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# Addition Doubles Game – Jump
Addition ‘Doubles’ and Two Times Table Game – Jump is an excellent FREE math game for learning the Doubles Basic Facts and exploring the 2 Times Table as repeated addition.
\$0.00
# Addition or Multiplication Game
## Doubles & 2 Times Table Game – Jump
Addition Doubles & 2 Times Table Game – Jump
Addition & Multiplication Games – Jump the Math Game is a fun Math lesson that teaches the Doubles Addition Facts and how these relate to the 2 Times Table. The exploration of these number facts is essential for efficient mental computations. This game aims to make the learning of these number facts enjoyable.
### Three Ways to Play:
1. Draw the game board up on the playground in chalk and students ‘jump’ to the answers. NB this is the version the students enjoy the most.
2. Draw up the game board on the classroom carpet in chalk. (This is heaps of fun too but from experience has been known to annoy the classroom cleaners just a little)
3. Print out the game boards and students place their fingers on the frog feet and jump to the answer with their fingers.
### Aims:
– to encourage ‘automaticity’ in the recall of Basic Addition & Multiplication Facts
– for students to see the relationship between Repeated Addition and Multiplication.
NB In my classrooms I do a lot of work on the need for efficiency in doing mental computations. We aim for, ‘SEE IT, SAY IT’. This addition game offers heaps of opportunities to discuss ‘efficient’ number strategies.
### How To Play:
1. Draw this grid on concrete using chalk. Each square should be about 30cm by 30cm.
2. Two players stand with their feet in the large feet facing the game board.
3. A ‘Caller’ asks a question from one of the cards.
4. When a player has worked out the answer they must jump from their spot onto the correct answer. NB Do not move until you know the answer.
5. Players then rotate positions but the winner stays to compete with the next player. NB The object of the game is to have a bit of fun while you learn some of the facts so make an effort to learn your ‘Doubles’ facts while you’re standing in line.
### Variations:
– After playing the games separately, combine the addition cards with the multiplication cards
### Before the Game:
– Safety is paramount – Students need to know this and need to know the game will be shut down as soon as there is any ‘silliness’. Make sure the students are well of this. I like to joke that I know this is a tough rule but I hate the paperwork of accident reports then as an aside mention the blood and pain hehe Know your audience 🙂
– Explore the Addition Doubles Facts and how they relate to the 2 Times Table as repeated addition. e.g. 2×3 = 3+3
– It is easy to get caught up in the fun of the competition but the essence of the learning experience is to focus on the facts ‘you need to work on’. In pairs generate ideas for strategies for improving your math while standing in line waiting for your next turn. Share these with the class and discuss them.
– Discuss ‘helpful thinking’ in your self-talk. e.g. ‘Ok I missed 9×2, I’ll repeat it to myself a few times, hey it is only 9+9, I know that so I just need to convert it to multiplication’. Unhelpful thinking – my granddad says he is not a ‘math person’ so I must have got it from him. This game sucks! This game is stupid! etc. Discuss implications of both types of thinking.
Multiplication Games – PowerPoint Version
### During the Game:
– Reminders about safety.
– Remind students in line to reflect on ‘What I need to practice’ and encourage them to do it.
### Possible After the Game Reflections:
– I enjoyed this Math game because…
– I think this game could be improved by…
– When standing on the feet I feel ……………. because ……………..
– Next time I play this game I need to focus on……. because…..
1 Addition Doubles Board
1 Two Times Table Board
13 Cards with the Addition Doubles Algorithms
13 Cards with the Multiplication Algorithms for the 2 Times Table
1 PowerPoint file with both boards and all the cards for easy discussion and game introduction
Enjoy |
# Function Application for the Real World
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#### Recommended Lessons and Courses for You
Lesson Transcript
Instructor: Laura Pennington
Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.
In this lesson, we'll recall what a function is and then look at applying functions in real life that involve both numbers and objects. We'll then learn how to solve problems using these functions.
## Functions
Do you have a different outfit that you wear each day of the week? If so, you might be surprised to learn that the relationship between the day of the week and your outfit represents a mathematical concept called a function.
Day Outfit
Sunday Plaid Skirt, Brown Blouse
Monday Jeans, Gray Sweater
Tuesday Jean Shorts, Green T-shirt
Wednesday Gray Slacks, Black Blouse
Thursday Flowered Dress
Friday Blue Shorts, White Sweater
Saturday Sweatpants and Sweatshirt
In mathematics, a function is a relationship between two sets of elements in which no element in the first set relates to more than one element in the second. We can think of a function as a rule that takes inputs from the first set and relates them to an element in the second set, which is the output.
In mathematics, we represent functions in many different ways; we can use words, tables, mappings, equations, and even graphs. Consider this example: If a state has a 6% sales tax, then we can use a function to calculate it. We can use the following equation to represent this function:
T = 0.06x
If we purchase a product for x dollars, then to calculate the tax, we would multiply x by 0.06, or the tax rate in decimal form. For instance, if you bought a shirt for \$25, then you can calculate the sales tax by plugging in 25 for x.
T = 0.06(25) = 1.5
We see the sales tax is \$1.50.
We can also represent this function using our other representations. Because our purchase price can be any number, we could never list all of the inputs and outputs! Using a table or mapping wouldn't be the best method in this situation, but we can easily represent it graphically by simply graphing the equation T = 0.06x.
We see that there are many ways to represent a function, and each one depends on its context. Now that we are reminded what a function is in mathematics, let's talk some more about using them in real-world settings!
## Functions in the Real World
When it comes to recognizing functions in the real world, it is useful to think of the term ''is a function of'' as ''is determined by.'' You see, in a function, no input can be related to two different outputs, otherwise, given the input, we couldn't determine the output. Therefore, the input determines the output, so the output is a function of the input.
For instance, think about those outfits again.
Day Outfit
Sunday Plaid Skirt, Brown Blouse
Monday Jeans, Gray Sweater
Tuesday Jean Shorts, Green T-shirt
Wednesday Gray Slacks, Black Blouse
Thursday Flowered Dress
Friday Blue Shorts, White Sweater
Saturday Sweatpants and Sweatshirt
If it's Tuesday, what outfit will you be wearing? Well, we look to see what outfit to which Tuesday is mapped, and we see that you will be wearing your jean shorts with your green t-shirt. That's not too hard, is it? You can see that your outfit ''is determined by'' the day of the week, so your outfit ''is a function of'' the day of the week.
Pretty neat, huh? You may have been familiar with functions involving numbers, but it's totally cool to see that the function concept can actually be applied to real world situations! Let's consider a couple more examples!
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# pipes and cistern aptitude questions
11. One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:
A. 81 min. B. 108 min. C. 144 min. D. 192 min.
Explanation:
Let the slower pipe alone fill the tank in x minutes.
Then, faster pipe will fill it in x minutes. 3
1 + 3 = 1 x x 36
4 = 1 x 36
x = 144 min.
12. A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half?
A. 15 min B. 20 min C. 27.5 min D. 30 min
Explanation:
Part filled by (A + B) in 1 minute = 1 + 1 = 1 . 60 40 24
Suppose the tank is filled in x minutes.
Then, x 1 + 1 = 1 2 24 40
x x 1 = 1 2 15
x = 30 min.
13. A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?
A. 3 hrs 15 min B. 3 hrs 45 min C. 4 hrs D. 4 hrs 15 min
Explanation:
Time taken by one tap to fill half of the tank = 3 hrs.
Part filled by the four taps in 1 hour = 4 x 1 = 2 . 6 3
Remaining part = 1 – 1 = 1 . 2 2
2 : 1 :: 1 : x 3 2
x = 1 x 1 x 3 = 3 hours i.e., 45 mins. 2 2 4
So, total time taken = 3 hrs. 45 mins.
14. Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:
A. 6 hours B.
6 2 hours 3
C. 7 hours D.
7 1 hours 2
Explanation:
(A + B)’s 1 hour’s work = 1 + 1 = 9 = 3 . 12 15 60 20
(A + C)’s hour’s work = 1 + 1 = 8 = 2 . 12 20 60 15
Part filled in 2 hrs = 3 + 2 = 17 . 20 15 60
Part filled in 6 hrs = 3 x 17 = 17 . 60 20
Remaining part = 1 – 17 = 3 . 20 20
Now, it is the turn of A and B and 3 part is filled by A and B in 1 hour. 20
Total time taken to fill the tank = (6 + 1) hrs = 7 hrs.
15. Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:
A. 10 B. 12 C. 14 D. 16
Explanation:
Part filled in 2 hours = 2 = 1 6 3
Remaining part = 1 – 1 = 2 . 3 3
(A + B)’s 7 hour’s work = 2 3
(A + B)’s 1 hour’s work = 2 21
C’s 1 hour’s work = { (A + B + C)’s 1 hour’s work } – { (A + B)’s 1 hour’s work }
= 1 – 2 = 1 6 21 14
C alone can fill the tank in 14 hours |
# 2. Area and Volume of Solids
Mind Map by , created almost 6 years ago
## Area and Volume of Solids Mind Map Exercise
54
2
0
Created by Jamie McLardy almost 6 years ago
FREQUENCY TABLES: MODE, MEDIAN AND MEAN
CUMULATIVE FREQUENCY DIAGRAMS
STEM AND LEAF DIAGRAMS
Cell Structure
GCSE Revision Tips
1. Significant Figures
New GCSE Maths required formulae
New GCSE Maths
GCSE Maths Symbols, Equations & Formulae
HISTOGRAMS
2. Area and Volume of Solids
1 Area of Triangles
1.1 Area = 1/2 b x h
1.1.1 h = The Hight of the Triangle = the distance between the top and bottom of the middle.
1.1.1.1 Can be called Altitude
1.1.2 b = The Width of the Triangle = the distance between the left and right corners of the bottom side.
1.2 The Area of a Triangle is basically the Width x Hight, and that would be devided by a half.
2 Area of a Parallelogram Area = b x h
3 Area of a Rhombus and Kite Area = 1/2 (b x h)
4 Area of a Trapezium
4.1 First off you would devide your Trapezium into two Triangles, making sure that the deviding line goes from the bottom left corner to the top right corner. Then you would use the Triangles Are Formulae
4.2 Triangle 1 Area = 1/2 (a x h)
4.2.1 A being the width for that Triangle.
4.3 Triangle 2 Area = 1/2 (b x h)
4.3.1 B being the width for that Triangle.
4.4 And to bring it all together to work out the area of the Original Trapezium, you just ass both of the answers together
4.4.1 Or you can just use the Formula
4.4.1.1 Trapezium Area = 1/2 (a+b) x h
4.4.1.1.1 A being the Width for Triangle 1
4.4.1.1.2 B being the Width for Triangle 2
5 Composite Areas
5.1 When it comes to the Area of Composite Shapes (Shapes that are not a set shape like a Square and are made up of 2 or more shapes)
5.1.1 The best bet would be to simpily break up the Composite Shape into the different base shapes, and then work out the Area by finding the Area of EACH shape and adding them all together.
6 Area of Squares
6.1 Area = b x h
6.2 Simple really..
7 Volume of Solids
7.1 Volume = Area of Cross Section x Length
7.2 Find the AREA of the shape and times that by the length of the Object, assuming that it is the same shape all the way down/across
8 Surface Area
8.1 The SURFACE area of lets say a cuboid is just simply the area of all the sides added together.
8.2 The Net for a cuboid is just each side's area, but placed down as if you were going to glue each side together to make a 3D object. |
# Continuity Theorems and Their Applications in Calculus
Theorems, related to the continuity of functions and their applications in calculus are presented and discussed with examples.
## Theorem 1
All polynomial functions and the functions sin x, cos x, arctan x and e x are continuous on the interval (-infinity , +infinity).
Example:Evaluate the following limits:
\lim_{x\to\0} \sin (x) \\\\ \lim_{x\to\pi} \cos (x) \\\\ \lim_{x\to\ -1} \arctan(x)
Solutions
If function f is continuous at x = a, then
\lim_{x\to\ a} \f(x) = f(a)
Since sin(x) is continuous
\lim_{x\to\0} \sin (x) = \sin(0) = 0
Since cos(x) is continuous
\lim_{x\to\pi} \cos (x) = \cos(\pi) = - 1
Since arctan(x) is continuous
\lim_{x\to\ -1} \arctan(x) = \arctan(-1) = - \pi / 4
## Theorem 2
If functions f and g are continuous at x = a, then
A. (f + g) is continuous at x = a,
B. (f - g) is continuous at x = a,
C. (f . g) is continuous at x = a,
D. (f / g) is continuous at x = a if g(a) is not equal to zero.
If g(a) = 0 then (f / g) is discontinuous at x = a.
Example:Let f(x) = sin x and g(x) = cos x. Where are the following functions (f + g), (f - g), (f . g) and (f / g) continuous?
Solutions:
Since both sin x and cos x are continuous everywhere, according to theorem 2 above (f + g), (f - g), (f . g) are continuous everywhere.
However (f / g) is continuous everywhere except at values of x for which make the denominator g(x) is equal to zero. These values are found by solving the trigonometric equation:
cos x = 0
The values which make cos x = 0 are given by:
x = ?/2 + k(?) , where k is any integer.
(f / g) is continuous everywhere except at x = ?/2 + k(?) , k integer.
## Theorem 3
A rational function is continuous everywhere except at the values of x that make the denominator of the function equal to zero.
Example:Find the values of x at which function f is discontinuous.
f(x) = \dfrac{x-2}{(2 x^2 + 2 x - 4)(x^4 + 5)}
Solutions:
The denominator of f is the product of two terms and is given by
(2 x^2 + 2 x - 4)(x^4 + 5)
The term x 4 + 5 is always positive hence never equal to zero. We now need to find the zeros of 2 x 2 + 2x - 4 by solving the equation:
2 x^2 + 2 x - 4 = 0
The solutions are: x = 1 and x = - 2
function f is discontinuous at x = 1 and x = -2.
## Theorem 4
If
\color{red}{\lim_{x\to\ a} g(x) = L}
and if f is a continuous function at x = L, then
\color{red}{\lim_{x\to\ a} f(g(x)) = f(\lim_{x\to\ a} g(x)) = f(L)}
Example:Evaluate the limit
\lim_{x\to\ a} \sin(2x + 5)
Solution: sin x is continuous everywhere and 2 x + 5 is a polynomial and also continuous everywhere. Hence
\lim_{x\to\ a} \sin(2x + 5) = \sin(\lim_{x\to\ a}(2x+5)) = \sin(2a + 5)
## Theorem 5
If g is a continuous function at x = a and function f is continuous at g(a), then the composition f o g is continuous at x = a.
Example:Show that any function of the form e ax + b is continuous everywhere, a and b real numbers.
f(x) = e x the exponential function and g(x) = ax + b a polynomial (linear) function are continuous everywhere. Hence the composition f(g(x)) = e ax + b is also continuous everywhere. More on Questions on Continuity with Solutions
Calculus Tutorials and Problems |
# What is the slope of the tangent line of (x^2/y-y^2)/(1+x)^2 =C , where C is an arbitrary constant, at (6,5)?
Apr 14, 2018
Slope is $\frac{3495}{648}$ and tangent is $y - 5 = \frac{3495}{648} \left(x - 6\right)$
#### Explanation:
As we are seeking tangent at point $\left(6 , 5\right)$, the curve $\frac{{x}^{2} / y - {y}^{2}}{1 + x} ^ 2 = C$ passes through it and hence
$C = \frac{{x}^{2} / y - {y}^{2}}{1 + x} ^ 2 = \frac{{6}^{2} / 5 - {5}^{2}}{1 + 5} ^ 2 = - \frac{89}{5 \cdot 36} = - \frac{89}{180}$
i.e. curve is $\frac{{x}^{2} / y - {y}^{2}}{1 + x} ^ 2 = - \frac{89}{180}$
or ${x}^{2} / y - {y}^{2} = - \frac{89}{180} {\left(1 + x\right)}^{2}$
Slope of tangent is given by value of $\frac{\mathrm{dy}}{\mathrm{dx}}$ at this point
and differentiating we get
$2 \frac{x}{y} - {x}^{2} / {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{89}{180} \left(2 x + 2\right)$
or at $\left(6 , 5\right)$ we have $\frac{12}{5} - \frac{36}{25} \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{89}{180} \cdot 14$
or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{25}{36} \left(\frac{12}{5} + \frac{483}{90}\right) = \frac{25}{36} \cdot \frac{699}{90} = \frac{3495}{648}$
and tangent is $y - 5 = \frac{3495}{648} \left(x - 6\right)$ |
# How to Find the Central Angle of a Circle?
A central angle is an angle formed by two arms whose vertex is at the center of the circle. In this step-by-step guide, you learn more about finding the central angle.
The central angle is useful for dividing a circle into sections. A slice of pizza is a good example of a central angle. A pie chart consists of several sections and helps to display different values.
## A step-by-step guide tofinding the central angleof a circle
The central angle is the angle subtended by an arc of a circle at the center of a circle. Radius vectors form the arms of the central angle.
In other words, it is an angle whose vertex is the center of a circle whose arms are two radii lines that intersect at two different points in the circle. When these two points are connected, they form an arc. The central angle is the angle subtended by this arc at the center of the circle.
### How to find the central angle of a circle?
The central angle is the angle between the two radii of the circle. To find the central angle, we need to find the length of the arc and the length of the radius. The following steps show how to calculate the central angle in radians.
There are three simple steps to finding the central angle:
• Identify the ends of the arc and the center of the circle (curve). $$AB$$ is the arc of the circle and $$O$$ is the center of the circle.
• Connect the end of the arc with the center of the circle. Also, measure the length of the arc and its radius. Here $$AB$$ is the arc length and $$OA$$ and $$OB$$ are the radii of the circle.
• Divide the length of the curve by the radius to get the central angle. Using the following formula, we will find the value of the central angle in radians.
$$\color{blue}{Central\:Angle=\frac{Length\:of\:the\:Arc}{Radius}}$$
### Finding the Central Angle of a Circle – Example 1:
If the arc length is $$8$$ inches and the central angle is $$120$$ degrees, find the radius of the arc.
Solution:
$$Radius\:of\:the\:arc= 8\space inches$$
$$Central\:angle= 120°$$
$$Central\:angle\:=\frac{\left(length\:of\:arc\:×\:360°\right)}{\left(2\pi \:×\:radius\right)}$$
$$radius\:=\frac{\left(length\:of\:arc\:×360°\right)}{\left(2\pi \:×\:Central\:angle\right)}$$
$$radius\:=\frac{\left(8\:×\:360°\right)}{\left(2\pi \:×120°\right)}$$
$$radius\:=\frac{12}{\pi \:}$$
## Exercises for Finding the Central Angleof a Circle
• From the diagram shown, find the $$m∠BOD$$ measure.
• From the diagram shown, find the $$m∠arc MNK$$ measure.
• If the arc length measurement is about $$21\space cm$$ and the length of the radius measures $$10\space cm$$, find the central angle.
• $$\color{blue}{60°}$$
• $$\color{blue}{234°}$$
• $$\color{blue}{120.38°}$$
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# Maryland - Grade 1 - Math - Operations and Algebraic Thinking - Addition and Subtraction Word Problems - 1.OA.1
### Description
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
• State - Maryland
• Standard ID - 1.OA.1
• Subjects - Math Common Core
### Keywords
• Math
• Operations and Algebraic Thinking
## More Maryland Topics
1.OA.4 Understand subtraction as an unknown-addend problem. For example, subtract 10 – 8 by finding the number that makes 10 when added to 8. Add and subtract within 20.
1.OA.5 Relate counting to addition and subtraction (e.g., by counting on 2 to add 2).
1.OA.6 Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1
Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used.
Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: A. 10 can be thought of as a bundle of ten ones — called a “ten.” B. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones). |
# The Extended Euclidean Algorithm
Explained step-by-step with examples.
Make sure that you have read the page about the Euclidean Algorithm (or watch the video instead). That page explains how to construct a table using the Euclidean Algorithm. In the Extended Euclidean Algorithm we're going to do the same, but with some extra columns in the table. So if you have no idea what we're talking about, this page is going to be confusing, while it really doesn't have to be. Just read that page about the Euclidean Algorithm first. Please
#### Text or video?
You can choose to read this page or watch the video at the bottom of this page. Both cover the same material, so there's no need to look at both. Reading this page might be quicker, but the video could feel a little more detailed.
#### Versions of the Extended Euclidean Algorithm
There are many versions of the Extended Euclidean Algorithm. You may have seen some already that require things like backwards substition or to write "gcd(...,...)=" over and over again while you're not really sure what you're doing. In our version, we don't need that kind of nonsense. We write down everything we need and skip all the things we don't need to write down. A table is perfect for this. We have already seen how to construct a table using the Euclidean algorithm, so now we're just going to add more columns.
#### Purpose
Why do we need more columns if the Euclidean Algorithm can already calculate the gcd? Why do we need the Extended Euclidean Algorithm at all?
Well, because it allows us to calculate some extra things.
Assume you know the two variables `a` and `b`.
Let's say we want to know the values of `s` and `t` such that:
`s × a + t × b = gcd(a, b)`
(This is called the Bézout identity, where `s` and `t` are the Bézout coefficients.)
The Euclidean Algorithm can calculate `gcd(a, b)`.
With the Extended Euclidean Algorithm, we can not only calculate `gcd(a, b)`, but also `s` and `t`.
That is what the extra columns are for.
#### Recap: the columns in the table of the Euclidean Algorithm
Remember when we were calculating the gcd of 36 (=a) and 10 (=b) with the Euclidean algorithm? Remember what the table looked like?
abqr
361036
10614
6412
4220
The table had 4 columns: a, b, q and r.
• q = the quotient of a/b
• r = the remainder of a/b
• On the first row a and b were just the a and b that we wanted to know the gcd of, so 36 and 10.
• On all other rows a = [b from the previous row] and b= [r from the previous row]
We can summarize this in an overview as follows:
ColumnValue on the first rowValue on other rows
athe input ab from the previous row
bthe input br from the previous row
qquotient of a/bquotient of a/b
rremainder of a/bremainder of a/b
#### The extra columns in the table of the Extended Euclidean Algorithm
In the Extended Euclidean Algorithm, we have all of the columns from the Euclidean Algorithm.
We also have some extra columns: s1, s2, s3, t1, t2 and t3.
What do we need to put in those columns? Have a look at this overview.
Overview of extra columns
ColumnValue on the first rowValue on other rows
s11s2 from the previous row
s20s3 from the previous row
s3s1 - q × s2 from the current row = 1s1 - q × s2 from the current row
t10t2 from the previous row
t21t3 form the previous row
t3t1 - q × t2 from the current rowt1 - q × t2 from the current row
So what do we see here?
• When creating the first row in the table, we don't need to calculate s1, s3, t1 and t2.
s1 and t2 are always 1 on the first row and s2 and t1 are always 0 on the first row.
• We also don't need to calculate s3 on the first row, because:
s1 - q × s2 = 1 - q × 0 = 1 - 0 = 1.
So for the first row in the table, we can always write down a 1 for s3. But for all other rows we actually do need to calculate s1 - q × s2.
• Just like we can copy b and r to a and b on the next row,
we can also copy s2 and s3 to s1 and s2 on the next row.
And t2 and t3 to t1 and t2 on the next row.
Does all of this sound confusing to you? Don't worry, just have a look at the example below where we're going to guide you through the algorithm step by step.
#### Example
a = 161 and b = 28. Calculate gcd(a, b) and s and t.
We begin with writing down the things that we would also write down in the Euclidean Algorithm:
abqr
16128521
If you didn't understand this step, go back to the explanation of the Euclidean Algorithm.
The Extended Euclidean algorithm uses extra columns, so we're going to add them to the right:
abqrs1s2s3t1t2t3
161285
As you can see in the overview of extra columns, some of the values we have to put on the first row are already known.
On the first row, the following is always the case: s1=1, s2=0, t1=0 and t2=1.
Usually we need to calculate s3, but not on the first row, since s3 = s1 - q × s2 = 1 - q × 0 = 1. So s3 is always 1 on the first row.
Now let's add these values to the table.
abqrs1s2s3t1t2t3
1612852110101
Next time when you create the first row, don't think to much. Just add 1 0 1 0 1 to the table after you wrote down the value of r.
Then the only thing left to do on the first row is calculating t3.
We look again at the overview of extra columns and we see that (on the first row) t3 = t1 - q × t2, with the values t1, q and t2 from the current row.
So t3 = t1 - q × t2 = 0 - 5 × 1 = -5. Let's add this value to the table to finish the first row:
abqrs1s2s3t1t2t3
1612852110101-5
Now the first row is done! We continue with the next row.
Let's start with copying some values from the previous row to this row. What values can we copy?
(This follows from the overview of columns in the Euclidean Algorithm and the overview of extra columns):
Copy this value from the previous rowAnd put it in this column on the current row
ba
rb
s2s1
s3s2
t2t1
t3t2
If you do this, then the table looks like this:
abqrs1s2s3t1t2t3
1612852110101-5
2821011-5
Here we have included colors to make it clear which values have been copied and where we put them.
Once you understand this, just ignore the colors. When you are creating a table, you don't have to color your own table. Unless you want to of course ;)
So now we only have to calculate q, r, s3 and t3 on this row.
First, we're looking for the greatest number below a, that's divisible by b.
On this row we have a=28 and b=21, so what's the biggest number below 28 that we can divide by 21? Yes, it's actually 21 itself.
q = [the greatest number below a, that's divisible by b]/b = 21/21 = 1
r = a - q × b = 28 - 1 × 21 = 7
If you find this confusing, look at the explanation of how to calculate q and r on the page of the Euclidean Algorithm.
Now let's add the values to the table:
abqrs1s2s3t1t2t3
1612852110101-5
282117011-5
From overview of extra columns we know that s3 = s1 - q × s2 and t3 = t1 - q × t2.
All values in this formulas are the values on the current row. So:
s3 = s1 - q × s2 = 0 - 1 × 1 = -1 and
t3 = t1 - q × t2 = 1 - 1 × -5 = 1 - -5 = 1 + 5 = 6
If we add these values to the table, we get:
abqrs1s2s3t1t2t3
1612852110101-5
28211701-11-56
Now we calculate the next row the same way, such that we get:
abqrs1s2s3t1t2t3
1612852110101-5
28211701-11-56
217301-14-56-23
As you can see, r = 0, so we are done! Now let's have a look at what we've actually calculated.
Results
Now, remember that we're calculating the value of s, t and gcd(a, b) in this equation:
`s × a + t × b = gcd(a, b)`
Since we have reached r=0, we are done and we can now find the answers in the last row of the table:
• gcd(a, b) = [b on the last row]
• s = [s2 on the last row]
• t = [t2 on the last row]
So in this case, we have found the following:
• gcd(161, 28) = 7
• s = -1
• t = 6
Executing the Extended Euclidean algorithm involves a lot of steps, so small mistakes are easily made.
Luckily we can check whether we have done it correctly.
Since we we're trying to find s, t and gcd(a, b) such that `s × a + t × b = gcd(a, b)`,
we can just check whether `s × a + t × b` is indeed equal to `gcd(a, b)`.
But there's a small problem: `s × a + t × b` can result in a negative number (e.g. if we use a negative a or b), while the gcd of two numbers is always positive.
In that case they are not equal, even though the calculation was correct. To so solve this, we can cheat a little bit:
we take the absolute value of `s × a + t × b` and compare that to the `gcd(a, b)`.
Example
In our earlier example, we used the Extended Euclidean algorithm on a=161 and b=28.
We found that s=-1, t=6 and gcd(161,28)=7.
So if we put these numbers in the formula, we get:
`|s × a + t × b|`= |-1 × 161 + 6 × 28| = |-161 + 168| = 7 = gcd(161,28) = `gcd(a, b)`
So as you can see, `|s × a + t × b|` is indeed equal to `gcd(a, b)`. So our calculation is correct.
#### What if b=0?
Then the gcd(a,b)=|a|, s=1 and t=0.
The gcd
Remember the property of the gcd that gcd(a,0)=|a|. That's why gcd(a,b)=|a| when b=0.
Why is s=1 and t=0?
We mentioned before that when r=0, you could add another row. If we do that with the earlier example, the table looks like this:
abqrs1s2s3t1t2t3
1612852110101-5
28211701-11-56
217301-14-56-23
70-14-6-23
This is exactly the same table as we saw before, but with an extra row.
In this extra row, b from the previous row has been copied into a again, r into b, s2 into s1, s3 into s2, t2 into t1 and t3 into t2.
So the answer now appears at two places: in the column of b where r=0 and in the column of a on the last row. The answer of s can now also be found in two places: column s2 on the row where r=0 and column s1 on the last row. Likewise, t can be found in column t2 (where r=0) and t1 (last row). This makes sense, because you can now see why a in the last row is the answer: b=0 and gcd(a,0)=|a|.
This extra row is actually part of the Extended Euclidean Algorithm, but we usually decide to skip it, since we can already tell the answers once we reach r=0.
But when b already equals 0 at the start, this extra row is actually the only row in the table. So if we decide to omit it, we have a table with no rows. This is what happens when you run our calculator.
Let's say a=15 and b=0. This is how the calculator displays the output:
abqr s1s2s3 t1t2t3
So as you can see, there are no rows.
If we wouldn't skip the extra row in our version of the Extended Euclidean algorithm and we run it again with a=15 and b=0, the table would look like this:
abqr s1s2s3 t1t2t3
150--10-01-
So here the "extra" row in the table is the only row in the table.
Which means that the answers we are looking for have already been shifted to the left. As if they have been copied from the previous row where r=0, although that row doesn't exist in this case.
So the answers that we are looking for can be found in the following columns again:
• gcd(a,b) in column a
• s in column s1
• t in column t1
Remember that in the first row we always have s1=1 and t1=0. Since this is the only row (whenever b=0 from the start), this is also the first row. That's why when b=0, we always have gcd(a,b)=|a|, s=1 and t=0.
#### Calculator
Do you want to see more examples? Do you want to practice more? Or are you just very lazy?
Then check out our awesome calculator that can do this entire calculation of the Extended Euclidean algorithm for you!
It shows all intermediate steps in the table, the final answers and also the verification of the answers.
→ Go the calculator now!
#### Multiplicative inverse
Interested to see how we can use the Extended Euclidean algorithm to calculate a multiplicative inverse? Check this page!
#### Video
For those who like to hear someone explain it instead of having to read a lot. |
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Home >> Rhombus >> Construction of Rhombus with Compass >>
## Construction of Rhombus with Compass
Area of Rhombus Difference & Similarity between Rhombus & Rectangle Difference & Similarity between Rhombus & Square Difference & Similarity between Rhombus & Parallelogram Construction of Rhombus with Compass
What is Rhombus and Diagonals of Rhombus ?
How to construct Perpendicular Bisector of Line Segment ?
Construction of Rhombus
Here you will understand how to construct a Rhombus; where only measure of its two diagonals are given.
Example: Construct Rhombus PQRS, where two diagonals PR and SQ are 5 cm and 7 cm respectively.
Solution: As per the given question
PR & SQ are diagonals of Rhombus PQRS
PR = 5 cm
SQ = 7 cm
Steps of construction are as follows:
Step 1: Use ruler and draw line segment SQ of 7 cm, which is also one of the diagonal (as shown below):
Step 2: Use compass and draw perpendicular bisector of line segment SQ (as shown below):
Note: We construct perpendicular bisector because in Rhombus, diagonals are perpendicular bisector to each other.
Step 3: Mark point of intersection (line segment and perpendicular bisector) as O (as shown below):
Step 4: Use compass and 2.5 cm wide open. With O as center, draw two arcs, cutting perpendicular bisector at point P and R (as shown below):
Note: Here we have chosen length of 2.5 cm because in Rhombus diagonals bisect each other at point of intersection. Since length of other diagonal is 5 cm (as given), so half of 5 cm is 2.5 cm
Step 5: Use ruler and join points P & S, P & Q, S & R and R & Q. And we get the resultant Rhombus PQRS (as shown below) |
# Point of Intersection Question6
In this page Point of Intersection Question6 we are going to see solution of second question in the worksheet point of intersection.
What does mean by point of intersection:
If two straight lines are not parallel then they will meet at a point.This common point for both straight lines is called the point of intersection.If the equations of two intersecting straight lines are given,then their intersecting point is obtained by solving equations simultaneously.
Find the intersection point of the straight lines 5 x + 6 y = 25 and
3 x - 4 y = 10
Solution:
5 x + 6 y = 25 ----- (1)
3 x - 4 y = 10 ------(2)
In the first equation coefficient of x is 5,in the second equation the coefficient of x is 3 and we have same signs for both equations. But the coefficient of y in the first equation is 6 and coefficient of y in the second equation is -4 and we have same signs.
To make the coefficient of y of the first equation as 12 we need to multiply the the whole equation by 2.To make the coefficient of y of the second equation as 12 we need to multiply the the whole equation by 3.
Then we are going to subtract the first equation from second equation since we have same signs.
(1) x 2 = > 10 x + 12 y = 50
(2) x 3 = > 9 x - 12 y = 30
10 x + 12 y = 50
9 x - 12 y = 30
--------------------
19 x = 80
--------------------
x = 80/19
Substituting x = 80/19 in the first equation
5 (80/19) + 6 y = 25
400/19 + 6 y = 25
(400+114y)/19 = 25
400 + 114 y = 25 x 19
400 + 114 y = 475
114 y = 475 - 400
114 y = 75
y = 75/114
y = 25/38
So the intersection point of the straight lines is (80/19,25/38) .Point of Intersection Question6 Point of Intersection Question6
Questions Solution (1) Find the intersection point of the straight lines x - 5y + 17 =0 and 2x + y + 1 = 0 Solution (2) Find the intersection point of the straight lines 5 x - 3 y - 8 = 0 and 2x - 3 y - 5 = 0 (3) Find the intersection point of the straight lines 4 x - 7 y =0 and 8 x - y - 26 = 0 Solution (4) Find the intersection point of the straight lines 3 x + 5 y - 6 =0 and 5 x - y - 10 = 0 Solution (5) Find the intersection point of the straight lines 2 x + 3 y =5 and 3 x + 4 y = 7 Solution (7) Find the intersection point of the straight lines 4 x - 3 y = 3 and 3 x + 2 y = 15 Solution (8) Find the intersection point of the straight lines 3 x + 2 y = 11 and 7 x - 3 y = 41 Solution (9) Find the intersection point of the straight lines 5 x + 3 y = 11 and 3 x + 5 y = -3 Solution (10) Find the intersection point of the straight lines 2 x - y = 15 and 5 x + 3 y = 21 Solution
Point of Intersection Question6 to Analytical Geometry |
# What operations can you associate with coefficients? What operations can you associate with exponents?
you can add, subtract, multiply, and divide coefficents, and exponents.
## Operations associated with coefficients:
1. Addition: You can add coefficients together. For example, if you have 2x + 3x, you can add the coefficients 2 and 3 to get 5x.
2. Subtraction: You can subtract coefficients. For example, if you have 4x - 2x, you can subtract the coefficients 2 from 4 to get 2x.
3. Multiplication: You can multiply coefficients together. For example, if you have 2x * 3, you can multiply the coefficient 2 by 3 to get 6x.
4. Division: You can divide coefficients. For example, if you have 6x / 2, you can divide the coefficient 6 by 2 to get 3x.
Operations associated with exponents:
1. Multiplication: You can multiply the same base with different exponents. For example, if you have x^2 * x^3, you can multiply the base x with exponents 2 and 3 to get x^(2+3) = x^5.
2. Division: You can divide the same base with different exponents. For example, if you have x^5 / x^2, you can divide the base x with exponents 5 by 2 to get x^(5-2) = x^3.
3. Exponentiation: You can raise a base with an exponent. For example, if you have x^3 ^ 2, you can raise the base x with exponent 3 to the exponent 2 to get x^(3*2) = x^6.
It's important to note that these operations can follow the rules of algebra and may involve simplification or manipulation of terms.
## To perform operations with coefficients, you can add, subtract, multiply, and divide them. When adding or subtracting coefficients, you simply combine the values. For example, if you have 2x + 3x, you add the coefficients of x (2 and 3) to get 5x.
When multiplying coefficients, you multiply their values together. For instance, if you have 2x multiplied by 3x, you multiply the coefficients of x (2 and 3), resulting in 6x^2.
When dividing coefficients, you divide their values. For example, if you have 6x divided by 2, you divide the coefficients (6 and 2), giving you 3x.
Now, for exponents, the operations associated are a bit different. When multiplying terms with exponents, you add the exponents. For instance, if you have x^2 multiplied by x^3, you add the exponents (2 and 3), resulting in x^5.
When dividing terms with exponents, you subtract the exponents. For example, if you have x^5 divided by x^3, you subtract the exponents (5 and 3), giving you x^2.
Finally, when raising a term with an exponent to another power, you multiply the exponents together. For instance, if you have (x^2)^3, you multiply the exponents (2 and 3), resulting in x^6.
Remember that these rules apply when the bases (the variables or terms) are the same. If the bases are different, you typically cannot perform operations directly on the exponents and coefficients. |
LCM of 27 and 36 is the smallest number among all common multiples of 27 and 36. The first few multiples of 27 and 36 are (27, 54, 81, 108, 135, 162, 189, . . . ) and (36, 72, 108, 144, 180, 216, 252, . . . ) respectively. There are 3 commonly used methods to find LCM of 27 and 36 - by prime factorization, by division method, and by listing multiples.
You are watching: What is the least common multiple of 27 and 36
1 LCM of 27 and 36 2 List of Methods 3 Solved Examples 4 FAQs
Answer: LCM of 27 and 36 is 108.
Explanation:
The LCM of two non-zero integers, x(27) and y(36), is the smallest positive integer m(108) that is divisible by both x(27) and y(36) without any remainder.
The methods to find the LCM of 27 and 36 are explained below.
By Listing MultiplesBy Division MethodBy Prime Factorization Method
### LCM of 27 and 36 by Listing Multiples
To calculate the LCM of 27 and 36 by listing out the common multiples, we can follow the given below steps:
Step 1: List a few multiples of 27 (27, 54, 81, 108, 135, 162, 189, . . . ) and 36 (36, 72, 108, 144, 180, 216, 252, . . . . )Step 2: The common multiples from the multiples of 27 and 36 are 108, 216, . . .Step 3: The smallest common multiple of 27 and 36 is 108.
∴ The least common multiple of 27 and 36 = 108.
### LCM of 27 and 36 by Division Method
To calculate the LCM of 27 and 36 by the division method, we will divide the numbers(27, 36) by their prime factors (preferably common). The product of these divisors gives the LCM of 27 and 36.
Step 3: Continue the steps until only 1s are left in the last row.
The LCM of 27 and 36 is the product of all prime numbers on the left, i.e. LCM(27, 36) by division method = 2 × 2 × 3 × 3 × 3 = 108.
### LCM of 27 and 36 by Prime Factorization
Prime factorization of 27 and 36 is (3 × 3 × 3) = 33 and (2 × 2 × 3 × 3) = 22 × 32 respectively. LCM of 27 and 36 can be obtained by multiplying prime factors raised to their respective highest power, i.e. 22 × 33 = 108.Hence, the LCM of 27 and 36 by prime factorization is 108.
☛ Also Check:
## FAQs on LCM of 27 and 36
### What is the LCM of 27 and 36?
The LCM of 27 and 36 is 108. To find the LCM of 27 and 36, we need to find the multiples of 27 and 36 (multiples of 27 = 27, 54, 81, 108; multiples of 36 = 36, 72, 108, 144) and choose the smallest multiple that is exactly divisible by 27 and 36, i.e., 108.
### Which of the following is the LCM of 27 and 36? 12, 2, 25, 108
The value of LCM of 27, 36 is the smallest common multiple of 27 and 36. The number satisfying the given condition is 108.
### How to Find the LCM of 27 and 36 by Prime Factorization?
To find the LCM of 27 and 36 using prime factorization, we will find the prime factors, (27 = 3 × 3 × 3) and (36 = 2 × 2 × 3 × 3). LCM of 27 and 36 is the product of prime factors raised to their respective highest exponent among the numbers 27 and 36.⇒ LCM of 27, 36 = 22 × 33 = 108.
### What is the Least Perfect Square Divisible by 27 and 36?
The least number divisible by 27 and 36 = LCM(27, 36)LCM of 27 and 36 = 2 × 2 × 3 × 3 × 3 ⇒ Least perfect square divisible by each 27 and 36 = LCM(27, 36) × 3 = 324 Therefore, 324 is the required number.
### If the LCM of 36 and 27 is 108, Find its GCF.See more: Dramatic Irony In Lord Of The Flies, What Is The Irony At The End Of Lord Of The Flies
LCM(36, 27) × GCF(36, 27) = 36 × 27Since the LCM of 36 and 27 = 108⇒ 108 × GCF(36, 27) = 972Therefore, the greatest common factor (GCF) = 972/108 = 9. |
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# The volume of an air bubble is doubled as it rises from the bottom of the lake to its surface. The atmospheric pressure is 75cm of mercury. The ratio of density of mercury to that of lake water is 40/3. The depth of the lake in meter is: (A) 10m (B) 15m (C) 20m (D) 25m
Last updated date: 13th Sep 2024
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Hint: Fluid pressure is the pressure at a point within a fluid rising due to the weight of the fluid.
Using following formula,
${{P}_{fluid}}=P+\rho gh$
P is pressure at the reference point, Here, P= ${{\text{P}}_{0}}$ , ${{\text{P}}_{0}}$ is atmosphere pressure , $\rho$ is density of fluid, g is acceleration due to gravity, h is height from reference point.
Complete step by step solution:
We have given, (V’) volume of air bubble $=2\times$ volume of air bubble at bottom (v).
i.e. $\text{{V}'=2V}$ ---------(1)
Also,
\begin{align} & \dfrac{(\rho ')\text{density of mercury}}{(\rho )\text{density of lake water}}=\dfrac{40}{3} \\ & \dfrac{\rho '}{\rho }=\dfrac{40}{3} \\ & \Rightarrow \rho '=\dfrac{40}{3}\rho \\ \end{align} ------(2)
Now, the pressure exerted at the surface is equal to 75cm of mercury.
From Idle gas equation,
\begin{align} & P=P'\left( \dfrac{V'}{V} \right) \\ & P=P'\left( \dfrac{2V}{V} \right) \\ & P=2P' \\ & \sin ce,P'={{P}_{0}} \\ \end{align}
$\therefore P=2{{P}_{0}}$ --------(3)
Pressure at h depth From eq. (3)
\begin{align} & {{P}_{0}}+\rho gh \\ & 2{{P}_{0}}={{P}_{0}}+\rho gh \\ & {{P}_{0}}=\rho gh \\ & \therefore \rho 'gh'=\rho gh \\ & h=\left( \dfrac{\rho '}{\rho } \right)\times h' \\ \end{align} ----(4)
Here, h is depth of lake h’ is depth = 75 cm
= 0.75m
Now, Put all the values in eq. (4)
\begin{align} & h=\left( \dfrac{40}{3} \right)\times 0.75 \\ & h=\dfrac{40}{3}\times \dfrac{75}{100} \\ & h=10m \\ \end{align}
Hence, option A is correct.
Note:
Factors that affect fluid Pressure:
There are two factors that affect fluid pressure. These two factors are the depth of the fluid and its density.
- Depth of the fluid: As the depth increases, the pressure exerted by the fluid becomes more.
- Density of the fluid: Denser fluids such as water exert more pressure than lighter fluids such as air. The molecules in the denser fluid are close to each other resulting in more collisions in the given area. |
# find in what ratio is the line joining (3,4) and (5,-7) cut by the co-ordinates axis
lfryerda | Certified Educator
To find the ratio of the line cut by the x-axis, we need to use the distance formula d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} between the two points (x_1,y_1) and (x_2,y_2).
The equation of the line between the end points is found by getting the slope of the line.
m={-7-4}/{5-3}
=-11/2
The equation of the line is then found using y=mx+b and solving for b.
4=-11/2\cdot 3+b
b={8+33}/2
b=41/2
so the equation of the line is y=-11/2 x +41/2 .
The x-intercept of this line is at y=0. That is, when 0=-11/2 x+41/2
which is at x=41/11.
Consider the three points A=(3,4), B=(41/11,0) and C=(5,-7). The point B is the x-intercept of the line.
The distance from A to B is
d_{AB}=\sqrt{(3-41/11)^2+(0-4)^2}
=\sqrt{64/121+16}
=\sqrt{2000}/11
and the distance from B to C is
d_{BC}=\sqrt{(5-41/11)^2+(-7-0)^2}
=\sqrt{196/121+49}
=\sqrt{6125}/11
This means that the ratio from AB to BC after cancelling out the common factor of 11 is
\sqrt{2000/6125}
=\sqrt{16/49}
=4/7
jeew-m | Certified Educator
The general equation for a line is y = mx+c
Our line goes through (3,4) and (5,-7)
So we can write;
4=m*3+c------------(1)
-7=m*5+c------------(2)
(1)-(2)
11 = -2m
m= -11/2
From (1) we can get 4=(-11/2)*3+c
c =41/2
Equation of the line joining (3,4) and (5,-7) is y=(-11/2)x+41/2
Since y values of two points changes 4 to -7 the line will cut x axis.
When y=0 then x=(41/2)(2/11) = 41/11
So the cutting point is (41/11,0)
Let legth AB=length (3,4) to ((41/11,0)
BC= length (5,-7) to ((41/11,0)
AB=sqrt[(3-41/11)^2+(4-0)^2] = sqrt[2000/121]
BC=sqrt[(5-41/11)^2+(-7-0)^2]= sqrt[6125/121]
So the ratio is AB/BC =sqrt[2000/121]/sqrt[6125/121]
=sqrt[2000/6125]
= 4/7
The the cordinate axis cut the line to the ratio of 4:7 |
## Classroom topics in focus: algebra as generalised arithmetic
Written by Kristen Tripet
Algebraic reasoning is foundational to all mathematical thinking. It is through algebra that we are able to explore and express mathematical structure, pattern and relationships. One of the big ideas in algebra is that of generalised arithmetic. The focus of arithmetic in the curriculum is not just about skill acquisition. Rather, it is about understanding the structure of our number system and exploring how computational methods can be expressed and generalised algebraically.
Algebra as generalised arithmetic was one of the focus topics used in the development of reSolve’s classroom resources. The two tasks below are taken from this topic and illustrate the development of algebraic reasoning.
## Year 4—Number maze
The properties of odd and even numbers are explored in Year 4.
• The maze starts at 5 and finishes at 14.
• You can move through the maze using horizontal, vertical and diagonal movements.
• Add together the numbers in each box that you pass through.
• The aim is to finish with a total that is an odd number.
There are multiple pathways through this maze. Through the course of this task, students are encouraged to look at how many odd and even numbers are in each pathway. They see that an odd number of odds is always required to give an odd total.
To explain this, students look at a visualisation of why two odd numbers always add to an even and similar results. They move from a specific case to the general case where they can see that the two unpaired parts of each number can be paired to make an even number.
This task is based on a mathematical ‘magic trick’. All these so called magic tricks have a basis in logic—they can be explained! The challenge of this task is to go beyond the entertaining. Students are asked to understand why it works and to see if they can go beyond to develop another one.
• Choose two numbers.
• Write down these numbers vertically aligned, and add them. Write the answer under the second number. Then add the last two numbers in the list, and continue until there are 10 numbers in the list. For example,
• Continue adding your chain of numbers until you have 10 numbers.
• Now add up the total of the 10 numbers in your list. (In the example the total will be 825).
Time for the teacher to perform some magic!
One student is asked to come to the board and record the numbers in their list. As the student writes the 10th number the teacher quickly tells them the total.
This ‘trick’ makes the teacher look as if they have added all 10 numbers very quickly, but in fact they have just multiplied the 7th number by 11 while the student has been writing the remaining numbers on the board. Generalising the addition chain reveals a beautiful Fibonacci pattern and allows the students to see what is happening.
If a and b are the starting numbers the seventh number is 5a + 8b and the total of the ten numbers is 55a + 88b, which can be factorised to 11(5a + 8b). The door is opened here to many more mathematical investigations. How can you quickly multiply by 11? Are there other numbers in the list that can be used to find the cumulative total? Will it work with any set of numbers? What if one of the starting numbers was 0 or a negative? Can you make up your own trick based on logic and represented algebraically? |
#### Please solve RD Sharma class 12 chapter Mean Value Theoram exercise 14.2 question 10 maths textbook solution
Answer: Point P $\left(\frac{a}{2 \sqrt{2}}, \frac{a}{2 \sqrt{2}}\right)$
Hint: Find $\frac{dy}{dx}$ and then use this formula for slope of tangent.
Given: $\mathrm{x}=a \cos ^{3} \theta, y=a \sin ^{3} \theta, \quad 0<\theta<\frac{\pi}{2} .$
The tangent to C is parallel to the chord joining the points (a, 0) and (0, a).
Solution:
\begin{aligned} &x=a \cos ^{3} \theta \\ &\frac{d x}{d \theta}=-3 a \cos ^{2} \theta \sin \theta \\ &y=a \sin ^{3} \theta \\ &\frac{d y}{d \theta}=3 a \sin ^{2} \theta \cos \theta \end{aligned}
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$
\begin{aligned} &\Rightarrow \frac{d y}{d x}=\frac{3 a \sin ^{2} \theta \cdot \cos \theta}{-3 a \cos ^{2} \theta \cdot \sin \theta} \\ &\Rightarrow \frac{d y}{d x}=-\tan \theta \end{aligned}
Now, slope of chord $=\frac{a-0}{0-a}$
Slope of chord $=\frac{a}{-a}=-1$
Slope of chord $=\frac{d y}{d x}$
\begin{aligned} &\Rightarrow-1=-\tan \theta \\ &\Rightarrow \tan \theta=1 \\ &\Rightarrow \tan \theta=\tan \frac{\pi}{4} \\ &\Rightarrow \theta=\frac{\pi}{4} \end{aligned}
Now, $x=a \cos ^{3} \theta$
\begin{aligned} &=a \cos ^{3} \frac{\pi}{4} \\ &=a\left(\frac{1}{\sqrt{2}}\right)^{2} \\ &=\frac{a}{2 \sqrt{2}} \quad\left[\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right] \end{aligned}
Now, $y=a \sin ^{3} \theta$
\begin{aligned} &=a \sin ^{3} \frac{\pi}{4} \\ &=a\left(\frac{1}{\sqrt{2}}\right)^{2} \\ &=\frac{a}{2 \sqrt{2}} \quad\left[\because \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right] \end{aligned}
Hence Point P is $\left(\frac{a}{2 \sqrt{2}}, \frac{a}{2 \sqrt{2}}\right)$ |
# point slope form
Your Search returned 8 results for point slope form
point slope form - show you how to find the equation of a line from a point on that line and the line's slope.
Formula: y - y1 = m(x - x1)
-11, -9, -7, -5, -3 What is the next number? What is the 200th term in this sequence?
-11, -9, -7, -5, -3 What is the next number? What is the 200th term in this sequence? We see that Term 1 is -11, Term 2 is -9, so we do a point slope equation of (1,-11)(2,-9) to get the [URL='https://www.mathcelebrity.com/search.php?q=%281%2C-11%29%282%2C-9%29']nth term of the formula[/URL] f(n) = 2n - 13 The next number is the 6th term: f(6) = 2(6) - 13 f(6) = 12 - 13 f(6) = [B]-1 [/B] The 200th term is: f(200) = 2(200) - 13 f(200) = 400 - 13 f(200) = [B]387[/B]
5, 14, 23, 32, 41....1895 What term is the number 1895?
5, 14, 23, 32, 41....1895 What term is the number 1895? Set up a point slope for the first 2 points: (1, 5)(2, 14) Using [URL='https://www.mathcelebrity.com/search.php?q=%281%2C+5%29%282%2C+14%29&x=0&y=0']point slope formula, our series function[/URL] is: f(n) = 9n - 4 To find what term 1895 is, we set 9n - 4 = 1895 and solve for n: 9n - 4 = 1895 Using our [URL='https://www.mathcelebrity.com/1unk.php?num=9n-4%3D1895&pl=Solve']equation solver[/URL], we get: n = [B]211[/B]
A circle has a center at (6, 2) and passes through (9, 6)
A circle has a center at (6, 2) and passes through (9, 6) The radius (r) is found by [URL='https://www.mathcelebrity.com/slope.php?xone=6&yone=2&slope=+2%2F5&xtwo=9&ytwo=6&pl=You+entered+2+points']using the distance formula[/URL] to get: r = 5 And the equation of the circle is found by using the center (h, k) and radius r as: (x - h)^2 + (y - k)^2 = r^2 (x - 6)^2 + (y - 2)^2 = 5^2 [B](x - 6)^2 + (y - 2)^2 = 25[/B]
Choose the equation of a line in standard form that satisfies the given conditions. perpendicular to
Choose the equation of a line in standard form that satisfies the given conditions. perpendicular to 4x + y = 8 through (4, 3). Step 1: Find the slope of the line 4x + y = 8. In y = mx + b form, we have y = -4x + 8. The slope is -4. To be perpendicular to a line, the slope must be a negative reciprocal of the line it intersects with. Reciprocal of -4 = -1/4 Negative of this = -1(-1/4) = 1/4 Using our [URL='https://www.mathcelebrity.com/slope.php?xone=4&yone=3&slope=+0.25&xtwo=3&ytwo=2&bvalue=+&pl=You+entered+1+point+and+the+slope']slope calculator[/URL], we get [B]y = 1/4x + 2[/B]
Find y if the line through (1, y) and (2, 7) has a slope of 4.
Find y if the line through (1, y) and (2, 7) has a slope of 4. Given two points (x1, y1) and (x2, y2), Slope formula is: slope = (y2 - y1)/(x2 - x1) Plugging in our coordinates and slope to this formula, we get: (7 - y)/(2 - 1) = 4 7 - y/1 = 4 7 - y = 4 To solve this equation for y, w[URL='https://www.mathcelebrity.com/1unk.php?num=7-y%3D4&pl=Solve']e type it in our search engine[/URL] and we get: y = [B]3[/B]
If the equation of a line passes through the points (1, 3) and (0, 0), which form would be used to w
If the equation of a line passes through the points (1, 3) and (0, 0), which form would be used to write the equation of the line? [URL='https://www.mathcelebrity.com/slope.php?xone=1&yone=3&slope=+&xtwo=0&ytwo=0&bvalue=+&pl=You+entered+2+points']Typing (1,3),(0,0) into the search engine[/URL], we get a point-slope form: [B]y - 3 = 3(x - 1)[/B] If we want mx + b form, we have: y - 3 = 3x - 3 Add 3 to each side: [B]y = 3x[/B]
Line Equation-Slope-Distance-Midpoint-Y intercept
Free Line Equation-Slope-Distance-Midpoint-Y intercept Calculator - Enter 2 points, and this calculates the following:
* Slope of the line (rise over run) and the line equation y = mx + b that joins the 2 points
* Midpoint of the two points
* Distance between the 2 points
* 2 remaining angles of the rignt triangle formed by the 2 points
* y intercept of the line equation
* Point-Slope Form
* Parametric Equations and Symmetric Equations
Or, if you are given a point on a line and the slope of the line including that point, this calculates the equation of that line and the y intercept of that line equation, and point-slope form.
Also allows for the entry of m and b to form the line equation
What is the slope of the line through (1,9) and (5,3)
What is the slope of the line through (1,9) and (5,3) [URL='https://www.mathcelebrity.com/slope.php?xone=1&yone=9&slope=+2%2F5&xtwo=5&ytwo=3&pl=You+entered+2+points']We run this through our slope calculator[/URL], and get an initial slope of 6/4. But this is not in simplest form. So we type 6/4 into our calculator, and s[URL='https://www.mathcelebrity.com/fraction.php?frac1=6%2F4&frac2=3%2F8&pl=Simplify']elect the simplify option[/URL]. We get [B]3/2[/B] |
If the length L of a rectangle is 3 meters more than twice its width and its perimeter is 300 meters, which of the following equations could be used to find L?
Aug 3, 2015
$L = \text{101 m}$
$w = \text{49 m}$
Explanation:
You have two equations to work with, the equation that describes the perimeter of the rectangle and the equation that describes the relationship between its length, $L$, and width, $w$.
The perimeter of a reactangle is always equal to
$P = 2 \cdot \left(L + w\right)$
$2 \left(L + w\right) = 300$
Now, you know that if you double the width and add three meters to the result, you get $L$. This means that
$L = 2 \cdot w + 3$
You now have a system of two equations
$\left\{\begin{matrix}2 L + 2 w = 300 \\ L = 2 w + 3\end{matrix}\right.$
Use the value of $L$ from the second equation to replace $L$ in the first equation. This will get you
$L = 2 w + 3$
$2 \cdot \left(2 w + 3\right) + 2 w = 300$
$4 w + 6 + 2 w = 300$
$6 w = 294 \implies w = \textcolor{g r e e n}{49}$
This means that $L$ is equal to
$L = 2 \cdot w + 3$
$L = 2 \cdot 49 + 3 = \textcolor{g r e e n}{101}$
The rectangle will thus be 49 m wide and 101 m long. |
# Factor $x^3-x^2+2$ in $\mathbb Z_{3}[x]$
Factor $x^3-x^2+2$ in $\mathbb Z_{3}[x]$ and explain why the factors are irreducible.
So the factor is supposed to be:
$x^3-x^2+2 = 2(x + 1)(2x^2 + 2x + 1)= (x + 1)(x^2 + x + 2)$.
But I don't really see how. What I see:
$x^3-x^2+2 \equiv x^3+2x^2+2$ in $\mathbb Z_{3}[x]$.
And I suppose it can factor to $x^2(x+2)+2$ ? I don't see where $2(x + 1)(2x^2 + 2x + 1)$ comes from.
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Since the degree is $3$ is enough to check for roots. – Sigur Nov 7 '12 at 10:39
Factoring means: Write your polynomial as a product of irreducible ones. $x^2(x+2) + 2$ is no factorisation. To find the linear factors of a polynomial, first find it's roots. So we plug in $0$, $1$ and $2$ into $p(x) = x^3 - x^2 + 2$, and get \begin{align*} p(0) &= 2\\ p(1) &= 2\\ p(2) &= 8 - 4 + 2 = 6 \equiv 0 \end{align*} So $2 \in \mathbb Z_3$ is a root. That means that $p$ has a factor $(X - 2) = (X+1) \in \mathbb Z_3[X]$, polynomial long division gives $X^3 - X^2 + 2 = (X+1) \cdot (X^2 + X + 2)$ as $q(X)= X^2 + X + 2$ has no zeros in $\mathbb Z_3$, it is (as a polynomial of degree two) irreducible and $p(X) = (X+1)\cdot (X^2 + X + 2)$ is the factorization of $p$ into irreducible factors you aimed for.
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Is there any reason why you choose $0,1$ and $2$ specifically? – Arvin Nov 7 '12 at 10:54
@Arvin: What else is there to choose? – Chris Eagle Nov 7 '12 at 10:58
@Arvin These are all the elements of $\mathbb Z_3$. – martini Nov 7 '12 at 10:58
ahhh of course.. oh me.. – Arvin Nov 7 '12 at 11:02
Let $P(x)=x^3-x^2+2$. Observe that $P(-1)=0$, hence $x-(-1)=x+1$ is a factor.
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The polynomial has $-1$ as root. If you divide by $x+1$ you obtain the factorization.
A polynomial $f(x) \in \mathbb{F}[x]$ ($\mathbb{F}$ field) with $\operatorname{deg}(f)\leq 3$ is irreducible iff it has no roots in $\mathbb{F}.$ If it has a root $a$ then is divisible by $x-a$.
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Notice that $$x^3-x^2+2=x^3+2x^2-1=x^3+x^2+x^2-1=x^2(x+1)+(x+1)(x-1)=(x+1)(x^2+x-1)$$
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You should really try to write mathematics in this site with LaTeX, otherwise things come up as above in your answer and it is very hard to understand them. – DonAntonio Nov 7 '12 at 10:47
@Amr all you need to do to make it a lot better is surround the whole equation with dollar signs (\\$) – Arvin Nov 7 '12 at 11:05 |
# Set Operations with Venn Diagrams
## Set Operations with Venn Diagrams
There are mainly four operations of sets. They are:
i. Union of sets
ii. Intersection of sets
iii. Difference of sets
iv. Complement of a set
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10 Math Problems officially announces the release of Quick Math Solver and 10 Math ProblemsApps on Google Play Store for students around the world.
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### i. Union of sets
The union of two sets A and B is the set of all elements that belongs to set A or set B or both the set A and B. Symbolically we write ‘A B’ and read as ‘A union B’.
Therefore, A B = {x:x A or x B or x both A and B}
For example:
i. A = {1, 2, 3, 4} and B = {3, 4, 5, 6}
A B = {1,2,3,4,5,6}
The shaded region represents the union of sets A and B.
ii. P = {a, b, c, d, e} and Q = {c, d}
P Q = {a,b,c,d,e}
The shaded region represents the union of sets P and Q.
### ii. Intersection of sets
The intersection of two sets A and B is the set of all common elements that belongs to both the sets A and B. Symbolically, we write ‘A B’ and read as ‘A intersection B’.
Therefore, A B = {x:x A and x B}
For example:
i. A = {1, 2, 3, 4} and B = {3, 4, 5, 6}
A B = {3, 4}
The shaded region represents the intersection of sets A and B.
ii. P = {a, b, c, d, e} and Q = {c, d}
P Q = { c, d }
The shaded region represents the intersection of sets P and Q.
### iii. Difference of sets
The difference of two sets A and B written as A – B is the set of all elements of set A only which are not in set B. Similarly, the difference B – A is the set of all elements of set B only which are not in set A.
Therefore, A B = {x:x A and x B} and B A = {x:x B and x A}
For example:
i. A = {1, 2, 3, 4} and B = {3, 4, 5, 6}
A B = {1, 2}
The shaded region represents the difference A B.
ii. A = {1, 2, 3, 4} and B = {3, 4, 5, 6}
B A = {5, 6}
The shaded region represents the difference B A.
### iv. Complement of a set
Let U be the universal set and A is any subset of U, then the complement of set A denoted Ac is the set of all the elements of set U which are not in set A.
Therefore, Ac = U – A ={x:x U and x A}
For example:
Let, U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {2, 4, 6, 8} then,
Ac = U – A = {1, 3, 5, 7, 9, 10}
The shaded region represents the complement of A.
### Workout Examples
Example 1: If U = {1, 2, 3, ……… 15}, A = {1, 2, 3, 4, 5, 6}, B = {2, 4, 6, 8, 10, 12} and C = {3, 6, 9, 12, 15} then find (A B) C and show in the venn diagram.
Solution: Here,
U = {1, 2, 3, ……… 15}
A = {1, 2, 3, 4, 5, 6}
B = {2, 4, 6, 8, 10, 12}
C = {3, 6, 9, 12, 15}
Now,
(A B) C = [{1, 2, 3, 4, 5, 6} {2, 4, 6, 8, 10, 12}] {3, 6, 9, 12, 15}
= {1, 2, 3, 4, 5, 6, 8, 10, 12} {3, 6, 9, 12, 15}
= {3, 6, 12}
Venn diagram,
The shaded region represents the the set (A B) C.
You can comment your questions or problems regarding the set operations and venn diagrams here. |
Case study Chapter 4(Determinant)
Case study 4:– Read the following and answer the question(Case study problem determinant 4)
  On the occasion of children’s day. Class teacher of class XII shri singh, decided to donate some money to students of class XII.
   If there were 8 studens less, every one would have got ₹10 more, however if there were 16 students more, everyone would have got ₹10 less.
(i) If number of students in class be x and Shri Singh has to decided to donate ₹y to each student, then the information provided in problem may be written in system of linear equation as
(a) 5x + 4y = 40, 5x – 8y = -80
(b)Â x – y = 40, 2x – 3y = 80
(c) 5x – 4y = 40, 5x – 8y = -80
(d) 8x + 10y = 40, 16x – 10y = 80
(ii) System of linear equations obtained in (i) may be written in matrix equation as
(a)
(b)
(c)
(d)
(iii) If AX = B, where A, X, B are matrix, then X should be
(a) X = ABÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b)
(c) Â Â Â Â Â Â Â Â Â (d)
(iv) If then is
(a)
(b)
(c)
(d)
(v) The number of students in class XII and amount distributed by Shri Singh respectively isÂ
(a) 30, ₹ 32             (b) 30, ₹30
(c) 30,₹32              (d) 32 ₹32
From question
(x – 8)(y + 10) = xy
⇒ x.y + 10x – 8y – 80 = x.y
⇒ 10x – 8y = 80
⇒ 5x – 4y = 40 —–(i)
Also,
(x + 16)(y – 10) = xy
⇒ x.y – 10x + 16y – 160 = x.y
⇒ 10x – 16y = -160
⇒ 5x – 8y = -80 —–(ii)
We have system of linear equations,
5x – 4y = 40 —–(i)
5x – 8y = -80 —–(ii)
We can write matrix system as
We have
AX = B
Pre multiplying both sides, we have
We have
⇒ |A| = -40 + 20 = -20 ≠0
NOW,
⇒ x = 32, and y = 30
Some other Case study problem
Case study 1:–Â Read the following and answer the questions(Case study problem determinant 1)
      Three friends Rahul, Ravi and Rakesh went to a vegetable market to purchase vegetable. From a vegetable shop Rahul purchased 1 kg of each potato, onion and Brinjal for a total of ₹ 21. Ravi purchased 4 kg of potato, 3 kg of onion and 2 kg of Brijal for ₹ 60 while Rakesh purchased 6 kg potato, 2 kg onion and 3 kg Brinjal for ₹70
Case study 2:- Read the following and answer the question(Case study problem determinant 2)
     Reshma wants to donate a rectangular plot of land for a school of her village. When she was asked by construction agency to give dimensions of the plot, she said that if its length is decreased by 50 m and breadth is increased by 50m, then its area will remain same, but if length is decreased by 10 m and breadth is decreased by 20m, then its area will decreased by 5300 m².
Case study 3:– Read the following and answer the question(Case study problem determinant 3 )
       The monthly incomes of two sister Reshma and Ritam are in the ratio 3:4 and their monthly expenditures are in the ratio 5;7. Each sister saves ₹ 15,000 per month.
Case study 5:– In coaching institutes, the students not only get academic guidance but also they get to know about career options and right goals as per their interest and academic record.(Case study problem determinant 5)
A coaching institute conduct classes in two sections A and Band fees for rich and poor children are different. In section A, there are 20 poor and 5 rich children and total monthly collection is ₹9,000, where as in section B, there are 5 poor and 25 rich children and total monthly collection is ₹ 26, 000.
For More question |
# 31 Rational Expressions and Equations
Jenna Lehmann
##### What is a Rational Expression?
A rational number is a number that can be written as a quotient of integers. In other words, it’s any number that can create a nice and neat fraction. A rational expression is also a quotient, it’s just made up of polynomials. A rational expression can be written in the form P/Q. For example:
or or
##### Evaluating Rational Expressions
Rational expressions have different numerical values depending on what values replace the variables. For example:
If then
If then
##### Identifying When a Rational Expression is Undefined
When the denominator of a rational expression is 0, then it is undefined. There are some equations in which the denominator is sometimes 0 and some in which the denominator is never 0. Take the following equation as an example.
When we set to equal 0, we find that when is 3, the denominator is 0. So when is 3, the rational expression is undefined.
The denominator of the above equation is never 0, and so it is never undefined.
Sometimes you will need to factor the denominator to find the variables which make it undefined, like in the following example.
and
So when equals 2 or 1, the expression is undefined.
##### Simplifying Rational Expressions
Sometimes a fraction made of polynomials can be simplified. This can be done by taking out the greatest common factor, like in the example below.
This may also be done by factoring the polynomials in both the numerator and denominator.
##### Multiplying Rational Expressions
Multiplying rational expressions follow the same rules as multiplying other kinds of fractions. The numerator of one fraction is multiplied by the numerator of the other, and the denominator of one fraction is multiplied by the other.
This goes for polynomials as well. Just be sure to FOIL when necessary.
##### Dividing Rational Expressions
Remember that dividing fractions is the same thing as flipping the second fraction and multiplying.
This is also true for polynomials. Again, don’t forget to FOIL when necessary.
##### Adding and Subtracting Rational Expressions with the Same Denominator
When adding and subtracting with rational expressions that have the same denominator, all you have to worry about is adding and subtracting what’s in the numerator.
This is true for any kind of term that may be in the numerator or denominator.
or
##### Finding the Least Common Denominator
Finding the least common denominator involves breaking each number up into its smallest variables and seeing how those numbers compare so that the smallest number that they both fit into appears. Take a look at the example below. 8 can be broken up into 2*2*2 and 6 can be broken up into 2*3. We need a number that satisfies the need for all of these numbers, even if a few overlap. The solution to this is 2*2*2*3. It incorporates the 8’s need to have 3 2’s and the 6’s need to have a 2 and a 3. 2*2*2*3 = 24 so our LCD is 24.
This also works with variables. We’re working with 5x and 15x^2. We need to find a number that fulfills the need of a 5 and an x as well as a 5, a 3, and 2 x’s. The equation 5*x*x*3 fulfills all of those needs. So our LCD is 15x^2.
##### Writing Equivalent Rational Expressions
When writing equivalent rational expressions, we are, in a sense, multiplying an expression of 1.
This is useful for when we want to translate one equation to keep the same value but use a different denominator.
In this case, we have to ask ourselves, “what can I multiply by in order to get ?”
##### Adding and Subtracting Rational Expressions with Different Denominators
Find the LCD and then multiply one rational expression by what’s missing from it to get that LCD in the denominator. Make sure to do this with the other fraction, if necessary, so that in the end, both fractions have the same denominator – the LCD.
##### Solving Equations Containing Rational Expressions
Sometimes, you will be asked to solve for a missing variable when there are equations containing rational expressions. I recommend doing what is needed so that the denominators of all the numbers are eliminated and you can work with whole numbers again.
Another, more complex example would be:
This chapter was originally posted to the Math Support Center blog at the University of Baltimore on October 7, 2019. |
# Section 9.6 What we are Learning:
## Presentation on theme: "Section 9.6 What we are Learning:"— Presentation transcript:
Section 9.6 What we are Learning:
To multiply a polynomial by a monomial To simplify expressions involving polynomials
Distributive Property:
For any real number a(b + c) = ab + ac This means: Multiply everything inside of the grouping symbol by what is on the outside of the grouping symbol.
Multiplying a Polynomial by a Monomial:
Combine any like terms first Use the distributive property: Multiply everything inside of the grouping symbol by what is on the outside of the grouping symbol. Use your Properties of Powers to multiply variables with exponents You can multiply vertically or horizontally
Example: -5x3(7x2 + 8x -3) Vertically Horizontally -5x3(7x2 + 8x -3)
-5x3 (7x2) + (-5x3)(8x) + (-5x3)(-3) -35x5 - 40x4 + 15x3 -5x3(7x2 + 8x -3) 7x2 + 8x -3 (x) x3 -35x5 - 40x4 + 15x3
Example: Find 3x2(6x2y + 4 – 2x2y)
Vertically Horizontally 3x2(6x2y + 4 – 2x2y) 3x2(4x2y + 4) 3x2(4x2y) + 3x2(4) 12x4y + 12x2 3x2(6x2y + 4 – 2x2y) 3x2(4x2y + 4) 4x2y + 4 (x) x2 12x4y + 12x2
Solving Equations That Contain Polynomials:
Combine any like terms Use the Distributive Property Use the Addition/Subtraction, Multiplication/Division Properties of Equality to solve for the variable. Remember: An equation is solved when a variable with a coefficient of one is on one side of the equation and the solution is on the other.
Example: 2x(x – 4) + 8x – 3 = 2x(5 + x) – 5x + 12
There are no like terms [2x(x) + 2x(-4)] + 8x – 3 = [2x(5) + 2x(x)] – 5x + 12 Distributive Property 2x2 – 8x + 8x – 3 = 10x + 2x2 – 5x + 12 Combine like terms 2x2 – 3 = 5x + 2x2 + 12 2x2 – 2x2 – 3 = 5x + 2x2 – 2x2 + 12 Subtraction Property of equality -3 = 5x + 12 -3 – 12 = 5x + 12 – 12 -15 = 5x -15/5 = 5x/5 Division Property of equality -3 = x Solution
Let’s Work These Together: Find each product
4a2(-8a3c + c -11) 11ab(2ab – 5a)
Let’s Work These Together:
Simplify Solve 8m(-9m2 + 2m – 6) + 11(2m2 – 4m + 12) y(y + 12) – 8y = 14 + y(y – 4)
Homework: Page 532 17 to 21 odd 27 to 31 odd 33 to 37 odd
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# How Do You Find the Perimeter and Area of a Trapezoid?
So far we have learned about the perimeter and area of a square, a rectangle, a rhombus and a parallelogram. Now we will look at how to find the perimeter and area of a trapezoid. But first, let’s take a look at one.
A trapezoid is a quadrilateral with two parallel sides of differing lengths. The other two sides can be at any length—and any angle!
## Perimeter of a Trapezoid
To find the perimeter you need to add the length of all four sides. In a trapezoid, the four sides can, and usually are, all different lengths, so all four sides must be measured out and added together.
Formula
### PerimeterofaTrapezoid
Two sides are parallel, but not the same length.
$P={\text{side}}_{1}+{\text{side}}_{2}+{\text{side}}_{3}+{\text{side}}_{4}$
Example 1
Find the perimeter of the following trapezoid:
You use the formula and add all four sides together: $\begin{array}{llll}\hfill P& =2\phantom{\rule{0.17em}{0ex}}\text{cm}+3\phantom{\rule{0.17em}{0ex}}\text{cm}+3\phantom{\rule{0.17em}{0ex}}\text{cm}+5\phantom{\rule{0.17em}{0ex}}\text{cm}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =13\phantom{\rule{0.17em}{0ex}}\text{cm}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
## Area of a Trapezoid
To find the area of a trapezoid you need to know the length of the two parallel sides and the height of the trapezoid. When you know this, you put the numbers into the following formula for the area of a trapezoid.
Formula
### AreaofaTrapezoid
$A=\frac{\left(a+b\right)\cdot h}{2}$
The height is always a right angle to both the parallel sides. When you know the two lengths and the height, you can use the formula.
Example 2
Find the area of the trapezoid
You find the area by using the formula: $\begin{array}{llll}\hfill A& =\frac{\left(4\phantom{\rule{0.17em}{0ex}}\text{cm}+2\phantom{\rule{0.17em}{0ex}}\text{cm}\right)\cdot 3\phantom{\rule{0.17em}{0ex}}\text{cm}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{6\phantom{\rule{0.17em}{0ex}}\text{cm}\cdot 3\phantom{\rule{0.17em}{0ex}}\text{cm}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{18\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{2}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =9\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ |
# Point
Point can be defined as that geometrical mark that does not have any dimensions and is represented using coordinates in the Cartesian system. But still it has much importance in math geometry. For defining a plane we must have at least three collinear points. From one point as many circles and lines can pass but when two points are considered, only one line and infinite number of circles can pass. When the number of points becomes three, only one circle can pass through it. For a point to lie on any curve or geometrical figure, it must satisfy the equation of that curve or figure.
For a point we define locus and also write the equation that will be specific for many such points. So, if we are defining a locus we must have all such points which satisfy certain conditions.
Example: Suppose we have a line 5y = 8x – 3, then find whether the points (2, 4) and (1, 1) satisfy it or not?
Solution: For the points (2, 4) and (1, 1) to lie on the line 5y = 8x – 3, they must satisfy the given equation. So, substituting them in the equation we can check whether they lie on the line or not:
For (2, 4),
5 y = 8 x – 3,
5 * 4 = 8 * 2 – 3,
Or 20 = 13 which is not possible,
For (1, 1),
5 y = 8 x – 3,
5 * 1 = 8 * 1 – 3,
Or 5 = 5,
Thus 2nd point lies on the line.
For (2, 4),
5 y = 8 x – 3,
This is how we make use of point geometry by using the coordinates of points which can be either in the form of rectangular coordinates or polar coordinates.
## Collinear Point
Mathematically collinear means to lie on the same line and this concept is important for solving the problems related to collinerarity of points. Two points are said to be collinear if they are supposed to lie on the same line. To be on a line means the points satisfy the equation of the line and even we can get the equation of line using two point form taking any two coll...Read More
## Theorem (If a Point Lies outside a Line, Then exactly One Plane contains the Line and the Point)
In math we have theorems based on various concepts that are actually possible to be proved. It is a kind of fact that is written in form of a statement. Likewise we have a suitable proof for the theorem that states if a Point Lies outside a line, then exactly one plane contains the line and point...Read More
Math Topics |
# Exercise 5.1 class 8 solution
### Exercise 5.1 class 8 solution-perfect square number
A perfect square is a number that can be expressed as the product of an integer with itself. In other words, if you take the square root of a perfect square, you will get an integer as the result. For example:
• 1 is a perfect square because 1 = 1 * 1 (the square root of 1 is 1).
• 4 is a perfect square because 4 = 2 * 2 (the square root of 4 is 2).
• 9 is a perfect square because 9 = 3 * 3 (the square root of 9 is 3).
• 16 is a perfect square because 16 = 4 * 4 (the square root of 16 is 4).
In general, if a positive integer “n” can be expressed as “n = a * a,” where “a” is also a positive integer, then “n” is a perfect square. Perfect squares are a subset of square numbers and have a special property of having integer square roots.
### Exercise 5.1 class 8 solution-properties of square number
Square numbers have several interesting properties and characteristics:
1. Definition: A square number is a number that can be expressed as the product of an integer with itself. In mathematical notation, a square number is often denoted as n^2, where n is an integer.
2. Perfect Squares: Square numbers are also known as perfect squares. These are the numbers that result from squaring an integer. For example, 1, 4, 9, 16, and 25 are all square numbers.
3. Integer Square Roots: The square root of a square number is an integer. For example, the square root of 9 is 3, and the square root of 16 is 4.
4. Patterns: When you list square numbers, you can observe patterns. The squares of consecutive integers increase by consecutive odd integers. For example, 1, 4, 9, 16, 25, and so on. The difference between consecutive squares is always an odd number (2, 5, 7, 9, …).
5. Sum of Consecutive Odd Numbers: Square numbers can also be expressed as the sum of consecutive odd numbers. For example, 16 is the sum of the first four odd numbers (1 + 3 + 5 + 7).
6. Pythagorean Triples: Square numbers are often related to Pythagorean triples, which are sets of three positive integers (a, b, c) that satisfy the Pythagorean theorem: a^2 + b^2 = c^2. The numbers 3, 4, and 5 form a Pythagorean triple since 3^2 + 4^2 = 5^2.
### Exercise 5.1 class 8 solution- square no. 1 to 30
Here are the square numbers from 1 to 30:
1. 1^2 = 1
2. 2^2 = 4
3. 3^2 = 9
4. 4^2 = 16
5. 5^2 = 25
6. 6^2 = 36
7. 7^2 = 49
8. 8^2 = 64
9. 9^2 = 81
10. 10^2 = 100
11. 11^2 = 121
12. 12^2 = 144
13. 13^2 = 169
14. 14^2 = 196
15. 15^2 = 225
16. 16^2 = 256
17. 17^2 = 289
18. 18^2 = 324
19. 19^2 = 361
20. 20^2 = 400
21. 21^2 = 441
22. 22^2 = 484
23. 23^2 = 529
24. 24^2 = 576
25. 25^2 = 625
26. 26^2 = 676
27. 27^2 = 729
28. 28^2 = 784
29. 29^2 = 841
30. 30^2 = 900
These are the squares of the numbers from 1 to 30. Each number on the left represents an integer, and the number on the right is the square of that integer.
### Exercise 5.1 class 8 solution- exercise preview
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### Exercise 5.1 class 8 solution-solution pdf
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Exercise 5.4 class 8 solution |
CBSE Class 10 Sample Paper for 2020 Boards - Maths Basic
Class 10
Solutions of Sample Papers for Class 10 Boards
## (iii) Which two students are equidistant from Gita.
### Transcript
Question 33 Read the following passage and answer the questions that follows: In a class room, four students Sita, Gita, Rita and Anita are sitting at A (3, 4), B (6, 7), C (9, 4), D (6, 1) respectively. Then a new student Anjali joins the class (i) Teacher tells Anjali to sit in the middle of the four students. Find the coordinates of the position where she can sit. Middle of the 4 students will be the mid point of BD and mid-point of AC So, it will be at point - (6, 4) (ii) Calculate the distance between Sita and Anita. Sita is point A (3, 4) , Anita is point D (6, 1) We need to calculate AD = β((π₯_2βπ₯_1 )^2+(π¦_2βπ¦_1 )^2 ) = β((0β3)^2+(0β(β4))^2 ) = β((β3)^2+(β4)^2 ) = β(3^2+4^2 ) = β(9+16) = β25 AD = β((π₯_2βπ₯_1 )^2+(π¦_2βπ¦_1 )^2 ) = β((6β3)^2+(1β4)^2 ) = β((3)^2+(β3)^2 ) = β(3^2+3^2 ) = β(2Γ3^2 ) = 3β2 Therefore, distance between Sita and Anita is πβπ units (iii) Which two students are equidistant from Gita. Here, Sita, Gita, Rita and Anita are sitting at A (3, 4), B (6, 7), C (9, 4), D (6, 1) respectively The point equidistant from point B are points A and C So, Sita and Rita are equidistant from Gita |
In this section, you will learn how to simplify radical expressions.
The radicals which are having same number inside the root and same index is called like radicals.
Unlike radicals don't have same number inside the radical sign or index may not be same.
The following steps will be useful to simply radical expressions
Step 1 :
Step 2 :
Take one number out of the radical for every two same numbers multiplied inside the radical sign, if the radical is a square root.
Take one number out of the radical for every three same numbers multiplied inside the radical sign, if the radical is a cube root.
Step 3 :
Simplify.
Examples :
√4 = √(2 2) = 2
√16 = √(2 2 2 2) = 2 2 = 2
3√27 = 3√(3 3 3) = 3
3√125 = 3√(5 5 5) = 5
Question 1 :
Simplify :
20 - 225 + 80
Solution :
Decompose 20, 225 and 80 into prime factors using synthetic division.
√20 = √2 2 5 = 2√5
√225 = √5 5 3 3 = 5 3 = 15
√225 = √2 2 2 2 5 = (2 2)5 = 4√5
Then, we have
20 - 225 + 80 = 2√5 - 15 + 4√5
20 - 225 + 80 = 6√5 - 15
20 - 225 + 80 = 6√5 - 15
20 - 225 + 80 = 3(2√5 - 5)
Question 2 :
Simplify :
√27 + √75 + √108 - √48
Solution :
Decompose 27, 75, 48 and 108 into prime factors using synthetic division.
√27 = √(3 3 3) = 3√3
√75 = √(5 5 ⋅ 3) = 5√3
√108 = √(3 3 3 ⋅ 2 ⋅ 2) = 3 ⋅ 2 ⋅ √3 = 6√3
√48 = √(2 2 2 2 3) = 2 ⋅ 2 ⋅ √3 = 4√3
Then, we have
√27 + √75 + √108 - √48 = 3√3 + 5√3 + 6√3 - 4√3
√27 + √75 + √108 - √48 = 10√3
Question 3 :
5√28 - √28 + 8√28
Solution :
5√28 - √28 + 8 √28
Because all the terms in the above radical expression are like terms, we can simplify as given below.
5√28 - √28 + 8√28 = 12√28
5√28 - √28 + 8√28 = 12√(2 ⋅ 7)
5√28 - √28 + 8√28 = 12 ⋅ 2√7
5√28 - √28 + 8√28 = 24√7
Question 4 :
9√11 - 6√11
Solution :
9√11 - 6√11
Because the terms in the above radical expression are like terms, we can simplify as given below.
9√11 - 6√11 = 3√11
Question 5 :
7√8 - 6√12 - 5√32
Solution :
7√8 - 6√12 - 5√32
Decompose 8, 12 and 32 into prime factors.
7√8 = 7√(2 2 2) = 7 ⋅ 2√2 = 14√2
6√12 = 6√(2 2 ⋅ 3) = 6 ⋅ 2√3 = 12√3
5√32 = √(2 2 2 ⋅ 2 ⋅ 2) = 5 ⋅ ⋅ 2 ⋅ √2 = 20√2
Then, we have
7√8 - 6√12 + 5√32 = 14√2 - 12√3 - 20√2
7√8 - 6√12 + 5√32 = 14√2 - 12√3 - 20√2
7√8 - 6√12 + 5√32 = -6√2 - 12√3
7√8 - 6√12 + 5√32 = -6(√2 + 2√3)
Question 6 :
2√99 + 2√27 - 4√176 - 3√12
Solution :
Decompose 99, 27, 176 and 12 into prime factors.
2√99 = 2√(3 3 11) = 2 ⋅ 3√11 = 6√11
2√27 = 2√(3 3 ⋅ 3) = 2 ⋅ 3√3 = 6√3
4√176 = √(2 2 2 ⋅ 2 ⋅ 11) = 4 ⋅ ⋅ 2√11 = 16√11
3√12 = 3√(2 2 3) = 3 ⋅ 2√3 = 6√3
Then, we have
2√99 + 2√27 - 4√176 - 3√12 = 6√11 + 6√3 - 16√11 - 6√3
2√99 + 2√27 - 4√176 - 3√12 = -10√11
Question 7 :
3√16 - 3√2 + 43√54
Solution :
Decompose 16 and 54 into prime factors.
3√16 = √(2 2 2 ⋅ 2) = 23√2
43√54 = 43√(3 3 3 ⋅ 2) = 4(33√2) = 123√2
Then, we have
3√16 - 3√2 - 43√54 = 23√2 - 3√2 + 123√2
3√16 + 3√2 - 43√54 = 133√2
Question 8 :
3√24 + 3√375 - 3√3
Solution :
Decompose 24 and 375 into prime factors.
3√24 = √(2 2 2 ⋅ 3) = 23√3
3√375 = 3√(5 5 5 ⋅ 3) = 53√3
Then, we have
3√24 + 3√375 - 3√3 = 23√3 + 53√3 - 3√3
3√24 + 3√375 - 3√3 = 63√3
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Data Handling Pictograph | Maths Grade 1 | Learning Concepts
Data handling
# Pictograph for Class 1 Math
This discussion is to inform the students about the data handling and the types of data handling. Pictograph is one of the simplest processes that is used to for data handling. From this concept the students will get to know about data and pictograph.
Also, the student will learn to
• Summarise and compare data in a planned way
• Express data using appropriate picture to make a pictograph
Each concept is explained to class 1 maths students with examples and illustrations, and a concept map is given at the end to summarise the idea. At the end of the page, two printable Pictograph worksheets with solutions are attached for students to practice. Download the worksheets and assess your knowledge.
## What Is Data?
• The information we have in which any particular objective is common through all the information.
• Or
• If we collect the information for any particular objective, then that information is called ‘Data’.
### Example :
Dani, Shubhrato, and Jiya each have some chocolates. Count the number of chocolates each one has.
In this case the information of the number of chocolates associated with each one is called Data. This data represents the number of chocolates each one has.
## What Is Data Handling?
• Data handling means studying the available data and representing it or sorting it in such a way that anyone can easily understand it.
### Example :
Suppose we have different colours of balls.
Now, we want to find out how many balls are there of each colour. So, we arrange the data like:
From arranged data in this manner, we can easily understand and get the number of balls of each colour.
So, the process of studying and arranging or sorting the data is known as ‘Data handling’.
We can arrange the data in the following manner:
1. Picture graph (Pictograph)
2. Table
## What Is A Pictograph?
• Picture graph or pictograph is the representation of the given data by using pictures or symbols.
• We can easily collect the information from the pictograph by studying it.
• The main advantage of the pictograph is, we can easily interpret a large amount of data.
## Interpreting Pictographs
• Title: It tells what the pictograph is about. It represents the main objective of the pictograph.
• Labels: It tells what kind of data is shown or it consists of the main objects related to which we collect the information.
• Pictures/Symbols: It tells the exact number of the objects.
• Key: It tells each picture or symbols is equal to.
### Example :
Kelvin and his friends are going on a picnic. There are a total of 12 children.
Each one brings his/her favourite food.
Now, if we want to know how many of the children are there whose favourite food is Pizza. We can make a pictograph here.
The pictograph looks like:
• Title of the pictograph – Favourite food. It tells that the pictograph is about a particular food is liked by how many children.
• Labels: In this pictograph, Pizza, juice, burger, chips and hot dog are the labels. They tell us that, we are collecting information about these foods.
• Pictures: Pictures tell us the number of foods.
• Key: Key helps us to count the number of objects.
From the above pictograph, we can easily find out how many likes to eat pizza.
Count each object.
Number of children who like pizza: 1
Number of children who like juice: 3
Number of children who like burgers: 2
Number of children who like chips: 2
Number of children who like hot dog: 4
In this way, we can easily get the information from a pictograph.
• - |
# Without actually calculating the cubes,
Question:
Without actually calculating the cubes, find the value of 36xy-36xy = 0
(i) $\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}$
(ii) $(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$
Thinking Process
In this method firstly check the values of a + b+ c, then . if a + b+c = Q, now use the identity a3 + b3 + c3 = 3abc.
Solution:
(i) Given, $\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}$ or $\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}+\left(-\frac{5}{6}\right)^{3}$
Here, we see that, $\frac{1}{2}+\frac{1}{3}-\frac{5}{6}=\frac{3+2-5}{6}=\frac{5-5}{6}=0$
$\therefore \quad\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}=3 \times \frac{1}{2} \times \frac{1}{3} \times\left(-\frac{5}{6}\right)=-\frac{5}{12}$
[using identity, if $a+b+c=0$, then $a^{3}+b^{3}+c^{3}=3 a b c$ ]
(ii) Given, $(02)^{3}-(0.3)^{3}+(0.1)^{3}$ or $(02)^{3}+(-0.3)^{3}+(0.1)^{3}$
Here, we see that, $02-0.3+0.1=0.3-0.3=0$
$\therefore \quad(02)^{3}+(-0.3)^{3}+(0.1)^{3}=3 \times(02) \times(-0.3) \times(0.1)$
[using identity, if $a+b+c=0$, then $a^{3}+b^{3}+c^{3}=3 a b c$ ]
$=-0.6 \times 0.03=-0.018$ |
# Algebra Chapter 1 Review
The material in this lesson is a review of previous algebra or pre-calculus courses. You
should be able to work through this material at a fairly rapid pace. If you find that much of
it is unfamiliar, or difficult, then you should consider reviewing at greater length. Any
college algebra text should have material similar to this lesson. At the University of Kansas,
the current text is College Algebra by Sullivan.
PART 1
By the end of this part of lesson 1, you should be able to
• distinguish between open, closed, and half-open intervals;
• understand and evaluate expressions involving absolute value; and
• work with exponents, including negative exponents and radical exponents , using the
Laws of Exponents.
Reading Assignment: Section 1.1 of Chapter 1
1. The symbol for absolute value is universal. Pay special attention to the definition of
the absolute value in the first highlighted box on page 6 of the text. The Absolute
Value Properties listed in the second highlighted box on page 6 are also very
important. Be sure to go over Example 3 at the bottom of page 6.
2. The Laws of Exponents (text, page 8; Table 1.4) have all the rules that you will need
when dealing with either integer or fractional exponents. It is important to remember
that an exponent is only attached to the obvious base; 2t2 is not the same as
3. The connection between nth roots of a number and fractional exponents is a very
important one (page 8; Table 1.3). You should be able to use these two concepts
interchangeably.
4. The distinction between multiplying or dividing two exponential expressions to the
same base—Laws 1 and 2—and taking powers—Law 3—is important. Multiplying
two exponential powers to the same base turns into addition or subtraction of
exponents, while raising an exponential to a power requires a product or division of
exponents.
5. Be careful when applying the Laws of Exponents. The exponent does NOT satisfy the
distributive property . Thus, to multiply (a + b)n one must use the binomial expansion
to expand this expression .
Practice Assignment
Section 1.1: Exercises 3, 7, 15, 19, 23, 25, 39, 41, 49, 55, 77, 81, 111, 113, 115, 119.
PART 2
By the end of this part of lesson 1, you should be able to
• multiply and factor algebraic expressions, especially expressions involving
polynomials;
• use the Quadratic Formula to find the roots of polynomials of degree two; and
simplify algebraic expressions (which may involve all arithmetic operations) and
arrive at a single fraction involving no negative exponents, and no common factors
between the numerator and the denominator.
Reading Assignment: Section 1.2 of Chapter 1
1. Multiplication of algebraic expressions is accomplished, in general, by using the
distributive rule illustrated in Example 2 (text, page 13). This method will always
provide the product, although there are some situations where a commonly used
product has a particular form; see Table 1.5 (page 13).
2. Factoring is the operation that attempts to un-multiply. This type of operation is
always more difficult than multiplication; see Table 1.6 (page 15). Factoring cannot
always be accomplished. Probably the best way to start is to attempt to group the
monomials
so that a common factor can be identified. Then, using the distributive law
in reverse, the original expression can be written as a product.
3. A root, sometimes called a zero, of a polynomial p is a number a such that p(a) = 0.
The roots of a polynomial are very difficult to find in general, and the problem of
finding roots is one that goes back to ancient Egypt and Mesopotamia. The
Babylonians and the Greeks could find roots of polynomials for all polynomials of
degree 1 and 2, but it was in the fourteenth century that roots of polynomials of degree
3 and 4 could be solved.
4. The Quadratic Formula gives a method for solving all polynomials of degree 2. This
formula was known by the Greek mathematicians of the B.C. era. While it may be
possible to solve such a polynomial by factoring and then applying the fact that a
product of real numbers is equal to zero if and only if at least one of the real numbers
is equal to zero, the Quadratic Formula gives a sure-fire and quick method for finding
the roots of a second degree polynomial.
5. Doing addition and subtraction with rational expressions (quotients of polynomials) is
always difficult. You must remember that these operations can only be accomplished
by first finding a common denominator, just as when trying to add or subtract rational
numbers. The explicit rules for this are in Table 1.8 (page 19). In contrast,
multiplication and division are relatively easy to perform. When in doubt, perform the
desired operations with rational numbers and compare. The rules for multiplying or
dividing are in Table 1.7 (page 19).
6. Sometimes an algebraic expression will contain several rational expressions, together
with some of the arithmetic operations . The process for dealing with such expressions
is the same as the process when rational numbers replace the rational expressions. In
example 10a (page 20), the problem is to find the quotient of two expressions. The only
way we can simplify a quotient is to get both the numerator and denominator in the
form of rational expressions and then use Table 1.7. In example 10b, we first need to
eliminate the negative exponents, and then proceed in a manner similar to 10a.
7. Cancellation between terms in the numerator and denominator can be very tricky. The
best way to determine whether cancellation can occur is by factoring both the
numerator and denominator before canceling. Cancellation can only occur if there are
common factors in both the numerator and denominator. See Example 8 on page 19 of
Practice Assignment
Section 1.2: Exercises 15, 19, 37, 41, 43, 49, 57, 63, 67, 73, 77, 81, 85, 87.
PART 3
By the end of this part of lesson 1 you should be able to
• compute the distance between two points;
• find the slope of a line ; and
• use the point-slope form to find the equation of a straight line.
Reading Assignment: Sections 1.3 and 1.4 of Chapter 1
1. Every point in the plane can be associated with a pair of real numbers and, conversely,
every pair of real numbers can be associated with a point in the plane. It is this
association—both ways—that allows algebra and geometry to flourish together: a
geometric concept can be given an algebraic interpretation, and an algebraic
expression can be associated with a geometric figure in the Cartesian plane. (The word
Cartesian is in honor of Rene Descartes, who with Pierre Fermat invented analytic
geometry.)
2. The Euclidean, or Cartesian, plane arises from having two intersecting number lines.
Their point of intersection is called the origin, and is denoted by O. Since the two lines
are number lines, each has a scale of units, although the scales are not necessarily the
same on both lines. Whenever we work with analytic geometry, for example in
drawing graphs , these scales must be clearly denoted. There are a few conventions
that have arisen concerning these number lines. For example, the number lines
intersect in a right angle; one horizontal—called the x-axis—and the other
vertical—called the y-axis. On the x-axis, positive reals are displayed to the right,
negatives to the left, while on the y-axis, positives go up and negatives down. Because
the x-axis and y-axis intersect at right angles to each other, we can use the Pythagorean
theorem in order to compute the distance between any two points. The reason for this
is that any vertical line and any horizontal line must also intersect in right angles.
Therefore, the line connecting the two points, together with a line parallel to the x-axis
and another parallel to the y-axis, form a right-angle triangle. (See Figure 1.7, page 25.)
3. Every line in the Euclidean plane, except vertical lines, has a slope. This slope is
computed by selecting two points on the line and using Equation 3 on page 33. Note
that it does not matter what two points you choose; the slope will always be the same.
There are two important slopes that you should remember. The slope of a vertical line
is undefined. Why? Horizontal lines have slopes equal to zero.
4. One of the two parts of calculus, differential calculus, deals with tangents to the graph
of a function. In the situations we will encounter, we will be given a point on the
graph and the slope of the tangent line. Thus, while there are several formulas for
determining the equation for a (tangent) line, the point-slope form is the most
important and you should use this form when doing the exercises.
5. Observe that in the distance formula, it is unimportant whether we use (a − b) or
(b − a) in computing the distance, but in computing the slope, confusing the
designation of the first versus the second point may result in an error in the sign of the
slope
. Thus, the slope of the line through the points (2, 4) and (−3, 6) is given by
6. Lines with positive slopes will rise from left to right, while lines with negative slopes
will fall from left to right. Parallel lines have equal slopes and two lines that are
perpendicular (intersect at right angles) will have slopes such that their product equals
(−1) (page 37).
7. In the slope intercept form y = mx + b, m is the slope and the point (0, b), is the
y-intercept—the point where the line intersects the y-axis. If the equation is given in
this form, or can be algebraically manipulated to get this form, then the slope of the
line will be given by the coefficient of x. However, it is crucial that the equation be
written so that the coefficient of y equals 1.
Practice Assignment
Section 1.3: Exercises 21, 23, 25, 27, 29, 35, 39.
Section 1.4: Exercises 13, 15, 17, 23, 27, 41, 51, 53, 69, 71.
Written Assigment 1
Section 1.1: Exercises 26, 56, 82, 90, 114, 116.
Section 1.2: Exercises 6, 10, 42, 58, 74, 84.
Section 1.3: Exercises 26, 28, 30, 34, 36, 38.
Section 1.4: Exercises 14, 16, 18, 42, 54, 70, 76.
Don’t forget:
• If an exercise asks you to graph a particular function, this graph should be included as
part of the solution to the exercise.
• Every graph must have a scale of units clearly marked on both axes.
• Be sure to show all necessary work, or explain completely the process used to find the |
# 5.4 - Inference for the Population Mean
5.4 - Inference for the Population Mean
## Overview
In this section, we discuss how to find confidence intervals for the population mean. The idea and interpretation of the confidence interval will be similar to that of the population proportion only applied to the population mean, $$\mu$$.
We start with the case where the population standard deviation, $$\sigma$$, is known. We continue to the more realistic case where $$\sigma$$ is not known. For the latter case, we need to recall the $$t$$-distribution. We end this section by presenting how to determine a sample size for a desired margin of error and confidence.
### Point Estimates for a Population Mean
The point estimate of the population mean, $$\mu$$ is:
Point Estimate of the Population Mean
$$\bar{x}=$$ sample mean
If one wants to know how accurate the sample mean is to estimate the population mean, we need some probability statement. We will want to know the sampling distribution of $$\bar{x}$$. From this distribution, we can get a confidence interval. Such an interval provides a range of values for which the parameter value is believed to fall. An interval is more likely to be "correct" than a point estimate.
# 5.4.1 - Construct and Interpret the CI
5.4.1 - Construct and Interpret the CI
## Constructing a Confidence Interval for the Population Mean
To construct a confidence interval for a population mean, we're going to apply the same three steps as with the population proportion, but first, let's look at the two possible cases.
## Case 1: $\sigma$ is known
In the previous lesson, we learned that if the population is normal with mean $$\mu$$ and standard deviation, $$\sigma$$, then the distribution of the sample mean will be Normal with mean $$\mu$$ and standard error $$\frac{\sigma}{\sqrt{n}}$$.
Following the similar idea to developing the confidence interval for $$p$$, the $$(1-\alpha)$$100% confidence interval for the population mean $$\mu$$ is...
$$P\left(\left|\dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\right|\le z_{\alpha/2}\right)=1-\alpha$$
A little bit of algebra will lead you to...
$$P\left(\bar{x}-z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\le \mu\le \bar{x}+z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\right)=1-\alpha$$
In other words, the $$(1-\alpha)$$100% confidence interval for $$\mu$$ is:
$$\bar{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$$
Notice for this case, the only condition we need is the population distribution to be normal.
#### Note!
The case where $$\sigma$$ is known is unrealistic. We explain it here briefly because it reinforces what we have previously learned. We do not present examples in this case.
## Case 2: $$\sigma$$ is unknown
When the population is normal or when the sample size is large then,
$$Z=\dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$$
where Z has a standard Normal distribution.
Usually, we don't know $$\sigma$$, so what can we do?
Recall that if X comes from a normal distribution with mean, $\mu$, and variance, $\sigma^2$, or if $n\ge 30$, then the sampling distribution will be approximately normal with mean $\mu$ and standard error, $$SE(\bar{X})=\frac{\sigma}{\sqrt{n}}$$
One way to estimate $$\sigma$$ is by $$s$$, the standard deviation of the sample, and replace $$\sigma$$ by $$s$$ in the above Z-equation. However, this new quotient no longer has a Z-distribution. Instead it has a t-distribution. We call the following a 'studentized' version of $$\bar{X}$$:
$$t=\dfrac{\bar{X}-\mu}{\dfrac{s}{\sqrt{n}}}$$
## Constructing the Confidence Interval
1. CHECK THE CONDITIONS
One of the following conditions need to be satisfied:
1. If the sample comes from a Normal distribution, then the sample mean will also be normal. In this case, $$\dfrac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$$ will follow a $$t$$-distribution with $$n-1$$ degrees of freedom.
2. If the sample does not come from a normal distribution but the sample size is large ($$n\ge 30$$), we can apply the Central Limit Theorem and state that $$\bar{X}$$ is approximately normal. Therefore, $$\dfrac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$$ will follow a $$t$$-distribution with $$n-1$$ degrees of freedom.
2. CONSTRUCT THE GENERAL FORM
$$(1-\alpha)$$100% Confidence Interval for the Population Mean, $$\mu$$
$$\bar{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$$
where the t-distribution has $$df = n - 1$$. This interval is also known as the one-sample t-interval for the population mean.
3. INTERPRET THE CONFIDENCE INTERVAL
We are $$(1-\alpha)100\%$$ confident that the population mean, $$\mu$$, is between $$\bar{x}-t_{\alpha/2}\frac{s}{\sqrt{n}}$$ and $$\bar{x}+t_{\alpha/2}\frac{s}{\sqrt{n}}$$.
## What if the conditions are not met?
What will you do if you cannot use the t-interval? What do we do when the above conditions are not satisfied?
1. If you do not know if the distribution comes from a normally distributed population and the sample size is small (i.e $$n<30$$), you can use the Normal Probability Plot to check if the data come from a normal distribution.
2. You may want to consider what is known as nonparametric statistical methods. A procedure such as the one-sample Wilcoxon procedure. Lesson 11 introduces nonparametric statistical methods.
# 5.4.2 - The t-distribution
5.4.2 - The t-distribution
In 1908, William Sealy Gosset from Guinness Breweries discovered the t-distribution. His pen-name was Student and thus it is called the "Student's t-distribution."
The t-distribution is different for different sample size, n. Thus, tables, as detailed as the standard normal table, are not provided in the usual statistics books. The graph below shows the t-distribution for degrees of freedom of 10 (blue) and 30 (red dashed).
### Properties of the t-distribution
1. t is symmetric about 0
2. t-distribution is more variable than the Standard Normal distribution
3. t-distributions are different for different degrees of freedom (d.f.).
4. The larger $n$ gets (or as $n$ goes to infinity), the closer the $t$-distribution is to the $z$.
5. The meaning of $t_\alpha$ is the $t$-value having the area "$\alpha$" to the right of it.
## Example 5-5: Finding t-values
Use this t-table or the one in your text to find following the example.
Find $$t_{0.05}$$ where the degree of freedom is 20.
In a t-distribution table below the top row represents the upper tail area, while the first column are the degrees of freedom.
The $$t_{0.05}$$ where the degree of freedom is 20 is 1.725 .
df 0.40 0.25 0.10 0.05 0.025 0.01 0.005 0.001 .0005
... ... ... ... ... ... ... ... ... ...
18 0.257 0.688 1.330 1.734 2.101 2.552 2.878 3.610 3.922
19 0.257 0.688 1.328 1.729 2.093 2.539 2.861 3.579 3.883
20 0.257 0.687 1.325 1.725 2.086 2.528 2.845 3.552 3.850
21 0.257 0.686 1.323 1.721 2.080 2.518 2.831 3.527 3.819
The graph shows that the $$\alpha$$ values at the top of this table are the upper tail areas of the distribution.
Note! When the corresponding degree of freedom is not given in the table, you can use the value for the closest degree of freedom that is smaller than the given one. We use this approach since it is better to err in a conservative manner (get a t-value that is slightly larger than the precise t-value).
Find $$t_{0.05}$$ where the degree of freedom is 34.
What do we do when the degrees of freedom are not on the table? The t-table degrees of freedom run continuously from 1 to 30, then go by intervals after 30 (e.g. after 30 we have 35). In such cases, we can use software such as Minitab to find a more exact value for the multiplier as opposed to using a degrees of freedom that is "close".
To find the t-value in Minitab...
1. From the Minitab Menu select Calc > Probability Distributions > t...
2. Choose inverse cumulative probability
3. Enter the degrees of freedom
4. Set the input constant as 0.95 (1 - 0.05).
5. Choose OK
The output from Minitab gives us $$t_{0.05}$$ with df= 34 as 1.69092.
P (X $$\le$$ x) x
0.95 1.69092
Find $$t_{0.05}$$ where the degree of freedom is 30.
The t-value for an $$\alpha$$ of .05 and df of 30 is 1.697.
df 0.40 0.25 0.10 0.05 0.025 0.01 0.005 0.001 .0005
... ... ... ... ... ... ... ... ... ...
27 0.256 0.684 1.314 1.703 2.052 2.473 2.771 3.421 3.690
28 0.256 0.683 1.313 1.701 2.048 2.467 2.763 3.408 3.674
29 0.256 0.683 1.311 1.699 2.045 2.462 2.756 3.396 3.659
30 0.256 0.683 1.310 1.697 2.042 2.457 2.750 3.385 3.646
Note! When the sample size is larger than 30, the t-values are not that different from the z-values. Thus, a crude estimate for $$t_{0.05}$$ with 34 degrees of freedom is $$z_{0.05} = 1.645$$. Although it is a crude estimate, when software is available, it is best to find the $t$ values rather than use the $z$.
# 5.4.3 - Example
5.4.3 - Example
## Example 5-6: Emergency Room Wait Time
You are interested in the average emergency room (ER) wait time at your local hospital. You take a random sample of 50 patients who visit the ER over the past week. From this sample, the mean wait time was 30 minutes and the standard deviation was 20 minutes. Find a 95% confidence interval for the average ER wait time for the hospital.
Is the population data normal? We don't know. However, the sample size is 50 which exceeds our minimum requirement of 30 in order to use the t-interval.
The population standard deviation is unknown; we only know the sample standard deviation. Having satisfied the conditions we proceed by finding the proper multiplier from the t-table. With n of 50 the d.f. are 49, and for 95% confidence, the alpha value is 5% or 0.05. From the t-table under the column 0.025 (remember we use α/2) and a d.f. of 40 (since 49 is not on the table), we arrive at a t-value of 2.021 Completing our confidence interval formula,\begin{align} &=\bar{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}\\ &=30\pm 2.021\dfrac{20}{\sqrt{50}}\\ &=30\pm 5.72\\ &=(24.28, 35.72) \end{align}
Note that, $$t_{0.025,49}\approx z_{0.025}$$ as the degrees of freedom is 49
We are 95% confident that mean emergency room wait time at our local hospital is from 24.28 minutes to 35.72 minutes.
Find the CI for a population mean in Minitab:
1. In Minitab choose Stat > Basic Statistics > 1-Sample t .
2. From the drop down box select the Summarized data option button. (If you have the raw data you would use the default drop down of One or more samples, each in a column.)
3. Enter the sample size, sample mean, and sample standard deviation in their respective text boxes.
4. Click the Options button. The default confidence level is 95. If your desire another confidence level edit appropriately.
5. Click OK and OK again.
#### Using Minitab: Emergency Room Wait Time Example
Referring to our prior example of average emergency room wait time from our discussion on confidence intervals for a population mean, our by-hand calculations produced a 95% confidence interval of 24.28 to 35.72 minutes. Recall the following for that example: sample size 50, sample mean 30, and sample standard deviation 20.
SampleText
In Minitab following the above steps, we get a 95% confidence interval:
N Mean StDev SE Mean 95% CI
50 30.00 20.00 2.84 (24.32, 35.68)
The slight discrepancy between the estimates is due to our by-hand calculation using the t-value associated with 40 degrees of freedom since the table did not include a d.f. of 49. Minitab used a t-value for the actually 49 degrees of freedom. With the larger degrees of freedom comes a smaller t-value. This would result in a smaller margin of error and a narrower interval - precisely what we have here.
The mean length of certain construction lumber is supposed to be 8.5 feet. A random sample of 81 pieces of such lumber gives a sample mean of 8.3 feet and a sample standard deviation of 1.2 feet.
What is the 95% CI for "mean length of such lumber?"
• Step 1: Check the conditions: The sample size is large ($n\ge 30$), so we may continue using the value from the t-distribution as our multiplier.
• Step 2: Construct the CI: The degrees of freedom are $n-1=80$. If we use the table, with d.f of 60, $t_{0.025}=2$.
The 95% confidence interval is \begin{align} &=\bar{x}\pm t_{0.025}\dfrac{s}{\sqrt{n}}\\ &=8.3\pm 2\dfrac{1.2}{\sqrt{81}}\\ &=8.3\pm 0.2667\\ &=(8.0333, 8.5667) \end{align}
What is the 99% CI for "mean length of such lumber?"
• Step 1: Check the conditions: The sample size is large ($n\ge 30$), so we may continue using the value from the t-distribution as our multiplier.
• Step 2: Construct the CI: The degrees of freedom are $n-1=80$. If we use the table, with d.f of 60, $t_{0.005}=2.66$. The 99% confidence interval is \begin{align} &=\bar{x}\pm t_{0.005}\dfrac{s}{\sqrt{n}}\\ &=8.3\pm 2.66\dfrac{1.2}{\sqrt{81}}\\ &=8.3\pm 0.3547\\ &=(7.9453, 8.6547) \end{align}
Reflecting back on interpretation of a proportion interval, we see the same basic structure: level of confidence, parameter of interest, lower and upper bounds.
# 5.4.4 - Checking Normality
5.4.4 - Checking Normality
## Using Normal Probability Plot to Check Normality
If the sample size is less than 30, one needs to use a Normal Probability Plot to check whether the assumption that the data come from a normal distribution is valid.
Normal Probability Plot
The Normal Probability Plot is a graph that allows us to assess whether or not the data comes from a normal distribution.
Note! This plot should be used as a guide for us to assess if the assumption that the data come from a normal distribution is valid or not. It should not be used to “test” an assumption.
## Example 5-7: Rattlesnake Lengths
It is very time consuming to find rattlesnakes and nerve racking to measure them (for obvious reasons). A scientist randomly finds 12 snakes from the central Pennsylvania area and measures their length. The following twelve measurements in inches are obtained:
40.2, 43.1, 45.5, 44.5, 39.5, 38.5, 40.2, 41.0, 41.6, 43.1, 44.9, 42.8
Using the above data, find a 90% confidence interval for the mean length of rattlesnakes in the central Pennsylvania area.
#### Step 1 Check Conditions
Think about what conditions you need to check. The sample size is only 12. The scenario does not give us an indication that the lengths follow a normal distribution. Therefore, let's do a normal probability plot to check whether the assumption that the data come from a normal distribution is valid.
##### Minitab: Creating a normal probability plot
To create a normal probability plot in Minitab:
1. Enter the 12 measurements into one column (name it length for this example) or upload the snakes.txt file.
2. Type or upload the data in the first column in Minitab.
3. Choose Graph > Probability Plot
Here is the normal probability plot for the rattlesnake data. What do you conclude about whether they may come from a normal distribution?
Since the points all fall within the confidence limits, it is reasonable to suggest that the data come from a normal distribution.
#### Step 2 Construct the CI
Now, we can proceed to find the 90% t-interval for the mean length of rattlesnakes in the central Pennsylvania area since even though the sample size is less than 30, the normality plot shows that the data may come from a normal distribution.
##### Minitab: Find the t-interval using Minitab
1. Enter the 12 measurements into one column (name it length for this example)
2. Choose Stat > Basic Statistics > 1-Sample t
3. Click on the variable (length for this example) and change to the desired confidence level
The Minitab output will provide the confidence interval. We get the following:
N Mean StDev SE Mean 90% CI
12 42.075 2.257 0.652 (40.905, 43.245)
View the video to see these steps within Minitab.
#### Step 3 Interpret the Interval
We are 90% confident that the population mean lengths of rattlesnakes is between 40.905 and 43.245 inches.
# 5.4.5 - Sample Size Computation
5.4.5 - Sample Size Computation
## Sample Size Computation for the Population Mean Confidence Interval
Recall that a $$(1-\alpha)$$100% confidence interval for $$\mu$$ is $$\bar{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$$ where the multiplier $$t$$ has a t-distribution with $$df = n - 1$$. Thus, the margin of error, E, is equal to:
$$E=t_{\alpha/2}\dfrac{s}{\sqrt{n}}$$
To determine the sample size, one first decides the confidence level and the half width of the interval one wants. Then we can find the sample size to yield an interval with that confidence level and with a half width not more than the specified one. The crude method to find the sample size: $$n=\left(\dfrac{z_{\alpha/2}\sigma}{E}\right)^2$$ Then round up to the next whole integer.
## Example 5-8: Spring Break
A marketing research firm wants to estimate the average amount a student spends during the Spring break. They want to determine it to within \$120 with 90% confidence. One can roughly say that it ranges from \$100 to \1700. How many students should they sample? #### Answer To use the formula, we need all the pieces for $$n=\left(\dfrac{z_{\alpha/2}\sigma}{E}\right)^2$$. We know that $$z_{\alpha/2}=1.645$$ (for 90%). The margin of error, $$E$$, is 120. The only piece missing is $$\sigma$$. Since the standard deviation is not given in the problem, we can estimate it using $$\dfrac{\text{range}}{4}$$ from Lesson 1. Therefore, $$\sigma=\dfrac{1700-100}{4}=400$$. So we have... \begin{align} n &=\left(\dfrac{1.645(400)}{120}\right)^2\\ &=30.07 \end{align} Therefore, a sample of size $$n=31$$ is required. Note! In homework and exams, it is fine if you simply use the cruder method. A more accurate method is provided in the following for your reference only. ### The Iterative Method A more accurate method to estimate the sample size: iteratively evaluate the formula since the t value also depends on n. $$n=\left(\dfrac{t_{\alpha/2}s}{E}\right)^2$$ Use the example above for illustration. Start with an initial guess forn\$, plug in the formula, and iteratively solve for $$n$$.
If the initial guess for $$n$$ is 20, $$t_{0.05} = 1.729$$ and degrees of freedom = 19,
$$n=\left(\dfrac{t_{\alpha/2}s}{E}\right)^2=n=\left(\dfrac{1.729(400)}{120}\right)^2=33.21$$
For $$n = 34$$, degree of freedom = 33, and $$t_{0.05} = 1.697$$
$$n=\left(\dfrac{t_{\alpha/2}s}{E}\right)^2=n=\left(\dfrac{1.697(400)}{120}\right)^2=31.99$$
If we use $$n = 32$$, the result is the same. Thus, the more accurate answer to the example is to sample 32 students.
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility |
Dimension of a Vector Space
# Dimension of a Vector Space
Before we precisely define what the dimension of a vector space is, we will first look at a very important theorem regarding bases that will give intuition to the subsequent definition.
Theorem 1: Let $V$ be a finite dimensional vector space. If $B_1$ and $B_2$ are bases of $V$, then the size of $B_1$ and $B_2$ are equal, that is $\rvert B_1 \rvert = \rvert B_2 \rvert$.
• Proof: Let $V$ be a finite dimensional vector space, and let $B_1$ and $B_2$ be bases of $V$. To prove this we will use the fact that a basis by definition is both a linearly independent set of vectors from $V$ and a spanning set of vectors from $V$.
• Since $B_1$ is a basis, $B_1$ by definition is also a linearly independent set of vectors. Also, since $B_2$ is a basis, $B_2$ by definition is a spanning set of $V$. By the Finite Dimensional Linearly Independent Set of Vectors Theorem the size of the linearly independent set of vectors $B_1$ must be less than or equal to the spanning set of vectors $B_2$, and so $\rvert B_1 \rvert ≤ \rvert B_2 \rvert$.
• Similarly, since $B_2$ is a basis, $B_2$ by definition is also a linearly independent set of vectors. Also, since $B_1$ is a basis, $B_1$ by definition is a spanning set of $V$. Once again by the Finite Dimensional Linearly Independent Set of Vectors Theorem, the size of the linearly independent set of vectors $B_2$ must be less than or equal to the spanning set of vectors $B_1$ and so $\rvert B_2 \rvert ≤ \rvert B_1 \rvert$.
• Since $\rvert B_1 \rvert ≤ \rvert B_2 \rvert$ and $\rvert B_2 \rvert ≤ \rvert B_1 \rvert$, it follows that $\rvert B_1 \rvert = \rvert B_2 \rvert$, and so both bases $B_1$ and $B_2$ have the same size. $\blacksquare$
We will now formally define the dimension of a vector space.
Definition: Let $V$ be a finite dimensional vector space over the field $\mathbb{F}$. The Dimension of $V$ denoted $\dim_{\mathbb{F}} V$ is the number of vectors in any basis of $V$. if $V$ is an infinite dimensional vector space over $\mathbb{F}$ then we write $\dim_{\mathbb{F}} V = \infty$.
We note from the theorem above, the dimension of a vector space is unique since the size of any two bases is of $V$ is always the same. Let's now look at some examples.
Consider the vector space $\mathbb{R}^n$ over the field of real numbers. We note that any basis of $\mathbb{R}^n$ will have $n$ vectors since the standard basis vectors form a basis, that is if $e_1 = (1, 0, 0, ..., 0)$, $e_2 = (0, 1, 0, ..., 0)$, …, $e_n = (0, 0, ..., 0, 1)$ then $V = \mathrm{span} (e_1, e_2, ..., e_n)$ and $\{ e_1, e_2, ..., e_n \}$ is linearly independent. Therefore $\dim_{\mathbb{R}} \mathbb{R}^n = n$.
For another example, consider the set of all polynomials of degree less than or equal to $n$, $\wp_n (\mathbb{F})$ over the real numbers. This is a finite dimensional vector space, and we know one such basis of this vector space to be $\{ 1, x, x^2, ..., x^n \}$ which contains $n + 1$ elements, so $\dim_{\mathbb{R}} \wp_n(\mathbb{F}) = n + 1$.
Yet another example is the vector space of complex numbers $\mathbb{C}$ over the set of real numbers. Once such basis for $\mathbb{C}$ is the set $\{ 1, i \}$ which spans $\mathbb{C}$ and is linearly independent. So $\dim_{\mathbb{R}} \mathbb{C} = 2$, which we should be familiar with since we can graph a complex number along a $2$-dimensional real-axis and imaginary-axis.
However, it is important to note that the underlying field is important when determining the dimension of a vector space. The vector space of complex numbers over the set of complex numbers has dimension $1$, that is $\dim_{\mathbb{C}} \mathbb{C} = 1$ as you should verify. |
# The Complex Plane
In the first post I explained why the mathematical “imaginary” number i is “real” (in more than one sense of the word). That weird number is just a stepping stone to the complex numbers, which are themselves stepping stones to the complex plane.
Which, in turn, is a big stepping stone to a fun fact about the Mandelbrot I want to write about. (But we all have to get there, first.) I think it’s a worthwhile journey — understanding the complex plane opens the door to more than just the Mandelbrot. (For instance, Euler’s beautiful “sonnet” also lives on the complex plane.)
As it turns out, the complex numbers cause this plane to “fly” a little bit differently than the regular X-Y plane does.
The two do have a lot in common, but the magic of i changes the math we do on points. (Mainly the multiplication of complex numbers.)
Recall that a complex number has the form:
a + bi
Where a and b are real numbers and i is the imaginary unit.
Since the two terms aren’t compatible (a plain number and one with an imaginary unit), we can’t carry out the addition. The two parts are forever separate. This makes complex numbers compound numbers.
Note that, if b=0, then the number is essentially the real number a. This is how the complex numbers include the real numbers (and therefore the integers and rationals). They are all values of a with b=0.
§
That the real and imaginary parts a and bi are incompatible means they are independent of each other, which means we can think of them as axes of a graph:
That’s all the complex plane is, an X-Y graph in which we use a complex number’s real component, a, as x and its imaginary component, bi, as y. This means the familiar real number line is the x-axis of the graph (where y is zero).
In the graph above, the blue points and numbers on the horizontal x-axis call out the integer points on the real number line. The red points and numbers on the vertical y-axis call out the integer multiples of i. The purple points and numbers show some combined real and imaginary values.
The “unit square” at the center separates an important behavior. The value 1.0 tends to be hugely significant in math — it’s the foundation of the integers and, more significantly here, the multiplicative identity. Points inside the square, when multiplied, tend to get closer to the origin (0,0i), and points outside, when multiplied, tend to get further from it.
The same basic behavior occurs with real numbers. Multiplying two numbers, both above 1.0, results in a number larger than either. Multiplying two numbers, both less than 1.0, results in a number smaller than either.
On a graph, smaller usually means closer to the center, and larger usually means further away. A multi-dimension graph provides the notation of magnitudethe distance from the center.
Note that when drawing the graph we ignore i and use only the a and b values. (We don’t ignore i when doing math on graph points. That’s where the magic kicks in.)
§ §
Let’s take a look at that math magic.
I’ll start by reversing the order of the real and imaginary components and swapping the letters:
ai + b
This doesn’t change anything. It’s still the same (complex) value. I want the imaginary part first, but I still want the letters to read from left to right just because that’s what we’re used to.
(Remember that what we use for ‘a’ and ‘b’ doesn’t matter. The letters are just placeholders for real numbers. We can use any letters we like.)
The reason for all the swapping is to highlight how the equation looks a lot like a low-order polynomial:
ax + b (= ax1 + bx0)
Which suggests that we treat complex numbers as polynomials when it comes to doing math with them. This is handy, because we already know how to do math that way — it’s basic algebra.
§
Adding the two complex numbers…
ai+b
ci+d
…(where a, b, c & d are all real numbers) is simple: We add the like parts (the two real parts and the two imaginary parts) and end up with a new complex number:
(ai+b)+(ci+d) = (ai+ci)+(b+d) = (a+c)i+(b+d)
The real numbers, a+c and b+d, add together while the i stands outside. Once we do those two additions, we end up with a complex number again.
Multiplying those two complex numbers is where it gets a little more interesting. We’re multiplying polynomials, which uses the inside-outside-left-right protocol from high school algebra class.
It ultimately boils down to this:
(ai+b) × (ci+d) = (bc+ad)i + (ac)i2+bd
And since i2 = -1, we can reduce it to:
Again, once we do the math on the real numbers (a, b, c & d), we end up with a new complex number.
If this section made your eyes glaze, that’s okay. The only important part is this:
(ai+b) × (ci+d) = (bc+ad)i+(bdac)
Multiplying two complex numbers together results in a certain specific result (which is a new complex number). As it turns out, this multiplication has certain interesting properties I’ll come back to.
§
Compare this with how we might think to multiply two ordinary X-Y coordinates.
In this case, both x and y are just real numbers, so there’s no need to treat them algebraically. We can just do a pair-wise multiplication:
[x, y] × [u, v] = [xu, yv]
Which is entirely legit. But it obviously delivers different results than we get with complex number multiplication.
To make the comparison apples-to-apples, here is the complex number — treated as a graph coordinate — multiplication result:
[x, yi] × [u, vi] = [(xuyv), (xv+yu)i]
Note that we’re back to the original (a,bi) form here. The real part (the x-axis value) comes first, and the imaginary part (the y-axis value) comes second. That’s to make it match the [x, y] form of coordinates.
Remember that the i in the equation just helps distinguish the real and imaginary parts and reminds us we’re working with complex numbers.
This different math matters because of what happens when we multiply points on the complex plane. A simple illustration is shown to the left.
The red and blue points are the two points we’re going to multiply. The red point is at [+3.2, +1.2i] and the blue point is at [+2.5, +1.9i].
The black point shows what we get if we multiply them simply as ordinary [x, y] points:
[3.2,1.2] × [2.5, 1.9] = [3.2×2.5, 1.2×1.9] = [8, 2.28]
The purple point shows the complex multiplication result:
[(3.2×2.5)-(1.2×1.9), ((3.2×1.9)+(1.2×2.5))i] = [5.72, 9.08i]
So the points end up in different places, but something a bit more important also happens.
§
If we look at the lines (vectors) drawn from origin to the points, those lines all have an angle with respect to the x-axis.
What might not be obvious (but is true) is that the angle of the purple line is the sum of the angles of the blue and red lines.
We can say either that the purple point is the result of rotating the blue point by the red point’s angle, or the result of rotating the red point by the blue point’s angle. It amounts to the same thing.
The crucial point is that multiplying complex points results in rotation.
Both the purple and black points are further away from the center because both the red and blue points are outside the unit square (the upper-right quadrant of which is the single square in the lower left).
§ §
This rotation aspect gives us a new way to find points on the unit circle.
We start with an easy point, the one at [+1, 0i]. It sits right on the x-axis, so it’s angle is zero. It’s the first point of the unit circle.
Then we decide how many points around the circle we want. The easiest approach is to pick a small angle and find a point for each increment of that angle around the circle. For instance, if we pick an angle of 1° there will be 360 points around the circle.
Now we find the point for that increment angle. That’s simply:
[cos(θ) , sin(θ)] = [cos(), sin()] = [+0.9998, +0.0175i]
We could have gotten the first point this way, too:
[cos(), sin()] = [+1.00, 0.00i]
But since it’s a “well-known” point, we didn’t bother.
Both of these are points on the unit circle, so by definition their magnitude is 1.0. Therefore, multiplying them will keep that magnitude constant.
If we multiply the first point (call it U for Unit circle) by the second point (R for Rotate), we end up with a new U, which — on this first iteration — will be the same as R.
[Recall how this works: We are either rotating U by R, which rotates it one degree, or we’re rotating R by U, which rotates it zero degrees. Either way, we end up with a point one-degree along the circle, which is where R is.]
Now we multiply U by R again which rotates it one more degree. We have a new U that is two degrees along the circle.
We continue to multiply U by R to give us 360 evenly spaced points along the unit circle.
If we want more points, we pick a smaller angle. A larger angle gives us fewer points. Below is an example that uses as the angle, which results in 120 points around the circle (360/3 = 120).
Points every 3° on the unit circle.
Included in the diagram is a line (in red) to the R(otation) point illustrating the 3° angle. The original U point is the one directly below it.
As you can see, each point is spaced by that angle. Each time we multiply a given point, the new point is rotated 3° counterclockwise.
§ §
That’s enough (possibly, for some, more than) for this time.
But now we finally have the groundwork to explore the heart of the Mandelbrot.
Stay on the plane, my friends!
## About Wyrd Smythe
The canonical fool on the hill watching the sunset and the rotation of the planet and thinking what he imagines are large thoughts. View all posts by Wyrd Smythe
#### 4 responses to “The Complex Plane”
• Wyrd Smythe
You know, for a “simple” thing I wanted to show you, this is taking a lot of explanation. Definitely feeling like I’ve bitten off more than I really want to chew right now.
And it’s the feeling that no one other than me even has any interest, so why exactly am I do this at all? [sigh]
• Wyrd Smythe
Simple: Because if it helps one person somewhere someday figure out one thing they wanted to understand, that’s a Good Thing.
I can’t begin to count the number of times some article, post, or comment, gave me that last little clue I needed to figure something out. I’m just hoping to return the favor.
• Wyrd Smythe
Here’s a great video explaining exactly why the complex plane is such a natural extension of the real number line.
• Sideband #73: Complex 4D | Logos con carne
[…] [For more, see: “Imaginary” Numbers and The Complex Plane.] […] |
DECIMALS – THE ART OF TENTHS
Introduction
The concepts of mathematics involve the division of two individuals who do not get divided completely. So, what was the strategy to solve this issue and come up with a solution for the same? This era of 1500-1650 AD was the solution to the question and gave rise to the element in the number system. The study to give the value of half and half of half is the finding to get the notation.
A step towards the decimals
Fraction, the word derived from the Latin word Fractio, meaning “break” was used from around 1800 BCE in Egypt to express parts of a whole. At first, the numbers were limited to fractions that are in the form of numerator and denominator where the top number in the fraction is the numerator and down is the denominator. The ancient Egyptians had symbols for 2 ⁄3 and 3 ⁄4 but other fractions were expressed as the sum of unit fractions, for example as 1 ⁄3 + 1 ⁄13 + 1 ⁄17. This system worked well for recording amounts but not for doing calculations.
The importance of tenth
Simon Stevin, a Flemish engineer, and mathematician in the late 16th and early 17th century, used many calculations in his work. He simplified the concept by using fractions with a base system of tenth powers. Stevin already predicted that a decimal system would eventually be universally accepted. In the old era of Rome, fractions were based on a system of twelfths, and written in words: 1 ⁄12 was called an uncia, 6 ⁄12 was semis, and 1 ⁄24 was semiuncia, but this system made it difficult for people to do any calculations.
Introducing decimals
Finding something that was conventional factors for both time-consuming and prone to errors, the decimal system began to use. The idea of decimal fraction as the denominator as 10 where had been used five centuries before Stevin, who made decimals for calculating and recording parts as a whole. In Stevin’s new notation, numbers that would previously have been written as the sum of fractions—for example, 32 + 5 ⁄10 + 6 ⁄100 + 7 ⁄1,000—could now be written as a single number. Stevin placed circles after each number; these were shorthand for the denominator of the original decimal fraction. The whole 32 would be followed by a 0, because 32 is an integer, whereas the 6 ⁄100, for example, was expressed as 6 and a 2 inside a circle. These 2 denoted the power of 10 of the original denominators, as 100 is 102. In the same vein, the 7 ⁄1,000 became a 7 followed by a 3 inside a circle which sums up to 32.567.
Conclusion
The decimals are denoted by the point which indicates the value of fractions and makes it easy to calculate. the numbers denoting the value of fraction which doesn’t divide completely. This gave a rise to the main invention in the times of this era. This got universally approved and mathematicians all over the world acknowledged this invention. |
# How do you simplify sqrt(500/720)?
Sep 24, 2015
=color(blue)(5/6
#### Explanation:
We first simplify both terms individually by prime factorisation:
$\sqrt{500} = \sqrt{5 \cdot 5 \cdot 5 \cdot 2 \cdot 2} = \sqrt{{5}^{2} \cdot 5 \cdot {2}^{2}} = 5 \cdot 2 \sqrt{5} = 10 \sqrt{5}$
$\sqrt{720} = \sqrt{{3}^{2} \cdot {2}^{2} \cdot {2}^{2} \cdot 5} = 3 \cdot 2 \cdot 2 \sqrt{5} = 12 \sqrt{5}$
Now our expression becomes:
$\frac{10 \sqrt{5}}{12 \sqrt{5}} = \frac{10 \cancel{\sqrt{5}}}{12 \cancel{\sqrt{5}}}$
$= \frac{10}{12}$
=color(blue)(5/6 |
# What is the length of the ladder if a ladder of length L is carried horizontally around a corner from a hall 3 feet wide into a hall 4 feet wide?
Mar 1, 2015
Consider a line segment running from $\left(x , 0\right)$ to $\left(0 , y\right)$ through the interior corner at $\left(4 , 3\right)$.
The minimum length of this line segment will be the maximum length of ladder that can be maneuvered around this corner.
Suppose that $x$ is beyond $\left(4 , 0\right)$ by some scaling factor, $s$, of 4, so
$x = 4 + 4 s = 4 \left(1 + s\right)$
[watch for the $\left(1 + s\right)$ showing up later as a value to be factored out of something.]
By similar triangles we can see that
$y = 3 \left(1 + \frac{1}{s}\right)$
By the Pythagorean Theorem, we can express the square of the length of the line segment as a function of $s$
${L}^{2} \left(s\right) = {3}^{2} \left({s}^{- 2} + 2 {s}^{- 1} + 1\right) + {4}^{2} \left(1 + 2 s + {s}^{2}\right)$
Normally we would take the derivative of L(s) to find the minimum but in this case it is easier to take the derivative of ${L}^{2} \left(s\right)$.
(Note that if $L \left(s\right)$ is a minimum as $s = {s}_{0}$, then ${L}^{2} \left(s\right)$ will also be a minimum at $s = {s}_{0}$.)
Taking the first derivative of ${L}^{2} \left(s\right)$ and setting it to zero, we get:
${3}^{2} \left(- 2 {s}^{- 3} - 2 {s}^{- 2}\right) + {4}^{2} \left(2 - 2 s\right) = 0$
Multiplying by ${s}^{3}$ and then factoring out $2 \left(1 + s\right)$
allows us to solve for $s$
$s = {\left(\frac{3}{4}\right)}^{\frac{2}{3}}$
Plugging this value back into the equation for ${L}^{2} \left(s\right)$ and taking the square root (I used a spreadsheet), we get
the maximum ladder length $= 9.87 f e e t$ (approx.) |
In mathematics, matrix addition is the operation of adding two matrices by adding the corresponding entries together. However, there are other operations which could also be considered as a kind of addition for matrices, the direct sum and the Kronecker sum.
## Entrywise sum
Two matrices must have an equal number of rows and columns to be added.[1] The sum of two matrices A and B will be a matrix which has the same number of rows and columns as do A and B. The sum of A and B, denoted A + B, is computed by adding corresponding elements of A and B:[2][3]
{\displaystyle {\begin{aligned}\mathbf {A} +\mathbf {B} &={\begin{bmatrix}a_{11}&a_{12}&\cdots &a_{1n}\\a_{21}&a_{22}&\cdots &a_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{m1}&a_{m2}&\cdots &a_{mn}\\\end{bmatrix}}+{\begin{bmatrix}b_{11}&b_{12}&\cdots &b_{1n}\\b_{21}&b_{22}&\cdots &b_{2n}\\\vdots &\vdots &\ddots &\vdots \\b_{m1}&b_{m2}&\cdots &b_{mn}\\\end{bmatrix}}\\&={\begin{bmatrix}a_{11}+b_{11}&a_{12}+b_{12}&\cdots &a_{1n}+b_{1n}\\a_{21}+b_{21}&a_{22}+b_{22}&\cdots &a_{2n}+b_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{m1}+b_{m1}&a_{m2}+b_{m2}&\cdots &a_{mn}+b_{mn}\\\end{bmatrix}}\\\end{aligned}}\,\!}
For example:
${\displaystyle {\begin{bmatrix}1&3\\1&0\\1&2\end{bmatrix}}+{\begin{bmatrix}0&0\\7&5\\2&1\end{bmatrix}}={\begin{bmatrix}1+0&3+0\\1+7&0+5\\1+2&2+1\end{bmatrix}}={\begin{bmatrix}1&3\\8&5\\3&3\end{bmatrix}}}$
We can also subtract one matrix from another, as long as they have the same dimensions. A B is computed by subtracting elements of B from corresponding elements of A, and has the same dimensions as A and B. For example:
${\displaystyle {\begin{bmatrix}1&3\\1&0\\1&2\end{bmatrix}}-{\begin{bmatrix}0&0\\7&5\\2&1\end{bmatrix}}={\begin{bmatrix}1-0&3-0\\1-7&0-5\\1-2&2-1\end{bmatrix}}={\begin{bmatrix}1&3\\-6&-5\\-1&1\end{bmatrix}}}$
## Direct sum
Another operation, which is used less often, is the direct sum (denoted by ⊕). Note the Kronecker sum is also denoted ⊕; the context should make the usage clear. The direct sum of any pair of matrices A of size m × n and B of size p × q is a matrix of size (m + p) × (n + q) defined as [4][2]
${\displaystyle \mathbf {A} \oplus \mathbf {B} ={\begin{bmatrix}\mathbf {A} &{\boldsymbol {0}}\\{\boldsymbol {0}}&\mathbf {B} \end{bmatrix}}={\begin{bmatrix}a_{11}&\cdots &a_{1n}&0&\cdots &0\\\vdots &\ddots &\vdots &\vdots &\ddots &\vdots \\a_{m1}&\cdots &a_{mn}&0&\cdots &0\\0&\cdots &0&b_{11}&\cdots &b_{1q}\\\vdots &\ddots &\vdots &\vdots &\ddots &\vdots \\0&\cdots &0&b_{p1}&\cdots &b_{pq}\end{bmatrix}}}$
For instance,
${\displaystyle {\begin{bmatrix}1&3&2\\2&3&1\end{bmatrix}}\oplus {\begin{bmatrix}1&6\\0&1\end{bmatrix}}={\begin{bmatrix}1&3&2&0&0\\2&3&1&0&0\\0&0&0&1&6\\0&0&0&0&1\end{bmatrix}}}$
The direct sum of matrices is a special type of block matrix, in particular the direct sum of square matrices is a block diagonal matrix.
The adjacency matrix of the union of disjoint graphs or multigraphs is the direct sum of their adjacency matrices. Any element in the direct sum of two vector spaces of matrices can be represented as a direct sum of two matrices.
In general, the direct sum of n matrices is:[2]
${\displaystyle \bigoplus _{i=1}^{n}\mathbf {A} _{i}=\operatorname {diag} (\mathbf {A} _{1},\mathbf {A} _{2},\mathbf {A} _{3},\ldots ,\mathbf {A} _{n})={\begin{bmatrix}\mathbf {A} _{1}&{\boldsymbol {0}}&\cdots &{\boldsymbol {0}}\\{\boldsymbol {0}}&\mathbf {A} _{2}&\cdots &{\boldsymbol {0}}\\\vdots &\vdots &\ddots &\vdots \\{\boldsymbol {0}}&{\boldsymbol {0}}&\cdots &\mathbf {A} _{n}\\\end{bmatrix}}\,\!}$
where the zeros are actually blocks of zeros; i.e., zero matrices.
## Kronecker sum
The Kronecker sum is different from the direct sum but is also denoted by ⊕. It is defined using the Kronecker product ⊗ and normal matrix addition. If A is n-by-n, B is m-by-m and ${\displaystyle \mathbf {I} _{k}}$ denotes the k-by-k identity matrix then the Kronecker sum is defined by:
${\displaystyle \mathbf {A} \oplus \mathbf {B} =\mathbf {A} \otimes \mathbf {I} _{m}+\mathbf {I} _{n}\otimes \mathbf {B} .}$ |
Question
# To save money for a video game, Ajay puts 1 rupee in an envelope. Each day for a total of 8 days he doubles the number of rupee from the day before. How much will he save on the eighth day?
Hint: In the question, it is given that on the first day Ajay puts 1 rupee and for the next coming days he doubles the amount of money everyday. So, start with 1 rupee and then double the amount of money everyday that he puts into the envelope and then add them all for eight days to get money collected on the 8th day.
In the question, it is given that:
Amount of money that Ajay put on first day = Re.1
And it is given in the question that everyday he doubles the money that he puts in the envelope. So, to calculate the amount of money that he puts on the next day we just multiply the previous day amount by 2.
So, the amount of money that he puts each of days is given as follow:
Amount of money that he puts on the first day $= {\text{Re}}{\text{.1}}$.
${\text{Amount of money that he puts on second day = 2}} \times {\text{amount of money that he puts on first day}} \\ {\text{ = 2}} \times 1{\text{Rupee = Rs}}{\text{.2}} \\ {\text{Amount of money that he puts on third day = 2}} \times {\text{amount of money that he puts on second day}} \\ {\text{ = 2}} \times {\text{Rs}}{\text{.2 = Rs}}{\text{.4}} \\ {\text{Amount of money that he puts on fourth day = 2}} \times {\text{amount of money that he puts on third day}} \\ {\text{ = 2}} \times {\text{Rs}}{\text{.4 = Rs}}{\text{.8}} \\ {\text{Amount of money that he puts on fifth day = 2}} \times {\text{amount of money that he puts on fourth day}} \\ {\text{ = 2}} \times {\text{Rs}}{\text{.8 = Rs}}{\text{.16}} \\ {\text{Amount of money that he puts on sixth day = 2}} \times {\text{amount of money that he puts on fifth day}} \\ {\text{ = 2}} \times {\text{Rs16 = Rs}}{\text{.32}} \\ {\text{Amount of money that he puts on seventh day = 2}} \times {\text{amount of money that he puts on sixth day}} \\ {\text{ = 2}} \times {\text{Rs}}{\text{.32 = Rs}}{\text{.64}} \\ {\text{Amount of money that he puts on eighth day = 2}} \times {\text{amount of money that he puts on seventh day}} \\ {\text{ = 2}} \times {\text{Rs64 = Rs}}{\text{.128}} \\$
So, total money that he collected till 8th day = sum of money collected on each day
= Re.1+Rs.2+Rs.4+Rs.8+Rs.16+Rs.32+Rs.64+Rs.128
= Rs.255.
Note:
In this type of question, the usual method is to calculate the amount for each day by doubling the amount of previous day and then add the amount of each day to get the final answer. But we can also ask this type of question using geometric progression. In this question, it is given that money doubles everyday. So, the common ratio(r)=2 and first term is given 1 and number of terms(n)=8. So, the money collected after 8th day is given by:
${\text{Sum of money = }}\dfrac{{a({r^n} - 1)}}{{r - 1}} = \dfrac{{1({2^8} - 1)}}{{2 - 1}} = {\text{Rs}}.255$ |
# Fraction calculator
This fraction calculator performs basic and advanced fraction operations, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. The calculator helps in finding value from multiple fractions operations. Solve problems with two, three, or more fractions and numbers in one expression.
## The result:
### 22/3 * 13/4 = 14/3 = 4 2/3 ≅ 4.6666667
Spelled result in words is fourteen thirds (or four and two thirds).
### How do we solve fractions step by step?
1. Conversion a mixed number 2 2/3 to a improper fraction: 2 2/3 = 2 2/3 = 2 · 3 + 2/3 = 6 + 2/3 = 8/3
To find a new numerator:
a) Multiply the whole number 2 by the denominator 3. Whole number 2 equally 2 * 3/3 = 6/3
b) Add the answer from the previous step 6 to the numerator 2. New numerator is 6 + 2 = 8
c) Write a previous answer (new numerator 8) over the denominator 3.
Two and two thirds is eight thirds.
2. Conversion a mixed number 1 3/4 to a improper fraction: 1 3/4 = 1 3/4 = 1 · 4 + 3/4 = 4 + 3/4 = 7/4
To find a new numerator:
a) Multiply the whole number 1 by the denominator 4. Whole number 1 equally 1 * 4/4 = 4/4
b) Add the answer from the previous step 4 to the numerator 3. New numerator is 4 + 3 = 7
c) Write a previous answer (new numerator 7) over the denominator 4.
One and three quarters is seven quarters.
3. Multiple: 8/3 * 7/4 = 8 · 7/3 · 4 = 56/12 = 14 · 4/3 · 4 = 14/3
Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD(56, 12) = 4. In the following intermediate step, cancel by a common factor of 4 gives 14/3.
In other words - eight thirds multiplied by seven quarters is fourteen thirds.
#### Rules for expressions with fractions:
Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
SymbolSymbol nameSymbol MeaningExample
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4)
The calculator follows well-known rules for the order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule.
Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right. |
# Internal Division of Line Segments
Internal Division an interval in a given ratio is that if the interval $(x_1,y_1)$ and $(x_2,y_2)$ is divided in the ratio $m:n$ then the coordinates are;
$$\large \Big(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\Big)$$
## Example 1: Basics
If $A$ and $B$ are the points $(-4,3)$ and $(2,-1)$ respectively, find the coordinates of $P$ such that $AP:PB=3:1$.
\begin{aligned} \displaystyle P(x,y) &= \Big(\frac{3 \times 2 + 1 \times -4}{3+1},\frac{3 \times -1 + 1 \times 3}{3+1}\Big) \\ &= \Big(\frac{1}{2},0\Big) \\ \end{aligned} \\
## Example 2: Negative Ratio of Internal Division of Line Segments
If $A$ and $B$ are the points $(-4,3)$ and $(2,-1)$ respectively, find the coordinates of $P$ such that $AP:PB=-4:5$.
\begin{aligned} \displaystyle P(x,y) &= \Big(\frac{-4 \times 2 + 5 \times -4}{-4+5},\frac{-4 \times -1 + 5 \times 3}{-4+5}\Big) \\ &= (-28,19) \\ \end{aligned} \\
## Example 3: Three Equal Parts using Internal Division of Line Segments
Divide the interval between $(-1,1)$ and $(5,10)$ into three equal parts.
\begin{aligned} \displaystyle (a,b) &= \Big(\frac{1 \times 5 + 2 \times -1}{1+2},\frac{1 \times 10 + 2 \times 1}{1+2}\Big) \\ &= (1,4) \\ (c,d) &= \Big(\frac{2 \times 5 + 1 \times -1}{2+1},\frac{2 \times 10 + 1 \times 1}{2+1}\Big) \\ &= (3,7) \\ \end{aligned} \\
## Example 4: Finding the Ratio of Internal Division of Line Segments
If the point $(-3,8)$ divides the interval between $(6,-4)$ and $(0,4)$ internally in the ratio $k:1$, find the value of $k$.
\begin{aligned} \displaystyle (-3,8) &= \Big(\frac{k \times 0 + 1 \times 6}{k+1},\frac{k \times 4 + 1 \times -4}{k+1}\Big) \\ &= \Big(\frac{6}{k+1},\frac{4k-4}{k+1}\Big) \\ \frac{6}{k+1} &= -3 \\ k+1 &= 6 \div -3 \\ k+1 &= -2 \\ \therefore k &= -3 \\ \end{aligned} \\ |
# Video: AQA GCSE Mathematics Higher Tier Pack 5 β’ Paper 3 β’ Question 3
Circle the equation of the line that is parallel to the π¦-axis. [A] π₯ = β2 [B] π¦ = 0 [C] π¦ β π₯ = 0 [D] π¦ + π₯ = 1
01:55
### Video Transcript
Circle the equation of the line that is parallel to the π¦-axis. Is it π₯ equals negative two, π¦ equals zero, π¦ minus π₯ equals zero, or π¦ plus π₯ equals one?
Letβs remind ourselves what a line is parallel to the π¦-axis would look like. This is one example. A common mistake here is to think that because the line travels in the same direction as the π¦-axis, it is π¦ is equal to some number. In fact, this is not true. The equation of this line is π₯ is equal to some number. And that number can be found by looking for the value at where the line crosses the π₯-axis. This line crosses the π₯-axis at π. So π₯ is equal to this number π. We could also have this line. This would have the equation π₯ is equal to negative π.
We can see in our list that the only equation that looks like these two is this one, π₯ is equal to negative two. In fact, the line π¦ equals zero is a horizontal line that passes through π¦ at zero. Itβs another way of describing the π₯-axis. The line π¦ minus π₯ equals zero is this one. This is because we can form an equation for π¦ in terms of π₯ by adding extra both sides. That tells us that π¦ is equal to π₯. So itβs the line that consists of all coordinates where the π¦-coordinate is equal to the π₯-coordinate.
And this diagonal line sloping downwards is the line π¦ plus π₯ equals one. We could rearrange this equation by subtracting π₯ from both sides. And we see that itβs the same as π¦ equals one minus π₯. This line has a π¦-intercept. It passes through the π¦-axis at one. And it has a gradient of negative one. It slopes downwards.
The equation of the line that is parallel to the π¦-axis is π₯ equals negative two. |
# 6.1 Understand percent
Page 1 / 6
By the end of this section, you will be able to:
• Use the definition of percent
• Convert percents to fractions and decimals
• Convert decimals and fractions to percents
Before you get started, take this readiness quiz.
1. Translate “the ratio of $33$ to $\text{5”}$ into an algebraic expression.
If you missed this problem, review Evaluate, Simplify and Translate Expressions .
2. Write $\frac{3}{5}$ as a decimal.
If you missed this problem, review Decimals and Fractions .
3. Write $0.62$ as a fraction.
If you missed this problem, review Decimals .
## Use the definition of percent
How many cents are in one dollar? There are $100$ cents in a dollar. How many years are in a century? There are $100$ years in a century. Does this give you a clue about what the word “percent” means? It is really two words, “per cent,” and means per one hundred. A percent is a ratio whose denominator is $100.$ We use the percent symbol $\text{%,}$ to show percent.
## Percent
A percent is a ratio whose denominator is $100.$
According to data from the American Association of Community Colleges $\left(2015\right)\text{,}$ about $\text{57%}$ of community college students are female. This means $57$ out of every $100$ community college students are female, as [link] shows. Out of the $100$ squares on the grid, $57$ are shaded, which we write as the ratio $\frac{57}{100}.$
Similarly, $\text{25%}$ means a ratio of $\frac{25}{100},\text{3%}$ means a ratio of $\frac{3}{100}$ and $\text{100%}$ means a ratio of $\frac{100}{100}.$ In words, "one hundred percent" means the total $\text{-100%-}$ is $\frac{100}{100},$ and since $\frac{100}{100}=1,$ we see that $\text{100%}$ means $1$ whole.
According to the Public Policy Institute of California $\left(2010\right)\text{,}\phantom{\rule{0.2em}{0ex}}\text{44%}$ of parents of public school children would like their youngest child to earn a graduate degree. Write this percent as a ratio.
## Solution
The amount we want to convert is 44%. $44%$ Write the percent as a ratio. Remember that percent means per 100. $\frac{44}{100}$
Write the percent as a ratio.
According to a survey, $\text{89%}$ of college students have a smartphone.
$\frac{89}{100}$
Write the percent as a ratio.
A study found that $\text{72%}$ of U.S. teens send text messages regularly.
$\frac{72}{100}$
In $2007,$ according to a U.S. Department of Education report, $21$ out of every $100$ first-time freshmen college students at $\text{4-year}$ public institutions took at least one remedial course. Write this as a ratio and then as a percent.
## Solution
The amount we want to convert is $21$ out of every $100.$
Write as a ratio. $\frac{21}{100}$ Convert the 21 per 100 to percent. $21%$
Write as a ratio and then as a percent: The American Association of Community Colleges reported that $62$ out of $100$ full-time community college students balance their studies with full-time or part time employment.
$\frac{62}{100},\text{62%}$
Write as a ratio and then as a percent: In response to a student survey, $41$ out of $100$ Santa Ana College students expressed a goal of earning an Associate's degree or transferring to a four-year college.
$\frac{41}{100},\text{41%}$
## Convert percents to fractions and decimals
Since percents are ratios, they can easily be expressed as fractions. Remember that percent means per $100,$ so the denominator of the fraction is $100.$
## Convert a percent to a fraction.
1. Write the percent as a ratio with the denominator $100.$
2. Simplify the fraction if possible.
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Berger describes sociologists as concerned with
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight? |
## Engage NY Eureka Math 4th Grade Module 6 Lesson 11 Answer Key
### Eureka Math Grade 4 Module 6 Lesson 11 Problem Set Answer Key
Question 1.
Plot the following points on the number line.
a. 0.2, $$\frac{1}{10}$$, 0.33, $$\frac{12}{100}$$, 0.21, $$\frac{32}{100}$$
The points are 0.2, 0.1, 0.33, 0.12, 0.21, 0.32.
Explanation:
In the above-given question,
given that,
the points on the number line are
0.2.
$$\frac{1}{10}$$ = 1/10 = 0.1.
0.33.
$$\frac{12}{100}$$ = 0.12.
0.21.
$$\frac{32}{100}$$ = 0.32.
b. 3.62, 3.7, 3$$\frac{85}{100}$$, $$\frac{38}{10}$$, $$\frac{364}{100}$$
The points are 3.62, 3.7, 3.85, 3.8, 3.64.
Explanation:
In the above-given question,
given that,
the points on the number line are
3.62.
3$$\frac{85}{100}$$ = 385/100 = 3.85.
3.7.
$$\frac{38}{10}$$ = 3.8.
$$\frac{364}{100}$$ = 3.64.
c. 6$$\frac{3}{10}$$, 6.31, $$\frac{628}{100}$$, $$\frac{62}{10}$$, 6.43, 6.40
The points are 6.3, 6.31, 6.28, 6.2, 6.43, 6.40.
Explanation:
In the above-given question,
given that,
the points on the number line are
6.31.
6$$\frac{3}{10}$$ = 63/10 = 6.3.
6.43.
$$\frac{628}{100}$$ = 6.28.
6.40.
$$\frac{62}{10}$$ = 6.2.
Question 2.
Arrange the following numbers in order from greatest to least using decimal form. Use the > symbol between each number.
a. $$\frac{27}{10}$$, 2.07, $$\frac{27}{100}$$, 2$$\frac{71}{100}$$, $$\frac{227}{100}$$, 2.72
b. 12$$\frac{3}{10}$$, 13.2, $$\frac{134}{100}$$, 13.02, 12$$\frac{20}{100}$$
c. 7$$\frac{34}{100}$$, 7$$\frac{4}{10}$$, 7$$\frac{3}{10}$$, $$\frac{750}{100}$$, 75, 7.2
a. The points are 2.7, 2.07, 0.27, 2.71, 2.27, 2.72.
Explanation:
In the above-given question,
given that,
the points on the number line are
27/10 = 2.7.
2.07.
0.27.
2.71.
2.27.
2.72.
the numbers from greatest to least are 2.72, 2.71, 2.7, 2.27, 2.07, 0.27.
b. The points are 12.3, 13.2, 1.34, 13.02, 12.2.
Explanation:
In the above-given question,
given that,
the points on the number line are 12$$\frac{3}{10}$$, 13.2, $$\frac{134}{100}$$, 13.02, 12$$\frac{20}{100}$$
123/10 = 12.3.
$$\frac{134}{100}$$ = 134/100= 1.34.
13.2.
12$$\frac{20}{100}$$ = 1220/100 = 12.2.
13.02.
the numbers from greatest to least are 13.2, 13.02, 12.3, 12.2, 1.34.
c. The points are 7.34, 7.4, 7.3, 7.5, 75, 7.2.
Explanation:
In the above-given question,
given that,
the points on the number line are 7$$\frac{34}{100}$$, 7$$\frac{4}{10}$$, 7$$\frac{3}{10}$$, $$\frac{750}{100}$$, 75, 7.2
75.
7$$\frac{34}{100}$$ = 734/100= 7.34.
7.2.
7$$\frac{4}{10}$$ = 74/10 = 7.4.
7$$\frac{3}{10}$$ = 73/10 = 7.3.
750/100 = 7.5.
the numbers from greatest to least are 75, 7.5, 7.4, 7.34, 7.3, 7.2..
Question 3.
In the long jump event, Rhonda jumped 1.64 meters. Mary jumped 1$$\frac{6}{10}$$ meters. Kerri jumped $$\frac{94}{100}$$ meter. Michelle jumped 1.06 meters. Who jumped the farthest?
Rhonda jumped the farthest = 1.64 meters.
Explanation:
In the above-given question,
given that,
Rhonda jumped 1.64 meters.
Mary jumped 16/10 = 1.6 meters.
Kerri jumped 0.94 meters.
Michelle jumped 1.06 meters.
the farthest jumped = 1.64 meters.
Question 4.
In December, 2$$\frac{3}{10}$$ feet of snow fell. In January, 2.14 feet of snow fell. In February, 2$$\frac{19}{100}$$ feet of snow fell, and in March, 1$$\frac{1}{10}$$ feet of snow fell. During which month did it snow the most? During which month did it snow the least?
The month in which it snows the most = December.
The month in which it snows the least = march.
Explanation:
In the above-given question,
given that,
In December, 2$$\frac{3}{10}$$ feet of snow fell = 23/10 = 2.3.
In January, 2.14 feet of snow fell.
In February, 2$$\frac{19}{100}$$ feet of snow fell = 2.19.
In March, 1$$\frac{1}{10}$$ feet of snow fell = 1.1.
the month in which it snows the most = December.
the month in which it snows the least = march.
### Eureka Math Grade 4 Module 6 Lesson 11 Exit Ticket Answer Key
Question 1.
Plot the following points on the number line using decimal form.
1 one and 1 tenth, $$\frac{13}{10}$$, 1 one and 20 hundredths, $$\frac{129}{100}$$, 1.11, $$\frac{102}{100}$$
The points are 1.1, 1.3, 1.20, 1.29, 1.11, 1.02.
Explanation:
In the above-given question,
given that,
1 one and 1 tenth = 1.1.
13/10 = 1.3.
1 one and 20 hundredths = 120/100 = 1.20.
129/100 = 1.29.
102/100 = 1.02.
Question 2.
Arrange the following numbers in order from greatest to least using decimal form. Use the > symbol between each number.
5.6, $$\frac{605}{100}$$, 6.15, 6$$\frac{56}{100}$$, $$\frac{516}{100}$$, 6 ones and 5 tenths
5.6 > 5.16 > 6.5 > 6.05 > 6.15 > 6.56.
Explanation:
In the above-given question,
given that,
the points are 605/100 = 6.05.
656/100 = 6.56.
516/100 = 5.16.
the points from least to greatest are
5.6 > 5.16 > 6.5 > 6.05 > 6.15 > 6.56.
### Eureka Math Grade 4 Module 6 Lesson 11 Homework Answer Key
Question 1.
Plot the following points on the number line using decimal form.
a. 0.6, $$\frac{5}{10}$$, 0.76, $$\frac{79}{100}$$, 0.53, $$\frac{67}{100}$$
The points on the number line = 0.6, 0.5, 0.76, 0.79, 0.53, 0.67.
Explanation:
In the above-given question,
given that,
the points on the number line are.
0.6.
5/10 = 0.5.
79/100 = 0.79.
67/100 = 0.67.
The points on the number line = 0.6, 0.5, 0.76, 0.79, 0.53, 0.67.
b. 8 ones and 15 hundredths, $$\frac{832}{100}$$, 8$$\frac{27}{100}$$, $$\frac{82}{10}$$, 8.1
The points on the number line = 8.15, 8.32, 8.27, 8.2, 8.1.
Explanation:
In the above-given question,
given that,
the points on the number line are.
8 ones and 15 hundredths = 8.15.
832/100 = 8.32.
827/100 = 8.27.
82/10 = 8.2.
The points on the number line = 8.15, 8.32, 8.27, 8.2, 8.1.
c. 13$$\frac{12}{100}$$, $$\frac{130}{10}$$, 13 ones and 3 tenths, 13.21, 13$$\frac{3}{100}$$
The points on the number line = 13.12, 13.0, 13.3, 13.21, 1.33.
Explanation:
In the above-given question,
given that,
the points on the number line are.
13 ones and 3 tenths = 13.3.
1312/100 = 13.12.
130/10 = 13.0.
133/100 = 13.3.
The points on the number line = 13.12, 13.0, 13.3, 13.21, 1.33.
Question 2.
Arrange the following numbers in order from greatest to least using decimal form. Use the > symbol between each number.
a. 4.03, 4 ones and 33 hundredths, $$\frac{34}{100}$$, 4$$\frac{43}{100}$$, $$\frac{430}{100}$$, 4.31
b. 17$$\frac{5}{10}$$, 17.55, $$\frac{157}{10}$$ 17 ones and 5 hundredths, 15.71, 15$$\frac{75}{100}$$
c. 8 ones and 19 hundredths, 9$$\frac{8}{10}$$, 81, $$\frac{809}{100}$$, 8.9, 8$$\frac{1}{10}$$
0.34 > 4.03 > 4.30 > 4.31 > 4.33 > 4.43.
Explanation:
In the above-given question,
given that,
the points on the number line are.
4 ones and 33 hundredths = 4.33.
34/100 = 0.34.
443/100 = 4.43.
430/100 = 4.30.
0.34 > 4.03 > 4.30 > 4.31 > 4.33 > 4.43.
Question 3.
In a paper airplane contest, Matt’s airplane flew 9.14 meters. Jenna’s airplane flew 9$$\frac{4}{10}$$ meters. Ben’s airplane flew $$\frac{904}{100}$$ meters. Leah’s airplane flew 9.1 meters. Whose airplane flew the farthest?
The airplane flew the farthest = Jenna airplane.
Explanation:
In the above-given question,
given that,
Matt’s airplane flew = 9.14 meters.
Jenna’s airplane flew = 94/10 = 9.4 meters.
Ben’s airplane flew = 904/100 = 9.04 meters.
Leah’s airplane flew = 9.1 meters.
jenna’s airplane flew the farthest.
Question 4.
Becky drank 1$$\frac{41}{100}$$ liters of water on Monday, 1.14 liters on Tuesday, 1.04 liters on Wednesday, $$\frac{11}{10}$$ liters on Thursday, and 1$$\frac{40}{100}$$ liters on Friday. Which day did Becky drink the most? Which day did Becky drink the least? |
# Question Video: Solving a System of Linear Equations Involving Percentages
You invested \$2300 in account 1, and \$2700 in account 2, which both pay simple interest annually. If the total amount of interest after one year is \$254, and account 2 has 1.5 times the interest of account 1, what are the interest rates?
03:24
### Video Transcript
You invested 2300 dollars in account one and 2700 dollars in account two, which both pay simple interest annually. If the total amount of interest after one year is 254 dollars and account two has 1.5 times the interest of account one, what are the interest rates?
In order to answer this question, we recall that in order to calculate the simple interest, we multiply 𝑃 by 𝑅 by 𝑇. 𝑃 is the principal amount or amount invested, 𝑅 is the interest rate written as a decimal, and 𝑇 is the time. In account one, the principal amount is 2300 dollars. We will let the interest rate for this account be 𝑥. As we are dealing with one year, the time is equal to one.
For account two, 𝑃 is equal to 2700 dollars. The time is also equal to one. As the interest rate is 1.5 times the interest of account one, 𝑅 is equal to 1.5𝑥. The amount of interest for account one is therefore equal to 2300 multiplied by 𝑥 multiplied by one. This is equal to 2300𝑥. For account two, we need to multiply 2700 by 1.5𝑥 by one. This is equal to 4050𝑥.
We are also told in the question that the total amount of interest is 245 dollars. This means that 2300𝑥 plus 4050𝑥 is equal to 254. The left-hand side simplifies to 6350𝑥. Dividing both sides of this equation by 6350 gives us 𝑥 is equal to 254 over 6350. This fraction simplifies to one over 25. Therefore, 𝑥 is equal to 0.04 as a decimal.
To convert from a decimal to a percentage, we multiply by 100. This means that the interest rate for account one is four percent. The interest rate for account two was 1.5 times this, and four multiplied by 1.5 is six. This means that the interest rate for account two is six percent. Four percent simple interest on 2300 dollars and six percent simple interest on 2700 dollars gives a total of 254 dollars of interest after one year. |
# ACT Math : How to find the intersection of a Venn Diagram
## Example Questions
### Example Question #1 : How To Find The Intersection Of A Venn Diagram
Fifty 6th graders were asked what their favorite school subjects were. Three students like math, science and English. Five students liked math and science. Seven students liked math and English. Eight people liked science and English. Twenty students liked science. Twenty-eight students liked English. Fourteen students liked math. How many students didn’t like any of these classes?
7
3
5
None of the answers are correct
10
5
Explanation:
Draw a Venn diagram with three subsets: Math, Science, and English. Start in the center with students that like all three subjects. Next, look at students that liked two subjects. Be sure to subtract out the ones already counted in the middle. Then, look at the students that only like one subject. Be sure to subtract out the students already accounted for. Once all of the subsets are filled, look at those students who don’t like any of these subjects. To find the students who don’t like any of these subjects add all of the students who like at least one subject from the total number of students surveyed, which is 50.
M = math
S = science
E = English
M∩S∩E = 3
M∩S = 5 (but 3 are already accounted for) so 2 for M and S ONLY
M∩E = 7 (but 3 are already accounted for) so 4 for M and E ONLY
S∩E = 8 (but 3 are already accounted for) so 5 for S and E ONLY
M = 14 (but 3 + 2 + 4 are already accounted for) so 5 for M ONLY
S = 20 (but 3 + 2 + 5 are already accounted for) so 10 for S ONLY
E = 28 (but 3 + 4 + 5 are already accounted for) so 16 for E ONLY
Therefore, the students already accounted for is 3 + 2 +4 + 5 + 5 + 10 + 16 = 45 students
So, those students who don’t like any of these subjects are 50 – 45 = 5 students
### Example Question #1 : Venn Diagrams
Set A contains the positive even integers less than 14. Set B contains the positive multiples of three less than 20. What is the intersection of the two sets?
A∩B = {6, 12}
A∩B = { }
A∩B = {6}
A∩B = {6, 12, 18}
A∩B = {4, 6, 8}
A∩B = {6, 12}
Explanation:
A = {2, 4, 6, 8, 10, 12}
B = {3, 6, 9, 12, 15, 18}
The intersection of a set means that the elements are in both sets: A∩B = {6, 12}
### Example Question #1 : How To Find The Intersection Of A Venn Diagram
Students at a local high school are given the option to take one gym class, one music class or one of each. Out of 100 students, 60 say that they are currently taking a gym class and 70 say that they are taking a music class. How many students are taking both?
Explanation:
This problem can be solved two ways, with a formula or with reason.
Using the formula, the intersection of the Venn diagram for which classes students take is:
By using reason, it is clear that 60 + 70 is greater than 100 by 30. It is assumed that this extra 30 students come from students who were counted twice because they took both classes.
### Example Question #1 : Intersection
In the Venn Diagram shown above, let the set be the set of all prime numbers less than . Let the set be the set of all odd numbers less than . What is the set , given by ? Use set notation to enumerate.
Explanation:
To find the intersection of a set of numbers, we need to find all numbers that are in both sets (definition of intersection). We can see that all the primes less than that are odd are also odd numbers less than , therefore they are in both sets and in the intersection. Thus the answer:
### Example Question #2 : How To Find The Intersection Of A Venn Diagram
In the Venn Diagram shown above, let , let and let . If the Universe is all integers less than or equal to 10, what is ? Use set notation to enumerate your answer.
Explanation:
means the complement of union - this is everything in the universe that is not in or . First we need to find :
- we find this by knowing anything in either set is in the union. Next we need to find things in the universe that are NOT in :
Since the universe is integers less than or equal to 10, we see that the only ones that fit that description that are not in are
### Example Question #1 : How To Find The Intersection Of A Venn Diagram
There are 75 juniors at a high school. 15 of the students are enrolled in Physics and 40 students are enrolled in Chemistry. 30 students are not enrolled in either Physics or Chemistry. How many students are enrolled in both Physics and Chemistry?
30
10
15
5
25 |
#### GCD (Greatest Common Division) of two numbers
Problem Definition: To find the Greatest Common Divisor (GCD)of two numbers.
Problem Analysis: GCD of two numbers is the greatest common factor of the given two numbers.
GCD of two numbers is obtained using the following method: Divide the larger of the two numbers by the smaller one.
Divide the divisor by the remainder.
Repeat this process till the remainder becomes zero.
The last divisor is the GCD of the two.
Solving by Example: Consider two numbers 188 and 423.
The smaller number is 188.Therefore divide 423 by 188.
Step 1
47(Remainder) becomes the divisor and 188 (Divisor) becomes the dividend.
Step 2
Since the remainder has become zero, 47 is the GCD of 188 and 423.
Algorithm Definition
Step 1: Start
Step 2: Read two numbers for which GCD is to be found, say (a,b).
Step 3: Let a be the larger of the two numbers.
Step 4: Divide a by b.
Step 5: Get the remainder of the division operation.
Step 6: Divide the divisor of the previous division operation with the remainder.
Step 7: Repeat steps 4,5,6 till the remainder become zero.
Step 8: The divisor of the last division operation performed is the GCD of the two numbers.
Step 9: Stop
Question. What is algorithm? Write the characteristics of an algorithm. Give an example for algorithm.
Definition of Algorithm
Algorithm is a step by step solution to a given problem. Sequence of steps written in order to carry out some particular task. Each step of an algorithm is written in english.
Various characteristics of an algorithm are
• The algorithm must have definite start and end.
• An algorithm may accept zero or more inputs
• An algorithm must produce at least one output
• The steps of an algorithm must be simple, easy to understand and unambiguous.
• Each step must be precise and accurate.
• The algorithm must contain finite number of steps.
Example:
Algorithm 1: Algorithm for finding of area of triangle
Step 1: start
Step 2: Read base and height of traingle
Step 3: Calulate area of traingle
Step 4: print area of traingle
Step 5: stop
Algorithm 2: Algorithm for finding of sum and average of given three numbers
Step 1: start
Step 2: Read three numbers i.e. A, B and C
Step 3: find sum of three numbers i.e. sum=A+B+C
Step 4: find average of three numbers i.e. average=sum/3
Step 4: print sum and average
Step 5: stop
Question : What is pseudocode? Give an example for pseudocode.
Definition of Pseudocode
Pseudocode is high level description of an algorithm that contains a sequence of steps written in combination of english and mathematical notations to solve a given problem.
Pseudocode is part english and part program logic. Pseudocodes are better than algorithm since it contains ordered steps and mathematical notations they are more closer to the statements of programming language.
This is essentially an intermediate-step towards the development of the actual code(program). Although pseudo code is frequently used, there are no set of rules for its exact writing.
Example:
Pseudocode 1: Pseudocode for finding area of traingle
Pseudocode Area_of_Traingle
BEGIN
CALCULATE Area_of_Triangle=(base*height)/2
PRINT Area_of_Triangle
END
Pseudocode 2:Pseudocode for finding sum and average of three numbers
Pseudocode SUM_AVG
BEGIN
CALCULATE sum=A+B+C
CALCULATE average=SUM/3
PRINT sum and average
END
Question : What is flowchart? List and define the purpose of symbols used to represent flowchart. Give an example for flowchart.
Definition of flowchart
Flowchart is a graphical or pictorial representation of an algorithm. Its a program "design tool in which standard graphical symbols are used to represent the logical flow of data through a function.
Flowchart is a combination of symbols. They show start and end points, the order and sequence of actions, and how one part of a flowchart is connected to another.
Example:
Finding the area of triangle |
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# A body cools from ${60^o}C$ to ${50^o}C$ in 5 minutes. If the temperature of the surrounding is ${30^o}C$ calculate the time period required by the body to cool to ${40^o}C$ .(A) 8.33 min(B) 5 min(C) 8.9 min(D) Less than 5 min
Last updated date: 22nd Feb 2024
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Hint Apply Newton’s Law of Cooling and find out the value of temperature changing coefficient for the first case of cooling. Newton’s law of cooling is used to calculate the value of the rate of change of temperature. Now use this value of temperature changing coefficient to find out the value of the time required for the body to cool down in the second case.
Formula used: $\dfrac{{dT}}{{dt}} = K\left[ {\dfrac{{\left( {{T_i} + {T_f}} \right)}}{2} - {T_o}} \right]$
Complete Step by step solution
Here we will use Newton’s Law of Cooling,
where rate of cooling, $\dfrac{{dT}}{{dt}} = K\left[ {\dfrac{{\left( {{T_i} + {T_f}} \right)}}{2} - {T_o}} \right]$.
Here,$t$ is the time taken,
$K$is the temperature changing coefficient,
${T_i}$ is the initial temperature,
${T_f}$ is the final temperature, and
${T_o}$ is the temperature of the surroundings.
For the first case,
${T_i} = {60^o}C,{T_f} = {50^o}C,{T_o} = {30^o}C,dt = 5\min$
Let us substitute these values in the above equation.
$\therefore \dfrac{{\left( {60 - 50} \right)}}{5} = K\left[ {\dfrac{{60 + 50}}{2} - 30} \right]$
$\Rightarrow \dfrac{{10}}{5} = K\left[ {\dfrac{{110}}{2} - 30} \right]$
$\therefore 2 = K(55 - 30)$
On further simplifying the equation we get,
$\Rightarrow 2 = K25$
$\therefore K = \dfrac{2}{{25}}$
In the second case we have the values as:
Initial temperature, ${T_i} = {50^o}C,$
Final temperature, ${T_f} = {40^o}C,$
Surrounding temperature, ${T_0} = {30^o}C,$
Temperature changing coefficient as found out from the first case, $K = \dfrac{2}{{25}}$
Substituting these values in the equation of Newton’s Law of Cooling we get,
$\dfrac{{\left( {50 - 40} \right)}}{{dt}} = \dfrac{2}{{25}}\left[ {\dfrac{{\left( {50 + 40} \right)}}{2} - 30} \right]$
$\Rightarrow \dfrac{{10}}{{dt}} = \dfrac{2}{{25}}\left( {\dfrac{{90}}{2} - 30} \right)$
$\Rightarrow \dfrac{{10}}{{dt}} = \dfrac{2}{{25}}\left( {45 - 30} \right)$
On further simplifying the equation we get,
$\Rightarrow \dfrac{{10}}{{dt}} = \dfrac{2}{{25}}\left( {15} \right)$
$\therefore dt = \dfrac{{25}}{3} = 8.33$
Therefore the time taken for the body to cool down from ${50^o}C$ to ${40^o}C$ , with a surrounding temperature of ${30^o}C$ is 8.33 minutes.
Hence option A. is the correct answer.
Note
We should be careful of the fact that the question mentions the cool down of the body from ${60^o}C$to ${50^o}C$ in the first case, and then further the body is cooled down from ${50^o}C$ to ${40^o}C$. Different ranges of temperature difference would yield different results. The unit of temperature cooling coefficient is ${s^{ - 1}}$ , hence we can see that the time taken is dependent on the temperature cooling coefficient. |
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# Limits and Continuity
Module - 2 Calculus
Limits and Continuity
Overview
In calculus, limits and continuity are important concepts that help us understand the behavior of functions as they approach certain values. A limit is a value that a function approaches as its input approaches a certain value. Continuity is the property of a function that describes whether it has any sudden jumps or breaks, or whether it can be drawn without lifting the pen from the paper.
Introduction to limits and continuity:
• Limits and continuity are fundamental concepts in calculus. They allow us to understand how functions behave as their inputs approach certain values and to make precise statements about the behavior of functions at certain points.
• Continuity refers to the property of a function where the output changes smoothly and gradually as the input changes. A function is continuous at a point in the event that the limit of the function at that point exists and is equal to the esteem of the function at that point. In the event that a function isn't continuous at a point, it is said to be discontinuous.
• Limits, on the other hand, allude to the behavior of a function as the input approaches a certain value/point. A limit can be utilized to portray what happens to a function if the input gets subjectively near to a certain value, if the function isn't defined at that point.
The definition of a limit:
The formal definition of a limit is given as follows:
For a function f(x), the limit of f(x) as x approaches a equals L, denoted as
lim f(x) = L
for every number ε > 0 there exists a corresponding number δ > 0 such that |f(x) - L| < ε whenever 0 < |x - a| < δ.
This definition can be hard to understand at first, but it basically says that as x gets arbitrarily close to a, f(x) gets arbitrarily close to L. The ε and δ values represent arbitrarily small positive numbers that determine the level of precision required for the limit.
For example, consider the function
f(x) = (x^2 - 1)/(x - 1)
If we try to evaluate f(1), we get an undefined expression (0/0). However, we can find the limit of f(x) as x approaches 1 by factoring the numerator and simplifying:
= f(x) = (x^2 - 1)/(x - 1)
= (x + 1)(x - 1)/(x - 1)
= x + 1
Now we can see that as x approaches 1, f(x) approaches 2. Therefore, we can write lim f(x) as x approaches 1 = 2.
Properties of limits:
There are several important properties of limits that are useful for evaluating and manipulating functions. These include:
• The limit of a whole is the sum of the limits:
lim (f(x) + g(x)) = lim f(x) + lim g(x)
• The limit of a product is the product of the limits:
lim (f(x) * g(x)) = lim f(x) * lim g(x)
• The limit of a quotient is the quotient of the limits:
lim (f(x) / g(x)) = lim f(x) / lim g(x) (given lim g(x) ≠ 0)
• The limit of a constant multiple is the constant times the restrain:
lim (k * f(x)) = k * lim f(x)
• The limit of a composite function is the restrain of the external function connected to the constrain of the inner function:
lim f(g(x)) = f(lim g(x))
In expansion to these properties, there are a few hypotheses that can be utilized to assess limits, such as the squeeze theorem, which states that in the event that
g(x) ≤ f(x) ≤ h(x)
for all x in a few interim containing a, and lim g(x) = lim h(x) = L, at that point lim f(x) = L. Another important theorem is the intermediate value theorem, which states that if f is a continuous function on the closed interval [a, b] and c is a number between f(a) and f(b), at that point there exists a number x between a and b such that f(x) = c.
Let's outline a few of these properties with a case. Consider the work
f(x) = x^2 - 3x + 2
We can use the properties of limits to evaluate lim f(x) as x approaches 2.
First, we can factor in the expression:
f(x) = x^2 - 3x + 2
= (x - 1)(x - 2)
Using the product property of limits, we can write:
lim f(x) as x approaches 2 = lim (x - 1)(x - 2) as x approaches 2
Then, we can use the limit laws to simplify this expression:
= lim (x - 1)(x - 2) as x approaches 2
= (lim (x - 2)) * (lim (x - 1)) as x approaches 2
= (2 - 2) * (2 - 1)
= 0
Therefore, we can conclude that lim f(x) as x approaches 2 = 0.
Evaluating limits:
Several methods for evaluating limits include direct substitution, factoring, and L'Hopital's rule. Direct substitution is the simplest method, and it involves substituting the input value directly into the function and evaluating the resulting expression.
For example, to evaluate
lim (x^2 - 4x + 3)/(x - 3)
as x approaches 3, we can simply substitute x = 3 into the expression to get:
lim (x^2 - 4x + 3)/(x - 3) = lim (3^2 - 4(3) + 3)/(3 - 3) = lim 0/0
Since this expression is indeterminate, we can use factoring or L'Hopital's rule to evaluate the limit further.
Factoring involves rewriting the function as a product of simpler expressions that can be canceled out.
For example, using the same limit as above, we can factor the numerator as
(x - 3)(x - 1) and cancel out the common factor of (x - 3) to get:
lim (x^2 - 4x + 3)/(x - 3) = lim (x - 1)
as x approaches 3
This limit evaluates to 2, so we can conclude that
lim (x^2 - 4x + 3)/(x - 3) as x approaches 3 = 2
L'Hopital's rule is another method for evaluating indeterminate limits. It involves taking the derivative of both the numerator and denominator of the function and evaluating the resulting expression. For example, to evaluate lim sin(x)/x as x approaches 0, we can apply L'Hopital's rule to get:
lim sin(x)/x = lim cos(x) as x approaches 0
This limit evaluates to 1, so we can conclude that lim sin(x)/x as x approaches 0 = 1.
Continuity and differentiability:
• Continuity and differentiability are closely related concepts in calculus. A function is said to be continuous at a point if its graph has no breaks or jumps at that point. A function is said to be differentiable at a point if it has a well-defined tangent line at that point.
• The differentiability implies continuity theorem states that if a function is differentiable at a point, then it is also continuous at that point. However, the converse is not necessarily true - a function can be continuous at a point but not differentiable at that point.
• For example, the function f(x) = |x| is continuous at x = 0, but it is not differentiable at that point because it has a sharp corner.
• The relationship between continuity and differentiability can also be communicated regarding the derivative. In the event that a function is differentiable at a point, at that point its derivative exists at that point.
• The derivative gives the slant of the tangent line at that point, and the slant of the tangent line can be utilized to decide whether the function is expanding or diminishing at that point.
Types of discontinuities:
• A discontinuity occurs when a function has a break in its graph, either due to a hole, jump, or asymptote.
• A removable discontinuity happens when a function has a gap in its graph, which can be filled in by rethinking the function at that point.
• For illustration, the function f(x) = (x^2 - 4)/(x - 2) includes a removable discontinuity at x = 2, where the function is vague. Be that as it may, we are able to redefine the work as f(x) = x + 2 for x ≠ 2 to fill within the gap and make the function nonstop.
• A jump discontinuity happens when a work has two diverse limits from the left and right sides of a point. For illustration, the work f(x) = {x on the off chance that x < 0, 1 in case x ≥ 0} includes a jump discontinuity at x = 0, since the left-hand restrain is and the right-hand constrain is 1.
• An infinite discontinuity occurs when a function approaches positive or negative infinity at a certain point. For example, the function f(x) = 1/x has an infinite discontinuity at x = 0 since the function grows without bound as x approaches 0.
Applications of limits and continuity:
Limits and continuity are important mathematical concepts that have a variety of applications in data science. Here are a few examples:
1. Derivatives and gradients: Limits are an essential component of calculus, which is the foundation of many data science techniques. Derivatives, which are based on limits, are used to calculate gradients, which are crucial for optimizing models using techniques such as gradient descent.
2. Probability distributions: Continuous probability distributions are often used in data science to model real-world phenomena. For example, the normal distribution is commonly used to model the distribution of heights or weights in a population.
3. Regression analysis: In regression analysis, a continuous variable is predicted based on other variables. The concept of continuity is essential here, as the prediction function must be continuous in order to be useful.
4. Time series analysis: Time series data is a sequence of values that are ordered by time. The continuity of time is an important aspect of time series analysis, as it allows us to model the relationship between past and future values.
5. Signal processing: Many signals, such as audio or video, are continuous in nature. Techniques from calculus, such as Fourier transforms, are used to analyze and process these signals.
Conclusion
In conclusion, limits and continuity are fundamental concepts in calculus that help us understand the behavior of functions as they approach certain values. They are important in many fields, such as physics, economics, and engineering, and are used to model and optimize various systems. The relationship between continuity and differentiability is also important, with differentiability implying continuity. Overall, understanding limits and continuity is crucial for a solid foundation in calculus.
Key takeaways
1. A limit is a value that a function approaches as its input approaches a certain value.
2. Continuity is the property of a function that describes whether it has any sudden jumps or breaks.
3. Evaluating limits can involve different methods, such as direct substitution, factoring, and L'Hopital's rule.
4. Types of discontinuities include removable, jump, and infinite discontinuities.
5. Differentiability implies continuity, but the converse is not necessarily true.
6. Limits and continuity are fundamental concepts in calculus and are important for understanding the behaviour of functions.
Quiz
1. What is the definition of a limit in calculus? a. The value that a function approaches as its input approaches a certain value b. The highest or lowest value of a function over a given interval c. The derivative of a function at a certain point d. The integral of a function over a certain interval
Answer: a. The value that a function approaches as its input approaches a certain value
2. What is a removable discontinuity? a. A discontinuity where the limit of the function approaches infinity b. A discontinuity where the function has a jump or break, but can be fixed by changing the value of the function at that point c. A discontinuity where the function is undefined at a certain point d. A discontinuity where the function is not defined over a certain interval
Answer: b. A discontinuity where the function has a jump or break, but can be fixed by changing the value of the function at that point
3. If a function is differentiable at a certain point, what does that imply about its continuity? a. The function is continuous at that point b. The function is discontinuous at that point c. The function is either continuous or discontinuous at that point, depending on the function d. The function may or may not be continuous at that point, depending on the function
Answer: a. The function is continuous at that point
4. What is L'Hopital's rule used for in calculus? a. To find the area under a curve b. To evaluate limits of indeterminate forms, such as 0/0 or infinity/infinity c. To find the slope of a tangent line to a curve d. To determine whether a function is increasing or decreasing at a certain point
Answer: b. To evaluate limits of indeterminate forms, such as 0/0 or infinity/infinity
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Geometry is branch of mathematics concerned with shapes, sizes and properties of figures. Geometry includes knowledge of angles, lines, triangles, quadrilaterals, circles and polygons.
### Angles
Based on the measurement, angles have been classified into different groups.
Complementary angles: Two angles taken together are said to be complementary if the sum of measurement of the angles equal to 90°. If ∠A + ∠B = 90°, then ∠A is complementary of ∠B and vice – versa.
Supplementary angles: Two angles are supplementary if sum of their measure is 180°. If ∠ A + ∠ B = 180°, then ∠A is supplementary of ∠B and vice – versa.
Linear Pair: Two angle drawn on a same point and have one arm common. If sum of their measure equals to 180°, then they are said to be liner pair of angles.
Adjacent angles: Two angles are adjacent if and only if they have one common arm between them.
### Lines
A line consists of infinite dots. A line is drawn by joining any two different points on a plane. Two different lines drawn can be either parallel or intersecting depending on their nature.
If two lines intersect at a point, then they form two pairs of opposite angles, which are known as vertically opposite angles and have same measure. In the figure, ∠PRQ and ∠SRT are vertically opposite angles. Also ∠QRS and ∠PRT are vertically opposite angles.
Also, ∠x + ∠y = 180° and are Linear pair angles.
Perpendicular Lines
An angle that has a measure of 90° is a right angle. If two lines intersect at right angels, the lines are perpendicular.
Parallel Lines
Two lines drawn on a plane are said to be parallel if they do not intersect each other.
Parallel Lines and Transverse
If a common line intersects two parallel lines L1 and L2, then that common line is known as transverse.
• Pair of corresponding angles = ∠1 & ∠5 and ∠4 & ∠6
• Pair of internal alternate angles = ∠2 & ∠5
• Pair of exterior alternate angles = ∠3 & ∠6
• Vertically opposite angles = ∠3 & ∠4
For parallel lines intersected by the transversal, the pair of corresponding angles, interior alternate angles and exterior alternate angles are equal.
∠1 = ∠5, ∠2 = ∠5, ∠3 = ∠6 and ∠3 = ∠4 |
# Chapter 2 Notes
```Bails – Math 094 Notes
Section 2.1 Introduction to Variables
Variable: letter that represents part of a rule that varies.
Constant: part of a rule that does not change.
Coefficient: the number in front of the variable.
Expression: combination of variables and constants terms separated by operations.
For the following expressions, identify the variable term, coefficient, and constant term.
1.
3x 7
variable term =
coefficient =
constant term =
2.
4y 11
variable term =
coefficient =
constant term =
3.
x 6
variable term =
coefficient =
constant term =
4.
y
variable term =
coefficient =
constant term =
5.
30
variable term =
coefficient =
constant term =
Review of Exponents:
Recall 32 = 3 3 = 9
so
3x2 = 3x x and
Rewrite each expression without exponents.
6.
7.
2x 3
8.
x4y 3
9.
x2y3 = x x y y y
xy 3
9x 2 yz3
Steps for evaluating an expression:
1.
Write the original expression
2.
Copy the original again, but replace the _______________ with a set of parenthesis
3.
Place given value of variable inside parenthesis
4.
Simplify
Evaluate each expression when a 3 , b 1, and c 2
3abc
10.
11.
a3
21
Chapter 2 Understanding Variables and Solving Equations
Evaluate each expression when a 3 , b 1, and c 2
12.
2a2 b2
13.
ab bc
14.
2a c
b
15.
3c 2 5b
3 a
16.
The expression for determining the cost per ounce is c z where c is the total cost
and z is the number of ounces. Evaluate the expression when the total cost of caviar
is \$48 for 16 oz.
17.
The expression for determining the perimeter of a rectangle is 2L + 2W, where L is
the length and W is the width. Evaluate the expression when
a.
The length is 15 centimeters (cm) and the width is 11 centimeters.
b.
The length is 20 feet (ft) and the width is 15 feet.
22
Bails – Math 094 Notes
Section 2.2 Simplifying Expressions
Similar (Like) Terms: have the same variable(s) and exponents. The number in front of the
variable (coefficient) can differ.
Like Terms Examples:
3x, 4x
2xy, -5xy
4y2, 10y2
Unlike Terms Examples:
3x, 3y
3x2, 4x
4x, 5xy
Steps for simplifying an expression:
1.
Rewrite expression without parenthesis
If necessary, make sure to distribute negative through parentheses
2.
Identify like ______________________
3.
Add/subtract the ______________________ (number in front of variable) & copy the variable
**Variables always stay the same in addition and subtraction
4.
Write each answer with the variables in alphabetical order and any constant term last.
Simplify (combine like terms) if possible.
3x 8x
1.
2.
xxx
3.
3y 7y y
4.
2x 2 5 x 2
5.
8 xy 9xy xy
6.
2x 3 4 x 1
7.
3x 5 9x 2
8.
4y 5 x 3y 7 x
9.
3 x 2 4 x x 2 11x
10.
6x 3 7x 2 10x
23
Chapter 2 Understanding Variables and Solving Equations
Review of Associative Property of Multiplication:
(a b) c = a (b c)
Simplify by using the associative property of multiplication.
11.
8 3x =
12.
9 3x 2 =
Review of Distributive Property: a b c ab ac
Note: If there is no number (coefficient) in front of the variable, the number is 1. Therefore, x = 1x
Use the distributive property to simplify each expression.
13.
2 6 x 1 =
14.
3 x 2 =
15.
4 4 x 1 =
16.
5 7 x 3 =
18.
3 x 2 15
Simplify each expression.
2 6x 5 2
17.
19.
4 6 9x 2
20.
6 10 x 1
21.
16 x 6 9 x 2 18
22.
3 5 2x 7 15 x
23.
2 3 x 7 4 8 x 1
24.
5 x 3 2x 9 4 3 x 2 11
24
Bails – Math 094 Notes
Section 2.3 Solving Equations Using Addition
If
A=B
Then
A+C=B+C
A, B, C are algebraic expressions
Adding/subtracting the same number to both sides of an equation keeps the equation balanced.
Note: A solution for an equation is a number that makes the equation a true statement.
Solve each equation and check the answer.
1.
x 7 11
2.
x 7 11
3.
x 7 11
4.
x 7 11
5.
33 27 x
6.
52 x 61
7.
x 17 21 28
8.
16 x 30 14
25
Chapter 2 Understanding Variables and Solving Equations
For each equation, simplify each side (if possible) then solve and check the answer.
9.
x 8 4 11 9
10.
6x 4 5x 3
11.
7 x 8 6 x 3 1
12.
3 6 9 9x 5 8x
13.
6 x 7 x 9 19 8
14.
x 3 7 x 9 x 4 7 9
26
Bails – Math 094 Notes
Section 2.4 Solving Equations Using Division
Division Property of Equality
If A = B, then
A B
C C
as long as C 0
Dividing by the same nonzero number on both sides of an equation keeps the equation balanced.
Solve each equation and check the answer.
1.
6 x 42
2.
5 x 40
3.
8 x 48
4.
121 11x
5.
27 x 0
6.
14 x
For each equation, simplify each side (if possible) then solve and check the answer.
7.
2 x 3 x 17 23
8.
8 x 15 x 71 15
27
Chapter 2 Understanding Variables and Solving Equations
For each equation, simplify each side (if possible) then solve and check the answer.
9.
8 2x 32
10.
60 2 15x
11.
14 x 10 x 36 44
12.
7 x 5 x 7 4 59
13.
82 96 5 4 x 8 2x 3 x
14.
36 36 41 50 x 3 2x 4 4 x
28
Bails – Math 094 Notes
Section 2.5 Solving Equations with Several Steps
Steps for solving equations with variables on both sides:
1.
Simplify each side of equation by removing parentheses and combining like terms
2.
Use _____________________ property to move all variables (letters) to one side.
3.
Use _____________________ property to move all constant terms (numbers) to the
opposite side of the equation.
4.
Use division property to get a _____________________ (the number in front of the
variable) of 1.
5.
Check the solution by going back to the original equation.
Solve each equation and show all work.
1.
5 x 10 55
2.
7 x 4 31
3.
3 x 6 36
4.
42 3 x 7
5.
3 x 2 x 40
6.
11x 4 x 8 15
29
Chapter 2 Understanding Variables and Solving Equations
Solve each equation and show all work.
7.
5 x 3 2x 3
8.
2x 7 4x 1
9.
5 x 10 19 8 x
10.
15 x 1 4 x 20
11.
8 2x 1 6 3 x 5
12.
7 x 8 2 x 13
13.
12 x 2 5 2x 1
14.
x 28 10 4 x 6 27
30
```
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# Quant Quiz on “COORDINATE GEOMETRY” for SSC CGL 2016
For the first time SSC 2016 examination will be held in Computer Based pattern and managing time will be one of the important factor while solving the questions. So in order to make students familiar with such a situations we are providing questions in a time based manner , which will help students to manage the time properly.
1. What is the distance of point of intersection of straight lines 2x+3y=6 and y=x+7 from origin ?
7
3
4
5
Solution:
2. The length intercepted by the straight line 12x-9y=108 between the coordinate axes is
12 unit
18 unit
15 unit
9 unit
Solution:
3. Area of triangle formed by straight lines 4x-3y+4=0,4x+3y-20=0 and x- axis is
3 sq. unit
6 sq. unit
12 sq. unit
24 sq. unit
Solution:
Area =1/2*(difference between x-intercept)*(y-coordinate of point of intersection)
Required Area =1/2 |5-(-1) |×4=12 square unit
4. Area of triangle formed by straight lines 4x-y=4,3x+2y=14 and y-axis is
11/2 sq. unit
11/4 sq. unit
22 sq. unit
11 sq. unit
Solution:
Required Area =1/2 |-4-7|×2=11 square unit
5. Ratio of area of triangle formed by straight lines 2x+3y=4 and 3x-y+5=0 with x-axis and y-axis is
1 : 2
2 : 1
4 : 1
None of these
Solution:
6. Area of quadrilateral formed by straight lines 2x=-5,2y=3,x+1=0 and y+2=0 is
21/2 sq. unit
21/4 sq. unit
21/8 sq. unit
21/16 sq. unit
Solution:
7. Area enclosed by equation |x|+|y|=4 is
16
32
24
48
Solution:
8. For what value of k system of equations 3x+4y=19, y-x=3 and 2x+3y=k has a solution ?
11
-11
14
-14
Solution:
Solving 3x+4y=19 and y-x=3
We get x=1,y=4
Putting (x,y)=(1,4) in 2x+3y=k
We have 2×1+3×4=k
⇒k=14
9. Which of the following pair represent equation of parallel straight lines.
2x+3y=4,4x+6y=9
x+2y=4,2x+y=4
y=3x+5,x=3y+5
None of these
Solution:
In option (a) a1/a2 =b1/b2 ≠c1/c2 . Hence lines given in alternative (a) shows parallel lines.
10. For what value of Ksystem of equation x+3y=K and 2x+6y=2K has infinitely many solution ?
K=1
K=2
for all real values of K
for no real value of K
Solution:
Here a1/a2 =b1/b2 =c1/c2 is always true. It has infinitely many solution for all real values of K.
11. Values of a andb so that system of equations 2x+3y=7 and 2ax+(a+b)y=28 has infinitely many solutions are
a=4,b=8
a=8,b=4
a=-4,b=-8
a=-8,b=-4
Solution:
Required condition is 2a/2=(a+b)/3=28/7
or, a=(a+b)/3=4
∴a=4,a+b=12
or,a=4,b=8
12. Value of k for which system of equations kx+2y=5, 3x+y=1 has unique solution is
k=1
k=2
k=3
All are true
Solution:
System has unique solution if a1/a2 ≠b1/b2
⇒k/3≠2/1
⇒k≠6
so options (a), (b), (c) are correct
13.
a
b
c
d
Solution:
14. Area of triangle formed by the straight line 8x-3y+24=0 and coordinate axes is
24 sq. unit
12 sq. unit
6 sq. unit
18 sq. unit
Solution:
15. Area of triangle formed by straight line y=mx+c with coordinate axes is
a
b
c
d
Solution:
×
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Before getting started, suppose you are blindfolded and asked to pick up a pen from a table. The only clue you are given is that the pen is at a distance of one arm from you. Would you be able to pick it up in one go?
Absolutely not!
Of course, you “may” pick it up but it’s uncertain. You may end up swinging your arm in one arm’s radius of yours.
So now, would you be able to pick up the pen having said that it’s at one arm’s distance in front of you?
Seems possible right?
This is exactly where the application of vectors in our real-life comes into the picture unknowingly. Wondering how? Let me explain.
In the first case when you were told that the pen is at one arm’s distance from you, you only know the magnitude of the distance which is why you end up puzzled in locating the pen.
But in the second case, along with the magnitude of the distance, you know the direction(in front of you) as well which made your task a cakewalk.
Let’s kick-start our learning and make more such problems easier to solve.
Vector
A vector in $\mathbb{R}^n$ is a column of $n$ real numbers. These real numbers are called the components or entries of the vector.
Ex: $\begin{bmatrix}5\\4\end{bmatrix}$ is a vector in $\mathbb{R}^2$
Fig 1: Vector $\begin{bmatrix}5\\4\end{bmatrix}$
%use s2
// define a vector
var v = DenseVector(arrayOf(5.0, 4.0))
println(v)
Output: [5.000000, 4.000000]
A vector in $\mathbb{R}^n$ , say $v$, is drawn as an arrow pointing from the first component in $v$ to the second component in $v$.
Norm of a vector: The norm of a vector $v$$\mathbb{R}^n$ is the length of the associated arrow when drawn as stated above, which may be calculated as the square root of the sum of the squares of $v$’s components. The norm of a vector $v$ is represented as $\lvert v\rvert$. A vector whose norm is 1 is called a unit vector.
Norm of a vector can be calculated as follows:
%use s2
//define a vector to find its norm
var v = DenseVector(arrayOf(3.0, 4.0))
//get the components of the vector
val x = v.get(1)
val y = v.get(2)
//calculate the square root of the sum of the squares of vector's components
val norm = (x.pow(2.0) + y.pow(2.0)).pow(0.5)
print(norm)
Output: 5.0
Vector Operations
Let us get familiar with the fundamental vector operations:
2. Scalar multiplication (multiplication of a real number and a vector).
This is the simple addition of two or more vectors whose summing concatenates them tail-to-head. A vector addition is denoted with “+” between two vectors.
$\begin{bmatrix}1\\2\end{bmatrix}+\begin{bmatrix}4\\2\end{bmatrix} = \begin{bmatrix}5\\4\end{bmatrix}$
%use s2
// define 2 vectors
var v1 = DenseVector(arrayOf(1.0, 2.0))
var v2 = DenseVector(arrayOf(4.0, 2.0))
println(vec_sum)
Output: [5.000000, 4.000000]
Scalar multiplication
Multiplying a vector by a real number is called scalar multiplication. Scalar multiplication is denoted by placing the scalar adjacent to the vector.
Ex: Multiplying a vector, $v = \begin{bmatrix}2\\1\end{bmatrix}$ with a real number, $s = 2$ is written as: $2\begin{bmatrix}2\\1\end{bmatrix} = \begin{bmatrix}4\\2\end{bmatrix}$
Fig 3: Scalar Multiplication
%use s2
// define a vector
var v = DenseVector(arrayOf(2.0, 1.0))
//define a scalar
val s = 2.0
// B = v*s
val B = v.scaled(s)
println(B)
Output: [4.000000, 2.000000]
Note:
1. Multiplying a vector by a positive real number k preserves its direction and multiplies its norm by k.
2. Multiplying a vector by a negative real reverses its direction and multiplies its norm by k.
3. Multiplying a vector by -1 reverses its direction and preserves its norm.
Let us illustrate an example implementing the operations involved in the concepts explained above.
Q: Simplify $4\begin{bmatrix}1.1\\2.2\end{bmatrix}-\begin{bmatrix}1.0\\2.4\end{bmatrix}$
%use s2
// define 1st vector
var m = DenseVector(arrayOf(1.1, 2.2))
//define scalar to multiply
val scalar = 4.0
//define 2nd vector
var n = DenseVector(arrayOf(1.0, 2.4))
// Scalar Multiplication
val B = m.scaled(scalar)
//Subtracting
val sol = B.minus(n)
println(sol)
Output: [3.400000, 6.400000]
Ever played games like Catapult Quest or Angry Birds? If you keenly observe, in games, vectors are used to store positions, directions and velocities just like the one below.
• The position vector indicates how far the object is.
• The velocity vector indicates the time taken or the amount of force that must be given.
• The direction vector indicates the direction in which the force should be applied.
Linear Combination
Let us define a list of vectors $v_1, v_2,..., v_k$. The linear combination of these vectors is expressed as:
$c_1v_1+c_2v_2+...+c_kv_k$
where $c_1, c_2,.., c_k$ are real numbers.
These $c$‘s are called the weights of the linear combination.
%use s2
// define vector v1
var v1 = DenseVector(arrayOf(1.0, 2.0))
//define c1
val c1 = 3.0
//Repeat the same for required number of terms
var v2 = DenseVector(arrayOf(3.0, 3.0))
val c2 = 1.0
var v3 = DenseVector(arrayOf(5.0, 3.0))
val c3 = 2.0
// Scalar Multiplication
val A = v1.scaled(c1)
val B = v2.scaled(c2)
val C = v3.scaled(c3)
println(sol)
Output: [16.000000, 15.00000]
Let us illustrate the same using graph for a better understanding.
Fig 4: Linear Combination of vectors
Note: An integer linear combination is a linear combination in which all the weights (c’s) are integers.
Span
The span of a given list of vectors is defined as the set of all vectors which can be written as the linear combination of the given list of vectors.
Ex: $\begin{bmatrix}x\\y\end{bmatrix}$ is said to be in the span of the vector-list $\left\{\begin{bmatrix}x_1\\y_1\end{bmatrix},\begin{bmatrix}x_2\\y_2\end{bmatrix}, \begin{bmatrix}x_3\\y_3\end{bmatrix}\right\}$ if:
$\begin{bmatrix}x\\y\end{bmatrix} = \alpha \begin{bmatrix}x_1\\y_1\end{bmatrix}+\beta \begin{bmatrix}x_2\\y_2\end{bmatrix}+\gamma \begin{bmatrix}x_3\\y_3\end{bmatrix}$
where $\alpha, \beta, \gamma$$\mathbb{R}$.
Let us solve a problem to improve our understanding.
Q: Determine whether the vector $\begin{bmatrix}19\\10\\-1\end{bmatrix}$ lies in the span of set of vectors: $S = \left\{\begin{bmatrix}3\\-1\\2\end{bmatrix}, \begin{bmatrix}-5\\0\\1\end{bmatrix}, \begin{bmatrix}1\\7\\-4\end{bmatrix}\right\}$.
Approach:
Let us first assume the equation: $c_1v_1+c_2v_2+c_3v_3 = v$, where $c_1, c_2, c_3$ are real numbers and try to find the solution $[c_1, c_2, c_3]$ if exists.
%use s2
// define vectors v1,v2,v3,v
var v1 = DenseVector(arrayOf(3.0, -1.0, 2.0))
var v2 = DenseVector(arrayOf(-5.0, 0.0, 1.0))
var v3 = DenseVector(arrayOf(1.0, 7.0, -4.0))
var v = DenseVector(arrayOf(19.0, 10.0, -1.0))
//we need to determine whether there exists a solution for the equation: c1v1+c2v2+c3v3 = v
//for real numbers c1,c2,c3.
// Create a 3x3 matrix for v1,v2,v3
val A = DenseMatrix(arrayOf
(doubleArrayOf(3.0, -5.0, 1.0),
doubleArrayOf(-1.0, 0.0, 7.0),
doubleArrayOf(2.0, 1.0, -4.0)))
//now we need to find whether there exists a solution for Ac=v where c = {c1, c2, c3}
//and if yes, print [c1, c2, c3]
// Create a solver for linear system
val precision = 1e-15
val solver = LinearSystemSolver(precision)
// Solve for Ac
val soln = solver.solve(A)
// solution for Ac = v
val c = soln.getParticularSolution(v)
println("soln is: $c") // verification val Ac = A.multiply(c) // Ac = v println("Ac =$Ac, same as \$v")
soln is: [4.000000, -1.000000, 2.000000]
Ac = [19.000000, 10.000000, -1.000000] , same as [19.000000, 10.000000, -1.000000] |
# Arc of a Circle
Concept
From our previous course on radian , we realise the radian = .We will need this information to find the arc of a circle. Let us look first at how to find the arc of a circle by using its radian.
Let us consider a circle with centre O and radius r.
We compare between the lengths and angles subtended of the arc and circle.
Hence the length of minor arc AB,
## Area of a Sector
We shall look at how to calculuate the area of a sector ,using radian and radius. The area of a sector is also directly proportional to the angle within the sector.
Example
Before the start of doing or looking through any of these questions, it would be good to draw labelled diagrams to give you a better understanding.
a) The radius of a circle with centre O is 5cm and the sector AOB subtends an angle of 2 radians. What is the length of its arc AB?
= 5(2) = 10 cm
The length of arc AB is 10 cm.
b) The circle with centre O has radius 3 cm and the angle which sector COD subtends is . Find the area of sector COD.
Remembering the formula for area of sector of a circle ,
A =
=
= 18cm
'Try it Out Yourself' Section
Figure out the area and arc of a circle's sector by trying the below exercises.
a) The length of arc AB is 12cm . Given that the centre is at O and that the radius of OA is 6cm , find the radian OAB.
b) The circle with centre O has radius 4 cm and the angle which sector COD subtends is . Find the area of sector COD. |
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# What is the difference between sine law and sine ratio?
## What is the difference between sine law and sine ratio?
The cosine rule relates the cosine of an angle of a triangle to the sides of the triangle. With its help , the angles of a triangle can be determined , if all its sides are known. The sine rules gives the ratio of the sine of two angles of a triangle, which equals to the ratio of the corresponding opposite sides.
How do you remember the law of sines and cosines?
How to Remember
1. think “abc”: a2 + b2 = c2,
2. then a 2nd “abc”: 2ab cos(C),
3. and put them together: a2 + b2 − 2ab cos(C) = c.
Can you use law of sines with law of cosines?
Use the law of cosines when you are given SAS, or SSS, quantities. For example: If you were given the lengths of sides b and c, and the measure of angle A, this would be SAS. SSS is when we know the lengths of the three sides a, b, and c. Use the law of sines when you are given ASA, SSA, or AAS.
### What is the difference between sine and cosine?
Sine and cosine — a.k.a., sin(θ) and cos(θ) — are functions revealing the shape of a right triangle. Looking out from a vertex with angle θ, sin(θ) is the ratio of the opposite side to the hypotenuse , while cos(θ) is the ratio of the adjacent side to the hypotenuse .
What is cosine law used for?
The cosine law is used to determine the third side of a triangle when we know the lengths of the other two sides and the angle between them.
What is the cosine law used for?
The Law of Cosines is used to find the remaining parts of an oblique (non-right) triangle when either the lengths of two sides and the measure of the included angle is known (SAS) or the lengths of the three sides (SSS) are known.
#### What do you use law of cosines for?
What do you use Law of Cosines for?
What is the phase difference between sine and cosine?
So the cosine wave is 90 degrees out of phase behind the sine wave or 270 degrees out of phase in front of the sine wave.
## What is the relationship between sine and cosine?
The sine of an angle is equal to the cosine of its complementary angle, and the cosine of an angle is equal to the sine of its complementary angle.
How to calculate law of sines?
– You only know the angle α and sides a and c; – Angle α is acute ( α < 90° ); – a is shorter than c ( a < c ); – a is longer than the altitude h from angle β, where h = c * sin (α) (or a > c * sin (α) ).
What are the rules of Sine and cosine?
The sine rule can be used to find an angle from 3 sides and an angle, or a side from 3 angles and a side. The cosine rule can find a side from 2 sides and the included angle, or an angle from 3 sides.
### How can one prove the law of cosines?
Proof of the Law of Cosines The Law of Cosines states that for any triangle ABC, with sides a,b,c For more see Law of Cosines. In the right triangle BCD, from the definition of cosine: or, Subtracting this from the side b, we see that In the triangle BCD, from the definition of sine: or In the triangle ADB, applying the Pythagorean Theorem
When to use the law of sines?
a/sin (A) = b/sin (B)
• 6/sin (45) = b/sin (75)
• 6sin (75) = bsin (45)[cross multiply to eliminate fractions]
• 6 (0.9659) = b√2/2[approximate value for sine of 75 degrees]
• 6 (0.9659)*2/√2 = b
• 8.196 = b |
# The Proof Course: Lecture 2
Many real-life situations lead us to considering a mathematical problem dealing with finding all possible numbers $$x$$ satisfying a certain formula. In most primitive cases, this formula is an equation involving basic arithmetic operations (like the one we considered in Lecture 1). As an example of a formula that does not fall in this category, consider the following one:
$$x<y^2$$ for every value of $$y$$ (Formula A)
In other words, the formula expresses the property that no matter what value of $$y$$ we pick, we will always have $$x<y^2$$. Let us write this purely symbolically as follows (so that it looks more like a formula!):
$$y\Rightarrow x<y^2$$ (symbolic form of Formula A)
In general, the symbol "$$\Rightarrow$$" describes logical implication of statements. Here the implication is: if $$y$$ has a specific value then $$x<y^2$$. In the symbolic form above, the assumption that $$y$$ has a specific value is expressed by just writing $$y$$ on the LHS (left-hand-side) of the implication symbol "$$\Rightarrow$$". Since we are not giving any further detail as to which specific value does $$y$$ have, the implication must not be dependent on such detail, and hence the RHS (right-hand-side), $$x<y^2$$, must hold for all values of $$y$$. Note however that this type of symbolic forms, where variables are allowed to be written on their own like in the LHS of the implication symbol above, is not a standard practice. We will nevertheless stick to it, as it makes understanding proofs easier.
So, what is the solution of Formula A? If $$x<y^2$$ needs to hold for every value of $$y$$, then in particular, it must hold for $$y=0$$, giving us $$x<0^2=0$$. This can be written out purely symbolically, as a proof:
1. $$y\Rightarrow x<y^2$$
2. $$x<0^2$$
3. $$x<0$$
However, as we know from Lecture 1 already, this proof only proves that if Formula A is true then $$x<0$$. In order for $$x<0$$ to be the solution of Formula A, we also need to prove that if $$x<0$$ then Formula A is true. Well, since $$0\leqslant y^2$$ is true for every $$y$$, combining $$x<0$$ with $$0\leqslant y^2$$ we will get $$x<y^2$$, as required in Formula A. So the proof is:
1. $$x<0$$
2. $$y\Rightarrow 0\leqslant y^2$$
3. $$y\Rightarrow x<y^2$$
Note that it seems as if this proof violates our requirement that in a basic proof, every line except the first one must be a logical conclusion of the previous one or several lines. Line 2 does not necessarily seem to be a conclusion of Line 1. Instead, it is simply a general true fact that does not seem to logically depend on Line 1 at all: it says that the square of every number is greater or equal to $$0$$. We can account for such situations by agreeing that "several" in "one or several lines" includes the case of "$$0$$ many". So in a basic proof we can also include lines that recall facts we know. If we had not done that in the above proof, we would have to skip from Line 1 directly to Line 3, and it may not have been so clear how does one logically conclude Line 3 from Line 1. So we allow inclusion of known facts as lines in a basic proof for the sake of clarity. Knowing this, we might want to make the first proof clearer by inserting one such line:
1. $$y\Rightarrow x<y^2$$
2. $$x<0^2$$
3. $$0^2=0$$
4. $$x<0$$
# The Proof Course: Lecture 1
In this blog-based lecture course we will learn how to build mathematical proofs.
Let us begin with something simple. You are most likely familiar with "solving an equation". You are given an "equation", say $x+2=2x-3$ with an "unknown" number $$x$$ and you need to find all possible values of $$x$$, so that the equation holds true. You then follow a certain process of creating new equations from the given one until you reach the solution: $2+3=2x-x$ $5=x$ This computation is in fact an example of a proof. To be more precise, there are two proofs here: one for proving that
if $$x+2=2x-3$$ then $$x=5$$ (Proposition A),
and the other proving that
if $$x=5$$ then $$x+2=2x-3$$ (Proposition B).
The first proof is the same as the series of equations above. The second proof is still the same series, but in reverse direction. The two Propositions A and B together guarantee that not only $$x=5$$ fulfills the original equation (Proposition B), but that there is no other value of $$x$$ that would fulfill the same equation (Proposition A). It is because of the presence of these two proofs in our computation that we can be sure that $$x=5$$ is indeed the solution of the equation $$x+2=2x-3$$.
In general, a proof is a series of mathematical formulas, like the equations above. However, in addition to a "vertical" structure of a proof, where each line displays a formula that has been derived from one or more previous lines, there is also a "horizontal" structure, where each line of a proof has a certain horizontal offset. This is, at least, according to a certain proof calculus formulated by someone by the name of Fitch. There are other ways of defining/describing proofs; in fact, there is an entire subject of proof theory, which studies these other ways. We will care little about those other ways and stick to the one we started describing, as it is closest to how mathematicians actually compose proofs in their everyday job.
So where were we? We were talking about "vertical" and "horizontal" structure of a proof. Not to complicate things too much at once, let us first get a handle on the vertical structure of proofs, illustrating it on various example proofs that have most primitive possible horizontal structure. We will then, slowly, complexify the horizontal structure as well.
For Proposition A, the proof goes like this:
1. $$x+2=2x-3$$
2. $$2+3=2x-x$$
3. $$5=x$$
The numbers at the start of each line are just for our reference purposes, they do not form part of the proof. Line 2 is a logical conclusion of Line 1: if $$x+2=2x-3$$ then it must be so that $$2+3=2x-x$$, since we could add $$3$$ to both sides of the equality and subtract $$x$$ as well – a process under which the equality will remain true if it were true at the start.
Line 3 is (again) a logical conclusion of Line 2: since $$5=2+3$$ and $$2x-x=x$$, so if the equality in Line 2 were true then the equality in Line 3 must be true as well.
A series of lines of mathematical formulas where every next line is a logical conclusion of the previous one or more lines, is a mathematical proof with simplest possible horizontal structure. We will call such proofs "basic".
Proposition B also has a basic proof:
1. $$5=x$$
2. $$2+3=2x-x$$
3. $$x+2=2x-3$$
Just as before, every next line is a logical conclusion of the previous one.
What about the first line (in each proof)? If the first line were to also satisfy the requirement that it is a logical conclusion of the previous lines, then, since there are no lines before the first line, it would appear that the first line is true on its own, without a need for justification. If course, in both proofs this is false: in the first proof, we cannot claim that Line 1 is true. Truth of Line 1 in the first proof depends on the value of $$x$$. Without knowing anything about the value of $$x$$, we cannot claim that $$x+2=2x-3$$, since if, say, $$x=0$$, then $$x+2=2x-3$$ is clearly false. The same for the second proof - we cannot claim that Line 1 is true. Instead, the role of the first line in each of the proofs is to "assume" they are true, and then see what conclusions can be drawn from such assumption. Recall that Proposition B, for instance, states that if $$x=5$$ then $$x+2=2x-3$$. It does not state that
$$x=5$$ and $$x+2=2x-3$$,
or that
$$x=5$$ or $$x+2=2x-3$$,
and so on. So in a basic proof the first line will always be an assumption, unlike the rest of the lines, which are conclusions from the previous one or several lines.
# Pure Mathematics: Job Description
What does a "pure mathematician" do? A shoemaker makes shoes, a musician makes music, an applied mathematician uses mathematics to solve some real-life problems... Each of these job descriptions have some sort of measurable output. What is such output for a pure mathematician?
Some will say that a pure mathematician solves problems in mathematics, i.e., mathematical problems that are not necessarily related to "real life". This does not do justice to the efforts of a pure mathematician: if you are keen to solve problems, rather solve real-life problems! The problem is that the language in which these "pure" mathematical problems are solved is such that it cannot (always) be used to solve the "real-life" problems. A pure mathematician wants to solve only those problems whose solutions are expressed in a pure mathematical language. This does not do justice to the efforts of a pure mathematician either: what a picky attitude! Besides, solve-a-problem style job description applies to every other job. Indeed, any job for which you expect to get paid requires some sort of problem-solving.
The job description of a pure mathematician is actually quite straightforward. A pure mathematician builds "proofs". A proof is a discussion that reaches a certain conclusion with a life-time guarantee of truthfulness of this conclusion. In no other discipline are you able to establish proofs with such a guarantee. Surely having a certainty in a certain fact is a useful thing in any area of life. Unfortunately though, as soon as your conclusions come close to describing how something in "real life" works, their certainty can no longer be guaranteed, i.e., they step out of the reach of pure mathematics. Still, pure mathematics is extremely useful in establishing the real-life-like close-to-certain conclusions, otherwise the disciplines such as applied mathematics, physics, chemistry, and many others, would hardly make any progress (for those who may not be aware of this, these disciplines, as well as many others, rely a lot on conclusions proved in pure mathematics).
The conclusions that a proof proves are called "theorems". Then there are "definitions", which are essentially shortcuts for building complex proofs. Now a proof starts with certain assumptions (always, in fact, for those who may have been deceived that unlike religion, science does not rely on unproved assumptions, but this is a topic for an entirely different discussion...). The universal assumptions, i.e., those that are used over and over in many different proofs, are called "axioms". Part of the task of a pure mathematician is coming up with appropriate definitions and axioms. In the end, they are to be used in a proof, otherwise, they are useless. Solving a pure mathematical problem is all about finding a proof: of a theorem, its negation, or if the theorem has not been precisely stated, finding a precise statement and then its proof. So fair and square, a pure mathematician is someone who builds proofs! |
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# How to Create a Spirallic Numbers Dataset
Do you want to make a (mysterious) spirallic numbers data set? This article will teach you how!
### Part 1 The Tutorial
1. 1
2. 2
Make the dataset by opening a new Excel worksheet and choosing a cell inwards and downwards about 10 rows and columns to start inputting the numbers into. You may start with 0 or 1 or any number; 1 is recommended.
3. 3
Proceed sequentially outwards in a spiral. If you want to save a little time, use a formula like "=J15+1" and copy it up, across or down off as many cells as need be. You may then find a like formula nearby as you go outwards, which will speed things up greatly. This will also allow you to use any originating number in the center, should you so choose..
4. 4
After you have built the data set up to 100 or 324 (as in the example), you're good to go!
5. 5
You will notice right away that the prime numbers are fairly evenly distributed in the 4 quadrants at this point. You might try blocking in non-primes in groups in patterns to see if you can predict primes!
6. 6
Notice there are a number of sequences to the columns and numbers which follow Summation Notation.
7. 7
Notice also that the first sequence is easy -- it's the diagonal from the center to the bottom right. Its pattern to produce the sequence 1,3,13,31,57,91,133,183,241,307, ... is -1*0+1=1, 1*2+1=3, 3*4+1 =13, 5*6+1 = 31, etc. You will find a similar pattern extending to the upper left.
• Notice that the second pattern is trickier: the vertical pattern that produces the sequence 1,4,15,34,61,96,139,190,249,316 is more involved. If one takes the difference of those numbers and then the differences of the differences, one gets 8's, so one knows that a constant value of 8 is being added in the sequence. Knowing that, a typical rate of change is produced by ((a*b)+c)+8 = result, with a being constant. Such is the case here as it turns out, for the sequence is produced by ((4*-2)+4+8)=4, ((4*1)+3+8)=15, ((4*6)+2+8)=34, ((4*13)+1+8) = 61, etc. with the increment to b being 3,5,7,... and the increment to c being -1, while a=4 and 8 are constants. There's an elegant way to write that in Summation Notation with N's and K's and i's and all but it's beyond the text capacity we're operating under. Actually, I just found another way to resolve the series: 1==((-1*0)-7)+8, 4=((1*2)-6)+8, 15=((3*4)-5)+8, 34=((5*6)-4)+8, 61=((7*8)-3)+8, 96=((9*10)-2)+8, etc.
8. 8
Notice the horizontal pattern is a bit trickier to discover. 8's were again a constant difference of the differences but it took me awhile to find the rest of the formula. The sequence of 11,28,53,86,127,176,233,298, ... is generated as far as I know by ((4*-5)+23)+8=11, ((4*-1)+24)+8 = 28, ((4*5)+25)+8 = 53, ((4*13)+26)+8 = 86, etc., with a=4 and 8 remaining constant while b increments by 4, 6, 8,... and c incrementing by 1. Again, there's a more elegant way to state that. If anyone happening to read this can help out, great. There's another way to resolve this sequence also: 2=((2*3)-12)+8, 11=((4*5)-17)+8, 28=((6*7)-22)+8, 53=((8*9)-27)+8, 86=((10*11)-32)+8, 127=((12*13)-37)+8, 176=((14*15)-42)+8, 233=((16*17)-47)+8, and 298 =((18*19)-52)+8, etc. with element c decrementing by -5 each time.
9. 9
Notice that the diagonals take the form of ((a*b)+c)+d = Sum wherein b_sub_0 is 1 greater than a_sub_zero, a_sub_1 was 2 greater than a_sub_0 and b_sub_1 was 2 greater than b_sub_0, i.e. (1*2) changed next to (3*4). This might be the general pattern instead of working with 4's as had been, The role of d=8, the difference of the differences, was still present, so then you just need to figure out which a and b pair best approximated the answer and adjust with c accordingly, by trial and error. It did not take long to hit upon the solutions.
10. 10
There are resolved 2 more sequences: 5=((2*3)-9)+8, 18=((4*5)-10)+8, 39=((6*7)-11)+8, etc. and also 8==((2*3)-6)+8, 23=((4*5)-5)+8, 46=((6*7)-4)+8, 77=((8*9)-3)+8, 116=((10*11)-2)+8,163=((12*13)-1)+8, 218=((14*15)-0)+8 and 281=((16*17)+1)+8. These were done by the analytical method described above in Step 9, though it's a good hunch the problem will only yield to greater finesse than the lazy mathematician possesses.
### Part 2 Explanatory Charts, Diagrams, Photos
1. 1
Review the spirallic dataset again to see if you can spot any other sequences in it -- how about the Fibonacci Series? Has anyone hi-lited that yet? Might be interesting! #
### Part 3 Helpful Guidance
1. 1
Make use of helper articles when proceeding through this tutorial:
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## Tips
• Nature uses spirals in building all sorts of forms, from organs to the abdomens of wasps, etc. -- all the way down to the stinger I imagine! That's the information I have from a Father (Priest) who lectured on Nature at a local college; I do not know so from firsthand microscopic observation however.
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Ex.3.2 Q6 Understanding Quadrilaterals Solution - NCERT Maths Class 8
Go back to 'Ex.3.2'
Question
(a) What is the minimum interior angle possible for a regular polygon? Why?
(b) What is the maximum exterior angle possible for a regular polygon?
Video Solution
Ex 3.2 | Question 6
Text Solution
What is Known?
We know polygons according to the number of sides (or vertices) they have.
What is Unknown?
Minimum interior angle possible for a regular polygon.
Maximum exterior angle possible for a regular polygon.
Reasoning:We know that the sum of measure of interior angle of triangle is $$={\rm{18}}0^\circ$$
Equilateral triangle is regular polygon having maximum exterior angle because it consists of least number of sides.
Steps:
(a) What is the minimum interior angle possible for a regular polygon? Why?
Consider a regular polygon having the least number of sides (i.e., an equilateral triangle).
We know Sum of all the angles of a triangle $$= {\rm{18}}0^\circ$$
\begin{align}x + x + x{\rm{ }}& = {\rm{ 18}}0^\circ \\{\rm{3}}x &= {\rm{18}}0^\circ \\x &= \frac{{{\rm{18}}0^\circ }}{3}\\x &= {\rm{6}}0^\circ\end{align}Thus, minimum interior angle possible for a regular polygon$$= 60^\circ$$
(b) What is the maximum exterior angle possible for a regular polygon?
We know that the exterior angle and an interior angle will always form a linear pair. Thus exterior angle will be maximum when interior angle is minimum.
Exterior angle
\begin{align}&= {\rm{18}}{0^{\rm{\circ}}} - \,\,{\rm{6}}0^\circ \\&={120^{\rm{\circ}}}.\end{align}
Therefore, maximum exterior angle possible for a regular polygon is $$120^\circ$$
Equilateral triangle is a regular polygon having maximum exterior angle because it consists of least number of sides.
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Lesson 6ª
PARENTHESIS, BRACKETS AND KEYS.
So far, we have used parentheses (...) and brackets [...] to separate specific operations. Remember that you have done several exercises with them: for example: 2[2+(3+6)-12]
For one parenthesis you open, there must be another one to close it. You also need to do this with brackets and keys.
Sometimes, we have the need to separate manu operations in the same exercise. We use keys {...} to cover these need. Keys work in the same way as parenthesis and brackets; they separate values.
If you have parentheses, brackets, and keys in the same exercise, you must start with the operations between parenthesis, then, solve the ones in brackets, and finally those operations inside keys. For example:
a) 2 {[3+2 (4-3)-2 (6-8)-5]} = 2 {[3+2 1-2 (-2)-5]} =
= 2 {[3+2+4-5]} = 2 {4} = 8
b) 3{4(2-7)-11-2[3(-5-1)]} = 3{4 (-5)-11-2[3 (-6)]}=
= 3{-20-11-2[-18]} =3{-31+36} = 3{5} = 15
Try to solve these three exercises on your own:
a) -3(4-6)-2{5[3-5(-7+5)-3]} the answer is: -94
b) -3{-4[-5(-6 x2-7)]} the answer is: 1140
c) 4-2{3[5-2(5-6)-7]+10}+17 = the answer is: 1
In case you've had any problems, here are the solutions:
a) -3 (-2)-2{5[3-5 (-2)-3]} = 6-2{5[3+10-3]} =
= 6-2{5 (10)]} = 6-2{50} = 6-100 = -94
b) -3{-4[-5(-6x 2-7)]} = -3{-4[-5(-12-7)]} =
= -3{-4[(-5 (-19)]} = -3{-4[95)]} =
= -3{-4 95} = -3{-380} = 1140
c) 4-2{3[5-2(5-6)-7]+10}+17 = 4-2{3[5-2(-1)-7]+10}+17 =
= 4-2{3[5+2-7]+10}+17 = 4-2{3 0+10}+17=
= 4-2{0+10}+17 = 4-2 10+17 = 4-20+17 = 21-20 = 1
Exercises
A) A person receives a monthly salary of 1400 € and another person, 48 € a day. What is the difference between the total amounts received by each person at the end of the year?
Solution:
The first person will receive in a year: 1400 x 12 = 16800
In a 365 day year, the second person will receive: 365 x 48 = 17520
The difference, in favour of the second person, is: 17520 – 16800 = 720
B) One car has ran 515 kilometres in 5 hours. Another car has traveled 749 kilometres in 7 hours. After having traveled for 10 hours, how many kilometres has each car covered?
Answer: The first one has traveled 1030 km. and the second, 1070km.
C) Two people work together in a construction site. The first one earns 5 € a day more than the second person. After several work days, the first one earned 450 € and the second person, 400 €. 1) How many days did they work? and 2) What is their daily wager?
Answers: 1) They worked for 10 days. 2) Their daily wagers are 45 and 40 €
Solution:
The exercise states that the first worker earns 5 € a day more than the second one.
After working for several days, the first one earns 450 € and the second one earns 400 €. This means that during the days they have worked, the first one earns: 450 – 400 = 50 € more than the second worker.
If the first one earns daily 5 € more than the second worker, we could deduct the amount of days they have worked. We simply divide:
If the first one has earned 450 € and the second 400 €, their wager (the total they have earned per day) will be:
D) Three friends decide to weight themselves. The first and second weighed 143 kilos combined; the second and third weighed 145 kilos combined; the first and third weighed 144 kilos combined. How much does each one weighs?
Answer: the first one 71 kilos, the second 72 kilos, and the third 73 kilos
This exercise isn't easy, so we will explain it. Try to understand it correctly.
Solution:
Before we begin solving it, lets explain a couple of things:
1) We can add and subtract anything that is homogeneous (of the same species), for example: bags of potatoes with bags of potatoes; kilos of onions with kilos of onions; Euros with Euros, etc.
Example: 5 bags of potatoes + 4 bags of potatoes = 9 bags of potatoes.
3 Euros + 9 Euros = 12 Euros.
2 first positions + 4 first positions = 9 first positions.
2 times the first one + 6 times the first one = 8 times the first one
2) If they ask us to divide ( 4 + 8 + 12 ) into two:
We have to divide each one of its segments, parts or terms:
4:2 +8:2 + 12:2
It is the same to add: 4 + 8 + 12 = 24 and then to divide it by 2; 24:2= 12;
to add the quotients of 4:2 +8:2 + 12:2 = 2 + 4 + 6 = 12
Following what was stated in the exercise, we can write:
the first one + the second one = 143
the second one + the third one = 145
the first one + the third one = 144
If we add all of them, we have:
the first one + the first one + the second one + the second one+ the third one + the third one = 143 + 145 + 144
We have: the first one two times + the second one two times + the third one two times = 432
The word "times" means multiply: for example, 4 times 3 = 12
The previous sum: the first one (1°) + the first one + the second one + the second one+ the third one + the third one = 143 + 145 + 144
We could write it:
We divide each term by 2 (because each segment is in the exercise twice) and we get:
We know that the first one + the second one + the third one = 216; and the first one + the second one = 143
If we subtract both expressions, we can calculate the weight of the third friend:
We find out that the third friend weighs 73 kilos.
We know that the first one + the second one + the third one = 216; and the second one + the third one = 145
If we subtract both expressions, we can calculate the weight of the first friend:
We find out the the first friend weighs 71 kilos.
We know that the first one + the second one + the third one = 216; and the first one + the third one = 144
If we subtract both expressions, we can calculate the weight of the second friend:
We find out that the second friend weighs 72 kilos.
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# Difference between a Rhombus and a Parallelogram
Updated on May 31, 2017
We may not normally use the words “rhombus” and “parallelogram,” but we actually see these shapes every day. Do you still remember what they are? For sure, these were discussed in school, but now we ask, “What exactly are they and how are they different from each other?” This article will explore the difference between a rhombus and a parallelogram.
## Descriptions
A rhombus, also known as “rhom” or “diamond,” is an equilateral quadrilateral, a term that refers to a figure with four parallel sides (the lines will never intersect even if they continue) with equal lengths. If you take a look at the picture above, you will notice that all sides are the same length (8 cm).
All opposing angles of a rhombus have equal lengths and its adjacent angles are supplementary angles (which means the sum of the two angles is 180°). Moreover, a rhombus’s diagonals are perpendicular to each other; they bisect each other at right angles.
A parallelogram, on the other hand, is a type of quadrilateral. Its opposing lines are parallel and of equal length. If you take a look at the picture above, you will notice that two opposing lines are 15 cm and the other two are 8 cm.
A parallelogram’s adjacent angles are supplementary and its consecutive angles are equal. Its diagonals bisect each other at an intersection. Examples of a parallelogram are rectangles, squares, and rhombuses.
## Rhombus vs Parallelogram
What, then, is the difference between a rhombus and a parallelogram?
Both a rhombus and a parallelogram are quadrilateral figures, which means both are four-sided figures. The main difference is that a parallelogram has two parallel opposing sides which are of equal lengths, whereas a rhombus has four sides that all have equal lengths (all four sides are parallel). In other words, a rhombus is a type of parallelogram with equal sides.
## Comparison Chart
Rhombus Parallelogram An equilateral quadrilateral; has four parallel sides with equal lengths Has parallel opposing lines that are of equal length
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# Ex.14.1 Q7 Statistics Solution - NCERT Maths Class 10
## Question
To find out the concentration of $$SO_2$$ in the air (in parts per million, i.e., ppm), the data was collected for $$30$$ localities in a certain city and is presented below:
Concentration of $$SO_2$$ (in ppm) Frequency $$0.00 - 0.04$$ $$4$$ $$0.04 - 0.08$$ $$9$$ $$0.08 - 0.12$$ $$9$$ $$0.12 - 0.16$$ $$2$$ $$0.16 - 0.20$$ $$4$$ $$0.20 - 0.24$$ $$2$$
Find the mean concentration of $$SO_2$$ in the air.
Video Solution
Statistics
Ex 14.1 | Question 7
## Text Solution
What is known?
The Concentration of $$SO_2$$ in the air (in parts per million, i.e., ppm) for $$30$$ localities in a certain city:
What is unknown?
The mean concentration of $$SO_2$$in the air
Reasoning:
We solve this question by step deviation method.
Mean, $$\overline x = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h$$
Steps:
We know that,
Class mark,$${x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}$$
Class size$$, h = 0.04$$
Taking assumed mean$$,a = 0.14$$
Content of SO2 Frequency $$(f_i)$$ $$x_i$$ $$d_i = x_i -a$$ \begin{align}{u_i}=\frac{ d_i}{h}\end{align} $$f_iu_i$$ $$0.00-0.04$$ $$4$$ $$0.02$$ $$-0.12$$ $$-3$$ $$-12$$ $$0.04-0.08$$ $$9$$ $$0.06$$ $$-0.08$$ $$-2$$ $$-18$$ $$0.08-0.12$$ $$9$$ $$0.10$$ $$-0.04$$ $$-1$$ $$-9$$ $$0.12-0.16$$ $$2$$ $$0.14(a)$$ $$0$$ $$0$$ $$0$$ $$0.16-0.20$$ $$4$$ $$0.18$$ $$0.04$$ $$1$$ $$4$$ $$0.20-0.24$$ $$2$$ $$0.22$$ $$0.08$$ $$2$$ $$4$$ $$\Sigma f_i=30$$ $$\Sigma f_iu_i=-31$$
From the table, we obtain
$\begin{array}{l}\sum {{f_i} = 30} \\\sum {{f_i}{u_i}} = - 31\end{array}$
\begin{align} \text { Mean }(\overline{{x}}) &={a}+\left(\frac{\sum {f}_{{i}} {u}_{{i}}}{{f}_{{i}}}\right) {h} \\ \overline{{x}} &=0.14+\left(\frac{-31}{30}\right) 0.04 \\ \overline{{x}} &=0.14-| 0.04133 \\ \overline{{x}} &=0.099 \end{align}
The mean concentration of $$SO_2$$ in the air is $$0.099$$.
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Geometry Mathematics 2 Set A 2014-2015 SSC (English Medium) 10th Standard Board Exam Question Paper Solution
Geometry Mathematics 2 [Set A]
Date: March 2015
Duration: 2h
[5] 1 | Solve any five sub-questions :
[1] 1.1
In the following figure seg AB ⊥ seg BC, seg DC ⊥ seg BC. If AB = 2 and DC = 3, find (A(triangleABC))/(A(triangleDCB))
Concept: Properties of Ratios of Areas of Two Triangles
Chapter: [0.01] Similarity
[1] 1.2
Find the slope and y-intercept of the line y = -2x + 3.
Concept: Intercepts Made by a Line
Chapter: [0.05] Co-ordinate Geometry
[1] 1.3
In the following figure, in ΔABC, BC = 1, AC = 2, ∠B = 90°. Find the value of sin θ.
Concept: Heights and Distances
Chapter: [0.06] Trigonometry
[1] 1.4
Find the diagonal of a square whose side is 10 cm.
Concept: Surface Area and Volume of Three Dimensional Figures
Chapter: [0.07] Mensuration
[1] 1.5
The volume of a cube is 1000 cm3. Find the side of a cube.
Concept: Surface Area and Volume of Three Dimensional Figures
Chapter: [0.07] Mensuration
[1] 1.6
If two circles with radii 5 cm and 3 cm respectively touch internally, find the distance between their centres.
Concept: Theorem of Touching Circles
Chapter: [0.03] Circle
[8] 2 | Solve any four sub-questions :
[2] 2.1
If sin θ =7/25, where θ is an acute angle, find the value of cos θ.
Concept: Trigonometric Ratios of Complementary Angles
Chapter: [0.06] Trigonometry
[2] 2.2
Draw ∠ABC of measure 120° and bisect it.
Concept: Basic Geometric Constructions
Chapter: [0.04] Geometric Constructions
[2] 2.3
Find the slope of the line passing through the points M(4,0) and N(-2,-3).
Concept: Slope of a Line
Chapter: [0.05] Co-ordinate Geometry
[2] 2.4
Find the area of the sector whose arc length and radius are 14 cm and 6 cm respectively.
Concept: Areas of Sector and Segment of a Circle
Chapter: [0.07] Mensuration
[2] 2.5
In the following figure, in Δ PQR, seg RS is the bisector of ∠PRQ.
PS = 11, SQ = 12, PR = 22. Find QR.
Concept: Similarity of Triangles
Chapter: [0.01] Similarity
[2] 2.6
In the following figure, if m(arc DXE) = 120° and m(arc AYC) = 60°. Find ∠DBE.
Concept: Areas of Sector and Segment of a Circle
Chapter: [0.07] Mensuration
[9] 3 | Solve any three sub-questions :
[3] 3.1
In the following figure, Q is the centre of a circle and PM, PN are tangent segments to the circle. If ∠MPN = 60°, find ∠MQN.
Concept: Number of Tangents from a Point on a Circle
Chapter: [0.03] Circle
[3] 3.2
Draw the tangents to the circle from the point L with radius 3 cm. Point ‘L’ is at a distance 8 cm from the centre ‘M’.
Concept: To Construct Tangents to a Circle from a Point Outside the Circle.
Chapter: [0.04] Geometric Constructions
[3] 3.3
The ratio of the areas of two triangles with the common base is 4 : 3. Height of the larger triangle is 2 cm, then find the corresponding height of the smaller triangle.
Concept: Properties of Ratios of Areas of Two Triangles
Chapter: [0.01] Similarity
[3] 3.4
Two buildings are in front of each other on either side of a road of width 10 metres. From the top of the first building which is 20 metres high, the angle of elevation to the top of the second is 45°. What is the height of the second building?
Concept: Heights and Distances
Chapter: [0.06] Trigonometry
[3] 3.5
Find the volume and surface area of a sphere of radius 8.4 cm.
(pi=22/7)
Concept: Surface Area and Volume of Three Dimensional Figures
Chapter: [0.07] Mensuration
[8] 4
[4] 4.1
Prove that "Opposite angles of a cyclic quadrilateral are supplementary".
Chapter: [0.03] Circle
[4] 4.2
Prove that sin6θ + cos6θ = 1 – 3 sin2θ. cos2θ.
Concept: Trigonometric Identities
Chapter: [0.06] Trigonometry
[4] 4.3
A test tube has diameter 20 mm and height is 15 cm. The lower portion is a hemisphere. Find the capacity of the test tube. (π = 3.14)
Concept: Surface Area and Volume of Different Combination of Solid Figures
Chapter: [0.07] Mensuration
[10] 5 | Solve any two sub-questions :
[5] 5.1
Prove that the angle bisector of a triangle divides the side opposite to the angle in the ratio of the remaining sides.
Concept: Similarity of Triangles
Chapter: [0.01] Similarity
[5] 5.2
Write down the equation of a line whose slope is 3/2 and which passes through point P, where P divides the line segment AB joining A(-2, 6) and B(3, -4) in the ratio 2 : 3.
Concept: Division of a Line Segment
Chapter: [0.04] Geometric Constructions [0.05] Co-ordinate Geometry
[5] 5.3
ΔRST ~ ΔUAY, In ΔRST, RS = 6 cm, ∠S = 50°, ST = 7.5 cm. The corresponding sides of ΔRST and ΔUAY are in the ratio 5 : 4. Construct ΔUAY.
Concept: Division of a Line Segment
Chapter: [0.04] Geometric Constructions [0.05] Co-ordinate Geometry
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Maharashtra State Board previous year question papers 10th Standard Board Exam Geometry Mathematics 2 with solutions 2014 - 2015
Maharashtra State Board 10th Standard Board Exam Geometry Maths 2 question paper solution is key to score more marks in final exams. Students who have used our past year paper solution have significantly improved in speed and boosted their confidence to solve any question in the examination. Our Maharashtra State Board 10th Standard Board Exam Geometry Maths 2 question paper 2015 serve as a catalyst to prepare for your Geometry Mathematics 2 board examination.
Previous year Question paper for Maharashtra State Board 10th Standard Board Exam Geometry Maths 2-2015 is solved by experts. Solved question papers gives you the chance to check yourself after your mock test.
By referring the question paper Solutions for Geometry Mathematics 2, you can scale your preparation level and work on your weak areas. It will also help the candidates in developing the time-management skills. Practice makes perfect, and there is no better way to practice than to attempt previous year question paper solutions of Maharashtra State Board 10th Standard Board Exam.
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## Where math comes alive
### Conquering Proportions, Part 2
In my first “Conquering Proportions” post, I showed how to save time by canceling terms horizontally as well as vertically. In this post you’ll learn how to save even more time with another shortcut. Let’s look at an example to refresh our memory.
Given a proportion such as this:
15 = 5
a 3
most people would do the traditional “cross-multiplying” step, to get:
5 x a = 15 x 3 (the x here is a true times sign; that’s why I’m using ‘a‘ as the variable, not ‘x.’)
If you follow the usual steps, the next thing would be to ÷ both sides by 5, to get:
a = (15 x 3) ÷ 5
But let’s look more closely at this answer expression: (15 x 3) ÷ 5
We can conceptualize this expression better if we think of the original proportion:
15 = 5
a 3
as containing two DIAGONALS.
One diagonal holds the 15 and the 3; the other diagonal holds the ‘a’ and the 5.
Let’s call the diagonal with the ‘a’ the ‘first diagonal.’ And since ‘5’ accompanies ‘a’ in that diagonal, we’ll call 5 the “variable’s partner.”
We’ll call the other diagonal just that, the “other diagonal.”
Now I know you’re getting ‘antsy’ for the shortcut, so just know it’s right around “the bend.”
Using our new terms, we can better understand the expression we got up above:
a = (15 x 3) ÷ 5
The (15 x 3) is the product (result of multiplication) of the “other diagonal,”
and ‘5’ is the “variable’s partner.
(15 x 3) ÷ 5
is simply (and here’s the shortcut):
(product of other diagonal) ÷ by (“variable’s partner.”)
We’ll call this the Proportion Shortcut Formula, or the PSF, for short.
The PSF saves a BIG STEP; using it, we no longer need to write out the cross-multiplication product the usual way, as:
5 x a = 15 x 3
Instead, using the PSF, we can go straight from the proportion to an expression for ‘a‘:
a = (15 x 3) ÷ 5
Let’s see how the PSF works in another proportion, such as:
9 = 45
13 a
What’s the “variable’s partner”? 9.
What’s in the “other diagonal”? 13 and 45.
So using PSF, the answer is this:
a = (13 x 45) ÷ 9
This simplifies to 65, of course. Isn’t it nice not to have to “cross-multiply” any more?
Another nice thing: the PSF works no matter where the variable is located in the original proportion. All you need to do is identify the “variable’s partner,” and the “other diagonal,” and then you’re all good go with the PSF.
Try a few of these to see how easy and convenient the PSF makes it to solve proportions.
PROBLEMS:
1) a = 15
12 36
2) 18 = a
24 4
3) 21 = 75
14 a
1) a = (12 x 15) ÷ 36
a = 5
2) a = (18 x 4) ÷ 24
a = 3
3) a = (14 x 75) ÷ 21
a = 50
### Common Core Math Workshop Saturday in Santa Fe
Anyone out there feeling frustrated or confused about the new Common Core math?
I’m hosting a free workshop at 4:00 – 5:30 this Saturday, May 2, at the LaFarge Library in Santa Fe.
The focus is how to understand the basics of Common Core math so that you can help your child understand the concepts being taught in school.
### Fraction Hack #2: The Size of the Smaller Number
Responding to my post about the fraction “hack” of using the gap between fraction numbers, Ivasallay wrote: “What if the numerator is smaller than the gap?”
High-Octane Boost for Math Ed
Good question, and thanks for sharing it. My answer: Yes, the numerator could be smaller than the gap, and if it is, that can help us simplify fractions, too.
Now we could have a fraction like 15/6, in which the lesser of the two numbers is the denominator, so to keep our discussion general I’m going to talk, not about the numerator, but rather about the “smaller fraction number,” whether numerator or denominator.
The way this matters is as follows: like the gap, the smaller fraction number provides an upper limit, a greatest possible value, for the GCF of the fraction’s two numbers. So if the fraction is 12/90 (smaller number being 12), that means that the GCF can be no larger than 12. If the fraction is 3/1011, with lesser number 3, the GCF can be no larger than 3.
The reason should be obvious, and when I say this I really mean it. Take the fraction 6/792, for example. Could a number larger than 6 go into both 6 and 792? Well there may be a number larger than 6 that goes into 792 evenly, but nothing larger than 6 can go into 6 itself, right? A large peg can’t go through a tiny hole, right? So there you go. Nothing larger than 6 can go into both 6 AND 792. QED.
So what does this mean for you, the math student, or parent of a math student, or the teacher of math students? … I means you want to keep in mind that in actuality two different numbers will help you nail down the size of the GCF. One is the gap between the fraction numbers, and the other is our “new friend,” the smaller of the two fraction numbers.
And here’s another … hack fact. (Whenever I say that, you know we’re heading into ‘nerd-land,’ right?) For both limiting numbers, the gap and the smaller fraction number, the only numbers that can possibly go into both fraction numbers are the FACTORS of those limiting numbers. So for example, if your fraction is 6/50, with the smaller number of 6, the only numbers that can possibly go into 6 and 50 are the factors of 6: i.e., 6, 3, or 2.
A nice rule of thumb: see which is smaller, the gap or the smaller fraction number. Then use that smaller number as your largest possible GCF. To nail this down, let’s do two example problems.
Example 1: 8/44. What’s smaller? 8 or the gap, 36. Obviously 8! So use 8. Test the factors of 8, which are 8, 4, 2. Notice that 8 doesn’t go into both 8 and 44. But 4 does, so 4 is the GCF, and using 4, the fraction simplifies down to 2/11.
Example 2: 22/36. What’s smaller? 22 or the gap, 14. Here the gap is smaller. So test the gap’s factors: 14, 7, 2. 14 doesn’t go into 22 and 36; nor does 7. But 2 does. So 2 is the GCF, and using 2, the fraction simplifies to 11/18.
Time for you all to try your hands at this fun practice, which catapults your “number sense” to new heights.
For each problem, 1) identify the fraction’s smaller number and the gap. 2) Say which of those two numbers is smaller. 3) Using that number’s factors, find the GCF. 4) Finally, using the GCF, simplify the fraction. Answers follow.
SIMPLIFY THE FRACTIONS:
a) 8/42
b) 12/20
c) 36/60
d) 18/96
e) 21/91
a) 8/42: 1) smaller # = 8; gap = 34. 2) 8 < 34. 3) GCF = 2. 4) 4/21
b) 12/20: 1) smaller # = 12; gap = 8. 2) 8 < 12. 3) GCF = 4. 4) 3/5
c) 36/60: 1) smaller # = 36; gap = 24. 2) 24 < 36. 3) GCF = 12. 4) 3/5
d) 18/96: 1) smaller # = 18; gap = 78. 2) 18 < 78. 3) GCF = 6. 4) 3/16
e) 21/91: 1) smaller # = 21; gap = 70. 2) 21 < 70. 3) GCF = 7. 4) 3/13
Josh Rappaport is the author of five math books, including the wildly popular Algebra Survival Guide and its trusty sidekick, the Algebra Survival Workbook. And FYI: the 2nd Edition of the Survival Guide was just released in March, so get it while it’s hot off the press! If you’d like to get tutored by Josh, you can. Josh and his remarkably helpful wife, Kathy, use Skype to tutor students in the U.S. and Canada in a wide range of subjects. They also prep students for the “semi-evil” ACT and SAT college entrance tests. If you’d be interested in seeing your ACT or SAT scores soar, shoot an email to Josh, sending it to: josh@SingingTurtle.com We’ll keep an eye out for your email, and in our office, our tutoring is always ON … except on Saturdays.
### “Hack” for Simplifying Fractions
So c’mon … everything that can be said about simplifying fractions has been said … right?
Not quite! Here’s something that might just be original … a hack to smack those fractions down to size.
Suppose you’re staring at an annoying-looking fraction: 96/104, and it’s annoying the heck out of you, particularly because it’s smirking at you!
But it won’t smirk for long. For you open up your bag of hacks (obtained @ mathchat.me) and …
1st) Subtract to get the difference between numerator and denominator. I also like to call this the gap between the numbers. Difference (aka, gap) = 104 – 96 = 8.
NOTE: Turns out that this gap, 8, is the upper limit for any numbers that can possibly go into BOTH 96 and 104. No number larger than 8 can go into both. And this is a … HACK FACT: The gap represents the largest number that could possibly go into BOTH numerator and denominator. In other words, the gap is the largest possible greatest common factor (GCF).
2nd) Try 8. Does 8 go into both 96 and 104? Turns out it does, so smack the numerator and denominator down to size: 96 ÷ 8 = 12, and 104 ÷ 8 = 13.
3rd) State the answer: 96/104 = 12/13.
Is it still smirking? I think … NOT!
Try another. Say you’re now puzzling over: 74/80.
1st) Subtract to get the gap. 80 – 74 = 6. So 6 is the largest number that can possibly go into BOTH 74 and 80.
2nd) So try 6. Does it go into both 74 and 80? No, in fact it goes into neither number.
NOTE: Turns out that even though 6 does NOT go into 74 OR 80, the fact that the gap is 6 still says something. It tells us that the only numbers that can possibly go into both 74 and 80 are the factors of 6: 6, 3 and 2. This, it turns out, is another … HACK FACT: Once you know the gap, the only numbers that can possibly go into the two numbers that make the gap are either the factors of the gap, or the gap number itself.
3rd) So now, try the next largest factor of 6, which just happens to be 3. Does 3 go into both 74 and 80? No. Like 6, 3 goes into neither 74 nor 80. But that’s actually a good thing because now there’s only one last factor to test, 2. Does 2 go into both 74 and 80? Yes! At last you’ve found a number that goes into both numerator and denominator.
4th) Hack the numbers down to size: 74 ÷ 2 = 37, and 80 ÷ 2 = 40.
5th) State the answer. 74/80 gets hacked down to 37/40, and that fraction, my dear friends, is the answer. 37/40 the final, simplified form of 74/80.
O.K., are you ready to smack some of those fractions down to size? I believe you are. So here are some problems that will let you test out your new hack.
As you slash these numbers down, remember this rule. In some of these problems the gap number itself is the number that divides into numerator and denominator. But in other problems, it’s not the gap number itself, but rather a factor of the gap number that slashes both numbers down to size. So if the gap number itself doesn’t work, don’t forget to check out its factors.
Ready then? Here you go … For each problem, state the gap and find the largest number that goes into both numerator and denominator. Then write the simplified version of the fraction.
a) 46/54
b) 42/51
c) 48/60
d) 45/51
e) 63/77
a) 46/54: gap = 8. Largest common factor (GCF) = 2. Simplified form = 23/27
b) 42/51: gap = 9. Largest common factor (GCF) = 3. Simplified form = 14/17
c) 48/60: gap = 12. Largest common factor (GCF) = 12. Simplified form = 4/5
d) 45/51: gap = 6. Largest common factor (GCF) = 3. Simplified form = 15/17
e) 63/77: gap = 14. Largest common factor (GCF) = 7. Simplified form = 9/11
Josh Rappaport is the author of five math books, including the wildly popular Algebra Survival Guide and its trusty sidekick, the Algebra Survival Workbook. Josh has been tutoring math for more years than he can count — even though he’s pretty good at counting after all that tutoring — and he now tutors students in math, nationwide, by Skype. Josh and his remarkably helpful wife, Kathy, use Skype to tutor students in the U.S. and Canada, preparing them for the “semi-evil” ACT and SAT college entrance tests. If you’d be interested in seeing your ACT or SAT scores rise dramatically, shoot an email to Josh, addressing it to: josh@SingingTurtle.com We’ll keep an eye out for your email, and our tutoring light will always be ON.
### FUNCTIONS: Notation Confusion & the Importance of Empathy
Working with one student on functions this morning, I was reminded of how much there is that students can fail to understand.
I was trying to explain to this student that the x-value is the input, and that the f(x) value is the output. But because of the repetition of x in both of these terms, he got confused.
Tales from the Tutoring Experience
I finally solved the problem by telling my student to view the parentheses around the x in (x) as like the slot that takes the coin in a vending machine. It sort of looks like a slot, too, right? So what goes inside it must be the “in”put. Now he at least understands clearly that what goes in the ( ) slot is the input.
Staying with the vending machine analogy, I told him that the f(x) is what the function machine (like a vending machine) gives you after you put the x-value in the slot.
I did need to clarify that when you’re working out the value of inputs and outputs, you must insert the x-value twice: once inside the ( ) slot, and secondly on the other side of the equation, in where the ‘x’ stands.
Obviously this student has a lot of trouble with processing the visual symbols of math. But working with him reminds me of something important. It shows how much students can get confused by math concepts and math notation. I feel that it’s important for us educators to keep this in mind as we teach. There’s so much that we take for granted in our understanding of math. But for students who struggle with notation and with the visual aspect of math, notation can be confusing.
One thing I try to do when I work with students comes from something I saw in a Great Courses class by Bruce Edwards, an excellent teacher of higher math. Mr. Edwards likes to say things like, “Now this is next part is a little bit tricky …” Just by saying this, Mr. Edwards shows that he understands that not everyone will get the concept, and that, I believe, helps students relax.
Ever since I saw Mr. Edwards use this way of talking, I’ve been using it in my tutoring work, too. And I find that it helps students. It makes them feel like no one will think badly of them for not understanding, since I, the teacher, have acknowledged that the concept is “tricky.” As a result, students relax, and that helps them be more relaxed in taking in what you’re going to tell them. A nice thing to learn from a master teacher, and another lesson in the importance of the way in which we talk to students to help them learn. There’s so much more to being a good math teacher than just being thorough and clear. The affective aspects of communicating, such as showing empathy, are very important as well.
### The Surge in Standardized Testing … Your Thoughts on the Controversy
This is a different kind of post; it deals not with math per se, but rather with the recent surge in standardized testing nationwide and with the controversy surrounding it.
As a tutor I work with a wide range of students, and lately I’ve noticed how much my high school students are talking about the standardized testing at their schools.
Tales from the Tutoring Experience
In my home state of New Mexico, many students grew so concerned about the new PARCC standardized tests that they walked out of school and protested, with placards and chants, the “whole nine yards.” I drove down to meet these students at the state capitol building, where the students were voicing their concerns.
Asked about the testing, the students told me that they’ve been told that if they don’t pass the PARCC, they’ll never get their diploma. They said they were told that directly by some of their teachers and counselors. (In the local newspaper this issue is debated, and it seems that there’s no clear consensus as to the implications of not passing the PARCC.)
As a math tutor I’m aware of the math content that is in my students’ curriculum, and I’ve noticed that some of the PARCC exam’s math problems are based on concepts that are not in students’ curriculum. For example, the PARCC’s math section has many problems on statistics, and my students have not been taught statistics through their curriculum.
My concern is that this exam is testing students on topics that are beyond their curriculum, and that the consequences of their failing might be not graduating.
The students told me that if instead of the PARCC they were given the SBA, the test that they’ve taken for many years now, they would not even consider protesting. They said their concern is that the district changed the test they would take rather abruptly from the SBA to the PARCC without giving teachers adequate time to teach students the PARCS’s content, which is largely aligned with Common Core State Standards for Math. Several teachers in Santa Fe have echoed this concern in the local media.
On the other hand, other students don’t seem concerned about the tests. They take it all in stride. And of course, local district officials say that the tests help establish a basis for evaluating both the progress of individual students, and the success of individual schools and districts. They argue that there needs to be a standardized “baseline” so that people and communities can make “apples-to-apples” comparisons of students, schools and districts.
I’m writing this post in hopes of starting a dialog to find out how people around the country feel about the many standardized tests dotting the educational landscape these days. In your opinion, are the tests a good thing, a bad thing? What are your thoughts and feelings on this topic? Please share. It would be great if this blog could be a forum for discussion of this issue.
### How to Find the LCM for 3+ Numbers — FAST!
Is there a quick-and-easy way to find the LCM for three or more numbers … WITHOUT prime factorizing?
Of course! We’ll demonstrate the technique by finding the LCM for 10, 14, 20.
High-Octane Boost for Math Ed
To begin, use the technique for finding the GCF for 10, 14, 20 that’s shown in my post: How to Find GCF for 3+ Numbers — FAST … no prime factorizing. If you don’t want to go to that post, no worries. I’ll re-show the technique here.
1st) Write the numbers from left to right:
………. 10 14 20
[The periods: …… are just to indent the lines. They have no mathematical meaning.]
2nd) If possible, rip out a factor common to all numbers. The factor 2 is common. So divide the three numbers by 2 [10 ÷ 2 = 5 and 20 ÷ 2 = 10] and show the result below:
2 | 10 14 20
………. 5 7 10
3rd) At this point, notice there’s no number that goes into the remaining numbers: 5, 7, 10. That means you’ve found that the GCF is the number pulled out, 2. At this point we’re at a crossroads. We’re done finding the GCF, but now we’re at the start of a new process, finding the LCM.
To proceed toward getting the LCM, see if there’s any number that goes into any pair of remaining numbers. Well, 5 goes into 5 and 10. So divide both those numbers by 5 [5 ÷ 5 = 1 and 10 ÷ 5 = 2] , and show the results below:
2 | 10 14 20
5 | 5 7 10
……….. 1 7 2
Notice that if there’s a number 5 doesn’t go into, you leave that number as is. So leave the 7 as 7.
4th) Repeat. See if there’s a number that goes into two of the remaining numbers. Since nothing goes into 1, 7, and 2, we’re done. To get the LCM, multiply all of the outer numbers. That means you multiply the numbers you pulled out on the left (2 and 5), and also multiply the numbers at the bottom (1, 7 and 2). Ignoring the meaningless 1, you have: 2 x 5 x 7 x 2 = 140, and that’s the LCM.
To see the process in more depth, let’s find the LCM for … not three, not four … but five numbers:
6, 12, 18, 30, 36.
1st) Write the numbers left to right:
……… 6 12 18 30 36
2nd) If possible, rip out a common factor. 2 is common, so divide all by 2 and show the results below:
2 | 6 12 18 30 36
………. 3 6 9 15 18
3rd) Repeat. See if there’s a number that goes into the five remaining numbers. 3 goes into all, so divide all by 3 and show the results below:
2 | 6 12 18 30 36
3 | 3 6 9 15 18
…….. 1 2 3 5 6
4th) Repeat. See if any number goes into the last remaining numbers. Nothing goes into all of them, so now you get the GCF by multiplying the left-hand column numbers. GCF = 2 x 3 = 6.
Proceeding to find the LCM, look for any number that goes into two or more of the remaining numbers. One such number is 3, which goes into the remaining 3 and 6. Divide those numbers by 3 and leave the other numbers as they are.
2 | 6 12 18 30 36
3 | 3 6 9 15 18
3 | 1 2 3 5 6
……… 1 2 1 5 2
5th) Interesting! Notice that 2 goes into the two remaining 2s, so pull out a 2 and show the results below:
2 | 6 12 18 30 36
3 | 3 6 9 15 18
3 | 1 2 3 5 6
2 | 1 2 1 5 2
…….. 1 1 1 5 1
6th) We’ve whittled the bottom row’s numbers so far down that finally there’s no number that goes into two or more of them (except 1, which doesn’t help). So we have all the numbers we need to find the LCM. Multiply them together. The left column gives us: 2 x 3 x 3 x 2. The bottom row gives us 1 x 1 x 1 x 5 x 1. Multiply all of those (non-1) numbers together, you get:
2 x 2 x 3 x 3 x 5 = 180, and that is the LCM! Pretty amazing, huh? And no prime factorizing, to boot.
Some people find that this process takes a bit of practice to get used to it. So here are a few problems to help you become an LCM-finding expert!
a) 12, 18, 30
b) 8, 18, 24
c) 15, 20, 30, 35
d) 16, 24, 40, 56
e) 16, 48, 64, 80, 112
And the answers. LCM for each set is:
a) 180
b) 72
c) 420
d) 1680
e) 6720 |
# Search by Topic
#### Resources tagged with Odd and even numbers similar to How Odd:
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Broad Topics > Numbers and the Number System > Odd and even numbers
### How Odd
##### Stage: 1 Challenge Level:
This problem challenges you to find out how many odd numbers there are between pairs of numbers. Can you find a pair of numbers that has four odds between them?
### Cube Bricks and Daisy Chains
##### Stage: 1 Challenge Level:
Daisy and Akram were making number patterns. Daisy was using beads that looked like flowers and Akram was using cube bricks. First they were counting in twos.
### Pairs of Legs
##### Stage: 1 Challenge Level:
How many legs do each of these creatures have? How many pairs is that?
### Lots of Biscuits!
##### Stage: 1 Challenge Level:
Help share out the biscuits the children have made.
### Diagonal Trace
##### Stage: 2 Challenge Level:
You can trace over all of the diagonals of a pentagon without lifting your pencil and without going over any more than once. Can the same thing be done with a hexagon or with a heptagon?
### Two Numbers Under the Microscope
##### Stage: 1 Challenge Level:
This investigates one particular property of number by looking closely at an example of adding two odd numbers together.
### Grouping Goodies
##### Stage: 1 Challenge Level:
Pat counts her sweets in different groups and both times she has some left over. How many sweets could she have had?
### The Set of Numbers
##### Stage: 1 Challenge Level:
Can you place the numbers from 1 to 10 in the grid?
### Odd Times Even
##### Stage: 1 Challenge Level:
This problem looks at how one example of your choice can show something about the general structure of multiplication.
### Number Round Up
##### Stage: 1 Challenge Level:
Arrange the numbers 1 to 6 in each set of circles below. The sum of each side of the triangle should equal the number in its centre.
### Multiplication Series: Number Arrays
##### Stage: 1 and 2
This article for teachers describes how number arrays can be a useful reprentation for many number concepts.
### Becky's Number Plumber
##### Stage: 2 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Seven Flipped
##### Stage: 2 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Always, Sometimes or Never?
##### Stage: 1 and 2 Challenge Level:
Are these statements relating to odd and even numbers always true, sometimes true or never true?
### Play to 37
##### Stage: 2 Challenge Level:
In this game for two players, the idea is to take it in turns to choose 1, 3, 5 or 7. The winner is the first to make the total 37.
### The Thousands Game
##### Stage: 2 Challenge Level:
Each child in Class 3 took four numbers out of the bag. Who had made the highest even number?
### Arrangements
##### Stage: 2 Challenge Level:
Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....?
### More Numbers in the Ring
##### Stage: 1 Challenge Level:
If there are 3 squares in the ring, can you place three different numbers in them so that their differences are odd? Try with different numbers of squares around the ring. What do you notice?
### Magic Vs
##### Stage: 2 Challenge Level:
Can you put the numbers 1-5 in the V shape so that both 'arms' have the same total?
### Crossings
##### Stage: 2 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
### Number Differences
##### Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Curious Number
##### Stage: 2 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
### What Do You Need?
##### Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Largest Even
##### Stage: 1 Challenge Level:
How would you create the largest possible two-digit even number from the digit I've given you and one of your choice?
### What Number?
##### Stage: 1 Short Challenge Level:
I am less than 25. My ones digit is twice my tens digit. My digits add up to an even number.
### Odd Tic Tac
##### Stage: 1 Challenge Level:
An odd version of tic tac toe
### Square Subtraction
##### Stage: 2 Challenge Level:
Look at what happens when you take a number, square it and subtract your answer. What kind of number do you get? Can you prove it?
### Red Even
##### Stage: 2 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### Always, Sometimes or Never? Number
##### Stage: 2 Challenge Level:
Are these statements always true, sometimes true or never true?
### What Could it Be?
##### Stage: 1 Challenge Level:
In this calculation, the box represents a missing digit. What could the digit be? What would the solution be in each case?
### Odds and Threes
##### Stage: 2 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### Part the Piles
##### Stage: 2 Challenge Level:
Try to stop your opponent from being able to split the piles of counters into unequal numbers. Can you find a strategy?
### Down to Nothing
##### Stage: 2 Challenge Level:
A game for 2 or more people. Starting with 100, subratct a number from 1 to 9 from the total. You score for making an odd number, a number ending in 0 or a multiple of 6.
### One of Thirty-six
##### Stage: 1 Challenge Level:
Can you find the chosen number from the grid using the clues?
### Break it Up!
##### Stage: 1 and 2 Challenge Level:
In how many different ways can you break up a stick of 7 interlocking cubes? Now try with a stick of 8 cubes and a stick of 6 cubes.
### Number Tracks
##### Stage: 2 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
### Numbers as Shapes
##### Stage: 1 Challenge Level:
Use cubes to continue making the numbers from 7 to 20. Are they sticks, rectangles or squares?
### Even and Odd
##### Stage: 1 Challenge Level:
This activity is best done with a whole class or in a large group. Can you match the cards? What happens when you add pairs of the numbers together?
### Take Three Numbers
##### Stage: 2 Challenge Level:
What happens when you add three numbers together? Will your answer be odd or even? How do you know?
### I Like ...
##### Stage: 1 Challenge Level:
Mr Gilderdale is playing a game with his class. What rule might he have chosen? How would you test your idea?
### Light the Lights
##### Stage: 1 Challenge Level:
Investigate which numbers make these lights come on. What is the smallest number you can find that lights up all the lights?
### Ring a Ring of Numbers
##### Stage: 1 Challenge Level:
Choose four of the numbers from 1 to 9 to put in the squares so that the differences between joined squares are odd.
### Sets of Numbers
##### Stage: 2 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
### Three Spinners
##### Stage: 2 Challenge Level:
These red, yellow and blue spinners were each spun 45 times in total. Can you work out which numbers are on each spinner?
### Sets of Four Numbers
##### Stage: 2 Challenge Level:
There are ten children in Becky's group. Can you find a set of numbers for each of them? Are there any other sets?
### Domino Sorting
##### Stage: 1 Challenge Level:
Try grouping the dominoes in the ways described. Are there any left over each time? Can you explain why?
### Number Detective
##### Stage: 2 Challenge Level:
Follow the clues to find the mystery number.
### Make 37
##### Stage: 2 and 3 Challenge Level:
Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37.
### A Mixed-up Clock
##### Stage: 2 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### Sorting Numbers
##### Stage: 1 Challenge Level:
Use the interactivity to sort these numbers into sets. Can you give each set a name? |
Home » Math Theory » Numbers » Number Change
# Number Change
## Definition
The topic of number changing is covered here. Number change is the knowledge that we either add to or subtract from one number to make another number, together with the understanding of how much to add or subtract.
Let us change the number 358 as an example, and let us see what happens if we change the number from 3 to 2 in the hundreds place. The number will be 158 since we took away 1 hundred.
Let us first review the idea of place value and how it affects the change that occurs to a number before we continue our discussion of number change and look at further instances.
## Place Value and Place Value Chart
In mathematics, each digit in a number has a place value. The value a digit in a number represents based on where it is in the number is known as place value.
The placement of each digit represents value. These positions begin at the unit place, often known as the ones position. Ones/unit, tens, hundreds, thousands, ten thousands, one hundred thousands, and so on are the place values of a number’s digits from right to left.
An illustration that enables us to locate and contrast the place values of the digits in numbers up to millions is known as a place value chart. As we move from left to right on the place value chart, a digit’s place value goes up by ten times and down by ten times.
Let us say, for example, in the number 654, 321
6 is in the hundred thousands place with a place value of 600,000.
5 is in the ten thousands place with a place value of 50,000.
4 is in the thousands place with a place value of 4,000.
3 is in the hundreds place with a place value of 300.
2 is in the tens place with a place value of 20.
1 is in the ones place with a place value of 1.
By multiplying the face value with the actual value of the number, one can get a number’s place value. A one-digit number has a place value equal to its face value. For instance, 1, 2, 3, 4, 5, 6, 7, 8, and 9 have the same place value and face value as 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. Any number’s place value for zero is always zero. Zero may appear anywhere in a number, but it will always have zero value.
Examples
Example 1
In the number 594, write down the place value of the digits.
The value of the digit 4, which is in the ones, is 4 x 1 = 4
The value of the digit 9, which is in the tens place, is 9 x 10 = 90
The value of the digit 5, which is in the hundreds place, is 5 x 100 = 500
Example 2
For the number 456 897, write down the place value of each digit.
The value of the digit 7, which is in the ones place, is 7 x 1 = 7.
The value of the digit 9, which is in the tens place, is 9 x 10 = 90.
The value of the digit 8, which is in the hundreds place, is 8 x 100 = 800.
The value of the digit 6, which is in the thousands place, is 6 x 1000 = 6000.
The value of the digit 5, which is in the ten thousands place, is 5 x 10000 = 50000.
The value of the digit 4, which is in the hundred thousands place, is 4 x 100000 = 400000.
Example 3
For the number 90 165 478, write down the place value of each digit.
The value of digit 8, which is in the ones place, 8 x 1 = 8.
The value of digit 7, which is in the tens place, 7 x 10 = 70.
The value of digit 4, which is in the hundreds place, 4 x 100 = 400.
The value of digit 5, which is in the thousands place, 5 x 1000 = 5000.
The value of digit 6, which is in the ten thousands place, is 6 x 10000 = 60000.
The value of digit 1, which is in the hundred thousands place, is 1 x 100000 = 100000.
The value of digit 0, which is in the millions place, is 0 x 1000000 = 0.
The value of digit 9, which is in the ten millions place, is 9 x 10000000 = 90000000.
Example 4
For the number 183 256 094, write down the place value of each digit.
The digit 4, which is in the ones place, is 4 x 1 = 4.
The digit 9, which is in the tens place, is 9 x 10 = 90.
The digit 0, which is in the hundreds place, is 0 x 100 = 0.
The digit 6, which is in the thousands place, is 6 x 1000 = 6000.
The digit 5, which is in the ten thousands place, is 5 x 10000 = 50000.
The digit 2, which is in the hundred thousands place, is 2 x 100000 = 200000.
The digit 3, which is in the millions place, is 3 x 1000000 = 0.
The digit 8, which is in the ten millions place, is 8 x 10000000 = 80000000.
The digit 1, which is in the hundred millions place, is 1 x 100000000 = 100000000.
Example 5
For the number 742 193 568, write down the place value of each digit.
The value of the digit 8, which is in the ones place, is 8 x 1 = 8.
The value of the digit 6, which is in the tens place, is 6 x 10 = 60.
The value of the digit 5, which is in the hundreds place, is 5 x 100 = 500.
The value of the digit 3, which is in the thousands place, is 3 x 1000 = 3000.
The value of the digit 9, which is in the ten thousands place, is 9 x 10000 = 90000.
The value of the digit 1, which is in the hundred thousands place, is 1 x 100000 = 100000.
The value of the digit 2, which is in the millions place, is 2 x 1000000 = 0.
The value of the digit 4, which is in the ten millions place, is 4 x 10000000 = 40000000.
The value of the digit 7, which is in the hundred millions place, is 7 x 100000000 = 700000000.
## Using Base Ten Blocks to Place Value
We can also write numbers in their expanded form using base-ten blocks to indicate the digit’s place value. The place value system allows for a range of values for each digit, from zero to nine. We re-start at zero but increase the value of the digit in the next highest place value by one when a digit’s value rises past nine. Let’s first learn what these base ten blocks stand for before using them to determine each digit’s place value.
Examples
Example 1:
To represent the number 89, use base ten blocks.
The value of the digit 9, which is in the ones place, is 9 x 1 = 9.
The value of the digit 8, which is in the tens place, is 8 x 10 = 80.
The illustration below shows the number 89 using base ten blocks.
Example 2:
To represent the number 453, use base ten blocks.
The value of the digit 3, which is in the ones place, is 3 x 1 = 3.
The value of the digit 5, which is in the tens place, is 5 x 10 = 50.
The value of the digit 4, which is in the hundreds place, is 4 x 100 = 400.
The illustration below shows the number 453 using base ten blocks.
Example 3:
To represent the number 142, use base ten blocks.
The value of the digit 2, which is in the ones place, is 2 x 1 = 2.
The value of the digit 4, which is in the tens place, is 4 x 10 = 40.
The value of the digit 1, which is in the hundreds place, is 1 x 100 = 100.
The illustration below shows the number 142 using base ten blocks.
Example 4:
To represent the number 1254, use base ten blocks.
The value of the digit 4, which is in the ones place, is 4 x 1 = 4.
The value of the digit 5, which is in the tens place, is 5 x 10 = 50.
The value of the digit 2, which is in the hundreds place, is 2 x 100 = 200.
The value of the digit 1, which is in the thousands place, is 1 x 100 = 1000.
The illustration below shows the number 1254 using base ten blocks.
## Examples:Number Change
Now that the concept of place value has been illustrated already let us talk about number change or how a number change if we change a digit.
Number change has something to do with either adding or subtracting from one number to make another number, together with the knowledge of how much to add or subtract.
Example 1
How much does the number 253 change if we take away 1 in the hundreds place?
Solution:
The value of the digit 3, which is in the ones place, is 3 x 1 = 3.
The value of the digit 5, which is in the tens place, is 5 x 10 = 50.
The value of the digit 2, which is in the hundreds place, is 2 x 100 = 200.
If we take away 1 in the hundreds place of the number 253, this means that we are subtracting 100 from 253. Therefore, we have 253 – 100, which is 153. Since we have subtracted 100, 153 is 100 less than 253.
253-100=153
Example 2
How much does the number 678 change if we take away 2 in the tens place?
Solution:
The value of the digit 8, which is in the ones place, is 8 x 1 = 8.
The value of the digit 7, which is in the tens place, is 7 x 10 = 70.
The value of the digit 6, which is in the hundreds place, is 6 x 100 = 600.
Taking away 2 in the tens place of 678 means that we have to subtract 20 from it. Hence, we have 678 – 20. Since we have subtracted 20 from 678, the new number is 658.
678-20=658
Example 3
How much does the number 5967 change if we take away 5 tens?
Solution:
The value of the digit 7, which is in the ones place, is 7 x 1 = 7.
The value of the digit 6, which is in the tens place, is 6 x 10 = 60.
The value of the digit 9, which is in the hundreds, place is 9 x 100 = 900.
The value of the digit 5, which is in the thousands place, is 5 x 1000 = 5000.
If we take away 5 tens from the number 5967, this means that we are subtracting 50 from 5967. Therefore, we have 5967 – 50, which is 5917. Since we have subtracted 50, the 5917 is 50 less than 5967.
5967-50=5917
Example 4
By how much does the number 62583 change if we add 4 hundreds?
Solution:
The value of the digit 3, which is in the ones place, is 3 x 1 = 3.
The value of the digit 8, which is in the tens place, is 8 x 10 = 80.
The value of the digit 5, which is in the hundreds place, is 5 x 100 = 500.
The value of the digit 2, which is in the thousands place, is 2 x 1000 = 2000.
The value of the digit 6, which is in the ten thousands place, is 6 x 10000 = 60000.
If we add 4 hundreds to the number 62583, this means that we are adding 400 to 62583. Thus, we have 62583 + 400, which is 62983.
62583+400=62983
Example 5
By how much does the number 12864 change if we add 7 hundreds?
Solution:
The value of the digit 4, which is in the ones place, is 4 x 1 = 4.
The value of the digit 6, which is in the tens place, is 6 x 10 = 60.
The value of the digit 8, which is in the hundreds place, is 8 x 100 = 800.
The value of the digit 2, which is in the thousands place, is 2 x 1000 = 2000.
The value of the digit 1, which is in the ten thousands place, is 1 x 10000 = 10000.
Adding 7 hundreds means adding 700 to the given number. Therefore we have 12864 + 700. Looking at the hundreds place, 800 + 700 is equal to 1500. This means we have 5 hundreds and add 1 to the thousands place. Since we need to add the whole 1 thousand to the number, the digit 2 in the thousands place will be 3. Hence, the new number is 13564.
12864+700=13564
Example 6
By how much does the number 678298 change if we add 5 thousands?
Solution:
The value of the digit 8, which is in the ones place, is 8 x 1 = 8.
The value of the digit 9, which is in the tens place, is 9 x 10 = 90.
The value of the digit 2, which is in the hundreds place, is 2 x 100 = 200.
The value of the digit 8, which is in the thousands place, is 8 x 1000 = 8000.
The value of the digit 7, which is in the ten thousands place, is 7 x 10000 = 70000.
The value of the digit 6, which is in the hundred thousands place, is 6 x 100000 = 600000.
Adding 5 thousands means adding 5000 to the given number. Thus, we have 678298 + 5000. Since the value in the thousands place is 8000 and we need to add 5000 to it, we have a total of 1300. This means that we have 3 in the hundreds place and add 1 to the ten thousands place. Since we need to add the whole 1 ten thousand to the number, then the digit 7 in the ten thousands place will be 8. Hence, the new number is 683298.
678298 +5000=683298
Example 7
Using number blocks, show how much the number 158 changes if we take away 2 tens?
Solution:
The value of the digit 8, which is in the ones place, is 8 x 1 = 8.
The value of the digit 5, which is in the tens place, is 5 x 10 = 50.
The value of the digit 1, which is in the hundreds place, is 1 x 100 = 100.
If we take away 2 tens from the number 158, this means that we are subtracting 20 from 158. Therefore, we have 158 – 20, which is 138. Since we have subtracted 20, the 138 is 20 less than 158.
158-20=138
Example 8
Using number blocks, show how much the number 3149 changes if we take away 2 thousands?
Solution:
The value of the digit 9, which is in the ones place, is 9 x 1 = 9.
The value of the digit 4, which is in the tens place, is 4 x 10 = 40.
The value of the digit 1, which is in the hundreds place, is 1 x 100 = 100.
The value of the digit 3, which is in the thousands place, is 3 x 1000 = 3000.
If we take away 2 thousands from the number 3149, this means that we are subtracting 2000 from 3149. Therefore, we have 3149 – 2000, which is 1149. Since we have subtracted 2000, the number 1149 is 2000 less than 3149.
3149-2000=1149
Example 8
Show how much the number 7286 changes if we add 2 hundreds.
Solution:
The value of the digit 6, which is in the ones place, is 6 x 1 = 6.
The value of the digit 8, which is in the tens place, is 8 x 10 = 80.
The value of the digit 2, which is in the hundreds place, is 2 x 100 = 200.
The value of the digit 7, which is in the thousands place, is 7 x 1000 = 7000.
If we add 2 hundreds to the number 7286, this means that we are adding 200 to 7286. Therefore, we have 7286 + 200, which is 7486. Since we have added 200, the number 7486 is 300 greater than 7286.
7286-200=7486
## Summary
• The value a digit in a number represents based on where it is in the number is known as place value.
• The placement of each digit represents value. These positions begin at the unit place, often known as the ones position. Ones/unit, tens, hundreds, thousands, ten thousand, one hundred thousand, and so on are the place values of a number’s digits from right to left.
• An illustration that enables us to locate and contrast the place values of the digits in numbers up to millions is known as a place value chart. As we move from left to right on the place value chart, a digit’s place value goes up by ten times and down by ten times.
• A one-digit number has a place value equal to its face value.
• Any number’s place value for zero is always zero. Zero may appear anywhere in a number, but it will always have zero value.
• The place value system allows for a range of values for each digit, from zero to nine. We re-start at zero but increase the value of the digit in the next highest place value by one when a digit’s value rises past nine. |
# Sides of a triangle are in the ratio of 12:17:25 and its perimeter is $540 \mathrm{~cm}$. Find its area.
Given:
The sides of a triangle are in the ratio of $12: 17: 25$ and its perimeter is $540\ cm$.
To find:
We have to find the area of the triangle.
Solution:
Let the common ratio between the sides of the triangle be $=\ a$
This implies,
The sides of the triangle as $12\ a, 17\ a$ and $25\ a$.
We know that,
Perimeter $P$ of a triangle with sides of length $a$ units, $b$ units and $c$ units
$P=(a+b+c)$units.
We have perimeter as $540\ cm$.
This implies,
$540\ cm=12\ a+17\ a+25\ a$
$540\ cm=54\ a$
This implies,
$a\ cm=\frac{540}{54}$
$a\ cm=10\ cm$
Therefore,
The sides of the triangle are $12\times10\ cm=120\ cm, 17\times\ 10\ cm=170\ cm$ and $25\times10\ cm=250\ cm$
By Heron's formula:
$A=\sqrt{s(s-a)(s-b)(s-c)}$
Since,
$S=\frac{a+b+c}{2}$
$S=\frac{120+170+250}{2}$
$S=\frac{540}{2}$
$S=270\ cm$
This implies,
$A=\sqrt{270(270-120)(270-170)(270-250)}$
$A=\sqrt{270(150)(100)(20)}$
$A=9000\ cm^2$
Therefore,
The area of the triangle is $9000\ cm^2$.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
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# How do you find the domain and range and intercepTs for R(x) = (x^2 + x - 12)/(x^2 - 4)?
Apr 7, 2017
see explanation.
#### Explanation:
To find any $\textcolor{b l u e}{\text{excluded values}}$ on the domain.
The denominator of R(x) cannot be zero as this would make R(x) undefined. Evaluating the denominator to zero and solving gives the values that x cannot be.
$\text{solve } {x}^{2} - 4 = 0 \Rightarrow {x}^{2} = 4 \Rightarrow x = \pm 2$
$\Rightarrow \text{domain is } x \in \mathbb{R} , x \ne \pm 2$
To find $\textcolor{b l u e}{\text{excluded values}}$ on the range.
divide all terms on numerator/denominator by the highest power of x, that is ${x}^{2}$
$R \left(x\right) = \frac{{x}^{2} / {x}^{2} + \frac{x}{x} ^ 2 - \frac{12}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{4}{x} ^ 2} = \frac{1 + \frac{1}{x} - \frac{12}{x} ^ 2}{1 - \frac{4}{x} ^ 2}$
as $x \to \pm \infty , \frac{1}{x} , \frac{12}{x} ^ 2 , \frac{4}{x} ^ 2 \to 0$
${\lim}_{x \to \pm \infty} , R \left(x\right) \to \frac{1 + 0 - 0}{1 - 0}$
$= \frac{1}{1} = 1 \leftarrow \textcolor{red}{\text{ excluded value}}$
$\text{range is } y \in \mathbb{R} , y \ne 1$
$\textcolor{b l u e}{\text{intercepts}}$
• " let x = 0, in equation, for y-intercept"
• " let y = 0, in equation, for x-intercepts"
$x = 0 \to y = \frac{- 12}{- 4} = 3 \leftarrow \textcolor{red}{\text{ y-intercept}}$
The numerator is the only part of R ( x ) that can equal zero when y = 0
$\Rightarrow y = 0 \to {x}^{2} + x - 12 = 0$
$\Rightarrow \left(x + 4\right) \left(x - 3\right) = 0$
$\Rightarrow x = - 4 , x = 3 \leftarrow \textcolor{red}{\text{ x-intercepts}}$
graph{(x^2+x-12)/(x^2-4) [-10, 10, -5, 5]} |
# Illustrative Mathematics Grade 7, Unit 1, Lesson 1: What are Scaled Copies?
Learning Targets:
• I can describe some characteristics of a scaled copy.
• I can tell whether or not a figure is a scaled copy of another figure.
Related Pages
Illustrative Math
#### Lesson 1: What are Scaled Copies?
Let’s explore scaled copies.
Illustrative Math Unit 7.1, Lesson 1 (printable worksheets)
#### Lesson 1 Summary
The following diagram describes some characteristics of a scaled copy and how to tell whether or not a figure is a scaled copy of another figure.
#### Lesson 1.1 Printing Portraits
Here is a portrait of a student. Move the slider under each image, A–E, to see it change.
Open Applet
1. Which images are most like the original? Which are least like the original? Explain your reasoning.
2. Some of the sliders make scaled copies of the original portrait. Which of the Portraits A–E do you think are scaled copies? Explain your reasoning.
3. What do you think “scaled copy” means?
#### Lesson 1.2 Scaling F
On the top left is the original drawing of the letter F. There are also several other drawings.
1. Identify all the drawings that are scaled copies of the original letter F drawing. Explain how you know.
2. Examine all the scaled copies more closely, specifically, the lengths of each part of the letter F. How do they compare to the original? What do you notice?
3. On the grid, draw a different scaled copy of the original letter F.
Open Applet
#### Lesson 1.3 Pairs of Scaled Polygons
Your teacher will give you a set of cards that have polygons drawn on a grid. Mix up the cards and place them all face up.
1. Take turns with your partner to match a pair of polygons that are scaled copies of one another. a. For each match you find, explain to your partner how you know it’s a match. b. For each match your partner finds, listen carefully to their explanation, and if you disagree, explain your thinking.
2. When you agree on all of the matches, check your answers with the answer key. If there are any errors, discuss why and revise your matches.
3. Select one pair of polygons to examine further. Use the grid below to produce both polygons. Explain or show how you know that one polygon is a scaled copy of the other.
Open Applet
#### Are you ready for more?
Is it possible to draw a polygon that is a scaled copy of both Polygon A and Polygon B? Either draw such a polygon, or explain how you know this is impossible.
#### Lesson 1 Practice Problems
1. Is it possible to draw a polygon that is a scaled copy of both Polygon A and Polygon B? Either draw such a polygon, or explain how you know this is impossible.
2. Tyler says that Figure B is a scaled copy of Figure A because all of the peaks are half as tall.
Do you agree with Tyler? Explain your reasoning.
3. Here is a picture of the Rose Bowl Stadium in Pasadena, CA.
Here are some copies of the picture. Select all the pictures that are scaled copies of the original picture.
4. Complete each equation with a number that makes it true.
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
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. |
# Discriminant of a Quadratic Equation – Formula and Examples
Quadratic equations are algebraic equations that have the form ax²+bx+c=0. Considering this form, the discriminant of a quadratic equation is the value b²-4ac. This value goes inside the square root of the general quadratic formula and determines the type of solutions that we will have.
In this article, we will learn everything related to the discriminant of a quadratic equation. In addition, we will use its formula to solve some practice problems.
Relevant for
See examples
Relevant for
See examples
## Formula for the discriminant of a quadratic equation and applications
When we have quadratic equations written in the form $latex a{{x}^2}+bx+c=0$, the discriminant is given by the following formula:
The value of the discriminant gives us information about the type of roots that we will get by solving a given quadratic equation. Namely, we have the following:
• When $latex b^2-4ac>0$, the quadratic equation has two real roots.
• When $latex b^2-4ac<0$, the quadratic equation has no real roots.
• When $latex b^2-4ac=0$, the quadratic equation has a repeated root.
We can understand this better if we remember that the general quadratic formula, which allows us to solve any quadratic equation, is the following:
$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
Therefore, we can see that the discriminant is the expression inside the square root of the general formula. This means that, when the value inside the square root is positive, we will have two real roots.
On the other hand, when the value inside the square root is negative, we won’t have any real roots (but we will have imaginary or complex roots). And finally, if the discriminant is equal to zero, we will have a single repeated root.
The discriminant of a quadratic equation indicates whether the graph of the quadratic equation intersects the x-axis at two different points, does not intersect the x-axis, or touches the x-axis at only one point, as we see in the following diagram:
The formula for the discriminant of a quadratic equation is used to solve the following examples. Each example has its respective solution, but try to solve the problems yourself.
### EXAMPLE 1
Find the discriminant of the equation $latex x^2+4x+5=0$.
Using the values $latex a=1$, $latex b=4$ and $latex c=5$ in the formula for the discriminant, we have:
$latex D=b^2-4ac$
$latex D=4^2-4(1)(5)$
$latex D=16-20$
$latex D=-4$
The discriminant of the equation is $latex D=-4$.
### EXAMPLE 2
Determine the discriminant of the equation $latex -2x^2+4x-2$.
We have the values $latex a=-2$, $latex b=4$, and $latex c=-2$. Using these values in the formula for the discriminant, we have:
$latex D=b^2-4ac$
$latex D=4^2-4(-2)(-2)$
$latex D=16-16$
$latex D=0$
The discriminant of the equation is $latex D=0$.
### EXAMPLE 3
Use the discriminant to show that the equation $latex 5x^2+4x+10=0$ has no real solutions.
Using the discriminant formula with the values $latex a=5$, $latex b=4$ and $latex c=10$, we have:
$latex D=b^2-4ac$
$latex D=4^2-4(5)(10)$
$latex D=16-200$
$latex D=-184$
The discriminant is negative, so the equation has no real roots. This is because we have a negative number inside the square root of the general quadratic formula.
### EXAMPLE 4
How many real roots does the equation $latex -3x^2-2x+5=0$ have?
We use the values $latex a=-3$, $latex b=-2$ and $latex c=5$ in the formula for the discriminant, and we have:
$latex D=b^2-4ac$
$latex D=(-2)^2-4(-3)(5)$
$latex D=4+60$
$latex D=64$
The discriminant is positive, so the equation has two real roots.
### EXAMPLE 5
Find the value of k in the equation $latex 3x^2-6x+k=0$, assuming that the equation has only one repeated root.
In this case, we have $latex a=3$, $latex b=-6$, and $latex c=k$. To find the value of k, we have to use the formula for the discriminant that has a value equal to zero, since the equation has only one root:
$latex D=b^2-4ac=0$
$latex (-6)^2-4(3)(k)=0$
$latex 36-12k=0$
$latex -12k=-36$
$latex k=3$
Therefore, if the quadratic equation has a single repeated root, the value of k is 3.
### EXAMPLE 6
Find the value of k in the equation $latex kx^2+5x-4=0$ if the equation has only one root.
We have the values $latex a=k$, $latex b=5$ and $latex c=-4$. We use the formula for the discriminant, set it equal to zero, and solve for k:
$latex D=b^2-4ac=0$
$latex (5)^2-4(k)(-4)=0$
$latex 25+16k=0$
$latex 16k=-25$
$latex k=-\frac{25}{16}$
Therefore, if the quadratic equation has a single repeated root, the value of k is 25/16.
### EXAMPLE 7
Prove that $latex y=2x^2+7x+7$ is always positive by using the discriminant.
Using the values $latex a=2$, $latex b=7$ and $latex c=7$ in the formula for the discriminant, we have:
$latex D=b^2-4ac$
$latex D=(7)^2-4(2)(7)$
$latex D=49-56$
$latex D=-7$
Since the discriminant is negative, the equation has no real roots. Then, the function does not cut the x-axis. Also, since the coefficient of $latex x^2$ is positive, the graph of the function opens up and is always above the x-axis, so the values are always positive.
## Discriminant of a quadratic equation – Practice problems
Use the formula for the discriminant of a quadratic equation to solve the following practice problems. |
Page 1 / 5
This module is part of a collection of modules that were developed to support laboratory activities in a Precalculus for PreEngineers (MATH 1508) at the University of Texas at El Paso. Contained in this module of applications of quadratic equations in various fields of engineering and science. These include the motion of an object under constant acceleration, quantitative management, and break-even analysis.
## Introduction
Quadratic equations play an important role in the modeling of many physical situations. Finding the roots of quadratic equations is a necessary skill. Being able to interpret these roots is an important ability that is important in understanding physical problems. In this module, we will present a number of applications of quadratic equations in several fields of engineering.
## Determining the roots of quadratic equations
A quadratic equation has the following form
${\text{ax}}^{2}+\text{bx}+c=0$
Because a quadratic equation involves a polynomial of order 2, it will have two roots. In general, a quadratic equation will either have two roots that are both real or have two roots that are both complex. For the present module, we will restrict our attention to quadratic equations that have two real roots.
There are three methods that are effective in solving for the roots of a quadratic equation. They are:
• Solution by factoring
• Solution by completing the square
• Solution by the quadratic formula
The applications that follow will include examples of each of these three methods of solution.
## Motion of an object under uniform acceleration
We will begin our study of quadratic equations by considering an application that you will likely encounter later in physics and mechanical engineering classes. Let us consider an object that is subject to a uniform acceleration. By uniform, we mean an acceleration that is constant. Such an object might be an automobile, an aircraft, a rocket, etc. The motion of an object subjected to uniform acceleration can be expressed mathematically by the following equation.
$s\left(t\right)=\frac{1}{2}a{t}^{2}+{v}_{0}t+{s}_{0}$
where s ( t ) represents the position of the object as function of time t ,
a represents the constant acceleration of the object,
v 0 represents the value of the object’s velocity at time t = 0, and
s 0 represents the position of the object at time t = 0.
An equation of this sort is called an equation of motion . We will illustrate its use in the following exercise.
Example 1: For our first example, let us consider a dragster on a drag strip of length one-quarter mile. For time t<0, the dragster is at rest at the starting line. At time = 0, the driver depresses his gas pedal to produce a uniform acceleration of 50 m/s 2 . Under these conditions, how far will the dragster travel in 1 second?
Because the dragster travels in a horizontal direction, we will represent its distance from the starting point as a fuction of time as x ( t ). We also know that the value for the acceleration ( a ) is 30 m / s 2 . We can incorporate these changes in equation (1) to produce a new equation of motion for the dragster.
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Can someone give me problems that involes radical expressions like area,volume or motion of pendulum with solution |
In other words, f: A!Bde ned by f: x7!f(x) is the full de nition of the function f. Determine whether or not the restriction of an injective function is injective. We will use the contrapositive approach to show that g is injective. A function $f: A \rightarrow B$ is surjective (onto) if the image of f equals its range. One example is $y = e^{x}$ Let us see how this is injective and not surjective. A function f from a set X to a set Y is injective (also called one-to-one) if distinct inputs map to distinct outputs, that is, if f(x 1) = f(x 2) implies x 1 = x 2 for any x 1;x 2 2X. Then in the conclusion, we say that they are equal! If $f(x_1) = f(x_2)$, then $2x_1 – 3 = 2x_2 – 3$ and it implies that $x_1 = x_2$. f: X → Y Function f is one-one if every element has a unique image, i.e. $f: R\rightarrow R, f(x) = x^2$ is not injective as $(-x)^2 = x^2$. Here's how I would approach this. For any amount of variables $f(x_0,x_1,…x_n)$ it is easy to create a “ugly” function that is even bijective. The function f: R … Therefore, fis not injective. You can find out if a function is injective by graphing it. f(x) = x3 We need to check injective (one-one) f (x1) = (x1)3 f (x2) = (x2)3 Putting f (x1) = f (x2) (x1)3 = (x2)3 x1 = x2 Since if f (x1) = f (x2) , then x1 = x2 It is one-one (injective) Let f: A → B be a function from the set A to the set B. Simplifying the equation, we get p =q, thus proving that the function f is injective. Next let’s prove that the composition of two injective functions is injective. B is bijective (a bijection) if it is both surjective and injective. It is a function which assigns to b, a unique element a such that f(a) = b. hence f -1 (b) = a. Informally, fis \surjective" if every element of the codomain Y is an actual output: XYf fsurjective fnot surjective XYf Here is the formal de nition: 4. The different mathematical formalisms of the property … So, to get an arbitrary real number a, just take x = 1, y = (a + 1)/2 Then f (x, y) = a, so every real number is in the range of f, and so f is surjective (assuming the codomain is the reals) Write the Lagrangean function and °nd the unique candidate to be a local maximizer/minimizer of f (x; y) subject to the given constraint. This means that for any y in B, there exists some x in A such that $y = f(x)$. Mathematics A Level question on geometric distribution? It is easy to show a function is not injective: you just find two distinct inputs with the same output. There can be many functions like this. This is equivalent to the following statement: for every element b in the codomain B, there is exactly one element a in the domain A such that f(a)=b.Another name for bijection is 1-1 correspondence (read "one-to-one correspondence).. Assuming the codomain is the reals, so that we have to show that every real number can be obtained, we can go as follows. Properties of Function: Addition and multiplication: let f1 and f2 are two functions from A to B, then f1 + f2 and f1.f2 are defined as-: f1+f2(x) = f1(x) + f2(x). De nition. Inverse Functions:Bijection function are also known as invertible function because they have inverse function property. Let f : A !B. f: X → Y Function f is one-one if every element has a unique image, i.e. Then f has an inverse. The differential of f is invertible at any x\in U except for a finite set of points. when f(x 1 ) = f(x 2 ) ⇒ x 1 = x 2 Otherwise the function is many-one. Use the gradient to find the tangent to a level curve of a given function. If it is, prove your result. As we established earlier, if $$f : A \to B$$ is injective, then the restriction of the inverse relation $$f^{-1}|_{\range(f)} : \range(f) \to A$$ is a function. (a) Consider f (x; y) = x 2 + 2 y 2, subject to the constraint 2 x + y = 3. A function is injective (one-to-one) if each possible element of the codomain is mapped to by at most one argument. Look for areas where the function crosses a horizontal line in at least two places; If this happens, then the function changes direction (e.g. This proves that is injective. If the function satisfies this condition, then it is known as one-to-one correspondence. Equivalently, a function is injective if it maps distinct arguments to distinct images. Transcript. Since f is both surjective and injective, we can say f is bijective. Problem 1: Every convergent sequence R3 is bounded. Whether functions are subjective is a philosophical question that I’m not qualified to answer. A function g : B !A is the inverse of f if f g = 1 B and g f = 1 A. Theorem 1. All injective functions from ℝ → ℝ are of the type of function f. Example 2.3.1. All injective functions from ℝ → ℝ are of the type of function f. If you think that it is true, prove it. Consider a function f (x; y) whose variables x; y are subject to a constraint g (x; y) = b. For many students, if we have given a different name to two variables, it is because the values are not equal to each other. A function $f: A \rightarrow B$ is bijective or one-to-one correspondent if and only if f is both injective and surjective. Proposition 3.2. There are two types of special properties of functions which are important in many di erent mathematical theories, and which you may have seen. BUT if we made it from the set of natural numbers to then it is injective, because: f(2) = 4 ; there is no f(-2), because -2 is not a natural number; So the domain and codomain of each set is important! X. Proof. POSITION() and INSTR() functions? As Q 2is dense in R , if D is any disk in the plane, then we must If you get confused doing this, keep in mind two things: (i) The variables used in defining a function are “dummy variables” — just placeholders. $f : N \rightarrow N, f(x) = x + 2$ is surjective. Functions find their application in various fields like representation of the computational complexity of algorithms, counting objects, study of sequences and strings, to name a few. This shows 8a8b[f(a) = f(b) !a= b], which shows fis injective. Still have questions? It also easily can be extended to countable infinite inputs First define $g(x)=\frac{\mathrm{atan}(x)}{\pi}+0.5$. Example 2.3.1. Prove that the function f: N !N be de ned by f(n) = n2 is injective. κ. If it isn't, provide a counterexample. Lv 5. Now suppose . The receptionist later notices that a room is actually supposed to cost..? Mathematical Functions in Python - Special Functions and Constants, Difference between regular functions and arrow functions in JavaScript, Python startswith() and endswidth() functions, Python maketrans() and translate() functions. Assuming m > 0 and m≠1, prove or disprove this equation:? surjective) in a neighborhood of p, and hence the rank of F is constant on that neighborhood, and the constant rank theorem applies. distinct elements have distinct images, but let us try a proof of this. Then , or equivalently, . Example. It's not the shortest, most efficient solution, but I believe it's natural, clear, revealing and actually gives you more than you bargained for. Then f is injective. So, $x = (y+5)/3$ which belongs to R and $f(x) = y$. Therefore . $f : R \rightarrow R, f(x) = x^2$ is not surjective since we cannot find a real number whose square is negative. Working with a Function of Two Variables. We have to show that f(x) = f(y) implies x= y. Ok, let us take f(x) = f(y), that is two images that are the same. Another exercise which has a nice contrapositive proof: prove that if are finite sets and is an injection, then has at most as many elements as . A function is injective if for every element in the domain there is a unique corresponding element in the codomain. We will de ne a function f 1: B !A as follows. The composition of two bijections is again a bijection, but if g o f is a bijection, then it can only be concluded that f is injective and g is surjective (see the figure at right and the remarks above regarding injections … Prove or disprove that if and are (arbitrary) functions, and if the composition is injective, then both of must be injective. Since the domain of fis the set of natural numbers, both aand bmust be nonnegative. Which of the following can be used to prove that △XYZ is isosceles? Erratic Trump has military brass highly concerned, 'Incitement of violence': Trump is kicked off Twitter, Some Senate Republicans are open to impeachment, 'Xena' actress slams co-star over conspiracy theory, Fired employee accuses star MLB pitchers of cheating, Unusually high amount of cash floating around, Flight attendants: Pro-Trump mob was 'dangerous', These are the rioters who stormed the nation's Capitol, 'Angry' Pence navigates fallout from rift with Trump, Late singer's rep 'appalled' over use of song at rally. In Mathematics, a bijective function is also known as bijection or one-to-one correspondence function. One example is $y = e^{x}$ Let us see how this is injective and not surjective. The formulas in this theorem are an extension of the formulas in the limit laws theorem in The Limit Laws. The rst property we require is the notion of an injective function. 6. In particular, we want to prove that if then . Proof. Suppose (m, n), (k, l) ∈ Z × Z and g(m, n) = g(k, l). 1 decade ago. Thus a= b. If given a function they will look for two distinct inputs with the same output, and if they fail to find any, they will declare that the function is injective. Instead, we use the following theorem, which gives us shortcuts to finding limits. Therefore, we can write z = 5p+2 and z = 5q+2 which can be thus written as: 5p+2 = 5q+2. 1.4.2 Example Prove that the function f: R !R given by f(x) = x2 is not injective. ... will state this theorem only for two variables. Show that A is countable. Proof. Assuming the codomain is the reals, so that we have to show that every real number can be obtained, we can go as follows. In this article, we are going to discuss the definition of the bijective function with examples, and let us learn how to prove that the given function is bijective. This is especially true for functions of two variables. On the other hand, multiplying equation (1) by 2 and adding to equation (2), we get , or equivalently, . Determine the gradient vector of a given real-valued function. If f: A ! Let f : A !B be bijective. Interestingly, it turns out that this result helps us prove a more general result, which is that the functions of two independent random variables are also independent. Relevance. To prove one-one & onto (injective, surjective, bijective) One One function. The term one-to-one correspondence should not be confused with the one-to-one function (i.e.) Last updated at May 29, 2018 by Teachoo. In other words there are two values of A that point to one B. This means a function f is injective if $a_1 \ne a_2$ implies $f(a1) \ne f(a2)$. QED. The value g(a) must lie in the domain of f for the composition to make sense, otherwise the composition f(g(a)) wouldn't make sense. Not Injective 3. encodeURI() and decodeURI() functions in JavaScript. Conversely, if the composition ∘ of two functions is bijective, it only follows that f is injective and g is surjective.. Cardinality. Prove that a function $f: R \rightarrow R$ defined by $f(x) = 2x – 3$ is a bijective function. Show that the function g: Z × Z → Z × Z defined by the formula g(m, n) = (m + n, m + 2n), is both injective and surjective. Step 2: To prove that the given function is surjective. Let f: R — > R be defined by f(x) = x^{3} -x for all x \in R. The Fundamental Theorem of Algebra plays a dominant role here in showing that f is both surjective and not injective. is a function defined on an infinite set . The inverse function theorem in infinite dimension The implicit function theorem has been successfully generalized in a variety of infinite-dimensional situations, which proved to be extremely useful in modern mathematics. 1.5 Surjective function Let f: X!Y be a function. Let f : A !B be bijective. 2. are elements of X. such that f (x. The equality of the two points in means that their coordinates are the same, i.e., Multiplying equation (2) by 2 and adding to equation (1), we get . Contrapositively, this is the same as proving that if then . As we have seen, all parts of a function are important (the domain, the codomain, and the rule for determining outputs). It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. Explanation − We have to prove this function is both injective and surjective. A Function assigns to each element of a set, exactly one element of a related set. (addition) f1f2(x) = f1(x) f2(x). When the derivative of F is injective (resp. Please Subscribe here, thank you!!! In general, you can tell if functions like this are one-to-one by using the horizontal line test; if a horizontal line ever intersects the graph in two di er-ent places, the real-valued function is not injective… A function f:A→B is injective or one-to-one function if for every b∈B, there exists at most one a∈A such that f(s)=t. How MySQL LOCATE() function is different from its synonym functions i.e. In mathematics, a bijective function or bijection is a function f : A → B that is both an injection and a surjection. An injective function must be continually increasing, or continually decreasing. Let a;b2N be such that f(a) = f(b). It takes time and practice to become efficient at working with the formal definitions of injection and surjection. We say that f is bijective if it is both injective and surjective. A more pertinent question for a mathematician would be whether they are surjective. Proving that a limit exists using the definition of a limit of a function of two variables can be challenging. The term bijection and the related terms surjection and injection … By definition, f. is injective if, and only if, the following universal statement is true: Thus, to prove . If not, give a counter-example. The French word sur means over or above, and relates to the fact that the image of the domain of a surjective function … Find stationary point that is not global minimum or maximum and its value . Using the previous idea, we can prove the following results. Solution We have 1; 1 2R and f(1) = 12 = 1 = ( 1)2 = f( 1), but 1 6= 1. So, to get an arbitrary real number a, just take, Then f(x, y) = a, so every real number is in the range of f, and so f is surjective. For functions of a single variable, the theorem states that if is a continuously differentiable function with nonzero derivative at the point a; then is invertible in a neighborhood of a, the inverse is continuously differentiable, and the derivative of the inverse function at = is the reciprocal of the derivative of at : (−) ′ = ′ = ′ (− ()).An alternate version, which assumes that is continuous and … Students can look at a graph or arrow diagram and do this easily. https://goo.gl/JQ8NysHow to prove a function is injective. Prove a two variable function is surjective? Equivalently, for all y2Y, the set f 1(y) has at most one element. Favorite Answer. atol(), atoll() and atof() functions in C/C++. The function f is called an injection provided that for all x1, x2 ∈ A, if x1 ≠ x2, then f(x1) ≠ f(x2). f. is injective, you will generally use the method of direct proof: suppose. Please Subscribe here, thank you!!! An injective (one-to-one) function A surjective (onto) function A bijective (one-to-one and onto) function A few words about notation: To de ne a speci c function one must de ne the domain, the codomain, and the rule of correspondence. Why and how are Python functions hashable? Prove that a composition of two injective functions is injective, and that a composition of two surjective functions is surjective. when f(x 1 ) = f(x 2 ) ⇒ x 1 = x 2 Otherwise the function is many-one. When f is an injection, we also say that f is a one-to-one function, or that f is an injective function. Example. https://goo.gl/JQ8Nys Proof that the composition of injective(one-to-one) functions is also injective(one-to-one) Determine the directional derivative in a given direction for a function of two variables. Please Subscribe here, thank you!!! If X and Y are finite sets, then there exists a bijection between the two sets X and Y if and only if X and Y have the same number of elements. https://goo.gl/JQ8Nys Proof that the composition of injective(one-to-one) functions is also injective(one-to-one) No, sorry. One example is the function x 4, which is not injective over its entire domain (the set of all real numbers). 2 W k+1 6(1+ η k)kx k −zk2 W k +ε k, (∀k ∈ N). For example, f(a,b) = (a+b,a2 +b) defines the same function f as above. Functions Solutions: 1. 5. the composition of two injective functions is injective 6. the composition of two surjective functions is surjective 7. the composition of two bijections is bijective Example $$\PageIndex{3}$$: Limit of a Function at a Boundary Point. Injective functions are also called one-to-one functions. Statement. The simple linear function f(x) = 2 x + 1 is injective in ℝ (the set of all real numbers), because every distinct x gives us a distinct answer f(x). How to check if function is one-one - Method 1 In this method, we check for each and every element manually if it has unique image Therefore fis injective. This concept extends the idea of a function of a real variable to several variables. The function … injective function. Injective 2. △XYZ is given with X(2, 0), Y(0, −2), and Z(−1, 1). Get your answers by asking now. It is clear from the previous example that the concept of difierentiability of a function of several variables should be stronger than mere existence of partial derivatives of the function. function of two variables a function $$z=f(x,y)$$ that maps each ordered pair $$(x,y)$$ in a subset $$D$$ of $$R^2$$ to a unique real number $$z$$ graph of a function of two variables a set of ordered triples $$(x,y,z)$$ that satisfies the equation $$z=f(x,y)$$ plotted in three-dimensional Cartesian space level curve of a function of two variables Misc 5 Show that the function f: R R given by f(x) = x3 is injective. Equivalently, for every $b \in B$, there exists some $a \in A$ such that $f(a) = b$. Then f(x) = 4x 1, f(y) = 4y 1, and thus we must have 4x 1 = 4y 1. Thus we need to show that g(m, n) = g(k, l) implies (m, n) = (k, l). This implies a2 = b2 by the de nition of f. Thus a= bor a= b. Step 1: To prove that the given function is injective. f(x, y) = (2^(x - 1)) (2y - 1) And not. Passionately Curious. Let b 2B. Thus fis injective if, for all y2Y, the equation f(x) = yhas at most one solution, or in other words if a solution exists, then it is unique. If a function is defined by an even power, it’s not injective. A function is said to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. They pay 100 each. Injective Functions on Infinite Sets. 1 Answer. Surjective (Also Called "Onto") A … That is, if and are injective functions, then the composition defined by is injective. This means a function f is injective if a1≠a2 implies f(a1)≠f(a2). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Consider the function g: R !R, g(x) = x2. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.. Visit Stack Exchange f(x,y) = 2^(x-1) (2y-1) Answer Save. 2. Join Yahoo Answers and get 100 points today. But then 4x= 4yand it must be that x= y, as we wanted. from increasing to decreasing), so it isn’t injective. 2 (page 161, # 27) (a) Let A be a collection of circular disks in the plane, no two of which intersect. Injective Bijective Function Deflnition : A function f: A ! Conclude a similar fact about bijections. $f: N \rightarrow N, f(x) = x^2$ is injective. (multiplication) Equality: Two functions are equal only when they have same domain, same co-domain and same mapping elements from domain to co-domain. There can be many functions like this. Theorem 3 (Independence and Functions of Random Variables) Let X and Y be inde-pendent random variables. (7) For variable metric quasi-Feje´r sequences the following re-sults have already been established [10, Proposition 3.2], we provide a proof in Appendix A.1 for completeness. $f: N \rightarrow N, f(x) = 5x$ is injective. 2 2A, then a 1 = a 2. Now as we're considering the composition f(g(a)). De nition 2. A function f: X!Y is injective or one-to-one if, for all x 1;x 2 2X, f(x 1) = f(x 2) if and only if x 1 = x 2. De nition 2.3. I'm guessing that the function is . A function $f: A \rightarrow B$ is injective or one-to-one function if for every $b \in B$, there exists at most one $a \in A$ such that $f(s) = t$. surjective) at a point p, it is also injective (resp. Explain the significance of the gradient vector with regard to direction of change along a surface. To prove injection, we have to show that f (p) = z and f (q) = z, and then p = q. In mathematical analysis, and applications in geometry, applied mathematics, engineering, natural sciences, and economics, a function of several real variables or real multivariate function is a function with more than one argument, with all arguments being real variables. For functions of more than one variable, ... A proof of the inverse function theorem. ... $\begingroup$ is how to formally apply the property or to prove the property in various settings, and this applies to more than "injective", which is why I'm using "the property". 1. and x. The inverse of bijection f is denoted as f -1 . Are all odd functions subjective, injective, bijective, or none? 2 2X. The term surjective and the related terms injective and bijective were introduced by Nicolas Bourbaki, a group of mainly French 20th-century mathematicians who, under this pseudonym, wrote a series of books presenting an exposition of modern advanced mathematics, beginning in 1935. x. 3 friends go to a hotel were a room costs $300. Prove … f . Write two functions isPrime and primeFactors (Python), Virtual Functions and Runtime Polymorphism in C++, JavaScript encodeURI(), decodeURI() and its components functions. You have to think about the two functions f & g. You can define g:A->B, so take an a in A, g will map this from A into B with a value g(a). See the lecture notesfor the relevant definitions. Example 99. 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The set f 1 ( y ) = x^2$ is surjective to distinct images but! And decodeURI ( ) functions in JavaScript... will state this theorem an. Is one-one if every element has a unique image, i.e. any x\in except. Can be thus written as: 5p+2 = 5q+2, to prove be with! Function of two surjective prove a function of two variables is injective is injective by graphing it a more question! Bor a= b we also say that they are equal: to one-one. Direction for a finite set of points U except for a mathematician would be whether they are equal state! 0 and m≠1, prove or disprove this equation: ) functions in C/C++ >! Only for two variables can be challenging 2: to prove that △XYZ is isosceles 8a8b [ f ( ). N, f ( x 2 Otherwise the function is many-one of fis the set f 1 ( y has. From ℝ → ℝ are of the type of function f. if you think that is... 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Bijection and the related terms surjection and injection … Here 's how would! Say that they are equal derivative in a given function find out a! Then a 1 = x 2 ) ⇒ x 1 ) = z, bijective ) one! Must prove a function of two variables is injective continually increasing, or continually decreasing x and y be a at. N \rightarrow N, f ( a bijection ) if it is easy to show a function two! A1≠A2 implies f ( x 1 = x 2 ) ⇒ x 1 = x 2 Otherwise the f... A given real-valued function Otherwise the function … Please Subscribe Here, you. Room costs$ 300 1+ η k ) kx k −zk2 W +ε! The notion of an injective function be that x= y, as we wanted: a → that..., thank you!!!!!!!!!!... Is bounded that I ’ m not qualified to answer theorem in the codomain is mapped by! Can write z = 5p+2 and z = 5q+2 f1f2 ( x 2 the! ], which is not injective over its entire domain ( the set of points )... 1.5 surjective function let f: R! R given by f ( a, b ) a=! ], which shows fis injective both aand bmust be nonnegative: 5p+2 = 5q+2 are injective from! Set of natural numbers, both aand bmust be nonnegative injective and surjective function … Please Subscribe Here thank! A philosophical question that I ’ m not qualified to answer universal statement is true: thus to. Theorem only for two variables term one-to-one correspondence should not be confused with the definitions! Be challenging we will use the method of direct proof: suppose sequence R3 is.... Convergent sequence R3 is bounded explanation − we have to prove that a limit exists using the of. A mathematician would be whether they are equal + 2 $is surjective generally the... It ’ s not injective and not and its value implies f ( x ) = f ( x 1... Distinct arguments to distinct images, but let us try a proof this. 3 } \ ): limit of a function is injective and$ f: N N! M≠1, prove it which gives us shortcuts to finding limits W +ε. F is a unique image, i.e. MySQL LOCATE ( ) and atof ( ) functions in JavaScript a+b! = 5p+2 and z = 5q+2 injective: you just find two distinct inputs with formal... = x^2 \$ is surjective answer Save an even power, it is an! Variable to several variables bijection and the related terms surjection and injection Here! The notion of an injective function extends the idea of a given direction for a f! For every element has a unique image, i.e. )! a= b f. thus a= bor b... Example \ ( \PageIndex { 3 } \ ): limit of function. One-To-One function ( i.e. 're considering the composition f ( p ) = x3 is if!
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Observing Student Representations
Introduction | Building Viewpoints | Student Work Reflection #1 | More Building Viewpoints | Student Work Reflection #2 | Observe a Classroom | Classroom Practice | Your Journal
Now let's look at how a middle school student solved the following problem:
This is the floor plan (or top view) of a structure made from cubes. The numbers in the squares indicate the number of cubes in that stack. Construct this building on the Building Mat.
Reflect on the problem and think about some questions you would ask this student. Then look at the questions listed below. For each question, think about an answer the student might provide, then select "Show Answer" to reveal a sample response.
As you read these questions and answers, analyze the student's representation. How does this representation help the student identify relationships among the different views of the building?
Teacher: Can you draw what you would see if you were in front of the structure, looking at it from eye level? Show Answer
Student: Yes:
Teacher: That is what the building would look like if you were looking down at it from the sky -- what we call a "bird's eye view." If you, get down and look at the front view of the building, at your eye level, can you draw what you see then? Show Answer
Student: (observes the building from eye level) Oh, okay:
Teacher: What would you see if you were in back of the structure, looking at it from your eye level? Can you draw that? Show Answer
Student: Yes:
Teacher: How is the view from the back different from the view from the front? Show Answer
Student: It is reversed.
Teacher: What would you see if you were on the right side of the structure, looking at it from your eye level? Can you draw that? Show Answer
Student: Yes:
Teacher: And now can you draw what you would see if you were on the left side of the structure, looking at it from your eye level? Show Answer
Student: Yes:
Teacher: How is the view from the right side different from the view from the left side? Show Answer
Student: It is reversed.
Teacher: Is there another name for this relationship? Show Answer
Student: It's a mirror image, or symmetrical.
Teacher: If you were under the building and looking straight up, what would you see? Show Answer
Student: It would be the same shape as from the top. (Note: The orientation of the shape would depend on where the student was standing: at the front of the building, at the back, at an angle, etc.)
Teaching Math Home | Grades 6-8 | Representation | Site Map | © | |
In elementary geometry, a polygon is a plane figure that is bounded by a finite chain of straight line segments closing in a loop to form a closed chain or circuit. Or simply a closed plane figure bounded by three or more line segment to form a closed loop is called a polygon.
Nomenclature:
• The line segments forming the polygon are called sides.
• The point of intersection of two line segments is called a vertex.
• Number of vertices of a polygon is equal to the number of line segments or sides.
Different types of Polygons: This is based on the number of sides that the polygon has. Here are few examples:
Diagonal of a Polygon: A line segment joining any two non-consecutive vertices is called
a diagonal of the polygon.
The dotted lines are diagonals of the shown polygons.
Interior and Exterior Angles of a Polygon: This is an important concept.
$\angle 1 \ and\ \angle 2$ are interior angles. These are made by the two sides of the polygon.
$\angle 3$ is called an exterior angle. This is formed by extending a side of the polygon as shown in the adjacent figure.
Just by looking at the figure you can tell that
$\angle 2 + \angle 3 = 180^{\circ}$
Hence we can say that:
Exterior Angle + Adjacent Interior Angle $= 180^{\circ}$
Convex Polygon vs Concave Polygon
If the interior angle of the polygon is less than $= 180^{\circ}$, then it is called convex polygon. If you look at any of the polygons shown above, you will see that all the interior angles are less than $= 180^{\circ}$.
But there can be cases where the interior angle of a polygon could be more than $= 180^{\circ}$. Take a look at the adjacent figure. Here you will see that $\angle 1 > 180^{\circ}$ (which is a reflex angle).
Regular Polygon: A polygon that satisfies the following condition is called a regular polygon.
1. All sides are equal
2. All interior angles are equal
3. All exterior angles are equal
For a regular polygon with $n$ sides we have the following:
$\displaystyle \text{Each Interior Angle }= \Bigg[\frac{(2n-4)\times 90^{\circ}}{n} \Bigg]^{\circ}$
Proof:
A polygon can be divided into $(n-2)$ triangles.
See a few examples in the adjacent figure.
We know that the sum of the angles of a triangle is $= 180^{\circ}$.
$\text{Therefore the sum of the interior angles }= (n-2) \times 180 = (2n - 4) \times 90^{\circ}$
$\Rightarrow \displaystyle \text{Interior Angle } = \frac{(2n-4)\times 90^{\circ}}{n}$
$\displaystyle \text{Each Exterior Angle } = \Big[\frac{360}{n} \Big]^{\circ}$
Proof:
$\text{Sum of all Interior Angles + Sum of all Exterior Angles } = n \times 180 = n \times (2 \times 90)^{\circ}$
$\text{Sum of all Exterior Angles }= n \times (2 \times 90)^{\circ} - (2n - 4) \times 90^{\circ}=360^{\circ}$
$\displaystyle \text{Each Exterior Angle } = \Big[\frac{360}{n} \Big]^{\circ}$
From the above point 2, we can also say that
$\displaystyle n = \Big[\frac{360}{\text{Each Exterior Angle}} \Big]^{\circ}$
$\text{Exterior Angle + Adjacent Interior Angle }= 180^{\circ}$
$\Rightarrow \text{Exterior Angle } =180^{\circ} - \text{Adjacent Interior Angle}$ |
# How do you multiply (2.4 * 10^-5)(4 * 10^-4)?
Mar 14, 2018
See a solution process below:
#### Explanation:
First, rewrite this expression as:
$\left(2.4 \cdot 4\right) \cdot \left({10}^{-} 5 \cdot {10}^{-} 4\right) \implies$
$9.6 \cdot \left({10}^{-} 5 \cdot {10}^{-} 4\right)$
Now, use this rule of exponents to multiply the 10s terms:
${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$
$9.6 \cdot \left({10}^{\textcolor{red}{- 5}} \cdot {10}^{\textcolor{b l u e}{- 4}}\right) \implies$
$9.6 \cdot {10}^{\textcolor{red}{- 5} + \textcolor{b l u e}{- 4}} \implies$
$9.6 \cdot {10}^{-} 9$ |
# What is the derivative of w =sqrt(x^2+y^2+z^2)?
Aug 30, 2015
$\frac{\partial w}{\partial x} = \frac{x}{\sqrt{{x}^{2} + {y}^{2} + {z}^{2}}}$
$\frac{\partial w}{\partial y} = \frac{y}{\sqrt{{x}^{2} + {y}^{2} + {z}^{2}}}$
$\frac{\partial w}{\partial z} = \frac{z}{\sqrt{{x}^{2} + {y}^{2} + {z}^{2}}}$
#### Explanation:
Since you're dealing with a multivariable function, you must treat $x$, $y$, and $z$ as independent variables and calculate the partial derivative of $w$, your dependent variable, with respect to $x$, $y$, and $z$.
When you differentiate with respect to $x$, you treat $y$ and $z$ as constants. LIkewise, when you differentiate with respect to $y$, you treat $x$ and $z$ as constants, and so on.
So, let's first differentiate $w$ with respect to $x$
$\frac{\partial w}{\partial x} = \frac{\partial}{\partial x} \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}$
$\frac{\partial w}{\partial x} = \frac{1}{2} \cdot {\left({x}^{2} + {y}^{2} + {z}^{2}\right)}^{- \frac{1}{2}} \cdot \frac{\partial}{\partial x} \left({x}^{2} + {y}^{2} + {z}^{2}\right)$
$\frac{\partial w}{\partial x} = \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \cdot \frac{1}{\sqrt{{x}^{2} + {y}^{2} + {z}^{2}}} \cdot \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x + 0 + 0\right)$
$\frac{\partial w}{\partial x} = \textcolor{g r e e n}{\frac{x}{\sqrt{{x}^{2} + {y}^{2} + {z}^{2}}}}$
You don't need to calculate the other two partial derivatives, all you have to do is recognize that the only thing that changes when you differentiate with respect to $y$ is that you get
$\frac{\partial}{\partial y} \left({x}^{2} + {y}^{2} + {z}^{2}\right) = 0 + 2 y + 0$
The same is true for the deivative with respect to $z$
$\frac{\partial}{\partial z} \left({x}^{2} + {y}^{2} + {z}^{2}\right) = 0 + 0 + 2 z$
This means that you have
$\frac{\partial w}{\partial y} = \textcolor{g r e e n}{\frac{y}{\sqrt{{x}^{2} + {y}^{2} + {z}^{2}}}}$
and
$\frac{\partial w}{\partial x} = \textcolor{g r e e n}{\frac{z}{\sqrt{{x}^{2} + {y}^{2} + {z}^{2}}}}$ |
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Fractions
Go back to 'Maths'
Fractions
Fractions are introduced around grade 3 (typically for students aged 8 or 9). They present a significant challenge for many students. Till now, children have only worked with whole numbers. Suddenly children are shown numbers that have two parts to them. There is a number at the top and another below. Difficult sounding terms like numerator and denominator are introduced. Children only know that each whole number has only one value. Now, two numbers separated by a symbol (½, ¾, etc.) represents a single number adding to the complexity.
What is a Fraction?
Fractions are a part of a whole.
Best ways to learn fractions
Visualisation is the best way to learn fractions. Visualisation builds an understanding of fractions as a concept and builds intuition. While teaching fractions emphasis should be on the situations where the object can easily be cut, folded, split or coloured in equal parts. Different manipulatives can also be used which helps in visualising fractions as a part of a whole.
Using shapes
Allow children to compare different visual representations where some part of the figure is shaded. Let children see that a whole has been divided into equal parts and a few of them have been shaded.
Using strips
Here we represent a strip divided into three equal parts. For the abstract representation, the total number of parts form the denominator. Now, by counting the number of shaded parts, the numerator of the fraction can be obtained.
Using number line
Ask children to divide the given number line into say three equal parts and then ask the represent each part as fractions. This is extremely helpful as this enables children to see that fractions are numbers, which can be placed on the number line.
Topics closely related to fractions
The chart given below shows the topics that are a pre-requisite to understanding fractions. It also shows topics that are related to fractions.
Sub Topics
Here are a few links that will take you through the journey that every Cuemath students undertake in the pursuit of understanding Fractions along with practice worksheets:
Simple ways to learn fractions
Here are a few time-tested and effective ways that you can use to help your child out at home:
Practice and more practice:
After understanding the concept then keep practising it. Nothing helps children learn faster than applying the concepts that they are taught. Research has shown that it allows them to cement their concepts and it leads to longer retention.
Address the common doubts:
Clear the common doubts that children usually have while comparing, ordering or operating on fractions.
1. Clear doubts such as when comparing fractions, one can only compare like fractions. Explain to the child this is primarily because fractions represent parts of the same whole, so we utilize the concepts of Equivalent Fractions to make them comparable.
Common mistakes or misconceptions
Misconception 1: Counting the shaded part is enough to identify the fraction.
Explanation: To identify a fraction, all parts must be equal.
For a part to be $$\frac{1}{4}$$th, 4 such parts should add up to make a whole. But if some parts are smaller, then depending on the shape we take the whole will be different.
Misconception 2: When finding fraction using area models, children incorrectly believe that the equal sized-piece must also look the same.
Children say this diagram does not show $$\frac{1}{2}$$ of the area of the square coloured because the pieces are “not the same (shape)”. They should know that if the areas of the parts are equal then they are identical. To address this use different geometrical shapes like circle, square, etc. and show how differently these shapes can be divided into the same number of parts.
Misconception 3: Children do not realise that while defined as parts of the whole, fractions are numbers. Children fail to understand that $$\frac{1}{2}$$, $$\frac{1}{4}$$, etc. are also numbers. To address this, use the number line model to show the children how fractions exist between whole numbers and how even whole numbers can be denoted using a fractional form. For instance, show 5/8 the following number line. Stress on how $$\frac{5}{8}$$ and $$\frac{7}{8}$$ lie between 0 and 1 as $$\frac{8}{8}$$ is nothing but 1 on the number line written in fractional form since the numbers between 0 and 1 are divided into 8 parts here.
Misconception 4: All fractions are less than 1. So, children may incorrectly believe that fractions are always less than 1. This will create a barrier to understand improper fractions. While it may be okay to not address this when fractions are just being introduced, do watch out for this misconception as children build fluency.
Tips and Tricks
1. Every time there’s an opportunity to talk about fractions, make use of it. When you open a pack of biscuits, ask children what fraction of the whole packet does each biscuit represent. When eating a pizza or cutting a cake or fruits this exercise can be repeated.
2. Referring to fractions such as $$\frac{3}{4}$$ as “three out of four” in addition to “three fourths” helps reiterate the idea that this fraction indicates 3 parts out of 4. Get students familiar with reading fractions in both these ways.
3. Don't introduce terms like numerator and denominator till students are completely comfortable with describing fractions in a simpler language like “three out of four”.
FAQ
Q1. Why $$3\frac14$$ is called a mixed fraction?
A whole number, when combined with a fraction, is known as a mixed fraction. Taking the above example is the whole number combined with a fraction $$\frac14$$.
Download Free Grade 4 Worksheets
Fractions-Consolidation of Basics
Grade4 | Worksheet 1
Fractions-Consolidation of Basics
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Fractions-A Consolidation
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Fractions-A Consolidation
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Fractions-A visual introduction
Grade3 | Worksheet 1
Fractions-A visual introduction
Grade3 | Worksheet 2
Understanding Fractions
Grade3 | Worksheet 3
Understanding Fractions
Grade3 | Worksheet 4
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A Consolidation of the Basics
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A Consolidation of the Basics
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# Negative exponents
How to calculate negative exponents.
## Negative exponents rule
The base b raised to the power of minus n is equal to 1 divided by the base b raised to the power of n:
b-n = 1 / bn
## Negative exponent example
The base 2 raised to the power of minus 3 is equal to 1 divided by the base 2 raised to the power of 3:
2-3 = 1/23 = 1/(2⋅2⋅2) = 1/8 = 0.125
## Negative fractional exponents
The base b raised to the power of minus n/m is equal to 1 divided by the base b raised to the power of n/m:
b-n/m = 1 / bn/m = 1 / (mb)n
The base 2 raised to the power of minus 1/2 is equal to 1 divided by the base 2 raised to the power of 1/2:
2-1/2 = 1/21/2 = 1/2 = 0.7071
## Fractions with negative exponents
The base a/b raised to the power of minus n is equal to 1 divided by the base a/b raised to the power of n:
(a/b)-n = 1 / (a/b)n = 1 / (an/bn) = bn/an
The base 2 raised to the power of minus 3 is equal to 1 divided by the base 2 raised to the power of 3:
(2/3)-2 = 1 / (2/3)2 = 1 / (22/32) = 32/22 = 9/4 = 2.25
## Multiplying negative exponents
For exponents with the same base, we can add the exponents:
a -na -m = a -(n+m) = 1 / a n+m
#### Example:
2-3 ⋅ 2-4 = 2-(3+4) = 2-7 = 1 / 27 = 1 / (2⋅2⋅2⋅2⋅2⋅2⋅2) = 1 / 128 = 0.0078125
When the bases are diffenrent and the exponents of a and b are the same, we can multiply a and b first:
a -nb -n = (a b) -n
#### Example:
3-2 ⋅ 4-2 = (3⋅4)-2 = 12-2 = 1 / 122 = 1 / (12⋅12) = 1 / 144 = 0.0069444
When the bases and the exponents are different we have to calculate each exponent and then multiply:
a -nb -m
#### Example:
3-2 ⋅ 4-3 = (1/9) ⋅ (1/64) = 1 / 576 = 0.0017361
## Dividing negative exponents
For exponents with the same base, we should subtract the exponents:
a n / a m = a n-m
#### Example:
26 / 23 = 26-3 = 23 = 2⋅2⋅2 = 8
When the bases are diffenrent and the exponents of a and b are the same, we can divide a and b first:
a n / b n = (a / b) n
#### Example:
63 / 23 = (6/2)3 = 33 = 3⋅3⋅3 = 27
When the bases and the exponents are different we have to calculate each exponent and then divide:
a n / b m
#### Example:
62 / 33 = 36 / 27 = 1.333
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# 16A: Newton’s Laws #3: Components, Friction, Ramps, Pulleys, and Strings
Having learned how to use free body diagrams, and then having learned how to create them, you are in a pretty good position to solve a huge number of Newton’s 2nd Law problems. An understanding of the considerations in this chapter will enable to you to solve an even larger class of problems. Again, we use examples to convey the desired information.
Example 16-1
A professor is pushing on a desk with a force of magnitude $$F$$ at an acute angle $$\theta$$ below the horizontal. The desk is on a flat, horizontal tile floor and it is not moving. For the desk, draw the free body diagram that facilitates the direct and straightforward application of Newton’s 2nd Law of motion. Give the table of forces.
Solution
While not a required part of the solution, a sketch often makes it easier to come up with the correct free body diagram. Just make sure you don’t combine the sketch and the free body diagram. In this problem, a sketch helps clarify what is meant by “at an acute angle $$\theta$$ below the horizontal.”
Pushing with a force that is directed at some acute angle below the horizontal is pushing horizontally and downward at the same time.
Here is the initial free body diagram and the corresponding table of forces
Table of Forces
Symbol=? Name Agent Victim
$F_N\notag$ Normal Force The Floor Desk
$F_{sf}\notag$ Static Friction Force The Floor Desk
$F_g=mg\notag$ Gravitational Force The Earth's Gravitational Field Desk
$F_p\notag$ Force of Professor Hands of Professor Desk
Note that there are no two mutually perpendicular lines to which all of the forces are parallel. The best choice of mutually perpendicular lines would be a vertical and a horizontal line. Three of the four forces lie along one or the other of such lines. But the force of the professor does not. We cannot use this free body diagram directly. We are dealing with a case which requires a second free body diagram.
Cases Requiring a Second Free Body Diagram in Which One of More of the Forces that was in the First Free Body Diagram is Replaced With its Components
Establish a pair of mutually perpendicular lines such that most of the vectors lie along one or the other of the two lines. After having done so, break up each of the other vectors, the ones that lie along neither of the lines, (let’s call these the rogue vectors) into components along the two lines. (Breaking up vectors into their components involves drawing a vector component diagram.) Draw a second free body diagram, identical to the first, except with rogue vectors replaced by their component vectors. In the new free body diagram, draw the component vectors in the direction in which they actually point and label them with their magnitudes (no minus signs).
For the case at hand, our rogue force is the force of the professor. We break it up into components as follows:
$\frac{F_{Px}}{F_P}=\cos\theta \quad \frac{|F_{Py}|}{F_P}=\sin\theta$
$F_{Px}=F_P\cos\theta \quad |F_{Py}|=F_P \sin \theta$
Then we draw a second free body diagram, the same as the first, except with $$F_P$$ replaced by its component vectors:
Example 16-2
A wooden block of mass m is sliding down a flat metal incline (a flat metal ramp) that makes an acute angle $$\theta$$ with the horizontal. The block is slowing down. Draw the directly-usable free body diagram of the block. Provide a table of forces.
Solution
We choose to start the solution to this problem with a sketch. The sketch facilitates the creation of the free body diagram but in no way replaces it.
Since the block is sliding in the down-the-incline direction, the frictional force must be in the up-the-incline direction. Since the block’s velocity is in the down-the-incline direction and decreasing, the acceleration must be in the up-the-incline direction.
Table of Forces
Symbol=? Name Agent Victim
$F_N\notag$ Normal Force The Ramp The Block
$F_{kf}=\mu_K F_N\notag$ Kinetic Friction Force The Ramp The Block
$F_g=mg\notag$ Gravitation Force The Earth's Gravitational Field The Block
No matter what we choose for a pair of coordinate axes, we cannot make it so that all the vectors in the free body diagram are parallel to one or the other of the two coordinate axes lines. At best, the pair of lines, one line parallel to the frictional force and the other perpendicular to the ramp, leaves one rogue vector, namely the gravitational force vector. Such a coordinate system is tilted on the page.
Cases Involving Tilted Coordinate Systems
For effective communication purposes, students drawing diagrams depicting phenomena occurring near the surface of the earth are required to use either the convention that downward is toward the bottom of the page (corresponding to a side view) or the convention that downward is into the page (corresponding to a top view). If one wants to depict a coordinate system for a case in which the direction “downward” is parallel to neither coordinate axis line, the coordinate system must be drawn so that it appears tilted on the page.
In the case of a tilted-coordinate system problem requiring a second free body diagram of the same object, it is a good idea to define the coordinate system on the first free body diagram. Use dashed lines so that the coordinate axes do not look like force vectors. Here we redraw the first free body diagram. (When you get to this stage in a problem, just add the coordinate axes to your existing diagram.)
Now we break up $$F_g$$ into its $$x$$, $$y$$ component vectors. This calls for a vector component diagram.
$\frac{F_{gx}}{F_g}=\sin\theta \quad \frac{|F_{gy}|}{F_g}=\cos \theta$
$F_{gx}=F_g \sin\theta \quad |F_{gy}|= F_g \cos\theta$
Next, we redraw the free body diagram with the gravitational force vector replaced by its component vectors.
Example 16-3
A solid brass cylinder of mass m is suspended by a massless string which is attached to the top end of the cylinder. From there, the string extends straight upward to a massless ideal pulley. The string passes over the pulley and extends downward, at an acute angle theta to the vertical, to the hand of a person who is pulling on the string with force $$F_T$$. The pulley and the entire string segment, from the cylinder to hand, lie in one and the same plane. The cylinder is accelerating upward. Provide both a free body diagram and a table of forces for the cylinder.
Solution
A sketch comes in handy for this one:
To proceed with this one, we need some information on the effect of an ideal massless pulley on a string that passes over the pulley.
Effect of an Ideal Massless Pulley
The effect of an ideal massless pulley on a string that passes over the pulley is to change the direction in which the string extends, without changing the tension in the string.
By pulling on the end of the string with a force of magnitude $$F_T$$, the person causes there to be a tension $$F_T$$ in the string. (The force applied to the string by the hand of the person, and the tension force of the string pulling on the hand of the person, are a Newton’s-3rd-law interaction pair of forces. They are equal in magnitude and opposite in direction. We choose to use one and the same symbol $$F_T$$ for the magnitude of both of these forces. The directions are opposite each other.) The tension is the same throughout the string, so, where the string is attached to the brass cylinder, the string exerts a force of magnitude $$F_T$$ directed away from the cylinder along the length of the string. Here is the free body diagram and the table of forces for the cylinder:
Table of Forces
Symbol=? Name Agent Victim
$F_T\notag$ Tension Force The String The Cylinder
$F_g=mg\notag$ Gravitational Force The Earth's Gravitational Field The Cylinder
Example 16-4
A cart of mass $$m_c$$ is on a horizontal frictionless track. One end of an ideal massless string segment is attached to the middle of the front end of the cart. From there the string extends horizontally, forward and away from the cart, parallel to the centerline of the track, to a vertical pulley. The string passes over the pulley and extends downward to a solid metal block of mass $$m_B$$. The string is attached to the block. A person was holding the cart in place. The block was suspended at rest, well above the floor, by the string. The person has released the cart. The cart is now accelerating forward and the block is accelerating downward. Draw a free body diagram for each object.
Solution
A sketch will help us to arrive at the correct answer to this problem.
Recall from the last example that there is only one tension in the string. Call it $$F_T$$. Based on our knowledge of the force exerted on an object by a string, viewed so that the apparatus appears as it does in the sketch, the string exerts a rightward force $$F_T$$ on the cart, and an upward force of magnitude FT on the block.
There is a relationship between each of several variables of motion of one object attached by a taut string, which remains taut throughout the motion of the object, and the corresponding variables of motion of the second object. The relationships are so simple that you might consider them to be trivial, but they are critical to the solution of problems involving objects connected by a taut spring.
The Relationships Among the Variables of Motion For Two Objects, One at One End and the Other at the Other
End, of an always-taut, Unstretchable String
Consider the following diagram.
Because they are connected together by a string of fixed length, if object $$1$$ goes downward $$5 cm$$, then object $$2$$ must go rightward $$5 cm$$. So if object $$1$$ goes downward at $$5 cm/s$$ then object $$2$$ must go rightward at $$5 cm/s$$. In fact, no matter how fast object $$1$$ goes downward, object $$2$$ must go rightward at the exact same speed (as long as the string does not break, stretch, or go slack). In order for the speeds to always be the same, the accelerations have to be the same as each other. So if object $$1$$ is picking up speed in the downward direction at, for instance, $$5 cm/s^2$$, then object $$2$$ must be picking up speed in the rightward direction at $$5 cm/s^2$$. The magnitudes of the accelerations are identical. The way to deal with this is to use one and the same symbol for the magnitude of the acceleration of each of the objects. The ideas relevant to this simple example apply to any case involving two objects, one on each end of an inextensible string, as long as each object moves only along a line collinear with the string segment to which the object is attached.
Let’s return to the example problem involving the cart and the block. The two free body diagrams follow:
Note that the use of the same symbol $$F_T$$ in both diagrams is important, as is the use of the same symbol $$a$$ in both diagrams. |
# 1.2 Solving Linear Systems by Substitution 9/19/12.
## Presentation on theme: "1.2 Solving Linear Systems by Substitution 9/19/12."— Presentation transcript:
1.2 Solving Linear Systems by Substitution 9/19/12
Solving a system of equations by substitution Step 1: Solve an equation for one variable. Step 2: Substitute Step 3: Solve the equation. Step 4: Plug back in to find the other variable. Step 5: Check your solution. Pick the easier equation. The goal is to get y= ; x= ; a= ; etc. Put the equation solved in Step 1 into the other equation. Get the variable by itself. Substitute the value of the variable into the equation. Substitute your ordered pair into BOTH equations.
1) Solve the system using substitution x + y = 5 y = 3 + x Step 1: Solve an equation for one variable. Step 2: Substitute The second equation is already solved for y! x + y = 5 x + (3 + x) = 5 Step 3: Solve the equation. 2x + 3 = 5 2x = 2 x = 1
1) Solve the system using substitution x + y = 5 y = 3 + x Step 4: Plug back in to find the other variable. x + y = 5 (1) + y = 5 y = 4 Step 5: Check your solution. (1, 4) (1) + (4) = 5 (4) = 3 + (1) The solution is (1, 4). What do you think the answer would be if you graphed the two equations?
Which answer checks correctly? 3x – y = 4 x = 4y - 17 1. (2, 2) 2. (5, 3) 3. (3, 5) 4. (3, -5)
2) Solve the system using substitution 3y + x = 7 4x – 2y = 0 Step 1: Solve an equation for one variable. Step 2: Substitute It is easiest to solve the first equation for x. 3y + x = 7 -3y x = -3y + 7 4x – 2y = 0 4(-3y + 7) – 2y = 0
2) Solve the system using substitution 3y + x = 7 4x – 2y = 0 Step 4: Plug back in to find the other variable. 4x – 2y = 0 4x – 2(2) = 0 4x – 4 = 0 4x = 4 x = 1 Step 3: Solve the equation. -12y + 28 – 2y = 0 -14y + 28 = 0 -14y = -28 y = 2
2) Solve the system using substitution 3y + x = 7 4x – 2y = 0 Step 5: Check your solution. (1, 2) 3(2) + (1) = 7 4(1) – 2(2) = 0
Solve the system using substitution. Checkpoint 1. Use Substitution 3 = y2x2x + 0 = y3x3x + ANSWER (), 9 3 –.
Solve the system using substitution. Tell which equation you chose to solve and use for the substitution. Explain. Checkpoint 2. Use Substitution 4 = 3y3y2x2x + 1 = 2y2yx + ANSWER () 5,5, 2 –.
Solve the system using substitution. Tell which equation you chose to solve and use for the substitution. Explain. Checkpoint 3. Use Substitution 10 = 2y2y4x4x + 5 = y3x3x – ANSWER () 2, 1.
Homework WS 1.2 all |
# Why does Partial Fraction Decomposition Result in Multiples of the Decomposed Fraction?
Let's say I have the rational function $\dfrac{x^3}{x^2 + x - 6}$. I use polynomial long division to get $x^3 = (x - 1)(x^2 + x - 6) + 7x -6 \implies \dfrac{x^3}{(x - 2)(x + 3)} = \dfrac{x^3}{x^2 + x - 6} = x - 1 + \dfrac{7x - 6}{x^2 + x - 6} = x - 1 + \dfrac{7x - 6}{(x - 2)(x + 3)}$.
$\dfrac{A}{x - 2} + \dfrac{B}{x + 3} = \dfrac{A(x + 3) + B(x - 2)}{(x - 2)(x + 3)} = \dfrac{Ax + 3A + Bx - 2B}{(x - 2)(x + 3)}$
$\therefore \dfrac{7x - 6}{(x - 2)(x + 3)} = \dfrac{Ax + 3A + Bx - 2B}{(x - 2)(x + 3)} \implies 7x - 6 = Ax + 3A + Bx - 2B$
$\implies 7x - 6 = (A + B)x + 3A - 2B$
$\therefore 7x = (A + B)x$ and $-6 = 3A - 2B$.
After solving this system of two linear equations, we get $B = \dfrac{27}{5}$ and $A = \dfrac{8}{5}$.
$\dfrac{7x - 6}{(x - 2)(x + 3)} = \dfrac{A}{x - 2} + \dfrac{B}{x + 3}$
$\therefore \dfrac{7x - 6}{(x - 2)(x + 3)} = \dfrac{8}{5x - 10} + \dfrac{27}{5x + 15}$
But when we expand $\dfrac{8}{5x - 10} + \dfrac{27}{5x + 15}$, we get $\dfrac{5(7x - 6)}{(x - 2)(x + 3)} \not = \dfrac{7x - 6}{(x - 2)(x + 3)}$.
Unless I'm consistently making the same error, I've noticed this phenomenon to be the same for other partial fraction decomposition problems: The result of the partial fraction decomposition is a constant multiple of the fraction that we were aiming to get.
I am confused as to why this happens. In doing the operations of partial fraction decomposition (as seen above), isn't our goal to get $\dfrac{7x - 6}{(x - 2)(x + 3)} = \dfrac{A}{x - 2} + \dfrac{B}{x + 3}$? Shouldn't our calculations (as seen above) generate exactly the result $\dfrac{7x - 6}{(x - 2)(x + 3)}$ rather than some multiple of it? What is the reasoning behind this? What am I misunderstanding?
I would greatly appreciate it if people could please take the time to explain this.
You made an arithmetic mistake; you lost track of the factor of $5$ in your common denominator:
$$\dfrac{8}{5(x-2)} + \dfrac{27}{5(x+3)} = \dfrac{8(x+3) + 27(x-2)}{\color{red}{\mathbf{5}}(x - 2)(x + 3)} = \dfrac{5(7x - 6)}{\color{red}{\mathbf{5}}(x - 2)(x + 3)}$$
• $x - 1 + \dfrac{8}{5x - 10} + \dfrac{27}{5x + 15} = x - 1 + \dfrac{40x + 120 + 135x - 270}{(5x - 10)(5x + 15)} = x - 1 + \dfrac{175x - 150}{5(x - 2)(x + 3)} = x - 1 + \dfrac{25(7x - 6)}{5(x - 2)(x + 3)} = x - 1 + \dfrac{5(7x - 6)}{(x - 2)(x + 3)}$ Can you see where I made an error? – The Pointer Mar 5 '17 at 3:23
• @ThePointer: If you calculate it that way, by multiplying the denominators rather than finding a least common multiple, the error is $(5x-10)(5x+15)=5(x-2) \cdot 5(x+3) = 25 (x-2)(x+3)$ – Hurkyl Mar 5 '17 at 3:27 |
# Difference between revisions of "2021 AMC 12A Problems/Problem 16"
The following problem is from both the 2021 AMC 10A #16 and 2021 AMC 12A #16, so both problems redirect to this page.
## Problem
In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200$.$$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200$$What is the median of the numbers in this list?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\textbf{(E)} ~167$
## Solution 1
There are $1+2+..+199+200=\frac{(200)(201)}{2}=20100$ numbers in total. Let the median be $k$. We want to find the median $k$ such that $$\frac{k(k+1)}{2}=20100/2,$$ or $$k(k+1)=20100.$$ Note that $\sqrt{20100} \approx 142$. Plugging this value in as $k$ gives $$\frac{1}{2}(142)(143)=10153.$$ $10153-142<10050$, so $142$ is the $152$nd and $153$rd numbers, and hence, our desired answer. $\fbox{(C) 142}$.
Note that we can derive $\sqrt{20100} \approx 142$ through the formula $$\sqrt{n} = \sqrt{a+b} \approx \sqrt{a} + \frac{b}{2\sqrt{a} + 1},$$ where $a$ is a perfect square less than or equal to $n$. We set $a$ to $19600$, so $\sqrt{a} = 140$, and $b = 500$. We then have $n \approx 140 + \frac{500}{2(140)+1} \approx 142$. ~approximation by ciceronii
## Solution 2
The $x$th number of this sequence is $\left\lceil\frac{-1\pm\sqrt{1+8x}}{2}\right\rceil$ via the quadratic formula. We can see that if we halve $x$ we end up getting $\left\lceil\frac{-1\pm\sqrt{1+4x}}{2}\right\rceil$. This is approximately the number divided by $\sqrt{2}$. $\frac{200}{\sqrt{2}} = 141.4$ and since $142$ looks like the only number close to it, it is answer $\boxed{(C) 142}$ ~Lopkiloinm
We can look at answer choice $C$, which is $142$ first. That means that the number of numbers from $1$ to $142$ is roughly the number of numbers from $143$ to $200$.
The number of numbers from $1$ to $142$ is $\frac{142(142+1)}{2}$ which is approximately $10000.$ The number of numbers from $143$ to $200$ is $\frac{200(200+1)}{2}-\frac{142(142+1)}{2}$ which is approximately $10000$ as well. Therefore, we can be relatively sure the answer choice is $\boxed{(C) \text{ } 142}.$
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Transcript
```Geometry Lesson 4-1: Apply Triangle Sum Properties
Triangle: a polygon with three sides
Interior angles: original angles (inside)
Exterior angles: Angles that are linear pairs to interior angles
Corollary to a theorem: The acute angles of a right triangle are complementary
CLASSIFYING TRIANGLES BY SIDES
Scalene Triangle
Isosceles Triangle
Equilateral Triangle
NO congruent
sides
At least 2
congruent sides
3 congruent sides
CLASSIFYING TRIANGLES BY ANGLES
Acute Triangle
Right
Triangle
Obtuse
Triangle
Equiangular
Triangle
*Notice that an
equilateral triangle is
also isosceles.
*An equiangular
triangle is also acute.
3 acute angles
1 right angel
1 obtuse angles
3 congruent
angles
Example 1: Classify ∆RST by its sides. Then determine if the triangle is a right triangle.
3-(-1) 4
-3-3
-6
-2
3
2-(-1) = 3
5-3
2
3 = -1
2
(Yes)
THEOREM 4.1: TRIANGLE SUM THEOREM
mA + mB + mC = 180
The sum of the measures of the interior angles of a triangle is 180.
THEOREM 4.2: EXTERIOR ANGLE THEOREM
m1 = m A + m B
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent
interior angles.
Example 2: Use the diagram at the right to find the measure of DCB and 1.
3x-9 = x + 73
2x = 82
X = 41
3x + 16 = 2x + 52
-16
-16
X = 36
COROLLARY TO THE TRIANGLE SUM THEOREM
mA + mB = 90
The acute angles of a right triangle are Complementary .
Example 3: The front face of the wheelchair ramp shown forms a right triangle. The measure of one
acute angle in the triangle is eight times the measure of the other. Find the measure of each acute angle.
8x + x = 90
X = 10
8 (10) = 80
```
Related documents |
# Factors of 30: Prime Factorization, Methods, Tree, and Examples
Factors of 30 are a set of integers that give zero as a remainder when 30 is divided from them. Not only do these numbers give zero as a remainder but they also yield a whole number quotient when 30 is divided from them.
Figure 1 – All factors of 30
In terms of multiplication, those numbers that when multiplied together give 30 as a product are termed as factors of 30. These two numbers which give 30 as the product are also termed as a Factor Pair. Factors for any number are the unique set of natural numbers which yield zero as a remainder whenever these numbers act as a divisor. There are multiple techniques for determining the factors of a number such as the division method, prime factorization, and the factor tree. For any number, the number 1 acts as the smallest factor, and the number itself acts as the largest factor. In the case of 30, the smallest factor is 1 and the greatest factor is the number itself, which is 30. This statement can be proved by the following multiplication of 1 and 30. This multiplication also proves that 1 and 30 act as a factor pair.
1 x 30 = 30
But 1 and 30 are not the only factors of 30. In this article, we will be diving into the details of the factors of 30 and the various techniques and methods which can be used to evaluate these factors.
## What Are the Factors of 30?
Factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. When these numbers act as the divisors, they produce zero as the reminder. The number 30 is an even composite number, meaning that it consists of more than 2 factors. Also, the number 30 has 8 factors in total.
## How To Calculate the Factors of 30?
You can calculate the factors of 30 through various techniques such as division. Let’s take a look at the division method first. The division method states that when a number acts as the divisor, it should produce a whole number quotient and zero as the remainder. If these two conditions for the number are met, only then the number can act as the factor. In the case of the number 30, since it is an even composite number so that means that the number is divisible by 2. Let’s take a look at its division from the number 2: $\frac{30}{2} = 15$ This division produced zero as a remainder and a whole number quotient which indicates that 2 is a factor of 30. Another rule of the division method is that for such divisors, which produce zero as the reminder, their quotient also acts as the factor. So in this case, 15 is also a factor of 30, since it is a quotient produced by the division of 2. Let’s take a look at the division of 30 by 15: $\frac{30}{15} = 2$ Hence, both 2 and 15 are factors of 30. Let’s take a look at some other factors of 30. $\frac{30}{3} = 10$ $\frac{30}{3} = 3$ So, both 3 and 10 act as the factors of 30. Similarly, consider the following division: $\frac{30}{5} = 6$ $\frac{30}{6} = 5$ So 5 and 6 are also the factors of 30. And lastly, let’s take a look at the following division: $\frac{30}{1} = 30$ $\frac{30}{30} = 1$ So, both 1 and 30 are also factors of 30. Hence in total, the number 30 has 8 factors and these factors are mentioned below: Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30
## Factors of 30 by Prime Factorization
Prime factorization is one of the unique ways to determine the factors of a number. In prime factorization, a number is broken down with the help of prime numbers and this division continues until 1 is achieved at the end. Prime factorization is the technique that is used to determine the prime factors of a number. Prime factors are those factors that are also prime numbers. In prime factorization, the division process continues until 1 is received as the end result. The prime factorization of the number 30 occurs in the following manner: $\frac{30}{2} = 15$ $\frac{15}{5} = 3$ $\frac{3}{3} = 1$ The prime factorization of the number 30 is also shown in figure 1 given below:
Figure 2 – Prime Factorization of 30
The prime factorization of 30 can be mathematically written as:
30 = 2 x 3 x 5
## Factor Tree of 30
A factor tree is a pictorial method of representing the prime factorization of a number. The unique aspect that distinguishes factor tree from prime factorization is that instead of ending the division process at 1, the division process ends at prime numbers. The factor tree begins with the number itself and then extends out its branches to possible divisors and quotients. At the end branches, prime numbers are obtained. The factor tree of the number 30 is shown below:
Figure 3 – Factor tree of 30
## Factors of 30 in Pairs
Factor pairs, as mentioned above, are the two possible numbers that when multiplied together give the original number as the product. The factor pairs for any number can be found by the multiplication method. A factor pair simply consists of a factor of a number and its whole number quotient. The factor pairs of 30 are given below:
2 x 15 = 30
1 x 30 = 30
3 x 10 = 30
5 x 6 = 30
Hence, the factor pairs of 30 are (1,30), (2,15), (3,10), and (5,6). These factor pairs can also consist of negative factors. They are pretty much the same as the positive factors only the reversed signs are different. The condition for negative factor pairs is that both the factors existing in the pair must have the negative sign. The negative factor pairs of 30 are (-1,-30), (-2,-15), (-3,-10), and (-5,-6).
## Factors of 30 Solved Examples
To further enhance the concept of the factors of 30, let’s take a look at some simple solved examples constituting the factors of 30.
### Example 1
Calculate the product of all the prime factors of 30.
### Solution
To calculate the product of all the factors of 30, let’s first list down the factors of 30. Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30 According to the prime factorization of 30, the following prime factors were obtained: Prime factors of 30 = 2, 3, 5 Now, to calculate the product of these prime factors, simply multiply them together. Their multiplication is shown below:
30 = 2 x 3 x 5
Hence the product obtained is 30.
### Example 2
Find the average of all the factors of 30.
### Solution
To find the average of all the factors of 30, let’s first note down the factors of 30. Following are the factors of 30: Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30 Calculating the average of these factors by using the following formula: $Average = \frac{\text{Sum of numbers}}{\text{Total numbers}}$ $Average = \frac{1+2+3+5+6+10+15+30}{8}$ $Average = \frac{72}{8}$
Average = 9
Hence, the average of all the factors of 30 is 9.
### Example 3
Find out the common factors between 30 and 15.
### Solution
To find out the common factors between 30 and 15, let’s first take a look at their total factors. The factors of 30 are given below: Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30 Similarly, the factors of 15 are given below: Factors of 15 = 1, 3, 5, 15 The common factors between two numbers are the factors that exist in the factor sets for both numbers. In this case, similar factors which exist both in the factor set of 30 and in the factor set of 15 are the common factors. So the common factors between 15 and 30 are 1, 3, 5, and 15.
### Example 4
List down the even and the odd factors of 30.
### Solution
To determine the even and the odd factors of 30, let’s first list down the factors of 30. Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30 The even factors would be the ones that are multiples of 2. So the even factors of the number 30 are 2, 6, 10, and 30. Similarly, the odd factors of the number 30 are the numbers that are not multiples of 30, so the odd factors of 30 are 1, 3, 5, and 15. Hence, these are the even and the odd factors of the number 30. All images/mathematical drawings are created with GeoGebra. |
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# Combining like terms review
A common technique for simplifying algebraic expressions. When combining like terms, such as 2x and 3x, we add their coefficients. For example, 2x + 3x = (2+3)x = 5x.
## What is combining like terms?
We call terms "like terms" if they have the same variable part. For example, $4x$ and $3x$ are like terms, but $4x$ and $3w$ are not like terms.
The cool thing about like terms is that we can combine them into a single term by adding their coefficients. For example:
$\begin{array}{rl}& 4x+3x\\ \\ =& \left(4+3\right)x\\ \\ =& 7x\end{array}$
## More examples
An example with more than two terms:
$\begin{array}{rl}& 2r+1+\left(-4r\right)+7\\ \\ =& 2r-4r+1+7\\ \\ =& \left(2-4\right)r+1+7\\ \\ =& -2r+8\end{array}$
An example with more than one variable:
$\begin{array}{rl}& 3x+2y+4x+7-y\\ \\ =& 3x+4x+2y-y+7\\ \\ =& \left(3+4\right)x+\left(2-1\right)y+7\\ \\ =& 7x+y+7\end{array}$
Problem 1
Combine the like terms to make a simpler expression.
$-3n-7+\left(-6n\right)+1$
Want more practice like this? Check out this exercise and this other exercise.
## Want to join the conversation?
• Why am i not getting this :(
• Suppose the temperature is a freezing -6 degrees.
It gets even colder than that by 16 degrees. (subtracting)
Now it's an even more freezing -22 degrees.
• uhhhhhhgg why have i failed like every single timeeeeeeeee
• what is 9t−3t+4
• That would be 6t+4 because 9t-3t=6t and 4 is just left alone so it equals 6t+4.
• the only thing i'm really confused about is how to do the fractions part
• You just have to know how to add, subtract, multiply and divide fractions and the simplifying algebraic expressions with fractions in them will become fairly easy :D
• I just don't understand the negatives! In one problem I must subtract and in an other one I must add... can some one help me?
• This is driving me crazy!
• How would I combine like terms if there is a variable as the numerator and not as the coefficient?
Like: w/60 + w/5
• Can someone help me with the problem 3x - 12 = 2x -4 + 3x + 6
I'm so confused on how to solve it, because I got 2 answers: one was -14/8 and one was 14/8. Which one is the right one?
(1 vote)
• Neither is correct. I confirmed this by substituting each value in the equation to see if the 2 sides are equal. In both cases, the 2 sides are unequal.
Here's correct solution:
1) Combine like terms on right side: 3x-12 = 5x+2
2) Subtract 3x from both sides: -12 = 2x + 2
3) Subtract 2 from both sides: -14 = 2x
4) Divide both sides by 2: -7 = x
Checking solution:
3(-7) - 12 = 2(-7) -4 + 3(-7) + 6
-21 - 12 = -14 -4 + -21 + 6
-33 = -18 + -15
-33 = -33
So x=-7 is the correct answer
Hope this helps.
• how can you subtract negative integers
(1 vote)
• If you have -2 - (-4) the subtraction sign and the negative sign for the second number cancel out so it becomes -2 + 4.
If you have -2 - 4 you simply add the 4 to 2 and put the negative sign in front |
# How do you describe the asymptote of a graph?
## How do you describe the asymptote of a graph?
An asymptote is a line that a graph approaches without touching. Similarly, horizontal asymptotes occur because y can come close to a value, but can never equal that value. In the previous graph, there is no value of x for which y = 0 ( ≠ 0), but as x gets very large or very small, y comes close to 0.
What are asymptotes give examples?
The asymptote (s) of a curve can be obtained by taking the limit of a value where the function does not get a definition or is not defined. An example would be \infty∞ and -\infty −∞ or the point where the denominator of a rational function is zero.
### What is a curved asymptote?
In analytic geometry, an asymptote (/ˈæsɪmptoʊt/) of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tends to infinity.
What are the three types of asymptote?
An asymptote is a line that the graph of a function approaches as either x or y go to positive or negative infinity. There are three types of asymptotes: vertical, horizontal and oblique. That is, as approaches from either the positive or negative side, the function approaches positive or negative infinity.
#### How do you define a horizontal asymptote?
The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator.
• Degree of numerator is less than degree of denominator: horizontal asymptote at y = 0.
• Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote.
What is a curvilinear asymptote?
A curvilinear asymptote is an asymptote that’s a curve. As x approaches ∞ (or -∞), the function will approach the curve. For example, we could have a function that gets closer to the parabola y = x2 as x approaches ∞ or -∞: © 2022 Shmoop University.
## How do you identify an asymptote?
The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator.
1. Degree of numerator is less than degree of denominator: horizontal asymptote at y = 0.
2. Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote. |
# How to solve improper fractions
This can be a great way to check your work or to see How to solve improper fractions. We can solve math problems for you.
## How can we solve improper fractions
College algebra students learn How to solve improper fractions, and manipulate different types of functions. In mathematics, a differential solver is a numerical method for solving differential equations. Differential equations are mathematical equations that describe how a function changes over time. They are used to model many real-world phenomena, such as the motion of a spring or the flow of a liquid. Differential solvers approximate the solution to a differential equation by calculating a series of values that get closer and closer to the true solution. There are many different types of differential solvers, each with its
To solve negative exponents fractions, we need to find the reciprocal of the base number. Then, we need to raise it to the power of the exponent. For example, to solve the fraction `2^-3`, we would take the reciprocal of 2 (`1/2`), and raise it to the power of -3 (`1/8`).
Some students prefer to start with the easiest problems first and work their way up to the more difficult ones. Others prefer to start with the most difficult problem and then work their way down to the easier ones. And still others prefer to work on all the problems at the same time. Ultimately, it is up to the individual student to decide what works best for them.
It can be incredibly satisfying to solve a hard math problem - the feeling of elation and accomplishment is like nothing else. When you're stuck on a problem, it can seem impossible, but when you finally crack the code and find the solution, it's all worth it.
Solving fractions can be a difficult task for students, but there are some tips and tricks that can help make the process easier. First, it is important to understand what a fraction is and how it works. A fraction is simply a number that represents a part of a whole. For example, if you have a pizza and you want to split it into equal pieces, each piece would be a fraction of the whole pizza. To solve a fraction, you need to find the amount of the
## Instant assistance with all types of math
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# 2003 AMC 10A Problems/Problem 22
## Problem
In rectangle $ABCD$, we have $AB=8$, $BC=9$, $H$ is on $BC$ with $BH=6$, $E$ is on $AD$ with $DE=4$, line $EC$ intersects line $AH$ at $G$, and $F$ is on line $AD$ with $GF \perp AF$. Find the length of $GF$.
$[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,8), C=(0,8), G=(-6,20), F=(-6,0); draw(D--A--B--C--D--F--G--Ep); draw(A--G); label("F",F,W); label("G",G,W); label("C",C,WSW); label("H",H,NNE); label("6",(6,8),N); label("B",B,NE); label("A",A,SW); label("E",Ep,S); label("4",(2,0),S); label("D",D,S);[/asy]$
$\mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30$
## Solutions
### Solution 1
$\angle GHC = \angle AHB$ (Vertical angles are equal).
$\angle F = \angle B$ (Both are 90 degrees).
$\angle BHA = \angle HAD$ (Alt. Interior Angles are congruent).
Therefore $\triangle GFA$ and $\triangle ABH$ are similar. $\triangle GCH$ and $\triangle GEA$ are also similar.
$DA$ is 9, therefore $EA$ must equal 5. Similarly, $CH$ must equal 3.
Because $GCH$ and $GEA$ are similar, the ratio of $CH\; =\; 3$ and $EA\; =\; 5$, must also hold true for $GH$ and $HA$. $\frac{GH}{GA} = \frac{3}{5}$, so $HA$ is $\frac{2}{5}$ of $GA$. By Pythagorean theorem, $(HA)^2\; =\; (HB)^2\; +\; (BA)^2\;...\;HA=10$.
$HA\: =\: 10 =\: \frac{2}{5}*(GA)$.
$GA\: =\: 25.$
So $\frac{GA}{HA}\: =\: \frac{GF}{BA}$.
$\frac{25}{10}\: =\: \frac{GF}{8}$.
Therefore $GF= \boxed{\mathrm{(B)}\ 20}$.
### Solution 2
Since $ABCD$ is a rectangle, $CD=AB=8$.
Since $ABCD$ is a rectangle and $GF \perp AF$, $\angle GFE = \angle CDE = \angle ABC = 90^\circ$.
Since $ABCD$ is a rectangle, $AD || BC$.
So, $AH$ is a transversal, and $\angle GAF = \angle AHB$.
This is sufficient to prove that $GFE \approx CDE$ and $GFA \approx ABH$.
Using ratios:
$\frac{GF}{FE}=\frac{CD}{DE}$
$\frac{GF}{FD+4}=\frac{8}{4}=2$
$GF=2 \cdot (FD+4)=2 \cdot FD+8$
$\frac{GF}{FA}=\frac{AB}{BH}$
$\frac{GF}{FD+9}=\frac{8}{6}=\frac{4}{3}$
$GF=\frac{4}{3} \cdot (FD+9)=\frac{4}{3} \cdot FD+12$
Since $GF$ can't have 2 different lengths, both expressions for $GF$ must be equal.
$2 \cdot FD+8=\frac{4}{3} \cdot FD+12$
$\frac{2}{3} \cdot FD=4$
$FD=6$
$GF=2 \cdot FD+8=2\cdot6+8=\boxed{\mathrm{(B)}\ 20}$
### Solution 3 (fastest)
We extend $BC$ such that it intersects $GF$ at $X$. Since $ABCD$ is a rectangle, it follows that $CD=8$, therefore, $XF=8$. Let $GX=y$. From the similarity of triangles $GCH$ and $GEA$, we have the ratio $3:5$ (as $CH=9-6=3$, and $EA=9-4=5$). $GX$ and $GF$ are the altitudes of $GCH$ and $GEA$, respectively. Thus, $y:y+8 = 3:5$, from which we have $y=12$, thus $GF=y+8=12+8=\boxed{\mathrm{(B)}\ 20}$
### Solution 4
Since $GF\perp AF$ and $AF\perp CD,$ we have $GF\parallel CD\parallel AB.$ Thus, $\triangle CDE\sim GFE.$ Suppose $GF=x$ and $FD=y.$ Thus, we have $\dfrac{x}{8}=\dfrac{y+4}{4}.$ Additionally, now note that $\triangle GAF\sim AHB,$ which is pretty obvious from insight, but can be proven by AA with extending $BH$ to meet $GF.$ From this new pair of similar triangles, we have $\dfrac{x}{8}=\dfrac{y+9}{6}.$ Therefore, we have by combining those two equations, $$\dfrac{y+9}{6}=\dfrac{y+4}{4}.$$ Solving, we have $y=6,$ and therefore $x=\boxed{\mathrm{(B)}\ 20}$
### Solution 5
Since there are only lines, you can resort to coordinate bashing. Let $FD=k$. Three lines, line $GF$, line $GE$, and line $GA$, intersect at $G$. Our goal is to find the y-coordinate of that intersection point.
Line $GF$ is $x=0$
Line $GE$ passes through $(k+4, 0)$ and $(k, 8)$. Therefore the slope is $-2$ and the line is $y-0=-2(x-k-4)$ which is $y=-2x+2k+8$
Line $GA$ passes through $(k+9, 0)$ and $(k+3, 8)$. Therefore the slope is $\frac{-4}{3}$ and the line is $y-0=-\frac{4}{3}(x-k-9)$ which simplifies to $y=-\frac{4}{3}x+\frac{4}{3}k+12$
We solve the system of equations with these three lines. First we plug in $x=0$
$y=2k+8$
$y=\frac{4}{3}k+12$
Next, we solve for k. $k=6$ Therefore $y=20$. The y-coordinate of this intersection point is indeed our answer. $\boxed{\mathrm{(B)}\ 20}$ ~superagh
### Solution 6 (simple coordinates)
Let $A$ be the origin of our coordinate system. Now line $GA$ has equation $-\frac{4}{3}x$. We can use point-slope form to find the equation for line $GE$. First, we know that its slope is $-2$, and we know that it passes through $E=(-5,0)$, so line $GE$ has equation $-2(x+5)$. Solving for the intersection by letting $-\frac{4}{3}x=-2(x+5)$, we get $x=-15$. Plugging this into our equation for line $GA$ gives us $G=(-15,20)$, so $GF= \boxed{\mathrm{(B)}\ 20}$ ~chrisdiamond10
### Solution 7 (system of equations through angle similarity)
First, using given information, we can find the values of some line segments in the figure. We find that $HA = 10$ (through Pythagorean Theorem), $CH = 3$, and $EA = 5$. Let Line $FD = x$ and let Line $FG = y$. We find that $\triangle FGE \sim \triangle CDE$ through some angle chasing (they both have a right angle, and they both share angle $\angle CED$. Using this information, we can write the equation $\frac{4}{8} = \frac{4+x}{y}$. Through simplifying this equation, we get that $y=2x+8$. Let point $I$ be the point on line $FG$ so that lines $CI$ and $FG$ are perpendicular, and we get that $GI = 2x$ and $FI =8$. Doing some more angle chasing, we can find that $\triangle GIH \sim \triangle GFA$, as they both share $\angle FGH$ and they both have a right angle.
With this information, we can write the equation $\frac{x+3}{y-8} = \frac{x+9}{y}.$ Simplifying this equation we get the equation $-8x+6y-72 = 0$. Plugging in $y=2x+8$ for $6y$, we get $4x-24 = 0$, so $x=6$. Lastly, to find the value of y, which is the value of Line $FG$, our desired value, we plug in $6$ for $x$ in the equation $y=2x+8$, we get $2(6)+8$, which, finally, we get our $y$ value of $20$, so therefore, our answer is $GF= \boxed{\mathrm{(B)}\ 20}$
### Solution 8 Line and Slope
Draw a coordinate plane with the y-axis centered on $CD$ and the x-axis centered on $AD$. From there, call the line passing through $CE$ $a$ and the line passing through $HA$ $b$. From there, you can find that the equations for these lines are $y=-2x+8$ and $y=-4/3 x+12$ respectively. We need to find the length of $GF$, so we are finding the y-value of the point $G$. Solving for this point, we get $-4/3 x+12=-2x+8$;$2/3x=-4$;$x=-6$. Now, plugging in the values, we find that $y=-2(-6)+8=20$, so our answer is $GF= \boxed{\mathrm{(B)}\ 20}$.
~iamcalifornia'sresidentidiot
2003 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 23 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
{{MAA Notice |
# Integers: Comparing and Ordering
## Presentation on theme: "Integers: Comparing and Ordering"— Presentation transcript:
Integers: Comparing and Ordering
Today’s Objective I will learn how to compare and order rational numbers.
Evaluation Questions What are rational numbers?
How do we compare and order rational numbers?
Rational Numbers
Whole Numbers Positive numbers that are not fractions or decimals.
Integers The set of whole numbers and their opposites.
Notes! The set of whole numbers and their opposites.
Positive Integers Notes! Integers greater than zero.
Negative Integers Notes! Integers less than zero.
Comparing Integers Notes! The further a number is to the right on the number line, the greater it’s value. Ex: -3 ___ -1 < . . -1 is on the right of -3, so it is the greatest.
Comparing Integers The farther a number is to the right on the number line, the greater it’s value. Ex: 2 ___ -5 > . . 2 is on the right of -5, so it is the greatest.
Comparing Integers The farther a number is to the right on the number line, the greater it’s value. Ex: 0 ___ -2 > . . 0 is on the right of -2, so it is the greatest.
Ordering Integers Notes! When ordering integers from least to greatest follow the order on the number line from left to right. Ex: 4, -5, 0, 2 . . . . Least to greatest: -5, 0, 2, 4
Ordering Integers Notes! When ordering integers from greatest to least follow the order on the number line from right to left. Ex: -4, 3, 0, -1 . . . . Greatest to least: 3, 0, -1, -4
Using Models If you are comparing tenths to hundredths, you can use a tenths grid and a hundredths grid. Here, you can see that 0.4 is greater than 0.36.
Notes! Real life Ordering Decimals! Line up the decimals. Think of all decimals like money!!! It works! Double Cheeseburger: \$ .99 Big Mac Value Meal: \$ 4.79 Chicken McNuggetts Meal: \$ 3.80 Small Drink: \$ .99 McFlurry: \$ 1.97 Salad: \$ 4.80 2 Cheeseburger Meal: \$ 3.70 Ice Cream Cone: \$ .87
Order Up! Least Expensive to Most Expensive
Ice Cream Cone Double Cheeseburger: Small Soft Drink: McFlurry: 2 Cheeseburger Meal: Chicken McNuggetts Meal: Big Mac Value Meal: Chicken Salad:
Comparing Fractions Strategy
Notes! Make sure the denominators are the same. Compare the numerators. If the denominators are not the same, then rewrite the fractions using a common denominator. The new fractions should be equivalent to the original fractions.
Writing Equivalent Fractions
One way to find a common denominator is to multiply the two original denominators. 5 3 > 4 6 6 x 4 = 24 20 > 18 x 4 24 x 6 20 18 24
Another way to compare fractions is to find the LCM of both denominators.
Use the LCM as the new denominator in the equivalent fractions. 7 5 < 9, 18, 27, 36, 45 9 12 12, 24, 36, 48, 60 20 < 21 x 3 x 4 20 21 36 36
Ordering Fractions Find the LCM of the denominators.
Notes! Find the LCM of the denominators. Use the LCM to write equivalent fractions. Put the fractions in order using the numerators.
Example - Order from Least to Greatest:
3 2 1 8 5 4 x 8 x 5 x 10 15 16 10 40 40 40 8, 16, 24, 32, 40, 48 1/4 , 3/8 , 2/5 5, 10, 15, 20, 25, 30, 35, 40 4, 8, 12, 16, 20, 24, 28, 32, 36, 40
HOMEWORK TIME!!!!
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Properties Questions on Arithmetic Mean
Here using the properties we will solve the different properties questions on arithmetic mean. Follow the steps to understand the explanations on how to solve the problems based on properties.
Worked-out examples on properties questions on arithmetic mean:
1. The mean of eight numbers is 25. If five is subtracted from each number, what will be the new mean?
Solution:
Let the given numbers be x1, x2, . . ., x8.
Then, the mean of these numbers = (x1 + x2 + ...+ x8)/8.
Therefore, (x1 + x2+...+x8)/8 = 25
⇒ (x1 + x2 + ... + x8) = 200 ……. (A)
The new numbers are (x1 - 5), (x2 - 5), …… ,(x8 - 5)
Mean of the new numbers = {(x1 - 5) + (x1 - 5) + …… + (x8 - 5)}/8
= [(x1 + x2 + ... + x8) - 40]/8
= (200 - 40)/8, [using (A)]
= 160/8
= 20
Hence, the new mean is 20.
2. The mean of 14 numbers is 6. If 3 is added to every number, what will be the new mean?
Solution:
Let the given numbers be x1, x2, x3, ….. x14.
Then, the mean of these numbers = x1 + x2 + x3+ ….. x14/14
Therefore, (x1 + x2 + x3 + ….. x14)/14 = 6
⇒ (x1 + x2 + x3 + ….. x14) = 84 ………………. (A)
The new numbers are (x1 + 3), (x2 + 3), (x3 + 3), …. ,(x14 + 3)
Mean of the new numbers
= (x1 + 3), (x2 + 3), (x3 + 3), …. ,(x14 + 3)/14
= (x1 + x2 + x3 + ….. x14) + 42
= (84 + 42)/14, [Using (A)]
= 126/14
= 9
Hence, the new mean is 9.
Statistics
Word Problems on Arithmetic Mean
Properties of Arithmetic Mean
Problems Based on Average
Properties Questions on Arithmetic Mean
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CBSE Class 7 Maths Notes Chapter 3 Data Handling Pdf free download is part of Class 7 Maths Notes for Quick Revision. Here we have given NCERT Class 7 Maths Notes Chapter 3 Data Handling.
## CBSE Class 7 Maths Notes Chapter 3 Data Handling
Data Handling Class 7 Notes
Collection and Organisation of data in a particular manner makes it easier for us to understand and interpret data.
Before collecting data, we need to know what we would use it for.
Examples
• Performance of your class in Mathematics.
• Performance of India in football or in cricket.
• Female literacy rate in a given area, or
• The number of children below the age of five in the families around you.
Average is a number that represents or shows the central tendency of a group of observations of data.
The arithmetic mean (AM) or simply mean is defined as follows:
$$Arithmeticmean=\frac { Sum\quad of\quad all\quad observations }{ Number\quad of\quad observations }$$
Mean always lies between the greatest and smallest observation of the data.
Range is the difference between the highest and the lowest observation of the data. i.e. Range = Highest observation – Lowest observation
Mode of a set of observation is the observation that occurs, the most often e.g. 2 is the mode of a set of numbers 1, 1, 2, 4, 3, 2, 1, 2, 2, 4.
Median refers to the value which lies in the middle of the data with half of the observations above it and the other half below it.
e.g. 24, 36,46,17,18, 25, 35 is given data.
Firstly, data is to arranged in ascending order i.e. 17,18, 24,25, 35,36, 46.
Since the median is the middle observation, therefore 25 is the median.
If the data has an odd number of items, then the median is the middle number.
If the data has an even number of items, then the median is mean of two middle numbers.
A bar graph is a representation of numbers using bars of uniform widths.
Mode of the data is the longest bar if the bar represents frequency.
Double bar graphs help to compare two collections of data at a glance.
The situation that may or may not happen, have a chance of happening.
$$Probabilityofanevent=\frac { Number\quad of\quad favourable\quad outcomes }{ Total\quad number\quad of\quad outcomes\quad in\quad the\quad experiment }$$
Probability of an event which has no chance of happening is ‘0’.
Probability of an event which is bound to happen is 1.
Data Handling Class 7 Notes Pdf
Collecting Data
A given collection of data may not give us a piece of the specific information related to that data. Before collecting data, we need to know what we would use it for.
Organisation of Data
After the collection of data, we have to record and organize it.
Many kinds of data we come across are put in tabular form. Our school rolls, progress report, index in the notebooks, temperature record and many others are all in tabular form. Similarly, census record, a record of the values of shares, the record of DA, the record of HRA, are all in tabular form.
When we put data in a proper table, it becomes more meaningful. We can then interpret the data and take some inferences from them.
Class 7 Data Handling Notes
Representative Values
In our day-to-day life, we come across many statements that involve the term ‘average’. Average is a number indicating the representative of the central value of a group of observations or data. This representative value or central value is known as measure of central tendency.
Different forms of data need different forms of representative or central value to describe it.
Notes On Data Handling Class 7
Arithmetic Mean
The most common representative value is the arithmetic mean or the mean.
Rule: To find the mean, we find the sum of all the observations and divide this by the number of observations.
Result:
(i) The mean of several observations is the value which is equally shared out among all the observations.
(ii) The mean lies in between the greatest and the smallest observation.
Data Handling Notes For Class 7
Range
Range = Highest observation – Lowest observation
Mode
The mode of a set of observations is the observation that occurs most often.
Mode of Large Data
Putting the same observations together and counting is not easy if the number of observations is large. In such cases, we tabulate the data. Tabulation can begin by putting tally marks and finding the frequency.
Class 7 Maths Chapter 3 Notes Pdf
More than one Mode
A set of numbers can have more than one mode.
For example: For numbers 2, 2, 2,3, 3, 4, 5, 5, 5, 6,6, 8 ; 2 and 5 both occur highest (three) times.
Therefore, they both are the modes of the data.
Median
Median is the middle most observation of the data arranged in ascending or descending order.
Class 7 Maths Data Handling Notes
Use of Bar Graphs with a Different Purpose
By looking at the bar graph, we can make deductions about the data.
For example, we can say that the mode is longest bar if the bar represents the frequency.
Choosing a Scale
We should choose a proper scale so that all the data may be represented on the available graph paper.
Class 7 Maths Chapter 3 Notes
Drawing a double bar graph
These are used when we have to make comparisons between two collections of data at a glance.
Chance
Chance predicts simply a possibility.
7th Standard Data Handling Notes
What is Probability?
Probability is the measure of the chance of a particular event. It is measured as
$$\frac { no.\quad of\quad favourable\quad outcomes }{ no.\quad of\quad all\quad possible\quad outcomes }$$
Note: Two outcomes are said to be equally likely if we cannot expect one in preference to the other.
We hope the given CBSE Class 7 Maths Notes Chapter 3 Data Handling Pdf free download will help you. If you have any query regarding NCERT Class 7 Maths Notes Chapter 3 Data Handling, drop a comment below and we will get back to you at the earliest. |
# Thread: Find number of ways
1. ## Find number of ways
There are 8 students from 4 different schools(2 from each school).In how many ways can these students be put in 4 different rooms(2 students in each room) so that no room contains students of the same school.
2. ## Re: Find number of ways
Originally Posted by pankaj
There are 8 students from 4 different schools(2 from each school).In how many ways can these students be put in 4 different rooms(2 students in each room) so that no room contains students of the same school.
There are $4!$ to assign the students from school $X$ to a different room.
What about the students from school $Y~?$
3. ## Re: Find number of ways
Hello, pankaj!
There are 8 students from 4 different schools (2 from each school).
In how many ways can these students be put in 4 different rooms (2 students in each room)
so that no room contains students of the same school.
Suppose the students are: . $A,A,B,B,C,C,D,D$
In how many ways can they partitioned into four unmatched pairs?
This is derangement of four objects.
There are $9$ derangements. **
Then the four pairs can be assigned rooms in $4!$ ways.
Answer: . $9\cdot4! \:=\:216$ ways.
** You can count them if you like . . . $\begin{array}{cccc}A&B&C&D \\ \hline B&A&D&C \\ C&A&D&B \\ D&A&B&C \\ B&D&A&C \\ C&D&A&B \\ D&C&A&B \\ B&C&D&A \\ C&D&B&A \\ D&C&B&A \end{array}$
4. ## Re: Find number of ways
Number of ways to put the $8$ students in $4$ rooms such that there are $2$ students in every room $=\dbinom{8}{2}\times\dbinom{6}{2}\times\binom{4}{2 }=28\times15\times6=2520.$
Let $A_i$ denote the property that the $i$th room contains students of the same school.
$n(A_1\cup A_2\cup A_3\cup A_4)=\sum n(A_{i})-\sum n(A_{i}\cap A_{j})+\sum n(A_{i}\cap A_{j}\cap A_{k})- n(A_{1}\cap A_{2}\cap A_{3}\cap A_{4})$.
$n(A_{1})=\binom{4}{1}\times\big( \binom{6}{2}\times \binom{4}{2}\times 1\big)=4\times 90=360$.
Therefore, $\sum n(A_{i})=360\times 4=1440$
$n(A_{1}\cap A_{2})=\dbinom{4}{2} \times 2! \times \dbinom{4}{2} =72$. Therefore, $\sum n(A_{i}\cap A_{j})=\dbinom{4}{2} \times 72=432$
$n(A_{1}\cap A_{2}\cap A_{3})=\dbinom {4}{3}\times 3!=24$.Therefore, $\sum n(A_{i}\cap A_{j}\cap A_{k})=24=24$
$n(A_{1}\cap A_{2}\cap A_{3}\cap A_{4})=4!=24$
$n(A_1\cup A_2\cup A_3\cup A_4)=1440-432+24-24=1008$
Thus required number of ways= $2520-n(A_1\cup A_2\cup A_3\cup A_4)=2520-1008=1512$
5. ## Re: Find number of ways
I want to make a slight correction
$\sum n(A_{i}\cap A_{j}\cap A_{k})=4\times 24=96$
Thus $\ n(A_{1}\cup A_{2}\cup A_{3}\cup A_{4})=1440-432+96-24=1080$
Finally,required number of ways= $2520-1080=1440$ |
## Introduction
Our word quadratic comes from the Latin word quadratus which means a square, so quadratics include squared numbers such as and . Something that contains only x and y is liniar, if it contains and it is cubic.
Here is the graph of y = + 5x +10 for -12<x<8
It is a quadratic curve called a parabola. (Circles and ellipses are also quadratic curves but their graphs include as well as x².) All parabolas are a similar shape, although they can be the other way up, and have an axis of symmetry. If you throw a ball across an open space it follows a parabolic trajectory as it first gets higher and then starts to fall again.
From the first graph you can see that x can have any value, and for every value of x there is only one value of y, but y has a minimum value (or a maximum in the second graph), and for every value of y above this value there are two values of x. Finding the value of y given the value of x is very simple and should take you only a few seconds even without a calculator, finding both values of x given the value of y is much more difficult.
If we are given y = + 3x + 10 and want to find the value of x when y is 6 we can write this down as + 3x + 10 = 6
First we turn it into a quadratic equation. A quadratic equation is always of the form
a + bx + c = 0
where a, b and c are constants. To do this we subtract 6 from both sides to give us the quadratic equation +3x + 4 = 0
There are three ways of solving a quadratic equation, that is, finding the value of x: factorisation, completing the square, and using the formula.
## Factorisation
If we have (x + p)(x + q) and multiply it out we have
+ qx + px + pq
or + (p + q)x +pq
So (x + 3)(x + 4) = + 7x + 12
To factorise + 5x + 6 we look for two numbers which multiplied together make 6 and added together make 5. Best to work through the factors of 6 starting with the smallest, that is 1: 6 = 1 × 6, no good, try 2 × 3 bingo!
If negative numbers are involved just remember 12 is +3 × +4 or -3 × -4, and -12 is +3 × -4 or -3 × +4
For example -2x -24 = (x + 4)(x - 6)
Once we have found the factors we can solve the equation. If (x + 2)(x - 3) = 0 then either (x + 2) = 0 or (x - 3) = 0 so x = -2 or x = +3
Factorisation is not always possible, so if you do not succeed after two or three attempts try the next method, completing the square.
If there is a number in front of the for example 6 + 19x + 15 factorisation is sometimes possible but only if you are very confident about what you are doing. Unless you are it is safer to use the formula.
## Completing the square
(x + n)² = +2n + n² so - 6x + 9 is (x - 3)²
If + 6x -11 = 0 then + 6x + 9 = 20 so (x +3)² = 20
so x + 3 = ±√(20) so x = -3 ±√(20) Don’t forget the plus or minus!
We complete the square by halving the number in front of the x and then squaring it. Then we add or subtract numbers to the right hand side of the equation to leave it balanced but with just the square on the left hand side.
Remenber that there is no (real - this is explained below) value of the square root of a negative number - look at the graph of the parabola at the top of this Page to see that for values of y below about 3 there is no real value of x. |
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# Tip #11 to Get a Top SAT Math Score
By McGraw-Hill Professional
Updated on Sep 10, 2011
Slope questions appear on every SAT. This topic could fill a college–level course, but the SAT tests only a few concepts.
Most importantly, to find the slope of two ordered pairs, use the slope formula:
You should also know these other groovy SAT slope facts:
• The slope of a line measures its steepness—the steeper the line, the bigger the slope.
• A line has a positive slope if it rises from left to right.
• A line has a negative slope if it falls from left to right.
• A horizontal line has a slope of 0.
Let's look at this question:
Solution: Plug the two points into the slope equation.
Once you have , you can simply "Use the Answers" and try each answer choice to see which one makes the equation work. Skill Preview: As we will see in Skill 20, you could also cross–multiply to solve this equation.
### Easy
1. If the slope of a line through the points (3, a) and (–2, –4) is , what is the value of a ?
1. A –2
2. B –1
3. C 0
4. D 1
5. E 2
2. What is the slope of the line through the point (2, –3) and the origin?
1. A –2.5
2. B –1.5
3. C –1
4. D 0.5
5. E 2.5
3. Which of the following is closest to the slope of the line in the figure below?
### Medium
1. In the figure below, the line containing (a, b) passes through the origin. What is the value of ?
1. A –1
2. B 0
3. C 1
4. D 3
5. E Undefined
### Hard
1. In the xy coordinate plane below, the indicated diameters of the two circles shown are parallel to the y axis. If the points M and N are centers of the circles, what is the slope of line MN (not shown)?
1. B Plug the two points into the slope equation.
2. Once you have , you can simply "Use the Answers" and try each answer choice to see which one makes the equation work. You could also cross–multiply (see Skill 20).
3. B This question is exactly like Question 1. You are given 2 points, and you simply plug them into the slope formula. The only twist is that the question refers to the origin as one of the two points so you have to know that "origin" means (0, 0), which is just vocab. Know it (which you do now) and you get it correct!
4. Notice how is not a choice. Don't be thrown. Be confident, unshakeable! After you finish this book, you will always know what to do; don't let the question intimidate you. So, which choice matches our ? Just enter it in your calculator, remember that means "–3 divided by 2," and you get – 1.5. Bingo!
5. C Use two points to count the , the change in y over the change in x.
6. C Remember that pictures on the SAT are drawn exactly to scale (we'll talk more about this excellent strategy in Skill 21). So just use the picture to determine the values of (a, b); it looks like (4, 4). The question just asks for , so the answer is choice C, since = 1.
7. B To get the slope of MN, we must know the coordinates for M and for N. Since these points are centers of the circles, they are halfway between the endpoints of each diameter shown. So M equals (2, 1) and N equals (7, 3). Now we plug these into the slope formula.
Go to: Tip #12 |
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# Composition of Functions
Two or more functions where the range of the first becomes the domain of the second.
%
Progress
Practice Composition of Functions
Progress
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Composition of Functions
If f(x) = x + 2, and g(x) = 2x + 4, what is f(g(x)) ?
A function can be conceptualized as a 'black box'. The input, or x value is placed into the box, and the box performs a specific set of operations on it. Once the operations are complete, the output (the " f(x) " or " y " value) is retrieved. Once the output is retrieved, the box is ready to work on the next input.
Using this idea, function composition can be seen as a box inside of a box. The input x value goes into the inner box, and then the output of the inner box is used as the input of the outer box.
This lesson is all about boxes inside of boxes. See if you can use what you learn to answer the question above before the review at the end.
Embedded Video:
### Guidance
Composition of Functions
Functions are often described in terms of “input” and “output.” For example, consider the function f ( x ) = 2 x + 3. When we input an x value, we output a y value, or a function value. We find the output by taking the input x , multiplying by 2, and adding 3. We can do this for any value of x . Now consider a second function g ( x ) = 5 x . For this function too, we can take an x value, input the x into g ( x ), and obtain an output. What happens if we take the output of g and use it as the input of f ?
#### Example A
Given the function definition above, g ( x ) = 5 x . Therefore if x = 4, then we have g (4) = 5(4) = 20. What happens if we then take the output of 20 and use it as the input of f ?
Solution:
Substituting 20 in for x in f ( x ) = 2 x + 3 gives: f (20) = 2(20) + 3 = 43.
The table below shows several examples of this same process:
x Output from g Output from f
2 10 23
3 15 33
4 20 43
5 25 53
Examining the values in the table, we can see a pattern: all of the final output values from f are 3 more than 10 times the initial input. We have created a new function called h ( x ) out of f ( x ) = 2 x + 3 in which g ( x ) = 5 x is the input:
h ( x ) = f (5 x ) = 2(5 x ) + 3 = 10 x + 3
When we input one function into another, we call this the composition of the two functions. Formally, we write the composed function as f ( g ( x )) = 10 x + 3 or write it as ( f o g ) x = 10 x + 3
#### Example B
Find f ( g ( x )) and g ( f ( x )):
a. f ( x ) = 3 x + 1 and g ( x ) = x 2 b. f ( x ) = 2 x + 4 and g ( x ) = (1/2) x - 2
Solution:
a. f ( x ) = 3 x + 1 and g ( x ) = x 2
f ( g ( x )) = f ( x 2 ) = 3( x 2 ) + 1 = 3 x 2 + 1
g ( f ( x )) = g (3 x + 1) = (3 x + 1) 2 = 9 x 2 + 6 x + 1
In both cases, the resulting function is quadratic.
b. f ( x ) = 2 x + 4 and g ( x ) = (1/2) x - 2
f ( g ( x )) = 2((1/2) x - 2) + 4 = (2/2) x - 4 + 4 = (2/2) x = x
g ( f ( x )) = g (2 x + 4) = (1/2)(2 x + 4) - 2 = x + 2 - 2 = x .
In this case, the composites were equal to each other, and they both equal x , the original input into the function. This means that there is a special relationship between these two functions. We will examine this relationship in Chapter 3. It is important to note, however, that f ( g ( x ) is not necessarily equal to g ( f ( x )).
#### Example C
Decompose the function f ( x ) = (3 x - 1) 2 - 5 into a quadratic function g ( x ) and a linear function h ( x ).
Solution:
When we compose functions, we are combining two (or more) functions by inputting the output of one function into another. We can also decompose a function. Consider the function f ( x ) = (2 x + 1) 2 . We can decompose this function into an “inside” and an “outside” function. For example, we can construct f ( x ) = (2 x + 1) 2 with a linear function and a quadratic function. If g ( x ) = x 2 and h ( x ) = (2 x + 1), then f ( x ) = g ( h ( x )). The linear function h ( x ) = (2 x + 1) is the inside function, and the quadratic function g ( x ) = x 2 is the outside function.
Let h ( x ) = 3 x - 1 and g ( x ) = x 2 - 5. Then f ( x ) = g ( h ( x )) because g ( h ( x )) = g (3 x - 1) = (3 x - 1) 2 - 5.
The decomposition of a function is not necessarily unique. For example, there are many ways that we could express a linear function as the composition of other linear functions.
Can you answer the question at the beginning of the lesson now?
If f(x) = x + 2, and g(x) = 2x + 4, what is f(g(x)) ?
f(g(x)) = f(2x + 4) = (2x + 4) + 2 = 2x + 6
Once you get the idea, composite functions aren't as difficult as they look!
### Vocabulary
A composite function is a function formed by using the output of one function as the input of another.
The input of a function is the value on which the function is performed (commonly the x value).
The output of a function is the result of the operations performed on x (commonly y or f(x) ).
### Guided Practice
Questions
1) Given:
$f(x) = 5x + 3$
$g(x) = 3x^2$
Find: $f(g(4))$
2) Given:
$h(n) = 7n +1 + 4(g(n))$
$g(t) = -t$
$f(x) = -2x + g(x)$
Find: $f(h(-5))$
3) Given: $g(x) = 5x^2$
$h(x)=5x^2 - 2x -4(g(x))$
Find $h(g(-1))$
Solutions
1) To find f ( g (4)), we need to know what $g(4)$ is, so we know what to substitute into $f(x)$ :
Substitute 4 for x for the function g ( x ), giving: $3 \cdot 4^2$
Simplify: $3 \cdot 16 = 48$
$\therefore g(4) = 48$
Substitute 48 for the x in the function $f(x)$ giving: $5(48) +3$
Simplify: $240 + 3 = 243$
$\therefore f(g(4)) = 243$
2) First, let's solve for the value of the inner function, $h(-5)$ . Then we'll know what to plug into the outer function.
$h(-5) = (7)(-5)+1+4(g(-5))$
To solve for the value of h , we need to solve $g(-5)$
$g(-5) = - (-5)$
$\therefore g(-5) = 5$
Now we have: $h(-5)=(7)(-5)+1+(4)(5)$
Simplify to get: $h(-5)=-14$
Now we know that $h(-5) = -14$ . That tells us that $f(h(-5))$ is $f(-14)$
Find $f(-14) = (-2)(-14)+g(-14)$
So to solve for the value of $f(-14)$ , we need to solve for the value of $g(-14)$
$g(-14) = - (-14)$
$\therefore g(-14) = 14$
Now we can finish up!
$f(-14) = (-2)(-14) + 14$
$\therefore f(-14) = 42$
3) First, solve for the value of the inner function $g(-1)$ to find what to plug into the outer function $h(g(-1))$
$g(-1) = 5(-1)^2$
$g(-1) = 5 \cdot 1$
$\therefore g(-1) = 5$
Next, solve for $h(g(-1))$ which we now know is: $h(5)$
$h(5) = 5(5^2) + (-2)(5) - 4(g(5))$
To solve for the value of h , we need to solve for the value of g (5).
$g(5) = 5(5^2)$
$g(5) = 125$
$\therefore h(5) = 5(5^2) + (-2)(5) + (-4)(125)$
Finally: $h(5) = -385$
### Practice
For problems 1-3:
$f(x) = 2x - 1$
$g(x) = 3x$
$h(x) = x^2 + 1$
1. Find: $f(g(-3))$
2. Find: $f(h(7))$
3. Find: $h(g(-4))$
4. Find: $f(g(h(2)))$
Evaluate each composition below:
1. Given: $f(x) = -5x + 2$ and $g(x) = \frac{1}{2}x + 4$ Find $f(g(12))$
2. Given: $g(x) = -3x + 6$ and $h(x) = 9x + 3$ Find $g(h(\frac{1}{3}))$
3. Given: $f(x) = -\frac{1}{5}x + 4$ and $g(x) = 4x^2$ Find $f(g(10))$
4. Given $g(x) = 3|x -4|+ 6$ and $h(x) = -x^3$ Find $h(g(4))$
5. Given $f(x) = \sqrt{x + 2}$ and $g(x) = |2x|$ Find $g(f(-7))$
6. Given $f(x) = -3x + 2$ and given $g(x) = 2x^2$ and given $h(x) = 4|7 - x| + 6$ Find $f(g(h(1)))$
7. Given $f(x) = (-3)$ and given $g(x) = \sqrt{2x}$ and given $h(x) = |4x| - 12$ Find $f(h(g(18)))$
8. Are compositions commutative? In other words, does $f(g(x)) = g(f(x))$ ?
9. Given: $f(x) = -2^2 - 5x$ and $h(x) = 3x + 2$ Find $f(h(x))$
10. Two functions are inverses of each other if $f(g(x)) = x$ and $g(f(x)) = x$ If $f(x) = x + 3$ , find its inverse: $g(x)$
11. A toy manufacturer has a new product to sell. The number of units to be sold, n , is a function of the price p such that: $n(p) = 30 - 25p$ The revenue r earned from the sales is a function of the number of units sold n such that: $r(n) = 1000 - \frac{1}{4}x^2$ Find the function for revenue in terms of price, p .
### Vocabulary Language: English
composite function
composite function
A composite function is a function $h(x)$ formed by using the output of one function $g(x)$ as the input of another function $f(x)$. Composite functions are written in the form $h(x)=f(g(x))$ or $h=f \circ g$.
domain
domain
The domain of a function is the set of $x$-values for which the function is defined.
Function
Function
A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$.
Function composition
Function composition
Function composition involves 'nested functions' or functions within functions. Function composition is the application of one function to the result of another function.
input
input
The input of a function is the value on which the function is performed (commonly the $x$ value).
Output
Output
The output of a function is the result of the operations performed on the independent variable (commonly $x$). The output values are commonly the values of $y$ or $f(x)$.
Range
Range
The range of a function is the set of $y$ values for which the function is defined. |
# What are the mathematical probabilities of rolling a certain number on a dice?
### The Basics of Dice Rolling Probability
Rolling a dice may seem like a game of pure chance, but there are mathematical probabilities to the outcome of the roll. Each dice has six sides, numbered 1 to 6, and the likelihood of rolling a certain number varies depending on the number of dice rolled and the desired outcome.
### The Probability of Rolling a Specific Number
To determine the probability of rolling a certain number on a single dice, you take the number of ways to roll the number you want, and divide it by the total number of outcomes. For example, if you want to roll a 3 on a single dice, there is only one way to do that, so the probability is 1/6 or approximately 16.67%.
### The Probability of Rolling a Certain Combination
If you want to roll a certain combination of numbers, such as rolling a 7 with two dice, you calculate the probability of rolling each number separately and then multiply those probabilities together. In the case of rolling a 7, there are six ways to roll a 7 with two dice (1+6, 2+5, 3+4, 4+3, 5+2, 6+1), so the probability is 6/36 or 1/6 or approximately 16.67%, which is the same as rolling a 3 on a single dice.
### The Probability of Rolling at Least One Certain Number
If you want to know the probability of rolling at least one certain number in multiple rolls of a dice, you can calculate the probability of not rolling that number and then subtract it from 1. For example, if you want to know the probability of rolling at least one 4 in three rolls of a dice, the probability of not rolling a 4 in one roll is 5/6, so the probability of not rolling a 4 in three rolls is (5/6)^3 or approximately 57.87%, and the probability of rolling at least one 4 is 1-(5/6)^3 or approximately 42.13%.
### The Importance of Understanding Probability
Understanding probabilities in dice rolling is important in games of chance such as craps, where betting on the outcome of dice rolls is a major component of the game. By understanding the probability of certain outcomes, players can make informed decisions about their bets and increase their chances of winning. |
INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith
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• ## Related Books
1 answerLast reply by: Dr Carleen EatonSun Apr 17, 2016 1:38 PMPost by Oscar Prado on April 12, 2016Why is the reason that we have to get the square root at the end of every problem?
### Pythagorean Theorem
• A right triangle is a triangle which has a right angle. The side opposite the right angle, called the hypotenuse, is the longest side of the triangle. The two sides forming the right angle are called the legs of the triangle.
• The Pythagorean Theorem states that if a right triangle has legs of lengths a and b, and hypotenuse of length c, then a2 + b2 = c2.
• This result can be used to find the length of any side of a right triangle if the other two sides are known.
• Some triples (a, b, c) of whole numbers, such as (3, 4, 5), satisfy the Pythagorean Theorem. Such triples are called Pythagorean triples. Note that any multiple of a Pythagorean triple is also a Pythagorean triple.
• The converse of the Pythagorean Theorem, which is also true, states that if the sides a, b, and c of a triangle satisfy the equation a2 + b2 = c2, then the triangle is a right triangle.
### Pythagorean Theorem
The legs of a right triangle have lengths 5 and 12. Find the length of the hypothenuse.
• a2 + b2 = c2
• 52 + 122 = c2
• 25 + 144 = c2
• 169 = c2
c = 13
The legs of a right triangle have lengths 10 and 8. Find the length of the hypothenuse.
• a2 + b2 = c2
• 82 + 102 = c2
• 64 + 100 = c2
• 164 = c2
c = 2√{41}
The legs of a right triangle have lengths 16 and 22. Find the length of the hypothenuse.
• 162 + 222 = c2
• 256 + 484 = c2
• 740 = c2
c = √{2 ×5 ×2 ×37} = 2√{185}
The hypothenuse of a right triangle has length 36 and the triangle is isoceles. Find the length of the legs.
• x2 + x2 = 362
• 2x2 = 1296
• x2 = 648
x = √{9 ×9 ×2 ×2 ×2} = 18√2
The hypothenuse of a right triangle has length 12 and the triangle is isoceles. Find the length of the legs.
• x2 + x2 = 122
• 2x2 = 144
• x2 = 72
x = √{2 ×2 ×2 ×3 ×3} = 6√2
The hypothenuse of a right triangle has length 16 and the triangle is isoceles. Find the length of the legs.
• x2 + x2 = 162
• 2x2 = 256
• x2 = 128
x = √{4 ×4 ×2 ×2 ×2} = 8√2
The hypothenuse of a right triangle is 18 and the length of one leg is 6. Find the area of the triangle.
• 62 + b2 = 182
• 36 + b2 = 324
• b2 = 288
• b = √{2 ×2 ×2 ×2 ×2 ×3 ×3} = 12√2
• area = [1/2]bh
• area = [1/2]( 6 )( 12√2 )
area = 36√2
The hypothenuse of a right triangle is 9 and the length of one leg is 4. Find the area of the triangle.
• 42 + b2 = 92
• 16 + b2 = 81
• b2 = 65
• b = √{65}
• area = [1/2]( 4 )( √{65} )
area = 2√{65}
The longest side of a triangle has length 10 and the shortest side has length 7. Find the length of the third side so that the triangle is a right triangle.
• 102 = a2 + 72
• 100 = a2 + 49
• a2 = 51
a = √{51}
The longest side of a triangle has length 30 and the shortest side has length 14. Find the length of the third side so that the triangle is a right triangle.
• 302 = a2 + 142
• 900 = a2 + 196
• a2 = 704
a = √{8 ×8 ×11} = 8√{11}
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
### Pythagorean Theorem
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Right Triangles 0:06
• Vertex
• Hypotenuse
• Legs
• Pythagorean Theorem 1:21
• Graphical Representation
• Example
• Pythagorean Triples 3:40
• Example
• Converse of the Pythagorean Theorem 4:36
• Example
• Example 1: Length of Hypotenuse 7:24
• Example 2: Length of Legs 9:02
• Example 3: Area of Triangle 12:00
• Example 4: Length of Side 14:59 |
Q:
# How do you teach algebra to beginners?
A:
Some students find algebra intimidating because they are uncomfortable replacing numbers with letters. However, basic algebra builds directly on things they already know from arithmetic. To teach beginning algebra, set up short exercises within 20-30 minute lessons using equations that are easy to prove.
## Keep Learning
Credit: altrendo images Altrendo Getty Images
1. Show how algebra works with basic addition
Set up an equation starting with single-digit numbers, such as 3 + 4 = ?. The student provides the answer 7. Increase one of the numbers, as in 3 + 5 = 8, to reinforce the pattern. Ask the student to find the missing number in a similar pattern, as in 3 + ? = 9. The student should answer 6. If the student has trouble, help them count up on their fingers. Let them know that by finding the missing number, they have solved an algebra problem.
2. Demonstrate converting addition to subtraction
By the time they begin algebra, most students are familiar with how addition and subtraction relate. Use that familiarity. Check the student's understanding. For example, write 8 + 7 = 15 and 15 - 8 = ?. The student should easily solve the second equation. Next, put the missing number in the middle of an operations, such as 11 + ? = 24. Ask the student to rewrite it as a subtraction problem like the example you showed them. The student should write 24 - 11 = ?.
3. Teach the student to solve on both sides
Write a new equation, such as 32 + ? = 51. Tell students they can solve the equation if they can get the question mark by itself, as they did in the other examples. Demonstrate how you can move the 32 to the other side by subtracting it from both sides. 32 + ? - 32 = 51 - 32 0 + ? = 51 - 32 19 = 51 - 32
4. Substitute x and repeat
Start using letters instead of question marks or empty boxes. Explain that the letter is just standing in for an unknown number. Repeat the lesson for different kinds of operations.
Sources:
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INTEGRATION TECHNIQUES:
# INTEGRATION TECHNIQUES:
## INTEGRATION TECHNIQUES:
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. INTEGRATION TECHNIQUES:
2. INTEGRATION… WHAT DO WE KNOW SO FAR? • In terms of getting integrals, we know how to do basic anti-differentiation with power functions, trig functions and exponential functions. • We started some techniques using “u-substitution” to solve integrals. More or less, “a reverse chain rule.” • We also did area, volume, and other physical applications using the definite integral. • This chapter is devoted to the indefinite integral.
3. 1) U-SUBSTITUTIONREVISITED • Just to freshen your minds, let’s do a u-substitution problem. There is a little catch to it… • For this problem, it is preferable to pick u = x+2, since du can never be in the denominator. • Use algebra to give a name for x+1. Since x=u-2, then x+1 would be u -2+1 or u-1. • After putting everything in, do the integration. • Note that in the final answer, the constant 2 does not need to be there, since C is more general.
4. PRODUCT RULE FLASHBACK • Remember back in Chapter 2, we mentioned the product rule? Looked like the following… In regular form…. Or differential form….
5. 2) INTEGRATION BY PARTS • If you play around with the differentials, you will get the following. • If you integrate both sides, you get what is known as the INTEGRATION BY PARTS FORMULA
6. EXAMPLE#1 • Integrate x*exdx. • STEP 1: • Pick your “u” and “dv.” • TRICK: L.I.P.E.T. This determines what u should be initially. Logarithm, Inverse, Power, Exponential, and Trig functions. • In this problem, a power function and an exponential function are present. Since power (P) comes before exponential (E), u should equal x. • From the u, find du. From the dv, find v. • Therefore u=x, then du=dx! Don’t forget that. • Simply plug this into the “parts formula.”
7. EXAMPLE#1 • I = integral to be solved. • Plug in u, v, du, and dv. • If the integral on the right looks easy to compute, then simply integrate it. • Don’t forget the +C!
8. EXAMPLE#2 • Integrate the function…
9. EXAMPLE#2 • Step 1: Find the u and dv: • An exponential and a trig function is present. • E (exponential) comes before T (trig). • Use the exponential function for u.
10. EXAMPLE#2 • Step#2: Plug u, du, v and dv into the formula • Simplify… • If the last integral is easy to compute, compute it. • However, it is not easy. In fact, we are in more mess than we started in. Looks like we have to use the integration by parts rule again.
11. EXAMPLE#2 • Use integration by parts again. Since there is an exponential function, call that “u.” After finding u, du, v, and dv, plug those values in the “Parts equation” and see what happens… • We have even more of a mess……but wait! The integral of exsin(x) is supposed to be equal to I (the integral we wanted to solve for in the first place!!). • So we can replace the integral of exsin(x) with I and add it to both sides. • You will see that it works out. • Always add the constant
12. RULE OF THUMB WITH INTEGRATION BY PARTS PROBLEMS • Always pick the right “u.” If the problem is getting really difficult, maybe you picked the wrong “u.” Just like in u-substitution. You had to pick the right “u” to work with the problem. • If the integral on the right does not look easy to compute, then do integration by parts for that integral only. • If you see the resulting integral looks like the integral you are asked to solve for in the first place, then simply combine the two like integrals and use algebra to solve for the integral.
13. 3) TRIGONOMETRIC INTEGRALS • You will always get the situation of funny combination of integrals. Trig functions are as such that you can translate from one function to a function with just sines and cosines. For example, you can always write tan, sec, csc, cot in terms of either sine or cosine. • Remember the following identities from PRE-CALCULUS!!!
14. IMPORTANT IDENTITIES
15. PROCEDURE • Well… I’m afraid to say it, but there is really no procedure or real template in attacking these problems except proper planning. • This takes a great deal of practice
16. EXAMPLE# 1 • Given: • Best thing to do is to break the cosine function down to a 2nd degree multiplied by a 1st degree cosine. • Since cos2x=1-sin2x, you can replace it. • Use u-substitution to solve the integral.
17. EXAMPLE#2 • Given • Break the 4th degree sine to two 2nd degre sines. • Use the sin2x theorem. • Expand the binomial squared. • For the cos22x, use the cos2x theorem. • Simplify • Don’t forget the +C!
18. EXAMPLE #3 • Whenever you see a tan, sec, csc, or cot, always convert them to sines and cosines. This way, you can cancel or combine whenever necessary. • In this case, tan2x=sin2x/cos2x. We are also lucky that the cos2x cancels. • Using the sin2x theorem, we can simply integrate.
19. NOTE ABOUT TRIGONOMETRIC INTEGRALS • There is no real rule for such integrals. But always remember: • 1) If there is a mix of sines and cosines, break them up until they resemble an easier form • 2) Use any trig theorem that would be relevant to make a problem simpler. • 3) Convert everything to sines or cosines.
20. 4) TRIGONOMETRIC SUBSTITUTION • Remember when we took derivatives of inverse trigonometric functions, we commonly dealt with sums or differences of squares. • Similarly, integrating sums of differences of square will lead us to the inverse trig functions. • However we need a stepping stone to integrate such functions.
21. GENERAL PROCEDURE Given that a is a constant, and u is a function, then follow the IF YOU HAVE THIS.. THEN USE THIS FORMULA
22. THETA? • Since we are working with a substitution, theta would be the variable to use subsitution… • Doesn’t make sense? Let’s do an example problem…
23. EXAMPLE #1 • Initially a very bad looking problem… • Focus on the denominator, inside the radical, you have 9-x2. In effect, that is a2-u2, a being 3 and u being x. If a2-u2 is used, then according to the table in the last slide, we would use u=a sinq. • You already found a name for x. You need to give a name for “dx.” Differentiate x with respect to q. Solve for dx.
24. EXAMPLE #1 • Here is the original problem… • With the substitutions of x and dx, here is the original problem… • Simplify a little… • Pull out constants when needed.
25. EXAMPLE #1 • The denominator is actually cos2x, according to the trig theorem. (Memorize them!) • Simplify • Integrate • We have our answer in terms of q!!! We need it in terms of x!
26. EXAMPLE #1 • Don’t forget what we said earlier. That x=3 sin q. We need to know what q is in order to find out the solution in terms of x.
27. EXAMPLE #1 • Simply replace all the q expressions with x expressions. • Simplify • Add constant!! • Sighs!! We’re finished!!!
28. That was a LOT of work!!! • Here are trig substitution steps… • 1) Find the correct equality statement using the table. • 2) Make the proper substitutions. Remember to have a substitute for x as well as dx. • 3) Integrate in terms of q. • 4) Convert all q terms to x terms.
29. 5) RATIONAL FUNCTIONS • Of course, there will always be functions in the form of a ratio of two functions. • Two integrate most rational functions, the method of partial fractions come into play. • <<Break from Calculus… entering Algebra Territory>>
30. PARTIAL FRACTIONS • This means you take a fraction and break it down into a sum of many fractions. • This way, we can add up the integrals of simpler easier fractions.
31. EXAMPLE • Given • The denominator could be factored to (x+5)(x-2). This way we could have new denominators for the two new fractions. • Add these new fractions and distribute. Make sure you bring all x terms together, as well as bringing all the constants together.
32. EXAMPLE • The coefficient of x on the right side is 1. In order to keep the equality true, the coefficient of x on the left side should also equal 1. • A+B=1 • Same thing with the constant. If the equality holds true, then -2A+5B must equal -9. • To solve for A and B, you use methods from algebra. <System of linear equations>. • If you multiply A+B=1 by 2, you will see that B=-1. Therefore A=2.
33. EXAMPLE • Since A=2 and B=-1, we can simply plug them in. • And integrate!!! • And the final answer!!
34. ANOTHER EXAMPLE • Given • Note: If the numerator has a higher degree than the denominator, then do long polynomial division. • If you actually do the long division, you will get x-1-1/(x+1). This is very easy to integrate.
35. POINTERS OF PARTIAL FRACTIONS • 1) Check if the top degree is bigger than the bottom. If so, perform long division • 2) If the denominator is factorable, then assume that the denominators of the new fractions will be those factors.
36. 6) QUADRATIC DENOMINATOR PROBLEMS • This is really no different than trigonometric substitution. • Strictly rational functions with a quadratic denominator that cannot be reduced. • To make the denominator easier to work with, you must complete the square
37. COMPLETING THE SQUARE • Given problem: • Look at the denominator. Take the coefficient of x and divide it by 2. • Take this result and square it. • 4/2 =22=4 • This result would form a perfect square when added to x2+4x. • The perfect square would be (x+2)2. • However, we have a 5. If you add and subtract 4, combining 5 and -4 will yield 1. • You have a form of u2+a2! Time for trig substitution!
38. EXAMPLE • Since we have u2+a2, we must use the fact of u=a*tan(q). • u=x+2 while a=1 • Substitute the values in appropriate spots.
39. EXAMPLE (work)
40. POINTERS • 1) Make the denominator into one of the three forms that allows trig substitution by the use of completing the square. • 2) Follow rules of trig substitution.
41. 7) IMPROPER INTEGRALS • This is not an integral evaluating technique. • An improper integral is basically an integral that has infinity as its limits or has a discontinuity within its limits.
42. IMPROPER INTEGRALS • Examples of improper integrals:
43. IMPROPER INTEGRALS • With limits of infinity, just use a letter to replace the infinity and treat as a limit. • And integrate as if nothing ever happened • Don’t forget to use the limit. • Amazing! As we start from 0 to infinity, we get closer to 1 square unit of area! We say that it converges to 1.
44. IMPROPER INTEGRALS • Since we have a discontinuity in this function at x=-2. To take this into account, we must split the integral into two parts. In addition, we cannot go exactly -2, but we have to get there pretty darn close. Therefore, we must use the one-sided limits from Chapter 1 to represent this. Two integrals: one from -3 to a little before -2, and a little after -2 to 2. • Other than that, simply integrate • Notice how we got an answer that don’t exist!! D.N.E (does not exist)! This means that this integral diverges. Also if an integral goes to infinity, it diverges.
45. POINTERS OF IMPROPER INTEGRALS • Remember to identify all the points of discontinuity. Remember to use limits before and after the points of discontinuity. • If you have infinity as your limit, remember to use infinity as your limit. • Other than that, use ALL of the previous techniques of integration mentioned.
46. FUNCTIONS WITHOUT AN ANTIDERIVATIVE • Besides three more chapters, this is the last of the single variable calculus. That is to say: y=f(x) in 2-dimensional x,y graph. • Before moving on, I must admit… even though all continuous functions have derivatives, not all continuous functions have simple integrals in terms of elementary functions. • Elementary functions are adding, subtracting, multiplication, division, power, rooting, exponential, logarithmic, trigonometric, all of their inverses as well as combinations or composition functions. Basically, all the functions you ever used were composed of elementary functions. • Some functions do not have elementary antiderivatives. For example… the classic (sin x)/x problem. • No matter what method you tried. Neither by u-substitution, integration by parts, trig substitution, partial fractions, or even guess and check will get you an antiderivative. • From my experience from differential equations class last year, the integral of sin(x)/x is Si(x) also known as the sine integral! • YOU DON’T NEED TO KNOW THAT!!!!
47. OUTTA THIS WORLD FUNCTIONS!!!!!!! • You will be dealing with functions like erf(x), Si(x), Ci(x), Shi(x), Chi(x), FresnelS(x), and FresnelC(x). Take their derivatives and you’ll get regular “sane” functions. AAHHH!!! HARI BOL!!!!
48. SUMMARY • Actually, for once, looking at the length and material of this chapter. I am quite amazed to say that I have no words to summarize this chapter. There has been so many methods of integration. Namely u-substitution, integration by parts, how to deal with trig integrals, trig substitution, partial fractions, quadratic denominators, and improper integrals. • All I can say is that review this material over again!!! • Like I said previously, there is no set way to do these problems. There are more than one way of doing it. • You have to know what to do when which problem arrives at you. |
# Determinants: What is Determinant of a Matrix, Calculating the Determinant (For CBSE, ICSE, IAS, NET, NRA 2022)
Glide to success with Doorsteptutor material for competitive exams : get questions, notes, tests, video lectures and more- for all subjects of your exam.
## What is Determinant of a Matrix?
• When a square matrix “A” of an order “n” is associated with a number, then it is titled as a determinant of the matrix.
• The determinant of a matrix is a special number that can be calculated from a square matrix.
• A Matrix is an array of numbers:
• The number involved in this square matrix can be a real number or a complex number.
• In the field of mathematics, Determinants can be used for a myriad of different calculations such as:
• Finding the area of a Triangle.
• Obtaining the solutions of linear equations in two or three variables.
• Solve Linear equations by using the inverse of a matrix
• Getting the adjoint and the inverse of a square matrix
## Calculating the Determinant
• First of all the matrix must be square (i.e.. have the same number of rows as columns) .
• Then it is just basic arithmetic.
## What Are the Properties of Determinants?
• The properties of determinants have its uses, in simplifying the evaluation by getting the maximum number of zeroes in a row or column.
• They hold true for determinants of any order.
• The determinant՚s value remains unchanged if the rows and columns are interchanged.
• For each element of a row or column which gets multiplied by a constant P, the value also gets multiplied by P.
• The sign of a determinant changes when any two rows or columns of a determinant are interchanged with each other.
• The value of a determinant is zero when any of the rows or columns of a determinant are identical to each other.
• The determinant obtained is a sum of two or more determinants if some or all elements of a column or a row are expressed as a sum of two or more terms
• When the equimultiples of corresponding elements are added, the value of the determinant remains the same, whenever the operation is multiplied.
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