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# What is the orthocenter of a triangle with corners at (3 ,1 ), (1 ,3 ), and (5 ,2 )#?
Oct 8, 2016
The orthocenter is at the point $\left(\frac{4}{3} , - \frac{17}{3}\right)$
#### Explanation:
Let's begin by writing the equation of the line that goes through point $\left(3 , 1\right)$ and perpendicular to the line going through points $\left(1 , 3\right)$ and $\left(5 , 2\right)$.
The slope, m, of the line going through points $\left(1 , 3\right)$ and $\left(5 , 2\right)$ is:
$m = \frac{3 - 2}{1 - 5} = - \frac{1}{4}$
The slope, n, of any line perpendicular is
$n = - \frac{1}{m} = 4$
Use point slope form of the equation of a line to obtain the equation of a line that we desire:
$y - {y}_{1} = n \left(x - {x}_{1}\right)$
$y - 1 = 4 \left(x - 3\right)$
$y - 1 = 4 x - 12$
$y = 4 x - 11$
Write the equation of a line that goes through point $\left(5 , 2\right)$ perpendicular to the line through the points $\left(3 , 1\right)$ and $\left(1 , 3\right)$:
The slope, m, of the line that goes through the points $\left(3 , 1\right)$ and $\left(1 , 3\right)$ is:
$m = \frac{3 - 1}{1 - 3} = \frac{2}{-} 2 = - 1$
The slope, n, of any line perpendicular is:
$n = - \frac{1}{m} = 1$
Again, Use the point slope form for the point $\left(5 , 2\right)$:
$y - {y}_{1} = n \left(x - {x}_{1}\right)$
$y - 2 = 1 \left(x - 5\right)$
$y = x - 7$
The orthocenter is at the intersection of these two lines:
$y = 4 x - 11$
$y = x - 7$
$x - 7 = 4 x - 11$
$4 = 3 x$
$x = \frac{4}{3}$
$y = \frac{4}{3} - 7$
$y = - \frac{17}{3}$
The orthocenter is at the point $\left(\frac{4}{3} , - \frac{17}{3}\right)$ |
Polar and pole of the ellipse
Polar and pole of the ellipse
If from a point A(x0, y0), exterior to the ellipse, drawn are tangents, then the secant line passing through the contact points, D1(x1, y1) and D2(x2, y2) is the polar of the point A. The point A is called the pole of the
polar, as shows the right figure.
Coordinates of the point A(x0, y0) must satisfy the equations of tangents, thus
t1 :: b2x0x1 + a2y0y1 = a2b2
t2 :: b2x0x2 + a2y0y2 = a2b2
and after subtracting t1 - t2
b2x0(x2 - x1) + a2y0(y2 - y1) = 0
obtained is
the slope of the secant line through points of contact D1and D2.
Thus, the equation of the secant line and after rearranging
b2x0x + a2y0y = b2x0x1 + a2y0y1, since b2x0x1 + a2y0y1 = a2b2
follows b2x0x + a2y0y = a2b2 the equation of the polar p of the point A(x0, y0).
Ellipse and line examples
Example: At a point A(-c, y > 0) where c denotes the focal distance, on the ellipse 16x2 + 25y2 = 1600 drawn is a tangent to the ellipse, find the area of the triangle that tangent forms by the coordinate axes.
Solution: Rewrite the equation of the ellipse to the standard form 16x2 + 25y2 = 1600 | ¸ 1600
We calculate the ordinate of the point A by plugging the abscissa into equation of the ellipse x = -6 => 16x2 + 25y2 = 1600, or, as we know that the point with the abscise x = - c has the ordinate equal to the value of the semi-latus rectum,
The area of the triangle formed by the tangent and the coordinate axes,
Example: Find a point on the ellipse x2 + 5y2 = 36 which is the closest, and which is the farthest from the line 6x + 5y - 25 = 0.
Solution: The tangency points of tangents to the ellipse which are parallel with the given line are, the
closest and the farthest points from the line.
Rewrite the equation of the ellipse to determine its axes,
Tangents and given line have the same slope, so Using the tangency condition, determine the intercepts c, therefore, the equations of tangents,
Solutions of the system of equations of tangents to the ellipse determine the points of contact, i.e., the
closest and the farthest point of the ellipse from the given line, thus
Example: Determine equation of the ellipse which the line -3x + 10y = 25 touches at the point P(-3, 8/5).
Solution: As the given line is the tangent to the ellipse, parameters, m and c of the line must satisfy the tangency condition, and the point P must satisfy the equations of the line and the ellipse, thus
Example: The line x + 14y - 25 = 0 is the polar of the ellipse x2 + 4y2 = 25. Find coordinates of the pole.
Solution: Intersections of the polar and the ellipse are points of contact of tangents drawn from the pole P to ellipse, thus solutions of the system of equations,
(1) x + 14y - 25 = 0 => x = 25 - 14y => (2) (2) x2 + 4y2 = 25 (25 - 14y)2 + 4y2 = 25 2y2 - 7y + 6 = 0, y1 = 3/2 and y2 = 2 y1 and y2 => x = 25 - 14y, x1 = 4 and x2 = -3. Thus, the points of contact D1(4, 3/2) and D2(-3, 2). The equations of the tangents at D1 and D2,
The solutions of the system of equations t1 and t2 are the coordinates of the pole P(1, 7/2).
Example: Find the equations of the common tangents of the curves 4x2 + 9y2 = 36 and x2 + y2 = 5.
Solution: The common tangents of the ellipse and the circle must satisfy the tangency conditions of these curves, thus
Pre-calculus contents H |
# What Is Half Of 13
What Is Half Of 13:
6.5 is the answer. Half of 13 would be 6.5. This is because 13 is an odd number and when you divide it by 2, the answer will always be a whole number with a remainder of 1. In this case, 6 is the whole number and .5 is the remainder. You can also think about it like this: 6 plus half of 13 is 7.5, so 6.5 must be half of 13.
It’s important to note that when working with halves, it’s sometimes easier to use fractions instead. For example, if you’re asked to find half of 8, you could say that 8 divided by 2 equals 4 with a remainder of 0. So 4 would be the whole number and .5 would be the fractional answer. This is because 8 is a even number and when you divide it by 2, the answer will always be a whole number with no remainder.
When it comes to halves, remember that if the number is even, the answer will be a whole number. If the number is odd, the answer will be a whole number plus a fractional part. Also, keep in mind that division by 2 always yields a result with a remainder of 1 or 0. As long as you remember these key concepts, you’ll be able to solve any half-problem!
### Why are fractions often easier than decimals when working with halves?
One reason why fractions can sometimes be easier than decimals when working with halves is that a fraction can be reduced to its lowest possible value while a decimal number cannot. This means it’s often easier to determine the whole number of a half-problem if you work with fractions instead of decimals.
Another reason why fractions are sometimes easier than decimals when working with halves is because dividing by 2 always yields an integer for an answer, whereas dividing by 2 can yield either an integer or decimal depending on the problem.
### What is the remainder when 4 is divided by 3?
There will be no remainder when four is divided by three, because four is evenly divisible (divisible without any remainder) by three; in other words, three goes into four zero times and leaves a remainder of zero. What is the remainder when 4 is divided by 7?
There will be a remainder of two when four is divided by seven, because four is not evenly divisible (divisible without any remainder) by seven; in other words, seven does not go into four two times and leave a remainder.
When working with halves, it is important to remember that division by two always yields a result with a remainder of 1 or 0. Keep this in mind when solving half-problems!
Additionally, if the number is even, the answer will be a whole number. If the number is odd, the answer will be a whole number plus a fractional part. Finally, fractions can sometimes be easier than decimals when working with halves.
### What is the remainder when 9 is divided by 5?
There will be a remainder of 4 when nine is divided by five, because nine is not evenly divisible (divisible without any remainder) by five; in other words, five does not go into nine four times and leave a remainder.
### What is half of 12?
Half of twelve would be six, because twelve divided by two would yield six as an answer with no fractional part left over. Remember that division resulting in an integer for the quotient always has a remainder of 1 or 0, and if it’s even (like this problem), then the whole number portion will be the answer. If it’s odd, like 15/2, then the whole number plus the fractional part (.5) will be the answer, making 15/2 equal to 7.5. |
1. ## Algebra
An electrical technician needs electrical tape and wire for his inventory. The tape costs $9 per package and the wire costs$6 per roll. Show the possible combinations of tape and wire he can purchase for less than $54. I have no clue how to solve this problem. 2. Originally Posted by Shinjiro An electrical technician needs electrical tape and wire for his inventory. The tape costs$9 per package and the wire costs $6 per roll. Show the possible combinations of tape and wire he can purchase for less than$54.
I have no clue how to solve this problem.
It's actually easier than you think.
First, you can set up an inequality:
x = tape
y = wire
$9x + 6y < 54$
Now, we can see that how much of one thing will affect how much of the other you can obtain, so, we can solve for each independently and see what we can come up with:
-First divide everything by 3:
$3x + 2y < 18$
Next, isolate either variable (I'm going to use y first).
$2y < 18 - 3x$
Divide by 2:
$y < 9 - \frac{3}{2}x$
This says that the number of packages of wire you can obtain is less than $9 - \frac{3}{2}x$, so it is dependent on how many packages of tape you get.
Let's go back to step one:
$3x + 2y < 18$
Now we isolate the other variable:
$3x < 18 - 2y$
Now we divide by 3:
$x < 6 - \frac{2}{3}y$
This tells you that the packages of tape must be less than $6 - \frac{2}{3}y$.
Therefore, all possible combinations of x and y are found in these two inequalities:
$x < 6 - \frac{2}{3}y$
$y < 9 - \frac{3}{2}x$
It all depends on what you buy first. |
# How do you calculate the power of an elevator motor
Contents
The total time of the trip can be calculated from the velocity of the elevator: t = = = 25 s. Thus the average power is given by: P = = = 3.9×104 Watts, or 39 kW.
## How do you calculate lifting power?
Let us calculate the work done in lifting an object of mass m through a height h. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh.
## How do you calculate the power of a machine?
Power is equal to work divided by time. In this example, P = 9000 J /60 s = 150 W . You can also use our power calculator to find work – simply insert the values of power and time.
## What type of motor is used in elevator lift?
Conventionally, there are, in general, three types of motors used in elevator systems: AC, DC and a hybrid between the two. The AC-2 motor is a primitive motor drive popular at least half a century ago for low-speed elevators. It is usually coupled with a worm gear to reduce speed and increase driving torque.
## How much power is required to lift a 30 N?
The power needed is 3 watts.
## What is the lift formula?
The lift equation states that lift L is equal to the lift coefficient Cl times the density r times half of the velocity V squared times the wing area A. For given air conditions, shape, and inclination of the object, we have to determine a value for Cl to determine the lift.
## What is the formula to calculate work done?
Work can be calculated with the equation: Work = Force × Distance. The SI unit for work is the joule (J), or Newton • meter (N • m). One joule equals the amount of work that is done when 1 N of force moves an object over a distance of 1 m.
## What is average power formula?
If the resistance is much larger than the reactance of the capacitor or inductor, the average power is a dc circuit equation of P=V2/R, where V replaces the rms voltage. An ac voltmeter attached across the terminals of a 45-Hz ac generator reads 7.07 V.
## What is the power output?
In physics, power is the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt, equal to one joule per second. … The output power of a motor is the product of the torque that the motor generates and the angular velocity of its output shaft.
## Which motor is used in fan?
In conventional ceiling fans, single phase induction motor is used. These motors consume minimum power and hence, are also known as fractional kilowatt motors. A single phase induction motor requires only one power phase for operating. It converts the electrical energy from the power input into mechanical energy.
IT IS INTERESTING: How can a universal motor be reversed
## How much weight can a 1 hp motor lift?
This means with 1 hp, you can lift about 550 lbs (250 kg) at a rate of 1 ft/sec, 1100 lbs at a rate of 0.5 ft/sec, 225 lb at a rate of 2 ft/sec, 1 lb at a rate of 550 ft/sec, and so on.
## What is the difference between an elevator and a lift?
Lifts are a design that conveys only one person. Lifts are the mechanisms that lift us to our desired level. An elevator is a type of vertical transport equipment that efficiently leads people or assets within a building’s floors. … The central contrast between Lift and Elevator is their usage.
## How much energy does it take to lift 1kg 1m?
On earth it takes about 10 Newton-meters (N-m) of energy to raise a 1 kilogram mass to a height of 1 meter. Since 1 N-m equals 1 Joule, that’s 10 Joules. If it takes 1 second to lift the weight 1 meter, than you have converted 10 Joules of energy to potential energy in one second. That’s 10 Watts of power.
## How much power is required to lift a?
If you lift a 2 kg weight 1 meter in 1 second, then the rate of energy conversion is 2 x 10 = 20 Joules per second, or 20 Watts of power. If you can keep lifting a 1 kg weight to a height of 1 meter every second, weight after weight after weight, then you will be steadily producing 10 Watts of power.
## How long will it take a 2.5 KW motor to lift a 80 kg block 30 m high?
Motor will take 9.4 s to lift block. Work done by motor to lift block is equal to change in gravitational potential energy of block.
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How To Teach Multiplication
Question
One of the tricks to learn the nine times table is
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Hi, I'm Donald Sinclair. I'm a science and math teacher with Greater London Tutors. Today, we are going to be looking at how to teach various topics. When teaching multiplication, it's always a good idea to tie it into the idea of addition, which the child should already be familiar with. Start off by saying that the times symbol simply means this many lots of. So, four times five means four lots of five. It can also be written as this: five plus five plus five plus five. Which the child should be able to add together and come together to get twenty. So, four times five equals twenty. Then you can ask things like “What happens if you multiply a number by one?” Well, if you multiply a number by one, that just means one lot of seven. So, if you multiply a number by one, it stays the same. Finally, what is a number times zero? Well, times zero means zero lots of eight, which means it's just zero. So, multiplying a number by zero is always equalled to zero. Still, the best way to learn multiplication is to learn the times tables. You can learn them off by heart, but some of them, you can teach simple tricks to remember the rest. For example, the two times tables are very simple, but you can remember it's just by counting and missing out every other number. So, starting at zero, you go missing the one you go to two, missing the three you go to four, missing the five you go to six. That way, it can be tiring of counting and make you learn very quickly. For the five times tables, you can teach that every number in the five times tables ends with either a zero or a five, as they go up in fives. Finally, the nine times tables, which a lot of people have problems with. There are two things to bear in mind with this. One, every number in the nine times table, the two digits will always add up to nine. It can also be learned from a simple trick in the fingers. If you hold up all ten fingers, to find, for example, three times nine, counting from left to right if you lower the third finger, then three times nine is going to be twenty seven. This can be done all the way up to nine times ten. It can be a very quick way to find out the value of the nine times table you're looking for. So these are some examples and tricks to help learn multiplication. . |
EASY
Math
# PROBLEM:
There are N children in the question playing hide and seek. In each round, one of them seeks and others hide.
You are given an array H of N numbers where H[i] represents the minimum number of rounds ith child have to hide to be satisfied. (1 based Indexing)
We aim to determine the least number of rounds that need to be played so that every child is satisfied.
# EXPLANATION:
Let’s say the number of rounds played is M. Now, the ith child can seek in at most (M - H[i]) rounds. ( 1 based Indexing) . And the summation of the number of rounds every child seeks has to be equal to M. So, we can have the following derivation based on upper observations.
\sum_{i = 1}^{N}(M - H_i) \geq M
(M*N - \sum_{i = 1}^{N} H_i ) \geq M
(M*N - \sum_{i = 1}^{N} H_i - M) \geq 0
(M*(N-1) - \sum_{i = 1}^{N} H_i ) \geq 0
(M*(N-1) - \sum_{i = 1}^{N} H_i ) \geq 0
M \geq \dfrac{\sum_{i = 1}^{N} H_i } { (N-1)}
Now, using ceiling function, we can have the minimum estimation of M, that is,
M_1 = ceil(\frac{ \sum_{i = 1}^{N} H_i }{N-1} )
But there has to be at least M2 rounds of play where,
M_2 = max(H_i) ; i \epsilon [1,n]
M = max(M_1,M_2)
# SOLUTIONS:
a.
N = 7
\sum_{i = 1}^{N} H_i = 65
M_1 = ceiling(65/6) = 11
M_2 = max(H[i]) = 15
M = max(M_1 , M_2) = 15
b.
N = 12
\sum_{i = 1}^{N} H_i = 100
M_1 = ceiling(100/11) = 10
M_2 = max(H[i]) = 9
M = max(M_1 , M_2) = 10
c.
N = 15
\sum_{i = 1}^{N} H_i = 2019
M_1 = ceiling(2019/14) = 145
M_2 = max(H[i]) = 140
M = max(M_1 , M_2) = 145
The formula should work for this also right ?
I dont understand why the formula is not working and we are again taking M2 and all
I understood everything else
Thanks a lot !!
2 Likes |
# Derivative of a Power Function
## Power Functions
A basic (non-transformed) power function has the form of a variable base x raised to a constant power n. In this activity we let n be a rational number. Therefore, n = p/q where p is an integer and q is a natural number. You should start by graphing several examples of power functions. Make sure that you can answer the following series of questions about the graphs of basic power functions before going on to examine their derivatives.
## Allways Present Quadrant
What quadrant is always present in the graph and what quadrant is always missing? Why?
## Common Point(s)
What point is on every graph of a basic power function, regardless of the power?
## First Quadrant Behavior
What does the size of the power, n, have to do with the shape of the graph of a basic power function in Quadrant I?
## Symmetry
What do the even or odd nature of p and q tell us about the graph of a basic power function? Consider what happens with a negative input to help answer this question.
## x = 0?
What happens for an input of 0?
## Derivatives
Now that you have a good feel for the properties of the graphs of basic power functions, manipulate the values of p and q to get the various cases of possible graph shapes for the original power function. For each of these try to figure out the appropriate shape of the derivative function, based upon your knowledge of the relationships between the graphs of a function and its derivative. When you think you have identified the basic shape of the derivative function, check your work by clicking on the checkbox to show the Derivative.
## Power Function Rule for Derivatives
The Power Function Rule for Derivatives is given above when you check the Derivative checkbox. To find the derivative of a power function, we simply bring down the original power as a coefficient and we subtract 1 from the power to get the new power. Therefore, the derivative of a power function is a constant times a basic power function. The graph of the derivative of a power function will also be the graph of a basic power function that has been modified by multiplying by a constant. Recall that multiplying a function by a constant results in a vertical stain (expansion |n|>1 or compression |n|<1 from the x-axis) and also a reflection over the x-axis if n is negative.
## Proof
One can use the definition of the derivative to prove this differentiation shortcut rule for integer powers of n. However, to prove the result in general requires that we first know the constant function rule, chain rule, exponential rule, logarithm rule, constant multiple rule, and sum rule, which we can apply to form a more general power/exponential rule. It turns out that the power function derivative rule works for any constant power. Note that this is for a variable base and a constant power. Do not confuse this with an exponential function, which has a constant base and variable exponent, or with a function with a variable base and a variable exponent. Be sure that you use derivative rules only where they apply.
## Warning
When I tried to use the derivative command in GeoGebra on these functions, it did not accurately graph the derivatives when x was negative in some cases. |
Content Blocks
## Learning Objectives
In this section, you will:
• Determine whether a relation represents a function.
• Find the value of a function.
• Determine whether a function is one-to-one.
• Use the vertical line test to identify functions.
• Graph the functions listed in the library of functions.
A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships.
## Determining Whether a Relation Represents a Function
A relation is a set of ordered pairs. The set consisting of the first components of each ordered pair is called the domain and the set consisting of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.
$$\{(1,\text{ }2),\text{ }(2,\text{ }4),\text{ }(3,\text{ }6),\text{ }(4,\text{ }8),\text{ }(5,\text{ }10)\}$$
The domain is $$\{1,\text{ }2,\text{ }3,\text{ }4,\text{ }5\}$$. The range is $$\{2,\text{ }4,\text{ }6,\text{ }8,\text{ }10\}$$.
Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter $$x$$. Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter $$y$$.
A function $$f$$ is a relation that assigns a single value in the range to each value in the domain. In other words, no x-values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, $$\{1, 2, 3, 4, 5\}$$, is paired with exactly one element in the range, $$\{2, 4, 6, 8, 10\}$$.
Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as
$$\{{(\text{odd},\text{ }1),\text{ }(\text{even},\text{ }2),\text{ }(\text{odd},\text{ }3),\text{ }(\text{even},\text{ }4),\text{ }(\text{odd},\text{ }5)}\}$$
Notice that each element in the domain, $$\{even, odd\}$$ is not paired with exactly one element in the range, $$\{1, 2, 3, 4, 5\}$$. For example, the term “odd” corresponds to three values from the range, $$\{1, 3, 5\}$$ and the term “even” corresponds to two values from the range, $$\{2, 4\}$$. This violates the definition of a function, so this relation is not a function.
Figure 1 compares relations that are functions and not functions.
## Functions
function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.”
The input values make up the domain, and the output values make up the range.
### Examples
Given a relationship between two quantities, determine whether the relationship is a function.
## Using Function Notation
Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions.
To represent “height is a function of age,” we start by identifying the descriptive variables $$h$$ for height and $$a$$ for age. The letters $$f$$, $$g$$, and $$h$$ are often used to represent functions just as we use $$x, y$$, and $$z$$ to represent numbers and $$A, B$$, and $$C$$ to represent sets.
$$\begin{array}{lcccc}h\text{ is }f\text{ of }a&&&&\text{We name the function }f;\text{ height is a function of age}.\\h=f(a)&&&&\text{We use parentheses to indicate the function input}\text{. }\\f(a)&&&&\text{We name the function }f;\text{ the expression is read as “}f\text{ of }a\text{.”}\end{array}$$
Remember, we can use any letter to name the function; the notation $$h(a)$$ shows us that $$h$$ depends on $$a$$. The value a a must be put into the function $$h$$ to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication.
We can also give an algebraic expression as the input to a function. For example $$f(a+b)$$ means "first add a and b, and the result is the input for the function f." The operations must be performed in this order to obtain the correct result.
### Function Notation
The notation $$y=f(x)$$ defines a function named $$f$$. This is read as "$$y is a function of x$$". The letter $$x$$ represents the input value, or independent variable. The letter $$y$$, or $$f(x)$$, represents the output value, or dependent variable.
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Content Blocks
## Learning Objectives
In this section, you will:
• Determine whether a relation represents a function.
• Find the value of a function.
• Determine whether a function is one-to-one.
• Use the vertical line test to identify functions.
• Graph the functions listed in the library of functions.
A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships.
## Determining Whether a Relation Represents a Function
A relation is a set of ordered pairs. The set consisting of the first components of each ordered pair is called the domain and the set consisting of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.
$$\{(1,\text{ }2),\text{ }(2,\text{ }4),\text{ }(3,\text{ }6),\text{ }(4,\text{ }8),\text{ }(5,\text{ }10)\}$$
The domain is $$\{1,\text{ }2,\text{ }3,\text{ }4,\text{ }5\}$$. The range is $$\{2,\text{ }4,\text{ }6,\text{ }8,\text{ }10\}$$.
Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter $$x$$. Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter $$y$$.
A function $$f$$ is a relation that assigns a single value in the range to each value in the domain. In other words, no x-values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, $$\{1, 2, 3, 4, 5\}$$, is paired with exactly one element in the range, $$\{2, 4, 6, 8, 10\}$$.
Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as
$$\{{(\text{odd},\text{ }1),\text{ }(\text{even},\text{ }2),\text{ }(\text{odd},\text{ }3),\text{ }(\text{even},\text{ }4),\text{ }(\text{odd},\text{ }5)}\}$$
Notice that each element in the domain, $$\{even, odd\}$$ is not paired with exactly one element in the range, $$\{1, 2, 3, 4, 5\}$$. For example, the term “odd” corresponds to three values from the range, $$\{1, 3, 5\}$$ and the term “even” corresponds to two values from the range, $$\{2, 4\}$$. This violates the definition of a function, so this relation is not a function.
Figure 1 compares relations that are functions and not functions.
## Functions
function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.”
The input values make up the domain, and the output values make up the range.
### Examples
Given a relationship between two quantities, determine whether the relationship is a function.
## Using Function Notation
Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions.
To represent “height is a function of age,” we start by identifying the descriptive variables $$h$$ for height and $$a$$ for age. The letters $$f$$, $$g$$, and $$h$$ are often used to represent functions just as we use $$x, y$$, and $$z$$ to represent numbers and $$A, B$$, and $$C$$ to represent sets.
$$\begin{array}{lcccc}h\text{ is }f\text{ of }a&&&&\text{We name the function }f;\text{ height is a function of age}.\\h=f(a)&&&&\text{We use parentheses to indicate the function input}\text{. }\\f(a)&&&&\text{We name the function }f;\text{ the expression is read as “}f\text{ of }a\text{.”}\end{array}$$
Remember, we can use any letter to name the function; the notation $$h(a)$$ shows us that $$h$$ depends on $$a$$. The value a a must be put into the function $$h$$ to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication.
We can also give an algebraic expression as the input to a function. For example $$f(a+b)$$ means "first add a and b, and the result is the input for the function f." The operations must be performed in this order to obtain the correct result.
### Function Notation
The notation $$y=f(x)$$ defines a function named $$f$$. This is read as "$$y is a function of x$$". The letter $$x$$ represents the input value, or independent variable. The letter $$y$$, or $$f(x)$$, represents the output value, or dependent variable.
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## Learning Objectives
In this section, you will:
• Determine whether a relation represents a function.
• Find the value of a function.
• Determine whether a function is one-to-one.
• Use the vertical line test to identify functions.
• Graph the functions listed in the library of functions.
A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships.
## Determining Whether a Relation Represents a Function
A relation is a set of ordered pairs. The set consisting of the first components of each ordered pair is called the domain and the set consisting of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.
$$\{(1,\text{ }2),\text{ }(2,\text{ }4),\text{ }(3,\text{ }6),\text{ }(4,\text{ }8),\text{ }(5,\text{ }10)\}$$
The domain is $$\{1,\text{ }2,\text{ }3,\text{ }4,\text{ }5\}$$. The range is $$\{2,\text{ }4,\text{ }6,\text{ }8,\text{ }10\}$$.
Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter $$x$$. Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter $$y$$.
A function $$f$$ is a relation that assigns a single value in the range to each value in the domain. In other words, no x-values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, $$\{1, 2, 3, 4, 5\}$$, is paired with exactly one element in the range, $$\{2, 4, 6, 8, 10\}$$.
Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as
$$\{{(\text{odd},\text{ }1),\text{ }(\text{even},\text{ }2),\text{ }(\text{odd},\text{ }3),\text{ }(\text{even},\text{ }4),\text{ }(\text{odd},\text{ }5)}\}$$
Notice that each element in the domain, $$\{even, odd\}$$ is not paired with exactly one element in the range, $$\{1, 2, 3, 4, 5\}$$. For example, the term “odd” corresponds to three values from the range, $$\{1, 3, 5\}$$ and the term “even” corresponds to two values from the range, $$\{2, 4\}$$. This violates the definition of a function, so this relation is not a function.
Figure 1 compares relations that are functions and not functions.
## Functions
function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.”
The input values make up the domain, and the output values make up the range.
### Examples
Given a relationship between two quantities, determine whether the relationship is a function.
## Using Function Notation
Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions.
To represent “height is a function of age,” we start by identifying the descriptive variables $$h$$ for height and $$a$$ for age. The letters $$f$$, $$g$$, and $$h$$ are often used to represent functions just as we use $$x, y$$, and $$z$$ to represent numbers and $$A, B$$, and $$C$$ to represent sets.
$$\begin{array}{lcccc}h\text{ is }f\text{ of }a&&&&\text{We name the function }f;\text{ height is a function of age}.\\h=f(a)&&&&\text{We use parentheses to indicate the function input}\text{. }\\f(a)&&&&\text{We name the function }f;\text{ the expression is read as “}f\text{ of }a\text{.”}\end{array}$$
Remember, we can use any letter to name the function; the notation $$h(a)$$ shows us that $$h$$ depends on $$a$$. The value a a must be put into the function $$h$$ to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication.
We can also give an algebraic expression as the input to a function. For example $$f(a+b)$$ means "first add a and b, and the result is the input for the function f." The operations must be performed in this order to obtain the correct result.
### Function Notation
The notation $$y=f(x)$$ defines a function named $$f$$. This is read as "$$y is a function of x$$". The letter $$x$$ represents the input value, or independent variable. The letter $$y$$, or $$f(x)$$, represents the output value, or dependent variable.
## Practice Quiz
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## Learning Objectives
In this section, you will:
• Determine whether a relation represents a function.
• Find the value of a function.
• Determine whether a function is one-to-one.
• Use the vertical line test to identify functions.
• Graph the functions listed in the library of functions.
A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships.
## Determining Whether a Relation Represents a Function
A relation is a set of ordered pairs. The set consisting of the first components of each ordered pair is called the domain and the set consisting of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.
$$\{(1,\text{ }2),\text{ }(2,\text{ }4),\text{ }(3,\text{ }6),\text{ }(4,\text{ }8),\text{ }(5,\text{ }10)\}$$
The domain is $$\{1,\text{ }2,\text{ }3,\text{ }4,\text{ }5\}$$. The range is $$\{2,\text{ }4,\text{ }6,\text{ }8,\text{ }10\}$$.
Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter $$x$$. Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter $$y$$.
A function $$f$$ is a relation that assigns a single value in the range to each value in the domain. In other words, no x-values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, $$\{1, 2, 3, 4, 5\}$$, is paired with exactly one element in the range, $$\{2, 4, 6, 8, 10\}$$.
Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as
$$\{{(\text{odd},\text{ }1),\text{ }(\text{even},\text{ }2),\text{ }(\text{odd},\text{ }3),\text{ }(\text{even},\text{ }4),\text{ }(\text{odd},\text{ }5)}\}$$
Notice that each element in the domain, $$\{even, odd\}$$ is not paired with exactly one element in the range, $$\{1, 2, 3, 4, 5\}$$. For example, the term “odd” corresponds to three values from the range, $$\{1, 3, 5\}$$ and the term “even” corresponds to two values from the range, $$\{2, 4\}$$. This violates the definition of a function, so this relation is not a function.
Figure 1 compares relations that are functions and not functions.
## Functions
function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.”
The input values make up the domain, and the output values make up the range.
### Examples
Given a relationship between two quantities, determine whether the relationship is a function.
## Using Function Notation
Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions.
To represent “height is a function of age,” we start by identifying the descriptive variables $$h$$ for height and $$a$$ for age. The letters $$f$$, $$g$$, and $$h$$ are often used to represent functions just as we use $$x, y$$, and $$z$$ to represent numbers and $$A, B$$, and $$C$$ to represent sets.
$$\begin{array}{lcccc}h\text{ is }f\text{ of }a&&&&\text{We name the function }f;\text{ height is a function of age}.\\h=f(a)&&&&\text{We use parentheses to indicate the function input}\text{. }\\f(a)&&&&\text{We name the function }f;\text{ the expression is read as “}f\text{ of }a\text{.”}\end{array}$$
Remember, we can use any letter to name the function; the notation $$h(a)$$ shows us that $$h$$ depends on $$a$$. The value a a must be put into the function $$h$$ to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication.
We can also give an algebraic expression as the input to a function. For example $$f(a+b)$$ means "first add a and b, and the result is the input for the function f." The operations must be performed in this order to obtain the correct result.
### Function Notation
The notation $$y=f(x)$$ defines a function named $$f$$. This is read as "$$y is a function of x$$". The letter $$x$$ represents the input value, or independent variable. The letter $$y$$, or $$f(x)$$, represents the output value, or dependent variable.
## Practice Quiz
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## Learning Objectives
In this section, you will:
• Determine whether a relation represents a function.
• Find the value of a function.
• Determine whether a function is one-to-one.
• Use the vertical line test to identify functions.
• Graph the functions listed in the library of functions.
A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships.
## Determining Whether a Relation Represents a Function
A relation is a set of ordered pairs. The set consisting of the first components of each ordered pair is called the domain and the set consisting of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.
$$\{(1,\text{ }2),\text{ }(2,\text{ }4),\text{ }(3,\text{ }6),\text{ }(4,\text{ }8),\text{ }(5,\text{ }10)\}$$
The domain is $$\{1,\text{ }2,\text{ }3,\text{ }4,\text{ }5\}$$. The range is $$\{2,\text{ }4,\text{ }6,\text{ }8,\text{ }10\}$$.
Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter $$x$$. Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter $$y$$.
A function $$f$$ is a relation that assigns a single value in the range to each value in the domain. In other words, no x-values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, $$\{1, 2, 3, 4, 5\}$$, is paired with exactly one element in the range, $$\{2, 4, 6, 8, 10\}$$.
Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as
$$\{{(\text{odd},\text{ }1),\text{ }(\text{even},\text{ }2),\text{ }(\text{odd},\text{ }3),\text{ }(\text{even},\text{ }4),\text{ }(\text{odd},\text{ }5)}\}$$
Notice that each element in the domain, $$\{even, odd\}$$ is not paired with exactly one element in the range, $$\{1, 2, 3, 4, 5\}$$. For example, the term “odd” corresponds to three values from the range, $$\{1, 3, 5\}$$ and the term “even” corresponds to two values from the range, $$\{2, 4\}$$. This violates the definition of a function, so this relation is not a function.
Figure 1 compares relations that are functions and not functions.
## Functions
function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.”
The input values make up the domain, and the output values make up the range.
### Examples
Given a relationship between two quantities, determine whether the relationship is a function.
## Using Function Notation
Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions.
To represent “height is a function of age,” we start by identifying the descriptive variables $$h$$ for height and $$a$$ for age. The letters $$f$$, $$g$$, and $$h$$ are often used to represent functions just as we use $$x, y$$, and $$z$$ to represent numbers and $$A, B$$, and $$C$$ to represent sets.
$$\begin{array}{lcccc}h\text{ is }f\text{ of }a&&&&\text{We name the function }f;\text{ height is a function of age}.\\h=f(a)&&&&\text{We use parentheses to indicate the function input}\text{. }\\f(a)&&&&\text{We name the function }f;\text{ the expression is read as “}f\text{ of }a\text{.”}\end{array}$$
Remember, we can use any letter to name the function; the notation $$h(a)$$ shows us that $$h$$ depends on $$a$$. The value a a must be put into the function $$h$$ to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication.
We can also give an algebraic expression as the input to a function. For example $$f(a+b)$$ means "first add a and b, and the result is the input for the function f." The operations must be performed in this order to obtain the correct result.
### Function Notation
The notation $$y=f(x)$$ defines a function named $$f$$. This is read as "$$y is a function of x$$". The letter $$x$$ represents the input value, or independent variable. The letter $$y$$, or $$f(x)$$, represents the output value, or dependent variable.
## Practice Quiz
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### Problems: Continuous Probability Distributions
1. There are very few continuous distributions that we can do much with without a lot of math, but the simple uniform distribution is one of them. A uniform distribution is one that is just that--uniform. If we imagine any possible number between 0 and 2 as equally likely, we have an infinite possible number of numbers, so the probability of any particular number is close to zero. Instead of focusing on particular numbers, we have to focus on intervals.
The probability that some number between zero and two will come up will be one. To find the probability of any interval, we find the length of the interval and multiply by .5. Hence, the probability of getting a number between zero and one is .5.
For the following, draw the picture and compute the probabilities. What is:
a) P(0<x<.8)
b) P(x>.8)
c) P(.5<x<1.5)
d) P(.5<x<1.2) or P(.8<x<1.3)
e) compute the probability that we will not be in the interval of .4 to 1.5.
f) compute the middle interval in which we will be 90% of the time.
g) compute the tails (outside intervals) that we will be in 5% of the time
2. The picture below shows a continuous probability distribution with the formula, y=.5x for the range, x=0 to x=2. In this case, area is probability.
To compute the probability of being between 0 and 1, for example, you need to know y when x equals 1. The formula says it is .5. Hence, to find the probability that x is between 0 and 1 Prob(0<x<1), we compute the area of the triangle with a length of 1 and a height of .5. The answer is .25 because .5HW = .5*.5*1.
a) Show that the area under the line is equal to 1.
b) Find the following probabilities:
Prob(0<x<.5)
Prob(0<x<1.5)
Prob(1.5<x<2)
Prob(.5<x<1.5)
Prob(x<.5 or x>1.5)
c) Suppose that we draw twice from this distribution and that probability of the second draw is independent of the first. What is the probability of getting greater than 1.5 on both draws? What is the probability of getting less than 1.5 on both draws? What is the probability of getting a result greater than 1.5 on one draw and less than 1.5 on the other draw?
3. Here is another continuous probability distribution for which we can work the problems without using complex math or tables. The range of possible outcomes is from zero to 1. The probability density function is y=2x. The probability histogram looks like this:
If you remember geometry, you can find the area of any interval using the formula for a triangle, which is .5(H*W) (one half height multiplied by width). We can find the height at any spot with the formula y=2x, where y is the height and x is the distance out on the x-axis. The total area under the triangle is 1, which it must be.
Find the following probabilities:
P(0<.5)
P(.5>x>1)
P(.2<x<.7)
Find the interval so there is a 5% chance of being below the interval and a 5% chance of being above it.
4. A uniform distribution that has a range from 0 to 4 and that is continuous would look like the picture below. Probability is area.
Find:
(0 < P(x) < 4) = ________
(P(x) > 1) = ________
(P(x) = 1) = ________
(P(x) < 1) = _________
What is the expected value of x? (You cannot do the math, but you should be able to see it.) |
## To what volume will a sample of gas expand if it is heated from 50.0°c and 2.33 l to 500.0°c?
Question
To what volume will a sample of gas expand if it is heated from 50.0°c and 2.33 l to 500.0°c?
in progress 0
1. The volume of the gas at [tex]500^{circ} text{C}[/tex] is [tex]fbox{begin 5.576,{text{L}}end{minispace}}[/tex].
Further Explanation:
Consider the pressure of the gas to be constant.
The change in the volume of the gas when the temperature of the gas is varied by keeping the pressure of the gas at a constant value is defined by the Charles’ Law.
Concept:
According to the Charles law, the volume of the gas is directly proportional to the temperature of the gas at constant pressure.
The Charles’ law can be stated as:
[tex]fbox{beginVpropto Tend{minispace}}[/tex]
The above expression can we written as.
[tex]dfrac{V_1}{T_1}=dfrac{V_2}{T_2}[/tex]
Convert the temperature of the gas into kelvin.
[tex]T=273+T^circtext{C}}}[/tex]
Here, T is the temperature in kelvin and [tex]T^circtext{C}}}[/tex] is the temperature in centigrade.
The initial temperature of the gas is [tex]50^circtext{C}[/tex]. The temperature of the gas in kelvin is.
[tex]begin{aligned}{T_1}&=273+50&=323,{text{K}}end{aligned}[/tex]
The final temperature of the gas is [tex]500^circtext{C}[/tex] . The temperature in kelvin is.
[tex]begin{aligned}{T_2}&=273+500&=773,{text{K}}end{aligned}[/tex]
Substitute the values of temperature and volume in the expression of the Charles’ Law.
[tex]begin{aligned}{V_2}&=frac{{{T_2}}}{{{T_1}}}{V_1}&=frac{{773,{text{K}}}}{{323,{text{K}}}}left({2.33,{text{L}}}right)&=5.576,{text{L}}end{aligned}[/tex]
Thus, the volume of the gas at [tex]500^circtext{C}}[/tex] will be [tex]fbox{begin 5.576,{text{L}}end{minispace}}[/tex].
1. Examples of wind and solar energy https://brainly.com/question/1062501
2. Stress developed in a wire https://brainly.com/question/12985068 |
HANDS-ON ACTIVITY 7.1: APPROXIMATING AREA WITH RIEMANN SUMS - Integration - AP CALCULUS AB & BC REVIEW - Master AP Calculus AB & BC
## Master AP Calculus AB & BC
Part II. AP CALCULUS AB & BC REVIEW
CHAPTER 7. Integration
HANDS-ON ACTIVITY 7.1: APPROXIMATING AREA WITH RIEMANN SUMS
There is an essential calculus tie between the integral and the area that is captured beneath a graph. You may, in fact, already know what it is. If you do, well, pin a rose on your nose. Those of you who don’t know will be kept in the dark for a couple of sections so that some suspense will build (I am nothing if not a showman). For now, let’s focus on using archaic and simplistic means to estimating the area “under” a curve. (The means are so simplistic that some students are actually disappointed. “This is calculus?” they ask, brows furrowed and tears forming in the corners of their eyes. My advice: embrace the easy. Just because a lot of calculus is tricky, not all of it has to be.) By the way, you may use your calculator freely on this activity.
1. Draw the graph of on the axes below.
2. We are going to approximate the area between f and the x-axis from x = 0 to x = 4 using rectangles (the method of Riemann sums). This is not the entire area in the first quadrant, just most of it. Draw four inscribed rectangles of width 1 on the interval [0,4] on your graph above.
3. What are the heights of each of the four rectangles? What is the total area of the rectangles? This area, although not the same as the area beneath the curve, is an approximation for that area called the lower sum.
4. The actual area between f and the x-axis on the interval [0,4] is 28/3. Why is one area greater?
________________________________________________________
________________________________________________________
NOTE. We are not really finding the area “under” a curve. For most graphs, there is infinite area under the curve. Instead, we will be calculating the area between the given curve and the x-axis. Keep that in the back of your mind.
5. How could you get a better approximation for the area beneath the curve if you still used inscribed rectangles?
________________________________________________________
________________________________________________________
6. On the axes below, graph/again. This time, draw four rectangles of width 1 that circumscribe the graph. Use the area of the circumscribed rectangles to approximate the area beneath the curve. This approximation is called the upper sum.
7. Compare the approximation you got in number 6 to the actual area.
________________________________________________________
________________________________________________________
NOTE. Consider only the endpoints’ heights when drawing the inscribed and circumscribed rectangles because you can compare their heights very easily. Choose the lower of the two heights for the inscribed and the larger of the two heights to draw circumscribed rectangles.
8. On the axes below, graph/again, and this time draw four rectangles of width one such that the height of each rectangle is given by the midpoint of each interval. The approximate area is called the midpoint sum. 9
9. Now you have found an inscribed sum, a circumscribed sum, and a midpoint sum. Explain what is likely meant by each of the remaining sums, and draw a sample rectangle on [a,b] that you would use with the technique to approximate the area beneath g.
10. Use 5 rectangles of equal width and the technique of midpoint sums to approximate the area beneath the curve h(x) = x3 — 2x2 — 5x + 7 on [0,4].
Hint: To figure out the width, ∆x, of n rectangles on the interval [a,b], use the formula
SELECTED SOLUTIONS TO HANDS-ON ACTIVITY 7.1
NOTE. Every rectangle below the x-axis counts as negative area.
2. The rectangles cannot cross the graph since they are inscribed. Thus, look at the endpoints of each interval, and choose the lower of the two endpoints’ heights. That will be the height of the inscribed rectangle for that interval.
3. The heights of the four rectangles are f(1) = 2.9, f(2) = 2.6, f(3) = 2.1, and f(4) = 1.4. The area of each rectangle is height ∙ width, and each of the widths is 1. Therefore, the total combined area is 2.9 ∙ 1 + 2.6 ∙ 1 + 2.1 ∙ 1 + 1.4 ∙ 1 = 9.
4. The actual area is Our area approximation is less than the actual area because it excludes little slivers of area between the curve and the rectangles. The most area is omitted on the interval [3,4]. Thus, the approximation is lower, hence the term lower sums.
5. If you used more rectangles, the approximation would be much better. In fact, the more rectangles you used, the less space would be omitted and the closer the approximation.
6. In order to draw rectangles that circumscribe the graph, look at each interval separately and choose the higher endpoint height—it will give the height for that rectangle.
The rectangles’ total area will be 3 ∙ 1 + 2.9 ∙ 1 + 2.6 ∙ 1 + 2.1 ∙ 1 = 10.6
7. This area is too large, which was expected (upper sum). The rectangles contain more area than the curve. In fact, the error in the circumscribed rectangle method was greater than the error in the inscribed rectangle method.
8. In this case, the heights of the rectangles will be given by the function values of the midpoints of the following intervals: and
The widths of the rectangles are still 1, so the total Riemann midpoint sum is 1 ∙ 2.975 + 1 ∙ 2.775 + 1 ∙ 2.375 + 1 ∙ 1.775 = 9.9.
TIP. The right-hand sum is not always the same as the lower sum and vice versa. That is only true if the graph is monotonic decreasing, as g is in problem 9.
9. The rectangle in a right-hand sum has the same height as the function at the right-hand endpoint of each interval. Similarly, a left-hand sum rectangle has the height of the function at the left-hand endpoint on each interval.
10. According to the given formula, each rectangle will have width as pictured in the below graph.
TIP. Because the width of each interval is 4/5 in problem 10, you can factor that value out to simplify your arithmetic.
The five intervals in this graph are and Because you’re doing midpoint sums, you should draw rectangles whose heights are given by the function values of the midpoints of those intervals: and The total area of these rectangles (and it’s no sin to use your calculator to help you out here) is
EXERCISE 2
Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.
YOU MAY USE YOUR GRAPHING CALCULATOR FOR ALL OF THESE PROBLEMS.
1. What type of Riemann sum is being applied in each of the following diagrams? If there is more than one correct answer, give both.
2. Approximate the area bounded by f(x) = sin x and the x-axis on the interval [0,π] using 4 rectangles and upper sums.
3. If g(x) is a continuous function that contains the values in the following table,
x 0 1 2 3 4 5 6 7 8 g(x) 2.1 3.6 5.1 6.3 6.9 7.2 7.3 4 3.2
approximate the area bounded by g(x) and the x-axis on [0,8] using
(a) 8 rectangles and right-hand sums
(b) 4 rectangles and midpoint sums
1. (a) Upper sums, circumscribed rectangles: On the first two rectangles, the right- hand endpoint is used to determine height, whereas the left-hand endpoint is being used for the third and fourth rectangles. Thus, it cannot be right- or left-hand sums.
(b) Lower sums, right-hand sums, inscribed rectangles: All these descriptions apply to this diagram since the right-hand endpoint of each interval forms the inscribed rectangles.
(c) Lower sums, inscribed rectangles: The lower of the two endpoints’ heights is chosen each time, not the right- or left-hand endpoint on a consistent basis.
(d) Midpoint sums: That one’s pretty clear from the diagram. The function value at each interval midpoint dictates the height of the rectangle there.
2. (a) If 4 rectangles are used, the width of each interval will be Therefore, the intervals will be and Because upper sums (circumscribed rectangles) are specified, the heights of the rectangles, from left to right, will be as these are the higher of the two endpoint function values for each interval.
The upper sum is
NOTE. g(0) is not used in 3(a) because it is not the right-hand endpoint of any interval.
3. (a) If 8 rectangles are used, the width of each will be 1. Because right-hand sums are specified, the function value at the right endpoint of each interval dictates the height. Thus, the right-hand sum will be
1 ∙ (g(1) + g(2) + g(3) + g(4) + g(5) + g(6) + g(7) + g(8))
3.6 + 5.1 + 6.3 + 6.9 + 7.2 + 7.3 + 4.0 + 3.2 = 43.6
(b) E ach rectangle will have width ∆x = 2, so the intervals are (0,2), (2,4), (4,6), and (6,8). The midpoints of these intervals are simple, and the heights come from their function values, so the midpoint sum is
2 ∙ (g(1) + g(3) + g(5) + g(7))
2(3.6 + 6.3 + 7.2 + 4.0) = 42.2
|
# Evaluate the limit of the function ln(1+x)/(sinx+sin3x) x-->0
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
We have to find the value of lim x--> 0 [ ln(1+x)/(sinx+sin3x)]
substituting x = 0, we get the indeterminate form 0/0. Therefore we can use l'Hopital's Rule and substitute the numerator and denominator with their derivatives.
=> lim x--> 0 [ (1/(1+x))/(cos x + 3*cos 3x)]
Substitute x = 0
[ (1/(1+x))/(cos x + 3*cos 3x)]
=> (1 / 1) / ( 1 + 3)
=> 1/4
The required value of lim x--> 0 [ln(1+x)/(sinx+sin3x)] = (1/4)
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
First, we'll substitute x by 0 into the function:
lim ln(1+x)/(sinx+sin3x) = ln 1/(sin 0 + sin 0) = 0/0
We've get an indetermination.
Since the trigonometric functions from denominator are matching, we'll transform the sum into a product:
sinx+sin3x = 2[sin (x+3x)/2][cos(x-3x)/2]
sinx+sin3x = 2sin 2x*cos(-x)
Since the cosine function is even, we'll put cos(-x) = cos x
sinx+sin3x = 2sin 2x*cos x
We'll re-write the limit:
lim ln(1+x)/(sinx+sin3x) = lim ln(1+x)/2sin 2x*cos x
We'll re-arrange the terms, crating remarcable limits:
(1/2)*lim ln(1+x)/sin 2x*lim (1/cos x)
But lim (1/cos x) = 1/cos 0 = 1
(1/2)*lim ln(1+x)/sin 2x= (1/2)lim [ln(1+x)/x]*(2x/sin 2x)*(x/2x)
(1/2)lim [ln(1+x)/x]*(2x/sin 2x)*(x/2x) = (1/2)*1*1*lim (x/2x)
(1/2)lim (x/2x) = 1/4
For x->0, lim ln(1+x)/(sinx+sin3x) = 1/4. |
Class 8 Maths Chapter 2 Linear Equations in One Variable MCQs (With Answers) - GMS - Learning Simply
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# Class 8 Maths Chapter 2 Linear Equations in One Variable MCQs (With Answers)
Linear Equations in One Variable MCQs
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# Class 8 Maths Chapter 2 Linear Equations in One Variable MCQs
Class 8 Maths Chapter 2 (Linear Equations in One Variable) MCQs are available online here for students. These objective questions are designed as per CBSE board syllabus and NCERT guidelines. The chapter-wise questions are provided here for students to make them understand each concept and help to score better marks in exams.
## MCQs on Class 8 Linear Equations in One Variable
Multiple choice questions (MCQs) are available for Class 8 Linear Equations in One Variable with each problem consisting of four options, out of which one is the correct answer. Students have to solve the problem and select the correct answer. Verify your solution with the answers provided here.
1. Which of the following is not a linear equation in one variable?
A. 33z+5
B. 33(x+y)
C. 33x+5
D. 33y+5
Explanation: In 33(x+y), x and y are two variables.
2. The solution of 2x-3=7 is:
A. 5
B. 7
C. 12
D. 11
Explanation: 2x-3=7
2x=7+3=10
x=10/2 = 5
3. The solution of 2y + 9 = 4 is:
A. 9/2
B. 4/9
C. -⅖
D. -5/2
Explanation: 2y+9 = 4
2y = 4-9 = -5
y=-5/2
4. The solution of y/5 = 10 is:
A. 15
B. 10
C. 50
D. 5
Explanation: y/5 = 10
y = 5×10 = 50
5. What should be added to -7/3 to get 3/7?
A. 21/58
B. 58/21
C. 47/21
D. 50/21
Explanation: Let the number be x
-7/3+x = 3/7
x=3/7+7/3 = (9+49)/21 = 58/21
6. The perimeter of the rectangle is 20cm. If the length of the rectangle is 6cm, then its breadth will be:
A. 4 cm
B. 6 cm
C. 10 cm
D. 14 cm
Explanation: Perimeter of rectangle = 2(Length+Breadth)
20 = 2(6+x)
6+x = 20/2
6+x = 10
x = 10-6
x=4 cm
7. The age of the father is three times the age of the son. If the age of the son is 15 years old, then the age of the father is:
A. 50 years
B. 55 years
C. 40 years
D. 45 years
Explanation: Let the age of the father is x
Given: x = 3 × (age of son) = 3 × (15) = 45 years
8. The difference between two whole numbers is 66. The ratio of the two numbers is 2: 5. The two numbers are:
A. 60 and 6
B. 100 and 33
C. 110 and 44
D. 99 and 33
Explanation: Let the two numbers be 2x and 5x since they are in the ratio of 2:5.
The difference between 5x and 2x = 66
5x – 2x = 66
3x = 66
x = 22
Hence, 2x = 2(22) = 44 and 5x = 5(22) = 110.
9. Three consecutive integers add up to 51. The integers are:
A. 16,17,18
B. 15,16,17
C. 17,18,19
D. 18,19,20
Explanation: Let the three consecutive integers be x, x+1, x+2
x+(x+1)+(x+2) = 51
3x+3 = 51
3x = 51 – 3
x = 48/3 = 16
x+1 = 16+1=17
x+2 = x+2 = 18
10. The solution for 3m = 5m – (8/5) is:
A. 8/5
B. ⅘
C. 5/4
D. 4/3
Explanation: 3m = 5m – (8/5)
8/5 = 5m – 3m
2m = 8/5
m = 8/10 = 4/5
At the helm of GMS Learning is Principal Balkishan Agrawal, a dedicated and experienced educationist. Under his able guidance, our school has flourished academically and has achieved remarkable milestones in various fields. Principal Agrawal’s visio… |
# DAV Class 5 Maths Chapter 4 Worksheet 6 Solutions
The DAV Class 5 Maths Book Solutions Pdf and DAV Class 5 Maths Chapter 4 Worksheet 6 Solutions of Fractional Numbers offer comprehensive answers to textbook questions.
## DAV Class 5 Maths Ch 4 Worksheet 6 Solutions
Question 1.
Multiply.
(a) $$\frac{1}{3}$$ × 2
Solution:
$$\frac{1}{3}$$ × 2
= $$\frac{1 \times 2}{3}$$
= $$\frac{2}{3}$$
(Product of whole number and numerator of a fractional number)
(b) $$\frac{5}{8}$$ × 9
Solution:
$$\frac{5}{8}$$ × 9
= $$\frac{5 \times 9}{8}$$
= $$\frac{45}{8}$$
= 5$$\frac{5}{8}$$
(Product of whole number and numerator of a fractional number)
(c) 4$$\frac{1}{2}$$ × 4
Solution:
4$$\frac{1}{2}$$ × 4
= $$\frac{9}{2}$$ × 4
= $$\frac{9 \times 4}{2}$$
= $$\frac{36}{2}$$
= 18
(d) 9$$\frac{1}{3}$$ × 27
Solution:
(Product of whole number and numerator of a fractional number)
(e) 10$$\frac{1}{10}$$ × 15
Solution:
(Product of whole number and numerator of a fractional number)
(f) 6 × $$\frac{4}{15}$$
Solution:
(Product of whole number and numerator of a fractional number)
(g) 100 × 3$$\frac{1}{10}$$
Solution:
(Product of whole number and numerator of fractional number)
(h) 52 × 2$$\frac{1}{13}$$
Solution:
(Product of whole number and numerator of fractional number)
(i) 49 × 7$$\frac{1}{7}$$
Solution:
(Product of whole number and numerator of fractional number)
(j) 3$$\frac{5}{8}$$ × 32
Solution:
(Product of whole number and numerator of fractional number)
(k) 45 × 2$$\frac{1}{9}$$
Solution:
(Product of whole number and numerator of fractional number)
(l) 50 × $$\frac{17}{15}$$
Solution:
(Product of whole number and numerator of fractional number)
DAV Class 5 Maths Chapter 4 Worksheet 6 Notes
Example 1.
Multiply 4 and $$\frac{1}{5}$$
Solution:
It means we have to find what is 4 times $$\frac{1}{5}$$
We know multiplication is repeated addition.
Therefore $$4 \times \frac{1}{5}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{4}{5}$$
Let us take rectangular strip of paper divided into 5 equal parts.
Therefore, 4 × $$\frac{1}{5}$$ or we can use the quick method
4 × $$\frac{1}{5}$$ = $$\frac{4 \times 1}{5}$$ = $$\frac{4}{5}$$
Example 2.
Multiply $$\frac{1}{2}$$ and 3.
Solution:
Let us consider a rectangle divided into 3 equal parts.
We further divide these 3 parts into 6 equal parts.
$$\frac{1}{2}$$ of 6 parts will be 3 equal parts (shaded portion)
Thus $$\frac{1}{2}$$ of 3 will be $$\frac{3}{2}$$ (three out of 2 equal parts)
Thus $$\frac{1}{2}$$ of 3 = $$\frac{3}{2}$$
or $$\frac{1}{2}$$ × 3 = $$\frac{1 \times 3}{2}$$ = $$\frac{3}{2}$$
Combining the two results we get
3 × $$\frac{1}{2}$$ = $$\frac{1}{2}$$ × 3 = $$\frac{3}{2}$$
Example 3.
Multiply $$\frac{2}{7}$$ and 3.
Solution:
3 × $$\frac{2}{7}$$ = $$\frac{2 \times 3}{7}$$ = $$\frac{6}{7}$$
(Product of whole number and numerator of a fractional number)
Remember: In order to get the product of whole number and a fractional number, we multiply the whole number and numerator of the fractional number. Denominator remains the same. |
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Table of Contents
# Convert Gallon to Litre
Table of Contents
Conversion of gallon to litre: Gallons and litres are used to measure the volume. A volume of 3.78541 litres makes a gallon. This article will help you to convert gallon to litre.
Gallons
gal
Litres
l
## What is Gallon?
A gallon is a unit of measurement that is used to measure liquids and it’s equal to 4.55 liters. It’s denoted by ‘gal.’
Common Usage:
• Used for measuring water in a tank.
• Used for measuring milk in a big can.
• Used for measuring oil in a big vessel.
## What is Litre?
Litre is the unit of volume in the metric system and it is equal to 1000 cubic centimetres. It’s denoted as ‘l.’
Common Usage:
• Used to measure the quantity of water in a glass.
• used to measure the capacity of a water bottle.
• Used to measure the capacity of milk in a vessel.
## How to convert Gallon to Litre?
To convert gallon to litre, multiply the given value in gallon by 3.78541 to get the value in litres. As
1 Gallon = 3.78541 Litres
The formula to convert gallon to litre is given below:
X gal = X × 3.78541 l
## Examples of the conversion of Gallon to Litre
Problem 1: Convert 40 gallons to litres.
Solution 1:
Step 1: The given value is 40 gallons.
Step 2: To convert gallons to litres, substitute the given values at the required places in the conversion table.
X gal = X × 3.78541 l
Hence,
40 gal = 40 × 3.78541 l
= 151.416 l
Therefore, 40 gallons is 151.416 litres.
Problem 2: Convert 6 gallons to litres.
Solution 2:
Step 1: The given value is 6 gallons.
Step 2: To convert gallons to litres, substitute the given values at the required places in the conversion table.
X gal = X × 3.78541 l
Hence,
6 gal = 6 × 3.78541 l
= 22.7125
Therefore, 6 gallons is 22.7125 litres.
Problem 3: Convert 20 gallons to litres.
Solution 3:
Step 1: The given value is 20 gallons.
Step 2: To convert gallons to litres, substitute the given values at the required places in the conversion table.
X gal = X × 3.78541 l
Hence,
20 gal = 20 × 3.78541 l
= 75.7082 l
Therefore, 20 gallons is 75.7082 litres.
Problem 4: Convert 15 gallons to litres.
Solution 4:
Step 1: The given value is 15 gallons.
Step 2: To convert gallons to litres, substitute the given values at the required places in the conversion table.
X gal = X × 3.78541 l
Hence,
15 gal = 15× 3.78541 l
= 56.7812 l
Therefore, 15 gallons is 56.7812litres.
Problem 5: Convert 36 gallons to litres.
Solution 5:
Step 1: The given value is 36 gallons.
Step 2: To convert gallons to litres, substitute the given values at the required places in the conversion table.
X gal = X × 3.78541 l
Hence,
36 gal =36 × 3.78541 l
= 136.275 l
Therefore, 36 gallons is 136.275 litres.
## FAQs on the Conversion of Gallon to Litre
Are 4 litres the same as 1 gallon?
Yes, 1 gallon is approximately 4 litres.
What is 1 gallon equal to in Litres?
1 Gallon = 3.78541 Litres
Are 3 litres the same as 1 gallon?
Yes, 1 gallon is 3.78541 litres.
What is a gallon?
A gallon is a unit of measurement that is used to measure liquids and it’s equal to 4.55 litres. It’s denoted by ‘gal.’
How to convert gallons to litres?
To convert gallon to litre, multiply the given value in gallon by 3.78541 to get the value in litres. As
1 Gallon = 3.78541 Litres
The formula to convert gallon to litre is given below:
X gal = X × 3.78541 l
Convert 5 gallons to litre.
To convert gallon to litre, multiply the given value in gallon by 3.78541 to get the value in litres. As
1 gallon = 3.78541 litres
5 gallons = 18.9271 litres
Convert 50 imperial gallons to litres.
To convert gallon to litre, multiply the given value in gallon by 3.78541 to get the value in litres. As
1 imperial gallon = 4.54609 litres
50 gallons = 227.304 litres
## Related conversion
References
The SI Base Units
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About Wiingy
Wiingy provides 1-to-1 online tutoring, instructor-led online technology courses, and web tutorials to school students, university students, and working professionals across the globe.
Wiingy works with top verified, qualified, and experienced instructors to deliver online lessons in Coding, Math, Science, and over 50 other subjects.
Parents and students have rated the teaching experience as 4.8/5 and above.
We are a community of over 20,000 students across 10+ countries growing daily.
Contact Us
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(M) +(1) 618-827-5802
Copyright Wiingy Pvt Ltd © 2021. All Rights Reserved |
# How to Calculate 4/6 Plus 1/6
Are you looking to work out and calculate how to add 4/6 plus 1/6? In this really simple guide, we'll teach you exactly what 4/6 + 1/6 is and walk you through the step-by-process of how to add two fractions together.
To start with, the number above the line in a fraction is called a numerator and the number below the line is called the denominator.
Why do you need to know this? Well, because the easiest way to add fractions together is to make sure both fractions have the same denominator.
Let's set up 4/6 and 1/6 side by side so they are easier to see:
4 / 6 + 1 / 6
In this calculation, some of the hard work has already been done for us because the denominator is the same in both fractions. All we need to do is add the numerators together and keep the denominator as it is:
4 + 1 / 6 = 5 / 6
You're done! You now know exactly how to calculate 4/6 + 1/6. Hopefully you understood the process and can use the same techniques to add other fractions together. The complete answer is below (simplified to the lowest form):
5/6
## Convert 4/6 plus 1/6 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:
5 / 6 = 0.8333
### Cite, Link, or Reference This Page
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "How to Calculate 4/6 plus 1/6". VisualFractions.com. Accessed on June 17, 2024. http://visualfractions.com/calculator/adding-fractions/what-is-4-6-plus-1-6/.
• "How to Calculate 4/6 plus 1/6". VisualFractions.com, http://visualfractions.com/calculator/adding-fractions/what-is-4-6-plus-1-6/. Accessed 17 June, 2024.
• How to Calculate 4/6 plus 1/6. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/adding-fractions/what-is-4-6-plus-1-6/.
### Preset List of Fraction Addition Examples
Below are links to some preset calculations that are commonly searched for: |
# 10th Class Mathematics Pair of Linear Equations in Two Variables Pair of Linear Equations in the Variables and Quadratic Equation
Pair of Linear Equations in the Variables and Quadratic Equation
Category : 10th Class
Pair of Linear Equations in two Variables and Quadratic Equation
Linear Equation in Two Variables
A linear equation in two variables is an equation which contains a pair of variables which can be graphically represented in xy-plane by using the coordinate system. For example ax + by=c and dx+ ey=f, is a pair of linear equations in two variables. Solutions of the linear equation in two variables are the pair of values of the variables that satisfies the given equation. In other words, we can say that a system of linear equation is nothing but two or more linear equations that are being solved simultaneously. Mostly, the system of equations are used in the business purposes by predicting their future events. They model a real life situation in two system of equations to find the solution and manage their business. We can make an accurate prediction by using system of equations. The solution of the system of equations in two variables is an ordered pair that satisfies each equation.
Graphical Representation of a Pair of Linear Equations in Two Variables
If ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$are a pair of linear equations in two variables such that:
• If$\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}$, then pair of linear equations is consistent with a unique solution.
• If$\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}$, then the pair of linear equations is consistent and dependent and having infinitely many solutions.
• If$\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}$, then the pair of linear equations is inconsistent and have no solution.
Unique Solution
If the lines represented by a pair of linear equations are intersecting each other at one point, then the system is said to have unique solution. The point at which the two lines intersect each other is called the solution of the system of equation.
No Solution
If the graph of the system of equation is parallel and does not intersect each other at any point, then it is said to have no solution.
Infinitely Many Solutions
If the lines represented by the pair of linear equations in two variables coincides each other, then it is said to have infinitely many solution.
Solving the System of Equations
There are different algebraic methods for solving the system of linear equations. The three different methods are:
• Elimination Method
• Substitution Method
• Cross Multiplication Method
• Example:
Find the relation between m and n for which the system of equations
$4x+6y=7and(m+n)x+(2m-n)y=21$, has unique solution.
(a) 2m=3n (b) m = 5n
(c)$2m\ne 3n$ (d) $m\ne 5n$
Explanation:
We have, the system of equations $4x+6y=7$and $(m+n)x+(2m-n)y=21$
For a unique solution, the required condition is
$\frac{4}{m+n}\ne \frac{6}{2m-n}$
$\Rightarrow 8m-4n\ne 6m+6n\Rightarrow 8m-6m\ne 6n+4n$$\Rightarrow 2m\ne 10n\Rightarrow m\ne 5n$
Quadratic equation is a type of polynomial of degree two. The general form of a quadratic equation is$a{{x}^{2}}+bx+c=0$, where a, b, c are the constants and$a\ne 0$. The quadratic equation which contains both second and first powers of the variable is called a complete quadratic equation and the equation in which first power is missing is called pure quadratic equation.
The values of variable x which satisfy the quadratic equation $a{{x}_{2}}+bx+c=0$$(a\ne 0)$is called the roots of a quadratic equation.
Nature of Roots of a Quadratic Equation
The general form of a quadratic equation is$a{{x}^{2}}+bx+c=0$. This equation can be solved by using the Discriminant method. In this method first we find the discriminant $D={{b}^{2}}-4ac$ of the given quadratic equation.
• If D > 0, then the given equation will have real and distinct roots and we can find the roots of the given equation.
• If D = 0, then the equation will have real and equal roots.
• If D < 0, then the given equation will have no real roots. In this case roots will be imaginary.
In case of real roots we can find the roots by using the formula, $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}or\frac{-b\pm \sqrt{D}}{2a}$
A quadratic equation can have maximum of two roots.
Relations between the Roots of a Quadratic Equation
If $\alpha$ and $\beta$are the roots of the equation $a{{x}^{2}}+bx+c=0$then the relation between the roots of the quadratic equation is given by:
Sum of the roots $=\alpha +\beta =-\frac{b}{a}$
Product of the roots $=\alpha \beta =\frac{c}{a}$
If $\alpha$ and $\beta$ are the roots of the quadratic equation, and S denotes its sum and P denotes its product, then the quadratic equation is given by: ${{x}_{2}}-Sx+P=0$
• Example:
The value of k for which the equation ${{k}^{2}}{{x}^{2}}-2(2k-1)x+4=0$have real and equal roots is:
(a) $\frac{1}{2}$ (b) $\frac{1}{8}$
(c) $\frac{1}{4}$ (d) $\frac{1}{6}$
Explanation: The given equation is ${{k}^{2}}{{x}^{2}}-2(2k-1)x+4=0$
Discriminant $(D)=4{{(2k-1)}^{2}}-4.4.{{k}^{2}}$
$=4(4{{k}^{2}}+1-4k)-16{{k}^{2}}$$=16{{k}^{2}}+4-16k-16{{k}^{2}}=4-16k$ |
# 2013 AMC 12A Problems/Problem 5
Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $$105$, Dorothy paid$$125$, and Sammy paid $$175$. In order to share the costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t-d$? $\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$ ## Solution 1 Simply write down two algebraic equations. We know that Tom gave $t$ dollars and Dorothy gave $d$ dollars. In addition, Tom originally paid $105$ dollars and Dorothy paid $125$ dollars originally. Since they all pay the same amount, we have: $$105 + t = 125 + d.$$ Rearranging, we have $$t-d = \boxed{\textbf{(B)} 20}.$$ Solution $\textcopyright 2018$ RandomPieKevin. All Rights Reserved. skrrt... ## Solution 2 Add up the amounts that Tom, Dorothy, and Sammy paid to get$$405$, and divide by 3 to get $$135$, the amount that each should have paid. Tom, having paid$$105$, owes Sammy $$30$, and Dorothy, having paid$$125$, owes Sammy \$$10$.
Thus, $t - d = 30 - 10 = 20$, which is $\boxed{\textbf{(B)}}$ |
Maximum Area of Rectangle in a Right Triangle - Problem with Solution
Maximize the area of a rectangle inscribed in a right triangle using the first derivative. The problem and its solution are presented
Problem with Solution
BDEF is a rectangle inscribed in the right triangle ABC whose side lengths are 40 and 30. Find the dimensions of the rectangle BDEF so that its area is maximum.
Solution to Problem:
Let the length BF of the rectangle be $$y$$ and the width BD be $$x$$. The area of the right triangle is given by $$\frac{1}{2} \times 40 \times 30 = 600$$. But the area of the right triangle may also be calculated as the sum of the areas of triangle BEC and BEA. Hence
$600 = \frac{1}{2} \times 40 \times y + \frac{1}{2} \times 30 \times x$
Let $$A$$ be the area of the rectangle. Hence
$A = y \times x$
We now use the first equation to express $$y$$ in terms of $$x$$ as follows
$y = \frac{600 - 15x}{20}$
Substitute in $$A$$ to obtain
$A(x) = \frac{x(600 - 15x)}{20}$
The graph of $$A(x)$$ as a function of $$x$$ is shown below. $$A(x)$$ has a maximum value for $$x = 20$$. This will be shown analytically as well
An expansion of $$A(x)$$ shows that $$A(x)$$ is a quadratic function with negative leading coefficient and therefore has a maximum value.
$A(x) = -\frac{3}{4}x^2 + 30x$
We now calculate the first derivative of $$A$$.
$A'(x) = -\frac{3}{2}x + 30$
Set $$A'(x) = 0$$ and solve for $$x$$.
$x = 20$
It is easy to check that $$A'(x)$$ is positive for $$x \lt 20$$ and negative for $$x > 20$$ and therefore $$A(x)$$ has a maximum at $$x = 20$$.
The maximum area is given by $$A(20)$$
$A(20) = -\frac{3}{4} \times 20^2 + 30 \times 20 = 300$
We now find $$y$$ as follows
$A = 300 = x \times y , \quad y = \frac{300}{20} = 15$
The dimensions of the rectangle that make its area maximum are $$x = 20$$ and $$y = 15$$. |
# 2004 AMC 12A Problems/Problem 15
The following problem is from both the 2004 AMC 12A #15 and 2004 AMC 10A #17, so both problems redirect to this page.
## Problem
Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?
$\mathrm{(A) \ } 250 \qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 350 \qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 500$
## Solutions
### Solution 1
Call the length of the race track $x$. When they meet at the first meeting point, Brenda has run $100$ meters, while Sally has run $\frac{x}{2} - 100$ meters. By the second meeting point, Sally has run $150$ meters, while Brenda has run $x - 150$ meters. Since they run at a constant speed, we can set up a proportion: $\frac{100}{x- 150} = \frac{\frac{x}{2} - 100}{150}$. Cross-multiplying, we get that $x = 350\Longrightarrow\boxed{\mathrm{(C)}\ 350}$.
## Solution by Alcumus
The total distance the girls run between the start and the first meeting is one half of the track length. The total distance they run between the two meetings is the track length. As the girls run at constant speeds, the interval between the meetings is twice as long as the interval between the start and the first meeting. Thus between the meetings Brenda will run $2\times100=200$ meters. Therefore the length of the track is $150 + 200 = 350$ meters $\Rightarrow\boxed{\mathrm{(C)}\ 350}$
## Video Solution
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CombinatoricsLesson1
# CombinatoricsLesson1 - Pure Math 30 Explained...
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Unformatted text preview: Pure Math 30: Explained! www.puremath30.com 323 Permutations & Combinations Lesson 1, Part One: The Fundamental Counting Principle The Fundamental Counting Principle: This is an easy way to determine how many ways you can arrange items. The following examples illustrate how to use it: Example 1: How many ways can you arrange the letters in the word MICRO? The basic idea is we have 5 objects, and 5 possible positions they can occupy. Example 2: How many ways can 8 different albums be arranged? We could approach this question in the same way as the last one by using the spaces and multiplying all the numbers, but there is a shorter way. The factorial function on your calculator will perform this calculation for you! 8! = 8 • 7 • 6 …. 2 •1 8! = 40320 Questions: 1) How many ways can the letters in the word PENCIL be arranged? TI-83 Info You can get the factorial function using: Math Prb ! 2) If there are four different types of cookies, how many ways can you eat all of them? 3) If three albums are placed in a multi-disc stereo, how many ways can the albums be played? 4) How many ways can you arrange all the letters in the alphabet? 5) How many ways can you arrange the numbers 24 through 28 (inclusive)? Answers: 1) 6! = 720 2) 4! = 24 3) 3! = 6 4) 26! = 4.03 • 1026 5) 5! = 120 Pure Math 30: Explained! www.puremath30.com 324 Permutations & Combinations Lesson 1, Part Two: Repetitions Not Allowed Repetitions Not Allowed: In many cases, some of the items we want to arrange are identical. For example, in the word TOOTH, if we exchange the places of the two O’s, we still get TOOTH. Because of this, we have to get rid of extraneous cases by dividing out repetitions. Example 1: How many ways can you arrange the letters in the word THESE? Do this as a fraction. Factorial the total number of letters and put this on top. Factorial the repeated letters and put them on the bottom. 5! 120 = = 60 2! 2 Example 2: How many ways can you arrange the letters in the word REFERENCE? 9! 362880 = = 7560 2!•4! 2•24 Questions: 1) How many ways can the letters in the word SASKATOON be arranged? TI-83 Info Make sure you put the denominator in brackets or you’ll get the wrong answer! 2) How many ways can the letters in the word MISSISSIPPI be arranged? 3) How many ways can the letters in the word MATHEMATICS be arranged? 4) If there are eight cookies (4 chocolate chip, 2 oatmeal, and 2 chocolate) in how many different orders can you eat all of them? Answers: 5) If a multiple choice test has 10 questions, of which one is answered A, 4 are answered B, 3 are answered C, and 2 are answered D, how many answer sheets are possible? 1) 9! = 45360 2!• 2!• 2! 2) 11! = 34650 4!• 4!• 2! 3) 11! = 4989600 2!• 2!• 2! 4) 8! = 420 4!• 2!• 2! 5) 10! = 12600 4!•3!• 2! Pure Math 30: Explained! www.puremath30.com 325 Permutations & Combinations Lesson 1, Part Three: Repetitions Are Allowed Repetitions Are Allowed: Sometimes we are interested in arrangements allowing the use of items more than once. Example 1: There are 9 switches on a fuse box. How many different arrangements are there? Each switch has two possible positions, on or off. Placing a 2 in each of the 9 positions, we have 29 = 512. Example 2: How many 3 letter words can be created, if repetitions are allowed? There are 26 letters to choose from, and we are allowed to have repetitions. There are 263 = 17576 possible three letter words. Questions: 1) If there are 4 light switches on an electrical panel, how many different orders of on/off are there? 2) How many 5 letter words can be formed, if repetitions are allowed? 3) How many three digit numbers can be formed? (Zero can’t be the first digit) There are ten digits in total, from zero to nine. 4) A coat hanger has four knobs. If you have 6 different colors of paint available, how many different ways can you paint the knobs? Answers: 1) 24 = 16 2) 265 = 11881376 3) 9 • 10 • 10 = 900 4) 64 = 1296 Pure Math 30: Explained! www.puremath30.com 326 Permutations & Combinations Lesson 1, Part Four: Arranging A Subset Arranging a subset of items: Sometimes you will be given a bunch of objects, and you want to arrange only a few of them: Example 1: There are 10 people in a competition. How many ways can the top three be ordered? Since only 3 positions can be filled, we have 3 spaces. Multiplying, we get 720. A fast way to do questions like this is to use the nPr feature on your calculator. n is the total number of items. r is the number of items you want to order. For Example 1, you would type: 10 Math PRB P n r enter 3 Example 2: There are 12 movies playing at a theater, in how many ways can you see two of them consecutively? OR You could use the spaces, but let’s try this question with the permutation feature. There are 12 movies, and you want to see 2, so type 12P2 into your calculator, and you’ll get 132. Example 3: How many 4 letter words can be created if repetitions are not allowerd? The answer is: 26P4 = 358800 OR Questions: 1) How many three letter words can be made from the letters of the word KEYBOARD 2) If there are 35 songs and you want to make a mix CD with 17 songs, how many different ways could you arrange them? 3) There are six different colored balls in a box, and you pull them out one at a time. How many different ways can you pull out four balls? 4) A committee is to be formed with a president, a vice-president, and a treasurer. There are 10 people to be selected from. How many different committees are possible? 5) A baseball league has 13 teams, and each team plays each other twice; once at home, and once away. How many games are scheduled? Answers: 1) 8P3 =336 2) 35P17 = 1.6 • 1024 3) 6P4 = 360 4) 10P3 = 720 5) 13P2 = 156 Pure Math 30: Explained! www.puremath30.com 327 Permutations & Combinations Lesson 1, Part Five: Specific Positions Specific Positions: Frequently when arranging items, a particular position must be occupied by a particular item. The easiest way to approach these questions is by analyzing how many possible ways each space can be filled. Example 1: How many ways can Adam, Beth, Charlie, and Doug be seated in a row if Charlie must be in the second chair? The answer is 6. Example 2: How many ways can you order the letters of KITCHEN if it must start with a consonant and end with a vowel? The answer is 1200. Example 3: How many ways can you order the letters of TORONTO if it begins with exactly two O’s? Exactly Two O’s means the first 2 letters must be O, and the third must NOT be an O. If the question simply stated two O’s, then the third letter could also be an O, since that case wasn’t excluded. Don’t forget repetitions! The answer from the left will be the numerator with repetitions divided out. 576 3!•2! = 48 Pure Math 30: Explained! www.puremath30.com 328 Permutations & Combinations Lesson 1, Part Five: Specific Positions Questions: 1) Six Pure Math 30 students (Brittany, Geoffrey, Jonathan, Kyle, Laura, and Stephanie) are going to stand in a line: How many ways can they stand if: a) Stephanie must be in the third position? b) Geoffrey must be second and Laura third? c) Kyle can’t be on either end of the line? d) Boys and girls alternate, with a boy starting the line? e) The first three positions are boys, the last three are girls? f) A girl must be on both ends? g) The row starts with two boys? h) The row starts with exactly two boys? i) Brittany must be in the second position, and a boy must be in the third? 2) How many ways can you order the letters from the word TREES if: a) A vowel must be at the beginning? b) It must start with a consonant and end with a vowel? c) The R must be in the middle? d) It begins with an E? e) It begins with exactly one E? f) Consonants & vowels alternate? Pure Math 30: Explained! www.puremath30.com 329 Permutations & Combinations Lesson 1, Part Five: Specific Positions Answers: 1) a) If Stephanie must be in the third position, place a one there to reserve her spot. You can then place the remaining 5 students in any position. b) Place a 1 in the second position to reserve Geoffrey’s spot, and place a 1 in the third position to reserve Laura’s spot. Place the remaining students in the other positions. c) Since Kyle can’t be on either end, 5 students could be placed on one end, then 4 at the other end. Now that 2 students are used up, there are 4 that can fill out the middle. d) Three boys can go first, then three girls second. Two boys remain, then two girls. Then one boy and one girl remain. e) Three boys can go first, then place the girls in the next three spots. f) Three girls could be placed on one end, then 2 girls at the other end. There are four students left to fill out the middle. g) Three boys could go first, then 2 boys second. Once those positions are filled, four people remain for the rest of the line. h) Three boys could go first, then two boys second. The third position can’t be a boy, so there are three girls that could go here. Then, three people remain to fill out the line. i) Place a 1 in the second position to reserve Brittany’s spot, then 3 boys could go in the third position. Now fill out the rest of the line with the four remaining people. 2) Note that since there are 2 E’s, all answers MUST be divided by 2! to eliminate repetitions. a) There are two vowels that can go first, then four letters remain to (Answer = 48 / 2! = 24) fill out the other positions. b) Three consonants could go first, and two vowels could go last. There are three letters to fill out the remaining positions. (Answer = 36 / 2! = 18) c) Place a 1 in the middle spot to reserve the R’s spot. Then fill out the rest of the spaces with the remaining 4 letters. (Answer = 24 / 2! = 12) d) Two E’s could go in the first spot, then fill the remaining spaces (Answer = 48 / 2! = 24) with the 4 remaining letters. e) Two E’s could go in the first spot, but the next letter must NOT be an E, so there are 3 letters that can go here. Fill out the three last spaces with the 3 remaining letters. (Answer = 36 / 2! = 18) f) Three consonants could go first, then 2 vowels, and so on. (Answer = 12 / 2! = 6) Pure Math 30: Explained! www.puremath30.com 330 Permutations & Combinations Lesson 1, Part Six: Adding Permutations More than one case (Adding): Given a set of items, it is possible to form multiple groups by ordering any 1 item from the set, any 2 items from the set, and so on. If you want the total arrangements from multiple groups, you have to ADD the results of each case. Example 1: How many words (of any number of letters) can be formed from CANS Since we are allowed to have any number of letters in a word, we can have a 1 letter word, a 2 letter word, a 3 letter word, and a 4 letter word. We can’t have more than 4 letters in a word, since there aren’t enough letters for that! The answer is 64 We could also write this using permutations: 4P1 + 4P2 + 4P3 + 4P4 = 64 Example 2: How many four digit positive numbers less than 4670 can be formed using the digits 1, 3, 4, 5, 8, 9 if repetitions are not allowed? We must separate this question into different cases. Numbers in the 4000’s have extra restrictions. Case 1 - Numbers in the 4000’s: There is only one possibility for the first digit {4}. The next digit has three possibilities. {1, 3, 5}. There are 4 possibilities for the next digit since any remaining number can be used, and 3 possibilities for the last digit. Case 2 - Numbers in the 1000’s and 3000’s: There are two possibilities for the first digit {1, 3}. Anything goes for the remaining digits, so there are 5, then 4, then 3 possibilities. Add the results together: 36 + 120 = 156 Answers: Questions: 1) How many one-letter, two-letter, or three-letter words can be formed from the word PENCIL? 3) 1) 6P1 + 6P2 + 6P3 = 156 2) 8P3 + 8P4 + 8P5 = 8736 There are two cases: The first case has five as the last digit, the second case has zero as the last digit. Remember the first digit can’t be 2) How many 3-digit, 4-digit, or 5-digit numbers can be made using the digits of 46723819? zero! 3) How many numbers between 999 and 9999 are divisible by 5 and have no repeated digits? Add the results to get the total: 952 Pure Math 30: Explained! www.puremath30.com 331 Permutations & Combinations Lesson 1, Part Seven: Items Always Together Always Together: Frequently, certain items must always be kept together. To do these questions, you must treat the joined items as if they were only one object. Example 1: How many arrangements of the word ACTIVE are there if C & E must always be together? There are 5 groups in total, and they can be arranged in 5! ways. The letters EC can be arranged in 2! ways. The total arrangements are 5! x 2! = 240 Example 2: How many ways can 3 math books, 5 chemistry books, and 7 physics books be arranged on a shelf if the books of each subject must be kept together? There are three groups, which can be arranged in 3! ways. The physics books can be arranged in 7! ways. The math books can be arranged in 3! ways. The chemistry books can be arranged in 5! ways. The total arrangements are 3! x 7! x 3! x 5! = 21772800 Questions: 1) How many ways can you order the letters in KEYBOARD if K and Y must always be kept together? 2) How many ways can the letters in OBTUSE be ordered if all the vowels must be kept together? 3) How many ways can 4 rock, 5 pop, & 6 classical albums be ordered if all albums of the same genre must be kept together? Answers: 1) 7! • 2! = 10080 2) 4! • 3! = 144 3) 3! • 4! • 5! • 6! = 12441600 . Pure Math 30: Explained! www.puremath30.com 332 Permutations & Combinations Lesson 1, Part Eight: Items Never Together Never Together: If certain items must be kept apart, you will need to figure out how many possible positions the separate items can occupy. Example 1: How many arrangements of the word ACTIVE are there if C & E must never be together? Method 1: You can place C & E in these spaces 5P2 ways. Get the answer by multiplying: First fill in the possible positions for the letters ATIV Next draw empty circles representing the positions C & E can go. 4! x 5P2 = 480 Method 2: First determine the number of ways ACTIVE can be arranged if C & E are ALWAYS together. (5! • 2!) Then subtract that from the total number of possible arrangements without restrictions (6!) The answer is 6! – (5! • 2!) = 480 ***This method does not work if there are more then two items you want to keep separate. Example 2: How many arrangements of the word DAUGHTER are there if none the vowels can ever be together? First fill in the possible positions for the consonants Next draw empty circles representing the positions the vowels can go. You can place the 3 vowels in the 6 spaces in 6P3 ways. 5! x 6P3 = 14400 Example 3: In how many ways can the letters from the word EDITOR be arranged if vowels and consonants alternate positions? First determine the number of arrangements with consonants first in the arrangement: Then determine the number of arrangements with vowels coming first: Add the results to get 72 possible arrangements. Questions: 1) How many ways can you order the letters in QUEST if the vowels must never be together? 2) If 8 boys and 2 girls must stand in line for a picture, how many line-up’s will have the girls separated from each other? 3) How many ways can you order the letters in FORTUNES if the vowels must never be together? 4) In how many ways can the letters AEFGOQSU be arranged if vowels and consonants alternate positions? Answers: 1) 3! • 4P2 = 72 or 5! – (4! • 2!) = 72 2) 8! • 9P2 = 2903040 or 10! – (9! • 2!) = 2903040 3) 5! • 6P3 =14400 4) Consonants first: 4×4×3×3×2×2×1×1 = 576 Vowels first: 4×4×3×3×2×2×1×1 = 576 Add results: 576 + 576 = 1152 Pure Math 30: Explained! www.puremath30.com 333 ...
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# What The Numbers Say About Hugh Grant (10/18/2019)
How will Hugh Grant perform on 10/18/2019 and the days ahead? Let’s use astrology to undertake a simple analysis. Note this is for entertainment purposes only – do not take this too seriously. I will first find the destiny number for Hugh Grant, and then something similar to the life path number, which we will calculate for today (10/18/2019). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology experts.
PATH NUMBER FOR 10/18/2019: We will analyze the month (10), the day (18) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. Here’s how it works. First, for the month, we take the current month of 10 and add the digits together: 1 + 0 = 1 (super simple). Then do the day: from 18 we do 1 + 8 = 9. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 1 + 9 + 12 = 22. This still isn’t a single-digit number, so we will add its digits together again: 2 + 2 = 4. Now we have a single-digit number: 4 is the path number for 10/18/2019.
DESTINY NUMBER FOR Hugh Grant: The destiny number will take the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Hugh Grant we have the letters H (8), u (3), g (7), h (8), G (7), r (9), a (1), n (5) and t (2). Adding all of that up (yes, this can get tedious) gives 50. This still isn’t a single-digit number, so we will add its digits together again: 5 + 0 = 5. Now we have a single-digit number: 5 is the destiny number for Hugh Grant.
CONCLUSION: The difference between the path number for today (4) and destiny number for Hugh Grant (5) is 1. That is smaller than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But this is just a shallow analysis! As mentioned earlier, this is not scientifically verified. If you want a forecast that we do recommend taking seriously, check out your cosmic energy profile here. Go see what it says for you now – you may be absolutely amazed. It only takes 1 minute.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. |
## 11.2 – Transformations of Functions
### Learning Objectives
• (11.2.1) – Graphing with transformations of functions
• Graph functions using vertical and horizontal shifts
• Graph functions using reflections about the $x$-axis and the $y$-axis
• Graph functions using compressions and stretches
• Combine transformations
• (11.2.2) – Transformations of quadratic functions
# (11.2.1) – Graphing with transformations of functions
We all know that a flat mirror enables us to see an accurate image of ourselves and whatever is behind us. When we tilt the mirror, the images we see may shift horizontally or vertically. But what happens when we bend a flexible mirror? Like a carnival funhouse mirror, it presents us with a distorted image of ourselves, stretched or compressed horizontally or vertically. In a similar way, we can distort or transform mathematical functions to better adapt them to describing objects or processes in the real world. In this section, we will take a look at several kinds of transformations.
Figure 1. (credit: “Misko”/Flickr)
### Graph functions using vertical and horizontal shifts
One simple kind of transformation involves shifting the entire graph of a function up, down, right, or left. The simplest shift is a vertical shift, moving the graph up or down, because this transformation involves adding a positive or negative constant to the function. In other words, we add the same constant to the output value of the function regardless of the input. For a function $g\left(x\right)=f\left(x\right)+k$, the function $f\left(x\right)$ is shifted vertically $k$ units.
Vertical shift by $k=1$ of the cube root function $f\left(x\right)=\sqrt[3]{x}$.
To help you visualize the concept of a vertical shift, consider that $y=f\left(x\right)$. Therefore, $f\left(x\right)+k$ is equivalent to $y+k$. Every unit of $y$ is replaced by $y+k$, so the $y\text{-}$ value increases or decreases depending on the value of $k$. The result is a shift upward or downward.
### A General Note: Vertical Shift
Given a function $f\left(x\right)$, a new function $g\left(x\right)=f\left(x\right)+k$, where $k$ is a constant, is a vertical shift of the function $f\left(x\right)$. All the output values change by $k$ units. If $k$ is positive, the graph will shift up. If $k$ is negative, the graph will shift down.
### Example: Adding a Constant to a Function
To regulate temperature in a green building, airflow vents near the roof open and close throughout the day. Figure 2 shows the area of open vents $V$ (in square feet) throughout the day in hours after midnight, $t$. During the summer, the facilities manager decides to try to better regulate temperature by increasing the amount of open vents by 20 square feet throughout the day and night. Sketch a graph of this new function.
### How To: Given a tabular function, create a new row to represent a vertical shift.
1. Identify the output row or column.
2. Determine the magnitude of the shift.
3. Add the shift to the value in each output cell. Add a positive value for up or a negative value for down.
### Example: Shifting a Tabular Function Vertically
A function $f\left(x\right)$ is given below. Create a table for the function $g\left(x\right)=f\left(x\right)-3$.
$x$ 2 4 6 8 $f\left(x\right)$ 1 3 7 11
### Try It
The function $h\left(t\right)=-4.9{t}^{2}+30t$ gives the height $h$ of a ball (in meters) thrown upward from the ground after $t$ seconds. Suppose the ball was instead thrown from the top of a 10-m building. Relate this new height function $b\left(t\right)$ to $h\left(t\right)$, and then find a formula for $b\left(t\right)$.
#### Identifying Horizontal Shifts
We just saw that the vertical shift is a change to the output, or outside, of the function. We will now look at how changes to input, on the inside of the function, change its graph and meaning. A shift to the input results in a movement of the graph of the function left or right in what is known as a horizontal shift.
Horizontal shift of the function $f\left(x\right)=\sqrt[3]{x}$. Note that $h=+1$ shifts the graph to the left, that is, towards negative values of $x$.
For example, if $f\left(x\right)={x}^{2}$, then $g\left(x\right)={\left(x - 2\right)}^{2}$ is a new function. Each input is reduced by 2 prior to squaring the function. The result is that the graph is shifted 2 units to the right, because we would need to increase the prior input by 2 units to yield the same output value as given in $f$.
### A General Note: Horizontal Shift
Given a function $f$, a new function $g\left(x\right)=f\left(x-h\right)$, where $h$ is a constant, is a horizontal shift of the function $f$. If $h$ is positive, the graph will shift right. If $h$ is negative, the graph will shift left.
### Example: Adding a Constant to an Input
Returning to our building airflow example from Example 2, suppose that in autumn the facilities manager decides that the original venting plan starts too late, and wants to begin the entire venting program 2 hours earlier. Sketch a graph of the new function.
### How To: Given a tabular function, create a new row to represent a horizontal shift.
1. Identify the input row or column.
2. Determine the magnitude of the shift.
3. Add the shift to the value in each input cell.
### Example: Shifting a Tabular Function Horizontally
A function $f\left(x\right)$ is given below. Create a table for the function $g\left(x\right)=f\left(x - 3\right)$.
$x$ 2 4 6 8 $f\left(x\right)$ 1 3 7 11
### Example: Identifying a Horizontal Shift of a Toolkit Function
This graph represents a transformation of the toolkit function $f\left(x\right)={x}^{2}$. Relate this new function $g\left(x\right)$ to $f\left(x\right)$, and then find a formula for $g\left(x\right)$.
### Example: Interpreting Horizontal versus Vertical Shifts
The function $G\left(m\right)$ gives the number of gallons of gas required to drive $m$ miles. Interpret $G\left(m\right)+10$ and $G\left(m+10\right)$.
### Try It
Given the function $f\left(x\right)=\sqrt{x}$, graph the original function $f\left(x\right)$ and the transformation $g\left(x\right)=f\left(x+2\right)$ on the same axes. Is this a horizontal or a vertical shift? Which way is the graph shifted and by how many units?
### Graph functions using reflections about the $x$-axis and the $y$-axis
Another transformation that can be applied to a function is a reflection over the $x$– or $y$-axis. A vertical reflection reflects a graph vertically across the $x$-axis, while a horizontal reflection reflects a graph horizontally across the $y$-axis. The reflections are shown in Figure 9.
Vertical and horizontal reflections of a function.
Notice that the vertical reflection produces a new graph that is a mirror image of the base or original graph about the x-axis. The horizontal reflection produces a new graph that is a mirror image of the base or original graph about the y-axis.
### A General Note: Reflections
Given a function $f\left(x\right)$, a new function $g\left(x\right)=-f\left(x\right)$ is a vertical reflection of the function $f\left(x\right)$, sometimes called a reflection about (or over, or through) the x-axis.
Given a function $f\left(x\right)$, a new function $g\left(x\right)=f\left(-x\right)$ is a horizontal reflection of the function $f\left(x\right)$, sometimes called a reflection about the y-axis.
### How To: Given a function, reflect the graph both vertically and horizontally.
1. Multiply all outputs by $-1$ for a vertical reflection. The new graph is a reflection of the original graph about the $x$-axis.
2. Multiply all inputs by $-1$ for a horizontal reflection. The new graph is a reflection of the original graph about the $y$-axis.
### Example: Reflecting a Graph Horizontally and Vertically
Reflect the graph of $s\left(t\right)=\sqrt{t}$ (a) vertically and (b) horizontally.
### Try It
Reflect the graph of $f\left(x\right)=|x - 1|$ (a) vertically and (b) horizontally.
### Example: Reflecting a Tabular Function Horizontally and Vertically
A function $f\left(x\right)$ is given. Create a table for the functions below.
1. $g\left(x\right)=-f\left(x\right)$
2. $h\left(x\right)=f\left(-x\right)$
$x$ 2 4 6 8 $f\left(x\right)$ 1 3 7 11
### Try It
$x$ −2 0 2 4 $f\left(x\right)$ 5 10 15 20
Using the function $f\left(x\right)$ given in the table above, create a table for the functions below.
a. $g\left(x\right)=-f\left(x\right)$
b. $h\left(x\right)=f\left(-x\right)$
### Graph functions using compressions and stretches
Adding a constant to the inputs or outputs of a function changed the position of a graph with respect to the axes, but it did not affect the shape of a graph. We now explore the effects of multiplying the inputs or outputs by some quantity.
We can transform the inside (input values) of a function or we can transform the outside (output values) of a function. Each change has a specific effect that can be seen graphically.
#### Vertical Stretches and Compressions
When we multiply a function by a positive constant, we get a function whose graph is stretched or compressed vertically in relation to the graph of the original function. If the constant is greater than 1, we get a vertical stretch; if the constant is between 0 and 1, we get a vertical compression. The graph below shows a function multiplied by constant factors 2 and 0.5 and the resulting vertical stretch and compression.
Vertical stretch and compression
### A General Note: Vertical Stretches and Compressions
Given a function $f\left(x\right)$, a new function $g\left(x\right)=af\left(x\right)$, where $a$ is a constant, is a vertical stretch or vertical compression of the function $f\left(x\right)$.
• If $a>1$, then the graph will be stretched.
• If $0 < a < 1$, then the graph will be compressed.
• If $a<0$, then there will be combination of a vertical stretch or compression with a vertical reflection.
### How To: Given a function, graph its vertical stretch.
1. Identify the value of $a$.
2. Multiply all range values by $a$.
3. If $a>1$, the graph is stretched by a factor of $a$.
If ${ 0 }<{ a }<{ 1 }$, the graph is compressed by a factor of $a$.
If $a<0$, the graph is either stretched or compressed and also reflected about the x-axis.
### Example: Graphing a Vertical Stretch
A function $P\left(t\right)$ models the number of fruit flies in a population over time, and is graphed below.
A scientist is comparing this population to another population, $Q$, whose growth follows the same pattern, but is twice as large. Sketch a graph of this population.
### How To: Given a tabular function and assuming that the transformation is a vertical stretch or compression, create a table for a vertical compression.
1. Determine the value of $a$.
2. Multiply all of the output values by $a$.
### Example: Finding a Vertical Compression of a Tabular Function
A function $f$ is given in the table below. Create a table for the function $g\left(x\right)=\frac{1}{2}f\left(x\right)$.
$x$ 2 4 6 8 $f\left(x\right)$ 1 3 7 11
### EXAMPLE
A function $f$ is given below. Create a table for the function $g\left(x\right)=\frac{3}{4}f\left(x\right)$.
$x$ 2 4 6 8 $f\left(x\right)$ 12 16 20 0
### Example: Recognizing a Vertical Stretch
The graph is a transformation of the toolkit function $f\left(x\right)={x}^{3}$. Relate this new function $g\left(x\right)$ to $f\left(x\right)$, and then find a formula for $g\left(x\right)$.
### EXAMPLE
Write the formula for the function that we get when we stretch the identity toolkit function by a factor of 3, and then shift it down by 2 units.
### Horizontal Stretches and Compressions
Now we consider changes to the inside of a function. When we multiply a function’s input by a positive constant, we get a function whose graph is stretched or compressed horizontally in relation to the graph of the original function. If the constant is between 0 and 1, we get a horizontal stretch; if the constant is greater than 1, we get a horizontal compression of the function.
Given a function $y=f\left(x\right)$, the form $y=f\left(bx\right)$ results in a horizontal stretch or compression. Consider the function $y={x}^{2}$. The graph of $y={\left(0.5x\right)}^{2}$ is a horizontal stretch of the graph of the function $y={x}^{2}$ by a factor of 2. The graph of $y={\left(2x\right)}^{2}$ is a horizontal compression of the graph of the function $y={x}^{2}$ by a factor of 2.
### A General Note: Horizontal Stretches and Compressions
Given a function $f\left(x\right)$, a new function $g\left(x\right)=f\left(bx\right)$, where $b$ is a constant, is a horizontal stretch or horizontal compression of the function $f\left(x\right)$.
• If $b>1$, then the graph will be compressed by $\displaystyle \frac{1}{b}$.
• If $0<b<1$, then the graph will be stretched by $\displaystyle \frac{1}{b}$.
• If $b<0$, then there will be combination of a horizontal stretch or compression with a horizontal reflection.
### How To: Given a description of a function, sketch a horizontal compression or stretch.
1. Write a formula to represent the function.
2. Set $g\left(x\right)=f\left(bx\right)$ where $b>1$ for a compression or $0<b<1$
for a stretch.
### Example: Graphing a Horizontal Compression
Suppose a scientist is comparing a population of fruit flies to a population that progresses through its lifespan twice as fast as the original population. In other words, this new population, $R$, will progress in 1 hour the same amount as the original population does in 2 hours, and in 2 hours, it will progress as much as the original population does in 4 hours. Sketch a graph of this population.
### Example: Finding a Horizontal Stretch for a Tabular Function
A function $f\left(x\right)$ is given below. Create a table for the function $\displaystyle g\left(x\right)=f\left(\frac{1}{2}x\right)$.
$x$ 2 4 6 8 $f\left(x\right)$ 1 3 7 11
### Example: Recognizing a Horizontal Compression on a Graph
Relate the function $g\left(x\right)$ to $f\left(x\right)$.
### Try It
Write a formula for the toolkit square root function horizontally stretched by a factor of 3.
### Combine transformations
Now that we have two transformations, we can combine them together. Vertical shifts are outside changes that affect the output ( $y\text{-}$ ) axis values and shift the function up or down. Horizontal shifts are inside changes that affect the input ( $x\text{-}$ ) axis values and shift the function left or right. Combining the two types of shifts will cause the graph of a function to shift up or down and right or left.
### How To: Given a function and both a vertical and a horizontal shift, sketch the graph.
1. Identify the vertical and horizontal shifts from the formula.
2. The vertical shift results from a constant added to the output. Move the graph up for a positive constant and down for a negative constant.
3. The horizontal shift results from a constant added to the input. Move the graph left for a positive constant and right for a negative constant.
4. Apply the shifts to the graph in either order.
### Example: Graphing Combined Vertical and Horizontal Shifts
Given $f\left(x\right)=|x|$, sketch a graph of $h\left(x\right)=f\left(x+1\right)-3$.
The function $f$ is our toolkit absolute value function. We know that this graph has a V shape, with the point at the origin. The graph of $h$ has transformed $f$ in two ways: $f\left(x+1\right)$ is a change on the inside of the function, giving a horizontal shift left by 1, and the subtraction by 3 in $f\left(x+1\right)-3$ is a change to the outside of the function, giving a vertical shift down by 3. The transformation of the graph is illustrated below.
Let us follow one point of the graph of $f\left(x\right)=|x|$.
• The point $\left(0,0\right)$ is transformed first by shifting left 1 unit: $\left(0,0\right)\to \left(-1,0\right)$
• The point $\left(-1,0\right)$ is transformed next by shifting down 3 units: $\left(-1,0\right)\to \left(-1,-3\right)$
Below is the graph of $h$.
### Try It
Given $f\left(x\right)=|x|$, sketch a graph of $h\left(x\right)=f\left(x - 2\right)+4$.
### Example: Identifying Combined Vertical and Horizontal Shifts
Write a formula for the graph shown below, which is a transformation of the toolkit square root function.
### EXAMPLE
Write a formula for a transformation of the toolkit reciprocal function $\displaystyle f\left(x\right)=\frac{1}{x}$ that shifts the function’s graph one unit to the right and one unit up.
### EXAMPLE
Given the toolkit function $f\left(x\right)={x}^{2}$, graph $g\left(x\right)=-f\left(x\right)$ and $h\left(x\right)=f\left(-x\right)$. Take note of any surprising behavior for these functions.
#### Combine Shifts and Stretches
When combining transformations, it is very important to consider the order of the transformations. For example, vertically shifting by 3 and then vertically stretching by 2 does not create the same graph as vertically stretching by 2 and then vertically shifting by 3, because when we shift first, both the original function and the shift get stretched, while only the original function gets stretched when we stretch first.
When we see an expression such as $2f\left(x\right)+3$, which transformation should we start with? The answer here follows nicely from the order of operations. Given the output value of $f\left(x\right)$, we first multiply by 2, causing the vertical stretch, and then add 3, causing the vertical shift. In other words, multiplication before addition.
Horizontal transformations are a little trickier to think about. When we write $g\left(x\right)=f\left(2x+3\right)$, for example, we have to think about how the inputs to the function $g$ relate to the inputs to the function $f$. Suppose we know $f\left(7\right)=12$. What input to $g$ would produce that output? In other words, what value of $x$ will allow $g\left(x\right)=f\left(2x+3\right)=12$? We would need $2x+3=7$. To solve for $x$, we would first subtract 3, resulting in a horizontal shift, and then divide by 2, causing a horizontal compression.
This format ends up being very difficult to work with, because it is usually much easier to horizontally stretch a graph before shifting. We can work around this by factoring inside the function.
$\displaystyle f\left(bx+p\right)=f\left(b\left(x+\frac{p}{b}\right)\right)$
Let’s work through an example.
$f\left(x\right)={\left(2x+4\right)}^{2}$
We can factor out a 2.
$f\left(x\right)={\left(2\left(x+2\right)\right)}^{2}$
Now we can more clearly observe a horizontal shift to the left 2 units and a horizontal compression. Factoring in this way allows us to horizontally stretch first and then shift horizontally.
### A General Note: Combining Transformations
When combining vertical transformations written in the form $af\left(x\right)+k$, first vertically stretch by $a$ and then vertically shift by $k$.
When combining horizontal transformations written in the form $f\left(bx+h\right)$, first horizontally shift by $h$ and then horizontally stretch by $\displaystyle \frac{1}{b}$.
When combining horizontal transformations written in the form $f\left(b\left(x+h\right)\right)$, first horizontally stretch by $\displaystyle \frac{1}{b}$ and then horizontally shift by $h$.
Horizontal and vertical transformations are independent. It does not matter whether horizontal or vertical transformations are performed first.
### Example: Finding a Triple Transformation of a Tabular Function
Given the table below for the function $f\left(x\right)$, create a table of values for the function $g\left(x\right)=2f\left(3x\right)+1$.
$x$ 6 12 18 24 $f\left(x\right)$ 10 14 15 17
### Example: Finding a Triple Transformation of a Graph
Use the graph of $f\left(x\right)$ to sketch a graph of $\displaystyle k\left(x\right)=f\left(\frac{1}{2}x+1\right)-3$.
# (11.2.2) – Transformations of Quadratic Functions
The standard form of a quadratic function presents the function in the form
$f\left(x\right)=a{\left(x-h\right)}^{2}+k$
where $\left(h,\text{ }k\right)$ is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function.
The standard form is useful for determining how the graph is transformed from the graph of $y={x}^{2}$. The figure below is the graph of this basic function.
#### Shift Up and Down by Changing the value of $k$
You can represent a vertical (up, down) shift of the graph of $f(x)=x^2$ by adding or subtracting a constant, $k$.
$f(x)=x^2 + k$
If $k>0$, the graph shifts upward, whereas if $k<0$, the graph shifts downward.
### Make Note
Write the equation for the graph of $f(x)=x^2$ that has been shifted up 4 units in the textbox below.
Now write the equation for the graph of $f(x)=x^2$ that has been shifted down 4 units in the textbox below.
Now check yourself!
#### Shift left and right by changing the value of $h$.
You can represent a horizontal (left, right) shift of the graph of $f(x)=x^2$ by adding or subtracting a constant, $h$, to the variable $x$, before squaring.
$f(x)=(x-h)^2$
If $h>0$, the graph shifts toward the right and if $h<0$, the graph shifts to the left.
### Make Note
Write the equation for the graph of $f(x)=x^2$ that has been shifted right 2 units in the textbox below.
Now write the equation for the graph of $f(x)=x^2$ that has been shifted left 2 units in the textbox below.
Now check yourself!
#### Stretch or compress by changing the value of $a$.
You can represent a stretch or compression (narrowing, widening) of the graph of $f(x)=x^2$ by multiplying the squared variable by a constant, a.
$f(x)=ax^2$
The magnitude of a indicates the stretch of the graph. If $|a|>1$, the point associated with a particular x-value shifts farther from the x-axis, so the graph appears to become narrower, and there is a vertical stretch. But if $|a|<1$, the point associated with a particular x-value shifts closer to the x-axis, so the graph appears to become wider, but in fact there is a vertical compression.
### Make Note
Write the equation for the graph of $f(x)=x^2$ that has been has been compressed vertically by a factor of $\displaystyle \frac{1}{2}$ in the textbox below.
Then, write the equation for the graph of $f(x)=x^2$ that has been vertically stretched by a factor of 3.
Now check yourself!
The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form.
$\begin{cases}a{\left(x-h\right)}^{2}+k=a{x}^{2}+bx+c\hfill \\ a{x}^{2}-2ahx+\left(a{h}^{2}+k\right)=a{x}^{2}+bx+c\hfill \end{cases}$
For the linear terms to be equal, the coefficients must be equal.
$\displaystyle -2ah=b,\text{ so }h=-\frac{b}{2a}$.
This is the axis of symmetry we defined earlier. Setting the constant terms equal:
$\large\begin{cases}a{h}^{2}+k=c\hfill \\ \text{ }k=c-a{h}^{2}\hfill \\ \text{ }=c-a-{\left(\frac{b}{2a}\right)}^{2}\hfill \\ \text{ }=c-\frac{{b}^{2}}{4a}\hfill \end{cases}$
In practice, though, it is usually easier to remember that $k$ is the output value of the function when the input is $h$, so $f\left(h\right)=k$.
#### Summary of Transformations
Vertical shift $g\left(x\right)=f\left(x\right)+k$ (up for $k>0$ ) Horizontal shift $g\left(x\right)=f\left(x-h\right)$ (right for $h>0$ ) Vertical reflection $g\left(x\right)=-f\left(x\right)$ Horizontal reflection $g\left(x\right)=f\left(-x\right)$ Vertical stretch $g\left(x\right)=af\left(x\right)$ ( $a>0$) Vertical compression $g\left(x\right)=af\left(x\right)$ $\left(01$ ) |
# Question Video: Finding the Value of an Unknown given Three Consecutive Terms of an Arithmetic Series Each in the Form of a Combination Mathematics
Given that 3 Ć š Cāā, 4 Ć š Cāā, 6 Ć š Cāā is an arithmetic sequence, find all possible values of š.
05:49
### Video Transcript
Given that three multiplied by š choose 10, four multiplied by š choose 11, six multiplied by š choose 12 is an arithmetic sequence, find all possible values of š.
We begin by recalling that in any arithmetic sequence, there is a common or constant difference between consecutive terms. This means that the difference between the third term, š sub three, and the second term, š sub two, must be equal to the difference between the second term and the first term. Substituting our three terms, we have six multiplied by š choose 12 minus four multiplied by š choose 11 is equal to four multiplied by š choose 11 minus three multiplied by š choose 10. We can rearrange and simplify this equation by adding four multiplied by š choose 11 and three multiplied by š choose 10 to both sides.
On the left-hand side, we have six multiplied by š choose 12 plus three multiplied by š choose 10. And on the right-hand side, we have eight multiplied by š choose 11. Next, we recall that when dealing with combinations, š choose š is equal to š factorial over š factorial multiplied by š minus š factorial. š choose 12 is therefore equal to š factorial over 12 factorial multiplied by š minus 12 factorial. And multiplying this by six, our first term becomes six multiplied by š factorial over 12 factorial multiplied by š minus 12 factorial.
We can rewrite the other two terms in our equation in the same manner. We notice that each of the three terms has a common factor of š factorial. So we can divide through by this. We can simplify this further by recalling that š factorial can be written as š multiplied by š minus one factorial. And this is also equal to š multiplied by š minus one multiplied by š minus two factorial, where š is greater than or equal to two. Using this property, we can rewrite the denominator of our first term as 12 multiplied by 11 multiplied by 10 factorial multiplied by š minus 12 factorial. In the same way, the second term can be rewritten as three over 10 factorial multiplied by š minus 10 multiplied by š minus 11 multiplied by š minus 12 factorial. And the sum of these two terms is equal to eight over 11 multiplied by 10 factorial multiplied by š minus 11 multiplied by š minus 12 factorial.
All three denominators have a common factor of 10 factorial multiplied by š minus 12 factorial. Clearing some space, our equation simplifies to six over 12 multiplied by 11 plus three over š minus 10 multiplied by š minus 11 is equal to eight over 11 multiplied by š minus 11. The first term has a common factor of six in the numerator and denominator. We can then multiply through by š minus 10 multiplied by š minus 11. After canceling the common factors, we can begin to distribute our parentheses. š minus 10 multiplied by š minus 11 is equal to š squared minus 21š plus 110. And eight multiplied by š minus 10 is equal to eight š minus 80.
We are now in a position to eliminate the denominators by multiplying through by 22. Our equation becomes š squared minus 21š plus 110 plus 66 is equal to 16š minus 160. And by subtracting 16š and adding 160 to both sides, we have the quadratic equation š squared minus 37š plus 336 equals zero. This can be solved using the quadratic formula or by factoring. Since negative 21 and negative 16 sum to negative 37 and the two values have a product of 336, our equation simplifies to š minus 21 multiplied by š minus 16 is equal to zero. And as at least one of the expressions in the parentheses must equal zero, we have two solutions š equals 21 and š equals 16. If the three given terms form an arithmetic sequence, then the two possible values of š are 21 and 16. We can check these by substituting them back in to the sequence. |
Did you recognize that 55 is the sum of the initially 10 herbal numbers i.e. 1+2+3+4+5+6+7+8+9+10=55 and also it is additionally the sum of the squares of the initially 5 herbal numbers i.e. 1+4+9+16+25=55? In this lesboy, we will certainly calculate the components of 55, the prime factors of 55, and the determinants of 55 in pairs. We will additionally go via a few addressed examples.
You are watching: What are all the factors of 55
Factors of 55: 1, 5, 11 and 55Prime Factorization of 55: 55 = 11 × 5
1 What Are the Factors of 55? 2 Tips and also Tricks 3 How to Calculate the Factors of 55? 4 Factors of 55 by Prime Factorization 5 Thinking Out of The Box! 6 Factors of 55 in Pairs 7 FAQs on Factors of 55
## What Are the Factors of 55?
The numbers through which 55 is divisible are the components of 55. For example, 55 is divisible by 11. Hence, 11 is a variable of 55.
Can you think of other components of 55?
Tips and also Tricks:
When you divide a number by its element, the quotient of the division is likewise a factor.An odd number cannot have also factors.Divisibility rules make the process of finding factors easy.
## How to Calculate the Factors of 55?
We have learnt that the factors of 55 have the right to be discovered making use of division. We can discover the factors of 55 using multiplication also. Can you think of two numbers whose product is 55? Can you think of all such possibilities?
The multiplicands of each such product are the factors of 55.
Hence, the determinants of 55 are 1, 5, 11 and also 55.
Explore factors utilizing illustrations and also interenergetic examples
## Factors of 55 by Prime Factorization
The prime factorization of 55 is expushing 55 as the product of prime numbers which provides the result as 55. Hence, the prime factorization of 55 is 55 = 5 × 11. From the prime factorization of 55, it is clear that 5 and 11 are the prime determinants of 55. We understand that 1 is the factor of eincredibly number.
Therefore, the components of 55 by prime factorization are 1, 5, 11, and 55.
Think Tank:
Can you uncover the prevalent components of 55, 60 and also 65?
## Factors of 55 in Pairs
The pair components of 55 are derived by composing 55 as a product of two numbers in all feasible means. In each product, both multiplicands form the pair determinants of 55.
The factors of 55 are 1, 5, 11, and also 55.
Product that Results in 55
Pair Factors of 55
1 × 55
(1, 55)
5 × 11
(5, 11)
The negative pair determinants of 55 are (-1, -55) and (-5, -11).
Example 1: Isabella, a dance teacher, wants to arselection 55 students of her dance course into teams for dance exercise. She wants each group to have equal variety of students. She also doesn"t want to form a group for either one student or all students. Based on this indevelopment, what can be the possible size(s) of the group?
Solution:
We already learned that the determinants of 55 are 1, 5, 11, and also 55. Because a team cannot have 1 or all the students, each team can have actually either 5 or 11 students.
Hence, the dimension of each group = 5 (or) 11.
See more: Determining Which Segments To Target For Firms Depends On An Interplay Between ______ Issues.
Example 2: William is stuck with finding the prevalent factors of 55 and 60. Can we assist him?
Solution:
We have actually currently learned that, the components of 55 are 1, 5, 11 and 55. The determinants of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. Hence, the components that are prevalent to both 55 and also 60 are 1 and 5. |
Section 4: Point-Slope Form of a Line
# Elementary and Intermediate Algebra: Graphs & Models (3rd Edition)
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Section 3.4 The Point-Slope Form of a Line 293 Version: Fall 2007 3.4 The Point-Slope Form of a Line In the last section, we developed the slope-intercept form of a line ( y = mx + b ). The slope-intercept form of a line is applicable when you’re given the slope and y -intercept of the line. However, there will be times when the y -intercept is unknown. Suppose for example, that you are asked to find the equation of a line that passes through a particular point P ( x 0 , y 0 ) with slope = m . This situation is pictured in Figure 1 . x y P ( x 0 ,y 0 ) Q ( x,y ) Figure 1. A line through ( x 0 , y 0 ) with slope m . Let the point Q ( x, y ) be an arbitrary point on the line. We can determine the equation of the line by using the slope formula with points P and Q . Hence, Slope = y x = y y 0 x x 0 . Because the slope equals m , we can set Slope = m in this last result to obtain m = y y 0 x x 0 . If we multiply both sides of this last equation by x x 0 , we get m ( x x 0 ) = y y 0 , or exchanging sides of this last equation, y y 0 = m ( x x 0 ) . This last result is the equation of the line. Copyrighted material. See: 1
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294 Chapter 3 Linear Functions Version: Fall 2007 The Point-Slope Form of a Line . If line L passes through the point ( x 0 , y 0 ) and has slope m , then the equation of the line is y y 0 = m ( x x 0 ) . (1) This form of the equation of a line is called the point-slope form . To use the point-slope form of a line, follow these steps. Procedure for Using the Point-Slope Form of a Line . When given the slope of a line and a point on the line, use the point-slope form as follows: 1. Substitute the given slope for m in the formula y y 0 = m ( x x 0 ) . 2. Substitute the coordinates of the given point for x 0 and y 0 in the formula y y 0 = m ( x x 0 ) . For example, if the line has slope 2 and passes through the point (3 , 4) , then substitute m = 2 , x 0 = 3 , and y 0 = 4 in the formula y y 0 = m ( x x 0 ) to obtain y 4 = 2( x 3) . l⚏ Example 2. Draw the line that passes through the point P ( 3 , 2) and has slope m = 1 / 2 . Use the point-slope form to determine the equation of the line. First, plot the point P ( 3 , 2) , as shown in Figure 2 (a). Starting from the point P ( 3 , 2) , move 2 units to the right and 1 unit up to the point Q ( 1 , 1) . The line through the points P and Q in Figure 2 (a) now has slope m = 1 / 2 . x y P ( - 3 , - 2) Q ( - 1 , - 1) Δ x =2 Δ y =1 x y R (0 , - 0 . 5) (a) The line through P ( 3 , 2) with slope m = 1 / 2 . (b) Checking the y -intercept. Figure 2.
Section 3.4 The Point-Slope Form of a Line 295 Version: Fall 2007 To determine the equation of the line in Figure 2 (a), we will use the point-slope form of the line y y 0 = m ( x x 0 ) . (3) The slope of the line is m = 1 / 2 and the given point is P ( 3 , 2) , so ( x 0 , y 0 ) = ( 3 , 2) . In equation (3) , set m = 1 / 2 , x 0 = 3 , and y 0 = 2 , obtaining y ( 2) = 1 2 ( x ( 3)) , or equivalently, y + 2 = 1 2 ( x + 3) . (4) This is the equation of the line in Figure 2 (a).
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Problem of the Week
Problem C and Solution
To the Next Level
Problem
Maddie Leet is participating in the first round of a math competition. She writes one contest a month in each of the first ten months of the school year. She can earn up to 100 points on each contest. On each of the first five contests she averaged 68 points. On the next three contests she averaged 80 points. In order to advance to the next round of the competition, she must obtain a minimum total of 750 points on the ten contests.
What is the minimum average Maddie requires on the final two contests in order to able to advance to the next round of the competition?
Solution
To determine an average of some numbers, we add the numbers together and divide the sum by the number of numbers.
$\mbox{Average}=\frac{\mbox{Sum of Numbers}}{\mbox{Number of Numbers}}$
Therefore, to determine the sum of the numbers we multiply their average by the number of numbers.
\begin{aligned} \mbox{Sum of Numbers}&=\mbox{Average}\times \mbox{Number of Numbers}\\ \mbox{Total Score for First 5 Contests}&=68\times 5=340\\ \mbox{Total Score for Next 3 Contests}&=80\times 3=240\\[3mm] \mbox{Total Score to Move On}&=750\\ \mbox{Total Score Needed for Last Two Contests}&=750-340-240=170\\ \mbox{Average Score for Final 2 Contests}&=170\div 2=85\\[-4mm]\end{aligned}
Maddie needs to average 85 points on her last two contests to have a total score of exactly 750 points. Therefore, the minimum average Maddie needs on the final two contests in order to advance to the next round of the math competition is 85 points. If she averages anything less than 85 points she will not move on. If she averages more than 85 points she will obtain a total score over 750 points and will move on to the next round. |
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# Class 10 NCERT Solutions- Chapter 14 Statistics – Exercise 14.4
• Last Updated : 03 Feb, 2023
### Question 1. The following distribution gives the daily income of 50 workers if a factory. Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.
Solution:
According to the question, we convert the given distribution to a less than type cumulative frequency distribution,
Now according to the table we plot the points that are corresponding to the ordered pairs (120, 12), (140, 26), (160, 34), (180, 40), and (200, 50) on a graph paper. Here x-axis represents the upper limit and y-axis represent the frequency. The curve obtained from the graph is known as less than type ogive curve.
### Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:
According to the given table, we get the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35). Now using these points we draw an ogive, where the x-axis represents the upper limit and y-axis represents the frequency. The curve obtained is known as less than type ogive.
Now, locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From this point, now we draw a perpendicular line to the x-axis and the intersection point which is perpendicular to x-axis is the median of the given data. After, locating point now we create a table to find the mode:
The class 46 – 48 has the maximum frequency, hence, this is the modal class
= 46, h = 2, f1 = 14, f0 = 5 and f2 = 4
Now we find the mode:
Mode =
On substituting the values in the given formula, we get
= 46 + 0.95 = 46.95
Hence, the mode is verified.
### Change the distribution to a more than type distribution and draw its ogive.
Solution:
According to the question, we change the distribution to a more than type distribution.
Now, according to the table we draw the ogive by plotting the points. Here, the a-axis represents the upper limit and y-axis represents the cumulative frequency. And the points are(50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on this graph paper. The graph obtained is known as more than type ogive curve.
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A cube is a three-dimensional geometric shape that is composed of six square faces, twelve edges, and eight vertices. In this article, we will focus on exploring the number of edges a cube has and delve into the properties and characteristics of this fascinating shape.
## The Definition of a Cube
Before we dive into the number of edges a cube possesses, let’s first establish what a cube is. A cube is a regular polyhedron, which means it has congruent faces and identical angles between faces. In the case of a cube, all six faces are squares, and each face meets at a right angle with its adjacent faces.
## The Anatomy of a Cube
A cube has several defining features that contribute to its unique properties. Understanding these features is crucial to determining the number of edges a cube has:
• Faces: A cube has six faces, each of which is a square. These faces are congruent and identical in size.
• Edges: A cube has twelve edges, which are the straight lines where two faces meet. Each edge is shared by two faces.
• Vertices: A cube has eight vertices, which are the points where three edges meet. Each vertex is shared by three faces.
• Diagonals: A cube has four space diagonals, which are the straight lines connecting opposite vertices of the cube.
## Calculating the Number of Edges
Now that we understand the basic components of a cube, let’s determine the number of edges it possesses. As mentioned earlier, a cube has twelve edges. To visualize this, imagine a cube and count the number of straight lines where two faces meet. Each of these lines represents an edge.
Alternatively, we can calculate the number of edges using a formula. The formula for determining the number of edges in a cube is:
Number of Edges = 12
Therefore, a cube always has twelve edges, regardless of its size or orientation.
## Real-World Examples
Cubes are not just abstract mathematical concepts; they have numerous real-world applications. Let’s explore a few examples where cubes play a significant role:
### 1. Rubik’s Cube
The Rubik’s Cube is a popular puzzle toy invented by Ernő Rubik in 1974. It consists of a 3x3x3 cube with colored stickers on each face. The cube can be twisted and turned to scramble the colors, and the objective is to solve it by returning each face to a single color. The Rubik’s Cube has twelve edges, which are crucial for its functionality and solving strategies.
### 2. Dice
A traditional six-sided die, commonly used in board games and gambling, is essentially a cube. Each face of the die represents a number from one to six, and the edges allow for randomization when rolled. A die has twelve edges, contributing to its fairness and unpredictability.
### 3. Building Blocks
Cubes are often used as building blocks in construction and architecture. For example, Lego bricks, which are widely popular among children and adults alike, are shaped like cubes. The edges of these cubes allow for interlocking and building various structures, fostering creativity and spatial awareness.
## Q&A
### Q1: Can a cube have more than twelve edges?
No, a cube cannot have more than twelve edges. The number of edges in a cube is fixed at twelve, regardless of its size or orientation.
### Q2: How many edges does a rectangular prism have?
A rectangular prism has twelve edges, just like a cube. However, unlike a cube, a rectangular prism has rectangular faces instead of square faces.
### Q3: What is the difference between an edge and a face?
An edge is a straight line where two faces of a cube meet, while a face is a flat surface of the cube. In other words, an edge is a line segment, and a face is a two-dimensional shape.
### Q4: Can a cube have curved edges?
No, a cube cannot have curved edges. By definition, a cube has straight edges that meet at right angles.
### Q5: How many edges does a triangular pyramid have?
A triangular pyramid, also known as a tetrahedron, has six edges. Each edge connects one vertex to another vertex.
## Summary
In conclusion, a cube has twelve edges, which are the straight lines where two faces meet. Understanding the anatomy of a cube, including its faces, edges, vertices, and diagonals, is essential to grasp its properties fully. Real-world examples such as the Rubik’s Cube, dice, and building blocks demonstrate the practical applications of cubes. Remember, a cube’s edges are fixed at twelve, making it a fascinating and well-defined geometric shape.
Ishaan Sharma is a tеch bloggеr and cybеrsеcurity analyst spеcializing in thrеat hunting and digital forеnsics. With еxpеrtisе in cybеrsеcurity framеworks and incidеnt rеsponsе, Ishaan has contributеd to fortifying digital dеfеnsеs. |
# 5.5 Divide Monomials
## Learning Objectives
By the end of this section, you will be able to:
• Simplify expressions using the Quotient Property for Exponents
• Simplify expressions with zero exponents
• Simplify expressions using the Quotient to a Power Property
• Simplify expressions by applying several properties
• Use the definition of a negative exponent
• Simplify expressions with integer exponents
• Simplify expressions by applying several properties with negative exponents
• Divide monomials
### Try It
Before you get started, take this readiness quiz:
1) Simplify: $\frac{8}{24}$
2) Simplify: $(2{m}^{3})^{5}$
3) Simplify: $\frac{12x}{12y}$
## Simplify Expressions Using the Quotient Property for Exponents
Earlier in this chapter, we developed the properties of exponents for multiplication. We summarize these properties below.
### Summary of Exponent Properties for Multiplication
If $a$ and $b$ are real numbers, and $m$ and $n$ are whole numbers, then
Product Property ${a}^{m}\times{{a}^{n}}={a}^{m+n}$ $({a}^{m})^{n}={a}^{m\times{n}}$ ${(ab)}^{m}={a}^{m}{b}^{m}$
Now we will look at the exponent properties for division. A quick memory refresher may help before we get started. You have learned to simplify fractions by dividing out common factors from the numerator and denominator using the Equivalent Fractions Property. This property will also help you work with algebraic fractions—which are also quotients.
### Equivalent Fractions Property
If $a$, $b$, and $c$ are whole numbers where $b\neq{0}, c\neq{0}$ then
$\displaystyle\frac{a}{b}=\frac{a\times{c}}{b\times{c}}$ and $\displaystyle\frac{a\times{c}}{b\times{c}}=\frac{a}{b}$
As before, we’ll try to discover a property by looking at some examples.
Example 1 Example 2 Consider $\displaystyle\frac{x^5}{x^2}$ $\displaystyle\frac{x^2}{x^3}$ What do they mean? $\displaystyle\frac{{x}\cdot {x}\cdot {x}\cdot {x}\cdot {x}}{{x}\cdot {x}}$ $\displaystyle\frac{{x}\cdot {x}}{{x}\cdot{x}\cdot{x}}$ Use the Equivalent Fractions Property. $\displaystyle\frac{\cancel x\cdot\cancel x\cdot x\cdot x\cdot x}{\cancel x\cdot\cancel x}$ $\displaystyle\frac{\cancel x\cdot\cancel x\cdot1}{\cancel x\cdot\cancel x\cdot x}$ Simplify $x^3$ $\displaystyle\frac{1}{x}$
Notice, in each case the bases were the same and we subtracted exponents.
When the larger exponent was in the numerator, we were left with factors in the numerator.
When the larger exponent was in the denominator, we were left with factors in the denominator—notice the numerator of 1.
We write:
$\displaystyle\frac{x^5}{x^2}$ $\displaystyle\frac{x^2}{x^3}$ $\displaystyle x^{5-2}$ $\displaystyle\frac{1}{x^{3-2}}$ $\displaystyle x^3$ $\displaystyle\frac{1}{x}$
This leads to the Quotient Property for Exponents.
### Quotient Property for Exponents
If $a$ is a real number, $a\neq{0}$, and $n$ are whole numbers, then
$\displaystyle\frac{a^m}{a^n}=a^{m-n}$, $m>n$ and $\displaystyle\frac{a^m}{a^n}=\frac{1}{a^{m-n}}$, $n>m$
A couple of examples with numbers may help to verify this property.
\begin{align*}\frac{3^4}{3^2}&=3^{4-2}\\[2ex]\frac{81}{9}&=3^{2}\\[2ex]9&=9\checkmark\end{align*} \begin{align*}\frac{5^2}{5^3}&=\frac1{5^{3-2}}\\[2ex]\frac{25}{125}&=\frac{1}{5^{1}}\\[2ex]\frac{1}{5}&=\frac{1}{5}\checkmark\end{align*}
### Example 5.5.1
Simplify:
a. $\frac{{x}^{9}}{{x}^{7}}$
b. $\frac{{3}^{10}}{{3}^{2}}$
Solution
To simplify an expression with a quotient, we need to first compare the exponents in the numerator and denominator.
a.
Step 1: Since $9 > 7$, there are more factors of $x$ in the numerator.
$\frac{x^9}{x^7}$
Step 2: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$
$\begin{eqnarray*}&=&x^{{\color{red}{9-7}}}\\\text{Simplify.}\;\;&=&x^2\end{eqnarray*}$
b.
Step 1: Since $10 > 2$, there are more factors of $x$ in the numerator.
$\frac{3^{10}}{3^2}$
Step 2: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$
$\begin{eqnarray*}&=&3^{\color{red}{{10\;-\;2}}}\\\text{Simplify.}\;\;&=&3^8 \end{eqnarray*}$
Notice that when the larger exponent is in the numerator, we are left with factors in the numerator.
### Try It
4) Simplify:
a. $\frac{{x}^{15}}{{x}^{10}}$
b. $\frac{{6}^{14}}{{6}^{5}}$
Solution
a. ${x}^{5}$
b. ${6}^{9}$
5) Simplify:
a. $\frac{{y}^{43}}{{y}^{37}}$
b. $\frac{{10}^{15}}{{10}^{7}}$
Solution
a. ${y}^{6}$
b. ${10}^{8}$
### Example 5.5.2
Simplify:
a. $\frac{{b}^{8}}{{b}^{12}}$
b. $\frac{{7}^{3}}{{7}^{5}}$
Solution
To simplify an expression with a quotient, we need to first compare the exponents in the numerator and denominator.
a.
Step 1: Since $12 > 8$, there are more factors of $b$ in the denominator.
$\frac{b^8}{b^{12}}$
Step 2: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}}$
$\begin{eqnarray*}&=&\frac{\color{red}{1}}{b^{\color{red}{{12\;-\;8}}}}\\\text{Simplify.}\;\;&=&\frac{1}{b^4}\end{eqnarray*}$
b.
Step 1: Since $5 > 3$, there are more factors of $3$ in the denominator.
$\frac{7^3}{7^5}$
Step 2: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}}$
$\begin{eqnarray*}&=&\frac{\color{red}{1}}{7^{\color{red}{5-3}}}\\[1ex]\text{Simplify.}\;\;&=&\frac{1}{7^2}\\[1ex] \text{Simplify.}\;\;&=&\frac{1}{49}\end{eqnarray*}$
Notice that when the larger exponent is in the denominator, we are left with factors in the denominator.
### Try It
6) Simplify:
a. $\frac{{x}^{18}}{{x}^{22}}$
b.
$\frac{{12}^{15}}{{12}^{30}}$
Solution
a. $\frac{1}{{x}^{4}}$
b. $\frac{1}{{12}^{15}}$
7) Simplify:
a. $\frac{{m}^{7}}{{m}^{15}}$
b. $\frac{{9}^{8}}{{9}^{19}}$
Solution
a. $\frac{1}{{m}^{8}}$
b. $\frac{1}{{9}^{11}}$
Notice the difference in the two previous examples:
• If we start with more factors in the numerator, we will end up with factors in the numerator.
• If we start with more factors in the denominator, we will end up with factors in the denominator.
The first step in simplifying an expression using the Quotient Property for Exponents is to determine whether the exponent is larger in the numerator or the denominator. Later in this chapter, we will explore this further while using negative exponents.
### Example 5.5.3
Simplify:
a. $\frac{{a}^{5}}{{a}^{9}}$
b. $\frac{{x}^{11}}{{x}^{7}}$
Solution
a. Is the exponent of $a$ larger in the numerator or denominator? Since $9 > 5$, there are more $a$‘s in the denominator and so we will end up with factors in the denominator.
Step 1: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}}$
$\begin{eqnarray*}&=&\frac{\color{red}{1}}{a^{\color{red}{{9\;-\;5}}}}\\\text{Simplify.}\;\;&=&\frac{1}{a^4}\end{eqnarray*}$
b. Notice there are more factors of $x$ in the numerator, since $11 > 7$. So we will end up with factors in the numerator.
Step 1: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}}$
$\begin{eqnarray*}&=&x^{{\color{red}{11\;-\;7}}}\\\text{Simplify.}\;\;&=&x^4\end{eqnarray*}$
### Try It
8) Simplify:
a. $\frac{{b}^{19}}{{b}^{11}}$
b. $\frac{{z}^{5}}{{z}^{11}}$
Solution
a. ${b}^{8}$
b. $\frac{1}{{z}^{6}}$
9) Simplify:
a. $\frac{{p}^{9}}{{p}^{17}}$
b. $\frac{{w}^{13}}{{w}^{9}}$
Solution
a. $\frac{1}{{p}^{8}}$
b. ${w}^{4}$
## Simplify Expressions with an Exponent of Zero
A special case of the Quotient Property is when the exponents of the numerator and denominator are equal, such as an expression like $\frac{{a}^{m}}{{a}^{m}}$. From your earlier work with fractions, you know that:
$\frac{2}{2}=1$ $\frac{17}{17}=1$ $\frac{-43}{-43}=1$
In words, a number divided by itself is $1$. So, $\frac{x}{x}=1$, for any $x$ $({x} \neq {0})$, since any number divided by itself is $1$.
The Quotient Property for Exponents shows us how to simplify $\frac{{a}^{m}}{{a}^{n}}$ when $m>n$ and when $n < m$ by subtracting exponents. What if $m=n$?
Consider $\frac{8}{8}$, which we know is $1$.
\begin{align*} &\;&\frac{8}{8}&=1\\[1ex] &\text{Write}\;8\;\text{as}\;2^3\;&\frac{{2}^{3}}{{2}^{3}}&=1\\[1ex] &\text{Subtract exponents}\;&{2}^{3-3}&=1\\ &\text{Simplify}\;&2^0&=1 \end{align*}
Now we will simplify $\frac{{a}^{m}}{{a}^{m}}$ in two ways to lead us to the definition of the zero exponent. In general, for ${a}\neq{0}$ :
$\begin{array}{cc}\displaystyle\frac{a^m}{a^m}\;\;&\displaystyle\frac{a^m}{a^m}\\[3ex]a^{m-m}\;\;&\displaystyle\frac{\overbrace{\cancel a\cdot \cancel a\cdot...\cdot\cancel a}^{{\color{blue}{m\text{ factors}}}}}{\underbrace{\cancel a\cdot \cancel a\cdot...\cdot\cancel a}_{{\color{blue}{m\text{ factors}}}}}\\[3ex]a^0\;\;&1\end{array}$
We see $\frac{{a}^{m}}{{a}^{m}}$ simplifies to ${a}^{0}$ and to $1$. So ${a}^{0}=1$
### Zero Exponent
If $a$ is a non-zero number, then ${a}^{0}=1$.
Any nonzero number raised to the zero power is $1$.
In this text, we assume any variable that we raise to the zero power is not zero.
### Example 5.5.4
Simplify:
a. ${9}^{0}$
b. ${n}^{0}$
Solution
The definition says any non-zero number raised to the zero power is $1$.
a.
Step 1: Use the definition of the zero exponent.
$\begin{eqnarray*}&=&9^0\\&=&1\end{eqnarray*}$
b.
Step 1: Use the definition of the zero exponent.
$\begin{eqnarray*}&=&n^0\\&=&1\end{eqnarray*}$
### Try It
10) Simplify:
a. ${15}^{0}$
b. ${m}^{0}$
Solution
a. 1
b. 1
11) Simplify:
a. ${k}^{0}$
b. ${29}^{0}$
Solution
a. 1
b. 1
Now that we have defined the zero exponent, we can expand all the Properties of Exponents to include whole number exponents.
What about raising an expression to the zero power? Let’s look at $(2x)^{0}$. We can use the product to a power rule to rewrite this expression.
\begin{align*} &\;&(2x)^{0}\\[1ex] &\text{Use the product to a power rule.}\;&{2}^{0}{x}^{0}\\[1ex] &\text{Use the zero exponent property.}\;&1\times{1}\\ &\text{Simplify}\;&1 \end{align*}
This tells us that any nonzero expression raised to the zero power is one.
### Example 5.5.5
Simplify:
a. $(5b)^{0}$
b. $(-4{a}^{2}b)^{0}$
Solution
a.
Step 1: Use the definition of the zero exponent.
$\begin{eqnarray*}&=&(5b)^0\\&=&1\end{eqnarray*}$
b.
Step 1: Use the definition of the zero exponent.
$\begin{eqnarray*}&=&(-4a^{2b})^0\\&=&1\end{eqnarray*}$
### Try It
12) Simplify:
a. $(11z)^{0}$
b. $(-11p{q}^{3})^{0}$
Solution
a. $1$
b. $1$
13) Simplify:
a. $(-6d)^{0}$
b. $(-8{m}^{2}{n}^{3})^{0}$
Solution
a. $1$
b. $1$
## Simplify Expressions Using the Quotient to a Power Property
Now we will look at an example that will lead us to the Quotient to a Power Property.
\begin{align*} &\;&\;&\left(\frac{x}{y}\right)^{3}\\[1ex] &\text{This means:}&\;&\frac{x}{y}\cdot\frac{x}{y}\cdot\frac{x}{y}\\[1ex] &\text{Multiply the fractions.}&\;&\frac{x\cdot{x}\cdot{x}}{y\cdot{y}\cdot{y}}\\[1ex] &\text{Write with exponents.}&\;&\frac{{x}^{3}}{{y}^{3}} \end{align*}
Notice that the exponent applies to both the numerator and the denominator.
\begin{align*} &\text{We write:}\;&\left(\frac{x}{y}\right)^{3}\\[1ex] &\;&\frac{{x}^{3}}{{y}^{3}}\;\;\;\end{align*}
This leads to the Quotient to a Power Property for Exponents.
### Quotient to a Power Property for Exponents
If $a$ and $b$ are real numbers, $b\neq{0}$, and $m$ is a counting number, then
$(\frac{a}{b})^{m}=\frac{{a}^{m}}{{b}^{m}}$
To raise a fraction to a power, raise the numerator and denominator to that power.
$\begin{eqnarray*}(\frac{2}{3})^3&=&\frac{2^3}{3^3}\\\frac{2}{3}\cdot\frac{2}{3}\cdot\frac{2}{3}&=&\frac{8}{27}\\\frac{8}{27}&=& \frac{8}{27}\checkmark\end{eqnarray*}$
### Example 5.5.6
Simplify:
a. $(\frac{3}{7})^{2}$
b. $(\frac{b}{3})^{4}$
c. $\left(\frac{k}{j}\right)^3$
Solution
a.
Step 1: Use the Quotient Property, $(\frac{a}{b})^{m}=\frac{{a}^{m}}{{b}^{m}}$
$\begin{eqnarray*}&=&{b}^{0}\\ \text{Simplify.}\;\;&=&1 \end{eqnarray*}$
b.
Step 1: Use the Quotient Property, $(\frac{a}{b})^{m}=\frac{{a}^{m}}{{b}^{m}}$
$\begin{eqnarray*}&=&\frac{b^{\color{red}{4}}}{3^{\color{red}{4}}}\\\text{Simplify.}\;\;&=&\frac{b^{4}}{81}\end{eqnarray*}$
c.
Step 1: Raise the numerator and denominator to the third power.
$\displaystyle\frac{k^{\color{red}{3}}}{j^{\color{red}{3}}}$
### Try It
14) Simplify:
a. $(\frac{5}{8})^{2}$
b. $(\frac{p}{10})^{4}$
c. $(\frac{m}{n})^{7}$
Solution
a. $\frac{25}{64}$
b. $\frac{{p}^{4}}{10,000}$
c. $\frac{{m}^{7}}{{n}^{7}}$
15) Simplify:
a. $(\frac{1}{3})^{3}$
b. $(\frac{-2}{q})^{3}$
c. $(\frac{w}{x})^{4}$
Solution
a. $\frac{1}{27}$
b. $\frac{-8}{{q}^{3}}$
c. $\frac{{w}^{4}}{{x}^{4}}$
## Simplify Expressions by Applying Several Properties
We’ll now summarize all the properties of exponents so they are all together to refer to as we simplify expressions using several properties. Notice that they are now defined for whole number exponents.
### Summary of Exponent Properties
If $a$ and $b$ are real numbers, and $m$ and $n$ are whole numbers, then
Product Property ${a}^{m}\times{a}^{n}={a}^{m+n}$ $({a}^{m})^{n}={a}^{m\times{n}}$ $(ab)^{m}={a}^{m}{b}^{m}$ $\frac{{a}^{m}}{{b}^{m}}={a}^{m-n}, a\neq{0} , m>n$ $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}},a\neq{0},n>m$ ${a}^{0}=1,a\neq{0}$ $(\frac{a}{b})^{m}=\frac{{a}^{m}}{{b}^{m}},b\neq{0}$
### Example 5.5.7
Simplify: $\frac{(y^4)^2}{y^6}$
Solution
Step 1: Multiply the exponents in the numerator.
$\frac{{y}^{8}}{{y}^{6}}$
Step 2: Subtract the exponents.
${y}^{2}$
### Try It
16) Simplify: $\frac{({m}^{5})^4}{m^7}$
Solution
${m}^{13}$
17) Simplify: $\frac{{({k}^{2})}^{6}}{{k}^{7}}$
Solution
${k}^{5}$
### Example 5.5.8
Simplify: $\frac{{b}^{12}}{{({b}^{2})}^{6}}$
Solution
Step 1: Multiply the exponents in the numerator.
$\frac{{b}^{12}}{{b}^{12}}$
Step 2: Subtract the exponents.
$\begin{eqnarray*}&=&{b}^{0}\\ \text{Simplify.}\;\;&=&1 \end{eqnarray*}$
### Try It
18) Simplify: $\frac{{n}^{12}}{{({n}^{3}})^{4}}$
Solution
$1$
19) Simplify: $\frac{{x}^{15}}{(x^3)^{5}}$
Solution
$1$
### Example 5.5.9
Simplify: $\left(\frac{{y}^{9}}{{y}^{4}}\right)^{2}$
Solution
Step 1: Remember parentheses come before exponents.
Notice the bases are the same, so we can simplify inside the parentheses.
Subtract the exponents.
$(y^5)^2$
Step 2: Multiply the exponents.
${y}^{10}$
### Try It
20) Simplify: $\left(\frac{{r}^{5}}{{r}^{3}}\right)^{4}$
Solution
${r}^{8}$
21) Simplify: $\left(\frac{{v}^{6}}{{v}^{4}}\right)^{3}$
Solution
${v}^{6}$
### Example 5.5.10
Simplify: $\left(\frac{{j}^{2}}{{k}^{3}}\right)^{4}$
Solution
Here we cannot simplify inside the parentheses first, since the bases are not the same.
Step 1: Raise the numerator and denominator to the third power using the Quotient to a Power Property, $(\frac{a}{b})^{m}=\frac{{a}^{m}}{{b}^{m}}$
$\left(\frac{{j}^{2}}{{k}^{3}}\right)^{4}$
Step 2: Use the Power Property and simplify.
$\frac{{j}^{8}}{{k}^{12}}$
### Try It
22) Simplify: $\left(\frac{a^3}{b^2}\right)^{4}$
Solution
$\frac{{a}^{12}}{{b}^{8}}$
23) Simplify: $\left(\frac{q^7}{r^5}\right)^3$
Solution
$\frac{{q}^{21}}{{r}^{15}}$
### Example 5.5.11
Simplify: $\left(\frac{2{m}^{2}}{5n}\right)^{4}$
Solution
Step 1: Raise the numerator and denominator to the fourth power, using the Quotient to a Power Property, $\left(\frac{a}{b}\right)^m=\;\frac{a^m}{b^m}$
$\frac{(2m^{2})^{4}}{{(5n)}^{4}}$
Step 2: Raise each factor to the fourth power.
$\frac{(2m^{2})^{4}}{{(5n)}^{4}}$
Step 3: Use the Power Property and simplify.
$\frac{16{m}^{8}}{625{n}^{4}}$
### Try It
24) Simplify: $\left(\frac{7x^3}{9y}\right)^2$
Solution
$\frac{49{x}^{6}}{81{y}^{2}}$
25) Simplify: $\left(\frac{3x^4}{7y}\right)^2$
Solution
$\frac{9{x}^{8}}{49{y}^{2}}$
### Example 5.5.12
Simplify: $\frac{\left(x^3\right)^4\times\left(x^2\right)^5}{\left(x^6\right)^5}$
Solution
Step 1: Use the Power Property, ${(a^m)}^{n}={a}^{mn}$
$\frac{\left(x^{12}\right)\times\left(x^{10}\right)}{\left(x^{30}\right)}$
Step 2: Add the exponents in the numerator.
$\frac{x^{22}}{x^{30}}$
Step 3: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}}$
$\frac{1}{{x}^{8}}$
### Try It
26) Simplify: $\frac{{({a}^{2})}^{3}{({a}^{2})}^{4}}{({a}^{4})^{5}}$
Solution
$\frac{1}{{a}^{6}}$
27) Simplify: $\frac{{({p}^{3})}^{4}{({p}^{5})}^{3}}{{({p}^{7})}^{6}}$
Solution
$\frac{1}{{p}^{15}}$
### Example 5.5.13
Simplify: $\frac{(10p^3)^2}{(5p)^3\cdot\left(2p^5\right)^4}$
Solution
Step 1: Use the Product to a Power Property, ${(ab)}^{m}={a}^{m}{b}^{m}$
$\frac{{(10)}^{2}{({p}^{3})}^{2}}{(5)^3(p)^{3}\cdot (2)^4(p^5)^4}$
Step 2: Use the Power Property, ${{a}^{m}}^{n}={a}^{m\times{n}}$
$\frac{100{p}^{6}}{125p^3\cdot 16{p}^{20}}$
Step 3: Add the exponents in the denominator.
$\frac{100{p}^{6}}{125\times{16}{p}^{23}}$
Step 4: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}}$
$\begin{eqnarray*}&=&\frac{100}{125\times{16{p}}^{17}}\\[1ex] \text{Simplify.}\;\;&=&\frac{1}{20{p}^{17}} \end{eqnarray*}$
### Try It
28) Simplify: $\frac{{(3{r}^{3}}^{2}{({r}^{3})}^{7}}{{({r}^{3})}^{3}}$
Solution
$9{r}^{18}$
29) Simplify: $\frac{{(2{x}^{4})}^{5}}{{(4{x}^{3})}^{2}{({x}^{3})}^{5}}$
Solution
$\frac{2}{x}$
## Use the Definition of a Negative Exponent
Up until this point, we have been careful to minimize our use of negative exponents. Before taking on division of polynomials, we will revisit some of the properties we’ve seen in the previous sections but, this time, with negative exponents.
### Quotient Property for Exponents
If $a$ is a real number, ${a}\neq 0$ , and $m$ and $n$ are whole numbers, then
$\frac{a^m}{a^n}=a^{m-n}$, for $m>n$ and $\frac{a^m}{a^n}=\frac{1}{a^{n-m}}$, for $n>m$.
What if we just subtract exponents regardless of which is larger?
Let’s consider $\frac{{x}^{2}}{{x}^{5}}$.
We subtract the exponent in the denominator from the exponent in the numerator.
$\begin{eqnarray*}&=&\frac{{x}^{2}}{{x}^{5}}\\&=&{x}^{2-5}\\&=&{x}^{-3}\end{eqnarray*}$
We can also simplify $\frac{{x}^{2}}{{x}^{5}}$ by dividing out common factors:
$\begin{eqnarray*}&=&\frac{{\color{red}{\cancel x}}\cdot{\color{red}{\cancel x}}}{{\color{red}{\cancel x}}\cdot{\color{red}{\cancel x}}\cdot x\cdot x\cdot x}\\&=&\frac1{x^3}\end{eqnarray*}$
This implies that ${x}^{-3}=\frac{1}{{x}^{3}}$ and it leads us to the definition of a negative exponent.
## Negative Exponents
If $n$ is an integer and $a\neq{0}$, then ${a}^{\text{−}n}=\frac{1}{{a}^{n}}$.
The negative exponent tells us we can re-write the expression by taking the reciprocal of the base and then changing the sign of the exponent.
Any expression that has negative exponents is not considered to be in simplest form. We will use the definition of a negative exponent and other properties of exponents to write the expression with only positive exponents.
For example, if after simplifying an expression we end up with the expression ${x}^{-3}$, we will take one more step and write $\frac{1}{{x}^{3}}$. The answer is considered to be in simplest form when it has only positive exponents.
### Example 5.5.14
Simplify:
a. ${4}^{-2}$
b. ${10}^{-3}$
Solution
a.
Step 1: Use the definition of a negative exponent, ${a}^{-n}=\frac{1}{{a}^{n}}$.
$\begin{eqnarray*}&=&\frac{1}{{4}^{2}}\\ \text{Simplify.}\;\;&=&\frac{1}{16} \end{eqnarray*}$
b.
Step 1: Use the definition of a negative exponent, ${a}^{-n}=\frac{1}{{a}^{n}}$.
$\begin{eqnarray*}&=&\frac{1}{{10}^{3}}\\ \text{Simplify.}\;\;&=&\frac{1}{1000}\end{eqnarray*}$
### Try It
30) Simplify:
a. ${2}^{-3}$
b. ${10}^{-7}$
Solution
a. $\frac{1}{8}$
b. $\frac{1}{{10}^{7}}$
31) Simplify:
a. ${3}^{-2}$
b. ${10}^{-4}$
Solution
a. $\frac{1}{9}$
b. $\frac{1}{10,000}$
In example 5.2.13 we raised an integer to a negative exponent. What happens when we raise a fraction to a negative exponent? We’ll start by looking at what happens to a fraction whose numerator is one and whose denominator is an integer raised to a negative exponent.
\begin{align*} &\;&\;&\frac{1}{a^{-n}}\\[1ex] &\text{Use the definition of a negative exponent,}\;{a}^{-n}=\frac{1}{a^n}&\;&\frac{1}{\frac{1}{{a}^{n}}}\\[1ex] &\text{Simplify the complex fraction.}&\;&1\cdot \frac{{a}^{n}}{1}\\[1ex] &\text{Multiply.}&\;&{a}^{n} \end{align*}
This leads to the Property of Negative Exponents.
### Property of Negative Exponents
If $n$ is an integer and ${a}\neq 0$, then $\frac{1}{{a}^{-n}}={a}^{n}$.
### Example 5.5.15
Simplify:
a. $\frac{1}{{y}^{-4}}$
b. $\frac{1}{{3}^{-2}}$
Solution
a.
Step 1: Use the property of a negative exponent, $\frac{1}{{a}^{-n}}={a}^{n}$.
${y}^{4}$
b.
Step 1: Use the property of a negative exponent, $\frac{1}{{a}^{-n}}={a}^{n}$.
$\begin{eqnarray*}&=&{3}^{2}\\\text{Simplify.}\;\;&=&9\end{eqnarray*}$
### Try It
32) Simplify:
a. $\frac{1}{{p}^{-8}}$
b. $\frac{1}{{4}^{-3}}$
Solution
a. ${p}^{8}$
b. $64$
33) Simplify:
a. $\frac{1}{{q}^{-7}}$
b. $\frac{1}{{2}^{-4}}$
Solution
a. ${q}^{7}$
b. $16$
Suppose now we have a fraction raised to a negative exponent. Let’s use our definition of negative exponents to lead us to a new property.
$(\frac{3}{4})^{-2}$ Use the definition of a negative exponent, ${a}^{-n}=\frac{1}{{a}^{n}}$. $\frac {1}{(\frac{3}{4})^{2}}$ Simplify the denominator. $\frac{1}{\frac{9}{16}}$ Simplify the complex fraction. $\frac{16}{9}$ But we know that $\frac{16}{9}$ is $(\frac{4}{3})^{2}$. $\frac{4^2}{3^2}$ This tells us that: $(\frac{3}{4})^{-2}=(\frac{4}{3})^{2}$
To get from the original fraction raised to a negative exponent to the final result, we took the reciprocal of the base—the fraction—and changed the sign of the exponent.
This leads us to the Quotient to a Negative Power Property.
### Quotient to a Negative Exponent Property
If $a$ and $b$ are real numbers, $a \neq 0$, $b\neq 0$, and $n$ is an integer, then $(\frac{a}{b})^{-n}= (\frac{b}{a})^{n}$.
### Example 5.5.16
Simplify:
a. $\left(\frac{5}{7}\right)^{-2}$
b. $\left(-\frac{2x}{y}\right)^{-3}$
Solution
a.
Step 1: Use the Quotient to a Negative Exponent Property, $(\frac{a}{b})^{-n}={(\frac{b}{a})}^{n}$.
Step 2: Take the reciprocal of the fraction and change the sign of the exponent.
$\begin{eqnarray*}&=&\left(\frac{7}{5}\right)^{2}\;\;\\\text{Simplify.}\;\;&=&\frac{49}{25} \end{eqnarray*}$
b.
Step 1: Use the Quotient to a Negative Exponent Property, $(\frac{a}{b})^{-n}={(\frac{b}{a})}^{n}$
Step 2: Take the reciprocal of the fraction and change the sign of the exponent.
$\begin{eqnarray*}&=&\left(-\frac{y}{2x}\right)^{3}\;\;\\[1ex]\text{Simplify.}\;\;&=&-\frac{{y}^{3}}{8{x}^{3}}\end{eqnarray*}$
### Try It
34) Simplify:
a. $\left(\frac{2}{3}\right)^{-4}$
b. $(-\frac{6m}{n})^{-2}$
Solution
a. $\frac{81}{16}$
b. $\frac{{n}^{2}}{36{m}^{2}}$
35) Simplify:
a. $\left(\frac{3}{5}\right)^{-3}$
b. $\left(-\frac{a}{2b}\right)^{-4}$
Solution
a. $\frac{125}{27}$
b. $\frac{16{b}^{4}}{{a}^{4}}$
When simplifying an expression with exponents, we must be careful to correctly identify the base.
### Example 5.5.17
Simplify:
a. $(-3)^{-2}$
b. $-3^{-2}$
c. $\left(-\frac{1}{3}\right)^{-2}$
d. $-\left(\frac{1}{3}\right)^{-2}$
Solution
a. Here the exponent applies to the base $-3$.
Step 1: Take the reciprocal of the base and change the sign of the exponent.
$\begin{eqnarray*}&=&\frac{1}{(-3)^2}\\[1ex] \text{Simplify.}\;\;&=&\frac{1}{9}\end{eqnarray*}$
b. The expression $-{3}^{-2}$ means “find the opposite of ${3}^{-2}$.”
Step 1: Rewrite as a product with $-1$.
$-1\times 3^{-2}$
Step 2: Take the reciprocal of the base and change the sign of the exponent.
$\begin{eqnarray*}&=&-1\cdot \frac{1}{{3}^{2}}\\[1ex]\text{Simplify.}\;\;&=&-\frac{1}{9}\end{eqnarray*}$
c. Here the exponent applies to the base $-\frac{1}{3}$.
Step 1: Take the reciprocal of the base and change the sign of the exponent.
$\begin{eqnarray*}&=&\left(-\frac{3}{1}\right)^2\\[1ex]\text{Simplify.}\;\;&=&9\end{eqnarray*}$
d. The expression $-(\frac{1}{3})^{-2}$ means “find the opposite of $(\frac{1}{3})^{-2}$.” Here the exponent applies to the base $\frac{1}{3}$.
Step 1: Rewrite as a product with $-1$.
$-1\cdot \left(\frac{3}{1}\right)^{2}$
Step 2: Take the reciprocal of the base and change the sign of the exponent.
$-9$
### Try It
36) Simplify:
a. $(-5)^{-2}$
b. ${−}{5}^{-2}$
c. $\left(-\frac{1}{5}\right)^{-2}$
d. ${−}\left(\frac{1}{5}\right)^{-2}$
Solution
a. $\frac{1}{25}$
b. $-\frac{1}{25}$
c. $25$
d. $-25$
37) Simplify:
a. $(-7)^{-2}$
b. ${−}{7}^{-2}$
c. $\left(-\frac{1}{7}\right)^{-2}$
d. $\left(-\frac{1}{7}\right)^{-2}$
Solution
a. $\frac{1}{49}$
b. $-\frac{1}{49}$
c. $49$
d. $-49$
We must be careful to follow the Order of Operations. In the next example, parts (a) and (b) look similar, but the results are different.
### Example 5.5.18
Simplify:
a. $4\times{2}^{-1}$
b. $(4\times{2})^{-1}$
Solution
a.
Step 1: Do exponents before multiplication.
$4\times{2}^{-1}$
Step 2: Use ${a}^{-n}=\frac{1}{{a}^{n}}$
$\begin{eqnarray*}&=&4\times \frac{1}{{2}^{1}}\\[1ex] \text{Simplify.}\;\;&=&2\end{eqnarray*}$
b.
Step 1: Simplify inside the parentheses first.
$(8)^{-1}$
Step 2: Use ${a}^{-n}=\frac{1}{{a}^{n}}$
$\begin{eqnarray*}&=&\frac{1}{{8}^{1}}\\[1ex] \text{Simplify.}\;\;&=&\frac{1}{8} \end{eqnarray*}$
### Try It
38) Simplify:
a. $6\times{3}^{-1}$
b. $(6\times 3)^{-1}$
Solution
a. $2$
b. $\frac{1}{18}$
39) Simplify:
a. $8\times {2}^{-2}$
b. $8\times {2}^{-2}$
Solution
a. $2$
b. $\frac{1}{256}$
When a variable is raised to a negative exponent, we apply the definition the same way we did with numbers. We will assume all variables are non-zero.
### Example 5.5.19
Simplify:
a. ${x}^{-6}$
b. $({u}^{4})^{-3}$
Solution
a.
Step 1: Use the definition of a negative exponent ${a}^{-n}=\frac{1}{{a}^{n}}$
$\frac{1}{{x}^{6}}$
b.
Step 1: Use the definition of a negative exponent ${a}^{-n}=\frac{1}{{a}^{n}}$.
$\begin{eqnarray*}&=&\frac{1}{(u^4)^3}\\[1ex] \text{Simplify.}\;\;&=&\frac{1}{u^{12}} \end{eqnarray*}$
### Try It
40) Simplify:
a. ${y}^{-7}$
b. $({z}^{3})^{-5}$
Solution
a. $\frac{1}{{y}^{7}}$
b. $\frac{1}{{z}^{15}}$
41) Simplify:
a. ${p}^{-9}$
b. $({q}^{4})^{-6}$
Solution
a. $\frac{1}{p}^{9}$
b. $\frac{1}{q}^{24}$
When there is a product and an exponent we have to be careful to apply the exponent to the correct quantity. According to the Order of Operations, we simplify expressions in parentheses before applying exponents. We’ll see how this works in the next example.
### Example 5.5.20
Simplify:
a. $5{y}^{-1}$
b. $(5y)^{-1}$
c. $(-5y)^{-1}$
Solution
a.
Step 1: Notice the exponent applies to just the base $y$.
Step 2: Take the reciprocal of $y$ and change the sign of the exponent.
$\begin{eqnarray*}&=&5\cdot \left(\frac{1}{y^1}\right)\\[1ex] \text{Simplify.}\;\;&=&\frac{5}{y} \end{eqnarray*}$
b.
Step 1: Here the parentheses make the exponent apply to the base $5y$.
Step 2: Take the reciprocal of $5y$ and change the sign of the exponent.
$\begin{eqnarray*}&=&\frac{1}{{5y}^1}\\[1ex] \text{Simplify.}\;\;&=&\frac{1}{5y} \end{eqnarray*}$
c.
Step 1: The base here is $−5y$.
Step 2: Take the reciprocal of $−5y$ and change the sign of the exponent.
$\begin{eqnarray*}&=&\frac{1}{-5y}\\[1ex]\text{Simplify.}\;\;&=&\left(-\frac{1}{5y}\right) \end{eqnarray*}$
### Try It
42) Simplify:
a. $(8{p}^{-1})$
b. ${(8p)}^{-1}$
c. ${(-8p)}^{-1}$
Solution
a. $\frac{8}{p}$
b. $\frac{1}{8p}$
c. $-\frac{1}{8p}$
43) Simplify:
a. ${11q}^{-1}$
b. ${(11q)}^{-1}$
c. ${(-11q)}^{-1}$
Solution
a. $\frac{1}{11q}$
b. $\frac{1}{11q}$
c. $-\frac{1}{11q}$
With negative exponents, the Quotient Rule needs only one form $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$, for $a\neq{0}$. When the exponent in the denominator is larger than the exponent in the numerator, the exponent of the quotient will be negative.
## Simplify Expressions with Integer Exponents
All of the exponent properties we developed earlier in the chapter with whole number exponents apply to integer exponents, too. We restate them here for reference.
## Summary of Exponent Properties
If $a$ and $b$ are real numbers, and $m$ and $n$ are integers, then
Product Property ${a^m}\times{a^n}=a^{m+n}$ $({a^m})^n={a}{mn}$ $(ab)^m=a^m b^m$ $\frac{a^m}{a^n}={a}^{m-n}$ $a^0=1 , {a}\neq{1}$ $(\frac {a}{b})^m=\frac{{a}^{m}}{{b}^{m}}, {b}\neq{0}$ ${a}^{-n}=\frac{1}{a^n}$ and $\frac{1}{a^{-n}}=a^{n}$ $(\frac{a}{b})^{-n}=(\frac{b}{a})^{n}$
### Example 5.5.21
Simplify:
a. ${x}^{-4}\times{x}^{6}$
b. ${y}^{-6}\times{y}^{4}$
c. ${z}^{-5}\times{z}^{-3}$
Solution
a.
Step 1: Use the Product Property, $a^m\cdot a^n=a^{m+n}$
$\begin{eqnarray*}&=&{x}^{-4+6}\\[1ex] \text{Simplify.}\;\;&=&x^2 \end{eqnarray*}$
b.
Step 1: Notice the same bases, so add the exponents.
$\begin{eqnarray*}&=&y^{-6+4}\\[1ex] \text{Simplify.}\;\;&=&y^{-2}\end{eqnarray*}$
Step 3: Use the definition of a negative exponent, $a^{-n}=\frac{1}{a^n}$
$\frac{1}{y^2}$
c.
Step 1: Add the exponents, since the bases are the same.
$\begin{eqnarray*}&=&z^{-5-3}\\[1ex] \text{Simplify.}\;\;&=&z^{-8}\end{eqnarray*}$
Step 3: Take the reciprocal and change the sign of the exponent, using the definition of a negative exponent.
$\frac{1}{z^{8}}$
### Try It
44) Simplify:
a. ${{x}^{-3}}\times{{x}^{7}}$
b. ${{y}^{-7}}\times{{y}^{2}}$
c. ${{z}^{-4}}\times{{z}^{-5}}$
Solution
a. ${x}^{4}$
b. $\frac{1}{{y}^{5}}$
c. $\frac{1}{{z}^{9}}$
45) Simplify:
a. ${{a}^{-1}}\times{{a}^{6}}$
b. $b^{-8}\times b^{4}$
c. ${c}^{-8}\times{{c}^{-7}}$
Solution
a. ${a}^{5}$
b. $\frac{1}{{b}^{4}}$
c. $\frac{1}{{c}^{15}}$
In the next two examples, we’ll start by using the Commutative Property to group the same variables together. This makes it easier to identify the like bases before using the Product Property.
### Example 5.5.22
Simplify: $({m}^{4}{n}^{-3})({m}^{-5}{n}^{-2})$
Solution
Step 1: Use the Commutative Property to get like bases together.
$m^4 m^{-5} n^{-2} n^{-3}$
Step 2: Add the exponents for each base.
$m^{-1} n^{-5}$
Step 3: Take reciprocals and change the signs of the exponents.
$\begin{eqnarray*}&=&\frac{1}{m^1} \frac{1}{n^5}\\[1ex] \text{Simplify.}\;\;&=&\frac{1}{mn^5}\end{eqnarray*}$
### Try It
46) Simplify: $({p}^{6}{q}^{-2})({p}^{-9}{q}^{-1})$
Solution
$\frac{1}{{p}^{3}{q}^{3}}$
47) Simplify: $({r}^{5}{s}^{-3})({r}^{-7}{s}^{-5})$
Solution
$\frac{1}{{r}^{2}{s}^{8}}$
If the monomials have numerical coefficients, we multiply the coefficients, just like we did earlier.
### Example 5.5.23
Simplify: $(2{x}^{-6}{y}^{8})(-5{x}^{5}{y}^{-3})$
Solution
Step 1: Rewrite with the like bases together
$2(-5)(x^{-6} x^{5}) (y^8 y^{-3})$
Step 2: Multiply the coefficients and add the exponents of each variable.
$-10x^{-1}y^5$
Step 3: Use the definition of a negative exponent, $a^{-n}=\frac{1}{a^n}$
$\begin{eqnarray*}&=&-10\cdot \left(\frac{1}{x^1}\right)\cdot y^5\\[1ex] \text{Simplify.}\;\;&=&-\frac{10y^5}{x}\end{eqnarray*}$
### Try It
48) Simplify: $(3{u}^{-5}{v}^{7})(-4{u}^{4}{v}^{-2})$
Solution
$-\frac{12{v}^{5}}{u}$
49) Simplify: $(-6{c}^{-6}{d}^{4})(-5{c}^{-2}{d}^{-1})$
Solution
$\frac{30{d}^{3}}{{c}^{8}}$
In the next two examples, we’ll use the Power Property and the Product to a Power Property.
### Example 5.5.24
Simplify: $(6{k}^{3})^{-2}$
Solution
Step 1: Use the Product to a Power Property, $(ab)^m=a^m b^m$
$(6)^{(-2)}(k^3)^{-2}$
Step 2: Use the Power Property, $(a^m)^n= {a}^{mn}$
$6^{-2}k^{-6}$
Step 3: Use the Definition of a Negative Exponent, $a^{-n}=\frac{1}{a^n}$
$\begin{eqnarray*}&=&\frac{1}{6^2}\cdot \frac{1}{k^6}\\[1ex]\text{Simplify.}\;\;&=&\frac{1}{36k^6}\end{eqnarray*}$
### Try It
50) Simplify: $(-4{x}^{4})^{-2}$
Solution
$\frac{1}{16{x}^{8}}$
51) Simplify: $\left(2{b}^{3}\right)^{-4}$
Solution
$\frac{1}{16{b}^{12}}$
### Example 5.5.25
Simplify: $(5{x}^{-3})^{2}$
Solution
Step 1: Use the Product to a Power property, $(ab)^m=a^m b^m$
$5^2(x^{-3})^2$
Step 2: Simplify $5^2$ and multiply the exponents of $x$ using the Power Property, $(a^m)^n=a^{mn}$
$25x^{-6}$
Step 3: Rewrite $x^{-6}$ by using the Definition of a Negative Exponent, $a^{-n}=\frac{1}{a^n}$
$\begin{eqnarray*}&=&25\cdot \left(\frac{1}{x^6}\right)\\[1ex] \text{Simplify.}\;\;&=&\frac{25}{x^6} \end{eqnarray*}$
### Try It
52) Simplify: $(8{a}^{-4})^{2}$
Solution
$\frac{64}{{a}^{8}}$
53) Simplify: $(2{c}^{-4})^{3}$
Solution
$\frac{8}{{c}^{12}}$
To simplify a fraction, we use the Quotient Property and subtract the exponents.
### Example 5.5.26
Simplify:$\frac{{r}^{5}}{{r}^{-4}}$
Solution
Step 1: Use the Quotient Property, $\frac{a^m}{a^n}=a^{m-n}$c
$\begin{eqnarray*}&=&r^{5-(-4)}\\[1ex]\text{Simplify.}\;\;&=&r^9 \end{eqnarray*}$
### Try It
54) Simplify: $\frac{{x}^{8}}{{x}^{-3}}$
Solution
${x}^{11}$
55) Simplify: $\frac{{y}^{8}}{{y}^{-6}}$
Solution
${y}^{14}$
## Divide Monomials
You have now been introduced to all the properties of exponents and used them to simplify expressions. Next, you’ll see how to use these properties to divide monomials. Later, you’ll use them to divide polynomials.
### Example 5.5.27
Find the quotient: ${56x^7}\div{8x^3}$
Solution
Step 1: Rewrite as a fraction.
$\frac{56{x}^{7}}{8{x}^{3}}$
Step 2: Use fraction multiplication.
$\frac{56}{8}\times\frac{x^7}{x^3}$
Step 3: Simplify and use the Quotient Property.
$7x^4$
### Try It
56) Find the quotient: $42{y}^{9}\div{6{y}^{3}}$
Solution
$7{y}^{6}$
57) Find the quotient: $48{z}^{8}\div{8{z}^{2}}$
Solution
$6{z}^{6}$
### Example 5.5.28
Find the quotient: $\frac{45{a}^{2}{b}^{3}}{-5a{b}^{5}}$
Solution
Step 1: Use fraction multiplication.
${\frac{45}{-5}}{\frac{{a}^{2}}{a}}{\frac{{b}^{3}}{{b}^{5}}}$
Step 2: Simplify and use the Quotient Property.
$\begin{eqnarray*}&=&-9\cdot {a}\cdot {\frac{1}{{b}^{2}}}\\[1ex] \text{Multiply.}\;\;&=&-\frac{9a}{{b}^{2}} \end{eqnarray*}$
### Try It
58) Find the quotient: $\frac{-72{a}^{7}{b}^{3}}{8{a}^{12}{b}^{4}}$
Solution
$-\frac{9}{{a}^{5}b}$
59) Find the quotient: $\frac{-63{c}^{8}{d}^{3}}{7{c}^{12}{d}^{2}}$
Solution
$\frac{-9d}{{c}^{4}}$
### Example 5.5.29
Find the quotient: $\frac{24{a}^{5}{b}^{3}}{48a{b}^{4}}$
Solution
Step 1: Use fraction multiplication.
$\frac{24}{48}\cdot \frac{a^5}{a}\cdot \frac{b^3}{b^4}$
Step 2: Simplify and use the Quotient Property.
$\begin{eqnarray*}&=&\frac{1}{2}\cdot {a}^{4}\cdot \frac{1}{b}\\[1ex] \text{Multiply.}\;\;&=&\frac{{a}^{4}}{2b} \end{eqnarray*}$
### Try It
60) Find the quotient:$\frac{16{a}^{7}{b}^{6}}{24a{b}^{8}}$
Solution
$\frac{2{a}^{6}}{3{b}^{2}}$
61) Find the quotient: $\frac{27{p}^{4}{q}^{7}}{-45{p}^{12}q}$
Solution
$-\frac{3{q}^{6}}{5{p}^{8}}$
Once you become familiar with the process and have practised it step by step several times, you may be able to simplify a fraction in one step.
### Example 5.5.30
Find the quotient: $\frac{14{x}^{7}{y}^{12}}{21{x}^{11}{y}^{6}}$
Solution
Be very careful to simplify $\frac{14}{21}$ by dividing out a common factor, and to simplify the variables by subtracting their exponents.
Step 1: Simplify and use the Quotient Property.
$\frac{2{y}^{6}}{3{x}^{4}}$
### Try It
62) Find the quotient: $\frac{28{x}^{5}{y}^{14}}{49{x}^{9}{y}^{12}}$
Solution
$\frac{4{y}^{2}}{7{x}^{4}}$
63) Find the quotient: $\frac{30{m}^{5}{n}^{11}}{48{m}^{10}{n}^{14}}$
Solution
$\frac{5}{8{m}^{5}{n}^{3}}$
In all examples so far, there was no work to do in the numerator or denominator before simplifying the fraction. In the next example, we’ll first find the product of two monomials in the numerator before we simplify the fraction. This follows the order of operations. Remember, a fraction bar is a grouping symbol.
### Example 5.5.31
Find the quotient:$\frac{(6{x}^{2}{y}^{3})(5{x}^{3}{y}^{2})}{(3{x}^{4}{y}^{5})}$
Solution
Step 1: Simplify the numerator.
$\begin{eqnarray*}&=&\frac{30{x}^{5}{y}^{5}}{3{x}^{4}{y}^{5}}\\[1ex]\text{Simplify.}\;\;&=&10x \end{eqnarray*}$
### Try It
64) Find the quotient: $\frac{(6{a}^{4}{b}^{5})(4{a}^{2}{b}^{5})}{12{a}^{5}{b}^{8}}$
Solution
$2a{b}^{2}$
65) Find the quotient: $\frac{(-12{x}^{6}{y}^{9})(-4{x}^{5}{y}^{8})}{-12{x}^{10}{y}^{12}}$
Solution
$-4x{y}^{5}$
Access these online resources for additional instruction and practice with dividing monomials:
### Key Concepts
• Quotient Property for Exponents:
• If $a$ is a real number, $a\neq{0}$, and $m$, $n$ are whole numbers, then:
$\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$, $m>n$ and $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{m-n}}$, $n>m$
• Zero Exponent
• If $a$ is a non-zero number, then ${a}^{0}=1$.
• Quotient to a Power Property for Exponents:
• If $a$ and $b$ are real numbers, $b\neq{0}$ and $m$ is a counting number, then:
$(\frac{a}{b})^{m}=\frac{{a}^{m}}{{b}^{m}}$
• To raise a fraction to a power, raise the numerator and denominator to that power.
• Summary of Exponent Properties
• If $a$, $b$ are real numbers and $m$, $n$ are whole numbers, then
Product Property $a^m a^n=a^{m+n}$ ${(a^m)^n}=a^{mn}$ $(ab)^m=a^m b^m$ $\frac{a^m}{b^m}=a^{m-n},a\neq{0} , m>n$ $\frac{a^m}{a^n}=\frac{1}{a^{n-m}},a\neq{0},n>n=m$ $a^{0}=1,a\neq{0}$ $(\frac{a}{b})^m\frac{a^m}{b^m},b\neq{0}$
Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this? |
# Adding Numbers in Expanded Form – Definition, Facts, Examples | How to do Expanded Form Addition?
Adding Numbers in Expanded Form can make the addition process simple. While moving to the large number addition, students may get confused while calculating addition. But if the students use Adding Numbers in Expanded Form, then they can easily solve the addition problems for large and small problems within a short period of time with accurate calculation. Get to know what is meant by expanded form, how to add numbers in expanded form, examples of expanded form addition in the later sections of the article.
Also, Refer to Similar Articles:
## How to Add Numbers in Expanded Form?
Adding Numbers in Expanded Form can possible as mentioned below. Carefully, read out the below process and follow it to find the Addition of Expanded Form of a Number.
(i) Note down the problems and write them in expanded form.
(ii) Write down the problem in columns.
(iii) Simplify the problem from the one’s column, ten’s column, and then hundred’s column.
(iv) If you get a carry while adding one’s column, then forward it to the ten’s column and add with the ten’s column digits.
(v) Similarly, if you get a carry while adding ten’s column, then forward it to the hundred’s column, and so on.
(vi) Finally, write down the answer.
Addition of 2-Digit Numbers in Expanded Form
Check out the below problems to know how to add 2-Digit Numbers in Expanded Form. We have given different examples on the Addition of 2-Digit Numbers in Expanded Form.
Example 1.
Solution:
Given numbers are 43 and 36.
Now, numbers are written in column form.
Explanation:
Firstly, add ones: 3 ones + 6 ones = 9 ones.
There is no carry. Therefore, now add tens.
Add Tens: 4 tens + 3 tens = 7 tens.
Write the 9 ones in one’s column and write the 7 tens in the tens column.
Example 2.
Add the expanded form of the numbers 23 + 16
Solution:
The given numbers are 23 and 16.
Write 23 in an expanded form. 23 = 20 + 3
Write 16 in an expanded form. 16 = 10 + 6
Now, add the digits of one’s place.
3 ones + 6 ones = 9 ones.
Again add the numbers of ten’s place
20 + 10 = 30
Therefore, the total sum = 30 + 9 = 39.
Example 3.
Add 52 and 24 in long form?
Solution:
The given numbers are 52 and 24.
Explanation:
Write 52 in an expanded form. 52 = 50 + 2
Write 24 in an expanded form. 24 = 20 + 4
Now, add the digits of one’s place.
Sum of digit of one’s place = 2 ones + 4 ones = 6 ones.
Again add the numbers of ten’s place.
Sum of digit of ten’s place = 5 ten’s + 2 ten’s = 7 ten’s
Therefore, the total sum = 70 + 6 = 76.
Example 4.
Add 27 and 50 in short form.
Solution:
The given numbers are 27 and 50.
Explanation:
Write 27 in an expanded form. 27 = 20 + 7
Write 50 in an expanded form. 50 = 50 +0
Now, add the digits of one’s place.
Sum of digit of one’s place = 7 ones + 0 ones = 7 ones.
Again add the numbers of ten’s place.
Sum of digit of ten’s place = 2 ten’s + 5 ten’s = 7 ten’s
Therefore, the total sum = 70 + 7 = 77.
Example 5.
Add the expanded form of the numbers 47 + 25
Solution:
The given numbers are 47 and 25.
Explanation:
Write 47 in an expanded form. 47 = 40 + 7
Write 25 in an expanded form. 25 = 20 + 5
Now, add the digits of one’s place.
Sum of digit of one’s place = 7 ones + 5 ones = 12 ones.
12 ones = 10 ones + 2 ones
10 ones = 1 ten.
Sum of digit of ten’s place = 4 ten’s + 2 ten’s = 6 ten’s.
6 ten’s + 1 ten = 7 ten’s
Therefore, the total sum = 60 + 10 + 2 = 72.
Addition of 3-Digit Numbers in Expanded Form:
Add 3-Digit Numbers in Expanded Form by following the below methods. Practice every problem given below to learn the addition of 3-Digit Numbers in Expanded Form.
Example 6.
Solution:
The given numbers are 425 and 341.
Explanation:
Write 425 in an expanded form. 425 = 400 + 20 + 5
Write 341 in an expanded form. 341 = 300 + 40 + 1
Now, add the digits of one’s place.
Sum of digit of one’s place = 5 ones + 1 ones = 6 ones.
Again add the numbers of ten’s place.
Sum of digit of ten’s place = 2 ten’s + 4 ten’s = 6 ten’s.
Again add the numbers of hundred’s place.
Sum of digit of hundred’s place = 4 hundred’s + 3 hundred’s = 7 hundred.
Therefore, the total sum = 700 + 60 + 6 = 766.
Example 7.
Add 472 and 416 by arranging the numbers in expanded form.
Solution:
The given numbers are 472 and 416.
Explanation:
Write 472 in an expanded form. 472 = 400 + 70 + 2
Write 416 in an expanded form. 416 = 400 + 10 + 6
Now, add the digits of one’s place.
Sum of digit of one’s place = 2 ones + 6 ones = 8 ones.
Again add the numbers of ten’s place.
Sum of digit of ten’s place = 7 ten’s + 1 ten’s = 8 ten’s.
Again add the numbers of hundred’s place.
Sum of digit of hundred’s place = 4 hundred’s + 4 hundred’s = 8 hundred.
Therefore, the total sum = 800 + 80 + 8 = 888.
Example 8.
Add 342, 513, and 466 by arranging the numbers in expanded form.
Solution:
The given numbers are 342, 513, and 466.
Explanation:
Write 342 in an expanded form. 342 = 300 + 40 + 2
Write 513 in an expanded form. 513 = 500 + 10 + 3
Write 466 in an expanded form. 466 = 400 + 60 + 6
Now, add the digits of one’s place.
Sum of digit of one’s place = 2 ones + 3 ones + 6 ones = 11 ones.
11 ones = 10 ones + 1 ones
10 ones = 1 ten
There is one carry after adding one’s digits.
Add the carry to the ten’s digits.
Again add the numbers of ten’s place.
Sum of digit of ten’s place = 4 ten’s + 1 ten’s + 6 ten’s = 11 ten’s.
11 ten’s = 10 ten’s + 1 ten.
10 ten’s = 1 hundred
There is one carry after adding ten’s digits.
Add the carry to the hundred’s digits.
Again add the numbers of hundred’s place.
Sum of digit of hundred’s place = 3 hundred’s + 5 hundred’s + 4 hundred’s = 12 hundred.
Therefore, the total sum = 1200 + 110 + 11 = 1321.
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# How do you simplify (-3-4i)^2?
Sep 18, 2016
$- 7 + 24 i$
#### Explanation:
${\left(- 3 - 4 i\right)}^{2}$
To square a binomial, rewrite the problem as the binomial multiplied by the binomial. Do NOT distribute the exponent!
$\left(- 3 - 4 i\right) \left(- 3 - 4 i\right)$
Distribute (FOIL)
$9 + 12 i + 12 i + 16 {i}^{2}$
Combine like terms. ${i}^{2} = - 1$
$9 + 24 i + 16 \cdot - 1$
$9 + 24 i - 16$
$- 7 + 24 i$ |
'
# Search results
Found 1335 matches
Cosine of the sum of three angles
Trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables. Geometrically, ... more
Tangent of the sum of three angles
Trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables. Geometrically, ... more
Secant of the sum of three angles
Trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables. Geometrically, ... more
Cosecant of the sum of three angles
Trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables. Geometrically, ... more
Worksheet 334
In a video game design, a map shows the location of other characters relative to the player, who is situated at the origin, and the direction they are facing. A character currently shows on the map at coordinates (-3, 5). If the player rotates counterclockwise by 20 degrees, then the objects in the map will correspondingly rotate 20 degrees clockwise. Find the new coordinates of the character.
To rotate the position of the character, we can imagine it as a point on a circle, and we will change the angle of the point by 20 degrees. To do so, we first need to find the radius of this circle and the original angle.
Drawing a right triangle inside the circle, we can find the radius using the Pythagorean Theorem:
Pythagorean theorem (right triangle)
To find the angle, we need to decide first if we are going to find the acute angle of the triangle, the reference angle, or if we are going to find the angle measured in standard position. While either approach will work, in this case we will do the latter. By applying the cosine function and using our given information we get
Cosine function
Subtraction
While there are two angles that have this cosine value, the angle of 120.964 degrees is in the second quadrant as desired, so it is the angle we were looking for.
Rotating the point clockwise by 20 degrees, the angle of the point will decrease to 100.964 degrees. We can then evaluate the coordinates of the rotated point
For x axis:
Cosine function
For y axis:
Sine function
The coordinates of the character on the rotated map will be (-1.109, 5.725)
Reference : PreCalculus: An Investigation of Functions,Edition 1.4 © 2014 David Lippman and Melonie Rasmussen
http://www.opentextbookstore.com/precalc/
Tangent of the sum of two angles (Bhāskara formula)
Trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables. Geometrically, ... more
Sine function
The trigonometric functions (also called the circular functions) are functions of an angle. They relate the angles of a triangle to the lengths of its ... more
Cosine function
The trigonometric functions (also called the circular functions) are functions of an angle. They relate the angles of a triangle to the lengths of its ... more
Tangent function
The trigonometric functions (also called the circular functions) are functions of an angle. They relate the angles of a triangle to the lengths of its ... more
Cotangent function
The trigonometric functions (also called the circular functions) are functions of an angle. They relate the angles of a triangle to the lengths of its ... more
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GRAPHING LINEAR EQUATIONS PARALLEL VS PERPENDICULAR LINES WARMUP
• Slides: 8
GRAPHING LINEAR EQUATIONS PARALLEL VS PERPENDICULAR LINES
WARM-UP: Find the slope and y-intercept of the following linear equations 1. Y + 3 = 2 x 2. Y + 3 x = -6 Y = 2 x - 3 Slope: 2 Y-int: -3 Y = -3 x - 6 Slope: -3 Y-int: -6 3. y - x - 21= 3 Y = x + 24 Slope: 1 Y-int: 24 4. 3 Y + 9 = 6 x 5. 2 y + 3 = 2 x 6. 4 y - x - 2= 3 Y = 2 x - 3 Y = x - 3/2 Y = x/4 +5/4 Slope: 2 Slope: 1/4 Y-int: -3/2 Y-int: 5/4
Parallel lines are linear equations which have the same slope Examples: Y=2 x+9 Y=2 x-23 Y=1/3 x +10 Y=1/3 x +5 Y=x +10 Y=7 +x Counter-examples: Y=-2 x+9 Y=2 x-23 Y=1/3 x +10 Y=3 x +5 Y=5 x +10 Y=7 +x
Example: 2 x + 6 y = 12 -2 x 6 y = -2 x + 12 6 6 6 y = -2/6 x + 2 y = -1/3 x + 2 Is parallel to y = -1/3 x + 5
REVIEW Find the reciprocal of the following numbers: 1. ⅔ 3/2 2. ⅜ 3. 8/3 7 4. 1/7 1/9 9 Find the negative reciprocal of the following numbers: 1. ½ -2 2. 4/9 -9/4 3. 8/7 -7/8 4. 3 -⅓
Perpendicular lines THE LINES ARE PERPENDICULAR IF THE PRODUCT OF THEIR SLOPES Examples: IS -1. Y=2 x+9 Y=-½x-23 Y=1/3 x +10 Y=-3 x +5 Y=x +10 Y=-1 X +21 Counter-examples: Y=-2 x+9 Y=-2 x-23 Y=⅓x +10 Y=3 x +10 Y=5 x +10 Y=7 +x
Example: Write an equation that is perpendicular to the following: 1. Y = 2/3 x + 7 Y = -3/2 x + 7 3. Y = -1/9 x + 29 Y = 9 x + 879 2. Y = 7 x -9 Y = -1/7 x + 25 4. Y = -5 x - 6 Y = -⅕x + 25
Practice re the equations parallel to each other? no no yes yes Which one is a parallel line? Which one is a perpendicular line? perpendicular parallel niether |
# Mathematics - Exponentiation (square, cube) - Power
Exponentiation is a binary operation involving two numbers:
1. the base (b) 1)
2. the exponent (n) (or index or power).
$$base^{exponent} = b^n = b_1 \times b_2 \times \dots \times b_n$$
## Example
$$2^3 = 2 \times 2 \times 2 = 8$$
## Text
In text notation or computer language, generally the exponentiation operator is noted ^
2^3 = 2 . 2 . 2 = 8
## Usage
### Permutation
The exponentiation is the formula to calculate the number of possible permutation (with repetition) in a set.
For instance, if you have a set of 3 elements {1,2,3}, if you can draw 2 elements, you have the following 9 possible selections (permutation):
• 1 1
• 1 2
• 1 3
• 2 1
• 2 2
• 2 3
• 3 1
• 3 2
• 3 3
that you can express with an exponentiation $$9 = 3^2 = b^n$$ because for each draw n, you have b choices. $$b_1 \times b_2 \times \dots \times b_n = b^n$$
### Base Calculation
You can even generalize the permutation calculation to base calculation (that's why b is called the base)
For instance, in a set of 10 digits { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 }, if you can draw 2 numbers, you will have 100 possibilities (permutation). $$b^n = 10^2 = 100$$
Note that:
• a set of 10 digits in mathematics is also known as the base 10 used in the decimal system
• 100 is the maximal number that you can get with the digits (ie 99+0)
therefore the exponentiation gives you also the maximum number in any base for any length.
For instance, in base 2 (binary) with 8 digits (octet/byte), the maximum (decimal) number is $$b^n = 2^8 = 256$$ You can then translate any base into decimal. For instance, to translate the binary string 10101 into decimal: $$10101 = 1 \times \text{max of 4 digits} + 1 \times \text{max of 2 digits} + 1 \times \text{max of 0 digits} = 1 \times 2^4 + 1 \times 2^2 + 1 \times 2^0 = 16 + 4 + 1 = 21$$
## Pronunciation
The exponentiation $b^n$ can be read as:
• b raised to the n-th power,
• b raised to the power of n,
• b raised by the exponent of n,
• most briefly as b to the n.
Example: The Exponent raises the first number, the base, to the power of the second number, the exponent.
Some exponents have their own pronunciation.
### Squared
$b^2$ is usually read as b squared.
### Cubed
$b^3$ is usually read as b cubed.
Cubing a number is the same as raising it to the third power.
## Law
$$base^a * base^b = base^{a+b}$$
## When the exponent (n) is
### a positive integer
When the exponent (n) is a positive integer, exponentiation corresponds to repeated multiplication.
$$b^n = \underbrace{b \times \dots \times b}_{n}$$
It's a product of n factors, each of which is equal to b (the product itself can also be called power):
### even
The exponents of a product are all even, the product is a perfect square.
Example: $2^2*3^6 = (2*3^2)^2$
## Inverse
• With the below exponentiation
$$b^n = a$$
$$b = \sqrt[n]{a}$$
• While the logarithm permits to get the exponent,
$$n = log_b(a)$$
## Javascript
Array(8).fill().map((element,index) => console.log(2 pow ${index} is${Math.pow(2,index)}));
## Documentation / Reference
Discover More
Byte Array (multi-byte word)
A byte array is any array of byte. Library represents them generally as a multi-byte word with the possibility to change the endianess. See for instance, the Array Buffer in Javascript To a integer...
Color - Luminance
Luminance is a photometric measure that is placed on a scale: that quantifies how a color is dark or light that goes from: 0 for darkest black to 1 for lightest white The ratio of luminance...
Function - Binary Function/Operation
A binary operation is an scalar operation with two arguments (arity of two) that produces one value. They are creating a binary relation. the addition operator, the multiplication operator ...
Machine Learning - (Univariate|Simple) Logistic regression
A Simple Logistic regression is a Logistic regression with only one parameters. For the generalization (ie with more than one parameter), see Logistic regression comes from the fact that linear regression...
Mathematics - Complex Exponential (Euler's formula)
Euler's formula provides a powerful connection between analysis and trigonometry, and provides an interpretation of the sine and cosine functions (the sinusoidal functions) as weighted sums of the exponential...
Mathematics - Exponential (Euler's number)
is the scientific constant, the exponential. Euler's number The number e is an important mathematical constant that is the base of the natural logarithm. The number e can be defined to be: the unique...
Mathematics - Inverse axiom
/ is inverse of - is inverse of + root is the inverse of exponentiation Functions f and g are functional inverses if: and are defined and are functions.
Mathematics - Logarithm Function (log)
is the number (#) of times you divide n by b until you get down to 1. trees With the base 2 where is the # of times you divide n by 2 until you get down to 1. You keep repeating dividing by two... |
## D9 Curve sketching
If you find some key points of a function such as: maxima, minima, or turning points; x and y axis intercepts; and regions where the gradient is positive or negative, you can put together a sketch of a curve. Read this section to find examples of this being done.
To sketch a curve it is helpful to find the
1. $$x$$ and $$y$$ intercepts
2. Maximum and minimum points.
This module describes how to do this.
### Definition of Maximum and Minimum
A stationary point is a point on a graph of a function $$y=f\left(x\right)$$ where the tangent to the curve is horizontal. At a stationary point the derivative function $$y=f'\left(x\right)=0.$$
A maximum stationary point occurs at $$x=a$$ if $$f'\left(a\right)=0$$ and $$f'\left(x\right)>0$$ for $$x<a$$ and $$f'\left(x\right)<0$$ for $$x>a$$ as shown below.
A minimum stationary point occurs at $$x=a$$ if $$f'\left(a\right)=0$$ and $$f'\left(x\right)$$< 0 for $$x<a$$ and $$f'\left(x\right)$$> 0 for $$x>a$$ as shown below.
### Example 1
Find the turning point of the parabola defined by $$y=x^{2}+4x+5$$ and determine if it is a maximum or minimum.
Solution
Let
\begin{align*} f\left(x\right)=x^{2}+4x+5 \end{align*} then \begin{align*} f'\left(x\right) & =2x+4. \end{align*} At a stationary point, $$f'\left(x\right)=0,$$ so \begin{align*} 2x+4 & =0\\ 2x & =-4\\ x & =-2 \end{align*} is the $$x-$$coordinate of a stationary point. When $$x=-2,$$ \begin{align*} f\left(-2\right) & =(-2)^{2}+4(-2)+5\\ & =1. \end{align*} Hence the coordinates of the stationary point are $$\left(-2,1\right).$$ Now we ascertain if it is a maximum or minimum.
A sign test can be used to determine whether the stationary point is minimum or a maximum by checking the slope of the tangent on each side of the stationary point. Consider the table below.
As we move from the left to the right of the stationary point at $$x=-2$$, the gradient changes from negative to positive. This indicates there is a minimum at $$\left(-2,1\right)$$.
Hence the turning point of the parabola is a minimum and occurs at $$\left(-2,1\right)$$.
### Example 2
Sketch the graph of $$y=x^{3}-x.$$
Solution strategy:
1. Find intercepts on $$x-$$axis.
2. Find stationary points.
3. Establish if stationary points are maximum or minimum values.
4. Plot intercepts and stationary points and sketch the graph.
Solution
For $$x-$$axis intercepts, we set $$y=0,$$ so \begin{align*} 0 & =x^{3}-x\\ & =x\left(x^{2}-1\right). \end{align*} Consequently, $$x=0$$ or \begin{align*} x^{2}-1 & =0\\ x & =1\text{ or -1. } \end{align*} Hence the $$x-$$axis intercepts are $$\left(-1,0\right),\ \left(0,0\right)\ \text{and \left(1,0\right) .}$$
For stationary points, \begin{align*} \frac{dy}{dx} & =0 \end{align*} that is \begin{align*} 3x^{2}-1 & =0\\ x^{2} & =\frac{1}{3}\\ x & =\pm\frac{1}{\sqrt{3}}\\ & \approx\pm0.58. \end{align*}
Substituting these $$x$$ values back in $$y=x-x^{3}$$ gives for $$x=1/\sqrt{3}$$, \begin{align*} y & =\left(\frac{1}{\sqrt{3}}\right)^{3}-\frac{1}{\sqrt{3}}\\ & \approx-0.39.\\ \end{align*} For $$x=-1/\sqrt{3}$$ \begin{align*} y & =\left(-\frac{1}{\sqrt{3}}\right)^{3}-\left(-\frac{1}{\sqrt{3}}\right)\\ & \approx0.38.\\ \end{align*} Hence the stationary points occur at approximately, $$\left(0.58,-0.39\right)$$ and $$\left(-0.58,0.38\right)$$. Now we decide if these points are maxima or minima. Consider the tables below.
For $$x\approx0.58$$ we have
and so the point $$x=1/\sqrt{3}\approx0.58$$ is a minimum.
For $$x\approx-0.58$$ we have
and so the point $$x=-1/\sqrt{3}\approx-0.58$$ is a maximum.
We can now graph $$y=x^{3}-x$$. First plot the $$x-$$intercepts and the stationary points:
We know $$E$$ is a maximum and $$D$$ is a minimum and so can graph $$y=x^{3}-x$$ as shown below.
### Exercise
Sketch the graphs of the following functions showing all intercepts and turning points
1. $$y=x^{2}-4x$$
2. $$y=x^{3}-2x^{2}+x$$
3. $$y=6-x-x^{2}$$
4. $$y=\left(x+1\right)^{4}$$ |
# Vector spaces: span, linear independence and basis
References: edX online course MIT 8.05.1x Week 3.
Sheldon Axler (2015), Linear Algebra Done Right, 3rd edition, Springer. Chapter 2.
Here, we investigate the ideas of the span of a vector space and see how this leads to the idea of linear independence of a set of vectors. I’ll summarize the main definitions and results here for future use; a more complete explanation together with some examples is given in Axler’s book, Chapter 2.
Span of a list of vectors
A list of vectors is just a subset of the vectors in a vector space, with the condition that the number of vectors in the subset is finite. The set of all linear combinations of the vectors ${\left(v_{1},\ldots,v_{m}\right)}$ in a list is called the span of that list. Since a general linear combination has the form
$\displaystyle v=\sum_{i=1}^{m}a_{i}v_{i} \ \ \ \ \ (1)$
where ${a_{i}\in\mathbb{F}}$ (recall that the field ${\mathbb{F}}$ is always taken to be either the real numbers ${\mathbb{R}}$ or the complex numbers ${\mathbb{C}}$), the span of a list itself forms a vector space which is a subspace of the original vector space. One result we can show is
Theorem 1 The span of a list of vectors in a vector space ${V}$ is the smallest subspace of ${V}$ containing all the vectors in the list.
Proof: Let the list be ${L\equiv\left(v_{1},\ldots,v_{m}\right)}$. Then ${S\equiv\mbox{span}\left(v_{1},\ldots,v_{m}\right)}$ is a subspace since it contains the zero vector if all ${a_{i}}$s are zero in 1, and since it contains all linear combinations of the list, it is closed under addition and scalar multiplication.
The span ${S}$ contains all ${v_{j}\in L}$ (just set ${a_{j}=\delta_{ij}}$ in 1). Now if we look at a subspace of ${V}$ that contains all the ${v_{i}}$s, it must also contain every vector in the span ${S}$, since a subspace must be closed under addition and scalar multiplication. Thus ${S}$ is the smallest subspace of ${V}$ that contains all the vectors in ${L}$. $\Box$
If ${S\equiv\mbox{span}\left(v_{1},\ldots,v_{m}\right)=V}$, that is, the span of a list is the same as the original vector space, then we say that ${\left(v_{1},\ldots,v_{m}\right)}$ spans ${V}$. This leads to the definition that a vector space is called finite-dimensional if it is spanned by some list of vectors. (Remember that all lists are finite in length!) A vector space that is not finite-dimensional is called (not surprisingly) infinite-dimensional.
Linear independence
Suppose a list ${\left(v_{1},\ldots,v_{m}\right)\in V}$ and ${v}$ is a vector such that ${v\in\mbox{span}\left(v_{1},\ldots,v_{m}\right)}$. This means that ${v}$ is a linear combination of ${\left(v_{1},\ldots,v_{m}\right)}$, so that 1 is true. However, using only the definitions above, there is no guarantee that there is only one choice for the scalars ${a_{i}}$ that satisfies 1. We might also have, for example
$\displaystyle v=\sum_{i=1}^{m}c_{i}v_{i} \ \ \ \ \ (2)$
where ${c_{i}\ne a_{i}}$. This means that we can write the zero vector as
$\displaystyle 0=\sum_{i=1}^{m}\left(a_{i}-c_{i}\right)v_{i} \ \ \ \ \ (3)$
Now, if the only way we can satsify this equation is to require that ${a_{i}=c_{i}}$ for all ${i}$, then we say that the list ${\left(v_{1},\ldots,v_{m}\right)}$ is linearly independent. (For completeness, the empty list (containing no vectors) is also declared to be linearly independent.) By reversing the above argument, we see that if the list ${\left(v_{1},\ldots,v_{m}\right)}$ is linearly independent, then there is only one set of scalars ${a_{i}}$ such that 1 is satisfied. In other words, any vector ${v\in\mbox{span}\left(v_{1},\ldots,v_{m}\right)}$ has only one representation as a linear combination of the vectors in the list.
A list that is not linearly independent is, again not surprisingly, defined to be linearly dependent. This leads to the linear dependence lemma:
Lemma 2 Suppose ${\left(v_{1},\ldots,v_{m}\right)}$ is a linearly dependent list in ${V}$. Then there exists some ${j\in\left\{ 1,2,\ldots,m\right\} }$ such that
(a) ${v_{j}\in\mbox{span}\left(v_{1},\ldots,v_{j-1}\right)}$;
(b) if ${v_{j}}$ is removed from the list ${\left(v_{1},\ldots,v_{m}\right)}$, the span of the remaining list, containing ${m-1}$ vectors, equals the span of the original list.
Proof: Because ${\left(v_{1},\ldots,v_{m}\right)}$ is linearly dependent, we can write
$\displaystyle \sum_{i=1}^{m}a_{i}v_{i}=0 \ \ \ \ \ (4)$
where not all of the ${a_{i}}$s are zero. Suppose ${j}$ is the largest index where ${a_{j}\ne0.}$ Then we can divide through by ${a_{j}}$ to get
$\displaystyle v_{j}=-\frac{1}{a_{j}}\sum_{i=1}^{j-1}a_{i}v_{i} \ \ \ \ \ (5)$
Thus ${v_{j}}$ is a linear combination of other vectors in the list, which proves part (a). Part (b) follows from the fact that we can represent any vector ${u\in\mbox{span}\left(v_{1},\ldots,v_{m}\right)}$ as
$\displaystyle u=\sum_{i=1}^{m}a_{i}v_{i} \ \ \ \ \ (6)$
We can replace ${v_{j}}$ in this sum by 5, so ${u}$ can be written as a linear combination of all the vectors in the list ${\left(v_{1},\ldots,v_{m}\right)}$ except for ${v_{j}}$. Thus (b) is true. $\Box$
We can use this lemma to prove the main result about linearly independent lists:
Theorem 3 In a finite-dimensional vector space ${V}$, the length of every linearly independent list is less than or equal to the length of every list that spans ${V}$.
Proof: Suppose the list ${A\equiv\left(u_{1},\ldots,u_{m}\right)}$ is linearly independent in ${V}$, and suppose another list ${B\equiv\left(w_{1},\ldots,w_{n}\right)}$ spans ${V}$. We want to prove that ${m\le n}$.
Since ${B}$ already spans ${V}$, if we add any other vector from ${V}$ to the list ${B}$, we will get a linearly dependent list, since this newly added vector can, by the definition of a span, be expressed a linear combination of the vectors in ${B}$. In particular, if we add ${u_{1}}$ from the list ${A}$ to ${B}$, then the list ${\left(u_{1},w_{1},\ldots,w_{n}\right)}$ is linearly dependent. By the linear independence lemma above, we can therefore remove one of the ${w_{i}}$s from ${B}$ so that the remaining list still spans ${V}$, and contains ${n}$ vectors. For the sake of argument, let’s say we remove ${w_{n}}$ (we can always order the ${w_{i}}$s in the list so that the element we remove is at the end). Then we’re left with the revised list ${B_{1}=\left(u_{1},w_{1},\ldots,w_{n-1}\right)}$.
We can repeat this process ${m}$ times, each time adding the next element ${u_{i}}$ from list ${A}$ and removing the last ${w_{i}}$. Because of the linear dependence lemma, we know that there must always be a ${w_{i}}$ that can be removed each time we add a ${u_{i}}$, so there must be at least as many ${w_{i}}$s as ${u_{i}}$s. In other words, ${m\le n}$ which is what we wanted to prove. $\Box$
This theorem can be used to show easily that any list of more than ${n}$ vectors in ${n}$-dimensional space cannot be linearly independent, since we know that we can span ${n}$-dimensional space with ${n}$ vectors (for example, the 3 coordinate axes in 3-d space). Conversely, since we can find a list of ${n}$ vectors in ${n}$-dimensional space that is linearly independent, any list of fewer than ${n}$ vectors cannot span ${n}$-dimensional space.
Basis of a finite-dimensional vector space
A basis of a finite-dimensional vector space is defined to be a list that is both linearly independent and spans the space. The dimension of the vector space is defined to be the length of a basis list. For example, in 3-d space, the list ${\left\{ \left(1,0,0\right),\left(0,1,0\right),\left(0,0,1\right)\right\} }$ is a basis, and since the length is 3, the dimension of the vector space is also 3. Any proper subset (that is, a subset with fewer than 3 members) of this basis is also linearly independent, but it does not span the space so is not a basis. For example, the list ${\left\{ \left(1,0,0\right),\left(0,1,0\right)\right\} }$ is linearly independent, but spans only the ${xy}$ plane.
A couple of examples of linear independence/dependence can be found here. |
The Remainder Theorem
The remainder theorem is a formula used to find the remainder when a polynomial is divided by a linear polynomial. In this step-by-step guide, you learn more about the remainder theorem.
When a certain number of things are divided into groups with an equal number of things in each group, the number of things left is known as the remainder. This is what “remains” after division.
Step by step guide to the remainder theorem
The remainder theorem is expressed as follows:
When a polynomial $$a(x)$$ is divided by a linear polynomial $$b(x)$$ whose zero is $$x = k$$, the remainder is obtained by $$r = a (k)$$. The remaining theorem enables us to compute the remainder of dividing any polynomial by a linear polynomial without performing the steps of the division algorithm.
Remainder theorem formula
The general formula of the remainder theorem is given as follows:
$$\color{blue}{p(x)=(x-c).q(x)+r(x)}$$
Important notes:
• When a polynomial $$a(x)$$ is divided by a linear polynomial $$b(x$$) whose zero is $$x = k$$, the remainder is given by $$r = a(k)$$
• The remainder theorem formula is: $$p(x)=(x-c).q(x)+r(x)$$
• The basic formula to check the division is: Dividend $$=$$ (Divisor $$×$$ Quotient) $$+$$ Remainder
The Remainder Theorem – Example 1:
Find the remainder when $$p(x)=3x^5-x^4+x^3-4x^2+2$$ is divided by $$q(x):x-1$$.
Replace the zero of $$q(x)$$ into the polynomial $$p(x)$$ to find the remainder $$r$$:
$$x-1=0 → x=1$$
$$p(1)=3(1)^5-(1)^4+(1)^3-4(1)^2+2$$
$$=3-1+1-4+2$$
$$=1$$
Therefore, the remainder is $$1$$.
The Remainder Theorem – Example 2:
Find the remainder when $$p(x)=x^3-x^2+x-1$$ is divided by $$q(x):x+1$$.
Replace the zero of $$q(x)$$ into the polynomial $$p(x)$$ to find the remainder $$r$$:
$$x+1=0 → x=-1$$
$$p(-1)=(-1)^3-(1)^2+(-1)-1$$
$$=-1-1-1-1$$
$$=-4$$
Therefore, the remainder is $$-4$$.
Exercises for the Remainder Theorem
1. Find the remainder after $$2x^2-5x-1$$ is divided by $$x-5$$.
2. Use the remainder theorem to evaluate $$f(x)=2x^5+4x^4-3x^3+8x^2+7$$ at $$x=2$$.
3. Find the remainder when $$4x^3-5x+1$$ is divided by $$2x-1$$.
4. Use the remainder theorem to find the remainder $$(x^6+4x^5+9x^3-4x^2+10) \div (x+1)$$.
1. $$\color{blue}{24}$$
2. $$\color{blue}{143}$$
3. $$\color{blue}{-1}$$
4. $$\color{blue}{-6}$$
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### Multiplication algorithm
The multiplication algorithm is a series of steps we use to multiply numbers that are too difficult to multiply mentally. In the early years of school you will practise this method with quite simple numbers.
The distributive property for multiplication is central to our multiplication algorithm because it enables us to calculate products one column at a time and then add the results together. The distributive property involves splitting the numbers into parts and then multiplying the parts.
For example, using the distributive law we calculate:
\begin{align*} 6\times14 &= 6\times10+6\times4\\\\ &= 60+24\\\\ &= 84\\\\ \end{align*}
We can show this as a picture, using areas to represent each number:
We can also use the multiplication algorithm:
1 4 × $$\ \ _2$$ 6 8 4
To multiply 23 × 17 we follow these steps:
• Set out the numbers so the digits line up according to their place value.
• Start with the ones. Multiply the ones digit in 17 by the ones digit in 23. Say '7 times 3 is 21'. Write '1' in the ones column and carry '2' to the tens column.
• Next, work with the tens. Multiply the ones digit in 17 by the tens digit in 23.
• Say '7 times 2 is 14'. Add the '2' carried from before. Write '6' in the tens column and a '1' in the hundreds column.
• Now multiply the tens digit in 17 by the ones digit in 23. This will give a certain number of tens, so start by writing '0' in the ones column. Say '1 times 3 is 3'. Write '3' in the tens column.
• Next, multiply the tens digit in 17 by the tens digit in 23. Say '1 times 2 is 2'. Write '2' in the hundreds column.
• Finally, add 161 to 230.
The product of 23 × 17 is 391.
Hundreds Tens Ones $$2_2$$ 3 × $$1\$$ 7 1 $$6\$$ 1 2 $$3\$$ 0 3 $$9\$$ 1 |
Introductory Statistics
# 9.2Outcomes and the Type I and Type II Errors
Introductory Statistics9.2 Outcomes and the Type I and Type II Errors
When you perform a hypothesis test, there are four possible outcomes depending on the actual truth (or falseness) of the null hypothesis H0 and the decision to reject or not. The outcomes are summarized in the following table:
ACTION H0 IS ACTUALLY ...
True False
Do not reject H0 Correct Outcome Type II error
Reject H0 Type I Error Correct Outcome
Table 9.2
The four possible outcomes in the table are:
1. The decision is not to reject H0 when H0 is true (correct decision).
2. The decision is to reject H0 when H0 is true (incorrect decision known as aType I error).
3. The decision is not to reject H0 when, in fact, H0 is false (incorrect decision known as a Type II error).
4. The decision is to reject H0 when H0 is false (correct decision whose probability is called the Power of the Test).
Each of the errors occurs with a particular probability. The Greek letters α and β represent the probabilities.
α = probability of a Type I error = P(Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true.
β = probability of a Type II error = P(Type II error) = probability of not rejecting the null hypothesis when the null hypothesis is false.
α and β should be as small as possible because they are probabilities of errors. They are rarely zero.
The Power of the Test is 1 – β. Ideally, we want a high power that is as close to one as possible. Increasing the sample size can increase the Power of the Test.
The following are examples of Type I and Type II errors.
## Example 9.5
Suppose the null hypothesis, H0, is: Frank's rock climbing equipment is safe.
Type I error: Frank thinks that his rock climbing equipment may not be safe when, in fact, it really is safe. Type II error: Frank thinks that his rock climbing equipment may be safe when, in fact, it is not safe.
α = probability that Frank thinks his rock climbing equipment may not be safe when, in fact, it really is safe. β = probability that Frank thinks his rock climbing equipment may be safe when, in fact, it is not safe.
Notice that, in this case, the error with the greater consequence is the Type II error. (If Frank thinks his rock climbing equipment is safe, he will go ahead and use it.)
## Try It 9.5
Suppose the null hypothesis, H0, is: the blood cultures contain no traces of pathogen X. State the Type I and Type II errors.
## Example 9.6
Suppose the null hypothesis, H0, is: The victim of an automobile accident is alive when he arrives at the emergency room of a hospital.
Type I error: The emergency crew thinks that the victim is dead when, in fact, the victim is alive. Type II error: The emergency crew does not know if the victim is alive when, in fact, the victim is dead.
α = probability that the emergency crew thinks the victim is dead when, in fact, he is really alive = P(Type I error). β = probability that the emergency crew does not know if the victim is alive when, in fact, the victim is dead = P(Type II error).
The error with the greater consequence is the Type I error. (If the emergency crew thinks the victim is dead, they will not treat him.)
## Try It 9.6
Suppose the null hypothesis, H0, is: a patient is not sick. Which type of error has the greater consequence, Type I or Type II?
## Example 9.7
It’s a Boy Genetic Labs claim to be able to increase the likelihood that a pregnancy will result in a boy being born. Statisticians want to test the claim. Suppose that the null hypothesis, H0, is: It’s a Boy Genetic Labs has no effect on gender outcome.
Type I error: This results when a true null hypothesis is rejected. In the context of this scenario, we would state that we believe that It’s a Boy Genetic Labs influences the gender outcome, when in fact it has no effect. The probability of this error occurring is denoted by the Greek letter alpha, α.
Type II error: This results when we fail to reject a false null hypothesis. In context, we would state that It’s a Boy Genetic Labs does not influence the gender outcome of a pregnancy when, in fact, it does. The probability of this error occurring is denoted by the Greek letter beta, β.
The error of greater consequence would be the Type I error since couples would use the It’s a Boy Genetic Labs product in hopes of increasing the chances of having a boy.
## Try It 9.7
“Red tide” is a bloom of poison-producing algae–a few different species of a class of plankton called dinoflagellates. When the weather and water conditions cause these blooms, shellfish such as clams living in the area develop dangerous levels of a paralysis-inducing toxin. In Massachusetts, the Division of Marine Fisheries (DMF) monitors levels of the toxin in shellfish by regular sampling of shellfish along the coastline. If the mean level of toxin in clams exceeds 800 μg (micrograms) of toxin per kg of clam meat in any area, clam harvesting is banned there until the bloom is over and levels of toxin in clams subside. Describe both a Type I and a Type II error in this context, and state which error has the greater consequence.
## Example 9.8
A certain experimental drug claims a cure rate of at least 75% for males with prostate cancer. Describe both the Type I and Type II errors in context. Which error is the more serious?
Type I: A cancer patient believes the cure rate for the drug is less than 75% when it actually is at least 75%.
Type II: A cancer patient believes the experimental drug has at least a 75% cure rate when it has a cure rate that is less than 75%.
In this scenario, the Type II error contains the more severe consequence. If a patient believes the drug works at least 75% of the time, this most likely will influence the patient’s (and doctor’s) choice about whether to use the drug as a treatment option.
## Try It 9.8
Determine both Type I and Type II errors for the following scenario:
Assume a null hypothesis, H0, that states the percentage of adults with jobs is at least 88%.
Identify the Type I and Type II errors from these four statements.
1. Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88% when that percentage is actually less than 88%
2. Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88% when the percentage is actually at least 88%.
3. Reject the null hypothesis that the percentage of adults who have jobs is at least 88% when the percentage is actually at least 88%.
4. Reject the null hypothesis that the percentage of adults who have jobs is at least 88% when that percentage is actually less than 88%. |
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# Exponential Decay
## Rational functions with x as an exponent in the denominator
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Exponential Decay
Suppose the amount of a radioactive substance is cut in half every 25 years. If there was originally 500 grams of the substance, could you write a function representing the amount of the substance after $x$ years? How much of the substance would there be after 100 years? Will the amount of the substance ever reach 0 grams? After completing this Concept, you'll be able to answer questions like these regarding exponential decay.
### Guidance
In the last Concept, we learned how to solve expressions that modeled exponential growth. In this Concept, we will be learning about exponential decay functions.
General Form of an Exponential Function: $y=a (b)^x$ , where $a=$ initial value and
$b = growth \ factor$
In exponential decay situations, the growth factor must be a fraction between zero and one.
$0
#### Example A
For her fifth birthday, Nadia’s grandmother gave her a full bag of candy. Nadia counted her candy and found out that there were 160 pieces in the bag. Nadia loves candy, so she ate half the bag on the first day. Her mother told her that if she continues to eat at that rate, it will be gone the next day and she will not have any more until her next birthday. Nadia devised a clever plan. She will always eat half of the candy that is left in the bag each day. She thinks that she will get candy every day and her candy will never run out. How much candy does Nadia have at the end of the week? Would the candy really last forever?
Solution: Make a table of values for this problem.
Day 0 1 2 3 4 5 6 7
# of Candies 160 80 40 20 10 5 2.5 1.25
You can see that if Nadia eats half the candies each day, then by the end of the week she has only 1.25 candies left in her bag.
Write an equation for this exponential function. Nadia began with 160 pieces of candy. In order to get the amount of candy left at the end of each day, we keep multiplying by $\frac{1}{2}$ . Because it is an exponential function, the equation is:
$y=160 \cdot \frac{1}{2}^x$
Graphing Exponential Decay Functions
#### Example B
Graph the exponential function $y=5 \cdot \left(\frac{1}{2}\right)^x$ .
Solution: Start by making a table of values. Remember when you have a number to the negative power, you are simply taking the reciprocal of that number and taking it to the positive power. Example: $\left(\frac{1}{2}\right)^{-2} = \left(\frac{2}{1}\right)^2 = 2^2$ .
$x$ $y=5 \cdot \left(\frac{1}{2}\right)^x$
–3 $y =5 \left(\frac{1}{2}\right)^{-3}=40$
–2 $y =5 \left(\frac{1}{2}\right)^{-2}=20$
–1 $y =5 \left(\frac{1}{2}\right)^{-1}=10$
0 $y =5 \left(\frac{1}{2}\right)^0=5$
1 $y =5 \left(\frac{1}{2}\right)^1=\frac{5}{2}$
2 $y =5 \left(\frac{1}{2}\right)^2=\frac{5}{4}$
Now graph the function.
Using the Property of Negative Exponents, the equation can also be written as $5 \cdot 2^{-x}$ .
#### Example C
Graph the functions $y=4^x$ and $y=4^{-x}$ on the same coordinate axes.
Solution:
Here is the table of values and the graph of the two functions.
Looking at the values in the table, we see that the two functions are “reverse images” of each other in the sense that the values for the two functions are reciprocals.
$x$ $y=4^x$ $y=4^{-\chi}$
–3 $y=4^{-3} = \frac{1}{64}$ $y=4^{-(-3)} = 64$
–2 $y=4^{-2} = \frac{1}{16}$ $y=4^{-(-2)} = 16$
–1 $y=4^{-1} = \frac{1}{4}$ $y=4^{-(-1)} = 4$
0 $y=4^0 = 1$ $y=4^{-(0)} = 1$
1 $y=4^1 = 4$ $y=4^{-(1)} =\frac{1}{4}$
2 $y=4^2 = 16$ $y=4^{-(2)} =\frac{1}{16}$
3 $y=4^3 = 64$ $y=4^{-(3)} = \frac{1}{64}$
Here is the graph of the two functions. Notice that these two functions are mirror images if the mirror is placed vertically on the $y-$ axis.
### Guided Practice
If a person takes 125 milligrams of a drug, and after the full dose is absorbed into the bloodstream, there is only 70% left after every hour, write a function that gives the concentration left in the bloodstream after $t$ hours. What is the concentration of the drug in the bloodstream after 3 hours?
Solution:
This will be a decay function in the form $f(t)=a\cdot b^t$ . We know the initial value is 125 milligrams. After one hour we multiply that by 0.70 to find 70% of 125. After the second hour, we multiply by 0.7 again and so on:
$\text{The initial dose.} && 125&=125 \\\text{After one hour.} && 125 \cdot 0.7&=87.5\\\text{After two hours.} && 125 \cdot 0.7 \cdot 0.7=125 \cdot 0.7^2&=61.25\\\text{After three hours.} && 125 \cdot 0.7 \cdot 0.7 \cdot 0.7=125 \cdot 0.7^3&=42.875$
Since we multiply by 0.7 to find the 70% that is left after each hour, the decay factor is 0.7. The function will be:
$f(t)=125(0.7)^t.$
We also found that after three hours the amount of the drug in the bloodstream will be 42.875 milligrams.
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Exponential Decay Functions (10:51)
1. Define exponential decay.
2. What is true about “b” in an exponential decay function?
3. Suppose $f(x)=a(b)^x$ . What is $f(0)$ ? What does this mean in terms of the $y-$ intercept of an exponential function?
Graph the following exponential decay functions.
1. $y=\frac{1}{5}^x$
2. $y=4 \cdot \left(\frac{2}{3}\right)^x$
3. $y=3^{-x}$
4. $y=\frac{3}{4} \cdot 6^{-x}$
1. The percentage of light visible at $d$ meters is given by the function $V(d)=0.70^d$ .
1. What is the growth factor?
2. What is the initial value?
3. Find the percentage of light visible at 65 meters.
1. A person is infected by a certain bacterial infection. When he goes to the doctor, the population of bacteria is 2 million. The doctor prescribes an antibiotic that reduces the bacteria population to $\frac{1}{4}$ of its size each day.
1. Draw a graph of the size of the bacteria population against time in days.
2. Find the formula that gives the size of the bacteria population in terms of time.
3. Find the size of the bacteria population ten days after the drug was first taken.
4. Find the size of the bacteria population after two weeks (14 days).
Mixed Review
1. The population of Kindly, USA is increasing at a rate of 2.14% each year. The population in the year 2010 is 14,578.
1. Write an equation to model this situation.
2. What would the population of Kindly be in the year 2015?
3. When will the population be 45,000?
1. The volume of a sphere is given by the formula $v=\frac{4}{3} \pi r^3$ . Find the volume of a sphere with a diameter of 11 inches.
2. Simplify $\frac{6x^2}{14y^3} \cdot \frac{7y}{x^8} \cdot x^0 y$ .
3. Simplify $3(x^2 y^3 x)^2$ .
4. Rewrite in standard form: $y-16+x=-4x+6y+1$ .
### Vocabulary Language: English Spanish
exponential decay
exponential decay
In exponential decay situations, the growth factor must be a fraction between zero and one. $0
Exponential Function
Exponential Function
An exponential function is a function whose variable is in the exponent. The general form is $y=a \cdot b^{x-h}+k$.
Model
Model
A model is a mathematical expression or function used to describe a physical item or situation. |
# Interactive Differential Equations: A Step-by-Step Approach to Methods & Modeling
## Section1.6Linear Terms
The most informative label you can place on a differential equation is whether it is linear or nonlinear. While linearity often refers to equations that graph as straight lines, the concept of linearity in differential equations is more nuanced. Understanding this distinction is crucial as it significantly influences how the equations are solved and interpreted.
Before jumping in, it may be helpful to revisit meaning of dependent variables, terms and coefficients as we will be referencing them throughout this discussion.
Now, let’s define what it means for a term in a differential equation to be linear.
### Definition11.Linear Term.
Linear Term
A term that has one of the forms
\begin{equation*} a(t)\ y,\ a(t)\ y',\ a(t)\ y'',\ a(t)\ y''',\ \ldots \end{equation*}
where $$y$$ is the dependent variable. The coefficient, $$a(t)\text{,}$$ can be a function of the independent variable, $$t\text{,}$$ but not the dependent variable.
### Note12.The linearity of a term depends entirely on the dependent variable.
The linearity of a term is determined by the presence of a single $$y$$ or derivative of $$y$$ to the power of $$1\text{.}$$ The coefficients, $$a(t)$$ play no part in the linearity.
### Example13.
$$\ \$$Given the differential equation,
\begin{equation*} e^{t}y^{(7)} + (t+1)y'y''' - t \ln y'' - y' \sin t - \tan y + \frac{4}{y} = \frac{3}{t}\text{,} \end{equation*}
identify each term and label the term as linear or nonlinear.
Solution.
To determine the linearity of a term we only need to consider the dependent variables (column $$3$$ below). We are looking for a single $$y$$ or derivative of $$y$$ raised to the power of $$1\text{.}$$ If we find such a term, we label it as linear. Otherwise, it is nonlinear. Let’s separate the different parts in the following table.
Term Coefficient Dependent Variables Linearity $$\ds e^{t}\ y^{(7)}$$ $$\ds e^{t}$$ $$\ds y^{(7)}$$ linear $$\ds (t+1)y'y'''$$ $$\ds (t+1)$$ $$\ds y'y'''$$ nonlinear $$\ds t \ln y''$$ $$\ds t$$ $$\ds \ln y$$ nonlinear $$\ds y' \sin t$$ $$\ds \sin t$$ $$\ds y'$$ linear $$\ds \tan y$$ $$\ds 1$$ $$\ds \tan y$$ nonlinear $$\ds \frac{4}{y}$$ $$\ds 4$$ $$\ds \frac{1}{y}$$ nonlinear $$\ds \uparrow$$ Linearity dependson this column only
Note, since constant terms do not contain a dependent variable, they are necessarily linear. So, in the context of differential equations, $$\frac{3}{t}$$ is also a linear term.
To help identify nonlinear terms, here are some common characteristics:
### Identifying Nonlinear Terms.
A term is nonlinear if it contains a dependent variable, $$y$$ or $$y^{(n)}\text{,}$$ that is
• Raised to a power other than 1, e.g., $$y^2$$ or $$(y')^{-4}\text{.}$$
• Inside another function, e.g., $$\ln(y)$$ or $$\sin(y)\text{.}$$
• Multiplied or divided by another $$y$$ or $$y^{(n)}\text{,}$$ e.g., $$y y'''$$ or $$\frac{y'}{y}\text{.}$$
Let’s practice this with one more example.
### Example14.
$$\ \$$Determine the linearity of each term in the following differential equations.
\begin{equation*} \frac{1}{t}y'' + y^2 + \ln(y') = e^t \end{equation*}
Solution.
\begin{align*} \underset{\text{linear}}{\underline{\frac{1}{t}{\color{blue} y'' }}} + \underset{\text{nonlinear}}{\underline{{\color{blue} y}^2}} + \underset{\text{nonlinear}}{\underline{\ln({\color{blue} y'})}} =\ \amp \underset{\text{linear}}{\underline{e^t}} \end{align*}
\begin{equation*} P^{(6)} + \frac{m P'}{P} = (m - 1)^2 \end{equation*}
Solution.
\begin{align*} \underset{\text{linear}}{\underline{P^{(6)}}} + \underset{\text{nonlinear}}{\underline{\frac{m {\color{blue} P'}}{{\color{blue} P}}}} =\ \amp \underset{\text{linear}}{\underline{(m - 1)^2}} \end{align*}
Assume that $$y$$ and $$t$$ are the dependent and independent variables, respectively.
#### 1.$$3t$$ is a linear term.
$$3t$$ is a linear term
• True
• Correct! The term $$3t$$ is linear because it is a function of the independent variable only.
• False
• Incorrect, review the examples of linear terms in the section on Linear Terms.
#### 2.$$y \sin(t)$$ is a linear term.
$$y \sin(t)$$ is a linear term
• True
• Correct! The term $$\sin(t)y$$ is linear.
• False
• Incorrect. The term $$\sin(t)y$$ is indeed linear.
#### 3.$$t\sin(y')$$ is a linear term.
$$t\sin(y')$$ is a linear term
• True
• Incorrect, review the rules for identifying nonlinear terms.
• False
• Correct! A term that includes the derivative of the dependent variable inside another function is nonlinear.
#### 4.$$e^{ty}$$ is a linear term.
$$e^{ty}$$ is a linear term
• False
• Correct! The term $$e^{ty}$$ is nonlinear because the dependent variable $$y$$ is inside an exponential function.
• True
• Incorrect. Review the rules for identifying nonlinear terms.
#### 5.Identify the linear term in the equation: $$\ds\frac{1}{t}y'' + y^2 + \ln(y') = e^t$$.
Identify the linear term in the equation: $$\ds\frac{1}{t}y'' + y^2 + \ln(y') = e^t$$
• $$\frac{1}{t}y''$$
• Correct! The term $$\frac{1}{t}y''$$ is linear because it involves a single derivative of the dependent variable.
• $$y^2$$
• Incorrect. This term is nonlinear.
• $$\ln(y')$$
• Incorrect. This term is nonlinear because it involves the dependent variable inside a logarithmic function.
• $$e^t$$
• Incorrect. This term is linear because it involves the independent variable only.
#### 6.Which of the following is a nonlinear term?
Which of the following is a nonlinear term?
• $$\ds 3y$$
• Incorrect. This is a linear term.
• $$\ds t\sin(y')$$
• I The term $$t\sin(y')$$ is nonlinear because it involves a derivative inside a trigonometric function.
• $$\ds \frac{d^2y}{dx^2}$$
• Incorrect. This is a linear term.
• $$\ds 5y'$$
• Incorrect. This is a linear term.
#### 7.A linear term can contain the dependent variable multiplied by the independent variable.
A linear term can contain the dependent variable multiplied by the independent variable
• True
• Correct! For example, $$t y$$ is a linear term.
• False
• Incorrect. Carefully review the examples above.
#### 8.Which of the following terms is linear?
Which of the following terms is linear?
• $$y^2$$
• Incorrect. $$y^2$$ is nonlinear because the dependent variable is raised to a power other than one.
• $$\sin(t)y'$$
• Correct! $$\sin(t)y'$$ is linear because it is a function of the independent variable multiplied by the first derivative of the dependent variable.
• $$\ln(y)$$
• Incorrect. $$\ln(y)$$ is nonlinear because it includes the dependent variable inside another function.
• $$y y'$$
• Incorrect. $$y \cdot y'$$ is nonlinear because it involves the product of the dependent variable and its derivative.
#### 9.Which of the following terms is linear?
Which of the following terms is linear?
• $$\ds \frac{1}{t}y''$$
• Correct! $$\frac{1}{t}y''$$ is linear because it is a function of the independent variable multiplied by the second derivative of the dependent variable.
• $$\ds y^3$$
• Incorrect. $$y^3$$ is nonlinear because the dependent variable is raised to a power other than one.
• $$\ds e^t y^2$$
• Incorrect. $$e^t y^2$$ is nonlinear because the dependent variable is squared.
• $$\ds y \cos(y)$$
• Incorrect. $$y \cdot \cos(y)$$ is nonlinear because it involves the product of the dependent variable and a function of the dependent variable.
#### 10.Which of the following is a characteristic of a nonlinear term?
Which of the following is a characteristic of a nonlinear term?
• It involves the dependent variable raised to the first power.
• Incorrect. This is a characteristic of a linear term.
• It involves the dependent variable only as a constant.
• Incorrect. This is a characteristic of a linear term.
• It includes the dependent variable inside another function.
• Correct! A nonlinear term includes the dependent variable inside another function.
• It involves the independent variable only.
• Incorrect. This is a characteristic of a linear term.
#### 11.Which term is an example of a nonlinear term?
Which term is an example of a nonlinear term?
• $$3$$
• Incorrect. $$3$$ is linear because it is a constant.
• $$3t$$
• Incorrect. $$3t$$ is linear because it is a function of the independent variable only.
• $$y^2$$
• Correct! $$y^2$$ is nonlinear because the dependent variable is squared.
• $$2t^2 y$$
• Incorrect. $$2t^2 y$$ is linear because it is a function of the independent variable multiplied by the dependent variable.
Hint.
Hint. |
# Ex.7.4 Q5 Triangles Solution - NCERT Maths Class 9
Go back to 'Ex.7.4'
## Question
In the given figure, $$PR \gt PQ$$ and $$PS$$
bisects $$\angle QPR$$. Prove that
$$\angle PSR \gt \angle PSQ$$.
Video Solution
Triangles
Ex 7.4 | Question 5
## Text Solution
What is Known?
$$\text{PR PQ}$$ and PS bisects $$\angle \text{QPR}$$
To prove:
$\angle \text{PSR }>\angle \text{PSQ}\text{.}$
Reasoning:
We can use exterior angle sum property to find the required inequality.
Steps:
As $$PR \gt PQ$$,
\begin{align} & \angle PQR \gt \angle PRQ\\&\left(\begin{array}{} \text {Angle opposite to larger}\\\text{ side is larger}) \ldots (1)\end{array}\right) \\\\ & PS \text { is the bisector of } \angle QPR \\ & \angle QPS = \angle RPS \ldots (2) \\\\ & \angle PSR \text{ is the exterior angle of } PQS \\ & \angle PSR = \angle PQR + \angle QPS \ldots (3)\\\\ & \angle PSQ \text{ is the exterior angle of } PRS \\ & \angle PSQ = \angle PRQ + \angle RPS \ldots (4)\\ \end{align}
Adding Equations ($$1$$) and ($$2$$), we obtain
\begin{align} &\angle PQR + \angle QPS \gt \angle PRQ + \angle RPS \\ & \angle PSR \gt \angle PSQ \\& \begin{bmatrix}\text{Using the values of}\\ \text{Equations (2),(3) and (4)}\end{bmatrix}\end{align}
Video Solution
Triangles
Ex 7.4 | Question 5
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### Dividing integer numbers
In algebra there are two important rules to remember when multiplying and dividing integer numbers:
• If both numbers are positive, the quotient is positive.
• If both numbers are negative, the quotient is positive.
• If the two numbers have different signs, the quotient is negative.
Typically the second bullet point gives algebra students a headache. They ask themselves, “WHY is the product or quotient of two negative numbers positive?!” Well, at the end of this article I have provided a little argument to show you why it’s true. If you don’t follow the argument, that’s okay. You can do some research and find plenty of other explanations of why the product or quotient of two negative numbers is positive.
For now, we will assume that the rule is true. Let’s look at two examples.
Example: Find the quotient $$- 14 \div 7$$
Solution: Here we simply divide $$- 14 \div 7$$, which is $$2$$. Then we notice in the original problem that one number is negative, and one is positive. So the answer will be negative. The solution is
$$\begin{gathered} - 14 \div 7 = - \left( {14 \div 7} \right) \\ = - 2 \\ \end{gathered}$$
Example: Find the quotient $$- 110 \div 10$$.
Solution: We have
$$- 110 \div 10 = - 11$$
Since one of the numbers is negative and one is positive.
Here is the argument I promised from above. Let a and b be two positive numbers. Then
$$\left( { - a} \right) + a = 0,$$ additive inverse property $$\left( { - b} \right)*\left[ {\left( { - a} \right) + a} \right] = \left( { - b} \right)*0,$$ multiply both sides by b $$\left( { - b} \right)\left( { - a} \right) + \left( { - b} \right)\left( a \right) = 0,$$ distributive property of multiplication $$\left( { - b} \right)\left( { - a} \right) + \left( { - ba} \right) = 0,$$ since $$\left( { - b} \right)\left( a \right) = \left( { - ba} \right)$$ $$\left( { - b} \right)\left( { - a} \right) + \left( { - ba} \right) + \left( {ba} \right) = 0 + \left( {ba} \right),$$ add ba to both sides $$\left( { - b} \right)\left( { - a} \right) + 0 = 0 + \left( {ba} \right),$$ additive inverse property $$\left( { - b} \right)\left( { - a} \right) = ba,$$ identity property
Chew on that!
3304 x
Find each quotient.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
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3392 x
Find each quotient.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
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3361 x
Find each quotient.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
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### Geometry
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Properties of Triangles
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Beginning Algebra
Beginning Trigonometry
Equations
Exponents
Factoring
Linear Equations and Inequalities
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How will Jonathan Adler perform on 11/12/2019 and the days ahead? Let’s use astrology to conduct a simple analysis. Note this is just for fun – don’t get too worked up about the result. I will first find the destiny number for Jonathan Adler, and then something similar to the life path number, which we will calculate for today (11/12/2019). By comparing the difference of these two numbers, we may have an indication of how well their day will go, at least according to some astrology practitioners.
PATH NUMBER FOR 11/12/2019: We will take the month (11), the day (12) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. How? Let’s walk through it. First, for the month, we take the current month of 11 and add the digits together: 1 + 1 = 2 (super simple). Then do the day: from 12 we do 1 + 2 = 3. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 2 + 3 + 12 = 17. This still isn’t a single-digit number, so we will add its digits together again: 1 + 7 = 8. Now we have a single-digit number: 8 is the path number for 11/12/2019.
DESTINY NUMBER FOR Jonathan Adler: The destiny number will calculate the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Jonathan Adler we have the letters J (1), o (6), n (5), a (1), t (2), h (8), a (1), n (5), A (1), d (4), l (3), e (5) and r (9). Adding all of that up (yes, this can get tedious) gives 51. This still isn’t a single-digit number, so we will add its digits together again: 5 + 1 = 6. Now we have a single-digit number: 6 is the destiny number for Jonathan Adler.
CONCLUSION: The difference between the path number for today (8) and destiny number for Jonathan Adler (6) is 2. That is lower than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t go jumping for joy yet! As mentioned earlier, this is of questionable accuracy. If you want to see something that we really strongly recommend, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. |
# How BigIntegers Work, part 2: Multiplication
In the last post, I gave an overview of how BigIntegers are stored and how addition and subtraction work. Now it's time to move up the hyperoperation sequence and and discuss multiplication.
#### Simple Multiplication
Probably the simplest way to think about multiplication is repeated addition. In other words, `4*5 = 5+5+5+5`. Using that definition, it would be trivial to write a multiplication function in terms of addition:
``````
// Assume we've defined "ZERO" and "ONE" constants and a "notZero(n)" function, which are all trivial.
// notZero can be as simple as "return n.sign !== 0".
function multiply(a, b) {
var product = ZERO;
while (notZero(a)) {
a = subtract(a, ONE);
}
return product;
}
``````
Something like the code above would work, but would be extremely slow. Imagine multiplying even relatively small numbers (for a big integer library) like 123,456,789 * 65,128,321. Even optimizing it by making sure a is smaller (remember, order doesn't matter in multiplication) we would have to run the loop 65,128,321 times, with an addition and subtraction each time. Now imagine two 100 digit numbers! Obviously this won't work.
The key thing to remember is that numbers are polynomials. So `123 * 456` is really `(1 * 100 + 2 * 10 + 3) * (4 * 100 + 5 * 10 + 6)`. Simplifying a little more, we get `(100 + 20 + 3) * (400 + 50 + 6)`. Lets leave the right side how it is and add up the terms on the left to get our original 123 back: `123 * (400 + 50 + 6)`. Now we can use the distributive property and get `123400 + 12350 + 123*6 = 49200 + 6150 + 738`.
All that is a great flashback to your high school algebra class, but how does it help us? Remember that to multiply by a factor of 10, you just add zeros to the end of the number. If we factor out the power of 10 from each term, we just have to do a couple single-digit multiplications, append some zeros, and do a couple additions. Easy! In fact, this is how you've been doing multiplication since grade school.
## `````` 123 x 456``````
``` 738 615 +492 ```
```56088 ```
So let's break down what we're doing. Starting from the right, take each digit from the bottom number, multiply it by each digit of the top number (also from right to left) and write down the result. I've glossed over it so far, but remember if you end up with a two digit number, you have to carry the first digit and add it to the next result. Then you move to the next digit, remembering to "multiply by 10" by shifting the numbers over to the left each time. Finally, add up all the "partial products" and you have the (hopefully correct) answer!
So, how do we do this in code? I'm glad you asked. Here's a straightforward implementation (we'll optimize it slightly by eliminating some of the intermediate steps later):
``````
function multiply(a, b) {
// To keep track of all the partial products
var partialProducts = [];
// For each digit of b
for (var i = 0; i < b.digits.length; i++) {
var carry = 0;
var digit;
var partial = {sign: 1, digits: []}; // Initialize a new partial product
// For each digit of a
for (var 0; j < a.digits.length; j++) {
digit = (b.digits[i] * a.digits[j]) + carry; // Multiply the digits, adding in any carry
carry = Math.floor(digit / 10); // Integer divide by 10 to get the first digit
partial.digits[j] = digit % 10; // Mod 10 gets the second digit
}
// Don't forget the final carry (if necessary)!
if (carry > 0) {
partial.digits[j] = carry;
}
// Shift the digits to the left to multiply by the next power of 10
for (var shift = 0; shift < i; shift++) {
// Unfortunately, JS's naming is backwards from ours.
// "array.unshift(0)" pushes a zero into the front of the array, which is what we want
partial.digits.unshift(0);
}
// Append this partial product to the list
partialProducts.push(partial);
}
// Now add up all the partial products
var product = ZERO;
for (var i = 0; i < partialProducts.length; i++) {
}
return product;
}
``````
Whew! That's pretty long, but fairly simple. For each digit of b, loop through each digit of a, multiplying the digits and storing them in the partial product, then add up the partial products at the end.
It's also still inefficient for several reasons. It's much better than the simple "loop and add" function at the beginning but we can do better, and actually shorten the code at the same time.
#### The Real Algorithm
The problem is all those partial products. Instead of building up a list (taking up a potentially large amount of memory) then adding them up at the end, we can just add in each product as we go. We can also avoid pushing all the zeros in by adjusting how we index into the arrays.
``````
function multiply(a, b) {
var partial = { sign: a.sign * b.sign, digits: [] }; // digits should be filled with zeros
// For each digit of b
for (var i = 0; i < b.digits.length; i++) {
var carry = 0;
var digit;
// For each digit of a
for (var j = i; j < a.digits.length + i; j++) {
// Multiply the digits, and add them to the current partial product, along with any carry
digit = partial.digits[j] + (b.digits[i] * a.digits[j - i]) + carry;
carry = Math.floor(digit / 10); // New carry
partial.digits[j] = digit % 10; // Put the result back into the partial product
}
// Don't forget the final carry (if necessary)!
if (carry) {
digit = partial.digits[j] + carry;
carry = Math.floor(digit / 10);
partial.digits[j] = digit % 10;
}
}
// That's it!
return partial;
}
``````
I didn't mention it before, but the nice thing about multiplication is that the sign is easy to get right. Because our `sign` property is a 1 (or zero) with the same sign as the number itself, the resulting sign is just `a.sign * b.sign`.
### Further Optimizations
The last function above, is basically how the BigInteger library does multiplication (except in base 10000000 and, as usual, I've skipped over some minor details for clarity). But there are actually more sophisticated algorithms to make multiplying huge numbers faster.
They are mostly based on Karatsuba multiplication, which is a recursive "divide and conquer" algorithm. The added complexity and overhead of recursive function calls makes it slower for "reasonably sized" numbers, so it's best to use simple long multiplication until you get to some threshold then switch to Karatsuba multiplication. I experimented with supporting it early on and the cutoff was somewhere in the several-thousand-digit range.
Tagged In: Calculator, Math |
# Calculation Short tricks: Calculate squares quickly!
By Jyoti Bisht|Updated : May 12th, 2021
Master your calculation tricks: Simplification is the most widely asked topic in almost every banking exam. The questions on simplification check your calculation speed. The speed of solving questions completely depends on your good command on topics like tables, squares, cubes, squares and cube roots, multiplication, etc. If you know the short tricks to solve all these topics, the time of solving question will be reduced and hence, accuracy also gets improved.
Master your calculation tricks: Simplification is the most widely asked topic in almost every banking exam. The questions on simplification check your calculation speed. The speed of solving questions completely depends on your good command on topics like tables, squares, cubes, squares and cube roots, multiplication, etc. If you know the short tricks to solve all these topics, the time of solving question will be reduced and hence, accuracy also gets improved.
### Unlock 360+ Mocks for all the bank and insurance exam
In the article, we are sharing the shortcut methods of solving squares quickly.
## Base Method of Calculating squares
Generally, we can calculate the square of a number by the formula (a2 + b2 + 2ab). This formula applies by splitting a number into a and b. For eg. (49)2 can be calculated using this formula by splitting into 40 and 9. But calculating with this method always is a lengthy process and not recommended to follow in the exam.
Base method of calculating squares is an easy method to square 2 and 3 digits numbers easily. The major thing is to find the Base here. The base is referred to n x 10x where n is 1, 2, 3, and so on. You have to take the value of base i.e, 10x which is nearer to the number.
Let us see the steps to find the square of a number using Base Method
Step 1: Find Base i.e, n x10x.
Step 2: Case 1: Square the (Number - Base) if Number > Base
Case 2: Square the (Base - Number) if Base > Number.
Make sure the number of digits in this part = number of zeroes in the base, add zeroes if it is of fewer digits or carry forward extra digits in case of more than the required digits.
Step 3: Case 1: Add (Number - Base) to Number if Number > Base and then multiply by n.
Case 2: Subtract (Base - Number) from Number if Base > Number and then multiply by n .
Step 4: Merge results of Step 2 and 3. Keep Step 2 result on the right side.
Suppose you have to find the square of 104.
Step 1: Now, here the base is 100 or 102.
Step 2: Here Number > Base or 104 > 100 so, (104 - 100)= 16
Step 3: 1 x (104 + 4) = 108
Step 4: 10816
In this way, you have to calculate the square using the base method. Let us take more examples to have a clear understanding.
Find the square of 99.
Step 1: Here, the base is 100 or 102.
Step 2: 100 > 99 (100- 99)= 01 (Here we added one zero to the left side of square number to make it a 2 digit number)
Step 3: 1 x (99 - 1) = 98
Step 4: 9801
Find the square of 119.
Step 1: Base is 100 or 102.
Step 2: (119- 100)= 3 61 (Here 3 will be carried forward as we need only 2 digits at this side)
Step 3: 1 x (119 + 19) = 138 + 3 carry = 141
Step 4: 14161
Find the square of 198
Step 1: Now, here the base is 200 or 2 x 102.
Step 2: (200- 198)= 04 (Here we added one zero to the left side of square number to make it a 2 digit number)
Step 3: 2 x (198- 2) = 392
Step 4: 39204
Find the square of 482
Step 1: Now, here the base is 500 or 5 x 102.
Step 2: (500- 482)= 3 24 (Here 3 will be carried forward as we need only 2 digits on this side)
Step 3: 5 x (482- 18) = 2320 + 3 carry = 2323 (Here we had multiplied it by 5 as the base was 5 x 10)
Step 4: 232324
Find the square of 1012
Step 1: Now, here the base is 1000 or 1 x 103.
Step 2: (1012- 1000)= 144 (Here we will keep three digits as the number of zeroes in the base is 3)
Step 3: (1012 + 12) = 1024
Step 4: 1024144
In the exam, you have don't have to write the individual steps. Just start calculating mentally and write the final value. After practising, you will be able to solve this quickly.
Calculate the following squares using the base method and answer in the comment section.
(A) (93)2 (B) (216)2 (C) (1001)2 (D) (3014)2
Base method is useful when numbers are nearer to the base as it will become more calculating when the difference between numbers and base becomes larger. To overcome such calculation, we are introducing one more method i.e., Duplex Method.
## Duplex Method to calculate squares
To understand this method, at first, you must know how to calculate duplex of the numbers.
Dup (a) = a2
Dup (ab) = 2 x a x b
Dup (abc) = 2 x a x c + b2
Dup (abcd) = 2 x a x d + 2 x b x c
Now let`s start calculating squares using this method.
(A) Method of calculating the square of a number ab.
It will be Dup a | Dup ab | Dup b
Calculate square of 42
Dup 4 | Dup 42 | Dup 2
42 |2 x 4 x 2| 22
16 | 1 6 | 4 (Keep only one digit in each part except first one)
1764
(B) Method of calculating the square of a number abc.
Dup a | Dup ab | Dup abc | Dup bc | Dup c
Find the square of 145
Dup 1 | Dup 14 | Dup 145 | Dup 45 | Dup 5
12 |2 x 1 x 4| 2 x 1 x 5 + 42 |2 x 4 x 5 | 52
1 | 8 | 2 6 | 4 0 | 2 5
21025
(C) Method of calculating the square of a number abcd.
Dup a | Dup ab | Dup abc | Dup abcd | Dup bcd | Dup cd | Dup d
Find the square of 1234
Dup 1 | Dup 12 | Dup 123 | Dup 1234 | Dup 234 | Dup 34 | Dup 4
12 |2 x 1 x 2| 2 x 1 x 3 + 22 |2 x 1 x 4 + 2 x 2 x 3 |2 x 2 x 4 + 32|2 x 3 x 4| 42
1 | 4 | 1 0 | 2 0 | 2 5 | 2 4 | 1 6
1522756
This is the way to solve squares using the duplex method. But in exams, don`t write steps. Just start calculating duplex from the right-hand side and write the final answer. Practice solving squares using this method without steps. You can calculate any number of digits from this method. The only need for this method is to know how to calculate duplex of the numbers.
Calculate the following squares using the duplex method and answer in the comment section.
(A) (87)2 (B) (529)2 (C) (1991)2 (D) (5018)2
Let us see some more tricks of solving squares of the numbers ending with digit 1, 5, 6 and 9.
## Trick to calculate squares of numbers ending with digit 5
Step 1: We all know 52 is 25. In the squares of numbers ending with 5, just write 25 on the right-hand side of the answer
Step 2: First digit of the number x Successive digit of the first number.
Calculate 252 = 2 x 3 | 25 = 625
Calculate 552 = 5 x 6 | 25 = 3025
Calculate 852 = 8 x 9 | 25 = 7225
Calculate 1252 = 12 x 13 | 25 = 15625
Calculate 3252= 32 x 33 | 25 = 105625
In the same way, you can calculate the square of other numbers ending with digit 5. You just need to strengthen your multiplication skills.
## Trick to calculate squares of numbers ending with digit 6
Suppose you have to find the square of 76.
Step : (Number - 1)2 + Number + (Number - 1)
(76)2= (75)2 + 75 + 76 = 5625 + 151 = 5776 (Calculate square of number ending with digit 5 using the above method.)
(146)2= (145)2 + 145 + 146 = 21025 + 291= 21316
(1006)2= (1005)2 + 1005 + 1006 = 1010025+ 2011= 1012036
To calculate squares of numbers ending with digit 6, you must know how to calculate squares of numbers ending with digit 5 and fast multiplication skills as it is the foremost requirement.
## Trick to calculate squares of numbers ending with digit 4
Suppose you have to find the square of 64.
Step : (Number + 1)2 - Number - (Number + 1)
(64)2= (65)2 - 65 - 64 = 4225 - 129 = 4096(Calculate square of number ending with digit 5 using the above method.)
(294)2= (295)2 - 295 - 294 = 87025 - 589= 86436
(3004)2= (3005)2 - 3005 - 3004 = 9030025 - 6009= 9024016
Again, you must know how to calculate squares of numbers ending with digit 5 with strong multiplication skills to calculate squares of numbers ending with digit 4.
## Trick to calculate squares of numbers ending with digit 1
Suppose you have to calculate the square of 91.
Step : (Number - 1)2 + Number + (Number + 1)
(91)2= (90)2 + 90 + 91 = 8100 + 181 = 8281
(161)2= (160)2 + 160 + 161 = 25600 + 321 = 25921
(921)2= (920)2 + 920 + 921 = 846400 + 1841 = 848241 (Calculate 922 mntally using base method)
(1201)2= (1200)2 + 1200 + 1201 = 1440000 + 2401 = 1442401
In the same way, you can calculate the square of other numbers ending with digit 1. For this, you must remember squares up to 20 and strong addition calculation mentally.
## Trick to calculate squares of numbers ending with digit 9
Suppose you have to calculate the square of 79.
Step : (Number + 1)2 - Number - (Number + 1)
(79)2= (80)2 - 80 - 79 = 6400 - 159 = 6241
(159)2= (160)2 - 160 - 1659 = 25600 - 319 = 25281
(579)2= (580)2 - 580 - 579 = 336400 - 1159 = 335241 (Calculate 922 mntally using base method)
(1239)2= (1240)2 - 1240 - 12039 = 1537600 - 2479 = 1535121 (Calculate 1242 using duplex method)
For this also, you must possess strong multiplication skills.
## Trick to calculate squares of numbers in the range of 51 to 70
Step : [25 + (Number - 50) | (Number - 50)2]
(51)2= 25 + 1 | 01 = 2601
(57)2= 25 + 7 | 49 = 3249
(66)2= 25 + 16 | 256 = 41 | 2 56 = 43 | 56 = 4356 (Keep only 2 digit on right side)
(69)2= 25 + 19 | 361 = 44 | 3 61 = 47 | 61 = 4761 (Keep only 2 digit on right side)
## Trick to calculate squares of numbers in the range of 31 to 50
Step : [25 - (50 - Number) | (50 - Number)2]
(49)2= 25 - 1 | 01 = 2401
(43)2= 25 - 7 | 49 = 1849
(37)2= 25 - 13 | 169 = 12 | 1 69 = 13 | 69 = 1369 (Keep only 2 digit on right side)
(31)2= 25 - 19 | 361 = 6 | 3 61 = 9 | 61 = 961 (Keep only 2 digit on right side)
### Another method: (a + b) (a - b) + b2
(88): it is 100 - 12 or a = 100 and b = 12
So, (88)= (88 + 12) (88 - 12) + 122 = 76 x 100 + 144 = 7744
(104)= (104 + 4) (104 - 4) + 42 = 108 x 100 + 146 = 10816
(438)(438 + 38) (438- 38) + 382 = 476 x 400 + 1444 = 191844
Calculate the following squares using any of the above mentioned method:
(A) (59)2 (B) (876)2 (C) (1441)2 (D) (995)2 (E) (1441)2 (F) (78)2
So, this was all about short tricks of squares. Practice these methods while calculating squares. In the upcoming articles, we will also cover the short tricks to calculate other topics like multiplication, cubes, square and cube roots, etc. Stay connected and keep going with practice.
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Member since Apr 2021
write a comment
Mathan KumarAug 8, 2020
thanks
Snehal MaliAug 11, 2020
What's this
Trick to calculate squares of numbers ending with digit 6
Suppose you have to find the square of 76.
Step : (Number - 1)<sup>2</sup> + Number + (Number - 1)
(76)<sup>2</sup>= (75)<sup>2</sup> + 75 + 76 = 5625 + 151 = 5776 (Calculate square of number ending with digit 5 using the above method.)
(146)<sup>2</sup>= (145)<sup>2</sup> + 145 + 146 = 21025 + 291= 21316
(1006)<sup>2</sup>= (1005)<sup>2</sup> + 1005 + 1006 = 1010025+ 2011= 1012036
In all these examples in last step u added 1 to the number but in formula u have written that number minus 1 in third step
Kunal KumarAug 23, 2020
In sq. Of no. ending with 1, it should be (number-1) at the end
Ak VermaAug 25, 2020
8649 46656 1002001 908596
Ak VermaAug 25, 2020
7569 279841 3964081 25180324
Naga DeviOct 22, 2020
These are very useful tricks to solve problems💯💯💯
Jayshri HatwarMay 12, 2021
.
Sanjay GiriMay 12, 2021
9053348687
ShitalJul 5, 2021
Hii
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Solving equations using additive and multiplicative inverses
9.2 Solving equations using additive and multiplicative inverses
To solve a linear equation, we do the same thing to both sides of the equation to get the variable on its own.
Consider the equation, $$3x + 2 = 23$$.
We can represent this equation in a flow diagram, where $$x$$ represents an unknown number:
When you reverse the process in the flow diagram, you start with the output number $$23$$, then subtract $$2$$ and then divide the answer by $$3$$:
We can write the steps in this reverse process as follows:
Subtract $$2$$ from both sides of the equation:
\begin{align} 3x + 2 &= 23 \\ 3x + 2 - 2 &= 23 - 2 \\ 3x &= 21 \end{align}
Divide both sides by $$3$$:
\begin{align} 3x &= 21 \\ 3x \div 3 &= 21 \div 3 \\ x &= 7 \end{align}
We say $$x = 7$$ is the solution of $$3x + 2 = 23$$, because $$3 \times 7 + 2 = 23$$.
We say that $$x = 7$$ makes the equation $$3x + 2 = 23$$ true.
The numbers $$+ 2$$ and $$- 2$$ are additive inverses of each other. When we add a number and its additive inverse, we always get $$0$$, so $$2 + ( - 2) = 0$$.
The numbers $$3$$ and $$\frac{1}{3}$$ are multiplicative inverses of each other. When we multiply a number and its multiplicative inverse, we always get $$1$$, so $$3 \times \frac{1}{3} = 1$$.
The additive and multiplicative inverses help us to isolate the unknown value or the input value.
a number that results in the answer $$0$$ when it is added to the original number
multiplicative inverse
a number that results in the answer $$1$$ when it is multiplied by the original number
Also remember:
• The multiplicative property of $$1$$: the product of any number and $$1$$ is that number.
• The additive property of $$0$$: the sum of any number and $$0$$ is that number.
Worked example 9.1: Using the additive inverse
Solve for $$b$$:
$- 2 = b + 4$
Get $$b$$ on its own.
To get $$b$$ on its own, we must do the opposite of what has been done to $$b$$. The $$4$$ has been added to $$b$$, so we must subtract $$4$$ from both sides of the equation.
\begin{align} - 2 &= b + 4 \\ 2 - 4 &= b + 4 - 4 \\ 6 &= b \\ \therefore b &= 6 \end{align}
“Doing the same thing to both sides” to get $$b$$ on its own is also called “applying the inverse operation”.
Worked example 9.2: Using the multiplicative inverse
Solve for $$b$$:
$- 7 = - 2b$
Get $$b$$ on its own.
We want the coefficient of $$b$$ to be $$1$$. We have $$-2b$$, which means that $$b$$ is being multiplied by $$-2$$. To get $$b$$ on its own with a coefficient of $$1$$, we must do the opposite of what has been done to $$b$$: we must divide by $$-2$$.
Do the same thing to the other side of the equation.
Whenever we solve equations, we must do the same thing to both sides to make sure that the value of the equation remains the same. So, we must divide both sides of the equation by $$-2$$.
\begin{align} - 7 &= - 2b \\ - 7 \div \left( - 2 \right) &= - 2b \div \left( - 2 \right) \\ \frac{- 7}{- 2} &= \frac{- 2b}{- 2} \\ \frac{7}{2} &= b \end{align}
$$1b$$ means the same thing as $$b$$ (because when you multiply any number by $$1$$, you get the same number). We usually write $$b$$ instead of $$1b$$ because it is simpler, and mathematicians like things to be as simple as possible.
When working with fractions in the equation, you will use the number called a reciprocal. When a fraction is multiplied by its reciprocal, the answer is $$1$$. For example, $$\frac{4}{5} \times \frac{5}{4} = 1$$. We say that $$\frac{4}{5}$$ and $$\frac{5}{4}$$ are reciprocals of each other.
reciprocal
the inverse or opposite of a value; the product of the number and its reciprocal is equal to $$1$$
Worked example 9.3: Solving equations with fractions
Solve for $$t$$:
$7 = \frac{3}{4}t$
Get $$t$$ on its own.
To get $$t$$ on its own, we need to get the coefficient of $$t$$ to be $$1$$. So, we need to think about what number to multiply $$\frac{3}{4}$$ by, so that is becomes $$1$$.
To get rid of the denominator, we need to multiply by $$4$$. To get rid of the numerator, we need to divide by $$3$$.
This is the same as multiplying by $$\frac{4}{3}$$. This number is called the reciprocal of $$\frac{3}{4}$$.
Multiply by the reciprocal.
We must do this on both sides of the equation.
\begin{align} 7 &= \frac{3}{4}t \\ 7 \times \frac{4}{3} &= \frac{3}{4}t \times \frac{4}{3} \\ \frac{28}{3} &= t \end{align}
You could also have done this in two steps. You could multiply both sides by the denominator first, and then divide both sides by the numerator.
temp text
Multi-step equations with inverses
Remember that to solve a linear equation, we do the same thing to both sides of the equation to get the variable on its own. To do this we must do the opposite of what has been done to that variable, but keep in mind the order of operations.
Worked example 9.4: Solving two-step linear equations
Solve for $$b$$:
$- 4 = - 2 + 4b$
Simplify the equation.
The $$2$$ has been subtracted from the side where $$b$$ is, so we must add $$2$$ to both sides of the equation.
\begin{align} - 4 &= - 2 + 4b \\ - 4 + 2 &= - 2 + 4b + 2 \\ - 2 &= 4b \end{align}
Get the variable on its own.
Now we can focus on getting the coefficient of $$b$$ to be $$1$$. To get $$b$$ on its own with a coefficient of $$1$$, we must do the opposite of what has been done to $$b$$. We must divide both sides of the equation by $$4$$.
\begin{align} - 2 &= 4b \\ - \frac{2}{4} &= \frac{4b}{4} \\ - \frac{1}{2} &= b \\ b &= - \frac{1}{2} \end{align}
You might have tried to solve this equation by first dividing both sides by the coefficient of $$b$$ (which is $$4$$ in this case). This would give you the correct answer, as long as you remembered to divide $$-2$$ by $$4$$. But this method can create fractions that are difficult to work with. (It is also easy to forget to divide $$-2$$ by $$4$$!) It is best to deal with the number first, and only divide by the coefficient of $$b$$ as the very last step.
Worked example 9.5: Solving two-step linear equations with decimals
Solve for $$m$$:
$2 + \text{0,2}m = - 3$
Use inverse operations to solve for $$m$$:
\begin{align} 2 + \text{0,2}m &= - 3 \\ 2 + \text{0,2}m - 2 &= - 3 - 2 \\ \text{0,2}m &= - 5 \\ \text{0,2}m \div \text{0,2} &= - 5 \div \text{0,2} \\ \therefore m &= - 25 \end{align}
If you don’t have a calculator, you might find it easier to convert the decimal into a common fraction:
\begin{align} \text{0,2}m &= - 5 \\ \text{0,2}m \div \text{0,2} &= - 5 \div \text{0,2} \\ m &= - 5 \div \frac{2}{10} \\ m &= - 5 \times \frac{10}{2} \\ m &= - \frac{50}{2} \\ m &= - 25 \end{align}
Worked example 9.6: Solving two-step linear equations with fractions
Solve for $$z$$:
$\frac{4}{5}z - 4 = 1$
Solve for $$z$$ using inverse operations.
To solve a linear equation, we do the same thing to both sides of the equation to get $$z$$ on its own. The $$4$$ has been subtracted from the $$z$$-term. So, we must add $$4$$ to both sides of the equation.
\begin{align} \frac{4}{5}z - 4 &= 1 \\ \frac{4}{5}z - 4 + 4 &= 1 + 4 \\ \frac{4}{5}z &= 5 \end{align}
Get $$z$$ on its own.
Now we need to get the coefficient of $$z$$ to be $$1$$. So, we need to think about what number to multiply $$\frac{4}{5}$$ by to get to $$1$$. To get rid of the denominator, we need to multiply by $$5$$. In the step after that, we will divide by $$4$$ to get rid of the numerator.
\begin{align} \frac{4}{5}z &= 5 \\ \frac{4}{5}z \times 5 &= 5 \times 5 \\ 4z &= 25 \\ \frac{4z}{4} &= \frac{25}{4} \\ z &= \frac{25}{4} \end{align}
You could have also multiplied both sides by the reciprocal, $$\frac{5}{4}$$.
\begin{align} \frac{4}{5}z \times \frac{5}{4} &= 5 \times \frac{5}{4} \\ z &= \frac{25}{4} \end{align}
Worked example 9.7: Using the distributive law to solve equations
Solve for $$b$$:
$- ( - a + 2) = 2$
Simplify the left hand side of the equation.
The first thing that we do when solving equations is simplify each side of the equation. We can use the distributive law to simplify the left-hand side of this equation. The right-hand side is already simplified, so we will leave it as is for now.
\begin{align} - ( - a + 2) &= 2 \\ - 1 \times ( - a) - 1 \times (2) &= 2 \\ a - 2 &= 2 \end{align}
Use inverses to solve for $$a$$.
Now we can solve the equation using inverse operations to isolate $$a$$.
\begin{align} a - 2 &= 2 \\ a - 2 + 2 &= 2 + 2 \\ a &= 4 \end{align}
You could have also solved this equation by first dividing both sides by $$-1$$.
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Factors
# Factors of 83 | Prime Factorization of 83 | Factor Tree of 83
Written by Prerit Jain
Updated on: 12 Jun 2023
## Factors of 83
Calculate Factors of
The Factors are
https://wiingy.com/learn/math/factors-of-83/
## What are the factors of 83
A factor of a given number is the number that divides the given number evenly into with zero remainder and no decimal points in the quotient. For example, 2 is a factor of 4 because 4 can be evenly divided by 2 (4 / 2 = 2).
To find the factors of a number, we can start by dividing the number by 2 and then working our way up to the number itself, dividing by each whole number along the way to see if it is a factor. If a number is a factor, we can add it to our list of factors.
For example, to find the factors of 83, we can do the following:
1. Divide 83 by 2: 83 / 2 = 41.5. Since 41.5 is not a whole number, 2 is not a factor of 83.
2. Divide 83 by 3: 83 / 3 = 27.67. Since 27.67 is not a whole number, 3 is not a factor of 83.
3. Continue dividing by the next whole number (in this case, 4) and check the result. We can continue this process until we reach the number itself.
Since 83 is a prime number, it has only two factors: 1 and itself. Therefore, the factors of 83 are 1 and 83.
## How to Find Factors of 83
To find the factors of 83, you can use one of the following methods:
1. Factors of 83 using the Multiplication Method
2. Factors of 83 using the Division Method
3. Prime Factorization of 83
4. Factor tree of 83
## Factors of 83 Using the Multiplication Method
To find the factors of a number using the multiplication method, we can start by writing down 1 and the number itself (83 in this case) and then finding all the other positive integers that can be multiplied together to equal the number.
For example, let’s say we want to find the factors of 83:
1. Write down 1 and the number itself (83). These will always be factors of the number.
2. Find two positive integers that, when multiplied together, equal the number (83).
In this case, the only two positive integers that can be multiplied together to equal 83 are 1 and 83.
1. The factors of the number are 1, the number itself (83), and any other positive integers that can be multiplied together to equal the number.
For 83, the factors using the multiplication method are 1 and 83.
## Factors of 83 Using the Division Method
To find the factors of a number using the division method, we can start by dividing the number by each whole number starting from 2 and working our way up to the number itself. If the result of the division is a whole number, then the divisor is a factor of the number.
For example, let’s say we want to find the factors of 83 using the division method:
1. Begin by dividing the number by 2. If the result is a whole number, then 2 is a factor of the number. For 83, 83 / 2 = 41.5, which is not a whole number. Therefore, 2 is not a factor of 83.
2. Divide the number by the next whole number (in this case, 3). If the result is a whole number, then 3 is a factor of the number. For 83, 83 / 3 = 27.67, which is not a whole number. Therefore, 3 is not a factor of 83.
3. Continue dividing the number by the next whole number (4 in this case) and checking the result. Repeat this process until you reach the number itself.
Since 83 is a prime number, it has only two factors: 1 and itself. Therefore, the factors of 83 using the division method are 1 and 83.
## Prime Factorization of 83
Calculate Prime Factors of
The Prime Factors of 83 =
83
https://wiingy.com/learn/math/factors-of-83/
The prime factorization of a number is the expression of the number as the product of its prime factors. A prime factor is a prime number that can be multiplied together to equal the original number.
For example,
1. The prime factorization of 12 is 2 x 2 x 3, because 2 x 2 x 3 = 12. In this case, 2 and 3 are the prime factors of 12.
2. The prime factorization of 83 is 83. This is because 83 is already a prime number, so it has no prime factors other than itself.
## Factor tree of 83
https://wiingy.com/learn/math/factors-of-83/
A factor tree is a graphical representation of the prime factorization of a number. To create a factor tree, we start with the number and then divide it by the smallest prime number which is a factor of the number. We repeat this process with each result until we reach a prime number.
In the case of 83, it is already a prime number, so its factor tree is simply a single branch with 83 at the end.
## Factor Pairs of 83
Calculate Pair Factors of
1 x 83=83
So Pair Factors of 83 are
(1,83)
https://wiingy.com/learn/math/factors-of-83/
A factor is a number that divides evenly into another number. To find the factors of a number, we can start by dividing the number by 2 and then working our way up to the number itself, dividing by each whole number along the way to see if it is a factor. If a number is a factor, we can add it to our list of factors.
For example, let’s say we want to find the factors of 83:
• Begin by dividing the number by 2. If the result is a whole number, then 2 is a factor of the number.
For 83, 83 / 2 = 41.5, which is not a whole number. Therefore, 2 is not a factor of 83.
• Divide the number by the next whole number (in this case, 3). If the result is a whole number, then 3 is a factor of the number.
For 83, 83 / 3 = 27.67, which is not a whole number. Therefore, 3 is not a factor of 83.
• Continue dividing the number by the next whole number (4 in this case) and checking the result. Repeat this process until you reach the number itself.
Since 83 is a prime number, it has only two factors: 1 and itself. Therefore, the factors of 83 are 1 and 83.
The factor pairs of 83 are (1, 83) and (83, 1). Factors of 83 – Quick Recap
• Factors of 83: 1 and 83.
• Negative Factors of 83: -1 and -83.
• Prime Factors of 83: 83.
• Prime Factorization of 83: 83.
## Factors of 83 – Fun Facts
• The factors of 83 are 1 and 83.
• 83 is a prime number, which means it has only two factors: 1 and itself.
• The sum of the factors of 83 is 84 (1 + 83 = 84).
• The product of the factors of 83 is 83 (1 x 83 = 83).
• The only even factor of 83 is 2, and 83 is not divisible by 2.
• The greatest common factor (GCF) of 83 and any other number is 1 since 1 is the only common factor.
## Solved Examples of Factor 83
Q.1: What is the prime factorization of 83?
Answer: The prime factorization of 83 is 83 = 83 x 1.
Q.2: How many factors does 83 have?
The number 83 is a prime number, which means it has only two factors: 1 and 83.
Q.3: Is 83 an abundant number?
No, 83 is not an abundant number. An abundant number is a number for which the sum of its proper divisors (excluding the number itself) is greater than the number itself. Since the only proper divisor of 83 is 1, and the sum of the proper divisors is 1, the sum is not greater than 83. Therefore, 83 is not an abundant number.
Q.4: How many pairs of factors equal to 83 can you find?
Solution:
There are 1 pair of factors that equal 83; (1,83).
Q.5: What is the greatest common factor for 73 and 63?
Solution:
The prime factorization of 85 is 5 * 17. Since 5 and 17 are both prime numbers and not perfect squares, it means that 85 does not have any square factors.
Q.6: What type of number is 83?
Solution:
The number 83 is a prime number since Since 83 cannot be divided evenly by any other number than 1 and 83, it is classified as prime number.
Q.7: What are the least common multiples between 63 and 73?
Solution:
The least common multiple between 63 and 73 is 1.
Q.8: Does 85 have any square factors?
Solution:
The prime factorization of 85 is: 5 * 17. Since 5 and 17 are both prime numbers and not perfect squares, it means that 85 does not have any square factors.
## Frequently Asked Questions on Factors of 83
### What is the greatest common factor between 83 and 30?
The greatest common factor (GCF) between 83 and 30 is 1.
### How many negative factors does the number 83 have?
There are 2 negative factors of 83; -1, -83.
### Is there a difference between the factors and multiples of 83?
While factors and multiples have similar definitions in that they both refer to groups or collections of related numbers generated by multiplying or dividing a given number, there is an important difference between them – factors refer to how many times the original number can be divided evenly while multiples make reference to how many times it has been multiplied by itself.
### Which number from 1-83 divides into it without any remainders?
Any number from 1-83 can divide into it with no remainders, but some will produce fractions or decimals instead of whole number results.
### What is the least common multiple (LCM) of 83?
The least common multiple (LCM) of 83 is 83 itself. Since 83 is a prime number, it does not have any other factors besides 1 and itself. Therefore, the LCM of 83 is simply 83.
### What is the sum of all positive integer divisors for 83?
The sum of the positive integer divisors for 83 is 84; 1+83= 84.
Written by by
Prerit Jain
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In Mathematics, the Taylor series is the most famous series that is utilized in several mathematical as well as practical problems. The Taylor theorem expresses a function in the form of the sum of infinite terms. These terms are determined from the derivative of a given function for a particular point. The standard definition of an algebraic function is provided using an algebraic equation. Likewise, transcendental functions are defined for a property that holds for them. A function may be well described by its Taylor series too. This series can also be used to determine various functions in plenty of areas of mathematics.
## Taylor’s Series Theorem
Assume that if f(x) be a real or composite function, which is a differentiable function of a neighbourhood number that is also real or composite. Then, the Taylor series describes the following power series :
$f(x) =f(a)\frac{f'(a)}{1!}(x-a)+\frac{f”(a)}{2!}(x-a)^{2}+\frac{f^{(3)}(a)}{3!}(x-a)^{3}+….$
In terms of sigma notation, the Taylor series can be written as
$\sum_{n=0}^{\infty }\frac{f^{n}(a)}{n!}(x-a)^{n}$
Where
f(n) (a) = nth derivative of f
n! = factorial of n.
## Taylor Series Formula and Proof
We know that the power series can be defined as
$f(x)= \sum_{n=0}^{\infty }a_{n}x^{n}=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+…$
When x = 0,
f(x)= a0
So, differentiate the given function, it becomes,
f’(x) = a1+ 2a2x + 2a3x2 + 4a4x3 +….
Again, when you substitute x = 0, we get
f’(0) =a1
So, differentiate it again, we get
f”(x) = 2a2 + 6a3x +12a4x2 + …
Now, substitute x=0 in second-order differentiation, we get
f”(0) = 2a2
Therefore, [f”(0)/2!] = a2
By generalising the equation, we get
f n (0) / n! = an
Now substitute the values in the power series we get,
$f(x)= f(0)+f'(0)x+\frac{f”(0)}{2!}x^{2}+\frac{f”‘(0)}{3!}x^{3}+….$
Generalise f in more general form, it becomes
f(x) = b + b1 (x-a) + b2( x-a)2 + b3 (x-a)3+ ….
Now, x = a, we get
bn = fn(a) / n!
Now, substitute bn in a generalised form
$f(x) =f(a)\frac{f'(a)}{1!}(x-a)+\frac{f”(a)}{2!}(x-a)^{2}+\frac{f^{(3)}(a)}{3!}(x-a)^{3}+….$
Hence, the Taylor series is proved.
## Taylor Series in Several Variables
The Taylor series is also represented in the form of functions of several variables. The general form of the Taylor series in several variables is
$T(x_{1}, x_{2}, x_{3},…x_{m})=f(a_{1}, a_{2}, a_{3},…a_{m})+\sum_{j=1}^{m}\frac{\partial f(a_{1}, a_{2}, a_{3},…a_{m})}{\partial x_{j}}(x_{j}-a_{j})+\frac{1}{2!\sum_{j=1}^{m}}$
## Maclaurin Series Expansion
If the Taylor Series is centred at 0, then the series is known as the Maclaurin series. It means that,
If a= 0 in the Taylor series, then we get
$f(x)= f(0)+f'(0)x+\frac{f”(0)}{2!}x^{2}+\frac{f”‘(0)}{3!}x^{3}+….$ is known as the Maclaurin series.
## Applications of Taylor Series
The uses of the Taylor series are:
• Taylor series is used to evaluate the value of a whole function in each point if the functional values and derivatives are identified at a single point.
• The representation of Taylor series reduces many mathematical proofs.
• The sum of partial series can be used as an approximation of the whole series.
• Multivariate Taylor series is used in many optimization techniques.
• This series is used in the power flow analysis of electrical power systems.
### Problems on Taylor’s Theorem
Question: Determine the Taylor series at x=0 for f(x) = ex
Solution: Given: f(x) = ex
Differentiate the given equation,
f’(x) = ex
f’’(x) =ex
f’’’(x) = ex
At x=0, we get
f’(0) = e0 =1
f’’(0) = e0=1
f’’’(0) = e0 = 1
When Taylor series at x= 0, then the Maclaurin series is
$f(x)= f(0)+f'(0)x+\frac{f”(0)}{2!}x^{2}+\frac{f”‘(0)}{3!}x^{3}+….$
ex = 1+ x(1) + (x2/2!)(1) + (x3/3!)(1) + …..
Therefore, ex = 1+ x + (x2/2!) + (x3/3!)+ ….. |
RS Aggarwal Class 9 Solutions Chapter 14 -Statistics Ex 14C (14.3)
RS Aggarwal Class 9 Ex 14C Chapter 14
Q.1: Find the arithmetic mean of:
(i) The first eight natural numbers
(ii) The first the odd numbers
(iii) The first five prime numbers
(iv) The first six even numbers
(v) The first seven multiples of 5
(vi) All the factors of 20
Sol:
(i) First eight natural numbers are:
1,2,3,4,5,6,7 and 8
Therefore, mean =SumofnumbersTotalnumbers$\frac{Sum \; of \; numbers}{Total \; numbers}$
=1+2+3+4+5+6+7+88=368=4.5$\frac{1+2+3+4+5+6+7+8}{8}=\frac{36}{8}=4.5$
= Therefore, Mean =4.5
(ii) First ten odd numbers are: 1, 3, 5, 7, 9, 11, 13, 15, 17 and 19
Therefore, mean =SumofnumbersTotalnumbers$\frac{Sum \; of \; numbers}{Total \; numbers}$
=1+3+5+7+9+11+13+15+17+1910=10010=10$\frac{1+3+5+7+9+11+13+15+17+19}{10}=\frac{100}{10}=10$
Therefore, Mean =10
(iii) First five prime numbers are : 2,3,5,7,11
Therefore, mean =SumofnumbersTotalnumbers$\frac{Sum \; of \; numbers}{Total \; numbers}$
=2+3+5+7+115=285=5.6$\frac{2+3+5+7+11}{5}=\frac{28}{5}=5.6$
Therefore, Mean =5.6
(iv) First six even numbers are: 2, 4, 6, 8, 10, 12
Therefore, mean =SumofnumbersTotalnumbers$\frac{Sum \; of \; numbers}{Total \; numbers}$
= 2+4+6+8+10+126=426=7$\frac{2+4+6+8+10+12}{6}=\frac{42}{6}=7$
Therefore, Mean = 7
(v) First seven multiples of 5 are: 5, 10,15, 20, 25, 30, 35
Therefore, mean = SumofnumbersTotalnumbers$\frac{Sum \; of \; numbers}{Total \; numbers}$
=5+10+15+20+25+30+357=1407=20$\frac{5+10+15+20+25+30+35}{7}=\frac{140}{7}=20$
Therefore, Mean = 20
(vi) Factors of 20 are: 1, 2, 4, 5, 10, 20
Therefore, mean =SumofnumbersTotalnumbers$\frac{Sum \; of \; numbers}{Total \; numbers}$
= 1+2+4+5+10+206=426=7$\frac{1+2+4+5+10+20}{6}=\frac{42}{6}=7$
Therefore, Mean = 7
Q.2: The number of children in 10 families of a taxably are: 2, 4, 3, 4, 2, 0, 3, 5, 1, 6. Find the mean number of children per family.
Sol:
Mean = SumofnumbersTotalnumbers$\frac{Sum \; of \; numbers}{Total \; numbers}$
= 2+4+3+4+2+0+3+5+1+610=3010=3$\frac{2+4+3+4+2+0+3+5+1+6}{10}=\frac{30}{10}=3$
Therefore, Mean = 3
Q.3: The following are the number of books issued in a school library during a week: 105, 216, 322, 167, 273, 405 and 346. Find the average number of books issued per day.
Sol:
Mean =SumofnumbersTotalnumbers$\frac{Sum \; of \; numbers}{Total \; numbers}$
=105+216+322+167+273+405+3467=18347=262$\frac{105+216+322+167+273+405+346}{7}=\frac{1834}{7}=262$
Therefore, Mean = 262
Q.4: The daily minimum temperature recorded (in degree F) at a place during a week was as under.
Monday Tuesday Wednesday Thursday Friday Saturday 35.5 30.8 27.3 32.1 23.8 29.9
Sol:
Mean Temperature = SumofTemperaturesNo.ofdays$\frac{Sum \; of \; Temperatures}{No. \; of \; days}$
=35.5+30.8+27.3+32.1+23.8+29.96=179.46=29.9$\frac{35.5+30.8+27.3+32.1+23.8+29.9}{6}=\frac{179.4}{6}=29.9$
Therefore, Mean = 29.9
Q.5: The percentages of mark obtained by 12 students of a class in mathematics are: 64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1. Find the mean percentage of marks.
Sol:
Mean = SumofmarksTotalnumberofstudents$\frac{Sum \; of \; marks}{Total \; number \; of \; students}$
=64+36+47+23+0+19+81+93+72+35+3+112=47412=39.5$\frac{64+36+47+23+0+19+81+93+72+35+3+1}{12}=\frac{474}{12}=39.5$
Therefore, Mean =39.5
Q.6: If the arithmetic mean of 7, 9, 11, 13, x, 21 is 13, find the value of x.
Sol:
Mean of the given number = 7+9+11+13+x+216$\frac{7+9+11+13+x+21}{6}$
Undefined control sequence \because$\left ( \because Mean=\frac{Sum\; of \; observation}{Number \; of \; observation} \right )$
And, mean = 13 (given)
Therefore, 13=61+x6$13 =\frac{61+x}{6}$
61+x=78$\Rightarrow 61+x =78$
x=7861=17$\Rightarrow x =78-61 =17$
Therefore, the value of x = 17
Q.7: The mean of 24 numbers is 35.If 3 is added to each number, what will be the new mean?
Sol:
Let, the given number be x1,x2,.x24$x_{1},x_{2},…….x_{24}$
Mean=x1+x2+.+x2424$\Rightarrow Mean =\frac{x_{1}+x_{2}+…….+x_{24}}{24}$
Therefore,x1+x2+.+x2424=35$\frac{x_{1}+x_{2}+…….+x_{24}}{24}=35$
x1+x2+.+x24=840$\Rightarrow x_{1}+x_{2}+…….+x_{24}=840$ . . . . . . . . (i)
The new number are (x1+3),(x2+3),.+(x24+3)$(x_{1}+3),(x_{2}+3),…….+(x_{24}+3)$.
Therefore, Mean of the new numbers
(x1+3)+(x2+3)+.+(x24+3)24=840+7224$\frac{(x_{1}+3)+(x_{2}+3)+…….+(x_{24}+3)}{24}=\frac{840+72}{24}$ (using (i) )
=91224=38$=\frac{912}{24} =38$
Therefore, Thenewmean=38$The new mean =38$
Q.8: The mean of 20 numbers is 43.If 6 is subtracted from each of the numbers, what will be the new mean?
Sol:
Let, the given numbers be x1,x2,.......x20$x_{1},x_{2}, . . . . . . . x_{20}$
Then, the mean of these numbers =
Therefore, x1+x2+.........+x2020=43$\frac{x_{1}+x_{2}+ . . . . . . . . . +x_{20}}{20}=43$
x1+x2+........+x20=860$\Rightarrow x_{1}+x_{2}+ . . . . . . . . +x_{20}=860$ . . . . . . . (i)
The new numbers are:
(x16)+(x26)+........+(x206)20=86012020$\Rightarrow \frac{(x_{1}-6)+(x_{2}-6)+ . . . . . . . . +(x_{20}-6)}{20}=\frac{860-120}{20}$ . . . . . . ( using (i) )
Therefore, 74020=37$\frac{740}{20}=37$
Q.9: The mean of 15 numbers is 27. If each number is multiplied by 4, what will be the mean of the new numbers?
Sol:
Let the given number be x1,x2,.,x15$x_{1},x_{2},……….,x_{15}$
Then, the mean of these numbers = x1+x2+.+x1515=27$\frac{x_{1}+x_{2}+……….+x_{15}}{15}=27$
x1+x2+.+x15=405$\Rightarrow x_{1}+x_{2}+……….+x_{15}=405$
The new numbers are (x1×4),(x2×4),.(x15×4)$(x_{1}\times 4),(x_{2}\times 4),…….(x_{15}\times 4)$
Therefore, mean of the new numbers = (x1×4)+(x2×4)+.+(x15×4)15=405×415=162015=108$\frac{(x_{1}\times 4)+(x_{2}\times 4)+…….+(x_{15}\times 4)}{15}=\frac{405\times 4}{15}=\frac{1620}{15}=108$
Therefore, the new mean = 108
Q.10: The mean of 12 numbers is 40. If each number is divided by 8, what will be the mean of the new numbers?
Sol:
Let the given number be x1,x2,.,x12$x_{1},x_{2},……….,x_{12}$
Then, the mean of these numbers = 40 x1+x2+.+x1212=40$\frac{x_{1}+x_{2}+……….+x_{12}}{12}=40$
x1+x2+.+x12=480$\Rightarrow x_{1}+x_{2}+……….+x_{12}=480$
The new numbers are (x1÷8),(x2÷8),.(x12÷8)$(x_{1}\div 8),(x_{2}\div 8),…….(x_{12}\div 8)$
Therefore, mean of the new numbers = (x1÷8)+(x2÷8)+.+(x12÷8)12=480÷812=6012=5$\frac{(x_{1}\div 8)+(x_{2}\div 8)+…….+(x_{12}\div 8)}{12}=\frac{480 \div 8}{12}=\frac{60}{12}=5$
Therefore, the new mean = 5
Q.11: The mean of 20 numbers is 18.If 3 is added to each of the first ten numbers; find the mean of the new set of 20 numbers.
Sol:
Let the given number be x1,x2,.,x20$x_{1},x_{2},……….,x_{20}$
Let the mean be X¯$\bar{X}$
Therefore,X¯=x1+x2+.+x2020=18$\bar{X}=\frac{x_{1}+x_{2}+……….+x_{20}}{20}=18$
x1+x2+.+x20=18×20=360$\Rightarrow x_{1}+x_{2}+……….+x_{20}=18\times 20=360$ . . . . (i)
But it is given that 3 is added to each of the first ten numbers.
Therefore, the first new ten numbers are
(x1+3),(x2+3),.,(x10+3)$(x_{1}+3),(x_{2}+3),……….,(x_{10}+3)$
Let, X¯$\bar {{X}’}$ be the mean of new numbers
=(x1+x2+.+x20)+3×1020$=\frac{(x_{1}+x_{2}+……….+x_{20})+3\times 10}{20}$
From (i), we know that =x1+x2+.+x20=360$=x_{1}+x_{2}+……….+x_{20}=360$
Therefore, mean of the new set of 20 numbers = 360+3020=39020=19.5$\frac{360+30}{20}=\frac{390}{20}=19.5$
Therefore, mean of the new set of 20 numbers = 19.5
Q.12: The mean weight of 6boys in a group is 48kg. The individual weights of five of them are 51kg, 45kg, 49kg, 46kg, and 44kg.Find the weight of the sixth boy.
Sol:
Mean weight of the boys =48 kg
Therefore, Mean weight = Sumoftheweightofsixboys6=48$\frac{Sum \; of \; the \;weight \; of \; six \; boys}{6}=48$
Thus Sum of the weight of 6 boys = (48×6)kg$(48\times 6)kg$ = 288 kg
Sum of the weight of 5 boys = (51 + 45 + 49 + 46 + 44) kg = 235 kg
Weight of the sixth boy = (Sum of Weight of 6 boys) – (Sum of the weight of 5 boys) = (288 – 235) = 53 kg
Q.13: The mean of the marks scored by 50students was found to be 39.Later on, it was discovered that a score of 43 was misread as 23.Find the correct mean.
Sol:
Calculated mean marks of 50 students = 39
Calculated sum of these marks = (39 \times 50) =1950
Corrected sum of these marks = [1950-(wrong number)+(correct number) ]
= (1950-23+43) = 1970
Therefore, the correct mean = 197050=39.4$\frac{1970}{50} =39.4$
Q.14: The mean of 100 items was found to be 64. Later on it was discovered that two items were misread as 26 and 9 instead of 36 and 90 respectively. Find the correct mean.
Sol:
Calculated mean of 100 items = 64
Sum of 100 items, as calculated = (100×64)=6400$(100\times 64)=6400$
Correct sum of these items = [6400 – (wrong items) + (correct items)] = [6400 – (26 + 9) + (36 + 90)] = [6400 – 35 + 126]
Therefore, the Correct mean = 6491100=64.91$\frac{6491}{100}=64.91$
Q.15: The mean of six numbers is 23. If one of the numbers is excluded, the mean of the remaining numbers is 20. Find the excluded number.
Sol:
Mean of 6 numbers = 23
Sum of 6 numbers = 23×6=138$23\times 6=138$
Again, mean of 5 numbers = 20
Sum of 5 number = 20×5=100$20 \times 5=100$
The excluded number = (Sum of 6 number) – (sum of 5 numbers) = (138-100) = 38
Therefore, the excluded number = 38
Practise This Question
Which of the following is not a greenhouse gas? |
# How to Calculate the Area of a Triangle?
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Students usually learn about triangles in their mathematics classes, including geometry and trigonometry. By learning everything about the types, area, and perimeter of a triangle, students will gain an understanding of this important shape and its applications in everyday life. Generally, the area of triangle is calculated using formula ½ × Base × Height where “Base” is any side of the triangle and “Height” is the length of the altitude drawn to the “Base” from its opposite vertex. But there are some more formulas used to find the area of different types of triangles. Let’s learn about types of triangles and ways to calculate their area.
## Types of Triangles
A triangle is a two-dimensional shape that has three sides and three vertices. The sum of all the angles of a triangle is 180 degrees. The triangles are further classified into different types based on the length of their sides and angels. There are mainly three types of triangles:- Equilateral Triangle, Isosceles Triangle, and Scalene Triangles.
An equilateral triangle is a triangle that has all the sides equal in length. The angle formed at each vertice of such a triangle is 60 degrees. The isosceles triangle is a triangle that has two equal sides and two equal angles. A scalene triangle is the most common type of triangle; it has no equal sides and no equal angles.
### Formula to Calculate Area of Triangle
Calculating the area of a triangle is an important math skill used by many professionals in different fields. The generic formula applied to find the area of a triangle is 0.5 * base * height, where ‘base’ represents the length of the base of the triangle, and height ‘represents’ the height/altitude of the triangle.
However, in certain cases, it’s difficult to find the height of the triangle. In these situations, many other equations can be applied, depending on the given values of the triangle:
### Area of Triangle With 3 Sides Formula
If you have the lengths of all three sides, use Herons formula to calculate the area of a triangle. Heron’s formula is applied to find the area of a triangle with three sides which says if a, b, and c are the three sides of a triangle, then its area is,
Area = √[s(s-a)(s-b)(s-c)]
Here, “s” is the semi-perimeter of the triangle. i.e., s = (a + b + c)/2. But there are 3 alternate formulas to find the same.
Area = 0.25 * √( (a + b + c) * (-a + b + c) * (a – b + c) * (a + b – c) )
### Area of Triangle With 2 Sides and Angle Between Them
If the length of the two sides and the angle between them is known, you can use the following formula to calculate the area of a triangle easily:
Area = 0.5 * a * b * sin(γ)
### Area of Triangle With 2 Angle and Side Between Them
There are different versions of the area of triangle formulas that can be used to calculate the area. For example, trigonometry or law of sines to derive it:
area = a² * sin(β) * sin(γ) / (2 * sin(β + γ))
### Area of an Equilateral Triangle
The total space enclosed within the sides of an equilateral triangle is called the area of an equilateral triangle. It is calculated by using the formula √3/4 × (side)2.
For example, if the length of a side of an equilateral triangle is 2 cm, then the area of this triangle can be √3/4 × (2)2 = √3 sq. cm.
#### Conclusion
Triangle is an important geometric shape vastly applied in fields like designing, graphics, architecture, construction, etc. It is helpful to relate to various designs and to build structures and artworks. It is crucial for students to have knowledge about triangles, their properties, and dimensions as it enables them to understand their applications. Cuemath offers interactive maths resources for students to efficiently learn an in-depth understanding of these concepts easily. To find some of the best math learning resources, visit the website and learn more. |
# Circumference Calculator
### Circumference
On this page, we will explain an important property of circles called the circumference. The topics we will cover are:
• What is the circumference of a circle?
• How can you calculate the circumference of a circle?
• An example of calculating the circumference of the Earth
What is the circumference of a circle?
The circumference of a circle is the distance around the outside of the circle. It is like the perimeter of other shapes like squares, but for shapes made up of straight lines, we use the word perimeter and for circles we use the word circumference.
This diagram shows the circumference of a circle.
There are also some other important distances on a circle, called the radius (r) and diameter (d) We will need these to calculate the circumference. The diameter is the distance from one side of the circle, to the other, passing through the center (middle of the circle). The radius is half of this distance.
This diagram shows the circumference, diameter, center and radius on a circle.
## How can you calculate the circumference of a circle?
If you know the diameter or radius of a circle, you can work out the circumference. To begin with, remember that pi is a number, written down with the symbol π. π is roughly equal to 3.14. Then, the formula for working out the circumference of the circle is:
Circumference of circle = π x Diameter of circle
which we usually write in the shortened form C = πd. This tells us that the circumference of the circle is three “and a bit” times as long as the diameter. We can see this on the graphic below:
You can also work out the circumference if you know the radius. Remember that the diameter is double the length of the radius. We already know that C = πd. If r is the radius of the circle, then d = 2r. So, C = 2πr.
Example 1
A circle has a diameter of 10cm, what is its circumference?
We know that C = πd. Since the diameter is 10cm, we have that C = π x 10cm = 31.42cm (to 2 decimal places).
Example 2
A circle has a radius of 3m, what is its circumference?
We know that C = 2πr. Since the radius is 3m, we have that C = π x 6m = C= 18.84 (to 2 decimal places).
Example 3
Find the missing length (marked with a ?) on the diagram below: |
# Antiderivatives Class 11 Mathematics Solutions | Exercise - 18.4
## Chapter - 18 Antiderivatives
Students in class 11 learn about antiderivatives, a type of calculus in which they learn how to find the area under the curve of a function and evaluate indefinite integrals, as well as solving differential equations. Antiderivatives can be calculated using integration by substitution, integration by parts, and trigonometric substitution, all of which are methods that students learn to use in order to calculate them.
As part of the course, students learn how to use the Fundamental Theorem of Calculus to calculate antiderivatives. Moreover, they learn how to identify a function's antiderivative as well as how to use the substitution rule for integrals in order to evaluate definite integrals.
The following sections provide a brief introduction to antiderivaties and some of the subtopics related to antiderivaties as well as some basic formulas related to antiderivatives before going into the solution section.
### Introduction
Antiderivatives, also known as indefinite integrals, are the inverse operation of derivatives. In other words, an antiderivative is the original function from which a derivative was derived.
For example, the derivative of f(x) = x² is f'(x) = 2x. Therefore, the antiderivative of 2x is f(x) = x². Antiderivatives can be used to find the area under a curve, and can also be used to solve differential equations.
### Integration by Parts
If a given function to be integrated is in the product form and it cannot be integrated either by reducing the integrand into the standard form or by substitution, we use the following rule known as the integration by parts.
This formula can be stated as follows:
The integral of the product of two functions = First function Integral of second
-Integral of (Derivative of first x Integral of second)
This is the formula of the integration of the product of two functions and is known as the "Integration by parts". The successfulness of the use of the above formula depends upon the proper choice of the first function. The first function must be chosen such that its derivative reduces to a simple form and second function should be easily integrable.
### Exercise - 18.4
This PDF contains all handwritten notes of class 11 Antiderivatives chapter, including solutions of Exercise 18.4. If you want the solutions of other exercises, click the button above.
NoteScroll the PDF to view all Solution
You are not allowed to post this PDF in any website or social platform without permission.
## Is Class 11 Mathematics Guide Helpful For Student ?
I have published this Notes for helping students who can't solve difficult maths problems. Student should not fully depend on this note for completing all the exercises. If you totally depend on this note and simply copy as it is then it may affect your study.
Student should also use their own will power and try to solve problems themselves. You can use this mathematics guide PDF as a reference. You should check all the answers before copying because all the answers may not be correct. There may be some minor mistakes in the note, please consider those mistakes.
## How to secure good marks in Mathematics ?
As, you may know I'm also a student. Being a student is not so easy. You have to study different subjects simultaneously. From my point of view most of the student are weak in mathematics. You can take me as an example, I am also weak in mathematics. I also face problems while solving mathematics questions .
If you want to secure good marks in mathematics then you should practise them everyday. You should once revise all the exercise which are already taught in class. When you are solving maths problems, start from easy questions that you know already. If you do so then you won't get bored.
Maths is not only about practising, especially in grade 11 you to have the basic concept of the problem. When you get the main concept of the problem then you can easily any problems in which similar concept are applied.
When your teacher tries to make the concept clear by giving examples then all students tries to remember the same example but you should never do that. You can create your own formula which you won't forget later.
If you give proper time for your practise with proper technique then you can definitely score a good marks in your examination.
Disclaimer: This website is made for educational purposes. If you find any content that belongs to you then contact us through the contact form. We will remove that content from our as soon as possible.
If you have any queries then feel free to comment down.
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# SSAT Upper Level Math : Properties of Triangles
## Example Questions
← Previous 1 3 4 5 6 7 8 9 17 18
### Example Question #1 : Perimeter Of A Triangle
What is the perimeter of a right triangle with hypotenuse and a leg of length ?
It cannot be determined from the information given.
Explanation:
Using the Pythagorean Theorem, the length of the second leg can be determined.
We are given the length of the hypotenuse and one leg.
The perimeter of the triangle is the sum of the lengths of the sides.
### Example Question #1 : How To Find The Perimeter Of A Right Triangle
Which of these polygons has the same perimeter as a right triangle with legs 6 feet and 8 feet?
A regular pentagon with sidelength one yard.
A regular hexagon with sidelength one yard.
A regular decagon with sidelength one yard.
A regular octagon with sidelength one yard.
None of the other responses is correct.
A regular octagon with sidelength one yard.
Explanation:
A right triangle with legs 6 feet and 8 feet has hypotentuse 10 feet, as this is a right triangle that confirms to the well-known Pythagorean triple 6-8-10. The perimeter is therefore feet, or 8 yards.
We are looking for a polygon with this perimeter. Each choice is a polygon with all sides one yard long, so we want the polygon with eight sides - the regular octagon is the correct choice.
### Example Question #2 : How To Find The Perimeter Of A Right Triangle
The lengths of the legs of a right triangle are units and units. What is the perimeter of this right triangle?
units
units
units
units
units
Explanation:
First, we need to use the Pythagorean Theorem to find the hypotenuse of the triangle.
Now, add up all three side lengths to find the perimeter of the triangle.
### Example Question #3 : How To Find The Perimeter Of A Right Triangle
A right triangle has leg lengths of and . Find the perimeter of this triangle.
Explanation:
First, use the Pythagorean Theorem to find the length of the hypotenuse.
Substituting in and for and (the lengths of the triangle's legs), we get:
Now, add up the three sides to find the perimeter:
### Example Question #4 : How To Find The Perimeter Of A Right Triangle
What is the perimeter of a right triangle with legs of length and , respectively?
Explanation:
In order to find the perimeter of the right triangle, we need to first find the missing length of the hypotenuse. In order to find the hypotenuse, use the Pythagorean Theorem:
, where and are the lengths of the legs of the triangle, and is the length of the hypotenuse.
Substituting in our known values:
Now that we have the lengths of all sides of the right triangle, we can calculate the perimeter:
### Example Question #5 : How To Find The Perimeter Of A Right Triangle
What is the perimeter of a right triangle with a hypotenuse of length and a leg of length ?
Not enough information provided
Explanation:
In order to find the perimeter of the right triangle, we need to first find the missing length of the second leg. In order to find the second leg, use the Pythagorean Theorem:
, where and are the lengths of the legs of the triangle, and is the length of the hypotenuse.
Substituting in our known values:
Subtracting from each side of the equation lets us isolate the variable for which we are solving:
Now that we have the lengths of all three sides of the right triangle, we can calculate the perimeter :
### Example Question #6 : How To Find The Perimeter Of A Right Triangle
Find the perimeter of a right triangle with two legs of length and , respectively.
Not enough information provided
Explanation:
In order to find the perimeter of the right triangle, we need to first find the missing length of the hypotenuse. In order to find the length of the hypotenuse, use the Pythagorean Theorem:
, where and are the lengths of the legs of the triangle, and is the length of the hypotenuse.
Substituting in our known values:
Now that we have all three sides of the right triangle, we can calculate the perimeter:
### Example Question #1 : Right Triangles
If a right triangle is similar to a right triangle, which of the other triangles must also be a similar triangle?
Explanation:
For the triangles to be similar, the dimensions of all sides must have the same ratio by dividing the 3-4-5 triangle.
The 6-8-10 triangle will have a scale factor of 2 since all dimensions are doubled the original 3-4-5 triangle.
The only correct answer that will yield similar ratios is the triangle with a scale factor of 4 from the 3-4-5 triangle.
The other answers will yield different ratios.
### Example Question #10 : Understand Categories And Subcategories Of Two Dimensional Figures: Ccss.Math.Content.5.G.B.3
What is the main difference between a right triangle and an isosceles triangle?
A right triangle has to have a angle and an isosceles triangle has to have equal, base angles.
An isosceles triangle has to have a angle and a right triangle has to have equal, base angles.
A right triangle has to have a angle and an isosceles triangle has to have equal, base angles.
A right triangle has to have a angle and an isosceles triangle has to have equal, base angles.
A right triangle has to have a angle and an isosceles triangle has to have equal, base angles.
A right triangle has to have a angle and an isosceles triangle has to have equal, base angles.
Explanation:
By definition, a right triangle has to have one right angle, or a angle, and an isosceles triangle has equal base angles and two equal side lengths.
### Example Question #1 : How To Find The Length Of The Side Of A Right Triangle
A right triangle has a hypotenuse of 39 and one leg is 36. What is the length of the other leg? |
# 27- Part 3/4 of the Solved problem 9-9-6, how to find LL?
## Part 3/4 for the Solved Problem 9-9-6, How To Find LL?
### The limiting coefficient Lp/ry for the built-up section.
As we can see from the beginning of our solved problem that the bracing distance=15′, we want to get the Nominal moment due to lateral-torsional buckling and compare the value obtained against Mn due to Local buckling.
How to get the ry? it is obtained from the relation of r^2y= Iy/A, then we estimate Lp/ry, for the limiting ratio Lp/ry=300/ sqrt(Fy).
In our part 3/4 for the solved problem where Fy=65 ksi, Lp/ry limiting ration =300/sqrt(65)=37.21, then the value is to be multiplied by ry.
After estimating its value, but we have a rather tough equation for Lr. First, we will estimate the ry value, for our built-up section where, r^2y=Iy/A, for Iy we estimate as we have a two flange and one web, then Iy can be estimated by the relation, Iy =parallel*perpendicular^3/12, for the flange=(5/8)*(16^3/12).
Multiply by 2 to include the lower flange as well. While for the web, Iy web=(26)*(5/16)^3/12. then Iy for the section=426.74 inch4.
### Check lateral torsional buckling, the value of Lp for the built-up section.
The total area At, we have estimated earlier as At/2=14.0625 inch2. Then At=2*14.0625= At=28.125 inch2.
ry =sqrt(Iy/At)=sqrt(426.74/28.125)=3.895 inch, that is the value of the ry. The radius of gyration for the Y-axis, then the plastic distance between bracings Lp limiting value=37.21*3.895=145.00 inch.
To convert to ft, divide /12=145/12=12.077 feet, Lp=12.10′. In our solved problem, part 3/4, the span is 45 feet, and the distance between bracings Lb=(L/3)=15″. L bracing=15′, the author aims to discuss the section in the second zone where Lb>Lp and <Lr.
### Part 3/4 Of The Solved Problem 9-9-6, the equation of Lr.
Let us estimate each item Cw= Iy*h^2/4c^2, Iy= 426.74 inch4. The Mn value is between Mnp=1728 ft-kips and Mr=1114 ft-kips, but for Lr relation, Lr=1.95*rts* (E/0.70Fy)*sqrt(J*c/Sx*ho*sqrt(1+sqrt(1+6.76Fy*(0.7*Fy*Sx*h0/(E*j*c)^2.
Let us estimate each item Cw= Iy*h^2/4c^2, Iy=426.74 inch4, ok, to estimate then ho=(27.25-5/8)=26.625 inch.
ho^2=(26.625)^2 inch2., taken as ho^2, to review the relation, ho is the distance between the CG of flanges.
The overall distance=total height-(tf/2)),Cw=426.74*(26.625)^2/ 4* c^2, where c=1 for doubly symmetrical sections.cw=75628 inch6.For r^2ts =sqrt(Iy*Cw/ry).
The equation, rts=(sqrt(Iy*Cw)/Sx), where Iy=426.74 inch4, Cw warping coefficient Cw=75628, I have estimated all these values, since we are not able to use the tables given by The AISC since our section is a built-up section.
r^2ts= sqrt(426.74*7.5628)/Sx, where Sx=294 inch3. then r^2ts=sqrt(426.74*7.5628)/294=19.32, then rts=4.395 inch
We will estimate Lr=1.95*rts*(E/0.70*Fy), ts=4.395, E=29000ksi, Sx=294 inch3, Fy=65 ksi. This bracket value(1.95*4.395*29000/0.70*65)*sqrt(J*C)/sqrt(Sx*ho) where J is the torsional constant first, C=1, Sx =294, ho=26.625″. The big square root, but first I have estimated the value of (1.954.39529000/0.7065)1/sqrt(29426.625), was considered as 61.7395, then multiplied by sqrt(J1)sqrt(1+sqrt(1+6.67(0.765)(29426.625^2/(29000J*1).
I have computed most of the items for the square roots until we estimate the J value. The square roots are re-written as sqrt(1+sqrt(1+6.76*12.815/ j^2).
### Review of J polar for non-round shapes.
From Prof. Beer’s handbook chapter -3 If we have a fixed circular at one end, the longitudinal axis due to the torsion occurring will be wrapped, and point A will move up to become point A’.
The torsion causes an angle phi φ to take place, and the rounded portion movement caused by the torsion=R*φ.
The shear strain gamma γ is the change of the length /original length=radial Angle=tan γ=R*φ/L.
As we remember the bending moment stress, f=M*Y/I, causes Compression stress at the top and tension stress at the bottom, but that equation is for the bending moment, now we have a torsion, τ the shear stress will be used instead of the bending stress f, τ max.
The torsion moment T will replace M, while y-max in the case of bending moment, will be replaced by R, since the section is round, where R is the radius, and instead of I, the polar moment of inertia I-polar will be used.
I polar is considered as Jp then the equation will be τmax=T*R/Jp, the strain gamma γ =R*φ/L, but as strain=τmax/G.
G instead of I to be used, but τmax=T*R/J. γmax= R*φ/L=T*R/Jp. Finally, we get the expression of φ=T*L/(Jp*G).
### Review of j polar for the non-round shapes.
For the non-round shapes, due to torsional, due to fixation on one side.
If we have a look at an element of length L, due to opposite Torsional moments, one at the fixed part, the other one acts at the free end, and both act against each other.
Check the τmax and φ in this case. By using the mathematical theory of elasticity for a straight bar is used for uniform rectangular cross-section, then τmax =T/c1*ab^2, where a is the longer side length, while b is the shorter distance, While φ=TL/Jpg will be replaced by TL/(c2ab^3)Jp as if Jp=c2ab^3, where c1 and c2 depend on a/b.
c1=c2=1/3(1-0.63b/a) for case of a/b>=5 only.
### J- value for the built-up section.
The built-up section, where the web height is > its width, can be considered as a group of rectangles.
For c1 =c2 to be used at infinity, where c1=c2=1/3 and Jp =1/3*a*b^3. Now refer to the next slide, where J =1/3ab^3, then φ= TL/((1/3)ab^3G) for the case of a rectangle. For the I beam, J will be the sum of 1/3ab^3, where a is the longer side while b is the shorter side.
Now refer back to Lr for the value of J, which is the sum of(1/3)*a*b^3 or 1/3*sum of (longer side)*(shorter side)^3.
For the flange at top and bottom, a=16″, b=5/8″, for the web a=26″ and b=5/16″, J=1/3(16*5/8^3*2+ 26*(5/16)^3). J=2.868 inch3.
For the Lr=61.7392*sqrt(2.868)*sqrt(1+sqrt(1+123.25/(2.868)^2.
This is a detailed calculation to find out the final value.Lr=104.575* (3.4897) divided by 12, Lr=30.41′.
Lb is>Lp and <Lr.
### Estimation of the coefficient of bending cb.
The next step is to estimate the Coefficient of bending Cb to continue to calculate the Mn due to lateral-torsional buckling. I have started with parts Ab and Cd these parts are similar. Estimate the moment values for four portions. Please refer to the first part of the next slide.
These are the CB values for These are the moment values for parts AB and CD.
The next post will include the Final Live Load value required.
This is the pdf file used in the illustration of this post.
This is the next post Part 4/4 of the Solved problem 9-9-6 How To Find LL?
This is a very useful external resource for steel beams. Chapter 8 – Bending Members.
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# What is the vertex form of the equation of the parabola with a focus at (12,22) and a directrix of y=11 ?
##### 1 Answer
Nov 24, 2017
$y = \frac{1}{22} {\left(x - 12\right)}^{2} + \frac{33}{2}$
#### Explanation:
$\text{the equation of a parabola in "color(blue)"vertex form}$ is.
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$
$\text{for any point "(x.y)" on a parabola}$
$\text{the focus and directrix are equidistant from } \left(x , y\right)$
$\text{using the "color(blue)"distance formula "" on "(x,y)" and } \left(12 , 22\right)$
$\Rightarrow \sqrt{{\left(x - 12\right)}^{2} + {\left(y - 22\right)}^{2}} = | y - 11 |$
$\textcolor{b l u e}{\text{squaring both sides}}$
$\Rightarrow {\left(x - 12\right)}^{2} + {\left(y - 22\right)}^{2} = {\left(y - 11\right)}^{2}$
${\left(x - 12\right)}^{2} \cancel{+ {y}^{2}} - 44 y + 484 = \cancel{{y}^{2}} - 22 y + 121$
$\Rightarrow {\left(x - 12\right)}^{2} = 22 y - 363$
$\Rightarrow y = \frac{1}{22} {\left(x - 12\right)}^{2} + \frac{33}{2} \leftarrow \textcolor{red}{\text{in vertex form}}$ |
## Tuesday, January 31, 2012
Young mathematicians’ first formal introduction to multiplication and division happens in 3rd Grade.
In our unit, Equal Groups, situations are presented in context. The situation usually requires students to identify the number of groups and the number of items within each group.
There are 5 tricycles. Each tricycle has three wheels. How many wheels are there in all?
Tricycles have three wheels. There are 5 tricycles. How many wheels are there in all?
Number of Groups: 5
Number of Items in Each Group: 3
Students are taught to represent the problem with both an addition equation and a multiplication equation to illustrate the connection, and use a variable for the missing piece of information.
Multiplication equation: 5x3=w (read 5 groups of 3)
Students generally begin solving the multiplication situations with their prior knowledge of repeated addition.
Repeated Addition: 3+3+3=9 and 3+3=6, so 9+6=15.
Then, some students move into skip counting if they can easily skip count by that number. If the situation allows counting skip counting by 2’s, 5’s, and 10’s, students will almost always start with this strategy. If however, the situation has them skip counting by 8’s, then we’ll teach students to use 10 as their anchor. Skip count by 10 and go back 2.
Skip Counting: 3,6,9,12,15.
One tool that we teach students to use to organize their thinking when skip counting is a ratio table. Ratio tables help students keep track of the number of groups as they skip count.
During the unit, students are introduced to multiplication situations using arrays, too. An array is a rectangular arrangement with rows and columns.
To create a common language with our students, we have them give us the dimensions of the row first and then the column. The array above would be a 5x3.
There are many ways to solve the array. One of the things students do is skip count the array.
They can skip count the array be either the rows or columns.
Students naturally begin to see and explore the commutative property of multiplication. Though the situation is 5x3, they realize they can applying their number knowledge properties, and solve 3x5 instead (5,10,15 or 5+5+5=15).
Students’ next level of understanding develops when they recognize that they can decompose an array into smaller arrays to help them solve problems.
In this situation, if a student didn’t know the product of 5x3 quickly, they could decompose the array into (5x2) + (5x1) = 15.
This understanding is extremely important as students move into problems that are more difficult when they begin multiplication like 6x8. When in the early stages of developing automaticity, they may not know 6x8, but if they know (3x8), they can solve 2(3x8), or, if they know 6x4, they can solve 2(6x4).
Students’ ability to think flexibly with decomposing arrays in multiple ways, builds a strong foundation for fluency in multiplication. The skill allows students to attack any multiplication equation for which they don’t automatically have a product, and leads into being able to solve more difficult equations like 14 x 12.
After the unit, Equal Groups, students have a solid conceptual foundation and can think about multiplication flexibly. But, if we stop there, they may never become fully fluent. We continue to practice fact fluency with our combination club flash cards, by playing multiplication bingo, doing fluency clicker reviews, and doing a timed fluency snapshot several times a week. The fluency snapshots are presented by similar facts. (5’s and 10’s together) (2’s and 4’s and 8’s together) (3’s, 6’s, 9’s, 12’s together) (7’s) and (11’s). Presentation with similar facts promotes the conceptual understanding we build throughout this unit.
Our goal is for every student to leave third grade knowing each of their multiplication facts within three seconds. This foundational knowledge creates automaticity and will help them be successful in fourth grade as they embark on more complex multiplication problems like 49 x 58.
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# Conditions for a Unique Triangle - AAS
Videos, examples, solutions, and lessons to help Grade 7 students learn how to draw triangles under different criteria to explore which criteria result in many, a few, or one triangle.
### New York State Common Core Math Grade 7, Module 6, Lesson 10
Worksheets for 7th Grade, Module 6, Lesson 10 (pdf)
The following figures give the Conditions for a Unique Triangle - Angle-Angle-Side (AAS). Scroll down the page for more examples and solutions.
### Lesson 10 Student Outcomes
• Students understand that two triangles are identical if two pairs of corresponding angles and one pair of corresponding sides are equal under some correspondence; two angle measurements and a given side length of a triangle determine a unique triangle.
• Students understand that the two angles and any side condition can be separated into two conditions:
(1) the two angles and included side condition and
(2) the two angles and the side opposite a given angle condition.
### Lesson 10 Summary
• The two angles and any side condition determines a unique triangle.
• Since the condition has two different arrangements, we separate it into two conditions: the two angles and included side condition and two angles and the side opposite a given angle condition.
• When drawing a triangle under the two angles and the side opposite a given angle condition, the angle opposite the given segment must be drawn on separate paper in order to locate the position of the third vertex.
Lesson 10 Classwork
Opening
In Lesson 6, we explored drawing triangles under the condition that two angles and a side length were provided.
• The arrangement of these parts was not specified, and a total of three non-identical triangles were drawn.
• In this lesson, we explore what happens when this condition is modified to take arrangement into consideration.
• Instead of drawing triangles given two angle measurements and a side length, we will draw triangles under the condition that two angles and the included side is provided and under the condition that two angles and the side opposite a given angle is provided.
Exploratory Challenge
1. A triangle XYZ has angles ∠X = 30° and ∠Y = 50° and included side XY = 6 cm. Draw triangle X’Y’Z' under the same condition as XYZ. Leave all construction marks as evidence of your work, and label all side and angle measurements.
Under what condition is X’Y’Z' drawn? Compare the triangle you drew to two of your peers' triangles. Are the triangles identical? Did the condition determine a unique triangle? Use your construction to explain why.
2. A triangle RST has angles ∠S = 90° and ∠T = 45° and included side ST = 7 cm. Draw triangle R’S’T' under the same condition. Leave all construction marks as evidence of your work, and label all side and angle measurements.
Under what condition is R’S’T' drawn? Compare the triangle you drew to two of your peers' triangles. Are the triangles identical? Did the condition determine a unique triangle? Use your construction to explain why.
3. A triangle JKL has angles ∠J = 60° and ∠L = 25° and side KL = 5 cm. Draw triangle J’K’L' under the same condition. Leave all construction marks as evidence of your work, and label all side and angle measurements.
Under what condition J’K’L' is drawn? Compare the triangle you drew to two of your peers' triangles. Are the triangles identical? Did the condition determine a unique triangle? Use your construction to explain why.
4. A triangle ABC has angles ∠C = 35° and ∠B = 105° and side AC = 7 cm. Draw triangle A’B’C' under the same condition. Leave all construction marks as evidence of your work, and label all side and angle measurements.
Under what condition is A’B’C' drawn? Compare the triangle you drew to two of your peers' triangles. Are the triangles identical? Did the condition determine a unique triangle? Use your construction to explain why.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
## Math Ia Type 2 Stellar Numbers.
This is an investigation about stellar numbers, it involves geometric shapes which form special number patterns. The simplest of these is that of the square numbers (1, 4, 9, 16, 25 etc…) The diagram below shows the stellar triangular numbers until the 6th triangle. The next three numbers after T5 would be: 21, 28, and 36. A general statement for nth triangular numbers in terms of n is: The 6-stellar star, where there are 6 vertices, has its first four shapes shown below:
The number of dots until stage S6: 1, 13, 37, 73, 121, 181 Number of dots at stage 7: 253 Expression for number of dots at stage 7: Since the general trend is adding the next multiple of 12 (12, 24, 36, 48 etc…) for each of the stars, so for S2 it would be 1+12=13, and for S3 it would be 13+24=37 General statement for 6-stellar star number at stage Sn in terms of n: For P=9: Since S1 must equal 1 then we can prove this formula by showing that:
So the first six terms are: 1, 19, 55, 109, 181, 271 Therefore the equation for the 9-Stellar star at For P=5: Since S1 must equal 1 then we can prove this formula by showing that: So the first six terms are: 1, 11, 31, 61, 101, 151 So the expression for 5-Stellar at General Statement for P-Stellar numbers at stage Sn in terms of P and = For P-Stellar number equals 10: For P-Stellar number equals 20: The General Statement works for all number fro 1 to positive infinity.
The equation was arrived at since the sum of arithmetic series can be found using , since the difference is always 2P then we can replace 2P with d, and since u1 is always equal to 1, we can replace it with 1 every time. The at the end of the equation serves the purpose of making the difference between the numbers in the series constant. This form of the equation will allow for only one variable to change, which will be . One of the things the student realized while solving this investigation was that the second term is always equal to , but the derived equation which is also works.
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# Group Theory Terminology
Understanding Group Theory Terminology in Further Mathematics is essential for developing advanced problem-solving skills and a deeper comprehension of various mathematical concepts. This branch of mathematics delves into the study of symmetry, structures, and operations within abstract systems and has numerous applications ranging from science and technology to finance and decision-making. By exploring the key concepts and characteristics of groups in mathematics, mastering basic and advanced terminology, and examining real-world examples, you can gain a better grasp of this important subject. Furthermore, understanding Group Theory's role in mathematical analysis and its connections to other fields can open up exciting future career opportunities for those who possess this critical knowledge. In this article, you will discover the many practical applications of Group Theory as they relate to further mathematics and various industries.
#### Create learning materials about Group Theory Terminology with our free learning app!
• Flashcards, notes, mock-exams and more
• Everything you need to ace your exams
## Explaining Group Theory Definition
Group Theory is a branch of mathematics that deals with the study of algebraic structures known as groups. A group is a set of elements accompanied by a binary operation that satisfies certain properties, which are essential for this topic. Understanding the key concepts and characteristics of groups in mathematics will help you develop a solid foundation in Further Mathematics.
### Key concepts in Group Theory
A group is an ordered pair $$(G, *)$$ consisting of a set $$G$$ and a binary operation $$*:GxG \rightarrow G$$ that satisfies the following properties:
1. Closure Property: For every $$a, b \in G$$, the result of the operation $$a * b$$ belongs to $$G$$.
2. Associative Property: For all $$a, b, c \in G$$, it holds that $$(a * b) * c = a * (b * c)$$.
3. Identity element: There exists an element $$e \in G$$ such that for every $$a \in G$$, $$a * e = e * a = a$$.
4. Inverse element: For each element $$a \in G$$, there exists an element $$b \in G$$ such that $$a * b = b * a = e$$, where $$e$$ is the identity element.
### Characteristics of groups in mathematics
There are several classifications and characteristics of groups that may be of interest:
• Abelian or commutative groups: Groups where the operation is commutative, meaning that for all $$a, b \in G$$, $$a * b = b * a$$.
• Finite and infinite groups: Groups with a finite number of elements are called finite, while groups with an infinite number of elements are called infinite groups.
• Subgroups: A subset of a group $$G$$ is a subgroup if it is non-empty and closed under the group operation.
• Order of a group: The number of elements in a finite group, denoted by |G|.
• Cyclic groups: Groups generated by a single element called a generator.
## Exploring Group Theory Terminology
There are many basic terms in Group Theory that you should be familiar with, including:
• Binary Operation: A function that takes in two elements from a set and returns another element of the same set.
• Identity Element: The element that, when combined with any other element in the group using the binary operation, returns the other element.
• Inverse Element: The element, when combined with another element using the binary operation, results in the identity element of the group.
• Generator: An element that generates all other elements of the group through repeated application of the binary operation.
• Homomorphism: A mapping between two groups that preserves the group structure.
• Isomorphism: A bijective homomorphism between two groups that implies the groups are structurally the same.
• Normal Subgroup: A subgroup which is invariant under conjugation by any element in the group.
Once you have mastered the basic terms, it's important to explore some advanced terminology in Group Theory:
• Group Action: A formal way to describe how a group acts on a set, preserving the set's structure.
• Permutation Groups: Groups formed by permutations of a finite set.
• Cosets: A subset of a group created by multiplying a fixed element with all elements of a subgroup.
• Factor Groups or Quotient Groups: The set of all cosets of a normal subgroup.$$N$$ of group $$G$$.
• Group Extensions: The process of constructing a new group using known groups and their properties.
• Representation Theory: The study of algebraic structures by representing their elements as linear transformations of vector spaces.
## Examining Group Theory Examples
Group Theory is applied in various real-world scenarios, such as:
• Cryptography: Group Theory plays a crucial role in encryption algorithms and secure communication.
• Physics: Understanding symmetry in particle physics and quantum mechanics often involves the use of Group Theory.
• Chemistry: Crystallographic groups, used to describe the symmetry of crystal structures, are based on Group Theory.
• Music Theory: Group Theory can aid in understanding the structures and symmetries in music scales and chords.
### Problem-solving using Group Theory techniques
Group Theory can be used to solve various mathematical problems, such as:
• Counting problems: Group actions can be used to search for fixed points in arrangements, as shown in Burnside's Lemma.
• Geometry: Symmetries of planar shapes can be analyzed and classified using Group Theory techniques.
• Combinatorics: Permutations and combination problems often involve the action of permutation groups on sets.
• Algebra: Solutions to algebraic equations can be discussed in terms of group actions and representations.
## The Importance of Group Theory in Decision Mathematics
Group Theory is essential in mathematical analysis as it allows for the systematic study of abstract algebraic structures, providing a deep understanding of various mathematical phenomena. It is intrinsically connected to many other branches of mathematics, serving as a unifying language and thus enabling a better comprehension of complex problems.
### Connections between Group Theory and other mathematical fields
Group Theory is interconnected with numerous other areas of mathematics, including:
• Linear Algebra: The study of vector spaces and linear transformations between them, which is often described using the language of Group Theory.
• Number Theory: The exploration of the properties and relationships of numbers, where Group Theory techniques play a significant role in understanding congruences and modular arithmetic.
• Combinatorics: Group Theory is instrumental in solving counting problems in permutations and combinations, such as Burnside's Lemma.
• Topology: The study of continuous deformations of geometric objects, which relies on Group Theory concepts in the analysis of transformation groups and symmetries.
• Geometry: In the field of geometric transformations and symmetries, Group Theory plays a central role in understanding planar shapes and their properties.
These connections facilitate the development of advanced mathematical tools and enable the creation of new approaches to analyze and solve complex problems that cross multiple disciplines.
### Applications of Group Theory in various decision-making situations
Group Theory has numerous applications in real-life decision-making situations such as:
• Game Theory: Group Theory can be applied to explore the symmetry of various games, which can lead to optimal strategies and decision-making.
• Optimization: In operations research, Group Theory techniques can be employed to analyze and solve optimization problems with a focus on symmetry and structure.
• Scheduling: Group Theory concepts can be used to efficiently organize and allocate resources, create timetables, and analyze rotation schemes.
• Network Analysis: Symmetries in network graphs can be studied using Group Theory methods, leading to better understanding and decision-making regarding network structures.
These applications demonstrate the significance of Group Theory within decision-making processes across diverse industries and disciplines.
## Advantages of Understanding Group Theory Importance
Developing a comprehensive understanding of Group Theory is valuable for students and professionals alike as it can lead to significant benefits in both educational and career pursuits.
### Benefits for students studying Group Theory
Students who learn Group Theory can gain the following advantages:
• A deeper understanding of abstract algebraic structures, which can enhance problem-solving skills and overall mathematical competency.
• Improved critical and logical thinking skills, as the study of Group Theory requires rigorous analytical thinking.
• Familiarity with a unifying language across multiple mathematical fields, enabling students to better communicate complex mathematical ideas and concepts.
• Access to a wide range of advanced topics and applications across various disciplines, broadening students' understanding and perspectives beyond core mathematical areas.
These benefits can open up diverse educational and research opportunities for students, greatly enriching their academic experience and progression.
### Future career opportunities requiring Group Theory knowledge
The knowledge of Group Theory can open up various career opportunities in different fields, including:
• Academia and Research: Being well-versed in Group Theory can lead to positions in research and education, exploring advanced topics in mathematics and its applications.
• Computer Science and Cryptography: Group Theory is essential to understanding and developing encryption algorithms, a vital aspect of computer security and data protection.
• Physics and Engineering: Engineers and physicists often rely on Group Theory concepts to study symmetries and structures in various applications such as particle physics and material sciences.
• Finance and Economics: Group Theory techniques can be applied to many economic and financial problems, including algorithmic trading, risk management, and economic modelling.
These career paths highlight the importance of mastering Group Theory, as it can lead to a wide array of exciting and fulfilling professional opportunities.
## Practical Applications of Group Theory in Further Mathematics
Group Theory has tremendously impacted both physics and engineering, thanks to its capability to describe symmetries, structural relationships, and transformations. In both disciplines, Group Theory has proven to be an invaluable tool, playing an essential role in a wide range of applications, some of which are listed below:
• Particle Physics: The study of fundamental particles, their interactions, and forces, relies heavily on Group Theory, especially to classify and comprehend their symmetry properties. Many theories and models, such as the Standard Model and gauge theories, depend on group representations and symmetries.
• Quantum Mechanics: Space-time symmetries, quantum states, and angular momentum are better understood using group representations. This understanding helps to unravel the underlying principles and behaviour of quantum systems.
• Solid State Physics and Material Science: Crystal structures, wave functions, and vibrational modes related to solids are effectively explained using Group Theory. Furthermore, it aids in investigating electronic behaviour and properties of materials by examining their symmetry groups.
• Control Theory and System Engineering: Group Theory can be leveraged to design and analyse engineering systems and control strategies based on their symmetries and invariances.
• Robotics and Kinematic Chains: Group Theory is helpful in defining the chain of transformations essential for controlling arm movements in robotics and understanding the complex joint systems accurately.
These applications demonstrate the significant role of Group Theory in shaping the advancement of science and technology, particularly in physics and engineering.
### Group Theory's impact on computer science and cryptography
Group Theory has made a substantial impact on computer science and cryptography, thanks to its distinctive properties and flexible framework for defining and analysing algebraic structures. Here are some prominent examples of how Group Theory is applied in these domains:
• Encryption Algorithms: Group Theory plays a vital role in designing and analysing secure communication algorithms such as the Diffie-Hellman key exchange, RSA, and elliptic curve cryptography.
• Algorithm Design: Many computer science algorithms utilise Group Theory concepts to provide efficient and transformative solutions. Certain divide-and-conquer strategies and algorithms for language recognition and parsing are based on group symmetries and invariance.
• Error-correcting codes: Group Theory is applied to develop error detection and correction techniques for reliable data transmission and storage, like Reed-Solomon and Hamming codes.
• Computer Graphics and Geometric Transformations: Group Theory techniques can be employed to study and implement transformations in computer graphics, image processing, and computer-aided design (CAD).
These examples illustrate the potential and impact of Group Theory on computer science and cryptography, offering robust and innovative solutions to various problems.
### Exploring Group Theory Applications in Business and Finance
In the realm of business and finance, Group Theory can be employed to devise decision-making strategies with a focus on symmetry and structure. By recognising and analysing patterns, one can find optimal solutions to various problems. Some practical applications of Group Theory in business decision-making are:
• Game Theory: Concepts from Group Theory can be applied to symmetric games, identifying optimal strategies, equilibria, and outcomes, subsequently aiding in effective decision-making.
• Portfolio Optimization: Group Theory can be utilised to develop algorithms for portfolio selection, particularly when considering symmetry in financial asset returns and correlations.
• Resource Allocation: Using Group Theory, resources can be efficiently allocated and managed based on symmetry properties and optimal combinations, ensuring operational effectiveness and cost minimisation.
• Market Analysis: Group Theory techniques can be used to study patterns within market data, leading to a better understanding of investment trends and allowing for more informed decisions.
These applications demonstrate how Group Theory can contribute to creating highly effective decision-making strategies in business and finance.
#### Assessing risk and optimising solutions through Group Theory techniques
Group Theory techniques can also be used in business and finance to evaluate the risk and to optimise solutions, ensuring profitability and stability. Key applications of Group Theory in risk assessment and solution optimisation are:
• Risk Management: Group Theory concepts can be useful in understanding and managing diverse financial instruments or systems based on their symmetries and invariances, particularly in evaluating risk and optimising trading strategies.
• Financial Modelling: Fitting financial models with underlying group structures can improve pricing and valuation, foresee potential issues with greater accuracy, and ultimately generate more viable economic projections.
• Credit Scoring: Group Theory helps in designing algorithms that recognise similarities in consumer behaviour, enabling more accurate credit scoring and better lending decisions.
• Optimisation Problems: Group Theory techniques can be used to solve many optimisation problems in finance and business, such as investment allocation, logistics, and supply chain management.
These applications highlight the significance of Group Theory in assessing risk and optimising solutions, facilitating more informed and effective decision-making in business and finance.
## Group Theory Terminology - Key takeaways
• Group Theory: branch of mathematics studying symmetry, structures, and operations within abstract systems; essential in developing advanced problem-solving skills and deeper mathematical understanding.
• Key concepts: Closure Property, Associative Property, Identity element, Inverse element; crucial for mastering Group Theory terminology.
• Characteristics of groups: Abelian/commutative, finite/infinite, subgroups, order, cyclic; various classifications and properties of mathematical groups.
• Real-world examples: cryptography, physics, chemistry, music theory; practical applications demonstrate Group Theory's importance and versatility.
• Connections to other fields: linear algebra, number theory, combinatorics, topology, geometry; understanding Group Theory's importance and applications can open up future career opportunities.
#### Flashcards inGroup Theory Terminology 12
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Who is the father of group theory?
The father of group theory is Évariste Galois, a French mathematician who made significant contributions to the development of the field in the early 19th century.
What is the goal of group theory?
The goal of group theory is to study the structure and properties of mathematical objects called groups, which are sets with a defined operation that satisfies specific rules. It aims to identify and classify different types of groups, understand their behaviour, and determine their applications in various areas of mathematics and other fields.
Why do we study group theory in mathematics?
We study group theory in mathematics because it provides a fundamental framework for understanding algebraic structures, symmetries and invariance. It has numerous applications across various fields, such as cryptography, chemistry, physics, and computer science. Additionally, group theory facilitates problem-solving and supplies essential tools for other branches of mathematics.
What is group theory example?
A group theory example is the set of integers Z under the operation of addition. In this group, the identity element is 0, and the inverse of any integer n is -n. These elements and operation satisfy group axioms: closure, associativity, identity, and inverse.
What is group theory?
Group theory is a branch of abstract algebra which studies mathematical objects called groups. These groups consist of a set of elements and a single binary operation that combines elements, satisfying certain properties like associativity, identity and inversibility. Group theory finds applications in various fields, including geometry, particle physics, and cryptography.
## Test your knowledge with multiple choice flashcards
What are the four main properties of a group in Group Theory?
What are some classifications and characteristics of groups in mathematics?
What are some basic terms in Group Theory that you should be familiar with?
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# To find Length or Breadth when Area of a Rectangle is given
Let’s discuss how to find length or breadth when area of a rectangle is given.
When we need to find length of a rectangle we need to divide area by breadth.
Length of a rectangle = Area ÷ breadth
ℓ = A ÷ b
Similarly, when we need to find breadth of a rectangle we need to divide area by length.
Breadth of a rectangle = Area ÷ length
b = A ÷ ℓ
Formulas:
For Example:
1. Find the length of a rectangle whose breadth is 5 cm and area is 90 sq. cm.
Solution:
Area (A) = 90 sq. cm.
Length (ℓ) = Area ÷ breadth
= 90 ÷ 5
= 18
Therefore, length of rectangle = 18 cm.
2. Find the breadth of a rectangle whose length is 200 cm and area is 10 m2.
Solution:
Length (l) = 200 cm,
Area (A) = 10 m2
100 cm = 1 m.
Length (ℓ) = Area ÷ breadth
Length (ℓ) = 200/100 m = 2 m.
Breadth (b) = A ÷ ℓ
= 10 ÷ 2 m.
= 5 m.
Therefore, breadth of rectangle = 5 m.
Area.
Area of a Rectangle.
Area of a Square.
To find Area of a Rectangle when Length and Breadth are of Different Units.
To find Length or Breadth when Area of a Rectangle is given.
Areas of Irregular Figures.
To find Cost of Painting or Tilling when Area and Cost per Unit is given.
To find the Number of Bricks or Tiles when Area of Path and Brick is given.
Worksheet on Area.
Practice Test on Area.
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
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## Logarithmic Differentiation
Suppose we wish to find $\displaystyle{\frac{dy}{dx}}$, where $\displaystyle{y = \frac{\sqrt[4]{x+1}}{(x+2)^6\sqrt{x+3}}}$.
At first blush, we might think we need to employ the quotient rule, the product rule, and a couple of chain rule applications involving derivatives of powers -- a task that is certianly doable, but likely to be algebraically tedious!
However, there is a simpler way!
For some complicated expressions involving product, quotients, and powers, we can use the properties of logarithms to make the expression more "differentiation-friendly".
Recall that $e^{\ln x} = x$
As such,
$${\large y} = {\large e}^{\left(\displaystyle{\ln \frac{\sqrt[4]{x+1}}{(x+2)^6\sqrt{x+3}}}\right)}$$
However, recalling $\ln AB = \ln A + \ln B$ and $\ln A/B = \ln A - \ln B$ under the appropriate conditions, we can break the exponent up into the sum/difference of some smaller pieces:
$${\large y} = {\large e}^{\left(\displaystyle{\ln \sqrt[4]{x+1} - \ln (x+2)^6 - \ln \sqrt{x+3}}\right)}$$
Then, recalling that $\ln A^p = p \ln A$ we can do even more to simplify these pieces.
$${\large y} = {\large e}^{\left(\displaystyle{\frac{1}{4}\ln (x+1) - 6\ln (x+2) - \frac{1}{2}\ln (x+3)}\right)}$$
Now, we can use the chain rule to find $dy/dx$:
$$\frac{dy}{dx} = {\large e}^{\left(\displaystyle{\frac{1}{4}\ln (x+1) - 6\ln (x+2) - \frac{1}{2}\ln (x+3)}\right)} \cdot \left(\frac{1}{4(x+1)} - \frac{6}{x+2} - \frac{1}{2(x+3)} \right)$$
As one final simplification, note that the power of $e$ above was just another way of writing the original $y$, and substitute this out for the formula given to us for $y$:
$$\frac{dy}{dx} = \frac{\sqrt[4]{x+1}}{(x+2)^6\sqrt{x+3}} \cdot \left(\frac{1}{4(x+1)} - \frac{6}{x+2} - \frac{1}{2(x+3)} \right)$$
Ta-Da! (That's a whole lot easier than using the quotient rule, product rule, and chain rule all together, don't you think?)
This process of rewriting $y$ as $e$ to a logarithm, and then using the properties of logarithms to simplify the exponent before differentiating (and finally making one last substitution) is called logarithmic differentiation and can be a real time saver under the right circumstances!
If you are wondering to whom you should direct your thanks, let's give a round of applause to Johann Bernoulli who came up with this technique in 1697! |
Question
# Construct the circumcircle and incircle of an equilateral $\Delta ABC$ with side 6cm and center $O$. Find the ratio of radii of circumcircle and incircle.
Hint: Draw a perpendicular bisector on the equilateral triangle which divides the sides of the equilateral triangle into two equal parts. Take their intersection point to draw a circumcircle and incircle. Now use trigonometric ratios in two different triangles for values of in-radius and circum-radius.
The pictorial representation of the given problem is shown above.
The equilateral triangle ABC with side 6 cm has a circumcircle and an incircle with center O and radii ${r_1}$ and ${r_2}$ respectively.
$\Rightarrow OB = {r_1}cm,{\text{ }}OD = {r_2}cm$
AE and CD are the perpendicular bisector of BC and AB respectively.
$\Rightarrow BE = CE = BD = AD = \dfrac{6}{2} = 3cm$
Since, FB is the bisector of $\angle ABC$
$\Rightarrow \angle FBC = \angle FBA = \dfrac{{{{60}^0}}}{2} = {30^0}$
Because in equilateral triangle all angles are equal which is ${60^0}$
Now, in $\Delta OBE,{\text{ cos3}}{{\text{0}}^0}{\text{ = }}\dfrac{{BE}}{{OB}} = \dfrac{3}{{{r_1}}}$
As we know ${\text{cos3}}{{\text{0}}^0} = \dfrac{{\sqrt 3 }}{2}$
$\Rightarrow {r_1} = \dfrac{3}{{\cos {{30}^0}}} = \dfrac{3}{{\dfrac{{\sqrt 3 }}{2}}} = \dfrac{6}{{\sqrt 3 }}cm$
Now in $\Delta OBD,{\text{ tan3}}{{\text{0}}^0}{\text{ = }}\dfrac{{OD}}{{BD}} = \dfrac{{{r_2}}}{3}$
As we know ${\text{tan3}}{{\text{0}}^0} = \dfrac{1}{{\sqrt 3 }}$
$\Rightarrow {r_2} = 3\tan {30^0} = 3\left( {\dfrac{1}{{\sqrt 3 }}} \right) = \sqrt 3 cm$
Now you have to calculate the ratio of radii circumcircle to incircle
$\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{\dfrac{6}{{\sqrt 3 }}}}{{\sqrt 3 }} = \dfrac{6}{{\sqrt 3 \times \sqrt 3 }} = \dfrac{6}{3} = 2cm$
So, the required ratio of the radii is 2 cm.
Note: In such types of question first draw the pictorial representation of the given problem, then draw the perpendicular bisectors on the triangle which divide its sides into two equal parts, then apply basic trigonometric property and calculate the radii of the two circles, then divide them we will get the required answer. |
# Trigonometric Substitution Integration
Have you ever experienced true pain with calculus? Want to relive those horrifying moments? Well look no further! We're going to be integrating some functions which will require trigonometric substitution! Yaaaay…
## The First Case:
Let's begin. The basic idea of integrating complex expressions is to just try a bunch of stuff and pray that one of those methods will lead you slightly closer to the solution than you started with. The first trick we'll use is the Pythagorean theorem. We're going to use x^2 + y^2 = z^2 as the values since "a" is already taken. Basically we're going to construct a right triangle, and that expression above is going to be the lengths of one of the sides. Then, we use the power of trigonometry to substitute in different functions that will actually be solvable. Plugging the above into the hypotenuse we get some nice values out of it:
It doesn't matter which variable we assign to which triangle leg. I'm setting it up like this. Now, use the triangle's properties to create equations that can be used to substitute values back in the original integral. We need a replacement for du and for that square root expression.
Alright, so now we just need to solve for that. Surprise, it's not easy. We'll have to pull out another trick: integration by parts. I'm not going to explain how integration by parts works. It turns out that sec^2(x) is easily integrable, so we'll use that as our "dv" value. the "u" value must be sec(x), then.
After using integration by parts and a trigonometric identity, we're left with this. We actually ended up with another copy of our starting place. This means we can combine that term with the left side of the equation and do some reduction, meaning that the last integral we need to care about is sec(theta). Just look this one up, honestly. It's another whole mess to figure this one out.
Now that no more integrals remain, it's time to resubstitute back the u and a variables! Remember the triangle we constructed from earlier to help us with this.
Note that near the end I used a property of logarithms to take out the (1/a) inside the log, then merging that value with C because they're both constants. And there's the first out of three forms. Oh yeah, there's two more problems in the set.
## The Second Case:
Do a similar process for this case. Set up the triangle, and do some substitutions.
Wait, that term looks familiar… we already solved for sec^3(theta)! We solved for sec(theta) as well. The only difference is that the value of the sec(theta) integral will be negative. Well, that's convenient.
Alright! One more to go.
## The Third Case:
The steps here are way easier: they just require different substitutions. For the last part, theta is by itself, meaning we'll have to go back to the triangle and find an equation that gets theta by itself.
Aaaaaand that's how you do it. |
# How do you calculate series error?
## How do you calculate series error?
Pollyanna, also known as “Secret Santa” is a Western Christmas tradition. In which groups, or community are randomly assigned a person they will give a gift to on Christmas. The identify of the gift giver is a secret not to be revealed.
## How do you find the summation error?
A white elephant gift exchange, Yankee swap or Dirty Santa is a party game where amusing and impractical gifts are exchanged during festivities. The goal of a white elephant gift exchange is To entertain party-goers rather than to gain a genuinely valuable or highly sought-after item.
## How do you find the maximum possible error in a series?
How to Determine Greatest Possible Error
1. Step1: Identify the last nonzero digit to the right of the decimal point in the given measurement.
2. Step 2: Determine the precision. …
3. Step 3: Divide the precision by 2 to determine the greatest possible error.
## How do you calculate alternating series?
How to Determine Greatest Possible Error
1. Step1: Identify the last nonzero digit to the right of the decimal point in the given measurement.
2. Step 2: Determine the precision. …
3. Step 3: Divide the precision by 2 to determine the greatest possible error.
## What is the remainder of a series?
Mathwords: Remainder of a Series. The difference between the nth partial sum and the sum of a series.
## How do you estimate differential errors?
We can also use differentials in Physics to estimate errors, say in physical measuring devices. In these problems, we’ll typically Take a derivative, and use the “dx” or “dy” part of the derivative as the error. Then, to get percent error, we’ll divide the error by the total amount and multiply by 100.
## What is the maximum possible error in a measurement?
The greatest possible error (GPE) is The largest amount a ballpark figure can miss the mark. It’s one half of the measuring unit you are using. For example: If measuring in centimeters, the GPE is 1/2 cm.
## How do you use alternating series error bound?
The greatest possible error (GPE) is The largest amount a ballpark figure can miss the mark. It’s one half of the measuring unit you are using. For example: If measuring in centimeters, the GPE is 1/2 cm.
## How do you evaluate a series?
The greatest possible error (GPE) is The largest amount a ballpark figure can miss the mark. It’s one half of the measuring unit you are using. For example: If measuring in centimeters, the GPE is 1/2 cm.
## How do you use taylor’s formula?
Taylor’s Formula: If f(x) has derivatives of all orders in a n open interval I containing a, then for each positive integer n and for each x ∈ I, F(x) = f(a) + f (a)(x − a) + f (a) 2! (x − a)2 + ··· + f(n)(a) n!
## What is estimation of error?
The difference between an estimated value and the true value of a parameter or, sometimes, of a value to be predicted.
## How do you calculate error and relative error in calculus?
To calculate relative error, Subtract the measured value by the real value and then divide the absolute of that number by the real value to get the relative error. We can then multiply by 100% to get the percent error.
## How do you calculate percentage error in statistics?
The margin of error can be calculated in two ways, depending on whether you have parameters from a population or statistics from a sample:
1. Margin of error (parameter) = Critical value x Standard deviation for the population.
2. Margin of error (statistic) = Critical value x Standard error of the sample.
## How do you calculate error in linear approximation?
This process can be summarized as: Linear Approximation Error: If the value of the x–variable is measured to be x = a with an “error” of ∆x units, then ∆f, the “error” in estimating f(x), is ∆f = f(x) – f(a) ≈ f ‘(a).
## How can we estimate the error in the sum and difference?
Solution : Let two physical quantities A and B has measured value `A+-DeltaA and B+- AB` where, <br> (1) For addition : <br> Let Z is quantity obtained by addition of A and B. <br> `:. Z=A+B` <br> Let error is Z be `DeltaZ` <br> `Z+-DeltaZ=(A+-DeltaA)+(B+-DeltaB)` <br> `A+B=Z` <br> `:.
## What are the different types of errors in measurement?
The errors that may occur in the measurement of a physical quantity can be classified into six types: Constant error, systematic error, random error, absolute error, relative error and percentage error. Each type of error in measurement are explained below.
## Why least count is maximum error?
Suppose the least count of a screw gauge is given 0.01 cm , so we can measure length up to 0.01 cm, below that we can’t measure the length. Hence maximum possible error is 0.01 cm. it is nothing but least count. So least count is considered as maximum possible error.
## What is summation error?
In words, this says that the error in the result of an addition or subtraction is The square root of the sum of the squares of the errors in the quantities being added or subtracted. |
# Lesson 5Flipped OutDevelop Understanding
## Learning Focus
Model a context using two different ways of thinking about the variables.
What are characteristics of inverse functions?
## Open Up the Math: Launch, Explore, Discuss
Chandler and Isaac both like to ride bikes for exercise. They are discussing whether or not they have a similar pace so that they can plan a time to bike together when they find that they both think about bike riding in a different way. Chandler says, “Sometimes I have a busy schedule and it’s hard to fit in a ride. I think about how much time I have to ride and then how many miles I can go.” Isaac says, “As I’m training for a race, I think about how many miles I need to ride and then how much time it will take.” They both look at each other blankly and decide to do a little mathematical modeling.
Chandler says, “If I have minutes to ride, I can go miles.”
### 1.
Model Chandler’s way of thinking about her ride using a table, graph, equation, and any other representation that you think demonstrates Chandler’s method. Be sure to label each representation.
### 2.
Explain the connections between the representations you found.
### 3.
How many miles will Chandler travel at minutes? What is Chandler’s speed?
Isaac explains that he plans his ride based on the number of miles he needs for his training regimen. He knows that for every miles he needs to go, he plans minutes.
### 4.
Model Isaac’s way of thinking about a ride, with miles as the input, using a table, graph, equation, and any other representation that you think demonstrates his method. Be sure to label each representation.
### 5.
What is Isaac’s speed? How is this different from how Chandler describes her speed? Who is faster?
### 6.
How many miles will Isaac travel at minutes?
### 7.
Using the equations, tables, and graphs for Isaac and Chandler, make a list of observations about how these situations compare to each other.
Using relationships that you have observed in this task, find the inverse of . Justify your solution using multiple representations.
## Takeaways
A function and its inverse:
## Lesson Summary
In this lesson, we learned about inverse functions using a context that had two different and useful ways to think of the relationship. Two functions are called inverse functions when their inputs and outputs have been switched.
## Retrieval
Solve for .
### 1.
Rewrite each expression. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 20.5: Calculating Free Energy Change (ΔG°)
Difficulty Level: At Grade Created by: CK-12
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Practice Calculating Free Energy Change
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#### Time for dessert!
When you are baking something, you heat the oven to the temperature indicated in the recipe. Then you mix all the ingredients, put them in the proper baking dish, and place them in the oven for a specified amount of time. If you had mixed the ingredients and left them out at room temperature, not much would change. The materials need to be heated to a given temperature for a set time in order for the ingredients to react with one another and produce a delicious final product.
### Calculating Free Energy(ΔG∘)\begin{align*}(\Delta G^\circ)\end{align*}
The free energy change of a reaction can be calculated using the following expression:
ΔG=ΔHTΔS\begin{align*}\Delta G^\circ=\Delta H^\circ - T\Delta S^\circ\end{align*}
where ΔG=free energy change (kJ/mol)\begin{align*}\Delta G = \text{free energy change (kJ/mol)}\end{align*}
ΔH=change in enthalpy (kJ/mol)\begin{align*}\Delta H = \text{change in enthalpy (kJ/mol)}\end{align*}
ΔS=change in entropy (J/Kmol)\begin{align*}\Delta S = \text{change in entropy (J/K} \cdot \text{mol)}\end{align*}
T=temperature (Kelvin)\begin{align*}T = \text{temperature (Kelvin)}\end{align*}
Note that all values are for substances in their standard state. In performing calculations, it is necessary to change the units for ΔS\begin{align*}\Delta S\end{align*} to kJ/K • mol, so that the calculation of ΔG\begin{align*}\Delta G\end{align*} is in kJ/mol.
#### Sample Problem: Gibbs Free Energy
Methane gas reacts with water vapor to produce a mixture of carbon monoxide and hydrogen according to the balanced equation below.
CH4(g)+H2O(g)CO(g)+3H2(g)\begin{align*}\text{CH}_4(g)+\text{H}_2\text{O}(g) \rightarrow \text{CO}(g)+3\text{H}_2(g)\end{align*}
The ΔH\begin{align*}\Delta H^\circ\end{align*} for the reaction is +206.1 kJ/mol, while the ΔS\begin{align*}\Delta S^\circ\end{align*} is +215 J/K • mol. Calculate the ΔG\begin{align*}\Delta G^\circ\end{align*} at 25°C and determine if the reaction is spontaneous at that temperature.
Step 1: List the known values and plan the problem.
Known
• ΔH=206.1 kJ/mol\begin{align*}\Delta H^\circ =206.1 \ \text{kJ/mol}\end{align*}
• ΔS=215 J/Kmol=0.215 kJ/Kmol\begin{align*}\Delta S^\circ = 215 \ \text{J/K}\cdot \text{mol}=0.215 \ \text{kJ/K}\cdot \text{mol}\end{align*}
• T=25C=298 K\begin{align*}T =25^\circ \text{C}=298 \ \text{K}\end{align*}
Unknown
• ΔG=? kJ/mol\begin{align*}\Delta G^\circ =? \ \text{kJ/mol}\end{align*}
Prior to substitution into the Gibbs free energy equation, the entropy change is converted to kJ/K • mol and the temperature to Kelvins.
Step 2: Solve.
ΔG=ΔHTΔS=206.1 kJ/mol298 K(0.215 kJ/Kmol)=+142.0 kJ/mol\begin{align*}\Delta G^\circ =\Delta H^\circ - T\Delta S^\circ = 206.1 \ \text{kJ/mol} - 298 \ \text{K}(0.215 \ \text{kJ/K}\cdot \text{mol})=+142.0 \ \text{kJ/mol}\end{align*}
The resulting positive value of ΔG\begin{align*}\Delta G\end{align*} indicates that the reaction is not spontaneous at 25°C.
The unfavorable driving force of increasing enthalpy outweighed the favorable increase in entropy. The reaction will be spontaneous only at some elevated temperature.
Available values for enthalpy and entropy changes are generally measured at the standard conditions of 25°C and 1 atm pressure. The values are slightly temperature dependent and so we must use caution when calculating specific ΔG\begin{align*}\Delta G\end{align*} values at temperatures other than 25°C. However, since the values for ΔH\begin{align*}\Delta H\end{align*} and ΔS\begin{align*}\Delta S\end{align*} do not change a great deal, the tabulated values can safely be used when making general predictions about the spontaneity of a reaction at various temperatures.
### Summary
• Calculations of free energy changes are described.
### Review
1. What would happen to ΔH\begin{align*}\Delta H\end{align*} if you forgot to change the units for ΔS\begin{align*}\Delta S\end{align*} to kJ/K • mol?
2. What are standard conditions for enthalpy and entropy changes?
3. At what temperature would the reaction become spontaneous?
### Explore More
Use the resource below to answer the questions that follow.
1. Why is ΔH\begin{align*}\Delta H\end{align*} negative in this example?
2. What would happen if you forgot to change the sign of the TΔS\begin{align*}T\Delta S\end{align*} value in the first calculation?
3. What indicates that the reaction is spontaneous?
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### Similar presentations
3 Certain classes of counting problems come up frequently, and it is useful to develop formulas to deal with them.
4 Example 1 – Casting Ms. Birkitt, the English teacher at Brakpan Girls High School, wanted to stage a production of R. B. Sheridans play, The School for Scandal. The casting was going well until she was left with five unfilled characters and five seniors who were yet to be assigned roles. The characters were Lady Sneerwell, Lady Teazle, Mrs. Candour, Maria, and Snake; while the unassigned seniors were April, May, June, Julia and Augusta. How many possible assignments are there?
5 Example 1 – Solution To decide on a specific assignment, we use the following algorithm: Step 1: Choose a senior to play Lady Sneerwell; 5 choices. Step 2: Choose one of the remaining seniors to play Lady Teazle; 4 choices. Step 3: Choose one of the now remaining seniors to play Mrs. Candour; 3 choices.
6 Example 1 – Solution Step 4: Choose one of the now remaining seniors to play Maria; 2 choices. Step 5: Choose the remaining senior to play Snake; 1 choice. Thus, there are 5 4 3 2 1 = 120 possible assignments of seniors to roles. contd
7 Permutations and Combinations What the situation in Example 1 has in common with many others is that we start with a sethere the set of seniors and we want to know how many ways we can put the elements of that set in order in a list. In this example, an ordered list of the five seniorssay, 1. May 2. Augusta 3. June
8 Permutations and Combinations 4. Julia 5. April corresponds to a particular casting: Cast Lady Sneerwell May Lady Teazle Augusta Mrs. Candour June Maria Julia Snake April
9 Permutations and Combinations We call an ordered list of items a permutation of those items. If we have n items, how many permutations of those items are possible? We can use a decision algorithm similar to the one we used in the Example 1 to select a permutation.
10 Permutations and Combinations Step 1: Select the first item; n choices. Step 2: Select the second item; n – 1 choices. Step 3: Select the third item; n – 2 choices.... Step n – 1: Select the next-to-last item; 2 choices. Step n: Select the last item; 1 choice. Thus, there are n (n – 1) (n – 2)... 2 1 possible permutations. We call this number n factorial, which we write as n!
11 Permutations and Combinations Permutations A permutation of n items is an ordered list of those items. The number of possible permutations of n items is given by n factorial, which is n! = n (n – 1) (n – 2)... 2 1 for n a positive integer, and 0! = 1.
12 Permutations and Combinations Visualizing Permutations Permutations of 3 colors in a flag:
13 Permutations and Combinations Quick Example The number of permutations of five items is 5! = 5 4 3 2 1 = 120. Sometimes, instead of constructing an ordered list of all the items of a set, we might want to construct a list of only some of the items. So, we can generalize our definition of permutation to allow for the case in which we use only some of the items, not all.
14 Permutations and Combinations Check that, if r = n below, this is the same definition we give for permutation of n items. Permutations of n items taken r at a time A permutation of n items taken r at a time is an ordered list of r items chosen from a set of n items. The number of permutations of n items taken r at a time is given by P(n, r ) = n (n – 1) (n – 2)... (n – r + 1) (there are r terms multiplied together). We can also write
15 Permutations and Combinations Quick Example The number of permutations of six items taken two at a time is P(6, 2) = 6 5 = 30 which we could also calculate as
16 Permutations and Combinations For ordered lists we used the word permutation; for unordered sets we use the word combination. Permutations and Combinations A permutation of n items taken r at a time is an ordered list of r items chosen from n. A combination of n items taken r at a time is an unordered set of r items chosen from n. Visualizing
17 Permutations and Combinations Note Because lists are usually understood to be ordered, when we refer to a list of items, we will always mean an ordered list. Similarly, because sets are understood to be unordered, when we refer to a set of items we will always mean an unordered set. In short: Lists are ordered. Sets are unordered.
18 Permutations and Combinations Quick Example There are six permutations of the three letters a, b, c taken two at a time: 1. a, b; 2. b, a; 3. a, c; 4. c, a; 5. b, c; 6. c, b. There are three combinations of the three letters a, b, c taken two at a time: 1. {a, b}; 2. {a, c}; 3. {b, c}. There are six lists containing two of the letters a, b, c. There are three sets containing two of the letters a, b, c.
19 Permutations and Combinations How do we count the number of possible combinations of n items taken r at a time? The number of permutations is P(n, r ), but each set of r items occurs r ! times because this is the number of ways in which those r items can be ordered. So, the number of combinations is P(n, r )/r !.
20 Permutations and Combinations Combinations of n items taken r at a time The number of combinations of n items taken r at a time is given by We can also write
21 Permutations and Combinations Quick Example The number of combinations of six items taken two at a time is which we can also calculate as
22 Permutations and Combinations Note There are other common notations for C(n, r ). Calculators often have n C r. In mathematics we often write which is also known as a binomial coefficient. Because C(n, r ) is the number of ways of choosing a set of r items from n, it is often read n choose r. |
Factors and Multiples Solved Examples
# Factors and Multiples Solved Examples | The Complete SAT Course - Class 10 PDF Download
## What are Factors and Multiples?
Factors and multiples are mathematical concepts that are closely related. A factor of a number is a number that can divide the given number exactly without leaving any remainder. In other words, a factor is a divisor of a given number. On the other hand, a multiple of a number is the product of that number and any other integer. In simpler terms, if you can obtain a particular number by multiplying two or more numbers, then those numbers are factors of that number, and the result is a multiple of those numbers.
For example, if we have the numbers 5 and 3, we can obtain the number 15 by multiplying them together, which means that 5 and 3 are factors of 15. Hence, 15 is a multiple of both 5 and 3. This relationship between factors and multiples is fundamental in mathematics and is used in many different contexts, including algebra and number theory.
### Factors and Multiples Questions with Solutions
Here are the steps to find the factors of a number:
• Choose two numbers that, when multiplied together, give the target number.
• Test whether these two numbers divide the target number without leaving a remainder.
• If they do, then they are factors of the target number.
• Write down the pair of factors.
• Repeat steps 1-4 with other number pairs until you have found all the factors.
• Remember that every number has at least two factors: 1 and the number itself.
For example, to find the factors of 35, we can start by trying 5 and 7 as the number pairs. We test whether they divide 35 exactly, and since they do, we know that 5 and 7 are factors of 35. Therefore, we can write 35 as the product of 5 and 7, i.e., 35 = 5 x 7.
1. What are the factors of 9?
The factors of 9 are the numbers that can divide 9 exactly without leaving any remainder. These numbers are 1, 3, and 9. We can see this by listing all the possible pairs of numbers that multiply to give 9:
1 × 9 = 9
3 × 3 = 9
9 × 1 = 9
Therefore, the factors of 9 are 1, 3, and 9.
2. What is the sum of factors of 12?
To find the sum of factors of 12, we need to first list all the factors of 12. The factors of 12 are 1, 2, 3, 4, 6, and 12.
To find the sum of these factors, we simply add them up:
1 + 2 + 3 + 4 + 6 + 12 = 28
Therefore, the sum of factors of 12 is 28.
3. Find the common factors of 30 and 45.
To find the common factors of 30 and 45, we need to list all the factors of each number and identify the factors that they have in common.
The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, and 30.
The factors of 45 are: 1, 3, 5, 9, 15, and 45.
So, the common factors of 30 and 45 are 1, 3, 5, and 15. These are the numbers that can divide both 30 and 45 without leaving a remainder.
4. What is the greatest common factor of 3 and 15.
To find the greatest common factor (GCF) of 3 and 15, we need to list all the factors of each number and identify the largest factor that they have in common.
The factors of 3 are: 1 and 3.
The factors of 15 are: 1, 3, 5, and 15.
So, the greatest common factor of 3 and 15 is 3, which is the largest number that can divide both 3 and 15 without leaving a remainder. Therefore, the GCF of 3 and 15 is 3.
5. Find the greatest common factor of 20 and 6?
To find the greatest common factor (GCF) of 20 and 6, we need to list all the factors of each number and identify the largest factor that they have in common.
The factors of 20 are: 1, 2, 4, 5, 10, and 20.
The factors of 6 are: 1, 2, 3, and 6.
The common factors of 20 and 6 are: 1 and 2.
Therefore, the greatest common factor of 20 and 6 is 2, which is the largest number that can divide both 20 and 6 without leaving a remainder.
### How to Find Multiples?
To find the multiples of a number, we can multiply the number by different whole numbers. The resulting numbers are known as the multiples of that number. For instance, if we take the number 35, then all the numbers obtained by multiplying 35 by another whole number are the multiples of 35, such as 35, 70, 105, 140, and so on. One of the simplest ways to find multiples is to use the skip counting method, where we add the number repeatedly to obtain its multiples.
6. Find the first 10 multiples of 10.
To find the first 10 multiples of 10, we can use the skip counting method and add 10 to the previous multiple to obtain the next one.
The first multiple of 10 is 10 itself.
To get the next multiple, we add 10 to 10, which gives us 20.
To get the third multiple, we add 10 to 20, which gives us 30.
We can continue this process to obtain the first 10 multiples of 10:
1 × 10 = 10
2 × 10 = 20
3 × 10 = 30
4 × 10 = 40
5 × 10 = 50
6 × 10 = 60
7 × 10 = 70
8 × 10 = 80
9 × 10 = 90
10 × 10 = 100
Therefore, the first 10 multiples of 10 are 10, 20, 30, 40, 50, 60, 70, 80, 90, and 100.
7. What are the first five multiples of 7?
To find the first five multiples of 7, we can use the skip counting method and add 7 to the previous multiple to obtain the next one.
Therefore, the first five multiples of 7 are 7, 14, 21, 28, and 35.
8. Find the least common multiple of 12 and 18.
To find the least common multiple (LCM) of 12 and 18, we can use different methods such as listing out their multiples or using prime factorization. Here, we'll use the prime factorization method:
Step 1: Find the prime factorization of each number.
12 = 22 x 3
18 = 2 x 32
Step 2: Write down the prime factors with their highest powers:
22 x 32
Step 3: Multiply the factors:
LCM of 12 and 18 = 22 x 32 = 36
Therefore, the least common multiple of 12 and 18 is 36.
9. Find the common multiples of 3 and 5.
To find the common multiples of 3 and 5, we can use the skip counting method and look for the numbers that appear in both the multiples of 3 and 5.
The multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, ...
The multiples of 5 are: 5, 10, 15, 20, 25, 30, 35, 40, ...
We can see that the common multiples of 3 and 5 are 15, 30, 45, 60, ... and so on.
Therefore, the common multiples of 3 and 5 are any numbers that are multiples of both 3 and 5, such as 15, 30, 45, 60, and so on.
10. What are the first three common multiples 4 and 8?
To find the common multiples of 4 and 8, we can use the skip counting method and look for the numbers that appear in both the multiples of 4 and 8.
The multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, ...
The multiples of 8 are: 8, 16, 24, 32, 40, ...
We can see that the common multiples of 4 and 8 are 8, 16, and 24.
Therefore, the first three common multiples of 4 and 8 are 8, 16, and 24.
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How do you convert a decimal to a percent?
How do you convert a decimal to a percent?
To convert a decimal to a percentage, multiply by 100 (just move the decimal point 2 places to the right). For example, 0.065 = 6.5% and 3.75 = 375%.
Why do you think you must learn to convert fraction decimal and percentage how is it important in business mathematics?
Being able to convert between fractions, decimals, and percentages is an essential skill at Key Stage 3 as it greatly develops a student’s concept of what the quantities mean. It also makes fractions and percentages of amounts much simpler to visualise and compare.
Why do we convert fractions to decimals?
It is a way to represent a fraction by a set of common denominators: 1/10, 1/100… The advantage of decimals is that because all of them are in common denominators already, addition, subtraction and comparing are quite straightforward.
When we change fraction to decimal form what is the main operation involved?
To convert a fraction to a decimal, we’ll just divide the numerator…by the denominator. In our example, we’ll divide 1 by 4. 1 divided by 4 equals 0. To keep dividing, we’ll add a decimal point and a zero after the 1.
What is 0.050 as a percentage?
Number Converter
1 in __ Decimal Percent
1 in 1,000 0.0010 0.10%
1 in 2,000 0.00050 0.050%
1 in 3,000 0.00033 0.033%
1 in 4,000 0.00025 0.025%
What is 0.007 as a percent?
Decimal to percent conversion table
Decimal Percent
0.04 4%
0.05 5%
0.06 6%
0.07 7%
What is the relationship between fractions decimals and percents?
The Relationship between Fractions, Decimals, and Percents – Making Conversions. Since a percent is a ratio a ratio can be written as a fraction, and a fraction can be written as a decimal. This means any of these forms can be converted to any of the others.
What is the relationship between fractions decimals and percentages?
What is .00001 as a percent?
Number Converter
1 in __ Decimal Percent
1 in 10,000 0.00010 0.010%
1 in 25,000 0.00004 0.004%
1 in 50,000 0.00002 0.002%
1 in 100,000 0.00001 0.001%
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# Graphs of Exponential Functions
## Growth and decay functions with varying compounding intervals
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Practice Graphs of Exponential Functions
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Exponential Functions
Have you ever been to a laboratory or conducted an experiment? Take a look at this dilemma.
“We have been given a dilemma by my friend Professor Smith,” Mr. Travis said upon the class’ return to the classroom.
“Here we go, see what you can do with this,” Mr. Travis wrote the following problem on the board.
In a laboratory, one strain of bacteria can double in number every 15 minutes. Suppose a culture starts with 60 cells, use your graphing calculator or a table of values to show the sample’s growth after 2 hours. Use the function b=602q\begin{align*}b=60 \cdot 2^q\end{align*} where b\begin{align*}b\end{align*} is the number of cells there are after q\begin{align*}q\end{align*} quarter hours.
To work on this problem, you have to understand exponential functions. Pay close attention during this Concept and you will know how to solve it by the end of it.
### Guidance
Let's think about exponential functions by looking at the following situation.
Two girls in a small town once shared a secret, just between the two of them. They couldn’t stand it though, and each of them told three friends. Of course, their friends couldn’t keep secrets, either, and each of them told three of their friends. Those friends told three friends, and those friends told three friends, and son on... and pretty soon the whole town knew the secret. There was nobody else to tell!
These girls experienced the startling effects of an exponential function.
If you start with the two girls who each told three friends, you can see that they told six people or 23\begin{align*}2 \cdot 3\end{align*}.
Those six people each told three others, so that 63\begin{align*}6 \cdot 3\end{align*} or 233\begin{align*}2 \cdot 3 \cdot 3\end{align*}—they told 18 people.
Those 18 people each told 3, so that now is 183\begin{align*}18 \cdot 3\end{align*} or 2333\begin{align*}2 \cdot 3 \cdot 3 \cdot 3\end{align*} or 54 people.
You can see how this is growing and you could show the number of people told in each round of gossip with a function: y=abx\begin{align*}y=ab^x\end{align*} where y\begin{align*}y\end{align*} is the number of people told, a\begin{align*}a\end{align*} is the two girls who started the gossip, b\begin{align*}b\end{align*} is the number of friends that they each told, and x\begin{align*}x\end{align*} is the number of rounds of gossip that occurred.
This is called an exponential function—any function that can be written in the form y=abx\begin{align*}y=ab^x\end{align*}, where a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} are constants, a0,b>0\begin{align*}a \ne 0,b>0\end{align*}, and b1\begin{align*}b \ne 1\end{align*}.
As we did with linear and quadratic functions, we could make a table of values and calculate the number of people told after each round of gossip. Use the function y=23x\begin{align*}y=2 \cdot 3^x\end{align*} where y\begin{align*}y\end{align*} is the number of people told and x\begin{align*}x\end{align*} is the number of rounds of gossip that occurred.
x rounds of gossip 012 345y people told 261854162486
How can you tell if a function is an exponential function?
If your function can be written in the form y=abx\begin{align*}y=ab^x\end{align*}, where a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} are constants, a0,b>0,\begin{align*}a \ne 0, b>0,\end{align*} and b1\begin{align*}b \ne 1\end{align*}, then it must be exponential. In quadratic equations, your functions were always to the 2nd\begin{align*}2^{nd}\end{align*} power. In exponential functions, the exponent is a variable. Their graphs will have a characteristic curve either upward or downward.
Exponential Functions
1. y=2x\begin{align*}y=2^x\end{align*}
2. c=410d\begin{align*}c=4 \cdot 10^d\end{align*}
3. \begin{align*}y=2 \cdot \left(\frac{2}{3}\right)^x\end{align*}
4. \begin{align*}t=4 \cdot 10^u\end{align*}
Not Exponential Functions
Exponential functions can be graphed by using a table of values like we did for quadratic functions. Substitute values for \begin{align*}x\end{align*} and calculate the corresponding values for \begin{align*}y\end{align*}.
Take a look at this one.
Graph \begin{align*}y=2^x\end{align*}.
Here is the table.
\begin{align*}x\end{align*} \begin{align*}y=2^x\end{align*} \begin{align*}y\end{align*}
\begin{align*}-3\end{align*} \begin{align*}y=2^{-3}\end{align*} \begin{align*} \frac{1}{8}\end{align*}
-2 \begin{align*}y=2^{-2}\end{align*} \begin{align*} \frac{1}{4}\end{align*}
-1 \begin{align*}y=2^{-1}\end{align*} \begin{align*} \frac{1}{2}\end{align*}
0 \begin{align*}y=2^0\end{align*} 0
1 \begin{align*}y=2^1\end{align*} 2
2 \begin{align*}y=2^2\end{align*} 4
3 \begin{align*}y=2^3\end{align*} 8
4 \begin{align*}y=2^4\end{align*} 16
5 \begin{align*}y=2^5\end{align*} 32
6 \begin{align*}y=2^6\end{align*} 64
Notice that the shapes of the graphs are not parabolic like the graphs of quadratic functions. Also, as the \begin{align*}x\end{align*} value gets lower and lower, the \begin{align*}y\end{align*} value approaches zero but never reaches it. As the \begin{align*}x\end{align*} value gets even smaller, the \begin{align*}y\end{align*} value may get infinitely close to zero but will never cross the \begin{align*}x\end{align*}-axis.
Identify each function.
#### Example A
\begin{align*}y=4^x\end{align*}
Solution: Exponential function
#### Example B
\begin{align*}f(x)=2x-1\end{align*}
Solution: Linear function
#### Example C
\begin{align*}y=ax^2-bx+c\end{align*}
Now let's go back to the dilemma from the beginning of the Concept.
First, we can create a t-table to go with the equation of the function. Here are the values in that table.
\begin{align*}q\end{align*} \begin{align*}b\end{align*}
0 60
1 120
2 240
3 480
4 960
5 1920
6 3840
7 7680
8 15360
Now here is our graph.
### Vocabulary
Exponential Functions
results that expand exponentially. The graph curves upward or downward.
### Guided Practice
Here is one for you to try on your own.
Graph \begin{align*}y=2 \cdot \left( \frac{2}{3}\right)^x\end{align*}
\begin{align*}x\end{align*} \begin{align*}y\end{align*}
\begin{align*}-3\end{align*} \begin{align*} \frac{27}{4}\end{align*}
-2 \begin{align*} \frac{9}{2}\end{align*}
-1 3
0 2
1 \begin{align*} \frac{4}{3}\end{align*}
2 \begin{align*} \frac{8}{9}\end{align*}
3 \begin{align*} \frac{16}{27}\end{align*}
4 \begin{align*} \frac{32}{81}\end{align*}
5 \begin{align*} \frac{64}{243}\end{align*}
6 \begin{align*} \frac{128}{729}\end{align*}
### Practice
Directions: Classify the following functions as exponential or not exponential. If it is not exponential, state the reason why.
1. \begin{align*}y=7^x\end{align*}
2. \begin{align*}c=-2 \cdot 10^d\end{align*}
3. \begin{align*}y=1^x\end{align*}
4. \begin{align*}y=4^x\end{align*}
5. \begin{align*}n=0 \cdot \left(\frac{1}{2}\right)^x\end{align*}
6. \begin{align*}y=5 \cdot \left(\frac{4}{3}\right)^x\end{align*}
7. \begin{align*}y=(-7)^x\end{align*}
8. Use a table of values to graph the function \begin{align*}y=3^x\end{align*}.
9. Use a table of values to graph the function \begin{align*}y=\left(\frac{1}{3}\right)^x\end{align*}.
10. What type of graph did you make in number 7?
11. What type of graph did you make in number 8?
12. Use a table of values to graph the function \begin{align*}y=-2^x\end{align*}.
13. Use a table of values to graph the function \begin{align*}y=5^x\end{align*}.
14. Use a table of values to graph the function \begin{align*}y=-5^x\end{align*}.
15. Use a table of values to graph the function \begin{align*}y=6^x\end{align*}.
### Vocabulary Language: English
Asymptotic
Asymptotic
A function is asymptotic to a given line if the given line is an asymptote of the function.
Exponential Function
Exponential Function
An exponential function is a function whose variable is in the exponent. The general form is $y=a \cdot b^{x-h}+k$.
grows without bound
grows without bound
If a function grows without bound, it has no limit (it stretches to $\infty$).
Horizontal Asymptote
Horizontal Asymptote
A horizontal asymptote is a horizontal line that indicates where a function flattens out as the independent variable gets very large or very small. A function may touch or pass through a horizontal asymptote.
Transformations
Transformations
Transformations are used to change the graph of a parent function into the graph of a more complex function. |
# SOLUTION: Solve each system by the addition method. 3/7x+5/9y=27 1/9x+2/7y=7 and 3x-2.5y=7.125 2.5x-3y=7.3125
Algebra -> -> SOLUTION: Solve each system by the addition method. 3/7x+5/9y=27 1/9x+2/7y=7 and 3x-2.5y=7.125 2.5x-3y=7.3125 Log On
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Question 175921: Solve each system by the addition method. 3/7x+5/9y=27 1/9x+2/7y=7 and 3x-2.5y=7.125 2.5x-3y=7.3125Answer by Alan3354(34679) (Show Source): You can put this solution on YOUR website!Solve each system by the addition method. 3/7x+5/9y=27 1/9x+2/7y=7 Since it says "linear systems" I'll assume 3/7x means (3/7)x (3x/7) + (5y/9) = 27 Do them like this to avoid ambiguities. (x/9) + (2y/7) = 7 Multiply both eqns by 63 to eliminate fractions. 27x + 35y = 1701 7x + 18y = 441 Multiply the 1st eqn by 7 and the 2nd by 27 to get 189x in both 189x + 245y = 11907 189x + 486y = 11907 Subtract eqn 2 from eqn 1 0x - 241y = 0 y = 0 x = 11907/189 x = 63 ------------- and 3x-2.5y=7.125 2.5x-3y=7.3125 -------------- Multiply both by 16 48x - 40y = 114 40x - 48y = 117 All integers now Multiply the 1st eqn by 5, the 2nd by 6 240x - 200y = 570 240x - 288y = 702 Subtract the 2nd from the 1st 0x + 88y = -132 y = -1.5 ------------- Sub for y into 2.5x-3y=7.3125 2.5x -3*(-1.5) = 7.3125 2.5x + 4.5 = 7.3125 2.5x = 2.8125 x = 1.125 |
# Range vs. Interquartile Range: What’s the Difference?
In statistics, the range and interquartile range are two ways to measure the spread of values in a dataset.
The range measures the difference between the minimum value and the maximum value in a dataset.
The interquartile range measures the difference between the first quartile (25th percentile) and third quartile (75th percentile) in a dataset. This represents the spread of the middle 50% of values.
## Example: How to Calculate Range & Interquartile Range
Suppose we have the following dataset:
Dataset: 1, 4, 8, 11, 13, 17, 19, 19, 20, 23, 24, 24, 25, 28, 29, 31, 32
We can use the following steps to calculate the range:
• Range = Maximum value – Minimum value
• Range = 32 – 1
• Range = 31
We can use the Interquartile Range Calculator to help us calculate the interquartile range:
• Interquartile Range = 3rd Quartile – 1st Quartile
• Interquartile Range = 26.5 – 12
• Interquartile Range = 14.5
The range tells us the spread of the entire dataset while the interquartile range tells us the spread of the middle half of the dataset.
## Range vs. Interquartile Range: Similarities & Differences
The range and interquartile range share the following similarity:
• Both metrics measure the spread of values in a dataset.
However, the range and interquartile range have the following difference:
• The range tells us the difference between the largest and smallest value in the entire dataset.
• The interquartile range tells us the spread of the middle 50% of values in the dataset.
## Range vs. Interquartile Range: When to Use Each
We should use the range when we’re interested in understanding the difference between the largest and smallest values in a dataset.
For example, suppose a professor administers an exam to 100 students. She can use the range to understand the difference between the highest score and the lowest score received by all of the students in the class.
Conversely, we should use the interquartile range when we’re interested in understanding the spread between the 75th percentile and 25th percentile of a dataset.
For example, if a professor administers an exam to 100 students, she can use the interquartile range to quickly understand the difference in exam score between a student who scored at the 75th percentile of scores and a student who scored at the 25th percentile.
It’s worth noting that we don’t have to choose between using the range or the interquartile range to describe the spread of values in a dataset.
We can use both metrics since they provide us with completely different information.
## The Drawback of Using the Range
The range suffers from one drawback: It is influenced by outliers.
To illustrate this, consider the following dataset:
Dataset: 1, 4, 8, 11, 13, 17, 19, 19, 20, 23, 24, 24, 25, 28, 29, 31, 32
The range of this dataset is 32 – 1 = 31.
However, consider if the dataset had one extreme outlier:
Dataset: 1, 4, 8, 11, 13, 17, 19, 19, 20, 23, 24, 24, 25, 28, 29, 31, 32, 378
The range of this dataset would now be 378 – 1 = 377.
Notice how the range changes dramatically as a result of one outlier.
Before calculating the range of any dataset, it’s a good idea to first check if there are any outliers that could cause the range to be misleading. |
Chapter 4, Exponential and Logarithmic Functions - Section 4.1 - Exponential Functions - 4.1 Exercises: 23
$\color{blue}{f(x)=\left(\dfrac{1}{4}\right)^x}$
Work Step by Step
The graph of the function $f(x)=a^x$ contains the point $(2,\frac{1}{16})$. This means that when $x=2$, $y = \frac{1}{16}$. Substitute $x$ and $y$ into $f(x) =a^x$ to obtain: $\begin{array}{ccc} \\&f(x) &= &a^x \\&f(2) &= &a^{2} \\&\dfrac{1}{16} &= &a^{2}\end{array}$ Note that $\dfrac{1}{16} = \left(\dfrac{1}{4}\right)^2$. Thus, the expression above is equivalent to: $\left(\dfrac{1}{4}\right)^2=a^2$ Use the rule "$a^m=b^m \longrightarrow a=b$" to obtain: $\dfrac{1}{4} = a$ With $a=\frac{1}{4}$, the function whose graph is given is $\color{blue}{f(x)=\left(\dfrac{1}{4}\right)^x}$.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
# AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.3
AP State Syllabus AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.3 Textbook Questions and Answers.
## AP State Syllabus 9th Class Maths Solutions 11th Lesson Areas Exercise 11.3
Question 1.
In a triangle ABC, E is the midpoint of median AD. Show that
i) ar ΔABE = ar ΔACE
Solution:
In ΔABC; AD is a median.
∴ ΔABD = ΔACD …………….. (1)
(∵ Median divides a triangle in two equal triangles)
Also in ΔABD; BE is a median.
∴ ΔABE = ΔBED = $$\frac{1}{2}$$ΔABD …………..(2)
Also in ΔACD; CE is a median.
∴ ΔACE = ΔCDE = $$\frac{1}{2}$$ΔACD …………….(3)
From (1), (2) and (3);
ΔABE = ΔACE
(OR)
ΔABD = ΔACD [∵ AD is median in ΔABC]
$$\frac{1}{2}$$ ΔABD = $$\frac{1}{2}$$ ΔACD
[Dividing both sides by 2]
ΔABE = ΔAEC
[∵ BE is median of ΔABD, CE is median of ΔACD]
Hence proved.
ii) arΔABE = $$\frac{1}{2}$$ ar(ΔABC)
Solution:
ΔABE = $$\frac{1}{2}$$ (ΔABD)
[From (i); BE is median of ΔABD]
ΔABE = $$\frac{1}{2}$$ [$$\frac{1}{2}$$ ΔABC]
[∵ AD is median of ΔABC]
= $$\frac{1}{4}$$ ΔABC
Hence,proved.
Question 2.
Show that the diagonals of a paral¬lelogram divide it into four triangles of equal area.
Solution:
□ABCD is a parallelogram.
The diagonals AC and BD bisect each other at ‘O’.
ΔABC and □ABCD lie on the same base AB and between the same parallels AB//CD.
∴ ΔABC = $$\frac{1}{4}$$ □ABCD
Now in ΔABC; BO is a median
[∵ O is the midpoint of both diagonals AC and BD]
∴ ΔAOB = ΔBOC ………….(1)
[ ∵ Median divides a triangle into two triangles of equal area]
Similarly ∵ABD and □ABCD lie on the same base AB and between the same parallels AB//CD.]
∴ ΔABD = $$\frac{1}{2}$$□ ABCD
Also ΔAOB = ΔAOD …………..(2)
[ ∵ AO is the median of AABD]
From (1) & (2)
ΔAOB = ΔBOC = ΔAOD
Similarly we can prove that
ΔAOD = ΔCOD [ ∵ OD is the median of ΔACD]
∴ ΔAOB = ΔBOC = ΔCOD = ΔAOD
Hence proved.
Question 3.
In the figure, ΔABC and ΔABD are two triangles on the same base AB. If line segment CD is bisected by $$\overline{\mathbf{A B}}$$ at O, show that ar (ΔABC) = ar (ΔABD).
Solution:
From the figure, in ΔAOC; ΔBOD
OA = OB [ ∵ given]
∠AOC = ∠BOD [Vertically opp. angles]
∴ ΔAOC ≅ ΔBOD (SAS congruence)
Thus AC = BD (CPCT)
∠OAC = ∠OBD (CPCT) .
But these are alternate interior angles for the lines AC, BD.
∴ AC // BD
As AC = BD and AC // BD;
□ABCD is a parallelogram.
AB is a diagonal of oABCD
⇒ ΔABC ≅ ΔABD
(∵ diagonal divides a parallelogram into two congruent triangles)
∴ ar(ΔABC) = ar (ΔABD)
Question 4.
In the figure ΔABC; D, E and F are the midpoints of sides BC, CA and AB respectively. Show that
i)BDEF is a parallelogram
ii) ar (ΔDEF) = $$\frac { 1 }{ 4 }$$ ar(ΔABC)
iii) ar (BDEF) = $$\frac { 1 }{ 2 }$$ ar(ΔABC)
Solution:
i) In ΔABC; D, E and F are the mid¬points of the sides.
∴ EF//BC FD//AC ED//AB
EF = $$\frac { 1 }{ 2 }$$BC FD = $$\frac { 1 }{ 2 }$$AC ED = $$\frac { 1 }{ 2 }$$AB
[ ∵ line joining the mid points of any two sides of a triangle is parallel to third side and equal to half of it]
∴ In □BDEF
BD = EF [ ∵ D is mid point of BC and $$\frac { 1 }{ 2 }$$ BC = EF]
DE = BF
∴ □BDEF is a parallelogram.
ii) □BDEF is a parallelogram (from (i))
Thus ΔBDF = ΔDEF
Similarly □CDFE; □AEDF are also parallelograms.
∴ ΔDEF = ΔCDE =ΔAEF
∴ ΔABC = ΔAEF+ ΔBDF + ΔCDF + ΔDEF
= 4ΔDEF
⇒ ΔDEF = $$\frac { 1 }{ 4 }$$ΔABC
iii) □BDEF = 2 ΔDEF …………..(1)
(from (ii))
ΔABC = 4 ΔDEF (2)
(from (ii))
From (1) & (2);
ΔABC = 2 (2ΔDEF) = 2 □BDEF
Hence proved.
Question 5.
In the figure D, E are points on the sides AB and AC respectively of ΔABC such that ar (ΔDBC) = ar (ΔEBC). Prove that DE // BC.
Solution:
ΔDBC = ΔEBC
The two triangles are on the same base BC and between the same pair of lines BC and DE.
As they are equal in area.
∴ BC // DE.
Question 6.
In the figure, XY is a line parallel to BC is drawn through A. If BE // CA and CF // BA are drawn to meet XY at E and F respectively. Show that ar (ΔABE) = ar (ΔACF).
Solution:
Given that XY//BC; BE//CA; CF//BA
In quad ABCF; AB//CF and BC//AF
Hence □ABCF is a parallelogram.
Also in □ACBE ; BC//AE and AC//BE
Hence □ACBE is a parallelogram.
Now in □ABCF and □ACBE
ΔABC = ΔACF …………..(1);
ΔABC = ΔABE …………..(2)
[∵ Diagonal divides a parallelogram into two congruent triangles]
∴ ΔACF = ΔABE [from (1) & (2)]
Hence proved.
Question 7.
In the figure, diagonals AC and BD of a trapezium ABCD with AB//DC inter¬sect each other at ‘O’. Prove that ar (ΔAOD) = ar (ΔBOC)
Solution:
Given that AB // CD
Now ΔADC and ΔBCD are on the same base and between the same parallels AB // CD.
∴ ΔADC = ΔBCD
⇒ ΔADC – ΔCOD = ΔBCD – ΔCOD
⇒ ΔAOD = ΔBOC [from the figure]
Question 8.
In the figure ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ΔACB) = ar (ΔACF)
(ii) ar (AEDF) = ar (ABCDE)
Solution:
ABCDE is a pentagon.
AC//BF
i) ΔACB and ΔACF are on the same base AC and between the same parallels AC//BF.
∴ ΔACB = ΔACF
ii) □AEDF = □AEDC + ΔACF
= □AEDC + ΔABC
[ ∵ ΔACF = ΔACB]
= area (ABCDE)
Hence proved.
Question 9.
In the figure, if ar (ΔRAS) = ar (ΔRBS) and ar (ΔQRB) = ar (ΔPAS) then show that both the quadrilaterals PQSR and RSBA are trapeziums.
ΔRAS = ΔRBS ………….. (1)
Both the triangles are on the same base RS and between the same pair of lines RS and AB.
As their areas are equal RS must be parallel to AB.
⇒ RS//AB
∴ □ABRS is a quadrilateral in which AB//RS.
∴ □ABRS (or) □RSBA is a trapezium.
Now AQRB = APAS (given)
⇒ ΔQRB – ΔRBS = ΔPAS – ΔRAS
[from (1) ΔRBS = ΔRAS]
⇒ ΔQRS = ΔPRS
These two triangles are on the same base RS and between the same pair of lines RS and PQ.
As these two triangles have same area RS must be parallel to PQ.
⇒ RS // PQ
Now in quad PQRS; PQ//RS.
Hence □PQRS is a trapezium.
Question 10.
A villager Ramayya has a plot of land in the shape of a quadrilateral. The grampanchayat of the village decided to take over some portion of his plot from one of the comers to construct a school. Ramayya agrees to the above proposal with the condition that he should be given equal amount of land in exchange of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will implemented. (Draw a rough sketch of the plot.)
Solution:
Let □ABCD is the plot of Ramayya.
School be constructed in the region ΔMCD where M is a point on BC such that □ABCD ≅ ΔADE
Draw the diagonal BD.
Draw a line parallel to BD through C which meets AB produced at E.
Join D, E
ΔADE is the required triangle.
Analysis:
ΔCED and ΔCEB are on the same base CE and between the same parallels CE and DB.
∴ ΔCED = ΔCEB [also from the figure]
ΔCEM + ΔCMD = ΔCEM + ΔBME
∴ ΔCMD = ΔBME
∴ ΔADE = □ABCD |
Multiplying fractions, decimals and percentages
# Multiplying fractions, decimals and percentages
Multiplying fractions, decimals and percentages
7.7
MICHAEL ANDERSON: Hello, and welcome to week 3.
10.6
PAULA KELLY: Last week, we saw how to express a fraction in many different ways, including as a decimal and a percentage. We also explored the different methods when converting between fractions, decimals, and percentages.
23.2
MICHAEL ANDERSON: This week we’ll be concentrating on multiplication. We’ll discover what it means to multiply a number by a fraction, decimal, or a percent. We’ll also work through a range of multiplication methods and the ideas behind them.
36.1
PAULA KELLY: We’ll look at different strategies that can be employed when multiplying with decimals and fractions, and how to find a percentage of an amount– for example, how to find 60% of 25.
47.6
MICHAEL ANDERSON: As we go through each step, you’ll find a range of examples, as well as questions for you to try yourself. We’ve also linked to some traditional classroom-based resources you may wish to explore in your own teaching.
59.1
PAULA KELLY: Don’t forget to join in the discussion, post your answers thoughts and questions in the sections below each step.
This week we’ll be looking at multiplying fractions, decimals and percentages. This allows us to answer questions such as:
• What is $$\frac{2}{3}$$ of $$\frac{4}{5}$$?
• What is 60% of 25?
There are many different ways to find the product of two or more numbers. Finding the product is what we mean by finding the answer when two or more numbers are multiplied together. Some methods are more efficient than other methods.
By the end of this week you’ll have practised a range of methods and will be able to choose methods for multiplying fractions, decimals and percentages. Importantly, we’ll be highlighting the issues with relying on ‘trick’ methods.
You can download this week’s problem sheet questions at the bottom of this first unit.
## Starter Activity
Without using a calculator, find the product of 62 and 31. In other words perform the calculation: $$62 \times 31$$.
When you have an answer you may like to check it using a calculator.
Now think of a different method of multiplying these numbers together.
Can you talk through the different methods you’ve used and explain each of the methods?
Now look back at the different methods used. Do you think the method you used can be extended to help you perform the calculation $$0.62 \times 0.31$$? |
Reducing Algebra Fractions by -1
• Aug 23rd 2008, 08:30 AM
cmf0106
Reducing Algebra Fractions by -1
The book is trying to show me that reducing a fraction or adding two fractions sometimes only requires that -1 be factored from one or more denominators. Here is the example they give:
$\displaystyle \frac{y-x}{x-y}=\frac{y-x}{-(-x+y)}=\frac{y-x}{-(y-x)}=\frac{1}{-1}=-1$
1.In the first step here I understand factoring out the "-" (the -1 value) because if the variable does not have a number coefficient we can assume its one so this part is fine.
$\displaystyle \frac{y-x}{-(-x+y)}$
2.Here$\displaystyle \frac{y-x}{-(-x+y)}$
once we distribute the "-" we get a negative y and a positive x, therefore we can rearrange the denominator to match the numerator in $\displaystyle \frac{y-x}{-(y-x)}$
3.Next we have $\displaystyle \frac{1}{-1}$ Now here is where I have a problem how can we know the values of both of these are 1 and -1? I am not certain how this is accomplished.
Does my logic up to 3's question look correct also?
• Aug 23rd 2008, 08:36 AM
Moo
Hi :)
Quote:
Originally Posted by cmf0106
The book is trying to show me that reducing a fraction or adding two fractions sometimes only requires that -1 be factored from one or more denominators. Here is the example they give:
$\displaystyle \frac{y-x}{x-y}=\frac{y-x}{-(-x+y)}=\frac{y-x}{-(y-x)}=\frac{1}{-1}=-1$
1.In the first step here I understand factoring out the "-" (the -1 value) because if the variable does not have a number coefficient we can assume its one so this part is fine.
$\displaystyle \frac{y-x}{-(-x+y)}$
2.Here$\displaystyle \frac{y-x}{-(-x+y)}$
once we distribute the "-" we get a negative y and a positive x, therefore we can rearrange the denominator to match the numerator in $\displaystyle \frac{y-x}{-(y-x)}$
What you said for 1. is not wrong. But I'd say they made these two transformations to show you that there is the common factor y-x.
But what you said in 2. looks incorrect to me. They don't distribute - again, they just use the commutativity of the addition : a+b=b+a. Here : -x+y=y+(-x)=y-x.
Quote:
3.Next we have $\displaystyle \frac{1}{-1}$ Now here is where I have a problem how can we know the values of both of these are 1 and -1? I am not certain how this is accomplished.
Relevant question for someone who begins ^^ It's so good seeing you trying to understand !
We have $\displaystyle \frac{y-x}{-(y-x)}$. This can be rewritten this way :
$\displaystyle \frac{1*{\color{red}(y-x)}}{(-1)*{\color{red}(y-x)}}$ and hence the result. Does it look correct to you ?
When you'll advance into maths lessons, maybe you will learn that most of these steps are only here to guide you to the solution :)
• Aug 23rd 2008, 08:40 AM
cmf0106
Thanks Moo! I will chew on that for awhile and let you know what the outcome is, I think it makes sense let me try a few more practice problems and we will see.
as for
"What you said for 1. is not wrong. But I'd say they made these two transformations to show you that there is the common factor y-x.
But what you said in 2. looks incorrect to me. They don't distribute - again, they just use the commutativity of the addition : a+b=b+a. Here : -x+y=y+(-x)=y-x."
Yah sorry I didnt mean to phrase it that way, it wouldnt make sense anyways because we factor out the "-", if we factored it out one step and then the next we redistributed it that would defeat the purpose of factoring (Itwasntme)
• Aug 23rd 2008, 09:02 AM
cmf0106
Alright moo update, this stuff is really putting me in a rut.
$\displaystyle \frac{3}{y-x}+\frac{x}{x-y}$
1.The book solves it $\displaystyle \frac{3}{y-x}+\frac{x}{x-y}=\frac{3}{-(-y+x)}+\frac{x}{x-y}=\frac{3}{-(x-y)}+\frac{x}{x-y}=\frac{-3}{x-y}+\frac{x}{x-y}=\frac{-3+x}{x-y}$
2. I started solving it like this, which looks different from the book $\displaystyle \frac{3}{y-x}+\frac{x}{x-y}=\frac{3}{-(-x+y)}+\frac{x}{x-y}=\frac{3+x}{(x-y)(x-y)}$
I used $\displaystyle \frac{3}{-(-x+y)}$ so it would yield the other fractions denominator $\displaystyle \frac{x}{x-y}$
Here $\displaystyle \frac{3+x}{(x-y)(x-y)}$ the two $\displaystyle (x-y)$ can merge into one but Im not understanding why the 3 is supposed to be -3 here, this is assuming my way of solving for it is correct in the first place.
• Aug 23rd 2008, 09:10 AM
Moo
Quote:
Originally Posted by cmf0106
Alright moo update, this stuff is really putting me in a rut.
Quote:
$\displaystyle \frac{3}{y-x}+\frac{x}{x-y}$
2. I started solving it like this, which looks different from the book $\displaystyle \frac{3}{y-x}+\frac{x}{x-y}=\frac{3}{-(-x+y)}+\frac{x}{x-y}=\frac{3+x}{(x-y)(x-y)}$
First problem in the first equality : y-x=-x+y, it's not equal to -(-x+y)
The purpose of this is to make appear the other fraction's denominator x-y. You can see that -(x-y)=-x+y=y-x.
So $\displaystyle \frac{3}{y-x}=\frac{3}{-(x-y)}$
Now if you want to know why it becomes $\displaystyle \frac{-3}{x-y}$ :
$\displaystyle \frac{3}{-(x-y)}=\frac{3}{-1} \cdot \frac{1}{x-y}=-3 \cdot \frac{1}{x-y}=\frac{-3}{x-y}$
See ? :)
Now, there is also a problem with the second equality. You should remember that :
$\displaystyle \frac ac+\frac bc=\frac{a+b}{c}$ and there is no multiplication involved. The multiplication occurs if there is not the same denominator ;)
• Aug 23rd 2008, 09:24 AM
cmf0106
But I could chose to make either both denominators appear as $\displaystyle y-x$ or $\displaystyle x-y$ correct?
• Aug 23rd 2008, 09:27 AM
Moo
Quote:
Originally Posted by cmf0106
But I could chose to make either both denominators appear as $\displaystyle y-x$ or $\displaystyle x-y$ correct?
Correct :)
It'd be nice if you could show how you would have done for y-x (Tongueout) (unless you have many more exercises ^^)
• Aug 23rd 2008, 10:07 AM
cmf0106
Quote:
Originally Posted by Moo
Correct :)
It'd be nice if you could show how you would have done for y-x (Tongueout) (unless you have many more exercises ^^)
Man this is probably the hardest thing Ive encountered so far, hard for me to wrap my brain around. Anyways lets give the other denominator a shot
so $\displaystyle \frac{3}{y-x}+\frac{x}{x-y}$
Lets Change$\displaystyle \frac{3}{y-x}=-(y-x)=-y+x$ this equals our other denominator $\displaystyle x-y$
next$\displaystyle =\frac{3}{-(y-x)}=\frac{-3+x}{x-y}$
• Aug 23rd 2008, 03:13 PM
masters
Quote:
Originally Posted by cmf0106
Man this is probably the hardest thing Ive encountered so far, hard for me to wrap my brain around. Anyways lets give the other denominator a shot
so $\displaystyle \frac{3}{y-x}+\frac{x}{x-y}$
Lets Change$\displaystyle {\color{red}\frac{3}{y-x}=\frac{3}{-(y-x)}=\frac{3}{-y+x}}$ this equals our other denominator $\displaystyle x-y$
next$\displaystyle =\frac{3}{-(y-x)}=\frac{-3+x}{x-y}={\color{red}\frac{x-3}{x-y}}$ |
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# Solving Inequalities
After studying this lesson, you will be able to:
• Solve inequalities.
The inequality symbols are:
> greater than
greater than or equal to
< less than
less than or equal to
The steps for solving inequalities are the same as those for solving equations:
1. Remove parentheses
2. Collect like terms on each side of the inequality symbol
3. Get the variables together on one side
4. Isolate the variable
5. Check
** There is one major difference in solving inequalities than equations: WHEN MULTIPLYING OR DIVIDING BY A NEGATIVE, REVERSE THE INEQUALITY SYMBOL.
Example 1
x + 8 > 10 x + 8 - 8 > 10 - 8 Subtract 8 from each side x > 2 This is the solution
Check by substituting a number greater than 2 into the original inequality:
3 + 8 >10
11>10 it checks!
Example 2
y - 4 6 y - 4 + 4 6 + 4 add 4 to each side y 10
Check by substituting a number less than or equal to 10 into the original inequality:
9 - 4 6
5 6
Example 3
6x - 10 26 6x - 10 + 10 26 + 10 Add 10 to each side 6x 36 Divide each side by 6Notice we divided by positive 6 so the inequality Symbol stays the same x 6 This is the solution...check by substituting into the original inequality |
# 11.6 Solving systems with gaussian elimination (Page 3/13)
Page 3 / 13
## Solving a $\text{\hspace{0.17em}}2×2\text{\hspace{0.17em}}$ System by gaussian elimination
Solve the given system by Gaussian elimination.
First, we write this as an augmented matrix.
We want a 1 in row 1, column 1. This can be accomplished by interchanging row 1 and row 2.
${R}_{1}↔{R}_{2}\to \left[\begin{array}{rrr}\hfill 1& \hfill -1& \hfill \\ \hfill 2& \hfill 3& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill \frac{1}{2}\\ \hfill & \hfill 6\end{array}\right]$
We now have a 1 as the first entry in row 1, column 1. Now let’s obtain a 0 in row 2, column 1. This can be accomplished by multiplying row 1 by $\text{\hspace{0.17em}}-2,$ and then adding the result to row 2.
$-2{R}_{1}+{R}_{2}={R}_{2}\to \left[\begin{array}{rrr}\hfill 1& \hfill -1& \hfill \\ \hfill 0& \hfill 5& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill \frac{1}{2}\\ \hfill & \hfill 5\end{array}\right]$
We only have one more step, to multiply row 2 by $\text{\hspace{0.17em}}\frac{1}{5}.$
$\frac{1}{5}{R}_{2}={R}_{2}\to \left[\begin{array}{rrr}\hfill 1& \hfill -1& \hfill \\ \hfill 0& \hfill 1& \hfill \end{array}|\begin{array}{cc}& \frac{1}{2}\\ & 1\end{array}\right]$
Use back-substitution. The second row of the matrix represents $\text{\hspace{0.17em}}y=1.\text{\hspace{0.17em}}$ Back-substitute $\text{\hspace{0.17em}}y=1\text{\hspace{0.17em}}$ into the first equation.
The solution is the point $\left(\frac{3}{2},1\right).$
Solve the given system by Gaussian elimination.
$\left(2,\text{\hspace{0.17em}}1\right)$
## Using gaussian elimination to solve a system of equations
Use Gaussian elimination to solve the given $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}×\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ system of equations .
Write the system as an augmented matrix .
Obtain a 1 in row 1, column 1. This can be accomplished by multiplying the first row by $\text{\hspace{0.17em}}\frac{1}{2}.$
Next, we want a 0 in row 2, column 1. Multiply row 1 by $\text{\hspace{0.17em}}-4\text{\hspace{0.17em}}$ and add row 1 to row 2.
The second row represents the equation $\text{\hspace{0.17em}}0=4.\text{\hspace{0.17em}}$ Therefore, the system is inconsistent and has no solution.
## Solving a dependent system
Solve the system of equations.
$\begin{array}{l}3x+4y=12\\ 6x+8y=24\end{array}$
Perform row operations on the augmented matrix to try and achieve row-echelon form .
$A=\left[\begin{array}{llll}3\hfill & \hfill & 4\hfill & \hfill \\ 6\hfill & \hfill & 8\hfill & \hfill \end{array}|\begin{array}{ll}\hfill & 12\hfill \\ \hfill & 24\hfill \end{array}\right]$
$\begin{array}{l}\hfill \\ \begin{array}{l}-\frac{1}{2}{R}_{2}+{R}_{1}={R}_{1}\to \left[\begin{array}{llll}0\hfill & \hfill & 0\hfill & \hfill \\ 6\hfill & \hfill & 8\hfill & \hfill \end{array}|\begin{array}{ll}\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\hfill \\ \hfill & 24\hfill \end{array}\right]\hfill \\ {R}_{1}↔{R}_{2}\to \left[\begin{array}{llll}6\hfill & \hfill & 8\hfill & \hfill \\ 0\hfill & \hfill & 0\hfill & \hfill \end{array}|\begin{array}{ll}\hfill & 24\hfill \\ \hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\hfill \end{array}\right]\hfill \end{array}\hfill \end{array}$
The matrix ends up with all zeros in the last row: $\text{\hspace{0.17em}}0y=0.\text{\hspace{0.17em}}$ Thus, there are an infinite number of solutions and the system is classified as dependent. To find the generic solution, return to one of the original equations and solve for $\text{\hspace{0.17em}}y.$
So the solution to this system is $\text{\hspace{0.17em}}\left(x,3-\frac{3}{4}x\right).$
## Performing row operations on a 3×3 augmented matrix to obtain row-echelon form
Perform row operations on the given matrix to obtain row-echelon form.
The first row already has a 1 in row 1, column 1. The next step is to multiply row 1 by $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ and add it to row 2. Then replace row 2 with the result.
$-2{R}_{1}+{R}_{2}={R}_{2}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -3& \hfill & \hfill 4& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill -3& \hfill & \hfill 3& \hfill & \hfill 4& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 3\\ \hfill & \hfill 0\\ \hfill & \hfill 6\end{array}\right]$
Next, obtain a zero in row 3, column 1.
$3{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -3& \hfill & \hfill 4& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill 0& \hfill & \hfill -6& \hfill & \hfill 16& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 3\\ \hfill & \hfill 0\\ \hfill & \hfill 15\end{array}\right]$
Next, obtain a zero in row 3, column 2.
$6{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -3& \hfill & \hfill 4& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 4& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 3\\ \hfill & \hfill 0\\ \hfill & \hfill 15\end{array}\right]$
The last step is to obtain a 1 in row 3, column 3.
Write the system of equations in row-echelon form.
$\left[\text{\hspace{0.17em}}\begin{array}{ccc}\text{\hspace{0.17em}}1& -\frac{5}{2}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{5}{2}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}1& 5\\ \text{\hspace{0.17em}}0& \text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}1\end{array}\text{\hspace{0.17em}}|\begin{array}{c}\frac{17}{2}\\ 9\\ 2\end{array}\right]$
## Solving a system of linear equations using matrices
We have seen how to write a system of equations with an augmented matrix , and then how to use row operations and back-substitution to obtain row-echelon form . Now, we will take row-echelon form a step farther to solve a 3 by 3 system of linear equations. The general idea is to eliminate all but one variable using row operations and then back-substitute to solve for the other variables.
## Solving a system of linear equations using matrices
Solve the system of linear equations using matrices.
$\begin{array}{c}\begin{array}{l}\hfill \\ \hfill \\ x\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}z=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8\hfill \end{array}\\ 2x\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3y\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}z=-2\\ 3x\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2y\text{\hspace{0.17em}}\text{\hspace{0.17em}}-9z=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}9\end{array}$
First, we write the augmented matrix.
Next, we perform row operations to obtain row-echelon form.
$\begin{array}{rrrrr}\hfill -2{R}_{1}+{R}_{2}={R}_{2}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 5& \hfill & \hfill -3& \hfill \\ \hfill 3& \hfill & \hfill -2& \hfill & \hfill -9& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -18\\ \hfill & \hfill 9\end{array}\right]& \hfill & \hfill & \hfill & \hfill -3{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 5& \hfill & \hfill -3& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -18\\ \hfill & \hfill -15\end{array}\right]\end{array}$
The easiest way to obtain a 1 in row 2 of column 1 is to interchange $\text{\hspace{0.17em}}{R}_{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{R}_{3}.$
$\text{Interchange}\text{\hspace{0.17em}}{R}_{2}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{R}_{3}\to \left[\begin{array}{rrrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill & \hfill 8\\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill & \hfill -15\\ \hfill 0& \hfill & \hfill 5& \hfill & \hfill -3& \hfill & \hfill -18\end{array}\right]$
Then
$\begin{array}{l}\\ \begin{array}{rrrrr}\hfill -5{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 57& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -15\\ \hfill & \hfill 57\end{array}\right]& \hfill & \hfill & \hfill & \hfill -\frac{1}{57}{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -1& \hfill & \hfill 1& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -12& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 1& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 8\\ \hfill & \hfill -15\\ \hfill & \hfill 1\end{array}\right]\end{array}\end{array}$
The last matrix represents the equivalent system.
Using back-substitution, we obtain the solution as $\text{\hspace{0.17em}}\left(4,-3,1\right).$
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
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test for convergence the series 1+x/2+2!/9x3
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100 meters
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e power cos hyperbolic (x+iy)
10y
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why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
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Solve 2cos x + 3sin x = 0.5 |
Advice 1: How to find the value of
Numeric expressions are constructed from numbers, arithmetic signs and brackets. If the expression contains variables, it will be called algebraic. Is a trigonometric expression that contains a variable under the signs of trigonometric functions. Tasks for the definition of numeric, trigonometric, algebraic expressions often found in school mathematics course.
Instruction
1
To find the value of a numeric expression, determine the order of actions in a given example. For convenience, label it with a pencil over the appropriate signs. Will follow all of these steps in a certain order: the actions in parentheses, exponentiation, multiplication, division, addition, subtraction. The resulting number will be the value of a numeric expression.
2
Example. Find the value of the expression (34∙10+(489-296)∙8):4-410. Determine the order of action. The first action is run in the inner brackets 489-296=193. Then, multiply 193∙8=1544 34∙10=340. Next activity: 340+1544=1884. Then follow the division 1884:4=461, and then subtract 461-410=60. You have found the value of this expression.
3
To find the value of trigonometric expressions at a known angle α, the pre - simplify the expression. To do this, apply the appropriate trigonometric formula. Calculate the values of trigonometric functions, substitute them in the example. Follow the steps.
4
Example. Find the value of the expression 2sin 30 ° ∙cos 30º∙tg 30º∙ctg 30º. Simplify this expression. To do this, use the formula tg α∙ctg α=1. Get: 2sin 30 ° ∙cos 30º∙1=2sin 30 ° ∙cos 30º. It is known that sin 30º=1/2 cos 30º=√3/2. Therefore, 2sin 30 ° ∙cos 30º=2∙1/2∙√3/2=√3/2. You have found the value of this expression.
5
The value of an algebraic expression depends on the value of the variable. To find the value of algebraic expressions with given variables, simplify the expression. Substitute variable values. Take the appropriate action. In the end, you will receive a number, and that is the value of algebraic expressions with given variables.
6
Example. Find the value of expression 7(a+y)-3(2a+3y) when a=21 and y=10. Simplify this expression, we get: a–2y. Substitute appropriate values for the variables and calculate: a–2y=21-2∙10=1. This is the value of the expression 7(a+y)-3(2a+3y) when a=21 and y=10.
Note
There are algebraic expressions that have no meaning for some values of the variables. For example, the expression x/(7–a) does not make sense if a=7, since the denominator becomes zero.
Advice 2 : How to simplify an expression in math
Learn how to simplify expressions in math just need to correctly and quickly solve the tasks, different equations. Simplify the expression implies a reduction in the number of actions, which facilitates the calculation and saves time.
Instruction
1
Learn how to calculate the degree with the natural indicators. When multiplying degrees with the same grounds get the degree number, the base of which remains the same, and the exponents are added b^m+b^n=b^(m+n). By dividing the degrees with the same bases receive the degree number, the base of which remains the same, and the exponents are subtracted, and the rate of the dividend is subtracted the index divisor of b^m:b^n=b^(m-n). During the construction of the degree in the degree obtained degree number, the base of which remains the same, and the two values are multiplied (b^m)^n=b^(mn)in the exponentiation of products of numbers in this degree is built every multiplier.(abc)^m=a^m*b^m*c^m
2
Play the polynomials into factors, i.e. imagine them as a product of multiple factors of polynomials and monomials. Take out a common factor of the brackets. Learn the basic formulas of reduced multiplication: the difference of squares, square sums, square differences, sum of cubes, difference of cubes, cube of sum and difference. For example, m^8+2*m^4*n^4+n^8=(m^4)^2+2*m^4*n^4+n^4)^2. These are the basic formulas to simplify expressions. Use the method of separating the complete square trinomial of the form ax^2+bx+c.
3
As often as possible, reduce fractions. For example, (2*a^2*b)/(a^2*b*c)=2/(a*c). But remember that you can cut only the multipliers. If the numerator and denominator of algebraic fractions to multiply by the same number other than zero, then the value of the fraction will not change. To convert a rational expression in two ways: chain and action. The second method is preferable because it is easier to check the results of the intermediate actions.
4
Often in expressions it is necessary to extract the roots. The roots of even degree is extracted from only non-negative expressions or numbers. The roots of an odd degree is retrieved from any expressions.
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# Binomial Probability & Binomial Experiments
Instructor: Betsy Chesnutt
Betsy teaches college physics, biology, and engineering and has a Ph.D. in Biomedical Engineering
Binomial probability can be used to determine the likelihood of a certain outcome in an experiment where there are only two possible outcomes (success and failure). In this lesson, learn how to apply binomial probability to a variety of situations.
## Binomial Experiments
If you flip a coin ten times, what is the probability that you will get heads exactly eight times? It doesn't seem very likely that you would get exactly eight heads in ten flips, but it's certainly not impossible. Binomial probability can help you determine exactly HOW likely it is to get ANY number of heads (or tails) in a coin flip experiment like this.
In order to use binomial probability to determine the likelihood that an event will occur, you first need to determine if the experiment is a binomial experiment or not. In order for an experiment to be considered a binomial experiment, each trial can only result in two possible outcomes. A coin flip is a great example of this, because every time you flip the coin, it has to land on either heads or tails. The prefix bi always means two, so think of this to remember that a binomial experiment is one in which there are two possible outcomes.
It's also important that the probability of success (represented by the symbol p) is the same for all the trials. For the coin flip, there is a 50% chance that it will land on heads every time you flip it, so the coin flip experiment meets this condition, too!
Finally, no matter how many trials are performed (n = the number of trials), each trial must not be affected by the outcome of any other trials. This is also true for the coin flip experiment, because each time you flip the coin, the outcome is not affected by the results of any of the previous trials.
Your coin flip experiment definitely meets all of these requirements, so now we know that it IS a binomial experiment.
## Binomial Probability
To determine the probability of getting heads exactly eight times in ten coin flips, you first need to know how many different ways there are to get exactly eight heads. You can do this by finding the total number of combinations that will give you number of successes that you want.
Next, determine the probability of success and the probability of failure. In this case, there is a 50% chance of success with each flip, so the probability of success is 0.5 and the probability of failure is 0.5. You can use these to calculate the probability of getting exactly eight heads in any one of the 45 possible combinations. Let's look at one of those possibilities to see how this works. One way to get eight heads is to have the first eight flips turn up heads and the last two turn up tails. What is the probability of this exact thing happening?
Since there are 45 different ways to get eight heads, you can find the probability of getting ANY one of these 45 different possibilities by multiplying the probability of getting one of these outcomes by the total number of combinations (45).
So, the probability of getting exactly eight heads when you flip a coin ten times is 0.0439 or 4.39%.
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# Quadratic Formula
##### Add yours
Algebra - Quadratic - Derive Quadratic Formula
Tip: This isn't the place to ask a question because the teacher can't reply.
1 of 3 videos by AJ Speller
## Key Questions
• Suppose you have $a {x}^{2} + b x + c = 0$ where a, b and c is any number. Plug these in the quadratic formula:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ Plug the values and solve for $x$.
• Since the quadratic formula is derived from the completing the square method, which always works. Note that factoring always works as well, but it is sometimes just very difficult to do it.
I hope that this was helpful.
• Your company is going to make frames as part of a new product they are launching.
The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be $28 c {m}^{2}$
The inside of the frame has to be 11 cm by 6 cm
What should the width x of the metal be?
Area of steel before cutting:
Area = (11 + 2x) × (6 + 2x) cm^2
Area = $66 + 22 x + 12 x + 4 {x}^{2}$
Area = $4 {x}^{2} + 34 x + 66$
Area of steel after cutting out the 11 × 6 middle:
Area = $+ 34 x + 66 - 66$
Area = $4 {x}^{2} + 34 x$
Let us solve this one graphically!
Here is the graph of 4x2 + 34x :
The desired area of 28 is shown as a horizontal line.
The area equals 28 cm2 when:
x is about -9.3 or 0.8
The negative value of x make no sense, so the answer is:
x = 0.8 cm (approx.)
You can try it out yourself and find the roots of the quadratic equation using the quadratic formula
x = [ -b ± sqrt(b^2 - 4ac) ] / (2a
There are many more examples at http://www.mathsisfun.com/algebra/quadratic-equation-real-world.html
• The quadratic formula is $x = \left(- b \pm \sqrt{{b}^{2} - 4 a c}\right) \div 2 a$.
1. Identify the values of $a$, $b$, and $c$ in $a {x}^{2} + b x + c = 0$.
2. Substitute each value into the formula $x = \left(- b \pm \sqrt{{b}^{2} - 4 a c}\right) \div 2 a$.
3. Simplify for the values of $- b$, ${b}^{2}$, $- 4 a$, and $2 a$.
4. Simplify for the value of $- 4 a c$.
5. Add product of $- 4 a c$ to ${b}^{2}$.
6. Simplify for the value of ${b}^{2} - 4 a c$.
7. Simplify for the value of $\sqrt{{b}^{2} - 4 a c}$. The value remains as it stands.
8. Simplify for the value of $x = \left(- b \pm \sqrt{{b}^{2} - 4 a c}\right) \div 2 a$. The value for $x$ should be determined using value .
Example:
$2 {x}^{2} - 16 x + 32 = 0$
1. $a = 2 , b = - 16 , c = 32$
2. $x = \left(- \left(- 16\right) \pm \sqrt{{\left(- 16\right)}^{2} - 4 \left(2\right) \left(32\right)}\right) \div 2 \left(2\right)$
3. $x = \left(16 \pm \sqrt{256 - 8 \left(32\right)}\right) \div 4$
4. $x = \left(16 \pm \sqrt{256 - 256}\right) \div 4$
5. $x = \left(16 \pm \sqrt{0}\right) \div 4$
6. $x = \left(16 \pm 0\right) \div 4$
7. $x = 16 \div 4$
8. $x = 4$
• This is a little bit tricky but also incredibly elegant!
You start from your general quadratic:
$a {x}^{2} + b x + c = 0$
Take $c$ to the right side:
$a {x}^{2} + b x = - c$
The idea is now to transform the left side in something like ${\left(a + b\right)}^{2}$;
Multiply by $a$;
${a}^{2} {x}^{2} + a b x = - a c$
Multiply by $4$;
$4 {a}^{2} {x}^{2} + 4 a b x = - 4 a c$
Add and subtract ${b}^{2}$ to the left side:
$4 {a}^{2} {x}^{2} + 4 a b x + {b}^{2} - {b}^{2} = - 4 a c$
Take the $- {b}^{2}$ to the right:
$4 {a}^{2} {x}^{2} + 4 a b x + {b}^{2} = {b}^{2} - 4 a c$
The left side can be written as:
${\left(2 a x + b\right)}^{2} = {b}^{2} - 4 a c$
And:
$2 a x + b = \pm \sqrt{{b}^{2} - 4 a c}$
And finally:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Poetry in algebra!
## Questions
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## The Power Of Negativity
###### By Mike O’Hern, Center Director of Mathnasium of West Knoxville
Remember that stuff last month about exponents? You might remember that we discovered that if you raise any number – any number at all – to the power of zero you end up with an answer of 1. We arrived at that conclusion by doing a series of divisions, and I think I mentioned that if we continued on that path we might discover something about negative exponents.
Perhaps one way to think about it is that the power of positivity unites, but the power of negativity divides! (Don’t take that one too far – it just popped into my head and I thought it had a ring to it.)
Now for a very brief review, 10^2 = 100, divide by 10, 10^1 = 10, divide by 10, 10^0 = 1. Each time we divide, on the left side we simply subtract 1 from the exponent, and on the right we divide by 10 just like we learned in fourth grade.
Now I know that after reading last month’s article, inquiring minds wondered what would come next. And perhaps some of you have already figured that out! But I can’t assume that it goes without saying (and I wouldn’t miss an opportunity to say something anyway), so we’ll take the next step here together.
Let’s just divide by 10 one more time. On the left, where you have 100, you simply subtract 1 from the exponent, just like we had done before. And what is 0 – 1? Negative one: -1. So the left side is 10^-1. And on the right side, what do we get when we divide 1 by 10? You might say 0.1, and you would be right, but for the moment this will be clearer if we use the fractional notation: 1/10. So 10^-1 = 1/10.
#### Perhaps one way to think about it is that the power of positivity unites, but the power of negativity divides!
One more step and this starts to get interesting. (I know you’re thinking “it’s about time!” – stop it.) Divide by 10 again. 10^-2 = 1/100. What’s so interesting about that? Notice that another way to write 1/100 is 1/10^2. Here’s the connection: 10^-2 = 1/10^2. 10^-3 = 1/10^3, and so on. And this turns out to be true in general!
STOP, MR. MIKE! Why in the world would we care about negative exponents? Glad you asked! One example comes to mind right away. In the scientific world we like to talk about distances in terms of meters, for example. So when we talk about the distance to the moon we can say it’s over 100,000,000 meters. That’s 10^8 meters.
So what if we’re talking about the distance between atoms? In space the distance between atoms can be up to a meter or so, but in a chemical compound here on earth it can be something like 1/10000000000 (or 0.0000000001 if you prefer) meters (it’s slightly more, but this is a math illustration!). That would be 10^-10 meters. If we were working with atoms regularly (that is, not just comparing them to space atoms!) we would refer to that as 1 angstrom, really, but the point is that when we want to work with very small numbers or compare things that have great differences, negative exponents can make things a lot simper for us.
That will have to suffice for this month. It’s been like 10^-6 years since I filled my coffee mug, so I’d best get back to that! |
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# APEX Calculus: for University of Lethbridge
## Section15.5Surface Area
In Section 7.4 we used definite integrals to compute the arc length of plane curves of the form $$y=f(x)\text{.}$$ We later extended these ideas to compute the arc length of plane curves defined by parametric or polar equations.
The natural extension of the concept of “arc length over an interval” to surfaces is “surface area over a region.”
Consider the surface $$z=f(x,y)$$ over a region $$R$$ in the $$xy$$-plane, shown in Figure 15.5.1.(a). Because of the domed shape of the surface, the surface area will be greater than that of the area of the region $$R\text{.}$$ We can find this area using the same basic technique we have used over and over: we’ll make an approximation, then using limits, we’ll refine the approximation to the exact value.
As done to find the volume under a surface or the mass of a lamina, we subdivide $$R$$ into $$n$$ subregions. Here we subdivide $$R$$ into rectangles, as shown in the figure. One such subregion is outlined in the figure, where the rectangle has dimensions $$\dx_i$$ and $$\dy_i\text{,}$$ along with its corresponding region on the surface.
In part Figure 15.5.1.(b) of the figure, we zoom in on this portion of the surface. When $$\dx_i$$ and $$\dy_i$$ are small, the function is approximated well by the tangent plane at any point $$(x_i,y_i)$$ in this subregion, which is graphed in part Figure 15.5.1.(b). In fact, the tangent plane approximates the function so well that in this figure, it is virtually indistinguishable from the surface itself! Therefore we can approximate the surface area $$S_i$$ of this region of the surface with the area $$T_i$$ of the corresponding portion of the tangent plane.
This portion of the tangent plane is a parallelogram, defined by sides $$\vec u$$ and $$\vec v\text{,}$$ as shown. One of the applications of the cross product from Section 11.4 is that the area of this parallelogram is $$\norm{\vec u\times \vec v}\text{.}$$ Once we can determine $$\vec u$$ and $$\vec v\text{,}$$ we can determine the area.
The vector $$\vec u$$ is tangent to the surface in the direction of $$x\text{,}$$ therefore, from Section 14.4, $$\vec u$$ is parallel to $$\la 1,0,f_x(x_i,y_i)\ra\text{.}$$ The $$x$$-displacement of $$\vec u$$ is $$\dx_i\text{,}$$ so we know that $$\vec u = \dx_i\la 1,0,f_x(x_i,y_i)\ra\text{.}$$ Similar logic shows that $$\vec v = \dy_i\la 0,1,f_y(x_i,y_i)\ra\text{.}$$ Thus:
\begin{align*} \text{surface area $$S_i$$} \amp \approx \,\text{area of $$T_i$$}\\ \amp = \norm{\vec u\times \vec v}\\ \amp = \norm{\dx_i\la 1,0,f_x(x_i,y_i)\ra\times\dy_i\la 0,1,f_y(x_i,y_i)\ra}\\ \amp =\sqrt{1+f_x(x_i,y_i)^2+f_y(x_i,y_i)^2}\dx_i\dy_i\text{.} \end{align*}
Note that $$\dx_i\dy_i = \Delta A_i\text{,}$$ the area of the $$i$$th subregion.
Summing up all $$n$$ of the approximations to the surface area gives
\begin{equation*} \text{surface area over $$R$$} \approx \sum_{i=1}^n \sqrt{1+f_x(x_i,y_i)^2+f_y(x_i,y_i)^2}\Delta A_i\text{.} \end{equation*}
Once again take a limit as all of the $$\dx_i$$ and $$\dy_i$$ shrink to 0; this leads to a double integral.
### Definition15.5.2.Surface Area.
Let $$z=f(x,y)$$ where $$f_x$$ and $$f_y$$ are continuous over a closed, bounded region $$R\text{.}$$ The surface area $$S$$ over $$R$$ is
\begin{align*} S \amp = \iint_R \, dS\\ \amp =\iint_R \sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\, dA\text{.} \end{align*}
We test this definition by using it to compute surface areas of known surfaces. We start with a triangle.
### Example15.5.3.Finding the surface area of a plane over a triangle.
Let $$f(x,y) = 4-x-2y\text{,}$$ and let $$R$$ be the region in the plane bounded by $$x=0\text{,}$$ $$y=0$$ and $$y=2-x/2\text{,}$$ as shown in Figure 15.5.4. Find the surface area of $$f$$ over $$R\text{.}$$
Solution.
We follow Definition 15.5.2. We start by noting that $$f_x(x,y) = -1$$ and $$f_y(x,y) = -2\text{.}$$ To define $$R\text{,}$$ we use bounds $$0\leq y\leq 2-x/2$$ and $$0\leq x\leq 4\text{.}$$ Therefore
\begin{align*} S \amp = \iint_R\, dS\\ \amp = \int_0^4\int_0^{2-x/2} \sqrt{1+(-1)^2+(-2)^2}\, dy\, dx\\ \amp = \int_0^4 \sqrt{6}\left(2-\frac x2\right)\, dx\\ \amp = 4\sqrt{6}\text{.} \end{align*}
Because the surface is a triangle, we can figure out the area using geometry. Considering the base of the triangle to be the side in the $$xy$$-plane, we find the length of the base to be $$\sqrt{20}\text{.}$$ We can find the height using our knowledge of vectors: let $$\vec u$$ be the side in the $$xz$$-plane and let $$\vec v$$ be the side in the $$xy$$-plane. The height is then $$\norm {\vec u - \proj{u}{v}} = 4\sqrt{6/5}\text{.}$$ Geometry states that the area is thus
\begin{equation*} \frac 12\cdot4\sqrt{6/5}\cdot\sqrt{20} = 4\sqrt{6}\text{.} \end{equation*}
We affirm the validity of our formula.
It is “common knowledge” that the surface area of a sphere of radius $$r$$ is $$4\pi r^2\text{.}$$ We confirm this in the following example, which involves using our formula with polar coordinates.
### Example15.5.5.The surface area of a sphere.
Find the surface area of the sphere with radius $$a$$ centered at the origin, whose top hemisphere has equation $$f(x,y)=\sqrt{a^2-x^2-y^2}\text{.}$$
Solution.
We start by computing partial derivatives and find
\begin{equation*} f_x(x,y) = \frac{-x}{\sqrt{a^2-x^2-y^2}} \text{ and } f_y(x,y) = \frac{-y}{\sqrt{a^2-x^2-y^2}}\text{.} \end{equation*}
As our function $$f$$ only defines the top upper hemisphere of the sphere, we double our surface area result to get the total area:
\begin{align*} S \amp = 2\iint_R \sqrt{1+ f_x(x,y)^2+f_y(x,y)^2}\, dA\\ \amp = 2\iint_R \sqrt{1+ \frac{x^2+y^2}{a^2-x^2-y^2}}\, dA\text{.} \end{align*}
The region $$R$$ that we are integrating over is bounded by the circle, centered at the origin, with radius $$a\text{:}$$ $$x^2+y^2=a^2\text{.}$$ Because of this region, we are likely to have greater success with our integration by converting to polar coordinates. Using the substitutions $$x=r\cos(\theta)\text{,}$$ $$y=r\sin(\theta)\text{,}$$ $$dA = r\, dr\, d\theta$$ and bounds $$0\leq\theta\leq2\pi$$ and $$0\leq r\leq a\text{,}$$ we have:
\begin{align} S \amp = 2\int_0^{2\pi}\int_0^a \sqrt{1+\frac{r^2\cos^2(\theta) +r^2\sin^2(\theta) }{a^2-r^2\cos^2(\theta) -r^2\sin^2(\theta) }}\, r\, dr\, d\theta\notag\\ \amp =2\int_0^{2\pi}\int_0^ar\sqrt{1+\frac{r^2}{a^2-r^2}}\, dr\, d\theta\notag\\ \amp =2\int_0^{2\pi}\int_0^ar\sqrt{\frac{a^2}{a^2-r^2}}\, dr\, d\theta.\tag{15.5.1}\\ \end{align}
Apply substitution $$u=a^2-r^2$$ and integrate the inner integral, giving
\begin{align} \amp = 2\int_0^{2\pi} a^2\, d\theta\notag\\ \amp = 4\pi a^2\text{.}\notag \end{align}
Our work confirms our previous formula.
### Example15.5.6.Finding the surface area of a cone.
The general formula for a right cone with height $$h$$ and base radius $$a$$ is
\begin{equation*} \ds f(x,y) = h-\frac{h}a\sqrt{x^2+y^2}\text{,} \end{equation*}
shown in Figure 15.5.7. Find the surface area of this cone.
Solution.
We begin by computing partial derivatives.
\begin{equation*} f_x(x,y) = -\frac{xh}{a\sqrt{x^2+y^2}} \text{ and } f_y(x,y)=-\frac{yh}{a\sqrt{x^2+y^2}}\text{.} \end{equation*}
Since we are integrating over the disk bounded by $$x^2+y^2=a^2\text{,}$$ we again use polar coordinates. Using the standard substitutions, our integrand becomes
\begin{equation*} \sqrt{1+ \left(\frac{hr\cos(\theta) }{a\sqrt{r^2}}\right)^2 + \left(\frac{hr\sin(\theta) }{a\sqrt{r^2}}\right)^2}\text{.} \end{equation*}
This may look intimidating at first, but there are lots of simple simplifications to be done. It amazingly reduces to just
\begin{equation*} \sqrt{1+\frac{h^2}{a^2}} = \frac1a\sqrt{a^2+h^2}\text{.} \end{equation*}
Our polar bounds are $$0\leq\theta\leq2\pi$$ and $$0\leq r\leq a\text{.}$$ Thus
\begin{align*} S \amp = \int_0^{2\pi}\int_0^ar\frac1a\sqrt{a^2+h^2}\, dr\, d\theta\\ \amp = \int_0^{2\pi} \left.\left(\frac12r^2\frac1a\sqrt{a^2+h^2}\right)\right|_0^ad\theta\\ \amp = \int_0^{2\pi} \frac12a\sqrt{a^2+h^2} \, d\theta\\ \amp = \pi a\sqrt{a^2+h^2}\text{.} \end{align*}
This matches the formula found in the back of this text.
### Example15.5.8.Finding surface area over a region.
Find the area of the surface $$f(x,y) = x^2-3y+3$$ over the region $$R$$ bounded by $$-x\leq y\leq x\text{,}$$ $$0\leq x\leq 4\text{,}$$ as pictured in Figure 15.5.9.
Solution.
It is straightforward to compute $$f_x(x,y) = 2x$$ and $$f_y(x,y) = -3\text{.}$$ Thus the surface area is described by the double integral
\begin{equation*} \iint_R \sqrt{1+(2x)^2+(-3)^2}\, dA = \iint_R \sqrt{10+4x^2}\, dA\text{.} \end{equation*}
As with integrals describing arc length, double integrals describing surface area are in general hard to evaluate directly because of the square-root. This particular integral can be easily evaluated, though, with judicious choice of our order of integration.
Integrating with order $$dx\, dy$$ requires us to evaluate $$\int \sqrt{10+4x^2}\, dx\text{.}$$ This can be done, though it involves Integration By Parts and $$\sinh^{-1}(x)\text{.}$$ Integrating with order $$dy\, dx$$ has as its first integral $$\int \sqrt{10+4x^2}\, dy\text{,}$$ which is easy to evaluate: it is simply $$y\sqrt{10+4x^2}+C\text{.}$$ So we proceed with the order $$dy\, dx\text{;}$$ the bounds are already given in the statement of the problem.
\begin{align*} \iint_R\sqrt{10+4x^2}\, dA \amp = \int_0^4\int_{-x}^x\sqrt{10+4x^2}\, dy \, dx\\ \amp = \int_0^4\left.\big(y\sqrt{10+4x^2}\big)\right|_{-x}^x dx\\ \amp =\int_0^4\big(2x\sqrt{10+4x^2}\big)\, dx.\\ \end{align*}
Apply substitution with $$u = 10+4x^2\text{:}$$
\begin{align*} \amp = \left.\left(\frac16\big(10+4x^2\big)^{3/2}\right)\right|_0^4\\ \amp = \frac13\big(37\sqrt{74}-5\sqrt{10}\big) \approx 100.825\,\text{units}^2\text{.} \end{align*}
So while the region $$R$$ over which we integrate has an area of 16 square units, the surface has a much greater area as its $$z$$-values change dramatically over $$R\text{.}$$
In practice, technology helps greatly in the evaluation of such integrals. High powered computer algebra systems can compute integrals that are difficult, or at least time consuming, by hand, and can at the least produce very accurate approximations with numerical methods. In general, just knowing how to set up the proper integrals brings one very close to being able to compute the needed value. Most of the work is actually done in just describing the region $$R$$ in terms of polar or rectangular coordinates. Once this is done, technology can usually provide a good answer.
We have learned how to integrate integrals; that is, we have learned to evaluate double integrals. In the next section, we learn how to integrate double integrals — that is, we learn to evaluate triple integrals, along with learning some uses for this operation.
### ExercisesExercises
#### Terms and Concepts
##### 1.
“Surface area” is analogous to what previously studied concept?
##### 2.
To approximate the area of a small portion of a surface, we computed the area of its plane.
##### 3.
We interpret $$\ds \iint_R\, dS$$ as “sum up lots of little .”
##### 4.
Why is it important to know how to set up a double integral to compute surface area, even if the resulting integral is hard to evaluate?
##### 5.
Why do $$z=f(x,y)$$ and $$z=g(x,y)=f(x,y)+h\text{,}$$ for some real number $$h\text{,}$$ have the same surface area over a region $$R\text{?}$$
##### 6.
Let $$f(x,y)$$ be a function defined over a region $$R$$ and let $$g(x,y)=2f(x,y)\text{.}$$ Why is the surface area of $$z=g(x,y)$$ over $$R$$ not twice the surface area of $$z=f(x,y)$$ over $$R\text{?}$$
#### Problems
##### Exercise Group.
In the following exercises, set up the iterated integral that computes the surface area of the graph of the given function over the region $$R\text{.}$$
###### 7.
$$f(x,y) = \sin(x) \cos(y)\text{;}$$$$R$$ is the rectangle with bounds $$0\leq x\leq 2\pi\text{,}$$ $$0\leq y\leq2\pi\text{.}$$
###### 8.
$$\ds f(x,y) = \frac{1}{x^2+y^2+1}\text{;}$$$$R$$ is bounded by the circle $$x^2+y^2=9\text{.}$$
###### 9.
$$\ds f(x,y) = x^2-y^2\text{;}$$$$R$$ is the rectangle with opposite corners $$(-1,-1)$$ and $$(1,1)\text{.}$$
###### 10.
$$\ds f(x,y) = \frac{1}{e^{x^2}+1}\text{;}$$$$R$$ is the rectangle bounded by
$$-5\leq x\leq 5$$ and $$0\leq y\leq 1\text{.}$$
##### Exercise Group.
In the following exercises, find the area of the given surface over the region $$R\text{.}$$
###### 11.
$$z = 3x-7y+2\text{;}$$ $$R$$ is the rectangle with opposite corners $$(-1,0)$$ and $$(1,3)\text{.}$$
###### 12.
$$z = 2x+2y+2\text{;}$$ $$R$$ is the triangle with corners $$(0,0)\text{,}$$ $$(1,0)$$ and $$(0,1)\text{.}$$
###### 13.
$$z = x^2+y^2+10\text{;}$$ $$R$$ is bounded by the circle $$x^2+y^2=16\text{.}$$
###### 14.
$$z = -2x+4y^2+7$$ over $$R\text{,}$$ the triangle bounded by $$y=-x\text{,}$$ $$y=x\text{,}$$ $$0\leq y\leq 1\text{.}$$
###### 15.
$$z = x^2+y$$ over $$R\text{,}$$ the triangle bounded by $$y=2x\text{,}$$ $$y=0$$ and $$x=2\text{.}$$
###### 16.
$$z = \frac23x^{3/2}+2y^{3/2}$$ over $$R\text{,}$$ the rectangle with opposite corners $$(0,0)$$ and $$(1,1)\text{.}$$
###### 17.
$$z = 10-2\sqrt{x^2+y^2}$$ over the region $$R$$ bounded by the circle $$x^2+y^2=25\text{.}$$ (This is the cone with height 10 and base radius 5; be sure to compare your result with the known formula.)
###### 18.
Find the surface area of the sphere with radius 5 by doubling the surface area of $$f(x,y) = \sqrt{25-x^2-y^2}$$ over the region $$R$$ bounded by the circle $$x^2+y^2=25\text{.}$$ (Be sure to compare your result with the known formula.)
###### 19.
Find the surface area of the ellipse formed by restricting the plane $$f(x,y) = cx+dy+h$$ to the region $$R$$ bounded by the circle $$x^2+y^2=1\text{,}$$ where $$c\text{,}$$ $$d$$ and $$h$$ are some constants. Your answer should be given in terms of $$c$$ and $$d\text{;}$$ why does the value of $$h$$ not matter? |
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# A path $1m$ wide is built inside a square park of side $30m$ along its sides. Find the area of the path. Calculate the cost of constructing the path at the rate of Rs.$70$ per ${m^2}$.
Last updated date: 14th Jun 2024
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Hint: Here is the problem of mensuration we have. We have to find the area of a particular shape. We are approaching the solution by subtracting the area of small Square from the area of the bigger square (Fig.1). Also here we have to find the cost of constructing the path as the rate is given. To find the cost we will use the unitary method.
Suppose there are two squares ABCD and PQRS also ABCD is the boundary of the square park and the area between the two squares ABCD and PQRS is the mentioned path in the question. We need to find the area of the path and the cost of constructing the path.
Area of the path = Area of square ABCD - Area of the square PQRS … equation (1)
To find the area of the two squares we need to know the length of the side of both squares.
Length of one side of square ABCD $= 30m$ [Given]
Length of one side of square PQRS $= 30 - 1 = 29m$ [ the width of the path is 1m]
We know that the area of the square is $A = {a^2}$ where $A$ stands for the area of the square and $a$ stands for the side of the square. Therefore,
Area of Square ABCD $= 4 \times {(30)^2} = 4 \times 900 = 3600{m^2}$
Area of Square PQRS $= 4 \times {(29)^2} = 4 \times 841 = 3364{m^2}$
Now, from equation (1),
Area of the path = Area of square ABCD - Area of the square PQRS
Area of the path$= 3600 - 3364 = 236{m^2}$
We have to also find the cost of constructing the path at the rate of Rs.$70$per${m^2}$.
Cost of constructing $1{m^2}$path $= 70Rs.$
$\therefore$ Cost of constructing $236{m^2}$path $= 236 \times 70 = 16,520Rs.$
Hence the area of the path is $236{m^2}$ and the cost of constructing the path at the rate of Rs.$70$per${m^2}$ is $16520Rs.$
Note:
Second method-
We can also find the area of the path by dividing the shape into rectangles as,
Area of Shape of the path$= Area(rect.AWZD) + Area(rect.SZRY) + Area(rect.YCBX) + Area(rect.XQPW)$ |
# 2.5. Independence#
Events $$A$$ and $$B$$ are independent if the information that one of them occurred does not change the chance of the other.
• If you don’t know whether $$A$$ has occurred, then the chance of $$B$$ is just $$P(B)$$.
• If you do know that $$A$$ occurred, then you have to update the chance of $$B$$ to $$P(B \mid A)$$, the conditional chance of $$B$$ given $$A$$.
The definition of independence says that $$A$$ and $$B$$ are independent if $$P(B \mid A) = P(B)$$.
For example, suppose you roll a die once and see an odd number. What does that tell you about the chance that the next roll is a 2? If your answer is, “Nothing – the chance that the next roll is a 2 is still 1/6,” then you have said that the two events “first roll is an odd number” and “second roll is 2” are independent.
## 2.5.1. Dice Versus Cards#
Suppose you are rolling a die twice. Then
\begin{split} \begin{align*} &P(2 \text{ on the second roll} \mid \text{odd number on the first roll}) \\ &= ~ P(2 \text{ on the second roll}) ~ = ~ \frac{1}{6} \end{align*} \end{split}
Independence is a natural assumption about successive rolls of a die. But when cards are dealt (at random without replacement, as we consistently assume) from a deck, then the situation is different.
Suppose two cards are dealt from a standard deck in which four out of 52 cards are aces. We know that by symmetry,
$P(\text{second card is an ace}) ~ = ~ \frac{4}{52}$
It’s worth reviewing the reason for this: since we have no other information, all 52 cards are equally likely to be appear on the second draw, and four of them are aces.
However,
\begin{split} \begin{align*} P(\text{second card is an ace} \mid \text{first card is an ace})~ &= ~ \frac{3}{51} \\ &\neq ~ P(\text{second card is an ace}) \end{align*} \end{split}
The information that the first card is an ace changes the probability that the second card is an ace. The events “the first card is an ace” and “the second card is an ace” are not independent. Knowing that the first card is an ace eliminates an ace from the deck, and probabilities have to be recalculated accordingly.
## 2.5.2. A Special Case of the Multiplication Rule#
Suppose you deal two cards. By the multiplication rule,
\begin{split} \begin{align*} & P(\text{both cards are aces}) \\ &= ~ P(\text{first card is an ace})P(\text{second card is an ace} \mid \text{first card is an ace}) \\ &= ~ \frac{4}{52} \times \frac{3}{51} \end{align*} \end{split}
If you roll two dice, then by the multiplication rule again,
\begin{split} \begin{align*} & P(\text{odd number on the first roll and } 2 \text{ on the second roll}) \\ &= ~ P(\text{odd number on the first roll})P(2 \text{ on the second roll} \mid \text{odd number on the first roll}) \\ &= ~ \frac{3}{6} \times \frac{1}{6} \end{align*} \end{split}
The second factor in the product is not affected by the condition that the first roll was an odd number.
What we have observed in the example above can be generalized as follows.
The multiplication rule says that for any two events $$A$$ and $$B$$,
$P(AB) ~ = ~ P(A)P(B \mid A)$
In the case of independent events $$A$$ and $$B$$ the second factor in the multiplication rule is the same as the unconditional chance of $$B$$:
$P(AB) ~ = ~ P(A)P(B) ~~ \text{if } A \text{ and } B \text{ are independent}$
Notice that if you are trying to find the chance of an intersection, then independence does not affect whether you multiply chances. Independence only affects what you multiply.
It is simplest to just remember the general multiplication rule. Independence is then an easy special case.
$P(\text{ odd number on the first roll and } 2 \text{ on the second roll}) ~ = ~ \frac{3}{6} \times \frac{1}{6} ~ = ~ \frac{3}{36}$
We can now check that the assumption of independence is consistent with an assumption that we have made all along: that all 36 outcomes of the two rolls are equally likely. Under that assumption, we would have solved the problem by saying that three pairs $$(1, 2)$$, $$(3, 2)$$, and $$(5, 2)$$ form the event and hence the chance is $$3/36$$. That’s the same as what we got by using independence.
The definition of independence extends to multiple events. A collection of events are mutually independent if knowing that any group of them has occurred doesn’t affect the chances of the others.
## 2.5.3. People v. Collins, 1968#
In 1964, the married couple Michael and Janet Collins were arrested for robbery in California. A jury found them guilty, based in part on a probability calculation.
Michael Collins appealed the conviction to the California Supreme Court. In 1968 the Court overturned the jury’s conviction.
In the statement of the majority opinion, Justice Raymond Sullivan wrote, “We deal here with the novel question whether evidence of mathematical probability has been properly introduced and used by the prosecution in a criminal case. While we discern no inherent incompatibility between the disciplines of law and mathematics and intend no general disapproval or disparagement of the latter as an auxiliary in the fact-finding processes of the former, we cannot uphold the technique employed in the instant case.”
To see what the Justices could not uphold, we start with their summary of the incident that led to the arrest. We apologize that the quote includes the term used to describe a black person in the statement as well as during the trial.
“On June 18, 1964, about 11:30 a.m. Mrs. Juanita Brooks, who had been shopping, was walking home along an alley in the San Pedro area of the City of Los Angeles. She was pulling behind her a wicker basket carryall containing groceries and had her purse on top of the packages. She was using a cane. As she stooped down to pick up an empty carton, she was suddenly pushed to the ground by a person whom she neither saw nor heard approach. She was stunned by the fall and felt some pain. She managed to look up and saw a young woman running from the scene. According to Mrs. Brooks the latter appeared to weigh about 145 pounds, was wearing “something dark,” and had hair “between a dark blond and a light blond,” but lighter than the color of defendant Janet Collins’ hair as it appeared at trial. Immediately after the incident, Mrs. Brooks discovered that her purse, containing between $35 and$40 was missing.
About the same time as the robbery, John Bass, who lived on the street at the end of the alley, was in front of his house watering his lawn. His attention was attracted by “a lot of crying and screaming” coming from the alley. As he looked in that direction, he saw a woman run out of the alley and enter a yellow automobile parked across the street from him. He was unable to give the make of the car. The car started off immediately and pulled wide around another parked vehicle so that in the narrow street it passed within 6 feet of Bass. The latter then saw that it was being driven by a male Negro, wearing a mustache and beard. At the trial Bass identified defendant as the driver of the yellow automobile. However, an attempt was made to impeach his identification by his admission that at the preliminary hearing he testified to an uncertain identification at the police lineup shortly after the attack on Mrs. Brooks, when defendant was beardless.
In his testimony Bass described the woman who ran from the alley as a Caucasian, slightly over 5 feet tall, of ordinary build, with her hair in a dark blonde ponytail, and wearing dark clothing. He further testified that her ponytail was “just like” one which Janet had in a police photograph taken on June 22, 1964.”
In the jury trial, the prosecutor had provided a table of probabilities:
Characteristic
Individual Probability
A. Partly yellow automobile
1/10
B. Man with mustache
1/4
C. Girl with ponytail
1/10
D. Girl with blond hair
1/3
E. Negro man with beard
1/10
F. Interracial couple in car
1/1000
The prosecutor had then multiplied all the probabilities together to get an answer of about 1 in 12,000,000. Quoting again from the Supreme Court’s statement, “Applying the product rule to his own factors the prosecutor arrived at a probability that there was but one chance in 12 million that any couple possessed the distinctive characteristics of the defendants.”
Though the jury had been swayed by this calculation, the Supreme Court understood the importance of checking assumptions.
• First, they objected to the fact that the prosecution had provided no basis for arriving at the probabilities in the table: “[W]e find the record devoid of any evidence relating to any of the six individual probability factors.”
• Second, they did not accept the assumption of independence of the factors. They pointed out the difficulty of believing that having a beard and having a mustache are independent. It is not hard to come up with other examples of this difficulty. Treating characteristics D, E, and F as independent is hard to justify.
When probability theory is used in practice, it is important to make sure that the assumptions of the theory are reasonable before doing calculations. Otherwise the analysis might be just as weak as the one in the Collins case, which “lacked an adequate foundation both in evidence and in statistical theory”.
## 2.5.4. Regina v. Clark, 2003#
The lessons learned in the Collins case have been widely broadcast and appear in many statistics textbooks – but clearly not the ones read by expert witnesses in the case of Sally Clark in the United Kingdom.
Two of Clark’s apparently healthy sons died in infancy. On both occasions, Clark was alone with her baby at the time of death, and there was no evidence of injury. After the second death, Clark was arrested for murder. Her defence was that the babies had died of Sudden Infant Death Syndrome (SIDS), referred to in the UK as Sudden Unexplained Death in Infancy (SUDI) or “cot death”. But she was found guilty.
Clark’s conviction was overturned on a second appeal, after she had served three years of her prison sentence. Among the reasons for the reversal were the acknowledgment by the court that there had been serious misuse of statistical reasoning.
In particular, a medical expert witness claimed that the chance of Clark’s two babies dying of SIDS was about 1 in 73 million. He arrived at this figure in two steps:
• First he estimated the chance of a cot death in a family like Clark’s. The estimate was based on a large-scale multidisciplinary analysis of SIDS that reported an overall SIDS death rate of 1 in 1303 live births in the UK. Clark was a lawyer and the family did not have financial struggles. After taking such factors into account, the expert set the rate at 1 in 8543, arguing that SIDS deaths were less likely in families like Clark’s than in the general population.
• Then he found the chance of two SIDs deaths in Clark’s family by the calculation
$\frac{1}{8543} \times \frac{1}{8543} = \frac{1}{72982849}$
That’s about 1 in 73 million.
The Royal Statistical Society (RSS) objected vigorously to this calculation on the grounds that the assumption of independence is hard to justify. Given that there is one SIDS death in a family, the risk to other babies in the family might be larger.
“There may well be unknown genetic or environmental factors that predispose families to SIDS, so that a second case within the family becomes much more likely,” the RSS wrote. “The well-publicised figure of 1 in 73 million thus has no statistical basis. Its use cannot reasonably be justified as a “ballpark” figure because the error involved is likely to be very large, and in one particular direction. The true frequency of families with two cases of SIDS may be very much less incriminating than the figure presented to the jury at trial.”
The Society also pointed out that the media had misinterpreted the erroneous figure of 1 in 73 million to be the chance that Clark was innocent. The error was an instance of the Prosecutor’s Fallacy in which the likelihood of the evidence given innocence was confused with the chance of innocence given the evidence.
Thirty-five years after the Collins case, another unjustifiable assumption of independence made by a prosecution witness resulted in a wrongful conviction for Clark. The Sally Clark case is regarded as a great miscarriage of justice and had devastating consequences. Though Clark was strongly supported by her husband throughout, she was in the end unable to cope with the loss of her sons and the injustice she suffered. She died in 2007.
Apart from statistical error, a serious flaw in the prosecution’s case was that another of its medical experts withheld blood test reports showing that the second baby had harmful bacteria in several places in his body including the cerebro-spinal fluid. After Clark’s conviction was overturned, the Journal of the Royal Society of Medicine (JRSM) addressed the questions the case raised about the conduct of medical expert witnesses.
The JRSM urged more appropriate investigation of SIDS deaths. It also made an important general recommendation. “[W]hen giving evidence to the police or to the court doctors would be wise to acknowledge the limitations in their understanding. They should present all relevant facts in a balanced manner, offer opinions only within their sphere of expertise and take care not to overstate their case. Wrong conclusions in either direction may be disastrous.” |
# How do you find the limit of (x-4)/(sqrt(x-3)-sqrt(5-x)) as x approaches 4?
Apr 13, 2015
The answer is: $1$.
This limit shows up in the $\frac{0}{0}$ indecision form, so we have to rationalise.
${\lim}_{x \rightarrow 4} \frac{x - 4}{\sqrt{x - 3} - \sqrt{5 - x}} =$
$= {\lim}_{x \rightarrow 4} \left[\frac{x - 4}{\sqrt{x - 3} - \sqrt{5 - x}} \cdot \frac{\sqrt{x - 3} + \sqrt{5 - x}}{\sqrt{x - 3} + \sqrt{5 - x}}\right] =$
$= {\lim}_{x \rightarrow 4} \frac{\left(x - 4\right) \left(\sqrt{x - 3} + \sqrt{5 - x}\right)}{x - 3 - 5 + x} =$
$= {\lim}_{x \rightarrow 4} \frac{\left(x - 4\right) \left(\sqrt{x - 3} + \sqrt{5 - x}\right)}{2 x - 8} =$
$= {\lim}_{x \rightarrow 4} \frac{\left(x - 4\right) \left(\sqrt{x - 3} + \sqrt{5 - x}\right)}{2 \left(x - 4\right)} =$
$= {\lim}_{x \rightarrow 4} \frac{\left(\sqrt{x - 3} + \sqrt{5 - x}\right)}{2} = \frac{\sqrt{4 - 3} + \sqrt{5 - 4}}{2} = 1$.
Apr 13, 2015
Rationalize the denominator by multiplying the fraction by $1$ in the form:
$\frac{\sqrt{x - 3} + \sqrt{5 - x}}{\sqrt{x - 3} + \sqrt{5 - x}}$
This works because $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$,
so $\left(\sqrt{a} - \sqrt{b}\right) \left(\sqrt{a} + \sqrt{b}\right) = a - b$. It looks like this:
$\frac{\left(x - 4\right)}{\left(\sqrt{x - 3} - \sqrt{5 - x}\right)} \frac{\left(\sqrt{x - 3} + \sqrt{5 - x}\right)}{\left(\sqrt{x - 3} + \sqrt{5 - x}\right)}$
=((x-4) (sqrt(x-3)+sqrt(5-x)))/((x-3)-(5-x)
$= \frac{\left(x - 4\right) \left(\sqrt{x - 3} + \sqrt{5 - x}\right)}{2 x - 8}$
$= \frac{\left(x - 4\right) \left(\sqrt{x - 3} + \sqrt{5 - x}\right)}{2 \left(x - 4\right)}$
$= \frac{\sqrt{x - 3} + \sqrt{5 - x}}{2}$
So, we have:
${\lim}_{x \rightarrow 4} \frac{x - 4}{\sqrt{x - 3} - \sqrt{5 - x}} = {\lim}_{x \rightarrow 4} \frac{\sqrt{x - 3} + \sqrt{5 - x}}{2}$
$= \frac{2}{2} = 1$ |
# How do you find the distance between (-2,4), (5,8)?
Dec 26, 2016
$\left\mid \left(- 2 , 4\right) : \left(5 , 8\right) \right\mid = \textcolor{g r e e n}{\sqrt{65}}$
#### Explanation:
The horizontal distance ($\Delta x$)between $\left(- 2 , 4\right)$ and $\left(5 , 8\right)$ is the absolute value of the difference between the $x$ components of the coordinates:
$\textcolor{w h i t e}{\text{XXX}} \Delta x = \left\mid 5 - \left(- 2\right) \right\mid = 7$
Similarly, the vertical distance ($\Delta y$)between $\left(- 2 , 4\right)$ and $\left(5 , 8\right)$ is the absolute value of the difference between the $y$ components of the coordinates:
$\textcolor{w h i t e}{\text{XXX}} \Delta x = \left\mid 8 - 4 \right\mid = 4$
Using the Pythagorean Theorem the distance between $\left(- 2 , 4\right)$ and $\left(5 , 8\right)$ is
$\textcolor{w h i t e}{\text{XXX}} d = \sqrt{{7}^{2} + {4}^{2}} = \sqrt{65}$ |
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ALL SIN TAN COS Rule
All Sin Tan Cos rule helps us to evaluate different trigonometric ratios in different quadrants.
The "all sin tan cos rule" can be remembered easily using the following phrases.
"All Sliver Tea Cups"
or
"All Students Take Calculus"
or
"All Students Take Coffee"
"All Sin Tan Cos rule" is also known as ASTC formula in trigonometry
ASTC formla has been explained clearly in the figure given below.
More clearly,
In the first quadrant (0° to 90°), all trigonometric ratios are positive.
In the second quadrant (90° to 180°), sin and csc are positive and other trigonometric ratios are negative.
In the third quadrant (180° to 270°), tan and cot are positive and other trigonometric ratios are negative.
In the fourth quadrant (270° to 360°), cos and sec are positive and other trigonometric ratios are negative.
Important Conversions
When we have the angles 90° and 270° in the trigonometric ratios in the form of
(90° + θ)
(90° - θ)
(270° + θ)
(270° - θ)
We have to do the following conversions,
sin θ <------> cos θ
tan θ <------> cot θ
csc θ <------> sec θ
For example,
sin (270° + θ) = - cos θ
cos (90° - θ) = sin θ
For the angles 0° or 360° and 180°, we should not make the above conversions.
(90° - θ) -------> I st Quadrant
(90° + θ) and (180° - θ) -------> II nd Quadrant
(180° + θ) and (270° - θ) -------> III rd Quadrant
(270° + θ), (360° - θ) and (θ) -------> IV th Quadrant
All Sin Tan Cos Rule - Practice Questions
Question 1 :
Evaluate : cos (270° - θ)
To evaluate cos (270° - θ), we have to consider the following important points.
(i) (270° - θ) will fall in the III rd quadrant.
(ii) When we have 270°, "cos" will become "sin"
(iii) In the III rd quadrant, the sign of "cos" is negative.
Considering the above points, we have
cos (270° - θ) = - sin θ
Question 2 :
Evaluate : sin (180° + θ)
To evaluate sin (180° + θ), we have to consider the following important points.
(i) (180° + θ) will fall in the III rd quadrant.
(ii) When we have 180°, "sin" will not be changed
(iii) In the III rd quadrant, the sign of "sin" is negative.
Considering the above points, we have
sin (180° + θ) = - sin θ
Based on the above two examples, we can evaluate the following trigonometric ratios.
sin (-θ) = - sin θcos (-θ) = cos θtan (-θ) = - tan θcsc (-θ) = - csc θsec (-θ) = sec θcot (-θ) = - cot θsin (90°-θ) = cos θcos (90°-θ) = sin θtan (90°-θ) = cot θcsc (90°-θ) = sec θsec (90°-θ) = csc θcot (90°-θ) = tan θsin (90°+θ) = cos θcos (90°+θ) = -sin θtan (90°+θ) = -cot θcsc (90°+θ) = sec θsec (90°+θ) = -csc θcot (90°+θ) = -tan θsin (180°-θ) = sin θcos (180°-θ) = -cos θtan (180°-θ) = -tan θ csc (180°-θ) = csc θsec (180°-θ) = -sec θcot (180°-θ) = -cot θsin (180°+θ) = -sin θcos (180°+θ) = -cos θtan (180°+θ) = tan θcsc (180°+θ) = -csc θsec (180°+θ) = -sec θcsc (180°+θ) = cot θsin (270°-θ) = -cos θcos (270°-θ) = -sin θtan (270°-θ) = cot θcsc (270°-θ) = -sec θsec (270°-θ) = -csc θcot (270°-θ) = tan θsin (270°+θ) = -cos θcos (270°+θ) = sin θtan (270°+θ) = -cot θcsc (270°+θ) = -sec θsec (270°+θ) = cos θcot (270°+θ) = -tan θ
Angles Greater than or Equal to 360°
If the angle is equal to or greater than 360°, we have to divide the given angle by 360 and take the remainder.
For example,
(i) Let us consider the angle 450°.
When we divide 450° by 360, we get the remainder 90°.
Therefore, 450° = 90°
(ii) Let us consider the angle 360°
When we divide 360° by 360, we get the remainder 0°.
Therefore, 360° = 0°
Based on the above two examples, we can evaluate the following trigonometric ratios.
sin (360° - θ) = sin (0° - θ) = sin (θ) = - sin θ
cos (360° - θ) = cos (0° - θ) = cos (θ) = cos θ
tan (360° - θ) = tan (0° - θ) = tan (θ) = - tan θ
csc (360° - θ) = csc (0° - θ) = csc (θ) = - csc θ
sec (360° - θ) = sec (0° - θ) = sec (θ) = sec θ
cot (360° - θ) = cot (0° - θ) = cot (θ) = - cot θ
Trigonometric Ratios of Angles Greater than or Equal to 360° - Practice Problems
Problem 1 :
Evaluate :
tan 735°
Solution :
The given 735° is greater than 360°.
So, we have to divide 735° by 360 and take the remainder.
When 735° is divided by 360, the remainder is 15°.
Therefore,
735° = 15° ------> tan 735° = tan 15°
Hence, tan 735° is equal to tan 15°.
Problem 2 :
Evaluate :
cos (-870°)
Solution :
Since the given angle (-870°) has negative sign, we have to assume it falls in the fourth quadrant.
In the fourth quadrant, "cos" is positive.
So, we have cos (-870°) = cos 870°.
The given 870° is greater than 360°.
So, we have to divide 870° by 360 and take the remainder.
When 870° is divided by 360, the remainder is 150°.
Therefore,
870° = 150° ------> cos 870° = cos 150°
cos 870° = cos (180° - 30°)
cos 870° = - cos 30°
cos 870 = - √3 / 2
Hence, cos 870° is equal to √3/2.
Problem 3 :
Find the value of :
(sin 780 sin 480° + cos 120° cos 60°)
Solution :
Let us find the value of each trigonometric ratio for the given angle.
sin 780° = sin 60° = √3 / 2
sin 480° = sin 120° = sin (180° - 60°) = sin 60° = √3/2
cos 120° = cos (180° - 60°) = - cos 60° = - 1/2
cos 60° = 1/2
sin780 sin480° + cos120° cos60° is
= (√3/2) x (√3/2) + (-1/2) x (1/2)
= (3/4) - (1/4)
= (3-1) / 4
= 2 / 4
= 1/2
Hence, the value of the given trigonometric expression is equal to 1/2.
Problem 4 :
Simplify :
cot (90°-θ) sin (180°+θ) sec(360°-θ) / tan(180°+θ) sec(-θ) cos(90°+θ)
Solution :
Using ASTC formula, we have
cot (90°-θ) = tan θ
sin (180°+θ) = - sin θ
sec(360°-θ) = sec θ
tan(180°+θ) = tan θ
sec(-θ) = sec θ
cos(90°+θ) = - sin θ
Then, the given expression is
= (tan θ x -sinθ x sec θ) / (tan θ x sec θ x -sin θ)
= 1
Hence, the simplification of the given trigonometric expression is equal to 1.
Problem 5 :
Simplify :
sec(360°-θ) tan(180°-θ) + cot(90°+θ) csc(270°-θ)
Solution :
Using ASTC formula, we have
sec (360°-θ) = sec θ
tan (180°-θ) = - tan θ
cot (90°+θ) = - tan θ
csc (270°-θ) = - sec θ
Then, the given expression is
= sec θ x (-tan θ) + (-tan θ) x (-sec θ)
= - sec θ x tan θ + sec θ x tan θ
= 0
Hence, the simplification of the given trigonometric expression is equal to 0.
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### Chapter 0
```Section 9.2
The Hyperbola
1
Objectives:
•
•
•
•
•
Locate a hyperbola’s vertices and foci.
Write equations of hyperbolas in standard form.
Graph hyperbolas centered at the origin.
Graph hyperbolas not centered at the origin.
Solve applied problems involving hyperbolas.
2
Definition of a Hyperbola
A hyperbola is the set of
points in a plane the
difference of whose
distances from two fixed
points, called foci,
is constant.
3
The Two Branches of a Hyperbola
The line through the foci
intersects the hyperbola at
two points, called the vertices.
The line segment that
joins the vertices is the
transverse axis.
The midpoint of the transverse
axis is the center of the hyperbola.
4
Standard Forms of the Equations of a Hyperbola
The standard form of the equation of a hyperbola with
center at the origin is x 2 y 2
y 2 x2
2 1 or 2 2 1.
2
a b
a b
5
Standard Forms of the Equations of a Hyperbola
(continued)
The vertices are a units from the center and the foci are c
units from the center.
For both equations, b2 = c2 – a2. Equivalently, c2 = a2 + b2.
6
Example: Finding Vertices and Foci from a Hyperbola’s
Equation
Find the vertices and locate the foci for the hyperbola with
the given equation: x 2 y 2
1.
25 16
a 2 25, a 5. The vertices are (–5, 0) and (5, 0).
c2 a 2 b2
c 2 25 16 41
c 41
The foci are at 41,0 and
41,0 .
7
Example: Finding Vertices and Foci from a Hyperbola’s
Equation (continued)
Find the vertices and locate the foci for the hyperbola with
the given equation:
2
2
x
y
1
25 16
The vertices are (–5, 0) and (5, 0).
The foci are at
41,0 and 41,0 .
8
Example: Finding Vertices and Foci from a Hyperbola’s
Equation
Find the vertices and locate the foci for the hyperbola
2
2
y
x
with the given equation:
1.
25 16
a 25, a 5. The vertices are (0, –5) and (0, 5).
2
c2 a 2 b2
c 2 25 16 41
c 41
The foci are at 0, 41 and 0, 41 .
9
Example: Finding the Equation of a Hyperbola from Its
Foci and Vertices (continued)
Find the vertices and locate the foci for the hypberbola
with the given equation:
2
2
y x
1
25 16
The vertices are (0, –5) and (0, 5).
The foci are at
0, 41 and 0, 41 .
10
Example: Finding the Equation of a Hyperbola from Its
Foci and Vertices
Find the standard form of the equation of a hyperbola
with foci at (0, –5) and (0, 5) and vertices (0, –3) and
(0, 3).
Because the foci are located at (0, –5) and (0, 5), the
transverse axis lies on the y-axis. Thus, the form of the
equation is y 2 x 2
2 1.
2
a b
The distance from the center to either vertex is 3, so
a = 3.
11
Example: Finding the Equation of a Hyperbola from Its
Foci and Vertices
Find the standard form of the equation of a hyperbola
with foci at (0, –5) and (0, 5) and vertices (0, –3) and
(0, 3).
The distance from the center to either focus is 5. Thus,
c = 5.
The equation is
c2 a 2 b2
y 2 x2
1
2
25 9 b
9 16
b 2 16
12
The Asymptotes of a Hyperbola Centered at the Origin
13
The Asymptotes of a Hyperbola Centered at the Origin
(continued)
As x and y get larger, the two branches of the graph of a hyperbola
approach a pair of intersecting straight lines, called asymptotes.
14
Graphing Hyperbolas Centered at the Origin
15
Example: Graphing a Hyperbola
2
2
x
y
Graph and locate the foci:
1.
36 9
What are the equations of the asymptotes?
Step 1 Locate the vertices.
x2 y 2
The given equation is in the form 2 2 1
a b
2
2
with a = 36 and b = 9.
a2 = 36, a = 6. Thus, the vertices are (–6, 0) and (6, 0).
16
Example: Graphing a Hyperbola (continued)
2
2
x
y
Graph and locate the foci:
1.
36 9
Step 2 Draw a rectangle
a2 = 36, a = 6.
b2 = 9, b = 3.
The rectangle passes
through the points
(–6, 0) and (6, 0)
and (0, –3) and (0, 3).
17
Example: Graphing a Hyperbola (continued)
2
2
x
y
Graph and locate the foci:
1.
36 9
Step 3 Draw extended
diagonals for the rectangle
to obtain the asymptotes.
b 3 1
a 6 2
The equations for the asymptotes
1
1
are y x and y x.
2
2
18
Example: Graphing a Hyperbola (continued)
2
2
x
y
Graph and locate the foci:
1.
36 9
Step 4 Draw the two branches
of the hyperbola starting at
each vertex and approaching
the asymptotes.
c 2 a 2 b 2 c 2 36 9 45
c 45 6.4
The foci are at 45,0 and
or (–6.4, 0) and (6.4, 0).
45,0 ,
19
Example: Graphing a Hyperbola (continued)
2
2
Graph and locate the foci: x y 1.
36 9
What are the asymptotes?
The foci are 45,0 and 45,0 ,
or (–6.7, 0) and (6.7, 0).
The equations of the asymptotes
1
1
are y x and y x.
2
2
20
Translations of Hyperbolas – Standard Forms of
Equations of Hyperbolas Centered at (h, k)
21
Translations of Hyperbolas – Standard Forms of
Equations of Hyperbolas Centered at (h, k) (continued)
22
Example: Graphing a Hyperbola Centered at (h, k)
Graph: 4 x2 24 x 9 y 2 90 y 153 0
We begin by completing the square on x and y.
4 x 24 x 9 y 90 y 153 0
2
2
(4 x 24 x) (9 y 90 y) 153
2
4 x2 6 x
2
9 y 2 10 y
153
4( x2 6 x 9) 9( y 2 10 y 25) 153 36 225
4( x 3)2 9( y 5)2 36
2
2
2
2
4( x 3) 9( y 5) 36
( x 3) ( y 5)
1
36
36
36
9
4
23
Example: Graphing a Hyperbola Centered at (h, k)
(continued)
Graph: 4 x2 24 x 9 y 2 90 y 153 0
( x 3)2 ( y 5) 2
( y 5)2 ( x 3) 2
1
1
9
4
4
9
We will graph
( y 5)2 ( x 3) 2
1
4
9
The center of the graph is (3, –5).
Step 1 Locate the vertices.
2
a 4, a 2.
The vertices are 2 units above and below the center.
24
Example: Graphing a Hyperbola Centered at (h, k)
(continued)
2
2
(
y
5)
(
x
3)
Graph:
1
4
9
Step 1 (cont) Locate the vertices.
The center is (3, –5). The vertices are 2 units above and 2
units below the center.
2 units above (3, –5 + 2) = (3, –3)
2 units below (3, –5 – 2) = (3, –7)
The vertices are (3, –3) and (3, –7).
25
Example: Graphing a Hyperbola Centered at (h, k)
(continued)
2
2
(
y
5)
(
x
3)
Graph:
1
4
9
Step 2 Draw a rectangle.
The rectangle passes through points that are 2 units above
and below the center - these points are the vertices, (3, –3)
and (3, –7). The rectangle passes through points that are 3
units to the right and left of the center.
3 units right (3 + 3, –5) = (6, –5)
3 units left (3 – 3, –5) = (0, –5)
26
Example: Graphing a Hyperbola Centered at (h, k)
(continued)
2
2
(
y
5)
(
x
3)
Graph:
1
4
9
Step 2 (cont) Draw a rectangle.
The rectangle passes through
(3, –3), (3, –7),
(0, –5), and (6, –5).
27
Example: Graphing a Hyperbola Centered at (h, k)
(continued)
2
2
(
y
5)
(
x
3)
Graph:
1
4
9
Step 3 Draw extended diagonals
of the rectangle to obtain the
asymptotes.
28
Example: Graphing a Hyperbola Centered at (h, k)
(continued)
2
2
(
y
5)
(
x
3)
Graph:
1
4
9
Step 3 (cont) Draw extended
diagonals of the rectangle
to obtain the asymptotes.
The asymptotes for the unshifted
2
hyperbola are y x.
3
The asymptotes for the shifted
hyperbola are
29
Example: Graphing a Hyperbola Centered at (h, k)
(continued)
2
2
(
y
5)
(
x
3)
Graph:
1
4
9
Step 3 (cont)
2
The asymptotes for the unshifted hyperbola are y x.
3
The asymptotes for the shifted hyperbola are
2
y 5 ( x 3).
3
2
2
2
y 5 ( x 3)
y5 x2
y x7
3
3
3
2
y 5 ( x 3)
3
2
y5 x2
3
2
y x3
3
30
Example: Graphing a Hyperbola Centered at (h, k)
(continued)
2
2
(
y
5)
(
x
3)
Graph:
1
4
9
Step 4 Draw the two branches
of the hyperbola by starting at
each vertex and approaching
the asymptotes.
31
Example: Graphing a Hyperbola Centered at (h, k)
(continued)
2
2
(
y
5)
(
x
3)
Graph:
1
4
9
The foci are at (3, –5 – c) and (3, –5 + c).
c2 a 2 b2
c 4 9 13
2
c 13
The foci are 3, 5 13 and 3, 5 13 .
32
Example: Graphing a Hyperbola Centered at (h, k)
(continued)
2
2
(
y
5)
(
x
3)
Graph:
1
4
9
The center is (3, –5).
The vertices are (3, –3)
and (3, –7). The foci are
3, 5 13 and 3, 5 13 .
The asymptotes are
2
y x7
3
2
and y x 3.
3
33
Example: An Application Involving Hyperbolas
An explosion is recorded by two microphones that are 2
miles apart. Microphone M1 receives the sound 3 seconds
before microphone M2. Assuming sound travels at 1100
feet per second, determine the possible locations of the
explosion relative to the location of the microphones.
We begin by putting the
microphones in a coordinate
system.
34
Example: An Application Involving Hyperbolas
(continued)
We know that M1 received the sound 3 seconds before M2.
Because sound travels at 1100 feet per second, the
difference between the distance from P to M1 and the
distance from P to M2 is 3300 feet. The set of all points P
(or locations of the explosion)
satisfying these conditions
fits the definition of a
hyperbola, with microphones
M1 and M2 at the foci.
35
Example: An Application Involving Hyperbolas
(continued)
We will use the standard form of the hyperbola’s equation.
P(x, y), the explosion point, lies on this hyperbola.
x2 y 2
2 1
2
a b
The differences between the distances is 3300 feet. Thus,
2a = 3300 and a = 1650.
x2
y2
2 1
2
1650 b
x2
y2
2 1
2,722,500 b
36
Example: An Application Involving Hyperbolas
(continued)
The distance from the center, (0, 0), to either focus
(–5280, 0) or (5280, 0) is 5280. Thus, c = 5280.
c2 a 2 b2
52802 16502 b 2
b2 52802 16502 25,155,900
The equation of the hyperbola with a microphone at each
focus is
x2
y2
1.
2,722,500 25,155,900
We can conclude that the explosion occurred somewhere
on the right branch (the branch closer to M1) of the
hyperbola given by this equation. |
# Geometry for Elementary School/Bisecting an angle
Geometry for Elementary School Why are the constructions not correct? Bisecting an angle Bisecting a segment
BISECT ANGLE $\angle ABC$
1. Use a compass to find points D and E, equidistant from the vertex, point B.
2. Draw the line $\overline{DE}$.
3. Construct an equilateral triangle on $\overline{DE}$ with third vertex F and get $\triangle DEF$. (Lines DF and EF are equal in length).
4. Draw the line $\overline{BF}$.
## Claim
1. The angles $\angle ABF$, $\angle FBC$ equal to half of $\angle ABC$.
## The proof
1. $\overline{DE}$ is a segment from the center to the circumference of $\circ B,\overline{BD}$ and therefore equals its radius.
2. Hence, $\overline{BE}$ equals $\overline{BD}$.
3. $\overline{DF}$ and $\overline{EF}$ are sides of the equilateral triangle $\triangle DEF$.
4. Hence, $\overline{DF}$ equals $\overline{EF}$.
5. The segment $\overline{BF}$ equals to itself
6. Due to the Side-Side-Side congruence theorem the triangles $\triangle ABF$ and $\triangle FBC$ congruent.
7. Hence, the angles $\angle ABF$, $\angle FBC$ equal to half of $\angle ABC$.
## Note
We showed a simple method to divide an angle to two. A natural question that rises is how to divide an angle into other numbers. Since Euclid's days, mathematicians looked for a method for trisecting an angle, dividing it into 3. Only after years of trials it was proven that no such method exists since such a construction is impossible, using only ruler and compass.
## Exercise
1. Find a construction for dividing an angle to 4.
2. Find a construction for dividing an angle to 8.
3. For which other number you can find such constructions? |
# [Solved] If 4 apples and 6 bananas cost \$1.56, and 9 apples and 7 bananas cost \$2.60, what is the cost of one apple and one banana?
Question: If 4 apples and 6 bananas cost \$1.56, and 9 apples and 7 bananas cost \$2.60. What are the cost of one apple and one banana?
We will solve the system of linear equations using the elimination method.
Step 1: Set up the equations
We are given the following information:
• 4 apples and 6 bananas cost \$1.56
• 9 apples and 7 bananas cost \$2.60
Let x be the cost of one apple, and y be the cost of one banana.
We can create the following system of linear equations:
1. 4x + 6y = 1.56
2. 9x + 7y = 2.60
Step 2: Eliminate one variable.
We’ll eliminate y by multiplying both equations by necessary multiples so that the coefficients of y in both equations are the same.
Multiply equation (1) by 7 and equation (2) by 6:
7(4x + 6y) = 7(1.56) 6(9x + 7y) = 6(2.60)
Which results in:
28x + 42y = 10.92 54x + 42y = 15.60
Step 3: Solve for x
Subtract equation (1) from equation (2) to eliminate the y variable:
54x + 42y – (28x + 42y) = 15.60 – 10.92 26x = 4.68
Now, divide by 26 to find the value of x:
x = 4.68 / 26 x ≈ 0.18
Step 4: Solve for y
Now that we have the value of x, we can find the value of y by plugging x back into either equation (1) or (2). We will use equation (1):
4x + 6y = 1.56 4(0.18) + 6y = 1.56 0.72 + 6y = 1.56
Subtract 0.72 from both sides:
6y = 0.84
Divide by 6 to find the value of y:
y ≈ 0.14
Step 5: Interpret the results
The cost of one apple (x) is approximately \$0.18, and the cost of one banana (y) is approximately \$0.14.
## Method 2
Solution: From the information given, we will form an equation.
We will convert the price to cents; it will make calculations easier.
The cost of 4 apples and 6 bananas cost \$1.56=156 ¢
Cost of 9 apples and 7 bananas \$2.60=260 ¢
Let a represent the number of apples.
Let b represent the number of bananas.
According to the question, 4 apples and 6 bananas cost a total of \$1.56, and 9 apples and 7 bananas cost a total of \$2.60
Now, we will combine these equations in such a way that it eliminates either a or b.
There are multiple ways of doing this; we will use the elimination method.
We will choose to first multiply equation 1 by 9 and equation 2 by 4; we get
9(4a + 6b = 156)
4(9a + 7b = 260)
Subtracting equation 4 from 3, we get
Dividing both sides of equation 5 by 26, we get
∴ Value of b = 14
Substituting the value of b in equation 1, we get
4a + 6(14) = 156
4a + 84 = 156
4a = 156 – 84
4a = 72
a = 18
∴Value of a = 18
Hence, the cost of an apple is 18¢ or \$0.18, and the cost of a banana is 14¢ or \$0.14
## Conclusion
If 4 apples and 6 bananas cost \$1.56, 9 apples and 7 bananas cost \$2.60. The cost of an apple is \$0.18, and the cost of a banana is \$0.14 |
Home | Rational Numbers
# Rational Numbers
When you want to count the number of books in your cupboard, you start with 1, 2, 3, … and so on.
Rational Numbers
When you want to count the number of books in your cupboard, you start with 1, 2, 3, … and so on. These counting numbers 1, 2, 3, … , are called Natural numbers. You know to show these numbers on a line (see Fig. 2.1).
We use N to denote the set of all natural numbers.
N= { 1, 2, 3, … }
Suppose there are 5 books in your cupboard and you remove them one by one; the number of books diminish step by step. You remove one, it becomes 4, remove one more, it becomes 3, again one more is removed leaving 2, once again remove one and you are left with 1. If this last one is also taken out, the cupboard is empty (since no books are there). To denote such a situation we use the symbol 0. It denotes absence of any quantity. Thus to say “there are no books”, you can write “the number of books is zero”. Including zero as a digit you can now consider the numbers 0, 1, 2, 3, … and call them Whole numbers. With this additional entity, the number line will look as shown below
We use W to denote the set of all Whole numbers.
W= { 0, 1, 2, 3, … }
Certain conventions lead to more varieties of numbers. Let us agree that certain conventions may be thought of as “positive” denoted by a ‘+’ sign. A thing that is ‘up’ or ‘forward’ or ‘more’ or ‘increasing’ is positive; and anything that is ‘down’ or ‘backward’ or ‘less’ or ‘decreasing’ is “negative” denoted by a ‘–’ sign.
For example, if I make a profit of Rs.1000 in my business, I would call that +1000, and if I lose Rs. 5000, that would be -5000. Why? Similarly, if a mountain’s base is 2 km below sea level and its peak is 3 km above sea level, then the altitude of its base is –2 and the altitude of its peak is +3. (What is its total height? Is it 5 km?).
With this understanding, you can treat natural numbers as positive numbers and rename them as positive integers; thereby you have enabled the entry of negative integers –1, –2, –3, … .
Note that –2 is “more negative” than –1. Therefore, among –1 and –2, you find that –2 is smaller and –1 is bigger. Are –2 and –1 smaller or greater than –3? Think about it.
The number line at this stage may be given as follows:
We use Z to denote the set of all Integers.
Z= { …, –3, –2, –1, 0, 1, 2, 3, … }.
Draw a copy of Fig 2.2. Hold your whole number line up to a mirror on zero. You will see the natural numbers reflected in the mirror. The reflected numbers attached with a minus sign are negative integers. So the numbers to the left of 0 are negative, and the numbers to the right of 0 are positive. But 0 is neither negative nor positive; 0 is just 0. It’s non-committal!
When you look at the figures (Fig. 2.2 and 2.3) above, you are sure to get amused by the gap between any pair of consecutive integers. Could there be some numbers in between?
How did you actually draw the number line N (Fig. 2.1) initially? Draw any line, mark a point 1 on it. From 1, choose another point on its right side at a preferred ‘unit’ distance and call it 2. Repeat this as many times as you desire. To get W (Fig. 2.2), from 1, go one unit on the left to get 0. Now Z is easier; just repeat the exercise on the left side.
You have come across fractions already. How will you mark the point that shows 1/2 on Z? It is just midway between 0 and 1. In the same way, you can plot 1/3, 1/4, 1/5,2 3/4.... etc. You may find that many different fractions are shown by the same point. Can you say ‘why’? Will 5/4 and 10/8 be represented by the same point? Do you think 7/9 and 35/55 represent the same point? You will now easily visualize similar fractions on the left side of zero. These are all fractions of the form a/b where a and b are integers with one restriction that b ≠ 0. (Why?) If a fraction is in decimal form, even then the setting is same.
Because of the connection between fractions and ratios of lengths, we name them as Rational numbers. Here is a rough picture of the situation:
Since a fraction can have many equivalent fractions , there are many possible names for the same rational number. Thus 1/3, 2/6, 8/24 8 all these denote the same rational number.
## 1. Denseness Property of Rational Numbers
Consider a, b where a > b and their AM(Arithmetic Mean) given by a+b / 2. Is this AM a rational number? Let us see.
Thus, for any two rational numbers, their average/midpoint is rational. We can repeat this process indefinitely to produce infinitely many rational numbers.
### Example 2.1
Find any two rational numbers between 1/2 and 2/3.
Solution 1
There is an interesting result that could help you to write instantly rational numbers between any two given rational numbers.
Let us take the same example: Find any two rational numbers between 1/2 and 2/3
Solution 2
Solution 3
Any more new methods to solve? Yes, if decimals are your favourites, then the above example can be given an alternate solution as follows:
Solution 4
Tags : Property, Solved Example Problems | Mathematics , 9th EM Mathematics : Real Numbers
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
9th EM Mathematics : Real Numbers : Rational Numbers | Property, Solved Example Problems | Mathematics |
## Successions
A succession is an ordered sequence of numbers, such as:
Bioprofe | Successions | 01
A succession of real numbers is an application of the set N (set of natural number excluding zero) in the set R of the real numbers.
It is called term of a succession to each one of the elements that make un the succession. To represent the different terms of a sequence, the same letter is used with different subscripts, which indicate the place which that therm occupies in the sequence.
For example:
• In the succession: a) 1, 2, 3, 4, 5, 6,… we have that: a5 = 5, because it is the term of the succession that occupies the fifth place.
• In the succession: b) 2, 4, 6, 8 , 10,… the third term would be written by b3 and it would corresponded to the value 6.
The really important thing when it comes to naming the terms of a sequence is the subscript because it denotes the place it occupies in the sequence. The letters with which the succession is designated are different for different sequences and are usually lowercase letters.
We called general term of a succession to the term that occupies the n-th and it is written with the letter that denotes the succession (for example a) with subscript n:(an).
If we focus on the values taken by the subscripts, we see that they are natural numbers, but the terms of the sequence do not have to be so, that is, the values taken by the sequence are real numbers. Therefore, we can affirm that a succession of real numbers is an application that corresponds to each natural number a real number.
## Ways to define a succession
There are several ways to define a succession:
• Giving its general term: A sequence has infinito terms and is frequently expressed by its general term an, that since an is a function that depends on n, it is enough to give natural values to the indeterminate n to obtain any term of the sequence.
For example: the sequence has as a general rule:
Bioprofe | Successions | 02
We can form the terms of the sequence by successively giving “n” los values 1, 2, 3…, having then:
Bioprofe | Successions | 03
• Giving a property that meets the terms of the succession: If we insure a property that meets the succession, it is easier to know the value of an.
– Ex: thanks to the following properties, get the successions:
1. Succession of pair numbers: 2, 4, 6, 8, …
2. Succession of prime numbers: 2, 3, 5, 7, 11, …
3. Succession of the natural numbers ending in 9: 9, 19, 29, 39, …
• By a law of recurrence: you can get a term from previous ones. – Ex: Knowing that the first term of a succession is 2, and each term, except the first one, is three times the previous term, write the first terms of the sequence.
Bioprofe | Successions | 04
It is said that a succession of real numbers is increasing when it is verified that each term is less than or equal to the next one, that is:
Bioprofe | Successions | 05
On the other side, it is said that a succession is decreasing if each term is greater than or equal to the next one:
Bioprofe | Successions | 06
## Arithmetic progressions
An arithmetic progression is a succession of real numbers in which the difference between two consecutive terms in the sequence is constant. This constant is called “difference of progression” and is usually denoted by the letter d.
In an arithmetic progression, where i is any natural number, it is verified that:
Bioprofe | Successions | 07
That is, each term is obtained by adding the difference to the previous one, d:
Bioprofe | Successions | 08
Depending on the value of the difference “d”, we can find different types of arithmetic sequences.
• If d>0, the succession is increasing, that is, each term is greater than the previous ones.
• If d<0, the succession is decreasing, each term being smaller than the previous ones.
• If d=0, the sequence is constant and all its terms are equal.
For example:
If a1=3 and d=2, What will be the first five terms of the arithmetic progression?
Bioprofe | Successions | 09
## General term of an arithmetic progression
As happens with successions, a progression is perfectly defined if we know its general term. But, if we wanted to find the term 50, its calculation would be very uncomfortable finding all the numbers, therefore, we are interested in finding the expression of the general term.
Knowing the values of a1 and d, and taking into account the definition of arithmetic progression, we can obtain the general term of the same.
Bioprofe | Successions | 10
Generally:
Bioprofe | Successions | 11
Therefore, the general term of an arithmetic progression is:
Bioprofe | Successions | 12
Generalizing this result, we can calculate the general term of an arithmetic progression knowing d and another term of the progression, not necessarily the first one.
More general, being ak the term of the progression that occupies place k, the general term of an arithmetic progression is:
Bioprofe | Successions | 13
Depending on the data we have, we can calculate the general term of an arithmetic progression in one way or another, because:
• If we know a1 and d, we can apply:
Bioprofe | Successions | 14
• If we know any term ai and d, we can use:
Bioprofe | Successions | 15
• If we know two any terms ar and as, by means of the equations:
Bioprofe | Successions | 16
Bioprofe | Successions | 17
we can clear d as a function of r, s, ar and as and therefore, we have:
Bioprofe | Successions | 18
For example:
Knowing that a10=24 y d=3; find the value of a40.
Bioprofe | Successions | 19
## Interpolation of n arithmetic means
We call interpolating “h arithmetic means” between two numbers given “p and q” to intercalate h terms between p and q so that they are in arithmetic progression, with p and q being the extremes.
p …………….. (h terms) ………………… q
For example:
In the case of the arithmetic progression: 2, a, b, c, 14
We can see that a1 = 2 and a5 =14 so the value of p is the first value and q will be the term h+2.
Bioprofe | Successions | 20
The difference of the progression is clearing d from the general expression.
Bioprofe | Successions | 21
Where:
Bioprofe | Successions | 22
Knowing the difference, the arithmetic means can be easily obtained, which will be:
Bioprofe | Successions | 23
## Sum of the consecutive terms of a limited arithmetic progression
Due to the arithmetic progressions have infinite terms, we can do the sum of the consecutive terms of a limited arithmetic progression in the following way:
Being the limited arithmetic progression:
Bioprofe | Successions | 24
The sum of these n terms will be:
Bioprofe | Successions | 25
And too:
Bioprofe | Successions | 26
Adding member to member the two previous equalities results:
Bioprofe | Successions | 27
Grouping and finally clearing, we have: 2Sn = (a1 + an)n
Bioprofe | Successions | 28
Therefore, the sum of the terms of a limited arithmetic progression equals the half-sum of the extremes by the number of terms.
## Geometric progressions
A geometric progression is a succession of real numbers in which the quotient between each term and the previous one is constant. This constant is called the ratio of progression and it is usually denoted by the r. That is, if i is a natural number and whenever ai is non-zero, the value of r is:
Bioprofe | Successions | 29
Or what is the same, each term is obtained by multiplying the previous one for the reason r:
Bioprofe | Successions | 30
## General term of a geometric progression
A geometric progression, due to it is a succession, is totally defined if we know its general term. We are going to obtain it without more than to apply the definition of geometric progression:
Bioprofe | Successions | 31
Therefore, the general term of a geometric progression is:
Bioprofe | Successions | 32
Generalizing this result, we can calculate the general term of a geometric progression knowing r and another term of the progression, not necessarily the first one. Being ak the term of the progression that occupies place k, the general term of a geometric progression is:
Bioprofe | Successions | 33
Depending on the value of r, we can find different types of geometric progressions:
• If r >1, the progression is growing, that is, each term is greater than the previous ones. {2, 4, 8, 16…}
• If 0< r <1, the progression is decreasing, that is, each term is smaller than the previous ones. {90, 30, 10,…}
• If r <0, the progression is alternate, that is, its terms change their sign according to the value of n. {-2, 4, -8, 16…}
• If r = 0, the progression is the progression formed by zeros from the second term. {7, 0, 0, 0,…}
• If r = 1, the progression is the constant progression formed by the first term.{2, 2, 2, 2,…}
Depending on the data we have, we can calculate the general term of a geometric progression in one way or another:
• If we know a1 and r, we can apply: an = a1 · r n-1
• If we know any term ak and r, we can use: an = ak · r n-k.
• If we know two any tems ap and aq, with ap no null, we need to know the reason r to be able to apply the previous reason. But we know that: an = ap · r n-p and an = aq · r n-q . We can clear r according to p, q, ap and aq and we have finally:
Bioprofe | Successions | 34
## Product of the terms of a geometric progression
In a geometric progression, the producto of two equidistante terms is constant. That is, if the natural subscripts p, q, t and s verify that p+q = t+s, then: ap · aq = at · as because:
ap·aq = a1 · rp-1·a1 · rq-1 = a12 ·rp-1 ·rq-1 = a12 · rp+q-2
at·as = a1 · rt-1·a1 · rs-1 = a12 ·rt-1 ·rs-1 = a12 · rt+s-2
And as: p+q = t+s, so: ap · aq= at · as
We can calculate the product of the n terms of a geometric progression, Pn. That is:
Pn= a1 · a2 · a3 ·……….· an-2 · an-1 · an
Applying the conmmtative property of the product, we have:
Pn = an · an-1 · an-2 · ……..· a3 · a2· a1
Multiplying these two equalities, we have:
Pn2 = (a1 · a2· a3·………· an-2· an-1·an) · (an· an-1· an-2·……· a3 · a2· a1)
Pn2 = (a1 · an) · (a2· an-1) · (a3· an-2)·……· (an-2 · a3) · (an1 · a2) · (an·a1)
As we can see, the subscripts corresponding to each pair of terms in parentheses add n + 1, so the product will always be the same in each factor, then: Pn2 = (a1 · an)n
The product of the first n terms of a geometric progression is given by:
Bioprofe | Successions | 35
## Sum of the terms of a geometric progression
The sum of the first n terms of a geometric progression, provided that r is different from 1, is given by:
Bioprofe | Successions | 36
According to the values of r, the value of Sn will be:
• If |r| > 1, the terms in absolute value grow indefinitely and the value of Sn s given by the previous formula.
• If |r| < 1, the sum of its terms when n is large, Sn approaches to a a/(1-r), since if in
Bioprofe | Successions | 37
we raise the reason |r|<1 at a power, the greater is the exponent, the smaller is the value of rn and if n is large enough, rn approaches 0. Therefore,
Bioprofe | Successions | 38
• If r = – 1, the consecutive terms are opposite: {a1, -a1, a1, -a1, ….} and Sn is equal to zero if n is even, and equal to a1 if n is odd. The sum of the series oscillates between these two values.
But could we add an unlimited number of consecutive terms of a geometric progression? Depending on the value of r it will be possible or not to obtain the sum of an unlimited number of terms:
• If r = 1, the progression is the constant progression formed by the first term: {a1, a1, a1, a1, …} and if a1 s positive, the sum of the terms will be increasing (if it were a1 negative it would be the increasing sum in absolute value, but negative). Therefore, if the number of terms is unlimited, this sum will be infinite.
• If |r| > 1, the terms grow indefinitely and the value of Sn for an unlimited number of terms, will also be infinite.
• If |r| < 1, the sum of its terms is approximated when n is large to a1/(1-r)
• If r = -1, the consecutive terms are opposite: {a1, -a1, a1, -a1, … } and Sn is equal to zero if n is even, and equal to a1 if n is odd. The sum of the series oscillates between these two values for a finite number of terms. For an unlimited number of terms we don not know if it is even or odd, so that the sum can not be realized unless that a1=0, case in which S = 0 = a1/(1-r). In the rest of the cases we say that the sum of infinite terms does not exist because its value is oscillating.
• If r < -1, the terms oscillate between positive and negative values, growing in absolute value. The sum of its infinite terms does not exist because its value is also oscillating.
Therefore, we can affirm that the sum of an unlimited number of terms of a geometric progression only takes a finite value if |r| < 1, and then it is given by:
Bioprofe | Successions | 39
In the rest of the cases, or it is infinite, or it does not exist as it oscillates.
For example:
Calculate the sum of all the terms of the geometric progression whose first term is 4 and the ratio is 1/2.
Bioprofe | Successions | 40
## Applications of the geometric progressions
### Generating fraction
An exacto or periodic number can be expressed as a fraction, “generating fraction” as follows:
• Calculation of the generation fraction of a pure periódico decimal expression.
We try to find a fracción such that dividing the numerator between the denominator gives us the starting number. That is:
Bioprofe | Successions | 41
We observe that it is the sum of the infinite terms of an unlimited geometric progression of ratio 1/100. Thus:
Bioprofe | Successions | 42
Calculation of the generativa fracción of a mixed periódico decimal.
In this case, the easiest is to decomise the expression as the sum of the two decimal expressions. That is:
Bioprofe | Successions | 43
### Compound capitalization
If we depositó in a Financial entity a quantity of money C0 during a time t and a credit r given in as much by one, we will obtain a benefit known as interes, and given by the formula:
Bioprofe | Successions | 44
The main characteristic of the compound capitalization is that the interest generated in a year, become part of the initial capital and produce interest in the following periods.
The final capital obtained after n years given an initial capital C0 and a credit r given in as much by one, is:
Bioprofe | Successions | 45 |
# Simplify a Ratio to the Form 1:n or m:1
In this worksheet, students will create and simplify ratios in the form 1:n or m:1 where the values of m and n can be decimals, converting values into the same units where required.
Key stage: KS 4
GCSE Subjects: Maths
GCSE Boards: AQA, Eduqas, Pearson Edexcel, OCR
Curriculum topic: Ratio, Proportion and Rates of Change
Curriculum subtopic: Ratio, Proportion and Rates of Change, Calculations with Ratio
Difficulty level:
### QUESTION 1 of 10
By now, you may have had a lot of practice in simplifying fractions and ratios into their simplest possible form by finding and applying the Highest Common Factor (HCF).
One of the things that you may have heard is that you cannot have decimals in the simplified versions of ratios.
There is, however, one important exception to this rule.
You could be asked to write a ratio in the form 1:n or m:1.
All this means is that either the first number must be 1 (1:n) or the second must be 1 (m:1).
This approach frequently means that you will have a decimal representing n or m.
In this situation alone, using a decimal in a ratio is allowed.
Let's look at this in action with some examples now.
e.g. Write 5:8 in the form 1:n.
As we are looking for 1:n, this means that the first number in our ratio must be 1 and the second can be a decimal or whatever we require.
Let's begin with our starting ratio:
5:8
In order to convert the first number to 1, we have to divide by 5 (as 5 ÷ 5 = 1).
So we need to do the same thing to the second number in the ratio:
5:8 ÷ 5 = 1 : 1.6
e.g. Write 25:10 in the form m:1.
As we are looking for m:1, this means that the second number in our ratio must be 1 and the first can be a decimal or whatever we require.
In order to convert the second number to 1, we have to divide by 10 (as 10 ÷ 10 = 1).
25:10 ÷ 10 = 2.5 : 1
In this activity, we will create and simplify ratios in the form 1:n or m:1 where the values of m and n can be decimals, converting values into the same units where required.
Type a number into the first space and a word into the second space to complete the sentence below.
A number in a ratio can only be a decimal if it is represented in the form 1:n.
Is the statement above true or false
True
False
Simplify the ratio 5:4 into the form 1:n.
True
False
Simplify 28:10 into the form m:1.
True
False
The ratios on the left below have been cancelled into the form m:1.
Match each uncancelled ratio with its correct simplified version.
## Column B
10:5
0.5 : 1
5:10
0.4 : 1
4:10
2:1
32:5
6.4 : 1
The ratios on the left below have been cancelled into the form 1:n.
Match each uncancelled ratio with its correct simplified version.
## Column B
10:5
1 : 0.15625
5:10
1 : 2.5
4:10
1:2
32:5
1 : 0.5
Write a ratio to compare £1 to 70 p in the form 1:n.
## Column B
10:5
1 : 0.15625
5:10
1 : 2.5
4:10
1:2
32:5
1 : 0.5
Write a ratio to compare 950 m to 1 km in the form m:1.
## Column B
10:5
1 : 0.15625
5:10
1 : 2.5
4:10
1:2
32:5
1 : 0.5
Which of these could be correct simplifications of 28:40?
Please note, where required, decimals have been rounded to two decimal places.
7:10
1.43 : 1
1 : 1.43
0.7 : 1
1 : 0.7
Simplify 28:35:42 into the form n:1:m.
7:10
1.43 : 1
1 : 1.43
0.7 : 1
1 : 0.7
• Question 1
Type a number into the first space and a word into the second space to complete the sentence below.
CORRECT ANSWER
EDDIE SAYS
The hardest thing to recall here is the exception to our usual rule - in this case only, one of the numbers in our ratio can be a decimal. To express a ratio in the form 1:n, we need to convert the first number into 1, then complete the same operation on the other side, which may result in the creation of a decimal.
• Question 2
A number in a ratio can only be a decimal if it is represented in the form 1:n.
Is the statement above true or false
CORRECT ANSWER
False
EDDIE SAYS
This was a bit of a trick question! We can only use decimals in ratios on two occasions - when we are expressing a ratio in the form 1:n, but also in the form m:1. If you can commit these situations to memory, then you will never use a decimal in a ratio in the wrong circumstances.
• Question 3
Simplify the ratio 5:4 into the form 1:n.
CORRECT ANSWER
EDDIE SAYS
As we are looking for 1:n, this means that the first number in our ratio must be 1 and the second can be a decimal (or whatever we require). Let's begin with our starting ratio: 5:4 In order to convert the first number to 1, we have to divide by 5 (as 5 ÷ 5 = 1). So we need to do the same thing to the second number in the ratio: 5:4 ÷ 5 = 1 : 0.8
• Question 4
Simplify 28:10 into the form m:1.
CORRECT ANSWER
EDDIE SAYS
As we are looking for m:1, this means that the second number in our ratio must be 1 this time. Let's begin with our starting ratio: 28:10 In order to convert the second number to 1, we have to divide by 10 (as 10 ÷ 10 = 1). So we need to do the same thing to the second number in the ratio: 28:10 ÷ 10 = 2.8 : 1
• Question 5
The ratios on the left below have been cancelled into the form m:1.
Match each uncancelled ratio with its correct simplified version.
CORRECT ANSWER
## Column B
10:5
2:1
5:10
0.5 : 1
4:10
0.4 : 1
32:5
6.4 : 1
EDDIE SAYS
As we are looking for m:1, this means that the second number in our ratio must be 1 in all these cases. Let's work through each starting ratio, one at a time. 10:5 > ÷ 5 = 2:1 5:10 > ÷ 10 = 0.5 : 1 4:10 > ÷ 10 = 0.4 : 1 32:5 > ÷ 5 = 6.4 : 1 Remember that our second number does not have to be a decimal, it is simply allowed to be if required.
• Question 6
The ratios on the left below have been cancelled into the form 1:n.
Match each uncancelled ratio with its correct simplified version.
CORRECT ANSWER
## Column B
10:5
1 : 0.5
5:10
1:2
4:10
1 : 2.5
32:5
1 : 0.15625
EDDIE SAYS
As we are looking for 1:n, this means that the first number in our ratio must be 1 in all these cases. Let's work through each starting ratio, one at a time. 10:5 > ÷ 10 = 1 : 0.5 5:10 > ÷ 5 = 1:2 4:10 > ÷ 4 = 1 : 2.5 32:5 > ÷ 32 = 1 : 0.15625 Can you see how the same ratios are converted into totally different amounts, depending on whether we are asked to apply 1:n or m:1?
• Question 7
Write a ratio to compare £1 to 70 p in the form 1:n.
CORRECT ANSWER
EDDIE SAYS
As we are looking for 1:n, this means that the first number in our ratio must be 1. The difficulty here is that we must find our starting ratio, using amounts which are expressed in different units (£ and p). So, to start with, we need to convert them into the same format. There are 100 p in £1, which leads us to the starting ratio (when we remove our units as they are now the same) of: 100:70 In order to convert the first number to 1, we have to divide by 100 (as 100 ÷ 100 = 1). So we need to do the same thing to the second number in the ratio: 100:70 ÷ 100 = 1 : 0.7
• Question 8
Write a ratio to compare 950 m to 1 km in the form m:1.
CORRECT ANSWER
EDDIE SAYS
As we are looking for m:1, this means that the second number in our ratio must be 1. Again, our units are different here, so we need to get them into the same format. There are 1000 m in 1 km, which leads us to the starting ratio (when we remove our units as they are now the same) of: 1000:950 1000:950 ÷ 1000 = 1 : 0.95
• Question 9
Which of these could be correct simplifications of 28:40?
Please note, where required, decimals have been rounded to two decimal places.
CORRECT ANSWER
7:10
1 : 1.43
0.7 : 1
EDDIE SAYS
It's important to notice here that the question doesn't specifically ask for 1:n or m:1, but rather a 'simplified version' in general. To test each option, we need to find the divisor and check it if works in both elements. e.g. For 7:10, we need to consider what 28:40 could have been divided by to reach this. The relationship between 28 and 7 would be '÷ 4'. Let's check that works in the other number: 40 ÷ 2 = 10 So the ratio 7:10 is an equivalent ratio to 28:40. How about using 1:n or m:1? If we use 1:n, then we need to convert our first number in the ratio to 1. To do this we divide by 28, which leads us to 1: 1.43 (to 2 decimal places). Can you find the equivalent ratio for m:1 independently? Did you spot that the three equivalent ratios present within these options?
• Question 10
Simplify 28:35:42 into the form n:1:m.
CORRECT ANSWER
EDDIE SAYS
This one may look confusing, but it is really just tying everything together. All this means is that the number in the middle has to be 1. To achieve this, we need to divide all parts of the ratio by 35: 28:35:42 ÷ 35 = 0.8 : 1 : 1.2 You can now create and simplify ratios in the form 1:n or m:1 where the values of m and n can be decimals, converting values into the same units where required. If you are feeling on a roll, why not try another activity to practise the concept of ratio more?
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# Calculus/Algebra
← Precalculus Calculus Trigonometry → Algebra
The purpose of this section is for readers to review important algebraic concepts. It is necessary to understand algebra in order to do calculus. If you are confident of your ability, you may skim through this section.
## Rules of arithmetic and algebra
The following laws are true for all ${\displaystyle a,b,c}$ in ${\displaystyle \mathbb {R} }$ whether these are numbers, variables, functions, or more complex expressions involving numbers, variable and/or functions.
• Commutative Law: ${\displaystyle a+b=b+a}$ .
• Associative Law: ${\displaystyle (a+b)+c=a+(b+c)}$ .
• Additive Identity: ${\displaystyle a+0=a}$ .
• Additive Inverse: ${\displaystyle a+(-a)=0}$ .
### Subtraction
• Definition: ${\displaystyle a-b=a+(-b)}$ .
### Multiplication
• Commutative law: ${\displaystyle a\times b=b\times a}$ .
• Associative law: ${\displaystyle (a\times b)\times c=a\times (b\times c)}$ .
• Multiplicative identity: ${\displaystyle a\times 1=a}$ .
• Multiplicative inverse: ${\displaystyle a\times {\frac {1}{a}}=1}$ , whenever ${\displaystyle a\neq 0}$
• Distributive law: ${\displaystyle a\times (b+c)=(a\times b)+(a\times c)}$.
### Division
• Definition: ${\displaystyle {\frac {a}{b}}=r+nb}$, where r is the remainder of a when divided by b, and n is an integer.
• Definition: ${\displaystyle {\frac {a}{b}}=a\times {\frac {1}{b}}}$ , whenever ${\displaystyle b\neq 0}$ .
Let's look at an example to see how these rules are used in practice.
${\displaystyle {\frac {(x+2)(x+3)}{x+3}}}$ ${\displaystyle =\left[(x+2)\times (x+3)\right]\times \left({\frac {1}{x+3}}\right)}$ (from the definition of division) ${\displaystyle =(x+2)\times \left[(x+3)\times \left({\frac {1}{x+3}}\right)\right]}$ (from the associative law of multiplication) ${\displaystyle =((x+2)\times (1)),\qquad x\neq -3}$ (from multiplicative inverse) ${\displaystyle =x+2,\qquad x\neq -3}$ (from multiplicative identity)
Of course, the above is much longer than simply cancelling ${\displaystyle x+3}$ out in both the numerator and denominator. However, it is important to know what the rules are so as to know when you are allowed to cancel. Occasionally people do the following, for instance, which is incorrect:
${\displaystyle {\frac {2\times (x+2)}{2}}={\frac {2}{2}}\times {\frac {x+2}{2}}=1\times {\frac {x+2}{2}}={\frac {x+2}{2}}}$ .
The correct simplification is
${\displaystyle {\frac {2\times (x+2)}{2}}=\left(2\times {\frac {1}{2}}\right)\times (x+2)=1\times (x+2)=x+2}$ ,
where the number ${\displaystyle 2}$ cancels out in both the numerator and the denominator.
## Interval notation
There are a few different ways that one can express with symbols a specific interval (all the numbers between two numbers). One way is with inequalities. If we wanted to denote the set of all numbers between, say, 2 and 4, we could write "all ${\displaystyle x}$ satisfying ${\displaystyle 2". This excludes the endpoints 2 and 4 because we use ${\displaystyle <}$ instead of ${\displaystyle \leq }$. If we wanted to include the endpoints, we would write "all ${\displaystyle x}$ satisfying ${\displaystyle 2\leq x\leq 4}$ ."
Another way to write these intervals would be with interval notation. If we wished to convey "all ${\displaystyle x}$ satisfying ${\displaystyle 2" we would write ${\displaystyle (2,4)}$. This does not include the endpoints 2 and 4. If we wanted to include the endpoints we would write ${\displaystyle [2,4]}$. If we wanted to include 2 and not 4 we would write ${\displaystyle [2,4)}$; if we wanted to exclude 2 and include 4, we would write ${\displaystyle (2,4]}$.
Thus, we have the following table:
Endpoint conditions Inequality notation Interval notation
Including both 2 and 4 all ${\displaystyle x}$ satisfying ${\displaystyle 2\leq x\leq 4}$
${\displaystyle [2,4]}$
Not including 2 nor 4 all ${\displaystyle x}$ satisfying ${\displaystyle 2
${\displaystyle (2,4)}$
Including 2 not 4 all ${\displaystyle x}$ satisfying ${\displaystyle 2\leq x<4}$
${\displaystyle [2,4)}$
Including 4 not 2 all ${\displaystyle x}$ satisfying ${\displaystyle 2
${\displaystyle (2,4]}$
In general, we have the following table, where ${\displaystyle a,b\in \mathbb {R} }$.
Meaning Interval Notation Set Notation
All values greater than or equal to ${\displaystyle a}$ and less than or equal to ${\displaystyle b}$ ${\displaystyle [a,b]}$ ${\displaystyle \{x:a\leq x\leq b\}}$
All values greater than ${\displaystyle a}$ and less than ${\displaystyle b}$ ${\displaystyle (a,b)}$ ${\displaystyle \{x:a
All values greater than or equal to ${\displaystyle a}$ and less than ${\displaystyle b}$ ${\displaystyle [a,b)}$ ${\displaystyle \{x:a\leq x
All values greater than ${\displaystyle a}$ and less than or equal to ${\displaystyle b}$ ${\displaystyle (a,b]}$ ${\displaystyle \{x:a
All values greater than or equal to ${\displaystyle a}$ ${\displaystyle [a,\infty )}$ ${\displaystyle \{x:x\geq a\}}$
All values greater than ${\displaystyle a}$ ${\displaystyle (a,\infty )}$ ${\displaystyle \{x:x>a\}}$
All values less than or equal to ${\displaystyle a}$ ${\displaystyle (-\infty ,a]}$ ${\displaystyle \{x:x\leq a\}}$
All values less than ${\displaystyle a}$ ${\displaystyle (-\infty ,a)}$ ${\displaystyle \{x:x
All values ${\displaystyle (-\infty ,\infty )}$ ${\displaystyle \{x:x\in \mathbb {R} \}}$
Note that ${\displaystyle \infty }$ and ${\displaystyle -\infty }$ must always have an exclusive parenthesis rather than an inclusive bracket. This is because ${\displaystyle \infty }$ is not a number, and therefore cannot be in our set. ${\displaystyle \infty }$ is really just a symbol that makes things easier to write, like the intervals above.
The interval ${\displaystyle (a,b)}$ is called an open interval, and the interval ${\displaystyle [a,b]}$ is called a closed interval.
Intervals are sets and we can use set notation to show relations between values and intervals. If we want to say that a certain value is contained in an interval, we can use the symbol ${\displaystyle \in }$ to denote this. For example, ${\displaystyle 2\in [1,3]}$ . Likewise, the symbol ${\displaystyle \notin }$ denotes that a certain element is not in an interval. For example ${\displaystyle 0\notin (0,1)}$ .
There are a few rules and properties involving exponents and radicals. As a definition we have that if ${\displaystyle n}$ is a positive integer then ${\displaystyle a^{n}}$ denotes ${\displaystyle n}$ factors of ${\displaystyle a}$ . That is,
${\displaystyle a^{n}=a\cdot a\cdot a\cdots a\qquad (n~{\mbox{times}})}$
If ${\displaystyle a\neq 0}$ then we say that ${\displaystyle a^{0}=1}$ .
If ${\displaystyle n}$ is a negative integer then we say that ${\displaystyle a^{-n}={\frac {1}{a^{n}}}}$ .
If we have an exponent that is a fraction then we say that ${\displaystyle a^{\frac {m}{n}}={\sqrt[{n}]{a^{m}}}=({\sqrt[{n}]{a}})^{m}}$ . In the expression ${\displaystyle {\sqrt[{n}]{a}}}$ , ${\displaystyle n}$ is called the index of the radical, the symbol ${\displaystyle {\sqrt {\;\;}}}$ is called the radical sign, and ${\displaystyle a}$ is called the radicand.
In addition to the previous definitions, the following rules apply:
Rule Example
${\displaystyle a^{n}\cdot a^{m}=a^{n+m}}$ ${\displaystyle 3^{6}\cdot 3^{9}=3^{15}}$
${\displaystyle (a^{n})^{m}=a^{n\cdot m}}$ ${\displaystyle (x^{4})^{5}=x^{20}}$
${\displaystyle (ab)^{n}=a^{n}b^{n}}$ ${\displaystyle (3x)^{5}=3^{5}x^{5}}$
We will use the following conventions for simplifying expressions involving radicals:
1. Given the expression ${\displaystyle a^{\frac {b}{c}}}$, write this as ${\displaystyle {\sqrt[{c}]{a^{b}}}}$
2. No fractions under the radical sign
3. No radicals in the denominator
4. The radicand has no exponentiated factors with exponent greater than or equal to the index of the radical
Example: Simplify the expression ${\displaystyle \left({\frac {1}{8}}\right)^{\frac {1}{2}}}$ Using convention 1, we rewrite the given expression as (1) ${\displaystyle \left({\frac {1}{8}}\right)^{\frac {1}{2}}={\sqrt[{2}]{\left({\frac {1}{8}}\right)^{1}}}={\sqrt {\frac {1}{8}}}}$ The expression now violates convention 2. To get rid of the fraction in the radical, apply the rule ${\displaystyle \left({\frac {a}{b}}\right)^{n}={\frac {a^{n}}{b^{n}}}}$ and simplify the result: (2) ${\displaystyle {\sqrt {\frac {1}{8}}}={\frac {\sqrt {1}}{\sqrt {8}}}={\frac {1}{\sqrt {8}}}}$ The resulting expression violates convention 3. To get rid of the radical in the denominator, multiply by ${\displaystyle {\frac {\sqrt {8}}{\sqrt {8}}}}$: (3) ${\displaystyle {\frac {1}{\sqrt {8}}}={\frac {1}{\sqrt {8}}}\cdot {\frac {\sqrt {8}}{\sqrt {8}}}={\frac {\sqrt {8}}{8}}}$ Notice that ${\displaystyle 8=2^{3}}$. Since the index of the radical is 2, our expression violates convention 4. We can reduce the exponent of the expression under the radical as follows: (4) ${\displaystyle {\frac {\sqrt {8}}{8}}={\frac {\sqrt {2^{3}}}{8}}={\frac {\sqrt {2^{2}\cdot 2}}{8}}={\frac {2\cdot {\sqrt {2}}}{8}}={\frac {\sqrt {2}}{4}}}$
#### Exercise
${\displaystyle 144^{\frac {5}{3}}=}$ ${\displaystyle {\sqrt[{3}]{}}}$
## Logarithms
Consider the equation
(5) ${\displaystyle y=b^{x}}$
${\displaystyle b}$ is called the base and ${\displaystyle x}$ is called the exponent. Suppose we would like to solve for ${\displaystyle x}$ . We would like to apply an operation to both sides of the equation that will get rid of the base on the right-hand side of the equation. The operation we want is called the logarithm, or log for short, and it is defined as follows:
Definition: (Formal definition of a logarithm)
${\displaystyle \log _{b}y=x}$ exactly if ${\displaystyle y=b^{x}}$ and ${\displaystyle x>0}$, ${\displaystyle b>0}$, and ${\displaystyle b\neq 1}$.
Logarithms are taken with respect to some base. What the equation is saying is that, when ${\displaystyle x}$ is the exponent of ${\displaystyle b}$, the result will be ${\displaystyle y}$.
### Example
Example: Calculate ${\displaystyle \log _{10}100000}$ ${\displaystyle \log _{10}100000}$ is the number ${\displaystyle x}$ such that ${\displaystyle 10^{x}=100000}$. Well ${\displaystyle 10^{5}=100000}$, so ${\displaystyle \log _{10}100000=5}$
### Common bases for logarithms
When the base is not specified, ${\displaystyle \log }$ is taken to mean the base 10 logarithm. Later on in our study of calculus we will commonly work with logarithms with base ${\displaystyle e\approx 2.718282}$ . In fact, the base ${\displaystyle e}$ logarithm comes up so often that it has its own name and symbol. It is called the natural logarithm, and its symbol is ${\displaystyle \ln }$ . In computer science the base 2 logarithm often comes up.
### Properties of logarithms
Logarithms have the property that ${\displaystyle \log _{b}x+\log _{b}y=\log _{b}(x\cdot y)}$ . To see why this is true, suppose that:
${\displaystyle \log _{b}x=r}$ and ${\displaystyle \log _{b}y=s}$
These assumptions imply that
${\displaystyle x=b^{r}}$ and ${\displaystyle y=b^{s}}$
Then by the properties of exponents
${\displaystyle x\cdot y=b^{r}\cdot b^{s}=b^{r+s}}$
According to the definition of the logarithm
${\displaystyle \log _{b}(x\cdot y)=r+s=\log _{b}x+\log _{b}y}$
Similarly, the property that ${\displaystyle \log _{b}x-\log _{b}y=\log _{b}\left({\frac {x}{y}}\right)}$ also hold true using the same method.
Historically, the development of logarithms was motivated by the usefulness of this fact for simplifying hand calculations by replacing tedious multiplication by table look-ups and addition.
#### Logarithmic powers and roots
Another useful property of logarithms is that ${\displaystyle \log _{b}a^{c}=c\cdot \log _{b}a}$ . To see why, consider the expression ${\displaystyle \log _{b}a^{c}}$ . Let us assume that
${\displaystyle \log _{b}a^{c}=x}$
By the definition of the logarithm
${\displaystyle a^{c}=b^{x}}$
Now raise each side of the equation to the power ${\displaystyle {\frac {1}{c}}}$ and simplify to get
${\displaystyle a=b^{\frac {x}{c}}}$
Now if you take the base ${\displaystyle b}$ log of both sides, you get
${\displaystyle \log _{b}a={\frac {x}{c}}}$
Solving for ${\displaystyle x}$ shows that
${\displaystyle x=c\cdot \log _{b}a}$
Similarly, the expression ${\displaystyle \log _{b}{\sqrt[{c}]{a}}={\frac {\log _{b}a}{c}}}$ holds true using the same methods.
### Converting between bases
Most scientific calculators have the ${\displaystyle \log }$ and ${\displaystyle \ln }$ functions built in, which do not include logarithms with other bases. Consider how one might compute ${\displaystyle \log _{b}a}$, where ${\displaystyle b}$ and ${\displaystyle a}$ are given known numbers, when we can only compute logarithms in some base ${\displaystyle \beta }$. First, let us assume that
${\displaystyle \log _{b}a=x}$
Then the definition of logarithm implies that
${\displaystyle a=b^{x}}$
If we take the base ${\displaystyle \beta }$ log of each side, we get
${\displaystyle \log _{\beta }a=\log _{\beta }b^{x}=x\cdot \log _{\beta }b}$
Solving for ${\displaystyle x}$ , we find that
${\displaystyle x={\frac {\log _{\beta }a}{\log _{\beta }b}}}$
For example, if we only use base 10 to calculate ${\displaystyle \log _{2}45}$, we get ${\displaystyle \log _{2}45={\frac {\log 45}{\log 2}}\approx {\frac {1.653212514}{0.301029996}}\approx 5.491853096}$ .
### Identities of logarithms summary
A table is provided below for a summary of logarithmic identities.
Formula Example
Product ${\displaystyle \log _{b}(x\cdot y)=\log _{b}x+\log _{b}y}$ ${\displaystyle \log _{2}32=\log _{2}(4\cdot 8)=\log _{2}4+\log _{2}8=2+3=5}$
Quotient ${\displaystyle \log _{b}\!{\frac {x}{y}}=\log _{b}x-\log _{b}y}$ ${\displaystyle \log _{2}16=\log _{2}\!{\frac {64}{4}}=\log _{2}64-\log _{2}4=6-2=4}$
Power ${\displaystyle \log _{b}\left(a^{c}\right)=c\log _{b}a}$ ${\displaystyle \log _{2}64=\log _{2}\left(2^{6}\right)=6\log _{2}2=6}$
Root ${\displaystyle \log _{b}{\sqrt[{c}]{a}}={\frac {\log _{b}a}{c}}}$ ${\displaystyle \log _{10}{\sqrt {1000}}={\frac {1}{2}}\log _{10}1000={\frac {3}{2}}=1.5}$
Change of base ${\displaystyle \log _{b}a={\frac {\log _{\beta }a}{\log _{\beta }b}}}$ ${\displaystyle \log _{9}243={\frac {\log _{3}243}{\log _{3}9}}={\frac {5}{2}}=2.5}$
## Factoring and roots
Given the expression ${\displaystyle x^{2}+3x+2}$ , one may ask "what are the values of ${\displaystyle x}$ that make this expression 0?" If we factor we obtain
${\displaystyle x^{2}+3x+2=(x+2)(x+1)}$
.
If ${\displaystyle x=-1,-2}$ , then one of the factors on the right becomes zero. Therefore, the whole must be zero. So, by factoring we have discovered the values of ${\displaystyle x}$ that render the expression zero. These values are termed "roots." In general, given a quadratic polynomial ${\displaystyle px^{2}+qx+r}$ that factors as
${\displaystyle px^{2}+qx+r=(ax+c)(bx+d)}$
then we have that ${\displaystyle x=-{\frac {c}{a}}}$ and ${\displaystyle x=-{\frac {d}{b}}}$ are roots of the original polynomial.
A special case to be on the look out for is the difference of two squares, ${\displaystyle a^{2}-b^{2}}$ . In this case, we are always able to factor as
${\displaystyle a^{2}-b^{2}=(a+b)(a-b)}$
For example, consider ${\displaystyle 4x^{2}-9}$ . On initial inspection we would see that both ${\displaystyle 4x^{2}}$ and ${\displaystyle 9}$ are squares of ${\displaystyle 2x}$ and ${\displaystyle 3}$, respectively. Applying the previous rule we have
${\displaystyle 4x^{2}-9=(2x+3)(2x-3)}$
### The AC method
There is a way of simplifying the process of factoring using the AC method. Suppose that a quadratic polynomial has a formula of
${\displaystyle x^{2}+qx+r}$
If there are numbers ${\displaystyle a}$ and ${\displaystyle b}$ that satisfy both
${\displaystyle a\cdot b=r}$ and ${\displaystyle a+b=q}$
Then, the result of factoring will be
${\displaystyle x^{2}+qx+r=(x+a)(x+b)}$
Given any quadratic equation ${\displaystyle ax^{2}+bx+c=0\ ,\ a\neq 0}$, all solutions of the equation are given by the quadratic formula:
${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$
Note that the value of ${\displaystyle b^{2}-4ac}$ will affect the number of real solutions of the equation.
If Then
${\displaystyle b^{2}-4ac>0}$ There are two real solutions to the equation
${\displaystyle b^{2}-4ac=0}$ There is only one real solution to the equation
${\displaystyle b^{2}-4ac<0}$ There are no real solutions to the equation
Example: Find all the roots of ${\displaystyle 4x^{2}+7x-2}$ Finding the roots is equivalent to solving the equation ${\displaystyle 4x^{2}+7x-2=0}$ . Applying the quadratic formula with ${\displaystyle a=4\ ,\ b=7\ ,\ c=-2}$ , we have: ${\displaystyle x={\frac {-7\pm {\sqrt {7^{2}-4(4)(-2)}}}{2(4)}}}$ ${\displaystyle x={\frac {-7\pm {\sqrt {49+32}}}{8}}}$ ${\displaystyle x={\frac {-7\pm {\sqrt {81}}}{8}}}$ ${\displaystyle x={\frac {-7\pm 9}{8}}}$ ${\displaystyle x={\frac {2}{8}}\ ,\ x={\frac {-16}{8}}}$ ${\displaystyle x={\frac {1}{4}}\ ,\ x=-2}$
The quadratic formula can also help with factoring, as the next example demonstrates.
Example: Factor the polynomial ${\displaystyle 4x^{2}+7x-2}$ We already know from the previous example that the polynomial has roots ${\displaystyle x={\frac {1}{4}}}$ and ${\displaystyle x=-2}$ . Our factorization will take the form ${\displaystyle C(x+2)\left(x-{\tfrac {1}{4}}\right)}$ All we have to do is set this expression equal to our polynomial and solve for the unknown constant C: ${\displaystyle C(x+2)\left(x-{\tfrac {1}{4}}\right)=4x^{2}+7x-2}$ ${\displaystyle C\left(x^{2}+\left(-{\tfrac {1}{4}}+2\right)x-{\tfrac {2}{4}}\right)=4x^{2}+7x-2}$ ${\displaystyle C\left(x^{2}+{\tfrac {7}{4}}x-{\tfrac {1}{2}}\right)=4x^{2}+7x-2}$ You can see that ${\displaystyle C=4}$ solves the equation. So the factorization is ${\displaystyle 4x^{2}+7x-2=4(x+2)\left(x-{\tfrac {1}{4}}\right)=(x+2)(4x-1)}$
### Vieta's formulae
Vieta's formulae relate the coefficients of a polynomial to sums and products of its roots. It is very convenient because under certain circumstances when the sums and products of the quadratic's roots are provided, one does not require to solve the whole quadratic polynomial.
Given any quadratic equation ${\displaystyle ax^{2}+bx+c=0\ ,\ a\neq 0}$, The roots ${\displaystyle x_{1},x_{2}}$ of the quadratic polynomial satisfy
${\displaystyle x_{1}+x_{2}=-{\frac {b}{a}},\quad x_{1}x_{2}={\frac {c}{a}}.}$
## Simplifying rational expressions
Consider the two polynomials
${\displaystyle p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}}$
and
${\displaystyle q(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots +b_{1}x+b_{0}}$
When we take the quotient of the two we obtain
${\displaystyle {\frac {p(x)}{q(x)}}={\frac {a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}}{b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots +b_{1}x+b_{0}}}}$
The ratio of two polynomials is called a rational expression. Many times we would like to simplify such a beast. For example, say we are given ${\displaystyle {\frac {x^{2}-1}{x+1}}}$ . We may simplify this in the following way:
${\displaystyle {\frac {x^{2}-1}{x+1}}={\frac {(x+1)(x-1)}{x+1}}=x-1,\qquad x\neq -1}$
This is nice because we have obtained something we understand quite well, ${\displaystyle x-1}$ , from something we didn't.
## Formulas of multiplication of polynomials
Here are some formulas that can be quite useful for solving polynomial problems:
${\displaystyle (a\pm b)^{2}=a^{2}\pm 2ab+b^{2}}$
${\displaystyle (a-b)(a+b)=a^{2}-b^{2}}$
${\displaystyle (a\pm b)^{3}=a^{3}\pm 3a^{2}b+3ab^{2}\pm b^{3}}$
${\displaystyle a^{3}\pm b^{3}=(a\pm b)(a^{2}\mp ab+b^{2})}$
## Polynomial Long Division
Suppose we would like to divide one polynomial by another. The procedure is similar to long division of numbers and is illustrated in the following example:
### Example
Divide ${\displaystyle x^{2}-2x-15}$ (the dividend or numerator) by ${\displaystyle x+3}$ (the divisor or denominator) Similar to long division of numbers, we set up our problem as follows: ${\displaystyle {\begin{array}{rl}\\x+3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}-2x-15\end{array}}\end{array}}}$ First we have to answer the question, how many times does ${\displaystyle x+3}$ go into ${\displaystyle x^{2}}$? To find out, divide the leading term of the dividend by leading term of the divisor. So it goes in ${\displaystyle x}$ times. We record this above the leading term of the dividend: ${\displaystyle {\begin{array}{rl}&~~\,x\\x+3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}-2x-15\end{array}}\\\end{array}}}$ , and we multiply ${\displaystyle x+3}$ by ${\displaystyle x}$ and write this below the dividend as follows: ${\displaystyle {\begin{array}{rl}&~~\,x\\x+3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}-2x-15\end{array}}\\&\!\!\!\!-{\underline {(x^{2}+3x)~~~}}\\\end{array}}}$ Now we perform the subtraction, bringing down any terms in the dividend that aren't matched in our subtrahend: ${\displaystyle {\begin{array}{rl}&~~\,x\\x+3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}-2x-15\end{array}}\\&\!\!\!\!-{\underline {(x^{2}+3x)~~~}}\\&\!\!\!\!~~~~~~-5x-15~~~\\\end{array}}}$ Now we repeat, treating the bottom line as our new dividend: ${\displaystyle {\begin{array}{rl}&~~\,x-5\\x+3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}-2x-15\end{array}}\\&\!\!\!\!-{\underline {(x^{2}+3x)~~~}}\\&\!\!\!\!~~~~~~-5x-15~~~\\&\!\!\!\!~~~-{\underline {(-5x-15)~~~}}\\&\!\!\!\!~~~~~~~~~~~~~~~~~~~0~~~\\\end{array}}}$ In this case we have no remainder.
### Application: Factoring Polynomials
We can use polynomial long division to factor a polynomial if we know one of the factors in advance. For example, suppose we have a polynomial ${\displaystyle P(x)}$ and we know that ${\displaystyle r}$ is a root of ${\displaystyle P}$ . If we perform polynomial long division using P(x) as the dividend and ${\displaystyle (x-r)}$ as the divisor, we will obtain a polynomial ${\displaystyle Q(x)}$ such that ${\displaystyle P(x)=(x-r)Q(x)}$ , where the degree of ${\displaystyle Q}$ is one less than the degree of ${\displaystyle P}$.
### Exercise
Use ^ to write exponents:
Factor ${\displaystyle x-1}$ out of ${\displaystyle 6x^{3}-4x^{2}+3x-5}$.
### Application: Breaking up a rational function
Similar to the way one can convert an improper fraction into an integer plus a proper fraction, one can convert a rational function ${\displaystyle P(x)}$ whose numerator ${\displaystyle N(x)}$ has degree ${\displaystyle n}$ and whose denominator ${\displaystyle D(x)}$ has degree ${\displaystyle d}$ with ${\displaystyle n\geq d}$ into a polynomial plus a rational function whose numerator has degree ${\displaystyle \nu }$ and denominator has degree ${\displaystyle \delta }$ with ${\displaystyle \nu <\delta }$ .
Suppose that ${\displaystyle N(x)}$ divided by ${\displaystyle D(x)}$ has quotient ${\displaystyle Q(x)}$ and remainder ${\displaystyle R(x)}$ . That is
${\displaystyle N(x)=D(x)Q(x)+R(x)}$
Dividing both sides by ${\displaystyle D(x)}$ gives
${\displaystyle {\frac {N(x)}{D(x)}}=Q(x)+{\frac {R(x)}{D(x)}}}$
${\displaystyle R(x)}$ will have degree less than ${\displaystyle D(x)}$ .
#### Example
Write ${\displaystyle {\frac {x-1}{x-3}}}$ as a polynomial plus a rational function with numerator having degree less than the denominator. ${\displaystyle {\begin{array}{rl}&~~\,1\\x-3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x-1\end{array}}\\&\!\!\!\!-{\underline {(x-3)~~~}}\\&\!\!\!\!~~~~~~~~~2~~~\\\end{array}}}$ so ${\displaystyle {\frac {x-1}{x-3}}=1+{\frac {2}{x-3}}}$
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# Find the Inverse Matrices if Matrices are Invertible by Elementary Row Operations
## Problem 552
For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A|I]$, where $I$ is the $3\times 3$ identity matrix.
(a) $A=\begin{bmatrix} 1 & 3 & -2 \\ 2 &3 &0 \\ 0 & 1 & -1 \end{bmatrix}$
(b) $A=\begin{bmatrix} 1 & 0 & 2 \\ -1 &-3 &2 \\ 3 & 6 & -2 \end{bmatrix}$.
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## Elementary Row Operations and Inverse Matrices
Recall the following procedure of testing the invertibility of $A$ as well as finding the inverse matrix if exists.
If the augmented matrix $[A|I]$ is transformed into a matrix of the form $[I|B]$, then the matrix $A$ is invertible and the inverse matrix $A^{-1}$ is given by $B$.
If the reduced row echelon form matrix for $[A|I]$ is not of the form $[I|B]$, then the matrix $A$ is not invertible.
## Solution.
### (a) $A=\begin{bmatrix} 1 & 3 & -2 \\ 2 &3 &0 \\ 0 & 1 & -1 \end{bmatrix}$
We apply the elementary row operations as follows.
We have
\begin{align*}
&[A|I]= \left[\begin{array}{rrr|rrr}
1 & 3 & -2 & 1 &0 & 0 \\
2 & 3 & 0 & 0 & 1 & 0 \\
0 & 1 & -1 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{R_2-2R_1}
\left[\begin{array}{rrr|rrr}
1 & 3 & -2 & 1 &0 & 0 \\
0 & -3 & 4 & -2 & 1 & 0 \\
0 & 1 & -1 & 0 & 0 & 1 \\
\end{array} \right]\6pt] &\xrightarrow{R_2\leftrightarrow R_3} \left[\begin{array}{rrr|rrr} 1 & 3 & -2 & 1 &0 & 0 \\ 0 & 1 & -1 & 0 & 0 & 1 \\ 0 & -3 & 4 & -2 & 1 & 0 \\ \end{array} \right] \xrightarrow{\substack{R_1-3R_2\\ R_3+3R_2}} \left[\begin{array}{rrr|rrr} 1 & 0& 1 & 1 &0 & -3 \\ 0 & 1 & -1 & 0 & 0 & 1 \\ 0 & 0 & 1 & -2 & 1 & 3 \\ \end{array} \right]\\[6pt] &\xrightarrow{\substack{R_1-R_3\\ R_2+R_3}} \left[\begin{array}{rrr|rrr} 1 & 0& 0 & 3 & -1 & -6 \\ 0 & 1 & 0 & -2 & 1 & 4 \\ 0 & 0 & 1 & -2 & 1 & 3 \\ \end{array} \right]. \end{align*} The left 3\times 3 part of the last matrix is the identity matrix. This implies that A is invertible and the inverse matrix is given by the right 3\times 3 matrix. Hence \[A^{-1}=\begin{bmatrix} 3 & -1 & -6 \\ -2 &1 &4 \\ -2 & 1 & 3 \end{bmatrix}.
### (b) $A=\begin{bmatrix} 1 & 0 & 2 \\ -1 &-3 &2 \\ 3 & 6 & -2 \end{bmatrix}$.
Now we consider the matrix $A=\begin{bmatrix} 1 & 0 & 2 \\ -1 &-3 &2 \\ 3 & 6 & -2 \end{bmatrix}$.
Applying elementary row operations, we obtain
\begin{align*}
&[A|I]= \left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
-1 & -3 & 2 & 0 & 1 & 0 \\
3 & 6 & -2 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{\substack{R_2+R_1\\ R_3-3R_1}}
\left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
0 & -3 & 4 & 1 & 1 & 0 \\
0 & 6 & -8 & -3 & 0 & 1 \\
\end{array} \right]\6pt] &\xrightarrow{R_3-2R_1} \left[\begin{array}{rrr|rrr} 1 & 0 & 2 & 1 &0 & 0 \\ 0 & -3 & 4 & 1 & 1 & 0 \\ 0 & 0 & 0 & -5 & -2 & 1 \\ \end{array} \right] \xrightarrow{\frac{-1}{3}R_2} \left[\begin{array}{rrr|rrr} 1 & 0 & 2 & 1 &0 & 0 \\ 0 & 1 & -4/3 & -1/3 & -1/3 & 0 \\ 0 & 0 & 0 & -5 & -2 & 1 \\ \end{array} \right]. \end{align*} The last matrix is in reduced row echelon form but the left 3\times 3 part is not the identity matrix I. It follows that the matrix A is not invertible. ## Related Question. Problem. Find the inverse matrix of \[A=\begin{bmatrix} 1 & 1 & 2 \\ 0 &0 &1 \\ 1 & 0 & 1 \end{bmatrix} if it exists. If you think there is no inverse matrix of $A$, then give a reason.
This is a linear algebra exam problem at the Ohio State University.
The solution is given in the post↴
Find the Inverse Matrix of a $3\times 3$ Matrix if Exists
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• Quiz 4: Inverse Matrix/ Nonsingular Matrix Satisfying a Relation (a) Find the inverse matrix of $A=\begin{bmatrix} 1 & 0 & 1 \\ 1 &0 &0 \\ 2 & 1 & 1 \end{bmatrix}$ if it exists. If you think there is no inverse matrix of $A$, then give a reason. (b) Find a nonsingular $2\times 2$ matrix $A$ such that $A^3=A^2B-3A^2,$ where […]
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