraction.
Step 1: Write $$3.2$$ as $$32$$ in the numerator.
Step 2: There is one digit after the decimal point. So add one zero in the denominator followed by $$1$$. The fraction is $\frac{32}{10}$.
Step 3: The common factor in the numerator and denominator is $$2$$. Divide numerator and denominator by $$2$$.
$\frac{32÷2}{10÷2}=\frac{16}{5}$
Therefore, the lowest fraction is $\frac{16}{5}$.
3. Write $$399.95$$ as lowest fraction.
Step 1: Write $$399.95$$ as $$39995$$ in the numerator.
Step 2: There is two digit after the decimal point. So add two zeros in the denominator followed by $$1$$. The fraction is $\frac{39995}{100}$.
Step 3: The common factor in the numerator and denominator is $$5$$. Divide numerator and denominator by $$5$$.
$\frac{39995÷5}{100÷5}=\frac{7999}{20}$
Therefore, the lowest fraction is $\frac{7999}{20}$.
|
# Question Video: Using a Given Arithmetic Sequence to Solve a System of Linear Equations Mathematics
Find the values of π₯ and π¦ given the arithmetic sequence (9π₯, 4π¦ + 1, 2π¦ β 6, 4π¦ + 2, ...).
04:35
### Video Transcript
Find the values of π₯ and π¦ given the arithmetic sequence nine π₯, four π¦ plus one, two π¦ minus six, four π¦ plus two.
Weβve been given four terms in an arithmetic sequence, and all four of these terms are given as expressions with a variable, either π₯ or π¦. In an arithmetic sequence, the terms are such that the difference between any two consecutive terms is the same amount, and we call that amount the common difference. This means if we take the first term and add the common difference, we get the second term. If we take the second term and add the common difference, we get the third term and so on. We could use this fact to set up some equations. For example, we could say that nine π₯ plus π equals four π¦ plus one. This is a true statement.
However, we have actually added an unknown variable. We now have three unknowns π₯, π, and π¦. We want to consider a strategy that would not require us to introduce a third variable. To do this, we can use what we know about arithmetic means. The terms between π sub one and π sub π in an arithmetic sequence are arithmetic means. This means that four π¦ plus one is an arithmetic mean. And two π¦ minus six is an arithmetic mean. Four π¦ plus one falls between π sub one and π sub three. And since four π¦ plus one is the arithmetic mean between π sub one and π sub three, then we combine π sub one and π sub three and divide by two to find π sub two. We say nine π₯ plus two π¦ minus six divided by two equals four π¦ plus one. And since two π¦ minus six is also an arithmetic mean, it falls between π sub two and π sub four.
We can create a second equation that says four π¦ plus one plus four π¦ plus two divided by two equals two π¦ minus six. In our first equation, we have two variables π₯ and π¦. In our second equation, we only have a π¦-variable, so itβs probably helpful for us to start there and solve for π¦. To do that, we can combine like terms in our numerator, which will give us eight π¦ plus three divided by two equals two π¦ minus six. We multiply both sides of our equation by two to give us eight π¦ plus three equals four π¦ minus 12. In our next step, we can subtract four π¦ from both sides and also subtract three from both sides of the equation. On the left, we will have our π¦-terms. Eight π¦ minus four π¦ equals four π¦.
And on the right side, weβll have the constants negative 12 minus three equals negative 15. We divide both sides by four, which leaves us with π¦ equals negative 15 over four. We can take what we found for π¦ and plug that into our first equation. But our calculations will be easier if we simplify before we plug in negative fifteen-fourths for π¦. Our first step to get that two out of the denominator, we multiply both sides of the equation by two, and we get nine π₯ plus two π¦ minus six equals eight π¦ plus two. We know weβre going to solve for π₯ here, so we want to get π¦ on the other side of the equation. To do that, weβll subtract two π¦ from both sides of the equation and weβll add six to both sides of the equation.
On the left side, weβre left with nine π₯. And on the right side, we have six π¦ plus eight. Then we substitute negative fifteen-fourths in for π¦. Six times negative fifteen-fourths plus eight will equal negative 29 over two. Dividing both sides of this equation by nine, we get that π₯ equals negative 29 over 18. We could go back and plug these values in to find the exact values of every term in this arithmetic sequence. However, this question was only asking us to identify the values of π₯ and π¦, which we have done here. π₯ equals negative 29 over 18, and π¦ equals negative fifteen-fourths.
|
# The Grouping Number Method
This is an excerpt from a previous post, but if I want to introduce someone to the grouping number method, they don’t need all of the commentary about the AC method.
Are you looking for a nice and intuitive method for factoring quadratics when the leading coefficient isn’t one? Most people were probably taught to break these these things down by trial and error. That isn’t awful once you’ve developed some intuition, but if you haven’t, trial and error quickly becomes tedious. What should we start with instead? The grouping number method works well, especially if you’re just starting out.
To set up the Grouping Number Method, let’s think about what happens in the following polynomial multiplication.
$(4x-1)(2x-3)$
$=2x(4x-1)-3(4x-1)$
$=8x^2 -2x -12x +3$
$=8x^2 -14x +3$
Normally, you’d just skip the second step, but it’s important to realize that what you’re really doing is using the distributive property twice. Additionally, let’s think about the -12 and the -2. It’s clear that they add up to the final coefficient of x, but it’s also true that their product (-12)(-2) = 24 is the same as the product of the leading coefficient and the constant term, (8)(3). Notice this has to be true. In either case, it’s the product of 4, -1, 2 and -3, merely in a different order.
So, how do we go in the other direction? That is, how do we factor instead of multiply? We’ll demonstrate the grouping number method on the polynomial $8x^2 - 14x + 3$.
The grouping number of a quadratic polynomial is the leading coefficient multiplied by the constant term.
So, in this case, it would be $8 times 3 = 24$.
Next, you want to look for two factors of your grouping number that add up to the coefficient of $x$. In this case, $(-12)(-2) = 24$ and $(-12)+(-2) = -14$. We use these two numbers to rewrite the middle term to get the following.
$8x^2 - 14x + 3 = 8x^2 - 2x - 12x + 3$.
In other words, to rewrite the polynomial in this way, you find two factors of your grouping number that add up to the coefficient of $x$ and use those to break down the middle term.
Why rewrite the polynomial in this way? Because it sets us up perfectly to factor by grouping.
Factoring by grouping is a technique for factoring a polynomial with four terms. In a nutshell, to factor by grouping, you remove the greatest common factor from the first two terms, then remove the greatest common factor from the last two terms. If the resulting binomial factors are the same you can factor this out to get the product of two binomials. Notice, that’s using the distributive property twice, just in the opposite direction as before.
Therefore, to finish factoring the polynomial we can factor by grouping.
$8x^2 - 14x + 3$
$= 8x^2 - 2x - 12x + 3$
$= 2x(4x - 1) - 3(4x - 1)$
$= (4x - 1)(2x - 3)$
Notice this exactly reverses the steps of the multiplication with which we started.
In short, as long as you have no common factors, the quadratic polynomials where the grouping number can be factored into two numbers that add up to the middle coefficient are exactly the ones that can be factored by grouping. All the others are prime.
# I’m Back
It’s been a rough couple of months. A heavy semester turned into a work-from-home marathon and that was followed by a shorter semester that was online from end-to-end. It was grueling. I’ve never been all that interested in teaching online; the investment of time seemed too extreme and I was not wrong. Still, there were pleasant surprises. We found out on a Friday that we’d have to start teaching on-line and I was able to figure out a lot of stuff over the weekend working with my colleagues in the mathematics department. Calculus II went on-line that Monday and although it wasn’t perfect, we barely missed a beat.
There’s still a lot to learn. My Science Fiction class in particular really drove home how much I depend on cues from students in the classroom. But it was still a rewarding albeit different experience from what I was used to. Having gone through the experience I’d be willing to try it again although hopefully not in such impromptu circumstances. It also has me pondering the possibility of doing parts of this blog with a “v” in front of it.
But I finally seem to be able to carve out some time for this. Later today, we’ll have the first in a series of posts on state flags in honor of flag day. There’s a post mortem on Mad Magazine in the works and I need to get back to these comics that seemed like they would be fun to write about.
I purchased these 60 years old to the month from their cover date but in August it will be 61 years from when they hit the stands. That should give you an idea of how long some of these things need to ruminate.
So, there’s more to come. Please stay tuned!
A couple of weeks ago I covered factoring in my College Algebra class, which we’ll mainly use for working with quadratic equations. This always makes me think about a method for factoring quadratics that I’d never seen before moving back to New York state. Years ago, I looked it up on what I think was an NYS Department of Education website which called it the “AC method” for factoring quadratics. When I tried to find this again, I only found it on YouTube called “slide and divide” and the “Berry Method.” Both of those names seem unnecessarily arcane to me, but for this post, I’ll call this process “slide and divide.” The references to an “AC Method” that I say today were actually talking about the “grouping number method” which we’ll come back to.
I hope that the fact that it’s become difficult to find references to what I think of as the AC Method is evidence that it’s going away. It needs to.
When I was in high school, I remember that factoring quadratics that looked like $ax^2 +bx + c$ was pretty straight forward when $a = 1$, but when $a$ was different from $1$ it got trickier. The only tool we were taught was trial and error, but you need to develop some intuition to use trial and error efficiently. Both the grouping number method and slide and divide attempt to give students a systematic process that builds on the experience of the $a = 1$ case to help them with the $a \neq 1$ case. If you’re a mathematician, and you’ve seen the slide and divide approach, you probably found it horrifying. I know I did. It goes something like this: say you want to factor the following polynomial.
$6 x^2 - x - 12$
You begin by replacing the leading coefficient with $1$ while replacing the constant term with the product of itself and the leading coefficient.
$x^2 - x - 72$
Why do we do this? I’m not sure it’s ever made clear. But this is a lot easier to factor since the leading coefficient is one. We just need to find two factors of $-72$ that add up to the coefficient of the middle term, $-1$, namely $-9$ and $8$ . These become the constant terms of the factors
$(x-9)(x+8)$
We then replace each $x$ with $6x$
$(6x-9)(6x+8)$
…and then remove the common factor from each binomial.
$(2x-3)(3x+4)$
With that, our quadratic polynomial is factored.
The single redeeming feature of this process is that it actually works, but think about these steps. Where did the $6$ in front of the $x^2$ go? Why does it make sense to multiply it into the $-12$? For that matter, why do the $6x$s magically reappear? And why is it okay to just cancel the common factors from the penultimate step? If I thought about it for a moment, I bet I’d actually have even more questions.
I’ve had a few students who wanted to use this method to factor, but none of them was ever able to explain why it works. That’s the ultimate problem with this method. It’s a list of meaningless steps that, once forgotten, will be gone forever because it doesn’t attach to any understanding. If you teach mathematics, it’s incumbent upon you to make your material meaningful to your students. Some people naively believe that this means you have to teach mathematics in a context that is directly relevant to the students’ interest, but in my opinion, that’s not true. You need to help them understand what’s happening and why it’s being done. Each step should have a clear and understandable rationale that helps to drive the process forward. Slide and divide does none of these things.
This process is not irredeemable, however. It is possible to work slide and divide in a way that gives it these characteristics. Let’s go through this in our original example.
$6 x^2 - x - 12$
The idea behind this method is that, if we can rewrite the polynomial with a leading coefficient of one, it will be easier to factor. To do this we need the leading term to be a perfect square. We can make that happen by multiplying the polynomial by $6$. Because we don’t want to change the polynomial, we’ll divide by $6$ at the same time.
$= \frac{36 x^2 - 6x - 72}{6}$
A simple substitution will give us a leading coefficient of $1$. We define $u = 6x$.
$= \frac{u^2 - u - 72}{6}$
Just like we did above, to factor our numerator we need two factors of $-72$ that add up to $-1$, that’s still $-9$ and $8$ .
$= \frac{(u - 9)(u+8)}{6}$
To get back to expressions that involve $x$s, we undo the substitution. That makes it clear why the $6x$s return.
$= \frac{(6x - 9)(6x+8)}{6}$
And now, we’d like to get rid of the $6$s since we don’t need them any longer. Where did the $6$ in the numerator go? It’s inside the common factors of the two binomials. We can factor those out…
$= \frac{3(2x - 3)2(3x+4)}{6}$
…and they clearly cancel with the $6$ in the denominator. That leaves us with our answer.
$= (2x - 3)(3x+4)$
That’s pretty cumbersome and so slide and divide leads you to a no-win-scenario. The quick way doesn’t engender understanding while the more rigorous approach isn’t quick.
So, what should we teach instead? The grouping number method that I mentioned earlier. To set it up, let’s think about what happens in the following polynomial multiplication.
$(4x-1)(2x-3)$
$=2x(4x-1)-3(4x-1)$
$=8x^2 -2x -12x +3$
$=8x^2 -14x +3$
Normally, you’d just skip the second step, but it’s important to realize that what you’re really doing is using the distributive property twice. Additionally, let’s think about the -12 and the -2. It’s clear that they add up to the final coefficient of x, but it’s also true that their product (-12)(-2) = 24 is the same as the product of the leading coefficient and the constant term, (8)(3). Notice this has to be true. In either case, it’s the product of 4, -1, 2 and -3, merely in a different order.
So, how do we go in the other direction? That is, how do we factor instead of multiply? We’ll demonstrate the grouping number method on the polynomial $8x^2 - 14x + 3$.
The grouping number of a quadratic polynomial is the leading coefficient multiplied by the constant term.
So, in this case, it would be $8 \times 3 = 24$.
Next, you want to look for two factors of your grouping number that add up to the coefficient of $x$. In this case, $(-12)(-2) = 24$ and $(-12)+(-2) = -14$. We use these two numbers to rewrite the middle term to get the following.
$8x^2 - 14x + 3 = 8x^2 - 2x - 12x + 3$.
In other words, to rewrite the polynomial in this way, you find two factors of your grouping number that add up to the coefficient of $x$ and use those to break down the middle term.
Why rewrite the polynomial in this way? Because it sets us up perfectly to factor by grouping.
Factoring by grouping is a technique for factoring a polynomial with four terms. In a nutshell, to factor by grouping, you remove the greatest common factor from the first two terms, then remove the greatest common factor from the last two terms. If the resulting binomial factors are the same you can factor this out to get the product of two binomials. Notice, that’s using the distributive property twice, just in the opposite direction as before.
Therefore, to finish factoring the polynomial we can factor by grouping.
$8x^2 - 14x + 3$
$= 8x^2 - 2x - 12x + 3$
$= 2x(4x - 1) - 3(4x - 1)$
$= (4x - 1)(2x - 3)$
Notice this exactly reverses the steps of the multiplication with which we started.
In short, as long as you have no common factors, the quadratic polynomials where the grouping number can be factored into two numbers that add up to the middle coefficient are exactly the ones that can be factored by grouping. All the others are prime.
|
```Algebra 1
Ch 2.6 – The Distributive Property
Objective
Students will use the distributive property
Before we begin…
The distributive property is a key algebraic
concept that looks something like this:
3(x + 4)
We will work with this property and its many
forms throughout this course…
It is expected that you are able to recognize
and know how to work with this property…
If you cannot recognize and work with the
distributive property you will not be
successful in this course…
The Distributive Property
The distributive property states:
To multiply a number by a sum or difference,
multiply each number inside the parentheses
by the number outside the parentheses
The distributive property can be used with
multiplication and addition or multiplication and
subtraction
Let’s see what it looks like…
Example 1
5(3 + 2)
15 + 10 = 25
Proof: 5(3+2) = 5(5) = 25
Algebraic Expressions
The distributive property can be used to
re-write algebraic expressions.
Use the same process…multiply what’s
on the outside of the parenthesis by
each term within the parenthesis
Let’s see what that looks like…
Example 2
3(x + 1)
3x + 3
Note: In this instance 3x and 3 are not like terms.
Therefore, you cannot combine them…so the expression
is simplified to just 3x + 3 – more on this later in the
lesson…
Distributive Property
There are 2 ways that you can see the
distributive property
With the multiplier on the left of the parenthesis
With the multiplier on the right of the
parenthesis
Example:
5(2 + 3)
OR
(b + 3)5
In either event you multiply what’s on the
out side of the parenthesis with EACH
term inside the parenthesis
Comments
Again…The distributive property is a
key algebraic concept…make no
mistake about it…you are REQUIRED
to be able to recognize and work with
the distributive property if you are to
pass Algebra 1!
Common Errors
The most common error that students
make when working the distributive
property is that they only multiply what
on the outside of the parenthesis with
the first term within the parenthesis
The other common error is that students
get the signs wrong…I do not give
partial credit for incorrect signs!
Example - Common Error
3(x - 1)
3x - 1
THIS IS INCORRECT!
Combining Like Terms
In this course you will be expected to
simplify expressions by combining like
terms…
In order to do that you have to be
familiar with the vocabulary and know
the definition of combining like terms…
Let’s take a look at that…
Vocabulary
Term – is the product of a number and a
variable. (product means to multiply)
Examples:
3x
3x2
-x
-xy2
Three times x
Three times x squared
Negative 1 times x
Negative 1 times x times y
squared
Vocabulary
Coefficient – the coefficient of a term is the
number in front of the variable.
If there is no number then the coefficient is
positive 1.
If there is no number and the variable is
negative then the coefficient is -1.
Examples:
-3x
x
-y
5y2
-3 is the coefficient
1 is the coefficient
-1 is the coefficient
5 is the coefficient
Vocabulary
Like terms – are terms that have the same variable
and exponent. They can be combined by adding or
subtracting.
Examples:
5x + 3x
same variable raised to the same power.
They can be combined by adding to get 8x
5x2 – 3x2
same variable raised to the same power. They
can be combined by subtracting to get 2x2
5x + 3y
Different variables raised to the same power –
they cannot be combined
5x2 – 3x4
Same variable raised to different powers – they
cannot be combined
Vocabulary
Constants – a number with no variable is
called a constant. Constant terms can be
combined by adding or subtracting.
Examples:
5x + 3 - 2
The constant terms are +3 and – 2.
They can be combined to get 5x + 1
- 7 + 6y - 2
The constant terms are – 7 and – 2.
They can be combined to get 6y – 9
Simplified Expressions
An expression is considered simplified if it has
no grouping symbols and all the like terms have
been combined
Example:
-x2 + 5x - 4 - 3x + 2
- x2 cannot be combined with anything because there is no other squared
term
+ 5x and – 3x can be combined because they have the same variable and
exponent to get +2x
- 4 and + 2 are constant terms and can be combined to get – 2.
The simplified expression is:
-x2 + 2x - 2
Comments
On the next couple of slides are some practice
problems…The answers are on the last slide…
Do the practice and then check your answers…If you
do not get the same answer you must question what
you did…go back and problem solve to find the
error…
If you cannot find the error bring your work to me and
I will help…
Your Turn
Use the distributive property to rewrite the
expression without parenthesis
1. 3(x + 4)
2. - (y – 9)
3. x(x + 1)
4. 2(3x – 1)
5. (2x – 4)(-3)
Your Turn
Simplify by combining like terms
6. 15x + (-4x)
7. 5 – x + 2
8. 4 + a + a
9. 8b + 5 – 3b
10. 9x3 – 2 – 4x3
Your Turn
Apply the distributive property then simplify
by combining like terms
11. (3x + 1)(-2) + y
12. 4(2 – a) – a
13. - 4(y + 2) – 6y
14. -x3 + 2x(x – x2)
15. 4w2 – w(2w – 3)
Your Turn Solutions
1. 3x + 12
8. 4 + 2a
2. -y + 9
9. 5b + 5
3. x2 + x
10. 5x3 – 2
4. 6x – 2
11. -6x – 2 + y
5. -6x + 12
12. 8 – 5a
6. 11x
13. -10y – 8
7. 7 - x
14. -x3 + x2
15. 2w2 + 3w
Summary
A key tool in making learning effective is being
able to summarize what you learned in a
lesson in your own words…
In this lesson we talked about the distributive
property… Therefore, in your own words
summarize this lesson…be sure to include key
concepts that the lesson covered as well as
any points that are still not clear to you…
I will give you credit for doing this
lesson…please see the next slide…
Credit
I will add 25 points as an assignment grade for you working on
this lesson…
To receive the full 25 points you must do the following:
Have your name, date and period as well a lesson
number as a heading.
Do each of the your turn problems showing all work
Have a 1 paragraph summary of the lesson in your own
words
Please be advised – I will not give any credit for work submitted:
Without a complete heading
Without showing work for the your turn problems
Without a summary in your own words…
```
|
# The appropriate direction of the graph for y = f ( x ) + 3 by shifting 3 units.
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
#### Solutions
Chapter 2.5, Problem 1E
(a)
To determine
## To fill: The appropriate direction of the graph for y=f(x)+3 by shifting 3 units.
Expert Solution
The graph of y=f(x)+3 is obtained from the graph of y=f(x) by shifting up 3 units.
### Explanation of Solution
Let the graph of f(x) is given by the equation,
y=f(x)
Add a constant c in f(x) ,
y=f(x)+c (1)
Add a constant to a function shifts its graph horizontally.
If c is positive then the graph will shifts upward by c unit at all points.
If c is negative then the graph will shifts downward by c unit at all points.
The given equation is,
y=f(x)+3 (2)
Compare equation (1) and (2) and obtain c=3 .
Since the value of c is 3, therefore the graph of f(x)+3 shifts upward by 3 unit.
Hence, the graph of y=f(x)+3 is obtained from the graph of y=f(x) by shifting up 3 units.
(b)
To determine
### To fill: The appropriate direction of the graph for y=f(x+3) by shifting 3 units .
Expert Solution
The graph of y=f(x+3) is obtained from the graph of y=f(x) by shifting left 3 units.
### Explanation of Solution
Let the graph of f(x) is given by the equation,
y=f(x)
Add a constant c in f(x) ,
y=f(x+c) (3)
Add a constant to a function shifts its graph horizontally.
If c is positive then the graph will shift to the left by c unit at all points.
If c is negative then the graph will shift to the right by c unit at all points.
The given equation is,
y=f(x+3) (4)
Compare equation (3) and (4) and obtain c=3 .
Since the value of c is 3, therefore the graph of f(x+3) shifts to the left by 3 unit.
Hence, the graph of y=f(x+3) is obtained from the graph of y=f(x) by shifting left 3 units.
### Have a homework question?
Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!
|
Home Lessons Calculators Worksheets Resources Feedback Algebra Tutors
# Calculator Output
```Simplifying
5j2 + -20 = 0
Reorder the terms:
-20 + 5j2 = 0
Solving
-20 + 5j2 = 0
Solving for variable 'j'.
Move all terms containing j to the left, all other terms to the right.
Add '20' to each side of the equation.
-20 + 20 + 5j2 = 0 + 20
Combine like terms: -20 + 20 = 0
0 + 5j2 = 0 + 20
5j2 = 0 + 20
Combine like terms: 0 + 20 = 20
5j2 = 20
Divide each side by '5'.
j2 = 4
Simplifying
j2 = 4
Reorder the terms:
-4 + j2 = 4 + -4
Combine like terms: 4 + -4 = 0
-4 + j2 = 0
Factor a difference between two squares.
(2 + j)(-2 + j) = 0
Subproblem 1Set the factor '(2 + j)' equal to zero and attempt to solve:
Simplifying
2 + j = 0
Solving
2 + j = 0
Move all terms containing j to the left, all other terms to the right.
Add '-2' to each side of the equation.
2 + -2 + j = 0 + -2
Combine like terms: 2 + -2 = 0
0 + j = 0 + -2
j = 0 + -2
Combine like terms: 0 + -2 = -2
j = -2
Simplifying
j = -2
Subproblem 2Set the factor '(-2 + j)' equal to zero and attempt to solve:
Simplifying
-2 + j = 0
Solving
-2 + j = 0
Move all terms containing j to the left, all other terms to the right.
Add '2' to each side of the equation.
-2 + 2 + j = 0 + 2
Combine like terms: -2 + 2 = 0
0 + j = 0 + 2
j = 0 + 2
Combine like terms: 0 + 2 = 2
j = 2
Simplifying
j = 2Solutionj = {-2, 2}```
Processing time: 1 ms. 76495489 equations since February 08, 2004. Disclaimer
# Equation Factoring Calculator
Equation: Variable: a A b B c C d D e E f F g G h H i I j J k K l L m M n N o O p P q Q r R s S t T u U v V w W x X y Y z Z AUTO Simplify Only
Hint: Selecting "AUTO" in the variable box will make the calculator automatically solve for the first variable it sees.
Home Lessons Calculators Worksheets Resources Feedback Algebra Tutors © Copyright 2001-2011 info @ algebrahelp.com
|
2
Q:
# A card is drawn from a pack of 52 cards. The probability of getting a queen of club or aking of heart is:
A) 2/13 B) 1/13 C) 1/26 D) 1/52
Explanation:
Here, n(S) = 52.
Let E = event of getting a queen of club or a king of heart.
Then, n(E) = 2.
P(E) =n(E)/n(S)=2/52=1/26.
Q:
When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 12?
A) 35/36 B) 17/36 C) 15/36 D) 1/36
Explanation:
When two dice are thrown simultaneously, the probability is n(S) = 6x6 = 36
Required, the sum of the two numbers that turn up is less than 12
That can be done as n(E)
= { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5) }
= 35
Hence, required probability = n(E)/n(S) = 35/36.
3 621
Q:
In a purse there are 30 coins, twenty one-rupee and remaining 50-paise coins. Eleven coins are picked simultaneously at random and are placed in a box. If a coin is now picked from the box, find the probability of it being a rupee coin?
A) 4/7 B) 2/3 C) 1/2 D) 5/6
Explanation:
Total coins 30
In that,
1 rupee coins 20
50 paise coins 10
Probability of total 1 rupee coins = 20C11
Probability that 11 coins are picked = 30C11
Required probability of a coin now picked from the box is 1 rupee = 20C11/30C11 = 2/3.
7 1002
Q:
In a box, there are 9 blue, 6 white and some black stones. A stone is randomly selected and the probability that the stone is black is ¼. Find the total number of stones in the box?
A) 15 B) 18 C) 20 D) 24
Explanation:
We know that, Total probability = 1
Given probability of black stones = 1/4
=> Probability of blue and white stones = 1 - 1/4 = 3/4
But, given blue + white stones = 9 + 6 = 15
Hence,
3/4 ----- 15
1 ----- ?
=> 15 x 4/3 = 20.
Hence, total number of stones in the box = 20.
10 1062
Q:
What is the probability of an impossible event?
A) 0 B) -1 C) 0.1 D) 1
Explanation:
The probability of an impossible event is 0.
The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.
The probability of a certain event is 1.
9 1484
Q:
In a box, there are four marbles of white color and five marbles of black color. Two marbles are chosen randomly. What is the probability that both are of the same color?
A) 2/9 B) 5/9 C) 4/9 D) 0
Explanation:
Number of white marbles = 4
Number of Black marbles = 5
Total number of marbles = 9
Number of ways, two marbles picked randomly = 9C2
Now, the required probability of picked marbles are to be of same color = 4C2/9C2 + 5C2/9C2
= 1/6 + 5/18
= 4/9.
9 1807
Q:
A bag contains 3 red balls, 5 yellow balls and 7 pink balls. If one ball is drawn at random from the bag, what is the probability that it is either pink or red?
A) 2/3 B) 1/8 C) 3/8 D) 3/4
Explanation:
Given number of balls = 3 + 5 + 7 = 15
One ball is drawn randomly = 15C1
probability that it is either pink or red =
14 1670
Q:
Two letters are randomly chosen from the word TIME. Find the probability that the letters are T and M?
A) 1/4 B) 1/6 C) 1/8 D) 4
Explanation:
Required probability is given by P(E) =
18 2376
Q:
14 persons are seated around a circular table. Find the probability that 3 particular persons always seated together.
A) 11/379 B) 21/628 C) 24/625 D) 26/247
Explanation:
Total no of ways = (14 – 1)! = 13!
Number of favorable ways = (12 – 1)! = 11!
So, required probability = $11!×3!13!$ = $39916800×66227020800$ = $24625$
|
# Difference of inequations
I'm currently studying for a class where the teacher's notes were given, but there are many errors here and there so I need to make sure that everything on it is correct. I'm given the following system of inequations:
$$\left\{\begin{matrix} 0 \leq t'_1 + t_2 \leq 10 \;\;\;\;\; (a) \\ 5\leq t_2 \leq 15 \;\;\;\;\;\;\;\;\;\;\;\; (b) \\ 12\leq t'_3+t_2\leq 22 \;\;\; (c) \end{matrix}\right.$$
Now I need to eliminate t2 form (a) and (c) so on the notes here is the resulting system:
$$\left\{\begin{matrix}-5 \leq t'_1 \leq 5 \;\;\;(a+(-b)) \\2\leq t'_3 \leq 17\;\;\; (c+(-b)) \\5\leq t_2\leq15 \;\;\;\;\;\;\;\;\;\;\;\;\;\;(b) \end{matrix}\right.$$
Now the question is: can someone guide me through the process of eliminating t2 ? Are the resulting inequations actually correct ?
To eliminate $t_2$, first we multiply the second inequation by $-1$ and then we add it to the first one:
$$5 \leq t_2 \leq 15 \implies -5 \geq -t_2 \geq -15$$ We need to flip the inequation signs since we're multiplying by a negative number.
Now, we have a problem here as we can't add 2 inequations if the inequation signs are opposite; we can only subtract them. Since subtracting them is equivalent to just adding them before multiplying by $-1$, we see that either way this won't work in eliminating $t_2$. The notes are definitely wrong; they forgot to flip the inequation signs. (Don't even ask me about the havoc they wreaked on the last inequation; for starters, $(b)$ isn't even the same inequation in the top than in the bottom!)
|
# Intermediate Algebra Questions With Solutions and Explanations - sample 4
Full explanations of solutions for intermediate algebra questions in sample 4 are presented. (True or False) x 2 and 2 x are like terms. Solution The statement "x 2 and 2 x are like terms" is FALSE because the two terms do not have the same power of x. (True or False) x-3 and -3x are unlike terms. Solution The statement "x-3 and -3x are unlike terms" is TRUE because the two terms do not have the same power of x. (True or False) 1 / (x - 9) = 0 for x = 9. Solution Substitute x by 9 in the expression 1 / (x - 9). 1 / (x - 9) = 1 / (9 - 9) = 1 / 0 = undefined The statement "1 / (x - 9) = 0 for x = 9" is FALSE. (True or False) The set of ordered pairs {(0,0),(2,0),(3,0),(10,0)} represents a function. Solution All values of the x coordinates are different and therefore the set of ordered pairs represents a function. The statement "The set of ordered pairs {(0,0),(2,0),(3,0),(10,0)} represents a function" is TRUE. (True or False) |a - b| = b - a if b - a < 0. Solution Recall that if x > 0, then | x | = x But if b - a < 0, then a - b > 0 and |a - b| = a - b The statement "|a - b| = b - a if b - a < 0" is FALSE. (True or False) |x2 + 1| = x2 + 1. Solution Recall that if x > 0, then | x | = x Since x2 + 1 is positive for all values of real x, then | x2 + 1 | = x2 + 1 The statement |x2 + 1| = x2 + 1" is TRUE. (True or False) √(x - 5) 2 = x - 5. Solution For the above statement to be true, it must be true for all values of x for which the expression are defined. Let x = - 4 and evaluate the right and left hand expressions. Left: √(x - 5) 2 = √(-4 - 5) 2 = √81 = 9 Right: x - 5 = - 4 - 5 = - 9 The above statement is not true for x = - 4 and therefore the statement "√(x - 5) 2 = x - 5" is FALSE. (True or False) (x - 2)(x + 2) = x2 - 4x - 4. Solution Expand (x - 2)(x + 2) (x - 2)(x + 2) = x2 + 2x - 2x - 4 Group like terms (x - 2)(x + 2) = x2 - 4 The statement "(x - 2)(x + 2) = x2 - 4x - 4" is FALSE. (True or False) sqrt(x + 9) = sqrt(x) + sqrt(9), for all x real. Solution For the above statement to be true, it must be true for all values of x for which the expression are defined. Let x = 16 and evaluate the right and left hand expressions. Left: sqrt(x + 9) = sqrt(16 + 9) = sqrt(25) = 5 Right: sqrt(x) + sqrt(9) = sqrt(16) + sqrt(9) = 4 + 3 = 7 The above statement is not true for x = 16 and therefore the statement "sqrt(x + 9) = sqrt(x) + sqrt(9), for all x real" is FALSE. (True or False) |x - 3| = |x| + |3|, for all x real and negative. Solution Start with |x - 3| |x - 3| = |3 - x| = |3 + (- x)| The absolute value of the sum of two positive numbers is equal to the sum of the numbers. Since x is negative - x is positive and hence |3 + (- x)| = |3| + |-x| = |3| + |x| The statement "|x - 3| = |x| + |3|, for all x real and negative" is TRUE. (True or False) (x + 2)3 = x3 + 23, for all x real. (True or False) If k = 4, then the equation x2 - kx = -4 has one solution only. (True or False) The discriminant of the equation: 2x2 - 4x + 9 = 0 is negative. (True or False) The degree of the polynomial P(x) = (x - 2)(-x + 3)(x - 4) is equal to -3. (True or False) The distance between the points (0 , 0) and (1 , 1) in a rectangular system of axes is equal to 1. (True or False) The slope of the line 2x + 3y = -2 is negative. (True or False) The relation 2y + x2 = 2 represents y as a function of x. (True or False) The relation 2y + x2 = 2 represents x as a function of y. (True or False) The relation |x| = |y| DOES NOT represent x as a function of y. (True or False) The relation |x| = |y| DOES NOT represent y as a function of x.
|
+0
# Algebra
0
1
4
+4
How many points of the form $$(x,y)$$, where both coordinates are positive integers, lie below the graph of the hyperbola $$xy=16$$?
Jul 24, 2024
#1
+593
0
I'm getting an answer of 42. I will post the solution later.
Jul 24, 2024
edited by AnswerscorrectIy Jul 24, 2024
#3
+1331
0
To determine the number of points of the form $$(x,y)$$, where both coordinates are positive integers, that lie below the graph of the hyperbola $$xy = 16$$, we need to find the integer pairs $$(x, y)$$ such that $$xy < 16$$.
### Step-by-Step Solution:
1. **Identify the Range for $$x$$**:
- For each $$x$$, we need $$y$$ to be an integer such that $$xy < 16$$.
- Since $$x$$ must be a positive integer, consider possible values of $$x$$ starting from 1 up to the point where $$x \cdot 1 = 16$$, so $$x$$ ranges from 1 to 15.
2. **Count $$y$$ Values for Each $$x$$**:
- For each $$x$$, find the largest integer $$y$$ such that $$y < \frac{16}{x}$$.
Here’s how this works for each $$x$$ from 1 to 15:
- $$x = 1$$: $$xy < 16 \implies y < \frac{16}{1} = 16$$. So, $$y$$ can be 1 to 15 (15 values).
- $$x = 2$$: $$xy < 16 \implies y < \frac{16}{2} = 8$$. So, $$y$$ can be 1 to 7 (7 values).
- $$x = 3$$: $$xy < 16 \implies y < \frac{16}{3} \approx 5.33$$. So, $$y$$ can be 1 to 5 (5 values).
- $$x = 4$$: $$xy < 16 \implies y < \frac{16}{4} = 4$$. So, $$y$$ can be 1 to 3 (3 values).
- $$x = 5$$: $$xy < 16 \implies y < \frac{16}{5} = 3.2$$. So, $$y$$ can be 1 to 3 (3 values).
- $$x = 6$$: $$xy < 16 \implies y < \frac{16}{6} \approx 2.67$$. So, $$y$$ can be 1 to 2 (2 values).
- $$x = 7$$: $$xy < 16 \implies y < \frac{16}{7} \approx 2.29$$. So, $$y$$ can be 1 to 2 (2 values).
- $$x = 8$$: $$xy < 16 \implies y < \frac{16}{8} = 2$$. So, $$y$$ can be 1 (1 value).
- $$x = 9$$ to $$x = 15$$: For these values, $$y < \frac{16}{x}$$ will always be less than 2, so $$y$$ can only be 1 (1 value each).
3. **Summarize the Counts**:
\begin{aligned} &15 \text{ values for } x = 1, \\ &7 \text{ values for } x = 2, \\ &5 \text{ values for } x = 3, \\ &3 \text{ values for } x = 4, \\ &3 \text{ values for } x = 5, \\ &2 \text{ values for } x = 6, \\ &2 \text{ values for } x = 7, \\ &1 \text{ value for } x = 8, \\ &1 \text{ value each for } x = 9 \text{ to } 15. \end{aligned}
$15 + 7 + 5 + 3 + 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 44$
Therefore, the number of points $$(x, y)$$, where both coordinates are positive integers, that lie below the graph of the hyperbola $$xy = 16$$ is:
$\boxed{44}$
Jul 24, 2024
#4
+2633
0
To determine how many points $$(x, y)$$ where both $$x$$ and $$y$$ are positive integers lie below the hyperbola $$xy = 16$$, we need to find the integer pairs $$(x, y)$$ such that $$xy < 16$$.
### Step-by-Step Solution:
1. **Consider values of $$x$$ and find corresponding $$y$$ values**:
For each $$x$$, $$y$$ must satisfy $$1 \leq y < \frac{16}{x}$$.
2. **Calculate pairs for each $$x$$**:
- $$x = 1$$:
$xy < 16 \implies y < \frac{16}{1} \implies y < 16 \implies y = 1, 2, 3, \ldots, 15 \quad (\text{15 values})$
- $$x= 2$$:
$xy < 16 \implies y < \frac{16}{2} \implies y < 8 \implies y = 1, 2, 3, \ldots, 7 \quad (\text{7 values})$
- $$x = 3$$:
$xy < 16 \implies y < \frac{16}{3} \implies y < 5.33 \implies y = 1, 2, 3, 4, 5 \quad (\text{5 values})$
- $$x = 4$$:
$xy < 16 \implies y < \frac{16}{4} \implies y < 4 \implies y = 1, 2, 3 \quad (\text{3 values})$
- $$x = 5$$:
$xy < 16 \implies y < \frac{16}{5} \implies y < 3.2 \implies y = 1, 2, 3 \quad (\text{3 values})$
- $$x = 6$$:
$xy < 16 \implies y < \frac{16}{6} \implies y < 2.67 \implies y = 1, 2 \quad (\text{2 values})$
- $$x = 7$$:
$xy < 16 \implies y < \frac{16}{7} \implies y < 2.29 \implies y = 1, 2 \quad (\text{2 values})$
- $$x = 8$$:
$xy < 16 \implies y < \frac{16}{8} \implies y < 2 \implies y = 1 \quad (\text{1 value})$
- $$x = 9$$ and higher:
$xy < 16 \implies y < \frac{16}{x} \implies y < \frac{16}{x} \implies y = 1 \quad (\text{1 value if } x \leq 15 \text{ else no values})$
3. **Count the total number of pairs**:
Summing all the valid $$y$$ values for each $$x$$:
$15 + 7 + 5 + 3 + 3 + 2 + 2 + 1 + 1 = 39$
Therefore, there are $$\boxed{39}$$ points of the form $$(x, y)$$ where both coordinates are positive integers and lie below the hyperbola $$xy = 16$$.
Jul 24, 2024
|
# Create a Polynomial with Given Degree and Real Zeros.
Welcome to Warren Institute! In this article, we will explore the fascinating world of Mathematics education. Today, our focus is on forming a polynomial with given real zeros and degree. This concept is crucial in understanding the behavior of polynomial functions. By analyzing the given real zeros and degree, we can construct a polynomial that accurately represents the information. Join us as we delve into the techniques and strategies involved in creating these polynomials. Let's unlock the mysteries of Mathematics education together! Stay tuned for more exciting content from Warren Institute.
## Finding the Real Zeros of a Polynomial
In this section, we will explore methods for finding the real zeros of a polynomial when its degree and some of its zeros are given. We will discuss techniques such as synthetic division, the Rational Root Theorem, and the Factor Theorem to simplify the process of finding these zeros.
## Constructing a Polynomial with Given Zeros
Here, we will learn how to construct a polynomial when its real zeros and degree are known. By using the concept of factoring, we can build a polynomial that satisfies the given conditions. We will also discuss the relationship between the zeros and factors of a polynomial.
## Applying the Fundamental Theorem of Algebra
The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros, counting multiplicities. In this section, we will explore how this theorem can help us determine the possible number of real zeros a polynomial can have based on its degree.
## Solving Real-World Problems with Polynomial Zeros
In this final section, we will apply our knowledge of finding polynomial zeros to solve real-world problems. By modeling these problems as polynomials, we can use the techniques discussed earlier to find the solutions. We will explore various examples that demonstrate the practical applications of polynomial zero-finding in Mathematics education.
## frequently asked questions
### How do I form a polynomial when the real zeros and degree are given?
To form a polynomial when the real zeros and degree are given, you can use the factor theorem. Start by finding the factors of the polynomial using the given real zeros. Then, multiply these factors together to obtain the polynomial expression.
### What steps are involved in forming a polynomial with given real zeros and degree?
The steps involved in forming a polynomial with given real zeros and degree are as follows:
1. Determine the given real zeros of the polynomial.
2. Use the given zeros to write the corresponding linear factors in the form (x - zero).
3. Multiply all the linear factors together to obtain the polynomial in factored form.
4. If necessary, simplify the polynomial by multiplying out the factors and combining like terms.
5. Verify that the degree of the polynomial matches the given degree.
Note: The given real zeros are the values of x for which the polynomial evaluates to zero.
### Can you provide an example of forming a polynomial using real zeros and degree?
Yes, I can provide an example. Let's say we have a polynomial of degree 3 and we know that its real zeros are x = -2, x = 1, and x = 3. To form the polynomial, we start by using the zero x = -2. This means that the polynomial has a factor of (x + 2). Similarly, we use the other two zeros to get factors of (x - 1) and (x - 3). Multiplying these factors together, we get the polynomial f(x) = (x + 2)(x - 1)(x - 3), which is a polynomial of degree 3 with real zeros at x = -2, x = 1, and x = 3.
### Are there any specific rules or guidelines to follow when forming a polynomial with real zeros and degree?
Yes, there are specific rules and guidelines to follow when forming a polynomial with real zeros and degree. One important rule is that the polynomial's degree must be equal to the number of real zeros. Additionally, the polynomial should have factors that correspond to each real zero, with each factor being raised to the power equal to its multiplicity.
### Are there any strategies or techniques that can be used to form a polynomial based on given real zeros and degree?
Yes, there are strategies and techniques to form a polynomial based on given real zeros and degree. One approach is to use the fact that if a real number 'a' is a zero of a polynomial, then (x - a) is a factor of the polynomial. By identifying all the given real zeros, we can write the polynomial as a product of linear factors. The degree of the polynomial will determine the number of real zeros and factors involved.
In conclusion, the ability to form a polynomial with given real zeros and degree is an essential skill in mathematics education. By understanding the concept of zeros and their corresponding factors, students are able to construct polynomials that accurately represent real-world situations or mathematical problems. This skill not only enhances their problem-solving abilities but also fosters a deeper understanding of algebraic concepts. With practice and guidance, students can master the art of forming polynomials and apply this knowledge to various mathematical contexts. By equipping students with this fundamental skill, educators can empower them to excel in their mathematical journey.
See also Exploring numbers from 31 to 40 in mathematics education
If you want to know other articles similar to Create a Polynomial with Given Degree and Real Zeros. you can visit the category General Education.
Michaell Miller
Michael Miller is a passionate blog writer and advanced mathematics teacher with a deep understanding of mathematical physics. With years of teaching experience, Michael combines his love of mathematics with an exceptional ability to communicate complex concepts in an accessible way. His blog posts offer a unique and enriching perspective on mathematical and physical topics, making learning fascinating and understandable for all.
Go up
|
# Practice Page
## Not the quiz
You must be logged in to take the quiz.
### 2002-E: Trigonometry 2: Trigonometric Functions
• Topic Cluster: Pure Mathematics
• Topic: Geometry
• Objective: Solve for unknown sides of a triangle using sine, cosine, and tangent, including problems in which the unknown side is in the denominator
• Content:
• Level: 2
We typically label angles by the Greek letter $$\theta$$, pronounced "theta." Once we have singled out a particular angle, each side has a new name:
• A is the opposite of $$\theta$$.
• B is the adjacent of $$\theta$$.
• C is the hypotenuse.
• $$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$$
• $$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$$
• $$\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$$
Whenever we know what an angle is, we can use the three trigonometric functions: the sine, the cosine, and the tangent. To remember these three formulas, we use the silly word "SOHCAHTOA," which stands for: "Sine is opposite over hypotenuse; cosine is adjacent over hypotenuse; tangent is opposite over adjacent."
#### Problems
In each case, use one of the trigonometric functions to solve for the unknown side. For full credit, you must write the formula before using it.
#### Variable in Denominator
In some trigonometry problems, the unknown variable will be located in the denominator! In this case, the most straightforward method is to first multiply both sides of the equation by the unknown variable, then solve for it:
##### Example:
$$\require{enclose} 48 = \frac{76}{X} \\ 48 \cdot X = \frac{76}{\enclose{horizontalstrike}{X}} \cdot \enclose{horizontalstrike}{X} \\ 48 \cdot X = 76 \\ X = \frac{76}{48} = 1.583$$
Solve for $$x$$ in each of the problems below. Please give all answers to 4 significant figures of accuracy.
1. $$49 = \frac{65}{x}$$
2. $$66 = \frac{82}{x}$$
3. $$42 = \frac{25}{x}$$
#### Saying the Names Right!
• Whenever we use the sine function, we write the abbreviation $$\sin$$, but when reading this out loud you always say "sine!"
• Whenever we use the cosine function, we write the abbreviation $$\cos$$, but when reading this out loud you always say "cosine!"
• Whenever we use the tangent function, we write the abbreviation $$\tan$$, but when reading this out loud you always say "tangent!"
1. How do you read out loud the following equation: $$\sin 90^{\circ} = 1$$.
2. How do you read out loud the following equation: $$\cos 90^{\circ} = 0$$.
3. How do you read out loud the following equation: $$\tan 45^{\circ} = 1$$.
#### Extra Practice
(Not required to receive full credit for this assignment.)
|
# How do you solve a problem like this? It says divide. then write each quotient in simplest form 5 divided by 3 1/3
75,816 results
1. ## Math
Write the power of 10 to multiply the divisor by to make it a whole number. Then, write the equivalent problem and find the quotient 16.65 ÷ 0.37
2. ## math
A=1/2h (b1 + b2) Solve for b1 A=1/2h (b1 + b2) Solve for b1 I assume this is A = (1/2)(h)(b1 + b2); i.e., the h is in the numerator and not the denominator Multiply both sides by 2. 2A = (2*1/2)h(b1 + b2). 2*1/2 on the right side is 1. 2A = h(b1 + b2)
3. ## Math - Another question to check, sorry... =/
7. 72 is what percent of 480? Write and solve an equation to solve the problem. Step 1. First, I will write out the equation. part = percent x whole Step 2. Now, fill in the parts… 72 = p x 480 Step 3. I will simply divide 72 by 480, which will be 0.15.
4. ## Math
1.Divide.Write your answer in simplest form. (1 point) 2/3 divided by 1/4 a.2 1/2 b.2 2/3 c.2 3/8 d.2 2.Find the product. (1 point) 1/2*3/7 a.4/14 b.3/14 c.7/6 d.2/7 3.Divide.Write the quotient in simplest form. (1 point) 3/5 divided by 2/10 a.6/10 b.3/5
5. ## math
108 is 36% of what number? Write and solve a proportion to solve the problem. First write the proportion to find the missing value, which I will replace it with n. 108/n = 36/100 Next write the cross products. So, 36n = 108 x 100 = 10,800 divide each side
6. ## ONE math question! Check please, thank you!
6) 108 is 36% of what number? Write and solve a proportion to solve the problem. Step 1. First, I will write the proportion to find the missing value, which I will replace it with n. 108/n = 36/100 Step 2. Next, I will write the cross products. So, 36n =
7. ## Math
Write a story problem to represent 6 divide 1/5
8. ## Math
I have to find the quotient of 6 divided by 1/5 So I write 1/6 x 1/5 to solve?
9. ## math
1. Divide. Write your answer in simplest form. 2/3 / 1/4 2. Find the product. 1/2 x 3/7 3. Divide. Write the quotient in simplest form. 3/5 / 2/10 4. Divide. Write the quotient in simplest form. 3/4 / 4/5 plz help me
10. ## Algebra A
write a short story problem that can be solved in one step. Write an equation that represents the problem. Name the property used to solve the problem, and then solve the problem. PLEASE HELP WITH THIS ASAP!!!
11. ## Math
1. What are the steps to solve this problem? 0.000027 ÷ 0.000009 Requirements: • Explain how to use scientific notation to solve the problem. • Describe how to divide numbers written in scientific notation. • Give you final answer in simplest form.
12. ## Algebra
I think that the goal is to get rid of the 5th root in this case. I know that I have to make whatever is under or in the root -in this case the number is X equal some number to the fifth power. I am supposed to use the rule a^(m/n)= n (the n-root) (don't
13. ## Algebra
Solve each equation by finding square roots. If the equation has no real-number solution, write no solution. 4g^2=25 In the equation when i divide 25 by 4 why do i keep it as 25/4 threw out solving the rest of the problem? why not change it to 4??? I'm
divide 891 by 40 and write the quotient as a mixed number
15. ## Math
What are the steps to show the quotient in simplest form? Be sure to include various key words. 0.000027 divided by 0.000009 My answer 0.000027 divided by 0.000009=3 Definition of a quotient The answer after you divide one number by another. The qutient in
16. ## Math
1. Write the place name for the 4 in the following number: 3,624,786 Answer: Thousand 2. Round the following number to the nearest ten: 53 Answer: 5 3. Find the sum of the following problem: 2,168 + 305 Answer: 2473 4. Estimate the sum of the following
17. ## algebra
Use synthetic division to divide the polynomial 2x3 – 45x + 28 by x + 5, and write the quotient polynomial and the remainder. [Be careful – notice that there is no x2 term.]. Show work.
18. ## Math
Write the following as an equation and then solve: "The quotient of a number and ten is thirteen."
19. ## Math
Write the steps to solve the following problem. Explain your steps as you solve it. What are the steps to show the quotient in simplest form? Be sure to include various key words you learned in this unit. 0.000027÷0.000009
20. ## Math
Divide 52728 by the smallest number so that the quotient is a perfect cube. Also find the cube root of the quotient
21. ## Math
Solve the system of equations by substitution and explain all your steps in words: 3x + y = 12 x = y − 8 . Is this a good way to explain this problem? Step 1: Since X is given to us. We can plug it into the x-value in the first equation. x= y-8 into
22. ## Algebra
Use synthetic division to divide the polynomial 2x^3 – 12x – 5 by x + 4, Write the quotient polynomial and the remainder. [Be careful – notice that there is no x2 term.]. Show work.
23. ## Algebra A
write a short story problem that can be solved in one step. Write an equation that represents the problem. Name the property used to solve the problem, and then solve the problem. can anyone give me the problem and the name of the property? ASAP HELP
24. ## Math
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided. 108 is 36% of what number? Write and solve a proportion to solve the problem. Note: Enter your answer and show all the steps that you use to solve this
25. ## Math
Explain how estimating the quotient helps you place the first digit in the quotient of a division problem
26. ## algebra 2
Would some one look at these problems and see if I got them right Thanks Just let me if they are wrong so I can go back over them 1.Write in simplest form 48/56 A. 6/7 2.Multiply 8x^3/5 *10/x^3 A. 16/x^4 3.Divide 4/x^7 / 12/x^3 A. 1/3x^2 4. divide
27. ## Math
divide 891 by 40 and write the quotient as a mixed number
28. ## Math
First find an estimate of the quotient. then find the exact quotient. 5 1/2 divide 3 4/7
29. ## Algebra (check my answers)
Please check these answers. Multiply -7(2a2 -4) = -14a2 + 28 Multiply 0.5(a + b -5) = 0.5a + 0.5b - 2.5 Factor bx - by + bz = b(x - y + z) Combine like terms (simplify) 12x + 4 - 11y - 6x - 9 - y = 6x - 12y - 5 Evaluate the expression 3x2y when x = 3 and y
30. ## algebra
How do you write the expression for 5 less than the quotient of a number and 2. Here is my answer but the quotient part is confusing me. 5
31. ## Math
You are supposed to come up with an equation to solve this problem. I have figured out how to do it, just not what the equation is supposed to look like. Anyways, here is the word problem. The taxi fare in Gotham City is \$2.40 for the first .5 miles, and
32. ## Math
1. First, write your own real-world percent problem. 2. Next, solve a classmate’s problem incorrectly. Include one or two mistakes in your solution. 3. Finally, respond to a different classmate’s post by finding the error in his or her solution.
33. ## Algebra
Solve for s: h=(square root of 3)times s/2 and solve for h V= (pi)r squared h / 3 Solve for s: h=(square root of 3)times s/2 Multiply both sides by 2. 2h = (sqrt 3)*s*2/2 which cancels the 2 on the right. 2h = (sqrt 3)*s Now divide the right side by
34. ## Math
How do you solve a problem like this? It says divide. then write each quotient in simplest form 5 divided by 3 1/3
35. ## math, correction,plz
Can someone correct these for me..plz... Problem #1 simplify. ((p^5)^4)/(p^6) MY answer: p^14 Problem #2 Write using positive exponents only. b^-6 MY answer: (1)/(b^6) Problem#3 Write 1.81 X 10^-2 in standard notation My answer. 0.0181 Problem #4 Subtract
37. ## College Algebra
Divide. (9x^3-10+19+6x^4)/(-2x^2+3x-3) Write your answer in the form Q(x)+ R(x)/ -2x^2+3x-3, where Q(x) is the quotient and R(x) is the remainder.
38. ## math,help
can someone help me with this i am lost now i've been doing math for the past 8 hrs. Problem # 1 solve: 4(x-10)-3x=x-40 Problem # 2 solve for t: a = p + prt Problem#3 translate to an algebraic equation 7 less than the quotient of a number and 3 is twice
39. ## MATH
20. A certain computer can perform 105 calculations per second. How many calculations can it perform in 10 seconds 21. Write the steps to solve the following problem. Explain your steps as you solve it. What are the steps to show the quotient in simplest
40. ## math, algebra
Problem: A formula for a football player's rushing average r with a total of y yards rushed in n carries of the ball is r=y/n. Solve for n. I have no other information this is how its in the book how do i even solve for n if theres no numbers. You don't
41. ## Algebra 1 HELP PLZ!!!
Use the Substitution method to solve the system of equations. x + y = -4 x - y = 2 PLz HELP!! I'M NOT BOB BUT I CAN HELP ITS REALLY SIMPLE ALGEBRA X + Y = -4......EQN 1 X - Y = 2.....EQN 2 From EQN 2, let x = 2 + y Substitute that into EQN 1 x - y = -4
42. ## Math
Can someone help me find the technique or techniques used in this problem? 146 divided by 68. I thought that 14 and 7 are easy to divide, so I used 140 divided by 70, which is 2. The quotient is about 2.
43. ## math
estimate quotient and divide (how do you estimate the quotient? 552 divided by 71,520 I don't understand the steps.
44. ## Math
Estimate each quotient 7 divide by 0.85=1.28 is that right 9.6 divide by 0.91= 10 divide by 1 =10 is this right
45. ## Calculus (Math 2A)
Solve the difference quotient for the given function f(x) = x^3, (f(a+h)-f(a))/h i forgot how to do this problem
46. ## Maths
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided. 108 is 36% of what number? Write and solve a proportion to solve the problem.
47. ## math,correction
can someone correct these for me i'll appreciate it thank. dirctions: solve 3x-7=2x+8 my answer: x = 15 directions: solve 2/5x=-10 my answer: x = -25 Directions: solve -3(x+1)=2(x-8)+3 my answer: x= -3.2 1 looks ok. Show your work on 2 and 3 and we can
48. ## math
. what are the steps to solve this problem? 0.000027/0.000009 requirements: explain how to use scientific notation to solve the problem describe how to divide written in scientific notation give your final answer in simplest form.
49. ## algebra
Use synthetic division to divide the polynomial 2x3 – 45x + 28 by x + 5, and write the quotient polynomial and the remainder. [Be careful – notice that there is no x2 term.]. Show work.
50. ## Math
1. What are the steps to solve this problem? 0.000027 ÷ 0.000009 • Explain how to use scientific notation to solve the problem. • Describe how to divide numbers written in scientific notation. • Give you final answer in simplest form. (5 points) I
51. ## Calculus
How do I use the chain rule to find the derivative of square root(1-x^2) also, are there any general hints or tips for determining when the chain rule and product or quotient rule should be used?? i'm having trouble discerning when both the chain rule and
52. ## Algebra
Can you please check my work? You have helped other students I know. Thank you:) Multiply -7(2a2 -4) = -14a2 + 28 Multiply 0.5(a + b -5) = 0.5a + 0.5b - 2.5 Factor bx - by + bz = b(x - y + z) Combine like terms (simplify) 12x + 4 - 11y - 6x - 9 - y = 6x -
53. ## writing expressions
I need to write algebraic expressions: for these 6 sentecnces 1) four more shirt than s shirts 2) the quotient of p and 5 3) the sum of t tv's and 11 tv's 4) five times your quiz score q 5) nine cards fewer than e cards 6) divide the total points p by 3
54. ## algebra help write expression
I need to write algebraic expressions: for these 6 sentecnces 1) four more shirt than s shirts 2) the quotient of p and 5 3) the sum of t tv's and 11 tv's 4) five times your quiz score q 5) nine cards fewer than e cards 6) divide the total points p by 3
55. ## Algebra
- 20> 8 +7a Solve inequalities like you would an equation, EXCEPT if you multiply or divide by a negative number, reverse the inequality sign. Let's look at your problem: - 20 > 8 + 7a First subtract 8 from both sides (whatever operation you do to one side
56. ## math
use the four digits 3 4 5 6 exactly once each to form a division problem which has a quotient of 0.08 write the appropriate division problem
57. ## Estimating DecimalsProducts and Quotients
Determine whether each product or quotient is resonable.If not resonable,find a resonable result. 1. 62.77(29.8)=187.0546 2.16.132/2.96=54.5 The numbers are correct except for the decimal places. The first one looks like 60*30 or 1800, so determine where
58. ## Math
Nathan can purchase 5 lemons for \$2. Which is one way to find the number of lemons Nathan can buy with \$20? A.add 6 and 2 and then multiply the sum by 20 B.add 2 and 20 and then multiply the sum by 6 C.divide 6 by 2 and then add the quotient to 20 D.divide
59. ## math
Will someone please show me how to do this problem. I have to write this ratios in simplest form. The ratio of 5 3/5 to 2 1/10. Change each of your mixed fractions to an improper fraction, then divide the first by the second. Remember to divide fraction #1
60. ## math
72 is what percent of 480? write and solve an equation to solve the problem. i finished the problem, but i can't figure out how to write the equation. PLEASE HELP!
61. ## math
Suppose you want to cover the backyard with decorative rock and plant some trees as the first phase of the project. You need 30 tons of rock to cover the area. If each ton cost \$60 and each tree is \$84, what is the maximum number of trees you can buy with
62. ## College Math II
Solve each problem. Find the quotient and remainder when x^2 – 5x + 9 is divided by x – 3. I don't understand how to divide and solve this problem. Can someone help me understand the steps on how to do it. Thanks.
63. ## Math
Are you trying to solve for x, n or p? solve for n Show me how to work this problem. 2n = 4xp - 6 what is the problem? what grade r u in i am in 6 grade To get n all by itself, you need to divide by 2. But if you divide by 2 on the left, you have to divide
64. ## programming
Your goal for this project is to write an unsigned 16-bit software implementation of both the multiplier and the divider architectures found in the book. Input/Output You are required to read two unsigned integers from the console. The valid range for both
65. ## help with solving
x^2+(x-2)^2 = (2x-6)^2 solve for x. thanks so much. Please keep in mind that tutors here won't simply do your work for you. I'm sure you'll get a better response if you show your work as far as you can get, so tutors will know where you get stuck. =) it
66. ## Need all the help i can get
Hi i'm pretty new to this and i don't know what to do?So i put these questions up im not sure if my answers will show up because i used color letters, but i need help checking and explaining steps N 108 is 36% of what number? Write and solve a proportion
67. ## Math
I am suppose to write 2/30 as a percent. My answer choices are a .o6%, b. .005% c. .6% d. .05% I thought I would divide 2 into 30 which would be .06 and then moving 2 spaces to the right equaling 6%. How do I solve this problem.
68. ## Math
What patterns help you divide?Draw a picture or make a chart to show how to divide 1,600 by 4.Give the quotient.
69. ## Algebra
please help me with this question and check my answers. Solve the problem. A rectangle with width 2x + 5 inches has an area of 2x4 + 9x3 - 12x2 - 79x - 60 square inches. Write a polynomial that represents its length. ( I don't know how to solve this
70. ## math
uhm basically this is the part 2 of applications of percent quiz (GA connections academy) 108 is 36% of what number? write and solve a proportion to solve the problem 72 is what percent of 480? write and solve an equation to solve the problem Explain how
71. ## elementary
write a problem write a word problem involving the number of pages in a book.then write the numerical expression that you would need to use to solve your problem.
72. ## math-volume
I'm confused with how to solve this problem, so help would be great! The volume of any spherical balloon can be found by using the formula V= 4/3 pi r^3. Write an equation for r in terms of V and pi All this is asking you to do is get r=something. First, I
73. ## Math...
please help....Note: Enter your answer and show all the steps that you use to solve this problem in the space provided. 108 is 36% of what number? Write and solve a proportion to solve the problem.
74. ## MATH...somewhat urgent
I have three problems I would like checked please.1: I need to graph and find the y intercept of 6y+5x=-18 First I want to solve for y therefore I will let x be 0: 6y+5(0)=-18 6y=-18...I will now divide y=-3 Therfore the first set to plot would be (0-3)
75. ## Algebra
Use synthetic division to divide the polynomial 2x^3 – 12x – 5 by x + 4, and write the quotient polynomial and the remainder. [Be careful – notice that there is no x^2 term.] Please Show work.
76. ## Math ,help
how can i simplify this more its for the following problem: Problem #22 solve by using the quadratic formual. 5x^2-4x+1=0 this is where i am but i do not know how to go further. x = (4 (+/-) sqrt (-4))/(10) Problem #23 Solve by completing the square.
77. ## Algebra
1. What are the steps to solve this problem? 0.000027 ÷ 0.000009 Requirements: Explain how to use scientific notation to solve the problem Describe how to divide numbers written is scientific notation Give your final answer in simplest form Plz help!!
78. ## math
how estimating the quotient helps you place the frist digit in the quotient of a divison problem
79. ## math
Explain how estimating the quotient helps you place the first digit in the quotient of division problem
80. ## Math
How is an estimating The quotient helps you place the first digit in the Quotient of a division problem
81. ## to bobpursley,math
This is what i have am i wrong or correct.Or how can word it or modify it to be with what the directions are asking? Directions: Create one statement in English representing an algebraic relationship. My answer: PROBLEM: A DOZEN EQQS COST \$1.20. WHAT IS
ok i need help with these questions i have been trying to figure them out and my brain isnt working. please help and thank you ahead of time. if you can help. 4. Q.write and algebraic expression for the product of 15 and c A. IS IS C-15 16. Q. define the
83. ## math
ok i need help with these questions i have been trying to figure them out and my brain isnt working. please help and thank you ahead of time. if you can help. 4. Q.write and algebraic expression for the product of 15 and c A. IS IS C-15 16. Q. define the
84. ## Math !-_-
Please help me out. Omg...i absolutley hate math. I try my best but...its just to hard for me. Ok so im working on algebraic expressions. Here i go. 1. The product of a number n and 9 2. The quotient of 21 and a number b I HAD MORE OF THOSE BUT I DID THEM.
85. ## math
I need some help 1. Write an algebraic expression to represent 5 plus the quotient of r and 14 2. Write an equation to represent the statement 4 less than the product of 3 and x is twenty-three 3. solve using mental math 4y + 3 = 19 4. solve using mental
86. ## MATH
Reiny, the problem with the answer x>-2, why divide by 1? Is it because you can't divide with variable? Reworked problem -7x>-2(x+15)-2(x+15) -7x + 2x > 30 -5x/-5 > 30/-5 x > -6 other side -2(x+15)
87. ## math
use a multiplication table to solve a division problem. in what order doo you find the quotient, the divisor, and the dividened.
88. ## a.p. calculus help please!
My a.p. calculus teacher doesn't know how to do this problem (like many others) so he's giving us 5 points extra credit on our test if we can figure it out, but I don't know how to do it: It's as the lim-->infinity (x+sinx)/(x+cosx) We know the answer is 1
89. ## Math 116
How do you write an inequality for this problem. You have a budget of \$2.500.00 and you want to buy 30 tons of rock for \$60.00 per ton. You also want to plant trees for \$84.00 per tree. Write an inequality that illustrates this problem and solve the
90. ## Math
Write an algebriac expression 1. the quotient of m and two Tell rather the statement is true or false: 2. quotient of 3 and 12 is 4
91. ## Math
How to solve this problem 8.3 was added to a certain number. This sum was then divided by 11.2. Finally, the quotient was multiplied by 8. This product came out to 16. Please help I'm in 5th gradethank you..
92. ## progamming JAVA
1.Write an algorithm, in English, to solve the following problem. Be sure to include pre- and post-conditions! 2.Problem: Solve X mod Y for all positive integers. Do so without using multiplication or division. 3.What is (7 / 3) + (4 % 5)?
93. ## Calculus
22. Divide the expression before differentiating. F(x) = (8x^3 - 1)/ (2x - 1) Would the way to solve this problem be... F(x) = [(8x^3) / (2x - 1)] - [(1)/(2x - 1)] = [(4x^2)/(1)] - [(1) / (2x - 1)] = (4x^2) - (2x-1)^-1 F'(x) = (8x + 1)
94. ## math
write a word problem involving the number of pages in a book. then write the numerical expression that you would need to use to solve your problem.
95. ## Math
How do you write the expression for...the sum of the quotient of p and 14, and the quotient of a q and 3
96. ## Math
Write a problem story that represents 5 divide by 1/6
97. ## math
Use compatible numbers to estimate the quotient. 256.1 divide by 82. My answer is 240 divide by 40 = 3.
98. ## math
1)how does the placement of the decimal point in the divisor and the dividend affect the decimal point in the quotient.2) when u divide a number by 0.03 ,will your quotient be larger or smaller than your dividend
99. ## Math
Write a story problem to represent six divide by 1/5
100. ## bryce
can someone help me write out these expressions? 4 more than p: product of c and 15: the quotient of n and e: The quotient of 17 and k: 23 less than x: the sum of v and 3: the sum of 9 and k minus 17: 7 more than 5 times n: 85 less than the product of t
|
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 8.7: Rectangle Facts
Difficulty Level: At Grade Created by: CK-12
Rectangle Facts – Apply Measurement Formulas
Teacher Notes
Each problem shows a rectangle separated into four smaller rectangles. The areas for three of the four rectangles are given. Students use the data provided and the formula for the area of a rectangle to figure out the dimensions of each of the three rectangles. Once they have found the dimensions of the three rectangles, they can figure out the dimensions of the fourth rectangle and determine its area.
Solutions:
1. 20 square inches
2. 32 square inches
3. 32 square inches
4. 6 square inches
5. 56 square inches
6. 8 square inches
Rectangle Facts – Apply Measurement Formulas
Rectangle ABCD\begin{align*}ABCD\end{align*} is divided into 4 smaller rectangles.
\begin{align*}& \mathbf{Describe:} && \text{The large rectangle contains} \ 4 \ \text{smaller rectangles. The areas of three of the}\\ &&& \text{rectangles are given.} \ A \ \text{is a square. All dimensions are whole numbers.}\\ \\ & \mathbf{My \ Job:} && \text{Use the given areas. Figure out the area of Rectangle} \ C.\\ \\ & \mathbf{Plan:} && \text{Find common factors of the areas.} \ A \ \text{is a square, so the dimensions can be easily}\\ &&& \text{determined. Figure out the dimensions of Rectangles} \ B \ \text{and} \ D. \ \text{This will give the}\\ &&& \text{dimensions for Rectangle} \ C. \ \text{Use the area formula to figure out the area of}\\ &&& \text{Rectangle} \ C.\\ \\ & \mathbf{Solve:} && A \ \text{is a square, so the dimensions are} \ 4 \ \text{in. by} \ 4 \ \text{in. That means that one of the}\\ &&& \text{dimensions of Rectangle} \ B \ \text{is} \ 4 \ \text{inches. The area is} \ 24 \ \text{square inches, so the other}\\ &&& \text{dimension must be} \ 24 \div 4, \ \text{or} \ 6 \ \text{inches. Likewise, one of the dimensions of}\\ &&& \text{Rectangle} \ D \ \text{is} \ 4, \ \text{so the other is} \ 20 \div 4, \ \text{or} \ 5 \ \text{inches. The dimensions of Rectangle}\\ &&& C \ \text{are} \ 6 \ \text{inches from the shared side with Rectangle} \ B \ \text{and} \ 5 \ \text{inches form the}\\ &&& \text{shared side with Rectangle} \ D. \ \text{The area of} \ C \ \text{is} \ 6 \times 5, \ \text{or} \ 30 \ \text{square inches.}\\ \\ & \mathbf{Check:} && A: 4 \ \text{by} \ 4 \ \text{inches with an area of} \ 16 \ \text{sq in.}\\ &&& B: 4 \ \text{by} \ 6 \ \text{inches with an area of} \ 24 \ \text{sq in.}\\ &&& C: 5 \ \text{by} \ 5 \ \text{inches with an area of} \ 30 \ \text{sq in.}\\ &&& D: 4 \ \text{by} \ 5 \ \text{inches with an area of} \ 20 \ \text{sq in.}\end{align*}
1. Rectangle \begin{align*}EFGH\end{align*} is separated into 4 smaller rectangles.
2. Rectangle \begin{align*}JKLM\end{align*} is separated into 4 smaller rectangles.
3. Rectangle \begin{align*}NPQR\end{align*} is separated into 4 smaller rectangles.
4. Rectangle \begin{align*}EFGH\end{align*} is separated into 4 smaller rectangles.
5. Rectangle \begin{align*}JKLM\end{align*} is separated into 4 smaller rectangles.
6. Rectangle \begin{align*}NPQR\end{align*} is separated into 4 smaller rectangles.
Extra for Experts: Rectangle Facts – Apply Measurement Formulas
1. Rectangle \begin{align*}ABCD\end{align*} is separated into 4 smaller rectangles.
2. Rectangle \begin{align*}EFGH\end{align*} is separated into 4 smaller rectangles.
3. Rectangle \begin{align*}JKLM\end{align*} is separated into 4 smaller rectangles.
4. Rectangle \begin{align*}NPQR\end{align*} is separated into 4 smaller rectangles.
5. Rectangle \begin{align*}STUV\end{align*} is separated into 4 smaller rectangles.
6. Rectangle \begin{align*}WXYZ\end{align*} is separated into 4 smaller rectangles.
Solutions:
1. 21 square inches
2. 24 square inches
3. 15 square inches
4. 80 square inches
5. 48 square inches
6. 108 square inches
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show Hide Details
Description
Tags:
Subjects:
Date Created:
Feb 23, 2012
|
# Polygon Definition
In Geometry, a polygon is a closed two-dimensional figure, which is made up of straight lines. Generally, from the name of the polygon, we can easily identify the number of sides of the shape. For example, a triangle is a polygon which has three sides. Here, let us discuss the polygon definition, types of a polygon, its formula, properties with an example.
## Polygon Definition in Maths
In Mathematics, a polygon is a closed two-dimensional shape having straight line segments. It is not a three-dimensional shape. A polygon does not have any curved surface. A polygon should have at least three sides. Each side of the line segment must intersect with another line segment only at its endpoint. Based on the number of sides of a polygon, we can easily identify the polygon shape. The list of polygon shapes with the number of sides is given below.
No. of Sides Polygon Shape 3 Triangle 4 Quadrilateral 5 Pentagon 6 Hexagon 7 Heptagon 8 Octagon 9 Nonagon 10 Decagon
## Types of Polygon
Based on the angle measure and the sides of a polygon, the polygon is classified into:
• Regular Polygon – All the interior angles and the sides are equal
• Irregular Polygon – All the interior angles and the sides are of different measure
• Convex polygon – All the interior angles of a polygon are strictly less than 180 degrees
• Concave Polygon – One or more interior angles of a polygon are more than 180 degrees
## Polygon Formula
The important polygon formulas are:
1. The sum of interior angles of a polygon with “n” sides =180°(n-2)
2. Number of diagonals of a “n-sided” polygon = [n(n-3)]/2
3. The measure of interior angles of a regular n-sided polygon = [(n-2)180°]/n
4. The measure of exterior angles of a regular n-sided polygon = 360°/n
### Polygon Properties
The important properties of the polygon are
• The sum of interior angles of all the quadrangles is equal to 360 degrees.
• If at least one of the interior angles is greater than 180 degrees, then it is called concave
• If a polygon does not cross over itself, and has only one boundary, it is called a simple polygon. Otherwise, it is a complex polygon
### Polygon Example
Question:
Find the sum of the interior angle of a pentagon
Solution:
We know that a pentagon has five sides.
The formula to find the sum of interior angles is given by:
Interior angle sum = 180°(n-2)
= 180°(5-2)
= 180° (3)
= 540°
Hence, the sum of the interior angles of a pentagon is 540°
|
# If Two Interior Angles on the Same Side of a Transversal Intersecting Two Parallel Lines Are in the Ratio 2:3, Then the Measure of the Larger Angle is - Mathematics
MCQ
If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2:3, then the measure of the larger angle is
• 54°
• 120°
• 108°
• 136°
#### Solution
Let us draw the following figure:
Here AB || CD with t as a transversal.
Also, ∠1and ∠2are the two angles on the same side of the transversal.
It is given that
∠1:∠2 = 2:3
Therefore, let
∠1 = 2x
and ∠2 = 3x
We also, know that, if a transversal intersects two parallel lines, then each pair of consecutive interior angles are supplementary.
Therefore,
∠1 + ∠2 = 180°
On substituting ∠1 =2x and ∠2 = 3x in equation above, we get:
2x +3x = 180°
5x = 180°
x = (180°) /5
x = 36°
Clearly, 3x >2x
Therefore,
∠2 > ∠1
Also,
∠2 = 3x
∠2 = 3(36°)
∠2 = 108°
Concept: Parallel Lines and a Transversal
Is there an error in this question or solution?
Chapter 10: Lines and Angles - Exercise 10.6 [Page 53]
#### APPEARS IN
RD Sharma Mathematics for Class 9
Chapter 10 Lines and Angles
Exercise 10.6 | Q 10 | Page 53
Share
|
# The Internet Is Completely Divided Over the Answer to This Simple Math Equation
The internet is completely divided over a simple algebraic equation that your middle schooler would likely solve in seconds. However, it's not that it's difficult to get to an answer; it's a matter of which answer you get. Some of those who solve the equation are getting an answer of 1 and others are landing at 16.
The confusion that's coming up is a result of the way people were taught the order of operations, or PEMDAS — remember that acronym? Here's a brief refresher as to how PEMDAS works, for those who need it:
P: solve the equation inside any parentheticals first.
E: solve any exponential equations (hint: there aren't any in this one, so we can ignore this)
M: solve any multiplication problems*
D: solve any division problems*
S: solve any subtraction problems*
*In most instances, you're taught to solve the MD and AS portions of PEMDAS from left to right, but some people were taught that if there's a number attached to a parenthetical (which implies that you multiply the two numbers), that you're still working on the parenthesis portion of PEMDAS. (Which is also known as BOMDAS in some parts of the world, but ultimately means the same thing.)
Depending on how you were taught PEMDAS, this is how you get both answers:
## How to Get to an Answer of 16
Here's the equation: 8 / 2(2+2)
First, add the numbers in the parenthesis, which equal 4. Then, multiply and divide from left to right. First, 8 divided by 2 is 4, and 4 times 4 is 16.
## How to Get to an Answer of 1
Here's the equation once again: 8 / 2(2+2)
First, add the numbers in the parenthesis, which equal 4. Then, to finish off the parenthesis, multiply 4 by the 2, which equals 8. Then, 8 divided by 8 is 1.
Now technically, even though the internet is torn, the correct answer is 16. BUT before you @ me, hear me out: say the equation was 8 / 2 x 4, thus eliminating the parenthesis component completely. You would always work from left to right, which would give you the answer of 16 (8 divided by 2 is 4, times 4 is 16).
What do you think?
|
# How do you solve 7+3x=5?
Oct 25, 2017
$x = - \frac{2}{3}$
#### Explanation:
$7 + 3 x = 5$
So the end goal is to find the value of $x$. First we want to isolate $x$. In this case we do thi by subtracting $7$ on both sides of the equation. Because what we do to one side of an equation, we do to the other side as well. Then simplify.
$7 - 7 + 3 x = 5 - 7$
$3 x = - 2$
After simplifying, we want to divide both sides of the equation by $3$ so that we get the variable by itself.
$\frac{3}{3} x = - \frac{2}{3}$
$x = - \frac{2}{3}$
Oct 25, 2017
$x = - \frac{2}{3}$
#### Explanation:
Given -
$7 + 3 x = 5$
Add $- 7$ to both sides
$7 + 3 x - 7 = 5 - 7$
$\cancel{7} + 3 x \cancel{- 7} = - 2$
Divide both sides by $3$
$\frac{3 x}{3} = \frac{- 2}{3}$
$\frac{\cancel{3} x}{\cancel{3}} = \frac{- 2}{3}$
$x = - \frac{2}{3}$
|
Suggested languages for you:
Americas
Europe
Problem 46
# Show that the zero mapping and the identity transformation are linear transformations.
Expert verified
The zero mapping, T₀, and the identity transformation, I, are both linear transformations. This is because they satisfy the two properties of linear transformations: For the zero mapping, T₀(u+v) = T₀(u) + T₀(v) and T₀(cu) = cT₀(u), as T₀ maps all vectors to the zero vector. For the identity transformation, I(u+v) = I(u) + I(v) and I(cu) = cI(u), since I maps each vector to itself.
See the step by step solution
## Step 1: Zero Mapping
Let T₀ : V → W be the zero mapping. We want to show that it satisfies the properties of a linear transformation. 1. Let u, v be arbitrary vectors in V. Then: T₀(u+v) = 0 (since T₀ maps all vectors to the zero vector) T₀(u) + T₀(v) = 0 + 0 = 0 (since T₀ also maps u and v to the zero vector) So, T₀(u+v) = T₀(u) + T₀(v). 2. Let u be an arbitrary vector in V and c be an arbitrary scalar. Then: T₀(cu) = 0 (since T₀ maps all vectors to the zero vector) cT₀(u) = c(0) = 0 (since T₀ maps u to the zero vector) So, T₀(cu) = cT₀(u). Since the zero-mapping satisfies both properties of a linear transformation, it is a linear transformation.
## Step 2: Identity Transformation
Let I : V → W be the identity transformation. We want to show that it satisfies the properties of a linear transformation. 1. Let u, v be arbitrary vectors in V. Then: I(u+v) = u+v (since I maps each vector to itself) I(u) + I(v) = u + v (since I maps u to u and v to v) So, I(u+v) = I(u) + I(v). 2. Let u be an arbitrary vector in V and c be an arbitrary scalar. Then: I(cu) = cu (since I maps each vector to itself) cI(u) = cu (since I maps u to u) So, I(cu) = cI(u). Since the identity transformation satisfies both properties of a linear transformation, it is a linear transformation.
We value your feedback to improve our textbook solutions.
## Access millions of textbook solutions in one place
• Access over 3 million high quality textbook solutions
• Access our popular flashcard, quiz, mock-exam and notes features
## Join over 22 million students in learning with our Vaia App
The first learning app that truly has everything you need to ace your exams in one place.
• Flashcards & Quizzes
• AI Study Assistant
• Smart Note-Taking
• Mock-Exams
• Study Planner
|
# How do you find the critical numbers for f(x) = x^(-2)ln(x) to determine the maximum and minimum?
Apr 15, 2017
$0$ is the minimum
${e}^{\frac{1}{2}}$ is the maximum
#### Explanation:
Take the derivative of $f \left(x\right)$. You will need to use the product rule. You also need to know that the derivative of $\ln \left(x\right)$ is $\frac{1}{x}$:
$f ' \left(x\right) = {x}^{-} 2 \left(\frac{1}{x}\right) + \ln \left(x\right) \left(- 2 {x}^{-} 3\right)$
$f ' \left(x\right) = {x}^{-} 3 - 2 \ln \left(x\right) {x}^{-} 3$
Factor out a ${x}^{-} 3$
$f ' \left(x\right) = {x}^{-} 3 \left(1 - 2 \ln \left(x\right)\right)$
Solve for $x$:
$x = 0 , {e}^{\frac{1}{2}}$
Plug in these numbers into the initial equation:
$f \left(0\right) = 0 \ln \left(0\right) = D N E$.
We'll need to use limits for this:
${\lim}_{x \rightarrow 0} {x}^{-} 2 \ln \left(x\right) = - \infty$. This is definitely a minimum
$f \left({e}^{\frac{1}{2}}\right) = \left({e}^{-} 1\right) \left(\ln \left({e}^{\frac{1}{2}}\right)\right) = \frac{1}{2 e}$. This is a maximum because plugging in anything before or after this will give a value less than this.
|
# Trigonometry
Check Problems from last lesson
The word trigonometry comes from Greek words that mean "triangle measurement."
When Egyptians first used a sundial around 1500 B.C., they were using trigonometry.
The picture to the right is a model of how a sundial works:
As the sun shines on the staff given by , it casts a shadow represented by . The staff is a fixed length, therefore the length of varies with the measure of angle , which changes as the sun moves throughout the day. The Egyptians looked at the ratio to determine the time. This ratio is called the tangent of .
A ratio of the lengths of two sides of a right triangle is called a tigonometric ratio. The three most common ratios are sine, cosine, and tangent. They are abbreviated sin,cos, and tan respectively.
Trigonometric Ratios Definition:
refer to the triangle to the right:
Trogonometric ratios are related to the acute angles of a triangle, not the right angle. The value of the trigonometric ratio, depends only on the angle. Consider the following three similar triangles (why are they similar?):
By the definition of the sin above we have , ,.
But since the triangles are similar, the ratio of their sides are equal, thus .
Activity: Open the GSP Sketch and use Calculate in the measure menu to calculate the sin,cos, and tangent of CAB by computing the ratios in the definition above.
Compare to the calculated sin,cos, and tangent.
Trigonometric ratios can be used to find the measures of acute angles in a right triangle when you know the measures of the two sides of the triangle. You must determine which ratio to use depending on which sides are given. Then, use the inverse trigonometric functions on your calculator to find the angle.
Example: Find the approximate measure of E.
We know the leg adjacent to E and the measure of the hypotenuse. We would therefore use the cosine, since it is defined in terms of the adjacent side and the hypontenuse:
Now on your calculator, make sure it is in degree mode and enter:
5 14 = INV COS
or on a TI-83 calculator: First hit mode key and select degree by using the arrows to put cursor over degree and pressing enter.
Now press 2nd COS 5 14 ) Enter
You should get 69.075168. So, E.
Click the lightbulb to practice what you've learned.
|
# Solve for x? 3(2x-2)=2(x+1)
Jun 5, 2018
$x = 2$
#### Explanation:
If you remember that a constant or whatever outside a parenthesis should be multiplied with each term inside the parenthesis.
Therefore,
$3 \left(2 x - 2\right) = 2 \left(x + 1\right)$
$6 x - 6 = 2 x + 2$
Move all the unknown x-s to the left of the equal sign, and the constants to the right of the equal sign:
$6 x - 2 x = 2 + 6$
$4 x = 8$
This gives $x = 2$
Jun 5, 2018
$x = 2$
#### Explanation:
Expand the brackets:
$3 \left(2 x - 2\right)$:
$3 \cdot 2 x = 6 x$ and $3 \cdot - 2 = - 6$
$2 \left(x + 1\right)$:
$2 \cdot x = 2 x$ and $2 \cdot 1 = 2$
simplify:
$6 x - 6 = 2 x + 2$
bring the 2x over the equals to make it -2x which then leads to
$6 x - 2 x - 6 = 2$ $\therefore$ $4 x - 6 = 2$
take the -6 over to make + 6 so you get:
$4 x = 6 + 2 \therefore 4 x = 8$
divide 8 by 4 which gives 2 therefore $x = 2$
|
# 8.1 - Compare/Order Real Numbers
8.1 The student will compare and order real numbers.
### BIG IDEAS
• I can determine sale costs while shopping, create a budget, dilute mixtures, understand scales on maps, interpret probabilities and odds, and convert among metric units.
• I will be able to interchange fraction, decimal, and percent values to interpret real world situations.
### UNDERSTANDING THE STANDARD
2016 VDOE Curriculum Framework - 8.1 Understanding
· Real numbers can be represented as integers, fractions (proper or improper), decimals, percents, numbers written in scientific notation, radicals, and p. It is often useful to convert numbers to be compared and/or ordered to one representation (e.g., fractions, decimals or percents).
· Proper fractions, improper fractions, and mixed numbers are terms often used to describe fractions. A proper fraction is a fraction whose numerator is less than the denominator. An improper fraction is a fraction whose numerator is equal to or greater than the denominator. An improper fraction may be expressed as a mixed number. A mixed number is written with two parts: a whole number and a proper fraction (e.g., 3 ). Fractions can have a positive or negative value.
· The density property states that between any two real numbers lies another real number. For example, between 3 and 5 we can find 4; between 4.0 and 4.2 we can find 4.16; between 4.16 and 4.17 we can find 4.165; between 4.165 and 4.166 we can find 4.1655, etc. Thus, we can always find another number between two numbers. Students are not expected to know the term density property but the concept allows for a deeper understanding of the set of real numbers.
· Scientific notation is used to represent very large or very small numbers.
· A number written in scientific notation is the product of two factors: a decimal greater than or equal to one but less than 10 multiplied by a power of 10 (e.g., 3.1 ´ 105 = 310,000 and 3.1 ´ 10–5 = 0.000031).
· Any real number raised to the zero power is 1. The only exception to this rule is zero itself. Zero raised to the zero power is undefined.
### ESSENTIALS
• How does the different ways rational numbers can be represented help us compare and order rational numbers?
Numbers can be represented as decimals, fractions, percents, and in scientific notation. It is often useful to convert numbers to be compared and/or ordered to one representation (e.g., fractions, decimals or percents).
• When are numbers written in scientific notation? Scientific notation is used to represent very large and very small numbers.
The student will use problem solving, mathematical communication, mathematical reasoning, connections, and representations to
· 8.11 Compare and order no more than five real numbers expressed as integers, fractions (proper or improper), decimals, mixed numbers, percents, numbers written in scientific notation, radicals, and π. Radicals may include both positive and negative square roots of values from 0 to 400. Ordering may be in ascending or descending order.
· 8.12 Use rational approximations (to the nearest hundredth) of irrational numbers to compare and order, locating values on a number line. Radicals may include both positive and negative square roots of values from 0 to 400 yielding an irrational number.
### KEY VOCABULARY
simplify, numerical expression, exponent, variable, rational number, positive, negative, whole number, fraction, decimal, integer, square root, perfect square, order of operations, real numbers, compare and order, percents, scientific notation, ascending order, descending order
Updated: Jul 30, 2019
|
Sei sulla pagina 1di 99
# Numerical Analysis Problems and Solutions
## CHAPTER 01: INTRODUCTION TO NUMERICAL CALCULATION
Fundamentals Of Numerical Methods
Principles Of Computer Operations
Number Representations
Fortran Rules And Vocabulary
## Fundamentals Of Numerical Methods
PROBLEM 01 – 0001: The digital computer is a basic tool used in arriving at the
solution of numerical problems that would otherwise be
extremely long and laborious. Give a brief explanation of
how the digital computer operates.
CONTENTS CHAPTER PREVIOUS NEXT PREP FIND
PROBLEM 01 – 0001: The digital computer is a basic tool used in arriving at the
## Explanation of the functions:
Input:
Magnetic tape, paper tape, and punched cards are the common
media for carrying information (data and program) from the outside world
## through, the input mechanism to the internal functions of the computer.
Figure 2 displays two punched cards and a piece of punched paper tape.
The input mechanism of the computer is able to receive information from
cards and tape by reading the punched holes. A particular computer may
have only the mechanism for reading cards; another computer may be
able to read from both cards and paper tape. Card (a) Fig. 2 is typical of a
data card containing two pieces of data. On the other hand, card (b)
## contains one instruction or one statement of a program.
Output:
Output information can be punched in cards and in paper tape. It is
paper.
Memory:
## information; however, the alphabetic information is stored in an equivalent
numerical form. Of course, the sign and the decimal associated with the
Control Unit:
## output, arithmetic and memory.
In this very brief look into how the digital computer operates, many
## intricate details have been avoided.
PROBLEM 01 – 0002:
In Algebra the expression N = N + 1 is meaningless, however, it is meaningful in Fortran. Explain.
Solution:
It means "take the value stored under the name N, add 1 to it, and store the result under the name N
again."
PROBLEM 01 – 0003:
Given a sequence of n numbers, x1, x2,…, xn, find the largest and the smallest number of this sequence and
write out the results.
Solution:
The figure shown is a flow chart for the solution of this problem. The READ statement reads the number n
into the location N and the numbers x1, x2,…, xn into the vector X. Locations XMAX and XMIN are next set
equal to X(1). The iteration statement sets up control for doing the operations that follow up to α for 1 = 2,
3,…,N. The first time through the loop XMAX is compared with X(2) (I = 2). If X(2) is greater than XMAX, then
X(2) is placed in XMAX; if it is equal to XMAX, control is sent directly to the end of the iteration loop. If X(2)
is less than XMAX, it is compared with XMIN, which contains X(1). If X(2) is less than XMIN, it is placed in
XMIN. If it is greater than or equal to XMIN , control is sent to the end of the iteration loop. The same
sequence of operations is then carried out using X(3). This continues for each I up to and including I = N.
Clearly when the iterations are completed, the contents of XMAX will have been compared with every
number and the largest retained in XMAX.
XMIN will likewise contain the smallest of the sequence. The
contents of XMAX and XMIN are then written out.
PROBLEM 01 – 0004: Construct a flow chart that will find the sum of a sequence of
numbers represented by a1, a2,…, an.
Solution:
This addition operation on the computer is done by finding first the partial sum of a1 + a2, then a1 + a2 + a3,
and so on, until each number has been added to the partial sum to obtain the total. Either of the two flow
charts in the accompanying figure will accomplish the necessary operations, the two solutions being the
same except for the control statements.
The first symbol in every flow chart is START. It indicates that appropriate control statements are to be
placed ahead of the program for proper initiation of the machine. The READ statement means that the
number n is read into a location labeled N and that the sequence of numbers a1, a2,…, an is read into the
locations labeled A(1), A(2),..., A(N).
The shorthand notation in this box will be used frequently. The
number in location N indicates the number of items in the set. A(1),…,
A(N) contain the numbers a1,…, an, respectively. The next statement is the
first arithmetic statement and places a floating-point zero in the contents of
a location called SUM.
The iteration statement in Figure (b) says to do the statements
between the box and the circle (in this case there is only one statement) N
times, the first time for I = 1, the second time for I = 2, etc., and the last
time for I = N. The box within the range of this iteration says to take the
current value of SUM, add the number in the location A(1), using the
current value of I, and store the result back into SUM. The first time
through the iteration this adds zero (SUM was initially set to zero) and A(1)
and stores the result, A(1), in SUM. The second time through the iteration,
this statement adds A(1), the current value of SUM, and A(2) , storing the
result, A(1) + A(2) , in SUM. The third time through the iteration, it adds
SUM [currently A(1) + A(2)] to A(3) and stores the result,
A(1) + A(2) + A(3) , back into SUM. Each time through the iteration, the
statement updates SUM by the next number in order from the set of
numbers A(1), A(2),…, A(N). At the conclusion of the iterations the
location SUM contains the sum of all the numbers A(1) through A(N).
In Fig. (a) the iteration is controlled by a logic statement and
illustrates the functions performed by the iteration statement. The integer 1
is placed into location I ahead of the loop. Then A(1) is added to the
contents of SUM, which is zero prior to this operation. SUM now contains
A(1). The logic statement tests the sign of I – N. If it is negative or equal to
zero, I is updated by 1, and control is returned to the arithmetic statement.
This sequence of operations is continued until I – N is greater than zero, at
which point SUM is written out.
The last box in each chart says to display the number in SUM on
the output. This is the conclusion to the problem.
The flow chart is completed with a STOP.
PROBLEM 01 – 0005: Find the sum of the first n terms in the Taylor-series
expansion for ex.
CONTENTS CHAPTER PREVIOUS NEXT PREP FIND
PROBLEM 01 – 0005: Find the sum of the first n terms in the Taylor-series
expansion for ex.
Solution: ex = 1 + x + (1/ 2!) x2 + (1/ 3!) x3 + … + [{xn–1} / {(n – 1)!}]
The arithmetic operations needed for evaluating this sum are simplified by
noting that the (i + 1)st term in the series is obtained from the ith term by
multiplying by (x/i). The second term is equal to 1 ∙ (x/1), and the (i + 1)st
term is equal to the product of
[{xi–1} / {(i – 1)!}] and (x/i)
The flow chart for the problem is shown in the figure. Numbers for
N and X are read into the machine. Locations TERM and EXPX, set to
contain floating point 1.0, will be used to store the results of the
computations. The iteration loop sets up control to repeat the operations to
a for I = 1, 2,..., N – 1. The first statement in the loop gives the next term in
the series by updating the previous term. The first time through the loop
TERM contains 1.0, which is multiplied by (x/1), and then stored back in
TERM, so that TERM now contains the second term of the series. Each
time through the loop TERM is updated, so that it contains the (i + 1)st
term in the series. The last statement in the loop adds the contents of
TERM to the contents of EXPX to give the partial sum of the first i + 1
terms of the series. The first time through the loop EXPX contains 1.0, and
TERM contains (x / 1.0) when this statement is executed. The updated
contents of EXPX are 1.0 + (x / 1.0), which is the sum of the first two terms
of the series. When the iteration has been completed n – 1 times, EXPX
contains the sum of the first n terms of the series. Note that the integer I is
used in a floating-point computation. Such statements may not be
acceptable in computer programs but are permitted here for convenience.
## PROBLEM 01 – 0006: Using
sin(x) = x – (x3 / 3!) + (x5 / 5!) – … + (– 1)n–1
[{x2n–1} / {(2n – 1)!}] + (– 1)n [{x2n+1} / {(2n + 1)!}] cos {ξx},
solve f(x) = 1∫0 sin (xt)dt.
CONTENTS CHAPTER PREVIOUS NEXT PREP FIND
## PROBLEM 01 – 0006: Using
sin(x) = x – (x3 / 3!) + (x5 / 5!) – … + (– 1)n–1
[{x2n–1} / {(2n – 1)!}] + (– 1)n [{x2n+1} / {(2n + 1)!}] cos {ξx},
solve f(x) = 1∫0 sin (xt)dt.
Solution: f(x) = 1∫0 [xt – {(x3t3) / (3!)} + … + (– 1)n–1 [{(xt)2n–1} / {(2n – 1)!}]
+ (– 1)n [{(xt)2n–1} / {(2n + 1)!}] cos {ξxt}] dt
= n∑j=1 (– 1)j–1 [{x2j–1} / {(2j – 1)! (2j)}]
+ (– 1)n [{x2n–1} / {(2n + 1)!}] 1∫0 t2n–1 cos {ξxt} dt
with {ξxt} between 0 and xt. The integral in the remainder is easily bounded
by [1 / (2n + 2)]; but one can also convert it to a simpler form. Although it
wasn’t proven, it can be shown that cos(ξxt) is a continuous function of t.
Then applying the integral mean value theorem,
∫0 sin(xt)dt = n∑j=1 (– 1)j–1 [{x2j–1} / {(2j – 1)! (2j)}]
1
## + (– 1)n [{x2n–1} / {(2n + 1)! (2n + 2)}] cos {ζx}
for some ζx between 0 and x.
PROBLEM 01 – 0007: Consider a routine for evaluating F(x); and, let F*(x) denote
the computed result obtained by the program as a
approximation to F(x). Evaluate by the experimental method
a probable error in the computed result.
## CONTENTS CHAPTER PREVIOUS NEXT PREP FIND
PROBLEM 01 – 0007: Consider a routine for evaluating F(x); and, let F*(x) denote
the computed result obtained by the program as a
approximation to F(x). Evaluate by the experimental method
a probable error in the computed result.
## Solution: It is more common to investigate by experimentation the accuracy
of computed results obtained with a function evaluation routine. The
simplest type of experimental testing consists of computing F*(x) for
selected values of x and checking against known values of F(x). Such
testing, which involves personal inspection of results, cannot be very
thorough. Fortunately the process of experimental testing can be made
more automatic and more thorough by writing a test program in the
following way.
With the aid of a random number generator compute a sequence
of, say, n test arguments, x1, x2,…, xn. For each xk compute (1) F*(xk), the
approximation to F(xk) produced by the function evaluation routine
being tested, and (2) F**(xk), another approximation to F(xk) that is
sufficiently accurate to be used as a check value for F*(xk). Compute
statistics that will provide a ready indication of the magnitude of absolute
error or relative error in the values of F*(xk). Commonly used test statistics
are the maximum relative error
max1≤k≤n |[{F*(xk) – F**(xk)} / {F**(xk)}]|
and the root-mean-square relative error
√[(1/n) n∑k=1 [{F*(xk) – F**(xk)} / {F**(xk)}]2]
The value of n, that is, the number of test arguments, can be very
large. Some programs have been tested with as many as n = 100,000
random arguments; but smaller values of n are normally used, such as
n = 5,000. A uniform random number generator can be used to generate
test arguments, with the range of values adapted to the argument range of
F(x). Thus, if the argument range for F(x) were a finite interval {a, b} , then
the distribution of xk could be uniform in {a, b}. An exponential random
number generator is sometimes useful when the argument range of F(x) is
infinite. For example, the distribution of xk might be exponential with mean
1 for testing of a In x routine for which the argument range was (0,∞ ).
Sometimes it is desirable to make several tests and generate test
arguments in different parts of the argument range of F(x) for different test
runs. In this way, for example, regions where F(x) is unstable can be given
special attention.
F**(xk) must be computed in such a way that it is a sufficiently
accurate approximation to F(xk) to be used as a check value for F*(xk).
Normally this means that F**(xk) must be computed in higher precision
than F*(xk) is. Thus, if F*(xk) is computed in single-precision arithmetic,
then F**(xk) might be computed in double-precision. A special- purpose,
high-precision routine for evaluating F(x) might have to be written just to
obtain F**(xk) with sufficient accuracy. The higher the precision in which
F*(xk) is obtained, of course, the more difficult it is to compute a suitably
accurate check value in still higher precision.
If computing F**(xk) does present difficulties on the computer in
which tests are being made, it may be desirable to compute check values
on another computer with which the necessary accuracy can be obtained
more readily. This is obviously more trouble than using just one computer
since test arguments and check values have to be transferred from one
computer to another in some fashion. Variable-precision computers with
which high-precision calculations can be performed with ease are
especially useful for the computation of check values in high precision.
(For example, Fortran programs can be run on the IBM 1620 with a
precision of 28 decimal digits.)
Principles Of Computer Operations
PROBLEM 01 – 0008: Computers are often imagined as giant electronic "brains"
which never make an error. This notion can be wrong.
Interpret this simple program to demonstrate this concept.
SUM = 0
DO 10 I = 1,10000
10 SUM = SUM + 0.0001
WRITE (3, 20) SUM
20 FORMAT (F15.8)
END
CONTENTS CHAPTER PREVIOUS NEXT PREP FIND
## PROBLEM 01 – 0008: Computers are often imagined as giant electronic "brains"
which never make an error. This notion can be wrong.
Interpret this simple program to demonstrate this concept.
SUM = 0
DO 10 I = 1,10000
10 SUM = SUM + 0.0001
WRITE (3, 20) SUM
20 FORMAT (F15.8)
END
Solution: Take a quantity SUM, make it equal to zero, and then add the
quantity 0.0001 to it 10,000 times, and the answer is 1,00000000, exactly.
If one runs this program on a typical computer, the answer is likely
to be
0.99935269
Granted that the error may be small– –only 0.00065 or so, or about 0.065
percent– –still, the answer is wrong. It certainly destroys the notion of a
computer's infallibility, and creates concern about the results of other,
more complicated calculations.
PROBLEM 01 – 0009: Solve the following two simultaneous equations for x1 and
x2, and formulate the equations for the computer.
A11 X1 + A12 X2 = C1
A21 X1 + A22 X2 = C2
CONTENTS CHAPTER PREVIOUS NEXT PREP FIND
PROBLEM 01 – 0009: Solve the following two simultaneous equations for x1 and
x2, and formulate the equations for the computer.
A11 X1 + A12 X2 = C1
A21 X1 + A22 X2 = C2
## Solution: The novice might be tempted to write these two equations
A11 X1 + A12 X2 = C1 (1)
A21 X1 + A22 X2 = C2 (2)
as Fortran substitution statements into a program. But, the substitution
statement is not an algebraic equation. One method of employing the
computer in Solving these two equations is to prepare algebraic
solutions– –using determinants, substitution method, or some other– –to
find solutions for X1 and X2 as shown in Eqs. (3) and (4)
X1 = [{A22 C1 – A12 C2} / {A11 A22 – A12 A21}] (3)
X2 = [{A11 C2 – A21 C1} / {A11 A22 – A12 A21}] (4)
By this method of solution, the computer is used only in performing the
numerical calculations shown in Eqs. (3) and (4) in order to ultimately find
the numerical value of X1 and X2. This might be a practical application of a
digital computer if one wishes to solve this pair of equations five hundred
times, each time for one set of values for the coefficients A11, A12, A21, A22,
C1, and C2.
Now to proceed with the problem one recognizes that Eqs. (3) and
(4) can be used as the basis for preparing the necessary Fortran program.
Obviously, in order to find one pair of solutions– –X1 and X2– –the
appropriate values of the six coefficients must be available in MEMORY.
After X1 and X2 have been calculated, then, it will be necessary either to
store these solutions in MEMORY or to output the appropriate values.
Because in Fortran alphabetic or numeric characters cannot be
used in a sub or super position, it is necessary to give Fortran variables
different names from those indicated in the above algebraic equations.
See the table for one acceptable list of variable names.
TABLE
Equivalent Symbols
Algebraic Fortran Algebraic Fortran
A11 A11 C1 C1
A12 A12 C2 C2
A21 A21 X1 X1
A22 A22 X2 X2
## The Fortran program will be:
C TWO EQUATIONS
1 READ 2, A11, A12, A21, A22, C1, C2
2 FORMAT (6F10.5)
D = A11 * A22 – A12 * A21
X1 = [(A22 * C1 – A12 * C2) / D]
X2 = [(A11 * C2 – A21 * C1) / D]
PUNCH 3, X1, X2
3 FORMAT (2E15.7)
GO TO 1
END
## PROBLEM 01 – 0010: One of the most commonly used control statements in
Fortran is the arithmetic IF statement; show how the
programmer uses the arithmetic IF statement.
## PROBLEM 01 – 0010: One of the most commonly used control statements in
Fortran is the arithmetic IF statement; show how the
programmer uses the arithmetic IF statement.
## Solution: The IF statement allows the programmer to change the sequence
of operations, depending upon whether the value of an expression is
negative, zero, or positive. For example, the factorial of an integer N may
be computed by means of the following program which contains an IF
statement.
•••
•••
I=0
NFACT = 1
10 I=I+1
NFACT = NFACT * I
IF (I – N) 10, 20, 20
20 •••
•••
The first two statements in the program set the initial values of I equal to 0
and of NFACT equal to 1. Statement 10 increases I by 1. The next
statement computes NFACT for the new value of I. The IF statement
checks whether I – N is less than, greater than, or equal to zero, i.e.,
whether I is less than, greater than, or equal to N.
## PROBLEM 01 – 0011: Write the corresponding Fortran statements to the following
mathematical expressions:
(a) X = A + B ∙ C (b) I = (J/2)4
(c) Z = √[sin x] (d) X ∙ (BD)
(e) [{DE} / F] – G (f) DE/(F–G)
Solution:
Mathematics FORTRAN
(a) X=A+B•C X=A+B*C
(b) I = (J/2)4 I = (J/2)** 4
(c) Z = √[sin x] Z = SQRT(SIN{X})
(d) X ∙ (B )
D
X * (B ** D)
(e) [{DE} / F] – G [{D**E} / {F–G}]
(f) DE/(F–G) D**(E / (F–G))
## PROBLEM 01 – 0011: Write the corresponding Fortran statements to the following
mathematical expressions:
(a) X = A + B ∙ C (b) I = (J/2)4
(c) Z = √[sin x] (d) X ∙ (BD)
(e) [{DE} / F] – G (f) DE/(F–G)
## CONTENTS CHAPTER PREVIOUS NEXT PREP FIND
PROBLEM 01 – 0011: Write the corresponding Fortran statements to the following
mathematical expressions:
(a) X = A + B ∙ C (b) I = (J/2)4
(c) Z = √[sin x] (d) X ∙ (BD)
(e) [{DE} / F] – G (f) DE/(F–G)
Solution:
Mathematics FORTRAN
(a) X=A+B•C X=A+B*C
(b) I = (J/2)4 I = (J/2)** 4
(c) Z = √[sin x] Z = SQRT(SIN{X})
(d) X ∙ (BD) X * (B ** D)
(e) [{D } / F] – G
E
[{D**E} / {F–G}]
(f) DE/(F–G) D**(E / (F–G))
Number Representations
PROBLEM 01 – 0012: In the computer the binary number system is used as
opposed to the decimal number system used by humans to
execute arithmetic operations. However, in both systems,
there are some similarities and differences. Contrast and
compare the two systems.
## PROBLEM 01 – 0012: In the computer the binary number system is used as
opposed to the decimal number system used by humans to
execute arithmetic operations. However, in both systems,
there are some similarities and differences. Contrast and
compare the two systems.
## Solution: Humans execute arithmetic operations in the decimal number
system, a system that has "ten" as its base; but when it comes to the
computer, the story is quite different. In the computer, internal calculations
are done in the binary system, a system that has "two" as its base.
The decimal number system has ten different digits, 0 through 9.
The binary number system has only two different digits, 0 and 1. As an
abbreviation, the two binary digits are usually called bits.
However, the binary number system is similar to the decimal
system in that the position of a particular digit in a number has a great
importance in both systems. Thus in both number systems the left digits of
a number are more important than the right digits since they have a
greater value.
In a decimal integer, each digit has 10 times the value of the digit to
its right. Thus the last digit on the right of an integer is the "units" digit, the
next is the "tens" digit, the next is the "hundreds" digit, and so on.
Therefore, the decimal number 532 can be interpreted as
5 hundreds
+ 3 tens
+ 2 units.
A more concise notation expressing the same thing is
532 = (5 × 102) + (3 × 101) + 2 × 100.
The same principle applies to non-integer numbers such as 14.37 below
14.37 = (1 × 101) + (4 × 100) + (3 × 10–1) + (7 × 10–2)
= 10 + 4 + 0.3 + 0.07.
Binary numbers can be expressed in exactly the same way, except that
they only contain the digits 0 and 1, and that adjacent digits differ in value
by a power of 2, rather than by a power of 10. For example, the binary
integer 1011 can be expressed as
1011 = (1 × 23) + (0 × 22) + (1 × 21) + (1 × 20).
Changing into decimal numbers, we see that the binary 1011 is equal to
8 + 2 + 1, or a decimal number 11. Because each digit of a binary number
can only be a 0 or a 1 and therefore can carry less information than the 10
digits of the decimal number system, binary numbers are generally much
longer than their decimal equivalents. For example, the decimal number
4094 is 111111111110 in binary.
Non-integer binary numbers can be expanded into powers of 2 and
converted into decimal almost as easily as integers. For example, the
binary number 110.11 can be written as
110.11 = (1 × 22) + (1 × 21) + (0 × 20) + (1 × 2–1) + (1 × 2–2).
where the exponent of 2 starts with 0 to the left of the decimal point and
increases to the left, and starts with – 1 to the right of the decimal point
and becomes more negative to the right. In the case of the binary 110.11,
the decimal equivalent is
4 + 2 + (1/2) + (1/4) = 6.75.
A decimal integer can always be exactly represented by a binary integer, for every integer can be
expressed as a sum of powers of 2. But this is not true of fractional numbers.
In general, it can easily be shown that a rational fraction can be exactly expressed with a finite
number of binary digits only if it can be expressed as the quotient of two integers p/q, where q is a power
of 2; that is, q = 2n for some integer n. Obviously only a small proportion of rational fractions will satisfy this
requirement.
Thus even some simple decimal fractions cannot be exactly expressed in the binary number system.
For example, the simple decimal fraction 0.1 is an infinitely long binary number 0.000110011 ... with two 1's
and two 0's repeating themselves forever. Every repeating decimal (such as 0.33333333...) is also repeating
in binary, but other numbers which are not repeating decimals in the decimal number system may become
repeating in binary.
The decimal number 0.0001 is a repeating binary fraction which begins with
0.000000000000011010001101101110001 ... and lasts for 104 bits before starting to repeat with the part
...11010001101101110001.... Every 104 bits from now on, this binary fraction starts to repeat itself.
Obviously an infinitely long binary fraction cannot be stored or used in a digital computer, and so some
finite number of bits must be used and the rest discarded. This automatically leads to a small error which,
by being repeated many times, can lead to a large error in the final answer.
## PROBLEM 01 – 0013: Perform the following conversions:
a) 1 1 0 1 1 1 0 1 12 into base 10
b) 4 5 78 into base 2
c) 7 B 316 into base 8
d) 1 2 4 2 510 into base 16
## PROBLEM 01 – 0013: Perform the following conversions:
a) 1 1 0 1 1 1 0 1 12 into base 10
b) 4 5 78 into base 2
c) 7 B 316 into base 8
d) 1 2 4 2 510 into base 16
Solution:
a) 1 1 0 1 1 1 0 1 12 may be converted into base 10 by considering this fact: Each digit in a base two number
may be thought of as a switch, a zero indicating "off", and a one indicating "on". Also note that each digit
corresponds, in base 10, to a power of two. To clarify, look at the procedure:
28 27 26 25 24 23 22 21 20
1 1 0 1 1 1 0 1 12
## (1 × 256) + (1 × 128) + (0 × 64) + (1 × 32) + (1 × 16) + (0 × 4) + (1 × 2) + (1 × 1) = 443.
If the switch is "on" , then you add the corresponding power of two. If not, you add a zero. Follow the next
conversion closely.
b) 4578 may be converted into base two by using the notion of triads.
Triads are three-bit groups of zeros and ones which correspond to octal (base 8) and decimal (base 10)
numbers. This table can, with some practice, be committed to memory:
000 0 0 0
001 1 1 1
010 2 2 2
011 3 3 3
100 4 4 4
101 5 5 5
110 6 6 6
111 7 7 7
## So, taking each digit of 4578 separately, one gets:
48 = 1002
58 = 1012
78 = 1112
which becomes 1001011112 = 4578
c) 7B316 is a hexadecimal number. Letters are needed to replace numbers, since the base here is 16. The
following chart will help:
1000 10 8 8
1001 11 9 9
1010 12 10 A
1011 13 11 B
1100 14 12 C
1101 15 13 D
1110 16 14 E
1111 17 15 F
10000 20 16 10
## 162 161 160
7 B 316
(7 × 256) + (11 × 16) + (3 × 1) = 197110
Now, convert to octal using this procedure: Divide 8, the base to be used, into 1971. Find the remainder
of this division and save it as the first digit of the new octal number. Then divide 8 into the quotient of
the previous division, and repeat the procedure. The following is an illustration:
8 1971
– 1968
3 → 3 246 × 8 = 1968
246
– 240
6 → 6 30 ×8 = 240
30
– 24
6 → 6 3 × 8 = 24
3 → 3 3 ÷ 8 ≠ an integer
## d) 1 2 4 2 510 may be converted to base 16 by this method:
Take the highest power of 16 contained in 12423. Then, subtract that value and try the next lowest
power. Continue the process until all the digits are accounted for.
12425
16 × 3 =
3
12288
137
161 × 8 = 128
9
160 × 9 = 9
0
## PROBLEM 01 – 0014: Perform the following conversions from
a) 1 0 0 1 1 0 1 0
b) 1 0 1 . 1 0 1
c) 0 . 0 0 1 1
base 2 to base 8, 10, and 16.
## PROBLEM 01 – 0014: Perform the following conversions from
a) 1 0 0 1 1 0 1 0
b) 1 0 1 . 1 0 1
c) 0 . 0 0 1 1
base 2 to base 8, 10, and 16.
Solution: a) To convert base 2 to base 8, use the triad method where three
digits are taken together and converted to equivalent decimal value; this
results in base 8 conversion
|0 1 0| |0 1 1| |0 1 0|2 = 2328
Similarly for hexadecimal conversion 4 bits are taken together. Zeros ar
assumed for most significant grouping.
|0 0 0 0| |1 0 0 1| |1 0 1 0| = 9A16.
For binary to decimal conversion
1 0 0 1 1 0 1 0
1×27 + 0×26 + 0×25 + 1×24 + 1×23 + 0×22 + 1×21 + 0×20
128 + 0 + 0 + 16 + 8 + 0 + 2 + 0 = 15410
b) i) 101 . 1012 = 5 . 58
ii) [{0101} / 5] . [{10102} / A] = 5 . A16
iii) 101 . 1012
(1 × 22 + 0 × 21 + 1 × 20) . (1 × 2–1 + 0 × 2–2 + 1 × 2–3)
= 4 + 0 + 1 + 0.5 + 0 + 0.125 = 5 . 62510
c) i) 0 . 0011 = |000| . |001| |100| = 0 . 148
ii) 0 . 0011 = |0000| . |0011| = 0 . 316
iii) 0 . 0011 = . (0 × 20 + 0 × 2–1 + 1 × 2–3 + 1 × 2–4)
= 0 + 0 + 0 + 0.125 + 0.0625 = 0.187510
## PROBLEM 01 – 0015: Describe the procedures involved in adding and subtracting
is almost identical to that of adding and subtracting Arabic digits. Actually,
there are a few ways in which these calculations may be performed. The
most efficient one is the method used in the decimal calculations. The
operations will be done from right to left. In using this method, each Hex
symbol will be translated to a decimal digit before the calculations, then,
upon completion of the calculations, each resulting decimal digit will be
retranslated to a corresponding Hex symbol.
Addition: In decimal calculations, as one goes from one column to the
next, one carries units of tens. For example:
## col. 2 col. 1 col. O
1 2 8
8 8
2 1 6
In Hex addition, instead of carrying units of ten, carry units of sixteen.
Example 1
6 E 8 8
+ 1 D 3
For col. O:
## Hex → Decimal → Decimal Sum → Hex Sum
8 8 8
3 3 3
11 → B
Note in this column the decimal sum is less than 16, and no units of
sixteen are carried.
For col. 1:
Hex → Decimal → Decimal Sum → Hex Sum
8 8 8
D 13 13
21 → 16 + 5
Here the decimal s-um is greater than 16. After factoring out the units of
sixteen (here only one), the remaining digit should be placed in this
column.
For col. 2:
Hex → Decimal → Decimal Sum → Hex Sum
E 14 14
1 1 1
15
+1 (form col.1)
16 → 16 + O
One unit of sixteen can be factored out, leaving zero as the remaining digit
for col. 2.
For col. 3:
Hex → Decimal → Decimal Sum → Hex Sum
6 6 (6/6)
+1 (form col.2)
7 → 7
Thus, the final result is:
6E88
+1D3
705B
Subtraction: In the subtraction procedure the units of sixteen are borrowed instead of being carried.
Example 2:
## col. 2 col. 1 col. Ø
3 5 E
– 2 B 8
For col. O
Hex → Decimal → Decimal → Hex Difference
Difference
E 14 14
8 8 –8
6 → 6
For col. 1:
Hex → Decimal → Decimal → Hex Difference
Difference
5 5 (5) + 16 (Borrowed unit from col.2)
B 11 11
10 → A
The difference in col.2 is zero since one unit was borrowed from 3. The
final result is:
2 + 16
3 5 E
– 2 B 8
A 6
## PROBLEM 01 – 0016: a) Convert 764 . 3018 to binary.
b) Convert 11011 . 101112 to octal.
## PROBLEM 01 – 0016: a) Convert 764 . 3018 to binary.
b) Convert 11011 . 101112 to octal.
Solution:
a) The binary groups are used to replace individual digits, as indicated:
7 6 4 3 0 1
↓ ↓ ↓ ↓ ↓ ↓
111 110 100 011 000 001
TABLE OP CONVERSION
Decimal Octal Binary
1 1 1 (or 001)
2 2 10(or 010)
3 3 11(or 011)
4 4 100
5 5 101
6 6 110
7 7 111
8 10 1000
9 11 1001
10 12 1010
11 13 1011
12 14 1100
13 15 1101
14 16 1110
15 17 1111
16 20 10000
The resulting number is the required binary number. Some additional insight into the process can be
obtained from the following diagram, which illustrates for the integral part the values of the numbers in
decimal form:
7 6 4
↓ ↓ ↓
7 × 82 6 × 81 4 × 80
↓ ↓ ↓
7 × 26 6 × 23 4 × 20
↓ ↓ ↓
(1×22+1×21+1×20)×26 (1×22+1×21+0×20)×23 (1×22+0×21+0×20)×20
↓ ↓ ↓
1×28+1×27+1×26 1×25+1×24+0×23 1×22+0×21+0×20
↓ ↓ ↓
111 110 100
b) Grouping the digits by threes, starting from the binary point, we have
11 011 . 101 11
It is necessary to introduce an additional zero in the first group and last group in order to have three
binary bits in each group, thus:
011 011 . 101 110
Each of these groups is now replaced by its octal equivalent, giving 33.568
## Solution: It is readily verified by division that
954 = 1 (3)6 + 0 (3)5 + 2(3)4 + 2 (3)3 + 1 (3)2 + 0 (3) + 0;
the successive divisions required are usually recorded in the following set
up.
3 |954
3 |318 + 0
3 |106 + 0
3 |35 + 1
3 |11 + 2
3 |3 + 2
1 +0
Thus the integral part of the given number is 1022100, in the scale of
three.
On the other hand, successive multiplications by 3 give
(4/10) = (12 / 30) = (1/3) + (2 / 30),
(2 / 30) = (6 / 90) = (0/9) + (6 / 90),
(6 / 90) = (18 / 270) = (1 / 27) + (8 / 270),
(8 / 270) = (24 / 810) = (2 / 81) + (4 / 810),
(4 / 810) = (12 / 2430) = (1 / 243) + (2 / 2430),
(2 / 2430) = (6 / 7290) = (0 / 729) + (6 / 7290),
…………………………………………………….
The process is now repeating and, collecting terms, one finds
954.4 (scale 10) = 1022100 . 101210121012 ... (scale 3).
## PROBLEM 01 – 0018: If a = 111010, b = 1011 (scale 2) , evaluate a + b, a – b, ab,
a/b.
CONTENTS CHAPTER PREVIOUS NEXT PREP FIND
## PROBLEM 01 – 0018: If a = 111010, b = 1011 (scale 2) , evaluate a + b, a – b, ab,
a/b.
Solution:
Fortran Rules And Vocabulary
PROBLEM 01 – 0019: Using FORTRAN rules what would be the value of sum in
each case ?
a) SUM = 3 – 1 + 5 * 2/2
b) SUM = (4 – 2) ** 2
c) SUM = 9 ** 2 + 9/2 ** 2
d) SUM = [[{(1 + 2) / (2 * 5)} * (10/2) + 7] / 3]
## CONTENTS CHAPTER PREVIOUS NEXT PREP FIND
PROBLEM 01 – 0019: Using FORTRAN rules what would be the value of sum in
each case ?
a) SUM = 3 – 1 + 5 * 2/2
b) SUM = (4 – 2) ** 2
c) SUM = 9 ** 2 + 9/2 ** 2
d) SUM = [[{(1 + 2) / (2 * 5)} * (10/2) + 7] / 3]
## Solution: The rules adopted by the creators of FORTRAN state that
evaluation of arithmetic operations is conducted from left to right, except
when the succeeding operation has a higher "binding power" than the one
currently being considered.
The binding power of "+" and "–" is the lowest in strength; operators
"*" and "/" are of middle strength, and raising to a power (**) has the
highest priority.
The concept of parenthesizing subexpressions was added to the
rules to achieve a sequence in operations different from the rules of
binding power of operators.
Therefore, using this knowledge the solutions are found to be:
a) First division 2/2, then multiplication by 5, followed by adding to – 1 and
to 3. The result is 7.
b) First, solving the expression inside the brackets, then using exponent 2.
The answer is 4. Note that the operation was done from left to right and
exponent binding power wasn't taken into consideration because of the
priority of parenthesis.
c) Using the binding power rule, 9**2 is 81 and 2**2 is 4. The third
operation division: (9/4) – which is 2.25. The last operation is addition. The
result is 83.25.
d) The result is 14.833.
## PROBLEM 01 – 0020: Each of the following is not a correct FORTRAN statement.
Give reasons in each case.
a) READ (2,1) K, L, M + 1
b) K + 5 = 37 * I – J * 13.
c) WRITE (3,13) I, J, K, L.
d) K = 2 × I
e) Z ** Y ** W
## PROBLEM 01 – 0020: Each of the following is not a correct FORTRAN statement.
Give reasons in each case.
a) READ (2,1) K, L, M + 1
b) K + 5 = 37 * I – J * 13.
c) WRITE (3,13) I, J, K, L.
d) K = 2 × I
e) Z ** Y ** W
## Solution: a) The only acceptable characters in a variable name in any
FORTRAN statement are the twenty-six letters of the alphabet and the ten
digits 0 to 9 (some systems also accept the \$ included in the variable
name). Algebraic signs cannot be included, and therefore M + 1 is an
invalid variable.
b) For FORTRAN statements involving arithmetic operations, use
the following symbols:
– for subtraction
** for exponentiation
* for multiplication
/ for division
No arithmetic operations are allowed on the left side of the "=" sign.
c) In the WRITE statement the first number (3) indicates the device
that is to be used to print the values obtained at the end of the program.
(Typewriter, pi*inter) . The second number (13) is the number of the
corresponding FORMAT statement. The listing of the variables should not
be followed by a period. This is an error.
d) FORTRAN has no symbol "x" (presumably we mean
multiplication, in which case the correct FORTRAN statement is K = 2 * I).
e) The expression is ambiguous. It could mean
(ZY)W or (Z)[(Y)W].
These are not always equal. For example, if Z = 2, Y = 3 W = 4:
(23)4 ≠ 2[(3)4]
## PROBLEM 01 – 0021: Find the mistakes in the following FORTRAN expressions.
a) (A – (B – (C + D (4. 7))
b) ((A/B)
d) V = 1.63 * / D
CONTENTS CHAPTER PREVIOUS NEXT PREP FIND
## PROBLEM 01 – 0021: Find the mistakes in the following FORTRAN expressions.
a) (A – (B – (C + D (4. 7))
b) ((A/B)
d) V = 1.63 * / D
## Solution: a) In FORTRAN any constant or variable in an expression, unless it
is the first constant or variable of that expression, must be preceded
immediately by one of the following:
a left parenthesis
one of the operators +, –, *, /, or ** .
Any constant or variable in an expression, unless it is the last
constant or variable of that expression, must be followed by one of the
following:
a right parenthesis
one of the operators +, – , *, /, or **.
The number of opened and closed parentheses must be equal. Therefore
the correct expression is (A – (B – (C + D(4.7)))).
b) By the same argument that was used in part (a), the correct
expressions are (A/B) or (A) / (B).
c) The statement is incorrectly formed, and has no meaning. Except
for the unary minus, each operator +, *, /, or ** must have a term or facto
both to its left and to its right in order for an expression to be correctly
formed.
PROBLEM 01 – 0022: Pick out the errors in the following FORTRAN statements
and explain briefly why they are incorrect:
a) RESULT = SUM / FLOAT (NUM) + 2ERR
b) IF (M* (N/M) = N) GO TO 20
c) INTEGER CAPITAL, COST, INCOME
d) AREA = LENGTH * WIDTH
e) TAU = BETA / – 3.0
## CONTENTS CHAPTER PREVIOUS NEXT PREP FIND
PROBLEM 01 – 0022: Pick out the errors in the following FORTRAN statements
and explain briefly why they are incorrect:
a) RESULT = SUM / FLOAT (NUM) + 2ERR
b) IF (M* (N/M) = N) GO TO 20
c) INTEGER CAPITAL, COST, INCOME
d) AREA = LENGTH * WIDTH
e) TAU = BETA / – 3.0
## Solution: a) FORTRAN variable names cannot begin with numbers, so 2ERR
is an invalid variable.
b) In an IF statement, comparisons are made by using relational
operators. An equality sign, which is used only in FORTRAN assignment
statements, should be replaced by .EQ. TERM.
c) CAPITAL is a seven-letter word. FORTRAN allows a maximum
of six characters in each variable name.
d) In FORTRAN, variables beginning with letters I through N are
integer unless specified otherwise. Since WIDTH and AREA are real
variable names, LENGTH cannot be used in order to avoid mixed mode
multiplication.
e) Two arithmetic operators may not be juxtaposed in a FORTRAN
assignment statement.
## CHAPTER 02: ERRORS AND APPROXIMATIONS IN NUMERICAL ANALYSIS
Significant Figures, Errors
Absolute And Relative Errors
Truncation And Round-Off Errors
Methods Of Approximation
## Significant Figures, Errors
PROBLEM 02 – 0023: The number 31.546824 is known to have a relative error no
worse than 1 part in 100,000. How many of the digits are
known to be correct?
## Solution: For this case the relative error is
[{ΔQ} / {|Q|}] ≤ [1 / {100,000}]
or
ΔQ ≤ [1 / {100,000}] |Q|
ΔQ is sought. In the below equation, Q is unknown and Q1 is known
(31.546824).
|Q| ≤ |Q1| + ΔQ
Hence
ΔQ ≤ [1 / {100,000}] (|Q| + ΔQ)
or
ΔQ ≤ .00031546824 + [1 / {100,000}] ΔQ
or
[{99,999} / {100,000}] ΔQ ≤ .00031546824
or
ΔQ ≤ .00032
Since ΔQ is less than half a unit in the thousandths place, the significant
digit in the thousandths place (the digit is correct, so the number has at
least five correct significant digits.
## PROBLEM 02 – 0024: Find the error in the evaluation of the fraction
cos 7° 10' ÷ log10 242.7, assuming that the angle may be in
error by 1' and that the number 242.7 may be in error by a
unit in its last figure.
Solution: Since this is a quotient of two functions, it is better to compute the
relative error from the formula Er ≤ Δu1/u1 + Δu2/u2 and then find the
absolute error from the relation Ea = NEr. Now write
N = [{cos 7° 10'} / {log10 242.7}] = [{cos x} / {log10 y}] = u1 / u2
and
Δu1 = Δ cos x = – sin xΔx,
Δu2 = Δ log10 y = 0.43429 (Δy / y).
∴ Er ≤ [{sin x} / {cos x}] Δx + [{0.43429} / {y log y}] Δy,
or
Er ≤ tan xΔx + [{0.435} / {y log y}] Δy.
Now taking x = 7° 10', Δx = 1' = 0.000291 radian, y = 242, Δy = 0.1,
and using a slide rule for the computation,
Er < 0.126 × 0.000291 + [{0.435 × 0.1] / {242 × 2.38}] = 0.00011.
Since N = cos 7° 10'/log 242.7 = 0.41599,
Ea = 0.00011 × 0.416 = 0.000046,
or
Ea < 0.00005.
The value of the fraction is therefore between 0.41604 and
0.41594, and the mean of these numbers is taken to four figures as the
best value of the fraction, or
N = 0.4160.
## PROBLEM 02 – 0025: The following quadratic approximations to f(x) = ex were
obtained by telescoping the Taylor series for ex:
g1(x) = 0.994571 + 1.130318x + 0.542990x2, – 1 ≤ x ≤ 1
g2(x) = 1.008129 + 0.860198x + 0.839882x2, 0 ≤ x ≤ 1
In each case a sufficient number of terms was employed in
the original Taylor series to ensure that the coefficients of
these approximations have essentially "converged." That is,
if additional terms were taken in the Taylor series, the
coefficients of g1(x) and g2(x) would not change significantly.
Note that gi(x) is a valid approximation over the interval
– 1 ≤ x ≤ 1 while g2(x) applies only over the interval
0 ≤ x ≤1. Compare the accuracy of these two approximations
on 0 ≤ x ≤ 1 where both are valid.
Solution: The errors in each of the approximations are plotted in the figure.
The errors in both approximations have nearly equal positive and negative
peaks. However, the error in g2 (x) is distributed much more uniformly over
the interval. (The error in g1(x) is distributed quite uniformly over the
interval – 1 ≤ x ≤ 1 for which it was con true ted, but g1(x) tends to
overestimate ex on much of 0 ≤ x ≤ 1). The magnitude of the maximum
error of g1(x) is approximately 0.054, while that of g2(x) is approximately
0.0101, or only about 1/5 that of g1(x). It is almost inevitable that the
quadratic g1(x) which must provide a good approximation over the interval
– 1 ≤ x ≤ 1 will have a larger maximum error than will an approximation of
the same type (g2(x)) which is constructed to serve as a good
approximation over an interval only half as large (0 ≤ x ≤ 1) . It would be
necessary to employ an economized approximation of higher degree than
2 in order to obtain the same accuracy on – 1 ≤ x ≤ 1 that g2(x) provides
on 0 ≤ x ≤ 1. It should be apparent that the simplest, most effective
approximations can be obtained by restricting the interval of approximation
to the absolute minimum size required.
## Absolute And Relative Errors
PROBLEM 02 – 0026: The following definitions apply to errors in a calculation:
The absolute error is defined as:
Errorabs = (Calculated value) – (True value)
The relative error is defined as:
Errorrel = [{(Calculated value) – (True value)}
/ {True value}]
The percentage error is defined as
Errorpct = Errorrel × 100
Assume the true value for a calculation should be 5.0, but
the calculated value is 4.0: Calculate the absolute error,
relative error and the percentage error.
## Solution: Errorabs = 4.0 – 5.0 = – 1.0.
Errorrel = [{4.0 – 5.0} / {5.0}] = [{– 1.0} / {5.0}] = – 0.2,
Errorpct = – 0.2 × 100 = – 20%.
## PROBLEM 02 – 0027: Show that the relative error in a quantity is approximately
equal to the absolute error in its natural logarithm, since
Δ(In y) ≈ d(ln y) = (dy / y) ≈ (Δy / y).
Solution: Using an exact formula, one has
Δ(In y) = In (y + Δy) – ln(y) = ln(1 + Δy / y). A comparison follows:
## (Δy / y) = 0.001 0.01 0.1 0.2 – 0.2 – 0.1
Δ(In y) = 0.00100 0.00995 0.095 0.182 – 0.223 – 0.105
## For logarithms to base 10 ("common logarithms") one has
Δ(log10 y) = log10 e ∙ Δ(ln y) ≈ 0.434 (Δy / y).
## PROBLEM 02 – 0028: Let x be a real number and F(x)it's floating–point k–digit
representation. For example, if x = (2/9) and k = 4, then
F(2/9) = .2222. The operations of addition, subtraction,
multiplication and division are defined as follows:
x (+) y = F[F(x) + F(y)]
x (–) y = F[F(x) – F(y)] (1)
x (×) y = F[F(x) × F(y)]
x (÷) y = F[{F(x)} / {F(y)}]
Compute the absolute error and relative error for the
operations defined in (1) for x = (1/3), y = (4/7). For
arithmetic calculations apply 4–digit chopping. Suppose
a = .5711, b = 4271, c = .1001 × 10–3,
compute
a (+) b
(a (–) y) (÷) c (2)
and the corresponding absolute errors and relative errors.
## Solution: With the help of a pocket calculator, we find
x = (1/3) F(x) = .3333
y = (4/7) F(y) = .5714
Then x (+) y = (1/3) (+) (4/7)
= F[F(x) + F(y)]
= F[. 3333 + .5714]
= F[.9047]
= .9047
The results of the operations are listed in Table 1.
Table 1
Operation Result Actual Value
x (+) y .9047 (19 / 21)
x (–) Y – .2381 – (5 / 21)
x (×) y .1904 (4 / 21)
x (÷) y .5833 (7 / 12)
## Let ω̃ be an approximate value of ω; then absolute error is defined by
|ω – ω̃| (3)
and the relative error is
[{|ω – ω̃|} / {|ω|}] (4)
assuming ω ≠ 0.
From Table 1 and (3), (4) we compute
Table 2
Operation Result Actual Value
x (+) y .6190 × 10–4 .6842 × 10–4
x (–) Y .4762 × 10–5 .2 × 10–4
x (×) y .7619 × 10–4 .4 × 10–3
x (÷) y .3333 × 10–4 .5714 × 10–4
## Note that the maximum relative error is .4 × 10–3.
If we apply, for example, 7-digit chopping then
F(x) = .3333333 F(y) = .5714286
and
x (+) y = .1904762.
The absolute error is .191 × 10–7, and the relative error is .131 × 10–6. The
accuracy has improved.
For a, b, c given in (2), we obtain
Table 3
Operation Result Actual Value Absolute Relative Error
Error
a (+) b 4271 4271.5711 .5711 .1123 × 10-3
(a (–) y) (÷) c – .003 – .000328 .00267 8.1445
Note that the absolute error in a (+) b is large, while the relative error is
small because we divide by a large actual value.
One should exercise caution in arranging the arithmetic operations
and avoiding multiplication by large numbers (or division by small
numbers) and subtraction of two numbers nearly equal.
## Truncation And Round-Off Errors
PROBLEM 02 – 0028:
Let x be a real number and F(x)it's floating–point k–digit representation. For example, if x = (2/9)
and k = 4, then F(2/9) = .2222. The operations of addition, subtraction, multiplication and division
are defined as follows:
x (+) y = F[F(x) + F(y)]
x (–) y = F[F(x) – F(y)] (1)
x (×) y = F[F(x) × F(y)]
x (÷) y = F[{F(x)} / {F(y)}]
Compute the absolute error and relative error for the operations defined in (1) for x = (1/3), y =
(4/7). For arithmetic calculations apply 4–digit chopping. Suppose
a = .5711, b = 4271, c = .1001 × 10–3, compute
a (+) b
(a (–) y) (÷) c (2)
and the corresponding absolute errors and relative errors.
## Solution: With the help of a pocket calculator, we find
x = (1/3) F(x) = .3333
y = (4/7) F(y) = .5714
Then x (+) y = (1/3) (+) (4/7)
= F[F(x) + F(y)]
= F[. 3333 + .5714]
= F[.9047]
= .9047
The results of the operations are listed in Table 1.
Table 1
Operation Result Actual Value
x (+) y .9047 (19 / 21)
x (–) Y – .2381 – (5 / 21)
x (×) y .1904 (4 / 21)
x (÷) y .5833 (7 / 12)
## Let ω̃ be an approximate value of ω; then absolute error is defined by
|ω – ω̃| (3)
and the relative error is
[{|ω – ω̃|} / {|ω|}] (4)
assuming ω ≠ 0.
From Table 1 and (3), (4) we compute
Table 2
Operation Result Actual Value
x (+) y .6190 × 10–4 .6842 × 10–4
x (–) Y .4762 × 10–5 .2 × 10–4
x (×) y .7619 × 10–4 .4 × 10–3
x (÷) y .3333 × 10–4 .5714 × 10–4
## Note that the maximum relative error is .4 × 10–3.
If we apply, for example, 7-digit chopping then
F(x) = .3333333 F(y) = .5714286
and
x (+) y = .1904762.
The absolute error is .191 × 10–7, and the relative error is .131 × 10–6. The
accuracy has improved.
For a, b, c given in (2), we obtain
Table 3
Operation Result Actual Value Absolute Relative Error
Error
a (+) b 4271 4271.5711 .5711 .1123 × 10-3
(a (–) y) (÷) c – .003 – .000328 .00267 8.1445
Note that the absolute error in a (+) b is large, while the relative error is
small because we divide by a large actual value.
One should exercise caution in arranging the arithmetic operations
and avoiding multiplication by large numbers (or division by small
numbers) and subtraction of two numbers nearly equal.
## PROBLEM 02 – 0029: Give a brief explanation of mathematical truncation error.
Solution: Mathematical truncation error refers to the error of approximation in
numerically solving a mathematical problem, and it is the error generally
associated with the subject of numerical analysis. It involves the
approximation of infinite process by finite ones, replacing non-computable
problems with computable ones. Following are some examples to make
the idea more precise.
(a) Using the first two terms of the Taylor series from
(1 + x)α = 1 + (α1)x + (α2)x2 + … + (αn)xn + (α n+1) [{xn+1} / {(1 + ξx)n+1–α}]
with
(αk) = [{α(α – 1)…(α – k + 1)} / {k!}] k = 1, 2, 3 …
for any real number a(for all cases the unknown ξx is located between x
and 0 ), one can write
√(1 + x) ≈ 1 + (1/2)x
which is a good approximation when x is small.
(b) For the differential equation problem
Y'(t) = f(t, Y(t)) Y(t0) = Y0
use the approximation of the derivative
Y'(t) ≈ [{Y(t + h) – Y(t)} / {h}]
for some small h. Let tj = t0 + jh for j ≥ 0, and define an approximate
solution function y(tj)by
[{y(tj +1) – y(tj)} / {h}] = f(tj, y(tj))
so one gets
y(tj + 1) = y(tj) + hf (tj, y(tj)) j≥0
This is Euler's method of solving a differential equation initial value
problem.
PROBLEM 02 – 0030: Use exact, chopping and rounding to evaluate the following:
f(x) = x3 – 4x2 + 2x – 2.2 at x = 2.41.
For chopping and rounding, apply three–digit arithmetic.
Compute the relative errors. To decrease their values, try to
carry out the same calculations for the same polynomial, but
written in a different form.
Solution: The numerical data necessary for the solution of the problem are
summarized in the following table.
x x2 x3 4x2 2x
exact 2.41 5.8081 13.997521 23.2324 4.82
3–digit 2.41 5.80 13.9 23.2 4.82
chopping
3–digit 2.41 5.81 14.0 23.2 4.82
rounding
Note that the three-digit chopped numbers retain the leading three digits
and may differ from the three-digit rounded numbers. For example, if
x = 14.7124, then Xchopped = 14.7, and Xrounded = 14.7. On the other hand, if
x = 14.7824, then Xchopped = 14.7. Xrounded = 14.8.
We obtain
f(2.41)exact = 13.997521 – 23.2324 + 4.82 – 2.2
= – 6.614879
f(2.41)chopped = 13.9 – 23.2 + 4.82 – 2.2
= – 6.68
f(2.41) rounded = 14.0 – 23.2 + 4.82 – 2.2
= – 6.58
The corresponding relative errors are:
relative error, 3-digit chopping = [{– 6.614879 + 6.68} / {– 6.614879}]
= .0098
relative error, 3-digit rounding = [{– 6.614879 + 6.58} / {– 6.614879}]
= .0052
In order to obtain smaller values for the relative errors we can write f(x) in
an equivalent form
f(x) = x[x(x – 4) + 2] – 2.2
Then, the exact value of f(x) at x = 2.41 is, of course, the same, but
f{2.41)chopped = 2.41[2.41(2.41 – 4) + 2] – 2.2
= 2.41[– 3.83 + 2] – 2.2
= – 4.41 – 2.2
= – 6.61
f(2.41)rounded = 2.41[2.41(2.41 – 4) + 2] – 2.2
= – 6.61
The relative error for both values is now
[{– 6.614879 + 6.61} / {– 6.614879}] = .00073
The relative error is now much smaller. The reason for this is that there is
a decrease in the number of error-producing computations. Note that in
x3 – 4x2 + 2x – 2.2, we had five multiplications and three additions, while in
x[x(x – 4) + 2] – 2.2, we had two multiplications and three additions.
PROBLEM 02 – 0031: When one gives the number of digits in a numerical value
one should not include zeros in the beginning of the number,
as these zeros only help to denote where the decimal point
should be. If one is counting the number of decimals, one
should of course include leading zeros to the right of the
decimal point.
For example, 0.001234 ± 0.000004 has five correct
decimals and three significant digits, while 0.001234
± 0.000006 has four correct decimals and two significant
digits.
The number of correct decimals gives one an idea of the
magnitude of the absolute error, while the number of
significant digits gives a rough idea of the magnitude of the
relative error.
Consider the following decimal numbers: 0.2397, – 0.2397,
0.23750, 0.23650, 0.23652. Shorten to three decimals by the
Round–off and Chopping methods.
## Solution: 0.2397 rounds to 0.240 (is chopped to 0.239),
– 0.2397 rounds to – 0.240 (is chopped to – 0.239),
0.23750 rounds to 0.238 (is chopped to 0.237),
0.23650 rounds to 0.236 (is chopped to 0.236),
0.23652 rounds to 0.237 (is chopped to 0.236).
Observe that when one rounds off a numerical value one produces
an error; thus it is occasionally wise to give more decimals than those
which are correct.
One consequence of these rounding conventions is that numerical
results which are not followed by any error estimations should often,
though not always, be considered as having an uncertainty of (1/2) unit in
the last decimal place.
## PROBLEM 02 – 0032: Assume that floating-point arithmetic with three significant
digits is used on the system
x + 400y = 801,
200x + 200y = 600.
Solve it by rounding off the errors.
Solution: Multiplying the first equation by – 200 and adding the result to the
second equation, one gets
– 7.98 × 104 y = – 15.9 × 104.
This gives the approximate value of 1.99 for y. Substitution of this
value in the first equation yields the approximate value x = 5. The relative
error in the value for x is 400% since the correct solution is (1, 2). The
problem here lies in the equation x = 801 – 400y. Any rounding error in y
(in this case 0.01) gets magnified by a factor of 400, which can be quite
significant because x is small.
PROBLEM 02 – 0033: Scaling is one way to get around errors that can be caused
by improportionate computations. Scale the system
x + 230y + 34602 = 20,000
30x + 5y + 0.1z = 300 (1)
0.001x + 0.002y + 0.003z = 7,
by columns.
Hint: Column scaling is similar to row scaling except that it
alters the solution. For example, if the first column is divided
by 10, then the new solution will be in terms of 10x1 instead
of x1.
## Solution: In order to scale the system (1) by columns, let x = x* / 102,
y = y* / 103, and z = z* / 104. This results in the system
0.01x* + 0.230y* + 0.3460z* = 20,000,
0.3x* + 0.005y* + 0.00001z* = 300, (2)
0.00001x* + 0.000002y* + 0.0000003z* = 7.
Of course, one must remember to replace x*, y*, and z* by their values in
terms of x, y, and z after solving the system (2).
## PROBLEM 02 – 0034: On the system
x + 400y = 801
200x + 200y = 600,
use the technique of partial pivoting to round off the
solution's error to three significant digits.
## Solution: Scaling the system by rows, one gets
0.001x + 0.4y = 0.801,
0.2x + 0.2y = 0.6.
The technique of partial pivoting requires that the second equation be
used as the pivot since 0.2 > 0.001. Carrying three significant digits, one
obtains the equation
0.399y = 0.798.
Thus y = 2.00 and x = 1.00. Note that a small round–off error in y would
have produced only a small error in x, since the second equation is less
sensitive than the first to an error in y.
Methods Of Approximation
## PROBLEM 02 – 0035: Approximate y = loge x = In x by polynomials of respective
max-degrees 1, 2, 3, in the neighborhood of the point (1, 0).
## Solution: Here, x0 = 1; and y' = x–1, y" = – x–2, y= 2x–3. Hence
f(1) = 0, f'(1) = 1, f"(1) = – 1, f"'(1) = 2,
and
y = x – 1,
y = (x–1) – (1/2) (x – 1)2,
y = (x – 1) – (1/2) (x – 1)2 + y (x – 1)3,
are the required approximating polynomials.
## PROBLEM 02 – 0036: Let f(x) = [1 / (1 + x)], then yk = [1 / (k + 1)], k = 0, 1, 2, ...
n – 1, and
n–1∫ f(x) dx = In n.
0
## (It is convenient to stop at the index n – 1 rather than the
index n.) Given the above, use the Euler Summation formula
n∑ yi = n∫0 f(x)dx + (1/2) (y0 + yn) + n∫0 P1(x)f'(x)dx
i=0
where
Pn(x) = a + a1(x – x0) + a2(x – x0)2 + ... + an(x – x0)
to derive the formula for the Euler's Constant.
Solution: n–1∑
k=0 [1 / {k + 1}] = In n + (1/2) [1 + (1/n)]
– n–1∫0 [{P1(x)} / {x + 1}2]dx
Here P1(x) is a polynomial used to approximate the function of x.
Since
|P1(x)| ≤ (1/2)
for every x and
k+1∫ [{P1(x)} / {x + 1}2]dx < 0,
k
then
0 > n–1∫0 [{P1(x)} / {x + 1}2]dx ≥ – (1/2) n–1∫0 [1 / {x + 1}2]dx
= – (1/2) [1 – (1/n)].
It follows that
∞∫ [{P1(x)} / {x + 1}2]dx = limn→∞ n–1∫0 [{P1(x)} / {x + 1}2]dx
0
## exists and is between 0 and – (1/2). Therefore the number C defined by
C = limn→∞ [n–1∑k=0 (1 / {k + 1}) – In n] = (1/2) – ∞∫0 [{P1(x)} / {x + 1}2]dx
exists. The number C is called Euler's constant; it can be shown that
C = 0.577216.
## PROBLEM 02 – 0037: If f(x) = (1/2) (x + |x|), find the parabola
y = a0 + a1x + a2x2
which minimizes the integral Is given by
Is = b∫a [f(x) – Pn(x)]2 dx
where
Pn(x) = y
in the interval [–1, 1].
## Solution: Thus, (the subscript s is omitted);
I = 1∫–1 [{(1/2) (x + |x|) – (a0 + a1x + a2x2)]2 dx
= 0∫–1 (a0 + a1x + a2x2)2 dx + 1∫0 [a0 + (a1 – 1) x + a2x2]2 dx
= (1/3) + 2a02 – a0 + (4/3) a0a2 + (2/3) a12 – (2/3)a1 + (2/5) a22 – (1/2)a2.
Now
I(a)0 = 4a0 – 1 + (4/3)a2,
I(a)1 = (4/3)a1 – (2/3),
I(a)2 = (4/3)a0 + (4/5)a2 – (1/2),
where
I(a)i = (∂I / ∂ai),
and
I(a)0 (a)0 = 4, I(a)0 (a)1 = I(a)1 (a)0 = 0,
I(a)1 (a)1 = (4/3), I(a)0 (a)2 = I(a)2 (a)0 = (4/3),
I(a)2 (a)2 = (4/3), I(a)1 (a)2 = I(a)2 (a)1 = 0,
where
I(a)i (a)j = [(∂2I) / (∂ai ∂aj)].
Sufficient conditions for a minimum are
I(a)0 = I(a)1 = I(a)2 = 0,
| I(a)0 (a)0 I(a)0 (a)1 I(a)0 (a)2|
| I(a)1 (a)0 I(a)1 (a)1 I(a)1 (a)2| > 0.
| I(a)2 (a)0 I(a)2 (a)1 I(a)2 (a)2|
These conditions are satisfied if
a0 = (3 / 32),
a1 = (16 / 32),
a2 = (15 / 32),
and therefore
y = (1 / 32) (3 + 16x + 15x2)
is the required polynomial. It is suggested to graph this parabola and
y = (1/2) (x + |x|)
on the same set of axes.
PROBLEM 02 – 0038: Approximate
y = esin x
at x = 0 by a function of type
C(x) = a0 + a1 sin (2π / p) (x – x0)
+ ... + an sin n (2π / p) (x – x0) + b1 cos (2π / p) (x – x0)
+ ... + bn cos m (2π / p) (x – x0),
where
a0, ... , an, b0, ... , bm and x0
are constants; C(x) is everywhere continuous and periodic
with period p and hence is suitable for approximating the
function y = f (x) at a point (x0, y0) if the function f(x) is itself
continuous and periodic with period p. As a rule, m is chosen
to be n, n – 1, or n + 1. The function C(x) has n + m + 1
arbitrary constants and hence can be determined by the
imposition of an equivalent number of conditions. Solve for
n = m = 1 and for n = m = 3.
## Solution: For n = m = 1, take x0 = 0 and put
C1(x) = a0 + a1 sin x + b1 cos x.
(The function to be approximated has the period p = 2π.) Determine the
three constants a0, a1, and b1 by equating the values of esin x and C1(x) and
the values of their first and second derivatives at x = x0. Obtain
C1(x) = 2 + sin x – cos x.
If the graphs of y = C1(x) and the given function were drawn, they would
show the periodicity of the two curves and indicate that C1(x) is a good
approximation to esin x from about – 2π / 9 to about 7π / 18 (roughly from
– 40° to 70°) in the period interval from – π to π.
C3(x) = a0 + a1 sin x + a2 sin 2x + a3 sin 3x + b1 cos x + b2 cos 2x
+ b3 cos 3x.
Since six derivatives are needed it is convenient to write esin x as a power
series in x, namely,
y = esin x = 1 + x + (x2 / 2!) – (3x4 / 4!) – (8x5 / 5!) – (3x6 / 6!)
+ (56x7 / 7!) + ... .
Obtain on substituting 0 for x, equating the values of corresponding derivatives, and solving,
C3(x) = (1 / 180) (200 + 210 sin x – 6 sin 2x – 6 sin 3x + 45 cos x
– 72 cos 2x + 7 cos 3x) .
The graphs of y = C3(x) and the given function would show that this time
the approximation is good from about – 4π / 9 to about 11π / 18 (– 80° to
110°) in the period interval from – π to π.
## PROBLEM 02 – 0039: Approximate y = esin x by a function of type
C(x) = a0 + a1 sin (2π / p) (x – x0)
+ ... + an sin n (2π / p) (x – x0) + b1 cos (2π / p) (x – x0)
+ ... + bn cos m (2π / p) (x – x0),
where
a0, ..., an, b0,..., bm and x0 are constants.
C(x) coincides with y = esin x at x = 0, π/3, π/2, π, 3π / 2.
## Solution: First determine the constants in
C2(x) = a0 + a1 sin x + a2 sin 2x + b1 cos x + b2 cos 2x
so that the graph of this equation passes through the five points
(0,1), (π/3, e(1/2)√3), (π/2, e), (π, 1), (3π/2, e–1) .
Substituting the coordinates of the points into C2(x), equating
corresponding values, and solving,
C2(x) = 1.272 + 1.175 sin x – 0.057 sin 2x – 0.272 cos 2x.
Its graph and the graph of y = esin x indicate that C2(x) is a good
approximation to y = esin x for all values of x.
Note that because of the periodicity of. the trigonometric functions, one is
more likely to run into a system of inconsistent linear equations (in the a's
and b's) than in polynomial approximation.
## PROBLEM 02 – 0040: Let Pn*(x) denote the polynomial of degree ≤ that
approximates F(x) = 16–x with minimax absolute error in
[0, a], and let Qn*(x) denote the polynomial of degree ≤ n that
approximates F(x) = 16–x with minimax relative error in [0, a].
Use the theorem:
Let Pn*(x) denote the polynomial of degree ≤ n that
approximates F(x) with minimax absolute error in [a, b].
Suppose that F(x) possesses an (n + 1) st derivative F(n+1)
(x) for x in [a, b] . If for two nonnegative numbers m and M
and for a ≤ x ≤ b either
m ≤ F(n+1)(x) ≤ M
or
m ≤ F(n+1)(x) ≤ M,
then
[{m(b – a)n+1} / {22n+1(n + 1)!}] ≤ max[a, b] |Pn*(x) – F(x)|
≤ [{M(b – a)n+1} / {22n+1(n + 1)!}];
to find bounds for both
max[0, a] |Pn*(x) – 16–x|
and
max[0, a] |{Qn*(x) – 16–x} / {16–x}|.
Solution: Since
F(n+1)(x) = (– 1)n+1(In 16)n+1 16–x,
one gets m ≤ ± F(n+1)(x) ≤ M for all x in [0, a],
where
m = 16–a(In 16)n+1
and
M = (In 16)n+1,
thus satisfying the condition of the theorem, it follows that
[{16–a(In 16)n+1 an+1} / {22n+1(n + 1)!}] ≤ max[0, a] |Pn*(x) – 16–x|
≤ [{(In 16)n+1 an+1} / {22n+1(n + 1)!}]
Applied directly, the theorem cannot give bounds for
max[0, a] |{Qn*(x) – 16–x} / {16–x}|.
With the additional analysis shown below, however, one can also obtain
such bounds.
max[0, a] |{Qn*(x) – 16–x} / {16–x}| ≥ max[0, a] |Qn*(x) – 16–x|
≥ max[0, a] |Pn*(x) – 16–x|
≥ [{16–a(In 16)n+1 an+1} / {22n+1(n + 1)!}]
Likewise,
max[0, a] |{Qn*(x) – 16–x} / {16–x}| ≤ max[0, a] |{Pn*(x) – 16–x} / {16–x}|
≤ 16a max[0, a] |Pn*(x) – 16–x|
≤ [{16a (In 16)n+1 an+1} / {22n+1(n + 1)!}]
Therefore
[{16–a(in 16)n+1 an+1} / {22n+1(n + 1)!}] ≤ max[0, a] |{Qn*(x) – 16–x} / {16–x}|
≤ [{16a (In 16)n+1 an+1} / {22n+1(n + 1)!}].
## PROBLEM 02 – 0041: Find the polynomial P2*(x) of degree ≤ 2 that approximates
F(x) = √x with minimax relative error in [(1 / 16), 1]. Use
Remez' method.
As initial estimates for the critical points of
[{P2*(x) – √x} / {√x}],
use
x1 = (1 / 16), x2 = .4, x3 = .8, and x4 = 1
Solution: Applying Reme z' method:
Step (1) Choose the initial estimates for the critical points.
Step (2) The system of linear equations
a0 + a1xk +…+ anxkn – (– 1)k μF (xk) = F(xk),
k = 1, 2,…, n + 2,
becomes
a0 + a1x1 + a2x12 + μ√x1 = √x1
a0 + a1x2 + a2x22 + μ√x2 = √x2
a0 + a1x3 + a2x32 + μ√x3 = √x3
a0 + a1x4 + a2x42 + μ√x4 = √x4
On each iteration this system of equations is solved numerically by means
of Gaussian elimination.
Step (3) On each iteration it is necessary to locate the extreme points in
[(1 / 16), 1]
of the relative-error function
[{P2(x) – √x} / {√x}],
where the coefficients of
P2(x) = a0 + a1x + a2x2
are obtained from step (2) . Two of the extreme points are
y1 = (1 / 16)
and
y4 = 1
on every iteration. The other two, y2 and y3, are the two roots of the
3a2x2 + a1x – a0 = 0
and are computed with the quadratic formula on every integration.
The quadratic equation in step (3) was derived as follows:
Let
E(x) = [{P2(x) – √x} / {√x}].
If one replaces P2(x) by a1 + a1x + a2x2, differentiates E(x), and set the
derivative equal to zero, then one obtains after some simplifications the
condition
3a2x2 + a1x – a0 = 0.
Since E(x) must have at least four extreme points in
[{1 / 16), 1],
whereas the quadratic equation has only two roots, it follows that y1 and
y4 must be the endpoints of the interval and that the two interior extreme
points, y2 and y3, must be the roots of the quadratic equation. Note,
incidentally, that this argument also establishes that
[{P2*(x) – √x} / {√x}]
must be a standard error function.
TABLE 1
After x2 x1 a0 a1 a2 μ
iteration
0 .400000 .800000 — — — —
1 .137582 .692428 .171896 1.339080 – .525123 – .014147
2 .146345 .588965 .170415 1.453820 – .659052 – .034816
3 .148921 .590636 .171509 1.442061 – .649966 – .036396
4 .148917 .590599 .171509 1.442106 – .650022 – .036407
Table 1 shows the results of four iterations with Remez' method. The
numbers shown in this table have been rounded to six decimal places.
Values of the iterates converged to six decimals in these four iterations.
## PROBLEM 02 – 0042: Find the polynomial P7*(x) of degree ≤ 7 that approximates
cos (1/4) πx
with minimax absolute error in [–1, 1] . Use the Remez'
method. The initial values are prescribed by
xk = (1/2) (b – a) cos [{n – k + 2} / {n + 1}] + (1/2) (b + a),
k = 1, 2, ..., n + 2. (1)
Solution: Since
cos (1/4) πx
is an even function and since *the approximation interval is symmetric
about the origin, P7*(x) contains only even-degree terms.
Let
P7*(x) = a0* + a2*x2 + a4*x4 + a6*x6.
The absolute-err or function P7*(x) – cos (1/4) πx is a standard error
function. It is also an even function, and its nine critical points are
therefore symmetrically located about the origin. That is,
– x1* = a9*, – x2* = x8*, – x3* = x7* – x4*, = x6*
and
x3* = 0.
If the starting values in step 1 of Remez' method are chosen so that
analogous conditions hold, then this symmetry be preserved throughout
subsequent iterations, and the odd- degree coefficients of P7(x) in step 2
will be zero on every iteration. Taking advantage of these facts, apply
Remez method in the following way:
Step 1) Because of the symmetries mentioned above, it is necessary to
use only x5, x6, x7 x8, and x9. For initial values use (1), that is,
xk = cos [{(9 – k)π} / {8}],
k = 5, 6, 7, 8, 9.
Step 2) The system of linear equations
a0 + a1xk + ... + anxkn – (– 1)kμ = F(xk),
k = 1, 2,..., n + 2,
for the n + 2 unknowns a0, a1,..., an, and μ,
simplifies to the followings
a0 + a2x52 + a4x54 + a6x56 + μ = cos (1/4) πx5
a0 + a2x62 + a4x64 + a6x66 – μ = cos (1/4) πx6
a0 + a2x72 + a4x74 + a6x76 + μ = cos (1/4) πx7
a0 + a2x82 + a4x84 + a6x86 – μ = cos (1/4) πx8
a0 + a2x92 + a4x94 + a6x96 + μ = cos (1/4) πx9.
On each iteration this system of equations is solved numerically ,by
means of Gaussian elimination.
Step 3) Since P7(x) – cos (1/4) πx is an even function, its extreme points
are distributed symmetrically about the origin. Therefore, as was said
earlier, it is necessary to locate only y5, y6, y7, y8, and y9. On every
iteration y5 = 0 and y9 = 1. To obtain y6, y7, and y8, differentiate the
absolute-error function P7(x) – cos (1/4) πx and compute the zeros of the
derivative. This can be done by isolating each yk by searching and then
computing it accurately by bisection. Since the starting values chosen for
step 1 are very good, yk is always close to xk and can be isolated easily by
searching near xk.
Because the initial choices for the xk's happen to be very accurate, after
only one iteration a0, a2, a4, a6, and μ become stabilized to ten decimal
places. The xk's do not converge quite so rapidly. The results after one
iteration are shown below, rounded to ten decimal places.
a0 = .9999999724 a6 = – .0003188805
a2 = – .3084242536 μ = .0000000276
a4 = .0158499153
PROBLEM 02 – 0043: Develop the power series expansion of cos (1/4) πx.
Solution: cos (1/4) πx = ∞∑k=0 [{(– 1)k {(1/4) π}2k} / {(2k)!}] x2k
= 1 – .3084251375x + .0158543442x4*
– .0003259919x6 + .0000035909x8 – .0000000246x10
+ ... ,
where the coefficients have been rounded to ten decimal places. Let P6(x)
denote the polynomial of degree six obtained by truncating this series after
the fourth term, and consider this polynomial as an approximation to cos
(1/4) πx for – 1 ≤ x ≤ 1. Graphs of the absolute-error function P6(x) – cos
(1/4) πx and the relative-error function
[{P6(x) – cos (1/4) πx} / {cos (1/4) πx}]
for 0 ≤ x ≤ 1 are shown in the figure. For both absolute error and relative
error, the magnitude of the error is greatest when x = 1. Simple
calculations show that the maximum absolute error is approximately .357
(10–5) and that the maximum relative error is approximately .504(10–5)
## PROBLEM 02 – 0044: Using Remez' method, determine the rational function
R5,2(x) in V5, 2[– 1, 1] that approximates cos (1/4)πx with
minimax absolute error in [– 1, 1]. Hint: Since cos (1/4)πx is
an even function and the approximation interval is symmetric
about the origin, the odd-degree coefficients in
Rm,n*(x) = [{p0* + p1*x + p2*x2 + ... + pm*xm}
/ {q0* + q1*x + q2*x2 + ... + qn*xn}], (1)
vanish so that it becomes just
R5,2* (X) = [{p0* + p2*x2 + p4*x4} / {1 + q2*x2}]
Solution: Approximate numerical values for p0*, p2*, p4*, and. Q2* need to be
computed. Since the absolute-error function for this approximation is an
even function, too, the nine critical points are symmetrically located about
the origin. That is,
– x1* = x9*, – x2* = x8*, – x3* = x7*, – x4* = x6*, and x5* = 0.
Choose the starting values xk, k = 1, 2, ..., 9, so that similar conditions hold
for them; then this will continue to be true on every iteration of Remez'
method, and the odd–degree coefficients of R5,2(x) in step 2 will vanish on
every iteration. Now apply the steps in Remez' method in the following
way:
Step 1) Because of the symmetries mentioned above, one needs to us
only x5, x6, x7, x8, and x9. Using
xk = (1/2) (b – a) cos [{(m + n + 2 – k)π} / {m + n + 1}] + (1/2) (b – a),
k = 1, 2, ..., m + n + 2,
to determine starting values, one gets
xk = cos [{(9 – k)π} / {8}], k = 5, 6, 7, 8, 9.
Step 2) The system of nonlinear equations
p0 + p1xk + ... + pmxkm + q1xk [– F(xk) – (–1)kμ] + q2xk2 [– F(xk)
– (–1)kμ] +...+ qnxkn [– F(xk) – (–1)kμ] – (– 1)k μ = F(xk),
k = 1, 2, 3,..., m + n + 2,
for the m + n + 2 unknowns p0, p1,..., pm, q1, q2,..., qn, and μ; becomes
p0 + p2x52 + p4x54 + q2x52 (– cos (1/4) πx5 + μ) + μ = cos (1/4)πx5
p0 + p2x62 + p4x64 + q2x62 (– cos (1/4) πx6 + μ) + μ = cos (1/4)πx6
p0 + p2x72 + p4x74 + q2x72 (– cos (1/4) πx7 + μ) + μ = cos (1/4)πx7
p0 + p2x82 + p4x84 + q2x82 (– cos (1/4) πx8 + μ) + μ = cos (1/4)πx8
p0 + p2x92 + p4x94 + q2x92 (– cos (1/4) πx9 + μ) + μ = cos (1/4)πx9
On each iteration with Remez' method, this system is solved for p0, p2,
p4, q2, and μ by the iterative method.
Step 3) Since the absolute–error function R5,2(x) – cos (1/4) πx is an even
function, it is sufficient to locate its extreme points in [0, 1], instead of
[– 1, 1]. The coefficients of R5,2(x) are the numbers obtained in step 2.
Always have y5 = 0 and y9 = 1. To obtain y6, y7, and y8, differentiate the
error function and compute the zeros of the derivative. This can be done
by isolating each yk by searching and then computing it accurately by
bisection. Since the starting values for the xk's chosen in step 1 happen to
be very good, yk is close to xk on every iteration and can easily be isolated
by searching in the neighborhood of xk.
Results after two iterations with Remez' method are shown below,
with constants rounded to ten decimal places. After two iterations the
numbers have stabilized to ten decimals.
p0 = 1.0000000241 q2 = .0209610796
p2 = – .2874648358 y = – .0000000241
p4 = .0093933390
## PROBLEM 02 – 0045: Using a rational approximation method, approximate tanh μx
for
– 1 ≤ x ≤ 1,
where
μ = (1/2) In 3.
Solution: First, find a rational approximation
R2,4(x) to
[{tan μx} / {x}]
by Maehly's method and use xR2,4(x) as an approximation to tanh μx. The object of using this
procedure (instead of applying Maehly's method to tanh μx itself) is to obtain an approximation to
tanh μx with low relative error around the origin, at which tanh μx is zero. Consider the equations
p0 = q0c0 + (1/2) n∑r=1 qrcr,
pk = q0ck + (1/2) qkc0 + (1/2) n∑r=1 qr(cr+k + c|r–k|),
k = 1,2, ..., m + n. (1)
In this case use m = 2 and n = 4. The ck's are the coefficients of the Chebyshev series for
[{tanh μx} / x];
all the odd–numbered ck's are zero because
[{tanh μx} / x]
is an even function. It is not difficult to show that the unknowns p1, q1, and q3 equal zero. If all the
zero terms are omitted and one simplifies the equations, then (1) becomes
p0 = q0c0 + (1/2) q2c2 + (1/2) q4c4
p2 = q0c2 + q2[c0 + (1/2)c4] + q4[(1/2)c2 + (1/2)c6]
0 = q0c4 + q2[(1/2)c2 + (1/2)c6] + q4[c0 + (1/2)c8]
0 = q0c6 + q2[(1/2)c4 + (1/2)c8] + q4[(1/2)c2 + (1/2)c10].
Approximate numerical values for the ck's appearing in the equations above can be computed by
the method of Chebyshev series expansion. In order to obtain a particular solution of the
equations, set q0 = 1 and solve for p0, p2, q2, and q4, computing approximate numerical values for
these unknowns. Then
P2(x) = .52320016427 T0(x) + .00739002644 T2(x)
= .51581013784 + .01478005287x2
and
Q4(x) = T0(x) + .06107956977 T2(x) + .00010081226 T4(x)
= .93902124249 + .12135264149x2 + .00080649804x4,
where the coefficients are rounded to 11 decimal places. Then
R2,4(x) = P2(x) / Q4(x) can be expressed in the form
R2,4(x) = [{.54930614399 + .01573984933x2}
/ {1 + .12923311636x2 + .00085887093x4}],
where the numerator and denominator have been scaled so that in thedenominator equals one
and the other coefficients are rounded to 11decimal places. The approximation to tanh μx for – 1 ≤
x ≤ 1 is xR2, 4(x).
CHAPTER 03: SERIES
Infinite Series (Test For Convergence And Divergence)
Taylor’s Series Expansion
Maclaurin’s Series Expansion
Power Series Expansion
Laurent Series Expansion
## Infinite Series (Test For Convergence And Divergence)
PROBLEM 03 – 0046: Define the term series with computational considerations.
Solution:
A series is a group of terms, often infinite, where the terms are formed according to some general
rule. Series are very useful for numerical computations of such constants as e, π, etc.; and for the
computations of terms such as log x, sin x, etc.
Note that a series can be used for computational purposes only if it converges, that is the only
kind of series that can be used for computation is a convergent series. The reason for this is that if
a series were not convergent but divergent, as one would add more and more terms, a continually
different result would be obtained. With a convergent series, however, the series approaches a
limit, and gives an ever more accurate result, the more terms added. Adding more terms does not
change the result substantially after the first several terms, it only makes the result more accurate.
Assuming, then, that a series converges, it can be differentiated, integrated, added to other series,
subtracted from other series, multiplied by a constant, etc.
For example, if the series for sin x is known, the series for cos x can be found by differentiating that
series term by term, since it is known that
(d / dx) (sin x) = cos x
In a computation, once convergence is established, the number of terms to be used in the
computation depends only on the accuracy desired.
## PROBLEM 03 – 0047: Show that:
a) ∞∑n=2 [(log n)2 / n3] is a convergent series.
b) ∞∑n=2 [1 / {n(log n)p}] is convergent if p > 1,
divergent for p ≤ 1.
Solution: a) To show
∞∑ [(log n)2 / n3]
n=2
is a convergent series apply the integral test, Hence, it must be shown that
limb→∞ b∫2 [(log x)2 / x3] dx < ∞.
To do this, use the method of integration by parts, Therefore, take
dv = x–3 dx;
v = [(– 1) / 2x2], u = (log x)2, du = [{2(log x)} / x]dx.
Then,
b∫ [(log x)2 / x3] dx = [{– (log x)2} / 2x2] |b 2
2
## + b∫2 [(2 log x) / 2x3] dx. (1)
Now, applying integration by parts again, let
dv = x–3 dx;
v = [(– 1) / 2x2], u = log x, du = (dx / x).
Then (1) becomes
b∫ [(log x)2 / x3] dx = [{– (log x)2} / 2x2] |b 2 – [(log x) / 2x2] |b 2
2
## + b∫2 (dx / 2x3).
Hence
limb→∞ b∫2 [(log x)2 / x3] dx
equals
limb→∞ ( [{– (log b)2} / 2b2] – [(log b) / 2b2] – (1 / 4b2) + [(log 2)2 / 8]
+ [(log 2) / 8] + [1 / 16])
Then, by L'Hospital's Rule the first three terms tend to 0 as b→∞.
Therefore
limb→∞ b∫2 [(log x)2 / x3] dx = [(log 2)2 / 8] + [(log 2) / 8] = [1 / 16] < ∞
Thus, the series is convergent.
b) To show
∞∑ [1 / {n(log n)p}]
n=2
## is convergent if p > 1, divergent for p ≤ 1, again apply the integral test.
Therefore set up the integral
b∫ [dx / {x(log x)p}].
2
This yields
log b∫ (du / up) = [u–p+1 / (– p + 1)] |log b log 2
log 2
Therefore,
limb→∞ b∫2 [dx / {x(log x)p}] = limb→∞ ( [(log b)–p+1 / (– p + 1)]
– [(log 2)–p+1 / (– p + 1)] )(p ≠ 1).
If p > 1,
the first term goes to 0 as b→∞ and the integral equals
[(log 2)–p+1 / (p – 1)] < ∞.
Thus, the series is convergent.
If p < 1, the first terra grows without bound as b→∞ and the integral is unbounded. Thus, the
series is divergent. For p = 1,
limb→∞ b∫2 [dx / {x(log x)p}] = limb→∞ b∫2 [dx / {x(log x)}]
= limb→∞ log(log b) – [log(log 2)]
Thus, the series is divergent. This shows that the series is convergent for p > 1;
and divergent for p ≤ 1.
## PROBLEM 03 – 0048: Determine if the series
(1/2) + (1/3) + (1 / 22) + (1 / 32) + (1 / 23) + (1 / 33) + …
is convergent or divergent, using the ratio test and the root test.
Solution:
The series can be rewritten as the sum of the sequence of numbers given by
an = [1 / 2{(n+1)/2}] if n is odd (n > 0)
and = [1 / 3(n/2)] if n is even (n > 0)
Now the ratio test states:
If ak > 0
and
limk→∞ (ak+1 / ak) = ℓ < 1,
then
∞∑ ak converges.
k=1
Similarly, if
limk→∞ (ak+1 / ak) = ℓ(1 < ℓ ≤ ∞)
then
∞∑ ak diverges.
k=1
## If ℓ = 1, the test fails. Therefore, applying this test gives:
If n is odd,
limn→∞ (an+1 / an) = limn→∞ [(1 / 3(n/2)) / (1 / 2{(n+1)/2})]
= limn→∞ (2{(n+1)/2} / 3(n/2))
= limn→∞ (2/3)(n/2) 2(1/2) = 0.
If n is even,
limn→∞ (an+1 / an) = limn→∞ [(1 / 2{(n+1)/2}) / (1 / 3(n/2))]
= limn→∞ (3/2)(n/2) 2(1/2)
and no limit exists.
Hence, the ratio test gives two different values, one < 1 and the other > 1, therefore the test fails to
determine if the series is convergent. Thus, another test, known as the root test is now applied.
This test states:
Let
∞∑ ak
k=1
## be a series of nonnegative terms, and let
limn→∞ (n√an) = S,
where
0 ≤ S ≤ ∞.
If:
1) 0 ≤ S < 1, the series converges
2) 1 < S ≤ ∞, the series diverges
3) S = 1, the series may converge or diverges.
Applying this test yields, if n is odd
limn→∞ n√an = limn→∞ n√[(1 / 2{(n+1)/2})
= limn→∞ n√(1 / 2(n/2)) n√[1 / {2(1/2)}]
= limn→∞ (1 / √2) (1 / 2(1/2n))
= (1 / √2) < 1.
If n is even
limn→∞ n√an = limn→∞ n√(1 / 3(n/2))
= (1 / √3) < 1.
Thus, since for both cases, the
limn→∞ n√an < 1,
the series converges.
PROBLEM 03 – 0049:
Determine if the series.
a) ∞∑k=1 [(k + 1)(1/2) / (k5 + k3 – 1)(1/3)]
converges or diverges. Use the limit test for convergence.
b) ∞∑k=1 [(k log k) / (7 + 11k – k2)]
converges or diverges. Use the limit test for divergence.
Solution:
a) To determine if the given series converges or diverges, the following test called the limit test for
convergence is used. This test states:
If
limk→∞ kp Uk = A for p > 1,
then
∞∑ Uk
k=1
converges absolutely.
To apply this test let p = (7/6) > 1 . Then since
Uk = [(k + 1)(1/2) / (k5 + k3 – 1)(1/3)],
limk→∞ kp Uk = limk→∞ [{k(7/6) (k + 1)(1/2)} / (k5 + k3 – 1)(1/3)]
= limk→∞ [{k(7/6) k(1/2)[1 + (1/k)](1/2)} / (k5 + k3 – 1)(1/3)]
= limk→∞ [{k(5/3) [1 + (1/k)](1/2)} / (k5 + k3 – 1)(1/3)]
= limk→∞ [{1 + (1/k)}(1/2) / {k–5(k5 + k3 – 1)}(1/3)]
= limk→∞ [{1 + (1/k)}(1/2) / {1 + (1/k2) – (1/k5)}(1/3)]
= [(1)(1/2) / (1)(1/3)]
=1
Therefore, the series converges absolutely.
b) For the series
∞∑ [(k log k) / (7 + 11k – k2)],
k=1
the following test called the limit test for divergence is used. This test states, if
limk→∞ k Uk = A ≠ 0 (or ± ∞)
then
∞∑ Uk
k=1
## diverges. If A = 0, the test fails.
Since
Uk = [(k log k) / (7 + 11k – k2)]
limk→∞ k Uk = limk→∞ [(k2 log k) / (7 + 11k – k2)]
= limk→∞ [(k2 log k) / {k2[(7/k2) + (11/k) – 1]}]
= limk→∞ [(log k) / {(7/k2) + (11/k) – 1}].
As k → ∞, log k → ∞
while
(7/k2) + (11/k) – 1 → – 1.
Thus
∞∑ [(k log k) / (7 + 11k – k2)] diverges.
k=1
## Taylor’s Series Expansion
PROBLEM 03 – 0050:
Find the Taylor series expansion of f(x) about x = x0 and show that a complete knowledge of the
behavior of f(x) at a point x0 determines the value of the function at any point x at which the series
converges.
Solution:
Let f(x) be a function with derivatives of all orders at x = x0, and assume that in a neighborhood of
x0, f(x) may be represented by the power series in x – x0:
f(x) ≡ c0 + c1(x – x0) + c2(x – x0)2 + ...
= ∞∑n=0 cn(x – x0)n. (a)
The derivatives of f(x) may be obtained by differentiating the series (a) term by term:
f'(x) = ∞∑n=1 ncn(x – x0)n–1
= c1 + 2c2(x – x0) + 3c3(x – x0)2 + ..., (b)
f"(x) = ∞∑n=2 n(n – 1) cn(x – x0)n–2
= 2 ∙ 1c2 + 3 ∙ 2c3(x – x0) + ..., (c)
.....................................................................................................
f(m)(x) = ∞∑n=m n(n – 1)(n – 2) ... (n – m + 1)cn(x – x0)n–m
= ∞∑n=m [n! / (n – m)!] cn(x – x0)n–m (d)
= m! cm + (m + 1)! cm+1(x – x0) + ...
Setting x = x0 on both sides of (a) , (b) , (c), and (d), one finds
f(x0) = c0 or c0 = f(x0),
f'(x0) = c1 or c1 = f'(x0),
f"(x0) = c2 or c2 = [f"(x0) / 2],
.....................................................................
f(m)(x0) = m! cm or cm = [{f(m)(x0)} / m!],
by means of which (a) becomes
f(x) = ∞∑n=0 [{f(n)(x0)} / n!] (x – x0)n, (e)
where
f(n)(x0) ≡ [(dnf) / (dxn)] |x=(x)0,
f(0)(x0) ≡ f(x0)
and 0! = 1.
Equation (e) is the Taylor series expansion. Note also that by setting x0 = 0 in (e), one gets the
Maclaurin series expansion:
f(x) = ∞∑n=0 [{f(n)(0)} / n!] xn.
PROBLEM 03 – 0051:
Evaluate √(1 + x) to two decimal places at x = .4 by Taylor series expansion. Consider the error
expression:
|Rm| ≤ |[{f(m)(x0)} / m!](x – x0)m|,
the mth derivative is computed at x0. |Rm| is the truncation error after the mth term.
## Solution: √(1 + x) = (1 + x)(1/2)
= 1 + (1/2)x – (1/8)x2 + (1 / 16)x3 – (5 / 128)x4 + ... .
To evaluate √(1 + .4) to two decimal places, at x = .4
1 + (1/2)(.4) – (1/8)(.4)2 = 1.18 [1.183];
|R3| < [(.4)3 / 16] = .004 [.003].
PROBLEM 03 – 0052:
Using the Taylor series expansion for ex about x = 0, find e0.5 to (e) (0.5)3. Bound the error using
the error expression
|[(dn+1f) / dxn+1]|max ∙ [(|x – a|)n+1 / (n + 1)!] (1)
(The subscript "max" denotes the maximum magnitude of the derivative on the interval from a to x
and (e) means error order), and compare the result with the actual error.
## Solution: From equation
f(x) = f(a) + (x – a)f'(a) + [(x – a)2 / 2!] f"(a) + [(x – a)3 / 3!] f'"(a) + ... + [(x – a)n / n!] f(n)(a) + ...
ex = e(0) + xe(0) + (x2 / 2!) e(0) + (x3 / 3!) e(0) + (x4 / 4!) e(0) + ...
or
ex = 1 + x + (x2 / 2!) + (x3 / 3!) + (x4 / 4!) + ...
and if x = 0.5,
e0.5 = 1 + 0.5 + [(0.5)2 / 2!] + [(0.5)3 / 3!] + [(0.5)4 / 4!] + ...
or, to (e)(0.5)3,
e0.5 = 1 + 0.5 + [(0.5)2 / 2!] = 1.625
Now according to (1), the error in this quantity should not be greater than
|[{d3 (ex)} / dt3]|max [(0.5)3 / 3!] = |ex|max (0.0208333)
where max denotes the maximum magnitude on 0 ≤ x ≤ 0.5.
|ex|max = e0.5 = 1.6487213,
so the error is no greater in magnitude than
(1.6487213) (0.0208333) = 0.0343831
The actual error is
e0.5 – 1.625 = 1.6487213 – 1.6250000 = 0.0237213
which lies within the error bound. Notice that in this case the error which was (e) (0.5) 3 was actually
PROBLEM 03 – 0053:
Show that f(x) = (x – 1)(1/2) cannot be expanded in a Taylor series about x = 0 or x = 1, but can be
expanded about x = 2. Carry out the expansion about x = 2.
## Solution: f(x) = (x – 1)(1/2)
f'(x) = (1/2) (x – 1)–(1/2)
f"(x) = – (1/4) (x – 1)–(3/2)
f"'(x) = (3/8) (x – 1)–(5/2)
For an expansion about x = 0, the quantities f(0), f'(0), f"(0), etc. are needed. These involve
noninteger powers of (– 1), e.g. (– 1)(1/2). These cannot be evaluated to give real values and thus this
expansion is impossible.
For an expansion about x = 1, one needs quantities such as
f(1) = (0)(1/2)
f'(1) = (1/2) (0)–(1/2)
f"(1) = – (1/4) (0)–(3/2)
While f(1) is bounded, all of the derivatives f'(1), f"(1), etc. are unbounded. Thus the expansion
about x = 1 is impossible.
The Taylor series expansion about x = 2 is
f(x) = (2 – 1)(1/2) + (x – 2)[(1/2)(2 – 1)–(1/2)] + [(x – 2)2 / 2][– (1/4)(2 – 1)–(3/2)]
+ [(x – 2)3 / 3!] [(3/8) (2 – 1)–(5/2)] + ...
All derivatives are bounded and finite and the series is
f(x) = 1 + [(x – 2) / 2] – [(x – 2)2 / 8] + [(x – 2)3 / 16] – ...
This series is convergent for |x – 2| < 1.
PROBLEM 03 – 0054: Write the Taylor expansion for ln x about the point x = 1. Use n terms.
Solution: f(x) = ln x f(1) = 0
f'(x) = (1/x) f'(1) = 1
f"(x) = [(– 1) / x2] f"(1) = – 1
f"'(x) = [(+ 2) / x3] f"'(1) = 2
fiv(x) = – 2 × (3 / x4) fiv(1) = – 2 × 3
∙ ∙
∙ ∙
∙ ∙
f(n–1)(x) = [{(– 1)n(n – 2)!} / (xn–1)] f(n–1)(1) = (– 1)n(n – 2)!
f(n)(x) = [{(– 1)n+1(n – 1)!} / xn] f(n) [1 + θ(x – 1)]
= [{(– 1)n+1(n – 1)!} / {1 + θ(x – 1)}]
Substituting these values into the expression
f(x) = f(a) + (x – a)f'(a) + [(x – a)2 / 2!]f"(a) + .... + [(x – a)n–1 / (n – 1)!]f(n–1)(x)
+ [(x – a)n / n!] f(n)[a + θ (x – a)],
0 < θ < 1,
and remembering that a = 1 in this case, one obtains
ln x = (x – 1) – (1 / 2!)(x – 1)2 + (2 / 3!)(x – 1)3
– [(2 × 3) / 4!] (x – 1)4 + ... + [{(– 1)n(n – 2)!(x – 1)n–1} / (n – 1)!]
+ [{(– 1)n+1(n – 1)! (x – 1)n} / {n![1 + θ (x – 1)]n}],
0<θ<1
or, simplifying the expression,
ln x = (x – 1) – [(x – 1)2 / 2] + [(x – 1)3 / 3] – [(x – 1)4 / 4] + ...
+ [{(– 1)n(x – 1)n–1} / (n – 1)] + [{(– 1)n+1(x – 1)n} / {n(1 – θ + θx)n}],
0<θ<1
PROBLEM 03 – 0055:
(a) Derive the first three (non-zero) terms of Taylor's series expansion for the function f(x) = sin x,
about the origin. Since it is about the origin, this series is also called a Maclaurin series.
(b) Use the above result to compute the value of sin x for
x = 0.2 rad, correctly to five decimal places.
Solution: (a) With the expansion about the origin the Taylor's series is:
f(x) = f(0) + xf'(0) + (x2 / 2!)f"(0) + (x3 / 3!)f(3)(0) + (x4 / 4!)f(4)(0)
+ (x5 / 5!)f(5)(0) + ...
For
f(x) = sin x f(0) = 0
f'(x) = cos x f'(0) = 1
f"(x) = – sin x f"(0) = 0
f(3)(x) = – cos x f(3)(0) = – 1
f(4)(x) = sin x f(4)(0) = 0
f(5)(x) = cos x f(5)(0) = 1
Therefore,
sin x = [(x – x3) / 3!] + (x5 / 5!) + ...
(b) For x = 0.2 rad,
sin 0.2 = 0.2 – [(0.2)3 / 6] + [(0.2)5 / 120] + ...
sin 0.2 = 0.2 – 0.0013333 + 0.0000027 + ...
To five decimal places,
sin 0.2 = 0.19867
It is noted that this is the correct answer for sin 0.2, to five decimal places. Actually, one can tell
from the above computations that this answer is correct to five places by using the theorem which
says that in a converging alternating series the error committed in stopping with any term is
always less than the first term neglected. The first term omitted in the calculation above for sin 0.2
is the third term, and it contributes only 3 in the sixth place.
## PROBLEM 03 – 0056: Obtain a first-degree polynomial approximation to
f(x) = (1 + x)(1/2)
by means of the Taylor expansion about x0 = 0.
Solution:
f'(x) = [1 / {2(1 + x)(1/2)}]
f"(x) = [(– 1) / {4[(1 + x)(1/2)]3}]
f(0) = 1
f'(0) = (1/2)
Thus the Taylor expansion with remainder term is
(1 + x)(1/2) = 1 + (x/2) – [x2 / {8[(1 + ξ)(1/2)]3}]
(Here ξ lies between 0 and x.).
Thus the first-degree polynomial approximation to (1 + x)(1/2) is [(1 + x) / 2].
The accuracy of this approximation depends on what set of values x is allowed to assume. If x ε [0,
.1] , for example, one gets
|(1 + x)(1/2) – [1 + (x/2)]| ≤ [.12 / {8[(1 + ξ)(1/2)]3}] ≤ (.12 / 8) = .00125
PROBLEM 03 – 0057:
Obtain a second-degree polynomial approximation to f(x) = e–(x)2 over [0, .1] by means of the
Taylor expansion about x0 = 0. Use the expansion to approximate f(.05), and bound the error.
## Solution: f(x) = e–(x)2 f'(x) = – 2xe–(x)2
f"(x) = (– 2 + 4x2)e–(x)2 f"'(x) = (12x – 8x3)e–(x)2
f(0) = 1 f'(0) = 0 f"(0) = – 2
Thus f(x) = 1 – x2 + [{f"'(t)x3} / 6]
where
0 < t < .1
Using f(x) ≅ 1 – x2, one gets f(.05) ≅ .9975. The truncation error is bounded by
|f(.05) – .9975| ≤ [(.05)3 / 6]maxtεI |f"'(t)|
where I = [0, .05]. To obtain a bound on |f'"(t)|, use the result that, for t ε I,
|f"'(t)| = |12 – 8t2| ∙ |t| ∙ |e–(t)2| < maxtεI |12 – 8t2| ∙ maxtεI |t| ∙ maxtεI |e–(t)2|
Hence
|f(0.5) – .9975| < [(.05)3 / 6] (12) (.05) (1.0) = 1.25 × 10–5
PROBLEM 03 – 0057:
Obtain a second-degree polynomial approximation to f(x) = e–(x)2 over [0, .1] by means of the
Taylor expansion about x0 = 0. Use the expansion to approximate f(.05), and
bound the error.
## Solution: f(x) = e–(x)2 f'(x) = – 2xe–(x)2
f"(x) = (– 2 + 4x2)e–(x)2 f"'(x) = (12x – 8x3)e–(x)2
f(0) = 1 f'(0) = 0 f"(0) = – 2
Thus f(x) = 1 – x2 + [{f"'(t)x3} / 6]
where
0 < t < .1
Using f(x) ≅ 1 – x2, one gets f(.05) ≅ .9975. The truncation error is bounded by
|f(.05) – .9975| ≤ [(.05)3 / 6]maxtεI |f"'(t)|
where I = [0, .05]. To obtain a bound on |f'"(t)|, use the result that, for t ε I,
|f"'(t)| = |12 – 8t2| ∙ |t| ∙ |e–(t)2| < maxtεI |12 – 8t2| ∙ maxtεI |t| ∙ maxtεI |e–(t)2|
Hence
|f(0.5) – .9975| < [(.05)3 / 6] (12) (.05) (1.0) = 1.25 × 10–5
## PROBLEM 03 – 0058: Obtain an approximate value of
(1/2) 1∫–1 [(sin x) / x] dx
by means of the Taylor expansion.
Solution: The expansion of sin x in a Taylor series about x0 = 0 through terms of degree 6 is
sin x = x – (x3 / 3!) + (x5 / 5!) – (x7 / 7!) cos ξ
[(sin x) / x] = 1 – (x2 / 6) + (x4 / 120) – (x6 / 7!) cos ξ
Thus
(1/2) 1∫–1 [(sin x) / x] dx = (1/2) [x – (x3 / 18) + (x5 / 600)] |1 –1
– (1/2) 1∫–1 (x6 / 7!) cos ξ dx
This simplifies to
(1/2) 1∫–1 [(sin x) / x] dx = (1,703 / 1,800) – (1/2) 1∫–1 (x6 / 7!) cos ξ dx
To bound the error, compute
|(1/2) 1∫–1 (x6 / 7!) cos ξ dx| ≤ (1/2) 1∫–1 (x6 / 7!) |cos ξ| dx
≤ (1/2) 1∫–1 (x6 / 7!) dx = [1 / {(7)(7!)}]
Hence
|(1/2) 1∫–1 [(sin x) / x] dx – (1,703 / 1,800)| ≤ [1 / {(7)(7!)}] < 3.0 × 10–5
## PROBLEM 03 – 0059: Find an approximation to the solution to x = cos x by use of
Taylor's series.
Solution: The general idea here is that the equation x = cos x is to be replaced by an
equation of the form
x = 1 – (x2 / 2!) + (x4 / 4!) – (x6 / 6!) ... + (– 1)n [x2n / (2n)!] (1)
Taking n = 1, for example, solve
x = 1 – (x2 / 2)
to obtain x = – 1 ±3(1/2). Since the only solution to x = cos x lies in the interval [0, 1], one obtains
– 1 + 3(1/2)
as an approximate solution. This value can be. used as an initial guess in a fixed-point iteration to
solve the equation. The iteration might be performed on the original equation x = cos x or on the
approximate equation (1) for some n > 1.
PROBLEM 03 – 0060: Find sin .1, using Taylor's series and t = 2 floating-point arithmetic.
## Solution: sin x = x – (x3 / 3!) + (x5 / 5!) – (x7 / 7!) ...
Hence
sin .1 = .1 – (.13 / 6) + (.15 / 120) – (.17 / 5,040) ...
Now in floating-point arithmetic
(.13 / 6) = [{(.10 × 100)(.10 × 100)(.10 × 100)} / (.60 × 101)]
= [(.10 × 10–2) / (.60 × 101)]
= .17 × 10–3
The sum of the first two terms of the series is
.10 × 100 – .17 × 10–3 = .10 × 100
Thus, the second term and all subsequent terms of the series do not affect the result.
## PROBLEM 03 – 0061: Expand the function cos x, in powers of x – a, where
a = – (π / 4),
and determine the interval of convergence.
## Solution: Expanding the given function in powers of x – a is equivalent to finding the
Taylor Series for the function. To find the Taylor Series determine f(x), f(a), f'(x), f'(a), f"(x), f"(a) , etc.
One finds:
f(x) = cos x; f(a) = f [– (π / 4)] = (√2 / 2).
f'(x) = – sin x; f'(a) = f' [– (π / 4)] = (√2 / 2).
This is a positive value because the value of sin x in the 2nd quadrant is negative, therefore – sin x
is positive.
f"(x) = – cos x; f"(a) = f" [– (π / 4)] = [(– √2) / 2].
f"'(x) = sin x; f"'(a) = f"' [– (π / 4)] = [(– √2) / 2].
Develop the series as follows:
f(x) = f(a) + f'(a)[x – a] + [{f"(a)} / 2!][x – a]2 + [{f"'(a)} / 3!][x – a]3 + ....
By substitution:
cos x = (√2 / 2) + (√2 / 2) [x + (π / 4)] – [(√2 / 2) / 2!] [x + (π / 4)]2
– [(√2 / 2) / 3!] [x + (π / 4)]3 + .....
To determine the law of formation, examine the terms of this series. The nth term of the series is
found to be:
[(√2 / 2) / (n – 1)!] [x + (π / 4)]n–1;
Then, the (n + 1)th term is:
[(√2 / 2) / n!] [x + (π / 4)]n.
Therefore the Taylor series is:
cos x = (√2 / 2) + (√2 / 2) [x + (π / 4)] – [(√2 / 2) / 2!] [x + (π / 4)]2
– [(√2 / 2) / 3!] [x + (π / 4)]3 + ... + [(√2 / 2) / (n – 1)!] [x + (π / 4)]n–1
+ [(√2 / 2) / n!] [x + (π / 4)]n + ...
To find the interval of convergence, use the Ratio Test. Set up the ratio
(Un+1 / Un),
obtaining:
[{√2[x + (π / 4)]n} / {2(n!)}] × [{2(n – 1)!} / {√2[x + (π / 4)]n–1}]
= [{x + (π / 4)}n / {n(n – 1)!}] × [(n – 1)! / {x + (π / 4)}n–1]
= [{x + (π / 4)} / n].
Now, one finds
limn→∞ |[{x + (π / 4)} / n]| = |0| = 0.
By the ratio test it is known that if
limn→∞ |(Un+1 / Un)| < 1
the series converges. Since 0 < 1, the series converges for all values of x.
## PROBLEM 03 – 0062: Give a Taylor expansion of f(x, y) = ex cos y on some
compact convex domain E containing (0, 0).
Solution: Assume f(x, y) = ɸ (x) Ψ(y) where ɸ (x) = ex and Ψ(y) = cos y. Also assume E =
Ex × Ey where Ex and Ey are compact convex subsets of R, i.e., closed and bounded intervals. Note
that
Supxε(E)x ||[{dkɸ (x)} / dxk]|| = emax(E)x
and
Supyε(E)y ||[{dkΨ(y)} / dxk]|| ≤ 1.
Hence ɸ (x) and Ψ(y) are real analytic on E. Furthermore, note that
ex = ∞∑i=0 (xi / i!) [(diex) / (dxi)] |x=0 = ∞∑i=0 (xi / i!) (1)
and
cos y = ∞∑i=0 (yi / i!)[(di cos y) / dyi] |y=0 = ∞∑i=0 (– 1)i [y2i / (2i)!] (2)
Since f = ɸ Ψ and ɸ and Ψ are real analytic on E, f is real analytic on E:
Supx, yεE ||[{∂kf(x, y)} / (∂xi∂yk–1)]|| ≤ emax(E)x.
hence, from (1) and (2)
ex cos y = [∞∑i=0 (xi / i!)] [∞∑j=0 (– 1)j {y2j / (2j)!}]
= ∞∑k=0 [∑(i+j=k)(i,j≥0) (– 1)j {(xiy2j) / [i!(2j)!]}].
The first three terms corresponding to k = 0, k = 1, and k = 2 give the approximation:
ex cos y ≈ 1 + x – (1/2)y2 + (1/2)x2 – (1/2)xy2 + (1 / 24)y4
which is known to be accurate near (0, 0).
n∑ αiti
i=0
1 + t + 3t4.
## Solution: To rewrite the polynomial
f(t) = n∑i=0 αiti
as a polynomial in x = t – 1, i.e., as
g(x) = m∑i=1 bixi,
use Taylor's Theorem. Note that f is C∞ and that
f'(t) = n∑i=0 iαiti–1,
f"(t) = n∑i=0 i(i – 1)αiti–2,...,
f(n)(t) = n∑i=0 i(i – 1) ... (i – n + 1)αiti–n,
and
0 = f(n+1)(t) = f(n+2)(t) = ....
To get a polynomial in t – 1, expand about 1. Hence
f(1) = n∑i=0 αi, f'(1) = n∑i=1 iαi, f"(1) = n∑i=2 i(i – 1)αi,...,
f(k)(1) = n∑i=k i(i – 1) ... (i – k + 1)αi, ...,
f(n)(1) = n! αn.
Therefore, the Taylor expansion
f(t) = n∑j=0 [{f(j)(1)} / j!] (t – 1)j
= n∑j=0 (1 / j!) n∑i=j i(i – 1) ... (i – j + 1) αi(t – 1)j
= n∑j=0 (1 / j!) n∑i=j [i! / (i – j)!] αi(t – 1)j
= n∑j=0 [n∑i=j {i! / [(i – j)!j!]} αi](t – 1)j
= n∑j=0 [n∑i=j (i j) αi](t – 1)j
= m∑j=0 bjxj = g(x).
Note that the degree of the new polynomial (1) is also n and the j-th coefficient is in terms of the
last n – j + 1 of the αi. For f(t) = 1 + t + 3t4, according to this method,
α0 = α1 = 1, α2 = α3 = 0
and
α4 = 3.
Hence
f(t) = 4∑j=0 [4∑i=j {i! / [(i – j)! j!]} αi](t – 1)j
= [{0! / (0!0!)}α0 + {1! / (1!0!)}α1 + {2! / (2!0!)}α2 + {3! / (3!0!)}α3
+ {4! / (4!0!)}α4](t – 1)0
+ [{1! / (0!1!)}α1 + {2! / (1!1!)}α2 + {3! / (2!1!)}α3 + {4! / (3!1!)}α4](t – 1)1
+ [{2! / (0!2!)}α2 + {3! / (1!2!)}α3 + {4! / (2!2!)}α4](t – 1)2
+ [{3! / (0!3!)}α3 + {4! / (1!3!)}α4](t – 1)3 + {4! / (0!4!)}α4 (t – 1)4
= (1 + 1 + 0 + 0 + 3) + (1 + 0 + 0 + 12)(t – 1) + (0 + 0 + 18)(t – 1)2
+ (0 + 12)(t – 1)3 + 3(t – 1)4
= 5 + 13(t – 1) + 18(t – 1)2 + 12(t – 1)3 + 3(t – 1)4. (2)
To show that (2) is actually equal to f(t); expand:
f(t) = 5 + 13(t – 1) + 18(t2 – 2t + 1) + 12(t3 – 3t2 + 3t – 1)
+ 3(t4 – 4t3 + 6t2 – 4t + 1)
= (5 – 13 + 18 – 12 + 3) + t(13 – 36 + 36 – 12) + t2(18 – 36 + 18)
+ t3(12 – 12) + t4(3)
= 1 + t + 3t4.
So, indeed,
1 + t + 3t4 = 5 + 13x + 18x2 + 12x3 + 3x4
where
x = t – 1.
PROBLEM 03 – 0064: A. Find the Taylor series expansion of f(x, y) about x0, y0.
B. Expand f(x, y) = cos xy about x0 = (π / 2), y0 = (1/2) up to
second-order terms.
## Solution: A. The Taylor series of a function z = f(x, y) of two independent variables x, y,
about the point (x0, y0) is obtained from
f(x0 + x) = ∞∑n=0 [{f(n)(x0)} / n!] xn (a)
by incrementing one of the variables at a time. Indicating the partial derivatives of z(x, y) by
fx = (∂z / ∂x), fy = (∂z / ∂y), fxx = (∂2z / ∂x2), fxy = [(∂2z / (∂x∂y)]
fyy = (∂2z / ∂y2),.., (b)
(a) applied to f(x0 + x, y0), i.e., to the function f considered as a function of x only while y0 is kept
constant, gives
f(x0 + x, y0) = f(x0, y0) + fx(x0, y0) (x / 1!) + fxx(x0, y0) (x2 / 2!) + ...,
from which, changing y0 into y0 + y on both sides of the equation,
f(x0 + x, y0 + y) = f(x0, y0 + y) + fx(x0, y0 + y) (x / 1!)
+ fxx(x0, y0 + y) (x2 / 2!) + ... . (c)
Expanding by means of (a) the coefficient of each power of x in (c) into a
power series in y about y0, while x0 is kept constant, yields
f(x0, y0 + y) = f(x0, y0) + fy(x0, y0)(y / 1!) + fyy(x0, y0)(y2 / 2!) + ...,
fx(x0, y0 + y) = fx(x0, y0) + fxy(x0, y0)(y / 1!) + fxyy(x0, y0)(y2 / 2!) + ... , (d)
fxx(x0, y0 + y) = fxx(x0, y0) + fxxy(x0, y0)(y / 1!) + fxxyy(x0, y0)(y2 / 2!) + ...
.................................................................................................................
Substituting (d) in (c), the Taylor series expansion of f(x, y) about x0, y0 becomes
f(x0 + x, y0 + y)
= f(x0, y0) + [fx(x0, y0)x + fy(x0, y0)y]
+ (1 / 2!)[fxx(x0, y0)x2 + 2fxy(x0, y0)xy + fyy(x0, y0)y2]
+ (1 / 3!) [fxxx(x0, y0)x3 + 3fxxy(x0, y0)x2y + 3fxyy(x0, y0)xy2
+ fyyy(x0, y0)y3] + ... .
B. Here
fx = – y sin xy; fy = – x sin xy; fxx = – y2 cos xy;
fxy = – sin xy – xy cos xy; fyy = – x2 cos xy;
f[(π / 2), (1/2)] = (√2 / 2);
fx[(π / 2), (1/2)] = – (1/2)(√2 / 2); fy[(π / 2), (1/2)] = – (π / 2)(√2 / 2);
fxx[(π / 2), (1/2)] = – (1/4)(√2 / 2);
fxy[(π / 2), (1/2)] = – (√2 / 2) [1 + (π / 4)];
fyy[(π / 2), (1/2)] = – (π2 / 4)(√2 / 2);
f[(π / 2) + x, (1/2) + y] = (√2 / 2){1 – [(1/2)x + (π / 2)y]
– [(1/8)x2 + {1 + (π / 4)}xy + (π2 / 8)y2]}.
## Solution: f(x) = ex f(0) = 1
f'(x) = ex f'(0) = 1
f"(x) = ex f"(0) = 1
f"'(x) = ex f"'(0) = 1
fiv(x) = ex fiv(0) = 1
f(n–1)(x) = ex f(n–1)(0) = 1
f(n)(x) = ex f(n)(θx) = eθx
Substituting these quantities in the expression for the Maclaurin series:
f(x) = f(0) + xf'(0) + (x2 / 2!) f"(0) + ... + [xn–1 / (n – 1)!] f(n – 1)(0)
+ (xn / n!) f(n)(θx),
0<θ<1
obtain
ex = 1 + x + (x2 / 2!) + (x3 / 3!) + ... + [xn–1 / (n – 1)!] + (xn / n!)eθx,
0<θ<1
## PROBLEM 03 – 0066: Evaluate [1 / (2e0.25)] correct to five decimal places by use of
the Maclaurin series for xe–(x)2.
Solution: The successive derivatives of this function soon become very unwieldy; hence
it is desirable to compute the coefficients of the power series and to estimate the magnitude of the
error committed by stopping at a certain term by other means. If in series
ex = 1 + (x / 1!) + (x2 / 2!) + ... + (xn / n!) + ... ,
R = ∞.
– x2 replaces x and then multiplying through by x, yields
xe–(x)2 = x – (x3 / 1!) + (x5 / 2!) – (x7 / 3!) ± ... + (– 1)n[x2n+1 / n!] + ... .
This is an alternating series and it can be shown that six terms are sufficient to yield, for
x = (1/2), [1 / (2e0.25)] = 0.38940,
a value which is correct to five decimal places.
## PROBLEM 03 – 0067: Find the Maclaurin series for the function:
(1/2)(ex + e–x),
and the interval of convergence.
Solution: Let
f(x) = (1/2)(ex + e–x).
To find the Maclaurin series, determine f(0), f'(x), f'(0), f"(x), f"(0), etc. One finds:
f(x) = (1/2)(ex + e–x) f(0) = 1
f'(x) = (1/2)(ex – e–x) f'(0) = 0
f"(x) = (1/2)(ex + e–x) f"(0) = 1
f"'(x) = (1/2)(ex – e–x) f"'(0) = 0
f4(x) = (1/2)(ex + e–x) f4(0) = 1, etc.
Now, the series is developed as follows:
f(x) = f(0) + f'(0)x + [{f"(0)} / 2!]x2 + [{f"'(0)} / 3!]x3 + ... .
By substitution:
(1/2)(ex + e–x) = 1 + 0 + (x2 / 2!) + 0 + (x4 / 4!) + ...
= 1 + (x2 / 2!) + (x4 / 4!) + ... .
To determine the law of formation examine the terms of this series. The nth term of the series is
found to be:
[x2n–2 / (2n – 2)!];
Then, the (n + 1)th term is:
[x2n / (2n)!].
Therefore the Maclaurin series is:
(1/2)(ex + e–x) = 1 + (x2 / 2!) + (x4 / 4!) + ... + [x2n–2 / (2n – 2)!] + [x2n / (2n)!].
To find the interval of convergence use the ratio test. Set up the ratio
(un+1 / un),
obtaining:
[x2n / (2n)!] ∙ [(2n – 2)! / x2n–2] = [x2n / {(2n)(2n – 2)!}] ∙ [(2n – 2)! / x2n–2]
= (x2 / 2n).
One finds:
limn→∞ |(x2 / 2n)| = |0| = 0.
By the ratio test it is known that, if
limn→∞ |(un+1 / un)| < 1,
the series converges. Since 0 is always less than 1, the series converges for all values of x.s
PROBLEM 03 – 0068: Find the Maclaurin series and the interval of convergence for
the function f(x) = cos x.
Solution: To find the Maclaurin series for the given function, determine f(0),
f'(x), f'(0), f"(x), f"(0), etc.
One finds:
f(x) = cos x f(0) = 1
f'(x) = – sin x f'(0) = 0
f"(x) = – cos x f"(0) = – 1
f"'(x) = sin x f"'(0) = 0
f4(x) = cos x f4(0) = 1
f5(x) = – sin x f5(0) = 0
f6(x) = – cos x f6(0) = – 1.
Develop the series as follows:
f(x) = f(0) + f'(0)x + [{f"(0)} / 2!]x2 + [{f"'(0)} / 3!]x3 + [{f4(0)} / 4!]x4
+ [{f5(0)} / 5!]x5 + [{f6(0)} / 6!]x6 + ...
By substitution:
cos x = 1 + 0 – (x2 / 2!) + 0 + (x4 / 4!) + 0 – (x6 / 6!) + ...
= 1 – (x2 / 2!) + (x4 / 4!) – (x6 / 6!) + ... .
To determine the law of formation, examine the terms of this series. The nth term of the series is
found to be
[x2n–2 / (2n – 2)!];
Then, the (n + 1)th term is:
[x2n / (2n)!].
Therefore, the Maclaurin series is:
cos x = 1 – (x2 / 2!) + (x4 / 4!) – (x6 / 6!) + ... ± [x2n–2 / (2n – 2)!] ± [x2n / (2n)!] ... .
To find the interval of convergence use the ratio test. Set up the ratio
(un+1 / un),
obtaining:
[x2n / (2n)!] × [(2n – 2)! / x2n–2] = [x2n / {(2n)(2n – 2)!}] × [(2n – 2)! / x2n–2] = (x2 / 2n).
Now, one finds
limn→∞ |(x2 / 2n)| = |0| = 0.
By the ratio test it is known that, if
limn→∞ |(un+1 / un)| < 1
the series converges.
Since 0 is always less than 1, the series converges for all values of x.
PROBLEM 03 – 0069: Using the Maclaurin expansion, find the value of the sine
function to eight significant figures.
Solution: The remainder term in the Maclaurin expansion of the sine function
sin x = x – (x3 / 3!) + (x5 / 5!) + ... , 0<θ<1
is
[{(– 1)p x2p+1 cos θx} / (2p + 1)!],
where p is the number of terms required. If one desires to compute the sine function to eight significant
figures, the relative error would be less than [1 / (2 × 108)]. Then, note that cosθx < 1, so it is required that
[1 / (sin x)] [x2p+1 / (2p + 1)!] < [1 / (2 × 108)]
The value of p, the number of terms required, depends on how large a value of x must be accommodated.
Because of the periodicity of the sine function, it is certainly sufficient to consider only x < 2π. Also, since
sin(π + x) = – sin x, values of x between π and 2π can be replaced by values between 0 and π. Further,
since sin(π – x) = sin x, angles between (π / 2) and π can be replaced by angles between 0 and (π / 2). For
values of x between 0 and (π / 2), the sin x in the denominator is no problem, since 1 ≤ [x / (sin x)] ≤ (π / 2)
for x in this range. The largest value of the left-hand side of the inequality occurs when x = (π / 2). For this
value of x, the inequality is satisfied by 2p + 1 = 15, or p = 7. Thus the value of the sine function will be
given in eight significant figures by the expression
sin x = x – (x3 / 3!) + (x5 / 5!) – (x7 / 7!) + (x9 / 9!) – (x11 / 11!) + (x13 / 13!)
U = X ∗Y
## Y = – [{([{(U ∗ A(13) – A(11)) ∗ U – A(9)} ∗ U – A(7)] ∗ U – A(5)) ∗ U
– A(3)} ∗ U – A(1)] ∗ X
## Power Series Expansion
PROBLEM 03 – 0070: Compute ln 2 correct to five decimal places by putting
x = (1/2) in the power series
ln [1 / (1 – x)] = x + (x2 / 2) + (x3 / 3) + (x4 / 4) + ...
+ (xn / n) +...,
R = 1.
## Solution: A simple calculation shows that
f(n+1)(x) = [n! / (1 – x)n+1].
Hence, the expression
E(x) = [(x – x0)n+1 / (n + 1)!] f(n+1)(X),
for the error term becomes (x0 is again 0),
[xn+1 / (n + 1)!] [n! / (1 – X)n+1] = [xn+1 / {(n + 1)(1 – X)n+1}].
We have x = (1/2); further, X must be chosen between 0 and (1/2) so that the magnitude of the error is as
large as possible, hence X is also (1/2).
The error term consequently reduces to [1 / (n + 1)] and the first n for which [1 / (n + 1)] < 0.000005 must be
found. The first n is n = 200,000.
On the other hand,
En = ln [1 / (1 – x)] – n∑i=1 (xi / i)
= [xn+1 / (n + 1)] + [xn+2 / (n + 2)] + ... < [xn+1 / (n + 1)]
+ [xn+2 / (n + 1)] + ...
= [xn+1 / (n + 1)] (1 + x + x2 + ...)
= [xn+1 / (n + 1)] [1 / (1 – x)].
If x = (1/2), the last expression becomes [1 / {2n(n + 1)}] and the first n for which this is less than 0.000005
must now be found. This time, n = 14. This estimate of the error gives a far better result in the shape of a
much
smaller n than the previous one. Taking the first fourteen terms of the series, one finds ln 2 = 0.693143+. If
the error committed in neglecting terms from (x15 / 15) onward is 0.000002–, or less, the value of ln 2 is
0.69314; but if the error is between 0.000002 and 0.000005, then ln 2 is 0.69315. Then one or two terms
more of the series must be computed. Since (1 / 15) ∙ 215 = 0.00002+,
ln 2 = 0.69315, correct to five decimal places.
PROBLEM 03 – 0071:
Compute sin 40° correct to five significant figures from
sin x = x – (x / 3!) + (x5 / 5!) ± ... + (– 1)n [x2n+1 / (2n + 1)!] + ...,
3
R = ∞. (1)
Solution: Equation (1) yields 40° = (2π / 9) = 0.6981317 rad. Thus, using
## [(0.7)n+1 / (n + 1)!] < 0.000005.
The first positive integral n for which this inequality holds (obtained by trial and error) is n = 7. Hence, a
polynomial of max-degree 7 is necessary to attain the desired precision. On the other hand, using
## [(0.7)2n+1 / (2n + 1)!] < 0.000005,
whence n = 4. That is, the first four terms of (1) are sufficient to yield sin 40° correct to five significant
figures; a result equivalent to the preceding one. We have
sin 40° 0.6981317 – [(0.6981317)3 / 3!] + [(0.6981317)5 / 5!] – [(0.6981317)7 / 7!] = 0.6427875,
## or, to five significant figures,
sin 40° = 0.64279.
## PROBLEM 03 – 0072: Compute, with at most 1 percent error, f"'(10), where
f(x) = (x3 + 1)–(1/2).
## Solution: x is large, but x–1 is fairly small. Expand in powers of x–1:
f(x) = (x3 + 1)–(1/2)
= x–(3/2) (1 + x–3)–(1/2)
= x–1.5 (1 – 0.5 ∙ x–3 + [(0.5 ∙ 1.5) / 2]x–6 – ...)
= x–1.5 – (1/2)x–4.5 + (3/8)x–7.5 – ...
Differentiate three times:
f"'(x) = – x–4.5 ([105 / 8] – [1,287 / 16]x–3 + ...).
For x = 10 the second term is less than 1 percent of the first; the terms after the second decrease quickly
and are negligible. One can show that the magnitude of each term is less than 8 ∙ x–3 of the previous term.
Hence one gets f"'(10) = – 4.14 ∙ 10–4 to the desired accuracy.
## PROBLEM 03 – 0073: Using power series representation, find the solution of
y" = x + y2 which passes through the point (0, 1).
## Solution: Assume the solution can be expressed as a power series
y = a0 + a1x + a2x2 + ... + anxn + ... ,
which converges in some interval about x0 = 0. Then
y' = a1 + 2a2x + ... + nanxn+1 + (n + 1) an+1xn + ... ,
and
y2 = a02 + (a0a1 + a1a0)x + (a0a2 + a1a1 + a2a0)x2
+ ... + (a0an + a1an–1 + ana0)xn + ... .
Therefore,
a1 + 2a2x + ... + (n + 1) an+1xn + ... = a02 + (a0a1 + a1a0 + 1)x
+ (a0a2 + a1a1 + a2a0)x2
+ ... + (a0an + ... + ana0)xn + ... .
Equate the coefficients of like powers of x and obtain the following system of simultaneous equations:
a1 = a02
2a2 = 2a0a1 + 1
3a3 = 2a0a2 + a12
..................................................................
(n + 1)an+1 = a0an + a1an–1 + ... + ana0.
Consequently,
a1 = a02
a2 = a03 + (1/2)
a3 = a04 + (1/3)a0
a4 = a05 + (5 / 12)a02
a5 = a06 + (1/2)a03 + (1 / 20)
...........................................................
Substituting these values for a1, a2, a3, ..., into the power series for y gives a general solution of the
differential equation. This might have been anticipated because no use has been made as yet of the point
(0, 1) through which the graph of the solution must pass. If this point is used, one finds, since x = 0, and y =
a0 = 1, that a1 = 1, a2 = (3/2), a3 = (4/3),
a4 = (17 / 12), a5 = (31 / 20), ... Consequently,
y = 1 + x + (3/2)x2 + (4/3)x3 + (17 / 12)x4 + (31 / 20)x5 + ... .
One can obtain the same result somewhat differently, and perhaps more briefly, recalling that
n! an = y0(n) = y(n)(x0) = [{dnf(x0)} / dxn],
where, to repeat, [{dnf(x0)} / dxn] is the nth derivative of f(x) evaluated at x0.
By successive differentiation (with respect to x) of the given differential
equation, y' = x + y2,
y" = 1 + 2yy',
y(3) = 2(yy" + y'2),
y(4) = 2(yy(3) + 3y'y"),
y(5) = 2(yy(4) + 4y'y(3) + 3y"2),
................................................................ ;
whence, at
(0, 1), y' = 1
and
y" = 3, y(3) = 8, y(4) = 34, y(5) = 186, ... .
Hence,
a0 = 1, a1 = 1, a2 = (3/2), a3 = (4/3), a4 = (17 / 12), a5 = (31 / 20) ... ,
as before.
## PROBLEM 03 – 0074: Find the radius of convergence of the power series
n
∑n=1 (xn / n2).
Then determine if the convergence is uniform for
– R ≤ x ≤ R.
## limn→∞ |(Un+1 / Un)| = limn→∞ |x| [n2 / (n + 1)2] = |x|,
so that R = 1. That is, the series converges absolutely for – 1 < x < 1 and diverges for |x| > 1. Note that this
result could have also been found by the relation,
R = (1 / α)
where
α = limn→∞ sup n√(|an|)
where
an = (1 / n2)
which yields
α = limn→∞ sup n√[|(1 / n2)|]
= limn→∞ sup n√(1 / n2)
= limn→∞ sup (1 / n(2/n))
= limn→∞ sup [1 / (e(2/n) log n)]
= [1 / {e limn→∞ sup (2/n) log n}]
= (1 / e0)
= 1,
therefore,
(1 / R) = 1,
so that as before R = 1. For x = ±1 the series converges by comparison with the harmonic series of order 2,
that is
|[(±1)n / n2]| ≤ (1 / n2)
Hence the series converges for – 1 ≤ x ≤ 1. To determine if the convergence is uniform in this interval, the
Weierstrass M-test for uniform convergence is needed. That is, one must find a convergent series of
constants
∑n=1 Mn
such that
|(xn / n2)| ≤ Mn
for all x in – 1 ≤ x ≤ 1 if one is to determine that
∑n=1 (xn / n2)
is uniformly convergent in this interval.
Since – 1 ≤ x ≤ 1,
|(xn / n2)| ≤ (1 / n2)
for all x in the range.
Therefore since
∑n=1 Mn
converges, this shows the given power series converges uniformly on – 1 ≤ x ≤ 1.
## PROBLEM 03 – 0075: Find a power series for small values of x, for:
a) tan x
b) [(sin x) / (sin 2x)] (x ≠ 0)
Solution: To find the power series of the given functions, one needs the following theorem:
Given the two power series
∑n=0 anxn = a0 + a1x + a2x2 + ... + anxn + ...
and
∑n=0 bnxn = b0 + b1x + b2x2 + ... + bnxn + ... ,
where b0 ≠ 0, and where both of the series are convergent in some interval |x| < R, let f be a function
defined by
f(x) = [(a0 + a1x + a2x2 + ... + anxn + ...) / (b0 + b1x + b2x2 + ... + bnxn + ...)].
Then for sufficiently small values of x the function f can be represented by the power series
f(x) = c0 + c1x + c2x2 + ... + cnxn + ... ,
where the coefficients c0, c1, c2, ... , cn, ... are found by long division or equivalently by solving the
following relations successively for each ci (i = 0 to ∞):
b0 c0 = a0
b0 c1 + b1 c0 = a1
b0 cn + b1cn–1 + ... + bnc0 = an
a) To find the power series expansion of tan x, Taylor's series for sin x and cos x are needed. That is
sin x = x – (x3 / 3!) + (x5 / 5!) – ...
and
cos x = 1 – (x2 / 2!) + (x4 / 4!) – ...
Then,
tan x = [(sin x) / (cos x)]
= [{x – (x3 / 3!) + (x5 / 5!) – ...} / {1 – (x2 / 2!) + (x4 / 4!) – ...}] (1)
Therefore, by the theorem the power series expansion of tan x can be found by dividing the numerator by
the denominator on the right side of (1). Hence, using long division one gets
x + (1/3)x3 + (2 / 15)x5 + ...
1 – (1/2)x2 + (1 / 24)x4 – ... √x – (1/6)x3 + (1 / 120)x5 – ...
x – (1/2)x3 + (1 / 24)x5 – ...
(1/3)x3 – (1 / 30)x5 + ...
(1/3)x3 – (1 / 6)x5 + ...
(2 / 15)x5 – ...
(2 / 15)x5 – ...
Thus
tan x = x + (1/3)x3 + (2 / 15)x5 + ...
b) Since
sin x = x – (x3 / 3!) + (x5 / 5!) – ...
one gets
sin (2x) = 2x – [(2x)3 / 3!] + [(2x)5 / 5!] – ...
so that
[(sin x) / (sin 2x)] = [{x – (x3 / 3!) + (x5 / 5!) – ...}
/ {2x – [(2x)3 / 3!] + [(2x)5 / 5!] – ...}]. (2)
Now multiplying the numerator and denominator on the right side of (2) by (1/x) yields
[(sin x) / (sin 2x)] = [{1 – (x2 / 6) + (x4 / 120) – ...}
/ {2 – (4/3)x2 + (4 / 15)x4 – ...}].
Now by long division
(1/2) + (1/4)x2 + (5 / 48)x4 + ...
2 – (4/3)x2 + (4 / 15)x4 √ 1 – (1/6)x2 + (1 / 120)x4 – ...
(1/2) – (4/6)x2 + (4 / 30)x4 – ...
(1/2)x2 – (15 / 120)x4 + ...
(1/2)x2 – (1 / 3)x4 + ...
(25 / 120)x4 – ...
(25 / 120)x4 – ...
Thus, for x ≠ 0 ,
[(sin x) / (sin 2x)] = (1/2) + (1/4)x2 + (5 / 48)x4 + ... .
## is valid for all values of x.
Solution: To do this problem make use of Taylor's formula. That is if
## f'(x), f"(x), ... , f(n)(x)
exist and are continuous in the interval a ≤ x ≤ b and if f(n+1)(x) exists in the interval a < x < b, then
f(x) = f(a) + f'(a)(x – a) + [{f"(a)(x – a)2} / 2!] + ... + [{f(n)(a)(x – a)n} / n!] + Rn (1)
## Rn = [{f(n+1)(ξ)} / (n + 1)!] (x – a)n+1
where a < ξ < x. Note that as n changes, ξ also changes in general. In addition, if for all x and ξ in [a, b],
limn→∞ Rn = 0,
## then (1) can be written in the form
f(x) = f(a) + f'(a)(x – a) + [{f"(a)} / 2!](x – a)2 + [{f"'(a)(x – a)3} / 3!] + ... (2)
## Note that (2) is the Taylor series or expansion of f(x).
For the given problem f(x) = ex, so that f(n)(x) = ex for all orders n. Now taking a = 0 in (1),
f(x) = f(0) + f'(0)x + [{f"(0)x2} / 2!] + ... + [{f(n)(0)xn} / n!] + Rn
which yields upon substitution of f(x) = ex,
ex = 1 + x + (x2 / 2!) + ... + (xn / n!) + Rn
where
Rn = [{f(n+1)(ξ)} / (n + 1)!] xn+1 = [eξ / (n + 1)!] xn+1
where
0 < ξ < x.
Now we must prove that
limn→∞ Rn = 0,
so that we will have the Taylor expansion for ex. However,
– |x| < ξ < |x|
and
0 < eξ < e|x|;
hence
## |Rn| = |[eξ / (n + 1)!] xn+1| ≤ |[{e|x||x|n+1} / (n + 1)!] (3)
Thus by (3) to prove
limn→∞ Rn = 0,
## limn→∞ [(|x|n) / n!] = 0.
To do this choose an integer N such that N ≥ 2 |x|. Then if n > N,
## ≤ [(|x|N) / N!] (1/2)n–N
Therefore
[(|x|n) / n!] ≤ [(|x|N) / N!] (1/2)n–N. (4)
Now keeping N fixed we have
limn→∞ (1/2)n–N = 0.
Therefore by (4)
limn→∞ [(|x|n) / n!] = 0.
This means that the series representation
ex = 1 + x + (x2 / 2!) + ... + (xn / n!) + ...
is valid for all values of x.
## PROBLEM 03 – 0077: a) Using power series, show that
[{d(sin x)} / dx] = cos x
and
[{d(cos x)} / dx] = – sin x.
b) Then show that sin a cos b + cos a sin b = sin (a + b) and
cos a cos b – sin a sin b = cos (a + b).
Solution: The power series expansion for sin x and cos x are for all values x
sin x = x – (x3 / 3!) + (x5 / 5!) – (x7 / 7!) + ... (1)
## cos x = 1 – (x2 / 2!) + (x4 / 4!) – (x6 / 6!) + ... . (2)
Since, by theorem, a power series can be differentiated term-by-term in any interval lying entirely within its
radius of convergence, one gets by (1) and (2), for all values x
[{d(sin x)} / dx] = 1 – (3x2 / 3!) + (5x4 / 5!) – (7x6 / 7!) + ...
= 1 – (x2 / 2!) + (x4 / 4!) – (x6 / 6!) + ... = cos x.
and
[{d(cos x)} / dx] = [(– 2x) / 2!] + (4x3 / 4!) – (6x5 / 5!) + ...
= – x + (x3 / 3!) – (x5 / 5!) + ... = – sin x
b) Using the result from part (a)
{sin x cos(h – x) + cos x sin(h – x)}'
= cos x cos(h – x) + sin x sin(h – x) – sin x sin(h – x) – cos x cos(h – x) = 0
Thus, sin x cos(h – x) + cos x sin(h – x) is constant and this constant equals the value sin h (this was found by
letting x = 0).
Hence
## sin x cos(h – x) + cos x sin(h – x) = sin h.
Now replacing x by a and h by a + b yields
sin a cos b + cos a sin b = sin(a + b).
From which differentiation with respect to a yields
cos a cos b – sin a sin b = cos(a + b).
Note that from this result one obtains
cos2x – sin2x = cos 2x.
PROBLEM 03 – 0078: Write the function f(z) = [z / (ez – 1)] in terms of a power series.
## Solution: The function
f(z) = [z / (ez – 1)] (1)
is analytic everywhere except at those points where ez – 1 vanishes and z does not, i.e., except at the points
±2πi, ±4πi, .... Substituting the power series
ex – 1 = (z / 1!) + (z2 / 2!) + ... + (zn / n!) +... .
into (1), and dividing both numerator and denominator by z, obtain
f(z) = [1 / {1 + (z / 2!) + ... + [zn / (n + 1)!] +...}]
(note that this expression is meaningful for z = 0) . The series in the denominator converges for all z and
does not vanish for z = 0, but otherwise has the same zeros as ez – 1. The two zeros closest to the origin are
at ±2πi, and therefore f(z) has a power series representation of the form
∑n=0 cn(z – zn)n
for z0 = 0
on the disk K: |z| < 2π. To find this representation, use the technique of division of power series. That is,
given an expression of the form
f(z) = [{g(z)} / {h(z)}]
## = [(∞∑n=0 an zn) / (∞∑n=0 bn zn)]
= ∞∑n=0 cn zn,
the coefficient cn may be found by using the coefficients c0, c1, c2, ..., cn–1 in formula
cn = [(an – c0bn – c1bn–1 – ... – cn–1b1) / b0] (2)
In the present case, the coefficients an, bn appearing in this problem are just
a0 = 1, an = 0 (n = 1, 2,...),
bn = [1 / (n + 1)!] (n = 0, 1, 2,...).
Therefore the first of the equations (2) gives
c0 = 1,
and the rest reduce to the recurrence relation
c0 [1 / (n + 1)!] + c1(1 / n!) + ... + cn = 0 (n = 1, 2,...), (3)
relating cn to the values eg, c0, c1, ... , cn–1.
The numbers cnn! are called the Bernoulli numbers and are denoted by Bn. To calculate Bn, use the
recurrence relation (3), which now takes the form
B0 [1 / {0!(n + 1)!}] + B1 [1 / (1! n!)] + ... + Bn [1 / (n! 1!)] = 0
(n = 1,2,...) (4)
(obviously B0 = c00! = 1) . Multiplying (4) by (n + 1)! and introducing the notation
[(n + 1)! / {k!(n + 1 – k)!}] = (n+1 k)
(the familiar binomial coefficient), find that
B0(n+1 0) + B1(n+1 1) + ... + Bn(n+1 n) = 0
(n = 1,2,...). (5)
Equation (5) can be written symbolically as
(1 + B)n+1 – Bn+1 = 0, (6)
where after raising 1 + B to the (n + 1)th power, every Bk is changed to Bk (k = 1, 2,...,n + 1). Using (6) and the
fact that B0 = 1, deduce step by step that
B0 + 2B1 = 0,
B0 + 3B1 + 3B2 = 0,
B0 + 4B1 + 6B2 + 4B3 = 0,
B0 + 5B1 + 10B2 + 10B3 + 5B4 = 0,
B1 = – (1/2)B0 = – (1/2),
B2 = – (1/3)B0 – B1 = (1/6),
B3 = – (1/4)B0 – B1 – (3/2)B2 = 0,
B4 = – (1/5)B0 – B1 – 2B2 – 2B3 = – (1 / 30)
B0 + 6B1 + 15B2 + 20B3 + 15B4 + 6B5 = 0,
B5 = – (1/6)B0 – B1 – (5/2)B2 – (10 / 3)B3 – (5/2)B4 = 0,
B0 + 7B1 + 21B2 + 35B3 + 35B4 + 21B5 + 7B6 = 0,
B6 = – (1/7)B0 – B1 – 3B2 – 5B3 – 5B4 – 3B5 = (1 / 42),
Collecting these results,
B0 = 1, B1 = – (1/2), B2 = (1/6) , B3 = 0, B4 = – (1 / 30), B5 = 0,
B6 = (1 / 42),... (7)
As suggested by (7), the Bernoulli numbers with odd indices greater than 1 vanish. To see this, write
f(z) = [z / (ez – 1)] = c0 + c1z + c2z2 + c3z3 + ... + cnzn + ... (8)
= B0 + (B1 / 1!)z + (B2 / 2!)z + (B3 / 3!)z + ... + (Bn / n!)zn + ...
2 3
## and then replace z by – z, obtaining
f(– z) = [(– z) / (e–z – 1)] = [(– zez) / {(e–z – 1)ez}] = [zez / (ez – 1)] (9)
= B0 – (B1 / 1!)z + (B2 / 2!)z2 – (B3 / 3!)z3 + ... + (Bn / n!)(– 1)nzn
Subtraction of (9) from (8) gives
f(z) – f(– z) = [z / (ez – 1)] – [zez / (ez – 1)] = – z
= 2(B1 / 1!)z + 2(B3 / 3!)z3 + ... + 2[B2m+1 / (2m + 1)!]z2m+1 + ...,
and hence, by the uniqueness of power series expansions,
2B1 = – 1, B3 = B5 = ... = B2m+1 = ... = 0,
as asserted. Using this fact, write the expansion (8) in the form
f(z) = [z / (ez – 1)] = 1 – (z/2) + ∞∑m=1 [B2m / (2m)!] z2m, (10)
where the series converges on the disk K: |z| < 2π. Moreover, the fact that f(z) becomes infinite for z = ±2πi
implies that K is the largest disk on which (10) converges.
## Laurent Series Expansion
PROBLEM 03 – 0079: Obtain series expansions of the function
f(z) = [(– 1) / {(z – 1)(z – 2)}]
about the point z0 = 0 in the regions
|z| < 1, 1 < |z| < 2, |z| > 2.
Solution: The function f has singular points, or points at which it is not analytic, at z 1 = 1, z2 = 2.
Taylor's theorem tells that f has a valid Taylor series expansion about z0 = 0 within the circle |z| = 1 since it is
analytic there. Hence
f(z) = ∞∑n=0 [{f(n)(z0)} / n!] (z – z0)n |z| < 1.
It is convenient, however, to use partial fractions to write
f(z) = [(– 1) / {(z – 1)(z – 2)}]
= [1 / (z – 1)] – [1 / (z – 2)]
= (1/2) [1 / (1 – z/2)] – [1 / (1 – z)]. (1)
For |z| < 1, we have |(z/2)| < 1 and note that these are the conditions under which each of the terms in (1)
can be represented by a geometric series. Hence, since
[1 / (1 – w)] = ∞∑n=0 wn, (|w| < 1),
f(z) = (1/2) ∞∑n=0 (z/2)n – ∞∑n=0 zn
= ∞∑n=0 [(1/2)(z/2)n – zn].
or
f(z) = ∞∑n=0 (2–n–1 – 1)zn (|z| < 1). (2)
Since the Taylor coefficients are unique this must be the Taylor expansion for f(z) in the region |z| < 1 and as
a by-product of the calculations, it has been deduced that
[{f(n)(0)} / n!]
must be equal to the coefficients in (2). Thus
f(n)(0) = n! (2–n–1 – 1).
In the region 1 < |z| < 2 one must use Laurent' s theorem which states that if f is analytic in the region
bounded by two concentric circles C1 and C2 centered at z0. then at each point z in that region f(z) is
represented by its Laurent series expansion
f(z) = ∞∑n=0 an (z – z0)n + ∞∑n=1 [bn / (z – z0)n] (3)
where the radius of C1 is assumed to be less than that of C2, and
an = (1 / 2πi) ∫(C)1 [{f(s)ds} / (s – z0)n+1] n = 0, 1, 2, ... (4)
bn = (1 / 2πi) ∫(C)2 [{f(s)ds} / (s – z0)–n+1] n = 1, 2, ... (5)
Formulas (4) and (5) are not very useful since the integrals are usually difficult to do. Therefore, note that in
the region 1 < |z| < 2, |(1/z)| < 1 and |(z/2)| < 1 so that one may use the geometric series expansion to write
f(z) = (1/z) [1 / (1 – 1/z)] + (1/2) [1 / (1 – z/2)]
= ∞∑n=0 (1 / zn+1) + (1/2) ∞∑n=0 (zn / 2n+1), (6)
which is valid for 1 < |z| < 2. Now the Laurent coefficients in (4) and (5) are unique, i.e., any representation
such as that in (6) must be the Laurent expansion regardless of how it is arrived at. Therefore, the
coefficients in the expansion must be equal to those of (4) and (5) so that as a by-product of the
calculations we have been able to find the values of the integrals there. E.g., for n = 1 in (5) see from (6) that
b1 = 1 so
∫|z|=2 f(z)dz = 2πi
Finally, for |z| > 2 one may write
f(z) = (1/z) [{1 / (1 – 1/z)} – {1 / (1 – 2/z)}] (7)
and |(1/z)| < 1 as well as |(2/z)| < 1 in the region |z| > 2.
Therefore, one may use the geometric series in this region to find from (7) that
f(z) = ∞∑n=0 (1 / zn+1) – ∞∑n=0 (2n / zn+1)
or
f(z) = ∞∑n=0 [(1 – 2n) / zn+1] (8)
Again, since the Laurent coefficients are unique it is deduced that the coefficients in (8) are the Laurent
coefficients. In particular since b1 = 0 in
(8) one finds from (4) that
∫C f(s)ds = 0,
where in this case, C can be any circle centered at z0 = 0 with radius r > 2.
PROBLEM 03 – 0080: Find the first three nonzero terms in the Laurent series
expansion of csc z about z0 = 0.
Solution: The Laurent series expansion of a function f(z) which is analytic in an annulus
bounded by the concentric circles C1 and C2 with centers at z0 is given by
f(z) = ∞∑n=0 an (z – z0)n + ∞∑n=1 [bn / (z – z0)n] (1)
where
an = (1 / 2πi) ∫(C)1 [{f(s)ds} / (s – z0)n+1] (n = 0, 1, 2, ...) (2)
bn = (1 / 2πi) ∫(C)2 [{f(s)ds} / (s – z0)–n+1] (n = 1, 2, ...) . (3)
Here the radius of C2 is assumed to be larger than the radius of C1. The integrals in (2) and (3) are usually
difficult and sometimes impossible to do by elementary methods so that a simpler way to proceed is to find
the Taylor series expansion for sin z and use long division to obtain a series representing csc z = (1 / sin z).
Then observing that the Laurent expansion is unique, it can be stated that the resulting series is indeed the
Laurent series.
Now since sin z is analytic Vz with |z| < ∞ its Taylor expansion can be written as
sin z = ∞∑n=0 [{f(n)(0)zn} / n!] = ∞∑n=1 [(– 1)n–1 / (2n – 1)!] z2n–1 (4)
Now sin z = 0 only for z = nπ, n = 0, ±1, ±2, ... .
Hence,
csc z = (1 / sin z)
is analytic in the annulus 0 < |z| < π can be represented by a series expansion there. Thus
csc z = (1 / sin z)
= [1 / {z – (z3 / 3!) + (z5 / 5!) – (z7 / 7!) + ...}]. (5)
Thus, by long division
z – (z3 / 3!) + (z5 / 5!) – (z7 / 7!) + ... [{(1/z) + (1/3!) + z3} / {3! ∙ 3! – 5!}] + …
√1 + 0 + 0 + 0 + 0
– [1 – (z / 3!) + (z / 5!) – (z7 / 7!) + ...]
3 5
## √[(z3 / 3!) – (z5 / 5!) + (z7 / 7!) – ...]
– [(z2/3!) – {z4/(3! ∙ 3!)} + {z6/(3! ∙ 5!)+ ...]
[z4 / {3! ∙ 3! – 5!}] – … .
Therefore, from equation (5),
csc z = (1/z) + (1 / 3!)z + [{1 / (3! ∙ 3!)} – (1 / 5!)]z3 + ...
or
csc z = (1/z) + (1/6)z + (7 / 360)z3 + ... (0 < |z| < π). (6)
As noted before, this must be the Laurent series for csc z since that series is unique. As a by-product of this
calculation, one may therefore equate the coefficients in (6) with the corresponding Laurent coefficients
given by (2) and (3) with C2 being the circle |z| = π, and C1 the degenerate circle |z| = 0. This provides a
useful way to calculate the integrals of equations (2) and (3).
PROBLEM 03 – 0081:
Find the principal part of the function f(z) = [(ez cos z) / z3] at its singular point and determine the type of
singular point it is.
Solution: The given function has an isolated singular point at z = 0. Such a function may be
represented by a Laurent series
f(z) = ∞∑n=0 an (z – z0)n + ∞∑n=1 [bn / (z – z0)n] (1)
where an and bn are the Laurent coefficients. Since a Laurent expansion is unique, one needn't calculate
these coefficients directly but may proceed as follows, noting that any series of the form (1) that are
obtained must be the Laurent series. First, expanding the numerator of f in a Taylor series about z0 = 0
yields
ez cos z = [1 + z + (z2 / 2!) + ...] [1 – (z2 / 2!) + ...]
= 1 + z – (z3 / 3!) + ... . |z| < ∞ (2)
This is valid by Taylor's theorem since ez cos z is analytic for all z. Hence
[(ez cos z) / z3] = (1 / z3) + (1 / z2) – (1 / 3) + ... 0 < |z| < ∞. (3)
As noted above, this must be the Laurent series for f(z). The principal part of a function f at a point z0 is
defined as the portion of its Laurent series involving negative powers of z – z0. Hence, the principal part of
f(z) = [(ez cos z) / z3]
at 0, its only singular point, can be seen from (3) to be
(1 / z3) + (1 / z2).
If the principal part of f at z0 contains at least one non-zero term but the number of such terms is finite, the
isolated singular point z0 is then called a pole of order m, where m is the largest of the powers
[bj / (z – z0)j]
in the principal part of f. Hence, in this case, it is said that f has a pole of order 3 at z0 = 0.
## at z0 = – 1. What type of singular point is z0?
Solution: Since the singular point z0 = – 1 is isolated, the function has a Laurent series
expansion about – 1 which is valid at every point except – 1 in the circular domain centered at – 1 with
radius r equal to the distance between – 1 and the next closest singularity, i.e.,
r = |2(1/3)|.
To find this expansion first expand
f1(z) = [z / (z3 + 2)]
in a Taylor series about z0 = – 1 to obtain
f1(z) = ∞∑n=0 [{f(0)(z0)} / n!] (z – z0)n. (1)
Now
f1(z) = [z / (z3 + 2)]
so
f1(1)(z) = [(2 – 2z3) / (z3 + 2)2]
and
f1(2) = [{6(z5 – 4z2)} / (z3 + 2)3]
etc. Using these results in (1) one finds
[z / (z3 + 2)] = – 1 + 4(z + 1) – 15(z + 1)2 + ... . (2)
## Now divide equation (2) by (z + 1)2 to obtain
f(z)[z / {(z + 1)2 (z3 + 2)}] = [(– 1) / (z + 1)2] + [4 / (z + 1)] – 15 + ... . (3)
The principal part of a function at a point z0 is defined as that part of its Laurent series involving negative
powers of (z – z0). Since (3) is the Laurent series representing
f(z) = [z / {(z + 1)2 (z3 + 2)}]
for 0 < |z + 1| < |2(1/3)|, it is concluded that the principal part of f at z0 = – 1 is
[(– 1) / (z + 1)2] + [4 / (z + 1)].
If the principal part of f at z0 contains at least one nonzero term but the number of such terms is
finite, then the isolated singular point z0 is called a pole of order m where m is the largest of
[bj / (z – z0)j]
in the principal part of f. Hence, in this case,
f(z) = [z / {(z + 1)2 (z3 + 2)}]
is said to have a pole of order 2 at z0 = – 1.
|
## What is the parametric equation of circle?
We have what’s called the parametric equation of the circle: x = rcosθ, y = rsinθ (where θ is a parameter). In other words, for all values of θ, the point (rcosθ, rsinθ) lies on the circle. Or, any point on the circle is (rcosθ, rsinθ), where θ is a parameter.
## What is a circular equation?
EQUATION OF A CIRCLE. The equation of a circle comes in two forms: 1) The standard form: (x – h)2 + (y-k)2 = r2. 2) The general form : x2 + y2 + Dx + Ey + F = 0, where D, E, F are constants. If the equation of a circle is in the standard form, we can easily identify the center of the circle, (h, k), and the radius, r .
## How do you parameterize a circle on a plane?
The way to derive a circle is to set up the identity of distance, i.e. the square of the distance of a point on the circle to its center (1 for unit circle) is equal to the sum of squares(Pythagorean theorem). If you are given the points on x+y=0 plane, then the setting of such points would be limited to (x,−x,z).
## What is the parametric equation of a straight line?
And this is the parametric form of the equation of a straight line: x = x1 + rcosθ, y = y1 + rsinθ.
## What is a Cartesian equation of a circle?
Cartesian coordinate system with a circle of radius 2 centered at the origin marked in red. The equation of a circle is (x − a)2 + (y − b)2 = r2 where a and b are the coordinates of the center (a, b) and r is the radius.
## How do you plot a circle?
follow these steps:Realize that the circle is centered at the origin (no h and v) and place this point there.Calculate the radius by solving for r. Set r-squared = 16. Plot the radius points on the coordinate plane. Connect the dots to graph the circle using a smooth, round curve.
You might be interested: Solving equation with rational expressions
## How do you write the standard form of an equation of a circle?
The standard form of a circle’s equation is (x-h)² + (y-k)² = r² where (h,k) is the center and r is the radius. To convert an equation to standard form, you can always complete the square separately in x and y.
## What is ellipse equation?
The standard form of the equation of an ellipse with center (0,0) and major axis parallel to the x-axis is. x2a2+y2b2=1. where. a>b. the length of the major axis is 2a.
## How do you sketch the equation of a curve?
The following steps are taken in the process of curve sketching:Domain. Find the domain of the function and determine the points of discontinuity (if any).Intercepts. Symmetry. Asymptotes. Intervals of Increase and Decrease. Local Maximum and Minimum. Concavity/Convexity and Points of Inflection. Graph of the Function.
## What is parameter circle?
A circle can be defined as the locus of all points that satisfy the equations. x = r cos(t) y = r sin(t) where x,y are the coordinates of any point on the circle, r is the radius of the circle and. t is the parameter – the angle subtended by the point at the circle’s center.
### Releated
#### Convert to an exponential equation
How do you convert a logarithmic equation to exponential form? How To: Given an equation in logarithmic form logb(x)=y l o g b ( x ) = y , convert it to exponential form. Examine the equation y=logbx y = l o g b x and identify b, y, and x. Rewrite logbx=y l o […]
#### H2o2 decomposition equation
What does h2o2 decompose into? Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts – either 2OHs or an H and HO2: If there are no other molecules to react with, the parts will form water and oxygen gas as these are more stable […]
|
For all $n,k$ in positive integers, there exists $m_1,...,m_k$ such that the following holds
Let $$n,k$$ be two positive integers. Prove that there exists $$m_1,...,m_k$$ in the positive integers such that
$$1+\frac{2^k-1}{n}=\prod_{i=1}^k\left(1+\frac{1}{m_i}\right)$$ Here is my attempt
We're going to construct a solution for the $$m_i$$'s, but rather than presenting it right away I will take time to go through the motivation behind it.
Assume that $$\exists m_1,...,m_k$$ such that $$1+\frac{2^k-1}{n}=\prod_{i=1}^k\left(1+\frac{1}{m_i}\right)$$ And consider \begin{align} \prod_{i=1}^{k+1}\left(1+\frac{1}{m_i}\right)&=\left(1+\frac{1}{m_{k+1}}\right)\prod_{i=1}^k\left(1+\frac{1}{m_i}\right) \\ &= \left(1+\frac{1}{m_{k+1}}\right)\left(1+\frac{2^k-1}{n}\right) \\ &= 1+\frac{2^{k+1}-1}{n} \end{align} Where the last step follows from the assumption. Hence $$\frac{1}{m_{k+1}}=\frac{2^k}{2^k+n-1} \text{ or } m_i=\frac{2^{i-1}+n-1}{2^{i-1}}$$ Now we have to prove this solution actually works. In other words, $$1+\frac{2^k-1}{n}=\prod_{i=1}^k\left(1+\frac{2^{i-1}}{2^{i-1}+n-1}\right)=\prod_{i=1}^k\left(\frac{2^{i}+n-1}{2^{i-1}+n-1}\right)$$ but expanding the RHS is not easy so let's go ahead and assume that the above is true $$\forall k\le s$$ we'll show that it's true for $$k=s+1$$
\begin{align}\prod_{i=1}^{s+1}\left(\frac{2^{i}+n-1}{2^{i-1}+n-1}\right)&=\frac{2^{s+1}+n-1}{2^{s}+n-1}\prod_{i=1}^{s}\left(\frac{2^{i}+n-1}{2^{i-1}+n-1}\right) \\&= \frac{2^{s+1}+n-1}{2^{s}+n-1}\left(1+\frac{2^s-1}{n}\right) \\&= 1+\frac{2^{s+1}-1}{n}\end{align}
Now my question is: Are the $$m_i$$'s integers? Since $$m_i=\frac{2^{i-1}+n-1}{2^{i-1}}$$ Then we should have $$2^{i-1}\mid n-1$$ but that's not true $$\forall n,i$$?
The $$m_i$$'s in your construction are not necessarily integers.
To illustrate the problem, consider the simple case of $$k = 2$$. You want to write $$\frac{n + 3}n$$ as a product $$\frac{m_1 + 1}{m_1} \cdot \frac{m_2 + 1}{m_2}$$.
Your strategy goes for $$m_1 = n$$ and $$m_2 = \frac{n + 1}2$$. However we see that $$\frac{n + 1}2$$ is not necessarily an integer.
In fact, the construction above works when $$n$$ is odd. For $$n$$ even, we can use instead $$m_1 = \frac n 2$$ and $$m_2 = n + 2$$.
This indicates how the solution would look like: you should separate the two cases of even/odd $$n$$.
We prove by induction on $$k$$. For $$k = 1$$, simply choosing $$m_1 = n$$ works. Now assume that the result is true for $$k$$ and we prove it for $$k + 1$$.
Thus we want to write $$\frac{n + 2^{k + 1} - 1}n$$ as a product $$\prod_{i = 1}^{k + 1} \frac{m_i + 1}{m_i}$$. As above, we will consider the parity of $$n$$.
If $$n$$ is odd, then we can write $$\frac{\frac{n + 1}2 + 2^k - 1}{\frac{n + 1}2} = \prod_{i = 1}^k \frac{m_i + 1}{m_i}$$ by induction hypothesis applied to $$\frac{n + 1}2$$. Choosing $$m_{k + 1} = n$$ gives us the willing identity.
If $$n$$ is even then we can write $$\frac{\frac n 2 + 2^k - 1}{\frac n 2} = \prod_{i = 1}^k \frac{m_i + 1}{m_i}$$ by induction hypothesis applied to $$\frac n 2$$. Choosing $$m_{k + 1} = n + 2^{k + 1} - 2$$ gives us the willing identity.
This finishes the induction step and hence the proof.
• I didn't understand is $m_i=n+2^i-2$ for all even $n$ and $m_i=n$ for all odd $n$?
– PNT
Jun 19 at 20:51
• It's an inductive construction which would be complicated to write down explicitly (probably would involve binary expansion of $n$). Do you understand proof by induction in general? Jun 19 at 20:53
• I do but you're inducting on $k$ not $n$ so how can you say that you're applying the induction hypothesis on $n+1/2$?
– PNT
Jun 19 at 21:16
• The statement for step $k$ is valid for any positive integer $n$, thus I apply it with $n$ replaced by $\frac{n + 1}2$ which is again a positive integer. Jun 19 at 21:23
• +1. In particular, if $k=2$ then $(3m_1-n)(3m_2-n)=n(n+1).$ So if $k=2=n$ then $(3m_1-2)(3m_2-2)=6$, which is impossible for integers $m_1,m_2$ as it would imply $4\equiv 6\mod 3.$ Jun 20 at 3:53
After thinking for some days, I would like to share my observations from the first answer and how it relates to binary notations and bit operations. This answer is based on the same recursion step and the same algorithm.
$$\frac{n_{k+1}+2^{k+1}-1}{n_{k+1}} = \frac{n_k+2^{k}-1}{n_{k}}\cdot \frac{m_{k+1}+1}{m_{k+1}} \tag 1$$
where $$n_k$$ is chosen depending on whether $$n_{k+1}$$ is even or odd,
\begin{align*} n_{k+1} \mapsto n_k &= \begin{cases} \dfrac {n_{k+1}}2,& n_{k+1}\equiv 0 \pmod 2\\ \dfrac {n_{k+1}+1}2,& n_{k+1}\equiv 1 \pmod 2\\ \end{cases}\\ &= \left\lfloor\frac{n_{k+1}+1}2\right\rfloor = \left\lceil\frac{n_{k+1}}2\right\rceil \tag2 \end{align*}
From $$n_{k+1}+2^{k+1}-1$$ to $$n_k+2^k-1$$ in the numerators of $$(1)$$ is actually simple: it's a floored division by $$2$$, or a right shift by $$1$$ bit:
\begin{align*} n_k+2^k-1 &= \left\lfloor\frac{n_{k+1}+1}2\right\rfloor + 2^k-1\\ &= \left\lfloor\frac{n_{k+1}+1}2 +2^k-1\right\rfloor\\ &= \left\lfloor\frac{n_{k+1}+2^{k+1}-1}2\right\rfloor \end{align*}
But from $$n_{k+1}$$ to $$n_k$$ in the denominators is less obvious. Applying $$(2)$$ is not simply a floored division nor a right shift by $$1$$ bit. Repeated application of $$(2)$$ to a positive $$n$$ will also never get to $$0$$. But there's still some pattern, by noting the parity of $$n \pmod 2$$ at each step, for example when applying to $$n=10$$,
$$\underbrace{10}_{\equiv 0} \mapsto \underbrace{5}_{\equiv1} \mapsto \underbrace{3}_{\equiv1}\mapsto \underbrace{2}_{\equiv0}\mapsto \underbrace{1}_{\equiv 1} \mapsto \underbrace{1}_{\equiv 1} \mapsto \ldots$$
Writing the remainders as bits from the least to the most significant, this gives an infinite binary string $$\ldots1111\ 0110_2$$. And one might notice this is the two's complement notation of $$-n = -10_{10}$$! In fact, considering how $$-n_{k+1}$$ and $$-n_k$$ are related by $$(2)$$,
$$-n_k = -\left\lceil\frac{n_{k+1}}2\right\rceil = \left\lfloor\frac{-n_{k+1}}2\right\rfloor$$
so applying arithmetic right shift by $$1$$ bit to $$-n_{k+1}$$ does give $$-n_k$$.
Hence both the numerators and denominators in $$(1)$$ are related by binary right shifts or floored divisions:
$$-\frac{n_{k+1}+2^{k+1}-1}{-n_{k+1}} \overset{>>1}{\underset{>>1}\longmapsto} -\frac{n_k+2^k-1}{-n_k}$$
On the left hand side, if both the numerators and denominators were even, i.e. with last bits $$0$$, then the two sides would be equal. But the numerators and denominators always have opposite parity, and the $$\frac{1+m_{k+1}}{m_{k+1}}$$ multiplier is here to round the odd one down to even.
If $$n_{k+1}$$ is even, then the numerator is odd, so choose $$m_{k+1}$$ to reduce the numerator by $$1$$:
\begin{align*} -\frac{n_{k+1}+2^{k+1}-1}{-n_{k+1}} &= -\frac{n_{k+1}+2^{k+1}-2}{-n_{k+1}}\cdot \frac{n_{k+1}+2^{k+1}-1}{n_{k+1}+2^{k+1}-2}\\ &= -\frac{n_k+2^k-1}{-n_k}\left(1+\frac{1}{n_{k+1}+2^{k+1}-2}\right) \end{align*}
If $$n_{k+1}$$ is odd, then the denominator is odd, so choose $$m_{k+1}$$ to reduce the denominator by $$1$$ (to be more negative):
\begin{align*} -\frac{n_{k+1}+2^{k+1}-1}{-n_{k+1}} &= -\frac{n_{k+1}+2^{k+1}-1}{-n_{k+1}-1}\cdot \frac{-n_{k+1}-1}{-n_{k+1}}\\ &= -\frac{n_k+2^k-1}{-n_k}\left(1+\frac{1}{n_{k+1}}\right) \end{align*}
This explains why the $$m_{k+1}$$ appears to be quite different depending on the parity of $$n_{k+1}$$:
$$m_{k+1} = \begin{cases} n_{k+1}+2^{k+1}-2,& n_{k+1}\equiv 0 \pmod 2\\ n_{k+1},& n_{k+1}\equiv 1 \pmod 2\\ \end{cases} \tag3$$
To conclude, by considering the recursion step $$(1)$$ with negated denominators, the choices of $$n_k$$ and $$m_{k+1}$$ are related to binary notations and bit shifts. Running the recursion down to $$k=0$$ gives the last $$m_1=n_1$$ and a fraction that can be eliminated: $$\dfrac{n_0+2^0-1}{n_0} = 1$$.
Omitted detail on two's complement, which is non-standard:
$$-10_{10} \overset{???}= \ldots1111\ 0110_2 = 2+4+16+32+64+\cdots$$
Either all mentions of bit shifts can be ignored and consider floored division only.
Or a fixed and finite number $$L$$ can be chosen to be the number of bits for two's complement. $$L$$ bits should be long enough to represent the numerator $$n+2^k-1$$ (and hence also the denominator $$\pm n$$).
|
# How do you solve 2(2+3)+4x=2(2x+2)+6?
May 28, 2018
All real numbers or $\left(- \infty , \infty\right)$ in interval notation.
#### Explanation:
$2 \left(2 + 3\right) + 4 x = 2 \left(2 x + 2\right) + 6$
First, simplify $2 \left(2 + 3\right)$:
$\textcolor{b l u e}{2 \left(2 + 3\right) = 2 \left(5\right) = 10}$
Put it back into the equation:
$10 + 4 x = 2 \left(2 x + 2\right) + 6$
Next, use the distributive property to simplify $2 \left(2 x + 2\right)$:
Following this image, we know that:
$\textcolor{b l u e}{2 \left(2 x + 2\right) = \left(2 \cdot 2 x\right) + \left(2 \cdot 2\right) = 4 x + 4}$
Put it back into the equation:
$10 + 4 x = 4 x + 4 + 6$
Add $4 + 6 = 10$:
$10 + 4 x = 4 x + 10$
Subtract $\textcolor{b l u e}{4 x}$ from both sides of the equation:
$10 + 4 x \quad \textcolor{b l u e}{- \quad 4 x} = 4 x + 10 \quad \textcolor{b l u e}{- \quad 4 x}$
$10 = 10$
Oh no! Our variables are gone now. Now we see if this equation is true. It is true that $10 = 10$, meaning that the answer is All real numbers or $\left(- \infty , \infty\right)$ in interval notation.
Hope this helps!
|
# How Many Tens Are In 740?
Tens are a basic unit of measurement that can be used to measure many things. When it comes to mathematics, tens are a type of counting or grouping system used to help simplify larger numbers. Knowing how many tens are in a certain number can make it easier to do calculations, such as addition and subtraction.
The number 740 is a three-digit number that can be broken down into hundreds, tens, and ones. The hundreds place is the 7, the tens place is the 4, and the ones place is the 0. So, the number 740 can be broken down into seven hundreds, four tens, and zero ones.
To determine how many tens there are in 740, we need to look at the tens place. The tens place of 740 is the 4. This means that there are four tens in 740. Four tens can also be expressed as forty, as there are four groups of ten in the number. Forty is the same as four tens.
To help remember that there are four tens in 740, you can use a strategy called chunking. Chunking is a strategy used to break down a number into smaller, easier to remember parts. In this case, you can break down 740 into 700 and 40. This helps make it easier to remember that there are four tens in 740.
In conclusion, the number 740 is made up of seven hundreds, four tens, and zero ones. To figure out how many tens there are in 740, you need to look at the tens place. In 740, the tens place is the 4, which means there are four tens in 740. This can also be expressed as forty, as there are four groups of ten in the number.
##
In summary, the number 740 is made up of seven hundreds, four tens, and zero ones. This means that there are four tens in 740. Knowing how many tens are in a number can be helpful when doing calculations like addition and subtraction. You can also use the strategy of chunking to help you remember how many tens are in a number.
|
Lesson Objectives
• Learn how to find arc length on a circle
• Learn how to find the area of a sector of a circle
## How to Find the Arc Length on a Circle
In our last lesson, we learned how to measure angles using radians. In general, if our angle θ is a central angle of a circle of radius r, and θ intercepts an arc of length s, then we can obtain our radian measure of θ as: $$θ=\frac{s}{r}$$ Therefore, if the arc length (s) is equal to the radius (r), the measure of the angle will be exactly 1 radian. $$s=r$$ $$θ=\frac{s}{r}$$ $$θ=\frac{r}{r}=1$$ Note: when working with angles, if the degree symbol "°" is not shown, our measure is given in terms of radians.
### Arc Length on a Circle
From our definition of our angle θ in radians above, we can obtain a formula for finding the length of an arc of a circle. We know that the measure of our angle θ in radians is represented by: $$θ=\frac{s}{r}$$ If we multiply both sides by r, we will obtain: $$\frac{s}{r}=θ$$ $$\require{cancel}\cancel{r}\cdot \frac{s}{\cancel{r}}=r \cdot θ$$ $$s=r θ$$ This leads us to the following formula:
### Formula for Arc Length
The length s of the arc intercepted on a circle of radius r by a central angle of measure θ radians is given by: $$s=rθ$$ where θ is in radians. Let's look at a few examples.
Example #1: Find the length of each arc. $$θ=\frac{π}{4}$$ $$r=12\hspace{.2em}\text{miles}$$ To find the length of our given arc, let's use our formula: $$s=r θ$$ Let's plug into our formula: $$s=12 \hspace{.2em}\text{miles}\cdot \frac{π}{4}$$ $$s=3\cancel{12}\hspace{.2em}\text{miles}\cdot \frac{π}{\cancel{4}}$$ $$s=3π \hspace{.2em}\text{miles}$$ Example #2: Find the length of each arc. $$θ=135°$$ $$r=12\hspace{.2em}\text{inches}$$ In this case, we can't immediately use our formula since our angle measure is given in degrees. We need to first convert θ to radians. $$135° \cdot \frac{π}{180°}=\frac{3π}{4}$$ Now that θ has been converted into radians, we can use our formula: $$s=r θ$$ $$s=12 \hspace{.2em}\text{inches}\cdot \frac{3π}{4}$$ $$s=3\cancel{12}\hspace{.2em}\text{inches}\cdot \frac{3π}{\cancel{4}}$$ $$s=9π \hspace{.2em}\text{inches}$$
## How to Find the Area of a Sector of a Circle
We have already seen that a complete circle or one full rotation forms an angle with a measure of 360° or $2π$ radians. The formula for the area of a complete circle with a radius r: $$\text{area of a circle}=π r^2$$ A sector of a circle is the portion of the interior of a circle intercepted by a central angle. Visually, it looks like a "piece of pie". If a central angle for a sector of a circle has a measure of θ radians, then the sector makes up the fraction: $\frac{θ}{2π}$ of a complete circle. To obtain the area of a sector, we will modify our area of a circle formula:
### Area of a Sector of a Circle with Radius r and Central Angle θ Measured in Radians
$$\frac{θ}{2π}(πr^2)=\frac{1}{2}r^2 θ$$ Let's look at an example.
Example #3: Find the area of each sector. $$θ=\frac{3π}{2}$$ $$r=9 \hspace{.1em}\text{meters}$$ Let's plug into our formula: $$\frac{1}{2}r^2 θ$$ $$\frac{1}{2}\cdot (9 \hspace{.1em}\text{meters})^2 \cdot \frac{3π}{2}$$ $$\frac{243 π}{4}\hspace{.1em}\text{square meters}$$ Example #4: Find the area of each sector. $$θ=60°$$ $$r=13 \hspace{.1em}\text{yards}$$ Since θ is given in degrees, our first step is to convert θ into radians. $$60° \cdot \frac{π}{180°}=\frac{π}{3}$$ Now, we can plug into our formula: $$\frac{1}{2}r^2 θ$$ $$\frac{1}{2}\cdot (13 \hspace{.1em}\text{yards})^2 \cdot \frac{π}{3}$$ $$\frac{169 π}{6}\hspace{.1em}\text{square yards}$$
#### Skills Check:
Example #1
Find the length of each arc. $$r=15\text{m}$$ $$θ=\frac{4π}{3}$$
A
$$150 π \hspace{.2em}\text{m}$$
B
$$\frac{27}{8}π \hspace{.2em}\text{m}$$
C
$$60 π \hspace{.2em}\text{m}$$
D
$$\frac{40}{3}π \hspace{.2em}\text{m}$$
E
$$20 π \hspace{.2em}\text{m}$$
Example #2
Find the length of each arc. $$r=17\text{ft}$$ $$θ=\frac{3π}{4}$$
A
$$\frac{51π}{4}\hspace{.2em}\text{ft}$$
B
$$\frac{26π}{5}\hspace{.2em}\text{ft}$$
C
$$24π \hspace{.2em}\text{ft}$$
D
$$169π \hspace{.2em}\text{ft}$$
E
$$\frac{56}{3}\hspace{.2em}\text{ft}$$
Example #3
Find the area of each sector. $$r=9\text{m}$$ $$θ=255°$$
A
$$\frac{20π}{3}\hspace{.2em}\text{m}^2$$
B
$$\frac{39π}{8}\hspace{.2em}\text{m}^2$$
C
$$\frac{152π}{3}\hspace{.2em}\text{m}^2$$
D
$$\frac{459π}{8}\hspace{.2em}\text{m}^2$$
E
$$225π \hspace{.2em}\text{m}^2$$
|
### Section 2.5 Part 2 Other tests for Zeros
```Section 2.5 Part Two
Other Tests for Zeros
Descartes’s Rule of Signs
Upper and Lower Bounds
Descartes’s Rule of Signs
• Let f(x) be a polynomial with real coefficients and
ao ≠ 0.
f ( x ) a n x a n 1 x
n
n 1
... a 2 x a1 x a 0
2
• The number of positive real zeros of f is either
equal to the number of variations in sign of f(x) or
less than that number by an even integer.
• The number of negative real zeros of f is either
equal to the number of variations in the sign of f(-x)
of less than that number by an even integer.
Apply Descarte’s Rule of Signs
• Consider f(x) = x2 + 8x + 15
• Since there are no variations in sign of f(x)
there are no positive roots.
• Since f(-x) = (-x)2 + 8(-x) + 15
= x2 – 8x + 15
has two variations, f(x) may have two or zero
negative roots.
x = {-3. -5}
Variation in Sign
• A variation in sign means that when the
polynomial is written in standard form that one
term has a different sign than the next.
• T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60
• For Variations of T(x) just look at the signs
T(x) has one variation in signs
• For Variations of T(-x) you must change the sign
of terms with odd numbered exponents
• T(-x) = -x5 + 9x4 – 19x3 – 21x2 + 92x – 60
T(-x) has four variation in signs
Find the Zeros of T(x)
• T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60
• 60 + 1, + 2 , + 3, + 4, + 5 , + 6 , + 10 ,
+ 12 , + 15 , + 20, + 30, + 60
• Since there is only one sign variation for
T(x) there is at most, one positive root
• SYN Program
Find the Zeros of T(x)
T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60
T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30)
+2
1
1
9
19
-21 -92 -60
2
22
82 122
11
41
61
30
60
0
Now that we have found the positive root we
know that any other real roots must be negative.
Find the Zeros of T(x)
T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60
T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30)
• 30 -1, -2 , -3, -4, -5 , -6 , -10 ,-15, - 30
T(x) = (x – 2)(x + 1)(x3 + 10x2 + 31x + 30)
-1
1
1
11 41
61
-1 -10
-31 -30
10
31
30
30
0
Find the Zeros of T(x)
T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60
T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30)
T(x) = (x – 2)(x + 1)(x3 + 10x2 + 31x + 30)
• 30 -1, -2 , -3, -4, -5 , -6 , -10 ,-15, - 30
T(x) = (x – 2)(x + 1)(x + 2)(x2 + 8x + 15)
-2
1
1
10 31
30
-2 -16
-30
8
15
0
Find the Zeros of T(x)
T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60
T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30)
T(x) = (x – 2)(x + 1)(x3 + 10x2 + 31x + 30)
T(x) = (x – 2)(x + 1)(x + 2)(x2 + 8x + 15)
T(x) = (x – 2)(x + 1)(x + 2)(x + 3)(x + 5)
x = {-5, -3, -2, -1, 2}
Homework 2.5
•
•
•
•
•
Zeros of Polynomial Functions
page 160
1 - 31 odd,
37 - 85 odd,
91 - 94 all
#5 Find all of the zeros of the
function.
F(x) = (x + 6)(x + i)(x – i)
x = {-6, -i, i}
#10 Use the rational zero test to list all of the
possible rational zeros of f. Verify that the
zeros of f shown are contained in the list.
f(x) = 4x5 – 8x4 – 5x3 + 10x2 + x – 2
4
Factors of 2
Factors of 4
2
1 2
1 2 4
1 2
-2
-4
-6
1
2
1
4
1 1
x { 1,
, , 1, 2}
2 2
#15 Find all of the real zeros of the
function.
h(t) = t3 + 12t2 + 21t + 10
Factors of 10
Factors of 1
1 2 5 10
1
Caution graph may be
6
With a cubic we expect
another turn down to the left
4
2
x { 10 , 1, 1}
-5
5
-2
#25 f(x) =
x3
Factors of 4
Factors of 1
+
x2
– 4x – 4
1 2 4
1
(a)
(b)
(c)
List the possible rational zeros of f,
sketch the graph of f so that some
possibilities can be eliminated
determine all the real zeros
1, 1, 2 , 2 , 4 , 4
4
Graph eliminates -4, 1, 4
f(x) = x3 + x2 – 4x – 4
2
f(x) = x2 (x +1)– 4(x + 1)
-5
5
f(x) = (x + 1)(x2– 4)
-2
f(x) = (x + 1)(x +2)(x – 2)
-4
-6
x { 2 , 1, 2}
#42. Find the polynomial function with
integer coefficients that has the given zeros.
5, 5, 1
Since imaginary solutions always
appear in conjugate pairs we
know to include
3i
1
f ( x ) ( x 5 )( x 5 )( x 1
3 i )( x 1
f ( x ) ( x 25 )( x 2 x 4 )
2
2
f ( x ) x 2 x 21 x 50 x 100
4
3
2
3i
3i )
```
|
## What is the position time equation?
Summary
1st equation v = v + at velocity-time
2nd equation s = s + vt + ½at2 position-time
3rd equation v2 = v2 + 2a(s − s) velocity-position
merton rule v = ½(v + v) average velocity
## What is a position vs time graph?
A position-time graph shows how far an object has traveled from its starting position at any given time since it started moving.
## What are the 3 equations of motion?
There are three equations of motion that can be used to derive components such as displacement(s), velocity (initial and final), time(t) and acceleration(a).Definition of Equations of MotionFirst Equation of Motion : v=u+at.Second Equation of Motion : s=ut+frac{1}{2}at^2.Third Equation of Motion : v^2=u^2+2as.
## What are the 5 equations of motion?
In circumstances of constant acceleration, these simpler equations of motion are usually referred to as the “SUVAT” equations, arising from the definitions of kinematic quantities: displacement (S), initial velocity (u), final velocity (v), acceleration (a), and time (t).
## What are the 5 kinematic equations?
Building on what you have learned so far and what Galileo presented, we have what my physics teacher, Glenn Glazier, liked to call the Five Sacred Equations of Kinematics for constant acceleration. In these equations, v is velocity, x is position, t is time, and a is acceleration. Remember, Δ means change in.
## What is the slope of a position vs time graph?
The principle is that the slope of the line on a position-time graph is equal to the velocity of the object. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s.
## How do we find speed?
Speed tells us how fast something or someone is travelling. You can find the average speed of an object if you know the distance travelled and the time it took. The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time.
## What does the slope of position time graph give?
What does the slope represent on a position graph? The slope of a position graph represents the velocity of the object. So the value of the slope at a particular time represents the velocity of the object at that instant.
## What are the 4 equations of motion?
In circumstances of constant acceleration, these simpler equations of motion are usually referred to as the SUVAT equations, arising from the definitions of kinematic quantities: displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t).
## What is the SI unit of velocity?
Velocity is a physical vector quantity; both magnitude and direction are needed to define it. The scalar absolute value (magnitude) of velocity is called speed, being a coherent derived unit whose quantity is measured in the SI (metric system) as metres per second (m/s) or as the SI base unit of (m⋅s1).
## What is motion formula?
Newton’s second law, which states that the force F acting on a body is equal to the mass m of the body multiplied by the acceleration a of its centre of mass, F = ma, is the basic equation of motion in classical mechanics.
## What is the position time graph?
In these graphs, the vertical axis represents the position of the object while the horizontal axis represents the time elapsed: the dependent variable, position, depends on the independent variable, time. In this way, the graph tells us where the particle can be found after some amount of time.
## How do you read a position vs time graph?
In a position-time graph, the velocity of the moving object is represented by the slope, or steepness, of the graph line. If the graph line is horizontal, like the line after time = 5 seconds in Graph 2 in the Figure below, then the slope is zero and so is the velocity. The position of the object is not changing.
### Releated
#### Equation for maximum height
What is the formula for height in physics? Determine how high the projectile traveled above its initial height by using the following formula where V is the initial vertical velocity and T is the time it takes to reach its peak: Height = V * T +1/2 * -32.2 ft/s^2 *T^2 For example, if you […]
#### Potential energy equation physics
What is potential energy physics? To summarize, potential energy is the energy that is stored in an object due to its position relative to some zero position. An object possesses gravitational potential energy if it is positioned at a height above (or below) the zero height. What is potential and kinetic energy? All forms of […]
|
USING OUR SERVICES YOU AGREE TO OUR USE OF COOKIES
# What Is The Prime Factorization Of 98
What is the prime factorization of 98? Answer: 2 * 7 * 7
The prime factorization of 98 has 3 prime factors. If you multiply all primes in the factorization together then 98=2 * 7 * 7. Prime factors can only have two factors(1 and itself) and only be divisible by those two factors. Any number where this rule applies can be called a prime factor. The biggest prime factor of 98 is 7. The smallest prime factor of 98 is 2.
## How To Write 98 As A Product Of Prime Factors
How to write 98 as a product of prime factors or in exponential notation? First we need to know the prime factorization of 98 which is 2 * 7 * 7. Next we add all numbers that are repeating more than once as exponents of these numbers.
Using exponential notation we can write 98=21*72
For clarity all readers should know that 98=2 * 7 * 7=21*72 this index form is the right way to express a number as a product of prime factors.
## Prime Factorization Of 98 With Upside Down Division Method
Prime factorization of 98 using upside down division method. Upside down division gives visual clarity when writing it on paper. It works by dividing the starting number 98 with its smallest prime factor(a figure that is only divisible with itself and 1). Then we continue the division with the answer of the last division. We find the smallest prime factor for each answer and make a division. We are essentially using successive divisions. This continues until we get an answer that is itself a prime factor. Then we make a list of all the prime factors that were used in the divisions and we call it prime factorization of 98.
2|98 We divide 98 with its smallest prime factor, which is 2
7|49 We divide 49 with its smallest prime factor, which is 7
7 The division of 7/49=7. 7 is a prime factor. Prime factorization is complete
The solved solution using upside down division is the prime factorization of 98=2 * 7 * 7. Remember that all divisions in this calculation have to be divisible, meaning they will leave no remainder.
## Mathematical Properties Of Integer 98 Calculator
98 is a composite figure. 98 is a composite number, because it has more divisors than 1 and itself. This is an even integer. 98 is an even number, because it can be divided by 2 without leaving a comma spot. This also means that 98 is not an odd digit. When we simplify Sin 98 degrees we get the value of sin(98)=-0.57338187199042. Simplify Cos 98 degrees. The value of cos(98)=-0.81928824529146. Simplify Tan 98 degrees. Value of tan(98)=0.69985365380953. When converting 98 in binary you get 1100010. Converting decimal 98 in hexadecimal is 62. The square root of 98=9.8994949366117. The cube root of 98=4.6104362920584. Square root of √98 simplified is 7√2. All radicals are now simplified and in their simplest form. Cube root of ∛98 simplified is 98. The simplified radicand no longer has any more cubed factors.
## Write Smaller Numbers Than 98 As A Product Of Prime Factors
Learn how to calculate factorization of smaller figures like:
## Express Bigger Numbers Than 98 As A Product Of Prime Factors
Learn how to calculate factorization of bigger amounts such as:
## Single Digit Properties For 98 Explained
• Integer 9 properties: 9 is odd and the square of 3. It is a composite, with the following divisors:1, 3, 9. Since the quantity of the divisors(excluding itself) is 4<9, it is a defective number. In mathematics nine is a perfect total, suitable and a Kaprekar figure. Any amount is divisible by 9 if and only if the quantity of its digits is. Being divisible by the count of its divisors, it is refactorizable. Each natural is the sum of at most 9 cubes. If any sum of the digits that compose it is subtracted from any natural, a multiple of 9 is obtained. The first odd square and the last single-digit quantity. In the binary system it is a palindrome. Part of the Pythagorean triples (9, 12, 15), (9, 40, 41). A repeated number in the positional numbering system based on 8. Nine is a Colombian digit in the numerical decimal system. If multiplied 9 always leads back to itself: 2×9=18 → 1+8=9, 3×9=27 → 2+7=9 in the same way if you add a number to 9, the result then refers to the initial digit: 7+9=16 → 1+6=7, 7+9+9=25 → 2+5=7, 7+9+9+9=34 → 3+4=7. If you put 111111111 in the square (ie 1 repeated 9 times) you get the palindrome 12345678987654321, also if you add all the numbers obtained: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 you get 81, and in turn 8 + 1 = 9.
• Integer 8 properties: Eight is even and a cube of 2. It is a composite, with the following 4 divisors:1, 2, 4, 8. Since the total of the divisors(excluding itself) is 7<8, it is a defective number. The sixth of the Fibonacci sequence, after 5 and before 13. It is the quantity of the twin primes 3 and 5. The first octagonal value. 8 is a Ulam, centered heptagonal and Leyland number. All amounts are divisible by 8 if and only if the result formed by its last three digits is. A refactorizable, being divisible by the count of its divisors. At the same time a highly totter and highly cototent quantity. It is the fourth term of the succession of Mian-Chowla. Any odd greater than or equal to 3, elevated to the square, to which subtract is subtracted 1 is divisible by 8 (example: 7²=49 49-1=48 divisible by 8). The sum of two squares, 8=2²+2². The sum of the digits of its cube: 8³=512, 5+1+2=8. The first 4-digit binary:1000. Part of the Pythagorean triples (6, 8, 10), (8, 15, 17). Eight is a repeated number in the positional numbering system based on 3 (22) and on the base 7 (11).
## Finding Prime Factorization Of A Number
The prime factorization of 98 contains 3 primes. The prime factorization of 98 is and equals 2 * 7 * 7. This answer was calculated using the upside down division method. We could have also used other methods such as a factor tree to arrive to the same answer. The method used is not important. What is important is to correctly solve the solution.
## List of divisibility rules for finding prime factors faster
Knowing these divisibility rules will help you find primes more easily. Finding prime factors faster helps you solve prime factorization faster.
Rule 1: If the last digit of a number is 0, 2, 4, 6 or 8 then it is an even integer. All even integers are divisible by 2.
Rule 2: If the sum of digits of a number is divisible by 3 then the figure is also divisible by 3 and 3 is a prime factor(example: the digits of 102 are 1, 0 and 2 so 1+0+2=3 and 3 is divisible by 3, meaning that 102 is divisible by 3). The same logic works also for number 9.
Rule 3: If the last two digits of a number are 00 then this number is divisible by 4(example: we know that 212=200+12 and 200 has two zeros in the end making it divisible with 4. We also know that 4 is divisible with 12). In order to use this rule to it's fullest it is best to know multiples of 4.
Rule 4: If the last digit of a integer is 0 or 5 then it is divisible by 5. We all know that 2*5=10 which is why the zero is logical.
Rule 5: All numbers that are divisible by both 2 and 3 are also divisible by 6. This makes much sense because 2*3=6.
## What Is Prime Factorization Of A Number?
In mathematics breaking down a composite number(a positive integer that can be the sum of two smaller numbers multiplied together) into a multiplication of smaller figures is called factorization. When the same process is continued until all numbers have been broken down into their prime factor multiplications then this process is called prime factorization.
Using prime factorization we can find all primes contained in a number.
|
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
Main content
### Course: 6th grade (Eureka Math/EngageNY)>Unit 6
Lesson 2: Topic B: Summarizing a distribution that is approximately symmetric using the mean and mean absolute deviation
# Calculating the mean
Learn how to calculate the mean by walking through some basic examples & trying practice problems.
The mean is used to summarize a data set. It is a measure of the center of a data set. Let's look at an example.
Claire has $5$ cookies, Brooke has $3$ cookies, Deandra has $6$ cookies, and Lucy has $2$ cookies. Find the mean number of cookies.
Let's start by drawing a picture to show each person and their cookies:
Imagine that the girls combined all of their cookies
and then each took the same number of cookies.
Each girl would have $4$ cookies. So, the mean is $4$ cookies.
Key idea: We can think of the mean as the number of cookies each girl would have if they were equally distributed among the four girls.
Find the mean number of bananas each of the monkeys has in the picture below.
bananas
## Calculating the mean
We don't need to draw a picture every time we want to calculate the mean. Instead, we can follow these steps:
Step 1: Add up all of the data points (this is like combining all of the cookies)
Step 2: Divide the total by the number of data points in the data set (this is like each girl taking the same number of cookies)
Let's do this for the data set $\left\{7,2,8,6,7\right\}$:
$\begin{array}{rlr}7+2+8+6+7& =30& \text{Step 1}\\ \\ \\ \\ \frac{30}{5}& =6& \text{Step 2}\end{array}$
The mean of this data set is $6$.
## Calculating the mean walkthrough
Let's find the mean of the data set $\left\{2,1,2,4,5,4\right\}$ together.
Add up all of the data points.
How many data points are in the data set?
Great! Now divide the total by the number of data points.
The mean of this data set is
.
Now it's time to try some practice on your own.
## Practice
Find the mean of the data set $\left\{5,23,8,12\right\}$.
Find the mean of the data set $\left\{2,7,5,4,6,3\right\}$.
Find the mean of the data set $\left\{4.5,5,3.5,2,2.5\right\}$.
## Want to join the conversation?
• how would you know when an outlier affects a data set?
(78 votes)
• An outlier is a number that is far from the data set. This could be the case such as in this set:
158, 156, 85, 145, 157, 159. 85 is the outlier. Without the 85 the mean would be 155, but with the 85 the mean is about 143. Just one number makes the mean decrease by 12. An outlier always affects a data set, because an outlier is a number that is nowhere near the current set of numbers.
(66 votes)
• How about the range? What is it?
(24 votes)
• The range of a numerical set is just the difference of the largest and smallest numbers. For example, lets say we have a set: {1, 4, 2, 9, 10} We will take the largest number, 10, and the smallest number, 1, and find the difference. 10-1=9. Therefore, the range of the set is 9. Hopefully this helps you understand range any better.
(77 votes)
• this are so hard sorry I'm not good with math
(11 votes)
• I would suggest keeping a notebook with all the math facts you have learned. Try to really get stuff pounded into your brain before you move on. For this you can use the rhyme: Hey diddle didle, the medians the middle, you add and divided for the mean, the mode is the one that appears the most, and the range is the difference between. Never skip stuff or guess randomly. Try to make sure you REALLY understand the answer. It might take a bit longer, but it's worth it in the long run. Also, don't be afraid to ask for help. I struggle with math, too. But just keep trying and I promise you'll get it. Also, just take a deep breath. It can get hard if you're super stressed. And I like to type out what I'm thinking, so if there is a problem, my teacher can figure out where I went wrong and correct it from there. Keep it up!
(46 votes)
• My teacher gives me this one like eighty times a day so I know all the answers but she still keeps giving it to me. >_<
(22 votes)
• Your teacher wants you to be very proficient in this I guess but now you can do it in a snap
(13 votes)
• Can there be more than one mode?
(10 votes)
• As you probably know, mode is the biggest set of ungrouped data. If two sets of data have the same amount, but the other data is less than the data that they have, then both of them are the modes. This is also true with 3 modes, 4 modes, 5 modes and so on.
(9 votes)
• A little challenging
(12 votes)
• How do I find a missing number when I know the mean?
(10 votes)
• what is median?
(5 votes)
• Median = Middle
(11 votes)
• so much easier now that I have practiced
(9 votes)
• 42 is the answer to life, the universe, and everything.
(7 votes)
|
Whether you are studying 6th grade math, are looking to get ahead and are in 4th or 5th grade, or are in 7th or 8th grade and math is a bit of a struggle, this course has it all!!! This course aligns with the Common Core 6th and 7th grade Math curricula.
Not a member yet? Click here to check out a bunch of free videos to see what makes our approach extraordinary!
### Chapter 1: Whole Numbers & Decimals
This chapter dives into and builds a solid foundation of your understanding of whole numbers and decimals. You will learn everything from how to read and compare whole numbers and decimals to how to perform the four operations with whole numbers and decimals.
### Chapter 2: Representing Data In Different Ways
In your world you are surrounded by data, or pieces of information that tell you a story about a certain situation or event. This chapter provides you with a thorough introduction into exploring and interpreting data using a variety of representations. You will learn about everything from measures of central tendency to a host of different kinds of graphs.
### Chapter 3: Variables, Algebraic Expressions, & Solving Equations
This Chapter is where you start to get your Algebra party on!!! You will explore how letters work with numbers (variables!) before learning the all-important translation skills where you take words and convert them into simpler Math expressions. You will then spend time investigating how to solve one-step equations before finishing off with the Distributive Property.
### Chapter 4: Divisibility, Factoring Numbers, & Fractions
This chapter takes you through many investigations involving properties of numbers. The journey begins with divisibility and exponents before going onto rocking how to find factors of numbers and using these factors to simplify fractions. You will then jump into some very cool lessons involving mixed numbers and improper fractions and using the Least Common Multiple to boost your fraction skills. The chapter rounds out nicely with seeing how to go between decimals and fractions before getting a small taste of the very radical concept of square roots!
### Chapter 5: Addition and Subtraction of Fractions
This chapter takes you through the ‘how’s’ and ‘why’s’ behind adding and subtracting fractions. Whether you are seeking to learn about estimating fraction addition and subtraction, want the basics about why you need a common denominator to add or subtract fractions, or are looking to add or subtract fractions with different denominators or mixed numbers, this chapter has you covered! A very neat way of solving one-step equations involving adding and subtracting fractions in the final lesson, too!
### Chapter 6: Multiplication and Division of Fractions
This chapter is all about mastering multiplying and dividing fractions. You will learn the processes involved in multiplying and dividing all kinds of fractions – mixed numbers included! Take the time to watch the videos explaining why the processes for multiplying and dividing fractions work. Your understanding of ‘why’ will pay off huge as you learn connected Math concepts in later Math courses! The final lesson in this chapter ties in how to solve one-step equations involving the multiplication of fractions.
### Chapter 7: Ratios, Proportions, and Percents
This chapter contains everything you need to know about ratios, proportions and percents. You will begin with a thorough view into ratios and units rates before navigating proportions and their uses in real-world word problems. The chapter rounds out nicely with lessons that show you how to convert between percents, fractions, and decimals and how to use percents to solve many classic Math problems you see in your Math class.
### Chapter 8: Fundamentals of Geometry
This chapter overviews some of the awesome foundations found in Geometry, a key field of Mathematics that is woven into all your Math courses from here through to the end of your Math studies! You will jump into lessons about the basics of everything from the building blocks of geometry before building to an understanding of major concepts like triangles, polygons, quadrilaterals, and congruence & similarity. The chapter concludes with looks into symmetry and the all-important ideas within transformations.
### Chapter 9: Unit Conversions, Perimeter, Area, & Volume
This chapter provides a strong overview of two- and three-dimensional geometric topics. The chapter kicks off with lessons on units of measurement and how to convert between them. You will then find yourself learning about perimeters and areas of all kinds of shapes. The chapter concludes with a tour through solids or three-dimensional figures and excellent videos demonstrating how to find surface areas and volumes of many of these solids.
### Chapter 10: Introduction to Probability
This chapter overviews the core ideas behind theoretical and experimental probability. The counting principle or calculating how many different ways a certain event can occur, will be expertly explained. You will roll through problems involving dice (I could not resist) and understand how to apply data from situations to make predictions about the likelihood of something occurring. The chapter concludes with a journey through independent and dependent events.
### Chapter 11: Integers – Properties, Operations, & Equations
One of the most important (and coolest!) chapters in all of your Math studies!! This chapter starts you off with an introduction into integers – a new number system! You will then master the rules for adding, subtracting, multiplying, and dividing integers by understanding why the rules make sense. The chapter rounds out with connecting your knowledge of solving one-step equations to doing so with integers and a bodacious introduction to the coordinate plane & graphing functions.
### Chapter 12: Two-Step Equations and Inequalities
This final chapter provides a brief link to some of the topics you will see covered in greater depth in Pre-Algebra. You will build on your understanding of solving one-step equations to learn how to rock solving two-step equations and why the solution processes work. You will close out the chapter with an exploration into inequalities involving variables and solving & graphing some one-step inequalities.
|
Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
# Simple Equations and Its Applications
In our day to day lives, we generally come across many situations where we observe simple equations and its applications.
## What is Simple Equation?
A mathematical equation which represents the relationship of two expressions on either side of the sign. It mostly has one variable and equal to symbol. Example: 2x – 4 = 2. In the given example, x is a variable.
Before understanding this, let us see an example, to visualize and interpret the meaning of simple equations.
Suppose, we have an equation, 12x + 40 = 100. What is the solution to this equation and what is the meaning? As we have already learned through the given image, how to solve simple equations, we would not go in details of solving. The solution of the above linear equation,(12x +40 =100) is x=5.
## Applications of Simple Equations
But now, the question is, how to interpret the meaning of the above equation. Let’s learn the applications of Simple Equation through an example.
Let us assume a situation, in which you go to the market to buy mangoes in a supermarket. Your mother gave you Rs.100 and said that you can keep Rs.40 for buying chocolates and for the remaining amount, you need to buy mangoes. When you went to the market, you found out that the price of one kg of mangoes was Rs.12. So, the final question is how much mangoes you should buy, so that everyone is satisfied.
Recall the concept of simple equations from previous classes and proceed. Let us assume that you have to purchased kg of mangoes, so the total amount of money that you will be spent on buying mangoes will be Rs.12x. Now, you have Rs.40 for yourself, so you have an equation:
12 x + 40 = 100
12x = 100 – 40
12x = 60
x = 60/12 = 5
This is the same equation as above. Now when you will solve this, you will get to know the value of 5kg of mangoes. It means that you can buy 5 kg mangoes and spend Rs.40 on buying chocolates.
This is one of the general applications of simple equations. We can consider any other situation, where any unknown is involved and we will surely find the application of simple equations in that particular situation.
### Transposition
The process of moving a term from one side of the equation to the other side is called transposing. It is the same as adding and subtracting a number from both sides of the equation.
Ex. Solve 5x + 9 = 19
-> Transposing 9 to the other side we get,
-> 5x = 19 – 9
-> 5x = 10
Transposing 5 to the other side we get,
x = 10/5
x = 2
Hence, it could be stated without doubt that the use of simple equations are applicable everywhere and not just restricted to classrooms. Â
|
# Strategies and Formulas for Tough GRE Sets
For some advanced Data Analysis and Probability questions, it will help you achieve better scores to know the logic and formulas behind set theory. Set theory hinges on two concepts: union and intersection. The union of sets is all elements from all sets. The intersection of sets is only those elements common to all sets.
Let’s call our sets A, B, and C, and use a Venn diagram to express their relationship.
If n = intersection and u = union, then we can describe the relationship between the sets thusly:
P(A u B u C) = P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)
To find the number of people in exactly one set: P(A) + P(B) + P(C) – 2P(A n B) – 2P(A n C) – 2P(B n C) + 3P(A n B n C)
To find the number of people in exactly two sets: P(A n B) + P(A n C) + P(B n C) – 3P(A n B n C)
To find the number of people in exactly three sets: P(A n B n C)
To find the number of people in two or more sets: P(A n B) + P(A n C) + P(B n C) – 2P(A n B n C)
To find the number of people in at least one set: P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)
To find the union of all set: (A + B + C + X + Y + Z + O)
Number of people in exactly one set: (A + B + C)
Number of people in exactly two of the sets: (X + Y + Z)
Number of people in exactly three of the sets: O
Number of people in two or more sets: (X + Y + Z + O)
If you’re like me, and formulas like these sometimes seem complicated and intimidating, let’s look at how making a Venn diagram and applying it to a tough GRE question can provide a little relief!
In 1997, N people graduated from college. If 1/3 of them received a degree in the applied sciences, and, of those, 1/4 graduated from a school in one of six northeastern states, which of the following expressions represents the number of people who graduated from college in 1997 who did not both receive a degree in the applied sciences and graduate from a school in one of six northeastern states?
(A) 11N/12
(B) 7N/12
(C) 5N/12
(D) 6N/7
(E) N/7
The key to understanding this question lies in the last sentence:
…who did not both receive a degree in the applied sciences and graduate from a school in one of six northeastern states?
We have two categories to sum: the people who ONLY received a science degree but NOT from one of the 6 schools, and the people who ONLY went to the 6 schools but did NOT receive a science degree. I made up variables for these categories (x and y).
If N = 12, there are 4 applied science students, 1 of which is both. That means x = 3. If 4 students are applied science, then 12-4 = 8 are from one of the six states but NOT applied science. y = 8.
3 + 8 = 11
So we are looking for an answer choice that gives us 11 when N = 12; the answer is A.
You aren’t likely to see many questions at this difficulty level on the actual GRE, but if you continue to challenge yourself, the medium GRE sets questions will soon look easy!
|
# Trigonometry : Graphing Tangent and Cotangent
## Example Questions
### Example Question #1 : Graphing Tangent And Cotangent
Which of the following best describes where the asymptotes are located on a tangent graph?
Angle measures where the cosine is 0, such as
Angle measures where the sine and cosine are equal, such as .
Angle measures where the sine is 0, such as .
Angle measures where the tangent cannot be calculated.
Angle measures where the cosine is 0, such as
Explanation:
In trigonometry,
.
It may also be thought of as .
This is because
and , so
This means that whenever cosine is 0, tangent is undefined, because it would be evaluated by dividing by 0.
### Example Question #2 : Graphing Tangent And Cotangent
Which of the following is not a solution to the following equation?
...
Explanation:
We can factor the original expression as follows:
So from this equation we conclude either that:
or
So any number that is not some integer multiple of away from these two solutions is not a solution to the original equation.
The only such choice is , which is ; n is not an integer, therefore it is not a solution.
### Example Question #3 : Graphing Tangent And Cotangent
The following is a graph of which function?
Explanation:
The graph looks to have infinite range, but multiple vertical asymptotes. That means we can limit our choices to tangent and cotangent graphs.
Furthermore, we observe that the graph starts at the bottom and increases from left to right, consistent with tangent graphs. So we narrow our focus to the choices involving tangents.
To decide between the remaining two graphs, observe that y-intercept (where x=0) of our graph is (0,1).
Now evaluated at is , which means that we need a vertical shift of unit.
Hence the best choice is:
### Example Question #4 : Graphing Tangent And Cotangent
Which of the following is the graph of ?
Explanation:
To derive the graph of , recall that . The graph of is
and the graph of is
Vertical asymptotes will occur in the graph of whenever . This is because the denominator of the tangent function will be equal to zero whenever the cosine function is equal to zero and then the entire function will be undefined at those points. Wherever cosine crosses the x-axis a vertical asymptote will occur. If we overlay the sine and cosine graphs we see the following:
So our tangent graph will follow the same form as the sine and cosine graphs when they are increasing, but will have vertical asymptotes wherever cosine crosses the x-axis.
And we are left with our graph of
### Example Question #5 : Graphing Tangent And Cotangent
Which of the following is the graph of ?
Explanation:
We will begin by considering the general graph of and apply transformations step by step to produce a graph of . The graph of is
The general equal of a tangent transformation equation is . A is the amplitude of the graph of tangent. Here, so we do not need to apply a transformation here. Next, we will consider the period. The period of the tangent function is equal to . So the period of our graph would be
Period =
Period =
So the period is shortened from to .
Now, we will consider is the phase shift of our graph. So we will shift our graph units to the left. This does not change our graph visually due to the period now being . Lastly, we will consider is the vertical shift of our graph, and so we must shift our graph 1 unit up.
And we are left with the graph of
### Example Question #6 : Graphing Tangent And Cotangent
Which of the following is the graph of ?
Explanation:
To derive the graph of recall that . So the tangent and cotangent graphs are reciprocals of one another. We will consider the tangent graph since it is one we are more familiar with:
Now we will simply invert the tangent graph to get the cotangent graph
And we are left with our cotangent graph
### Example Question #7 : Graphing Tangent And Cotangent
Which of the following is the graph of .
Explanation:
First, we will consider the graph of and apply transformations step-by-step. The graph of is
The general form of a cotangent transformation function is . For our function , and so we need to increase the amplitude 4 units.
here so we do not need to make any changes to the period of this graph. , giving us a negative phase shift of units.
This leaves us with our graph of .
### Example Question #1 : Graphing Tangent And Cotangent
True or False: The period of tangent and cotangent function is .
|
# Class 6 Maths Chapter 3 Exercise 3.1 Solution – Playing With Numbers NCERT
NCERT Class 6 Math Exercise 3.1 Solution of Chapter 3 Playing with Numbers with explanation. Here we provide Class 6 Maths all Chapters in Hindi for cbse, HBSE, Mp Board, UP Board and some other boards.
Also Read : – Class 6 Maths NCERT Solution
NCERT Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.1 Solution in english Medium.
## Class 6 Math Chapter 3 Exercise 3.1 Solution
Q.1. Write all the factors of the following numbers:
A. 24
Ans. 24 = 1×24
24 = 2×12
24 = 3×8
24 = 4×6
24 = 6×4
Stop here since 4 and 6 have occurred earlier.
Hence, the factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24
B. 15
Ans.
15 = 1×15
15 = 3×5
15 = 5×3
Stop here since 3 and 5 have occurred earlier.
Hence, the factors of 15 are 1, 3, 5, 15.
C. 21
Ans.
21 = 1×21
21 =3×7
21 =7×3
Stop here since 3 and 7 have occurred earlier.
Hence, the factors of 21 are 1, 3, 7, 21.
D. 27
Ans.
27 = 1×27
27 =3×9
27 =9×3
Stop here since 3 and 9 have occurred earlier.
Hence, the factors of 27 are 1, 3, 9, 27.
E. 12
Ans.
12 = 1×12
12 = 2×6
12 = 3×4
12 = 4×3
Stop here since 3 and 4 have occurred earlier.
Hence, the factors of 12 are 1, 2, 3, 4, 6, 12
F. 20
Ans.
20 = 1×20
20 = 2×10
20 = 4×5
20 = 5×4
Stop here since 4 and 5 have occurred earlier.
Hence, the factors of 20 are 1, 2, 4, 5, 10, 20.
G. 18
Ans.
18 = 1×18
18 = 2×9
18 = 3×6
18 = 6×3
Stop here since 3 and 6 have occurred earlier.
Hence, the factors of 18 are 1, 2, 3, 6, 9, 18.
H. 23
Ans.
23 = 1×23
23 = 23×1
Stop here since 1 and 23 have occurred earlier.
Hence, the factors of 23 are 1, 23.
I. 36
Ans.
36= 1×36
36= 2×18
36= 3×12
36= 4×9
36= 6×6
Stop here since 6 and 6 are same.
Hence, the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.
Q.2. Write first five multiples of:
A. 5
Ans. The required multiples are :
5×1 = 5
5×2 = 10
5×3 = 15
5×4 = 20
5×5 = 25
Hence, the first five multiples of 5 are 5, 10, 15, 20, 25.
B. 8
Ans. The required multiples are :
8×1 = 8
8×2 = 16
8×3 = 24
8×4 = 32
8×5 = 40
Hence, the first five multiples of 8 are 8, 16, 24, 32, 40.
C. 9
Ans. The required multiples are :
9×1 = 9
9×2 = 18
9×3 = 27
9×4 = 36
9×5 = 45
Hence, the first five multiples of 9 are 9, 18, 27, 36, 45 .
Q.3. Match the items in column 1 with the items in column 2.
Column 1 1. 35 2. 15 3. 16 4. 20 5. 25 Column 2 A. Multiple of 8 B. Multiple of 7 C. Multiple of 70 D. Factor of 30 E. Factor of 50 F. factor of 20
Ans. 1 – B
2 – D
3 – A
4 – F
5 – E
Q.4. Find all the multiples of 9 upto 100.
Ans. All the multiples of 9 upto 100 are as :
9×1 = 9
9×2 = 18
9×3 = 27
9×4 = 36
9×5 = 45
9×6 = 54
9×7 = 63
9×8 = 72
9×9 = 81
9×10 = 90
9×11 = 99
Hence, all the multiples of 9 upto 100 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.
error:
|
# Division with Remainder
When we divided whole numbers, we gave such examples that result is whole number.
But in general it is not true.
What if we try to divide 11 by 4?
There are two 4s in 11 and something extra: 11=4+4+3 or 11=4*2+3.
This is called division with remainder. It occurs when result of division is not whole number.
In general if m is dividend, n is divisor, q is quotient and r is remainder then color(green)(m=n*q+r). Note, that n should be greater than r. All numbers should be whole numbers.
In the above example we could write 11=4*1+7. Although it is correct, but here 4<7, so remainder is found incorrectly.
When remainder equals 0 then result of division is whole number.
For example, since 12/3=4 we can write that 12=4*3+0 or simply 12=4*3.
In general, remainder can be found using the same technique, which we saw when divided whole numbers.
Let's go through a couple of examples.
Example 1. Find remainder after division of 87 by 2.
Write in special form:
$$\begin{array}{r} 2\hspace{1pt})\overline{\hspace{1pt}87}\\ \end{array}$$$First let's divide 8 by 2. How many 2s are in 8? 2+2+2+2=8. There are four 2s, so 8=2*4. Write down 4. Multiply 2 by 4. Result is 8. Write it down. $$\begin{array}{r}\color{green}{4}\phantom{7}\\\color{green}{2}\hspace{1pt})\overline{\hspace{1pt}87}\\-\underline{\color{green}{8}}\phantom{7}\\ \end{array}$$$
Now subtract 8 from 8. Result is 0.
$$\begin{array}{r}4\phantom{7}\\2\hspace{1pt})\overline{\hspace{1pt}\color{purple}{8}7}\\-\underline{\color{purple}{8}}\phantom{7}\\\color{purple}{0}\phantom{7}\end{array}$$$Drag 7 down. $$\begin{array}{r}4\phantom{7}\\2\hspace{1pt})\overline{\hspace{1pt}8\color{red}{7}}\\-\underline{8}\phantom{7}\\0\color{red}{7}\end{array}$$$
So, what have we done?
We've first done division, then multiplication, then subtraction.
Let's proceed in the same way.
Divide 7 by 2. There are three 2s and additional one: 7=2*3+1.
Now, multiply 2 and 3. Result is 6.
$$\begin{array}{r}4\color{blue}{3}\\\color{blue}{2}\hspace{1pt})\overline{\hspace{1pt}87}\\-\underline{8}\phantom{7}\\07\\-\underline{\phantom{0}\color{blue}{6}}\end{array}$$$Now, subtract 7 from 6. Result is 1. $$\begin{array}{r}43\\ 2\hspace{1pt})\overline{\hspace{1pt}87}\\-\underline{8}\phantom{7}\\0\color{purple}{7}\\-\underline{\phantom{0}\color{purple}{6}}\\\color{cyan}{1}\end{array}$$$
We are done, because 1<2, i.e. remainder is less than divisor.
So, remainder is 1:87=2*43+1 .
Let's do another example.
Example 2. Find remainder after division of 74 by 3.
Write in special form:
$$\begin{array}{r}3\hspace{1pt})\overline{\hspace{1pt}74}\\ \end{array}$$$How many 3s are in 7? 7=3+3+1. There are two 3s and something extra. Write down 2. Now, multiply 2 by 3. Result is 6. Write it down. $$\begin{array}{r}\color{green}{2}\phantom{4}\\ \color{green}{3}\hspace{1pt})\overline{\hspace{1pt}74}\\-\underline{\color{green}{6}}\phantom{4}\\ \end{array}$$$
Now subtract 6 from 7. Result is 1.
$$\begin{array}{r}2\phantom{4}\\3\hspace{1pt})\overline{\hspace{1pt}\color{purple}{7}4}\\-\underline{\color{purple}{6}}\phantom{4}\\\color{purple}{1}\phantom{4}\end{array}$$$Drag 4 down. $$\begin{array}{r}2\phantom{4}\\ 3\hspace{1pt})\overline{\hspace{1pt}7\color{red}{4}}\\-\underline{6}\phantom{4}\\1\color{red}{4}\end{array}$$$
Next, continue with 14.
How many 3s are in 14? Four 3s and something extra: 14=3+3+3+3+2. There are four 3s.
Now, multiply 3 and 4. Result is 12.
$$\begin{array}{r}2\color{blue}{4}\\\color{blue}{3}\hspace{1pt})\overline{\hspace{1pt}74}\\-\underline{6}\phantom{4}\\14\\-\underline{\color{blue}{12}}\end{array}$$$Subtract 14 from 12. Result is 2. $$\begin{array}{r}24\\3\hspace{1pt})\overline{\hspace{1pt}74}\\-\underline{6}\phantom{4}\\\color{purple}{14}\\-\underline{\color{purple}{12}}\\\color{cyan}{2}\end{array}$$$
We are done, because we number that is less than 3 (divisor).
So, remainder is 2,72=3*24+2 .
Finally, let's work through a slightly harder example.
Example 3. Find remainder after division of 2194 by 12.
$$\begin{array}{r} 12\hspace{1pt})\overline{\hspace{1pt}2194}\\ \end{array}$$$How many 12s are in 2? Zero! 12 is greater than 2. So, we just add next digit (take 21 instead of 2): how many 12s are in 21? 12+9=21. There is one. Write down 1. Multiply 12 by 1. Result is 12. Write it down. $$\begin{array}{r}\color{green}{1}\phantom{94}\\\color{green}{12}\hspace{1pt})\overline{\hspace{1pt}2194}\\-\underline{\color{green}{12}}\phantom{94}\\ \end{array}$$$
Subtract 12 from 21. Result is 9.
$$\begin{array}{r}1\phantom{94}\\ 12\hspace{1pt})\overline{\hspace{1pt}\color{purple}{21}94}\\-\underline{\color{purple}{12}}\phantom{94}\\\color{purple}{9}\phantom{94}\end{array}$$$Drag 9 down. $$\begin{array}{r}1\phantom{94}\\ 12\hspace{1pt})\overline{\hspace{1pt}21\color{red}{9}4}\\-\underline{12}\phantom{94}\\9\color{red}{9}\phantom{4}\end{array}$$$
Continue with 99.
How many 12s are there in 99? 99=12+12+12+12+12+12+12+12+3. There are eight 12s.
Multiply 12 and 8. Result is 96.
$$\begin{array}{r}1\color{blue}{8}\phantom{4}\\ \color{blue}{12}\hspace{1pt})\overline{\hspace{1pt}2194}\\-\underline{12}\phantom{94}\\99\phantom{4}\\-\underline{\color{blue}{96}}\phantom{4}\end{array}$$$Subtract 99 from 96. Result is 3. $$\begin{array}{r}18\phantom{4}\\12\hspace{1pt})\overline{\hspace{1pt}2194}\\-\underline{12}\phantom{94}\\\color{green}{99}\phantom{4}\\-\underline{\color{green}{96}}\phantom{4}\\\color{green}{3}\phantom{4}\end{array}$$$
Drag down 4.
$$\begin{array}{r}18\phantom{4}\\ 12\hspace{1pt})\overline{\hspace{1pt}219\color{red}{4}}\\-\underline{12}\phantom{94}\\99\phantom{4}\\-\underline{96}\phantom{4}\\3\color{red}{4}\end{array}$$$Finally, determine how many 12s in 34? 24=12+12+10. There are two 12s. Multiply 12 by 2. Result is 24. Subtract 34 from 24. Result is 10. $$\begin{array}{r}18\color{purple}{2}\\\color{purple}{12}\hspace{1pt})\overline{\hspace{1pt}2194}\\-\underline{12}\phantom{94}\\99\phantom{4}\\-\underline{96}\phantom{4}\\34\\-\underline{\color{purple}{24}}\\\color{cyan}{10}\end{array}$$$
We are done, because 10<12.
So, remainder is 10, 2184=12*182+10 .
Now, it is your turn. Take pen and paper and solve following problems:
Exercise 1. Find remainder after division of 97 by 3.
Answer: 1 (97=3*32+1).
Exercise 2. Find remainder after division of 67 by 4.
Answer: 3 (67=4*16+3).
Exercise 3. Find remainder after division of 125 by 5.
Answer: 0 (125=5*25).
Exercise 4. Find remainder after division of 1653 by 24.
Answer: 21 (1653=24*68+21).
Exercise 5. Find remainder after division of 59540 by 447.
Answer: 446 (59450=447*132+446).
|
Tambahkan ke favorit Tautkan di sini
Beranda
MatematikaRia.com
# Triangles
## Equilateral, Isosceles and Scalene
There are three special names given to triangles that tell how many sides (or angles) are equal.
There can be 3, 2 or no equal sides/angles:
### Equilateral Triangle
Three equal sides
Three equal angles, always 60°
Two equal sides
Two equal angles
No equal sides
No equal angles
## What Type of Angle?
Triangles can also have names that tell you what type of angle is inside:
### Acute Triangle
All angles are less than 90°
### Right Triangle
Has a right angle (90°)
### Obtuse Triangle
Has an angle more than 90°
## Combining the Names
Sometimes a triangle will have two names, for example:
### Right Isosceles Triangle
Has a right angle (90°), and also two equal angles
Can you guess what the equal angles are?
## Area
The area is half of the base times height. "b" is the distance along the base "h" is the height (measured at right angles to the base) Area = ½bh
The formula works for all triangles.
Another way of writing the formula is bh/2
### Example: What is the area of this triangle?
Height = h = 12
Base = b = 20
Area = bh/2 = 20 × 12 / 2 = 120
Just make sure that the "h" is measured at right angles to the "b".
## Why is the Area "Half of bh"?
Imagine you "doubled" the triangle (flip it around one of the upper edges) to make a square-like shape (it would be a "parallelogram" actually), THEN the whole area would be bh (that would be for both triangles, so just one is ½bh), like this:
You can also see that if you sliced the new triangle and placed the sliced part on the other side you get a simple rectangle, whose area is bh.
|
Categories :
## How do you find the area of a triangle with 3 sides?
What is the Area of Triangle with 3 Sides Equal? If a triangle has 3 equal sides, it is called an equilateral triangle. The area of an equilateral triangle can be calculated using the formula, Area = a2(√3/4), where ‘a’ is the side of the triangle.
## Can you solve a triangle with 3 sides?
“SSS” is when we know three sides of the triangle, and want to find the missing angles. To solve an SSS triangle: use The Law of Cosines first to calculate one of the angles. then use The Law of Cosines again to find another angle.
What is a triangle with 3 equal side lengths?
A triangle with all sides equal is called an equilateral triangle, and a triangle with no sides equal is called a scalene triangle. An equilateral triangle is therefore a special case of an isosceles triangle having not just two, but all three sides and angles equal.
### How do we find area of a triangle?
So, the area A of a triangle is given by the formula A=12bh where b is the base and h is the height of the triangle. Example: Find the area of the triangle. The area A of a triangle is given by the formula A=12bh where b is the base and h is the height of the triangle.
### How do you find the area of an irregular shape with 4 sides?
The area of any irregular quadrilateral can be calculated by dividing it into triangles. Example: Find the area of a quadrilateral ABCD whose sides are 9 m, 40 m, 28 m and 15 m respectively and the angle between the first two sides is a right angle. The area of the quadrilateral ABCD =(180+126)=306 square meters.
How do you calculate an uneven area?
How to use irregular area calculator?
1. Step 1: Measure all sides of the area in one unit (Feet, Meter, Inches or any other).
2. Step 2: Enter length of horizontal sides into Length 1 and Length 2. And Width of the vertical sides into Width 1 and Width 2.
3. Step 3: Press calculate button.
4. Our Formula: Area = b × h.
#### How do you find the side of a triangle given two sides?
Boundless Algebra
1. The Pythagorean Theorem, a2+b2=c2, a 2 + b 2 = c 2 , can be used to find the length of any side of a right triangle.
2. The side opposite the right angle is called the hypotenuse (side c in the figure).
#### What is the shortest side of a 30 60 90 triangle?
And so on. The side opposite the 30° angle is always the smallest, because 30 degrees is the smallest angle. The side opposite the 60° angle will be the middle length, because 60 degrees is the mid-sized degree angle in this triangle.
How do you classify a triangle by side lengths?
Classifying Triangles by Side Lengths
1. An equilateral triangle has side lengths that are the same. Here is an example.
2. A scalene triangle is a triangle where the lengths of all three sides are different.
3. An isosceles triangle has two side lengths that are the same and one side length that is different.
## How do you classify a triangle using side lengths?
Equilateral triangle: A triangle with three sides of equal length. Isosceles triangle: A triangle with at least two sides of equal length. Line of symmetry: A line through a figure that creates two halves that match exactly. Obtuse angle: An angle with a measure greater than 90 degrees but less than 180 degrees.
## What are the formulas for triangles?
The area formulas for all the different types of triangles like the equilateral triangle, right-angled triangle, and isosceles triangle are given below….Area of an Isosceles Triangle:
Given Dimensions Area of a Triangle Formula
When base and height is given. Area of a right triangle = 1/2 × Base × Height
|
10 Pre-Calculus Missteps to Avoid - dummies
# 10 Pre-Calculus Missteps to Avoid
Here are ten pitfalls that normally trip up the pre-calculus student. Make sure that you understand the order of operations, how to properly multiply binomials and monomials with exponents, how to break up and flip fractions, how not to combine terms, and how to keep from disregarding the negative sign.
## Going out of order (of operations)
Operations in an expression or an equation aren’t all meant to be done from left to right. For example, 3 – 7(x – 2) doesn’t equal –4(x – 2) or –4x + 8. You’re supposed to do multiplication first, which means distributing the –7 first 3 – 7(x – 2) = 3 – 7x + 14. And then you combine like terms to get –7x + 17.
Remember your order of operations (PEMDAS) all the time, every time:
Parentheses (and other grouping devices)
Exponents (and roots)
Multiplication and Division, from left to right in order as you find them
Addition and Subtraction, also from left to right
## Multiplying binomials incorrectly
When multiplying binomials, always remember to multiply them in the correct order. You remember FOIL — First, Outside, Inside, Last. This includes when squaring any binomial. The biggest mistake made in these situations is something like when squaring a binomial: (x – 4)2 and getting x2 + 16. That’s forgetting a whole lot of multiplying, and it’s not correct. It should look like this:
(x – 4)2 = x2 – 4x – 4x + 16 = x2 – 8x + 16.
You may use other orders, but it pays to be careful and consistent.
Don’t fall for a big trap and break a fraction up incorrectly.
doesn’t equal .
If you’re skeptical, just pick a value for x and plug it into both expressions and see whether you get the same answer twice. You won’t, because it doesn’t work. On the other hand,
.
See the difference?
When reducing fractions, each term in the fraction has to be divided. The division bar is a grouping symbol, and you have to simplify the numerator and denominator separately before doing the division. For example,
.
Instead, you factor the denominator, first, and then reduce:
## Combining terms that can’t be combined
Yet another frequent mistake occurs when terms that aren’t meant to be combined are combined. 4x – 1 suddenly becomes 3x, which it’s not. 4x – 1is simplified, meaning that it’s an expression that doesn’t contain any like terms. 3a4b5 + 2a5b4 is also simplified. (It can be factored, but it’s still simplified.) Those exponents are close, but close only counts in horseshoes and hand grenades. When counting in the real world, you can’t combine apples and bananas. Four apples plus three bananas is still four apples and three bananas. It’s the same in algebra: 4a + 3b is simplified.
## Forgetting to flip the fraction
When dealing with complex fractions, remember that you’re dealing with division of fractions.
doesn’t become .
Remember that a division bar is division. To divide a fraction, you must multiply by the reciprocal of the denominator, so
.
## Losing the negative (sign)
It’s true that in life you’re not supposed to be negative, but in math, don’t disregard a negative sign — especially when subtracting polynomials or raising to powers.
(4x3 – 6x + 3) – (3x3 – 2x + 4) isn’t the same thing as 4x3 – 6x + 3 – 3x3 – 2x + 4. If you do it that way, you’re not subtracting the whole second polynomial, only its first term. The right way to do it is 4x3 – 6x + 3 – 3x3 – (–2x) – 4 = 4x3 – 6x + 3 – 3x3 + 2x – 4, which simplifies to x3 – 4x –1. The issue here is a special case of the failure to correctly apply the distributive law; it frequently occurs with other coefficients as well (not just –1). Failure to write the parentheses often directly contributes to these errors.
Similarly, when subtracting rational functions, take care of that negative sign.
What happened? Someone forgot to subtract the whole second polynomial on the top. Instead, this is the way to do it:
And one other caution about negative signs is when raising negative expressions to powers. There’s a difference between –26 and (–2)6. In the expression –26, the order of operations says to raise the 2 to the sixth power first, and then find the opposite: –26 = –64. The expression (–2)6 says to raise the number –2 to the sixth power: (–2)6 = 64.
## Oversimplifying roots
When it comes to roots, there are all kinds of errors that can occur. For instance, suddenly becomes 5 when that’s not even close!
You know that .
Don’t add or subtract roots that aren’t like terms, either.
isn’t , now or ever. They’re not like terms, so you can’t add them.
You can rewrite it as and then subtract the like terms getting .
## Executing exponent errors
When multiplying monomials, you don’t multiply the exponents.
x4 × x3 = x7, not x12. Don’t mistake the multiplication of exponential expressions with powers of exponential expressions.
Raising .
Also, when there’s more than one term being raised to the power, you have to multiply the whole expression the number of times shown with the exponent. Raising (x2 – 2x +3)3 does not mean
.
## Ignoring extraneous
Sometimes, when solving radical or rational equations, you have to perform an operation (such as multiplying all the terms by a variable) which can introduce an extraneous or false answer.
For example, when told to solve
you dutifully subtract x from each side and then square both sides of the equation:
or 4x + 1 = 1 – 2x + x2.
Simplifying, you get 0 = x2 – 6x = x(x – 6). This new equation has two solutions: x = 0 and x = 6. When substituting back into the original equation, the x = 0 works just fine:
becomes .
But that’s not the case with x = 6:
The solution x = 6 is extraneous.
## Misinterpreting trig notation
The trig functions are great to work with, and they come with some specialty notation that makes writing about them quicker and easier. You just need to be careful and interpret the notation correctly.
The two cases in point have to do with squaring the function or finding the inverse of the function.
When you see sin2x, this means to find the sine and then square the result.
|
How do you write the equation for a circle where the points passes through the points (1,1), (-2, 2), and (-5,1)?
Nov 13, 2016
${\left(x - - 2\right)}^{2} + {\left(y - - 3\right)}^{2} = {5}^{2}$
Explanation:
Use the standard form for the equation of a circle:
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
And the points, $\left(1 , 1\right) , \left(- 2 , 2\right) , \mathmr{and} \left(- 5 , 1\right)$
To write 3 equations:
${\left(1 - h\right)}^{2} + {\left(1 - k\right)}^{2} = {r}^{2} \text{ [1]}$
${\left(- 2 - h\right)}^{2} + {\left(2 - k\right)}^{2} = {r}^{2} \text{ [2]}$
${\left(- 5 - h\right)}^{2} + {\left(1 - k\right)}^{2} = {r}^{2} \text{ [3]}$
Temporarily eliminate the variable r by setting the left side of equation [1] equal to the left side of equation [2] and the same for equation [1] and equation [3]:
${\left(1 - h\right)}^{2} + {\left(1 - k\right)}^{2} = {\left(- 2 - h\right)}^{2} + {\left(2 - k\right)}^{2}$
${\left(1 - h\right)}^{2} + {\left(1 - k\right)}^{2} = {\left(- 5 - h\right)}^{2} + {\left(1 - k\right)}^{2} \text{ [4]}$
Please notice that equation [4] has ${\left(1 - k\right)}^{2}$ on both sides so we can subtract this form both sides:
${\left(1 - h\right)}^{2} + {\left(1 - k\right)}^{2} = {\left(- 2 - h\right)}^{2} + {\left(2 - k\right)}^{2}$
${\left(1 - h\right)}^{2} = {\left(- 5 - h\right)}^{2}$
Use the pattern ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$ to expand all of the squares:
$1 - 2 h + {h}^{2} + 1 - 2 k + {k}^{2} = 4 + 4 h + {h}^{2} + 4 - 4 k + {k}^{2}$
$1 - 2 h + {h}^{2} = 25 + 10 h + {h}^{2}$
The ${h}^{2} \mathmr{and} {k}^{2}$ terms cancel:
$1 - 2 h + 1 - 2 k = 4 + 4 h + 4 - 4 k \text{ [5]}$
$1 - 2 h = 25 + 10 h \text{ [6]}$
Use equation 6 to solve for h:
$- 12 h = 24$
$h = - 2$
Combine like tems of equation [5] with k terms on the left and constant and h terms on the right:
$2 k = 6 h + 6$
$k = 3 h + 3$
Substitute -2 for h:
$k = 3 \left(- 2\right) + 3$
$k = - 3$
Substitute the values for h and k into one of the first three equations (I will use equation [2]):
${\left(- 2 - - 2\right)}^{2} + {\left(2 - - 3\right)}^{2} = {r}^{2} \text{ [2]}$
${\left(0\right)}^{2} + {\left(5\right)}^{2} = {r}^{2}$
Solve for r:
$r = 5$
The equation of the circle is:
${\left(x - - 2\right)}^{2} + {\left(y - - 3\right)}^{2} = {5}^{2}$
|
Search
# Bearing and Distances
Welcome to Class !!
We are eager to have you join us !!
In today’s Mathematics class, We will be discussing Bearing and Distances. We hope you enjoy the class!
CONTENT:
i. Compass bearing
ii Three figure bearing
iii. Finding the bearing of a point from another
### COMPASS BEARING
A bearing gives the direction between two points in terms of an angle in degrees. The two types of bearing are compass bearing and three-figure bearings.
The four major compass directions are North (N) South (S) East (E) and West (W)
Apart from the four main points or directions, there are also four main secondary directions i.e. NE (northeast), SE (south-east), SW (south-west), NW (north-west). The angle between each point is 45o
Worked examples
Draw a sketch to show each of these bearings marketing the angles clearly.
a) N35oW (B) N70oE (C.) S58oW
Solution
1. N35oW means from N measure 35o toward the W or 35oW of N
Solution
In a), the direction start from a wrong point (W) instead of N, therefore,
90 – 18 = 72o
i.e. N72oW
In b), the direction starts from a wrong point (E) instead of S therefore:
90 – 55 = 35o i.e. S35oE
Evaluation: Class Work
Find the compass direction of point A from point O in these diagrams.
Reading Assignment
NGM BK CHAPTER 23, page 185 – 187
Essential Mathematics for JSS BK 2, CHAPTER 24, pg 246-247
##### THREE-FIGURE BEARINGS
Three-figure bearings are given as the number of degrees from north, measured in a clockwise direction. Any direction can be given as a three-figure bearing. Three-digit are always given but angles less than 100o need extra zero to be written in front of the digits e.g. 008o, 060o, 070o up to 099o
Worked Example
Find the three-figure bearings of A, B, C, and D from X
Solution
1. The arrow N shows the direction N, NXA = 63o. the bearing of A from X is 063o
2. NXB = 180 – 35 = 145o. The bearing of B from X is 145o
3. NXC clockwise = 180 + 75 = 255o. The bearing of C from X is 255o
4. NXD clockwise = 360 – 52 = 308o. The bearing of D from X is 308o.
Evaluation:
In the figure below, find the bearings of A, B, C and D from X.
Reference
NGM Bk. 2 Chapter 23, page 180 – 19
Evaluation
In each diagram below, calculate
i) the bearing of B from A
ii) the bearing of A from B
GENERAL EVALUATION
From a point P, the bearing of a house is 060o. From a point Q 100m due east of P, the bearing is 330o.
Draw a labelled sketch to show the positions of P, Q and the house.
REVISION QUESTION
1. A girl is facing East. If she turns clockwise through 2 right angles, then the direction she would be facing is ……………………..
2. A student is facing South-East. If he turns anticlockwise through 1800, then the direction he would be facing is …………………..
WEEKEND ASSIGNMENT
1. The bearing of X from Y is 196o. The bearing of Y from X is 016o B. 074o C. 106o D. 196o
2. A boat sails on a bearing of 225o. Using compass bearing, in what direction is the boat sailing? South East B. North-East C. South West D. North West
3. The bearing of point A from B is 058o. Find the bearing of point B from point 058o B. 122o C. 3020 D. 238o
4. Which of the following statements is not true when we specify a direction with bearing? Measure the angle from North B. Measure anticlockwise C. Measure clockwise D. Always use three digits
5. In the diagram below, which of the following angles is the bearing of P from Q? 065o B. 2450 C. 295o D. 115o
We have come to the end of this class. We do hope you enjoyed the class?
Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.
In our next class, we will be talking about Statistics: Data Presentation. We are very much eager to meet you there.
Get more class notes, videos, homework help, exam practice etc on our app [CLICK HERE]
Upgrade your teaching with ready-made & downloadable class notes on our app [CLICK HERE]
### 7 thoughts on “Bearing and Distances”
1. Shomoye Mariam
The bearing of sokoto from Kaduna is 050 what Is the bearing of Kaduna from sokoto
1. Deborah Temitope Babalola
If the bearing is less than 180 degrees, add 180. And if it is more than 180, subtract 180 degrees in other to get the back or reciprocal bearing
To your question 050+180 =230. Hence the bearing of Kaduna from sokoto is 230 degrees
Don`t copy text!
|
# Derivative of e^3x by first principle and chain rule
The derivative of e3x is 3e3x. The function e^3x is an exponential function with an exponent 3x. In this note, we will find the derivative of e to the power 3x by the first principle of derivatives and by the chain rule of derivatives.
## Derivative of e^3x using first principle
As we know that the derivative of a function $f(x)$ by first principle is the below limit
$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h},$
so taking $f(x)=e^{3x}$ in the above equation, the derivative of $e^{3x}$ from first principle is
$\dfrac{d}{dx}(e^{3x})$ $=\lim\limits_{h \to 0} \dfrac{e^{3(x+h)}-e^{3x}}{h}$
$=\lim\limits_{h \to 0} \dfrac{e^{3x+3h}-e^{3x}}{h}$
$=\lim\limits_{h \to 0} \dfrac{e^{3x} \cdot e^{3h}-e^{3x}}{h}$
$=\lim\limits_{h \to 0} \dfrac{e^{3x} (e^{3h}-1)}{h}$
$=e^{3x} \lim\limits_{h \to 0} \dfrac{e^{3h}-1}{3h}$ $\times 3$
Let $t=3h$. Thus $t \to 0$ when $h \to 0.$
So from above, we get $=e^{3x} \lim\limits_{t \to 0} \dfrac{e^{t}-1}{t}$ $\times 3$
$=e^{3x} \times 1 \times 3$ as the limit of $(e^t-1)/t$ is one when t tends to zero.
$=3e^{3x}$
Thus the derivative of e3x is 3e3x and this is obtained by the first principle of derivatives.
Now we will find the derivative of e to the power 3x by the chain rule of derivatives.
## Derivative of e^3x by Chain Rule
Let $z=3x$. Therefore, we have $dz/dx=3.$ By the chain rule of derivatives, we have
$\dfrac{d}{dx}(e^{3x})$ $=\dfrac{d}{dz}(e^{z}) \cdot \dfrac{dz}{dx}$
$=e^z \cdot 3$ as the derivative of ez with respect to z is ez, and $dz/dx=3.$
$=3e^z$
$=3e^{3x}$ as $z=3x$
So the derivative of e^3x is 3e^3x and this is obtained by the chain rule of derivatives.
## Question Answer on Derivative of e^3x
Question 1: Find the derivative of e^3.
Note that e3 is a constant number as the number e is a constant. We know that the derivative of a constant is zero (see the page on Derivative of a constant is 0). Thus we can say that the derivative of e cube is zero.
|
Contemporary Mathematics
# 6.4Compound Interest
Contemporary Mathematics6.4 Compound Interest
Figure 6.6 The impact of compound interest (credit: "English Money" by Images Money/Flickr, CC BY 2.0)
## Learning Objectives
After completing this section, you should be able to:
1. Compute compound interest.
2. Determine the difference in interest between simple and compound calculations.
3. Understand and compute future value.
4. Compute present value.
5. Compute and interpret effective annual yield.
For a very long time in certain parts of the world, interest was not charged due to religious dictates. Once this restriction was relaxed, loans that earned interest became possible. Initially, such loans had short terms, so only simple interest was applied to the loan. However, when loans began to stretch out for years, it was natural to add the interest at the end of each year, and add the interest to the principal of the loan. After another year, the interest was calculated on the initial principal plus the interest from year 1, or, the interest earned interest. Each year, more interest was added to the money owed, and that interest continued to earn interest.
Since the amount in the account grows each year, more money earns interest, increasing the account faster. This growth follows a geometric series (Geometric Sequences). It is this feature that gives compound interest its power. This module covers the mathematics of compound interest.
## Understand and Compute Compound Interest
As we saw in Simple Interest, an account that pays simple interest only pays based on the original principal and the term of the loan. Accounts offering compound interest pay interest at regular intervals. After each interval, the interest is added to the original principal. Later, interest is calculated on the original principal plus the interest that has been added previously.
After each period, the interest on the account is computed, then added to the account. Then, after the next period, when interest is computed, it is computed based on the original principal AND the interest that was added in the previous periods.
The following example illustrates how compounded interest works.
## Determine the Difference in Interest Between Simple and Compound Calculations
It is natural to ask, does compound interest make much of a difference? To find out, we revisit Abena’s CD.
## Example 6.40
### Comparing Simple to Compound Interest on a 3-Year CD
Abena invested $1,000 in a CD that earned 4% compounded annually, and the CD was worth$1,124.86 after 3 years. Had Abena invested in a CD with simple interest, how much would the CD have been worth after 3 years? How much more did Abena earn using compound interest?
1.
Oksana deposits $5,000 in a CD that earned 3% compounded annually and was worth$5,627.54 after 4 years. Had Oksana invested in a CD with simple interest, how much would the CD have been worth after 4 years? How much more did Oksana earn using compound interest?
## Understand and Compute Future Value
Imagine investing for 30 years and compounding the interest every month. Using the method above, there would be 360 periods to calculate interest for. This is not a reasonable approach. Fortunately, there is a formula for finding the future value of an investment that earns compound interest.
## FORMULA
The future value of an investment, $AA$, when the principal $PP$ is invested at an annual interest rate of $rr$ (in decimal form), compounded $nn$ times per year, for $tt$ years, is found using the formula $A=P(1+rn)ntA=P(1+rn)nt$. This is also referred to as the future value of the investment.
## Checkpoint
Note, sometimes the formula is presented with the total number of periods, $nn$, and the interest rate per period, $rr$. In that case the formula becomes $A=P(1+r)nA=P(1+r)n$.
## Example 6.41
### Computing Future Value for Compound Interest
In the following, compute the future value of the investment with the given conditions.
1. Principal is $5,000, annual interest rate is 3.8%, compounded monthly, for 5 years. 2. Principal is$18,500, annual interest rate is 6.25%, compounded quarterly, for 17 years.
In the following, compute the future value of the investment with the given conditions.
1.
Principal is $7,600, annual interest rate is 4.1%, compounded monthly, for 10 years. 2. Principal is$13,250, annual interest rate is 2.79%, compounded quarterly, for 25 years.
## FORMULA
The money invested in an account bearing an annual interest rate of $rr$ (in decimal form), compounded $nn$ times per year for $tt$ years, is called the present value, $PVPV$, of the account (or of the money) and found using the formula $PV=A(1+rn)n×tPV=A(1+rn)n×t$, where $AA$ is the value of the account at the investment’s end. Always round this value up to the nearest penny.
## Example 6.44
### Computing Present Value
Find the present value of the accounts under the following conditions.
1. $AA$ = $250,000, invested at 6.75 interest, compounded monthly, for 30 years. 2. $AA$ =$500,000, invested at 7.1% interest, compounded quarterly, for 40 years.
Find the present value of the accounts under the following conditions.
1.
$A$ = $1,000,000, invested at 5.75% interest, compounded monthly, for 40 years. 2. $A$ =$175,000, invested at 3.8% interest, compounded quarterly, for 20 years.
## Compute and Interpret Effective Annual Yield
As we’ve seen, quarterly compounding pays interest 4 times a year or every 3 months; monthly compounding pays 12 times a year; daily compounding pays interest every day, and so on. Effective annual yield allows direct comparisons between simple interest and compound interest by converting compound interest to its equivalent simple interest rate. We can even directly compare different compound interest situations. This gives information that can be used to identify the best investment from a yield perspective.
Using a formula, we can interpret compound interest as simple interest. The effective annual yield formula stems from the compound interest formula and is based on an investment of $1 for 1 year. Effective annual yield is $Y=(1+rn)n-1Y=(1+rn)n-1$ where $YY$ = effective annual yield, $rr$ = interest rate in decimal form, and $nn$ = number of times the interest is compounded in a year. $YY$ is interpreted as the equivalent annual simple interest rate. ## Example 6.46 ### Determine and Interpret Effective Annual Yield for 6% Compounded Quarterly Suppose you have an investment paying a rate of 6% compounded quarterly. Determine and interpret that effective annual yield of the investment. ## Your Turn 6.46 1. Calculate and interpret the effective annual yield for an investment that pays at a 7% interest compounded quarterly. ## Example 6.47 ### Determine and Interpret Effective Annual Yield for 5% Compounded Daily Calculate and interpret the effective annual yield on a deposit earning interest at a rate of 5% compounded daily. ## Your Turn 6.47 1. Calculate and interpret the effective annual yield on a deposit earning 2.5% compounded daily. ## Example 6.48 ### Choosing a Bank Minh has a choice of banks in which he will open a savings account. He will deposit$3,200 and he wants to get the best interest he can. The banks advertise as follows:
Bank Interest Rate
ABC Bank 2.08% compounded monthly
123 Bank 2.09% compounded annually
XYZ Bank 2.05% compounded daily
Which bank offers the best interest?
1.
Isabella decides to deposit $5,500 in a CD but needs to choose between banks that offer CDs. She identifies four banks and finds out the terms of their CDs. Her findings are in the table below. Bank Interest Rate Smith Bank 3.08% compounded quarterly Park Bank 3.11% compounded annually Town Bank 3.09% compounded daily Community Bank 3.10% compounded monthly Which bank has the best yield? ## Check Your Understanding 19. What is compound interest? 20. Which yields more money, simple interest or compound interest? 21. Find the future value after 15 years of$8,560.00 deposited in an account bearing 4.05% interest compounded monthly.
22.
$10,000 is deposited in an account bearing 5.6% interest for 5 years. Find the difference between the future value when the interest is simple interest and when the interest is compounded quarterly. 23. Find the present value of$75,000 after 28 years if money is invested in an account bearing 3.25% interest compounded monthly.
24.
What can be done to compare accounts if the rates and number of compound periods per year are different?
25.
Find the effective annual yield of an account with 4.89% interest compounded quarterly.
## Section 6.4 Exercises
1 .
What is the difference between simple interest and compound interest?
2 .
What is a direct way to compare accounts with different interest rates and number of compounding periods?
3 .
Which type of account grows in value faster, one with simple interest or one with compound interest?
How many periods are there if interest is compounded?
4 .
Daily
5 .
Weekly
6 .
Monthly
7 .
Quarterly
8 .
Semi-annually
In the following exercises, compute the future value of the investment with the given conditions.
9 .
Principal = $15,000, annual interest rate = 4.25%, compounded annually, for 5 years 10 . Principal =$27,500, annual interest rate = 3.75%, compounded annually, for 10 years
11 .
Principal = $13,800, annual interest rate = 2.55%, compounded quarterly, for 18 years 12 . Principal =$150,000, annual interest rate = 2.95%, compounded quarterly, for 30 years
13 .
Principal = $3,500, annual interest rate = 2.9%, compounded monthly, for 7 years 14 . Principal =$1,500, annual interest rate = 3.23%, compounded monthly, for 30 years
15 .
Principal = $16,000, annual interest rate = 3.64%, compounded daily, for 13 years 16 . Principal =$9,450, annual interest rate = 3.99%, compounded daily, for 25 years
In the following exercises, compute the present value of the accounts with the given conditions.
17 .
Future value = $250,000, annual interest rate = 3.45%, compounded annually, for 25 years 18 . Future value =$300,000, annual interest rate = 3.99%, compounded annually, for 15 years
19 .
Future value = $1,500,000, annual interest rate = 4.81%, compounded quarterly, for 35 years 20 . Future value =$750,000, annual interest rate = 3.95%, compounded quarterly, for 10 years
21 .
Future value = $600,000, annual interest rate = 3.79%, compounded monthly, for 17 years 22 . Future value =$800,000, annual interest rate = 4.23%, compounded monthly, for 35 years
23 .
Future value = $890,000, annual interest rate = 2.77%, compounded daily, for 25 years 24 . Future value =$345,000, annual interest rate = 2.99%, compounded daily, for 19 years
In the following exercises, compute the effective annual yield for accounts with the given interest rate and number of compounding periods. Round to three decimal places.
25 .
Annual interest rate = 2.75%, compounded monthly
26 .
Annual interest rate = 3.44%, compounded monthly
27 .
Annual interest rate = 5.18%, compounded quarterly
28 .
Annual interest rate = 2.56%, compounded quarterly
29 .
Annual interest rate = 4.11%, compounded daily
30 .
Annual interest rate = 6.5%, compounded daily
The following exercises explore what happens when a person deposits money in an account earning compound interest.
31 .
Find the present value of $500,000 in an account that earns 3.85% compounded quarterly for the indicated number of years. 1. 40 years 2. 35 years 3. 30 years 4. 25 years 5. 20 years 6. 15 years 32 . Find the present value of$1,000,000 in an account that earns 6.15% compounded monthly for the indicated number of years.
1. 40 years
2. 35 years
3. 30 years
4. 25 years
5. 20 years
6. 15 years
33 .
In the following exercises, the number of years can reflect delaying depositing money. 40 years would be depositing money at the start of a 40-year career. 35 years would be waiting 5 years before depositing the money. Thirty years would be waiting 10 years before depositing the money, and so on. What do you notice happens if you delay depositing money?
34 .
For each 5-year gap for exercise 32, compute the difference between the present values. Do these differences remain the same for each of the 5-year gaps, or do they differ? How do they differ? What conclusion can you draw?
35 .
Daria invests $2,500 in a CD that yields 3.5% compounded quarterly for 5 years. How much is the CD worth after those 5 years? 36 . Maurice deposits$4,200 in a CD that yields 3.8% compounded annually for 3 years. How much is the CD worth after those 3 years?
37 .
Georgita is shopping for an account to invest her money in. She wants the account to grow to $400,000 in 30 years. She finds an account that earns 4.75% compounded monthly. How much does she need to deposit to reach her goal? 38 . Zak wants to create a nest egg for himself. He wants the account to be valued at$600,000 in 25 years. He finds an account that earns 4.05% interest compounded quarterly. How much does Zak need to deposit in the account to reach his goal of $600,000? 39 . Eli wants to compare two accounts for their money. They find one account that earns 4.26% interest compounded monthly. They find another account that earns 4.31% interest compounded quarterly. Which account will grow to Eli’s goal the fastest? 40 . Heath is planning to retire in 40 years. He’d like his account to be worth$250,000 when he does retire. He wants to deposit money now. How much does he need to deposit in an account yielding 5.71% interest compounded semi-annually to reach his goal?
41 .
Jo and Kim want to set aside some money for a down payment on a new car. They have 6 years to let the money grow. If they want to make a $15,000 down payment on the car, how much should they deposit now in an account that earns 4.36% interest compounded monthly? 42 . A newspaper’s business section runs an article about savings at various banks in the city. They find six that offer accounts that offer compound interest. Bank A offers 3.76% compounded daily. Bank B offers 3.85% compounded annually. Bank C offers 3.77% compounded weekly. Bank D offers 3.74% compounded daily. Bank E offers 3.81% compounded semi-annually. To earn the most interest on a deposit, which bank should a person choose? 43 . Paola reads the newspaper article from exercise 32. She really wants to know how different they are in terms of dollars, not effective annual yield. She decides to compute the future value for accounts at each bank based on a principal of$100,000 that are allowed to grow for 20 years. What is the difference in the future values of the account with the highest effective annual yield, and the account with the second highest effective annual yield?
44 .
Paola reads the newspaper article from exercise 32. She really wants to know how different they are in terms of dollars, not effective annual yield. She decides to compute the future value for accounts at each bank based on a principal of $100,000 that are allowed to grow for 20 years. What is the difference in the future values of the account with the highest effective annual yield, and the account with the lowest effective annual yield? 45 . Jesse and Lila need to decide if they want to deposit money this year. If they do, they can deposit$17,400 and allow the money to grow for 35 years. However, they could wait 12 years before making the deposit. At that time, they’d be able to collect $31,700 but the money would only grow for 23 years. Their account earns 4.63% interest compounded monthly. Which plan will result in the most money, depositing$17,400 now or depositing $31,700 in 12 years? 46 . Veronica and Jose are debating if they should deposit$15,000 now in an account or if they should wait 10 years and deposit $25,000. If they deposit money now, the money will grow for 35 years. If they wait 10 years, it will grow for 25 years. Their account earns 5.25% interest compounded weekly. Which plan will result in the most money, depositing$15,000 now or depositing \$25,000 in 10 years?
Order a print copy
As an Amazon Associate we earn from qualifying purchases.
|
The basic concept of finding the missing number in addition sum of one-digit number. These give us support and idea about the basic addition facts, as well as help us to understand the addition's relationship to subtraction. It also assists as an introduction to learning algebra (as the missing number could be represented by a variable or a letter). While solving the addition problems children will complete the missing addend to find the correct calculation. Now we will learn how to find missing number in addition.
What is the missing number in these addition sums?
(i) 3 + ? = 8
Follow the procedure to find the missing addend. The question given is ‘3 and how many more make 8?’ The answer is 5.
Therefore, when the number is put in the place of the question mark (?), the sum is written as 3 + 5 = 8.
(ii) ? + 2 = 6
To find the missing addend we read the question given as ‘how many more and 2 make 6?’ The answer is 4.
Therefore, when the number is put in the place of the question mark (?), the sum is written as 4 + 2 = 6.
(iii) 5 + ? = 9
To find the missing addend we read the question given as ‘5 and how many more make 9?’ The answer is 4.
Therefore, when the number is put in the place of the question mark (?), the sum is written as 5 + 4 = 9.
(iv) ? + 6 = 7
To find the missing addend we read the question given as ‘how many more and 6 make 7?’ The answer is 1.
Therefore, when the number is put in the place of the question mark (?), the sum is written as 1 + 6 = 7.
`
|
# NCERT Class 8 Mathematics Solutions: Chapter 6 – Squares and Square Roots Exercise 6.4 Part 3
Get top class preparation for IMO-Level-2 right from your home: fully solved questions with step-by-step explanation- practice your way to success.
Image of square root formula to find square root digit
Question 2: Find the number of digits in the square root of each of the following numbers (without any calculation):
(i) 64
(ii) 144
(iii) 4489
(v) 390625
(iv) 27225
(i) 64
Here, 64 contain two digits which are even.
Therefore, number of digits in square root
Or
Let us place bars and pair (two digits) the digits from right to left, we get:
There is only one bar.
So, the square root of 64 has only one digit.
(ii) 144
Here, 144 contain three digits which are odd.
Therefore, number of digit in square root
Or
Let us place bars and pair (two digits) the digits from right to left, we get:
There are two bars.
So, the square root of 144 has two digits.
(iii) 4489
Here, 4489 contains four digits which are even.
Therefore, number of digits in square root
Or
Let us place bars and pair (two digits) the digits from right to left, we get:
There are two bars.
So, the square root of 4489 has two digits.
(iv) 390625
Here, 390625 contain six digits which are even.
Therefore, number of digits in square root
Or
Let us place bars and pair (two digits) the digits from right to left, we get: 390625 =
There are three bars.
So, the square root of 390625 has three digits.
(v) 27225
Here, 27225 contain five digits which are odd.
Therefore, number of digits in square root
Or
Let us place bars and pair (two digits) the digits from right to left, we get:
There are three bars.
So, the square root of 144 has three digits.
|
Question
# Differentiate $secx$ by the first principle.
Open in App
Solution
## Let us consider$secx=\frac{1}{\mathrm{cos}x}$By quotient rule derivative of $secx=\frac{1}{\mathrm{cos}x}$[The quotient rule of differentiation is defined as the ratio of two functions (1st function / 2nd function ), is equal to the ratio of (differentiation of 1st function $×$the 2nd function $-$ differentiation of the second function $×$the first function ) to the square of the 2nd function]so,$\frac{\frac{d}{dx}×1-\frac{d}{dx}\mathrm{cos}x×1}{{\left(\mathrm{cos}x\right)}^{2}}\phantom{\rule{0ex}{0ex}}$by simplifying the equation$=\frac{0×\mathrm{cos}x-\frac{d}{dx}\mathrm{cos}x×1}{{\left(\mathrm{cos}x\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}x}{\mathrm{cos}x×\mathrm{cos}x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{sin}x}{\mathrm{cos}x}×\frac{1}{\mathrm{cos}x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=tanx×secx$Thus, differentiation of $secx$ by the first principle is$\mathbf{}\mathbit{t}\mathbit{a}\mathbit{n}\mathbit{x}\mathbf{×}\mathbit{s}\mathbit{e}\mathbit{c}\mathbit{x}$.
Suggest Corrections
4
Related Videos
Derivative of Simple Functions
MATHEMATICS
Watch in App
|
}
# How to convert the rectangle to square?
1985/12/01 Goñi, Jesus Mari | Etxeberria, J. Iturria: Elhuyar aldizkaria
How to convert the rectangle to square? The way forward to find the answer is divided into several stages.
The way forward to find the answer is divided into several stages.
At the foot of a rectangle: if we call and at height the value of a surface b will be given by the formula axb. Then, the value to be given on each side to form a square with that surface will be a.b.
• Therefore, we have moved from the problem we had before to a new problem. If values a and b are known, how to complete the value a.b.
Let's try.
Figure 1:
OA=1 unit AB = value a (with compass) OA'=b (with compass) A'B'= x AB and A'B' are perpendicular to the OM line
Next:
B'D = unit C, midpoint of segment A'D with center A'C'D in C B'C', perpendicular to line A'D' from B'C', being the height of the right triangle A'C'D = B'C' A'B'
B'C'= a.b
That was the value we were looking for.
It is enough, therefore, to repeat the work we have done to convert a rectangle into square, replacing the generals a and b with the foot and the concrete heights of that rectangle. Then using the B'C' segment as side, to complete the square.
This evolution is interesting in itself, but the most interesting thing is how easy it is to rectify other polygons. Thus, the rectangle plays the role of mediator.
Squaring a polygon, that is, turning it into a square equivalent to the surface, is not easy if we want to do it in a correct way, but if we use the rectangle as an intermediate auxiliary element everything is easier.
As the square a polygon and calculate its surface are equivalent, the interest of this operation seems evident.
In this issue, changing a little theme, we want to propose a new game.
A and B are wagons of a train. L, the locomotive. T is a tunnel. Along this tunnel you can reach a locomotive but without cars.
The double line in Figure 2 is the railroad.
The essence of the game is how to move from the situation that appears in figure 2 to the one that appears in figure 3 moving as little as possible the locomotive?
eu es fr en ca gl
## Gai honi buruzko eduki gehiago
Elhuyarrek garatutako teknologia
##### Mosaikoak
2018-10-29Zientzia
1989-1-1Zientzia
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.