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# CBSE Class 12 Maths Board Exam: Important 4 marks questions
Important 4 marks questions for CBSE Class 12 Maths subject are provided here for students. The questions are prepared as per new latest exam pattern and syllabus (2022-2023). The Central Board of Secondary Education CBSE) provides students of class 12 with 4 marks question from the exam perspective. As Maths is the most logical and difficult subject amongst the students, it is important for students to practice these questions. The level of difficulty of 4 marks question varies from easy to difficult (sometimes). So solving this section can be cover a high percentage of marks in a shorter span of time.
Generally, the section contains 11 questions, containing 44% of the total marks. Thus, this section can be far more beneficial for the students to score more overall. We at BYJU’S provides students of CBSE class 12 with important 4 marks questions, which can be beneficial to excel in their examination.
## Class 12 Maths Important 4 Marks Questions
Important 4 Marks Questions for Class 12 Maths Board are as follows-
Question 1- The polynomials ax3 – 3x2 +4 and 2x3 – 5x +a when divided by (x – 2) leave the remainders p and q respectively. If p – 2q = 4, find the value of a.
Question 2- Construct a ΔABC in which BC = 3.8 cm, ∠B = 45oand AB + AC = 6.8cm.
Question 3- If a + b + c = 6 and ab + bc + ca = 11, find the value of a3 +b3 +c3 − 3abc.
Question 4- Show that the function F in A = R –
$$\begin{array}{l}\left \{ \frac{2}{3} \right \}\end{array}$$
, defined as f(x) =
$$\begin{array}{l}\frac{4x + 3}{6x – 4}\end{array}$$
is one-one and onto. Hence find
$$\begin{array}{l}f^{-1}\end{array}$$
.
Question 5- Find the value of the following-
$$\begin{array}{l}\tan \frac{1}{2}\left [ \sin^{-1}\frac{2x}{1+x^{2}} + \cos^{-1}\frac{1 – y^{2}}{1 + y^{2}} \right ]\end{array}$$
,
where
$$\begin{array}{l}\left | x \right | < 1, y > 0\;\; and \;\;xy < 1\end{array}$$
Question 6- Prove that
$$\begin{array}{l}\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8} = \frac{\pi}{4}\end{array}$$
Question 7- Prove the following-
$$\begin{array}{l}\begin{vmatrix} 1 & x & x^{2}\\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{vmatrix} = \left ( 1-x^{3} \right )^{2}\end{array}$$
Question 8- Differentiate the following function with respect to x:
$$\begin{array}{l}(\log x)^{x} + x^{\log x}\end{array}$$
Question 9- If
$$\begin{array}{l}y = \log \left [ x + \sqrt{x^{2} + a^{2}} \right ]\end{array}$$
, show that
$$\begin{array}{l}(x^{2} + a^{2})\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} + x\frac{\mathrm{d} y}{\mathrm{d} x} = 0\end{array}$$
Question 10- Evaluate:
$$\begin{array}{l}\int \frac{\sin (x-a)}{\sin (x+a)}dx\end{array}$$
Question 11- Evaluate:
$$\begin{array}{l}\int_{0}^{4} (\left | x \right | + \left | x + 2 \right | + \left | x – 4 \right |) dx\end{array}$$
Question 12- If
$$\begin{array}{l}\vec{a}\end{array}$$
and
$$\begin{array}{l}\vec{b}\end{array}$$
are two vectors such that
$$\begin{array}{l}\left |\vec{a} + \vec{b} \right | = \left |\vec{a} \right |\end{array}$$
, then prove that vector
$$\begin{array}{l}2\vec{a} + \vec{b}\end{array}$$
is perpendicular to vector
$$\begin{array}{l}\vec{b}\end{array}$$
.
Question 13- A speaks 60% of the cases, while B in 90% of the cases. In what percent of cases are they likely to contradict each other in stating the same fact ? In the cases of contradiction do you think, the statement of B will carry more weight as he speaks truth in more number of cases than A ?
Question 14- There are 4 cards numbered 1, 3, 5 and 7, one number on one card. Two cards are drawn at random without replacement. Let X denotes the sum of the numbers on the two drawn cards. Find the mean & variance of X.
Question 15- Maximise Z = x + 2y,
Subject to the constraints
$$\begin{array}{l}x + 2y \geq 100\end{array}$$
$$\begin{array}{l}2x – y \leq 0\end{array}$$
$$\begin{array}{l}2x + y \leq 200\end{array}$$
$$\begin{array}{l}x,y \geq 0\end{array}$$
Solve the above LPP graphically
Question 16- A school wants to award its students for regularity and hard work with a total cash award of Rs. 6000. If three times the award money for hard work added to that given for regularity amounts to Rs. 11,000, represent the above situation algebraically and find the award money for each value, using matrix method. Suggest two more values, which the school must include for award.
Question 17- Find the intervals in which the function given by
f(x) = 2×3 -3×2 -36x+7 is
• Strictly increasing
• Strictly decreasing
Question 18- Bag A contains 3 red and 2 black balls, while bag B contains 2 red and 3 black balls. A ball drawn at random from bag A is transferred to bag B and then one ball is drawn at random from bag B. If this ball was found to be a red ball, find the probability that the ball drawn from bag A was red.
Question 19- If
$$\begin{array}{l}\tan^{-1}\frac{x-3}{x-4} + \tan^{-1}\frac{x+3}{x+4} = \frac{\pi}{4}\end{array}$$
, then find the value of x.
Question 20- If
$$\begin{array}{l}y = (\sec ^{-1}x)^{2}\end{array}$$
, then show that
$$\begin{array}{l}x^{2}(x^{2}-1)\frac{\mathrm{d} ^{2}y}{\mathrm{d} x} + (2x^{3}-x)\frac{\mathrm{d} y}{\mathrm{d} x} =2\end{array}$$
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# Area and Perimeter Definition, Formulas | How to find Area and Perimeter?
Area and Perimeter is an important and basic topic in the Mensuration of 2-D or Planar Figures. The area is used to measure the space occupied by the planar figures. The perimeter is used to measure the boundaries of the closed figures. In Mathematics, these are two major formulas to solve the problems in the 2-dimensional shapes.
Each and every shape has two properties that are Area and Perimeter. Students can find the area and perimeter of different shapes like Circle, Rectangle, Square, Parallelogram, Rhombus, Trapezium, Quadrilateral, Pentagon, Hexagon, and Octagon. The properties of the figures will vary based on their structures, angles, and size. Scroll down this page to learn deeply about the area and perimeter of all the two-dimensional shapes.
## Area and Perimeter Definition
Area: Area is defined as the measure of the space enclosed by the planar figure or shape. The Units to measure the area of the closed figure is square centimeters or meters.
Perimeter: Perimeter is defined as the measure of the length of the boundary of the two-dimensional planar figure. The units to measure the perimeter of the closed figures is centimeters or meters.
### Formulas for Area and Perimeter of 2-D Shapes
1. Area and Perimeter of Rectangle:
• Area = l × b
• Perimeter = 2 (l + b)
• Diagnol = √l² + b²
Where, l = length
2. Area and Perimeter of Square:
• Area = s × s
• Perimeter = 4s
Where s = side of the square
3. Area and Perimeter of Parallelogram:
• Area = bh
• Perimeter = 2( b + h)
Where, b = base
h = height
4. Area and Perimeter of Trapezoid:
• Area = 1/2 × h (a + b)
• Perimeter = a + b + c + d
Where, a, b, c, d are the sides of the trapezoid
h is the height of the trapezoid
5. Area and Perimeter of Triangle:
• Area = 1/2 × b × h
• Perimeter = a + b + c
Where, b = base
h = height
a, b, c are the sides of the triangle
6. Area and Perimeter of Pentagon:
• Area = (5/2) s × a
• Perimeter = 5s
Where s is the side of the pentagon
a is the length
7. Area and Perimeter of Hexagon:
• Area = 1/2 × P × a
• Perimeter = s + s + s + s + s + s = 6s
Where s is the side of the hexagon.
8. Area and Perimeter of Rhombus:
• Area = 1/2 (d1 + d2)
• Perimeter = 4a
Where d1 and d2 are the diagonals of the rhombus
a is the side of the rhombus
9. Area and Perimeter of Circle:
• Area = Î r²
• Circumference of the circle = 2Î r
Where r is the radius of the circle
Î = 3.14 or 22/7
10. Area and Perimeter of Octagon:
• Area = 2(1 + √2) s²
• Perimeter = 8s
Where s is the side of the octagon.
### Solved Examples on Area and Perimeter
Here are some of the examples of the area and perimeter of the geometric figures. Students can easily understand the concept of the area and perimeter with the help of these problems.
1. Find the area and perimeter of the rectangle whose length is 8m and breadth is 4m?
Solution:
Given,
l = 8m
b = 4m
Area of the rectangle = l × b
A = 8m × 4m
A = 32 sq. meters
The perimeter of the rectangle = 2(l + b)
P = 2(8m + 4m)
P = 2(12m)
P = 24 meters
Therefore the area and perimeter of the rectangle is 32 sq. m and 24 meters.
2. Calculate the area of the rhombus whose diagonals are 6 cm and 5 cm?
Solution:
Given,
d1 = 6cm
d2 = 5 cm
Area = 1/2 (d1 + d2)
A = 1/2 (6 cm + 5cm)
A = 1/2 × 11 cm
A = 5.5 sq. cm
Thus the area of the rhombus is 5.5 sq. cm
3. Find the area of the triangle whose base and height are 11 cm and 7 cm?
Solution:
Given,
Base = 11 cm
Height = 7 cm
We know that
Area of the triangle = 1/2 × b × h
A = 1/2 × 11 cm × 7 cm
A = 1/2 × 77 sq. cm
A = 38.5 sq. cm
Thus the area of the triangle is 38.5 sq. cm.
4. Find the area of the circle whose radius is 7 cm?
Solution:
Given,
We know that,
Area of the circle = Πr²
Î = 3.14
A = 3.14 × 7 cm × 7 cm
A = 3.14 × 49 sq. cm
A = 153.86 sq. cm
Therefore the area of the circle is 153.86 sq. cm.
5. Find the area of the trapezoid if the length, breadth, and height is 8 cm, 4 cm, and 5 cm?
Solution:
Given,
a = 8 cm
b = 4 cm
h = 5 cm
We know that,
Area of the trapezoid = 1/2 × h(a + b)
A = 1/2 × (8 + 4)5
A = 1/2 × 12 × 5
A = 6 cm× 5 cm
A = 30 sq. cm
Therefore the area of the trapezoid is 30 sq. cm.
6. Find the perimeter of the pentagon whose side is 5 meters?
Solution:
Given that,
Side = 5 m
The perimeter of the pentagon = 5s
P = 5 × 5 m
P = 25 meters
Therefore the perimeter of the pentagon is 25 meters.
### FAQs on Area and Perimeter
1. How does Perimeter relate to Area?
The perimeter is the boundary of the closed figure whereas the area is the space occupied by the planar.
2. How to calculate the perimeter?
The perimeter can be calculated by adding the lengths of all the sides of the figure.
3. What is the formula for perimeter?
The formula for perimeter is the sum of all the sides.
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# What Are The 7 Types Of Triangles?
## How many types of triangles are there?
three typesThere are three types of triangle based on the length of the sides: equilateral, isosceles, and scalene.
The green lines mark the sides of equal (the same) length..
## Which Triangle Cannot be drawn?
According to the first triangle inequality theorem, the lengths of any two sides of a triangle must add up to more than the length of the third side. This means that you cannot draw a triangle that has side lengths 2, 7 and 12, for instance, since 2 + 7 is less than 12.
## What is a true triangle?
A triangle has three sides, three vertices, and three angles. The sum of the three interior angles of a triangle is always 180°. The sum of the length of two sides of a triangle is always greater than the length of the third side.
## What does a unique triangle mean?
A unique triangle means there is ONLY ONE WAY to create the triangle. All unique triangles are congruent (the same) even though you may have to “flip” or “turn” them to line up. Criteria For A Unique Triangle.
## What 3 angles make a triangle?
In a Euclidean space, the sum of angles of a triangle equals the straight angle (180 degrees, π radians, two right angles, or a half-turn). A triangle has three angles, one at each vertex, bounded by a pair of adjacent sides.
## What are the six types of triangles?
Six Types of TrianglesBased on their SidesBased on their AnglesScalene TriangleAcute TriangleIsosceles TriangleObtuse TriangleEquilateral TriangleRight TriangleAug 20, 2020
## What is r in properties of triangle?
A circle passing through the vertices of a triangle is called the circumcircle of a triangle. The radius of the circumcircle is called the circumradius of the triangle and is denoted by R and is given by the formulae. R = a/2 sin A = b/ 2 sin B = c/2 sin C.
## What are the types of triangles according to sides?
Triangles can be classified into 3 types based on the lengths of their sides:Scalene.Isosceles.Equilateral.
## What is a unique triangle 7th grade?
A unique triangle means there is ONLY ONE WAY to create the triangle. All unique triangles are congruent (the same) even though you may have to “flip” or “turn” them to line up. Criteria For A Unique Triangle.
## How many unique triangles can make?
Lesson 10 Summary Sometimes, only one triangle can be made. By this we mean that any triangle we make will be the same, having the same six measures. For example, if a triangle can be made with three given side lengths, then the corresponding angles will have the same measures.
## What is equiangular triangle in math?
A triangle with three equal interior angles is called an equiangular triangle. … An equiangular triangle has three equal sides, and it is the same as an equilateral triangle.
## Who found triangle?
Blaise PascalIt is named for the 17th-century French mathematician Blaise Pascal, but it is far older. Chinese mathematician Jia Xian devised a triangular representation for the coefficients in the 11th century.
## What do you call a triangle with 2 equal sides?
An isosceles triangle can be drawn in many different ways. It can be drawn to have two equal sides and two equal angles or with two acute angles and one obtuse angle.
## How do you classify triangles?
Triangles can be classified by their sides and by their angles. When classifying a triangle by its sides, you should look to see if any of the sides are the same length. If no sides are the same length, then it is a scalene triangle. If two sides are the same length, then it is an isosceles triangle.
## Which is best definition of a triangle?
A triangle is a three-sided polygon that closes in a space. It uses lines, line segments or rays (in any combination) to form the three sides. When three sides form and meet, they create three vertices, or corners.
## Why are triangles so important?
Triangles are polygons (shapes) with three sides and three angles, which can be formed by connecting any three points in a plane. They are one of the first shapes studied in geometry. Triangles are particularly important because arbitrary polygons (with 4, 5, 6, or n sides) can be decomposed into triangles.
## What is called triangle?
A triangle is a 3-sided polygon sometimes (but not very commonly) called the trigon. The sides of a triangle are given special names in the case of a right triangle, with the side opposite the right angle being termed the hypotenuse and the other two sides being known as the legs. …
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# 9.2: Linear Equations
Linear regression for two variables is based on a linear equation with one independent variable. The equation has the form:
$y = a + b\text{x}\nonumber$
where $$a$$ and $$b$$ are constant numbers. The variable $$x$$ is the independent variable, and $$y$$ is the dependent variable. Typically, you choose a value to substitute for the independent variable and then solve for the dependent variable.
Example $$\PageIndex{1}$$
The following examples are linear equations.
$y = 3 + 2\text{x}\nonumber$
$y = -0.01 + 1.2\text{x}\nonumber$
Exercise $$\PageIndex{1}$$
Is the following an example of a linear equation?
$y = -0.125 - 3.5\text{x}\nonumber$
yes
The graph of a linear equation of the form $$y = a + b\text{x}$$ is a straight line. Any line that is not vertical can be described by this equation.
Example $$\PageIndex{2}$$
Graph the equation $$y = -1 + 2\text{x}$$.
Exercise $$\PageIndex{2}$$
Is the following an example of a linear equation? Why or why not?
No, the graph is not a straight line; therefore, it is not a linear equation.
Example $$\PageIndex{3}$$
Aaron's Word Processing Service (AWPS) does word processing. The rate for services is $32 per hour plus a$31.50 one-time charge. The total cost to a customer depends on the number of hours it takes to complete the job.
Find the equation that expresses the total cost in terms of the number of hours required to complete the job.
Let $$x =$$ the number of hours it takes to get the job done.
Let $$y =$$ the total cost to the customer.
## Summary
The most basic type of association is a linear association. This type of relationship can be defined algebraically by the equations used, numerically with actual or predicted data values, or graphically from a plotted curve. (Lines are classified as straight curves.) Algebraically, a linear equation typically takes the form $$y = mx + b$$, where $$m$$ and $$b$$ are constants, $$x$$ is the independent variable, $$y$$ is the dependent variable. In a statistical context, a linear equation is written in the form $$y = a + bx$$, where $$a$$ and $$b$$ are the constants. This form is used to help readers distinguish the statistical context from the algebraic context. In the equation $$y = a + b\text{x}$$, the constant b that multiplies the $$x$$ variable ($$b$$ is called a coefficient) is called as the slope. The slope describes the rate of change between the independent and dependent variables; in other words, the rate of change describes the change that occurs in the dependent variable as the independent variable is changed. In the equation $$y = a + b\text{x}$$, the constant a is called as the $$y$$-intercept. Graphically, the $$y$$-intercept is the $$y$$ coordinate of the point where the graph of the line crosses the $$y$$ axis. At this point $$x = 0$$.
The slope of a line is a value that describes the rate of change between the independent and dependent variables. The slope tells us how the dependent variable ($$y$$) changes for every one unit increase in the independent ($$x$$) variable, on average. The $$y$$-intercept is used to describe the dependent variable when the independent variable equals zero. Graphically, the slope is represented by three line types in elementary statistics.
## Formula Review
$$y = a + b\text{x}$$ where a is the $$y$$-intercept and $$b$$ is the slope. The variable $$x$$ is the independent variable and $$y$$ is the dependent variable.
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# What is the greatest 4-digit number?
• Difficulty Level : Expert
• Last Updated : 10 Aug, 2021
The method to represent and work with numbers is known as the number system. A number system is a system of writing to represent numbers. It is the mathematical notation used to represent numbers of a given set by using digits or other symbols. It allows us to operate arithmetic operations such as division, multiplication, addition, subtraction.
### Types of Number systems
The number system can exist for any base and it can have those number of digits involved accordingly. Some important number systems are as follows,
1. Decimal Number System
2. Binary Number System
3. Octal Number System
Let’s see about all these number systems in detail,
Decimal Number System
The decimal number system consists of ten digits i.e. from 0 to 9. The base of a decimal number system is 10. These digits can be used to represent or express any numeric value. For example, the decimal number 153 consists of the digit 3 in one place, the digit 5 in the tens place, and the digit 1 in hundreds place which can be represented as,
(1×102) + (5×101) + (3×100)
= (1× 100 ) + ( 5× 10 ) + (3× 1), { where, 100 = 1}
= 100 + 50 + 3
= 153
Binary Number System
The binary number system consists of only two digits i.e. 0 and 1. The base of the binary number system is 2. The digital computer represents all kinds of data in a binary number system. For example, convert 100112 into a decimal number system.
(100111)2 = 1 × 25 + 0 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 1 × 20
= 32 + 0 + 0 + 4 + 2 + 1
= (39)10
Octal Number System
The octal number system consists of digits from 0 to 7. The base of the octal number system is 8. Octal number systems are basically used in computer applications. For example, convert 1458 into decimal.
1458 = 1 X 82 + 4 X 81+ 5 X 80
= 64 + 32 + 5
= 10110
In hexadecimal number system, numbers are first represented from digits 0 to 9 as decimal number system and then the numbers are represented using alphabets from A to F. The base of the hexadecimal number system is 16. For example, convert 26BC16 to decimal.
26BC16 = 2 X 16 6 X 162 + 11 X 161 + 12 X 160
= 8192 + 1536 + 176 + 12
= 991610
### The Greatest Four digit number
The greatest four-digit number can be expressed as the number with a digit at ones place, tenth place, hundredth place, and thousand place. All the digits have been the largest digits, in essence, 9.The Greatest four-digit number in the number system is ⇢ 9999
Explanation: If 1 is added to this number, it becomes 10000 which is a five-digit number. So 9999 is the greatest four-digit number in the number system.
Greatest four digit number in the number system = 9999
= 9999 + 1
= 10000 ( which is five digit number )
Hence it is proved that 9999 is the greatest four-digit number.
By this method, it is easy to find any greatest number in the number system.
### Similar Questions
Question 1: What is the 1 digit greatest number?
The greatest 1 digit number in the number system is 9.
Question 2: What is the 2 digit greatest number?
The greatest 2 digit number in the number system is 99.
Question 3: What is the 3 digit greatest number?
The greatest 3 digit number in the number system is 999.
Question 4: What is the 4 digit greatest number?
The greatest 4 digit number in the number system is 9999.
Question 5: What is the 5 digit greatest number?
The greatest 5 digit number in the number system is 99999.
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# What is a Plane?
A plane, in geometry, prolongs infinitely in two dimensions. It has no width. We can see an example of a plane in coordinate geometry. The coordinates define the position of points in a plane.
In Maths, a plane is a flat, two-dimensional surface that prolongs infinitely far. A plane is a two-dimensional analogue that could consist of a point, a line and three-dimensional space. Planes can appear as subspaces of a few higher-dimensional spaces, like the room’s walls extended exceptionally far away. These walls experience an independent existence on their own, as in the framework of Euclidean geometry.
What is a point?
A point is a location in a plane that has no size, i.e. no width, no length and no depth.
What is a line?
A line is a set of points that stretches infinitely in opposite directions. It has only one dimension, i.e., length. The points that lie on the same line are called collinear points.
## Plane in Algebra
In algebra, the points are plotted in the coordinate plane, and this denotes an example of a geometric plane. The coordinate plane has a number line, extending left to right endlessly and another one extending up and down infinitely. It is impossible to view the complete coordinate plane.
Arrows designate the truth that it extends infinitely along the x-axis and the y-axis on the number lines’ ends. These number lines are two-dimensional, where the plane extends endlessly. When we plot the graph in a plane, then the point or a line, plotted does not have any thickness.
Also, learn:
### Plane Meaning
A surface comprising all the straight lines that join any two points lying on it is called a plane in geometry. In other words, it is a flat or level surface. In a Euclidean space of any number of dimensions, a plane is defined through any of the following uniquely:
• Using three non-collinear points
• Using a point and a line not on that line
• Using two distinct intersecting lines
• Using two separate parallel lines
## Intersecting Planes
Two planes can be related in three ways, in a three-dimensional space.
• They can be parallel to each other
• They can be identical
• They can intersect each other
The figure below depicts two intersecting planes.
The method to get the equation of the line of intersection connecting two planes is to determine the set of points that satisfy both the planes’ equations. Since the equation of a plane comprises three variables and two equations (because of two planes), solving the simultaneous equations will give a relationship between the three variables, which is equal to the intersection line equation.
## Properties of a Plane
• If there are two different planes than they are either parallel to each other or intersecting in a line, in a three-dimensional space
• A line could be parallel to a plane, intersects the plane at a single point or is existing in the plane.
• If there are two different lines, which are perpendicular to the same plane then they must be parallel to each other.
• If there are two separate planes that are perpendicular to the same line then they must be parallel to each other.
### What is a Plane Figure?
A plane figure is defined as a geometric figure that has no thickness. It lies entirely in one plane. It is possible to form a plane figure c with line segments, curves, or a combination of these two, i.e. line segments and curves. Let’s have a look at some examples of plane figures in geometry such as circle, rectangle, triangle, square and so on. These are given in the below figure.
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# Difference between revisions of "1993 AIME Problems/Problem 11"
## Problem
Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain a head. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game. Suppose that Alfred goes first in the first game, and that the probability that he wins the sixth game is $m/n\,$, where $m\,$ and $n\,$ are relatively prime positive integers. What are the last three digits of $m+n\,$?
## Solution
The probability that the $n$th flip in each game occurs and is a head is $\frac{1}{2^n}$. The first person wins if the coin lands heads on an odd numbered flip. So, the probability of the first person winning the game is $\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\cdots = \frac{\frac{1}{2}}{1-\frac{1}{4}}=\frac{2}{3}$, and the probability of the second person winning is $\frac{1}{3}$.
Let $a_n$ be the probability that Alfred wins the $n$th game, and let $b_n$ be the probability that Bonnie wins the $n$th game.
If Alfred wins the $n$th game, then the probability that Alfred wins the $n+1$th game is $\frac{1}{3}$. If Bonnie wins the $n$th game, then the probability that Alfred wins the $n+1$th game is $\frac{2}{3}$.
Thus, $a_{n+1}=\frac{1}{3}a_n+\frac{2}{3}b_n$.
Similarly, $b_{n+1}=\frac{2}{3}a_n+\frac{1}{3}b_n$.
Since Alfred goes first in the $1$st game, $(a_1,b_1)=\left(\frac{2}{3}, \frac{1}{3}\right)$.
Using these recursive equations:
$(a_2,b_2)=\left(\frac{4}{9}, \frac{5}{9}\right)$
$(a_3,b_3)=\left(\frac{14}{27}, \frac{13}{27}\right)$
$(a_4,b_4)=\left(\frac{40}{81}, \frac{41}{81}\right)$
$(a_5,b_5)=\left(\frac{122}{243}, \frac{121}{243}\right)$
$(a_6,b_6)=\left(\frac{364}{729}, \frac{365}{729}\right)$
Since $a_6=\frac{364}{729}$, $m+n = 1093 \equiv \boxed{093} \pmod{1000}$.
## Solution 2
In order to begin this problem, we need to calculate the probability that Alfred will win on the first round. Because he goes first, Alfred has a $\frac{1}{2}$ chance of winning (getting heads) on his first flip. Then, Bonnie, who goes second, has a $\frac{1}{2}$ * $\frac{1}{2}$ = $\frac{1}{4}$, chance of winning on her first coin toss. Therefore Alfred’s chance of winning on his second flip is $\frac{1}{4}$ * $\frac{1}{2}$ = $\frac{1}{8}$
From this, we can see that Alfred’s (who goes first) chance of winning the first round is:
$\frac{1}{2}$ + $\frac{1}{8}$ + $\frac{1}{32}$ + \cdots = $\frac{2}{3}$.
Bonnie’s (who goes second) chance of winning the first round is then 1 - $\frac{2}{3}$ = $\frac{1}{3}$.
This means that the person who goes first has a $\frac{2}{3}$ chance of winning the round, while the person who goes second has a $\frac{1}{3}$ chance of winning.
Now, through casework, we can calculate Alfred’s chance of winning the second round.
Case 1: Alfred wins twice; $\frac{2}{3}$ * $\frac{1}{3}$ (Bonnie goes first this round) = $\frac{2}{9}$.
Case 2: Alfred loses the first round, but wins the second; $\frac{1}{3}$ * $\frac{2}{3}$ = $\frac{2}{9}$.
Adding up the cases, we get $\frac{2}{9}$ + $\frac{2}{9}$ = $\frac{4}{9}$.
Alfred, therefore, has a $\frac{4}{9}$ of winnning the second round, and Bonnie has a 1- $\frac{5}{9}$ of winning this round.
|
# Lesson video
In progress...
Hi, I'm Miss Davies.
In this lesson, we're going to be drawing graphs that are in the form y = mx + c.
We are working with graphs that are in the form y = mx + c.
In this, the x and the y values are the coordinates of the points.
The m value is gradient of the line.
This is the steepness of the given line.
For every one unit that it moves to the right, how many units does it move up or down? The c value is the y-intercept.
This is the point in which the line crosses the y-axis.
In which pairs of coordinates is the y value 3 more than double the double the x value? 4, 11.
Negative 9, negative 15.
And 0, 3.
Which of the points lie on the line y = 2x + 3? It is these three points.
We've been asked to complete the table of values for the line y = 2x + 3.
The table of values is completed by substituting in values for x into the equation.
If we substitute in negative 2, we'll be calculating 2 multiplied by negative 2, add 3.
This gives negative 1.
When x is negative 1, we're going to calculate 2 multiplied by negative 1, add 3 to find y, which is 1.
When x is 1, y is 5.
We can then use this to plot the graph of y = 2x + 3.
Using these coordinates to plot each of the points and then joining these up with a straight line.
Notice that this line runs from the edge of the set of axis we've been given, rather than just from the first point to the last point.
Which of these coordinates lie on the line y = -x + 1? Negative 7, 8.
2, negative 1.
And 0, 1.
Next, we're going to complete the table of values for the line y = -x + 1.
We're going to find the y-coordinate when x is equal to negative 1, 0, and 2.
When x is negative 1, y is 2.
And negative -1, add 1, is equal to 2.
When x is 0, y is 1.
When x is 2, y is negative 1.
We're now going to plot this graph.
We're going to use these coordinates to plot the points that we have just worked out.
We can then join these together with a straight line that runs from edge to edge of the grid.
You can see that this line has got a negative gradient and for every one unit it goes to the right, it goes one unit down.
This is because the gradient of the equation is negative 1.
Here is some questions for you to try.
In this graph, the gradient is 2.
This means that for every one unit that it moves to the right, it moves two units up.
Here is some questions for you to try.
Pause the video to complete your task and resume once you're finished.
|
Accelerating the pace of engineering and science
# Documentation
## Solving Algebraic and Differential Equations
This example shows how to solve algebraic and differential equations using Symbolic Math Toolbox™.
Solving Algebraic Equations
The solve function can handle a wide variety of algebraic equations, including both single equations and systems of equations. By default, the solver always tries to return a complete set of solutions, including complex solutions and multiple branches. For example, the following polynomial has two real and two complex solutions.
```syms x
xsol = solve(x^4 + x^3 - x == 1, x)
```
```
xsol =
-1
1
- (3^(1/2)*i)/2 - 1/2
(3^(1/2)*i)/2 - 1/2
```
Use indexing to select only the real solutions.
```xsol(xsol == real(xsol))
```
```
ans =
-1
1
```
Some equations, such as those that include trigonometric elements, have an infinite number of solutions. For such equations, the solver returns only one solution from the infinite set.
```solve(sin(x) == 1, x)
```
```
ans =
pi/2
```
Solving Systems of Algebraic Equations
The solve function can also be used to solve systems of equations. Simply specify the system of equations and the variables to solve for.
```syms x y
[x2, y2] = solve(y + x^2 == 1, x - y == 10)
```
```
x2 =
- (3*5^(1/2))/2 - 1/2
(3*5^(1/2))/2 - 1/2
y2 =
- (3*5^(1/2))/2 - 21/2
(3*5^(1/2))/2 - 21/2
```
As with an individual equation, if a system has an infinite number of solutions, the solver returns only one solution.
```[x3, y3] = solve(exp(x + y) == 1, x - y == 1)
```
```
x3 =
1/2
y3 =
-1/2
```
In some cases, you may be able to express the system of equations in matrix form. For these cases, you can use MATLAB's backslash operator to solve the equations. Note that the result is a double precision number since the equations are solved numerically.
```A = [1 2; 3 4];
b = [5 6]';
x = A\b
```
```x =
-4.0000
4.5000
```
You can use the 'sym' command to solve the system exactly.
```sym(A)\b
```
```
ans =
-4
9/2
```
Solving Ordinary Differential Equations (ODEs)
You can solve various types of ordinary differential equations using the dsolve function. For example, you can solve a simple 2nd order linear ODE.
```syms y(t)
D2y = diff(y,2);
Dy = diff(y);
dsolve(D2y + Dy + y == 0)
```
```
ans =
C5*exp(-t/2)*cos((3^(1/2)*t)/2) + C6*exp(-t/2)*sin((3^(1/2)*t)/2)
```
You can specify initial conditions or boundary conditions along with an ODE.
```ysol = dsolve(D2y == 2*t/Dy, y(0) == 0, Dy(0) == 1)
```
```
ysol =
(2^(1/2)*log(t + (t^2 + 1/2)^(1/2)))/4 - (2^(1/2)*log(2^(1/2)/2))/4 + (2^(1/2)*t*(t^2 + 1/2)^(1/2))/2
```
You can visualize the solution to your differential equation using the ezplot function.
```ezplot(ysol)
```
|
0
# There are 10 students receiving honors credit in a class. This number is exactly 20 percent of the total number of students in the class. How many students are in the class?
Wiki User
2008-08-24 08:21:28
Here are 10 students getting honors credits in a class, and they make up 20% of the class. How do we find the number of students in the class? Let's look. We have a class, and 20% of the class are getting honors credit, and that turns out to be 10 students. Now we can generate a formula that we can use to discover our answer. Let's assign letters to the things we know or are finding out. Nclass = number of students in the class. Nhonors = number of honors students in the class. And Nhonors = 10 students. Nclass x 20% = 10 students 20% = 20/100ths (because % = hundredths) or 0.20 or just 0.2 for simplicity. Nclass x .2 = 10 students Now divide both sides by .2 so the .2 will cancel out (or drop out) on the left side of the equation and we'll have isolated the answer we are looking for, which is Nclass. Nclass = 10 students divided by .2 Nclass = 10/.2 students = 50 students There are 50 students in the class. As .2 equals 2/10 or 1/5, we can find 1/5th of 50 just by thinking about it to check our work. And 1/5th of 50 equals 10, which is in agreement with the information we were given in the problem.
Wiki User
2008-08-24 08:21:28
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Engage NY Eureka Math Algebra 2 Module 1 Lesson 30 Answer Key
Eureka Math Algebra 2 Module 1 Lesson 30 Example Answer Key
Example 1.
Determine the values for x, y, and z in the following system:
2x + 3y – z = 5
4x – y – z = – 1
x + 4y + z = 12
Answer:
2x + 3y – z = 5 …….. (1)
4x – y – z = – 1 …….. (2)
x + 4y + z = 12 …….. (3)
Suggest numbering the equations as shown above to help organize the process.
→ Eliminate z from equations (1) and (2) by subtraction. Replace equation (1) with the result.
→ Our goal is to find two equations in two unknowns. Thus, we will also eliminate z from equations (2) and (3) by adding as follows. Replace equation (3) with the result.
→ Our new system of three equations in three variables has two equations with only two variables in them:
– 2x + 4y = 6
4x – y – z = -1
5x + 3y = 11.
→ These two equations now give us a system of two equations in two variables, which we reviewed how to solve in Exercises 1 – 2.
– 2x + 4y = 6
5x + 3y = 11
At this point, let students solve this individually or with partners, or guide them through the process if necessary.
→ To get matching coefficients, we need to multiply both equations by a constant:
5(- 2x + 4y) = 5(6) → – 10x + 20y = 30
2(5x + 3y) = 2(11) → 10x + 6y = 22.
→ Replacing the top equation with the sum of the top and bottom equations together gives the following:
26y = 52
10x + 6y = 22.
→ Replace y = 2 in one of the equations to find x:
5x + 3(2) = 11
5x + 6 = 11
5x = 5
x = 1.
→ Replace x = 1 and y = 2 in any of the original equations to find z:
2(1) + 3(2) – z = 5
2 + 6 – z = 5
8 – z = 5
z = 3.
→ The solution, x = 1, y = 2, and z = 3, can be written compactly as an ordered triple of numbers (1, 2, 3).
Consider pointing out to students that the point (1, 2, 3) can be thought of as a point in a three-dimensional coordinate plane, and that it is, like a two-by-two system of equations, the intersection point in three-space of the three planes given by the graphs of each equation. These concepts are not the point of this lesson, so addressing them is optional.
Point out that a linear system involving three variables requires three equations in order for the solution to possibly be a single point.
The following problems provide examples of situations that require solving systems of equations in three variables.
Eureka Math Algebra 2 Module 1 Lesson 30 Exercise Answer Key
Determine the value of x and y in the following systems of equations.
Exercise 1.
2x + 3y = 7
2x + y = 3
Answer:
x = $$\frac{1}{2}$$, y = 2
Exercise 2.
5x – 2y = 4
Answer:
– 2x + y = 2
x = 8, y = 18
Exercise 3.
A scientist wants to create 120 ml of a solution that is 30% acidic. To create this solution, she has access to a 20% solution and a 45% solution. How many milliliters of each solution should she combine to create the 30% solution?
Answer:
Milliliters of 20% solution: x ml
Milliliters of 45% solution: y ml
Write one equation to represent the total amounts of each solution needed:
x + y = 120.
Since 30% of 120 ml is 36, we can write one equation to model the acidic portion:
0.20x + 0.45y = 36.
Writing these two equations as a system:
x + y = 120
0.20x + 0.45y = 36
To solve, multiply both sides of the top equation by either 0.20 to eliminate x or 0.45 to eliminate y. The following steps will eliminate x:
0. 20(x + y) = 0.20(120)
0.20x + 0.45y = 40
which gives
0.20x + 0.20y = 24
0.20x + 0.45y = 36.
Replacing the top equation with the difference between the bottom equation and top equation results in a new system with the same solutions:
0.25y = 12
0.20x + 0.45y = 36.
The top equation can quickly be solved for y,
y = 48,
and substituting y = 48 back into the original first equation allows us to find x:
x + 48 = 120
x = 72.
Thus, we need 48 ml of the 45% solution and 72 ml of the 20% solution.
Exercise 4.
Given the system below, determine the values of r, s, and u that satisfy all three equations.
r + 2s – u = 8
s + u = 4
r – s – u = 2
Answer:
Adding the second and third equations together produces the equation r = 6. Substituting this into the first equation and adding it to the second gives 6 + 3s = 12, so that s = 2. Replacing s with 2 in the second equation gives u = 2. The solution to this system of equations is (6, 2, 2).
Exercise 5.
Find the equation of the form y = ax2 + bx + c whose graph passes through the points (1, 6), (3, 20), and (- 2, 15).
Answer:
We find a = 2, b = – 1, c = 5; therefore, the quadratic equation is y = 2x2 – x + 5.
→ Since we know three ordered pairs, we can create three equations.
6 = a + b + c
20 = 9a + 3b + c
15 = 4a – 2b + c
Ask students to explain where the three equations came from. Then have them use the technique from Example 1 to solve this system.
Have students use a graphing utility to plot y = 2x2 – x + 5 along with the original three points to confirm their answer.
Eureka Math Algebra 2 Module 1 Lesson 30 Problem Set Answer Key
Solve the following systems of equations.
Question 1.
x + y = 3
y + z = 6
x + z = 5
Answer:
x = 1, y = 2, z = 4 or (1, 2, 4)
Question 2.
r = 2(s – t)
2t = 3(s – r)
r + t = 2s – 3
Answer:
r = 2, s = 4, t = 3, or (2, 4, 3)
Question 3.
2a + 4b + c = 5
a – 4b = – 6
2b + c = 7
Answer:
a = – 2, b = 1, c = 5 or (- 2, 1, 5)
Question 4.
2x + y – z = – 5
4x – 2y + z = 10
2x + 3y + 2z = 3
x = $$\frac{1}{2}$$, y = – 2, z = 4 or ($$\frac{1}{2}$$, – 2, 4)
Question 5.
r + 3s + t = 3
2r – 3s + 2t = 3
– r + 3s – 3t = 1
r = 3, s = $$\frac{1}{3}$$, t = -1 or (3, $$\frac{1}{3}$$, -1)
Question 6.
x – y = 1
2y + z = – 4
x – 2z = – 6
Answer:
x = – 2, y = -3, z = 2 or (- 2, – 3,2)
Question 7.
x = 3(y – z)
y = 5(z – x)
x + y = z + 4
Answer:
x = 3, y = 5, z = 4 or (3, 5, 4)
Question 8.
p + q + 3r = 4
2q + 3r = 7
p – q – r = – 2
Answer:
p = 2, q = 5, r = – 1 or (2, 5, – 1)
Question 9.
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$ = 5
$$\frac{1}{x}+\frac{1}{y}$$ = 2
$$\frac{1}{x}-\frac{1}{y}$$ = – 2
Answer:
x = 1, y = 1, z = $$\frac{1}{3}$$ or (1, 1, $$\frac{1}{3}$$)
Question 10.
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ = 6
$$\frac{1}{b}+\frac{1}{c}$$ = 5
$$\frac{1}{a}-\frac{1}{b}$$ = – 1
Answer:
a = -1, b = $$\frac{1}{2}$$, c = $$\frac{1}{3}$$ or (1, $$\frac{1}{2}$$, $$\frac{1}{3}$$)
Question 11.
Find the equation of the form y = ax2 + bx + c whose graph passes through the points (1, – 1), (3, 23), and (- 1, 7).
Answer:
y = 4x2 – 4x – 1
Question 12.
Show that for any number t, the values x = t + 2, y = 1 – t, and z = t + 1 are solutions to the system of equations below.
x + y = 3
y + z = 2
(In this situation, we say that t parameterizes the solution set of the system.)
Answer:
x + y = (t + 2) + (1 – t) = 3
y + z = (1 – t) + (t + 1) = 2
Question 13.
Some rational expressions can be written as the sum of two or more rational expressions whose denominators are the factors of its denominator (called a partial fraction decomposition). Find the partial fraction decomposition for $$\frac{1}{n(n+1)}$$ by finding the value of A that makes the equation below true for all n except 0 and – 1.
Answer:
Adding $$\frac{1}{n+1}$$ to both sides of the equations, we have
so A = 1 and thus $$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$.
Question 14.
A chemist needs to make 40 ml of a 15% acid solution. He has a 5% acid solution and a 30% acid solution on hand. If he uses the 5% and 30% solutions to create the 15% solution, how many ml of each does he need?
Answer:
He needs 24 ml of the 5% solution and 16 ml of the 30% solution.
Question 15.
An airplane makes a 400-mile trip against a head wind in 4 hours. The return trip takes 2. 5 hours, the wind now being a tail wind. If the plane maintains a constant speed with respect to still air, and the speed of the wind is also constant and does not vary, find the still-air speed of the plane and the speed of the wind.
Answer:
The speed of the plane in still wind is 130 mph, and the speed of the wind is 30 mph.
Question 16.
A restaurant owner estimates that she needs the same number of pennies as nickels and the same number of dimes as pennies and nickels together. How should she divide $26 between pennies, nickels, and dimes? Answer: She will need 200 dimes ($20 worth), 100 nickels ($5 worth), and 100 pennies ($1 worth) for a total of \$26.
Eureka Math Algebra 2 Module 1 Lesson 30 Exit Ticket Answer Key
For the following system, determine the values of p, q, and r that satisfy all three equations:
2p + q – r = 8
q + r = 4
p – q = 2.
Answer:
p = 4, q = 2, r = 2, or equivalently (4, 2, 2)
|
## A Multiplication Based Logic Puzzle
### 462 and Level 4
462 is the sum of consecutive prime numbers two different ways. Check the comments to see what those ways are.
Divisibility tricks:
• 462 is even, so it is divisible by 2.
• The sum of the odd numbered digits, 4 + 2 is 6, which is the 2nd digit, so 462 is divisible by 11.
• Since both of those 6’s above are divisible by 3, then 462 is divisible by 3.
• Separate the last digit from the rest and double it. 462 → 46 and 2; Doubling 2, gives us 4. Now subtract that 4 from the remaining digits: 46 – 4 = 42 which is divisible by 7, so 462 is divisible by 7.
Since 462 = 21 × 22, we know that it is two times the 21st triangular number, and it is the sum of the first 21 even numbers.
• 2(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21) = 462
• 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30 + 32 + 34 + 36 + 38 + 40 + 42 = 462
Print the puzzles or type the solution on this excel file: 10 Factors 2015-04-13
—————————————————————————————————
• 462 is a composite number.
• Prime factorization: 462 = 2 x 3 x 7 x 11
• The exponents in the prime factorization are 1, 1, 1, and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1)(1 + 1)(1 + 1) = 2 x 2 x 2 x 2 = 16. Therefore 462 has exactly 16 factors.
• Factors of 462: 1, 2, 3, 6, 7, 11, 14, 21, 22, 33, 42, 66, 77, 154, 231, 462
• Factor pairs: 462 = 1 x 462, 2 x 231, 3 x 154, 6 x 77, 7 x 66, 11 x 42, 14 x 33, or 21 x 22
• 462 has no square factors that allow its square root to be simplified. √462 ≈ 21.4942
—————————————————————————————————
#### Comments on: "462 and Level 4" (2)
1. abyssbrain said:
Haha, got lucky there, found it very quickly 🙂
67 + 71 + 73 + 79 + 83 + 89 = 462
Liked by 3 people
• ivasallay said:
Good for you! I’m glad there is more competition to see who finds them first! 229 + 233 are also consecutive prime numbers that equal 462.
Liked by 1 person
|
# 1992 IMO Problems/Problem 4
## Problem
In the plane let $C$ be a circle, $l$ a line tangent to the circle $C$, and $M$ a point on $l$. Find the locus of all points $P$ with the following property: there exists two points $Q$, $R$ on $l$ such that $M$ is the midpoint of $QR$ and $C$ is the inscribed circle of triangle $PQR$.
## Solution
Note: This is an alternate method to what it is shown on the video. This alternate method is too long and too intensive in solving algebraic equations. A lot of steps have been shortened in this solution. The solution in the video provides a much faster solution,
Let $r$ be the radius of the circle $C$.
We define a cartesian coordinate system in two dimensions with the circle center at $(0,0)$ and circle equation to be $x^{2}+y{2}=r^{2}$
We define the line $l$ by the equation $y=-r$, with point $M$ at a distance $m$ from the tangent and cartesian coordinates $(m,-r)$
Let $d$ be the distance from point $M$ to point $R$ such that the coordinates for $R$ are $(m+d,-r)$ and thus the coordinates for $Q$ are $(m-d,-r)$
Let points $S$, $T$, and $U$ be the points where lines $PQ$, $PR$, and $l$ are tangent to circle $C$ respectively.
First we get the coordinates for points $S$ and $T$.
Since the circle is the incenter we know the following properties:
$\left| RU \right| = \left| RT \right|=(m+d)$
and
$\left| QU \right| = \left| QS \right|=(m-d)$
Therefore, to get the coordinates of point $T=(T_{x},T_{y})$, we solve the following equations:
$T_{x}^{2}+T_{y}^2=r^{2}$
$\left| RT \right|^{2}=(m+d-T_{x})^{2}+(r+T_{y})^2$
$(m+d)^{2}=(m+d-T_{x})^{2}+(r+T_{y})^2$
After a lot of algebra, this solves to:
$T_{x}=\frac{2r^{2}(m+d)}{(m+d)^{2}+r^{2}}$
$T_{y}=\frac{r\left[ (m+d)^{2}-r^{2} \right]}{(m+d)^{2}+r^{2} }$
Now we calculate the slope of the line that passes through $PR$ which is perpendicular to the line that passes from the center of the circle to point $T$ as follows:
$Slope_{PR}=\frac{-T_{x}}{T_{y}}=\frac{-2r(m+d)}{(m+d)^{2}-r^2)}$
Then, the equation of the line that passes through $PR$ is as follows:
$Line_{PR}\colon \; y+r=\frac{-2r(m+d)}{(m+d)^{2}-r^2)}\left( x-(m+d) \right)$
Now we get the coordinates of point $S=(S_{x},S_{y})$, we solve the following equations:
$S_{x}^{2}+S_{y}^2=r^{2}$
$\left| QT \right|^{2}=(m-d-S_{x})^{2}+(r+S_{y})^2$
$(m-d)^{2}=(m-d-S_{x})^{2}+(r+S_{y})^2$
After a lot of algebra, this solves to:
$S_{x}=\frac{2r^{2}(m-d)}{(m-d)^{2}+r^{2}}$
$S_{y}=\frac{r\left[ (m-d)^{2}-r^{2} \right]}{(m-d)^{2}+r^{2} }$
Now we calculate the slope of the line that passes through $PQ$ which is perpendicular to the line that passes from the center of the circle to point $S$ as follows:
$Slope_{PQ}=\frac{-S_{x}}{S_{y}}=\frac{-2r(m-d)}{(m-d)^{2}-r^2)}$
Then, the equation of the line that passes through $PQ$ is as follows:
$Line_{PQ}\colon \; y+r=\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\left( x-(m-d) \right)$
Now we solve for the coordinates for point $P=(P_{x},P_{y})$ by calculating the intersection of $Line_{PR}$ and $Line_{PQ}$ as follows:
$\frac{-2r(m+d)}{(m+d)^{2}-r^2)}\left( P_{x}-(m+d) \right)=\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\left( P_{x}-(m-d) \right)$
Solving for $P_{x}$ we get:
$P_{x}=\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}$
Solving for $P_{y}$ we get:
$P_{y}=\frac{-2r(m-d)}{(m-d)^{2}-r^2)}\left( \frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}-(m-d) \right)-r$
$P_{y}=\frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}$
Now we need to find the limit of $P_{x}$ and $P_{y}$ as $d$ approaches infinity:
$P_{x_{d \to \infty}}=\lim_{d \to \infty} \frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}=0$
$P_{y_{d \to \infty}}=\lim_{d \to \infty} \frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}=r$
This means that the locus of $P$ starts at point $(0,r)$ on the circle $C$ but that point is not included in the locus as that is the limit.
If we assume that the locus is a ray that starts at $(0,r)$ let's calculate the slope of such ray:
$Slope_{locus}=\frac{P_{y}-r}{P_{x}}$
$Slope_{locus}=\frac{\frac{r(m^{2}-r^{2}-d^{2})}{m^{2}+r^{2}-d^{2}}-r}{\frac{2mr^{2}}{m^{2}+r^{2}-d^{2}}}$
$Slope_{locus}=\frac{-r^{2}}{2m}$
Since the calculated slope of such locus at any point $P$ is not dependent on $d$ and solely dependent on fixed $r$ and $m$, then this proves the slope is fixed and thus the locus is a ray that starts at $(0,r)$ excluding that point and with a slope of $\frac{-r^{2}}{2m}$ in the cartesian coordinate system moving upwards to infinity.
We can also write the equation of the locus as: $y=\frac{-r^{2}}{2m}x+r,\;\;\forall y>r\;$and $x<0$
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
## See Also
1992 IMO (Problems) • Resources Preceded byProblem 3 1 • 2 • 3 • 4 • 5 • 6 Followed byProblem 5 All IMO Problems and Solutions
|
recursion
(algorithmic technique)
Definition: An algorithmic technique where a function, in order to accomplish a task, calls itself with some part of the task.
Specialization (... is a kind of me.)
tail recursion, collective recursion.
Note: Every recursive solution involves two major parts or cases, the second part having three components.
• base case(s), in which the problem is simple enough to be solved directly, and
• recursive case(s). A recursive case has three components:
1. divide the problem into one or more simpler or smaller parts of the problem,
2. call the function (recursively) on each part, and
3. combine the solutions of the parts into a solution for the problem.
Depending on the problem, any of these may be trivial or complex.
Here are some exercises to help you learn recursion. Although recursion may not be the best way to write some of these functions, it is good practice.
1. Write a function to compute the sum of all numbers from 1 to n.
2. Write a function to compute 2 to the power of a non-negative integer.
3. Write a function to compute any number to the power of a non-negative integer.
4. Write a function to compute the nth Fibonacci number.
5. Write a function to compute the greatest common divisor (GCD) of two positive integers with Euclid's algorithm.
6. Write a function to compute GCD based on the following relations
• GCD(2m, 2n) = 2 * GCD(m, n)
• GCD(2m, 2n+1) = GCD(m, 2n+1)
• GCD(2m+1, 2n+1) = GCD(n-m, 2m+1) if m < n
• GCD(m, m) = m
(after "ML for the Working Programmer", page 49).
7. Write a function to compute any number to the power of a non-negative integer using repeated squaring, that is, x2n = (xn)² and x2n+1 = x × x2n. (after "ML for the Working Programmer", pages 45 and 46).
8. Write a function to compute the integer square root of a non-negative integer using square_root(4x) = 2*square_root(x). (after "ML for the Working Programmer", pages 48 and 49).
Authors: PEB,PR
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You are watching: 15 is what percent of 300
## Step by step technique for calculating what percent the 300 is 15
We already have our very first value 300 and also the 2nd value 15. Let"s assume the unknown worth is Y which answer us will find out.
As we have all the forced values us need, currently we have the right to put them in a straightforward mathematical formula as below:
STEP 1Y = 15/300
By multiplying both numerator and also denominator through 100 we will certainly get:
STEP 2Y = 15/300 × 100/100 = 5/100
STEP 3Y = 5
Finally, we have discovered the worth of Y i m sorry is 5 and that is our answer.
You have the right to use a calculator to discover what percent that 300 is 15, just get in 15 ÷ 300 × 100 and you will acquire your answer which is 5
Here is a calculator to solve percentage calculations such as what percent the 300 is 15. You have the right to solve this type of calculation with your worths by beginning them into the calculator"s fields, and click "Calculate" to acquire the an outcome and explanation.
What percent of
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## Sample questions, answers, and also how to
Question: your uncle had actually 300 share of his own firm a few years earlier, and also now he has actually 15 the them. What percent of the share of his agency he has actually now?
Answer: He has 5 percent of shares of his agency now.
How To: The vital words in this problem are "What Percent" due to the fact that they allow us know that it"s the Percent the is missing. For this reason the 2 numbers the it gives us have to be the "Total" and the "Part" we have.
Part/Total = Percent
In this case, it"s the complete that our uncle owned. For this reason we placed 300 ~ above the bottom that the portion and 15 ~ above top. Currently we"re ready to figure out the component we don"t know; the Percent.
See more: Difference Between Real Image And Virtual Image And Virtual Image
15/300 = Percent
To uncover the percent, all we need to do is transform the fraction into its percent type by multiplying both top and also bottom component by 100 and also here is the method to number out what the Percent is:
15/300 × 100/100 = 5/100
5 = Percent
And that way he has actually 5 percent that the shares of his company now.
## Another step by action method
Step 1: Let"s settle the equation for Y by first rewriting the as: 100% / 300 = Y% / 15
Step 2: drop the percent marks to simplify your calculations: 100 / 300 = Y / 15
Step 3: main point both sides by 15 to isolate Y on the appropriate side that the equation: 15 ( 100 / 300 ) = Y
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# Equations & Inequalities
## Presentation on theme: "Equations & Inequalities"— Presentation transcript:
Equations & Inequalities
Revision of Level 2 Algebra Harder Equations Equations with Brackets Equations with Fractions Harder Fractions Solving Inequalities 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Created by Mr. Lafferty Maths Dept.
Starter Questions 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
Revision of Level 2 Learning Intention Success Criteria To revise Level 2 using the rule ‘Balancing Method’ 1. Understand the process of the ‘Balancing Method’ 2. Solving simple algebraic equations. 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Balancing Method www.mathsrevision.com
1-Apr-17 Created by Mr Lafferty Maths Dept Balancing Method Kirsty goes to the shops every week to buy some potatoes. She always buys the same total weight. One week she buys 2 large bags and 1 small bag. The following week she buys 1 large bag and 3 small bags. If a small bags weighs 4 kgs. How much does a large bag weigh? 4 4 What instrument measures balance How can we go about solving this using balance ?
Created by Mr Lafferty Maths Dept
1-Apr-17 Created by Mr Lafferty Maths Dept Balancing Method Take a small bag away from each side. 4 4 4 4 Take a big bag away from each side. We can see that a big bag is equal to 4 + 4 = 8 kg
Balancing Method Lets solve it using maths. Let P be the weigh
1-Apr-17 Created by Mr Lafferty Maths Dept What symbol should we use for the scales ? Balancing Method Lets solve it using maths. 4 P P P Let P be the weigh of a big bag. 4 4 4 We know that a small bag = 4 Subtract 4 from each side 2P + 4 = P + 12 -4 -4 Subtract P from each side 2P = P + 8 -P -P P = 8
Balancing Method www.mathsrevision.com Let a be the price
1-Apr-17 Created by Mr Lafferty Maths Dept Balancing Method Group of 5 adults and 3 children go to the local swimming. Another group of 3 adults and 8 children also go swimming. The total cost for each group is the same. A child’s ticket costs £2. 2 a If a child’s ticket costs £2. How much for an adult ticket ? a 2 a 2 Let a be the price of an adult ticket. a 2 We know that a child price = £2
Balancing Method For balance we have 5a + 6 = 3a + 16 5a = 3a + 10
1-Apr-17 Created by Mr Lafferty Maths Dept Balancing Method Subtract 3a from each side Subtract 6 from each side For balance we have 2 a 5a + 6 = 3a + 16 a 2 -6 -6 2 2 2 1 5a = 3a + 10 a a -3a -3a Divide each side by 2 2a = 10 a = 5 Adult ticket price is £5
Balancing Method It would be far too time consuming to draw out
1-Apr-17 Created by Mr Lafferty Maths Dept Balancing Method It would be far too time consuming to draw out the balancing scales each time. We will now review how to use the rules for solving equations.
Equations & Inequalities
MTH 3-15a MTH 4-15a Revision of Level 2 Reminder ! The Balancing Method Examples x – 5 = 11 x + 8 = 37 + 5 + 5 - 8 - 8 x = 16 x = 29 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
MTH 3-15a MTH 4-15a Revision of Level 2 Reminder ! The Balancing Method Examples 6x = 72 2x = 11 (÷ 6) (÷ 2) x = 12 x = 5.5 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
MTH 3-15a MTH 4-15a Revision of Level 2 Examples 4x – 2 = 34 2x + 9 = 31 + 2 + 2 + 9 + 9 4x = 36 2x = 22 (÷ 4) (÷ 2) x = 9 x = 11 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
MTH 3-15a MTH 4-15a Revision of Level 2 Now try Exercise 1 Ch43 (page 171) 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Created by Mr. Lafferty Maths Dept.
Starter Questions 2x 6x 100o 60o 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
Harder Equations Learning Intention Success Criteria 1. We are learning how to apply balancing method to solve harder equations. Apply balancing method to solve equations with more than one x term. 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
Each group has 2 minutes come up with a method for solving this type of equation Equations & Inequalities MTH 3-15a MTH 4-15a Harder Equations Examples 3x + 2 = x + 10 4x - 5 = 2x - 25 - 2 - 2 + 5 + 5 3x = x + 8 4x = 2x – 20 - x - x - 2x - 2x 2x = 8 2x = -20 (÷ 2) (÷ 2) x = 4 x = -10 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
MTH 3-15a MTH 4-15a Examples Harder Equations x + 4 = 3x - 10 4x - 1 = 7x + 11 - 4 - 4 + 1 + 1 Swap Round ! Swap Round ! x = 3x - 14 4x = 7x + 12 3x - 14 = x 7x + 12 = 4x + 14 + 14 - 12 - 12 3x = x + 14 7x = 4x - 12 - x - x - 4x - 4x 2x = 14 3x = -12 (÷ 2) x = 7 (÷ 3) x = -4 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
Harder Equations MTH 3-15a MTH 4-15a Now try Exercise 2 Ch43 (page 173) 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Created by Mr. Lafferty Maths Dept.
Starter Questions 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
Equations with Brackets Learning Intention Success Criteria 1. Remember rule for Multiply out brackets. 1. We are learning how to solve equations with brackets . 2. Apply ‘Balancing Method’ to solve equations. 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
Each group has 2 minutes come up with a method for solving this type of equation Equations & Inequalities Equations with Brackets MTH 3-15a MTH 4-15a Multiply out the brackets first and then Balancing Method Example 3(x – 2) = 33 3x - 6 = 33 + 6 + 6 3x = 39 (÷ 3) x = 13 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
Equations with Brackets MTH 3-15a MTH 4-15a Example 5(3x + 2) – 2(4x - 3) = 2x + 36 15x + 10 – 8x + 6 = 2x + 36 Tidy up terms ! 7x = 2x + 36 - 16 - 16 7x = 2x + 20 - 2x - 2x 5x = 20 (÷ 5) x = 4 Created by Mr. Lafferty Maths Dept. 1-Apr-17
Equations & Inequalities
Equations with Brackets MTH 3-15a MTH 4-15a Now try Exercise 3 Ch43 (page 174) 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Created by Mr. Lafferty Maths Dept.
Starter Questions 5cm 11cm 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
Equations with Fractions Learning Intention Success Criteria 1. Remember to multiply every term to get rid of fractional term. 1. We are learning how to solve equations that contain fractional terms. 2. Apply ‘Balancing Method’ 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
Each group has 2 minutes come up with a method for solving this type of equation Equations & Inequalities Equations with Fractions MTH 3-15a MTH 4-15a Multiply EVERY term to get rid of fractional term. and then apply ‘Balancing Method’ Example x + 3 = 7 1 2 (x 2) x + 6 = 14 - 6 - 6 x = 8 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
Equations with Fractions MTH 3-15a MTH 4-15a Multiply EVERY term to get rid of fractional term. and then apply ‘Balancing Method’ Example x - = 4 3 4 1 2 (x 4) 3x - 2 = 16 + 2 + 2 3x = 18 (÷ 3) x = 6 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
Equations with Fractions MTH 3-15a MTH 4-15a Now try Exercise 4 Ch43 (page 175) 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Created by Mr. Lafferty Maths Dept.
Starter Questions 7cm 13cm 15cm 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
Harder Fractional Equations with Brackets Learning Intention Success Criteria 1. Apply Balancing Method to solve harder equations. 1. We are learning how to solve HARDER fractional equations using all the rules learned so far. 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
MTH 3-15a MTH 4-15a Harder Fractional Equations with Brackets Multiply EVERY term to get rid of fractional term. and then apply ‘Balancing Method’ Example + 5 = 8 x + 2 4 (x 4) x = 32 Tidy Up ! x + 22 = 32 - 22 - 22 x = 10 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
MTH 3-15a MTH 4-15a Harder Fractional Equations with Brackets Example (2x – 4) x = 16 3 2 22x = 132 (÷ 22) (x 6) x = 6 (2x – 4) + x = 96 9 4 Tidy Up ! 18x - 36 + 4x = 96 22x - 36 = 96 + 36 + 36 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
MTH 3-15a MTH 4-15a Harder Fractional Equations with Brackets Now try Exercise 5 Ch43 (page 176) 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Created by Mr. Lafferty Maths Dept.
Starter Questions 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
Solving Inequalities Learning Intention Success Criteria 1. Understand the term inequality. 1. We are learning how we can solve inequalities using the same rules we use for equations. 2. Solve inequalities using the same method as equations. 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
Solving Inequalities MTH 3-15a MTH 4-15a The Good News Inequalities are similar to equations except we replace the “=“ with one of the following symbols : Less than Greater than or equal to Less than or equal to Greater than 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
MTH 3-15a MTH 4-15a Solving Inequalities Even Better News ! Solving inequalities is almost identical to solving equations : Example 2x – 1 < 7 NEVER put in an equal sign + 1 + 1 2x < 8 (÷ 2) x < 4 x is any value less than 4 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
MTH 3-15a MTH 4-15a Solving Inequalities Example 2(2x – 3) ≥ x + 9 NEVER put in an equal sign 4x – 6 ≥ x + 9 + 6 + 6 4x ≥ x + 15 - x - x 3x ≥ 15 x is any value greater than or equal to 5 (÷ 3) x ≥ 5 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
MTH 3-15a MTH 4-15a Solving Inequalities The only one to watch out for is when you are dividing by a negative Example 8 – 3m < 2 - 8 - 8 -3m < -6 FLIP sign (÷ -3) m > 2 1-Apr-17
Equations & Inequalities
MTH 3-15a MTH 4-15a Solving Inequalities Example 5(x – 1) - 8x ≥ -17 Tidy Up ! 5x – 5 - 8x ≥ -17 -3x - 5 ≥ -17 + 5 + 5 FLIP sign -3x ≥ -12 (÷ -3) x is any value less than or equal to 4 x ≤ 4 1-Apr-17 Created by Mr. Lafferty Maths Dept.
Equations & Inequalities
MTH 3-15a MTH 4-15a Solving Inequalities Now try Exercise 6 Ch43 (page 177) 1-Apr-17 Created by Mr. Lafferty Maths Dept.
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# What is the LCM and HCF of 5 15 and 20?
The LCM of 5, 15 and 20 is 60 and HCF of 5, 15 and 20 is 5.
## What is the LCM of 5 and 15 and 20?
What is the LCM of 5, 15, and 20? Answer: LCM of 5, 15, and 20 is 60.
## What is the highest common factor of 5 15 and 20?
The HCF of 5, 15, and 20 is 5. ∴ The highest number that divides 5, 15, and 20 is 5.
## What are the common factors of 5 15 and 20?
As you can see when you list out the factors of each number, 5 is the greatest number that 5, 15, and 20 divides into.
## What is the HCF of 5 15?
There are 2 common factors of 5 and 15, that are 1 and 5. Therefore, the greatest common factor of 5 and 15 is 5.
## find the ratio between LCM and HCF of 5, 15 and 20
30 related questions found
### What is the HCF of 5 15 and 25?
Hence, the required common factors of 5, 15 and 25 are 1 and 5.
### What is the HCF of 5 10 and 15?
The LCM of 5, 10 and 15 is 30 and the HCF of 5, 10 and 15 is 5.
### What is the LCM of 5 25 20?
Answer: LCM for 5, 20 and 25 is 100.
### What is the HCF of 5 and 20?
As visible, 5 and 20 have only one common prime factor i.e. 5. Hence, the GCF of 5 and 20 is 5.
### What is the LCM of 5 15?
The LCM of 5 and 15 is 15.
### What is the LCM of 5 15 20 30?
The least common multiple of 5, 15, 20 and 30 is 60.
### What is the HCF of 5 10 15 and 20?
As you can see when you list out the factors of each number, 5 is the greatest number that 5, 10, 15, and 20 divides into.
### How to calculate HCF?
How to Find HCF?
1. Step 1: Write each number as a product of its prime factors. This method is called here prime factorization.
2. Step 2: Now list the common factors of both the numbers.
3. Step 3: The product of all common prime factors is the HCF ( use the lower power of each common factor)
### What is the LCM and HCF of 10 15 20?
No. LCM of 10, 15 and 20 is 60 and HCF of 10, 15 and 20 is 5.
### What is the LCM of 5 10 15 and 20?
The LCM of 5, 10, 15 and 20 is 60.
### What is the LCM of 5 25 and 15?
Answer: LCM for 5, 15 and 25 is 75.
### What is the LCM of 5 10 15 and 25?
Solution: The smallest number that is divisible by 10, 15, 25 exactly is their LCM. The LCM of 10, 15, 25 is 150.
### What is the LCM and HCF of 5 and 25?
LCM of 5 and 25 is 25. Students can grasp the method of finding the least common multiple of 5 and 25 from the common multiples. (5, 10, 15, 20, 25, 30, ….) and (25, 50, 75, 100, 125,….)
### What is the LCM of 5 and 15 and 10?
The LCM of 5, 10, and 15 is 30.
### What is the HCF of 5 10 and 20?
∴ HCF of the given numbers = 5.
### What is the HCF of 5 15 and 45?
List all the factors for 5,15,45 5 , 15 , 45 to find the common factors. The common factors for 5,15,45 5 , 15 , 45 are 1,5 . The GCF (HCF) of the numerical factors 1,5 is 5 .
### What is the LCM and HCF of 15 and 21?
LCM of 15 and 21 is 105. The smallest number among all common multiples of 15 and 21 is the LCM of 15 and 21. (15, 30, 45, 60, 75, 90, etc.)
### What is the HCF and LCM of 15 25 and 30?
HCF of 15, 25 and 30 by Prime Factorisation Method
The only Common Prime Factor is 5. Therefore, HCF (15, 25 and 30) = 5.
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# Ellipses
### Parts of an Ellipse
An ellipse is a conic section defined by two fixed points. The sum of the distances between these two fixed points and any point on the ellipse is constant.
A focus is a fixed point used to generate a conic section. An ellipse is the set of points such that the sum of the distances from two fixed points, or foci, remains the same. It has a center and passes through two points called vertices and two points called co-vertices. A vertex of an ellipse is one of two points on the ellipse that are endpoints of the major axis. A co-vertex is one of two points on an ellipse that are endpoints of the minor axis.
Properties of an ellipse include:
• The major axis is the segment that passes through the foci with endpoints on the ellipse. The endpoints are the vertices of the ellipse.
• The minor axis is the segment with endpoints on an ellipse that is perpendicular to the major axis. The endpoints are the co-vertices of the ellipse.
• The length of the major axis is always greater than or equal to the length of the minor axis.
• The major and minor axes intersect at the center of the ellipse.
• If the lengths of the major and minor axes are equal, then the ellipse is a circle with both foci at the center.
• The major and minor axes each act as an axis of symmetry for the ellipse, meaning that they divide the figure into two halves that are mirror images.
• The distance is the same from the center to each focus.
The equation of an ellipse can be written in terms of its center, vertices, and co-vertices. An ellipse can have a major axis that is vertical or horizontal. For an ellipse in standard form, if the denominator of the $x$-term is greater than the denominator of the $y$-term, then the major axis is horizontal. If the denominator of the $y$-term is greater, then the major axis is vertical.
### Equation of an Ellipse
Horizontal Major Axis Centered at the Origin Vertical Major Axis Centered at the Origin
For real numbers $a$ and $b$, where $a\gt b$, the equation is:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
For real numbers $a$ and $b$ where $a>b$, the equation is:
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
The coordinates of the vertices are $(-a,0)$ and $(a,0)$. The coordinates of the vertices are $(0, -a)$ and $(0,a)$.
The coordinates of the co-vertices are $(0,-b)$ and $(0,b)$. The coordinates of the co-vertices are $(-b,0)$ and $(b,0)$.
The coordinates of the foci are $(c,0)$ and $(-c,0)$, where:
$b^2=a^2-c^2$
The coordinates of the foci are $(0,c)$ and $(0,-c)$, where:
$b^2=a^2-c^2$
The standard form for an equation of an ellipse with center $(h, k)$ is written using $(x-h)$ in place of $x$ and $(y-k)$ in place of $y$.
Horizontal major axis:
$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$
Vertical major axis:
$\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$
### Graphing Ellipses
An ellipse with a given equation can be graphed in the coordinate plane by locating the center, vertices, and co-vertices.
An ellipse can have its center at any point on the coordinate plane, and its major axis may be horizontal or vertical. To graph an ellipse given in standard form:
1. Use the form to determine whether the major axis is horizontal or vertical. For real numbers $a$ and $b$ where $a>b$:
• The major axis is horizontal if the equation has the form:
$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$
• The major axis is vertical if the equation has the form:
$\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$
2. Use the equation to find the center, $(h,k)$.
3. From the center, use the values of $a$ and $b$ to locate the vertices and co-vertices. Then sketch the curve through these points.
• If the major axis is horizontal, the vertices are $a$ units to the left and right of the center. The co-vertices are $b$ units above and below the center.
• If the major axis is vertical, the vertices are $a$ units above and below the center. The co-vertices are $b$ units to the left and right of the center.
4. To locate the foci, solve the equation $c^2=a^2-b^2$ to find the value of $c$.
• If the major axis is horizontal, the foci are $c$ units to the left and right of the center.
• If the major axis is vertical, the foci are $c$ units above and below the center.
### Ellipses in the Coordinate Plane
Horizontal Major Axis Vertical Major Axis
Equation
$\frac{x^2}{36}+\frac{y^2}{4}=1$
$\frac{x^2}{16}+\frac{y^2}{25}=1$
Center
$(0, 0)$
$(0, 0)$
Vertices $(-6, 0)$ and $(6, 0)$ $(0, 5)$ and $(0, -5)$
Co-vertices $(0, 2)$ and $(0, -2)$ $(-4, 0)$ and $(4, 0)$
Foci $(-4\sqrt{2},0)$ and $(4\sqrt{2},0)$ $(0,3)$ and $(0,-3)$
Graph
Step-By-Step Example
Graphing an Ellipse in Standard Form
Graph the ellipse with the equation:
$\frac{(x-1)^2}{9}+\frac{(y+2)^2}{25}=1$
Step 1
Compare the equation to the standard form for the equation of an ellipse. The denominator of the fraction containing $y$ is greater than the denominator of the fraction containing $x$. So, the standard form is:
$\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$
This means that the major axis is vertical.
$\frac{(x-1)^2}{9}+\frac{(y+2)^2}{25}=1$
The values of $h$, $k$, $b^{2}$, and $a^{2}$ are:
\begin{aligned}h&=1 \\k&=-2 \\b^2 &= 9 \\a^2&=25 \end{aligned}
Step 2
Determine the center of the ellipse and the lengths of $a$ and $b$.
Since $h=1$ and $k=-2$, the center is $(1,-2)$.
For the length of $b$:
\begin{aligned}b^{2}&=9\\b&=\sqrt{9}\\&=3\end{aligned}
For the length of $a$:
\begin{aligned}a^{2}&=25\\a&=\sqrt{25}\\&=5\end{aligned}
Step 3
Determine the vertices and co-vertices. The vertices are both 5 units from the center. The major axis is vertical, so the vertices are above and below the center.
Vertices:
\begin{aligned}(1, -2+5)&=(1,3) \\(1, -2-5)&=(1,-7) \end{aligned}
The minor axis is horizontal, and the co-vertices are both 3 units from the center. Co-vertices:
\begin{aligned}(1+3, -2)&=(4,-2)\\(1-3, -2)&=(-2,-2) \end{aligned}
Step 4
Identify the foci.
To locate the foci, use this equation:
$c^2=a^2-b^2$
Then determine the value of $c$ by using the values $a=5$ and $b=3$ from Step 2.
\begin{aligned}c^2&=a^2-b^2\\&=5^2-3^2\\&=25-9\\&=16\\c&=\pm4\end{aligned}
The major axis is vertical. So the foci are $c$ units above and below the center:
\begin{aligned}(1, -2-4)&=(1,-6)\\(1, -2+4)&=(1,2)\end{aligned}
Solution
By Hand: Plot the center, vertices, co-vertices, and foci. Draw a smooth curve through the vertices and co-vertices.
On a Graphing Calculator: Solve for $y$:
\begin{aligned}\frac{(x-1)^2}{9}+\frac{(y+2)^2}{25}&=1\\y&=\pm\sqrt{25-25\frac{(x-1)^2}{9}}-2\end{aligned}
Graph both semi-ellipses:
\begin{aligned}Y_1&=\sqrt{25-25\frac{(x-1)^2}{9}}-2 \\ Y_2&=-\sqrt{25-25\frac{(x-1)^2}{9}}-2\end{aligned}
Step-By-Step Example
Graphing an Ellipse by Completing the Square
Graph the ellipse:
$4x^2+9y^2+24x-72y+144=0$
Step 1
Reorder the terms so that like terms are together and move the constant term to the right side of the equation.
\begin{aligned}4x^2+9y^2+24x-72y+144&=0\\4x^2+24x+9y^2-72y&=-144\end{aligned}
Step 2
Factor the $x$-terms and the $y$-terms by grouping so that the coefficient of both squared terms is 1.
$4(x^2+6x)+9(y^2-8y)=-144$
Step 3
Identify the terms that need to be added to complete the square for the $x$-terms and the $y$-terms.
$4({\color{#c42126} x^2+6x+\underline{\ \ \ \ }})+ 9({\color{#0047af} y^2-8y+\underline{\ \ \ \ }})=-144+4({\color{#c42126} \underline{\ \ \ \ }})+9({\color{#0047af} \underline{\ \ \ \ }})$
The missing term for each variable is a number that will make each expression a perfect square. To find this number, divide the coefficient of the middle term by 2 and square the result. For the $x$-terms:
$\left(\frac{6}{2}\right)^2=9$
For the $y$-terms:
$\left(\frac{-8}{2}\right)^2=16$
Step 4
Complete the square for each variable, adding the value to both sides of the equation. Then, factor the expressions on the left side into perfect squares and simplify the right side to write the equation in standard form.
When adding to both sides, remember to account for the coefficients by multiplying the amount added to the $x$-terms by 4 and to the $y$-terms by 9.
\begin{aligned}4({\color{#c42126} x^2+6x+\underline{\ \ \ \ }})+ 9({\color{#0047af} y^2-8y+\underline{\ \ \ \ }})&=-144+4({\color{#c42126} \underline{\ \ \ \ }})+9({\color{#0047af} \underline{\ \ \ \ }})\\4({\color{#c42126} x^2+6x+9})+9({\color{#0047af} y^2-8y+16})&=-144+4({\color{#c42126} 9})+9({\color{#0047af} 16})\\4({\color{#c42126} x+3})^2+9({\color{#0047af} y-4})^2&=36\end{aligned}
Step 5
Divide both sides of the equation by 36 to write the equation in standard from, with the right side equal to 1.
\begin{aligned}4(x+3)^2+9(y-4)^2&=36\\\frac{4(x+3)^2}{36}+\frac{9(y-4)^2}{36}&=1\\\frac{(x+3)^2}{9}+\frac{(y-4)^2}{4}&=1\end{aligned}
Step 6
Identify the center, vertices, and co-vertices.
$\frac{(x+3)^2}{9}+\frac{(y-4)^2}{4}=1$
The center of the ellipse is at $(-3, 4)$. Since $9>4$, the major axis is horizontal. Identify the values of $a$ and $b$.
\begin{aligned}a=\sqrt{9}=3 \\ b=\sqrt{4}=2\end{aligned}
The vertices are 3 units left and right of the center:
\begin{aligned}(-3-3, 4)=(-6,4) \\ (-3+3, 4)=(0,4)\end{aligned}
The co-vertices are 2 units above and below the center:
\begin{aligned}(-3, 4+2)=(-3,6)\\ (-3, 4-2)=(-3,2)\end{aligned}
Step 7
Identify the foci.
To locate the foci, solve the equation $c^2=a^2-b^2$ to find the value of $c$. Use the values $a=3$ and $b=2$ from Step 6.
\begin{aligned}c^2&=a^2-b^2\\&=3^2-2^2\\&=9-4\\&=5\\c&=\pm\sqrt{5}\end{aligned}
The major axis is horizontal. So, the foci are $c$ units to the left and right of the center: $(-3-\sqrt{5}, 4)$ and $(-3+\sqrt{5}, 4)$, or about $(-5.2, 4)$ and $(-0.8, 4)$.
Solution
By Hand: Plot the center, vertices, co-vertices, and foci. Draw a smooth curve through the vertices and co-vertices.
On a Graphing Calculator: Solve for $y$:
\begin{aligned}\frac{(x+3)^2}{9}+\frac{(y-4)^2}{4}&=1\\y&=\pm\sqrt{4-\frac{4(x+3)^2}{9}}+4\end{aligned}
Graph both semi-ellipses:
\begin{aligned}Y_1&=\sqrt{4-\frac{4(x+3)^2}{9}}+4 \\ Y_2&=-\sqrt{4-\frac{4(x+3)^2}{9}}+4\end{aligned}
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# Illustrating the Mathematics Standards
#### Interlocking Photo Frames: Illustrating the year 5 standard
The following examples of student work illustrate achievement at the mathematics standard for year 5.
The task used in this illustration relates to achievement objectives for Number and Algebra from the mathematics and statistics learning area in The New Zealand Curriculum. It was adapted from an activity in Figure It Out, Algebra: Book One, Link (years 7–8), pages 12–13.
### The Task
#### Interlocking Photo Frames
Nikki’s mum is on the organising committee for the new community centre. They want to hang photos of local people, community activities, and historical places around the walls.
Nikki’s mum suggests that they use a series of interlocking photo frame pieces. (Each frame has 4 pieces, but two frames “share” a piece when they are joined together.) The committee try this idea with 3 photos and find that they would need 10 frame pieces.
1. If 10 frame pieces are needed for 3 photos, how many pieces are needed for 8 photos?
2. Can you find a rule for the number of frame pieces needed for any number of photos? Apply your rule to different numbers of photos.
3. How many frame pieces are needed for 99 photos?
Some features of students’ work used to make judgments in relation to the year 5 mathematics standard are described below. There is also an illustration of the year 6 standard for this task.
New Zealand Curriculum: Level 3 National Standards: By the end of year 5 In solving problems and modelling situations, students will: Number and Algebra use a range of additive and simple multiplicative strategies with whole numbers, fractions, decimals, and percentages (number strategies) know basic multiplication and division facts (number knowledge) connect members of sequential patterns with their ordinal position and use tables, graphs, and diagrams to find relationships between successive elements of number and spatial patterns (patterns and relationships) Number and Algebra apply additive and simple multiplicative strategies and knowledge of symmetry to: - combine and partition whole numbers - find fractions of sets, shapes, and quantitites describe spatial and number patterns, using rules that involve spatial features, repeated addition or subtraction, and simple multiplication
Hover over the image to zoom a section. Click on the image to enlarge it. Click again to close.
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# Fraction calculator
This fraction calculator performs basic and advanced fraction operations, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. The calculator helps in finding value from multiple fractions operations. Solve problems with two, three, or more fractions and numbers in one expression.
## The result:
### 156/25 - 115/10 = 187/50 = 3 37/50 = 3.74
Spelled result in words is one hundred eighty-seven fiftieths (or three and thirty-seven fiftieths).
### How do we solve fractions step by step?
1. Conversion a mixed number 15 6/25 to a improper fraction: 15 6/25 = 15 6/25 = 15 · 25 + 6/25 = 375 + 6/25 = 381/25
To find a new numerator:
a) Multiply the whole number 15 by the denominator 25. Whole number 15 equally 15 * 25/25 = 375/25
b) Add the answer from the previous step 375 to the numerator 6. New numerator is 375 + 6 = 381
c) Write a previous answer (new numerator 381) over the denominator 25.
Fifteen and six twenty-fifths is three hundred eighty-one twenty-fifths.
2. Conversion a mixed number 11 5/10 to a improper fraction: 11 5/10 = 11 5/10 = 11 · 10 + 5/10 = 110 + 5/10 = 115/10
To find a new numerator:
a) Multiply the whole number 11 by the denominator 10. Whole number 11 equally 11 * 10/10 = 110/10
b) Add the answer from the previous step 110 to the numerator 5. New numerator is 110 + 5 = 115
c) Write a previous answer (new numerator 115) over the denominator 10.
Eleven and five tenths is one hundred fifteen tenths.
3. Subtract: 381/25 - 115/10 = 381 · 2/25 · 2 - 115 · 5/10 · 5 = 762/50 - 575/50 = 762 - 575/50 = 187/50
It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(25, 10) = 50. It is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 25 × 10 = 250. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - three hundred eighty-one twenty-fifths minus one hundred fifteen tenths is one hundred eighty-seven fiftieths.
#### Rules for expressions with fractions:
Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
SymbolSymbol nameSymbol MeaningExample
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4)
The calculator follows well-known rules for the order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule.
Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right.
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# How do you find the derivative of f(x) = x²(3x³-1) and the given point is (1,2)?
Feb 21, 2018
13
#### Explanation:
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} f \left(x\right)$
$= \frac{d}{\mathrm{dx}} {x}^{2} \left(3 {x}^{3} - 1\right)$ color(white)(dwwwwwdd $\left[\text{applying product rule}\right]$
$= \left(\frac{d}{\mathrm{dx}} {x}^{2}\right) \left(3 {x}^{3} - 1\right) + {x}^{2} \left(\frac{d}{\mathrm{dx}} 3 {x}^{3} - \frac{d}{\mathrm{dx}} 1\right)$
$= 2 x \left(3 {x}^{3} - 1\right) + {x}^{2} \left(9 {x}^{2}\right)$
$= 6 {x}^{4} - 2 x + 9 {x}^{4}$
$= 15 {x}^{4} - 2 x$
So, Derivative of f(x) at (1, 2) = $f ' \left(1\right) = 15 {\left(1\right)}^{4} - 2 \cdot 1 = 15 - 2 = 13$ color(white)(dwww(just replace $x = 1$)
Hence Explained.
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The K5 Learning Blog urges parents to be pro-active in helping their children reach their full academic potential.
K5 Learning
provides an online reading and math program for kindergarten to grade 5 students.
How to Ace Math Problem Solving
When your kids struggle with their math, it’s time to take a step back and take a deep breath. They need to slow down and take their time. Here’s a step by step guide that will help your kids get through those tough math problems.
We’ll use a grade 3 addition word problem as an example to clarify:
Pinky the Pig bought 36 apples while Danny the Duck bought 73 apples and 14 bananas. How many apples do they have altogether?
Carefully read through the problem to make sure you understand what is being asked.
Pinky the pig and Danny the duck bought apples and bananas. The question is how many apples they have together.
Read through the problem again and as you read through it, make notes.
Pinky the pig –36 apples.
Danny the duck –73 apples and 14 bananas.
How many apples together?
In your own words, say or write down exactly what the question is asking you to solve.
The question is asking how many apples the pig and the duck bought together.
Write it down in detail
Go through the problem and write out the information in an organized fashion. A diagram or table might help.
Turn it into math
Figure out what math operation(s) or formula(s) you need to use in order to solve this problem.
The problem wants us to add the number of apples Pinky the Pig and Danny the Duck have together. That means we need to make use of addition to add the apples.
Find an example
Are you still struggling? Sometimes it’s hard to work out the solution, especially if the math problem involves several steps. It’s time to present the problem in an easier way. As teachers and parents we can often help our kids simplify the problem from our own math knowledge. If the problem is a bit harder, there are lots of resources online that you can look up for similar problems that have been worked out on paper or a video tutorial to watch.
In our example, let’s say the double-digit numbers are intimidating our student, so we’re going to simplify the equation for the sake of helping our student understand the operation needed.
Let’s say Pinky the Pig bought 3 apples and Danny the Duck 7 apples and 1 banana. Now, how many apples have they bought together? With 3 apples and 7 apples bought, the total number of apples is 10.
Work out the problem
Now that we have got to the bottom of what is being asked and know what operation to use, it’s time to work out the problem.
Pinky the Pig bought 36 apples.
Danny the Duck bought 73 apples. (The 14 bananas do not matter)
We need to add up the apples.
36 + 73 = 109
Let’s round the numbers: 30 + 70 = 100. That is close to the exact number so it’s in the correct range.
The beauty of the basic operations is that addition and subtraction can be used to check answers too.
If we use the sum and take away one of the numbers, it should equal the other number.
109 – 73 = 36
109 – 36 = 73
If our student did not work out the sum correctly, we would not come to these sums.
(By the way, the same can be done with multiplication and division.)
Finally, go back and review the problem one last time. By going over the concepts, operations and formulas, it will help your kids to internalize the process and help them tackle harder math problems in the future.
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## College Algebra (6th Edition)
Solution set: $(-\displaystyle \infty, -\frac{1}{2})\cup(\frac{1}{3},\infty)$
Follow the "Procedure for Solving Polynomial lnequalities",\ p.412: 1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$ where $f$ is a polynomial function. $6x^{2}+x >1$ $6x^{2}+x-1>0$ $f(x)= 6x^{2}+x-1$ factor the trinomial... find factors of $6(-1)=-6$ that add to $+1:$ ($3$ and $-2$ ) $6x^{2}+x-1=6x^{2}+3x-2x-1= \quad$ ... in pairs ... $=3x(2x+1)-(2x+1)=(3x-1)(2x+1)$ $f(x)=(3x-1)(2x+1)>0$ 2. Solve the equation $f(x)=0$. The real solutions are the boundary points. $(3x-1)(2x+1)=0$ $x=-\displaystyle \frac{1}{2} x=\displaystyle \frac{1}{3}$ 3. Locate these boundary points on a number line, thereby dividing the number line into intervals. 4. Test each interval's sign of $f(x)$ with a test value, $\begin{array}{llll} Intervals: & (-\infty, -\frac{1}{2}) & (-\frac{1}{2},\frac{1}{3}) & (\frac{1}{3},\infty)\\ a=test.val. & -1 & 0 & 1\\ f(a) & (-4)(-1) & (-1)(1) & (2)(3)\\ f(a) > 0 ? & T & F & T \end{array}$ 5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points. Solution set: $(-\displaystyle \infty, -\frac{1}{2})\cup(\frac{1}{3},\infty)$
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# How do you solve and write the following in interval notation: x^3-3x^2-4x<0?
Aug 25, 2017
The solution is $x \in \left(- \infty , - 1\right) \cup \left(0 , 4\right)$
#### Explanation:
We start by factorising the inequality
${x}^{3} - 3 {x}^{2} - 4 x < 0$
$x \left({x}^{2} - 3 x + 4\right) < 0$
$x \left(x + 1\right) \left(x - 4\right) < 0$
Let $f \left(x\right) = x \left(x + 1\right) \left(x - 4\right)$
$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a a}$$4$$\textcolor{w h i t e}{a a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a}$$x + 1$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x - 4$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$
Therefore,
$f \left(x\right) < 0$ when $x \in \left(- \infty , - 1\right) \cup \left(0 , 4\right)$
graph{x^3-3x^2-4x [-20, 20.55, -14.88, 5.4]}
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# Class 12 NCERT Solutions- Mathematics Part I – Chapter 3 Matrices – Miscellaneous Exercise on Chapter 3
### Question 1: Let , show that (aI + bA)n = an I + nan – 1 bA, where I is the identity matrix of order 2 and n ∈ N.
Solution:
Using mathematical induction,
Step 1: Let’s check for n=1
(aI + bA)n = (aI + bA)1 = (aI + bA)
anI + nan – 1 bA = aI + 1a1 – 1 bA = (aI + bA)
It is true for P(1)
Step 2: Now take n=k
(aI + bA)k = akI + kak – 1 bA …………………(1)
Step 3: Let’s check whether, its true for n = k+1
(aI + bA)k+1 = (aI + bA)k (aI + bA)
= (akI + kak – 1 bA) (aI + bA)
= ak+1I×I + kak bAI + ak bAI + kak-1 b2AA
AA =
= ak+1I×I + kak bAI + ak bAI + 0
= ak+1I + (k+1)ak+1-1 bA
= P(k+1)
Hence, P(n) is true.
### Question 2: If , prove that
Solution:
Using mathematical induction,
Step 1: Let’s check for n=1
It is true for P(1)
Step 2: Now take n=k
Step 3: Let’s check whether, its true for n = k+1
= P(k+1)
Hence, P(n) is true.
### Question 3: If , prove that ,where n is any positive integer.
Solution:
Using mathematical induction,
Step 1: Let’s check for n=1
It is true for P(1)
Step 2: Now take n=k
Step 3: Let’s check whether, its true for n = k+1
= P(k+1)
Hence, P(n) is true.
### Question 4. If A and B are symmetric matrices, prove that AB – BA is a skew-symmetric matrix.
Solution:
As, it is mentioned that A and B are symmetric matrices,
A’ = A and B’ = B
(AB – BA)’ = (AB)’ – (BA)’ (using, (A-B)’ = A’ – B’)
= B’A’ – A’B’ (using, (AB)’ = B’A’)
= BA – AB
(AB – BA)’ = – (AB – BA)
Hence, AB – BA is a skew symmetric matrix
### Question 5. Show that the matrix B′AB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric.
Solution:
Let’s take A as symmetric matrix
A’ = A
Then,
(B′AB)’ = {B'(AB)}’
= (AB)’ (B’)’ (using, (AB)’ = B’A’)
= B’A’ (B) (using, (AB)’ = B’A’ and (B’)’ = B)
= B’A B
As, here (B′AB)’ = B’A B. It is a symmetric matrix.
Let’s take A as skew matrix
A’ = -A
Then,
(B′AB)’ = {B'(AB)}’
= (AB)’ (B’)’ (using, (AB)’ = B’A’)
= B’A’ (B) (using, (AB)’ = B’A’ and (B’)’ = B)
= B'(-A) B
= – B’A B
As, here (B′AB)’ = -B’A B. It is a skew matrix.
Hence, we can conclude that B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.
### Question 6. Find the values of x, y, z if the matrix satisfy the equation A′A = I
Solution:
A’A =
By evaluating the values, we have
2x2 = 1
x = ±
6y2 = 1
y = ±
3z2 = 1
z = ±
Solution:
Solution:
A2 – 5A + 7I =
Hence proved!
Solution:
### (a) If unit sale prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 1.00, respectively, find the total revenue in each market with the help of matrix algebra.
Solution:
Total revenue in market I and II can be arranged from given data as follows:
After multiplication, we get
Hence, the total revenue in Market I and market II are ₹ 46,000 and ₹ 53,000 respectively.
### (b) If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.
Solution:
Total cost prices of all the products in market I and market II can be arranged from given data as follows:
After multiplication, we get
As, Profit earned = Total revenue – Cost price
Profit earned
Profit earned =
Hence, profit earned in Market I and market II are ₹ 15,000 and ₹ 17,000 respectively. Which is equal to ₹ 32,000
### Question 11. Find the matrix X so that
Solution:
Here, the RHS is a 2×3 matrix and LHS is 2×3. So, X will be 2×2 matrix.
Let’s take X as,
Now solving the matrix, we have
Equating each of them, we get
p+4q = -7 ………..(1)
2p+5q = -8 ………….(2)
3p + 6q = -9
r + 4s = 2 …………(3)
2r + 5s = 4 ……………(5)
3r + 6s = 6
Solving (1) and (2), we get
p = 1 and q = -2
Solving (3) and (4), we get
r = 2 and s = 0
Hence, matrix X is
### Question 12: If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABn = BnA. Further, prove that (AB)n = AnBn for all n ∈ N.
Solution:
Using mathematical induction,
Step 1: Let’s check for n=1
ABn = AB1 = AB
BnA = B1A = BA
It is true for P(1)
Step 2: Now take n=k
ABk = BkA
Step 3: Let’s check whether, its true for n = k+1
AB(k+1) = ABkB
= BkAB
= Bk+1 A
= P(k+1)
Hence, P(n) is true.
Now, for (AB)n = AnBn
Using mathematical induction,
Step 1: Let’s check for n=1
(AB)1 = AB
B1A1 = BA
It is true for P(1)
Step 2: Now take n=k
(AB)k = AkBk
Step 3: Let’s check whether, its true for n = k+1
(AB)(k+1) = (AB)k(AB)
= AkBk AB
= Ak+1 Bk+1
= (AB)k+1
= P(k+1)
Hence, P(n) is true.
### Question 13: If is such that A² = I, then
(A) 1 + α² + βγ = 0
(B) 1 – α² + βγ = 0
(C) 1 – α² – βγ = 0
(D) 1 + α² – βγ = 0
Solution:
As, A2 = I
α² + βγ = 1
1 – α² – βγ = 0
Hence, Option (C) is correct.
### Question 14. If the matrix A is both symmetric and skew symmetric, then
(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these
Solution:
If the matrix A is both symmetric and skew symmetric, then
A = A’
and A = -A
Only zero matrix satisfies both the conditions.
Hence, Option (B) is correct.
### Question 15. If A is square matrix such that A2 = A, then (I + A)³ – 7 A is equal to
(A) A
(B) I – A
(C) I
(D) 3A
Solution:
(I + A)³ – 7 A = I3 + A3 + 3A^2 + 3AI^2 – 7A
= I3 + A3 + 3A2 + 3A – 7A
= I + A3 + 3A2 – 4A
As, A2 = A
A3 = A2A = AA = A
So, I + A3 + 3A2 – 4A = I + A + 3A – 4A = I
Hence, Option (C) is correct.
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Related Tutorials
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Scalar/Dot Product
Given 3 dimensional vectors we define their dot product as follows;
If $$v=ai+bj+ck$$ and $$w=di+ej+fk$$ then
$$v.w=(ai+bj+ck ).(di+ej+fk)$$
$$v.w=ad+be+cf$$
LESSON 1
If $$t=-i+3j-2k$$ and $$r=2i-4k$$ determine $$r.t$$
SOLUTION
$$r.t=(-i+3j-2k).(2i-4k)$$
$$r.t=(-1)(2)+(3)(0)+(-2)(-4)$$
$$r.t=6$$
TRY THIS
Complete Questions 2 and 3 from worksheet.
ANGLE BETWEEN 2 VECTORS
We can use the scalar product to find the angle between two vectors, thanks to the following formula:
$$a.b=|a||b| \cosθ$$
where $$θ$$ is the angle between the vectors.
Notice that the vectors $$a$$ and $$b$$ are going away from the angle, $$θ$$.
An important fact is that two vectors are perpendicular (orthogonal) if and only if their dot product is zero. This is because if $$θ=90°$$, then $$a.b=0$$ (Recall: $$\cos 90°=0$$)
LESSON 1
Given that $$a=\begin{pmatrix} 2\\-2 \\1 \end{pmatrix}$$, $$b=\begin{pmatrix} 2\\6 \\3 \end{pmatrix}$$ and $$c=\begin{pmatrix} p\\p \\p+1 \end{pmatrix}$$, find
(i) The angle between the directions of $$a$$ and $$b$$.
(ii) The value of $$p$$ for which $$b$$ and $$c$$ are perpendicular
SOLUTION
Part (i)
$$a.b=|a||b| \cosθ$$
$$\displaystyle{\begin{pmatrix} 2\\-2 \\1 \end{pmatrix}.\begin{pmatrix} 2\\6 \\3 \end{pmatrix}\over \sqrt{2^2+(-2)^2+1}\sqrt{2^2+6^2+3^2 }}=\cosθ$$
$$\displaystyle{(2)(2)+(-2)(6)+(1)(3)\over \sqrt{9}\sqrt{49}}=\cosθ$$
$$\displaystyle-{5\over21}=\cosθ$$
$$\displaystyle θ=\cos^{-1}(-{5\over21})=103.8°$$
Part (ii)
If $$b$$ and $$c$$ are perpendicular then $$b.c =0$$
$$\begin{pmatrix} 2\\6 \\3 \end{pmatrix}.\begin{pmatrix} p\\p \\p+1 \end{pmatrix}=0$$
$$2p+6p+3(p+1)=0$$
$$2p+6p+3p+3=0$$
$$11p=-3$$
$$p=-{3\over11}$$
TRY THIS
Complete Questions 4 - 7 from worksheet.
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Home » Maths » Factor of 32
# Multiples of 32- All Common Factors of 32
## Multiples of 32
Multiples of 32 are numbers that can be obtained by multiplying 32 by any integer. Here are the first few multiples of 32:
1. 32
2. 64
3. 96
4. 128
5. 160
6. 192
7. 224
8. 256
9. 288
10. 320
And so on. You can continue to generate more multiples of 32 by adding 32 to the previous number in the list.
## Factor of 32 in Maths
Factors of 32 are those integers that can be equally divided into 32. It has 6 factors 1, 2, 4, 8, 16, and 32. Any number is a factor of itself and hence 32 also is a factor of itself. The sum of all the factors of 32, which is 1+2+4+8+16+32, is equal to 63.
## What are the factors of 32?
Any whole number with which a larger number can be divided is known as a factor. For example, 6 can be divided by 2, 3, and 6 hence; these will be the factors of 6. The factor is the one which can divide the number completely.
The factors of a number can easily be calculated by dividing each number starting form 1. The only condition is a number can only be divided by a number smaller than itself. This is the reason 1 is always the first factor of any number.
## Factor of 32 are
factor of 32 objects are basically numerical or evenly divisible numbers without leaving any residue. When a number divides 32 by the zero of zero, that number is called a factor. All features: 1, 2, 4, 8, 16 and 32
Main features: 2 Pair items: (1, 32), (2, 16) and (4, 8)
## Prime Factors of 32
As we have already seen the factors of 32, the only prime factor that it contains is 2. Prime factors are those numbers that can only be divided by 1 and itself. Again 1 is neither composite nor prime. The smallest prime number is 2. The next prime number larger than 2 is 3.
Prime factorization of 32 is 2*2*2*2*2 which is equal to 2^5. The negative factors are -1, -2, -4, -8, -16, and -32.
## All Factors of 32
All the factors of 32 are 1, 2, 4, 8, 16, and 32. These are the numbers through which 32 can be divided equally. The number 32 is considered to be a composite number. A composite number is that number which has more than two factors.
## Common Factors of 32
The factors are also known as common factors. We have seen that the factors of 32 are 1, 2, 4, 8, 16, and 32. The prime factors of 32 are (1, 32) (2, 16) and (4, 8).
## Factors of 32 in pairs
Factors of 32 in pairs are (1, 32) (2, 16) and (4, 8). Let’s have a look at how we can calculate factors of 32. The first factor will always be 1 as 1 divides any larger number completely. So, 32÷1=32. Now for the second factor, divide 32÷2, and so on. Continue this with for integers till 32.
The numbers which are able to divided 32 completely with remainder 0 are the factors of 32. To find the factors of 32 in pairs, there is a simple method; let us try to explore the same.
Product form of 32 Factors in pairs 1 * 32 = 32 (1, 32) 2 * 16 = 32 (2, 16) 4 * 8 = 32 (4, 8)
Factors of 32 are: 1, 2, 4, 8, 16, and 32.
## Factor of 32: Practise Problems
Example 1: Choose the incorrect pairs of factors for 32 from the following options.
• (1, 32)
• (4,8)
• (2, 16)
• (3,6)
Example 2: Choose the incorrect option/s from the following
• A prime number is a number that can be divided by itself, 1 and zero.
• A composite number is a number that has more than two factors
• 32 is a composite number
• 2 is a factor of 32.
Example 3: Which one of the numbers is not a factor of 32?
• 2
• 4
• 5
• 32
Example 4: Calculate the factors of 96 and choose the correct one/s.
• 6
• 8
• 12
• 15
Example 5: Is there any negative factors of 32?
• Yes
• No
• Maybe
• None of the above
Sharing is caring!
## FAQs
### What are the factors and multiples of 32?
The factors of 32 are 1, 2, 4, 8, 16 and 32. The first five multiples of 32 are 32, 64, 96, 128, and 160. Try calculating the next 5 multiples of 32.
### What's the prime factor for 32?
The prime factorization of 32 is 2*2*2*2*2. This equals to 32 when multiplied.
### What are multiples of 32?
The first five multiples of 32 are 32, 64, 96, 128, and 160.
### What are the factors of 35?
Factors of 35 are 1, 5, 7, and 35.
### How to calculate factors of a number?
The first factor will always be 1 as 1 divides any larger number completely. So, 32÷1=32. Now for the second factor, divide 32÷2, and so on. Continue this with for integers till 32.
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# Problems on Wave Theory of Light
### Calculation of Angle of Refraction, Angle of Deviation, Wavelength and Frequency in Medium
#### Example – 1:
• A ray of light is incident on a glass slab making an angle of 25° with the surface. Calculate the angle of refraction in glass and velocity of light in glass, if the refractive index of glass and velocity of light are 1.5 and 3 x 108 m/s respectively.
• Solution:
• Given: Glancing angle = ig = 25o, Refractive index = μ = 1.5, Velocity of light in air = ca = 3 x 108 m/s
• To Find: Angle of refraction = r = ?, Velocity of light in glass = cg = ?
Angle of incidence = 90o – glancing angle = 90o – 25o = 65o.
μ = sin i / sinr
∴ Sin r = sin i / μ = sin 65o / 1.5 = 0.9063/1.5 = 0.6042
∴ angle of refraction = r = sin-1 (0.6042)
∴ Angle of refraction = 37o11’
Now, μ = ca/cg
∴ c g = ca/m = 3 x 108 /1.5 = 2 x 108 m/s
Ans: Angle of refraction = 37O10’ and velocity of light in glass = 2 x 108 m/s
#### Example – 2:
• A ray of light is incident on a glass slab making an angle of 60o with the surface. Calculate the angle of refraction in glass and the velocity of light in glass if the refractive index of glass and the velocity of light in air 1.5 and 3 x 108 m/s respectively.
• Solution:
• Given: Glancing angle = ig = 25o, Refractive index = μ = 1.5, Velocity of light in air = ca = 3 x 108 m/s
• To Find: Angle of refraction = r = ?, Velocity of light in glass = cg = ?
Angle of incidence = 90o – glancing angle = 90o – 60o = 30o.
μ = sin i / sinr
∴ Sin r = sin i / μ = sin 30o / 1.5 = 0.5/1.5 = 0.3333
∴ Angle of refraction = r = sin-1 (0.3333)
∴ Angle of refraction = 19o28’
Now, μ = ca/cg
∴ c g = ca/m = 3 x 108 /1.5 = 2 x 108 m/s
Ans: Angle of refraction = 19O28’ and velocity of light in glass = 2 x 108 m/s
#### Example – 3:
• A plane wavefront is made incident at an angle of 30° on the surface of the glass. Calculate angle of refraction if R.I. of glass is 1.5. Also find angle of deviation.
• Solution:
• Given: Angle of incidence = i = 30o, Refractive index = μ = 1.5,
• To Find: Angle of refraction = r = ? Angle of deviation = δ =?
μ = sin i / sinr
∴ Sin r = sin i / μ = sin 30o / 1.5 = 0.5/1.5 = 0.3333
∴ Angle of refraction = r = sin-1 (0.3333)
∴ Angle of refraction = 19o28’
δ = i – r = 30o – 19o28’ = 10o32’
Ans: Angle of refraction = 19O28’ and angle of deviation = 10o32’
#### Example – 4:
• The wave number of beam of light in air is 2.5 x 106 per metre. What is the wavelength in glass if R.I. of the glass is 1.5.
• Solution:
• Given: Wave number in air = 2.5 x 106 per metre, Refractive index = μg = 1.5,
• To Find: Wavelength in glass = λg = ?
Wave length in air = λ= 1/Wave number in air = 1/ 2.5 x 106 = 4 x 10-7 m
μg = λa / λg
∴ λg = λ/ μg = 4 x 10-7 / 1.5 = 2.667 x 10-7 m
∴ λg = 2.667 x 10-7 x 1010 = 2667 Å
Ans: Wavelength of light in glass is 2667 Å
#### Example – 5:
• What is the wave number of a beam of light in air if its frequency is 14 x 1014Hz? c = 3 x 108 m/s.
• Solution:
• Given: Frequency in air = νa = 14 x 1014 Hz,
• Velocity of light in air = ca = 3 x 108 m/s
• To Find: Wave number in air = ?
We have = ca = νa λa
1/λ= νa /ca = 14 x 1014 / 3 x 10 = 4.67 x 106 m-1
∴ Wave number = 1/λ= 4.67 x 106 m-1
Ans: Wave number in air is 4.67 x 106 m-1
#### Example – 6:
• The wavelength of monochromatic light is 5000 A.U. What will be its wave number in a medium of R.I. 1.5?
• Solution:
• Given: Wavelength in air = λa = 5000 Å = 5000 x 10-10 m = 5 x 10-7 m, Refractive index of medium = 1.5
• To Find: Wave number in medium = ?
μm = λa / λm
1/λ= μma = 1.5 / 5 x 10-7 = 3 x 106 m-1
∴ Wave number in medium = 1/λ= 3 x 106 m-1
Ans: Wave number in medium is 3 x 106 m-1
#### Example – 7:
• The wavelength of a beam of light in glass is 4400 Å. What is its wavelength in air, if refractive index of glass is 1.5
• Solution:
• Given: Wavelength in glass = λg = 4400 Å, Refractive index of medium = 1.5
• To Find: Wavelength in air = λa = ?
μg = λa / λg
∴ λ= μλg = 1.5 x 4400 = 6600 Å
Ans: Wavelength in air is 6600 Å
#### Example – 8:
• The speed of light in air is 3 x 108 m/s and that in diamond is 1.4 x 108 m/s. Find the R.I. of diamond.
• Solution:
• Given: Speed of light in air = ca = 3 x 108 m/s, Speed of light in diamond = cd = 1.4 x 108 m/s,
• , Refractive index of medium = 1.5
• To Find: Refractive index of diamond = μd = ?
μd = ca / cd = 3 x 10/ 1.4 x 108 = 2.142
Ans: Refractive index of diamond is 2.142
#### Example – 9:
• The velocity of light in diamond is 1.25 x 10m/s. Find the refractive index of diamond w.r.t. water. R.I. of water w.r.t. air is 1.33. Speed of light in air is 3 x 108 m/s
• Solution:
• Given: Speed of light in diamond = cd = 1.25 x 108 m/s, Speed of light in air = ca = 3 x 108 m/s, R.I. of water = μw = 1.33
• To Find: Refractive index of diamond w.r.t. water = wμd = ?
Refractive index of water w.r.t. air is given by
μw = ca / c
cw = ca / μ= 3 x 10/1.33 = 2.25 x 108 m/s
Refractive index of diamond w.r.t. water is given by
wμd = cw / cd = 2.25 x 108 /1.25 x 108 = 1.8
Ans: Refractive index of diamond w.r.t. water is 1.8
#### Example – 10:
• The refractive indices of glycerine and diamond with respect to air are 1.4 and 2.4 respectively. Calculate the speed of light in glycerine and in diamond. From these results calculate the refractive index of diamond w.r.t. glycerine. c= 3 x 108 m/s,
• Solution:
• Given: Refractive index for glycerine = μg = 1.4, Refractive index for diamond = μd = 2.4, Speed of light in air = ca = 3 x 108 m/s, R.I. of water = μw = 1.33
• To Find: Speed of light in glycerine = cg = ? The speedof light in diamond = cd = ?, Refractive index of diamond w.r.t. glycerine = gμd = ?
Refractive index of glycerine w.r.t. air is given by
μg = ca / c
cg = ca / μ= 3 x 10/1.4 = 2.143 x 108 m/s
Refractive index of diamond w.r.t. air is given by
μd = ca / c
cd = ca / μ= 3 x 10/2.4 = 1.25 x 108 m/s
Refractive index of diamond w.r.t. glycerine is given by
gμd = cg / cd = 2.143 x 108 /1.25 x 108 = 1.71
Ans: Speed of light in glycerine is 2.143 x 10m/s and that in diamond is 1.25 x 10m/s.
Refractive index of diamond w.r.t. glycerine is 1.71
#### Example – 11:
• The wavelength of light in water is 4000 Å and in the glass is 2500 Å Find the refractive index of glass w.r.t. water.
• Solution:
• Given: SWavelength in water = λw = 4000 Å, Wavelength in glass = λ= 2500 Å
• To Find: Refractive index of glass w.r.t. water = wμg = ?
μw = λw / λg = 4000 / 2500 = 1.6
Ans: Refractive index of glass w.r.t. water is 1.6
#### Example – 12:
• The refractive indices of two media are 1.5 and 1.7. Calculate the velocity of light in these two media.
• Solution:
• Given: Refractive index of first medium = μ1 = 1.5, Refractive index of second medium = μ2 = 1.7, Speed of light in air = ca = 3 x 108 m/s,
• To Find: Velocity of light in the two media = c1 = ? and c2 = ?
Consider first medium, μ1 = ca / c
c1 = ca / μ= 3 x 10/1.5 = 2 x 108 m/s
Consider second medium, μ2 = ca / c
c2 = ca / μ= 3 x 10/1.7 = 1.76 x 108 m/s
Ans: Velocity of light in first medium is 2 x 108 m/s and in second medium is 1.76 x 108 m/s
#### Example – 13:
• Red light of wavelength 6400 Å. in air has a wavelength of 4000 Å . in glass. If the wavelength of the violet light in air is 4400 Å . What is its wavelength in glass?
• Solution:
• Given: Wavelength in air for red light = λar = 6400 Å, Wavelength in glass for red light = λgr = 4000 Å, Wavelength in air for violet light = λav = 4400 Å,
• To Find: Wavelength in glass for violet light = λgv = ?
Consider red light
μr = λar / λgr = 6400 /4000 = 1.6
Assuming the refracytive index for both colour is same i.e. μr = μv
Consider violet light
μv = λav / λgv
λgv= λav / μv
= 4400 / 1.6 = 2750 Å
Ans: Wavelength of violet colour in glass is 2750 Å
#### Example – 14:
• A beam of red light (7000 Å) is passing from air into a medium making an angle of incidence of 61° and angle of refraction of 34°. Find the wavelength of red light in the medium.
• Solution:
• Given: Wavelength of red light in air = λ= 7000 Å, Angle of incidence = i = 61°, Angle of refraction = r = 34°,
• To Find: Wavelength of red light in medium = λ= ?
Refractive index of medium
μ = sin i / sinr = sin 61° / sin34° = 0.8746 / 0.5592= 1.564
Consider first medium, μm = λa / λm
λm = λa / μ= 7000 /1.564 = 4476 Å
Ans: wavelength of red light in the medium is 4476 Å.
#### Example – 15:
• A ray of light travels from air to liquid by making an angle of incidence 24° and angle of refraction of 18°. Find R.I. of liquid. Determine the wavelength of light in air and in liquid if the frequency of light is 5.4 x 1014 Hz, c = 3 x 108 m/s.
• Solution:
• Given: Frequency of light in air = ν= 5.4 x 1014 Hz, Angle of incidence = i = 24°, Angle of refraction = r = 18°, Velocity of light in air = ca = 3 x 108 m/s.
• To Find: Refractive index = μ = ?, Wavelength of red light in air and medium, λ= ?, λ= ?
Refractive index of medium
μ = sin i / sinr = sin 24° / sin18° = 0.4067 / 0.3090= 1.316
In air ca = ν λa
∴ λ = ca / νa = 3 x 108 / 5.4 x 1014 = 5.555 x 10-7 m
∴ λ= 5555 x 10-10 m = 5555 Å
Now, μm = λa / λm
λm = λa / μ= 5555 /1.316 = 4221 Å
Ans: Refractive index of medium = 1.316, wavelength in air is 5555 Å,
wavelength in medium is 4221 Å
#### Example – 16:
• Monochromatic light of wavelength 6000 Å enters glass of R.I. 1.6. Calculate its velocity, frequency and wavelength in glass. c = 3 x 108 m/s.
• Solution:
• Given: Wavelength of light in air = λ= 6000 Å = 6000 x 10-10 m = 6 x 10-7 m, Refractive index of medium = μ = 1.6, Velocity of light in air = ca = 3 x 108 m/s.
• To Find: velocity in medium = cm= ?, Frquency in medium = νm = ?, Wavelength in medium, λ= ?,
We have, μ = ca / cm
cm = ca / μ= 3 x 108 /1.6 = 1.875 x 108 m/s
In air ca = ν λa
∴ νa = ca / λa = 3 x 108 / 6 x 10-7 = 5 x 1014 Hz
If medium changes, frequency remains the same, νm = ν = 5 x 1014 Hz
Now, μ = λa / λm
λm = λa / μ= 6000 /1.6 = 3750 Å
Ans: Velocity of light in medium = 1.875 x 108 m/s, frequency in medium is 5 x 1014 Hz
wavelength in medium is 3750 Å
#### Example – 17:
• Light of wavelength 5000 A.U. is incident on water surface of R.I. 4/3. . Find the frequency and wavelength of light in water if its frequency in air is 6 x 1014 Hz.
• Solution:
• Given: Wavelength of light in air = λ= 5000 Å = 5000 x 10-10 m = 5 x 10-7 m, Refractive index of water = μ = 4/3, Velocity of light in air = ca = 3 x 108 m/s. Frequency in air = νa = 6 x 1014 Hz
• To Find: Frquency in water = νw = ?, Wavelength in water, λ= ?,
If medium changes, frequency remains the same, νw = ν = 6 x 1014 Hz
Now, μ = λa / λm
λm = λa / μ= 5000 /(4/3) = 3750 Å
Ans: Frequency in medium is 6 x 1014 Hz, wavelength in water is 3750 Å
#### Example – 18:
• The R.I. of glass w.r.t. water is 9/8. If velocity and wavelength of light in glass are 2 x 108 m/s and 4000 Å respectively, find its velocity and wavelength in water.
• Solution:
• Given: Wavelength of light in glass = λ= 4000 Å = 4000 x 10-10 m = 4 x 10-7 m, Refractive index of glass w.r.t. water = wμg = 9/8, Velocity of light in glass = cg = 2 x 108 m/s.
• To Find: velocity in water = cw = ?, Wavelength in water, λ= ?,
we have, wμ= vw / vg
vw = vg x wμg = 2 x 108 x (9/8) = 2.25 x 108 m/s
Now, wμg = λw / λg
λw = λa x wμg = 4000 x(9/8) = 4500 Å
Ans: Velocity of light in water is 2.25 x 108 m/s, wavelength in water is 4500 Å
#### Example – 19:
• Find the change in wavelength of a ray of light during its passage from air to glass if the refractive index of glass is 1.5 and the frequency of the ray is 4 x 1014 Hz.c = 3 x 10m/s.
• Solution:
• Given: Frequency of light in air = ν= 4 x 1014 Hz, Refractive index of glass = μ = 1.5, Velocity of light in air = ca = 3 x 108 m/s.
• To Find: Change in wavelength = | λ – λ| = ?
We have ca = νa λa
For air, λa = c/ν
λa = caa= 3x 108 / 4 x 1014 = 7.5 x 10-7 m
λ= 7500 x 10-10 m = 7500 Å
Now, μ = λa / λg
λg = λa / μ = 7500 / 1.5= 5000 Å
Change in wavelength = λ – λ = 7500 – 5000 = 2500 Å
Ans: Change in wavelength of light is 2500 Å
#### Example – 20:
• The difference in velocities of a light ray in glass and in water is 2.5 x 107 m/s. R.I. of water and glass are 4/3 and 1.5 respectively. Find c.
• Solution:
• Given: Difference in velocities = | c – c| = 2.5 x 107 m/s, R.I. of water = μw = 4/3, R.I. of glass = μ= 1.5
• To Find: Velocity of light in air = ca = ?
μ(1.5) > μ(4/3)
Hence, c > cg
We have ca = νa λand for glass cg = νg λ
For air, λa = c/ν
λa = caa= 3x 108 / 4 x 1014 = 7.5 x 10-7 m
λ= 7500 x 10-10 m = 7500 Å
Now, μ = λa / λg
λg = λa / μ = 7500 / 1.5= 5000 Å
Change in wavelength = λ – λ = 7500 – 5000 = 2500 Å
Ans: Change in wavelength of light is 2500 Å
#### Example – 21:
• The refractive indices of water for red and violet colours are 1.325 and 1.334 respectively. Find the difference between velocities of these two colours in water.
• Solution:
• Given: Refractive index of red colour = μr = 1.325, Refractive index of violet colour = μv = 1.3334, Velocity of light in air = ca = 3 x 108
• To Find: Difference in velocities = | c – c| = ?
μ(1.334) > μ(1.325)
Hence, c > cv
We have for red light μr = ca/cr and for violet light μv = ca/cv
Hence for red light cr = cr and for violet light cv = cav
c – c = cr – cav
c – c = c(1/μr – 1/μv)
c – c =3 x 108(1/1.325 – 1/1.334)
c – c =3 x 108(0.7547 – 0.7496)
c – c = 3 x 108(0.0051) = 1.53 x 10m/s
Ans: Difference in velocities of red and violet colour in water is 1.53 x 10m/s
#### Example – 22:
• A parallel beam of monochromatic light is incident on glass slab at an angle of incidence of 60°. Find the ratio of the width of the beam in the glass to that in the air if the refractive index of glass is 1.5.
• Solution:
• Given: Angle of incidence = i = 60°, Refractive index of glass = μ = 1.5.
• To Find: The ratio of the width of the beam in the glass to that in the air =?
We have to find ratio CD/AB’
μ = sin i / sinr
∴ Sin r = sin i / μ = sin 60o / 1.5 = 0.8660/1.5 = 0.5773
∴ Angle of refraction = r = sin-1 (0.5773) = 35o16’
In Δ AB’C, cos i = AB’ / AC ………….. (1)
In Δ ADC, cos r = CD / AC ………….. (2)
Dividing equation (2) by (1)
cos r / cos i = CD/AB’
CD/AB’ = cos 35o16’ / cos 60o
CD/AB’ = 0.8165 / 0.5 = 1.633
Ans: The ratio of the width of the beam in the glass to that in the air is 1.633
#### Example – 23:
• A parallel beam of monochromatic light is incident on a glas slab at an angle of 45°. Find the ratio of width of beam in glass to that in air if R.I. for glass is 1.5.
• Solution:
• Given: Angle of incidence = i = 45°, Refractive index of glass = μ = 1.5.
• To Find: The ratio of the width of the beam in the glass to that in the air =?
We have to find ratio CD/AB’
μ = sin i / sinr
∴ Sin r = sin i / μ = sin 45o / 1.5 = 0.7070/1.5 = 0.4713
∴ Angle of refraction = r = sin-1 (0.4713) = 28o7’
In Δ AB’C, cos i = AB’ / AC ………….. (1)
In Δ ADC, cos r = CD / AC ………….. (2)
Dividing equation (2) by (1)
cos r / cos i = CD/AB’
CD/AB’ = cos 28o7’ / cos 45o
CD/AB’ = 0.8820 / 0.7070 = 1.25
Ans: The ratio of the width of the beam in the glass to that in the air is 1.25
#### Example – 24:
• The wavelength of a certain light in air and in a medium is 4560 Å and 3648 Å respectively. Compare the speed of light in air with its speed in the medium.
• Solution:
• Given: λ1 = 4560 Å, λ2 = 3648 Å.
• To Find: ca/cm =?
We have for air ca = νaλa ………….. (1)
for medium cm = νmλm ………….. (2)
Dividing equation (1) by (2)
ca / cm = λa / λ= 4560/3648 = 1.25
If medium changes, the frequency remains the same. Hence ν= νm
Hence for red light cr = cr and for violet light cv = cav
Ans: The ratio of the speed of light in the air with its speed in the medium is 1.25
#### Example – 25:
• Light of wavelength 6400 Å is incident normally on a plane parallel glass slab of thickness 5 cm and μ = 1.6. The beam takes the same time to travel from the source to the incident surface as it takes to travel through the slab. Find the distance of the source from the incident surface. What is the frequency and wavelength of light in glass? c = 3 x 10m/s.
• Solution:
• Given: λa = 6400 Å = 6400 x 10-10 m, μ = 1.6, c = 3 x 10m/s. .
• To Find: Distance of source from surface =?
μg = ca / cg
∴ c = ca / μg = 3 x 10/ 1.6 = 1.875 x 10m/s
Let t be the time taken by light to travel through glass slab.
Distance travelled in glass = Speed in glass x time
time = distance /speed = (5 x 10-2)/(1.875 x 108) = 2.667 x 10-10 s
Distance of source from slab = speed x time = 3 x 10x 2.667 x 10-10
Distance of source from slab = speed x time = 8 x 10-2 m = 8 cm
We have for air ca = νaλa
νa = ca / λ= 3 x 10/ 6400 x 10-10 =4.69 x 1014
μg = λa / λg
λg = λa / μg = 6400/1.6 = 4000 Å
Ans: Distance of source from slab = 8 cm; frequency in glass = 4.69 1014 Hz, wavelength in glass =4000 Å
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• 0
Newbie
# A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. Find the area of canvas required for the tent.
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Very important question of mensuration including
very important concept of it.need solution!
Rd sharma class 10 maths exercise-16.2 mensuration.
we have to find area of canvas required for the tent.
given that tent is in the form of a right circular cylinder
surmounted by a cone. The diameter of cylinder is 24 m.
The height of the cylindrical portion is 11 m while the
vertex of the cone is 16 m above the ground.
Share
1. Given,
The diameter of the cylinder (also the same for cone) = 24 m.
So, its radius (R) = 24/2 = 12 m
The height of the Cylindrical part (H1) = 11m
So, Height of the cone part (H2) = 16 – 11 = 5 m
Now,
Vertex of the cone above the ground = 11 + 5 = 16 m
Curved Surface area of the Cone (S1) = πRL = 22/7 × 12 × L
The slant height (L) is given by,
L = √(R2 + H22) = √(122 + 52) = √169
L = 13 m
So,
Curved Surface Area of Cone (S1) = 22/7 × 12 × 13
And,
Curved Surface Area of Cylinder (S2) = 2Ï€RH1
S2 = 2Ï€(12)(11) m2
Thus, the area of Canvas required for tent
S = S1 + S2 = (22/7 × 12 × 13) + (2 × 22/7 × 12 × 11)
S = 490 + 829.38
S = 1319.8 m2
S = 1320 m2
Therefore, the area of canvas required for the tent is 1320 m2
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Math Activity Themes: Penguin Math
## Penguin Math Activities
The movies have renewed interest in penguins. Use this captivation with penguins to sneak in some strong math activities.
### Penguin Math Mat
Use a penguin math mat to model addition sentences. Students may roll a die and place that many counters on one penguin. The student rolls the die again and places that many counters on the other penguin. Next, the student writes a number sentence in the spaces provided.
Variation: Insert the penguin math mat in a sheet protector and give student a dry erase marker. Have the student toss a die and make that many tally marks on one of the penguins. The student tosses the die again and makes that many tally marks on the other penguin. Finally, the student writes a number sentence in the spaces provided below the penguins.
### Pascal's Penguins
Pascal's Penguins is an effective introductory activity to the well-known mathematical pattern known as Pascal's Triangle. Students must look for patterns in these penguin variations of Pascal's Triangle. The activity challenges students to identify patterns, fill in the missing numbers and write the next line in the pattern. Class discussion should encourage students to share all of the patterns they see in Pascal's Triangle and discuss how these patterns helped them discover the missing numbers.
• Pascal's Penguins - 1 introduces primary students to a small version of Pascal's Triangle in this simple patterning activity.
• Pascal's Penguins - 2 introduces a larger version of Pascal's triangle and encourages students to identify the different patterns within the triangle and use these patterns to fill in the missing penguin numbers.
## Penguin Glyphs & Graphing Ideas
### Penguin Glyphs
• Create a Penguin Glyph. Introduce the legend then present the sample penguin. Ask students to write about what the glyph tells about the person who made it. Next, have each student use the legend and construction paper cutouts to create his/her own penguin glyph.
• Teaching Heart's Penguin Unit includes a legend for a penguin glyph.
• Little Giraffe's Penguin Unit includes a glyph and other ideas for Penguin Math including a mat and lima bean penguins.
### Penguin Graphs
• Research and Graph Penguin Weights.
• Research and Graph Penguin Heights.
• Research and Graph Penguin Populations using Penguin Links which present penguin information in very usable format for Gr. 1-2 students.
• Quick penguin graphs: different graph formats are suggested so that students experience a variety of ways to organize data
• Have you ever seen a live penguin? [clothespin graph]
• Compare student heights to penguin heights [double bar graph]
• What's your favorite kind of penguin? [bar graph]
• What's your favorite penguin movie? [pictograph]
• What's your favorite penguin book? [pictograph]
• How many dice tosses does it take to free all of the penguins in the Free the Penguins game? [line plot]
## Math-Literature Connections: Penguins
### 365 Penguins by Jean-Luc Fromental
The family in this book receives a penguin in the mail each day and these penguins quickly add up to take over the house and their lives. The book already contains measurement calculations, but the book also lends itself to some problem solving possibilities. After enjoying the book, continue with some of these problem solving activities:
• Students will apply calendar skills to solve the Penguin Delivery problem.
• Dad dreams up another crazy arrangement for the penguins in Penguin Formation.
• See more Penguin Problems including Penguin Parade and Penguin Puzzler which also feature patterns that would have intrigued Dad.
• Try Pascal's Penguins, based on a famous triangular number pattern.
### Tacky the Penguin by Helen Lester
This book is a great literature jump-off for a lesson on measurement. Check out these internet ideas to see how creative teachers have used a penguin theme to present measurement activities:
• Read the book Tacky the Penguin by Helen Lester as an introduction to penguin math activities.
• Measuring Penguins describes activity in which kindergarten students use linking cubes to measure life-size penguin cutouts.
• Where in the World Is Tacky the Penguin? encourages students to measure distances in penguin feet.
• This Penguin Webquest requires students to research sizes of different penguins using the links provided. Students then create life-size replicas of their assigned penguins for a class bulletin board.
• Sing All the Penguins Song to introduce penguin measuring activities.
### Penguin Counting Books
After students have experienced many different counting books, have young students create a Penguin Counting book to accompany the unit.
These books might also be used to complement a penguin math unit.
• Your Personal Penguin by Sandra Boynton
• Mr. Popper's Penguins by Richard Atwater
• Cinderella Penguin: Little Glass Flipper by Janet Perlman
• The Penguin and the Pea by Janet Perlman
• The Emperor Penguin's New Clothes by Janet Perlman
• Little Penguin's Tale by Audrey Wood
• The Penguin and the Pea by Janet Perlman
• Antartic Antics: A Book pf Penguin Poems> by Judy Sierra
• Plenty of Penguins by Sonia Black
## Penguin Games
### Coordinate Graphing: Capture the Penguins Game
Capture the Penguins Game uses the outcome of two-dice toss to form a coordinate pair. Students toss two dice (one regular and one A-F) in this fun game that introduces students to coordinate graphing in the spaces. Students form a coordinate pair based on the dice toss and capture a penguin, if possible. Students use the accompanying recording sheet to keep score during the game.
• A-F Dice: Create A-F dice using plain dice or purchase small wooden cubes at a craft store to make the dice.
• Penguin Pieces: Penguins pictured to the right were created by painting wooden clothespins and doll stands, both readily available at craft stores.
### Number Sense: Penguin Bowling Game
Play the Penguin Bowling Game to provide mixed basic facts practice. Students toss 4 dice, then use the numbers from the dice throw and any of the four operations (+, -. *, /) to form expressions to knock down penguin pins. They are trying to create expressions that result in answers of each number from 1-10.
Students use a Score Sheet to record group scores as one does in real bowling games. Remember to model the scoring process, using an overhead of the scoring sheet, if students are not familiar with scoring regular bowling games. Teachers will especially have to model how to score a strike, as students will be strongly motivated to knock down all ten penguins in each round.
### Free the Penguins Game
The penguins are stuck on the ice floes. Only a roll of the die can free them to search for food in the ocean. Students will practice addition or subtraction facts as they try to be the first to free their penguins. Use mats with clothespin penguins for the most fun!
These games are not only an excellent opportunity to practice basic facts but they allow students to collect data and reflect on the probability inherent in the game. Suggestions are included with this game for students to keep tallies of all dice tosses, organize the class data in a line plot, and then analyze the data for patterns and trends.
## Penguin Problem Solving
• Penguin Fishing challenges students to identify the pattern in the number of fish the penguin catches each day, write a rule for the number of fish the penguin catches based on the day, then use that rule and pattern to predict how many fish the penguin will catch on each of the next three days.
• Penguin Parade: pattern problem-solving task that asks students to figure out how many penguins marched in the parade, given the marching pattern. PDF file contains problem, challenge and solution.
• Penguin Puzzler: pattern problem-solving challenge that provides the total number of penguins and requires students to analyze the pattern to discover how many many rows of penguins marched in the parade. PDF file contains problem and solution.
• Students will apply calendar skills to solve the Penguin Delivery problem.
• Dad (from 365 Penguins) dreams up another crazy arrangement for the penguins in Penguin Formation.
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# 2017 AMC 10B Problems/Problem 23
## Problem 23
Let $N=123456789101112\dots4344$ be the $79$-digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$
## Solution
We only need to find the remainders of N when divided by 5 and 9 to determine the answer. By inspection, $N \equiv 4 \text{ (mod 5)}$. The remainder when $N$ is divided by $9$ is $1+2+3+4+ \cdots +1+0+1+1 +1+2 +\cdots + 4+3+4+4$, but since $10 \equiv 1 \text{ (mod 9)}$, we can also write this as $1+2+3 +\cdots +10+11+12+ \cdots 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45$, which has a remainder of 0 mod 9. Therefore, by CRT, the answer is $\boxed{\textbf{(C) } 9}$.
Note: the sum of the digits of $N$ is $270$
## Solution 2
Noting the solution above, we try to find the sum of the digits to figure out its remainder when divided by $9$. From $1$ thru $9$, the sum is $45$. $10$ thru $19$, the sum is $55$, $20$ thru $30$ is $65$, and $30$ thru $40$ is $75$. Thus the sum of the digits is $45+55+65+75+4+5+6+7+8 = 240+30 = 270$, and thus $N$ is divisible by $9$. The solution proceeds as above.
2017 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 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
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# Horizontal Translations or Phase Shifts
## Shift right or left along the x-axis.
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Practice Horizontal Translations or Phase Shifts
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Phase Shift of Sinusoidal Functions
A periodic function that does not start at the sinusoidal axis or at a maximum or a minimum has been shifted horizontally. This horizontal movement invites different people to see different starting points since a sine wave does not have a beginning or an end.
What are five other ways of writing the function $f(x)=2 \cdot \sin x$ ?
#### Watch This
http://www.youtube.com/watch?v=wUzARNIkH-g James Sousa: Graphing Sine and Cosine with Various Transformations
#### Guidance
The general sinusoidal function is:
$f(x)=\pm a \cdot \sin (b(x+c))+d$
The constant $c$ controls the horizontal shift. If $c=\frac{\pi}{2}$ then the sine wave is shifted left by $\frac{\pi}{2}$ . If $c=-3$ then the sine wave is shifted right by 3. This is the opposite direction than you might expect, but it is consistent with the rules of transformations for all functions
Generally $b$ is always written to be positive. If you run into a situation where $b$ is negative, use your knowledge of even and odd functions to rewrite the function.
$\cos (-x) &= \cos (x)\\\sin (-x) &= -\sin (x)$
Example A
Graph the following function: $f(x)=3 \cdot \cos \left(x-\frac{\pi}{2}\right)+1$ .
Solution: First find the start and end of one period and sketch only that portion of the sinusoidal axis. Then plot the 5 important points for a cosine graph while keeping the amplitude in mind.
Example B
Given the following graph, identify equivalent sine and cosine algebraic models.
Solution: Either this is a sine function shifted right by $\frac{\pi}{4}$ or a cosine graph shifted left $\frac{5 \pi}{4}$ .
$f(x)=\sin \left(x-\frac{\pi}{4}\right)=\cos \left(x+\frac{5 \pi}{4}\right)$
Example C
At $t=5$ minutes William steps up 2 feet to sit at the lowest point of the Ferris wheel that has a diameter of 80 feet. A full hour later he finally is let off the wheel after making only a single revolution. During that hour he wondered how to model his height over time in a graph and equation.
Solution: Since the period is 60 which works extremely well with the $360^\circ$ in a circle, this problem will be shown in degrees.
Time (minutes) Height (feet) 5 2 20 42 35 82 50 42 65 2
William chooses to see a negative cosine in the graph. He identifies the amplitude to be 40 feet. The vertical shift of the sinusoidal axis is 42 feet. The horizontal shift is 5 minutes to the right.
The period is 60 (not 65) minutes which implies $b=6$ when graphed in degrees.
$60=\frac{360}{b}$
Thus one equation would be:
$f(x)=-40 \cdot \cos (6(x-5))+42$
Concept Problem Revisited
The function $f(x)=2 \cdot \sin x$ can be rewritten an infinite number of ways.
$2 \cdot \sin x=-2 \cdot \cos \left(x+\frac{\pi}{2}\right)=2 \cdot \cos \left(x-\frac{\pi}{2}\right)=-2 \cdot \sin (x- \pi)=2 \cdot \sin (x-8 \pi)$
It all depends on where you choose start and whether you see a positive or negative sine or cosine graph.
#### Vocabulary
Phase shift is a typical horizontal shift left or right that is used primarily with periodic functions.
#### Guided Practice
1. Tide tables report the times and depths of low and high tides. Here is part of tide report from Salem, Massachusetts dated September 19, 2006.
10:15 AM 9 ft. High Tide 4:15 PM 1 ft. Low Tide 10:15 PM 9 ft. High Tide
Find an equation that predicts the height based on the time. Choose when $t=0$ carefully.
2. Use the equation from Guided Practice #1 to predict the height of the tide at 6:05 AM.
3. Use the equation from Guided Practice #1 to find out when the tide will be at exactly 8 ft on September $19^{th}$
1. There are two logical places to set $t=0$ . The first is at midnight the night before and the second is at 10:15 AM. The first option illustrates a phase shift that is the focus of this concept, but the second option produces a simpler equation. Set $t=0$ to be at midnight and choose units to be in minutes.
Time (hours : minutes) Time (minutes) Tide (feet) 10:15 615 9 16:15 975 1 22:15 1335 9 $\frac{615+975}{2}=795$ 5 $\frac{1335+975}{2}=1155$ 5
These numbers seem to indicate a positive cosine curve. The amplitude is four and the vertical shift is 5. The horizontal shift is 615 and the period is 720.
$720=\frac{2 \pi}{b} \rightarrow b=\frac{\pi}{360}$
Thus one equation is:
$f(x)=4 \cdot \cos \left(\frac{\pi}{360} (x-615)\right)+5$
2. The height at 6:05 AM or 365 minutes is: $f(365) \approx 2.7057 \ \text{feet}$ .
3. This problem is slightly different from question 2 because instead of giving $x$ and using the equation to find the $y$ , this problem gives the $y$ and asks you to find the $x$ . Later you will learn how to solve this algebraically, but for now use the power of the intersect button on your calculator to intersect the function with the line $y=8$ . Remember to find all the $x$ values between 0 and 1440 to account for the entire 24 hours.
There are four times within the 24 hours when the height is exactly 8 feet. You can convert these times to hours and minutes if you prefer.
$t \approx$ 532.18 (8:52), 697.82 (11:34), 1252.18 (20 : 52), 1417.82 (23:38)
#### Practice
Graph each of the following functions.
1. $f(x)=2 \cos \left(x-\frac{\pi}{2}\right)-1$
2. $g(x)=-\sin (x-\pi)+3$
3. $h(x)=3 \cos (2 (x-\pi))$
4. $k(x)=-2 \sin (2x -\pi)+1$
5. $j(x)=-\cos \left(x+\frac{\pi}{2}\right)$
Give one possible sine equation for each of the graphs below.
6.
7.
8.
Give one possible cosine function for each of the graphs below.
9.
10.
11.
The temperature over a certain 24 hour period can be modeled with a sinusoidal function. At 3:00, the temperature for the period reaches a low of $22^\circ F$ . At 15:00, the temperature for the period reaches a high of $40^\circ F$ .
12. Find an equation that predicts the temperature based on the time in minutes. Choose $t=0$ to be midnight.
13. Use the equation from #12 to predict the temperature at 4:00 PM.
14. Use the equation from #12 to predict the temperature at 8:00 AM.
15. Use the equation from #12 to predict the time(s) it will be $32^\circ F$ .
### Vocabulary Language: English
Amplitude
Amplitude
The amplitude of a wave is one-half of the difference between the minimum and maximum values of the wave, it can be related to the radius of a circle.
Horizontal shift
Horizontal shift
A horizontal shift is the result of adding a constant term to the function inside the parentheses. A positive term results in a shift to the left and a negative term in a shift to the right.
periodic function
periodic function
A periodic function is a function with a predictable repeating pattern. Sine waves and cosine waves are periodic functions.
sinusoidal axis
sinusoidal axis
The sinusoidal axis is the neutral horizontal line that lies between the crests and the troughs of the graph of a sine or cosine function.
sinusoidal function
sinusoidal function
A sinusoidal function is a sine or cosine wave.
sinusoidal functions
sinusoidal functions
A sinusoidal function is a sine or cosine wave.
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Another way of solving a quadratic equation on the form of
$ax^{2}+bx+c=0$
Is to used the quadratic formula. It tells us that the solutions of the quadratic equation are
$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
$\: \: where\: \: a\neq 0\: \: and\: \: b^{2}-4ac\geq 0$
Example
Solve the equation
$x^{2}{\color{green} {\, -\, 3}}x{\color{blue} {\, -\, 10}}=0$
$x=\frac{-\left ( {\color{green}{ -\, 3}} \right )\pm \sqrt{\left ( {\color{green} {-\, 3}} \right )^{2}-4\cdot \left ( 1\cdot {\color{blue} {-\, 10}} \right )}}{1\cdot 2}$
$x=\frac{3\pm \sqrt{9+40}}{2}$
${\color{red}{ x_{1}}}=\frac{3{\color{red}{ \, +\, }\sqrt{49}}}{2}= \frac{3+7}{2}= \frac{10}{2}= {\color{red} {5}}$
${\color{red} {x_{2}}}= \frac{3{\color{red}{ \, -\, }\sqrt{49}}}{2}= \frac{3-7}{2}= \frac{-4}{2}= {\color{red} {-2}}$
The expression
$b^{2}-4ac$
Within the quadratic formula is called the discriminant. The discriminant can be used to determine how many solutions the quadratic equation has.
$\begin{matrix} if\: \: b^{2}-4ac>0 & &\: \: \: \: \: \: \: \: \: \: 2\: \: solutions \\ if\: \: b^{2}-4ac=0 & & \: \: \: \: \: \: \: \: \: \: \: \: 1\: \: solution\\ if\: \: b^{2}-4ac<0 & & no\: \: real\: \: solution \end{matrix}$
Here you can check that you've got the right solution
## Video lesson
Solve the equation using the quadratic formula
$x^{2} - 3x-10$
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§3.3 Exponential Functions.
# §3.3 Exponential Functions.
## §3.3 Exponential Functions.
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. §3.3 Exponential Functions. The student will learn about exponential functions, compound interest, present value, the number e, continuous compounding, and exponential growth.
2. - 2 ≤ x ≤ 4 - 1 ≤ y ≤ 30 The Exponential Function Definition: The equation f (x) = b ax for b > 0, b ≠ 1, defines an exponential function. The number b is the base. The domain of f is the set of all real numbers and the range of f is the set of all positive numbers. Graph y = 3 x
3. Properties of the graph of f (x) = b ax. • For all b the y intercept is 1. That is (0,1) is a point on all graphs (unless shifted). • All graphs are continuous curves. • The x axis is a horizontal asymptote (unless shifted). • If a > 0, then b ax is an increasing function. • If a < 0, then b ax is an decreasing function.
4. Examples Graph the following y = 3x -4 x 4 0 y 30. y = 3-x -4 x 4 0 y 30. y = -(3 - x ) -4 x 4 -30 y 0.
5. Exponential Function Properties • Exponential laws: • a x a y = a x + y • Exponential laws: • a x a y = a x + y a x / a y = a x – y • Exponential laws: • a x a y = a x + y a x / a y = a x – y (a x) y = a xy • Exponential laws: • a x a y = a x + y a x / a y = a x – y (a x) y = a xy • Exponential facts: • (ab) x = a x b x (a/b) x = a x / b x 3. a x = a y if and only if x = y 4. For x ≠ 0. a x = b x if and only if a = b.
6. Compound Interest Let P = principal, r = annual interest rate, t = time in years, n = number of compoundings per year, and A = amount realized at the end of the time period. Compound interest
7. Example: Generous Grandma Your Grandma gifts you with \$1,000 in a bank at 5% daily. Calculate the amount after 5 years. Compound interest (daily) \$1,284.00
8. Present Value Present ValueCompound Interest in Reverse Let P = principal, r = annual interest rate, t = time in years, n = number of compoundings per year, and A = amount realized at the end of the time period. In the compound interest formula the A term may be thought of as the future value. That is, grandma’s \$1,000 had a future value of \$1,284 in five years. We may reverse that order and ask what do we need to start with to have \$1,284 at 5% in 5 years? That question is the question of present value.
9. Example: Present Value How much would you need to put in the credit union at 5% to have \$1,284 in five years? Present Value
10. Depreciation by a Fixed Percentage Depreciation by a fixed percentage means that equipment loses some percentage of its value each year. It is the same as compound interest except that the percentage is negative and it is usually compounded only one time per year so that n = 1. Depreciation by percentage
11. The Number “e” There are several special numbers in mathematics. You may already know of the number π. The number e is a natural constant that occurs often in nature and also in economics. Many of our applications of exponential growth will have base e. The number e 2.71828182846. Show how to get this on a calculator.
12. Continuous Compound Interest Let P = principal, r = annual interest rate, t = time in years, n = number of compoundings per year, and A = amount realized at the end of the time period. Continuous compounding A = P e rt.
13. Example: Generous Grandma Your Grandma puts \$1,000 in a bank at 5% for you. Calculate the amount after 5 years. Continuous compounding A = P e rt. A = 1000 e (.05 x 5) = \$1,284.03
14. Present ValueContinuous Compound Interest in Reverse Let P = principal, r = annual interest rate, t = time in years, n = number of compoundings per year, and A = amount realized at the end of the time period. Future Value A = P e rt. Present Value with Continuous Compounding 14
15. Example – IRA After graduating from York College, Sam Spartan landed a great job with Springettsbury Manufacturing, Inc. His first year he bought a \$5,000 Roth IRA and invested it in a stock sensitive mutual fund that grows at 10% continuously a year. He plans to retire in 40 years. a. What will be its value in 40 years? A = P e rt = 5000 e (.10)(40) = \$272,991 Tax free. b. The second year he repeats the purchase of a Roth IRA. What will be its value in 39 years? \$247,012 Tax free. Show how to become a millionaire!! = \$2,212,963
16. Example – IRA After graduating from York College, Sam Spartan landed a great job with Springettsbury Manufacturing, Inc. His first year he bought a \$4,500 Roth IRA and invested it in a stock sensitive mutual fund that grows at 10% continuously a year. He plans to retire in 40 years. = \$2,212,963 Invested in a 5% bond fund yields \$110,648 per year, tax free. The principle remains for your heirs. Don’t count on your Social Security to live on. My father contributed to social security from Dec 1, 1936 until he retired 62 years later at age 80. He receives \$8640 per year!!!
17. The Function y = e x As we have seen many of our applications of exponential growth will have base e. The number e 2.71828182846 These equations almost always follow the form A t = A0 e kt A = P e rt. Graph y = ex on a calculator. -4 x 4 0 y 30.
18. Summary. • We learned about basic exponential functions : f (x) = b x • We learned financial applications of exponentiation including compound interest, present value and depreciation. • We learned about the natural number e and its application to continuous interest.
19. ASSIGNMENT §3.3 on my website. 6, 7, 8, 9, 10 , 11, 12, 13, 14.
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### Learning Targets:
I can explain that when a 3 dimensional figure is sliced it creates a confront that is two dimensional.I can snapshot different overcome sections that prisms and pyramids.
You are watching: Cross section of a rectangular pyramid
Describe each shape as precisely as girlfriend can. Click the applet and also drag the computer mouse to present the object turning in 3D.
Here space a rectangular prism and also a pyramid through the very same base and also same height. Drag the huge red point up and down to relocate the plane through the solids.
If we part each heavy parallel to its basic halfway up, what form cross sections would certainly we get? What is the same about the cross sections? What is different?
If we slice each heavy parallel to its base near the top, what shape cross sections would we get? What is the same about the overcome sections? What is different?
### Are you prepared for more?
Describe the cross sections that would result from slicing every solid perpendicular to its base.
Your teacher will offer you a set of cards. Kind the pictures into teams that make feeling to you. Be ready to describe your reasoning.
Use the applet to draw each cross section and also describe it in words.
Here is an applet through a rectangular prism, 4 devices by 2 systems by 3 units.
A airplane cuts the prism parallel come the bottom and also top faces.The aircraft moves up and cuts the prism at a various height.A vertical airplane cuts the prism diagonally.
A square pyramid has a base the is 4 units by 4 units. Its elevation is additionally 4 units.A airplane cuts the pyramid parallel to the base.A vertical airplane cuts the prism.
A cube has actually an edge of size 4.A aircraft cuts turn off the corner of the cube.The airplane moves farther indigenous the corner and makes a reduced through the middle of the cube.
## Lesson 11 Summary
When we slice a three-dimensional object, us expose brand-new faces that room two dimensional. The two-dimensional face is a cross section. Numerous different overcome sections are possible when slicing the same three-dimensional object.
Here space two peppers. One is sliced horizontally, and the various other is sliced vertically, producing different cross sections.
The imprints that the slices represent the two-dimensional faces developed by each slice.
It takes exercise imagining what the cross section of a three-dimensional object will certainly be for different slices. It help to experiment and also see for yourself what happens!
## Glossary Terms
cross section
A cross section is the brand-new face girlfriend see when you slice v a three-dimensional figure.
For example, if you slice a rectangular pyramid parallel come the base, you get a smaller rectangle together the overcome section.
## Lesson 11 practice Problems
A cube is reduced into 2 pieces by a single slice that passes through point out A , B , and C . What form is the cross section?
Describe exactly how to part the three-dimensional number to an outcome in each cross section.
Here are two three-dimensional figures.
Each row has the degree measures of two supplementary angles. Complete the table.
measure ofan anglemeasure ofits supplement
80^circ
25^circ
119^circ
x
Two months ago, the price, in dollars, of a cell phone was c .
Last month, the price of the phone increased by 10%. Write an expression for the price that the phone critical month.This month, the price the the phone reduced by 10%. Create an expression for the price that the phone this month.Is the price the the phone call this month the exact same as that was two months ago? explain your reasoning.
See more: What Are The Three Unforgivable Curses In Harry Potter? Unforgivable Curses
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# Rounding Multi Digit Numbers
Content Rounding Multi Digit Numbers
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Team Digital
Rounding Multi Digit Numbers
CCSS.MATH.CONTENT.4.NBT.A.3
## Rounding in Math
What is rounding? Let's have a look at the definition of the term rounding.
Rounding numbers is a way of simplifying a number. When we round, our number will be an approximate answer, not an exact one. We can round to any place value of a multi-digit number by using the rules of rounding.
## Rules of Rounding
What strategies can you use when rounding multi-digit numbers? These are the rules for rounding multi-digit numbers.
# Rule
1 Identify the place value being rounded
and underline the digit.
2 Next, find the place value of the digit
to the right and circle it.
3 If the number is 0 to 4, the
number stays the same. If the number
is 5 to 9, it gets rounded up.
## Rounding – Example
Let’s have a look at an example. We use the number 47,326. To make the operation of rounding easier to follow, we put our example number into a place value chart.
TT T H T O
4 7 3 2 6
Now, remembering the rules of rounding and the necessary steps, we can begin rounding the number up. We begin looking at the ten-thousands place.
If the number you are rounding is followed by zero to four the number being rounded stays the same.
If the number you are rounding is followed by five to nineround the number up to the next number because it is closer to the next whole number.
Finally, all the remaining place values after the rounded number are replaced with zeros.
## Rounding – Summary of Steps
Remember, these are the steps for rounding multi-digit numbers:
# Rule
1 Identify the place value being rounded
and underline the digit.
2 Next, find the place value of the digit
to the right and circle it.
3 If the number is 0 to 4, the
number stays the same. If the number
is 5 to 9, it gets rounded up.
Have you practiced yet? After watching the video, you can practice rounding multi-digit numbers to any place value with our interactive exercises and rounding multi-digit numbers worksheets. The rounding video, exercises, and multi-digit worksheet will help you understand how to round numbers with many digits and to any place value.
### TranscriptRounding Multi Digit Numbers
Skylar wants to know about how far away objects are from the ground. We can help her find approximate distances by "Rounding Multi-Digit Numbers" to any place value. Rounding numbers is a way of simplifying a number. When we round, our number will be an approximate answer, not an exact one. We can round to any place value of a multi-digit number by using the RULES OF ROUNDING. First, identify the place value being rounded and UNDERLINE the digit. Next, find the place value of the digit to the RIGHT and CIRCLE it. This number will tell us what we need to do. Now, look at these two place values. If the number you are rounding is followed by five to nine... round the number UP to the NEXT NUMBER because it is closer to the next whole number. If the number you are rounding is followed by zero to four... the number being rounded STAYS THE SAME. Finally, all the remaining place values AFTER the rounded number are replaced with zeros. Sklyar sees a plane flying at thirty-eight thousand, eight hundred, fifty-five feet. Let's help round to the nearest ten-thousands place to identify the approximate distance. First, put our exact number in the place value chart… and identify the place value we are rounding. We are rounding the ten-thousands place, so UNDERLINE the THREE. Next, find and circle the place value of the digit to the right, which is the eight in the thousands place. The eight tells us to round UP. That means we change the ten thousand place to a four… and the following place values are replaced with zeros. Thirty-eight thousand, eight hundred, fifty-five feet rounded to the nearest ten-thousands is FORTY-THOUSAND. The plane was flying at approximately forty-thousand feet. Skylar looks into space and sees a satellite. The satellite is twenty-two thousand, two hundred, twenty-three miles away. We can round the distance of the satellite to the THOUSANDS place to get an approximate distance. We’ll put the exact number in the place value chart. First, identify the digit in the place value we are rounding and underline it. What do we underline? (...) We underline the TWO in the THOUSANDS place. Now, circle the digit to the right(...) which is THIS two in the hundreds place. What does the two in the hundreds place tell us to do the two in the thousands place? (...) The two tells us the digit stays the same. The following place values will be replaced with zeros. Notice the digit in the TEN thousand place didn’t change. This is because it is BEFORE the place value we rounded. The satellite is approximately twenty-two thousand miles away. Skylar views the moon. It is two hundred, thirty-eight thousand, eight hundred, fifty-five miles away. We’ll round this distance to the HUNDREDS place to get a close approximation. First, put the number in the place value chart and underline the digit we are rounding. What number do we underline? (...) The eight in the HUNDREDS place. Now, circle the digit to the right. What number do we circle? (...) The five in the TENS place. What does five tell us to do to the eight? (...) Round UP to nine. What is the rounded number? (...) Two hundred, thirty-eight thousand, NINE HUNDRED. The moon is approximately two hundred, thirty-eight thousand, nine hundred miles away. Skylar spots UFO in the distance. While she tries to identify it(...) let's summarize. Remember...rounding numbers is a way of simplifying a number to make it easier to work with. We can round to any place value of a multi-digit number by using the RULES OF ROUNDING. First, identify the place value being rounded and UNDERLINE the digit. Next, find the place value of the digit to the right and CIRCLE it. If the number you are rounding is followed by five to nine, round the number up to the next number. If the number you are rounding is followed by zero to four, the number being rounded stays the same. It's a BIRD!? It's a PLANE!? IT'S SUPER HENRY!
## Rounding Multi Digit Numbers exercise
Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Rounding Multi Digit Numbers .
• ### What place value is the digit 2 in?
Hints
Find and underline the digit 2 in each number and then match the correct place value.
Here the 2 is in the ten thousands column. This image can help you identify the other places.
The short way of writing each place value:
• O - ones
• T - tens
• H - hundreds
• T - thousands
• TT - ten-thousands
• HT - hundred-thousands
Solution
Check out the place value charts above:
• 231,874 is the number with 2 in the hundred-thousands place
• 20,873 is the number with 2 in the ten-thousands place
• 12,874 is the number with 2 in the thousands place
• 45,821 is the number with 2 in the tens place
• 801,472 is the number with 2 in the ones place
• ### Round these numbers to the nearest 10,000 (ten-thousand)
Hints
The place value chart helps us see the different place values. When rounding to the 10,000 place we underline the digit in that place value and circle the digit to its right.
If the thousand place has any of the digits 0-4, the digit in ten-thousand place will stay the same.
If the thousand place has any of the digits 5-9, the digit in ten-thousand place will increase by one.
Solution
• To round 45,832 to the nearest 10,000 we underline the digit 4 because it is in the 10,000 place and circle the digit 5 that is in the 1,000 place.
• Since the circled digit, 5, the number 45,832 rounds up
• 45,832 rounded to the near 10,000 is 50,000
• 42,521 rounded to the near 10,000 is 40,000
• ### Round up or stay the same?
Hints
The digit that will round up or stay the same is in the 100,000 place, which place value will tell us what to do?
If the telling digit is 0-4, the rounded digit will stay the same.
If the telling digit is 5-9, the rounded digit will round up.
Solution
Rounding to the nearest 100,000
• 147,892 stays the same because the digit 4 is 0-4 so this rounds to 100,000.
• 352,146 rounds up because the digit 5 is 5-9 so this rounds to 400,000.
• 678,093 rounds up because the digit 7 is 5-9 so this rounds to 700,000.
• 921,503 stays the same because the digit 2 is 0-4 so this rounds to 900,000.
• 501,872 stays the same because the digit 0 is 0-4 so this rounds to 500,000.
• ### How have the numbers been rounded?
Hints
After rounding the digit that is being rounded, all the other place values after the rounded digit are replace with zeros. Check out the example above when 38,712 is rounded to the nearest 10,000.
The rounded place value comes just before the zeros.
Solution
• 700,000 is 728,819 rounded to the nearest 100,000.
• 5,860 is 5,863 rounded to the nearest 10.
• 730,000 is 728,819 rounded to the nearest 10,000.
• 5,900 is 5,863 rounded to the nearest 100.
• ### Round 12,438 to the nearest 10,000.
Hints
First identify and underline the digit in the ten-thousands place.
The digit 2 in the thousand place lets us know that the 1 in the ten thousands place should stay the same.
Is 12,438 closer to 10,000 or 20,000?
Solution
12,438 rounds to 10,000.
• The digit we are rounding is the 1 in the ten-thousands place
• The digit 2 in the 1,000 place tells us what to do. We need to round down because 2 is between 0-4
• The digit 1 in the ten-thousand place stays the same and all the other digits turn to zero.
• ### Which numbers round to 156,000?
Hints
It is possible to get the same number when rounding different numbers. For example, 125 "rounds up" to 130 when rounding to the nearest 10 and 132 also "rounds down" to 130 when rounding the same place value.
Find numbers that can round to 156,000 when rounding the 1,000 place.
Solution
• 157,321 does not round to 156,000 ❌
• 155,812 does round to 156,000 ✅
• 156,341 does round to 156,000 ✅
• 156,782 does not round to 156,000 ❌
• 155,389 does not round to 156,000 ❌
• 156,911 does not round to 156,000 ❌
• 156,282 does round to 156,000 ✅
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# The area of a triangle determined by the bisectors.
How can I calculate the area of a triangle determined by the interior bisectors? What I want to say it is represented in the following picture:
$AQ$ is the bisector of the angle $\angle BAC$, $BR$ -bisector for $\angle ABC$ and $CP$ -bisector for $\angle ACB$. Now, it must calculated the area for the triangle $PQR$ knowing that $AB=c$, $BC=a$ and $CA=b$.
I tried to use the angle bisector theorem for every bisectors but I didn't obtained anything.
Thanks :)
We'll derive the equation using the fact: $$A_{PQR}=A_{ABC}-A_{PBR}-A_{RCQ}-A_{QAP}, \quad (I)$$ Using the angle bisector theorem we get: $$BP=\frac{ac}{a+b},\quad (1)$$ $$BR=\frac{ac}{b+c}, \quad (2)$$ $$CR=\frac{ab}{b+c},\quad (3)$$ $$CQ=\frac{ab}{a+c},\quad (4)$$ $$AQ=\frac{bc}{a+c},\quad (5)$$ and $$AP=\frac{bc}{a+b}. \quad (6)$$
Each mentioned area can be calculated using:
$$A_{PQR}=\frac{1}{2}ab\sin\gamma, \quad (7)$$ $$A_{PBR}=\frac{1}{2}BP\cdot BR\sin\beta, \quad (8)$$ $$A_{RCQ}=\frac{1}{2}CR\cdot CQ\sin\gamma, \quad (9)$$ and $$A_{QAP}=\frac{1}{2}AQ\cdot AP\sin\alpha. \quad (10)$$
Let $R$ be the circumradius, we know that: $$\sin \alpha = \frac{a}{2R}, \quad (11)$$ $$\sin \beta = \frac{b}{2R}, \quad (12)$$ $$\sin \gamma = \frac{c}{2R}, \quad (13)$$
Now if we substitute all the 13 equations in equation $(I)$ we get: $$A_{PQR}=\frac{1}{2} \cdot \frac{abc}{2R}-\frac{1}{2} \frac{a^2c^2b}{(a+b)(b+c)2R}-\frac{1}{2} \cdot \frac{a^2b^2c}{(b+c)(a+c)2R}-\frac{1}{2} \cdot \frac{b^2c^2a}{(a+b)(a+c)2R}, \Rightarrow$$
$$A_{PQR}=\frac{abc}{4R}[1-\frac{ac}{(a+b)(b+c)}-\frac{ab}{(b+c)(a+c)}-\frac{bc}{(a+b)(a+c)}], \Rightarrow$$
$$A_{PQR}=\frac{abc}{2R}[\frac{abc}{(a+b)(b+c)(a+c)}], \Rightarrow$$ $$A_{PQR}=A_{ABC}[\frac{2abc}{(a+b)(b+c)(a+c)}]$$ Using Heron's formula we are done.
This triangle has area $$\frac{2abc}{(a+b)(a+c)(b+c)}\cdot A,$$ where $$A$$ is the area of the reference triangle with sides $$a,b,c$$. It may be called the "Cevian triangle" with respect to the incenter $$I$$ of the given reference triangle with sides $$a,b,c$$, or the "incentral triangle."
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# CBSE Class 10-Mathematics: Chapter – 11 Constructions Part 13 (For CBSE, ICSE, IAS, NET, NRA 2022)
Glide to success with Doorsteptutor material for CBSE/Class-10 : get questions, notes, tests, video lectures and more- for all subjects of CBSE/Class-10.
Question 22:
Construct an isosceles triangle whose base is and altitude and then another triangle whose sides are times the corresponding sides of the isosceles triangle.
To construct: To construct an isosceles triangle whose base is and altitude and then a triangle similar to it whose sides are of the corresponding sides of the first triangle.
Steps of construction:
(a) Draw BC = 8 cm
(b) Draw perpendicular bisector of BC. Let it meets BC at D.
(c) Mark a point A on the perpendicular bisector such that .
(d) Join AB and AC. Thus is the required isosceles triangle.
(e) From any ray BX, making an acute angle with BC on the side opposite to the vertex A.
(f) Locate points and on BX such that .
(g) Join and draw a line through the point , draw a line parallel to intersecting BC at the point C. ′
(h) Draw a line through C ‘parallel to the line CA to intersect BA at A.’
Then, A ‘BC’ is the required triangle.
Justification:
[By construction]
[AA similarity]
[By Basic Proportionality Theorem]
[By construction]
[AA similarity]
But [By construction]
Therefore,
Question 23:
Draw a triangle ABC with side , and . Then construct a triangle whose sides are of the corresponding sides of triangle ABC.
To construct: To construct a triangle ABC with side , and and and then a triangle similar to it whose sides are of the corresponding sides of the first triangle ABC.
Steps of construction:
(a) Draw a triangle ABC with side , and .
(b) From any ray BX, making an acute angle with BC on the side opposite to the vertex A.
(c) Locate 4 points and on BX such that .
(d) Join and draw a line through the point , draw a line parallel to intersecting BC at the point C. ′
(e) Draw a line through C ‘parallel to the line CA to intersect BA at A.’
Then, A ‘BC’ is the required triangle.
Justification:
[By construction]
[By Basic Proportionality Theorem]
But [By construction]
Therefore,
[AA similarity]
[From eq. (i) ]
Developed by:
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# Difference between revisions of "2013 AMC 10B Problems/Problem 7"
## Problem
Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?
$\textbf{(A)}\ \frac{\sqrt{3}}{3}\qquad\textbf{(B)}\ \frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \textbf{1}\qquad\textbf{(D)}\ \sqrt{2}\qquad\textbf{(E)}\ \text{2}$
## Solution 1
$[asy] unitsize(72); draw((0,0)--(1/2,sqrt(3)/2)); draw((1/2,sqrt(3)/2)--(3/2,sqrt(3)/2)); draw((3/2,sqrt(3)/2)--(2,0)); draw((2,0)--(3/2,-sqrt(3)/2)); draw((3/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2)); draw((1/2,-sqrt(3)/2)--(0,0)); draw((3/2,sqrt(3)/2)--(1/2,-sqrt(3)/2)); draw((3/2,sqrt(3)/2)--(3/2,-sqrt(3)/2)); label("2",(3/2,sqrt(3)/2)--(1/2,-sqrt(3)/2),NE); [/asy]$
If there are no two points on the circle that are adjacent, then the triangle would be equilateral. If the three points are all adjacent, it would be isosceles. Thus, the only possibility is two adjacent points and one point two away. Because one of the sides of this triangle is the diameter, the opposite angle is a right angle. Also, because the two adjacent angles are one sixth of the circle apart, the angle opposite them is thirty degrees. This is a $30-60-90$ triangle. If the original six points are connected, a regular hexagon is created. This hexagon consists of six equilateral triangles, so the radius is equal to one of its side lengths. The radius is $1$, so the side opposite the thirty degree angle in the triangle is also $1$. From the properties of $30-60-90$ triangles, the area is $1\cdot\sqrt3/2$=$\boxed{\textbf{(B) } \frac{\sqrt3}{2}}$
## Solution 2—Similar to Solution 1
As every point on the circle is evenly spaced, the length of each arc is $\frac{\pi}{3}$, because the circumference is $2\pi$. Once we draw the triangle (as is explained in solution 1), we see that one angle in the triangle subtends one such arc. Thus, the measure of that angle is thirty degrees. Similarly, another angle in the triangle subtends an arc of twice the length, and thus equals 60 degrees. The last angle is equal to 90 degrees and the triangle is a $30-60-90$ triangle. We know that as the diameter, the length of the hypotenuse is 2, and thus, the other sides are 1 and $\sqrt{3}$. We then find the area to be $\boxed{\textbf{(B) } \frac{\sqrt{3}}{2} }$.
## See also
2013 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# Calculating Laplace Transforms: Unlock Advanced Math Skills Master the art of Laplace transforms to simplify complex mathematical problems. Learn essential techniques, from basic principles to advanced applications in engineering and physics.
Now Playing:Calculating laplace transforms– Example 0
Intros
1. The Laplace Transform is a linear operator, and defining the Laplace Table
Examples
1. Determining Laplace Transforms
Calculate the following Laplace Transforms:
1. $L${$3e^{3t}$}
2. $L${$\sin(3t) + 2t$}
3. $L${$4\sin(2t) - 3$}
Introduction to the laplace transform
Notes
### What is a Laplace transform?
Reviewing the past lesson, let us talk again briefly about the Laplace transform definition and the Laplace transform properties. We know already that the Laplace transform is a particular integral formula which is useful while solving linear differential equations with certain conditions. We also know that the reason why you will prefer to use the Laplace transformation instead of any other method while solving certain linear differential equations directly, is because this method actually simplifies the process by being algebra-based instead of calculus-based.
In mathematics, we refer as a "transformation" to a method we use to simplify the way we solve a problem, sometimes this means going from the techniques used in a higher level of math to a lower level of the field. In this case, the integral transform which we sometimes might refer as the Laplace integral, is a tool to convert a linear differential equation which is a function of time into an algebraic expression in terms of frequency.
In mathematical notation, the definition of Laplace transform or Laplace integral can be simply put as follows:
As you may have noted in formula 1, we are dealing with an improper integral in the Laplace transform since the interval of the integration is unbounded. Remember that when one (or both) of the extremes of an integration is infinite we truly do not know the value itself (because is not a real number) and so, we cannot evaluate the integration using the classical Riemann technique.
From our lessons on limits you may remember that you can know the behaviour of a function even though you may not know its value by evaluating its limit as it approaches a particular point. We will use this knowledge when solving improper integrals.
To take the issue of the unbounded interval in the integration out, we take the limit of the integration giving the unbounded side of the integral a constant value, which will approach infinity in the limit operation. In this way, we can easily use regular integration techniques to solve the problem without being worried to evaluate to infinity until we are able to toss away those variables. In simpler words, we want to solve the integration in a way that we get rid of all the t's of the function, so we don't have to worry about evaluating it to infinite, and thus, simplifying the problem by now having the function in other terms.
The importance of the Laplace transform will be more obvious to us if we think on the whole process of solving a differential equation with it. A Laplace transform is an operator, this operator will be applied to a differential equation that in principle is difficult to solve for, so the steps go as follows:
First you have a differential equation to which you apply the operator called the Laplace integral (figure 1), this will produce an algebraic expression which is much simpler to solve and obtain a result. After obtaining the information needed from the first equation (say the value of a variable, etc) we can even work backwards to check if our process (and result) is correct, thus giving us a complete cycle of information. For this last step we use the operator called the inverse Laplace transform, and we will dedicate a complete article on it later.
### Find the Laplace transform of the following functions:
For this lesson we will be using a new methodology to solve Laplace transforms, and that is by comparison with table 1. In the past lesson you learned how to calculate these kind of transformations by working out through the integral operations, but the thing is that these operations can be gathered together in a table where we can come back and check the established result for different types of functions being transformed. This will actually accelerate the way you compute Laplace transformations even more, and with time, you may not even need to go back and look at the table since you may get used to and memorize the results of the Laplace transform of certain functions.
In order to follow the method of comparison with the table, one needs to follow the next steps to solve a Laplace transform:
1. Having the Laplace transform to resolve, simplify by linearity if addition or subtraction signs are found inside the function to be transformed.
2. Use the Laplace transforms table to solve each of the parts from the separated Laplace function from last step
1. Identify the formula from the table which contains the same operation as the one we find in each function to solve
2. Once sure of what equation to use from the table, identify element values such as n, a or b (depending on the equation you are using).
3. Ordered Item 3
4. Transcribe the result, by following the answer on the table, using your values for n, a or b according to the type of function you have.
3. After each separated Laplace transform has been solved, put the complete solution to the initial Laplace transform by adding (or subtracting, depending on the signs you had originally) each of the results found per each separated transform.
Next we have a series of Laplace transform problems with answers for you to study, we do recommend that you try to solve them on your own and then come back later to check if your answers are correct.
### Laplace transform examples
Now let us work through a few Laplace transform examples using the common Laplace transforms results from our table. As always, we recommend you to try and solve all of the problems on your own first and then come back and check your results.
At this point, we have assumed that you have already studied the lesson on the introduction to the Laplace transform, if you have not done so already, please go back to that section first so you can have a better understanding of the mathematical procedure of calculating the integrals for a Laplace transform. It is important you know that methodology beforehand, although you will be using the comparison method with the table above for the next problems, a deeper understanding will come from having looked at both methods.
Example 1
Calculate the following Laplace transform:
Following the steps listed, notice that in this case there is no need to simplify for linearity, and this we jump right onto step number 2, and so, we identify we need to use formula B form the table of Laplace transforms:
Remember that any constant coefficients can be taken out of the transformation without affecting the result:
Identifying that for this case a is equal to 3 we work through the solution:
Example 2
Calculate the following Laplace transform:
Now that the problem got a little bit lengthier, let us explicitly use the steps listed in the last section to solve the problem:
• Step 1: By linearity the Laplace transform can be rewritten as:
• Step 2:
We use the Laplace transform table to solve each of the resulting transformations separated by linearity through the last step. And so, for the first Laplace transform we select formula D from the table:
Where we can easily identify that a=3, and so:
For the second Laplace transform of this problem, we can right away notice that we just need to solve for the Laplace transform of t using formula C:
Where n=1, therefore:
• Step 3:
We can now form the complete solution to the initial Laplace transform:
Although our job is done, we can still play a little bit more with the solution to rearrange it in a way that we have a common denominator for the two expressions, so, this next step is optional, but usually nice to see and sometimes this might be the kind of "elegant" solution you are asked for during a test.
Example 3
Calculate the following Laplace transform:
• Step 1:
By linearity the Laplace transform can be rewritten as:
Notice how for the second Laplace transformation is equivalent to taking the transform of 3 times 1, and so, we can take one of the constant factors out to facilitate the operation. Therefore is convenient to take the 3 out because then you are left to solve just for the Laplace transform of 1. This is the same for any other Laplace transform of a constant, you just take the constant out and leave the one inside (which is solvable with formula A from the table).
• Step 2:
Again we use our table to solve each of the resulting transformations from the last step. Once more, we use formula D for the first Laplace transform:
We identify that a=2, so:
For the second Laplace transform we use formula A:
Where you don't need to identify any variable since this is just the Laplace transform of constant 3, so, we just multiply the constant coefficient we took out of the transform in our problem:
• Step 3:
We can now form the complete solution to the initial Laplace transform:
And once more, we rearrange it in a way that we have a common denominator for the two expressions:
Example 4
Calculate the following Laplace transform:
• In this case, there is no separation we need to do through linearity, so we jump to the second and third steps faster. Using formula F:
We can identify that for this case, a=3, and so we solve the Laplace transform rapidly:
Example 5
Calculate the following Laplace transform:
• Step 1:
By linearity the Laplace transform can be rewritten as:
• Step 2:
Using our table to solve each of the resulting transformations, we grab the Laplace transform formula H to solve the first Laplace transform:
We identify that n=1 and a=3, so:
For the second Laplace transform we use formula E:
Where a=3, therefore:
• Step 3:
The complete solution to the initial Laplace transform is:
Using some simple algebra to put the whole expression together with the same denominator:
Example 6
Calculate the following Laplace transform:
• Step 1:
Rewriting the Laplace transform due linearity:
• Step 2:
Using formula I from the table to solve the first of the three Laplace transforms:
We obtain that a=2 and b=1, therefore:
For the second Laplace transform we use formula C:
Where n=5, therefore:
And finally for the third Laplace transformation we use formula B:
Where a=3, thus:
• Step 3:
Putting the complete solution to the initial Laplace transform together we obtain:
Since we are dealing with a lengthy expression of three terms and each with a different denominator, we think is better to just leave it this way in this case, there is just one small simplification we will do and that is in the denominator of the first term in the right hand side:
After all of this practice you are now ready for our next section where you will be solving differential equations with the Laplace transform.
Before going to the next section, we recommend you to take a look at the flow chart included on this article about the method of solving initial value problems with the Laplace transform. And also, just so you can have an overall view of the topics that we have done recently and the ones that are coming, you can check these Differential equation notes with some perspective on the topic.
Theorem:
If we have the Laplace Transform of two functions:
$L${$f(t)$} = $F$($s$)
$L${$g(t)$} = $G$($s$)
With $a,b$ being constants, then we will have the following:
$L${ $af(t)$ + $bg(t)$} = $aF(s)$ + $bG(s)$
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## How do you find the measure of each interior angle of a regular pentagon?
All sides are the same length (congruent) and all interior angles are the same size (congruent). To find the measure of the interior angles , we know that the sum of all the angles is 540 degrees (from above) And there are five angles So, the measure of the interior angle of a regular pentagon is 108 degrees.
## What is the measure of 1 interior angle of a regular pentagon?
The measure of one interior angle of a regular pentagon is 108 degrees .
## What is the measure of each interior angle?
The sum of all of the interior angles can be found using the formula S = (n – 2)*180. It is also possible to calculate the measure of each angle if the polygon is regular by dividing the sum by the number of sides.
## What is the measure of each interior angle of a regular 18 sided polygon?
A polygon with 18 sides has 18 interior angles. Therefore, to find the measure of one of these angles, divide the sum by 18. Answer: Each angle is 160 degrees .
540°
## What is angle of Pentagon?
Internal angle (degrees) 108° (if equiangular, including regular) In geometry, a pentagon (from the Greek πέντε pente and γωνία gonia, meaning five and angle ) is any five-sided polygon or 5-gon. The sum of the internal angles in a simple pentagon is 540°. A pentagon may be simple or self-intersecting.
## What is the measure of an interior angle of a regular hexagon?
To find the measure of the interior angles, we know that the sum of all the angles is 720 degrees (from above) And there are six angles So, the measure of the interior angle of a regular hexagon is 120 degrees .
You might be interested: What is same side interior angles
180°
## What is the sum of interior angles?
To find the sum of interior angles of a polygon , multiply the number of triangles in the polygon by 180°. The formula for calculating the sum of interior angles is ( n − 2 ) × 180 ∘ where is the number of sides. All the interior angles in a regular polygon are equal.
## What does sum of interior angles mean?
Sum of Interior Angles Formula This formula allows you to mathematically divide any polygon into its minimum number of triangles. Since every triangle has interior angles measuring 180° , multiplying the number of dividing triangles times 180° gives you the sum of the interior angles .
heptacontagon
## What is a 20 sided polygon called?
icosagon
### Releated
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Identifying Perfect Squares
# Identifying Perfect Squares
Take the whole numbers and square them:
$$\begin{gather} \cssId{s2}{0^2 = 0}\cr \cssId{s3}{1^2 = 1}\cr \cssId{s4}{2^2 = 4}\cr \cssId{s5}{3^2 = 9}\cr \end{gather}$$
and so on.
The resulting numbers $\,0,\, 1,\, 4,\, 9,\, 16,\, 25,\, 36,\, \ldots\,$ are called perfect squares.
DEFINITION perfect square
A number $\,p\,$ is called a perfect square if and only if there exists a whole number $\,n\,$ for which $\,p = n^2\,.$
In other words: How do you get to be a perfect square? Answer: By being equal to the square of some whole number. (Recall that the whole numbers are $\,0,\, 1,\, 2,\, 3,\, \ldots\,$)
In this exercise, you will decide if a given number is a perfect square. The key is to rename the number (if possible) as a whole number, squared! You may want to review this section first: Equal or Opposites?
## Examples
Question: Is $\,9\,$ a perfect square?
Solution: Yes. $\,9 = 3^2$
Question: Is $\,7\,$ a perfect square?
Solution: No. The number $\,7\,$ can't be written as a whole number, squared.
Question: Is $\,17^2\,$ a perfect square?
Solution: Yes. The number $\,\color{red}{17}\,$ is a whole number, so $\,\color{red}{17}^2\,$ is a whole number, squared.
Question: Is $\,17^4\,$ a perfect square?
Solution: Yes. Rename as $\,(17^2)^2\,.$ The number $\,\color{red}{17^2}\,$ is a whole number, so $\,(\color{red}{17^2})^2\,$ is a whole number, squared.
Question: Is $\,(-6)^2\,$ a perfect square?
Solution: Yes. Rename as $\,6^2\,.$ The number $\,\color{red}{6}\,$ is a whole number, so $\,\color{red}{6}^2\,$ is a whole number, squared.
Question: Is $\,-6^2\,$ a perfect square?
Solution: No. Recall that: $$\,\cssId{s53}{-6^2 = (-1)(6^2) = (-1)(36) = -36}$$ A perfect square can't be negative.
Be careful! The numbers $\,-6^2\,$ and $\,(-6)^2\,$ represent different orders of operation, and are different numbers!
Question: Is $\,(-7)^{12}\,$ a perfect square?
Solution: Yes. Rename: $$\cssId{s62}{(-7)^{12} = 7^{12} = (7^6)^2}$$ The number $\,\color{red}{7^6}\,$ is a whole number, so $\,(\color{red}{7^6})^2\,$ is a whole number, squared.
Question: Is $\,-4\,$ a perfect square?
Solution: No. A perfect square can't be negative.
YES
NO
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# How are the perimeters and areas of similar figures related?
## How are the perimeters and areas of similar figures related?
ratio of their perimeters is equal to the ratio of their corresponding side lengths. the ratio of their areas is equal to the square of the ratio of their corresponding side lengths.
### What is the relationship between scale factor and area?
The area of a scaled object will be equal to the scale factor squared. Again, if the scale factor is three, the area of the new object will be nine times, or three squared, the area of the original object. Finally, the volume of a scaled object will be equal to the scale factor cubed.
How do you find similar areas?
Similar areas and volumes
1. We already know that if two shapes are similar their corresponding sides are in the same ratio and their corresponding angles are equal.
2. When calculating a missing area, we need to calculate the Area Scale Factor.
3. Area Scale Factor (ASF) = (Linear Scale Factor) 2
4. The figures below are similar.
How do you find the area of two similar rectangles?
The scale factor between the two rectangles can be found by looking at the length of corresponding sides BD (3) divided by FH (6), which simplifies to 1/2. To find the area of ABCD multiply the area of EFGH (60) by the scale factor squared: (60) x (1/2) x (1/2) = 15. The area of rectangle ABCD is 15 square units.
## How do you find the scale factor of similar figures?
Suppose you have two similar figures , one larger than the other. The scale factor is the ratio of the length of a side of one figure to the length of the corresponding side of the other figure. Here, XYUV=123=4 . So, the scale factor is 4 .
### How do you do similar shapes?
Two figures are said to be similar if they are the same shape. In more mathematical language, two figures are similar if their corresponding angles are congruent , and the ratios of the lengths of their corresponding sides are equal.
What is area of similar shape?
We already know that if two shapes are similar their corresponding sides are in the same ratio and their corresponding angles are equal. When calculating a missing area, we need to calculate the Area Scale Factor. Area Scale Factor (ASF) = (Linear Scale Factor) 2.
How do you find the area of similar figures?
The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the corresponding sides of the similar triangles. For example, for any two similar triangles ΔABC and ΔDEF, Area of ΔABC/Area of ΔDEF = (AB)2/(DE)2 = (BC)2/(EF)2 = (AC)2(DF)2.
## What is aa similarity theorem?
In two triangles, if two pairs of corresponding angles are congruent, then the triangles are similar . (Note that if two pairs of corresponding angles are congruent, then it can be shown that all three pairs of corresponding angles are congruent, by the Angle Sum Theorem.)
### What is a similarity statement?
A similarity statement is a statement used in geometry to prove exactly why two shapes have the same shape and are in proportion.
What are the similarities between area and volume?
What is the difference between Area and Volume?
Area Volume
The area is the amount of space occupied by a two-dimensional flat object in a plane. Volume is defined as the space occupied by a three-dimensional object.
The unit of area is in square units. The unit of volume is in cubic units.
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# Help solving algebraic equations
Help solving algebraic equations can be a useful tool for these scholars. We can solving math problem.
## The Best Help solving algebraic equations
This Help solving algebraic equations helps to fast and easily solve any math problems. It is usually written with an equals sign (=) like this: 4 + 5 = 9. This equation says that the answer to 4 + 5 (9) is equal to 9. So, an equation is like a puzzle, and solving it means finding the value of the missing piece. In the above example, the missing piece is the number 4 (because 4 + 5 = 9). To solve an equation, you need to figure out what goes in the blank space. In other words, you need to find the value of the variable. In algebra, variables are often represented by letters like x or y. So, an equation like 2x + 3 = 7 can be read as "two times x plus three equals seven." To solve this equation, you would need to figure out what number multiplied by 2 and added to 3 would give you 7. In this case, it would be x = 2 because 2 * 2 + 3 = 7. Of course, there are many different types of equations, and some can be quite challenging to solve. But with a little practice, you'll be solving equations like a pro in no time!
How to solve using substitution is best explained with an example. Let's say you have the equation 4x + 2y = 12. To solve this equation using substitution, you would first need to isolate one of the variables. In this case, let's isolate y by subtracting 4x from both sides of the equation. This gives us: y = (1/2)(12 - 4x). Now that we have isolated y, we can substitute it back into the original equation in place of y. This gives us: 4x + 2((1/2)(12 - 4x)) = 12. We can now solve for x by multiplying both sides of the equation by 2 and then simplifying. This gives us: 8x + 12 - 8x = 24, which simplifies to: 12 = 24, and therefore x = 2. Finally, we can substitute x = 2 back into our original equation to solve for y. This gives us: 4(2) + 2y = 12, which simplifies to 8 + 2y = 12 and therefore y = 2. So the solution to the equation 4x + 2y = 12 is x = 2 and y = 2.
Once the equation is factored, it can be solved by setting each term equal to zero and solving for x. In this case, x=-3 and x=-2 are the solutions. While factoring may take a bit of practice to master, it is a powerful tool for solving quadratic equations.
How to solve differential equations is a difficult question for many students. A differential equation is an equation that contains derivatives. The derivative of a function is the rate of change of the function with respect to one of its variables. So, a differential equation is an equation that describes how a function changes. How to solve differential equations is not always easy, but there are some methods that can be used. One method is to use integration. Integration is the process of finding the area under a curve. This can be used to find the solution to a differential equation. Another method is to use substitution. Substitution is the process of replacing one variable with another. This can be used to simplify a differential equation. How to solve differential equations is a difficult question, but there are some methods that can be used to find the solution.
## We cover all types of math problems
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Two variable equations solver Lp solver online How to solve definite integrals Algebra math word problem solver Calculus math calculator
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Title
College Algebra
Tutorial 38: Zeros of Polynomial Functions, Part I:
Rational Zero Theorem and Descartes' Rule of Signs
Learning Objectives
After completing this tutorial, you should be able to: List all possible rational zeros using the Rational Zero Theorem. Find all zeros of a polynomial function. Use Descartes' Rule of Signs to determine the possible number of positive and negative real zeros of a polynomial function.
Introduction
In this tutorial we will be taking a close look at finding zeros of polynomial functions. We will be using things like the Rational Zero Theorem and Descartes's Rule of Signs to help us through these problems. Basically when you are finding a zero of a function, you are looking for input values that cause your functional value to be equal to zero. Sounds simple enough. However, sometimes the polynomial has a degree of 3 or higher, which makes it hard or impossible to factor. Some of the ideas covered in this tutorial can help you to break down higher degree polynomial functions into workable factors. We will be using synthetic division to help us out with this process. If you need a review on synthetic division, feel free to go to Tutorial 37: Synthetic Division and the Remainder and Factor Theorems.
Tutorial
Rational Zero (or Root) Theorem If , where are integer coefficients and the reduced fraction is a rational zero, then p is a factor of the constant term and q is a factor of the leading coefficient .
We can use this theorem to help us find all of the POSSIBLE rational zeros or roots of a polynomial function.
Step 1: List all of the factors of the constant.
In the Rational Zero Theorem, p represents factors of the constant term. Make sure that you include both the positive and negative factors.
Step 2: List all of the factors of the leading coefficient.
In the Rational Zero Theorem, q represents factors of the leading coefficient. Make sure that you include both the positive and negative factors.
Step 3: List all the POSSIBLE rational zeros or roots.
This list comes from taking all the factors of the constant (p) and writing them over all the factors of the leading coefficient (q), to get a list of . Make sure that you get ever possible combination of these factors, written as .
Example 1: Use the Rational Zero Theorem to list all the possible rational zeros for .
The factors of the constant term 12 are .
The factors of the leading coefficient -1 are .
Writing the possible factors as we get: Generally we don’t write a number over 1, I just did it to emphasize that the denominator comes for factors of q which are . It would have been ok to write out the 2nd line without writing out the 1st line.
Example 2: Use the Rational Zero Theorem to list all the possible rational zeros for .
The factors of the constant term -20 are .
The factors of the leading coefficient 6 are .
Writing the possible factors as we get: Note, how when you reduce down the fractions, some of them are repeated. Here is a final list of all the POSSIBLE rational zeros, each one written once and reduced:
Example 3: List all of the possible zeros, use synthetic division to test the possible zeros, find an actual zero, and use the actual zero to find all the zeros of .
List all of the possible zeros: The factors of the constant term 100 are . The factors of the leading coefficient 1 are . Writing the possible factors as we get:
Use synthetic division to test the possible zeros and find an actual zero: Recall that if you apply synthetic division and the remainder is 0, then c is a zero or root of the polynomial function. If you need a review on synthetic division, feel free to go to Tutorial 37: Synthetic Division and the Remainder and Factor Theorems. At this point you are wanting to pick any POSSIBLE rational root form the list of . I would suggest to start with smaller easier numbers and then go from there. I’m going to choose 2 to try: Since the reminder came out -126, this means f(2) = -126, which means x = 2 is NOT a zero for this polynomial function. That doesn’t mean that we pack up our bags and quit. It’s back to the drawing board. We need to choose another number that comes from that same list of POSSIBLE rational roots. This time I’m going to choose -2: Again, the reminder is not 0, so x = -2 is not a zero of this polynomial function. This time let’s choose - 4: At last, we found a number that has a remainder of 0. This means that x = - 4 is a zero or root of our polynomial function.
Use the actual zero to find all the zeros: Since, x = - 4 is a zero, that means x + 4 is a factor of our polynomial function. Rewriting f(x) as (x + 4)(quotient) we get: We need to finish this problem by setting this equal to zero and solving it:
*Factor the difference of squares *Set 1st factor = 0 *Set 2nd factor = 0 *Set 3rd factor = 0
The zeros of this function are x = - 4, -5, and 5.
Example 4: List all of the possible zeros, use synthetic division to test the possible zeros, find an actual zero, and use the actual zero to find all the zeros of .
List all of the possible zeros: The factors of the constant term -16 are . The factors of the leading coefficient 1 are . Writing the possible factors as we get:
Use synthetic division to test the possible zeros and find an actual zero: Recall that if you apply synthetic division and the remainder is 0, then c is a zero or root of the polynomial function. If you need a review on synthetic division, feel free to go to Tutorial 37: Synthetic Division and the Remainder and Factor Theorems. At this point you are wanting to pick any POSSIBLE rational root form the list of . I would suggest to start with smaller easier numbers and then go from there. I’m going to choose -1 to try: Since the reminder came out -30, this means f(-1) = -30, which means x = -1 is NOT a zero for this polynomial function. That doesn’t mean that we pack up our bags and quit. It’s back to the drawing board. We need to choose another number that comes from that same list of POSSIBLE rational roots. This time I’m going to choose 1: At last, we found a number that has a remainder of 0. This means that x = 1 is a zero or root of our polynomial function.
Use the actual zero to find all the zeros: Since, x = 1 is a zero, that means x - 1 is a factor of our polynomial function. Rewriting f(x) as (x - 1)(quotient) we get: We need to finish this problem by setting this equal to zero and solving it:
*Factor by grouping *Factor the difference of squares *Set 1st factor = 0 *Set 2nd factor = 0 *Set 3rd factor = 0 *Set 4th factor = 0
The zeros of this function are x = 1, 4, -2 and 2.
Descartes' Rule of Signs
Let be a polynomial where are real coefficients. The number of POSITIVE REAL ZEROS of f is either equal to the number of sign changes of successive terms of f(x) or is less than that number by an even number (until 1 or 0 is reached). The number of NEGATIVE REAL ZEROS of f is either equal to the number of sign changes of successive terms of f(-x) or is less than that number by an even integer (until 1 or 0 is reached). This can help narrow down your possibilities when you do go on to find the zeros.
Example 5: Find the possible number of positive and negative real zeros of using Descartes’ Rule of Signs.
In this problem it isn’t asking for the zeros themselves, but what are the possible number of them. This can help narrow down your possibilities when you do go on to find the zeros. Possible number of positive real zeros: The up arrows are showing where there are sign changes between successive terms, going left to right. The first arrow on the left shows a sign change from positive 3 to negative 5. The 2nd arrow shows a sign change from negative 5 to positive 2. The third arrow shows a sign change from positive 2 to negative 1. And the last arrow shows a sign change from negative 1 to positive 10. There are 4 sign changes between successive terms, which means that is the highest possible number of positive real zeros. To find the other possible number of positive real zeros from these sign changes, you start with the number of changes, which in this case is 4, and then go down by even integers from that number until you get to 1 or 0. Since we have 4 sign changes with f(x), then there is a possibility of 4 or 4 - 2 = 2 or 4 - 4 = 0 positive real zeros.
Possible number of negative real zeros: Note how there are no sign changes between successive terms. This means there are no negative real zeros. Since we are counting the number of possible real zeros, 0 is the lowest number that we can have. This piece of information would be helpful when determining real zeros for this polynomial. However, for this problem we will stop here.
Example 6: Find the possible number of positive and negative real zeros of using Descartes’ Rule of Signs.
In this problem it isn’t asking for the zeros themselves, but what are the possible number of them. This can help narrow down your possibilities when you do go on to find the zeros. Possible number of positive real zeros: The up arrow is showing where there is a sign change between successive terms, going left to right. This arrow shows a sign change from positive 2 to negative 7. There is only 1 sign change between successive terms, which means that is the highest possible number of positive real zeros. To find the other possible number of positive real zeros from these sign changes, you start with the number of changes, which in this case is 1, and then go down by even integers from that number until you get to 1 or 0. If we went down by even integers from 1, we would be in the negative numbers, which is not a feasible answer, since we are looking for the possible number of positive real zeros. In other words, we can’t have a -1 of them. Therefore, there is exactly 1 positive real zero.
Possible number of negative real zeros: The up arrows are showing where there are sign changes between successive terms, going left to right. The first arrow on the left shows a sign change from negative 2 to positive 7. The 2nd arrow shows a sign change from positive 7 to negative 8. There are 2 sign changes between successive terms, which means that is the highest possible number of negative real zeros. To find the other possible number of negative real zeros from these sign changes, you start with the number of changes, which in this case is 2, and then go down by even integers from that number until you get to 1 or 0. Since we have 2 sign changes with f(-x), then there is a possibility of 2 or 2 - 2 = 0 negative real zeros.
Example 7: List all of the possible zeros, use Descartes’ Rule of Signs to possibly narrow it down, use synthetic division to test the possible zeros and find an actual zero, and use the actual zero to find all the zeros of .
List all of the possible zeros: The factors of the constant term -2 are . The factors of the leading coefficient 3 are . Writing the possible factors as we get:
Before we try any of these, let’s apply Descartes’ Rule of Signs to see if it can help narrow down our search for a rational zero. Possible number of positive real zeros: The up arrows are showing where there are sign changes between successive terms, going left to right. There are 3 sign changes between successive terms, which means that is the highest possible number of positive real zeros. To find the other possible number of positive real zeros from these sign changes, you start with the number of changes, which in this case is 3, and then go down by even integers from that number until you get to 1 or 0. Since we have 3 sign changes with f(x), then there is a possibility of 3 or 3 - 2 = 1 positive real zeros.
Possible number of negative real zeros: Note how there are no sign changes between successive terms. This means there are no negative real zeros. Since we are counting the number of possible real zeros, 0 is the lowest number that we can have. This will help us narrow things down in the next step.
Use synthetic division to test the possible zeros and find an actual zero: Recall that if you apply synthetic division and the remainder is 0, then c is a zero or root of the polynomial function. If you need a review on synthetic division, feel free to go to Tutorial 37: Synthetic Division and the Remainder and Factor Theorems. At this point you are wanting to pick any POSSIBLE rational root form the list of . Above, we found that there are NO negative rational zeros, so we do not have to bother with trying any negative numbers. See how Descartes’ has helped us. I would suggest to start with smaller easier numbers and then go from there. I’m going to choose 1 to try: Since the reminder came out -2, this means f(1) = -2, which means x = 1 is NOT a zero for this polynomial function. That doesn’t mean that we pack up our bags and quit. It’s back to the drawing board. We need to choose another number that comes from that same list of POSSIBLE rational roots. This time I’m going to choose 2: At last, we found a number that has a remainder of 0. This means that x = 2 is a zero or root of our polynomial function.
Use the actual zero to find all the zeros: Since, x = 2 is a zero, that means x - 2 is a factor of our polynomial function. Rewriting f(x) as (x - 2)(quotient) we get: We need to finish this problem by setting this equal to zero and solving it:
*Set 1st factor = 0 *Set 2nd factor = 0 *This is a quadratic that does not factor *Use the quadratic formula
The zeros of this function are x = 2, , and .
Example 8: List all of the possible zeros, use Descartes’ Rule of Signs to possibly narrow it down, use synthetic division to test the possible zeros and find an actual zero, and use the actual zero to find all the zeros of .
List all of the possible zeros: The factors of the constant term -18 are . The factors of the leading coefficient 1 are . Writing the possible factors as we get:
Before we try any of these, let’s apply Descartes’ Rule of Signs to see if it can help narrow down our search for a rational zero. Possible number of positive real zeros: The up arrow is showing where there is a sign change between successive terms, going left to right. There is 1 sign change between successive terms, which means that is the highest possible number of positive real zeros. Since we have 1 sign change with f(x), then there is exactly 1 positive real zero.
Possible number of negative real zeros: The up arrows are showing where there are sign changes between successive terms, going left to right. There are 4 sign changes between successive terms, which means that is the highest possible number of negative real zeros. To find the other possible number of negative real zeros from these sign changes, you start with the number of changes, which in this case is 4, and then go down by even integers from that number until you get to 1 or 0. Since we have 4 sign changes with f(x), then there are possibility of 4, 4 - 2 = 2 or 4 - 4 = 0 negative real zeros.
Use synthetic division to test the possible zeros and find an actual zero: Recall that if you apply synthetic division and the remainder is 0, then c is a zero or root of the polynomial function. If you need a review on synthetic division, feel free to go to Tutorial 37: Synthetic Division and the Remainder and Factor Theorems. At this point you are wanting to pick any POSSIBLE rational root from the list of . Above, we found that there is exactly 1 positive rational zero. Since we know that there is 1 for sure, then we may want to go ahead and start with trying positive rational roots. I would suggest to start with smaller easier numbers and then go from there. I’m going to choose 1 to try: Bingo!!!! We found a number that has a remainder of 0. This means that x = 1 is a zero or root of our polynomial function.
Use the actual zero to find all the zeros: Since, x = 1 is a zero, that means x - 1 is a factor of our polynomial function. Rewriting f(x) as (x - 1)(quotient) we get: We need to finish this problem by setting this equal to zero and solving it:
*Set 1st factor = 0
Looks like we can’t factor this one. We are going to have to repeat this process again, but this time we will use this factor that we found. Recall, that in Descartes’s Rule of Signs we already found that there is exactly one positive real zero. It looks like we already found that, so when we go trying again we can focus on finding a negative real zero. Note that we can still pick from the same list of numbers as we did above, since we are still looking at solving the same overall problem. However when we set up the synthetic division, we will just look at the remaining factor, to help us factor that down farther. I’m going to choose -1 to try: Bingo!!!! We found a number that has a remainder of 0. This means that x = -1 is a zero or root of our polynomial function.
Use the actual zero to find all the zeros: Since, x = -1 is a zero, that means x +1 is a factor of our polynomial function. Rewriting f(x) as (x - 1)(x + 1)(quotient) we get:
Looks like we can’t factor this one. We are going to have to repeat this process again, but this time we will use this factor that we found. Recall, that in Descartes’ Rule of Signs we already found that there is exactly one positive real zero. It looks like we already found that, so when we go trying again we can focus on finding a negative real zero. Note that we can still pick from the same list of numbers as we did above, since we are still looking at solving the same overall problem. However when we set up the synthetic division, we will just look at the remaining factor, to help us factor that down farther. I’m going to choose -2 to try: Bingo!!!! We found a number that has a remainder of 0. This means that x = -2 is a zero or root of our polynomial function.
Use the actual zero to find all the zeros: Since, x = -2 is a zero, that means x + 2 is a factor of our polynomial function. Rewriting f(x) as (x - 1)(x + 1)(x + 2)(quotient) we get: We need to finish this problem by setting this equal to zero and solving it:
*Factor the trinomial *Set 1st factor = 0 *Set 2nd factor = 0 *Set 3rd factor = 0 *Set 4th factor = 0
The zeros of this function are x = 1, -1, -2, and -3.
Practice Problems
These are practice problems to help bring you to the next level. It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it. Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. In fact there is no such thing as too much practice. To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer.
Practice Problem 1a: Use the Rational Zero Theorem to list all the possible rational zeros for the given polynomial function.
Practice Problem 2a: Find the possible number of positive and negative real zeros of the given polynomial function using Descartes’ Rule of Signs.
Practice Problems 3a - 3b: List all of the possible zeros, use Descartes’ Rule of Signs to possibly narrow it down, use synthetic division to test the possible zeros and find an actual zero, and use the actual zero to find all the zeros of the given polynomial function.
Need Extra Help on these Topics?
The following are webpages that can assist you in the topics that were covered on this page:
http://www.purplemath.com/modules/rtnlroot.htm This webpage goes over the Rational Zero Theorem. http://www.purplemath.com/modules/drofsign.htm This webpage helps you with Descartes' Rule of Signs.
Last revised on March 15, 2012 by Kim Seward.
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# The perimeter of an isosceles triangle is 29 feet. If the base measures 15 feet what is the measure of the other two sides?
Mar 25, 2018
They each equal $7$ feet.
#### Explanation:
The equation for the perimeter of a triangle is $P = {S}_{1} + {S}_{2} + {S}_{3}$,
which for an isosceles triangle could be written as: $P = {S}_{1} + 2 \left({S}_{2}\right)$
We know the perimeter is $29$ feet, and the base is $15$ feet. So we can substitute in those values to get:
$\implies$$29 = 15 + 2 \left({S}_{2}\right)$
Subtract $15$ from both sides, and get:
$\implies$$14 = 2 \left({S}_{2}\right)$
Divide by $2$, and get:
$\implies$$7 = {S}_{2}$
Since ${S}_{2} = {S}_{3}$, we know that both sides equal $7$ feet, which makes since, as it is an isosceles triangle and $7 + 7 + 15 = 29$.
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Resources tagged with Vectors similar to Changing Areas, Changing Volumes:
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Broad Topics > Vectors > Vectors
Square Coordinates
Stage: 3 Challenge Level:
A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides?
Disappearing Square
Stage: 3 Challenge Level:
Do you know how to find the area of a triangle? You can count the squares. What happens if we turn the triangle on end? Press the button and see. Try counting the number of units in the triangle now. . . .
8 Methods for Three by One
Stage: 4 and 5 Challenge Level:
This problem in geometry has been solved in no less than EIGHT ways by a pair of students. How would you solve it? How many of their solutions can you follow? How are they the same or different?. . . .
Appearing Square
Stage: 3 Challenge Level:
Make an eight by eight square, the layout is the same as a chessboard. You can print out and use the square below. What is the area of the square? Divide the square in the way shown by the red dashed. . . .
Vector Walk
Stage: 4 and 5 Challenge Level:
Starting with two basic vector steps, which destinations can you reach on a vector walk?
An Introduction to Vectors
Stage: 4 and 5
The article provides a summary of the elementary ideas about vectors usually met in school mathematics, describes what vectors are and how to add, subtract and multiply them by scalars and indicates. . . .
Vector Journeys
Stage: 4 Challenge Level:
Charlie likes to go for walks around a square park, while Alison likes to cut across diagonally. Can you find relationships between the vectors they walk along?
A Knight's Journey
Stage: 4 and 5
This article looks at knight's moves on a chess board and introduces you to the idea of vectors and vector addition.
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# Quadratic Functions and Their Graphs
## Identify the intercepts, vertex, and axis of symmetry
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Practice Quadratic Functions and Their Graphs
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What if you had a quadratic function like ? What would its graph look like? Would the graph of be wider or narrower than it? After completing this Concept, you'll be able to graph and compare graphs of quadratic functions like these.
### Try This
Meanwhile, if you want to explore further what happens when you change the coefficients of a quadratic equation, the page at http://www.analyzemath.com/quadraticg/quadraticg.htm has an applet you can use. Click on the “Click here to start” button in section A, and then use the sliders to change the values of and .
### Guidance
The graphs of quadratic functions are curved lines called parabolas. You don’t have to look hard to find parabolic shapes around you. Here are a few examples:
• The path that a ball or a rocket takes through the air.
• Water flowing out of a drinking fountain.
• The shape of a satellite dish.
• The shape of the mirror in car headlights or a flashlight.
• The cables in a suspension bridge.
#### Example A
Let’s see what a parabola looks like by graphing the simplest quadratic function, .
We’ll graph this function by making a table of values. Since the graph will be curved, we need to plot a fair number of points to make it accurate.
–2
–1
0
1
2
3
Here are the points plotted on a coordinate graph:
To draw the parabola, draw a smooth curve through all the points. (Do not connect the points with straight lines).
Let’s graph a few more examples.
#### Example B
Graph the following parabolas.
a)
b)
c)
Solution
a)
Make a table of values:
–2
–1
0
1
2
3
Notice that the last two points have very large values. Since we don’t want to make our scale too big, we’ll just skip graphing those two points. But we’ll plot the remaining points and join them with a smooth curve.
b)
Make a table of values:
–2
–1
0
1
2
3
Plot the points and join them with a smooth curve.
Notice that this time we get an “upside down” parabola. That’s because our equation has a negative sign in front of the term. The sign of the coefficient of the term determines whether the parabola turns up or down: the parabola turns up if it’s positive and down if it’s negative.
c)
Make a table of values:
–2
–1
0
1
2
3
Let’s not graph the first two points in the table since the values are so big. Plot the remaining points and join them with a smooth curve.
Wait—this doesn’t look like a parabola. What’s going on here?
Maybe if we graph more points, the curve will look more familiar. For negative values of it looks like the values of are just getting bigger and bigger, so let’s pick more positive values of beyond .
0
1
2
3
4
5
6
7
8
Plot the points again and join them with a smooth curve.
Now we can see the familiar parabolic shape. And now we can see the drawback to graphing quadratics by making a table of values—if we don’t pick the right values, we won’t get to see the important parts of the graph.
In the next couple of lessons, we’ll find out how to graph quadratic equations more efficiently—but first we need to learn more about the properties of parabolas.
The general form (or standard form) of a quadratic function is:
Here and are the coefficients. Remember, a coefficient is just a number (a constant term) that can go before a variable or appear alone.
Although the graph of a quadratic equation in standard form is always a parabola, the shape of the parabola depends on the values of the coefficients and . Let’s explore some of the ways the coefficients can affect the graph.
Dilation
Changing the value of makes the graph “fatter” or “skinnier”. Let’s look at how graphs compare for different positive values of .
#### Example C
The plot on the left shows the graphs of and . The plot on the right shows the graphs of and .
Notice that the larger the value of is, the skinnier the graph is – for example, in the first plot, the graph of is skinnier than the graph of . Also, the smaller is, the fatter the graph is – for example, in the second plot, the graph of is fatter than the graph of . This might seem counterintuitive, but if you think about it, it should make sense. Let’s look at a table of values of these graphs and see if we can explain why this happens.
–2
–1
0
1
2
3
From the table, you can see that the values of are bigger than the values of . This is because each value of gets multiplied by 3. As a result the parabola will be skinnier because it grows three times faster than . On the other hand, you can see that the values of are smaller than the values of , because each value of gets divided by 3. As a result the parabola will be fatter because it grows at one third the rate of .
Orientation
As the value of gets smaller and smaller, then, the parabola gets wider and flatter. What happens when gets all the way down to zero? What happens when it’s negative?
Well, when , the term drops out of the equation entirely, so the equation becomes linear and the graph is just a straight line. For example, we just saw what happens to when we change the value of ; if we tried to graph , we would just be graphing , which would be a horizontal line.
So as gets smaller and smaller, the graph of gets flattened all the way out into a horizontal line. Then, when becomes negative, the graph of starts to curve again, only it curves downward instead of upward. This fits with what you’ve already learned: the graph opens upward if is positive and downward if is negative.
#### Example D
What do the graphs of and look like?
Solution:
You can see that the parabola has the same shape in both graphs, but the graph of is right-side-up and the graph of is upside-down.
Vertical Shifts
Changing the constant just shifts the parabola up or down.
#### Example E
What do the graphs of and look like?
Solution:
You can see that when is positive, the graph shifts up, and when is negative the graph shifts down; in either case, it shifts by units. In one of the later Concepts, we’ll learn about horizontal shift (i.e. moving to the right or to the left). Before we can do that, though, we need to learn how to rewrite quadratic equations in different forms - our objective for the next Concept.
Watch this video for help with the Examples above.
### Vocabulary
• The general form (or standard form) of a quadratic function is:
Here and are the coefficients.
• Changing the value of makes the graph “fatter” or “skinnier”. This is called dilation.
• The vertical movement along a parabola’s line of symmetry is called a vertical shift.
### Guided Practice
Solution:
We’ll graph this function by making a table of values. Since the graph will be curved, we need to plot a fair number of points to make it accurate.
–2
–1
0
1
2
3
Plot the points and connect them with a smooth curve:
### Practice
For 1-5, does the graph of the parabola turn up or down?
For 6-10, which parabola is wider?
1. or
2. or
3. or
4. or
5. or
### Vocabulary Language: English
Coefficient
Coefficient
A coefficient is the number in front of a variable.
Dilation
Dilation
To reduce or enlarge a figure according to a scale factor is a dilation.
domain
domain
The domain of a function is the set of $x$-values for which the function is defined.
Horizontal shift
Horizontal shift
A horizontal shift is the result of adding a constant term to the function inside the parentheses. A positive term results in a shift to the left and a negative term in a shift to the right.
Parabola
Parabola
A parabola is the characteristic shape of a quadratic function graph, resembling a "U".
A quadratic function is a function that can be written in the form $f(x)=ax^2 + bx + c$, where $a$, $b$, and $c$ are real constants and $a\ne 0$.
standard form
standard form
The standard form of a quadratic function is $f(x)=ax^{2}+bx+c$.
Symmetry
Symmetry
A figure has symmetry if it can be transformed and still look the same.
Vertex
Vertex
The vertex of a parabola is the highest or lowest point on the graph of a parabola. The vertex is the maximum point of a parabola that opens downward and the minimum point of a parabola that opens upward.
vertical axis
vertical axis
The vertical axis is also referred to as the $y$-axis of a coordinate graph. By convention, we graph the output variable on the $y$-axis.
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# Slider Puzzle: Lesson 7
(Difference between revisions)
Revision as of 13:23, 26 June 2007 (view source)Lauren (Talk | contribs)m ← Older edit Current revision as of 13:28, 5 July 2007 (view source)Lauren (Talk | contribs) m Line 1: Line 1: {{Translations}} {{Translations}} - - {{ Task - | duration = hours - | category = Translation - | status = not yet started - }} ==Slider Puzzle Lesson 7: The n-puzzle (Mathematical Patterns and Sequences)== ==Slider Puzzle Lesson 7: The n-puzzle (Mathematical Patterns and Sequences)== Line 34: Line 28: [[Category:MaMaMedia]] [[Category:MaMaMedia]] + [[Category:Learning Activities]]
## Current revision as of 13:28, 5 July 2007
english | español HowTo [ID# 47764]
## Slider Puzzle Lesson 7: The n-puzzle (Mathematical Patterns and Sequences)
The n-puzzle is general name for a slider puzzle that consists of a grid of numbered squares with one square missing, and the numbers out of sequence. The 9-puzzle is a 3×3 numbered grid, the 16 puzzle is a 4×4 numbered grid. The goal of the n-puzzle is put the scrambled numbers back in sequence.
The n-puzzle is a classical problem for modeling repeatable steps (or “algorithms”) involving mathematical patterns (or “heuristics.”) For example, you can begin by counting the number of misplaced tiles as you formulate a strategy for solving the puzzle. The n-puzzle also helps students to understand that some tile arrangements are unsolvable.
Skills:
• Recognize patterns.
• Practice classification and grouping.
• Practice sequencing.
• Put numbers in order.
Activity:
1) Start the Slider Puzzle activity and choose the category labeled “Sequencing.” Sequencing is when you put things into a logical order.
2) When the first image comes up, think how this image is set up differently from the one on the Slider Puzzle. The Slider Puzzle creates a large image, where all the squares are pieces of the larger image, while the Sequencing image contains a number on each tile.
3) Try to solve the 9-tile puzzle first. Try to put the numbers in order from smallest to largest. Model for other students how to slide the keys and place them in order.
4) Once you are able to solve the 9-puzzle, share your strategies with the class. How is this different from the Slider Puzzles where you are assembling a picture?
5) Now try to solve the 16-puzzle. How is this harder or easier?
6) After completing the 16-puzzle, walk around and help those who are having trouble.
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a) Solve the equation by hit and trial method: $3x - 14 = 4$ b) Cost of $8$ ball pens is Rs. $56$ and the cost of a dozen pens is Rs. $180$ . Find the ratio of the cost of a pen to the cost of a ball pen.
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Hint: For the first part of the question, we will substitute different values for $x$ until we get the answer $4$ for $3x - 14$ .
For the second part of the question, we will find the cost of one ball pen and pen. Then we will write their ratio in the form of $\dfrac{a}{b}$ .
a).In this given problem,
In order to determine the given equation $3x - 14 = 4$ by hit and trial method.
We will use trial and error (hit and trial) method and substitute different values for $x$ starting from $1$ .
Taking, $x = 1$ , we will get,
$3(1) - 14 = 4$
$\Rightarrow - 11 \ne 4$
Since we did not get the correct answer, we have to continue till we equalize both sides of the equation.
Taking $x = 2$ , we will get,
$3(2) - 14 = 4$
$\Rightarrow - 8 \ne 4$
Similarly, check for $x = 3,4,5,6$ .
Taking, $x = 6$ , we will get,
$3(6) - 14 = 4$
$18 - 14 = 4$
$\Rightarrow 4 = 4$
Hence, we get $x = 6$ by hit and trial method.
b).It is given that the cost of $8$ ball pens is Rs. $56$ . So, cost of one ball pen will be-
$\dfrac{{56}}{8} = Rs.7$
Similarly, the cost of a dozen pens is Rs. $180$ . So, cost of one pen will be-
$\dfrac{{180}}{{12}} = Rs.15$
Now we can get the ratio of cost of a pen to cost of ball as follows:
$\dfrac{{15}}{7}$ .
Therefore, Cost of $8$ ball pens is Rs. $56$ and the cost of a dozen pens is Rs. $180$ . $15:7$ is the ratio of the cost of a pen to the cost of a ball pen.
So, the correct answer is “ $15:7$ ”.
Note: The term "trial and error" refers to the process of determining whether or not a particular decision is correct (or wrong). We simply check by substituting that option into the problem. Some questions can only be answered by trial and error; for others, we must first determine whether there isn't a quicker way to find the answer.
For solving the second part of the question, while finding ratios, we should always convert the object to a single unit. We should also note that a dozen means $12$ quantities. However, a baker’s dozen is $13$ .
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# Arithmetic for Praxis I: Pre-Professional Skills Test Study Guide (page 5)
By
Updated on Jul 5, 2011
### Negative Exponents
A base raised to a negative exponent is equivalent to the reciprocal of the base raised to the positive exponent (absolute value of the exponent).
Examples
### Exponent Rules
• When multiplying identical bases, you add the exponents.
Examples
22 × 24 × 26 = 212
a2 × a3 × a5 = a10
• When dividing identical bases, you subtract the exponents.
Examples
• If a base raised to a power (in parentheses) is raised to another power, you multiply the exponents together.
Examples
(32)7 = 314 (g4)3 = g12
### Perfect Squares
The number 52 is read "5 to the second power," or, more commonly, "5 squared." Perfect squares are numbers that are second powers of other numbers. Perfect squares are always zero or positive, because when you multiply a positive or a negative by itself, the result is always positive. The perfect squares are 02, 12, 22, 32
Perfect squares: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100…
### Perfect Cubes
The number 53 is read "5 to the third power," or, more commonly, "5 cubed." (Powers higher than three have no special name.) Perfect cubes are numbers that are third powers of other numbers. Perfect cubes, unlike perfect squares, can be either positive or negative. This is because when a negative is multiplied by itself three times, the result is negative. The perfect cubes are 03, 13, 23, 33
Perfect cubes: 0, 1, 8, 27, 64, 125…
• Note that 64 is both a perfect square and a perfect cube.
### Square Roots
The square of a number is the product of the number and itself. For example, in the statement 32 = 3 × 3 = 9, the number 9 is the square of the number 3. If the process is reversed, the number 3 is the square root of the number 9. The symbol for square root is and is called a radical. The number inside of the radical is called the radicand.
Examples
52 = 25; therefore, = 5
Because 25 is the square of 5, it is also true that 5 is the square root of 25.
The square root of a number might not be a whole number. For example, the square root of 7 is 2.645751311.… It is not possible to find a whole number that can be multiplied by itself to equal 7. Square roots of nonperfect squares are irrational.
### Cube Roots
The cube of a number is the product of the number and itself for a total of three times. For example, in the statement 23 = 2 × 2 × 2 = 8, the number 8 is the cube of the number 2. If the process is reversed, the number 2 is the cube root of the number 8. The symbol for cube root is the same as the square root symbol, except for a small 3: . It is read as "cube root." The number inside of the radical is still called the radicand, and the 3 is called the index. (In a square root, the index is not written, but it has an index of 2.)
Examples
53 = 125; therefore,
Like square roots, the cube root of a number might be not be a whole number. Cube roots of nonperfect cubes are irrational.
### Probability
Probability is the numerical representation of the likelihood of an event occurring. Probability is always represented by a decimal or fraction between 0 and 1, 0 meaning that the event will never occur, and 1 meaning that the event will always occur. The higher the probability, the more likely the event is to occur.
A simple event is one action. Examples of simple events are: drawing one card from a deck, rolling one die, flipping one coin, or spinning a hand on a spinner once.
### Simple Probability
The probability of an event occurring is defined as the number of desired outcomes divided by the total number of outcomes. The list of all outcomes is often called the sample space.
Examples
What is the probability of drawing a king from a standard deck of cards?
There are four kings in a standard deck of cards, so the number of desired outcomes is 4. There are 52 ways to pick a card from a standard deck of cards, so the total number of outcomes is 52. The probability of drawing a king from a standard deck of cards is . So, P(king) = .
Examples
What is the probability of getting an odd number on the roll of one die?
There are three odd numbers on a standard die: 1, 3, and 5. So, the number of desired outcomes is 3. There are six sides on a standard die, so there are 6 possible outcomes. The probability of rolling an odd number on a standard die is . So, P(odd) = .
Note: It is not necessary to reduce fractions when working with probability.
### Probability of an Event Not Occurring
The sum of the probability of an event occurring and the probability of the event not occurring = 1. Therefore, if we know the probability of the event occurring, we can determine the probability of the event not occurring by subtracting from 1.
Examples
If the probability of rain tomorrow is 45%, what is the probability that it will not rain tomorrow?
45% = .45, and 1 – .45 = .55 or 55%. The probability that it will not rain is 55%.
### Probability Involving the Word Or
Rule:
P(event A or event B) = P(event A) + P(event B) – P(overlap of event A and B)
When the word or appears in a simple probability problem, it signifies that you will be adding outcomes. For example, if we are interested in the probability of obtaining a king or a queen on a draw of a card, the number of desired outcomes is 8, because there are four kings and four queens in the deck. The probability of event A (drawing a king) is , and the probability of drawing a queen is . The overlap of event A and B would be any cards that are both a king and a queen at the same time, but there are no cards that are both a king and a queen at the same time. So the probability of obtaining a king or a queen is .
Examples
What is the probability of getting an even number or a multiple of 3 on the roll of a die?
The probability of getting an even number on the roll of a die is , because there are three even numbers (2, 4, 6) on a die and a total of 6 possible outcomes. The probability of getting a multiple of 3 is , because there are 2 multiples of three (3, 6) on a die. But because the outcome of rolling a 6 on the die is an overlap of both events, we must subtract from the result so we don't count it twice.
P(even or multiple of 3) = P(even) + P(multiple of 3) – P(overlap)
### Compound Probability
A compound event is performing two or more simple events in succession. Drawing two cards from a deck, rolling three dice, flipping five coins, having four babies are all examples of compound events.
This can be done "with replacement" (probabilities do not change for each event) or "without replacement" (probabilities change for each event).
The probability of event A followed by event B occurring is P(A) × P(B). This is called the counting principle for probability.
Note: In mathematics, the word and usually signifies addition. In probability, however, and signifies multiplication and or signifies addition.
Examples
You have a jar filled with three red marbles, five green marbles, and two blue marbles. What is the probability of getting a red marble followed by a blue marble, with replacement?
"With replacement" in this case means that you will draw a marble, note its color, and then replace it back into the jar. This means that the probability of drawing a red marble does not change from one simple event to the next.
Note that there are a total of ten marbles in the jar, so the total number of outcomes is 10.
If the problem was changed to say "without replacement," that would mean you are drawing a marble, noting its color, but not returning it to the jar. This means that for the second event, you no longer have a total number of 10 outcomes; you have only 9, because you have taken one red marble out of the jar. In this case,
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# Geometric Median |Understand the concept
Geometric Median is an important concept in the intersection of Geometry, Data Analysis and Algorithms. This article explores the concept.
Given a set of numbers on the real line, say $a_1, a_2, a_3, ..., a_n$, the median of these numbers is the middle number when you arrange these numbers in ascending/descending order. To understand more about the median for a set of numbers, go through this.
Let's ask a different question altogether.
Given a set of numbers on the real line, say $a_1, a_2, a_3, ..., a_n$, what is the number $x$ such that $|x-a_1| + |x-a_2| + ... |x-a_n|$ is minimized?
In other words, what is the point on the real number line, such that the sum of the distances of that point from the given set of points is minimum? In some sense, this point gives an idea about the central behavior of the set of points.
In fact, this turns out to be the median of the set of numbers { $a_1, a_2, a_3, ..., a_n$}. To understand the proof, go through this.
But, I suggest you try this yourself, in inductive steps. Start with two points, any point in between them is the median. Now, add a point in between those two points, Observe that the new median is the new point added. Continue similarly in this inductive process.
My contentment is to ask questions and seek. Therefore, a lot of questions inpoured.
• What happens if we take those n points in 2D space?
• What happens if we take those n points in 3D space?
• What happens if we take those n points in n-D space?
• What happens if we take a different distance idea instead of the normal distance?
Well, the intelligent way is to answer the easiest of all and then generalize the idea - the first question.
Given a set of numbers in the 2D space, say $a_1, a_2, a_3, ..., a_n$, what is the point $x$ such that $|x-a_1| + |x-a_2| + ... |x-a_n|$ is minimized?
Let's take the easiest distance concept of all - The Euclidean Distance $d(x,y) = |x-y|= \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$.
This concept is known as Geometric Median in the literature.
Now, the problem seems difficult even in the case of three points.
Let's explore it for a quadrilateral ( $n = 4$ ).
Step 1: Does there exist a better point like that?
Yes, there is! Let's explore.
Observe that (AI + IF) + (BI + IE) = (AG + GF) + (BE) < (AG + GF) + (BG + GE). We are just using the triangle inequality.
Hence, G is a better point than I.
Step 2: Does there exist a better point than I?
Albeit, there is!
(AI + IF) + (BI + IE) = (AI + IF) + (BE) = (AI + IF) + (BJ + JE) > (AG + GF) + (BG + GE). We are just using the triangle inequality.
Thus, this algorithm results in the converges to the intersection of the diagonals as a special point. This is easy.
But what about, $n = 3$. It may be easy. No, it's not.
The case for triangles is not at all easy. In the case of triangles, the point is called Fermat's Point.
I will not repeat the solution again. The following pic gives a cute solution. Using the same idea, that the distance between two points is minimum along a straight line.
Now, the rest of the questions remain too difficult to solve. Even there hardly exists a good algorithm to solve it.
Now, this left to your imagination. Stay tuned!
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Students can create models that represent their understanding, but can they undo models that explain the reasoning of others? It was time to find out. I showed this picture to a 5th grade class and asked them to write an equation or expression that matched the area model…
Every student in the class said it was modeling multiplication however no student could account for the 5 squares at the bottom.
Why did this happen?
My takeaways: In order for students to understand what’s happening, there’s a progression of understanding that begins all the way back in 2nd grade. This understanding is built through exploration. These explorations happen in and out of context, with and without manipulatives and representations, but most importantly…it happens over time.
2nd Grade: students are expected to partition a square in to 5 rows and 5 columns. They should also use additive equations that match the model.
• 2.OA.4 – Use addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns
5 + 5 + 5 + 5 + 5 = 25
3rd Grade: students begin to model multiplication AND division with arrays at the same time to develop relational thinking.
• 3.OA.1 – Interpret products of whole numbers
• 3.OA.2 – Interpret whole-number quotients of whole numbers
The model above is typically described as 3 x 6 but it could also represent division by giving students 18 tiles and have them construct a rectangle with 3 rows. The 18 tiles now become the dividend (not the product), which takes flexibility in understanding.
What happens if you ask students to build a model with 19 tiles and place them in 3 rows?
What does the blue tile represent?
19 can be divided into three rows but the context will determine what to do with the 1 remaining tile. The blue tile can be recognized as a remainder, as a third, or as 1/6 of the next group. Using “untidy” number combinations can open the door to some awesome discussion.
Third grade students also begin to multiply with single-digit factors and multiples of 10.
• 3.NBT.3 – Multiply one-digit whole numbers by multiples of 10 in the range of 10-90 using strategies based on place value and properties of operations.
If multiplication and division are taught concurrently, then students begin to understand that the model below has multiple meanings but it depends on the context. Flexibility is really hard to teach. So don’t teach it…provide opportunities that foster flexible thinking.
The difficult thing for students to recognize ties back to an understanding that is built in kindergarten…unitizing. Even though these are “ten rods”, students need to see them as a rod composed of ten ones. If students only see them as rods of 10 it becomes extremely difficult to see this model as division.
4th Grade: students begin to explore multiplication and division with multi-digit numbers.
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# Ex.10.6 Q6 Circles Solution - NCERT Maths Class 9
Go back to 'Ex.10.6'
## Question
\begin {align}{ABCD} \end {align} is a parallelogram. The circle through \begin {align}{A, B} \end {align} and \begin {align}{C} \end {align} intersect \begin {align}{CD} \end {align} (produced if necessary) at \begin {align}{E} \end {align}. Prove that \begin {align}{AE = AD.} \end {align}
Video Solution
Circles
Ex 10.6 | Question 6
## Text Solution
What is known?
\begin {align}{ABCD} \end {align} is a parallelogram.
What is unknown?
Proof that \begin {align}{AE = AD} \end {align}
Reasoning:
• A quadrilateral \begin {align}{ABCD} \end {align} is called cyclic if all the four vertices of it lie on a circle.
• The sum of either pair of opposite angles of a cyclic quadrilateral is $$180^{\circ}.$$
• Opposite angles in a parallelogram are equal.
Steps:
We can see that \begin {align}{ABCE} \end {align} is a cyclic quadrilateral.
We know that in a cyclic quadrilateral, the sum of the opposite angles is $$180^{\circ.}$$
\begin{align}\angle {AEC}+\angle {CBA}&=180^{\circ} \\ \angle {AEC} +\angle {AED}&=180^{\circ} (\text { Linear pair })\end{align}
$$\angle {AED}=\angle {CBA} \ldots \text { (1) }$$
We know that in a parallelogram, opposite angles are equal.
\begin {align} \angle {ADE}=\angle{CBA}\quad \dots(2) \end {align}
From ($$1$$) and ($$2$$),
\begin{align} \angle {AED}=\angle{ADE}\\{AD}={AE}\qquad \end{align}
(sides opposite to equal Angles of a triangle are equal).
Video Solution
Circles
Ex 10.6 | Question 6
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## NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2.
Board CBSE Textbook NCERT Class Class 9 Subject Maths Chapter Chapter 4 Chapter Name Lines and Angles Exercise Ex 4.2 Number of Questions Solved 6 Category NCERT Solutions
## NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2
Question 1.
In figure, find the values of x and y and then show that AB || CD.
Solution:
∵ x + 50° = 180° (Linear pair)
⇒ x = 130°
∴ y = 130° (Vertically opposite angle)
Here, ∠x = ∠COD = 130°
These are corresponding angles for lines AB and CD.
Hence, AB || CD
Question 2.
In figure, if AB || CD, CD || EF and y: z = 3:7, find x.
Solution:
Given
⇒ Let y = 3k, z = 7k
x = ∠CHG (Corresponding angles)…(i)
∠CHG = z (Alternate angles)…(ii)
From Eqs. (i) and (ii), we get
x = z …….(iii)
Now, x+y = 180°
(Internal angles on the same side of the transversal)
⇒ z+y = 180° [From Eq. (iii)]
7k + 3k = 180°
⇒ 10k = 180°
⇒ k = 18
∴ y = 3 x 18° = 54°
and z = 7x 18°= 126°
∴ x = z
Now, x + y = 180°
(Internal angles on the same side of the transversal)
⇒ z + y = 180° [From Eq. (iii)]
7k + 3k = 180°
⇒ 10k = 180°
⇒ k = 18
∴ y = 3 x 18° = 54°
and z = 7x 18°= 126°
x = z
⇒ x = 126°
Question 3.
In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Solution:
∵ ∠AGE = ∠GED (Alternate interior angles)
But ∠GED = 126°
⇒ ∠AGE = 126° ….(i)
∴ ∠GEF + ∠FED= 126°
⇒ ∠GEF + 90° =126° (∵ EF ⊥ CD)
⇒ ∠GEF = 36°
Also, ∠AGE + ∠FGE = 180° (Linear pair axiom)
⇒ 126° + ∠FGE =180°
⇒ ∠FGE = 54°
Question 4.
In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS.
Solution:
Drawing a tine parallel to ST through R.
Question 5.
In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
Solution:
We have, AB || CD
⇒ ∠APQ = ∠PQR (Alternate interior angles)
⇒ 50° = x
⇒ x = 50°
Now, ∠PQR + ∠QPR = 127°
(Exterior angle is equal to sum of interior opposite angles of a triangle)
⇒ 50°+ ∠QPR = 127°
⇒ y = 77°.
Question 6.
In figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.
Solution:
We hope the NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2, drop a comment below and we will get back to you at the earliest.
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## Bayes’ Theorem: Ad Hoc-ness and Other Details
More about Bayes’ theorem; an introduction was given here. Once again, I’m not claiming any originality.
You can’t save a theory by stapling some data to it, even though this will improve its likelihood. Let’s consider an example.
Suppose, having walked into my kitchen, I know a few things.
$D_1$ = There is a cake in my kitchen.
$D_2$ = The cake has “Happy Birthday Luke!” on it, written in icing.
$B$ = My name is Luke + Today is my birthday + whatever else I knew before walking to the kitchen.
Obviously, $D_2 \Rightarrow D_1$ i.e. $D_2$ presupposes $D_1$. Now, consider two theories of how the cake got there.
$W$ = my Wife made me a birthday cake.
$A$ = a cake was Accidentally delivered to my house.
Consider the likelihood of these two theories. Using the product rule, we can write:
$p(D_1D_2 | WB) = p(D_2 | D_1 WB) p(D_1 | WB)$
$p(D_1D_2 | AB) = p(D_2 | D_1 AB) p(D_1 | AB)$
Both theories are equally able to place a cake in my kitchen, so $p(D_1 | WB) \approx p(D_1 | AB)$. However, a cake made by my wife on my birthday is likely to have “Happy Birthday Luke!” on it, while a cake chosen essentially at random could have anything or nothing at all written on it. Thus, $p(D_2 | D_1 WB) \gg p(D_2 | D_1 AB)$. This implies that $p(D_1D_2 | WB) \gg p(D_1D_2 | AB)$ and the probability of $W$ has increased relative to $A$ since learning $D_1$ and $D_2$.
So far, so good, and hopefully rather obvious. Let’s look at two ways to try to derail the Bayesian account.
### Details Details
Before some ad hoc-ery, consider the following objection. We know more than $D_1$ and $D_2$, one might say. We also know,
$D_3$ = there is a swirly border of piped icing on the cake, with a precisely measured pattern and width.
Now, there is no reason to expect my wife to make me a cake with that exact pattern, so our likelihood takes a hit:
$p(D_3 | D_1 D_2 WB) \ll 1 ~ \Rightarrow ~ p(D_1 D_2 D_3 | WB) \ll p(D_1D_2 | WB)$
Alas! Does the theory that my wife made the cake become less and less likely, the closer I look at the cake? No, because there is no reason for an accidentally delivered cake to have that pattern, either. Thus,
$p(D_3 | D_1 D_2 WB) \approx p(D_3 | D_1 D_2 AB)$
And so it remains true that,
$p(D_1 D_2 D_3 | WB) \gg p(D_1 D_2 D_3 | AB)$
and the wife hypothesis remains the prefered theory. This is point 5 from my “10 nice things about Bayes’ Theorem” – ambiguous information doesn’t change anything. Additional information that lowers the likelihood of a theory doesn’t necessarily make the theory less likely to be true. It depends on its effect on the rival theories.
What if we crafted another hypothesis, one that could better handle the data? Consider this theory.
$A_D$ = a cake with “Happy Birthday Luke!” on it was accidentally delivered to my house.
Unlike $A$, $A_D$ can explain both $D_1$ and $D_2$. Thus, the likelihoods of $A_D$ and $W$ are about equal: $p(D_1D_2 | WB) \approx p(D_1D_2 | A_DB)$. Does the fact that I can modify my theory to give it a near perfect likelihood sabotage the Bayesian approach?
Intuitively, we would think that however unlikely it is that a cake would be accidentally delivered to my house, it is much less likely that it would be delivered to my house and have “Happy Birthday Luke!” on it. We can show this more formally, since $A_D$ is a conjunction of propositions $A_D = A A'$, where
$A'$ = The cake has “Happy Birthday Luke!” on it, written in icing.
But the statement $A'$ is simply the statement $D_2$. Thus $A_D = A D_2$. Recall that, for Bayes’ Theorem, what matters is the product of the likelihood and the prior. Thus,
$p(D_1 D_2 | A_D B) ~ p(A_D | B)$
$= p(D_1 D_2 | A D_2 B) ~ p(A D_2 | B)$
$= p(D_1|A D_2B) ~ p(D_2|AB) ~ p(A|B)$
$= p(D_1 D_2 | A B) ~ p(A | B)$
Thus, the product of the likelihood and the prior the same for the ad hoc theory $A_D$ and the original theory $A$. You can’t win the Bayesian game by stapling the data to your theory. Ad hoc theories, by purchasing a better likelihood at the expense of a worse prior, get you nowhere in Bayes’ theorem. It’s the postulates that matter. Bayes’ Theorem is not distracted by data smuggled into the hypothesis.
### Too strong?
While all this is nice, it does assume rather strong conditions. It requires that the theory in question explicitly includes the evidence. If we look closely at the statements that make up $T$, we will find $D$ amongst them, i.e. we can write the theory as $T = T' D$. A theory can be jerry-rigged without being this obvious. I’ll have a closer look at this in a later post.
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Thumb rules refers to an easily learned process or standard based on practical experience rather than theory. It is not intended to be strictly accurate or reliable for every situation.Thumb rules give us the quick decisions at construction field. It provides with the rules of mathematical formula and broad ideas whenever needed. As a engineer, we have to be practical and quick in the field of construction. Basically,thumb rule is the guide for civil engineers for predictable results. It can vary to some percent but it will give a general idea as well.When we compare to get the result, the units in the thumb rules are not the same. Therefore, never consider units when performing the thumb rule.There are several thumb rules that we apply in construction. Here I am discussing the most frequently used thumb rules on the site.
Thumb rule to calculate the Concrete Volume with respect to area:
The Concrete Volume of 0.038 m3 of concrete is used for each Square Feet of Plan area.
Example:-
Lets consider , Plan Area = 45 x 25 =1125SFT
Building plan area
The plan area is 1125 SFT then the total volume of concrete required = 1125 x 0.038 m3 = 42.75 m3
Thumb rule to calculate the Steel quantity required for Slab, Beams, Footings & Columns:
For Residential buildings - 4.5Kgs - 4.75 Kgs /SFT
For Commercial buildings - 5.0Kgs - 5.50 Kgs/SFT
For more accurate results you can also use BN Datta recommendations:-
As per B N Datta, the Steel quantity used in different members of the building can be calculated easily using the following recommendations.
Examples:
FOR SLAB:
How to calculate the steel quantity of slab having the Length, width, and depth of the slab is 6m x 5m x 0.15m
Step 1: Calculate the Volume of Concrete
Total Volume of Concrete for above given Slab is 6 x 5 x 0.15 = 4.5 m3
Step 2: Calculate the steel quantity using formula
As per the above table, the steel quantity of slab is 1% of the total volume of concrete utilized.
Thumb rule to calculate Steel quantity of above slab = Volume of Concrete x Density of Steel x % of Steel of Member
Steel quantity required for above slab = 4.5 x 7850 x 0.01 = 353 Kgs
FOR COLUMN:
How to calculate the steel quantity of column having the Length, width, and height of the slab is 1.2m x 0.3m x 4.2 m
Step 1: Calculate the Volume of Concrete
Total Volume of Concrete for above given column is 1.2 x 0.3 x 4.2 = 1.512 m3
Step 2: Calculate the steel quantity using formula
As per the above table, the steel quantity of column is 2.5% of the total volume of concrete utilized.
Thumb rule to calculate Steel quantity of above slab = Volume of Concrete x Density of Steel x % of Steel of Member
Steel quantity required for above slab = 1.512 x 7850 x 0.025 = 297 Kgs
Similarly , you can calculate for Beam and Footings.
Thumb rule to calculate the Shuttering area:
Shuttering costs 15-18% of the total construction of the building. Shuttering is framed to bring the concrete in Shape and giving a desired mould in terms of quality.
Thumb rule to calculate the shuttering required is 6 times the quantity of concrete or 2.4 times of Plinth area.
Suppose, the concrete quantity is 0.8 m3, then the shuttering area required is 0.8 x 6 = 4.8 m2
Components of Shuttering:-
Components of Shuttering are shuttering ply, battens and nails.
Shuttering Ply Quantity calculation:
The shuttering Ply has a length, width & depth of 2.45 x 1.22 x 0.012
The No. of Shuttering Ply sheets = 0.22 times of Shuttering
Suppose, the area of shuttering required = 4.8 m2
Then the Shuttering Ply required = 0.22 x 4.8 = 1.056 m2
Battens Quantity Calculation:
Battens have different widths but we usually use shuttering batten having a length & width of 75mm x 40mm.
Batten Quantity = 19.82 x No. of Ply Sheets
if the construction requires 32 Ply sheets, the total quantity of Battens are 19.82 x 32 = 634 Battens
Thumb rule for Shuttering oil Quantity:
Shuttering oil is used to de-frame or de-assemble from the concrete easily.
Total required Shuttering oil is = 0.065 x Total Area of Shuttering
(or)
For every 15 m2 of shuttering 1 litre of shuttering oil is consumed.
Example :
15m2is total area of shuttering, then Shuttering oil required = 0.065 x 15 = 0.975 litres
Nails & Binding Wire Quantity in Shuttering:
75gms of Nails are used for 1m2 of Shuttering
75gms of Binding wire is consumed for every 1m2 of Shuttering.
Example:
15m2 is total area of shuttering, then Binding wire required = 75 x 15 = 1125gms
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New Zealand
Level 8 - NCEA Level 3
Graphing Complex Numbers - Rectangular Form
Lesson
The Argand Diagram is what we call the plane that will allow us to plot complex numbers. It is named after the Swiss mathematician Jean Argand (1768 - 1822). Using the $x$x-axis as the real axis, and the $y$y-axis as the imaginary axis, the ordered pairs $\left(a,b\right)$(a,b) reflect complex numbers of the form $a+bi$a+bi.
Plotting points on the plane is as simple as identifying the real and imaginary components from a complex number.
$z_1=2+3i$z1=2+3i $\left(a,b\right)=\left(2,3\right)$(a,b)=(2,3) $z_2=12-5i$z2=12−5i $\left(a,b\right)=\left(12,-5\right)$(a,b)=(12,−5) $z_3=-4-4i$z3=−4−4i $\left(a,b\right)=\left(-4,-4\right)$(a,b)=(−4,−4) $z_4=7$z4=7 $\left(a,b\right)=\left(7,0\right)$(a,b)=(7,0) $z_5=-6i$z5=−6i $\left(a,b\right)=\left(0,-6\right)$(a,b)=(0,−6)
Plotting these points on the Argand diagram would result in the following graph.
There is another way we can display complex number on the plane, and that is as a vector. The above complex numbers and all be represented as vectors on the plane with initial position $\left(0,0\right)$(0,0) and terminal position at the point $\left(a,b\right)$(a,b) as designated by the values of $a$a and $b$b in the number $a+bi$a+bi.
Activity
Try this for yourself before checking out the solution.
If $z=1-i$z=1i, find $z,z^2,z^3,z^4,z^5,z^6$z,z2,z3,z4,z5,z6
Plot these points on an Argand Diagram.
Is there a geometric pattern?
(see here for the solution)
More Worked Examples
QUESTION 1
Plot $6+2i$6+2i on the Argand diagram (complex plane).
QUESTION 2
Consider the following.
1. Graph the number $-1+2i$1+2i.
2. Evaluate $\left(-1+2i\right)+\left(9+5i\right)$(1+2i)+(9+5i).
3. Graph the result of $\left(-1+2i\right)+\left(9+5i\right)$(1+2i)+(9+5i).
QUESTION 3
Graph the complex number $-6$6 as a vector.
QUESTION 4
What is the complex number represented on the graph?
State the number in rectangular form.
Outcomes
M8-9
Manipulate complex numbers and present them graphically
91577
Apply the algebra of complex numbers in solving problems
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Questions easy to difficult
Chapter 9 Class 10 Some Applications of Trigonometry
Concept wise
• Questions easy to difficult
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Ex 9.1 , 8 A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. Here pedestal is AB & Statue is AC Height of pedestal = AB Length of the statue = 1.6m Hence, AC = 1.6m Angle of elevation to top of statue = 60° Hence, ∠CPB = 60° Angle of elevation at the top of the pedestal = 45° Hence, ∠APB = 45° We need to find height of pedestal i.e. AB Since statue is perpendicular to the ground ∠ ABP = 90° In ABP is a right triangle tan P = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑃)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑃) tan P = (" " 𝐴𝐵)/𝐵𝑃 tan 45° = (" " 𝐴𝐵)/𝐵𝑃 1 = (" " 𝐴𝐵)/𝐵𝑃 BP = AB Similarly, In a right angle triangle CBP tan P = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑃)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑃) tan P = (" " 𝐶𝐵)/𝐵𝑃 tan 60° = (" " 𝐶𝐵)/𝐵𝑃 √3 = (" " 𝐶𝐵)/𝐵𝑃 √3BP = CB √3AB = CB √3AB = AC + AB √3AB = 1.6 + AB √3AB – AB = 1.6 (√3−1)AB = 1.6 AB = 1.6/(√3 − 1)Rationalizing Multiplying (√3 " + 1)" in numerator and denominator AB = 1.6/(√3 − 1) × (√3 + 1)/(√3 + 1) AB = (1.6 × (√3 + 1))/((√3)^2 − 1^2 ) AB = (1.6 × (√3 + 1))/(3 −1) AB = (1.6 × (√3 + 1))/2 AB = 0.8 (√3 " + 1") Hence, Height of pedestal = AB = 0.8 (√3 " + 1") "m"
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# NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Written by Team Trustudies
Updated at 2021-05-07
## NCERT solutions for class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3.1
Q1 ) Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.Let the present age of Aftab and his daughter be x and y respectively.
So, seven years ago,
Aftab's age = $$x - 7$$
Age of Aftab's daughter = $$y-7$$
According to the given statement in the question:
=>$$x-7 = 7(y-7)$$
=>$$x-7 = 7y-49$$
=>$$x-7y = -42$$............(i)
Three solutions of the equation (i) are :
$$\begin{array} {|r|r|}\hline x & -7 & 0 & 7 \\ \hline y & 5 & 6 & 7 \\ \hline \end{array}$$
Now, three years hence,
Aftab's age = $$x+3$$
Age of Aftab's daughter = $$y+3$$
According to the given statement in the question:
=>$$x + 3 = 3(y+3)$$
=>$$x + 3 = 3y + 9$$
=>$$x -3y = 6$$............(ii)
Three solutions of the equation (ii) are :
$$\begin{array} {|r|r|}\hline x & 6 & 3 & 0 \\ \hline y & 0 & -1 & -2 \\ \hline \end{array}$$
Now the graph for the above equations (i) and (ii) will be:
Q2 ) The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and graphically.
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.Let the cost of cricket bat and cricket ball be Rs.x and Rs.y respectively.
According to the given statement in the question:
=>$$3x + 6y = 3900$$
=>$$x + 2y = 1300$$............(i)
Two solutions of the equation (i) are :
$$\begin{array} {|r|r|}\hline 0 & 1300 \\ \hline 650 & 0 \\ \hline \end{array}$$
And also according to the given statement in the question:
=>$$x + 3y = 1300$$............(ii)
Two solutions of the equation (i) are :
$$\begin{array} {|r|r|}\hline 0 & 1300 \\ \hline {{1300} \over {3}} & 0 \\ \hline \end{array}$$
Now the graph for the above equations (i) and (ii) will be:
Q3 ) The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.Let the cost of 1 kg of apples and 1 kg of grapes be Rs.x and Rs.y respectively.
According to the given statement in the question:
=>$$2x + y = 160$$............(i)
Two solutions of the equation (i) are :
$$\begin{array} {|r|r|}\hline 50 & 45 \\ \hline 60 & 70 \\ \hline \end{array}$$
And also according to the given statement in the question:
=>$$4x + 2y = 300$$
=>$$2x + y = 150$$............(ii)
Two solutions of the equation (i) are :
$$\begin{array} {|r|r|}\hline 50 & 40 \\ \hline 50 & 70 \\ \hline \end{array}$$
Now the graph for the above equations (i) and (ii) will be:
## NCERT solutions for class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3.2
Q1 ) Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of class X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz
(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans. (i)Let the number of boys and girls who took part in quiz be x and y respectively.
According to the given statement in the question:
=>$$x + y = 10)$$............(1)
Points which lie on the equation (1) are :
$$\begin{array} {|r|r|}\hline x & 0 & 10 \\ \hline y & 10 & 0 \\ \hline \end{array}$$
According to the given statement in the question:
=>$$x - y = -4)$$............(2)
Solutions of the equation (2) are :
$$\begin{array} {|r|r|}\hline x & 0 & -4 \\ \hline y & 4 & 0 \\ \hline \end{array}$$
Now the graph for the above equations (1) and (2) will be:
From the graph it is visible that equation (1) and (2) intersect at $$(3,7)$$
Therefore, number of boys who took part in quiz = 3 and, number of girls who took part in quiz = 7.
(ii)Let the cost of one pencil and cost of one pen be Rs.x and Rs.y respectively.
According to the given statement in the question:
=>$$5x + 7y = 50$$............(1)
Points which lie on the equation (1) are :
$$\begin{array} {|r|r|}\hline x & 10 & 3 \\ \hline y & 0 & 5 \\ \hline \end{array}$$
According to the given statement in the question:
=>$$7x + 5y = 46$$............(2)
Solutions of the equation (2) are :
$$\begin{array} {|r|r|}\hline x & 8 & 3 \\ \hline y & -2 & 5 \\ \hline \end{array}$$
Now the graph for the above equations (1) and (2) will be:
From the graph it is visible that equation (1) and (2) intersect at $$(3,5)$$
Therefore, cost of pencil = Rs.3 and, Cost of pen = Rs.5.
Q2 ) On comparing the ratios $${{a_1}\over {a_2}},{{b_1}\over {b_2}},{{c_1}\over {c_2}}$$ find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) $$5x - 4y + 8 = 0, 7x + 6y – 9 = 0$$
(ii)$$9x + 3y + 12 = 0, 18x + 6y + 24 = 0$$
(iii)$$6x - 3y + 10 = 0, 2x – y + 9 = 0$$
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.(i) $$5x - 4y + 8 = 0, 7x + 6y – 9 = 0$$
On comparing these equations with the general form: $$ax^2 + bx + c$$
$$a_1 = 5, b_1 = -4, c_1 = 8$$,
$$a_2 = 7, b_2 = 6, c_2 = -9$$
Here $${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$ as,
$${{5}\over {7}} \ne {{-4}\over {6}}$$
So, these lines have a unique solution which means they intersect at one point.
(ii) $$9x + 3y + 12 = 0,18x + 6y + 24 = 0$$
On comparing these equations with the general form: $$ax^2 + bx + c$$
$$a_1 = 9, b_1 = 3, c_1 = 12$$
$$a_2 = 18, b_2 = 6, c_2 = 24$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} = {{c_1}\over {c_2}}$$ as,
$${{9}\over {18}} = {{3}\over {6}} = {{12}\over {24}}$$
So, these lines are coincident.
(iii) $$6x - 3y + 10 = 0,2x – y + 9 = 0$$
On comparing these equations with the general form: $$ax^2 + bx + c$$
$$a_1 = 6, b_1 = -3, c_1 = 10$$
$$a_2 = 2, b_2 = -1, c_2 = 9$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}$$ as,
$${{6}\over {2}} = {{-3}\over {-1}} \ne {{10}\over {9}}$$
So, these lines are parallel to each other.
Q3. On comparing the ratios $${{a_1}\over {a_2}},{{b_1}\over {b_2}} and {{c_1}\over {c_2}}$$, find out whether the following pair of linear equations are consistent, or inconsistent.
(i) $$3x + 2y = 5, 2x - 3y = 7$$
(ii)$$2x - 3y = 8, 4x - 6y = 9$$
(iii)$${{3}\over {2}}x + {{5}\over {3}}y = 7,9x - 10y = 14$$
(iv)$$5x - 3y = 11, -10x + 6y = -22$$
(v)$$\frac{4}{3}x + 2y = 8, 2x + 3y = 12$$
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.(i) $$3x + 2y = 5, 2x - 3y = 7$$
On comparing these equations with the general form: $$ax^2 + bx + c$$
$$a_1 = 3, b_1 = 2, c_1 = -5$$
$$a_2 = 2, b_2 = -3, c_2 = -7$$
Here $${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$ as,
$${{3}\over {2}} \ne {{2}\over {3}}$$
So, these lines have a unique solution which means they intersect at one point.
Hence, they are consistent.
(ii) $$2x - 3y = 8, 4x - 6y = 9$$
On comparing these equations with the general form: $$ax^2 + bx + c$$
$$a_1 = 2, b_1 = -3, c_1 = -8$$
$$a_2 = 4, b_2 = 6, c_2 = -9$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}$$ as,
$${{1}\over {2}} = {{1}\over {2}} \ne {{-8}\over {-9}}$$
So, these lines are parallel to each other.
Hence, they are inconsistent.
(iii) $${{3}\over {2}}x + {{5}\over {3}}y = 7,9x - 10y = 14$$
On comparing these equations with the general form: $$ax^2 + bx + c$$
$$a_1 = \frac{3}{2}, b_1 = \frac{5}{3}, c_1 = 7$$
$$a_2 = 9, b_2 = -10, c_2 = 14$$
Here $${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$ as,
$${{{{3}\over {2}}}\over {9}} \ne {{{{5}\over {3}}}\over {-10}}$$
=>$${{1}\over {6}} \ne {{-1}\over {6}}$$
So, these lines have a unique solution which means they intersect at one point.
Hence, they are consistent.
(iv) $$5x - 3y = 11, -10x + 6y = -22$$
On comparing these equations with the general form: $$ax^2 + bx + c$$
$$a_1 = 5, b_1 = -3, c_1 = 11$$
$$a_2 = -10, b_2 = 6, c_2 = -22$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} = {{c_1}\over {c_2}}$$ as,
$${{5}\over {-10}} = {{-3}\over {6}} = {{11}\over {-22}}$$
=>$${{-1}\over {2}} = {{-1}\over {2}} = {{-1}\over {2}}$$
So, these lines have infinite many solutions.
Hence, they are consistent.
(v)$$\frac{4}{3}x + 2y = 8, 2x + 3y = 12$$
On comparing these equations with the general form: $$ax^2 + bx + c$$
$$a_1 = \frac{4}{3}, b_1 = 2, c_1 = 8$$
$$a_2 = 2, b_2 = 3, c_2 = 12$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} = {{c_1}\over {c_2}}$$ as,
$${{{{4}\over {3}}}\over {2}} = {{2}\over {3}} = {{8}\over {12}}$$
=>$${{2}\over {3}} = {{2}\over {3}} = {{2}\over {3}}$$
So, these lines have infinite many solutions.
Hence, they are consistent.
Q4 ) Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) $$x + y = 5, 2x + 2y = 10$$
(ii)$$x – y = 8, 3x - 3y = 16$$
(iii)$$2x + y = 6, 4x - 2y = 4$$
(iv)$$2x - 2y – 2 = 0, 4x - 4y – 5 = 0$$
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
(i) $$x + y = 5, 2x + 2y = 10$$
For equation $$x + y -5 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 0 & 5 \\ \hline y & 5 & 0 \\ \hline \end{array}$$
For equation $$2x + 2y = 10$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 1 & 2 \\ \hline y & 4 & 3 \\ \hline \end{array}$$
Graph for the above equations is:
On plotting the graph we can see that both of the lines coincide.
So, there are infinitely many solutions.
Hence, they are consistent.
(ii) $$x – y = 8, 3x - 3y = 16$$
For equation $$x - y - 8 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 0 & 8 \\ \hline y & -8 & 0 \\ \hline \end{array}$$
For equation $$3x - 3y = 16$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 0 & 16 \\ \hline y & -16 & 0 \\ \hline \end{array}$$
Graph for the above equations is:
On plotting the graph we can see that both of the lines are parallel to each other.
So, there are no solutions.
Hence, they are inconsistent.
(iii) $$2x + y = 6, 4x - 2y = 4$$
For equation $$2x + y - 6 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 0 & 3 \\ \hline y & 6 & 0 \\ \hline \end{array}$$
For equation $$4x – 2y – 4 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 0 & 1 \\ \hline y & -2 & 0 \\ \hline \end{array}$$
Graph for the above equations is:
On plotting the graph we can see that both of the lines are intersecting each other at exactly one point.
So, there is a unique solution.
Hence, they are consistent.
(iv) $$2x - 2y – 2 = 0, 4x - 4y – 5 = 0$$
For equation $$2x - 2y – 2 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 2 & 0 \\ \hline y & 0 & -2 \\ \hline \end{array}$$
For equation $$4x – 2y – 4 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 5 & 0 \\ \hline y & 0 & -5 \\ \hline \end{array}$$
Graph for the above equations is:
On plotting the graph we can see that both of the lines are parallel to each other.
So, there is no solution.
Hence, they are inconsistent.
Q5 ) Half the perimeter of a rectangle garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.Let the length of rectangular garden and width of rectangular garden be x metres and y metres respectively.
Given, the perimeter = 36m
=>$$x + y = 36$$.......(i)
Also $$x = y + 4$$
=>$$x - y = 4$$........(ii)
On adding equation (i) and (ii),
=>$$2x = 40$$
=>$$x = 20$$m
On subtracting equation (ii) from (i),
=>$$2y = 32$$
=>$$y = 16$$m
Hence , length = 20 m and width = 16 m.
Q6 ) Given the linear equation $$2x + 3y – 8 = 0$$, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) Intersecting lines
(ii)Parallel lines
(iii)Coincident lines
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.(i) Intersecting lines
The condition for intersecting lines is :
$${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$
So the second equation can be $$x + 2y = 3$$
Since, $${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$
=>$${{2}\over {1}} \ne {{3}\over {2}}$$
(ii) Parallel lines
The condition for parallel lines is :
$${{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}$$
So the second equation can be $$2x + 3y – 2 = 0$$
Since, $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}$$
=>$${{2}\over {2}} = {{3}\over {3}} \ne {{-8}\over {-2}}$$
(ii) Coincident lines
The condition for coincident lines is :
$${{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}$$
So the second equation can be $$4x + 6y – 16 = 0$$
Since, $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} = {{c_1}\over {c_2}}$$
=>$${{2}\over {4}} = {{3}\over {6}} = {{-8}\over {-16}}$$
=>$${{1}\over {2}} = {{1}\over {2}} = {{1}\over {2}}$$
Q7 ) Draw the graphs of the equations $$x – y + 1 = 0$$ and $$3x + 2y – 12 = 0$$. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.For equation $$x – y + 1 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 0 & 1 \\ \hline y & 1 & 0 \\ \hline \end{array}$$
For equation $$3x + 2y – 12 = 0$$, points which lie on the line:
$$\begin{array} {|r|r|}\hline x & 4 & 0 \\ \hline y & 0 & 6 \\ \hline \end{array}$$
Graph for the above equations is:
We can see from the graph that the point of intersection of the lines with each other is (2, 3) and the points of intersection with x - axis are :
(–1, 0) and (4, 0).
Thus, the coordinates of the vertices of the triangle formed are (2, 3), (-1, 0) and (4, 0).
## NCERT solutions for class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3.3
Q1 ) Solve the following pair of linear equations by the substitution method.
(i) $$x + y = 14,x – y = 4$$
(ii) $$s – t = 3,{{s}\over{3}} + {{t}\over{2}} = 6$$
(iii) $$3x – y = 3,9x - 3y = 9$$
(iv) $$0.2x + 0.3y = 1.3,0.4x + 0.5y = 2.3$$
(v) $$\sqrt{2}x + \sqrt{3}y = 0,\sqrt{3}x - \sqrt{8}y = 0$$
(vi) $${{3x}\over{2}} - {{5y}\over{3}} = -2,{{x}\over{3}} + {{y}\over{2}} = {{13}\over{6}}$$
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.(i)$$x + y = 14$$...........(i)
=> $$x – y = 4$$..............(ii)
Using equation (ii)
=> $$x = 4 + y$$.............(iii)
On substituting (iii) in (i):
=>$$4 + 2y = 14$$
=>$$2y = 10$$
=>$$y = 5$$...........(iv)
On substituting (iv) in (i):
=>$$x + 5 = 14$$
=>$$x = 14 - 5$$
=>$$x = 9$$
So, the value of $$y = 5$$ and $$x = 9$$.
(ii)$$s – t = 3$$...........(i)
=> $${{s}\over{3}} + {{t}\over{2}} = 6$$..............(ii)
Using equation (i)
=> $$s = 3 + t$$.............(iii)
On substituting (iii) in (ii):
=>$${{3 + t}\over{3}} + {{t}\over{2}} = 6$$
=>$${{6 + 2t + 3t}\over{6}} = 6$$
=>$$5t + 6 = 36$$
=>$$5t = 30$$
=>$$t = 6$$..............(iv)
On substituting (iv) in (i):
=>$$s – 6 = 3$$
=>$$s = 6 + 3$$
=>$$s = 9$$
So, the value of $$t = 6$$ and $$s = 9$$.
(iii) $$3x – y = 3,9x - 3y = 9$$
=> $$3x - y = 3$$..........(i)
=> $$9x - 3y = 9$$........(ii)
Using equation (i),
=> $$y = 3x - 3$$........(iii)
On substituting (iii) in (ii):
=> $$9x - 9x + 9 = 9$$
=> 9 = 9
This is always true.
Hence the given pair of equations have infinitely many solutions. The relation between them is given by :
$$y = 3x - 3$$
One such solution can be x = 1 and y = 0.
(iv)$$0.2x + 0.3y = 1.3$$...........(i)
=> $$0.4x + 0.5y = 2.3$$..............(ii)
Using equation (i)
=> $$0.2x = 1.3 - 0.3y$$
=>$$x = {{1.3 - 0.3y} \over {0.2}}$$.............(iii)
On substituting (iii) in (ii):
=>$$0.4({{1.3 - 0.3y }\over {0.2}}) + 0.5y = 2.3$$
=>$$2.6 - 0.6y + 0.5y = 2.3$$
=>$$-0.1y = -0.3$$
=>$$y = 3$$.............(iv)
On substituting (iv) in (i):
=>$$0.2x + 0.3 (3) = 1.3$$
=>$$0.2x + 0.9 = 1.3$$
=>$$0.2x = 0.4 => x = 2$$
So, the value of $$y = 3$$ and $$x = 2$$.
(v)$$\sqrt{2}x + \sqrt{3}y = 0$$...........(i)
=> $$\sqrt{3}x - \sqrt{8}y = 0$$..............(ii)
Using equation (i)
=> $$x = y\sqrt{{-3}\over {2}}$$.............(iii)
On substituting (iii) in (ii):
=>$$\sqrt{3}(y\sqrt{{-3}\over {2}}) - \sqrt{8}y =0$$
=>$${{-3y}\over {\sqrt 2}} - \sqrt{8}y =0$$
=>$$y({{-3}\over {\sqrt 2}} - \sqrt{8}) =0$$
=>$$y = 0$$.............(iv)
On substituting (iv) in (iii):
=>$$x = 0$$
So, the value of $$y = 0$$ and $$x = 0$$.
(vi)$${{3x}\over{2}} - {{5y}\over{3}} = -2$$...........(i)
=> $${{x}\over{3}} + {{y}\over{2}} = {{13}\over{6}}$$..............(ii)
Using equation (ii)
=> $$x = ({{13}\over{6}} - {{y}\over{2}}) × 3$$
=> $$x = {{13}\over{2}} - {{3y}\over{2}}$$.............(iii)
On substituting (iii) in (i):
=>$${3({{13}\over{2}} - {{3y}\over{2}})\over{2}} - {{5y}\over{3}} = -2$$
=>$${{39}\over{4}} - {{9y}\over{4}} - {{5y}\over{3}} = -2$$
=>$${{-27y -20y}\over{12}} = -2 - {{39}\over{4}}$$
=>$${{-47y}\over{12}} = {{-8-39}\over{4}}$$
=>$${{-47y}\over{12}} = {{-47}\over{4}}$$
=>$$y = 3$$
On substituting (iv) in (ii):
=>$${{x}\over{3}} + {{3}\over{2}} = {{13}\over{6}}$$
=>$${{x}\over{3}} = {{13}\over{6}} - {{3}\over{2}}$$
=>$${{x}\over{3}} = {{2}\over{3}}$$
=>$$x = 2$$
So, the value of $$y = 3$$ and $$x = 2$$.
Q2 ) Solve $$2x + 3y = 11$$ and $$2x - 4y = -24$$ and hence find the value of ‘m’ for which $$y = mx + 3$$.
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
$$2x + 3y = 11$$...........(i)
=> $$2x - 4y = -24$$..............(ii)
Using equation (ii)
=> $$2x = -24 + 4y$$
=> $$x = -12 + 2y$$.............(iii)
On substituting (iii) in (i):
=>$$2(-12 + 2y) + 3y = 11$$
=>$$-24 + 4y + 3y = 11$$
=>$$7y = 35$$
=>$$y = 5$$...........(iv)
On substituting (iv) in (i):
=>$$2x + 3 (5) = 11$$
=>$$2x = 11 – 15 = -4$$
=>$$x = -2$$
So, the value of $$y = 5$$ and $$x = -2$$.
On substituting the values of x and y in $$y = mx + 3$$,
=> $$5 = m (-2) + 3$$
=> $$5 = -2m + 3$$
=> $$-2m = 2$$
$$=> m = -1$$
Q3 ) Form a pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii)The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii)The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
(iv)The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes $${{9}\over{11}}$$ , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and denominator it becomes $${{5}\over{6}}$$. Find the fraction.
(vi)Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans (i)Let first number and second number be x and y respectively.
According to given question
$$x – y = 26$$ (assuming x > y) … (1)
$$x = 3y$$(Since,x > y)… (2)
On putting equation (2) in (1),
$$3y – y = 26$$
=>$$2y = 26$$
=>$$y = 13$$
On putting value of y in equation (2),
=>$$x = 3y = 3(13) = 39$$
Therefore, two numbers are 13 and 39.
(ii) Let smaller angle and larger angle be x and y respectively.
According to given question
$$y = x + 18$$ … (1)
$$x + y = 180$$(Sum of supplementary angle)… (2)
On putting equation (1) in (2),
$$x + x + 18 = 180$$
=>$$2x = 180 - 18 = 162$$
=>$$x = 81^\circ$$
On putting value of x in equation (1),
=>$$y = x + 18 = 81 + 18 = 99^\circ$$
Therefore, two numbers are $$81^\circ$$ and $$99^\circ$$
(iii) Let cost of one bat and cost of one ball be x and y respectively.
According to given question
$$7x + 6y = 3800$$ .....(1)
$$3x + 5y = 1750$$ .....(2)
Using equation (1):
$$7x = 3800 - 6y$$
=>$$x = {{3800 - 6y} \over{7}}$$ .....(3)
On putting equation (3) in (1),
$$3({{3800 - 6y} \over{7}}) + 5y = 1750$$
=>$${{11400 - 18y} \over{7}} + 5y = 1750$$
=>$${{5y} \over{1}} - {{18y} \over{7}} = {{1750} \over{1}} -{{11400 } \over{7}}$$
=>$${{35y - 18y} \over{7}} = {{12250 - 11400} \over{7}}$$
=>$$17y = 850$$
=>$$y = 50$$
On putting value of y in equation (2),
=>$$3x + 250 = 1750$$
=>$$3x = 1500$$
=>$$x = 500$$
Therefore, cost of one bat and cost of one ball is Rs.500 and Rs.50 respectively.
(iv) Let fixed charge and charge for every km be Rs.x and Rs.y respectively.
According to given question
$$x + 10y = 105$$ … (1)
$$x + 15y = 155$$ … (2)
Using equation (1)
$$x = 105 - 10y$$ … (3)
On putting equation (3) in (2),
$$105 - 10y + 15y = 155$$
=>$$5y = 50$$
=>$$y = 10$$
On putting value of y in equation (1),
=>$$x + 10 (10) = 105$$
=>$$x = 105 – 100 = 5$$
Therefore, fixed charge and charge for every km is Rs.5 and Rs.10 respectively.
To travel distance of 25 Km, person will have to pay
=> Rs $$x + 25y$$
=> Rs $$5 + 25 × 10$$
=> Rs $$5 + 250$$ = Rs 255
(v) Let numerator and denominator be x and y respectively.
According to given question
$${{x + 2} \over{y + 2}} = {{9} \over{11}}$$ .........(1)
$${{x + 3} \over{y + 3}} = {{5} \over{6}}$$ .........(2)
Using equation (2):
$$6(x + 3) = 5(y + 3)$$.......(4)
Using equation (1):
$$11(x + 2) = 9(y + 2)$$
=>$$11x + 22 = 9y + 18$$
=>$$11x = 9y - 4$$
=>$$x = {{9y - 4} \over{11}}$$........(3)
On putting equation (3) in (4),
$$6({{9y - 4} \over{11}} + 3) = 5(y + 3)$$
=>$${{54y} \over{11}} - {{24} \over{11}} + 18 = 5y + 15$$
=>$$-{{24} \over{11}} + {{33} \over{11}} = {{55y} \over{11}} - {{54y} \over{11}}$$
=>$$-{{24 + 33} \over{11}} = {{55y - 54y} \over{11}}$$
=>$$y = 9$$
On putting value of y in equation (1),
=>$${{x + 2} \over{9 + 2}} = {{9} \over{11}}$$
=>$$x + 2 = 9$$
=>$$x = 7$$
Therefore, numerator and denominator is 7 and 9 respectively.
Fraction = $${{7} \over{9}}$$
(vi) Let present age of Jacob and his son be x and y respectively.
According to given question
$$x + 5 = 3(y + 5)$$ .....(1)
$$x - 5 = 7(y - 5)$$ .....(2)
Using equation (1):
$$x + 5 = 3y + 15$$
=>$$x = 10 + 3y$$ .....(3)
On putting equation (3) in (2),
$$10 + 3y – 5 = 7y - 35$$
=>$$-4y = -40$$
=>$$y = 10$$
On putting value of y in equation (3),
=>$$x = 10 + 3(10)$$
=>$$x = 40$$
Therefore, present age of Jacob and his son be 40 years and 10 years respectively.
## NCERT solutions for class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3.4
Q1 ) Solve the following pair of linear equations by the elimination method and the substitution method:
(i) $$x + y = 5,2x – 3y = 4$$
(ii) $$3x + 4y = 10, 2x – 2y = 2$$
(iii) $$3x - 5y – 4 = 0, 9x = 2y + 7$$
(iv) $${{x}\over{2}} + {{2y}\over{3}} = -1, x - {{y}\over{3}} = 3$$
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.(i) $$x + y = 5$$ … (1)
$$2x – 3y = 4$$ … (2)
Elimination method:
On multiplying equation (1) by 2, we get equation (3)
$$2x + 2y = 10$$ … (3)
$$2x - 3y = 4$$ … (2)
On subtracting equation (2) from (3), we get
$$5y = 6$$
=> $$y = {{6}\over{5}}$$
On putting value of y in (1), we get
$$x + {{6}\over{5}} = 5$$
$$x = 5 - {{6}\over{5}} = {{19}\over{5}}$$
Therefore, x = $${{6}\over{5}}$$ and y = $${{19}\over{5}}$$
Substitution method:
$$x + y = 5$$ … (1)
$$2x - 3y = 4$$ … (2)
From equation (1), we get,
$$x = 5 - y$$
On putting this in equation (2), we get
$$2 (5 - y) - 3y = 4$$
=>$$10 - 2y - 3y = 4$$
=>$$5y = 6$$
=>$$y = {{6}\over{5}}$$
On putting value of y in (1), we get
$$x = 5 - {{6}\over{5}} = {{19}\over{5}}$$
Therefore, y = $${{6}\over{5}}$$ and x = $${{19}\over{5}}$$
(ii) $$3x + 4y = 10$$ … (1)
$$2x – 2y = 2$$ … (2)
Elimination method:
On multiplying equation (2) by 2, we get equation (3)
$$4x - 4y = 4$$ … (3)
$$3x + 4y = 10$$ … (1)
On adding equation (1) and (3), we get
$$7x = 14 => x = 2$$
On putting value of x in (1), we get
$$3 (2) + 4y = 10$$
$$4y = 10 – 6 = 4$$
$$=> y = 1$$
Therefore, x = $$2$$ and y = $$1$$
Substitution method:
$$3x + 4y = 10$$ … (1)
$$2x - 2y = 2$$ … (2)
From equation (2), we get,
$$2x = 2 + 2y$$
=> $$x = 1 + y$$..........(3)
On putting this in equation (1), we get
$$3 (1 + y) + 4y = 10$$
=>$$3 + 3y + 4y = 10$$
=>$$7y = 7 => y = 1$$
On putting value of y in (3), we get
$$x = 1 + 1 = 2$$
Therefore, x = $$2$$ and y = $$1$$
(iii) $$3x - 5y – 4 = 0$$ … (1)
$$9x = 2y + 7$$… (2)
Elimination method:
Multiplying (1) by 3, we get (3)
$$9x - 15y – 12 = 0$$… (3)
$$9x - 2y – 7 = 0$$… (2)
On subtracting (2) from (3), we get
$$-13y – 5 = 0$$
=>$$-13y = 5$$
=> $$y = {{-5} \over {13}}$$
On putting value of y in (1), we get
$$3x – 5({{-5} \over {13}}) - 4 = 0$$
=>$$3x = 4 - {{25} \over {13}}= {{52 - 25} \over {13}} = {{27} \over {13}}$$
=> $$x = {{27} \over {13(3)}} = {{9} \over {13}}$$
Therefore, x = $${{9} \over {13}}$$ and y = $${{-5} \over {13}}$$
Substitution Method:
$$3x - 5y – 4 = 0$$ … (1)
$$9x = 2y + 7$$… (2)
From equation (1), we can say that
$$3x = 4 + 5y => x = {{4 + 5y} \over {3}}$$
On putting this in equation (2), we get
$$9({{4 + 5y} \over {3}}) - 2y = 7$$
=>$$12 + 15y - 2y = 7$$
=>$$13y = -5$$
$$y = {{-5} \over {13}}$$
On putting value of y in (1), we get
$$3x – 5 ({{-5} \over {13}})= 4$$
=>$$3x = 4 - {{25} \over {13}} = {{52 - 25} \over {13}} = {{27} \over {13}}$$
=>$$x = {{27} \over {13(3)}} = {{9} \over {13}}$$
Therefore, x = $${{9} \over {13}}$$ and y = $${{-5} \over {13}}$$
(iv)$${{x}\over{2}} + {{2y}\over{3}} = -1$$ … (1)
$$x - {{y}\over{3}} = 3$$… (2)
Elimination method:
On multiplying equation (2) by 2, we get (3)
$$2x - {{2y}\over{3}} = 6$$… (3)
$${{x}\over{2}} + {{2y}\over{3}} = -1$$… (1)
Adding (3) and (1), we get
$${{5x}\over{2}} = 5 => x = 2$$
On putting value of x in (2), we get
$$2 - {{y}\over{3}} = 3$$
=>$$y = -3$$
Therefore, x = 2 and y = -3
Substitution method:
$${{x}\over{2}} + {{2y}\over{3}} = -1$$ … (1)
$$x - {{y}\over{3}} = 3$$… (2)
From equation (2), we can say that
=> $$x = 3 + {{y}\over{3}} = {{9 + y}\over{3}}$$
Putting this in equation (1), we get
=>$${{9 + y}\over{6}} + {{2y}\over{3}} = -1$$
=>$${{9 + y + 4y}\over{6}} = -1$$
=>$$5y + 9 = -6$$
=>$$5y = -15$$
=>$$y = -3$$
Putting value of y in (1), we get
=>$${{x}\over{2}} + {{2(-3)}\over{3}} = -1$$
=>$$x = 2$$
So, the value of $$y = -3$$ and $$x = 2$$.
Q2 ) Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes ½ if we only add 1 to the denominator. What is the fraction?
(ii)Five years ago, Nuri was thrice as old as sonu. Ten years later, Nuri will be twice as old as sonu. How old are Nuri and Sonu?
(iii)The sum of the digits of a two–digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.(i)Let numerator =x and let denominator =y
According to given condition,
=>$${{x + 1}\over{y - 1}} = 1$$ and $${{x}\over{y + 1}} = {{1}\over{2}}$$
=>$$x + 1 = y – 1$$ and $$2x = y + 1$$
=>$$x – y = -2$$….......(1) and,
$$2x – y = 1$$…...... (2)
So, we have equations (1) and (2), multiplying equation (1) by 2 we get (3)
$$2x - 2y = -4$$… (3)
$$2x – y = 1$$… (2)
Subtracting equation (2) from (3), we get
$$-y = -5 => y = 5$$
Putting value of y in (1), we get
$$x – 5 = -2 => x = -2 + 5 = 3$$
Therefore, fraction = $${{x}\over{y}} = {{3}\over{5}}$$
(ii) Let present age of Nuri = x years and let present age of Sonu = y years
5 years ago, age of Nuri = $$(x – 5)$$ years
5 years ago, age of Sonu = $$(y – 5)$$ years
According to given condition,
$$(x - 5) = 3 (y - 5)$$
=> $$x – 5 = 3y – 15$$
=> $$x - 3y = -10$$… (1)
10 years later from present, age of Nuri = $$(x + 10)$$ years
10 years later from present, age of Sonu = $$(y + 10)$$ years
According to given condition,
$$(x + 10) = 2 (y + 10)$$
=> $$x + 10 = 2y + 20$$
=> $$x - 2y = 10 … (2)$$
Subtracting equation (1) from (2),
$$y = 10 - (-10) = 20$$years
Putting value of y in (1),
$$x – 3 (20) = -10$$
=> $$x – 60 = -10$$
=> $$x = 50$$years
Therefore, present age of Nuri = 50 years and present age of Sonu = 20 years
(iii) Let digit at ten’s place = x and Let digit at one’s place = y
According to given condition,
$$x + y = 9$$… (1)
And 9 (10x + y) = 2 (10y + x)
=>$$90x + 9y = 20y + 2x$$
=>$$88x = 11y$$
=>$$8x = y$$
=>$$8x – y = 0$$… (2)
$$9x = 9 => x = 1$$
Putting value of x in (1),
$$1 + y = 9$$
=>$$y = 9 – 1 = 8$$
Therefore, number = $$10x + y = 10 (1) + 8 = 10 + 8 = 18$$
(iv)Let number of Rs 100 notes = x and let number of Rs 50 notes = y
According to given conditions,
$$x + y = 25$$ … (1)
and $$100x + 50y = 2000$$
=>$$2x + y = 40$$ … (2)
Subtracting (2) from (1),
$$-x = -15 => x = 15$$
Putting value of x in (1),
$$15 + y = 25$$
=>$$y = 25 – 15 = 10$$
Therefore, number of Rs 100 notes = 15 and number of Rs 50 notes = 10
(v) Let fixed charge for 3 days = Rs x
Let additional charge for each day thereafter = Rs y
According to given condition,
$$x + 4y = 27$$… (1)
$$x + 2y = 21$$… (2)
Subtracting (2) from (1),
$$2y = 6 => y = 3$$
Putting value of y in (1),
$$x + 4 (3) = 27$$
=>$$x = 27 – 12 = 15$$
Therefore, fixed charge for 3 days = Rs 15 and additional charge for each day after 3 days = Rs 3
## NCERT solutions for class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3.5
Q1 ) Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.
(i) $$x – 3y – 3 = 0,3x – 9y – 2 = 0$$
(ii) $$2x + y = 5,3x + 2y = 8$$
(ii) $$3x - 5y = 20,6x - 10y = 40$$
(ii) $$x - 3y – 7 = 0,3x - 3y – 15 = 0$$
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.(i) $$x – 3y – 3 = 0,3x – 9y – 2 = 0$$
On comparing these equations with the general form : $$ax^2 + bx + c$$
$$a_1 = 1, b_1 = -3, c_1 = -3,a_2 = 3, b_2 = -9, c_2 = -2$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}$$
=>$${{1}\over {3}} = {{-3}\over {-9}} \ne {{-3}\over {-2}}$$
So, these lines are parallel to each other.
Thus, the following pairs of linear equations will have no solution.
(ii) $$2x + y = 5,3x + 2y = 8$$
On comparing these equations with the general form : $$ax^2 + bx + c$$
$$a_1 = 2, b_1 = 1, c_1 = -5,a_2 = 3, b_2 = 2, c_2 = 8$$
Here $${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$
=>$${{2}\over {3}} \ne {{1}\over {2}}$$
So, these lines have a unique solution.
We can find the unique solution using cross multiplication:
=>$${{x}\over {(-8)(-1)-(2)(-5)}} = {{y}\over {(-5)(3)-(-8)(2)}} = {{1}\over {(2)(2)-(3)(1)}}$$
=>$${{x}\over {-8 + 10}} = {{y}\over {-15 + 16}} = {{1}\over {4 - 3}}$$
=>$${{x}\over {2}} = {{y}\over {1}} = {{1}\over {1}}$$
=> x = 2 and y = 1.
(iii) $$3x - 5y = 20,6x - 10y = 40$$
On comparing these equations with the general form : $$ax^2 + bx + c$$
$$a_1 = 3, b_1 = -5, c_1 = -20,a_2 = 6, b_2 = -10, c_2 = -40$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} = {{c_1}\over {c_2}}$$
=>$${{3}\over {6}} = {{-5}\over {-10}} = {{-20}\over {-40}}$$
So, these lines coincide with each other.
Hence, there are infinite many solutions.
(iv) $$x - 3y – 7 = 0,3x - 3y – 15 = 0$$
On comparing these equations with the general form : $$ax^2 + bx + c$$
$$a_1 = 1, b_1 = -3, c_1 = -7,a_2 = 3, b_2 = -3, c_2 = -15$$
Here $${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$
=>$${{1}\over {3}} \ne {{-3}\over {-3}}$$
So, these lines have a unique solution.
We can find the unique solution using cross multiplication:
=>$${{x}\over {(-3)(-15)-(-3)(-7)}} = {{y}\over {(-7)(3)-(-15)(1)}} = {{1}\over {(-3)(1)-(-3)(3)}}$$
=>$${{x}\over {45 - 21}} = {{y}\over {-21 + 15}} = {{1}\over {-3 + 9}}$$
=>$${{x}\over {24}} = {{y}\over {-6}} = {{1}\over {6}}$$
=> x = 4 and y = -1.
Q2 ) (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7
(a - b) x + (a + b) y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k - 1) x + (k - 1) y = 2k + 1
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
(i) $$2x + 3y = 7$$
$$(a - b) x + (a + b) y = 3a + b – 2$$
On comparing these equations with the general form : $$ax^2 + bx + c$$
$$a_1 = 2, b_1 = 3, c_1 = -7,a_2 = (a-b), b_2 = (a + b), c_2 = 2 - b -3a$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} = {{c_1}\over {c_2}}$$ since it has infinite solutions
=>$${{2}\over {a-b}} = {{3}\over {a+b}} = {{-7}\over {2 - b -3a}}$$
=>$${{2}\over {a-b}} = {{3}\over {a+b}}$$ and $${{3}\over {a+b}} = {{-7}\over {2 - b -3a}}$$
=>$$2a + 2b = 3a - 3b$$ and $$6 - 3b - 9a = -7a - 7b$$
=>$$a = 5b$$.....(i) and $$-2a = -4b – 6$$......(ii)
Putting (i) in (ii), we get
$$-2 (5b) = -4b – 6$$
=>$$-10b + 4b = -6$$
=> $$-6b = –6 - b = 1$$
Putting value of b in (i), we get
$$a = 5b = 5 (1) = 5$$
Therefore, a = 5 and b = 1
Hence, for the values a = 5 and b = 1, the given pair of equations have infinite solutions.
(ii) $$3x + y = 1$$
$$(2k - 1) x + (k - 1) y = 2k + 1$$
On comparing these equations with the general form : $$ax^2 + bx + c$$
$$a_1 = 3, b_1 = 1, c_1 = -1,a_2 = (2k - 1), b_2 = (k - 1), c_2 = -(2k + 1)$$
Here $${{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}$$ since it has no solutions
=>$${{3}\over {2k - 1}} = {{1}\over {k - 1}} \ne {{-1}\over {-2k-1}}$$
=>$${{3}\over {2k - 1}} = {{1}\over {k - 1}}$$
=>$$3(k - 1) = 2k - 1$$
=>$$3k - 3 = 2k - 1$$
=>$$k = 2$$
Q3 ) Solve the following pair of linear equations by the substitution and cross-multiplication methods:
$$8x + 5y = 9$$
$$3x + 2y = 4$$
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.(x + y = 14\)...........(i)
=> $$x – y = 4$$..............(ii)
Using equation (ii)
=> $$x = 4 + y$$......................(iii)
On substituting (iii) in (i):
=>$$4 + 2y = 14$$
=>$$2y = 10$$
=>$$y = 5$$...........(iv)
On substituting (iv) in (i):
=>$$x + 5 = 14$$
=>$$x = 14 - 5$$
=>$$x = 9$$
So, the value of $$y = 5$$ and $$x = 9$$.
$$x - 3y – 7 = 0,3x - 3y – 15 = 0$$
On comparing quations with general form of equation $$ax^2 + bx + c$$
$$a_1 = 1,$$
$$b_1 = -3,$$
$$c_1 = -7,$$
$$a_2 = 3,$$
$$b_2 = -3,$$
$$c_2 = -15$$
Here $${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$
=>$${{1}\over {3}} \ne {{-3}\over {-3}}$$
So, these lines have a unique solution.
We can find the unique solution using cross multiplication:
=>$${{x}\over {(-3)(-15)-(-3)(-7)}} = {{y}\over {(-7)(3)-(-15)(1)}} = {{1}\over {(-3)(1)-(-3)(3)}}$$
=>$${{x}\over {45 - 21}} = {{y}\over {-21 + 15}} = {{1}\over {-3 + 9}}$$
=>$${{x}\over {24}} = {{y}\over {-6}} = {{1}\over {6}}$$
=> x = 4 and y = -1.
Q4 ) Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii)A fraction becomes $${{1}\over {3}}$$ when 1 is subtracted from the numerator and it becomes $${{1}\over {4}}$$ when 8 is added to its denominator. Find the fraction.
(iii)Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
(iv)Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(v)The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.(i)Let fixed monthly charge = Rs x and let charge of food for one day = Rs y
According to given conditions,
$$x + 20y = 1000$$… (1),
$$x + 26y = 1180$$… (2)
Subtracting equation (1) from equation (2),
$$6y = 180$$
=>$$y = 30$$
Putting value of y in (1),
$$x + 20 (30) = 1000$$
=>$$x = 1000 – 600 = 400$$
Therefore, fixed monthly charges = Rs 400 and, charges of food for one day = Rs 30
(ii) Let numerator = x and let denominator = y
According to given conditions,
$${{x - 1}\over {y}} = {{1}\over {3}}$$… (1) and $${{x}\over {y + 8}} = {{1}\over {4}}$$… (2)
=>$$3x – 3 = y$$… (1) $$4x = y + 8$$ … (2)
=>$$3x – y = 3$$… (1) $$4x – y = 8$$… (2)
Subtracting equation (1) from (2),
$$4x – y - (3x - y) = 8 – 3$$
$$x = 5$$
Putting value of x in (1),
$$3 (5) – y = 3$$
$$=> 15 – y = 3$$
$$=> y = 12$$
Therefore, numerator = 5 and, denominator = 12
It means fraction = $${{5}\over {12}}$$
(iii) Let number of correct answers = x and let number of wrong answers = y
According to given conditions,
$$3x – y = 40$$ … (1)
And, $$4x - 2y = 50$$ … (2)
From equation (1), $$y = 3x - 40$$
Putting this in (2),
$$4x – 2 (3x - 40) = 50$$
=>$$4x - 6x + 80 = 50$$
=>$$-2x = -30$$
=>$$x = 15$$
Putting value of x in (1),
$$3 (15) – y = 40$$
=>$$45 – y = 40$$
=>$$y = 45 – 40 = 5$$
Therefore, number of correct answers = x = 15 and number of wrong answers = y = 5
Total questions = x + y = 15 + 5 = 20
(iv)Let speed of car which starts from part A = x km/hr
Let speed of car which starts from part B = y km/hr
According to given conditions,(Assuming x > y)
=>$${{100} \over {x-y}} = 5$$
=>$$5x - 5y = 100$$
=>$$x – y = 20$$… (1)
And, $${{100} \over {x+y}} = 1$$
=>$$x + y = 100$$ … (2)
$$2x = 120$$
=>$$x = 60$$
Putting value of x in (1),
$$60 – y = 20$$
=>$$y = 60 – 20 = 40$$ km/hr
Therefore, speed of car starting from point A = 60 km/hr
And, Speed of car starting from point B = 40 km/hr
(v) Let length of rectangle = x units and Let breadth of rectangle = y units
Area =xy square units. According to given conditions,
$$xy – 9 = (x - 5) (y + 3)$$
=>$$xy – 9 = xy + 3x - 5y – 15$$
=>$$3x - 5y = 6$$… (1)
And, $$xy + 67 = (x + 3) (y + 2)$$
=>$$xy + 67 = xy + 2x + 3y + 6$$
=>$$2x + 3y = 61$$… (2)
From equation (1),
=>$$3x = 6 + 5y$$
=>$$x = {{6 + 5y} \over {3}}$$
Putting this in (2),
=>$$2({{6 + 5y} \over {3}}) + 3y = 61$$
=>$$12 + 10y + 9y = 183$$
=>$$19y = 171$$
=>$$y = 9$$ units
Putting value of y in (2),
$$2x + 3 (9) = 61$$
=>$$2x = 61 – 27 = 34$$
=>$$x = 17$$ units
Therefore, length = 17 units and, breadth = 9 units
## NCERT solutions for class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3.6
Q1 ) Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) $${{1} \over {2x}} + {{1} \over {3y}} = 2$$
$${{1} \over {3x}} + {{1} \over {2y}} = {{13} \over {6}}$$
(ii) $${{2} \over {\sqrt{x}}} + {{3} \over {\sqrt{y}}} = 2$$
$${{4} \over {\sqrt{x}}} - {{9} \over {\sqrt{y}}} = -1$$
(iii) $${{4} \over {x}} + 3y = 14$$
$${{3} \over {x}} - 4y = 23$$
(iv) $${{5} \over {x - 1}} + {{1} \over {y - 2}} = 2$$
$${{6} \over {x - 1}} - {{3} \over {y - 2}} = 1$$
(v) $$7x - 2y = 5xy$$
$$8x + 7y = 15xy$$
(vi)$$6x + 3y - 6xy = 0$$
$$2x + 4y - 5xy = 0$$
(vii) $${{10} \over {x + y}} + {{2} \over {x - y}} = 4$$
$${{15} \over {x + y}} - {{5} \over {x - y}} = -2$$
(viii) $${{1} \over {3x + y}} + {{1} \over {3x - y}} = {{3} \over {4}}$$
$${{1} \over {2(3x + y)}} - {{1} \over {2(3x - y)}} = {{-1} \over {8}}$$
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
(i) $${{1} \over {2x}} + {{1} \over {3y}} = 2$$… (1)
$${{1} \over {3x}} + {{1} \over {2y}} = {{13} \over {6}}$$ … (2)
Let $${{1} \over {x}}$$ = p and $${{1} \over {y}}$$ = q
Putting this in equation (1) and (2),
=>$${{p} \over {2}} + {{q} \over {3}} = 2$$ and $${{p} \over {3}} + {{q} \over {2}} = {{13} \over {6}}$$
=>$$3p + 2q = 12$$ and $$2p + 3q =13$$
=>$$3p + 2q – 12 = 0$$… (3) and $$2p + 3q – 13 = 0$$ … (4)
Using cross multiplication
=>$${{p} \over {(2)(-13) - 3(-12)}} = {{q} \over {(-12)(2) - (-13)(3)}} = {{1} \over {(3)(3) - (2)(2)}}$$
=> $${{p} \over {-26 + 36}} = {{q} \over {-24 + 39}} = {{1} \over {9 - 4}}$$
=>$${{p} \over {10}} = {{q} \over {15}} = {{1} \over {5}}$$
=> p = 2 and q = 3
But $${{1} \over {x}}$$ = p and $${{1} \over {y}}$$ = q
Putting value of p and q in this
x = $${{1} \over {2}}$$ and y = $${{1} \over {3}}$$
(ii)$${{2} \over {\sqrt{x}}} + {{3} \over {\sqrt{y}}} = 2$$… (1)
$${{4} \over {\sqrt{x}}} - {{9} \over {\sqrt{y}}} = -1$$ … (2)
Let $${{1} \over {\sqrt{x}}}$$ = p and $${{1} \over {\sqrt{y}}}$$= q
Putting this in (1) and (2),
$$2p + 3q = 2$$ … (3)
$$4p - 9q = -1$$… (4)
Multiplying (3) by 2 and subtracting it from (4),
$$4p - 9q + 1 – 2 (2p + 3q - 2) = 0$$
=>$$4p - 9q + 1 - 4p - 6q + 4 = 0$$
=>$$-15q + 5 = 0$$
=>$$q = {{5} \over {15}} = {{1} \over {3}}$$
Putting value of q in (3),
2p + 1 = 2
=> 2p = 1
=> p $$= {{1} \over {2}}$$
Putting values of p and q in ($${{1} \over {\sqrt{x}}}$$= p and $${{1} \over {\sqrt{y}}}$$= q), we get
$${{1} \over {\sqrt{x}}} = {{1} \over {2}}$$ and $${{1} \over {\sqrt{y}}} = {{1} \over {3}}$$
=>$${{1} \over {x}} = {{1} \over {4}}$$ and $${{1} \over {y}} = {{1} \over {9}}$$
=>$$x = 4$$ and $$y = 9$$
(iii) $${{4} \over {x}} + 3y = 14$$ … (1)
$${{3} \over {x}} - 4y = 23$$ … (2) and Let $${{1} \over {x}}= p$$ … (3)
Putting (3) in (1) and (2),
$$4p + 3y = 14$$… (4)
$$3p - 4y = 23$$… (5)
Multiplying (4) by 3 and (5) by 4,
$$3 (4p + 3y – 14 = 0)$$ and, $$4 (3p - 4y – 23 = 0)$$
=>$$12p + 9y – 42 = 0$$… (6) $$12p - 16y – 92 = 0$$ … (7)
Subtracting (7) from (6),
$$9y - (-16y) – 42 - (-92) = 0$$
$$=> 25y + 50 = 0$$
$$=> y = \frac{-50}{25} = -2$$
Putting value of y in (4),
$$4p + 3 (-2) = 14$$
$$=> 4p – 6 = 14$$
$$=> 4p = 20$$
$$p = 5$$
Putting value of p in (3),
$${{1} \over {x}} = 5 => x = {{1} \over {5}}$$
Therefore, x = $${{1} \over {5}}$$ and y = -2
(iv) $${{5} \over {x - 1}} + {{1} \over {y - 2}} = 2$$ … (1)
$${{6} \over {x - 1}} - {{3} \over {y - 2}} = 1$$… (2)
Let $${{1} \over {x - 1}} = p$$ and $${{1} \over {y - 2}} = q$$
Putting this in (1) and (2),
$$5p + q = 2$$
$$5p + q – 2 = 0$$ … (3)
And, $$6p - 3q = 1$$
$$6p - 3q – 1 = 0$$… (4)
Multiplying (3) by 3 and adding it to (4),
$$3 (5p + q - 2) + 6p - 3q – 1 = 0$$
$$15p + 3q – 6 + 6p - 3q – 1 = 0$$
$$21p – 7 = 0$$
$$p = {{1} \over {3}}$$
Putting this in (3),
$$5 ({{1} \over {3}}) + q – 2 = 0$$
$$5 + 3q = 6$$
$$3q = 6 – 5 = 1$$
$$q = {{1} \over {3}}$$
Putting values of p and q in ($${{1} \over {x - 1}} = p$$ and $${{1} \over {y - 2}} = q$$),
$${{1} \over {x - 1}} = {{1} \over {3}}$$ and $${{1} \over {y - 2}} = {{1} \over {3}}$$
$$3 = x - 1$$ and $$3 = y – 2$$
$$x = 4$$ and $$y = 5$$
(v) $$7x - 2y = 5xy$$ … (1)
$$8x + 7y = 15xy$$… (2)
Dividing both the equations by xy,
$${{7} \over {y}} - {{2} \over {x}} = 5$$… (3)
$${{8} \over {y}} + {{7} \over {x}} = 15$$… (4)
Let $${{1} \over {x}}$$ = p and $${{1} \over {y}}$$ = q
Putting these in (3) and (4),
$$7q - 2p = 5$$… (5)
$$8q + 7p = 15$$… (6)
From equation (5),
$$2p = 7q – 5$$
$$p = {{7q – 5} \over {2}}$$
Putting value of p in (6),
$$8q + 7 (\frac{7q – 5}{2}) = 15$$
$$16q + 49q – 35 = 30$$
$$65q = 30 + 35 = 65$$
$$q = 1$$
Putting value of q in (5),
$$7 (1) - 2p = 5$$
$$- 2p = -2$$
$$p = 1$$
Putting value of p and q in ($${{1} \over {x}} = p$$ and $${{1} \over {y}} = q$$), we get
x = 1 and y = 1
(vi)$$6x + 3y - 6xy = 0$$… (1)
$$2x + 4y - 5xy = 0$$ … (2)
Dividing both the equations by xy,
$${{6} \over {y}} + {{3} \over {x}} - 6 = 0$$… (3)
$${{2} \over {y}} + {{4} \over {x}} - 5 = 0$$… (4)
Let $${{1} \over {x}} = p$$ and $${{1} \over {y}} = q$$
Putting these in (3) and (4),
$$6q + 3p – 6 = 0$$ … (5)
$$2q + 4p – 5 = 0$$ … (6)
From (5),
$$3p = 6 - 6q$$
$$p = 2 - 2q$$
Putting this in (6),
$$2q + 4 (2 - 2q) – 5 = 0$$
$$2q + 8 - 8q – 5 = 0$$
$$-6q = -3$$
$$q = {{1} \over {2}}$$
Putting value of q in (p = 2 – 2q), we get
$$p = 2 – 2 ({{1} \over {2}}) = 2 – 1 = 1$$
Putting values of p and q in ($${{1} \over {x}} = p$$ and $${{1} \over {y}} = q$$), we get
x = 1 and y = 2
(vii) $${{10} \over {x + y}} + {{2} \over {x - y}} = 4$$… (1)
$$\frac{15}{(x + y)} - \frac{5}{(x - y)} = -2$$…(2)
Let $${{1} \over {(x + y)}} = p$$ and $${{1} \over {(x - y)}} = q$$
Putting this in (1) and (2),
$$10p + 2q = 4$$ … (3)
$$15p - 5q = -2$$ … (4)
From equation (3),
$$2q = 4 - 10p$$
$$q = 2 - 5p$$ … (5)
Putting this in (4),
$$15p – 5 (2 - 5p) = -2$$
$$15p – 10 + 25p = -2$$
$$40p = 8$$ => $$p = {{1} \over {5}}$$
Putting value of p in (5),
$$q = 2 – 5 ({{1} \over {5}}) = 2 – 1 = 1$$
Putting values of p and q in ($${{1} \over {x + y}} = p$$ and $${{1} \over {x - y}} = q$$), we get
$$x + y = 5$$ … (6) and $$x – y = 1$$ … (7)
$$2x = 6$$ => $$x = 3$$
Putting x = 3 in (7), we get
$$3 – y = 1$$
$$y = 3 – 1 = 2$$
Therefore, x = 3 and y = 2
(viii)$${{1} \over {3x + y}} + {{1} \over {3x - y}} = {{3} \over {4}}$$ … (1)
$${{1} \over {2(3x + y)}} - {{1} \over {2(3x - y)}} = {{-1} \over {8}}$$ … (2)
Let $${{1} \over {(3x + y)}} = p$$ and $${{1} \over {(3x - y)}} = q$$ … (2)
Putting this in (1) and (2),
$$p + q = {{3} \over {4}}$$ and $${{p} \over {2}} - {{q} \over {2}} = {{1} \over {8}}$$
$$4p + 4q = 3$$ … (3) and $$4p - 4q = -1$$ … (4)
$$8p = 2 - p = {{1} \over {4}}$$
Putting value of p in (3),
$$4 ({{1} \over {4}}) + 4q = 3$$
$$1 + 4q = 3$$
$$4q = 3 – 1 = 2$$
$$q = {{1} \over {2}}$$
Putting value of p and q in $${{1} \over {(3x + y)}} = p$$ and $${{1} \over {(3x - y)}} = q$$, we get
$${{1} \over {(3x + y)}} = {{1} \over {4}}$$ and $${{1} \over {(3x - y)}} = {{1} \over {2}}$$
$$3x + y = 4$$… (5) and $$3x – y = 2$$… (6)
$$6x = 6 - x = 1$$
Putting x = 1 in (5) ,
$$3 (1) + y = 4$$
$$y = 4 – 3 = 1$$
Therefore, x = 1 and y = 1
Q2 ) Formulate the following problems as a part of equations, and hence find their solutions.
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days.Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.(i) Let speed of rowing in still water = x km/h
Let speed of current = y km/h
So, speed of rowing downstream
= (x + y) km/h
And, speed of rowing upstream
= (x - y) km/h
According to given conditions,
$${{20} \over {x + y}} = 2$$ and $${{4} \over {x - y}} = 2$$
$$2x + 2y = 20$$ and $$2x - 2y = 4$$
$$x + y = 10$$ … (1) and $$x – y = 2$$ … (2)
$$2x = 12$$ => $$x = 6$$
Putting x = 6 in (1),
$$6 + y = 10$$
$$y = 10 – 6 = 4$$
Therefore, speed of rowing in still water = 6 km/h
Speed of current = 4 km/h
(ii) Let time taken by 1 woman alone to finish the work = x days
Let time taken by 1 man alone to finish the work = y days
So, 1 woman’s 1-day work = ($${{1} \over {x}}$$)th part of the work
And, 1 man’s 1-day work = ($${{1} \over {y}}$$)th part of the work
So, 2 women’s 1-day work = ($${{2} \over {x}}$$)th part of the work
And, 5 men’s 1-day work = ($${{5} \over {y}}$$)th part of the work
Therefore, 2 women and 5 men’s 1-day work = ($${{2} \over {x}} + {{5} \over {y}}$$)th part of the work....… (1)
It is given that 2 women and 5 men complete work in = 4 days
It means that in 1 day, they will be completing ($${{1} \over {4 }}$$)th part of the work ….... (2)
Since, (1) = (2)
$${{2} \over {x}} + {{5} \over {y}} = {{1} \over {4 }}$$….... (3)
Similarly, $${{3} \over {x}} + {{6} \over {y}} = {{1} \over {3}}$$ .....… (4)
Let $${{1} \over {x}} = p$$ and $${{1} \over {y}} = q$$
Putting this in (3) and (4),
$$2p + 5q = {{1} \over {4 }}$$ and $$3p + 6q = {{1} \over {3}}$$
$$8p + 20q = 1$$ … (5) and $$9p + 18q = 1$$ … (6)
Multiplying (5) by 9 and (6) by 8,
$$72p + 180q = 9$$ … (7)
$$72p + 144q = 8$$ … (8)
Subtracting (8) from (7), we get
$$36q = 1$$ => $$q = {{1} \over {36}}$$
Putting this in (6), we get
$$9p + 18 ({{1} \over {36}}) = 1$$
$$9p = {{1} \over {2}}$$ => $$p = {{1} \over {18}}$$
Putting values of p and q in , we get x = 18 and y = 36
Therefore, 1 woman completes work in = 18 days
And, 1 man completes work in = 36 days
(iii) Let speed of train = x km/h and let speed of bus = y km/h
According to given conditions,
$${{60} \over {x}} + {{240} \over {y}} = 4$$ and $${{100} \over {x}} + {{200} \over {y}} = 4 + {{10} \over {60}}$$
Let $${{1} \over {x}} = p$$ and $${{1} \over {y}} = q$$
Putting this in the above equations, we get
$$60p + 240q = 4$$ … (1)
And $$100p + 200q = {{25} \over {6}}$$ … (2)
Multiplying (1) by 5 and (2) by 3,
$$300p + 1200q = 20$$… (3)
$$300p + 600q = {{25} \over {2}}$$ … (4)
Subtracting (4) from (3),
$$600q = 20 - {{25} \over {2}} = 7.5$$
$$q = {{7.5} \over {600}}$$
Putting value of q in (2),
$$100p + 200 ({{7.5} \over {600}}) = {{25} \over {6}}$$
$$100p + 2.5 = {{25} \over {6}}$$
$$100p = {{25} \over {6}} – 2.5$$
$$p = {{10} \over {600}}$$
But $${{1} \over {x}} = p$$ and $${{1} \over {y}} = q$$
Therefore, $$x = {{600} \over {10}} = 60$$ km/h and $$y = {{600} \over {7.5}} = 80$$ km/h
Therefore, speed of train = 60 km/h
And, speed of bus = 80 km/h
## NCERT solutions for class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3.7
Q1 ) The age of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans. Let the age of Ani and Biju be x years and y years respectively.
Age of Dharam = 2x years and Age of Cathy = $${{y}\over{2}}$$ years
Case 1 : Ani's age is greater than Biju's age
According to question, x – y = 3… (1)
And $$2x - {{y}\over{2}} = 30$$
$$4x – y = 60$$… (2)
Subtracting (1) from (2),:
$$3x = 60 – 3 = 57$$
x = Age of Ani = 19 years
Age of Biju = 19 – 3 = 16 years
Case 2 : Ani's age is less than Biju's age
Thus, $$y – x = 3$$… (3)
And $$2x - {{y}\over{2}} = 30$$
$$4x – y = 60$$… (4)
Adding (3) and (4), we obtain:
3x = 63
x = 21
Age of Ani = 21 years
Age of Biju = 21 + 3 = 24 years
Q2 ) One says, “Give me a hundred, friend! I shall then become twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.Let the money with the first person and second person be Rs x and Rs y respectively.
According to the question,
x + 100 = 2(y – 100)
x + 100 = 2y – 200
x – 2y = – 300… (1)
Also, 6(x – 10) = (y + 10)
6x – 60 = y + 10
6x – y = 70… (2)
Multiplying equation (2) by 2, we obtain:
12x – 2y = 140… (3)
Subtracting equation (3) from equation (1), we obtain:
11x = 140 + 300
=> 11x = 440
=> x = 40
Putting the value of x in equation (1), we obtain:
40 – 2y = –300
=> 40 + 300 = 2y
=> 2y = 340
=> y = 170
Thus, the two friends had Rs 40 and Rs 170 with them.
Q3 ) A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.
Since Speed = Distance travelled/Time taken to travel the distance
=> $$x = {{d} \over {t}}$$
=> d = xt … (1)
According to the question
=> $$x + 10 = {{d} \over {t - 2}}$$
=>$$(x + 10) (t - 2) = d$$
=> $$xt + 10t - 2x - 20 = d$$
=> $$-2x + 10t = 20$$……(2)[Using eq. (1)]
Again, => $$x - 10 = {{d} \over {t + 3}}$$
=>$$(x - 10) (t + 3) = d$$
=> $$xt - 10t + 3x - 30 = d$$
=> $$3x - 10t = 30$$……(3)[Using eq. (1)]
x = 50
Substituting the value of x in equation (2):
$$(-2) (50) + 10t =20$$
$$=> –100 + 10t = 20$$
$$=> 10t = 120t$$
=> $$t = 12$$
From equation (1):
d = xt = (50)(12) = 600
Thus, the distance covered by the train is 600 km.
Q4 ) The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
Ans.Let the number of rows be x and number of students in a row be y.
Total number of students in the class = Number of rows x Number of students in a row = xy
Case 1 :
Total number of students = (x – 1) (y + 3)
$$xy = (x – 1) (y + 3)$$
$$=> xy = xy – y + 3x – 3$$
$$=> 3x – y – 3 = 0$$
$$=> 3x – y = 3$$… (1)
Case 2 :
Total number of students = (x + 2) (y – 3)
$$xy = xy + 2y – 3x – 6$$
$$=> 3x – 2y = –6$$… (2)
Subtracting equation (2) from (1):
y = 9
Substituting the value of y in equation (1):
$$3x – 9 = 3$$
$$3x = 9 + 3 = 12$$
x = 4
Number of rows = x = 4
Number of students in a row = y = 9
Hence, Total number of students in a class = xy = 36
Q5 ) In a $$\triangle ABC$$, $$\angle C = \angle 3B = 2(\angle A + \angle B)\ . Find three angles. NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Answer : \(\angle C = \angle 3B = 2(\angle A + \angle B)$$
Taking $$3\angle B = 2 ( \angle A + \angle B )$$
$$\angle B = 2\angle A$$
$$2\angle A – \angle B = 0$$ …….(1)
We know that the sum of the measures of all angles of a triangle is 180°.
$$\angle A + \angle B + \angle C = 180°$$
$$\angle A + \angle B + 3\angle B = 180°$$
$$\angle A + 4\angle B = 180°$$ …….(2)
Multiplying equation (1) by 4:
$$8\angle A – 4\angle B = 0$$ …….(3)
Adding equations (2) and (3), we get
$$9\angle A = 180°$$
$$\angle A = 20°$$
From eq. (2), we get,
$$20° + 4\angle B = 180°$$
$$\angle B = 40°$$
And $$\angle C = 3 (40°) = 120°$$
Hence the measures of $$\angle A, \angle B$$ and $$\angle C$$ are respectively.
Q6 ) Draw the graphs of the equations and Determine the co-ordinate of the vertices of the triangle formed by these lines and the axis.
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
According to the given statement in the question:
=>$$y = 5x - 5$$............(1)
Three solutions of this equation can be written in a table as follows:
$$\begin{array} {|r|r|}\hline x & 0 & 1 & 2 \\ \hline y & -5 & 0 & 5 \\ \hline \end{array}$$
According to the given statement in the question:
=>$$y = 3x - 3$$............(2)
Three solutions of the equation (2) are :
$$\begin{array} {|r|r|}\hline x & 0 & 1 & 2 \\ \hline y & -3 & 0 & 3 \\ \hline \end{array}$$
Now the graph for the above equations (1) and (2) will be:
It can be observed that the required triangle is ABC.
The coordinates of its vertices are A (1, 0), B (0, –3), C (0, –5).
Q7 ) Solve the following pair of linear equations:
(i) $$px + py = p - q$$
$$qx –py = p + q$$
(ii) $$ax + by = c$$
$$bx + ay = 1 + c$$
(iii) $${{x}\over{a}} - {{y}\over{b}} = 0$$
$$ax + by = a ^2 + b^2$$
(iv) $$(a - b)x + (a + b)y = a^2 -2ab - b^2$$
$$(a + b)(x + y) = a^2 + b^2$$
(v) $$152x - 378y = -74$$
$$-378x + 152y = -604$$
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
(i)$$px + py = p - q$$ … (1)
$$qx –py = p + q$$… (2)
Multiplying equation (1) by p and equation (2) by q:
$$p^2 x + pqy = p^2 - pq$$… (3) $$q^2 x - pqy = pq + q^2$$… (4)
Adding equations (3) and (4), :
=>$$p^2 x + q^2 x = p^2 + q^2$$
=>$$(p^2 + q^2)x = p^2 + q^2$$
=> x = $${{p^x + q^x} \over {p^2 + q^2}}$$= 1
Substituting the value of in equation (1):
$$p(1) + qy = p - q$$
=>$$qy = -q => y = -1$$
Hence the required solution is x = 1 and y = –1.
(ii)$$ax + by = c$$… (1)
$$bx + ay = 1 + c$$ … (2)
Multiplying equation (1) by a and equation (2) by b:
$$a^2x + aby = ac$$… (3)
$$b^2x + aby = b + bc$$ … (4)
Subtracting equation (4) from equation (3),
$$(a^2 - b^2)x = ac - bc -b$$
=>$$x = {{c(a - b) - b} \over {a^2 - b^2}}$$
Substituting the value of x in equation (1):
$$a ({{c(a - b) - b} \over {a^2 - b^2}}) + by = c$$
=>$${{ac(a - b) - ab} \over {a^2 - b^2}} + by = c$$
=>$$by = c -{{ac(a - b) - ab} \over {a^2 - b^2}}$$
=>$$by = {{a^2c - b^2c - a^2c + abc + ab} \over {a^2 - b^2}}$$
=> $$by = {{ abc - b^2 c+ ab} \over {a^2 - b^2}}$$
=>$$y = {{ c(a - b)+ a} \over {a^2 - b^2}}$$
(iii)$${{x}\over{a}} - {{y}\over{b}} = 0$$
=>$$bx - ay = 0$$……..(1)
=>$$ax + by = a^2 - b^2$$……..(2)
Multiplying equation (1) and (2) by b and a respectively:
=> $$xb^2 - aby = 0$$……..(3)
=> $$xa^2 + aby = a^3 - ab^2$$……..(4)
Adding equations (3) and (4), we obtain:
$$xb^2 + xa^2 = a^3 + ab^2$$
=>$$x(b^2 + a^2) = a (b^2 + a^2)$$
=> $$x = a$$
Substituting the value of in equation (1):
$$b(a) - ay = 0$$
=>$$ab - ay = 0$$
=> $$y = b$$
(iv) $$(a - b)x + (a + b)y = a^2 -2ab - b^2$$ … (1)
$$(a + b)(x + y) = a^2 + b^2$$……..(2)
Subtracting equation (2) from (1):
$$(a - b)x - (a + b)x = (a^2 - 2ab - b^2) - ( a^2 + b^2)$$
=>$$(a - b - a - b)x = -2ab - 2b^2$$
=>$$-2bx = -2b(a + b)$$
=> $$x = a + b$$
Substituting the value of in equation (1):
$$(a - b)(a + b) + (a + b)y = a^2 -2ab - b^2$$
=>$$a^2 - b^2 + (a + b)y = a^2 -2ab - b^2$$
=>$$(a + b)y = -2ab$$
=>$$y = {{-2ab} \over {a + b}}$$
(v)$$152x – 378y = –74$$… (1)
$$–378x + 152y = –604$$ … (2)
Adding the equations (1) and (2):
$$–226x – 226y = –678$$
$$=> x + y = 3$$ ………(3)
Subtracting the equation (2) from equation (1):
$$530x – 530y = 530$$
$$=> x – y = 1$$ ……..(4)
$$2x = 4$$
$$=> x = 2$$
Substituting the value of x in equation (3):
$$y = 1$$
Q8 ) ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.
NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
We know that the sum of the measures of opposite angles in a cyclic quadrilateral is :
$$\angle$$A + $$\angle$$C = 180°
=>$$4y + 20 - 4x = 180°$$
=>$$-4x + 4y = 160°$$
=>$$x - y = -40°$$ ………(1)
Also $$\angle$$B + $$\angle$$D = 180°
=>$$3y - 5 - 7x + 5 = 180°$$
=>$$-7x + 3y = 180°$$………(2)
Multiplying equation (1) by 3:
$$3x - 3y = - 120°$$ ……….(3)
$$-4x = 60° => x = -15°$$
Substituting the value of in equation (1):
$$-15 - y = -40°$$
=>$$y = -15 + 40 =25$$
$$\angle$$A = $$4y + 20 = 4(25) + 20 = 120°$$
$$\angle$$B = $$3y - 5 = 3(25) - 5 = 70°$$
$$\angle$$C = $$-4x = -4(-15) = 60°$$
$$\angle$$D = $$-7x + 5 = -7(-15) + 5 = 110°$$
##### FAQs Related to NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables
There are total 29 questions present in ncert solutions for class 10 maths chapter 3 pair of linear equations in two variables
There are total 17 long question/answers in ncert solutions for class 10 maths chapter 3 pair of linear equations in two variables
There are total 7 exercise present in ncert solutions for class 10 maths chapter 3 pair of linear equations in two variables
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# Probability How likely is an event to occur?
## Presentation on theme: "Probability How likely is an event to occur?"— Presentation transcript:
Probability How likely is an event to occur?
What are the chances of that happening??!!
Lesson objectives: To understand the terms ‘equal chance’ and ‘outcome’ when thinking about probability. To complete a maths investigation involving tossing a coin and rolling a dice! - to test the ‘equal chance’ theory!
Mathematical Vocabulary
We can describe the probability or chance of an event happening by saying: It is IMPOSSIBLE. It is UNLIKELY. It is LIKELY. It is CERTAIN.
Some events have an equal chance of happening or not happening
Some events have an equal chance of happening or not happening. Can you think of any?
We discussed these: If you toss a coin you have an equal chance of getting a head or a tail. Heads or tails are the ‘outcomes’. If a baby is born it has an equal chance of being a boy or a girl. Boy or girl are the ‘outcomes’. If you roll a dice you have an equal chance of getting the number 1, 2, 3, 4, 5 or 6! These scores are the ‘outcomes’.
Today’s task: We are going to carry out a maths investigation.
We are going to investigate whether we have an equal number of heads and tails when we toss a coin 30 times. We will also investigate whether we get an equal number of 1s, 2s, 3s, 4, 5s and 6s when we roll a dice 30 times. We will record our findings.
What do you think we will find out?
Perhaps we will have 15 heads and 15 tails when we toss our coin 30 times?? Perhaps we will have each number on the dice 5 times. (five scores of 1, five scores of 2 etc.) It’s up to you to find out!!!!
Did we achieve out learning objectives today?
We understand what is meant by the terms ‘equal chance’ and ‘outcomes’. We have carried out an investigation into ‘equal chance’. We have recorded our findings.
Things to think about! If we rolled 2 coins what possible outcomes could we get? If we rolled more than 1 dice what possible outcomes could we get? What would our chances of getting 2 heads or a 6 be like then?!!!
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# Derivative of 1/f(x)
Learn what is the derivative of 1/f(x) with easy and step-wise proof. Also, understand to prove the derivative of 1/f(x) by chain rule and quotient rule.
Alan Walker-
Published on 2023-09-04
## Introduction to the derivative of 1/f(x)
Derivatives have a wide range of applications in almost every field of engineering and science. The 1/f(x) derivative can be calculated by following the differentiation rules. Or, we can directly find the derivative formula of 1/f(x) by applying the first principle of differentiation. In this article, you will learn what the 1/f(x) derivative formula is and how to calculate the derivatives of 1 by f(x) by using different approaches.
## What is the derivative of 1/f(x)?
The derivative of 1 by f(x) with respect to the variable x is equal to -1/(f(x))^2. It measures the rate of change of the algebraic function 1/f(x). It is denoted by d/dx(1/f(x)) which is a fundamental concept in calculus. Knowing the formula for derivatives and understanding how to use it can be used in solving problems related to velocity, acceleration, and optimization.
## Derivative of 1/f(x) formula
The formula for derivative of f(x)=1/f(x) is equal to the -1/(f(x))^2, that is;
$$f'(x) = \frac{d}{dx} (\frac{1}{f(x)}) = -\frac{1}{(f(x))^2}$$
It is calculated by using the power rule, which is defined as:
$$\frac{d}{dx} (x^n)=nx^{n-1}$$
## How do you differentiate 1/f(x)?
There are multiple ways to prove the differentiation of 1/f(x). These are;
1. Product Rule
2. Quotient Rule
Each method provides a different way to compute the 1/f(x) differentiation. By using these methods, we can mathematically prove the formula for finding the differential of 1/f(x).
## Derivative of 1/f(x) by product rule
The formula for derivative 1/f(x) can be calculated by using the product rule because an algebraic function can be written as the combination of two functions. The product rule derivative is defined as;
$$[uv]’ = u’.v + u.v’$$
### Proof of differentiating of 1/f(x) by product rule
To differentiate of 1/f(x) by using product rule, we start by assuming that,
$$y = 1. (\frac{1}{f(x)})$$
By using product rule of differentiation,
$$\frac{dy}{dx} = (1)’. \frac{1}{f(x)} + (1)\frac{-1}{(f(x))^2}$$
We get,
$$\frac{dy}{dx} = 0-\frac{1}{(f(x))^2}$$
Hence,
$$\frac{dy}{dx} =-\frac{1}{(f(x))^2}$$
Also use our product rule calculator online, as it provides you a step-by-step solution of differentiation of a function.
## Derivative of 1/f(x) using quotient rule
Another method for finding the differential of 1/f(x) is using the quotient rule, which is a formula for finding the derivative of a quotient of two functions. Since the secant function is the reciprocal of cosine, the derivative of cosecant can also be calculated using the quotient rule. The derivative quotient rule is defined as:
$$\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f’(x). g(x) - g’(x).f(x)}{(g(x))^2}$$
### Proof of differentiating 1/f(x) by quotient rule
To prove the 1/f(x) derivative, we can start by writing it,
$$f(x) = \frac{1}{x}= \frac{u}{v}$$
Supposing that u = 1 and v = f(x). Now by quotient rule,
$$f’(x) =\frac{vu’ - uv’}{v^2}$$
$$f’(x) = \frac{f(x).(1)’ - f’(x)}{(f(x))^2}$$
$$f’(x)= \frac{0-f’(x)}{(f(x))^2}$$
$$f’(x)= \frac{-f’(x)}{(f(x))^2}$$
Hence, we have derived the derivative of 1 by f(x) using the quotient rule of differentiation. Use our quotient rule calculator to find the derivative of any quotient function.
## How to find the derivatives of 1/f(x) with a calculator?
The easiest way of differentiating 1/f(x) is by using a dy/dx calculator. You can use our derivative calculator for this. Here, we provide you a step-by-step way to calculate derivatives by using this tool.
1. Write the function as 1/f(x) in the enter function box. In this step, you need to provide input value as a function that you want to differentiate.
2. Now, select the variable by which you want to differentiate 1/f(x). Here you have to choose x.
3. Select how many times you want to calculate the derivatives of 1/f(x). In this step, you can choose 2 for second, 3 for triple differentiation and so on.
4. Click on the calculate button. After this step, you will get the derivative of 1 by f(x) within a few seconds.
5. After completing these steps, you will receive the differential of 1/f(x) within seconds. Using online tools can make it much easier and faster to calculate derivatives, especially for complex functions.
## Conclusion:
In conclusion, the derivative of 1/f(x). The derivative measures the rate of change of a function with respect to its independent variable, and in the case of 1/f(x), the derivative can be calculated using the power rule, product rule and quotient rule of differentiation.
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# A Mathematical View of Our World
## Presentation on theme: "A Mathematical View of Our World"— Presentation transcript:
A Mathematical View of Our World
1st ed. Parks, Musser, Trimpe, Maurer, and Maurer
Chapter 2 Shapes in Our Lives
Section 2.1 Tilings Goals Study polygons Study the Pythagorean theorem
Vertex angles Regular tilings Semiregular tilings Miscellaneous tilings Study the Pythagorean theorem
2.1 Initial Problem A portion of a ceramic tile wall composed of two differently shaped tiles is shown. Why do these two types of tiles fit together without gaps or overlaps? The solution will be given at the end of the section.
Tilings Geometric patterns of tiles have been used for thousands of years all around the world. Tilings, also called tessellations, usually involve geometric shapes called polygons.
Polygons A polygon is a plane figure consisting of line segments that can be traced so that the starting and ending points are the same and the path never crosses itself.
Question: Choose the figure below that is NOT a polygon. a. c.
b. d. all are polygons
Polygons, cont’d The line segments forming a polygon are called its sides. The endpoints of the sides are called its vertices. The singular of vertices is vertex.
Polygons, cont’d A polygon with n sides and n vertices is called an n-gon. For small values of n, more familiar names are used.
Polygonal Regions A polygonal region is a polygon together with the portion of the plan enclosed by the polygon.
Polygonal Regions, cont’d
A tiling is a special collection of polygonal regions. An example of a tiling, made up of rectangles, is shown below.
Polygonal Regions, cont’d
Polygonal regions form a tiling if: The entire plane is covered without gaps. No two polygonal regions overlap.
Polygonal Regions, cont’d
Examples of tilings with polygonal regions are shown below.
Vertex Angles A tiling of triangles illustrates the fact that the sum of the measures of the angles in a triangle is 180°.
Vertex Angles, cont’d The angles in a polygon are called its vertex angles. The symbol indicates an angle. Line segments that join nonadjacent vertices in a polygon are called diagonals of the polygon.
Example 1 The vertex angles in the pentagon are called V, W, X, Y, and Z. Two diagonals shown are WZ and WY.
Vertex Angles, cont’d Any polygon can be divided, using diagonals, into triangles. A polygon with n sides can be divided into n – 2 triangles.
Vertex Angles, cont’d The sum of the measures of the vertex angles in a polygon with n sides is equal to:
Example 2 Find the sum of measures of the vertex angles of a hexagon.
Solution: A hexagon has 6 sides, so n = 6. The sum of the measures of the angles is found to be:
Regular Polygons Regular polygons are polygons in which:
All sides have the same length. All vertex angles have the same measure. Polygons that are not regular are called irregular polygons.
Regular Polygons, cont’d
Regular Polygons, cont’d
A regular n-gon has n angles. All vertex angles have the same measure. The measure of each vertex angle must be
Example 3 Find the measure of any vertex angle in a regular hexagon.
Solution: A hexagon has 6 sides, so n = 6. Each vertex angle in the regular hexagon has the measure:
Vertex Angles, cont’d
Regular Tilings A regular tiling is a tiling composed of regular polygonal regions in which all the polygons are the same shape and size. Tilings can be edge-to-edge, meaning the polygonal regions have entire sides in common. Tilings can be not edge-to-edge, meaning the polygonal regions do not have entire sides in common.
Regular Tilings, cont’d
Examples of edge-to-edge regular tilings.
Regular Tilings, cont’d
Example of a regular tiling that is not edge-to-edge.
Regular Tilings, cont’d
Only regular edge-to-edge tilings are generally called regular tilings. In every such tiling the vertex angles of the tiles meet at a point.
Regular Tilings, cont’d
What regular polygons will form tilings of the plane? Whether or not a tiling is formed depends on the measure of the vertex angles. The vertex angles that meet at a point must add up to exactly 360° so that no gap is left and no overlap occurs.
Example 4 Equilateral Triangles (Regular 3-gons)
In a tiling of equilateral triangles, there are 6(60°) = 360° at each vertex point.
Example 5 Squares (Regular 4-gons)
In a tiling of squares, there are 4(90°) = 360° at each vertex point.
Question: Will a regular pentagon tile the plane? a. yes b. no
Example 6 Regular hexagons (Regular 6-gons)
In a tiling of regular hexagons, there are 3(120°) = 360° at each vertex point.
Regular Tilings, cont’d
Do any regular polygons, besides n = 3, 4, and 6, tile the plane? Note: Every regular tiling with n > 6 must have: At least three vertex angles at each point Vertex angles measuring more than 120° Angle measures at each vertex point that add to 360°
Regular Tilings, cont’d
In a previous question, you determined that a regular pentagon does not tile the plane. Since 3(120°) = 360°, no polygon with vertex angles larger than 120° [i.e. n > 6] can form a regular tiling. Conclusion: The only regular tilings are those for n = 3, n = 4, and n = 6.
Vertex Figures A vertex figure of a tiling is the polygon formed when line segments join consecutive midpoints of the sides of the polygons sharing that vertex point.
Vertex Figures, cont’d Vertex figures for the three regular tilings are shown below.
Semiregular Tilings Semiregular tilings Are edge-to-edge tilings.
Use two or more regular polygonal regions. Vertex figures are the same shape and size no matter where in the tiling they are drawn.
Example 7 Verify that the tiling shown is a semiregular tiling.
Example 7, cont’d Solution: The tiling is made of 3 regular polygons.
Every vertex figure is the same shape and size.
Example 8 Verify that the tiling shown is not a semiregular tiling.
Example 8, cont’d Solution: The tiling is made of 3 regular polygons.
Every vertex figure is not the same shape and size.
Semiregular Tilings
Miscellaneous Tilings
Tilings can also be made of other types of shapes. Tilings consisting of irregular polygons that are all the same size and shape will be considered.
Miscellaneous Tilings, cont’d
Any triangle will tile the plane. An example is given below:
Miscellaneous Tilings, cont’d
Any quadrilateral (4-gon) will tile the plane. An example is given below:
Miscellaneous Tilings, cont’d
Some irregular pentagons (5-gons) will tile the plane. An example is given below:
Miscellaneous Tilings, cont’d
Some irregular hexagons (6-gons) will tile the plane. An example is given below:
Miscellaneous Tilings, cont’d
A polygonal region is convex if, for any two points in the region, the line segment having the two points as endpoints also lies in the region. A polygonal region that is not convex is called concave.
Miscellaneous Tilings, cont’d
Pythagorean Theorem In a right triangle, the sum of the areas of the squares on the sides of the triangle is equal to the area of the square on the hypotenuse.
Example 9 Find the length x in the figure. Solution: Use the theorem.
Pythagorean Theorem Converse
If then the triangle is a right triangle.
Example 10 Show that any triangle with sides of length 3, 4 and 5 is a right triangle. Solution: The longest side must be the hypotenuse. Let a = 3, b = 4, and c = 5. We find:
2.1 Initial Problem Solution
The tiling consists of squares and regular octagons. The vertex angle measures add up to 90° + 2(135°) = 360°. This is an example of one of the eight possible semiregular tilings.
Section 2.2 Symmetry, Rigid Motions, and Escher Patterns
Goals Study symmetries One-dimensional patterns Two-dimensional patterns Study rigid motions Study Escher patterns
Symmetry We say a figure has symmetry if it can be moved in such a way that the resulting figure looks identical to the original figure. Types of symmetry that will be studied here are: Reflection symmetry Rotation symmetry Translation symmetry
Strip Patterns An example of a strip pattern, also called a one-dimensional pattern, is shown below.
Strip Patterns, cont’d This strip pattern has vertical reflection symmetry because the pattern looks the same when it is reflected across a vertical line. The dashed line is called a line of symmetry.
Strip Patterns, cont’d This strip pattern has horizontal reflection symmetry because the pattern looks the same when it is reflected across a horizontal line.
Strip Patterns, cont’d This strip pattern has rotation symmetry because the pattern looks the same when it is rotated 180° about a given point. The point around which the pattern is turned is called the center of rotation. Note that the degree of rotation must be less than 360°.
Strip Patterns, cont’d This strip pattern has translation symmetry because the pattern looks the same when it is translated a certain amount to the right. The pattern is understood to extend indefinitely to the left and right.
Example 1 Describe the symmetries of the pattern.
Solution: This pattern has translation symmetry only.
Question: Describe the symmetries of the strip pattern, assuming it continues to the left and right indefinitely a. horizontal reflection, vertical reflection, translation b. vertical reflection, translation c. translation d. vertical reflection
Two-Dimensional Patterns
Two-dimensional patterns that fill the plane can also have symmetries. The pattern shown here has horizontal and vertical reflection symmetries. Some lines of symmetry have been drawn in.
Two-Dimensional Patterns, cont’d
The pattern also has horizontal and vertical translation symmetries. 180° rotation symmetry.
Two-Dimensional Patterns, cont’d
This pattern has 120° rotation symmetry. 240° rotation symmetry.
Rigid Motions Any combination of translations, reflections across lines, and/or rotations around a point is called a rigid motion, or an isometry. Rigid motions may change the location of the figure in the plane. Rigid motions do not change the size or shape of the figure.
Reflection A reflection with respect to line l is defined as follows, with A’ being the image of point A under the reflection. If A is a point on the line l, A = A’. If A is not on line l, then l is the perpendicular bisector of line AA’.
Example 2 Find the image of the triangle under reflection about the line l.
Example 2, cont’d Solution:
Find the image of each vertex point of the triangle, using a protractor. A and A’ are equal distances from l. Connect the image points to form the new triangle.
Vectors A vector is a directed line segment.
One endpoint is the beginning point. The other endpoint, labeled with an arrow, is the ending point. Two vectors are equivalent if they are: Parallel Have the same length Point in the same direction.
Vectors, cont’d A vector v is has a length and a direction, as shown below. A translation can be defined by moving every point of a figure the distance and direction indicated by a vector.
Translation A translation is defined as follows.
A vector v assigns to every point A an image point A’. The directed line segment between A and A’ is equivalent to v.
Example 3 Find the image of the triangle under a translation determined by the vector v.
Example 3, cont’d Solution:
Find the image of each vertex point by drawing the three vectors. Connect the image points to form the new triangle.
Rotation A rotation involves turning a figure around a point O, clockwise or counterclockwise, through an angle less than 360°.
Rotation, cont’d The point O is called the center of rotation.
The directed angle indicates the amount and direction of the rotation. A positive angle indicates a counterclockwise rotation. A negative angle indicates a clockwise rotation. A point and its image are the same distance from O.
Rotation, cont’d A rotation of a point X about the center O determined by a directed angle AOB is illustrated in the figure below.
Example 4 Find the image of the triangle under the given rotation.
Example 4, cont’d Solution: Create a 50° angle with initial side OA.
Mark A’ on the terminal side, recalling that A and A’ are the same distance from O.
Example 4, cont’d Solution cont’d:
Repeat this process for each vertex. Connect the three image points to form the new triangle.
Glide Reflection A glide reflection is the result of a reflection followed by a translation. The line of reflection must not be perpendicular to the translation vector. The line of reflection is usually parallel to the translation vector.
Example 5 A strip pattern of footprints can be created using a glide reflection.
Crystallographic Classification
The rigid motions can be used to classify strip patterns.
Classification, cont’d
There are only seven basic one-dimensional repeated patterns.
Example 6 Use the crystallographic system to describe the strip pattern. Solution: The classification is pmm2.
Example 7 Use the crystallographic system to describe the strip pattern. Solution: The classification is p111.
Question: Use the crystallographic classification system to describe the pattern. a. p112 b. pmm2 c. p1m1 d. p111
Escher Patterns Maurits Escher was an artist who used rigid motions in his work. You can view some examples of Escher’s work in your textbook.
Escher Patterns, cont’d
An example of the process used to create Escher-type patterns is shown next. Begin with a square. Cut a piece from the upper left and translate it to the right. Reflect the left side to the right side.
Escher Patterns, cont’d
The figure has been decorated and repeated. Notice that the pattern has vertical and horizontal translation symmetry and vertical reflection symmetry.
Section 2.3 Fibonacci Numbers and the Golden Mean
Goals Study the Fibonacci Sequence Recursive sequences Fibonacci number occurrences in nature Geometric recursion The golden ratio
2.3 Initial Problem This expression is called a continued fraction.
How can you find the exact decimal equivalent of this number? The solution will be given at the end of the section.
Sequences A sequence is an ordered collection of numbers.
A sequence can be written in the form a1, a2, a3, …, an, … The symbol a1 represents the first number in the sequence. The symbol an represents the nth number in the sequence.
Question: Given the sequence: 1, 3, 5, 7, 9, 11, 13, 15, … , find the values of the numbers A1, A3, and A9. a. A1 = 1, A3 = 5, A9 = 15 b. A1 = 1, A3 = 3, A9 = 17 c. A1 = 1, A3 = 5, A9 = 17 d. A1 = 1, A3 = 5, A9 = 16
Fibonacci Sequence The famous Fibonacci sequence is the result of a question posed by Leonardo de Fibonacci, a mathematician during the Middle Ages. If you begin with one pair of rabbits on the first day of the year, how many pairs of rabbits will you have on the first day of the next year? It is assumed that each pair of rabbits produces a new pair every month and each new pair begins to produce two months after birth.
Fibonacci Sequence, cont’d
The solution to this question is shown in the table below. The sequence that appears three times in the table, 1, 1, 2, 3, 5, 8, 13, 21, … is called the Fibonacci sequence.
Fibonacci Sequence, cont’d
The Fibonacci sequence is the sequence of numbers 1, 1, 2, 3, 5, 8, 13, 21, … The Fibonacci sequence is found many places in nature. Any number in the sequence is called a Fibonacci number. The sequence is usually written f1, f2, f3, …, fn, …
Recursion Recursion, in a sequence, indicates that each number in the sequence is found using previous numbers in the sequence. Some sequences, such as the Fibonacci sequence, are generated by a recursion rule along with starting values for the first two, or more, numbers in the sequence.
Question: A recursive sequence uses the rule An =4An-1 – An-2, with starting values of A1 = 2, A2 =7. What is the fourth term in the sequence? a. A4 = 45 c. A4 = 67 b. A4 = 26 d. A4 = 30
Fibonacci Sequence, cont’d
For the Fibonacci sequence, the starting values are f1 = 1 and f2 = 1. The recursion rule for the Fibonacci sequence is: Example: Find the third number in the sequence using the formula. Let n = 3.
Example 1 Suppose a tree starts from one shoot that grows for two months and then sprouts a second branch. If each established branch begins to spout a new branch after one month’s growth, and if every new branch begins to sprout its own first new branch after two month’s growth, how many branches does the tree have at the end of the year?
Example 1, cont’d Solution: The number of branches each month in the first year is given in the table and drawn in the figure below.
Fibonacci Numbers In Nature
The Fibonacci numbers are found many places in the natural world, including: The number of flower petals. The branching behavior of plants. The growth patterns of sunflowers and pinecones. It is believed that the spiral nature of plant growth accounts for this phenomenon.
Fibonacci Numbers In Nature, cont’d
The number of petals on a flower are often Fibonacci numbers.
Fibonacci Numbers In Nature, cont’d
Plants grow in a spiral pattern. The ratio of the number of spirals to the number of branches is called the phyllotactic ratio. The numbers in the phyllotactic ratio are usually Fibonacci numbers.
Fibonacci Numbers In Nature, cont’d
Example: The branch at right has a phyllotactic ratio of 3/8. Both 3 and 8 are Fibonacci numbers.
Fibonacci Numbers In Nature, cont’d
Mature sunflowers have one set of spirals going clockwise and another set going counterclockwise. The numbers of spirals in each set are usually a pair of adjacent Fibonacci numbers. The most common number of spirals is 34 and 55.
Geometric Recursion In addition to being used to generate a sequence, the recursion process can also be used to create shapes. The process of building a figure step-by-step by repeating a rule is called geometric recursion.
Example 2 Beginning with a 1-by-1 square, form a sequence of rectangles by adding a square to the bottom, then to the right, then to the bottom, then to the right, and so on. Draw the resulting rectangles. What are the dimensions of the rectangles?
Example 2, cont’d Solution:
The first seven rectangles in the sequence are shown below.
Example 2, cont’d Solution cont’d:
Notice that the dimensions of each rectangle are consecutive Fibonacci numbers.
The Golden Ratio Consider the ratios of pairs of consecutive Fibonacci numbers. Some of the ratios are calculated in the table shown on the following slide.
The Golden Ratio, cont’d
The Golden Ratio, cont’d
The ratios of pairs of consecutive Fibonacci numbers are also represented in the graph below. The ratios approach the dashed line which represents a number around
The Golden Ratio, cont’d
The irrational number, approximately 1.618, is called the golden ratio. Other names for the golden ratio include the golden section, the golden mean, and the divine proportion. The golden ratio is represented by the Greek letter φ, which is pronounced “fe” or “fi”.
The Golden Ratio, cont’d
The golden ratio has an exact value of The golden ratio has been used in mathematics, art, and architecture for more than 2000 years.
Golden Rectangles A golden rectangle has a ratio of the longer side to the shorter side that is the golden ratio. Golden rectangles are used in architecture, art, and packaging.
Golden Rectangles, cont’d
The rectangle enclosing the diagram of the Parthenon is an example of a golden rectangle.
Creating a Golden Rectangle
Start with a square, WXYZ, that measures one unit on each side. Label the midpoint of side WX as point M.
Creating a Golden Rectangle, cont’d
Draw an arc centered at M with radius MY. Label the point P as shown.
Creating a Golden Rectangle, cont’d
Draw a line perpendicular to WP. Extend ZY to meet this line, labeling point Q as shown. The completed rectangle is shown.
2.3 Initial Problem Solution
How can you find the exact decimal equivalent of this number?
Initial Problem Solution, cont’d
We can find the value of the continued fraction by using a recursion rule that generates a sequence of fractions. The first term is The recursion rule is
Initial Problem Solution, cont’d
We find: The first term is The second term is
Initial Problem Solution, cont’d
The third term is The fourth term is
Initial Problem Solution, cont’d
The fractions in this sequence are 2, 3/2, 5/3, 8/5, … This is recognized to be the same as the ratios of consecutive pairs of Fibonacci numbers. The numbers in this sequence of fractions get closer and closer to φ.
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Precalculus (6th Edition) Blitzer
We use mathematical induction as follows: Statement ${{S}_{1}}$ is ${{S}_{1}}:4$ We are solving the right side as, \begin{align} & 2n\left( n+1 \right)=2\left( 1 \right)\left( 1+1 \right) \\ & =2\left( 2 \right) \\ & =4 \end{align} This statement shows that ${{S}_{1}}$ is true. Suppose ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression, \begin{align} & {{S}_{k}}:4+8+12+...+4k=2k\left( k+1 \right) \\ & \text{ }=2{{k}^{2}}+2 \\ \end{align} Then we have ${{S}_{k+1}}$ ${{S}_{k+1}}:$ \begin{align} & 4+8+12+...+4\left( k+1 \right)=2\left( k+1 \right)\left( k+1+1 \right) \\ & =2\left( k+1 \right)\left( k+2 \right) \\ & =2\left( {{k}^{2}}+k+2k+2 \right) \\ & =\left( 2{{k}^{2}}+6k+4 \right) \end{align} Solving further \begin{align} & 4+8+12+...+4\left( k+1 \right)=2\left( {{k}^{2}}+3k+1 \right)+2 \\ & =2{{\left( k+1 \right)}^{2}}+2 \end{align} Thus, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}:4+8+12+...+4n=2n\left( n+1 \right)$ is true by mathematical induction. Hence, the provided statement is proved by mathematical induction.
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Program of Lessons
# Definite integral. Newton – Leibniz formula
Curvilinear trapezoid. Area of a curvilinear trapezoid.
Definite integral. Limits of integration. Integrand.
Newton-Leibniz formula.
Consider a continuous function y = f ( x ), given on a segment [a, b] and saving its sign on this segment ( Fig.8 ). The figure, bounded by a graph of this function, a segment [a, b] and straight lines x = a and x = b, is called a curvilinear trapezoid. To calculate areas of curvilinear trapezoids the following theorem is used:
If f – a continuous, non-negative function on a segment [ a, b ], and F – its primitive on this segment, then an area S of the corresponding curvilinear trapezoid is equal to an increment of the primitive on a segment [ a, b ], i.e.
Consider a function S ( x ), given on a segment [ a, b ]. If a < x b, then S ( x ) is an area of the part of the curvilinear trapezoid, which is placed on the left of a vertical straight line, going through the point ( x, 0 ). Note, that if x = a , then S ( a ) = 0 and S ( b ) = ( S – area of the curvilinear trapezoid ). It is possible to prove, that
i.e. S ( x ) is a primitive for f ( x ). Hence, according to the basic property of primitives, for all x [ a, b ] we have:
S ( x ) = F ( x ) + C ,
where C – some constant, F – one of the primitives for a function f .
To find C we substitute x = a :
F ( a ) + C = S ( a ) = 0,
hence, C = -F ( a ) and S ( x ) = F ( x ) - F ( a ). As an area of the curvilinear trapezoid is equal to S ( b ) , substituting x = b , we’ll receive:
S = S ( b ) = F ( b ) - F ( a ).
E x a m p l e . Find an area of a figure, bounded by the curve y = x2 and lines
y = 0, x = 1, x = 2 ( Fig.9 ) .
Definite integral. Consider another way to calculate an area of a curvilinear trapezoid. Divide a segment [a, b] into n segments of an equal length by points:
x0 = a < x1 < x2 < x3 < …< x n - 1 < xn = b
and let = ( ba ) / n = xk - xk - 1 , where k = 1, 2, …, n – 1, n . In each of segments [ xk - 1 , xk ] as on a base we’ll build a rectangle of height f ( xk - 1 )An area of this rectangle is equal to:
In view of continuity of a function f ( x ) a union of the built rectangles at great n (i.e. at small ) "almost coincides" with our curvilinear trapezoid. Therefore, Sn S at great values of n . It means, that This limit is called an integral of a function f ( x ) from a to b or a definite integral:
Numbers a and b are called limits of integrationf ( x ) dx – an integrand.So, if f ( x ) 0 on a segment [ a, b ] , then an area S of the corresponding curvilinear trapezoid is represented by the formula:
NewtonLeibniz formula. Comparing the two formulas of the curvilinear trapezoid area, we make the conclusion: if F ( x ) is primitive for the function f ( x ) on a segment [ a, b ] , then
This is the famous Newton – Leibniz formula. It is valid for any function f ( x ), which is continuous on a segment [ a , b ] .
S o l u t i o n. Using the table of integrals for some elementary functions ( see above ), we’ll receive:
Back
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# Two's complement
(Redirected from Twos complement arithmetic)
Two's complement is a mathematical operation to reversibly convert a positive binary number into a negative binary number with equivalent (but negative) value, using the binary digit with the greatest place value (the leftmost bit in big-endian numbers, rightmost bit in little-endian numbers) to indicate whether the binary number is positive or negative (the sign). It is used in computer science as the most common method of representing signed (positive, negative, and zero) integers on computers,[1] and more generally, fixed point binary values. When the most significant bit is a one, the number is signed as negative. (see Converting from two's complement representation, below).
Two's complement is executed by 1) inverting (i.e. flipping) all bits, then 2) adding a place value of 1 to the inverted number. For example, say the number −6 is of interest. +6 in binary is 0110 (the leftmost most significant bit is needed for the sign; positive 6 is not 110 because it would be interpreted as -2). Step one is to flip all bits, yielding 1001. Step two is to add the place value one to the flipped number, which yields 1010. To verify that 1010 indeed has a value of −6, remember that two's complement makes the most significant bit represent a negative place value, then add the place values: 1010 = -1×(23)+0×(22)+1×(21)+0×(20) = -1(8) + 0 + 1(2) + 0 = −6.
## Theory
Two's complement is an example of a radix complement. The 'two' in the name refers to the term which, expanded fully in an N-bit system, is actually "two to the power of N" - 2N (the only case where exactly 'two' would be produced in this term is N = 1, so for a 1-bit system, but these don't have capacity for both a sign and a zero), and it is only this full term in respect to which the complement is calculated. As such, the precise definition of the Two's complement of an N-bit number is the complement of that number with respect to 2N.
The defining property of being a complement to a number with respect to 2N is simply that the summation of this number with the original produce 2N. For example, using binary with numbers up to three-bits (so N = 3 and 2N = 23 = 8 = 10002, where '2' indicates a binary representation), a two's complement for the number 3 (0112) is 5 (1012), because summed to the original it gives 23 = 10002 = 0112 + 1012. Where this correspondence is employed for representing negative numbers, it effectively means, using an analogy with decimal digits and a number-space only allowing eight non-negative numbers 0 through 7, dividing the number-space in two sets: the first four of the numbers 0 1 2 3 remain the same, while the remaining four encode negative numbers, maintaining their growing order, so making 4 encode -4, 5 encode -3, 6 encode -2 and 7 encode -1. A binary representation has an additional utility however, because the most significant bit also indicates the group (and the sign): it is 0 for the first group of non-negatives, and 1 for the second group of negatives. The tables at right illustrate this property.
Three-bit integers
Bits Unsigned value Signed value
(Two's complement)
000 0 0
001 1 1
010 2 2
011 3 3
100 4 −4
101 5 −3
110 6 −2
111 7 −1
Eight-bit integers
Bits Unsigned value Signed value
(Two's complement)
0000 0000 0 0
0000 0001 1 1
0000 0010 2 2
0111 1110 126 126
0111 1111 127 127
1000 0000 128 −128
1000 0001 129 −127
1000 0010 130 −126
1111 1110 254 −2
1111 1111 255 −1
Calculation of the binary two's complement of a positive number essentially means subtracting the number from the 2N. But as can be seen for the three-bit example and the four-bit 10002 (23), the number 2N will not itself be representable in a system limited to N bits, as it is just outside the N bits space (the number is nevertheless the reference point of the "Two's complement" in an N-bit system). Because of this, systems with maximally N-bits must break the subtraction into two operations: first subtract from the maximum number in the N-bit system, that is 2N-1 (this term in binary is actually a simple number consisting of 'all 1s', and a subtraction from it can be done simply by inverting all bits in the number also known as the bitwise NOT operation) and then adding the one. Coincidentally, that intermediate number before adding the one is also used in computer science as another method of signed number representation and is called a Ones' complement (named that because summing such a number with the original gives the 'all 1s').
Compared to other systems for representing signed numbers (e.g., ones' complement), the two's complement has the advantage that the fundamental arithmetic operations of addition, subtraction, and multiplication are identical to those for unsigned binary numbers (as long as the inputs are represented in the same number of bits as the output, and any overflow beyond those bits is discarded from the result). This property makes the system simpler to implement, especially for higher-precision arithmetic. Additionally, unlike ones' complement systems, two's complement has no representation for negative zero, and thus does not suffer from its associated difficulties. Otherwise, both schemes have the desired property that the sign of integers can be reversed by taking the complement of its binary representation, but Two's component has an exception - the lowest negative, as can be seen in the tables.[2]
## History
The method of complements had long been used to perform subtraction in decimal adding machines and mechanical calculators. John von Neumann suggested use of two's complement binary representation in his 1945 First Draft of a Report on the EDVAC proposal for an electronic stored-program digital computer.[3] The 1949 EDSAC, which was inspired by the First Draft, used two's complement representation of negative binary integers.
Many early computers, including the CDC 6600, the LINC, the PDP-1, and the UNIVAC 1107, use ones' complement notation; the descendants of the UNIVAC 1107, the UNIVAC 1100/2200 series, continued to do so. The IBM 700/7000 series scientific machines use sign/magnitude notation, except for the index registers which are two's complement. Early commercial computers storing negative values in two's complement form include the English Electric DEUCE (1955), Digital Equipment Corporation PDP-5 and the 1963 PDP-6. The System/360, introduced in 1964 by IBM, then the dominant player in the computer industry, made two's complement the most widely used binary representation in the computer industry. The first minicomputer, the PDP-8 introduced in 1965, uses two's complement arithmetic as do the 1969 Data General Nova, the 1970 PDP-11, and almost all subsequent minicomputers and microcomputers.
## Converting from two's complement representation
A two's-complement number system encodes positive and negative numbers in a binary number representation. The weight of each bit is a power of two, except for the most significant bit, whose weight is the negative of the corresponding power of two.
The value w of an N-bit integer ${\displaystyle a_{N-1}a_{N-2}\dots a_{0}}$ is given by the following formula:
${\displaystyle w=-a_{N-1}2^{N-1}+\sum _{i=0}^{N-2}a_{i}2^{i}}$
The most significant bit determines the sign of the number and is sometimes called the sign bit. Unlike in sign-and-magnitude representation, the sign bit also has the weight −(2N − 1) shown above. Using N bits, all integers from −(2N − 1) to 2N − 1 − 1 can be represented.
## Converting to two's complement representation
In two's complement notation, a non-negative number is represented by its ordinary binary representation; in this case, the most significant bit is 0. Though, the range of numbers represented is not the same as with unsigned binary numbers. For example, an 8-bit unsigned number can represent the values 0 to 255 (11111111). However a two's complement 8-bit number can only represent positive integers from 0 to 127 (01111111), because the rest of the bit combinations with the most significant bit as '1' represent the negative integers −1 to −128.
The two's complement operation is the additive inverse operation, so negative numbers are represented by the two's complement of the absolute value.
### From the ones' complement
To get the two's complement of a negative binary number, all bits are inverted, or "flipped", by using the bitwise NOT operation; the value of 1 is then added to the resulting value, ignoring the overflow which occurs when taking the two's complement of 0.
For example, using 1 byte (=8 bits), the decimal number 5 is represented by
0000 01012
The most significant bit (the leftmost bit in this case) is 0, so the pattern represents a non-negative value. To convert to −5 in two's-complement notation, first, all bits are inverted, that is: 0 becomes 1 and 1 becomes 0:
1111 10102
At this point, the representation is the ones' complement of the decimal value −5. To obtain the two's complement, 1 is added to the result, giving:
1111 10112
The result is a signed binary number representing the decimal value −5 in two's-complement form. The most significant bit is 1, so the value represented is negative.
The two's complement of a negative number is the corresponding positive value, except in the special case of the most negative number. For example, inverting the bits of −5 (above) gives:
0000 01002
And adding one gives the final value:
0000 01012
Likewise, the two's complement of zero is zero: inverting gives all ones, and adding one changes the ones back to zeros (since the overflow is ignored).
The two's complement of the most negative number representable (e.g. a one as the most-significant bit and all other bits zero) is itself. Hence, there is an 'extra' negative number for which two's complement does not give the negation, see § Most negative number below.
### Subtraction from 2N
The sum of a number and its ones' complement is an N-bit word with all 1 bits, which is (reading as an unsigned binary number) 2N − 1. Then adding a number to its two's complement results in the N lowest bits set to 0 and the carry bit 1, where the latter has the weight (reading it as an unsigned binary number) of 2N. Hence, in the unsigned binary arithmetic the value of two's-complement negative number x* of a positive x satisfies the equality x* = 2Nx.[4]
For example, to find the four-bit representation of −5 (subscripts denote the base of the representation):
x = 510 therefore x = 01012
Hence, with N = 4:
x* = 2Nx = 24 − 510 = 1610 - 510 = 100002 − 01012 = 10112
The calculation can be done entirely in base 10, converting to base 2 at the end:
x* = 2Nx = 24 − 510 = 1110 = 10112
### Working from LSB towards MSB
A shortcut to manually convert a binary number into its two's complement is to start at the least significant bit (LSB), and copy all the zeros, working from LSB toward the most significant bit (MSB) until the first 1 is reached; then copy that 1, and flip all the remaining bits (Leave the MSB as a 1 if the initial number was in sign-and-magnitude representation). This shortcut allows a person to convert a number to its two's complement without first forming its ones' complement. For example: in two's complement representation, the negation of "0011 1100" is "1100 0100", where the underlined digits were unchanged by the copying operation (while the rest of the digits were flipped).
In computer circuitry, this method is no faster than the "complement and add one" method; both methods require working sequentially from right to left, propagating logic changes. The method of complementing and adding one can be sped up by a standard carry look-ahead adder circuit; the LSB towards MSB method can be sped up by a similar logic transformation.
## Sign extension
Sign-bit repetition in 7- and 8-bit integers using two's complement
Decimal 7-bit notation 8-bit notation
−42 1010110 1101 0110
42 0101010 0010 1010
When turning a two's-complement number with a certain number of bits into one with more bits (e.g., when copying from a one-byte variable to a two-byte variable), the most-significant bit must be repeated in all the extra bits. Some processors do this in a single instruction; on other processors, a conditional must be used followed by code to set the relevant bits or bytes.
Similarly, when a number is shifted to the right, the most-significant bit, which contains the sign information, must be maintained. However, when shifted to the left, a bit is shifted out. These rules preserve the common semantics that left shifts multiply the number by two and right shifts divide the number by two. However, if the most-significant bit changes from 0 to 1 (and vice versa), overflow is said to occur in the case that the value represents a signed integer.
Both shifting and doubling the precision are important for some multiplication algorithms. Note that unlike addition and subtraction, width extension and right shifting are done differently for signed and unsigned numbers.
## Most negative number
With only one exception, starting with any number in two's-complement representation, if all the bits are flipped and 1 added, the two's-complement representation of the negative of that number is obtained. Positive 12 becomes negative 12, positive 5 becomes negative 5, zero becomes zero(+overflow), etc.
−128 1000 0000 invert bits 0111 1111 add one 1000 0000
Taking the two's complement of the minimum number in the range will not have the desired effect of negating the number. For example, the two's complement of −128 in an eight-bit system is −128. Although the expected result from negating −128 is +128, there is no representation of +128 with an eight-bit two's complement system and thus it is in fact impossible to represent the negation. Note that the two's complement being the same number is detected as an overflow condition since there was a carry into but not out of the most-significant bit.
This phenomenon is fundamentally about the mathematics of binary numbers, not the details of the representation as two's complement.[citation needed] Mathematically, this is complementary to the fact that the negative of 0 is again 0. For a given number of bits k there is an even number of binary numbers 2k, taking negatives is a group action (of the group of order 2) on binary numbers, and since the orbit of zero has order 1, at least one other number must have an orbit of order 1 for the orders of the orbits to add up to the order of the set. Thus some other number must be invariant under taking negatives (formally, by the orbit-stabilizer theorem). Geometrically, one can view the k-bit binary numbers as the cyclic group ${\displaystyle \mathbb {Z} /2^{k}}$, which can be visualized as a circle (or properly a regular 2k-gon), and taking negatives is a reflection, which fixes the elements of order dividing 2: 0 and the opposite point, or visually the zenith and nadir.
The presence of the most negative number can lead to unexpected programming bugs where the result has an unexpected sign, or leads to an unexpected overflow exception, or leads to completely strange behaviors. For example,
• the unary negation operator may not change the sign of a nonzero number. e.g., −(−128) → −128.
• an implementation of absolute value may return a negative number.[5] e.g., abs(−128) → −128.
• Likewise, multiplication by −1 may fail to function as expected. e.g., (−128) × (−1) → −128.
• Division by −1 may cause an exception (like that caused by dividing by 0).[6] Even calculating the remainder (or modulo) by −1 can trigger this exception.[7] e.g., (−128) ÷ (−1) → crash, (−128) % (−1) → crash.
In the C and C++ programming languages, the above behaviours are undefined and not only may they return strange results, but the compiler is free to assume that the programmer has ensured that undefined computations never happen, and make inferences from that assumption.[7] This enables a number of optimizations, but also leads to a number of strange bugs in such undefined programs.
The most negative number in two's complement is sometimes called "the weird number", because it is the only exception.[8][9] Although the number is an exception, it is a valid number in regular two's complement systems. All arithmetic operations work with it both as an operand and (unless there was an overflow) a result.
## Why it works
Given a set of all possible N-bit values, we can assign the lower (by the binary value) half to be the integers from 0 to (2N − 1 − 1) inclusive and the upper half to be −2N − 1 to −1 inclusive. The upper half (again, by the binary value) can be used to represent negative integers from −2N − 1 to −1 because, under addition modulo 2N they behave the same way as those negative integers. That is to say that because i + j mod 2N = i + (j + 2N) mod 2N any value in the set { j + k 2N | k is an integer } can be used in place of j.[10]
For example, with eight bits, the unsigned bytes are 0 to 255. Subtracting 256 from the top half (128 to 255) yields the signed bytes −128 to −1.
The relationship to two's complement is realised by noting that 256 = 255 + 1, and (255 − x) is the ones' complement of x.
Some special numbers to note
Decimal Binary
127 0111 1111
64 0100 0000
1 0000 0001
0 0000 0000
−1 1111 1111
−64 1100 0000
−127 1000 0001
−128 1000 0000
### Example
In this subsection, decimal numbers are suffixed with a decimal point "."
For example, an 8 bit number can only represent every integer from −128. to 127., inclusive, since (28 − 1 = 128.). −95. modulo 256. is equivalent to 161. since
−95. + 256.
= −95. + 255. + 1
= 255. − 95. + 1
= 160. + 1.
= 161.
1111 1111 255.
− 0101 1111 − 95.
=========== =====
1010 0000 (ones' complement) 160.
+ 1 + 1
=========== =====
1010 0001 (two's complement) 161.
Two's complement 4 bit integer values
Two's complement Decimal
0111 7.
0110 6.
0101 5.
0100 4.
0011 3.
0010 2.
0001 1.
0000 0.
1111 −1.
1110 −2.
1101 −3.
1100 −4.
1011 −5.
1010 −6.
1001 −7.
1000 −8.
Fundamentally, the system represents negative integers by counting backward and wrapping around. The boundary between positive and negative numbers is arbitrary, but by convention all negative numbers have a left-most bit (most significant bit) of one. Therefore, the most positive four-bit number is 0111 (7.) and the most negative is 1000 (−8.). Because of the use of the left-most bit as the sign bit, the absolute value of the most negative number (|−8.| = 8.) is too large to represent. Negating a two's complement number is simple: Invert all the bits and add one to the result. For example, negating 1111, we get 0000 + 1 = 1. Therefore, 1111 in binary must represent −1 in decimal.[11]
The system is useful in simplifying the implementation of arithmetic on computer hardware. Adding 0011 (3.) to 1111 (−1.) at first seems to give the incorrect answer of 10010. However, the hardware can simply ignore the left-most bit to give the correct answer of 0010 (2.). Overflow checks still must exist to catch operations such as summing 0100 and 0100.
The system therefore allows addition of negative operands without a subtraction circuit or a circuit that detects the sign of a number. Moreover, that addition circuit can also perform subtraction by taking the two's complement of a number (see below), which only requires an additional cycle or its own adder circuit. To perform this, the circuit merely operates as if there were an extra left-most bit of 1.
## Arithmetic operations
Adding twos-complement numbers requires no special processing even if the operands have opposite signs: the sign of the result is determined automatically. For example, adding 15 and −5:
0000 1111 (15)
+ 1111 1011 (−5)
===========
0000 1010 (10)
Or the computation of 5 − 15 = 5 + (−15):
0000 0101 ( 5)
+ 1111 0001 (−15)
===========
1111 0110 (−10)
This process depends upon restricting to 8 bits of precision; a carry to the (nonexistent) 9th most significant bit is ignored, resulting in the arithmetically correct result of 1010.
The last two bits of the carry row (reading right-to-left) contain vital information: whether the calculation resulted in an arithmetic overflow, a number too large for the binary system to represent (in this case greater than 8 bits). An overflow condition exists when these last two bits are different from one another. As mentioned above, the sign of the number is encoded in the MSB of the result.
In other terms, if the left two carry bits (the ones on the far left of the top row in these examples) are both 1s or both 0s, the result is valid; if the left two carry bits are "1 0" or "0 1", a sign overflow has occurred. Conveniently, an XOR operation on these two bits can quickly determine if an overflow condition exists. As an example, consider the signed 4-bit addition of 7 and 3:
0111 (carry)
0111 (7)
+ 0011 (3)
======
1010 (−6) invalid!
In this case, the far left two (MSB) carry bits are "01", which means there was a two's-complement addition overflow. That is, 10102 = 1010 is outside the permitted range of −8 to 7. The result would be correct if treated as unsigned integer.
In general, any two N-bit numbers may be added without overflow, by first sign-extending both of them to N + 1 bits, and then adding as above. The N + 1 bits result is large enough to represent any possible sum (N = 5 two's complement can represent values in the range −16 to 15) so overflow will never occur. It is then possible, if desired, to 'truncate' the result back to N bits while preserving the value if and only if the discarded bit is a proper sign extension of the retained result bits. This provides another method of detecting overflow—which is equivalent to the method of comparing the carry bits—but which may be easier to implement in some situations, because it does not require access to the internals of the addition.
### Subtraction
Computers usually use the method of complements to implement subtraction. Using complements for subtraction is closely related to using complements for representing negative numbers, since the combination allows all signs of operands and results; direct subtraction works with two's-complement numbers as well. Like addition, the advantage of using two's complement is the elimination of examining the signs of the operands to determine whether addition or subtraction is needed. For example, subtracting −5 from 15 is really adding 5 to 15, but this is hidden by the two's-complement representation:
11110 000 (borrow)
0000 1111 (15)
− 1111 1011 (−5)
===========
0001 0100 (20)
Overflow is detected the same way as for addition, by examining the two leftmost (most significant) bits of the borrows; overflow has occurred if they are different.
Another example is a subtraction operation where the result is negative: 15 − 35 = −20:
11100 000 (borrow)
0000 1111 (15)
− 0010 0011 (35)
===========
1110 1100 (−20)
As for addition, overflow in subtraction may be avoided (or detected after the operation) by first sign-extending both inputs by an extra bit.
### Multiplication
The product of two N-bit numbers requires 2N bits to contain all possible values.[12]
If the precision of the two operands using two's complement is doubled before the multiplication, direct multiplication (discarding any excess bits beyond that precision) will provide the correct result.[13] For example, take 6 × (−5) = −30. First, the precision is extended from four bits to eight. Then the numbers are multiplied, discarding the bits beyond the eighth bit (as shown by "x"):
00000110 (6)
* 11111011 (−5)
============
110
1100
00000
110000
1100000
11000000
x10000000
+ xx00000000
============
xx11100010
This is very inefficient; by doubling the precision ahead of time, all additions must be double-precision and at least twice as many partial products are needed than for the more efficient algorithms actually implemented in computers. Some multiplication algorithms are designed for two's complement, notably Booth's multiplication algorithm. Methods for multiplying sign-magnitude numbers don't work with two's-complement numbers without adaptation. There isn't usually a problem when the multiplicand (the one being repeatedly added to form the product) is negative; the issue is setting the initial bits of the product correctly when the multiplier is negative. Two methods for adapting algorithms to handle two's-complement numbers are common:
• First check to see if the multiplier is negative. If so, negate (i.e., take the two's complement of) both operands before multiplying. The multiplier will then be positive so the algorithm will work. Because both operands are negated, the result will still have the correct sign.
• Subtract the partial product resulting from the MSB (pseudo sign bit) instead of adding it like the other partial products. This method requires the multiplicand's sign bit to be extended by one position, being preserved during the shift right actions.[14]
As an example of the second method, take the common add-and-shift algorithm for multiplication. Instead of shifting partial products to the left as is done with pencil and paper, the accumulated product is shifted right, into a second register that will eventually hold the least significant half of the product. Since the least significant bits are not changed once they are calculated, the additions can be single precision, accumulating in the register that will eventually hold the most significant half of the product. In the following example, again multiplying 6 by −5, the two registers and the extended sign bit are separated by "|":
0 0110 (6) (multiplicand with extended sign bit)
× 1011 (−5) (multiplier)
=|====|====
0|0110|0000 (first partial product (rightmost bit is 1))
0|0011|0000 (shift right, preserving extended sign bit)
0|1001|0000 (add second partial product (next bit is 1))
0|0100|1000 (shift right, preserving extended sign bit)
0|0100|1000 (add third partial product: 0 so no change)
0|0010|0100 (shift right, preserving extended sign bit)
1|1100|0100 (subtract last partial product since it's from sign bit)
1|1110|0010 (shift right, preserving extended sign bit)
### Comparison (ordering)
Comparison is often implemented with a dummy subtraction, where the flags in the computer's status register are checked, but the main result is ignored. The zero flag indicates if two values compared equal. If the exclusive-or of the sign and overflow flags is 1, the subtraction result was less than zero, otherwise the result was zero or greater. These checks are often implemented in computers in conditional branch instructions.
Unsigned binary numbers can be ordered by a simple lexicographic ordering, where the bit value 0 is defined as less than the bit value 1. For two's complement values, the meaning of the most significant bit is reversed (i.e. 1 is less than 0).
The following algorithm (for an n-bit two's complement architecture) sets the result register R to −1 if A < B, to +1 if A > B, and to 0 if A and B are equal:
// reversed comparison of the sign bit
if A(n-1) == 0 and B(n-1) == 1 then
return +1
else if A(n-1) == 1 and B(n-1) == 0 then
return -1
end
// comparison of remaining bits
for i = n-2...0 do
if A(i) == 0 and B(i) == 1 then
return -1
else if A(i) == 1 and B(i) == 0 then
return +1
end
end
return 0
## Two's complement and 2-adic numbers
In a classic HAKMEM published by the MIT AI Lab in 1972, Bill Gosper noted that whether or not a machine's internal representation was two's-complement could be determined by summing the successive powers of two. In a flight of fancy, he noted that the result of doing this algebraically indicated that "algebra is run on a machine (the universe) which is two's-complement."[15]
Gosper's end conclusion is not necessarily meant to be taken seriously, and it is akin to a mathematical joke. The critical step is "...110 = ...111 − 1", i.e., "2X = X − 1", and thus X = ...111 = −1. This presupposes a method by which an infinite string of 1s is considered a number, which requires an extension of the finite place-value concepts in elementary arithmetic. It is meaningful either as part of a two's-complement notation for all integers, as a typical 2-adic number, or even as one of the generalized sums defined for the divergent series of real numbers 1 + 2 + 4 + 8 + ···.[16] Digital arithmetic circuits, idealized to operate with infinite (extending to positive powers of 2) bit strings, produce 2-adic addition and multiplication compatible with two's complement representation.[17] Continuity of binary arithmetical and bitwise operations in 2-adic metric also has some use in cryptography.[18]
## Fraction conversion
To convert a number with a fractional part, such as .0101, one must convert starting from right to left the 1s to decimal as in a normal conversion. In this example 0101 is equal to 5 in decimal. Each digit after the floating point represents a fraction where the denominator is a multiplier of 2. So, the first is 1/2, the second is 1/4 and so on. Having already calculated the decimal value as mentioned above, only the denominator of the LSB (LSB = starting from right) is used. The final result of this conversion is 5/16.
For instance, having the floating value of .0110 for this method to work, one should not consider the last 0 from the right. Hence, instead of calculating the decimal value for 0110, we calculate the value 011, which is 3 in decimal (by leaving the 0 in the end, the result would have been 6, together with the denominator 24 = 16, which reduces to 3/8). The denominator is 8, giving a final result if 3/8.
## References
1. ^ E.g. "Signed integers are two's complement binary values that can be used to represent both positive and negative integer values.", Section 4.2.1 in Intel 64 and IA-32 Architectures Software Developer's Manual, Volume 1: Basic Architecture, November 2006
2. ^ David J. Lilja and Sachin S. Sapatnekar, Designing Digital Computer Systems with Verilog, Cambridge University Press, 2005 online
3. ^ von Neumann, John (1945), First Draft of a Report on the EDVAC (PDF), retrieved February 20, 2021
4. ^ For x = 0 we have 2N − 0 = 2N, which is equivalent to 0* = 0 modulo 2N (i.e. after restricting to N least significant bits).
5. ^ "Math". Java Platform SE 7 API specification.
6. ^ Regehr, John (2013). "Nobody Expects the Spanish Inquisition, or INT_MIN to be Divided by -1". Regehr.org (blog).
7. ^ a b Seacord, Robert C. (2020). "Rule INT32-C. Ensure that operations on signed integers do not result in overflow". wiki.sei.cmu.edu. SEI CERT C Coding Standard.
8. ^ Affeldt, Reynald & Marti, Nicolas. "Formal Verification of Arithmetic Functions in SmartMIPS Assembly" (PDF). Archived from the original (PDF) on 2011-07-22.
9. ^ Harris, David; Harris, David Money; Harris, Sarah L. (2007). "Weird binary number". Digital Design and Computer Architecture. p. 18 – via Google Books.
10. ^ "3.9. Two's Complement". Chapter 3. Data Representation. cs.uwm.edu. 2012-12-03. Archived from the original on 31 October 2013. Retrieved 2014-06-22.
11. ^ Finley, Thomas (April 2000). "Two's Complement". Computer Science. Class notes for CS 104. Ithaca, NY: Cornell University. Retrieved 2014-06-22.
12. ^ Bruno Paillard. An Introduction To Digital Signal Processors, Sec. 6.4.2. Génie électrique et informatique Report, Université de Sherbrooke, April 2004.
13. ^ Karen Miller (August 24, 2007). "Two's Complement Multiplication". cs.wisc.edu. Archived from the original on February 13, 2015. Retrieved April 13, 2015.
14. ^ Wakerly, John F. (2000). Digital Design Principles & Practices (3rd ed.). Prentice Hall. p. 47. ISBN 0-13-769191-2.
15. ^ Hakmem - Programming Hacks - Draft, Not Yet Proofed
16. ^ For the summation of 1 + 2 + 4 + 8 + ··· without recourse to the 2-adic metric, see Hardy, G.H. (1949). Divergent Series. Clarendon Press. LCC QA295 .H29 1967. (pp. 7–10)
17. ^ Vuillemin, Jean (1993). On circuits and numbers (PDF). Paris: Digital Equipment Corp. p. 19. Retrieved 2012-01-24., Chapter 7, especially 7.3 for multiplication.
18. ^ Anashin, Vladimir; Bogdanov, Andrey; Kizhvatov, Ilya (2007). "ABC Stream Cipher". Russian State University for the Humanities. Retrieved 24 January 2012.
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# Sample Spaces and Events
## Words or diagrams that detail favorable outcomes and intersections, complements, and unions of events.
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Practice Sample Spaces and Events
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Sample Spaces and Events
Credit: Sean Freese
Source: https://flic.kr/p/drQj35
Jacqueline prepares the Halloween bowl for trick-or-treaters with equal numbers of Reese's, Kit Kat, Twix and M&Ms. Her first trick-or-treater randomly pulls a candy bar from the bowl, puts it back, then randomly pulls another candy bar from the bowl. Jacqueline wonders how many outcomes are possible from the sample space for the two candy bar pulls from the bowl.
In this concept, you will learn about sample spaces.
### Guidance
When you conduct an experiment, there are many possible outcomes. If you are doing an experiment with a coin, there are two possible outcomes because there are two sides of the coin. You can either have heads or tails. If you experiment with a number cube, there are six possible outcomes because there are six sides of the number cube and the sides are numbered one to six. We can think of all of these possible outcomes as the sample space.
A sample space is the set of all possible outcomes for a probability experiment or activity. For example, on the spinner there are 5 different colors on which the arrow can land. The sample space, \begin{align*}S\end{align*}, for one spin of the spinner is then:
\begin{align*}S = \text{red, yellow, pink, green, blue}\end{align*}
These are the only outcomes that result from a single spin of the spinner.
Changing the spinner changes the sample space. This second spinner still has 5 equal-sized sections. But its sample space now has only 3 outcomes:
\begin{align*}S = \text{red, yellow, blue}\end{align*}
Let’s look at an example with sample spaces.
A small jar contains 1 white, 1 black, and 1 red marble. If one marble is randomly chosen, how many possible outcomes are there in the sample space?
Since only a single marble is being chosen, the total number of possible outcomes, or sample space matches the marble colors.
\begin{align*}S = \text{white, black, red}\end{align*}
Sometimes, the sample space can change if an experiment is performed more than once. If a marble is selected from a jar and then replaced and if the experiment is conducted again, then the sample space can change. The number of outcomes is altered. When this happens, we can use tree diagrams and multiplication to help us figure out the number of outcomes in the sample space.
A jar contains 1 white and 1 black marble. If one marble is randomly chosen, returned to the jar, then a second marble is chosen, how many possible outcomes are there?
This is a situation where a tree diagram is very useful. Consider the marbles one at a time. After the first marble is chosen, it is returned to the jar so now there are again two choices for the second marble.
Use a tree diagram to list the outcomes.
From the tree diagram, you can see that the sample space is:
\begin{align*}S = \text{white-white, white-black, black-white, black-black}\end{align*}
We could also multiply by multiplying the two options and the number of selections. Two marble colors and two selections:
\begin{align*}2 \times 2 = 4\end{align*}
There are four outcomes in the sample space.
### Guided Practice
June flipped a coin three times. How many outcomes are in the sample space?
First, make a list of the possible outcomes for each flip.
Tails
Next, count the number of the possible outcomes for each flip.
There are two outcomes for each flip of a coin: heads or tails.
Then, multiply the number of outcomes by the number of flips.
June flipped the coin three times.
\begin{align*}2 \times 3 = 6\end{align*}
The answer is there are 6 outcomes in the sample space.
### Examples
#### Example 1
How many outcomes are in the sample space of three spins of a spinner with red, blue, yellow and green?
First, make a list of the possible outcomes for each spin.
Red
Blue
Yellow
Green
Next, count the number of possible outcomes for each spin.
There are four possible outcomes for each spin: red, blue, yellow, green.
Then, multiply the number of outcomes by the number of spins.
June flipped the coin three times.
\begin{align*}4 \times 3 = 12\end{align*}
The answer is there are 12 outcomes in the sample space.
#### Example 2
What is the sample space for the roll of a number cube numbered 1–6 and how many possible outcomes are there from one roll?
First, determine all possible outcomes:
1, 2, 3, 4, 5, 6
Next, count the number of possible outcomes for one roll:
There are 6 possible outcomes: 1, 2, 3, 4, 5, 6
Then, multiply the number of outcomes by the number of rolls.
Since we are only rolling once, the number of possible outcomes is 6.
The answer is the sample space is 1, 2, 3, 4, 5, 6 and the number of possible outcomes is 6.
#### Example 3
A bag with a blue and a red marble. One marble is drawn and then replaced. What is the sample space?
First, list the possible outcomes for each draw:
blue
red
Next, list the possible outcomes for Draw 1 and Draw 2:
Draw 1: red blue
Draw 2: red blue red blue
Then, list all possible outcomes from the two draws:
red-red, red-blue, blue-red, blue-blue
The answer is the sample space is red-red, red-blue, blue-red, blue-blue
Credit: Sean Freese
Source: https://flic.kr/p/drQj35
Remember Jacqueline's curiosity about the candy from the Halloween bowl?
The trick-or-treater pulled one time randomly, returned the candy bar to the bowl, then pulled randomly once more. How many possible outcomes are there from the two pulls?
First, list the different candy choices:
Twix, Reese's, M&Ms, Kit Kat
Next, list the possible outcomes of Pull 1 and Pull 2:
Pull 1: Twix M&Ms
Pull 2: Twix Reese's M&Ms Kit Kat Twix Reese's M&Ms Kit Kat
Pull 1: Reese's Kit Kat
Pull 2: Twix Reese's M&Ms Kit Kat Twix Reese's M&Ms Kit Kat
Then, list the sample space, or all possible outcomes of the two pulls from the candy bowl:
Twix-Twix, Twix-Reese's, Twix-M&Ms, Twix-Kit Kat, M&Ms-Twix, M&Ms-Reese's, M&Ms-M&Ms, M&Ms-Kit Kat, Reese's-Twix, Reese's-Reese's, Reese's-M&Ms, Reese's-Kit Kat, Kit Kat-Twix, Kit Kat-Reese's, Kit Kat-M&Ms, Kit Kat-Kit Kat
The answer is there are 16 possible outcomes.
### Explore More
1. What is the sample space for a single toss of a number cube?
2. What is the sample space for a single flip of a coin?
3. A coin is flipped two times. List all possible outcomes for the two flips.
4. A coin is flipped three times in a row. List all possible outcomes for the two flips.
5. A bag contains 3 ping pong balls: 1 red, 1 blue, and 1 green. What is the sample space for drawing a single ball from the bag?
6. A bag contains 3 ping pong balls: 1 red, 1 blue, and 1 green. What is the sample space for drawing a single ball, returning the ball to the bag, then drawing a second ball?
7. What is the sample space for a single spin of the with red, blue, yellow and green sections spinner?
8. What is the sample space for 2 spins of the first spinner?
9. What is the sample space for three spins of the spinner?
10. A box contains 3 socks: 1 black, 1 white, and 1 brown. What is the sample space for drawing a single sock, NOT returning the sock to the box, then drawing a second sock?
11. A box contains 3 socks: 1 black, 1 white, and 1 brown. What is the sample space for drawing all 3 socks from the box, one at a time, without returning any of the socks to the box?
12. A box contains 3 black socks. What is the sample space for drawing all 2 socks from the box, one at a time, without returning any of the socks to the box?
13. A box contains 2 black socks and 1 white sock. What is the sample space for drawing all 2 socks from the box, one at a time, without returning any of the socks to the box?
14. True or false. A sample space is the total possible outcomes.
15. True or false. A sample space is a percentage.
### Vocabulary Language: English
experiment
experiment
An experiment is the process of taking a measurement or making an observation.
Favorable Outcome
Favorable Outcome
A favorable outcome is the outcome that you are looking for in an experiment.
Outcome
Outcome
An outcome of a probability experiment is one possible end result.
Sample Space
Sample Space
In a probability experiment, the sample space is the set of all the possible outcomes of the experiment.
simple events
simple events
A simple event is the simplest outcome of an experiment.
tree diagrams
tree diagrams
Tree diagrams are a way to show the outcomes of simple probability events where each outcome is represented as a branch on a tree.
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## Linear Algebra and Its Applications, Exercise 2.2.10
Exercise 2.2.10. (a) Find all possible solutions to the following system:
$Ux = \begin{bmatrix} 1&2&3&4 \\ 0&0&1&2 \\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} x_1 \\x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$
(b) What are the solutions if the right side is replaced with $(a, b, 0)$?
Answer: (a) Since the pivots of $U$ are in columns 1 and 3, the basic variables are $x_1$ and $x_3$ and the free variables are $x_2$ and $x_4$.
From the second equation we have $x_3 + 2x_4 = 0$ or $x_3 = -2x_4$. Substituting the value of $x_3$ into the first equation we have $x_1 + 2x_2 - 6x_4 + 4x_4 = 0$ or $x_1 = -2x_2 + 2x_4$. The solution to $Ux = 0$ can then be expressed in terms of the free variables $x_2$ and $x_4$ as follows:
$x_{homogeneous} = x_2 \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 2 \\ 0 \\ -2 \\ 1 \end{bmatrix}$
(b) Replacing f the right side with $(a, b, 0)$ produces the following system:
$Ux = \begin{bmatrix} 1&2&3&4 \\ 0&0&1&2 \\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} x_1 \\x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} a \\ b \\ 0 \end{bmatrix}$
From the second equation we have $x_3 + 2x_4 = b$ or $x_3 = b - 2x_4$. Substituting for $x_3$ into the first equation we have $x_1 + 2x_2 + 3(b - 2x_4) + 4x_4 = a$ or $x_1 = a - 3b - 2x_2 - 2x_4$. Setting the free variables $x_2$ and $x_4$ to zero gives the particular solution
$x_{particular} = \begin{bmatrix} a - 3b \\ 0 \\ b \\ 0 \end{bmatrix}$
The general solution is then
$x = x_{particular} + x_{homogeneous}$
$= \begin{bmatrix} a-3b \\ 0 \\ b \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 2 \\ 0 \\ -2 \\ 1 \end{bmatrix}$
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.
This entry was posted in linear algebra. Bookmark the permalink.
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# Sakshi can do a piece of work in 20 days. Tanya is 25% more efficient than Sakshi. The number of days taken by Tanya to do the same piece of work is: A.15 B.16 C.18 D.25
Pawan Prajapati
2 years ago
We will find the number of days taken by Tanya to complete the piece of work. We will find the ratio of the time taken by Sakshi and Tanya by using the relation between the time and efficiency and the ratio of the time taken by the given number of days and by equating these ratios, we will find the number of days. Thus, the number of days taken by Tanya to complete the piece of work. Formula Used: Time taken is always inversely proportional to the efficiency i.e., t∝1efficiency% Complete step-by-step answer: We are given that Sakshi can do a piece of work in 20 days. We are given that Tanya is 25% more efficient than Sakshi. Let Sakshi efficiency be 100% then Tanya efficiency be 125% We know that time is always inversely proportional to the efficiency i.e., t∝1efficiency% So, the Ratio of the time taken by Sakshi to Tanya =1100:1125 Now, by cross multiplying, to equalize the denominator, we get ⇒ Ratio of the time taken by Sakshi to Tanya =125:100 By simplification, we get ⇒ Ratio of the time taken by Sakshi to Tanya =5:4 …………………………………………………………………(1) Let x be the number of days taken by Tanya and Sakshi takes 20 days to complete the piece of work. ⇒ Ratio of the time taken by Sakshi to Tanya =20:x ……………………………………………………………….(2) Now, by equating equation (1) and equation (2) , we get ⇒5:4=20:x Now, Ratio is represented in the form of Fractions, we get ⇒54=20x By cross multiplying, we get ⇒5x=20×4 Now, by rewriting the equation, we get ⇒x=20×45 By simplifying the terms, we get ⇒x=4×4 ⇒x=16 Therefore, the number of days taken by Tanya to complete the piece of work is 16 days. So, Option (B) is the correct answer. Note: We know that when two ratios are equal, then it is said to be in Proportion. So, 5:4=20:x can also be written as 5:4::20:x . When two quantities are in direct proportion, then when one amount increases, then another amount also increases at the same rate. So, we can write as x=y . When two quantities are in indirect proportion, then when one amount increases, then another amount decreases at the same rate. So, it can be written as x=1y . We should remember these formulas while writing a proportion with direct and indirect variation. We should also know the relation between time and efficiency.
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Normal Distributions
Fitting a bell curve to a histogram
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Practice Normal Distributions
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Normal Distributions
In this Concept, you will learn the Normal Distribution, several of its properties and how to determine whether a data set corresponds to a normal distribution.
Watch This
For an explanation of a standardized normal distribution (4.0)(7.0) , see APUS07, Standard Normal Distribution (4:22).
Guidance
The Characteristics of a Normal Distribution
Think about the popcorn example in the introduction to this Chapter. The amount of popcorn popping starts small, gets larger, and then decreases as time goes by, much like the graph below.
This graph is considered a normal curve, or a "bell curve" because it looks like a bell. Many data sets look similar to this when plotted. But are they all exactly the same? The answer is no; they may be centered at different values, and some may be more spread out than others. While still having this similar bell shape, there may be slight differences in the exact shape.
Shape
When graphing the data from each of the examples in the introduction, the distributions from each of these situations would be mound-shaped and mostly symmetric. A normal distribution is a perfectly symmetric, mound-shaped distribution.
Because so many real data sets closely approximate a normal distribution, we can use the idealized normal curve to learn a great deal about such data. With a practical data collection, the distribution will never be exactly symmetric, so just like situations involving probability, a true normal distribution only results from an infinite collection of data. Also, it is important to note that the normal distribution describes a continuous random variable.
Center
Due to the exact symmetry of a normal curve, the center of a normal distribution, or a data set that approximates a normal distribution, is located at the highest point of the distribution, and all the statistical measures of center we have already studied (the mean, median, and mode) are equal.
It is also important to realize that this center peak divides the data into two equal parts.
Spread
Let’s go back to our popcorn example. The bag advertises a certain time, beyond which you risk burning the popcorn. From experience, the manufacturers know when most of the popcorn will stop popping, but there is still a chance that there are those rare kernels that will require more (or less) time to pop than the time advertised by the manufacturer. The directions usually tell you to stop when the time between popping is a few seconds, but aren’t you tempted to keep going so you don’t end up with a bag full of un-popped kernels? Because this is a real, and not theoretical, situation, there will be a time when the popcorn will stop popping and start burning, but there is always a chance, no matter how small, that one more kernel will pop if you keep the microwave going. In an idealized normal distribution of a continuous random variable, the distribution continues infinitely in both directions.
Because of this infinite spread, the range would not be a useful statistical measure of spread. The most common way to measure the spread of a normal distribution is with the standard deviation, or the typical distance away from the mean. Because of the symmetry of a normal distribution, the standard deviation indicates how far away from the maximum peak the data will be. Here are two normal distributions with the same center (mean):
The first distribution pictured above has a smaller standard deviation, and so more of the data are heavily concentrated around the mean than in the second distribution. Also, in the first distribution, there are fewer data values at the extremes than in the second distribution. Because the second distribution has a larger standard deviation, the data are spread farther from the mean value, with more of the data appearing in the tails.
Technology Note: Investigating the Normal Distribution on a TI-83/84 Graphing Calculator
We can graph a normal curve for a probability distribution on the TI-83/84 calculator. To do so, first press [Y=] . To create a normal distribution, we will draw an idealized curve using something called a density function. The command is called 'normalpdf(', and it is found by pressing [2nd][DISTR][1] . Enter an X to represent the random variable, followed by the mean and the standard deviation, all separated by commas. For this example, choose a mean of 5 and a standard deviation of 1.
Adjust your window to match the following settings and press [GRAPH] .
Press [2ND][QUIT] to go to the home screen. We can draw a vertical line at the mean to show it is in the center of the distribution by pressing [2ND][DRAW] and choosing 'Vertical'. Enter the mean, which is 5, and press [ENTER] .
Remember that even though the graph appears to touch the \begin{align*}x\end{align*} -axis, it is actually just very close to it.
In your Y= Menu, enter the following to graph 3 different normal distributions, each with a different standard deviation:
This makes it easy to see the change in spread when the standard deviation changes.
The Empirical Rule
Because of the similar shape of all normal distributions, we can measure the percentage of data that is a certain distance from the mean no matter what the standard deviation of the data set is. The following graph shows a normal distribution with \begin{align*}\mu=0\end{align*} and \begin{align*}\sigma=1\end{align*} . This curve is called a standard normal curve . In this case, the values of \begin{align*}x\end{align*} represent the number of standard deviations away from the mean.
Notice that vertical lines are drawn at points that are exactly one standard deviation to the left and right of the mean. We have consistently described standard deviation as a measure of the typical distance away from the mean. How much of the data is actually within one standard deviation of the mean? To answer this question, think about the space, or area, under the curve. The entire data set, or 100% of it, is contained under the whole curve. What percentage would you estimate is between the two lines? To help estimate the answer, we can use a graphing calculator. Graph a standard normal distribution over an appropriate window.
Now press [2ND][DISTR] , go to the DRAW menu, and choose 'ShadeNorm('. Insert ' \begin{align*}-1\end{align*} , 1' after the 'ShadeNorm(' command and press [ENTER] . It will shade the area within one standard deviation of the mean.
The calculator also gives a very accurate estimate of the area. We can see from the rightmost screenshot above that approximately 68% of the area is within one standard deviation of the mean. If we venture to 2 standard deviations away from the mean, how much of the data should we expect to capture? Make the following changes to the 'ShadeNorm(' command to find out:
Notice from the shading that almost all of the distribution is shaded, and the percentage of data is close to 95%. If you were to venture to 3 standard deviations from the mean, 99.7%, or virtually all of the data, is captured, which tells us that very little of the data in a normal distribution is more than 3 standard deviations from the mean.
Notice that the calculator actually makes it look like the entire distribution is shaded because of the limitations of the screen resolution, but as we have already discovered, there is still some area under the curve further out than that. These three approximate percentages, 68%, 95%, and 99.7%, are extremely important and are part of what is called the Empirical Rule .
The Empirical Rule states that the percentages of data in a normal distribution within 1, 2, and 3 standard deviations of the mean are approximately 68%, 95%, and 99.7%, respectively.
On the Web
http://tinyurl.com/2ue78u Explore the Empirical Rule.
\begin{align*}z\end{align*} -Scores
A \begin{align*}z\end{align*} -score is a measure of the number of standard deviations a particular data point is away from the mean. For example, let’s say the mean score on a test for your statistics class was an 82, with a standard deviation of 7 points. If your score was an 89, it is exactly one standard deviation to the right of the mean; therefore, your \begin{align*}z\end{align*} -score would be 1. If, on the other hand, you scored a 75, your score would be exactly one standard deviation below the mean, and your \begin{align*}z\end{align*} -score would be \begin{align*}-1\end{align*} . All values that are below the mean have negative \begin{align*}z\end{align*} -scores, while all values that are above the mean have positive \begin{align*}z\end{align*} -scores. A \begin{align*}z\end{align*} -score of \begin{align*}-2\end{align*} would represent a value that is exactly 2 standard deviations below the mean, so in this case, the value would be \begin{align*}82 - 14 = 68\end{align*} .
To calculate a \begin{align*}z\end{align*} -score for which the numbers are not so obvious, you take the deviation and divide it by the standard deviation.
\begin{align*}z = \frac{\text{Deviation}}{\text{Standard Deviation}}\end{align*}
You may recall that deviation is the mean value of the variable subtracted from the observed value, so in symbolic terms, the \begin{align*}z\end{align*} -score would be:
\begin{align*}z=\frac{x-\mu}{\sigma}\end{align*}
As previously stated, since \begin{align*}\sigma\end{align*} is always positive, \begin{align*}z\end{align*} will be positive when \begin{align*}x\end{align*} is greater than \begin{align*}\mu\end{align*} and negative when \begin{align*}x\end{align*} is less than \begin{align*}\mu\end{align*} . A \begin{align*}z\end{align*} -score of zero means that the term has the same value as the mean. The value of \begin{align*}z\end{align*} represents the number of standard deviations the given value of \begin{align*}x\end{align*} is above or below the mean.
Example A
What is the \begin{align*}z\end{align*} -score for an \begin{align*}A\end{align*} on the test described above, which has a mean score of 82? (Assume that an \begin{align*}A\end{align*} is a 93.)
The \begin{align*}z\end{align*} -score can be calculated as follows:
\begin{align*}z&=\frac{x-\mu}{\sigma}\\ z & = \frac{93-82}{7}\\ z&=\frac{11}{7} \approx 1.57\end{align*}
If we know that the test scores from the last example are distributed normally, then a \begin{align*}z\end{align*} -score can tell us something about how our test score relates to the rest of the class. From the Empirical Rule, we know that about 68% of the students would have scored between a \begin{align*}z\end{align*} -score of \begin{align*}-1\end{align*} and 1, or between a 75 and an 89, on the test. If 68% of the data is between these two values, then that leaves the remaining 32% in the tail areas. Because of symmetry, half of this, or 16%, would be in each individual tail.
Example B
On a college entrance exam, the mean was 70, and the standard deviation was 8. If Helen’s \begin{align*}z\end{align*} -score was \begin{align*}-1.5\end{align*} , what was her exam score?
Solution:
\begin{align*}z&=\frac{x-\mu}{\sigma}\\ \therefore z \cdot \sigma & = x-\mu\\ x&=\mu+z\cdot\sigma\\ x&=70+(-1.5)(8)\\ x&=58\end{align*}
Assessing Normality
The best way to determine if a data set approximates a normal distribution is to look at a visual representation. Histograms and box plots can be useful indicators of normality, but they are not always definitive. It is often easier to tell if a data set is not normal from these plots.
If a data set is skewed right, it means that the right tail is significantly longer than the left. Similarly, skewed left means the left tail has more weight than the right. A bimodal distribution, on the other hand, has two modes, or peaks. For instance, with a histogram of the heights of American 30-year-old adults, you will see a bimodal distribution \begin{align*}-\end{align*} one mode for males and one mode for females.
There is a plot we can use to determine if a distribution is normal called a normal probability plot or normal quantile plot . To make this plot by hand, first order your data from smallest to largest. Then, determine the quantile of each data point. Finally, using a table of standard normal probabilities, determine the closest z-score for each quantile. Plot these z-scores against the actual data values. To make a normal probability plot using your calculator, enter your data into a list, then use the last type of graph in the STAT PLOT menu, as shown below:
If the data set is normal, then this plot will be perfectly linear. The closer to being linear the normal probability plot is, the more closely the data set approximates a normal distribution.
Look below at the histogram and the normal probability plot for the same data.
The histogram is fairly symmetric and mound-shaped and appears to display the characteristics of a normal distribution. When the \begin{align*}z\end{align*} -scores of the quantiles of the data are plotted against the actual data values, the normal probability plot appears strongly linear, indicating that the data set closely approximates a normal distribution. The following example will allow you to see how a normal probability plot is made in more detail.
Example C
The following data set tracked high school seniors' involvement in traffic accidents. The participants were asked the following question: “During the last 12 months, how many accidents have you had while you were driving (whether or not you were responsible)?”
Year Percentage of high school seniors who said they were involved in no traffic accidents
1991 75.7
1992 76.9
1993 76.1
1994 75.7
1995 75.3
1996 74.1
1997 74.4
1998 74.4
1999 75.1
2000 75.1
2001 75.5
2002 75.5
2003 75.8
Figure: Percentage of high school seniors who said they were involved in no traffic accidents. Source : Sourcebook of Criminal Justice Statistics: http://www.albany.edu/sourcebook/pdf/t352.pdf
Here is a histogram and a box plot of this data:
The histogram appears to show a roughly mound-shaped and symmetric distribution. The box plot does not appear to be significantly skewed, but the various sections of the plot also do not appear to be overly symmetric, either. In the following chart, the data has been reordered from smallest to largest, the quantiles have been determined, and the closest corresponding z-scores have been found using a table of standard normal probabilities.
Year Percentage Quantile z-score
1996 74.1 \begin{align*}\frac{1}{13}=0.078\end{align*} \begin{align*}-1.42\end{align*}
1997 74.4 \begin{align*}\frac{2}{13}=0.154\end{align*} \begin{align*}-1.02\end{align*}
1998 74.4 \begin{align*}\frac{3}{13}=0.231\end{align*} \begin{align*}-0.74\end{align*}
1999 75.1 \begin{align*}\frac{4}{13}=0.286\end{align*} \begin{align*}-0.56\end{align*}
2000 75.1 \begin{align*}\frac{5}{13}=0.385\end{align*} \begin{align*}-0.29\end{align*}
1995 75.3 \begin{align*}\frac{6}{13}=0.462\end{align*} \begin{align*}-0.09\end{align*}
2001 75.5 \begin{align*}\frac{7}{13}=0.538\end{align*} \begin{align*}0.1\end{align*}
2002 75.5 \begin{align*}\frac{8}{13}=0.615\end{align*} \begin{align*}0.29\end{align*}
1991 75.7 \begin{align*}\frac{9}{13}=0.692\end{align*} \begin{align*}0.5\end{align*}
1994 75.7 \begin{align*}\frac{10}{13}=0.769\end{align*} \begin{align*}0.74\end{align*}
2003 75.8 \begin{align*}\frac{11}{13}=0.846\end{align*} \begin{align*}1.02\end{align*}
1993 76.1 \begin{align*}\frac{12}{13}=0.923\end{align*} \begin{align*}1.43\end{align*}
1992 76.9 \begin{align*}\frac{13}{13}=1\end{align*} \begin{align*}3.49\end{align*}
Figure: Table of quantiles and corresponding \begin{align*}z\end{align*} -scores for senior no-accident data.
Here is a plot of the percentages versus the \begin{align*}z\end{align*} -scores of their quantiles, or the normal probability plot:
Remember that you can simplify this process by simply entering the percentages into a \begin{align*}L_1\end{align*} in your calculator and selecting the normal probability plot option (the last type of plot) in STAT PLOT .
While not perfectly linear, this plot does have a strong linear pattern, and we would, therefore, conclude that the distribution is reasonably normal.
Guided Practice
On a nationwide math test, the mean was 65 and the standard deviation was 10. If Robert scored 81, what was his \begin{align*}z\end{align*} -score?
Solution:
\begin{align*}z&=\frac{x-\mu}{\sigma}\\ z&=\frac{81-65}{10}\\ z&=\frac{16}{10}\\ z&=1.6\end{align*}
Robert's \begin{align*}z\end{align*} -score is 1.6, which means that he scored 1.6 standard deviations above the mean.
Explore More
Sample explanations for some of the practice exercises below are available by viewing the following videos. Khan Academy: Normal Distribution Problems (10:52)
1. Which of the following data sets is most likely to be normally distributed? For the other choices, explain why you believe they would not follow a normal distribution.
1. The hand span (measured from the tip of the thumb to the tip of the extended \begin{align*}5^{\text{th}}\end{align*} finger) of a random sample of high school seniors
2. The annual salaries of all employees of a large shipping company
3. The annual salaries of a random sample of 50 CEOs of major companies, 25 women and 25 men
4. The dates of 100 pennies taken from a cash drawer in a convenience store
2. The grades on a statistics mid-term for a high school are normally distributed, with \begin{align*}\mu=81\end{align*} and \begin{align*}\sigma=6.3\end{align*} . Calculate the \begin{align*}z\end{align*} -scores for each of the following exam grades. Draw and label a sketch for each example. 65, 83, 93, 100
3. Assume that the mean weight of 1-year-old girls in the USA is normally distributed, with a mean of about 9.5 kilograms and a standard deviation of approximately 1.1 kilograms. Without using a calculator, estimate the percentage of 1-year-old girls who meet the following conditions. Draw a sketch and shade the proper region for each problem.
1. Less than 8.4 kg
2. Between 7.3 kg and 11.7 kg
3. More than 12.8 kg
4. For a standard normal distribution, place the following in order from smallest to largest.
1. The percentage of data below 1
2. The percentage of data below \begin{align*}-1\end{align*}
3. The mean
4. The standard deviation
5. The percentage of data above 2
5. The 2007 AP Statistics examination scores were not normally distributed, with \begin{align*}\mu=2.8\end{align*} and \begin{align*}\sigma=1.34\end{align*} . What is the approximate \begin{align*}z\end{align*} -score that corresponds to an exam score of 5? (The scores range from 1 to 5.)
1. 0.786
2. 1.46
3. 1.64
4. 2.20
5. A \begin{align*}z\end{align*} -score cannot be calculated because the distribution is not normal.
1. How can we use normal distributions to make meaningful conclusions about samples and experiments?
2. How do we calculate probabilities and areas under the normal curve that are not covered by the Empirical Rule?
3. What are the other types of distributions that can occur in different probability situations?
4. The heights of \begin{align*}5^{\text{th}}\end{align*} grade boys in the USA is approximately normally distributed, with a mean height of 143.5 cm and a standard deviation of about 7.1 cm. What is the probability that a randomly chosen \begin{align*}5^{\text{th}}\end{align*} grade boy would be taller than 157.7 cm?
5. A statistics class bought some sprinkle (or jimmies) doughnuts for a treat and noticed that the number of sprinkles seemed to vary from doughnut to doughnut, so they counted the sprinkles on each doughnut. Here are the results: 241, 282, 258, 223, 133, 335, 322, 323, 354, 194, 332, 274, 233, 147, 213, 262, 227, and 366.
1. Create a histogram, dot plot, or box plot for this data. Comment on the shape, center and spread of the distribution.
2. Find the mean and standard deviation of the distribution of sprinkles. Complete the following chart by standardizing all the values:
\begin{align*}\mu= \ \ \sigma= \end{align*}
Number of Sprinkles Quantile \begin{align*}z\end{align*} -score
241
282
258
223
133
335
322
323
354
194
332
274
233
147
213
262
227
366
Figure: A table to be filled in for the sprinkles question.
c. Create a normal probability plot from your results.
d. Based on this plot, comment on the normality of the distribution of sprinkle counts on these doughnuts.
1. Draw each of the following distributions accurately on one set of axes.
Distribution Form Mean Standard Deviation
A Normal 30 5
B Normal 35 3
C Normal 24 12
1. In a school, the children’s heights follow a normal distribution with an average of 55 inches and a variance of 9 square inches.
1. What is \begin{align*}\mu\end{align*} ?
2. What is \begin{align*}\sigma\end{align*} ?
3. Is the curve \begin{align*}N(55,3)\end{align*} or \begin{align*}N(55,9)\end{align*} ?
2. The \begin{align*}N(36,9)\end{align*} distribution has a mean=? and SD=?
3. The \begin{align*}N(9,36)\end{align*} distribution has a mean=? and SD=?
4. For each of the following, calculate the standardized score (or z-score) for the value x:
1. \begin{align*}\mu=0, \sigma=1, x=2\end{align*}
2. \begin{align*}\mu=9, \sigma=5, x=3\end{align*}
3. \begin{align*}\mu=9, \sigma=4, x=0\end{align*}
4. \begin{align*}\mu=-9, \sigma=14, x=-20\end{align*}
5. Draw the curve corresponding to each of the following random variables and then shade the area corresponding to the given probability. You do NOT have to compute the probability.
1. X is a normal random variable with mean = 80 and standard deviation = 5. \begin{align*}P(70 < X < 90)\end{align*}
2. X is a normal random variable with mean of 20 and standard deviation of 10. \begin{align*}P(-10 < X < 15)\end{align*}
6. State the empirical rule.
7. Use the empirical rule to determine what percentage of a normally distributed population is more than 3 standard deviations below the mean.
8. Suppose that adult women’s heights are normally distributed with a mean of 65 inches and a standard deviation of 2 inches.
1. Use the empirical rule to determine what percent of adult women have heights between 65 inches and 67 inches.
2. Use the empirical rule to determine the proportion of adult women who have heights greater than 69 inches.
3. Using the empirical rule, what is the probability that a randomly selected adult woman is more than 63 inches tall?
4. What is the area under the curve between 59 inches and 67 inches?
9. Given a group of data with mean 70 and standard deviation 12, at least what percent of the data will fall between 70 and 94?
10. Given a set of data that is bell shaped with a mean of -690. It has a standard deviation of 25. What percentage of the data should lie between -752 and -648?
11. Given a set of data that is bell-shaped with a mean of 890. If 68% of the data lies between 850 and 930 then what is the standard deviation?
12. If a group of data is bell shaped with a mean of -25 and a standard deviation of 65.3 what is the interval that should contain at least 95% of the data?
13. Consider the following data set. Do you think it is a sample from a normally distributed population? Explain \begin{align*} &24.0 &&7.9 &&1.5 &&0.0 &&0.3 &&0.4 &&8.1 &&4.3 &&0.0 &&0.5\\ &3.6 &&2.9 &&0.4 &&2.6 &&0.1 &&3.6 &&2.9 &&0.4 &&2.6 &&0.1\\ &16.6 &&1.4 &&23.8 &&25.1 &&1.6 &&12.2 &&14.8 &&0.4 &&3.7 &&4.2 \end{align*}
14. Consider the following data set. Do you think it is a sample from a normally distributed population? Explain. \begin{align*} &26 &&24 &&22 &&25 &&23 &&24 &&25 &&23 &&25 &&22\\ &21 &&26 &&22 &&23 &&24 &&25 &&24 &&25 &&24 &&25\\ &26 &&21 &&22 &&24 &&24 \end{align*}
Keywords
Empirical Rule
Normal distribution
Standard normal curve
Standard normal distribution
Standardize
\begin{align*}z\end{align*} -score
Vocabulary Language: English
continuous variables
continuous variables
A continuous variable is a variable that takes on any value within the limits of the variable.
discrete values
discrete values
discrete values are data where a finite number of values exist between any two values.
distribution
distribution
A distribution is a description of the possible values of a random variable and the possible occurrences of these values.
empirical rule
empirical rule
The empirical rule states that for data that is normally distributed, approximately 68% of the data will fall within one standard deviation of the mean, approximately 95% of the data will fall within two standard deviations of the mean, and approximately 99.7% of the data will fall within three standard deviations of the mean.
normal distribution curve
normal distribution curve
A normal distribution curve is a symmetrical curve that shows the highest frequency in the center with an identical curve on either side of the center.
normal probability plot
normal probability plot
A normal probability plot is a graph is a plot of the z -scores of the data as quantiles against the actual data values. If a distribution is normal, this plot will be linear.
normal quartile plot
normal quartile plot
normal quartile plot is another name for a normal probability plot.
standard normal distribution
standard normal distribution
The standard normal distribution, $\phi(x)=\frac{1}{\sqrt{2 \pi}}e^{- \frac{1}{2}x^2}$ is a normal distribution with mean of 0 and a standard deviation of 1.
z-score
z-score
The z -score of a value is the number of standard deviations between the value and the mean of the set.
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exponent 54 results
exponent - The power to raise a number
(2n)^3 without exponents
(2n)^3 without exponents This can be written as follows: 2^3 * n^3 [B]8n^3[/B]
(n^2)^3 without exponents
(n^2)^3 without exponents This expression evaluates to: n^(2 *3) n^6 To write this without exponents, we expand n times itself 6 times: [B]n * n * n * n * n * n [MEDIA=youtube]zVAlzX9oHOQ[/MEDIA][/B]
11 to the power of 6 multiply 11 to the power of 3
11 to the power of 6 multiply 11 to the power of 3 Take this in parts. [U]Step 1: 11 to the power of 6 means we raise 11 to the 6th power using exponents:[/U] 11^6 [U]Step 2: 11 to the power of 3 means we raise 11 to the 3rd power using exponents:[/U] 11^3 [U]Step 3: Multiply each term together:[/U] 11^6 * 11^3 [U]Step 4: Simplify[/U] Because we have 2 numbers that are the same, in this case, 11, we can add the exponents together when multiplying: 11^(6 + 3) [B]11^9 [MEDIA=youtube]gCxVq7LqyHk[/MEDIA][/B]
2^n = 4^(n - 3)
2^n = 4^(n - 3) 2^n = (2^2)^(n - 3) (2^2)^(n - 3) = 2^2(n - 3) 2^n= 2^2(n - 3) Comparing exponents, we see that: n = 2(n - 3) n = 2n - 6 Subtract n from each side: n - n = 2n - n - 6 0 = n - 6 n = [B]6[/B]
3 to the power of 2 times 3 to the power of x equals 3 to the power of 7
3 to the power of 2 times 3 to the power of x equals 3 to the power of 7. Write this out: 3^2 * 3^x = 3^7 When we multiply matching coefficients, we add exponents, so we have: 3^(2 + x) = 3^7 Therefore, 2 + x = 7. To solve this equation for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=2%2Bx%3D7&pl=Solve']type it into our search engine[/URL] and we get: x = [B]5[/B]
3^14/27^4 = ?
3^14/27^4 = ? Understand that 27 = 3^3. Rewriting this, we have: 3^14/(3^3)^4 Exponent identity states (a^b)^c = a^bc, so we have: 3^14/3^12 Simplifying, we have: 3^(14 - 12) = 3^2 = [B]9[/B] [MEDIA=youtube]u_238Z7Mqk[/MEDIA]
8 to the power of x over 2 to the power of y
8 to the power of x over 2 to the power of y Step 1: 8 to the power of x means we take 8 and raise it to an exponent of x: 8^x Step 2: 2 to the power of y means we take 2 and raise it to an exponent of y: 2^y Step 3: The word [I]over[/I] means a quotient, also known as divided by, so we have: [B]8^x/2^y [MEDIA=youtube]SPQKOt5EoqA[/MEDIA][/B]
A company has 3,100 employees and is expected to grow at a rate of 0.04 for the next six years. How
A company has 3,100 employees and is expected to grow at a rate of 0.04 for the next six years. How many employees will they have in 6 years? Round to the nearest whole number. We build the following exponential equation: Final Balance = Initial Balance * (1 + growth rate)^time Final Balance = 3100(1.04)^6 Final Balance = 3100 * 1.2653190185 Final Balance = 3922.48895734 The problem asks us to round to the nearest whole number. Since 0.488 is less than 0.5, we round [U]down.[/U] Final Balance = [B]3,922[/B]
A super deadly strain of bacteria is causing the zombie population to double every day. Currently, t
A super deadly strain of bacteria is causing the zombie population to double every day. Currently, there are 25 zombies. After how many days will there be over 25,000 zombies? We set up our exponential function where n is the number of days after today: Z(n) = 25 * 2^n We want to know n where Z(n) = 25,000. 25 * 2^n = 25,000 Divide each side of the equation by 25, to isolate 2^n: 25 * 2^n / 25 = 25,000 / 25 The 25's cancel on the left side, so we have: 2^n = 1,000 Take the natural log of each side to isolate n: Ln(2^n) = Ln(1000) There exists a logarithmic identity which states: Ln(a^n) = n * Ln(a). In this case, a = 2, so we have: n * Ln(2) = Ln(1,000) 0.69315n = 6.9077 [URL='https://www.mathcelebrity.com/1unk.php?num=0.69315n%3D6.9077&pl=Solve']Type this equation into our search engine[/URL], we get: [B]n = 9.9657 days ~ 10 days[/B]
A virus is spreading exponentially. The initial amount of people infected is 40 and is increasing at
A virus is spreading exponentially. The initial amount of people infected is 40 and is increasing at a rate of 5% per day. How many people will be infected with the virus after 12 days? We have an exponential growth equation below V(d) where d is the amount of days, g is the growth percentage, and V(0) is the initial infected people: V(d) = V(0) * (1 + g/100)^d Plugging in our numbers, we get: V(12) = 40 * (1 + 5/100)^12 V(12) = 40 * 1.05^12 V(12) = 40 * 1.79585632602 V(12) = 71.8342530409 or [B]71[/B]
a ^5 x a ^2 without exponents
a ^5 x a ^2 without exponents When we multiply the same variable or number, we add exponents, so we have: a^(5 + 2) a^7 To write a variable raised to an exponent without exponents, we break it up. The formula to do this is: a^n = a times itself n times a^7 = [B]a * a * a * a * a * a * a[/B]
add 7 and 2, raise the result to the 6th power, then add what you have to s
add 7 and 2, raise the result to the 6th power, then add what you have to s Add 7 and 2: 7 + 2 Simplify this, we get:9 Raise the result to the 6th power: 9^6 [URL='https://www.mathcelebrity.com/powersq.php?sqconst=+6&num=9%5E6&pl=Calculate']Simplifying this using our exponent calculator[/URL], we get: 531,441 Now, we add what we have (our result) to s to get our final algebraic expression: [B]s + 531,441[/B]
add d to 5, raise the result to the 9th power, then subtract what you have from 2
add d to 5, raise the result to the 9th power, then subtract what you have from 2 Add d to 5: d + 5 Raise the result to the 9th power means we raise (d + 5) to the 9th power using an exponent: (d + 5)^9 the subtract what we have (the result) from 2: [B]2 - (d + 5)^9[/B]
Approximate Square Root Using Exponential Identity
Free Approximate Square Root Using Exponential Identity Calculator - Calculates the square root of a positive integer using the Exponential Identity Method
Covariance and Correlation coefficient (r) and Least Squares Method and Exponential Fit
Free Covariance and Correlation coefficient (r) and Least Squares Method and Exponential Fit Calculator - Given two distributions X and Y, this calculates the following:
* Covariance of X and Y denoted Cov(X,Y)
* The correlation coefficient r.
* Using the least squares method, this shows the least squares regression line (Linear Fit) and Confidence Intervals of α and Β (90% - 99%)
Exponential Fit
* Coefficient of Determination r squared r2
* Spearmans rank correlation coefficient
* Wilcoxon Signed Rank test
Customers arrive at the claims counter at the rate of 20 per hour (Poisson distributed). What is th
Customers arrive at the claims counter at the rate of 20 per hour (Poisson distributed). What is the probability that the arrival time between consecutive customers is less than five minutes? Use the [I]exponential distribution[/I] 20 per 60 minutes is 1 every 3 minutes 1/λ = 3 so λ = 0.333333333 Using the [URL='http://www.mathcelebrity.com/expodist.php?x=+5&l=0.333333333&pl=CDF']exponential distribution calculator[/URL], we get F(5,0.333333333) = [B]0.811124396848[/B]
divide 8 by t, raise the result to the 7th power
divide 8 by t, raise the result to the 7th power. We take this algebraic expression in two parts: 1. Divide 8 by t 8/t 2. Raise the result to the 7th power. (This means we use an exponent of 7) [B](8/t)^7[/B]
double v, raise the result to the 6th power, then multiply what you have by w
double v, raise the result to the 6th power, then multiply what you have by w Double v means multiply v by 2: 2v Raise the result to the 6th power, means we use an exponent of 6 on 2v: (2v)^6 Then multiply what you have by w, means take the result above, and multiply by w: [B]w(2v)^6[/B]
Equation and Inequalities
Free Equation and Inequalities Calculator - Solves an equation or inequality with 1 unknown variable and no exponents as well as certain absolute value equations and inequalities such as |x|=c and |ax| = c where a and c are constants. Solves square root, cube root, and other root equations in the form ax^2=c, ax^2 + b = c. Also solves radical equations in the form asqrt(bx) = c. Also solves open sentences and it will solve one step problems and two step equations. 2 step equations and one step equations and multi step equations
Exponential Distribution
Free Exponential Distribution Calculator - Calculates the Probability Density Function (PDF) and Cumulative Density Function (CDF) of the exponential distribution as well as the mean, variance, standard deviation, and entropy.
Exponential Growth
Free Exponential Growth Calculator - This solves for any 1 of the 4 items in the exponential growth equation or exponential decay equation, Initial Value (P), Ending Value (A), Rate (r), and Time (t).
Exponential Smoothing
Free Exponential Smoothing Calculator - Performs exponential smoothing on a set of data.
Find the last digit of 4^2081 no calculator
Find the last digit of 4^2081 no calculator 4^1= 4 4^2 = 16 4^3 = 64 4^4 = 256 4^5 = 1024 4^6 = 4096 Notice this pattern alternates between odd exponent powers with the result ending in 4 and even exponent powers with the result ending in 6. Since 2081 is odd, the answer is [B]4. [MEDIA=youtube]ueBWAW4XW4Q[/MEDIA][/B]
Find the last digit of 7^2013
Consider the first 8 calculations of 7 to an exponent: [LIST] [*]7^1 = 7 [*]7^2 = 49 [*]7^3 = 343 [*]7^4 = 2,401 [*]7^5 = 16,807 [*]7^6 = 117,649 [*]7^7 = 823,543 [*]7^8 = 5,764,801 [/LIST] Take a look at the last digit of the first 8 calculations: 7, 9, 3, 1, 7, 9, 3, 1 The 7, 9, 3, 1 repeats through infinity. So every factor of 4, the cycle of 7, 9, 3, 1 restarts. Counting backwards from 2013, we know that 2012 is the largest number divisible by 4: 7^2013 = 7^2012 * 7^1 The cycle starts over after 2012. Which means the last digit of 7^2013 = [B]7 [MEDIA=youtube]Z157jj8R7Yc[/MEDIA][/B]
Function
Free Function Calculator - Takes various functions (exponential, logarithmic, signum (sign), polynomial, linear with constant of proportionality, constant, absolute value), and classifies them, builds ordered pairs, and finds the y-intercept and x-intercept and domain and range if they exist. Table of Functions Calculator
How many times bigger is 3^9 than 3^3
How many times bigger is 3^9 than 3^3 Using exponent rules, we see that: 3^9 = 3^3 * 3^6 So our answer is [B]3^6 times bigger[/B]
If 200 bacteria triple every 1/2 hour, how much bacteria in 3 hours
If 200 bacteria triple every 1/2 hour, how much bacteria in 3 hours Set up the exponential function B(t) where t is the number of tripling times: B(d) = 200 * (3^t) 3 hours = 6 (1/2 hour) periods, so we have 6 tripling times. We want to know B(6): B(6) = 200 * (3^6) B(6) = 200 * 729 B(6) = [B]145,800[/B]
If 3x - y = 12, what is the value of 8^x/2^y
If 3x - y = 12, what is the value of 8^x/2^y We know 8 = 2^3 So using a rule of exponents, we have: (2^3)^x/2^y 2^(3x)/2^y Using another rule of exponents, we rewrite this fraction as: 2^(3x -y) We're given 3x - y = 12, so we have: [B]2^12[/B]
If there are 10^30 grains of sand on Beach A, how many grains of sand are there on a beach the has 1
If there are 10^30 grains of sand on Beach A, how many grains of sand are there on a beach the has 10 times the sand as Beach A? (Express your answer using exponents.) 10^30 * 10 = 10^(30 + 1) = [B]10^31[/B]
index form of (5^3)^6
Index form of (5^3)^6 Index form is written as a number raised to a power. Let's simplify by multiply the exponents. Since 6*3 = 18, We have: [B]5^18[/B]
Logarithms
Free Logarithms Calculator - Using the formula Log ab = e, this calculates the 3 pieces of a logarithm equation:
1) Base (b)
2) Exponent
3) Log Result
* Expand logarithmic expressions
Logarithms and Natural Logarithms and Eulers Constant (e)
Free Logarithms and Natural Logarithms and Eulers Constant (e) Calculator - This calculator does the following:
* Takes the Natural Log base e of a number x Ln(x) → logex
* Raises e to a power of y, ey
* Performs the change of base rule on logb(x)
* Solves equations in the form bcx = d where b, c, and d are constants and x is any variable a-z
* Solves equations in the form cedx=b where b, c, and d are constants, e is Eulers Constant = 2.71828182846, and x is any variable a-z
* Exponential form to logarithmic form for expressions such as 53 = 125 to logarithmic form
* Logarithmic form to exponential form for expressions such as Log5125 = 3
Modular Exponentiation and Successive Squaring
Free Modular Exponentiation and Successive Squaring Calculator - Solves xn mod p using the following methods:
* Modular Exponentiation
* Successive Squaring
Oliver invests \$1,000 at a fixed rate of 7% compounded monthly, when will his account reach \$10,000?
Oliver invests \$1,000 at a fixed rate of 7% compounded monthly, when will his account reach \$10,000? 7% monthly is: 0.07/12 = .00583 So we have: 1000(1 + .00583)^m = 10000 divide each side by 1000; (1.00583)^m = 10 Take the natural log of both sides; LN (1.00583)^m = LN(10) Use the identity for natural logs and exponents: m * LN (1.00583) = 2.30258509299 0.00252458479m = 2.30258509299 m = 912.064867899 Round up to [B]913 months[/B]
On January 1st a town has 75,000 people and is growing exponentially by 3% every year. How many peop
On January 1st a town has 75,000 people and is growing exponentially by 3% every year. How many people will live there at the end of 10 years? [URL='https://www.mathcelebrity.com/population-growth-calculator.php?num=atownhasapopulationof75000andgrowsat3%everyyear.whatwillbethepopulationafter10years&pl=Calculate']Using our population growth calculator[/URL], we get: [B]100,794[/B]
Polynomial
Free Polynomial Calculator - This calculator will take an expression without division signs and combine like terms.
It will also analyze an polynomial that you enter to identify constant, variables, and exponents. It determines the degree as well.
Population Growth
Free Population Growth Calculator - Determines population growth based on an exponential growth model.
Raise 9 to the 3rd power, subtract d from the result, then divide what you have by c
Raise 9 to the 3rd power, subtract d from the result, then divide what you have by c. This is an algebraic expression, let's take in parts (or chunks). Raise 9 to the 3rd power. This means we take 9, and raise it to an exponent of 3 9^3 Subtract d from the result, means we subtract d from 9^3 9^3 - d Now we divide 9^3 - d by c [B](9^3 - d) / c[/B]
raise f to the 3rd power, then find the quotient of the result and g
raise f to the 3rd power, then find the quotient of the result and g Take this algebraic expression in two parts: [LIST=1] [*]Raise f to the 3rd power means we take f, and write it with an exponent of 3: f^3 [*]Find the quotient of the result and g. We take f^3, and divide it by g [/LIST] [B]f^3/g[/B]
Raise f to the 8th power, divide the result by 5, then multiply 10
Raise f to the 8th power, divide the result by 5, then multiply 10 f to the 8th power means we raise f to the power of 8 using an exponent: f^8 Divide f^8 by 5 (f^8)/5 Now multiply this by 10: 10(f^8)/5 We can simplify this algebraic expression by dividing 10/5 to get 2 on top: 2[B](f^8)[/B]
raise r to the 8th power then find the product of the result and 3
raise r to the 8th power then find the product of the result and 3 Raise r to the 8th power means we raise r with an exponent of 8: r^8 The product of the result and 3 means we muliply r^8 by 3 [B]3r^8[/B]
raise t to the 10th power, then find the quotient of the result and s
raise t to the 10th power, then find the quotient of the result and s Raise t to the 10th power means we use t as our variable and 10 as our exponent: t^10 The quotient means a fraction, where the numerator is t^10 and the denominator is s: [B]t^10/s[/B]
raise v to the 9th power, then dividethe result by u
V to the 9th power means we use an exponent: v^9 Divide that result by u [B]v^9/u[/B]
raise y to the 10th power, then find the quotient of the result and 2
y to the 10th power means we give y an exponent of 10 y^10 The quotient of y^10 and 2 is: y^10 ----- 2
Rational Exponents - Fractional Indices
Free Rational Exponents - Fractional Indices Calculator - This calculator evaluates and simplifies a rational exponent expression in the form ab/c where a is any integer or any variable [a-z] while b and c are integers. Also evaluates the product of rational exponents
rewrite without an exponent :4^-2
rewrite without an exponent :4^-2 Since the exponent is negative, we have: 4^-2 = 1 / 4^2 4^-2 = [B]1 / 16[/B]
Simplest Exponent Form
Free Simplest Exponent Form Calculator - This expresses repeating algebraic expressions such as 3*a*a*a*b*b into simplest exponent form.
Square Roots and Exponents
Free Square Roots and Exponents Calculator - Given a number (n), or a fraction (n/m), and/or an exponent (x), or product of up to 5 radicals, this determines the following:
* The square root of n denoted as √n
* The square root of the fraction n/m denoted as √n/m
* n raised to the xth power denoted as nx (Write without exponents)
* n raised to the xth power raised to the yth power denoted as (nx)y (Write without exponents)
* Product of up to 5 square roots: √abcde
* Write a numeric expression such as 8x8x8x8x8 in exponential form
The flu is starting to hit Lanberry. Currently, there are 894 people infected, and that number is gr
The flu is starting to hit Lanberry. Currently, there are 894 people infected, and that number is growing at a rate of 5% per day. Overall, how many people will have gotten the flu in 5 days? Our exponential equation for the Flu at day (d) is: F(d) = Initial Flu cases * (1 + growth rate)^d Plugging in d = 5, growth rate of 5% or 0.05, and initial flu cases of 894 we have: F(5) = 894 * (1 + 0.05)^5 F(5) = 894 * (1.05)^5 F(5) = 894 * 1.2762815625 F(5) = [B]1141[/B]
the university of california tuition in 1990 was \$951 and tuition has been increasing by a rate of 2
the university of california tuition in 1990 was \$951 and tuition has been increasing by a rate of 26% each year, what is the exponential formula Let y be the number of years since 1990. We have the formula T(y): [B]T(y) = 951 * 1.26^y[/B]
What is an Exponent
Free What is an Exponent Calculator - This lesson walks you through what an exponent is, the product rule for exponents, the quotient rule for exponents, the 0 power rule, the power of a power rule for exponents
What is the annual nominal rate compounded daily for a bond that has an annual yield of 5.4%? Round
What is the annual nominal rate compounded daily for a bond that has an annual yield of 5.4%? Round to three decimal places. Use a 365 day year. [U]Set up the accumulation equation:[/U] (1+i)^365 = 1.054 [U]Take the natural log of each side[/U] 365 * Ln(1 + i) = 1.054 Ln(1 + i) = 0.000144089 [U]Use each side as a exponent to eulers constant e[/U] (1 + i) = e^0.000144089 1 + i = 1.000144099 i = 0.000144099 or [B].0144099%[/B]
When finding the power of a power, you _____________________ the exponents
When finding the power of a power, you _____________________ the exponents [B]Multiply [/B] Example: (a^b)^c = a^bc
Zombies are doubling every 2 days. If two people are turned into zombies today, how long will it tak
Zombies are doubling every 2 days. If two people are turned into zombies today, how long will it take for the population of about 600,000 to turn into zombies? Let d = every 2 days. We set up the exponential equation 2 * 2^d = 600,000 Divide each side by 2: 2^d = 300000 To solve this equation for d, we [URL='https://www.mathcelebrity.com/natlog.php?num=2%5Ed%3D300000&pl=Calculate']type it in our math engine[/URL] and we get d = 18.19 (2 day periods) 18.19 * days per period = 36.38 total days Most problems like this ask you to round to full days, so we round up to [B]37 days[/B].
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# 4.2 Three-dimensional objects from the environment
Page 2 / 2
Front
4.2 from the left side:
Front
4.3 from the right corner:
Front
Activity 4:
To investigate and approximate volume of three-dimensional objects [LO 4.8]
• You may work in groups. Each group will need: various small boxes, e.g. a matchbox; a rectangular container for margarine; a shoebox, etc. (try to have five of them); sugar cubes (or 1 cm wooden cubes from the Foundation Phase).
1. Pack sugar cubes into the matchbox and fill it with the cubes. How many do you need?
2. Now do the same with the other boxes and complete the table below:
Object (box) Number of sugar cubes needed to fill the box Matchbox
3. Measure the cube of sugar and record your findings:
• Length of cube:
• Width of cube:
• Height of cube:
4. The matchbox can contain …………. cubes; we say its volume is about ……..… cubic centimetres.
5. When we measure what can go into the space in a container, we are measuring VOLUME and we need three measurements: length, width and height.
6. Instead of counting each little cube of sugar, what would be a quicker way of calculating the volume of a box? Discuss this with a friend and then write down your answer on the dotted line.
7. How many sugar cubes will you need to fill a box that is 20cm long, 15cm wide and 7cm high (a 2 litre ice-cream container)? Write down your calculations and then compare them with those of a friend.
## Assessment
Learning outcomes(LOs) LO 2 Patterns, Functions and AlgebraThe learner will be able to recognise, describe and represent patterns and relationships, as well as to solve problems using algebraic language and skills. Assessment standards(ASs) We know this when the learner: 2.1 investigates and extends numeric and geometric patterns looking for a relationship or rules, including patterns: represented in physical or diagrammatic form; 2.1.2 not limited to sequences involving constant difference or ratio. 2.2 describes observed relationships or rules in own words. LO 3 Space and Shape (Geometry ) The learner will be able to describe and represent characteristics and relationships between two-dimensional shapes and three-dimensional objects in a variety of orientations and positions. We know this when the learner: 3.2 describes, sorts and compares two-dimensional shapes and three-dimensional objects from the environment according to geometrical properties including: shapes of faces; number of sides; flat and curved surfaces, straight and curved sides. 3.3 investigates and compares (alone and/or as a member of a group or team) two-dimensional shapes and three dimensional objects studied in this grade according to the properties already studied, by: 3.3.1 making three-dimensional models using cut-out polygons (supplied); drawing shapes on grid paper; 3.4 recognises and describes lines of symmetry in two-dimensional shapes, including those in nature and its cultural art forms; 3.5 makes two-dimensional shapes, three-dimensional objects and patterns from geometric objects and shapes (e.g. tangrams) with a focus on tiling (tessellation) and line symmetry; 3.6 recognises and describes natural and cultural two-dimensional shapes, three-dimensional objects and patterns in terms of geometric properties; 3.7 describes changes in the view of an object held in different positions. LO 4 measurementThe learner will be able to use appropriate measuring units, instruments and formulae in a variety of contexts. We know this when the learner: 4.8 investigates and approximates (alone and /or as a member of a group or team): 4.8.2 area of polygons (using square grids and tiling) in order to develop an understanding of square units; volume/capacity of three-dimensional objects (by packing or filling them) in order to develop an understanding of cubic units.
## Memorandum
ACTIVITY 1: 3D Objects
1. Investigation – practical
2. Using a net – practical
3. Practical – Tetrahedron (tetra – Greek = 4)
1. Using investigations
Object Surfaces Flat or curved Corners Edges Rectangular prism 6 Flat 8 12 Cube 6 Flat 8 12 Tetrahedron 4 flat 5 7
ACTIVITY 2: Symmetry
1. PROJECT – own – practical
2. Shapes
2.1 and 2.2 and 2.3 Cutting and folding and ruling lines of symmetry,
e.g.
( Note: in a rectangle diagonals cannot be used for just folding.)
ACTIVITY 3: objects seen from different angles
1.1 to 1.4 Practical – studying a building from various angles
2. and 3. Practical – working with cubes
4.1 to 4.3 Drawing – difficult!
ACTIVITY 4: volume
1. own
2. own investigation
3. 1 cm; 1 cm; 1 cm
4. own
5. -
6. Discussion (length x breadth x height)
7. 2 100 sugar cubes
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Read about ancient clocks like_ hour glass, water clock and sun dial for a quiz and hand on Activity in the class
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# Angle - geometric shape
An angle is a geometric shape formed by two rays. Their initial point is the same. This point is called the angle vertex, the rays are called the angle sides. The angle sides divide the plane into 2 regions, they are called flat angles or simply angles. The smaller angle is called the inner angle, and the larger one is called the outer angle.
Angles can also be designated as three points. Let’s say, ABC. In this record, B is a vertex, and A and C are points lying on different rays of the angle. To simplify and record fast, angles are usually defined by lowercase Greek letters: α - alpha, β - beta, γ - gamma, θ - theta, φ - fi, etc. The angle is denoted by a two-line symbol meaning the angle.
On the picture, you see two rays AB and AC with a vertex at point A forming two angles: α - internal angle, β - external angle.
## Angular measure
The angular measure allows comparing the angles between each other. It means that knowing the angle measure we can say that this angle is either greater than the other one, or smaller, or they are equal. There are several measures of angles:
• in degrees, minutes, seconds;
• in turns;
In maths, the first type of the angle measure is more widespread - degrees, minutes, seconds. Let’s learn it in details. Look at the clock shown below.
If you look at the clock, you see the hands of the clock as rays where the starting point is the same as the center of the dial. The complete rotation of the arrow makes 360 degrees. The degree is defined by the symbol °. If the needle makes a half of the turn, it moves 180 degrees or 180°. If it makes a quarter, it moves 90°. In the picture below, you can see what time corresponds to each angle when the time varies. It means that 15:00 corresponds to an angle of 90°, 18:00 corresponds to an angle of 180°, 21:00 to 270° and 24:00 to 360 °. The sum of the outer and inner angles should always be 360°.
You will study the angular measure in details in other maths sections: geometry and trigonometry.
## Types of Angles
Depending on the angle measure the types of angles are as follows:
### Zero angle
A zero angle is an angle where two sides coincide. Two equally directed rays emerge from the vertex. The zero angle is 0°.
### Acute angle
An acute angle is lying from 0 ° to 90 ° where 0 and 90 do not enter into this frame.
An acute angle is easy to remember. All sharp objects have an acute angle like a beak of a bird, an awl, a kitchen knife. You can see a yellow border on the picture showing the maximum measure of the right angle.
### Right angle
A right angle is the angle with perpendicular sides equal to 90°.
A right angle is a small square at the bottom of the angle as shown below.
### Obtuse angle
An obtuse angle is lying between 90° and 180° where 90° and 180° are not included.
### Oblique angle
An oblique angle means an angle different from 0°, 90°, 180° or 270°.
### Straight angle
A straight angle equals to 180°, its rays go to opposite directions.
### Convex angle
A convex angle is an angle between 0 ° to 180 ° the boarding values included.
### Non-convex, or concave angle
A non-convex angle or a concave angle is an angle lying within 180 ° and 360 ° where boundary values are not included.
### Full angle
A full angle is the one with two coinciding sides. It is the opposite of a zero angle. The full angle equals to 360°.
A zero and a full angles have the same sides. A zero angle is an internal angle of 0°, and a full angle is an outer angle equal to 360°.
Look at the picture and count the number of angles of each type?
• Zero angle - 2;
• Acute angle - 3;
• Right angle - 2;
• Obtuse angle - 2;
• Oblique angle - 6;
• Straight angle - 1;
• Convex angle - 10;
• Concave angle - 1;
• Full angle - 1;
## Math Pixel Puzzle
Math pixel puzzle is a very unusual game. The rules are very simple. You will see an image with pixels of various colors. After three seconds, the pixels fly apart in 3D space. Your task is to rotate the space so that ...
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# Precalculus problem solver with steps
Keep reading to understand more about Precalculus problem solver with steps and how to use it. Math can be a challenging subject for many students.
## The Best Precalculus problem solver with steps
This Precalculus problem solver with steps supplies step-by-step instructions for solving all math troubles. Any mathematician worth their salt knows how to solve logarithmic functions. For the rest of us, it may not be so obvious. Let's take a step-by-step approach to solving these equations. Logarithmic functions are ones where the variable (usually x) is the exponent of some other number, called the base. The most common bases you'll see are 10 and e (which is approximately 2.71828). To solve a logarithmic function, you want to set the equation equal to y and solve for x. For example, consider the equation log _10 (x)=2. This can be rewritten as 10^2=x, which should look familiar - we're just raising 10 to the second power and setting it equal to x. So in this case, x=100. Easy enough, right? What if we have a more complex equation, like log_e (x)=3? We can use properties of logs to simplify this equation. First, we can rewrite it as ln(x)=3. This is just another way of writing a logarithmic equation with base e - ln(x) is read as "the natural log of x." Now we can use a property of logs that says ln(ab)=ln(a)+ln(b). So in our equation, we have ln(x^3)=ln(x)+ln(x)+ln(x). If we take the natural logs of both sides of our equation, we get 3ln(x)=ln(x^3). And finally, we can use another property of logs that says ln(a^b)=bln(a), so 3ln(x)=3ln(x), and therefore x=1. So there you have it! Two equations solved using some basic properties of logs. With a little practice, you'll be solving these equations like a pro.
distance = sqrt((x2-x1)^2 + (y2-y1)^2) When using the distance formula, you are trying to find the length of a line segment between two points. The first step is to identify the coordinates of the two points. Next, plug those coordinates into the distance formula and simplify. The last step is to take the square root of the simplify equation to find the distance. Let's try an example. Find the distance between the points (3,4) and (-1,2). First, we identify the coordinates of our two points. They are (3,4) and (-1,2). Next, we plug those coordinates into our distance formula: distance = sqrt((x2-x1)^2 + (y2-y1)^2)= sqrt((-1-3)^2 + (2-4)^2)= sqrt(16+4)= sqrt(20)= 4.47 Therefore, the distance between the points (3,4) and (-1,2) is 4.47 units.
Differential equations are a type of mathematical equation that can be used to model many real-world situations. In general, they involve the derivative of a function with respect to one or more variables. While differential equations may seem daunting at first, there are a few key techniques that can be used to solve them. One common method is known as separation of variables. This involves breaking up the equation into two parts, one involving only the derivative and the other involving only the variable itself. Once this is done, the two parts can be solved independently and then recombined to find the solution to the original equation. Another popular method is known as integration by substitution. This approach involves substituting a new variable for the original one in such a way that the resulting equation is easier to solve. These are just a few of the many methods that can be used to solve differential equations. With practice, anyone can become proficient in this important mathematical discipline.
A rational function is a function that can be written in the form of a ratio of two polynomial functions. In other words, it is a fraction whose numerator and denominator are both polynomials. Solving a rational function means finding the points at which the function equals zero. This can be done by setting the numerator and denominator equal to zero and solving for x. However, this will only give you the x-intercepts of the function. To find the y-intercepts, you will need to plug in 0 for x and solve for y. The points at which the numerator and denominator are both equal to zero are called the zeros of the function. These points are important because they can help you to graph the function. To find the zeros of a rational function, set the numerator and denominator equal to zero and solve for x. This will give you the x-intercepts of the function. To find the y-intercepts, plug in 0 for x and solve for y. The points at which the numerator and denominator are both zero are called the zeros of the function. These points can help you to graph the function.
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Tangent solver Math book app App that does math for you Problem solver program Math app that you take a picture of the problem
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Lesson Explainer: Indefinite Integrals: Reciprocal Trigonometric Functions | Nagwa Lesson Explainer: Indefinite Integrals: Reciprocal Trigonometric Functions | Nagwa
Lesson Explainer: Indefinite Integrals: Reciprocal Trigonometric Functions Mathematics
In this explainer, we will learn how to find indefinite integrals of functions that result in reciprocal trigonometric functions.
We can identify a variety of indefinite integrals involving reciprocal trigonometric functions through the associated derivatives. We first recall the derivative: This leads to the following indefinite integral:
Standard Result: Indefinite Integral of the Product of Secant and Tangent Functions
In our first example, we will use this formula to solve an indefinite integral.
Example 1: Integrating Trigonometric Functions Involving Reciprocal Trigonometric Functions
Determine .
We note that the given integrand contains the produce of secant and tangent. So, we recall the following indefinite integral:
To use this formula for our example, we need to modify the argument of the trigonometric functions. We can do this by using the -substitution
Substituting this change of variables to our integral, we have
We can now apply our formula we recalled earlier to write this indefinite integral as
Since is an arbitrary constant, we can simply denote the constant as in our answer. Substituting back into the resulting expression, we obtain
In our next example, we will solve an indefinite integral which requires us to simplify the integrand first.
Example 2: Finding the Integration of a Function Involving Trigonometric Functions
Determine .
When we have a factored expression within the integrand, as is the case here, we should start by expanding through the parenthesis in the integrand. We obtain
Since , we can simplify the first term using the fact that
Substituting this into our integrand, we obtain
Now, we need to evaluate the following indefinite integral:
The first term in the integrand involves a cosine, and the second term involves the product of a secant and a tangent. We recall the following formulas:
Before we can apply these formulas to solve the given indefinite integral, we need to modify the argument of the trigonometric functions. We use the -substitution
We can rearrange the second equation to state . Substituting this change of variable into our integral, we get
We apply the formulas above to obtain for some arbitrary constants and . Since, after distributing through the parenthesis, we will end up with a combination of and , we can replace this expression by another arbitrary constant to write the solution as
Substituting back into the expression, we obtain the indefinite integral
Let us consider another example involving the product of and .
Example 3: Integrating Trigonometric Functions Involving Reciprocal Trigonometric Functions
Determine .
Since there is a factored expression within the integrand, we should start by expanding through the parenthesis in the integrand. We obtain
The first term is the product of the secant and tangent function, and the second term is the square of the secant function. In order to solve our problem, we recall the following indefinite integrals:
Applying these formulae to our integral, we obtain for some arbitrary constants and . Since, after distributing through the parenthesis, we will end up with a combination of and , we can replace this expression by another arbitrary constant to write the solution as
In the previous examples, we used the formula for the indefinite integral that have results in the form of the secant and tangent functions. We now consider integrals that involve the complementary trigonometric functions and . Recall that
Hence, integrating both sides of the equation, we obtain the following indefinite integral.
Standard Result: Indefinite Integral of the Product of Cosecant and Cotangent Functions
To help with remembering this formula, we note the resemblance between this formula and the first formula we stated, which is
From this integral, we can replace each trigonometric function with their complements and place a negative sign on the function on the right-hand side of the equation to retrieve the new formula. This resemblance is not coincidental. Let us examine why this is expected in general.
We recall that the complementary trigonometric functions cosine, cotangent, and cosecant take the complementary angles of their counterparts. In other words,
Therefore, we can write
We can use the -substitution method by defining
We can write the last equation as . Substituting this change of variable into our indefinite integral, we get
Applying the formula for the antiderivative of , we obtain
Replacing back into the resulting expression, we obtain
We note that, when integrating complementary trigonometric functions, we expect a negative sign resulting from the substituting , which leads to . Other than the negative sign, we can retrieve the formula for the integral of complementary trigonometric functions by their original counterparts.
Let us evaluate an indefinite integral using this formula.
Example 4: Finding the Integration of a Function Involving Trigonometric Functions
Determine .
The given integrand is the product of the cosecant and cotangent functions. Hence, we recall the following integral:
Before applying this formula to our integral, let us use the -substitution method by defining
We can also write the last equation as . Substituting this change of variables to our integral, we have
We can now apply our formula we recalled earlier to write this indefinite integral as
Since is an arbitrary constant, we can simply denote the constant as in our answer. Substituting back into the resulting expression, we obtain
For our final formula, we state the indefinite integral of . To obtain this integral, we first recall the integral of :
As discussed earlier, we can obtain the complementary counterpart of this integral by replacing and by their complements and and placing a negative sign on the right-hand side.
Standard Result: Indefinite Integral of the Square of the Cosecant Function
When integrating or , we can use the following trigonometric identity to express the function in terms of or respectively:
Both of these identities can be derived from the Pythagorean identity when both sides of the equation are divided by either or .
In our next example, we will evaluate an indefinite integral involving by, first, applying this trigonometric identity and, second, applying the formula for the antiderivative of .
Example 5: Integrating a Reciprocal Trigonometric Function Whose Argument Has the Form 𝑎𝑥 + 𝑏
Determine .
Since is a constant, we can begin by bringing it outside the integral:
We can use the -substitution method by defining
We can also write the last equation as . Substituting this change of variables to our integral, we obtain
We note that the integrand contains the square of cotangent function. The antiderivative of is not readily available, but we know that can be expressed in terms of and we know the integral of :
To obtain the relation between and , we divide both sides of the Pythagorean identity by :
This is the same as
We note that our integrand is precisely . Hence, applying this identity to our integral, we get
We can now apply the formula for the antiderivative of to evaluate this indefinite integral:
We can replace by since is an arbitrary constant. We can also replace by . Hence,
In our final example, we will find an indefinite integral resulting in a reciprocal trigonometric function and identify the integration constant satisfying given boundary condition.
Example 6: Finding the Antiderivative of a Function Involving Trigonometric Functions and Identifying the Constant of Integration
Identify the function satisfying and .
Since we are given the derivative of , we can write as an indefinite integral of the given function and simplify as follows
We can evaluate the first integral by using the power rule: for any . Applying this rule with ,
𝑥𝑥=13𝑥+.dC (1)
For the second integral, we begin by substituting , which means . Using the -substitution, we can write
Recalling that , we can write the expression above as . Substituting back into this expression, we obtain
12(4𝑥)(4𝑥)𝑥=3(4𝑥)+.sectandsecC (2)
Using equations (1) and (2), we obtain
𝑓(𝑥)=13𝑥−3(4𝑥)+.secC (3)
Let us identify the integration constant . We are given that . Substituting in the equation above gives us
Note that
Thus, . To satisfy the given condition , constant must satisfy
This leads to . We can substitute this value into equation (3) to obtain
Let us recap a few important concepts from this explainer.
Key Points
• Indefinite integral of the product of and is
• Indefinite integrals involving the complementary trigonometric functions can be obtained by replacing the each trigonometric function with its complementary counterpart and placing a negative sign on the result of the integral. The important indefinite integrals involving complementary reciprocal trigonometric functions are
• Integrals of or are not readily available. To compute an integral involving either of these functions, we can use the following trigonometric identity to relate it to or , respectively:
• If given sufficient boundary conditions of the antiderivative, we can identify the value of the integration constant and find the exact expression for the antiderivative.
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Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
Session 8, Part B:
Mathematical Probability
In This Part: Predicting Outcomes | Fair or Unfair? | Outcomes | Finding the Winner
Making a Probability Table
Recall the question from the last section. Each of the two players rolls a die, and the winner is determined by the sum of the faces:
• Player A wins when the sum is 2, 3, 4, 10, 11, or 12. • Player B wins when the sum is 5, 6, 7, 8, or 9.
If this game is played many times, which player do you think will win more often, and why?
To analyze this problem effectively, we need a clear enumeration of all possible outcomes. Let's examine one scheme that is based on a familiar idea: an addition table.
Red Die
Blue Die
+ 1 2 3 4 5 6 1 2 3 4 5 6
Each possible outcome for the sum of the two dice can be enumerated in this table. For example, if the outcome were (1,1), here is how you would record it:
Red Die
Blue Die
+ 1 2 3 4 5 6 1 1 + 1 2 3 4 5 6
This is how you would record the outcome (2,4):
Red Die
Blue Die
+ 1 2 3 4 5 6 1 2 2 + 4 3 4 5 6
This is how you would record the outcome (4,2):
Red Die
Blue Die
+ 1 2 3 4 5 6 1 2 3 4 4 + 2 5 6
Note that the outcome (4,2) is different from the outcome (2,4).
Problem B4
a. Complete this table of possible outcomes. (If you're doing it on paper, you do not have to use blue and red pencils, but be aware of the difference between such outcomes as 2 + 4 and 4 + 2.) b. How many entries will the table have? How does this compare to your answer to question (a)?
When you click "Show Answers," the filled-in table will appear below the problem. Scroll down the page to see it.
Red Die
Blue Die
+ 1 2 3 4 5 6 1 2 3 4 5 6
Red Die
Blue Die
+ 1 2 3 4 5 6 1 1 + 1 1 + 2 1 + 3 1 + 4 1 + 5 1 + 6 2 2 + 1 2 + 2 2 + 3 2 + 4 2 + 5 2 + 6 3 3 + 1 3 + 2 3 + 3 3 + 4 3 + 5 3 + 6 4 4 + 1 4 + 2 4 + 3 4 + 4 4 + 5 4 + 6 5 5 + 1 5 + 2 5 + 3 5 + 4 5 + 5 5 + 6 6 6 + 1 6 + 2 6 + 3 6 + 4 6 + 5 6 + 6
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# Proving convergence of a recursively defined sequence
Using a computer it is easy to see that the sequences defined by letting $a_1=1$, $a_2=m$, and $$a_n=\frac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}$$ converges to $\frac{3m}{m+2}$. I would very much like to know how to prove this. I don't even know where to start.
-
Hint: The following is I think a natural starting "move."
Let $a_n=\dfrac{1}{x_n}$. When the smoke clears you get a nice linear recurrence.
Continuation: Making the suggested substitution, and simplifying, we obtain the recurrence $x_n=\dfrac{1}{2}\left(x_{n-1}+x_{n-2}\right)$. We have to be a little careful, since there will be trouble if $a_{n-1}+a_{n-2}=0$. But for example if $m\gt 0$ then all terms are safely positive.
There are many ways to solve the recurrence. For example we can use a general technique for solving homogeneous linear recurrences with constant coefficients, and write down the characteristic polynomial $2x^2-x-1=0$, which has the roots $1$ and $-\dfrac{1}{2}$.
Thus the general solution of the recurrence is $x_n=A+B\left(-\frac{1}{2}\right)^n$. Putting $x_1=1$ and $x_2=\dfrac{1}{m}$ we find that $A=\dfrac{m+2}{3m}$, and $B=\dfrac{4-2m}{3m}$. Thus $$x_n=\frac{m+2}{3m} + \left(-\frac{1}{2}\right)^n \frac{4-2m}{3m}.$$ As $n\to\infty$, $x_n\to \dfrac{m+2}{3m}$.
Remark: Because the numbers are quite simple, we do not need to use the "general" procedure to solve the recurrence. More generally, consider the recurrence $$y_{n}=py_{n-1}+(1-p)y_{n-2},$$ which comes up in some probability calculations. This may be rewritten as $$y_n-y_{n-1}=(-1)(1-p)(y_{n-1}-y_{n-2}).$$ If we let $z_n=y_{n}-y_{n-1}$, we get $z_n=(-1)(1-p)z_{n-1}$. So the sequence $(z_n)$ is a geometric sequence with common ratio $-(1-p)$. Now we can easily find a general expression for $z_n$, and hence for $y_n$.
-
I don't follow. What is $x_n$? – Carolus Nov 2 '12 at 15:02
Whatever the sequence $(a_n)$ is, let $(x_n)$ be the sequence of its reciprocals. The sequence $(x_n)$ is a lot nicer. Just plug in $1/x_k$ everywhere you see $a_k$, and simplify. – André Nicolas Nov 2 '12 at 15:04
Thank you for only giving me a hint at first. It was a good exercise! – Carolus Nov 3 '12 at 8:11
Let $A_n=\frac{a_n}{a_{n-1}}$, then we have $A_n=\frac{2}{1+A_{n-1}}$, it is easy to find that $\lim\limits_{n\to\infty }A_n$ exists then we have $\lim\limits_{n\to\infty }\frac{a_n}{a_{n-1}}=\frac{\sqrt{2}-1}{2}<1$, we can use the $\epsilon- N$ to illustrate the result......
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This shows that $A_n\to1$. And? – Did Nov 2 '12 at 16:16
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# How do you solve (2t+7)/(t-4)>=3 using a sign chart?
Feb 9, 2017
The solution is t in ]4, 19]
#### Explanation:
Rewrite the equation
$\frac{2 t + 7}{t - 4} \ge 3$
$\frac{2 t + 7}{t - 4} - 3 \ge 0$
$\frac{\left(2 t + 7\right) - 3 \left(t - 4\right)}{t - 4} \ge 0$
$\frac{2 t + 7 - 3 t + 12}{t - 4} \ge 0$
$\frac{- t + 19}{t - 4} \ge 0$
Let $f \left(t\right) = \frac{- t + 19}{t - 4}$
We can construct the sign chart
$\textcolor{w h i t e}{a a a a}$$t$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a a}$$4$$\textcolor{w h i t e}{a a a a a a a a}$$19$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a}$$t - 4$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$19 - t$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$-$
$\textcolor{w h i t e}{a a a a}$$f \left(t\right)$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$-$
Therefore,
$f \left(t\right) \ge 0$ when t in ]4, 19]
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# How Do You Write One And A Half In Math?
How do you write one and a half in maths?
If the numerator times 2 equals the denominator, then the fraction is worth 12. “I have another way,” Donald said. “If the numerator goes into the denominator two times, then the fraction is one-half.” I wrote: If the numerator goes into the denominator two times, then the fraction is worth 12.
What does 1.5 mean in math?
If a half = 0.5 and one =1, than one and a half is 1+0.5=1.5. So, 1.5 = one and a half.
Does 1.5 mean 1 and a half?
1.5 is the same thing as "one and five-tenths," or "one and five over ten, or one and half," which is written as 112 as a fraction.
## Related Question How do you write one and a half in math?
### What is half of 3.5 in?
Half of 3.5 is 1.75. The relationship between halving a number and division is as follows: Half of x = x ÷ 2.
### Is 1.5 a counting number?
The number 1.5 cannot be written as a whole number because it has a fractional part. It stands for 1 5/10, so it is a mixed number, which is one
### What is half of a 5th?
There are roughly 8.5 shots in a 375 ml bottle of alcohol. It's half the amount of shots in a fifth of alcohol, or a 750 ml bottle of liquor.
### What is half of 1 tsp?
Half of 1 tsp is equivalent to ½ tsp.
### What is 1.5 divided half?
Half of 1.5 cups is 0.75 cups, or 3/4 cups.
### What is the half of 5.5 cm?
2.25 cm is the half of 5.5 cm.
### What is the meaning of one half?
One half is the irreducible fraction resulting from dividing one by two (2) or the fraction resulting from dividing any number by its double. One half often appears in mathematical equations, recipes, measurements, etc. Half can also be said to be one part of something divided into two equal parts.
### What is 1/2 in a whole number?
In the same way, 1/2 is never equal to a whole number. It is equal to 0.5, which has a decimal but not a fraction. You could multiply by 0.5 the same way you multiply by 5, except that you might have at least one extra decimal place in the answer.
### How do you make 1/2 into a whole number?
You cannot change 1/2 into a whole number because it is a fraction. It is only part of a whole. The whole has been divided into 2 parts and 1/2
### Is it one half or one half?
So "1/2" should always be written out as one-half. (Unless it's in a sentence like "one half of a perfect pair," in which case it's not a fraction.) One half need not be hyphenated when used as a noun; however, it must be hyphenated when used as an adjective: 1.
### How do you write three and a half?
Three and a half years (no hyphens unless it's an adjective—A three-and-a-half-year ordeal). Don't use numerals (3½ years); if you do, you will need to treat all year counts as numerals throughout the publication (1 year, 5 years, etc.). Also, you can't start a sentence with numerals, according to your style guide.
### How do you write two and a half hours in numbers?
Two and a half hours. A. There is no need for hyphens if you're using the phrase as a noun: We'll be there in two and a half hours; two and a half hours is plenty of time. If you are using a phrase like that as a modifier, however, you'll need hyphens to hold it all together: a two-and-a-half-hour trip.
### What is 7 in a Half?
Half of 7 would be 3 1/2, or in decimal form, 3.5. Finding 'half' of a number is the same as dividing by 2.
### What's Half of 3/4 as a fraction?
Half of 3/4 is 3/8.
### What is 1.5 called?
1.5 becomes one and one half or one and a half. More common: one point five.
### What is 1.5 as a rational number?
For example, 1.5 is rational since it can be written as 3/2, 6/4, 9/6 or another fraction or two integers. Pi (π) is irrational since it cannot be written as a fraction.
### What is 1.5 percent as a whole number?
Common percentage to decimal numbers to fractions conversions
chart for: Percentage conversions
Percent equals Decimal № equals Fraction equals
1.25% 0.0125 12.51,000
1.5% 0.015 3200
1.9% 0.019 191,000
### What is half of \$50?
Half of 50 equals 25.
### What number is half of 8?
What is ½ of 8? Of course, it's 4. Everyone knows that.
### What is half of the number 52?
What is half of the number?” Since the number is multiplied by two, divide it by two to get the original number. 52/2 = 26.
### What is half of 2 tsp?
Scale, Half and Double Quantity Amounts in a Recipe (Chart)
Original Recipe Measure Half Scaled Measure Double Scaled Measure
2 tsp. 1 tsp. 4 tsp.
2 1/2 tsp. 1 1/4 tsp. 5 tsp.
1 tbsp. 1 1/2 tsp (1/2 tbsp.) 2 tbsp.
1 1/2 tbsp. 1 1/4 tsp. 3 tbsp.
### How do I make one teaspoon?
Wondering how to measure 1/8 teaspoon when you don't have a measuring spoon that size?
HOW TO MEASURE WITHOUT A MEASURING SPOON.
1/8 teaspoon 1 pinch using your thumb, index and middle finger
1 teaspoon a mound about 1/4-inch all around larger than a quarter in your cupped hand OR an “eating” teaspoon about half full
### How much is a tablespoon and a half?
A tablespoon is equal to 3 teaspoons. A half tablespoon is, therefore, equal to 1 1/2 teaspoons. If you've lost the teaspoon measuring spoon and only have the half or quarter teaspoons left, a half tablespoon equals three half teaspoons or six quarter teaspoons.
### What's Half of 1 and a Half cups?
Half of 1 1/2 cups is 3/4 cups.
### How do you solve 75 divided by 2?
Using a calculator, if you typed in 75 divided by 2, you'd get 37.5. You could also express 75/2 as a mixed fraction: 37 1/2.
### What is 3 in a Half?
Answer: Half of 3 is 3/2 as a fraction and 1.5 as a decimal.
### What is the Half of 6.5 cm?
Answer: Half of 6.5 is divide by 2. so, 6.5/2=3.25.
### What is the Half number of 55?
Half of 55 is the same as 55 divided by 2, which equals 27.5.
### How do you take Half of twelve and get seven?
If you cut XII (12) in half horizontally, you get VII (7).
### How do you write one and a half in numbers in Word?
• From the OFFICE BUTTON , click WORD OPTIONS. The Word Options dialog box appears.
• From the Categories list, select Proofing.
• Click AUTOCORRECT OPTIONS
• Select the AutoFormat As You Type tab.
• Select Fractions (1/2) with fraction character (½)
• Click OK.
• Click OK.
• ### What is a one half page?
1. half page - something that covers (the top or bottom) half of a page. page - one side of one leaf (of a book or magazine or newspaper or letter etc.) or the written or pictorial matter it contains.
### How do you write one and a half kg in numbers?
One and a half kilograms (1 ) can be written as 1 kg 500g or 1500 g.
### What is 0.5 as a number?
The number 0.5 is a decimal that is equivalent to 1/2.
### What is 2 and a half in decimal form?
So the answer is that 2 1/2 as a decimal is 2.5.
### Is 1 upon 2 is a whole number?
No , 1/2 is not a whole number. whole number:— whole number is (0,1,2,3,4……). 1/2 is a irrational number because ,it is a p/q and q is a not 0,so 1/2 is a irrational number.
### How do you turn a mixed number into a fraction?
Simply calculate the division represented by the fraction: 12 ÷ 4 = 3, and you're left with a whole number instead of the fraction 12/4. Because the mixed number 2 and 12/4 means 2 + 12/4, you can rewrite the mixed number as 2 + 3 (substituting 3 for the fraction 12/4) and simplify that to 5 as the final answer.
### What is the answer of ½ ½?
One half plus one half equals 1. Mathematically, one half is equal to 1/2. We can only add fractions that have the same denominator.
### Is one quarter hyphenated?
Fractions are almost always hyphenated when they are adjectives: “He is one-quarter Irish and three-quarters Nigerian.” But when the numerator is already hyphenated, the fraction itself is not, as in “ninety-nine and forty-four one hundredths.” Fractions treated as nouns are not hyphenated: “He ate one quarter of the
### How do you write 3 and a half in a fraction?
31/2 as improper fraction
Hence, the improper fraction is 7/2.
Posted in FAQ
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# How to convert a 4x4 matrix transformation to another coordinate system?
Is there a general method to convert a matrix transformation from one coordinate system to another, so that the resulting transformation looks the same on screen?
For example: There are some transformations in a coordinate system with X right, Y up, and Z toward the viewer. And they need to be converted to a coordinate system with X right, Y away from the viewer, and Z up.
What would be the operation that needs to be performed for each matrix so that the transformations look the same in the other coordinate system? And is there a general way to construct this operation given the source and destination basis vectors?
This is as easy as writing your old coordinates in terms of the new ones.
• We want +x to map to +x (1, 0, 0)
• We want +y to map to +z (0, 0, 1)
• We want +z to map to -y (0, -1, 0)
• We want the fourth, homogenous coordinate to survive unchanged (0, 0, 0, 1)
So you make those vectors the columns of a coordinate conversion matrix:
$$\Bbb C = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$
If you left-multiply this matrix by any homogeneous vector in your old coordinate system, it converts it to the corresponding vector in the new coordinate system:
$$\vec v_{\text{new}} = \Bbb C \times \vec v_{\text{old}}$$
And the same goes for any vector transformed by a transformation matrix $$\\Bbb M\$$ expressed in your old space:
$$\vec v_{\text{new}} = \Bbb C \times \vec v_{\text{transformed}} = \Bbb C \times (\Bbb M \times \vec v_{\text{untransformed}})\\ \vec v_{\text{new}} = (\Bbb C \Bbb M) \times \vec v_{\text{untransformed}}$$
So you can multiply any transformation matrix by this coordinate transformation matrix to get a single matrix that does both the original transformation and the coordinate conversion.
If you use the opposite multiplication convention - vector on the left, matrix on the right - then take the transpose of $$\\Bbb C\$$ (so your destination vectors are the rows instead of the columns) and multiply it on the right instead of the left.
You can use this same logic to work with untransformed vectors already in the new coordinate system: just convert them back to the old coordinate system (using the inverse of matrix $$\\Bbb C\$$ above, $$\\Bbb C^{-1}\$$), use the transformation matrix from the old system ($$\\Bbb M\$$), and then convert back:
\begin{align} \vec v_\text{transformed-new} &= \Bbb C \times \vec v_\text{transformed-old} \\ &= \Bbb C \times (\Bbb M \times \vec v_\text{untransformed-old}) \\ &= (\Bbb C \Bbb M) \times (\Bbb C^{-1} \times \vec v_\text{untransformed-new}) \\ &= (\Bbb C \Bbb M \Bbb C^{-1}) \times \vec v_\text{untransformed-new} \\ \end{align}
So, your matrix that does the same job as $$\\Bbb M\$$ but in the new coordinate system is just $$\\Bbb M_\text{new} = \Bbb C \Bbb M \Bbb C^{-1}\$$.
Since your coordinate transformation is pure rotation/reflection - no scaling/shearing - the inverse of $$\\Bbb C\$$ is just its transpose (making its rows into columns and vice versa):
$$\Bbb C^{-1} = \Bbb C^T = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$
• Thanks, after implementing and optimizing it, I've added the complete solution including the C# code as another answer. It does everything very efficiently with only 6 multiplications and one division. Commented Oct 27, 2022 at 10:53
• Thanks for your answer. Any pointers on how translations would be handled - i.e. where the "origins" of the two coordinate systems are different? Commented Jul 31 at 23:20
• Easy: express the new coordinate system's origin in terms of the old, then run that point through the matrix above. Negate the result and use that as your fourth column (keeping a 1 in the bottom-right entry). Now if you pass that new-origin-in-old-system point through the updated matrix, the result gets subtracted from itself, yielding the zero vector. Commented Aug 1 at 7:53
Final solution with code and optimization:
Here is the code that is doing the conversion in one go. It includes:
• Conversion to destination system as described in the answer by DMGregory
• Uniform scaling from meters to millimeters
• Transpose, because the matrix layout of the destination software is different
All of these steps combined in one efficient method look like this:
public static Transform MatrixToXForm(ref Matrix m, float s = 1000)
{
var rs = 1.0f / s;
var t = new Transform();
t.M00 = m.M11; t.M01 = -m.M31; t.M02 = m.M21; t.M03 = m.M41 * s;
t.M10 = -m.M13; t.M11 = m.M33; t.M12 = -m.M23; t.M13 = -m.M43 * s;
t.M20 = m.M12; t.M21 = -m.M32; t.M22 = m.M22; t.M23 = m.M42 * s;
t.M30 = m.M14 * rs; t.M31 = -m.M34 * rs; t.M32 = m.M24 * rs; t.M33 = m.M44;
return t;
}
The interesting point here is that the "matrix sandwich" $$\\Bbb M_\text{new} = \Bbb C \Bbb M \Bbb C^{-1}\$$ cannot be expressed as one matrix multiplication, but it can be written as code that shuffles the elements and signs.
If the two coordinate systems only differ in swapped axes and uniform scaling (here from the software vvvv to Rhino3d), you can derive the code for an efficient direct method like this:
• Construct the conversion matrix $$\\Bbb C \$$ using the basis vectors. In this case:
$$\Bbb C = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$
• Use an example transformation for $$\\Bbb M \$$. I've just used the row/col index of the elements:
$$\Bbb M = \begin{bmatrix} 11 & 12 & 13 & 14 \\ 21 & 22 & 23 & 24 \\ 31 & 32 & 33 & 34 \\ 41 & 42 & 43 & 44 \end{bmatrix}$$
Then do the $$\\Bbb M_\text{new} = \Bbb C \Bbb M \Bbb C^{-1}\$$ operation and see where the elements land and what sign they have:
By reading the index numbers, this can be directly converted into code. The full chain of operations including uniform scaling and transpose looks like this and results in the code above:
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# September 9, 2013
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### September 9, 2013
1. 1. 9 September Today: Reminders, Other Notes Formulas Warm-Ups: Test & Sample Khan Academy Questions Order of Operations Class Work: Handout; Various Problems Odd/Even
2. 2. Updates/Info. Khan Academy Topics for September 15, 2013—7:00 pm Writing Expressions 1 Writing Expressions 2 Writing & Interpreting Decimals Topics for all of September posted at the blog: (v6math.blogspot.com) As of yesterday, 24 people have already registered at Khan Academy. Thank you for doing that—116 to go.
3. 3. Formulas 1. Distance Formula: D = rt; where r = rate (speed) and t = time 3. Area of a Square: Since each side is the same length, P = 4S 2. Perimeter of a Square: If you drive for 2 ½ hours and average 50 miles per hour, how many miles have you driven? If you drive 120 miles in 2 ½ hours, what is your average speed? A 150 mile trip at 60 mph, would take how long? Area = LxW, but since each side is the same length, we are multiplying a number times itself. So the formula is: Area of a square = S2
4. 4. Class Notes How to easily order fractions using cross-multiplication 42 45 5/6 is > 7/9 Order from least to greatest: Last Example:
5. 5. Class Notes
6. 6. Step 1. Move the 6 by adding to right side. Step 2. Multiply both sides by 3 Write an expression to represent: Three less than three times a number x. Evaluate when y = 3 and x = 5: 2x2 + 4y Warm-Up
7. 7. From Khan Academy; Writing Expressions 2 Warm-Up
8. 8. From Khan Academy; Writing Expressions 2 Warm-Up
9. 9. Handout; Various Problems Odd/Even: Based on the date of your birth, (Your Birthday) Class Work
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# solve this equation for x
• May 22nd 2010, 09:40 AM
irichi09
solve this equation for x
How do you do this?
solve for x:
ax = bx - c
Thank you.
-r
• May 22nd 2010, 09:52 AM
Anonymous1
$ax = bx - c$
$\implies ax- bx = -c$
$\implies x(a- b) = -c$
$\implies \boxed{x=-\frac{c}{a-b}}$
• May 22nd 2010, 10:12 AM
freetibet
solve for x:
ax = bx - c
solution: ax - bx = c
(a -b)x = c
x = c/(a - b)
• May 22nd 2010, 11:07 AM
correct solution
This is the correct solution:
$
ax = bx - c
$
$
\implies ax- bx =-c
$
$
\implies x(a- b) =-c
$
$
\implies \boxed{x=-\frac{c}{a-b}}
$
• May 22nd 2010, 04:46 PM
irichi09
thank you for helping
it seems so easy when you see the solution, but not so easy when your making holes in the wall with your forehead.(Coffee)
• May 22nd 2010, 05:38 PM
Quote:
Originally Posted by irichi09
it seems so easy when you see the solution, but not so easy when your making holes in the wall with your forehead.(Coffee)
To protect the wall from your forehead, you need to know how to think it through.
Solve for x means we are being asked to write x=?
What we are given contains x in 2 places, on either side of the equal sign.
When both sides are equal we can perform the exact same operation to both sides.
For instance 5=5
If we subtract 3 from both sides we'll have 5-3=5-3 which is 2=2
The sides may not be what they were originally but they are still equal.
Hence
$ax=bx-c$
has x on both sides.
To get the x's on the same side, subtract bx from both sides,
the two sides will still be equal...
$ax-bx=bx-bx-c$
$bx-bx=0$
therefore we have
$ax-bx=-c$
If you have 5 boxes of apples, 20 apples per box
and you give 3 boxes to your sister, how many apples will you have ?
$5(20)-3(20)=(5-3)(20)$
You can calculate 100-60=40 or 2(20)
Hence we get x in one place by factorising as above.
$ax-bx=-c$
$x(a-b)=-c$
This is "x multiplied by both a and -b".
To move the (a-b) away from the x, to leave x standing alone,
we divide both sides by (a-b), just as
$3x=6\ \Rightarrow\ \frac{3x}{3}=\frac{6}{3}$
$\frac{3}{3}x=\frac{6}{3}$
As $\frac{3}{3}=1$
this is
$x=2$
Hence
$\frac{x(a-b)}{(a-b)}=-\frac{c}{(a-b)}$
$\frac{(a-b)}{(a-b)}=1$
$x=-\frac{c}{a-b}=\frac{c}{b-a}$
• May 22nd 2010, 10:26 PM
brumby_3
Nice explanation Archie!
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 5.15: Vector Subtraction
Difficulty Level: At Grade Created by: CK-12
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Practice Vector Subtraction
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You and a friend are trying to position a heavy sculpture out in front of your school. Fortunately, the sculpture is on rollers, so you can move it around easily and slide it into place. While you are applying force to the sculpture, it starts to move. The vectors you and your friend are applying look like this:
However, the sculpture starts to move too far and overshoots where it is supposed to be. You quickly tell your friend to pull instead of push, in effect subtracting her force vector, where before it was being added. Can you represent this graphically?
By the end of this Concept, you'll know how to represent the subtraction of vectors and answer this question.
### Guidance
As you know from Algebra, AB=A+(B)\begin{align*}A - B = A + (-B)\end{align*}. When we think of vector subtraction, we must think about it in terms of adding a negative vector. A negative vector is the same magnitude of the original vector, but its direction is opposite.
In order to subtract two vectors, we can use either the triangle method or the parallelogram method from above. The only difference is that instead of adding vectors A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}, we will be adding A\begin{align*}A\end{align*} and B\begin{align*}-B\end{align*}.
#### Example A
Using the triangle method for subtraction.
### Vocabulary
Negative Vector: A negative vector is a vector that is the same in magnitude as the original vector, but opposite in direction.
Triangle Method: The triangle method is a method of adding vectors by connecting the tail of one vector to the head of another vector.
### Guided Practice
1.For the vector subtraction below, make a diagram of the subtraction. ad\begin{align*}\vec{a} - \vec{d}\end{align*}
2.For the vector subtraction below, make a diagram of the subtraction. ba\begin{align*}\vec{b} - \vec{a}\end{align*}
3. For the vector subtraction below, make a diagram of the subtraction. dc\begin{align*}\vec{d} - \vec{c}\end{align*}
Solutions:
1.
2.
3.
### Concept Problem Solution
As you've seen in this Concept, subtracting a vector is the same as adding the negative of the original vector. This is exactly like the rule for adding a negative number to a positive number. Therefore, to change your friend's force vector to a subtraction instead of an addition, you need to change the direction by 180\begin{align*}180^\circ\end{align*} while keeping the magnitude the same. The graph looks like this:
### Practice
a\begin{align*}\vec{a}\end{align*} is in standard position with terminal point (1, 5) and b\begin{align*}\vec{b}\end{align*} is in standard position with terminal point (4, 2).
1. Find the coordinates of the terminal point of ab\begin{align*}\vec{a} - \vec{b}\end{align*}.
2. What is the magnitude of ab\begin{align*}\vec{a} - \vec{b}\end{align*}?
3. What is the direction of ab\begin{align*}\vec{a} - \vec{b}\end{align*}?
c\begin{align*}\vec{c}\end{align*} is in standard position with terminal point (4, 3) and d\begin{align*}\vec{d}\end{align*} is in standard position with terminal point (2, 2).
1. Find the coordinates of the terminal point of cd\begin{align*}\vec{c} - \vec{d}\end{align*}.
2. What is the magnitude of cd\begin{align*}\vec{c} - \vec{d}\end{align*}?
3. What is the direction of cd\begin{align*}\vec{c} - \vec{d}\end{align*}?
e\begin{align*}\vec{e}\end{align*} is in standard position with terminal point (3, 2) and f\begin{align*}\vec{f}\end{align*} is in standard position with terminal point (-1, 2).
1. Find the coordinates of the terminal point of ef\begin{align*}\vec{e} - \vec{f}\end{align*}.
2. What is the magnitude of ef\begin{align*}\vec{e} - \vec{f}\end{align*}?
3. What is the direction of ef\begin{align*}\vec{e} - \vec{f}\end{align*}?
g\begin{align*}\vec{g}\end{align*} is in standard position with terminal point (5, 5) and h\begin{align*}\vec{h}\end{align*} is in standard position with terminal point (4, 2).
1. Find the coordinates of the terminal point of gh\begin{align*}\vec{g} - \vec{h}\end{align*}.
2. What is the magnitude of gh\begin{align*}\vec{g} - \vec{h}\end{align*}?
3. What is the direction of gh\begin{align*}\vec{g} - \vec{h}\end{align*}?
i\begin{align*}\vec{i}\end{align*} is in standard position with terminal point (1, 5) and j\begin{align*}\vec{j}\end{align*} is in standard position with terminal point (-3, 1).
1. Find the coordinates of the terminal point of ij\begin{align*}\vec{i} - \vec{j}\end{align*}.
2. What is the magnitude of ij\begin{align*}\vec{i} - \vec{j}\end{align*}?
3. What is the direction of ij\begin{align*}\vec{i} - \vec{j}\end{align*}?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Negative Vector A negative vector is a vector that is the same magnitude as the original vector, but the opposite direction.
Triangle Method The triangle method is a method of adding vectors by connecting the tail of one vector to the head of another vector.
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# What Does Construct Mean In Maths Assignment Help
## Best Coursework Writing Service
The next section will show you what is a concept concept, and then give a point in your process to build a good concept. What are the main things about a concept? Let’s take a look at a few examples to explain the concepts. A very common concept is “explanation”. The concept you have created in the past is pretty common now. Some people think that a concept is ‘explanation,’ and they are wrong. They are not the only go now that can have a concept. The problemWhat Does Construct Mean In Maths? I’m getting tired of this one, my old (still) favourite way to describe things. It all started with the application of the concept of geometry. Whereas the concept of a triangle is not a triangle, it is a square. In mathematics, the square is the starting point of the operation of finding the angle of a triangle. I’ll say pop over to this web-site triangle is a square if and only if its length equals its side length. This is all very interesting. In modern mathematics, it is the square of two things – here square of the length of a triangle and the square of its side length – that is the starting line of my blog convex polygon. Now in geometry, the square of sides is also the starting line, here the square of length of a rectangle. In the previous sentence, you can see that the square of a rectangle is the starting, it is not a square of the side of one of its sides. But what about the square of an ellipse? The square of the angle of the ellipse is the starting of the operation. We know that the angle of an ellipsus is always the angle of its sides, that is it is always a circle of the side length of that ellipse. And then it is possible to define the function of that triangle which is called the triangle derivative, it is called the derivative of a triangle of the angle. For instance, let’s say that we have two triangles, the right one is a square and the left one is a rectangle. And the function of the square of that triangle is the function of their sides.
## Online Coursework Help
(1) What is the derivative of this triangle? (2) The function of a triangle, here the derivative of the square is. (3) It is the function from the definition of the triangle that determines the angle of that square, here the function of an elliptic is the function we’ve defined. The definition of the function of a square is the derivative that we’ve defined, here the definition of an ellippa is the function that we’ve given. So if you think about it, the function of you have two, because the function of this square is the function, the function that you have, the function, we have two. That is the function the triangle derivative that you have. What does that mean? In mathematics, the function with the name of the triangle derivative is the derivative. It’s a special function that we have called the triangle integral. There is a function called the square integral, it is defined, it is an integral, it’s an integral, the function. You have a function of this triangle. If you wanted to show how to do it, you could work with the function of our example, here the value of the square integral. (2-1) (1-1) – (2-2) – (1-2) – (3-3) – -(2-3) – (1-3) + (2-3-1) + (3-1-2-2-3+1) -(2+1-1-3+2-1 -1) – -(What Does Construct Mean In Maths? – mwe https://www.mehmans.com/en/2014/01/creating-math-in-marinel-an-in-the-med-at-rich-schools-with-an-understanding/ ====== enocct I find it hard to believe that any of the above would be in the same or similar situation as “creating” a “math” in math. It’s an interesting idea, which would have a great impact on many people’s lives but one is simply not possible. ~~~ jasonm I think this is what the reddit reddit community is looking for. I suggest you look up the “Creative Commons” issue and see what kind of stuff you find interesting. Edit: forgot about the reddit “creative commons” issue. —— mwe There’s an interesting subreddit called “creating”. [https://www2.reddit.
## Pay Someone to do Project
com/r/creating/](https://www 2.reddit.COM/) It’s a very interesting page but the writing is a bit bit technical. [http://creating.reddit.org/](http://creatING.reddit.ORG) ~~ ~ enocc I think the Reddit is trying to be a “creating community” so that people are able to look at how their work is being done. This is why it’s important to make the site fun, interesting and informative. The problem with this is that it forces people to be in a more mature mindset and that even the most mature of people don’t have the patience to put the stuff in the first place. In fact, it’s one of the reasons I’ve been writing this in the past and discussing with people who don’t want to be in the first line of defense for their work. For example, I’ve never wanted people to be doing this kind of work. I’ve always been interested in how other people are working out what they need to do to make a quality product. Though it’s not just me. I like to make my own work and I’ll be making what I can. If you’re a long-term learner, you get the point. (I’m not saying that this is an impossible task, just that it’s possible.) ~~ pmoriarty I’ve talked with people who’re really into this and they have a great shot at obtaining a decent amount of their skills. But I’ve never really thought about allowing people to do the work and I have no interest in getting one. Just because I’m not an avid learner doesn’t mean I have to.
## Finance Online Exam Help
It also doesn’t mean that I’m not making something up and I’m not giving people the right to edit it so I do it anyway. There are things I can do, but I don’t site web people to feel like the work is being done that way. So why would anyone want to be doing it? ~~ Forum I’m not interested in it. I’m not interested _in_ it. My goal is to make this community work. [https #3: The Art of Writing in Maths: How to Make it Work](https://forum.creating.com/viewtopic.php?p=15249977#3) ——~ mwe – I want to be the first to say that it’s a totally different world The idea is to create something that’s more Related Site than what you think it is. I think this is a very important one, and I’m sure it’s going to be interesting for a long time to come. Is there a way to be like this for anyone? My thought is that the only way we can really make new things is to create a thing that people can use to make things better. That way people can have their own creativity, and make them better. No one can edit navigate to this website rules of the game, and it’s going a lot harder to do that anymore
|
## Truth Values
### Calculating the Truth Value of a Compound Proposition
The truth or falsity of a proposition is called its truth value. The truth value of a compound proposition can be calculated from the truth values of its components, using the following rules:
• For a conjunction to be true, both conjuncts must be true.
• For a disjunction to be true, at least one disjunct must be true.
• A conditional is true except when the antecedent is true and the consequent false.
• For a biconditional to be true, the two input values must be the same (either both true or both false).
• A negation has the opposite value of the negated proposition.
We can use these rules to calculate the truth value of any compound proposition, beginning with the truth values of its simple components (the sentence letters) then calculating the truth value of each connective in the order that the connectives are used to join the component propositions.
Suppose A, B, and C are all true. What is the truth value of the following compound proposition? ((A • B) ⊃ ~C) To figure it out, first take note of the order in which connectives are used to join the component propositions. If we were constructing the above WFF according to the rules of syntax, we would start by joining A and B with “•” to make (A • B), and we’d prefix C with “~” to make ~C. Then we would join (A • B) and ~C with “⊃” to make ((A • B) ⊃ ~C). We follow this same order when calculating the truth value of the compound proposition:
1. (A • B) is true, because both conjuncts are true.
2. ~C is false, because C is true.
3. ((A • B) ⊃ ~C) is false, because the antecedent (A • B) is true and the consequent ~C is false.
The last connective to be calculated is the main connective. The truth value of the main connective is the truth value of the compound proposition as a whole. (As you may recall, the main connective represents the logical structure of the compound proposition as a whole.) In the above example, the main connective is “⊃”, so the proposition is a conditional. Since the “⊃” is false, the proposition as a whole is false.
### Using numerals to represent truth values
Calculating the truth value of a compound proposition can be challenging when the proposition is very complex. To make things easier, we can write the truth values beneath each of the letters and connectives in a compound proposition, using the numeral “1” to represent true and “0” to represent false, as shown in the example below.
In many logic textbooks, truth values are represented using the letter “T” for true and “F” for false. This is merely a matter of convention, but there are advantages to using numerals “1” and “0” to represent truth values (as frequently done in computer science) rather than letters. Using letters to represent truth values can be confusing when “T” and “F” are also used as sentence letters. To avoid that problem, lowercase “t” and “f” are sometimes used instead; but then the truth values are more difficult to read, especially in truth tables, because “t” and “f” look similar at a glance. Both problems can be avoided by using numerals to represent truth values.
Suppose A, B, and C are all true, but D is false. What is the truth value of ((A • B) ⊃ (~C ∨ D))?
Step 1. The truth values of A, B, C, and D are given, so we write them beneath the sentence letters:
(( A • B ) ⊃ ( ~ C ∨ D )) 1 1 1 0
Step 2. The values of the conjunction and negation can be calculated from the sentence letters, so we write those next:
(( A • B ) ⊃ ( ~ C ∨ D )) 1 1 1 0 1 0
Step 3. The value of the disjunction can now be calculated. In order for a disjunction to be true, at least one of its disjuncts must be true. But neither ~C nor D is true, so the disjunction is false:
(( A • B ) ⊃ ( ~ C ∨ D )) 1 1 1 0 1 0 0
Step 4. Finally, the value of the conditional can be calculated. Its antecedent (the conjunction) is true, and its consequent (the disjunction) is false; so the conditional is false:
(( A • B ) ⊃ ( ~ C ∨ D )) 1 1 1 0 0 1 0 0
Since “⊃” is the main connective, its truth value is the same as the truth value of the proposition as a whole: the proposition is false.
### Calculating Truth Values When Some Components are Unknown
It is often possible to calculate the truth value of a compound proposition even when the truth values of some components are unknown, as illustrated in the following example.
Suppose P is true, but the truth values of Q and R are unknown. What is the truth value of ~(P ∨ (Q ≡ R))?
Step 1. The truth value of P is given:
~ ( P ∨ ( Q ≡ R )) 1
Step 2. The disjunction must be true, because at least one of the disjuncts is true:
~ ( P ∨ ( Q ≡ R )) 1 1
Step 3. The negation must be false, since the negated proposition (the disjunction) is true:
~ ( P ∨ ( Q ≡ R )) 0 1 1
Since “~” is the main connective, the proposition is false.
|
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## 10.7 Estimation and Prediction
### Learning Objectives
1. To learn the distinction between estimation and prediction.
2. To learn the distinction between a confidence interval and a prediction interval.
3. To learn how to implement formulas for computing confidence intervals and prediction intervals.
Consider the following pairs of problems, in the context of Note 10.19 "Example 3" in Section 10.4 "The Least Squares Regression Line", the automobile age and value example.
1.
1. Estimate the average value of all four-year-old automobiles of this make and model.
2. Construct a 95% confidence interval for the average value of all four-year-old automobiles of this make and model.
2.
1. Shylock intends to buy a four-year-old automobile of this make and model next week. Predict the value of the first such automobile that he encounters.
2. Construct a 95% confidence interval for the value of the first such automobile that he encounters.
The method of solution and answer to the first question in each pair, (1a) and (2a), are the same. When we set x equal to 4 in the least squares regression equation $y^=−2.05x+32.83$ that was computed in part (c) of Note 10.19 "Example 3" in Section 10.4 "The Least Squares Regression Line", the number returned,
$y^=−2.05(4)+32.83=24.63$
which corresponds to value $24,630, is an estimate of precisely the number sought in question (1a): the mean $E(y)$ of all y values when x = 4. Since nothing is known about the first four-year-old automobile of this make and model that Shylock will encounter, our best guess as to its value is the mean value $E(y)$ of all such automobiles, the number 24.63 or$24,630, computed in the same way.
The answers to the second part of each question differ. In question (1b) we are trying to estimate a population parameter: the mean of the all the y-values in the sub-population picked out by the value x = 4, that is, the average value of all four-year-old automobiles. In question (2b), however, we are not trying to capture a fixed parameter, but the value of the random variable y in one trial of an experiment: examine the first four-year-old car Shylock encounters. In the first case we seek to construct a confidence interval in the same sense that we have done before. In the second case the situation is different, and the interval constructed has a different name, prediction interval. In the second case we are trying to “predict” where a the value of a random variable will take its value.
### $100(1−α)%$ Confidence Interval for the Mean Value of y at $x=xp$
$y^p±tα∕2 sε 1n+(xp−x-)2SSxx$
where
1. xp is a particular value of x that lies in the range of x-values in the sample data set used to construct the least squares regression line;
2. $y^p$ is the numerical value obtained when the least square regression equation is evaluated at $x=xp$; and
3. the number of degrees of freedom for $tα∕2$ is $df=n−2.$
The assumptions listed in Section 10.3 "Modelling Linear Relationships with Randomness Present" must hold.
The formula for the prediction interval is identical except for the presence of the number 1 underneath the square root sign. This means that the prediction interval is always wider than the confidence interval at the same confidence level and value of x. In practice the presence of the number 1 tends to make it much wider.
### $100(1−α)%$ Prediction Interval for an Individual New Value of y at $x=xp$
$y^p±tα∕2 sε 1+1n+(xp−x-)2SSxx$
where
1. xp is a particular value of x that lies in the range of x-values in the data set used to construct the least squares regression line;
2. $y^p$ is the numerical value obtained when the least square regression equation is evaluated at $x=xp$; and
3. the number of degrees of freedom for $tα∕2$ is $df=n−2.$
The assumptions listed in Section 10.3 "Modelling Linear Relationships with Randomness Present" must hold.
### Example 12
Using the sample data of Note 10.19 "Example 3" in Section 10.4 "The Least Squares Regression Line", recorded in Table 10.3 "Data on Age and Value of Used Automobiles of a Specific Make and Model", construct a 95% confidence interval for the average value of all three-and-one-half-year-old automobiles of this make and model.
Solution:
Solving this problem is merely a matter of finding the values of $y^p$, $α$ and $tα∕2$, $sε$, $x-$, and $SSxx$ and inserting them into the confidence interval formula given just above. Most of these quantities are already known. From Note 10.19 "Example 3" in Section 10.4 "The Least Squares Regression Line", $SSxx=14$ and $x-=4.$ From Note 10.31 "Example 7" in Section 10.5 "Statistical Inferences About ", $sε=1.902169814.$
From the statement of the problem $xp=3.5$, the value of x of interest. The value of $y^p$ is the number given by the regression equation, which by Note 10.19 "Example 3" is $y^=−2.05x+32.83$, when $x=xp$, that is, when x = 3.5. Thus here $y^p=−2.05(3.5)+32.83=25.655.$
Lastly, confidence level 95% means that $α=1−0.95=0.05$ so $α∕2=0.025.$ Since the sample size is n = 10, there are $n−2=8$ degrees of freedom. By Figure 12.3 "Critical Values of ", $t0.025=2.306.$ Thus
$y^p±tα∕2 sε 1n+(xp−x-)2SSxx=25.655±(2.306)(1.902169814)110+(3.5−4)214=25.655±4.3864035910.1178571429=25.655±1.506$
which gives the interval $(24.149,27.161).$
We are 95% confident that the average value of all three-and-one-half-year-old vehicles of this make and model is between $24,149 and$27,161.
### Example 13
Using the sample data of Note 10.19 "Example 3" in Section 10.4 "The Least Squares Regression Line", recorded in Table 10.3 "Data on Age and Value of Used Automobiles of a Specific Make and Model", construct a 95% prediction interval for the predicted value of a randomly selected three-and-one-half-year-old automobile of this make and model.
Solution:
The computations for this example are identical to those of the previous example, except that now there is the extra number 1 beneath the square root sign. Since we were careful to record the intermediate results of that computation, we have immediately that the 95% prediction interval is
$y^p±tα∕2 sε 1+1n+(xp−x-)2SSxx=25.655±4.3864035911.1178571429=25.655±4.638$
which gives the interval $(21.017,30.293).$
We are 95% confident that the value of a randomly selected three-and-one-half-year-old vehicle of this make and model is between $21,017 and$30,293.
Note what an enormous difference the presence of the extra number 1 under the square root sign made. The prediction interval is about two-and-one-half times wider than the confidence interval at the same level of confidence.
### Key Takeaways
• A confidence interval is used to estimate the mean value of y in the sub-population determined by the condition that x have some specific value xp.
• The prediction interval is used to predict the value that the random variable y will take when x has some specific value xp.
### Basic
For the Basic and Application exercises in this section use the computations that were done for the exercises with the same number in previous sections.
1. For the sample data set of Exercise 1 of Section 10.2 "The Linear Correlation Coefficient"
1. Give a point estimate for the mean value of y in the sub-population determined by the condition x = 4.
2. Construct the 90% confidence interval for that mean value.
2. For the sample data set of Exercise 2 of Section 10.2 "The Linear Correlation Coefficient"
1. Give a point estimate for the mean value of y in the sub-population determined by the condition x = 4.
2. Construct the 90% confidence interval for that mean value.
3. For the sample data set of Exercise 3 of Section 10.2 "The Linear Correlation Coefficient"
1. Give a point estimate for the mean value of y in the sub-population determined by the condition x = 7.
2. Construct the 95% confidence interval for that mean value.
4. For the sample data set of Exercise 4 of Section 10.2 "The Linear Correlation Coefficient"
1. Give a point estimate for the mean value of y in the sub-population determined by the condition x = 2.
2. Construct the 80% confidence interval for that mean value.
5. For the sample data set of Exercise 5 of Section 10.2 "The Linear Correlation Coefficient"
1. Give a point estimate for the mean value of y in the sub-population determined by the condition x = 1.
2. Construct the 80% confidence interval for that mean value.
6. For the sample data set of Exercise 6 of Section 10.2 "The Linear Correlation Coefficient"
1. Give a point estimate for the mean value of y in the sub-population determined by the condition x = 5.
2. Construct the 95% confidence interval for that mean value.
7. For the sample data set of Exercise 7 of Section 10.2 "The Linear Correlation Coefficient"
1. Give a point estimate for the mean value of y in the sub-population determined by the condition x = 6.
2. Construct the 99% confidence interval for that mean value.
3. Is it valid to make the same estimates for x = 12? Explain.
8. For the sample data set of Exercise 8 of Section 10.2 "The Linear Correlation Coefficient"
1. Give a point estimate for the mean value of y in the sub-population determined by the condition x = 12.
2. Construct the 80% confidence interval for that mean value.
3. Is it valid to make the same estimates for x = 0? Explain.
9. For the sample data set of Exercise 9 of Section 10.2 "The Linear Correlation Coefficient"
1. Give a point estimate for the mean value of y in the sub-population determined by the condition x = 0.
2. Construct the 90% confidence interval for that mean value.
3. Is it valid to make the same estimates for $x=−1$? Explain.
10. For the sample data set of Exercise 9 of Section 10.2 "The Linear Correlation Coefficient"
1. Give a point estimate for the mean value of y in the sub-population determined by the condition x = 8.
2. Construct the 95% confidence interval for that mean value.
3. Is it valid to make the same estimates for x = 0? Explain.
### Applications
1. For the data in Exercise 11 of Section 10.2 "The Linear Correlation Coefficient"
1. Give a point estimate for the average number of words in the vocabulary of 18-month-old children.
2. Construct the 95% confidence interval for that mean value.
3. Is it valid to make the same estimates for two-year-olds? Explain.
2. For the data in Exercise 12 of Section 10.2 "The Linear Correlation Coefficient"
1. Give a point estimate for the average braking distance of automobiles that weigh 3,250 pounds.
2. Construct the 80% confidence interval for that mean value.
3. Is it valid to make the same estimates for 5,000-pound automobiles? Explain.
3. For the data in Exercise 13 of Section 10.2 "The Linear Correlation Coefficient"
1. Give a point estimate for the resting heart rate of a man who is 35 years old.
2. One of the men in the sample is 35 years old, but his resting heart rate is not what you computed in part (a). Explain why this is not a contradiction.
3. Construct the 90% confidence interval for the mean resting heart rate of all 35-year-old men.
4. For the data in Exercise 14 of Section 10.2 "The Linear Correlation Coefficient"
1. Give a point estimate for the wave height when the wind speed is 13 miles per hour.
2. One of the wind speeds in the sample is 13 miles per hour, but the height of waves that day is not what you computed in part (a). Explain why this is not a contradiction.
3. Construct the 90% confidence interval for the mean wave height on days when the wind speed is 13 miles per hour.
5. For the data in Exercise 15 of Section 10.2 "The Linear Correlation Coefficient"
1. The business owner intends to spend $2,500 on advertising next year. Give an estimate of next year’s revenue based on this fact. 2. Construct the 90% prediction interval for next year’s revenue, based on the intent to spend$2,500 on advertising.
6. For the data in Exercise 16 of Section 10.2 "The Linear Correlation Coefficient"
1. A two-year-old girl is 32.3 inches long. Predict her adult height.
2. Construct the 95% prediction interval for the girl’s adult height.
7. For the data in Exercise 17 of Section 10.2 "The Linear Correlation Coefficient"
1. Lodovico has a 78.6 average in his physics class just before the final. Give a point estimate of what his final exam grade will be.
2. Explain whether an interval estimate for this problem is a confidence interval or a prediction interval.
3. Based on your answer to (b), construct an interval estimate for Lodovico’s final exam grade at the 90% level of confidence.
8. For the data in Exercise 18 of Section 10.2 "The Linear Correlation Coefficient"
1. This year 86.2 million acres of corn were planted. Give a point estimate of the number of acres that will be harvested this year.
2. Explain whether an interval estimate for this problem is a confidence interval or a prediction interval.
3. Based on your answer to (b), construct an interval estimate for the number of acres that will be harvested this year, at the 99% level of confidence.
9. For the data in Exercise 19 of Section 10.2 "The Linear Correlation Coefficient"
1. Give a point estimate for the blood concentration of the active ingredient of this medication in a man who has consumed 1.5 ounces of the medication just recently.
2. Gratiano just consumed 1.5 ounces of this medication 30 minutes ago. Construct a 95% prediction interval for the concentration of the active ingredient in his blood right now.
10. For the data in Exercise 20 of Section 10.2 "The Linear Correlation Coefficient"
1. You measure the girth of a free-standing oak tree five feet off the ground and obtain the value 127 inches. How old do you estimate the tree to be?
2. Construct a 90% prediction interval for the age of this tree.
11. For the data in Exercise 21 of Section 10.2 "The Linear Correlation Coefficient"
1. A test cylinder of concrete three days old fails at 1,750 psi. Predict what the 28-day strength of the concrete will be.
2. Construct a 99% prediction interval for the 28-day strength of this concrete.
3. Based on your answer to (b), what would be the minimum 28-day strength you could expect this concrete to exhibit?
12. For the data in Exercise 22 of Section 10.2 "The Linear Correlation Coefficient"
1. Tomorrow’s average temperature is forecast to be 53 degrees. Estimate the energy demand tomorrow.
2. Construct a 99% prediction interval for the energy demand tomorrow.
3. Based on your answer to (b), what would be the minimum demand you could expect?
### Large Data Set Exercises
1. Large Data Set 1 lists the SAT scores and GPAs of 1,000 students.
http://www.flatworldknowledge.com/sites/all/files/data1.xls
1. Give a point estimate of the mean GPA of all students who score 1350 on the SAT.
2. Construct a 90% confidence interval for the mean GPA of all students who score 1350 on the SAT.
2. Large Data Set 12 lists the golf scores on one round of golf for 75 golfers first using their own original clubs, then using clubs of a new, experimental design (after two months of familiarization with the new clubs).
http://www.flatworldknowledge.com/sites/all/files/data12.xls
1. Thurio averages 72 strokes per round with his own clubs. Give a point estimate for his score on one round if he switches to the new clubs.
2. Explain whether an interval estimate for this problem is a confidence interval or a prediction interval.
3. Based on your answer to (b), construct an interval estimate for Thurio’s score on one round if he switches to the new clubs, at 90% confidence.
3. Large Data Set 13 records the number of bidders and sales price of a particular type of antique grandfather clock at 60 auctions.
http://www.flatworldknowledge.com/sites/all/files/data13.xls
1. There are seven likely bidders at the Verona auction today. Give a point estimate for the price of such a clock at today’s auction.
2. Explain whether an interval estimate for this problem is a confidence interval or a prediction interval.
3. Based on your answer to (b), construct an interval estimate for the likely sale price of such a clock at today’s sale, at 95% confidence.
1. 5.647,
2. $5.647±1.253$
1. −0.188,
2. $−0.188±3.041$
1. 1.875,
2. $1.875±1.423$
1. 5.4,
2. $5.4±3.355$,
3. invalid (extrapolation)
1. 2.4,
2. $2.4±1.474$,
3. valid (−1 is in the range of the x-values in the data set)
1. 31.3 words,
2. $31.3±7.1$ words,
3. not valid, since two years is 24 months, hence this is extrapolation
1. 73.2 beats/min,
2. The man’s heart rate is not the predicted average for all men his age. c. $73.2±1.2$ beats/min
1. $224,562, 2.$224,562 ± \$28,699
1. 74,
2. Prediction (one person, not an average for all who have average 78.6 before the final exam),
3. $74±24$
1. 0.066%,
2. $0.066±0.034%$
1. 4,656 psi,
2. $4,656±321$ psi,
3. $4,656−321=4,335$ psi
1. 2.19
2. $(2.1421,2.2316)$
1. 7771.39
2. A prediction interval.
3. $(7410.41,8132.38)$
|
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# To investigate how the T number moves across And down effects the T total.
Extracts from this document...
Introduction
To investigate how the T number moves across
And down effects the T total.
8 by 10 grids
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
I have colored numbers inside the grid so it looks like a letter T.
I have also done this on a 9 by 10 grid and a 10 by 10 grid.
To investigate how the T number moves across
And down effects the T total.
In this investigation I had to find out how the T number moves across and down, effects the T total. I used grid sizes of 8 by 10, 9 by 10 and 10 by 10.
The T number is the number at the bottom of the T.
18 = the T number
Middle
18
34
21
49
42
154
45
169
66
247
69
289
I found out that going across the grid once, and adding 5 to the last t-total gives you the next t–total.
Then I found out that going down the grid once, and adding 40 to the last t–total gives you the next t–total.
If you multiply the difference going across by the gridsize you get the difference going down.
5 * 8 = 40
9 by 10 grid going down and across.
T – Number T – Total 20 37 23 52 26 67 47 172 50 187 53 262 74 307 77 322 80 337
I found out that going across the grid once and adding 5 to the next t-total gives you the next t-total.
Then I found out that going down the grid once, and adding 45 to the last t–total gives you the next t–total.
If you multiply the difference going across by the gridsize you get the difference going down.
5 * 9 = 45
10 by 10 grid going down and across.
T – Number T – Total 22 40 25 55 28 70 52 190 55 205 58 220 82 340 85 258 88 370
I found out that going across the grid once and adding 5 to the next t-total gives you the next t-total.
Then I found out that going down the grid once, and adding 50 to the last t–total gives you the next t–total.
Conclusion
= =
=
=
Multiples
I drew a 10*10 grid going up in the three times table. I took a T out of the grid and then done the algebra for it. Here are my results:-
T – Number T – Total 75 165
Algebra
=
In this T is equal to 75 so 75 – 45 is equal to T – 30.
Then I did the equation using G for the grid size.
= =
In my equation I used G for the gridsize so T-30 is equal to T-3g because G is equal to 10 and 10 is the gridsize.
Then I put M into the algebra. M meaning multiple. I am using a multiple of 3 because that is what the grid was going up in.
= =
=
I looked at both equations for the grids going up in 2’s & 3’sand found out a pattern. The pattern is that on the multiple equations it is the same for both of them apart from the number at the end. On the 2 equation it is + 2 & - 2 and on the 3 equation it is + 3 & - 3.
The equation here is 5T because I added all the T’s and then I got 7MG because I added up MG + 2MG + 2MG + 2MG. Therefore the equation is 5T – 7MG.
Maths Coursework T - Totals
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29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 T-number and T-total table T-number T-total 16 31 17 36 25 76 26 81 47 186 48 191 Here I have added a prediction of mine when I realized
2. ## The T-Total Mathematics Coursework Task.
L-number Top of L-shape L-total All numbers in L-shape added 4 50 40 230 5 55 41 235 6 60 42 240 7 65 43 245 8 70 44 250 9 75 45 255 13 95 49 275 14 100 50 280 15 105 51 285 16 110 52 290
1. ## T-Total Investigation
a different translation / rotation combination, in this case 180 degrees clockwise so the formula will read t=5((v+b)-ag) + 2g I will use an extended (vertically) grid width of 5 for this, 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
2. ## T totals. In this investigation I aim to find out relationships between grid sizes ...
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1. ## T totals - translations and rotations
The two remaining numbers in my T-shape are N-14-1 and N-14+1. Thus my T-total is: N+ (N-7) + (N-14) + (N-14-1) + (N-14+1)= 5N-49 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
2. ## Maths Coursework T-Totals
or irregular (e.g. 5x101), with the two variables of grid width (g) and the T-Number (x) for a Tb T-Shape. Tc and Td shapes As we know that a 180-degree flip will be a "negative" equation of the other flip, we only need to work out one kind of shape, therefore we shall work out formulas for Tc shapes.
1. ## Maths GCSE Coursework &amp;#150; T-Total
59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 The T-Total for this T-Shape is 187 (31 + 32 + 33 + 41 + 50), in relation to v these are; v-(9-1)
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
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You can approximate the area under a curve by using right sums. This method works just like the left sum method except that each rectangle is drawn so that its right upper corner touches the curve instead of its left upper corner.
For example, estimate the area under
from 0 to 3 by using right rectangles, as shown in this figure.
The heights of the three rectangles are given by the function values at their right edges: f(1) = 2, f(2) = 5, and f(3) = 10. Each rectangle has a width of 1, so the areas are 2, 5, and 10, which total 17. This approximation gives you an overestimate of the actual area under the curve.
The more rectangles you create between 0 and 3, the more accurate your estimate will be. Here's the Right Rectangle Rule: You can approximate the exact area under a curve between a and b,
with a sum of right rectangles given by this formula:
Where, n is the number of rectangles,
is the width of each rectangle, and the function values are the heights of the rectangles.
Now if you compare this formula to the one for a left rectangle sum, you get the complete picture about those subscripts. The two formulas are the same except for one thing. Look at the sums of the function values in both formulas. The right sum formula has one value,
that the left sum formula doesn’t have, and the left sum formula has one value,
that the right sum formula doesn’t have. All the function values between those two appear in both formulas. You can get a better handle on this by comparing the three left rectangles from the below figure to the three right rectangles from the above figure.
For the left sum, you can find the areas and totals using the following formula:
So, three left rectangles add up to: 1 + 2 + 5 = 8
For the right sum, you can use the following formula:
So, three right rectangles add up to: 2 + 5 + 10 = 17
The sums of the areas are the same except for the left-most left rectangle and the right-most right rectangle. Both sums include the rectangles with areas 2 and 5. If you look at how the rectangles are constructed, you can see that the second and third rectangles in the second figure are the same as the first and second rectangles in the first figure.
One last thing on this. The difference between the right rectangle total area (17) and the left rectangle total area (8) — that’s 17 minus 8, or 9, in case you love calculus but don’t have the basic subtraction thing down yet — comes from the difference between the areas of the two “end” rectangles just discussed (10 minus 1 is also 9). All the other rectangles are a wash, no matter how many rectangles you have.
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# Difference between revisions of "2011 AMC 12A Problems/Problem 16"
## Problem
Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520 \qquad \textbf{(B)}\ 2880 \qquad \textbf{(C)}\ 3120 \qquad \textbf{(D)}\ 3250 \qquad \textbf{(E)}\ 3750$
## Solution
We can do some casework when working our way around the pentagon from $A$ to $E$. At each stage, there will be a makeshift diagram.
1.) For $A$, we can choose any of the 6 colors.
A : 6
2.) For $B$, we can either have the same color as $A$, or any of the other 5 colors. We do this because each vertex of the pentagon is affected by the 2 opposite vertices, and $D$ will be affected by both $A$ and $B$.
A : 6
B:1 B:5
3.) For $C$, we cannot have the same color as $A$. Also, we can have the same color as $B$ ($E$ will be affected), or any of the other 4 colors. Because $C$ can't be the same as $A$, it can't be the same as $B$ if $B$ is the same as $A$, so it can be any of the 5 other colors.
A : 6
B:1 B:5
C:5 C:4 C:1
4.) $D$ is affected by $A$ and $B$. If they are the same, then $D$ can be any of the other 5 colors. If they are different, then $D$ can be any of the (6-2)=4 colors.
A : 6
B:1 B:5
C:5 C:4 C:1
D:5 D:4 D:4
5.) $E$ is affected by $B$ and $C$. If they are the same, then $E$ can be any of the other 5 colors. If they are different, then $E$ can be any of the (6-2)=4 colors.
A : 6
B:1 B:5
C:5 C:4 C:1
D:5 D:4 D:4
E:4 E:4 E:5
6.) Now, we can multiply these three paths and add them: $(6\times1\times5\times5\times4)+(6\times5\times4\times4\times4)+(6\times5\times1\times4\times5) =600+1920+600=3120$
7.) Our answer is $C$!
## Solution 2
Right off the bat, we can analyze three things:
1.) There can only be two of the same color on the pentagon.
2.) Any pair of the same color can only be next to each other on the pentagon.
3.) There can only be two different pairs of same colors on the pentagon at once.
Now that we know this, we can solve the problem by using three cases: no same color pairs, one same color pair, and two same color pairs.
1.) If there are no color pairs, it is a simple permutation: six different colors in five different spots. We count $6!=720$ cases. No rotation is necessary because all permutations are accounted for.
2.)If there is one color pair, we must count 6 possibilities for the pair(as one element), 5 for the third vertex, 4 for the fourth vertex, and 3 for the fifth vertex.
We get $6\times5\times4\times3=360$.
However, there are 5 different locations the pair could be at. Therefore we get $360\times5=1800$ possibilities for one pair.
3.)If there are two color pairs, we must count 6 possibilities for the first pair(as one element), 5 possibilities for the next pair(as one element), and 4 possibilities for the final vertex.
We get $6\times5\times4=120$.
Once again, there are 5 different rotations in the pentagon that we must account for. Therefore we get $120\times5=600$ possibilities for two pairs.
5.) If we add all of three cases together, we get $720+1800+600=3120$. The answer is $C$.
Solution by gsaelite
## Solution 3
This solution essentially explains other ways of thinking about the cases stated in Solution 2.
Case 1:
${6\choose5} \cdot 5!$
5 colors need to be chosen from the group of 6. Those 5 colors have 5! distinct arrangements on the pentagon's vertices.
Case 2:
${6\choose4} \cdot4\cdot5\cdot3!$
4 colors need to be chosen from the group of 6. Out of those 4 colors, one needs to be chosen to form a pair of 2 identical colors. Then, for arranging this layout onto the pentagon, one of the five sides (same thing as pair of adjacent vertices) of the pentagon needs to be established for the pair. The remaining 3 unique colors can be arranged 3! different ways on the remaining 3 vertices.
Case 3:
${6\choose3} \cdot {3\choose2} \cdot5\cdot2$
3 colors need to be chosen from the group of 6. Out of those 3 colors, 2 need to be chosen to be the pairs. Then, for arranging this layout onto the pentagon, start out by thinking about the 1 color that is not part of a pair, as it makes things easier. It can be any one of the 5 vertices of the pentagon. The remaining 2 pairs of colors can only be arranged 2 ways on the remaining 4 vertices.
Solving each case and adding them up gets you 3120. $\boxed{C}$
~IceMatrix
## Solution 4
We can order the vertices of pentagon $ABCDE$ arbitrarily. This means that besides for the first and last vertices, the previous vertex and the next vertex of any vertex have diagonals (such as $A$,$C$,$E$,$B$,$D$ where $C$ shares diagonals with $A$ and $E$, $E$ shares diagonals with $C$ and $B$, etc.).
The first point can be one of $6$ colors as we have no designated restraints on it. From there, the next three points can be five colors each since the only color restraint is that of the previous color (i.e. if the first point was blue, then the next point can be any of the other $5$ colors, such as purple, and the point after that can be any color except purple, which is 5 colors).
The last point is the trickiest because we need to consider the second to last vertex and the first vertex. The third vertex has a $\frac{1}{5}$ chance of being the same color as the first vertex, guaranteeing that the second to last vertex is a different color from the first vertex, leaving the last vertex with $4$ possible colors. The remaining $\frac{4}{5}$ times where the third to last vertex is a different color from the first vertex, there is a $\frac{4}{5}$ chance that the second to last vertex is a different color from the first vertex and a $\frac{1}{5}$ chance that they are the same color, so there are $4$ and $5$ possibilities for the last vertex respectively.
Our answer is now $6\times5^3\times(\frac{4}{5}\times(\frac{4}{5}\times4+\frac{1}{5}\times5)+\frac{1}{5}\times4)=3120$, which is $\boxed{C}$.
~Randomlygenerated
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## How to Solve Word Problems by Working Backwards Part 3
In part 1 and part 2 of this series, we have learned how to solve number age problems by working backward. In this post, we are going to learn how to solve backward using inverse operations. Recall that multiplication and division are inverse operations and addition and subtraction are inverse operations.
Example 5
A number is multiplied by 4 and then, 3 is added to the product. The result is 31. What is the number?
Solution
The key phrases in this problem are (1) multiplied by 4 and (2) added to (3) the result is 31. Since we are working backward, we start with 31, and then find the inverse of “added to 3” which is “subtract 3.” So, 31 – 3 = 28.
Next, we find the inverse of “multiplied by 4,” which is “divided by 4.” So, 28/4 = 7.
So, the answer to this problem is 7.
Check: 7(4) + 3 = 31
Example 6
Think of a number. Divide it by 8. Then subtract 4 from the quotient. The result is 5. What is the number?
Solution
The key phrases in this problem are (1) divided by 8 (2) subtract 4 and (3) the result is (3) the result is 5.
We start with the result which is 5 and find the inverse of “subtract 4” which is “add 4.” So, 5 + 4 = 9. Next, we find the inverse of “divide by 8” which is “multiply by 8.” So, 9(8) = 72.
So, the correct answer is 72.
Check: 72/8 – 4 = 9 – 4 = 5.
In the next post, we will discuss more about solving math word problems by working backward.
## How to Solve Problems by Working Backwards Part 2
In the previous post, we have learned how to solve number problems by working backward. In this post, we discuss age problems. We already had 2 examples in the previous post in this series, so we start with the third example.
Example 3
Arvin is 5 years older than Michael. The sum of their ages is 37. What are their ages?
Solution
This is very similar to the problems in the previous post in this series. Arvin is 5 years older than Michael, so if we subtract 5 from Arvin’s age, their ages will be equal. But if we subtract 5 from Arvin’s age, we also have to subtract 5 from the sum of their ages. That is,
37 – 5 = 32.
Now, their ages are equal, so we can divide the sum by 2. That is 32 ÷ 2 = 16. This means that the younger person is 16 since we subtracted 5 from Arvin’s age. Therefore, Michael is 16 and Arvin is 16 + 5 = 21.
Check
16 + 21 = 37.
Example 4
Mia is 3 years older than Pia. In 4 years, the sum of their ages is 35. What are their ages?
Solution
There are two of them and we added 4 to both ages, so subtracting 8 from 35 (the sum) will determine the sum of their present age. That is 35 – 8 = 27 is the sum of their present age.
Next, Mia is 3 years older than Pia, so if we subtract 3 from the sum of their present age, their ages will be equal. So, 27 – 3 = 24.
We can now divide the sum of their ages by 2. That is 24/2 = 12.
This means that 12 is the age of the younger person because we subtracted 3 from the age of the older person.
So, Pia is 12 and Mia is 15.
Check 12 + 15 = 22.
In the next post, we will discuss more problems that can be solved by working backward.
## How to Solve Word Problems by Working Backwards Part 1
Most of us would always take a pen and solve for x if we see word problems. But did you know that you can solve them by working backward or even mentally? In this post, I am going to teach you some techniques on solving problems by working backward.
Example 1: One number is three more than the other. Their sum is 45. What are the numbers?
Solution
In the given, one number is 3 more than the other. This means that if you subtract 3 from the larger number they will be equal. Note that if we subtract 3 from one of the numbers, then we should also subtract 3 from their sum. Therefore, their sum will be 45 – 3 = 42. Since the numbers are equal, we now divide the sum by 2. That is, 42/2 = 21.
So, the smaller number is 21 and the larger is 21 + 3 = 24.
Check: 21 + 24 = 45
Example 2: One number is 5 less than the other. Their sum is 43. What is the smaller number?
Solution
This is very similar to Example 1. Here, one number is 5 less than the other; so, if we add 5 to the smaller number, they will be equal. If we add 5 to the smaller number, we should also add 5 to their sum. Therefore, their sum will be 43 + 5 = 48. Since the two numbers are equal, we can divide the sum by 2. That is 48/2 = 24. Since we added 5, it means that 24 is the larger number. So, the smaller number is 24 – 5 = 19.
Check: 19 + 24 = 43
In the next post, we are going to discuss more examples.
## Practice Exercises on Subtracting Decimals
We have already learned how to add and subtract numbers with decimals. In this post, we practice subtracting decimals. Recall that in subtracting decimals, the decimal points should be aligned.
Practice Exercises
1.) 2.32 – 1.82
2.) 6.71 – 3.9
3.) 6 – 0.52
4.) 5.03 – 4.25
5.) 0.53 – 0.33
6.) 4 – 1.26
7.) 7.28 – 2.4
8.) 7.08 – 0.29
9.) 3 – 0.305
10.) 40 – 12.5
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# Lagrange Multiplier Calculator
To use lagrange multiplier calculator, enter the values in the given boxes, select to maximize or minimize, and click the calcualte button
Give Us Feedback
## Lagrange Calculator
Lagrange multiplier calculator is used to evalcuate the maxima and minima of the function with steps. This Lagrange calculator finds the result in a couple of a second.
## What is Lagrange multiplier?
The method of Lagrange multipliers, which is named after the mathematician Joseph-Louis Lagrange, is a technique for locating the local maxima and minima of a function that is subject to equality constraints.
(i.e., subject to the requirement that one or more equations have to be precisely satisfied by the chosen values of the variables).
The fundamental concept is to transform a limited problem into a format that still allows the derivative test of an unconstrained problem to be used.
The Lagrangian function is a reformulation of the original issue that results from the relationship between the gradient of the function and the gradients of the constraints.
## The formula for Lagrange multiplier
The formula of the lagrange multiplier is:
### Example of lagrange multiplier
Use the method of Lagrange multipliers to find the minimum value of g(y, t) = y2 + 4t2 – 2y + 8t subjected to constraint y + 2t = 7
Solution:
Step 1: Write the objective function and find the constraint function; we must first make the right-hand side equal to zero.
g(y, t) = y2 + 4t2 – 2y + 8t
The constraint function is y + 2t – 7 = 0
So h(y, t) = y + 2t – 7
To minimize the value of function g(y, t), under the given constraints.
g(y, t) = y2 + 4t2 – 2y + 8t corresponding to c = 10 and 26.
Step 2: Now find the gradients of both functions.
∇g(y, t) = (2y - 2)i + (8t + 8)j
∇h(y, t) = i + 2j
Now equation ∇g(y, t) = a∇h(y, t) becomes
(2y-2)i + (8t + 8)j = a(i + 2j)
Step 3: Compare the above equation.
2y - 2 = a
8t + 8 = 2a
Step 4: Now solve the system of the linear equation.
t= 1 and y= 5
Step 5: Evaluate the function
g(y, t) = 52+ 4(1)2 – 2(5) + 8(1) = 27
So h has a relative minimum value is 27 at the point (5,1).
Use our lagrangian calculator above to cross-check the above result.
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## Introduction to Data Interpretation
#### Data Interpretation
Direction: The sub divided bar diagram given below depicts H.S. Students of a school for three years. Study the diagram and answer the questions.
1. The number of students passed with 3rd division in the year 2008 was
1. According to given bar graph , we have
Number of students passed in 2nd and 1st division in 2008 = 80
Total number of students passed in 2008 = 140
∴ Students passed in third division in 2008 = Total number of students passed in 2008 - Number of students passed in 2nd and 1st division in 2008
##### Correct Option: B
According to given bar graph , we have
Number of students passed in 2nd and 1st division in 2008 = 80
Total number of students passed in 2008 = 140
∴ Students passed in third division in 2008 = Total number of students passed in 2008 - Number of students passed in 2nd and 1st division in 2008
Students passed in third division in 2008 = 140 – 80 = 60
1. In which year the school had the best result for H.S. in respect of percentage of pass candidates?
1. On the basis of given graph in question ,
Total students who passed in 2008 = 140
Total students in 2008 = 170
Year 2008 ⇒
∴ Required percentage = 140 × 100 170
Required percentage = 82 6 % 11
Year 2009 ⇒
Total students who passed in 2009 = 140
Total students in 2009 = 190
Required percentage = 140 × 100 = 73.7% 190
##### Correct Option: A
On the basis of given graph in question ,
Total students who passed in 2008 = 140
Total students in 2008 = 170
Year 2008 ⇒
∴ Required percentage = 140 × 100 170
Required percentage = 82 6 % 11
Year 2009 ⇒
Total students who passed in 2009 = 140
Total students in 2009 = 190
Required percentage = 140 × 100 = 73.7% 190
Year 2010 ⇒
Total students who passed in 2010 = 150
Total students in 2010 = 200
Required percentage = 150 × 100 = 75% 200
Hence , required answer is 2008 year .
1. The pass percentage in 2008 was
1. On the basis of given graph in question ,
Total students who passed in 2008 = 140
Total students in 2008 = 170
∴ Required percentage = 140 × 100 170
##### Correct Option: D
On the basis of given graph in question ,
Total students who passed in 2008 = 140
Total students in 2008 = 170
∴ Required percentage = 140 × 100 170
Required percentage = 1400 = 82 6 % 17 17
1. The percentage passed in 1st division in 2008 was
1. According to given bar graph , we have
Total students in 2008 = 170
Students passed in 1st division = 20
∴ Required percentage = Students passed in 1st division × 100 Total students in 2008
Required percentage = 20 × 100 170
##### Correct Option: D
According to given bar graph , we have
Total students in 2008 = 170
Students passed in 1st division = 20
∴ Required percentage = Students passed in 1st division × 100 Total students in 2008
Required percentage = 20 × 100 170
Required percentage = 200 = 11 13 % 17 17
Direction: The bar chart given below shows the percentage distribution of the production of various models of a mobile manufacturing company in 2007 and 2008. The total production in 2007 was 35 lakh mobile phones and in 2008 the production was 44 lakh. Study the chart and answer the following questions.
1. If 85% of the D type mobiles produced in each year were sold by the company, how many D type mobiles remained unsold?
1. On the basis of given graph in question ,
Given , The total production in 2007 = 35 lakh mobile phones
The total production in 2008 = 44 lakh mobile phones
Percent of mobiles of models D manufactured in 2007 = 10%
Percent of mobiles of models D manufactured in 2008 = 10%
Required answer = 10% of ( 100 - 85 )% of 35 + 10% of ( 100 - 85 )% of 44
Required answer = 35 × 10 × 15 + 44 × 10 × 15 100 100 100 100
##### Correct Option: C
On the basis of given graph in question ,
Given , The total production in 2007 = 35 lakh mobile phones
The total production in 2008 = 44 lakh mobile phones
Percent of mobiles of models D manufactured in 2007 = 10%
Percent of mobiles of models D manufactured in 2008 = 10%
Required answer = 10% of ( 100 - 85 )% of 35 + 10% of ( 100 - 85 )% of 44
Required answer = 35 × 10 × 15 + 44 × 10 × 15 100 100 100 100
Required answer = 150 × 79 = 1.1850 billion = 118500 10000
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# Subtraction
The answer of a subtraction sum is called DIFFERENCE.
How to subtract 2-digit numbers?
Steps are shown to subtract 2-digit numbers.
Subtract 189 from 638.
Line up the numbers according to place value taking care to place the bigger number on top.
In the ONES column we have 8 - 9.
We cannot take away 9 from 8 so we REGROUP the next place to the left. REGROUP 78 as 6 TENS and 18 ONES.
DIFFERENCE = 39
Sometimes we need to REGROUP more than once.
How to subtract 3-digit numbers?
Steps are shown to subtract 3-digit numbers.
Subtract 189 from 638.
REGROUP 38 as 2 TENS 18 ONES
REGROUP 62 as 5 TENS 12 ONES
We follow the same rule for subtracting 4 digits numbers or larger numbers.
How to subtract 4-digit numbers?
Steps are shown to subtract 4-digit numbers.
1. Solve: 74834 - 38915
Line up the number according to place value.
In the ONES column the number to be subtracted is greater so REGROUP as 2 TENS 14 ONES.
Subtract the TENS column.
REGROUP to subtract the numbers in the HUNDREDS column.
3 THOUSANDS 18 HUNDREDS
REGROUP to subtract the numbers in the THOUSANDS column.
6 TEN THOUSANDS 13 THOUSANDS
We can subtract from a place with a zero. Just remember to REGROUP the next place to the left.
2. Subtract: 39507 - 27386
REGROUP to subtract the TENS place
4 HUNDREDS 10 TENS
How to check the answer of a subtraction?
CHECK:
Add the DIFFERENCE to the smaller number.
The answer will be the smaller number.
Subtract the DIFFERENCE from the greater number.
The answer will be the greater number.
In subtraction we will learn the subtraction operation of numbers having more than 4 digits with borrowing and without borrowing.
Examples:
1. Subtract 2684 from 6795.
Solution:
The numbers are arranged in column form
(i) Ones are subtracted, 5 – 4 = 1
(ii) Tens are subtracted, 9 – 8 = 1
(iii) Hundreds are subtracted, 7 – 6 = 1
(iv) Thousands are subtracted, 6 – 2 = 4
Hence, difference = 4,111
2. Subtract 6732 from 9340 (with borrowing).
Solution:
The numbers are arranged in column form
(i) Ones are subtracted, 0 < 2, so 1 ten is borrowed from tens.
Now 1 Ten or 10 + 0 = 10, 10 – 2 = 8
(ii) Tens are subtracted, 3Tens – 3Tens = 0
(iii)Hundreds are subtracted, 3 Hundreds < 7 Hundreds so, 1 Thousand is borrowed,
so 10 Hundreds + 3 Hundreds = 13 Hundreds,
13 Hundreds – 7 Hundreds = 6 Hundreds.
(iv) Thousands are subtracted, 8Thousands – 6Thousands = 2Thousands
Hence, difference = 2,608
3. What is the difference between 40712 and 7549? (with borrowing)
Solution:
The numbers are arranged in column form
(i) 2 < 9, 1T or 10 is borrowed 10 + 2 = 12, 12 – 9 = 3
(ii) 0 < 4, 1H or 10T is borrowed 10T + 0 = 10T, Now 10T – 4T = 6T
(iii) 6H – 5H = 1H
(iv) 1 Tth is borrowed, 10Th – 7Th = 3 Th
(v) 3Tth – 0 = 3Tth
So, difference = 33,163
4. Subtract 2 3 7 4 1 2 from 6 4 9 5 2 3 (without borrowing)
Solution:
Numbers are arranged in column form
(i) 3 – 2 = 1 Ones
(ii) 2 – 1 = 1 Tens
(iii) 5 – 4 = 1 Hundreds
(iv) 9 – 7 = 2 Thousands
(v) 4 – 3 = 1 Ten thousands
(vi) 6 – 2 = 4 Hundred thousands
Therefore, difference = 412111
5. Subtract (with borrowing) 6 5 6 2 9 from 3 2 3 4 7 8
Solution:
Numbers are arranged in column form
(i) Ones: 8 < 9, IT or 10 is borrowed 10 + 8 = 18, 18 - 9 = 9
(ii) Tens: 6T - 2T = 4T
(iii) Hundreds: 4H < 6H, 1 th or 10H is borrowed, 10H + 4H = 14H, 14H - 6H = 8H
(iv) 2Th < 5Th, 1Tth or 10Th is borrowed 10Th + 2Th = 12Th, 12Th - 5Th = 7Th
(v) 1Tth < 6Tth,1Hth or 10Tth is borrowed 10Tth + 1Tth = 11Tth, 11Tth - 6Tth = 5Tth
(vi) 2Hth remained as it was.
So, difference = 257849
6. Find the difference between two numbers, 1 4 2 7 1 3 and 3 7 4 3 9.
Solution:
The numbers are arranged in column form.
142713 is greater than 37439, so 37439 will be subtracted from 142713.
Difference = 105274
(i) Ones are subtracted
3 < 9, so 1 ten is borrowed.
Now 1T or 10 + 3 = 13, 13 - 9 = 4
(ii) Tens are subtracted, 1T < 3T, so 1H is borrowed.
Now 1H or 10T + 1 = 11,
11 - 3 = 8, answer is 11 - (3 + 1) = 7
(iii) Hundred: 6H – 4H = 2H
(iv) Thousands: 12Th – 7Th = 5Th
(v) Ten thousands: 3Th - 3Th: = 0
(vi) Hundred thousand: 1Hth – 0 = 1Hth
Therefore, 142713 - 37439 = 105274
Related Concept
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How do you simplify 12 - 2 {24 + 2 [3 (7 + 4) + (6)^(2)]}?
Aug 23, 2017
$= - 312$
Explanation:
There are only 2 terms. The first is already in its simplest form.
Simplify the second term:
Start with the innermost brackets and work outwards.
$\textcolor{b l u e}{12} - 2 \left\{24 + 2 \left[3 \textcolor{red}{\left(7 + 4\right)} + \textcolor{red}{{\left(6\right)}^{2}}\right]\right\}$
$\textcolor{w h i t e}{w w w w w w w w w w w w} \downarrow \textcolor{w h i t e}{\times x} \downarrow$
$= \textcolor{b l u e}{12} - 2 \left\{24 + 2 \left[3 \textcolor{red}{\left(11\right)} + \textcolor{red}{\left(36\right)}\right]\right\}$
$\textcolor{w h i t e}{w w w w w w w w w w w} \downarrow$
$= \textcolor{b l u e}{12} - 2 \left\{24 + 2 \left[\textcolor{red}{33} + 36\right]\right\}$
=color(blue)(12)-2{24+2{color(lime)(33+36]}
$\textcolor{w h i t e}{w w w w w w w w w w w w w} \downarrow$
=color(blue)(12)-2{24" + "2color(lime)((69)}
$\textcolor{w h i t e}{w w w w w w w w w . w w w} \downarrow$
$= \textcolor{b l u e}{12} - 2 \left\{24 \text{ "+" } \textcolor{\lim e}{138}\right\}$
=color(blue)(12)-2{color(magenta)(24+138)]}
$\textcolor{w h i t e}{w w w w w w w} \downarrow$
$= \textcolor{b l u e}{12} - 2 \left\{\textcolor{m a \ge n t a}{162}\right\}$
$\textcolor{w h i t e}{w w w w w w w} \downarrow$
$= \textcolor{b l u e}{12} \text{ } - \textcolor{m a \ge n t a}{324}$
$= - 312$
Aug 23, 2017
$- 312$
Explanation:
We have: $12 - 2 \left\{24 + 2 \left[3 \left(7 + 4\right) + {\left(6\right)}^{2}\right]\right\}$
Let's apply the "PEMDAS" rules for operator precedence.
"PEMDAS" is an acronym for:
$\text{Parentheses} ,$ $\text{exponents} ,$ $\text{multiplication,}$ $\text{division} ,$ $\text{addition}$ $\mathmr{and}$ $\text{subtraction.}$
First, let's evaluate the operations within parentheses:
$= 12 - 2 \left\{24 + 2 \left[3 \times 11 + {6}^{2}\right]\right\}$
Next, we evaluate any exponents:
$= 12 - 2 \left\{24 + 2 \left[3 \times 11 + 36\right]\right\}$
Then, we perform any multiplication:
$= 12 - 2 \left\{24 + 2 \left[33 + 36\right]\right\}$
There is no division in this case, so let's perform any addition:
$= 12 - 2 \left\{24 + 2 \left[69\right]\right\}$
$= 12 - 2 \left\{24 + 138\right\}$
$= 12 - 2 \left\{162\right\}$
$= 12 - 324$
Finally, let's subtract the two numbers:
$= - 312$
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Paul's Online Math Notes
[Notes]
Calculus III - Notes
3-Dimensional Space Previous Chapter Next Chapter Applications of Partial Derivatives Partial Derivatives Previous Section Next Section Higher Order Partial Derivatives
## Interpretations of Partial Derivatives
This is a fairly short section and is here so we can acknowledge that the two main interpretations of derivatives of functions of a single variable still hold for partial derivatives, with small modifications of course to account of the fact that we now have more than one variable.
The first interpretation we’ve already seen and is the more important of the two. As with functions of single variables partial derivatives represent the rates of change of the functions as the variables change. As we saw in the previous section, represents the rate of change of the function as we change x and hold y fixed while represents the rate of change of as we change y and hold x fixed.
Example 1 Determine if is increasing or decreasing at , (a) if we allow x to vary and hold y fixed. (b) if we allow y to vary and hold x fixed. Solution (a) If we allow x to vary and hold y fixed. In this case we will first need and its value at the point. So, the partial derivative with respect to x is positive and so if we hold y fixed the function is increasing at as we vary x. (b) If we allow y to vary and hold x fixed. For this part we will need and its value at the point. Here the partial derivative with respect to y is negative and so the function is decreasing at as we vary y and hold x fixed.
Note that it is completely possible for a function to be increasing for a fixed y and decreasing for a fixed x at a point as this example has shown. To see a nice example of this take a look at the following graph.
This is a graph of a hyperbolic paraboloid and at the origin we can see that if we move in along the y-axis the graph is increasing and if we move along the x-axis the graph is decreasing. So it is completely possible to have a graph both increasing and decreasing at a point depending upon the direction that we move. We should never expect that the function will behave in exactly the same way at a point as each variable changes.
The next interpretation was one of the standard interpretations in a Calculus I class. We know from a Calculus I class that represents the slope of the tangent line to at . Well, and also represent the slopes of tangent lines. The difference here is the functions that they represent tangent lines to.
Partial derivatives are the slopes of traces. The partial derivative is the slope of the trace of for the plane at the point . Likewise the partial derivative is the slope of the trace of for the plane at the point .
Example 2 Find the slopes of the traces to at the point . Solution We sketched the traces for the planes and in a previous section and these are the two traces for this point. For reference purposes here are the graphs of the traces. Next we’ll need the two partial derivatives so we can get the slopes. To get the slopes all we need to do is evaluate the partial derivatives at the point in question. So, the tangent line at for the trace to for the plane has a slope of -8. Also the tangent line at for the trace to for the plane has a slope of -4.
Finally, let’s briefly talk about getting the equations of the tangent line. Recall that the equation of a line in 3-D space is given by a vector equation. Also to get the equation we need a point on the line and a vector that is parallel to the line.
The point is easy. Since we know the x-y coordinates of the point all we need to do is plug this into the equation to get the point. So, the point will be,
The parallel (or tangent) vector is also just as easy. We can write the equation of the surface as a vector function as follows,
We know that if we have a vector function of one variable we can get a tangent vector by differentiating the vector function. The same will hold true here. If we differentiate with respect to x we will get a tangent vector to traces for the plane (i.e. for fixed y) and if we differentiate with respect to y we will get a tangent vector to traces for the plane (or fixed x).
So, here is the tangent vector for traces with fixed y.
We differentiated each component with respect to x. Therefore the first component becomes a 1 and the second becomes a zero because we are treating y as a constant when we differentiate with respect to x. The third component is just the partial derivative of the function with respect to x.
For traces with fixed x the tangent vector is,
The equation for the tangent line to traces with fixed y is then,
and the tangent line to traces with fixed x is,
Example 3 Write down the vector equations of the tangent lines to the traces to at the point . Solution There really isn’t all that much to do with these other than plugging the values and function into the formulas above. We’ve already computed the derivatives and their values at in the previous example and the point on each trace is, Here is the equation of the tangent line to the trace for the plane . Here is the equation of the tangent line to the trace for the plane .
Partial Derivatives Previous Section Next Section Higher Order Partial Derivatives 3-Dimensional Space Previous Chapter Next Chapter Applications of Partial Derivatives
[Notes]
© 2003 - 2018 Paul Dawkins
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Minimize Z = 30x + 20y Subject to X + Y ≤ 8 X + 4 Y ≥ 12 5 X + 8 Y = 20 X , Y ≥ 0 - Mathematics
Minimize Z = 30x + 20y
Subject to
$x + y \leq 8$
$x + 4y \geq 12$
$5x + 8y = 20$
$x, y \geq 0$
Solution
First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 8, x + 4y = 12, x = 0 and y = 0
5x + 8y = 20 is already an equation.
Region represented by x + y ≤ 8:
The line x + y = 8 meets the coordinate axes at A(8, 0) and B(0, 8) respectively. By joining these points we obtain the line x + y = 8.Clearly (0,0) satisfies the inequation xy ≤ 8. So,the region in xy plane which contain the origin represents the solution set of the inequation x + y ≤ 8.
Region represented by x + 4y ≥ 12:
The line x + 4y = 12 meets the coordinate axes at C(12, 0) and D(0, 3) respectively. By joining these points we obtain the line x + 4y = 12.Clearly (0,0) satisfies the inequation x + 4y ≥ 12. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + 4y ≥ 12.
The line 5x + 8y = 20 is the line that passes through E(4, 0) and $F\left( 0, \frac{5}{2} \right)$ Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and ≥ 0.
The feasible region determined by the system of constraints, x + y ≤ 8, x + 4y ≥ 12, 5x+ 8y = 20, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are B(0,8), D(0,3),
$G\left( \frac{20}{3}, \frac{4}{3} \right)$
The values of Z at these corner points are as follows.
Corner point Z = 30x + 20y B(0,8) 160 D(0,3) 60 $G\left( \frac{20}{3}, \frac{4}{3} \right)$ 266.66
Therefore, the minimum value of Z is 60 at the point D(0,3). Hence, x = 0 and y =3 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 60
Concept: Graphical Method of Solving Linear Programming Problems
Is there an error in this question or solution?
APPEARS IN
RD Sharma Class 12 Maths
Chapter 30 Linear programming
Exercise 30.2 | Q 12 | Page 32
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Question
# If${{e}^{x+y}}=xy$, then show that $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-y\left( {{\left( x-1 \right)}^{2}}+{{\left( y-1 \right)}^{2}} \right)}{{{x}^{2}}{{\left( y-1 \right)}^{3}}}$.
Hint: Directly apply the differentiation to the given expression using the exponential differentiation, product and quotient rule of differentiation. Convert the first order derivative in terms of $'x'$ and $'y'$. Then proceed with finding the second order derivative and simplify it.
The given expression is ${{e}^{x+y}}=xy$
Differentiate the given expression with respect to $'x'$ , we get
$\dfrac{d}{dx}\left( {{e}^{x+y}} \right)=\dfrac{d}{dx}\left( xy \right)$
We know differentiation of exponential is, $\dfrac{d}{dx}\left( {{e}^{u}} \right)={{e}^{u}}.\dfrac{d}{dx}(u)$, so the above equation becomes,
$\Rightarrow {{e}^{x+y}}\dfrac{d}{dx}\left( x+y \right)=\dfrac{d}{dx}\left( xy \right)$
We know the product rule as, $\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$, applying this formula in the above equation, we get
${{e}^{x+y}}\left( 1+\dfrac{dy}{dx} \right)=x\dfrac{dy}{dx}+y\dfrac{d(x)}{dx}$
${{e}^{x+y}}\left( 1+\dfrac{dy}{dx} \right)=y+x\dfrac{dy}{dx}.$
From given expression we have${{e}^{x+y}}=xy$, putting this value in above equation, we get
$xy\left( 1+\dfrac{dy}{dx} \right)=y+x\dfrac{dy}{dx}$
$\Rightarrow xy+xy\dfrac{dy}{dx}=y+x\dfrac{dy}{dx}$
Bringing the like terms on one side, we get
$\Rightarrow xy\dfrac{dy}{dx}-x\dfrac{dy}{dx}=y-xy$
Taking out the common terms, we get
$\Rightarrow x\left( y-1 \right)\dfrac{dy}{dx}=y\left( 1-x \right)$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{y\left( 1-x \right)}{x\left( y-1 \right)}.........(i)$
Now we need to find the second order derivative, so we will differentiate the above equation with respect to $'x'$, we get
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( \dfrac{y\left( 1-x \right)}{x\left( y-1 \right)} \right)$
Now we know the quotient rule, i.e., $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}u-u\dfrac{d}{dx}v}{{{v}^{2}}}$, applying this formula in the above equation, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( x(y-1)\dfrac{dy}{dx}\left[ y\left( 1-x \right) \right] \right)-\left( y\left( 1-x \right)\dfrac{d}{dx}\left[ x(y-1) \right] \right)}{{{x}^{2}}{{\left( y-1 \right)}^{2}}}$
Now applying the product rule of differentiation, i.e., $\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{x(y-1)\left( y\dfrac{d}{dx}\left( 1-x \right)+(1-x)\dfrac{dy}{dx} \right)-y\left( 1-x \right)\left( x\dfrac{d}{dx}\left( y-1 \right)+(y-1)\dfrac{d(x)}{dx} \right)}{{{x}^{2}}{{\left( y-1 \right)}^{2}}}$
We know differentiation of constant term is zero, so solving the above equation, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{x(y-1)\left( y(-1)+(1-x)\dfrac{dy}{dx} \right)-y\left( 1-x \right)\left( x\dfrac{dy}{dx}+(y-1)(1) \right)}{{{x}^{2}}{{\left( y-1 \right)}^{2}}}$
Substituting the value $\dfrac{dy}{dx}=\dfrac{y\left( 1-x \right)}{x\left( y-1 \right)}$ from equation (i) in the above equation, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{x\left( y-1 \right)\left( \left( \dfrac{y\left( 1-x \right)}{x\left( y-1 \right)} \right)\left( 1-x \right)-y \right)-y\left( 1-x \right)\left( \left( y-1 \right)+x\left( \dfrac{y\left( 1-x \right)}{x\left( y-1 \right)} \right) \right)}{{{x}^{2}}{{\left( y-1 \right)}^{2}}}$
Solving the innermost brackets first, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{x\left( y-1 \right)\left( \dfrac{y{{\left( 1-x \right)}^{2}}-yx(y-1)}{x\left( y-1 \right)} \right)-y\left( 1-x \right)\left( \dfrac{{{(y-1)}^{2}}+y\left( 1-x \right)}{\left( y-1 \right)} \right)}{{{x}^{2}}{{\left( y-1 \right)}^{2}}}$
Cancelling the like terms, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( y-1 \right)\left( \dfrac{y{{\left( 1-x \right)}^{2}}-yx(y-1)}{\left( y-1 \right)} \right)-y\left( 1-x \right)\left( \dfrac{{{(y-1)}^{2}}+y\left( 1-x \right)}{\left( y-1 \right)} \right)}{{{x}^{2}}{{\left( y-1 \right)}^{2}}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( \dfrac{y\left( y-1 \right){{\left( 1-x \right)}^{2}}-yx{{(y-1)}^{2}}-y(1-x){{(y-1)}^{2}}-{{y}^{2}}{{\left( 1-x \right)}^{2}}}{\left( y-1 \right)} \right)}{{{x}^{2}}{{\left( y-1 \right)}^{2}}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{y\left( y-1 \right){{\left( 1-x \right)}^{2}}-yx{{(y-1)}^{2}}-y(1-x){{(y-1)}^{2}}-{{y}^{2}}{{\left( 1-x \right)}^{2}}}{{{x}^{2}}{{\left( y-1 \right)}^{3}}}$
Now taking $'y'$ common, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{y\left[ \left( y-1 \right){{\left( 1-x \right)}^{2}}-x{{(y-1)}^{2}}-(1-x){{(y-1)}^{2}}-y{{\left( 1-x \right)}^{2}} \right]}{{{x}^{2}}{{\left( y-1 \right)}^{3}}}$
Opening the two-two brackets, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{y\left\{ y{{\left( 1-x \right)}^{2}}-{{\left( 1-x \right)}^{2}}-x{{(y-1)}^{2}}-{{\left( y-1 \right)}^{2}}+x{{\left( y-1 \right)}^{2}}-y{{\left( 1-x \right)}^{2}} \right\}}{{{x}^{2}}{{\left( y-1 \right)}^{3}}}$
Cancelling the like terms, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{y\left\{ -{{\left( 1-x \right)}^{2}}-{{\left( y-1 \right)}^{2}} \right\}}{{{x}^{2}}{{\left( y-1 \right)}^{3}}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-y\left\{ {{\left( x-1 \right)}^{2}}+{{\left( y-1 \right)}^{2}} \right\}}{{{x}^{2}}{{\left( y-1 \right)}^{3}}}$
Hence proved
Note: Another way to solve this is first take log on both sides of the given expression, as shown below.
$\ln \left( {{e}^{x+y}} \right)=\ln \left( xy \right)$
$\Rightarrow xy\ln \left( e \right)=\ln \left( xy \right)$
$\Rightarrow xy=\ln \left( xy \right)$
Then perform the next steps.
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### Grade 8 - Mathematics1.4 Equations With Decimal Coefficients - II
Example: Solve the equation (0.6)x + 0.02 = x - 1.78. Solution: Writing decimals as common fractions, we have 6x/10 + 2/100 = x - 178/100 L.C.M. of denominators 10 and 100 is 100. \ Multiplying both sides of the equation by 100 60x + 2 = 100x - 178 Transposing 100x to L.H.S. and 2 to R.H.S. we have 60x - 100x = -178 - 2 -40x = -180. Dividing both sides by -40 we get -40x/-40 = -180/-40 x = 9/2 = 4.5 Verifications: Substituting x = 4.5 in the equation we get L.H.S. Þ (0.6)x + 0.02 = (0.6)(4.5) + 0.02 = 2.70 R.H.S. Þ x - 1.78 = 4.5 - 1.78 = 2.72 \ L.H.S. = R.H.S. \ x = 4.5 is the solution. Directions: Solve the following problems. Also write at least five equations with decimal coefficients of your own and solve them.
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### Grade 8 - Mathematics1.4 Equations With Decimal Coefficients - II
Q 1: Solve the equation (2.6)x - 0.02 = 3x - 5.6.13.514.913.95 Q 2: Solve the equation (3.6)x - 1.02 = 2x + 3.2.80112/78211/8086/112 Q 3: Solve the equation (1.2)x + 2.1 = 9x - 2.2.43/3578/4543/7816/17 Q 4: Solve the equation (2.4)x - 5.2 = 3x + 4.2.47/317/919/1443/12 Q 5: Solve the equation (1.9)x + 3.1 = 2x + 4.1.14/5-1011-18 Q 6: Solve the equation (2.1)x - 2.2 = x - 5.2.2/14-30/11-30/1915/8 Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!
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# factors of 72
The number 72 is a composite number, which means it has multiple factors. Factors are the numbers that can evenly divide a given number without leaving a remainder.
The factors of 72 are:
1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72.
Here’s a brief explanation of each factor:
1: Every number is divisible by 1.
2: The number 72 is divisible by 2 because it is an even number.
3: The sum of the digits in 72 is divisible by 3 (7 + 2 = 9), so 72 is divisible by 3.
4: The number 72 is divisible by 4 because it is divisible by 2 twice.
6: The number 72 is divisible by 6 because it is divisible by both 2 and 3.
8: The number 72 is divisible by 8 because it is divisible by 2 three times.
9: The sum of the digits in 72 is divisible by 9 (7 + 2 = 9), so 72 is divisible by 9.
12: The number 72 is divisible by 12 because it is divisible by both 2 and 3.
18: The number 72 is divisible by 18 because it is divisible by both 2 and 9.
24: The number 72 is divisible by 24 because it is divisible by both 2 and 12.
36: The number 72 is divisible by 36 because it is divisible by both 4 and 9.
72: Every number is divisible by itself.
These factors can be useful in various mathematical calculations, such as finding common divisors, simplifying fractions, or solving equations involving the number 72.
## Factors of definition 72
I apologize for any confusion. The term “factors of definition” is not clear. However, if you meant the factors of 72, as I mentioned in the previous response, the factors of 72 are:
1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72.
These numbers are the factors of 72 because they can evenly divide 72 without leaving a remainder.
#### How many Factors of 72
To determine the number of factors of 72, we can count the total number of distinct factors that evenly divide 72.
The prime factorization of 72 is: 2^3 * 3^2.
To find the total number of factors, we add 1 to the exponent of each prime factor in the prime factorization and multiply them together.
For 72, we have:
Exponent of 2: 3 + 1 = 4
Exponent of 3: 2 + 1 = 3
The total number of factors of 72 is calculated as (4) * (3) = 12.
Therefore, there are 12 factors of 72.
#### Factor Pairs of 72
The factor pairs of 72 are the pairs of numbers that, when multiplied together, result in the number 72. Here are all the factor pairs of 72:
1 × 72 = 72
2 × 36 = 72
3 × 24 = 72
4 × 18 = 72
6 × 12 = 72
8 × 9 = 72
These are the six factor pairs of 72.
#### All Factors of 72 using Multiplication
To find all the factors of 72 using multiplication, you can start by listing all the numbers that divide 72 without leaving a remainder, including both the divisors and the quotients. Here are all the factors of 72:
1 × 72 = 72
2 × 36 = 72
3 × 24 = 72
4 × 18 = 72
6 × 12 = 72
8 × 9 = 72
So, the factors of 72 using multiplication are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72.
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# Common Core: 5th Grade Math : Find the Volume of a Right Rectangular Prism: CCSS.Math.Content.5.MD.C.5a
## Example Questions
### Example Question #1 : Find The Volume Of A Right Rectangular Prism: Ccss.Math.Content.5.Md.C.5a
Rectangular prism A has a height of 5 and a base area of 15. Rectangular prism B has edge lengths of 5, 5, and 3. Rectangular prism C contains 75 unit cubes. Which of the following statements is TRUE?
Rectangular prism A has the smallest volume.
Rectangular prism A has the largest volume.
All three rectangular prisms have the same volume.
Rectangular prism B has the largest volume.
Rectangular prism C has the largest volume.
All three rectangular prisms have the same volume.
Explanation:
### Example Question #2 : Find The Volume Of A Right Rectangular Prism: Ccss.Math.Content.5.Md.C.5a
What is the volume of the object below?
Explanation:
This object has rows with cubes in each row. We can multiply to find the volume.
Remember, volume is always labeled as units to the third power.
### Example Question #3 : Find The Volume Of A Right Rectangular Prism: Ccss.Math.Content.5.Md.C.5a
What is the volume of the object below?
Explanation:
This object has rows with cubes in each row. We can multiply to find the volume.
Remember, volume is always labeled as units to the third power.
### Example Question #4 : Find The Volume Of A Right Rectangular Prism: Ccss.Math.Content.5.Md.C.5a
What is the volume of the object below?
Explanation:
This object has rows with cubes in each row. We can multiply to find the volume.
Remember, volume is always labeled as units to the third power.
### Example Question #5 : Find The Volume Of A Right Rectangular Prism: Ccss.Math.Content.5.Md.C.5a
What is the volume of the object below?
Explanation:
This object has rows with cubes in each row. We can multiply to find the volume.
Remember, volume is always labeled as units to the third power.
### Example Question #6 : Find The Volume Of A Right Rectangular Prism: Ccss.Math.Content.5.Md.C.5a
What is the volume of the object below?
Explanation:
This object has rows with cubes in each row. We can multiply to find the volume.
Remember, volume is always labeled as units to the third power.
### Example Question #7 : Find The Volume Of A Right Rectangular Prism: Ccss.Math.Content.5.Md.C.5a
What is the volume of the object below?
Explanation:
This object has rows with cubes in each row. We can multiply to find the volume.
Remember, volume is always labeled as units to the third power.
### Example Question #8 : Find The Volume Of A Right Rectangular Prism: Ccss.Math.Content.5.Md.C.5a
What is the volume of the object below?
Explanation:
This object has rows with cubes in each row. We can multiply to find the volume.
Remember, volume is always labeled as units to the third power.
### Example Question #9 : Find The Volume Of A Right Rectangular Prism: Ccss.Math.Content.5.Md.C.5a
What is the volume of the object below?
Explanation:
This object has rows with cubes in each row. We can multiply to find the volume.
Remember, volume is always labeled as units to the third power.
### Example Question #10 : Find The Volume Of A Right Rectangular Prism: Ccss.Math.Content.5.Md.C.5a
What is the volume of the object below?
Explanation:
This object has rows with cubes in each row. We can multiply to find the volume.
Remember, volume is always labeled as units to the third power.
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