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## Want to keep learning?
This content is taken from the UNSW Sydney's online course, Maths for Humans: Linear, Quadratic & Inverse Relations. Join the course to learn more.
2.27
## UNSW Sydney
Parabola through three points
# A quadratic function through three points
Another application of quadratic functions is to curve fitting, also called the theory of splines. Since a parabola $$\normalsize{y=ax^2+bx+c}$$ is specified by three numbers, it is reasonable to suppose that we could fit a parabola to three points in the plane. This is indeed the case, and it is a useful idea.
In this step we see how to algebraically fit a parabola to three points in the Cartesian plane. This involves recalling, or learning, how to solve three equations in three unknowns. This is a useful skill on its own right.
## The unique circle through three non-collinear points
A line is determined by two points. A circle, on the other hand, is determined by three points —as long as these points are not collinear (all three points cannot lie on the same line). The construction of a circle which passes through three points is a standard exercise in Euclidean geometry: we construct the perpendicular bisectors of the line segments determined by these three points, and then these three lines meet at the circumcenter of the triangle $$\normalsize{ABC}$$, namely the centre of the unique circle which passes through all three points.
Here is a GeoGebra image of this construction:
What about a parabola? Well, if we restrict ourselves to parabolas given by quadratic equations, that is curves which the form $$\normalsize{y=ax^2+bx+c}$$, then since there are three unknown quantities, we can expect that we can fit such a curve to three distinct points. Let’s see how this works in practice.
## Fitting a parabola to three points
Suppose we want to find a parabola with equation $$\normalsize{y=ax^2+bx+c}$$ which passes through the points $$\normalsize{[0,1], [1,5]}$$ and $$\normalsize{[2,3]}$$. Substituting each of the three points into the equation, we get
$\Large{1=c}$ $\Large{5=a+b+c}$ $\Large{3=4a+2b+c}$
This is three equations in three unknowns. This is more complicated, but in this particular case, things are simpler since the first equation already tells us that $$\normalsize{c=1}$$, so the other two equations become $$\normalsize{a+b=4}$$ and $$\normalsize{2a+b=1}$$.
Solving these equations, yields $$\normalsize{a=-3}$$, $$\normalsize{b=7}$$ and also $$\normalsize{c=1}$$, so the required parabola is $$\normalsize{y=-3x^2+7x+1}$$. Here is the graph:
## Solving three linear equations in three unknowns
Since two linear equations represent two lines in the plane, their common solution corresponds to the geometric meet of the two lines. For three linear equations in three unknowns, the situation actually corresponds to the common intersection point of three planes in three-dimensional space!
Fortunately the ancient Chinese were able to develop a general technique for solving such systems of equations. Here we just try to find a simple practical method.
Suppose we want to find the equation of the quadratic function $$\normalsize{y=ax^2+bx+c}$$ which passes through the points $$\normalsize{[1,3], [2,-1]}$$ and $$\normalsize{[4,1]}$$. It means we have three equations, one for each of the points – since we know the points given must satisfy the unknown equation. The three equations are
$\Large{3=a+b+c} \label{b1p} \tag {1}$ $\Large{-1=4a+2b+c} \label{b2p} \tag {2}$ $\Large{1=16a+4b+c}. \label{b3p} \tag 3$
What is the strategy? It is simple: we try to eliminate one of the variables, leaving us with two equations in two unknowns. This we know how to solve.
To get two equations in two variables, let’s eliminate $$\normalsize{c}$$ from the first two equations. We do that by subtracting one from the other, say subtract the first from the second:
$\Large{-4=3a+b} \label{b4p} \tag{4}$
Please make sure you understand how we got that. Now we do the same with the last two equations: take the third equation and subtract the second:
$\Large{2=12a+2b}$
We can make that a bit simpler by dividing all the coefficients by $$\normalsize{2}$$ to get
$\Large{1=6a+b}. \label{b5p} \tag {5}$
Now we treat $$\normalsize{(\ref{b5p})}$$ and $$\normalsize{(\ref{b4p})}$$ as new equations, and use them to find $$\normalsize{a}$$ and $$\normalsize{b}$$.
If we take $$\normalsize{(\ref{b5p})}$$–$$\normalsize{(\ref{b4p})}$$ we get $$\normalsize{5=3a}$$, so that $$\normalsize{a=5/3}$$, and then plugging back into either $$\normalsize{(\ref{b3p})}$$ or $$\normalsize{(\ref{b4p})}$$ gives $$\normalsize{b=-9}$$. Then putting both of these back into say $$\normalsize{(\ref{b1p})}$$ gives $$\normalsize{3=5/3-9+c}$$ so that $$\normalsize{c=31/3}$$.
The parabola is therefore $$\normalsize{y=5/3x^2-9x+31/3}$$. Here is its graph:
Q1 (M): Find the equation of the quadratic function $$\normalsize{y=ax^2+bx+c}$$ which passes through the points $$\normalsize{[-1,9], [2,3]}$$ and $$\normalsize{[5,15]}$$.
Q2 (C): Find a quadratic relation of the form $$\normalsize{x=ay^2 + by +c}$$ which passes through the points $$\normalsize{[0,1], [1,5]}$$ and $$\normalsize{[2,3]}$$.
A1. The answer is the parabolic function $$\normalsize{y=x^2-3x+5}$$.
A2. The answer is the curve $$x=- \frac{1}{8}\left(3y^2-20y+17\right)$$. Note this is not a function.
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# How do you multiply (4sqrt18)(2sqrt14)?
Jun 18, 2015
$48 \sqrt{7}$
#### Explanation:
You can multiply the numbers outside the roots and them try to manipulate the arguments of the square roots:
$4 \cdot 2 \sqrt{18} \sqrt{14} =$
$= 8 \sqrt{9 \cdot 2} \sqrt{7 \cdot 2} =$
$= 8 \cdot 3 \sqrt{2} \sqrt{7 \cdot 2} =$ take one big root:
$= 24 \sqrt{2 \cdot 7 \cdot 2} =$
$= 24 \sqrt{7 \cdot 4} = 48 \sqrt{7}$
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# Basic facts and techniques of Boats and Streams of Quantitative Aptitude
Boats and Streams is a part of the Quantitative aptitude section. This is just a logical extension of motion in a straight line. One or two questions are asked from this chapter in almost every exam. Today I will tell you some important facts and terminologies which will help you to make better understanding about this topic.
## Basic Concept
If direction of boat is same as direction of the stream, then it is known as DOWNSTREAM
and if directions are opposite, then it is known as UPSTREAM. Following figure is representing the same:
i.e. if boat is moving with stream then it is known as Downstream and if opposite to stream, then it is Downstream.
### Downstream Speed and Upstream Speed
In case of Downstream, as you can see the direction is same, speeds of stream and boat will be added to get Downstream speed.
If Speed of boat in still water = u km/hr
Speed of stream = v km/hr, then
Downstream Speed = (u+v) km/hr
Similarly, if I talk about upstream speed, as the direction of boat and stream is opposite, speed of both will be subtracted.
i.e. Upstream Speed = (u-v) km/hr
Study the following figure, notice the directions and try to remember this i.e.
If directions are same then speeds will be added and
If directions are opposite then speeds will be subtracted
### Speeds of Boat and Stream if Downstream and Upstream Speeds are given
Speed of Boat = 1/2 (Downstream Speed + Upstream Speed)
Speed of Stream = 1/2 (Downstream Speed - Upstream Speed)
## Problems with Solution
Example1: Speed of boat in still water is 5 km/hr and speed of stream is 1 km/hr. Find the downstream speed and upstream speed.
Solution: Given that, u = 5 km/hr
v = 1 km/hr
Downstream speed = u+ v km/hr
⇒ 5+ 1= 6 km/hr
Upstream speed = u -v km/hr
⇒ 5- 1 = 4 km/hr
Example2: A man takes 3 hours to row a boat 15km downstream of river and 2 hours 30 min to cover a distance of 5 km upstream. Find speed of river or stream.
Solution: We need to find speed of stream from downstream speed and upstream speed. See how I calculate it:
As You know, Speed = Distance/ Time
So, Downstream Speed = (15)/3 = 5 km/hr
Upstream Speed = 5/2.5 = 2 km/hr
Now, As i have discussed, Speed of stream = 1/2 (Downstream Speed - Upstream Speed)
⇒Speed of stream = 1/2 (5-2)
⇒3/2 = 1.5 km/hr
Example3: A man can row 7km/hr in still water. If in a river running at 2 km/hr, it takes him 50 minutes to row to his place and back, how far off is the place?
*Important Question*
Solution: Given, u = 7km/hr
v = 2km/hr
From u and v , we can calculate Downstream speed and upstream speed.
Downstream Speed = (u + v) = 7+2 = 9 km/hr
Upstream Speed = (u-v) = 7-2 = 5 km/hr
Now, we need to find DISTANCE and time is given,
Time = Distance / Speed
Let required distance = x km
Time taken in downstream + Time taken in upstream = Total Time
⇒ (x/9) + (x/5) = (50)/(60)........................ 50 minutes = (50)/(60) hrs
⇒Calculating the above equation: x = 2.68km
I hope You have better understanding of this topic now. Also check the following topics:
#### What's trending in BankExamsToday
Smart Prep Kit for Banking Exams by Ramandeep Singh - Download here
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# How do you differentiate f(x)= 3x^2 (4x - 12)^2 using the product rule?
Nov 12, 2015
Taken you to a point where you should be able to complete the calculation.
#### Explanation:
Let $u = 3 {x}^{2} \text{ }$ then $\frac{\mathrm{du}}{\mathrm{dx}} = 6 x$ ..............(1)
Let $v = {\left(4 x - 12\right)}^{2} \text{ }$ then $\frac{\mathrm{dv}}{\mathrm{dx}} = 8 \left(4 x - 12\right) = 32 x - 96$.....(2)
Product $\left(v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}\right) \text{ }$
$\textcolor{b l u e}{\text{~~~~~~~~~Foot Note showing How to obtain (2) ~~~~~~~~~~~~~~}}$
$\textcolor{b r o w n}{\text{Let } w = 4 x - 12}$
$\textcolor{b r o w n}{\frac{\mathrm{dw}}{\mathrm{dx}} = 4}$
$\textcolor{b r o w n}{\text{But "v=w^2 " so } \frac{\mathrm{dv}}{\mathrm{dw}} = 2 w}$
$\textcolor{b r o w n}{\text{But } \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{\mathrm{dv}}{\mathrm{dw}} \times \frac{\mathrm{dw}}{\mathrm{dx}}}$
color(brown)("(dv)/(dx)= 2w times 4 = 8w = 8(4x-12))
$\textcolor{b l u e}{\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$
By substitution we have:
$\left\{{\left(4 x - 12\right)}^{2} \times 6 x\right\} + \left\{3 {x}^{2} \times \left(32 x - 96\right)\right\}$
$\textcolor{b l u e}{\text{I have left the final calculation and simplification for you to complete}}$
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# Thearrangement in increasing order for the given numbers.
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
#### Solutions
Chapter 2, Problem 29RE
To determine
## To find: Thearrangement in increasing order for the given numbers.
Expert Solution
The number can be arranged in increasing order as f''5<0<f'(5)<f'(2)<1<f'(3) .
### Explanation of Solution
Given information:
The given numbers are 01f'(2)f'(3)f'(5)f''(5) .
The given graph is shown in figure (1).
Figure (1)
Calculation:
Using the given graph it observe that, the graph of f is a concave downword at the point 5 .
This implies that f''(x)<0 on the interval (3,9) .
So, f''(5)<0 .
The derivative of a function is the slope of the tangent line at that point.
So, draw the tangent line at the point 2,3 and 5 .
Figure (2)
From the figure (2) observe that the graph f is increases at the point 2,3 and 5 .
So, f'(x)>0 at the point 2,3 and 5 .
That is, f'(2)>0,f'(3)>0, and f'(5)>0
The slope of the tangent line increases when the slope becomes sleeper.
The slope of the tangent line at x=3 is larger than the slope at x=2,5 , because the slope is sleeper at x=3 .
The slope of the tangent line at x=2 is larger than the slope at x=5 , because the slope is sleeper at x=2 .
Finally, The slope of the tangent line at x=5 is is closely appeared at horizontal line, the tangent line at x=5 has a small positive slope (nearly zero).
From the above discussion the number can be arranged as.
0<f'(5)<f'(2)<1<f'(3)
At f''5<0 .
The number are arranged as.
f''5<0<f'(5)<f'(2)<1<f'(3)
Therefore, the number can be arrangedin increasing order as f''5<0<f'(5)<f'(2)<1<f'(3) .
### Have a homework question?
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Lesson Video: Converting Lengths: Meters and Centimeters | Nagwa Lesson Video: Converting Lengths: Meters and Centimeters | Nagwa
# Lesson Video: Converting Lengths: Meters and Centimeters Mathematics
In this video, we will learn how to measure length using mixed units, convert between lengths using partitioning, and compare lengths given in meters and centimeters.
13:01
### Video Transcript
Converting Lengths: Meters and Centimeters
In this video, we’re going to learn how to measure length using mixed units, to convert between lengths using partitioning, and also to compare lengths given in meters and centimeters. Which of these two snakes is longer, the one that measures 100 centimeters or the one that measures one meter? Well, this is a trick question to test whether you’re awake. They’re both the same length. One meter is exactly the same length as 100 centimeters, and it’s important that we remember this fact because we’re going to be using it again and again in this video.
This is because we’re going to be learning how to convert measurements so we can write them in three different ways: in centimeters and then, depending on the measurement, either meters or meters and centimeters. We’ll see when we need to use this in a moment. But for now, because we know that one meter is the same as 100 centimeters, we can start to use this to help us convert between the two. This snake is two meters long. How many centimeters long is it? We know that one meter is worth 100 centimeters, and so two meters would be two lots of 100 centimeters or 200 centimeters. It’s quite easy to convert between the two when we’re thinking of whole numbers of meters.
Three meters is 300 centimeters. Four meters is 400 centimeters and so on. And we can even work the other way, changing centimeters into meters. This snake is 600 centimeters long. And what’s the fact that we need to use to help us here? We know 100 centimeters are the same as one meter. So we need to ask ourselves, how many lots of 100 are there in 600? There are six lots of 100 in 600. And if each of these is worth one meter, then we know the snake is six meters long. 600 centimeters equals six meters. They’re just different ways of saying the same length.
So far, we’ve been looking at measurements that are whole numbers of meters. But what about this snake? It’s 228 centimeters long. Because 228 isn’t a multiple of 100, we’re going to need to give the measurement in meters and some centimeters. And to do this, we need to start with 228 and partition it so we take the 100s out. There are two 100s in 228, so we can partition this number into 200 and then the 28 that’s left over. Now because we’ve split up this number or partitioned it so that we could see the 100s, we can convert them. We know that 200 centimeters are exactly the same as two meters, so at least we’ve changed part of our measurement into meters. We can’t change it all though; we’ve still got those 28 centimeters left over. This snake’s length is two meters and 28 centimeters.
You know, we can also convert in the opposite direction. We can start with a measurement in meters and centimeters and then convert it so that we write it just in centimeters. This snake is one meter and 15 centimeters long. Now we can partition this length into a part worth one meter and another part worth 15 centimeters. A part of the length is already in centimeters, so we don’t need to worry about this for now. Let’s just concentrate on the part that isn’t in centimeters. How many centimeters are the same as one meter? 100 centimeters. And we know if we put our two parts back together again — 100 plus another 15 is 115 centimeters. One meter 15 centimeters is the same as 115 centimeters.
Let’s have a go at answering some questions now where we have to convert between meters and centimeters.
Look at the ruler. One meter equals 100 centimeters. How many centimeters are in three meters? Three meters equals what centimeters.
To begin within this question, we’re told to look at the picture of the ruler. So, we better start by doing that. What can we see? Looks a little bit like a number line, doesn’t it? It starts at zero. And we can see each multiple of 10 is labeled: 10, 20, 30, and so on all the way up to 100. Now if we look at the start of the ruler, we can see the letters cm. This shows us the unit of measurement that our ruler shows. cm stands for centimeters. And so when we’re counting 10, 20, 30, and so on, we’re counting in centimeters. This ruler goes from zero to 100 centimeters.
But there’s something else interesting about our ruler. Can you spot it? At the other end of the ruler, there’s an arrow labeled one m. So if we count all the way up to 100 centimeters, it’s the same as one m. Do you know what one m stands for? Of course, we’re told in the fact above the ruler, aren’t we? One meter. One meter is the same as 100 centimeters, and our ruler helps to show this. Now, if one meter is the same as 100 centimeters, we can use this to help answer our question “How many centimeters are in three meters?” Let’s use a bar model to help us.
If we know one meter is the same as 100 centimeters, then two meters must be the same as two lots of 100 centimeters or 200 centimeters. And so to answer our question, three meters will be the same as three lots of 100 centimeters or 300 centimeters. The number of centimeters that are in three meters is 300. Three meters equals 300 centimeters.
Write in centimeters: eight meters and five centimeters.
In this question, we’re given a measurement. Now we might think at first glance that this looks like two measurements because not only can we see a number of meters, but also some centimeters too. But this is one single measurement. It just means slightly more than eight meters, eight meters and five centimeters. Now at the moment, this measurement has been written, as we’ve said already, in meters and also centimeters. But we can convert this measurement so that we write it just in centimeters, which is exactly what this question asks us to do. To help us, we can split up or partition eight meters five centimeters. And we can split it into eight meters and five centimeters.
Now our question asks us to write the whole measurement in centimeters. And at the moment, part of it already is in centimeters, so we can ignore this for a moment. Let’s concentrate on the part that isn’t in centimeters. How can we convert eight meters into centimeters? Is there a fact we can use to help us? We know that one meter is the same as 100 centimeters, and so eight meters must be the same as eight lots of 100 centimeters or 800 centimeters. Now that we’ve converted our number of meters into centimeters, both parts are written in centimeters. So we just need to add the two parts back together again. 800 centimeters plus another five centimeters equals 805 centimeters. We can write eight meters and five centimeters in centimeters as 805 centimeters.
Complete: What centimeters equals four meters and 70 centimeters.
To help us answer this question, we’re shown a part–whole model. Now, normally, if we wanted to find the missing number here, that’s the missing whole, with a part–whole model, we just add the two parts together. But in this particular question, we can’t just add four and 70 to find the answer. Can you see why? In this question, the numbers four and 70 represent two measurements: four meters and 70 centimeters. We can’t just add four and 70 together because they’re different units. If we look at the missing number in our part–whole model, we can see that we’re looking for a number of centimeters. And we can see this in the question, too.
For us to write this whole measurement in centimeters, we’re going to need to change those four meters into centimeters, aren’t we? Is there a fact We know that can help us? We know that one meter is the same as 100 centimeters, and so four meters is the same as four lots of 100 centimeters or 400 centimeters. Should we write this on our part–whole model to help us? Instead of four meters, let’s write 400 centimeters instead. They’re exactly the same. Now we can add our two parts together. 400 plus another 70 equals 470. And so we know that four meters and 70 centimeters is exactly the same distance as 470 centimeters. Our missing number is 470.
A red car is 411 centimeters long, a blue car is 500 centimeters long, and a black car is three meters long. Which car is the longest?
In this question, we need to compare the lengths of three cars. Let’s have a look at the measurements: 411 centimeters, 500 centimeters, three meters. What do you notice about these measurements? They’re not all measured in the same unit, are they? We’ve got two measurements in centimeters and one in meters. This means that we can’t just compare the numbers and look at them and see which one is largest.
Instead, we’re going to have to convert the measurements so that they’re all in the same unit of measurement. Should we change them all into meters or centimeters? It doesn’t matter which we choose, but with questions like this, we need to use a bit of common sense. Two of our measurements are already in centimeters, so the quickest thing to do would be to change the one that’s in meters into centimeters, and then they’re all in centimeters. We know that one meter is the same as 100 centimeters, and so three meters is the same as three lots of 100 centimeters or 300 centimeters.
Now that all our measurements are in centimeters, we could show them all on a number line. A red car is 411 centimeters. That’s about here on the number line. A blue car is 500 centimeters long. And although we said a black car was three meters, we worked out this was exactly the same as 300 centimeters. We made sure all our measurements were in the same unit so that we could then compare them. The longest of the three cars is the blue car.
So what have we learned in this video? We’ve learned how to measure length using mixed units, convert between lengths using partitioning, and compare lengths given in meters and centimeters.
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# 125% of 16 is what number?
Apr 2, 2018
See a solution process below:
#### Explanation:
"Percent" or "%" means "out of 100" or "per 100", Therefore 125% can be written as $\frac{125}{100}$.
When dealing with percents the word "of" means "times" or "to multiply".
Finally, lets call the number we are looking for "n".
Putting this altogether we can write this equation and solve for $n$ while keeping the equation balanced:
$n = \frac{125}{100} \times 16$
$n = \frac{2000}{100}$
$n = 20$
125% of 16 is $\textcolor{red}{20}$
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# Geometric Sequences
Module by: First Last. E-mail the author
Summary: In this section, you will:
• Find the common ratio for a geometric sequence.
• List the terms of a geometric sequence.
• Use a recursive formula for a geometric sequence.
• Use an explicit formula for a geometric sequence.
Note: You are viewing an old style version of this document. The new style version is available here.
Many jobs offer an annual cost-of-living increase to keep salaries consistent with inflation. Suppose, for example, a recent college graduate finds a position as a sales manager earning an annual salary of $26,000. He is promised a 2% cost of living increase each year. His annual salary in any given year can be found by multiplying his salary from the previous year by 102%. His salary will be$26,520 after one year; $27,050.40 after two years;$27,591.41 after three years; and so on. When a salary increases by a constant rate each year, the salary grows by a constant factor. In this section, we will review sequences that grow in this way.
## Finding Common Ratios
The yearly salary values described form a geometric sequence because they change by a constant factor each year. Each term of a geometric sequence increases or decreases by a constant factor called the common ratio. The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term.
### A General Note: Definition of a Geometric Sequence:
A geometric sequence is one in which any term divided by the previous term is a constant. This constant is called the common ratio of the sequence. The common ratio can be found by dividing any term in the sequence by the previous term. If a 1 a 1 is the initial term of a geometric sequence and r r is the common ratio, the sequence will be
{ a 1 , a 1 r, a 1 r 2 , a 1 r 3 ,...}. { a 1 , a 1 r, a 1 r 2 , a 1 r 3 ,...}.
### How To:
Given a set of numbers, determine if they represent a geometric sequence.
1. Divide each term by the previous term.
2. Compare the quotients. If they are the same, a common ratio exists and the sequence is geometric.
### Example 1
#### Problem 1
##### Finding Common Ratios
Is the sequence geometric? If so, find the common ratio.
1. 1,2,4,8,16,... 1,2,4,8,16,...
2. 48,12,42,... 48,12,42,...
##### Solution
Divide each term by the previous term to determine whether a common ratio exists.
1. 2 1 =2 4 2 =2 8 4 =2 16 8 =2 2 1 =2 4 2 =2 8 4 =2 16 8 =2
The sequence is geometric because there is a common ratio. The common ratio is 2.
2. 12 48 = 1 4 4 12 = 1 3 2 4 = 1 2 12 48 = 1 4 4 12 = 1 3 2 4 = 1 2
The sequence is not geometric because there is not a common ratio.
##### Analysis
The graph of each sequence is shown in Figure 1. It seems from the graphs that both (a) and (b) appear have the form of the graph of an exponential function in this viewing window. However, we know that (a) is geometric and so this interpretation holds, but (b) is not.
### Q&A:
If you are told that a sequence is geometric, do you have to divide every term by the previous term to find the common ratio?
No. If you know that the sequence is geometric, you can choose any one term in the sequence and divide it by the previous term to find the common ratio.
### Try It:
#### Exercise 1
Is the sequence geometric? If so, find the common ratio.
5,10,15,20,... 5,10,15,20,...
##### Solution
The sequence is not geometric because 10 5 15 10 10 5 15 10 .
### Try It:
#### Exercise 2
Is the sequence geometric? If so, find the common ratio.
100,20,4, 4 5 ,... 100,20,4, 4 5 ,...
##### Solution
The sequence is geometric. The common ratio is 1 5 1 5 .
## Writing Terms of Geometric Sequences
Now that we can identify a geometric sequence, we will learn how to find the terms of a geometric sequence if we are given the first term and the common ratio. The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. For instance, if the first term of a geometric sequence is a 1 =2 a 1 =2 and the common ratio is r=4, r=4, we can find subsequent terms by multiplying 24 24 to get 8 8 then multiplying the result 84 84 to get 32 32 and so on.
a 1 =2 a 2 =(24)=8 a 3 =(84)=32 a 4 =(324)128 a 1 =2 a 2 =(24)=8 a 3 =(84)=32 a 4 =(324)128
The first four terms are {–2–8–32–128}. {–2–8–32–128}.
### How To:
Given the first term and the common factor, find the first four terms of a geometric sequence.
1. Multiply the initial term, a 1 , a 1 , by the common ratio to find the next term, a 2 . a 2 .
2. Repeat the process, using a n = a 2 a n = a 2 to find a 3 a 3 and then a 3 a 3 to find a 4, a 4, until all four terms have been identified.
3. Write the terms separated by commons within brackets.
### Example 2
#### Problem 1
##### Writing the Terms of a Geometric Sequence
List the first four terms of the geometric sequence with a 1 =5 a 1 =5 and r=–2. r=–2.
##### Solution
Multiply a 1 a 1 by 2 2 to find a 2 . a 2 . Repeat the process, using a 2 a 2 to find a 3 , a 3 , and so on.
a 1 =5 a 2 =2 a 1 =10 a 3 =2 a 2 =20 a 4 =2 a 3 =40 a 1 =5 a 2 =2 a 1 =10 a 3 =2 a 2 =20 a 4 =2 a 3 =40
The first four terms are { 5,–10,20,–40 }. { 5,–10,20,–40 }.
### Try It:
#### Exercise 3
List the first five terms of the geometric sequence with a 1 =18 a 1 =18 and r= 1 3 . r= 1 3 .
##### Solution
{ 18,6,2, 2 3 , 2 9 } { 18,6,2, 2 3 , 2 9 }
## Using Recursive Formulas for Geometric Sequences
A recursive formula allows us to find any term of a geometric sequence by using the previous term. Each term is the product of the common ratio and the previous term. For example, suppose the common ratio is 9. Then each term is nine times the previous term. As with any recursive formula, the initial term must be given.
### A General Note: Recursive Formula for a Geometric Sequence:
The recursive formula for a geometric sequence with common ratio r r and first term a 1 a 1 is
a n =r a n1 ,n2 a n =r a n1 ,n2
(6)
### How To:
Given the first several terms of a geometric sequence, write its recursive formula.
1. State the initial term.
2. Find the common ratio by dividing any term by the preceding term.
3. Substitute the common ratio into the recursive formula for a geometric sequence.
### Example 3
#### Problem 1
##### Using Recursive Formulas for Geometric Sequences
Write a recursive formula for the following geometric sequence.
{6913.520.25...} {6913.520.25...}
##### Solution
The first term is given as 6. The common ratio can be found by dividing the second term by the first term.
r= 9 6 =1.5 r= 9 6 =1.5
Substitute the common ratio into the recursive formula for geometric sequences and define a 1 . a 1 .
a n =r a n1 a n =1.5 a n1 for n2 a 1 =6 a n =r a n1 a n =1.5 a n1 for n2 a 1 =6
##### Analysis
The sequence of data points follows an exponential pattern. The common ratio is also the base of an exponential function as shown in Figure 2
### A General Note:
Do we have to divide the second term by the first term to find the common ratio?
No. We can divide any term in the sequence by the previous term. It is, however, most common to divide the second term by the first term because it is often the easiest method of finding the common ratio.
### Try It:
#### Exercise 4
Write a recursive formula for the following geometric sequence.
{2 4 3 8 9 16 27 ...} {2 4 3 8 9 16 27 ...}
##### Solution
a 1 =2 a n = 2 3 a n1 for n2 a 1 =2 a n = 2 3 a n1 for n2
## Using Explicit Formulas for Geometric Sequences
Because a geometric sequence is an exponential function whose domain is the set of positive integers, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms.
a n = a 1 r n1 a n = a 1 r n1
Let’s take a look at the sequence {183672144288...}. {183672144288...}. This is a geometric sequence with a common ratio of 2 and an exponential function with a base of 2. An explicit formula for this sequence is
a n =18· 2 n1 a n =18· 2 n1
The graph of the sequence is shown in Figure 3.
### A General Note: Explicit Formula for a Geometric Sequence:
The nth term of a geometric sequence is given by the explicit formula:
a n = a 1 r n1 a n = a 1 r n1
(13)
### Example 4
#### Problem 1
##### Writing Terms of Geometric Sequences Using the Explicit Formula
Given a geometric sequence with a 1 =3 a 1 =3 and a 4 =24, a 4 =24, find a 2 . a 2 .
##### Solution
The sequence can be written in terms of the initial term and the common ratio r. r.
3,3r,3 r 2 ,3 r 3 ,... 3,3r,3 r 2 ,3 r 3 ,...
Find the common ratio using the given fourth term.
a n = a 1 r n1 a 4 =3 r 3 Write the fourth term of sequence in terms of α 1 and r 24=3 r 3 Substitute 24 for a 4 8= r 3 Divide r=2 Solve for the common ratio a n = a 1 r n1 a 4 =3 r 3 Write the fourth term of sequence in terms of α 1 and r 24=3 r 3 Substitute 24 for a 4 8= r 3 Divide r=2 Solve for the common ratio
Find the second term by multiplying the first term by the common ratio.
a 2 =2 a 1 =2(3) =6 a 2 =2 a 1 =2(3) =6
##### Analysis
The common ratio is multiplied by the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by multiplying the first term by the common ratio nine times or by multiplying by the common ratio raised to the ninth power.
### Try It:
#### Exercise 5
Given a geometric sequence with a 2 =4 a 2 =4 and a 3 =32 a 3 =32 , find a 6 . a 6 .
##### Solution
a 6 =16,384 a 6 =16,384
### Example 5
#### Problem 1
##### Writing an Explicit Formula for the nth Term of a Geometric Sequence
Write an explicit formula for the nth nth term of the following geometric sequence.
{21050250...} {21050250...}
##### Solution
The first term is 2. The common ratio can be found by dividing the second term by the first term.
10 2 =5 10 2 =5
The common ratio is 5. Substitute the common ratio and the first term of the sequence into the formula.
a n = a 1 r (n1) a n =2 5 n1 a n = a 1 r (n1) a n =2 5 n1
The graph of this sequence in Figure 4 shows an exponential pattern.
### Try It:
#### Exercise 6
Write an explicit formula for the following geometric sequence.
{–13–927...} {–13–927...}
##### Solution
a n = (3) n1 a n = (3) n1
## Solving Application Problems with Geometric Sequences
In real-world scenarios involving arithmetic sequences, we may need to use an initial term of a 0 a 0 instead of a 1 . a 1 . In these problems, we can alter the explicit formula slightly by using the following formula:
a n = a 0 r n a n = a 0 r n
### Example 6
#### Problem 1
##### Solving Application Problems with Geometric Sequences
In 2013, the number of students in a small school is 284. It is estimated that the student population will increase by 4% each year.
1. Write a formula for the student population.
2. Estimate the student population in 2020.
##### Solution
1. The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04.
Let P P be the student population and n n be the number of years after 2013. Using the explicit formula for a geometric sequence we get
P n =284 1.04 n P n =284 1.04 n
2. We can find the number of years since 2013 by subtracting.
20202013=7 20202013=7
We are looking for the population after 7 years. We can substitute 7 for n n to estimate the population in 2020.
P 7 =284 1.04 7 374 P 7 =284 1.04 7 374
The student population will be about 374 in 2020.
### Try It:
#### Exercise 7
A business starts a new website. Initially the number of hits is 293 due to the curiosity factor. The business estimates the number of hits will increase by 2.6% per week.
1. Write a formula for the number of hits.
2. Estimate the number of hits in 5 weeks.
##### Solution
1. P n = 2931.026 a n P n = 2931.026 a n
2. The number of hits will be about 333.
### Media:
Access these online resources for additional instruction and practice with geometric sequences.
## Key Equations
recursive formula for nthnth term of a geometric sequence a n =r a n−1 ,n≥ 2 a n =r a n−1 ,n≥ 2 explicit formula for nth nth term of a geometric sequence a n = a 1 r n−1 a n = a 1 r n−1
## Key Concepts
• A geometric sequence is a sequence in which the ratio between any two consecutive terms is a constant.
• The constant ratio between two consecutive terms is called the common ratio.
• The common ratio can be found by dividing any term in the sequence by the previous term. See Example 1.
• The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. See Example 2 and Example 4.
• A recursive formula for a geometric sequence with common ratio r r is given by a n =r a n1 a n =r a n1 for n2 n2 .
• As with any recursive formula, the initial term of the sequence must be given. See Example 3.
• An explicit formula for a geometric sequence with common ratio r r is given by a n = a 1 r n1 . a n = a 1 r n1 . See Example 5.
• In application problems, we sometimes alter the explicit formula slightly to a n = a 0 r n . a n = a 0 r n . See Example 6.
## Section Exercises
### Verbal
#### Exercise 8
What is a geometric sequence?
##### Solution
A sequence in which the ratio between any two consecutive terms is constant.
#### Exercise 9
How is the common ratio of a geometric sequence found?
#### Exercise 10
What is the procedure for determining whether a sequence is geometric?
##### Solution
Divide each term in a sequence by the preceding term. If the resulting quotients are equal, then the sequence is geometric.
#### Exercise 11
What is the difference between an arithmetic sequence and a geometric sequence?
#### Exercise 12
Describe how exponential functions and geometric sequences are similar. How are they different?
##### Solution
Both geometric sequences and exponential functions have a constant ratio. However, their domains are not the same. Exponential functions are defined for all real numbers, and geometric sequences are defined only for positive integers. Another difference is that the base of a geometric sequence (the common ratio) can be negative, but the base of an exponential function must be positive.
### Algebraic
For the following exercises, find the common ratio for the geometric sequence.
#### Exercise 13
1,3,9,27,81,... 1,3,9,27,81,...
#### Exercise 14
0.125,0.25,0.5,1,2,... 0.125,0.25,0.5,1,2,...
##### Solution
The common ratio is 2 2
#### Exercise 15
2, 1 2 , 1 8 , 1 32 , 1 128 ,... 2, 1 2 , 1 8 , 1 32 , 1 128 ,...
For the following exercises, determine whether the sequence is geometric. If so, find the common ratio.
#### Exercise 16
6,12,24,48,96,... 6,12,24,48,96,...
##### Solution
The sequence is geometric. The common ratio is 2.
#### Exercise 17
5,5.2,5.4,5.6,5.8,... 5,5.2,5.4,5.6,5.8,...
#### Exercise 18
1, 1 2 , 1 4 , 1 8 , 1 16 ,... 1, 1 2 , 1 4 , 1 8 , 1 16 ,...
##### Solution
The sequence is geometric. The common ratio is 1 2 . 1 2 .
#### Exercise 19
6,8,11,15,20,... 6,8,11,15,20,...
#### Exercise 20
0.8,4,20,100,500,... 0.8,4,20,100,500,...
##### Solution
The sequence is geometric. The common ratio is 5. 5.
For the following exercises, write the first five terms of the geometric sequence, given the first term and common ratio.
#### Exercise 21
a 1 =8, r=0.3 a 1 =8, r=0.3
#### Exercise 22
a 1 =5, r= 1 5 a 1 =5, r= 1 5
##### Solution
5,1, 1 5 , 1 25 , 1 125 5,1, 1 5 , 1 25 , 1 125
For the following exercises, write the first five terms of the geometric sequence, given any two terms.
#### Exercise 23
a 7 =64, a 10 =512 a 7 =64, a 10 =512
#### Exercise 24
a 6 =25, a 8 =6.25 a 6 =25, a 8 =6.25
##### Solution
800,400,200,100,50 800,400,200,100,50
For the following exercises, find the specified term for the geometric sequence, given the first term and common ratio.
#### Exercise 25
The first term is 2, 2, and the common ratio is 3. 3. Find the 5th term.
#### Exercise 26
The first term is 16 and the common ratio is 1 3 . 1 3 . Find the 4th term.
##### Solution
a 4 = 16 27 a 4 = 16 27
For the following exercises, find the specified term for the geometric sequence, given the first four terms.
#### Exercise 27
a n ={ 1,2,4,8,... }. a n ={ 1,2,4,8,... }. Find a 12 . a 12 .
#### Exercise 28
a n ={ 2, 2 3 , 2 9 , 2 27 ,... }. a n ={ 2, 2 3 , 2 9 , 2 27 ,... }. Find a 7 . a 7 .
##### Solution
a 7 = 2 729 a 7 = 2 729
For the following exercises, write the first five terms of the geometric sequence.
#### Exercise 29
a 1 =486, a n = 1 3 a n1 a 1 =486, a n = 1 3 a n1
#### Exercise 30
a 1 =7, a n =0.2 a n1 a 1 =7, a n =0.2 a n1
##### Solution
7,1.4,0.28,0.056,0.0112 7,1.4,0.28,0.056,0.0112
For the following exercises, write a recursive formula for each geometric sequence.
#### Exercise 31
a n ={ 1,5,25,125,... } a n ={ 1,5,25,125,... }
#### Exercise 32
a n ={ 32,16,8,4,... } a n ={ 32,16,8,4,... }
##### Solution
a = 1 32, a n = 1 2 a n1 a = 1 32, a n = 1 2 a n1
#### Exercise 33
a n ={ 14,56,224,896,... } a n ={ 14,56,224,896,... }
#### Exercise 34
a n ={ 10,3,0.9,0.27,... } a n ={ 10,3,0.9,0.27,... }
##### Solution
a 1 =10, a n =0.3 a n1 a 1 =10, a n =0.3 a n1
#### Exercise 35
a n ={ 0.61,1.83,5.49,16.47,... } a n ={ 0.61,1.83,5.49,16.47,... }
#### Exercise 36
a n ={ 3 5 , 1 10 , 1 60 , 1 360 ,... } a n ={ 3 5 , 1 10 , 1 60 , 1 360 ,... }
##### Solution
a 1 = 3 5 , a n = 1 6 a n1 a 1 = 3 5 , a n = 1 6 a n1
#### Exercise 37
a n ={ 2, 4 3 , 8 9 , 16 27 ,... } a n ={ 2, 4 3 , 8 9 , 16 27 ,... }
#### Exercise 38
a n ={ 1 512 , 1 128 , 1 32 , 1 8 ,... } a n ={ 1 512 , 1 128 , 1 32 , 1 8 ,... }
##### Solution
a 1 = 1 512 , a n =4 a n1 a 1 = 1 512 , a n =4 a n1
For the following exercises, write the first five terms of the geometric sequence.
#### Exercise 39
a n =4 5 n1 a n =4 5 n1
#### Exercise 40
a n =12 ( 1 2 ) n1 a n =12 ( 1 2 ) n1
##### Solution
12,6,3, 3 2 , 3 4 12,6,3, 3 2 , 3 4
For the following exercises, write an explicit formula for each geometric sequence.
#### Exercise 41
a n ={ 2,4,8,16,... } a n ={ 2,4,8,16,... }
#### Exercise 42
a n ={ 1,3,9,27,... } a n ={ 1,3,9,27,... }
##### Solution
a n = 3 n1 a n = 3 n1
#### Exercise 43
a n ={ 4,12,36,108,... } a n ={ 4,12,36,108,... }
#### Exercise 44
a n ={ 0.8,4,20,100,... } a n ={ 0.8,4,20,100,... }
##### Solution
a n =0.8 (5) n1 a n =0.8 (5) n1
#### Exercise 45
a n ={1.25,5,20,80,...} a n ={1.25,5,20,80,...}
#### Exercise 46
a n ={ 1, 4 5 , 16 25 , 64 125 ,... } a n ={ 1, 4 5 , 16 25 , 64 125 ,... }
##### Solution
a n = ( 4 5 ) n1 a n = ( 4 5 ) n1
#### Exercise 47
a n ={ 2, 1 3 , 1 18 , 1 108 ,... } a n ={ 2, 1 3 , 1 18 , 1 108 ,... }
#### Exercise 48
a n ={ 3,1, 1 3 , 1 9 ,... } a n ={ 3,1, 1 3 , 1 9 ,... }
##### Solution
a n =3 ( 1 3 ) n1 a n =3 ( 1 3 ) n1
For the following exercises, find the specified term for the geometric sequence given.
#### Exercise 49
Let a 1 =4, a 1 =4, a n =3 a n1 . a n =3 a n1 . Find a 8 . a 8 .
#### Exercise 50
Let a n = ( 1 3 ) n1 . a n = ( 1 3 ) n1 . Find a 12 . a 12 .
##### Solution
a 12 = 1 177,147 a 12 = 1 177,147
For the following exercises, find the number of terms in the given finite geometric sequence.
#### Exercise 51
a n ={ 1,3,9,...,2187 } a n ={ 1,3,9,...,2187 }
#### Exercise 52
a n ={ 2,1, 1 2 ,..., 1 1024 } a n ={ 2,1, 1 2 ,..., 1 1024 }
##### Solution
There are 12 12 terms in the sequence.
### Graphical
For the following exercises, determine whether the graph shown represents a geometric sequence.
#### Exercise 54
##### Solution
The graph does not represent a geometric sequence.
For the following exercises, use the information provided to graph the first five terms of the geometric sequence.
#### Exercise 55
a 1 =1, r= 1 2 a 1 =1, r= 1 2
#### Exercise 56
a 1 =3, a n =2 a n1 a 1 =3, a n =2 a n1
#### Exercise 57
a n =27 0.3 n1 a n =27 0.3 n1
### Extensions
#### Exercise 58
Use recursive formulas to give two examples of geometric sequences whose 3rd terms are 200. 200.
##### Solution
Answers will vary. Examples: a 1 =800, a n =0.5a n1 a 1 =800, a n =0.5a n1 and a 1 =12.5, a n =4a n1 a 1 =12.5, a n =4a n1
#### Exercise 59
Use explicit formulas to give two examples of geometric sequences whose 7th terms are 1024. 1024.
#### Exercise 60
Find the 5th term of the geometric sequence {b,4b,16b,...}. {b,4b,16b,...}.
##### Solution
a 5 =256b a 5 =256b
#### Exercise 61
Find the 7th term of the geometric sequence {64a(b),32a(3b),16a(9b),...}. {64a(b),32a(3b),16a(9b),...}.
#### Exercise 62
At which term does the sequence {10,12,14.4,17.28, ...} {10,12,14.4,17.28, ...} exceed 100? 100?
##### Solution
The sequence exceeds 100 100 at the 14th term, a 14 107. a 14 107.
#### Exercise 63
At which term does the sequence { 1 2187 , 1 729 , 1 243 , 1 81 ... } { 1 2187 , 1 729 , 1 243 , 1 81 ... } begin to have integer values?
#### Exercise 64
For which term does the geometric sequence a n =36 ( 2 3 ) n1 a n =36 ( 2 3 ) n1 first have a non-integer value?
##### Solution
a 4 = 32 3 a 4 = 32 3 is the first non-integer value
#### Exercise 65
Use the recursive formula to write a geometric sequence whose common ratio is an integer. Show the first four terms, and then find the 10th term.
#### Exercise 66
Use the explicit formula to write a geometric sequence whose common ratio is a decimal number between 0 and 1. Show the first 4 terms, and then find the 8th term.
##### Solution
Answers will vary. Example: Explicit formula with a decimal common ratio: a n =400 0.5 n1 ; a n =400 0.5 n1 ; First 4 terms: 400,200,100,50; a 8 =3.125 400,200,100,50; a 8 =3.125
#### Exercise 67
Is it possible for a sequence to be both arithmetic and geometric? If so, give an example.
## Glossary
common ratio:
the ratio between any two consecutive terms in a geometric sequence
geometric sequence:
a sequence in which the ratio of a term to a previous term is a constant
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# 5 Parallelogram Problem Solving Techniques for Mastering Geometry
쉬운 목차
## Introduction to Parallelogram Properties
A parallelogram, with its opposite sides parallel and equal in length, serves as a cornerstone of geometric theory, captivating both mathematicians and students. Grasping these characteristics is crucial for delving into geometric problem solving.
## Equations Central to Parallelograms
True to its shape, a parallelogram holds specific equations based on its sides and angles, which permit a range of calculations like deducing the area, the diagonals’ lengths, or solving for variables such as ‘x’.
## Identifying Variables within Parallelogram Problems
To efficiently address parallelogram problem solving, one must discern what ‘x’ signifies, be it an angle, a side’s dimension, or another element like height or diagonal.
## Approaches to Solve for X with Angles
Using angles can unlock many answers in a parallelogram. With opposite angles being congruent and adjacent ones supplementing to 180 degrees, equations formulating ‘x’ are easily established.
### Example Calculation Using Angles
An example is a parallelogram with angles (2x + 30) degrees and (x + 50) degrees as supplementary:
[2x + 30 + x + 50 = 180]
Finding ‘x’ from this equation allows precise determination of the angles.
## Calculating Side Lengths and X Variables
Sides of a parallelogram often relate to ‘x’, leading to ratios that reveal the unknown when compared with measured sides.
### Example Calculation of Side Lengths
If a side is (4x + 5) units and its opposite counterpart measures 13 units, the equation would be:
[4x + 5 = 13]
Solving this uncovers ‘x’ and the parallelogram’s dimensions.
## Determining Diagonal Lengths and Their Relation to X
The bisecting diagonals in a parallelogram, while not necessarily equal, present patterns that aid in calculating ‘x’.
### Example Calculation Using Diagonals
With diagonals intersecting at E, if one is (3x + 7) units and a segment from a corner to E is (x + 9) units, the formula is then:
[(3x + 7) = 2(x + 9)]
The segments’ equality facilitates ‘x’ determination.
## Applying Sine and Cosine Laws in Parallelogram Problems
Complex situations involving sides and angles may require the sine and cosine laws, valuable trigonometric tools for resolving ‘x’.
## Systems of Equations for Advanced Problem-Solving
Complicated cases with several unknowns call for systems of equations, offering a robust approach to discover ‘x’ values.
## Relevance of Area and Perimeter in Parallelogram Calculations
Formulas for area and perimeter give rise to algebraic problems involving ‘x’ where substitution and resolution lead to the sought-after value.
### Area-Based Example Calculation
When a parallelogram’s area is defined as (A = base × height) and the height is (3x + 4) units with a known base, it results in:
[A = base × (3x + 4)]
Fulfilling the area allows us to solve for ‘x’.
## The Influence of Geometric Theorems on X Values
Geometric theorems, like those on parallel lines intersected by a transversal, are instrumental in creating equations to resolve for ‘x’ in parallelogram challenges.
## Practical Applications of Parallelogram Theory
Beyond academia, the skills in parallelogram problem solving have tangible benefits in architecture, engineering, and digital graphics where exactness is paramount.
## Conclusion on Parallelogram Calculations Mastery
Achieving proficiency in parallelogram problem solving is a significant step in advancing one’s grasp of geometry, whether academically engaged or professionally active with geometric forms.
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## New Maths Curriculum: Year 2 Multiplication and Division
Year 2 Multiplication and Division
It’s time to start learning those times tables off by heart; beginning with the 2x, 5x and 10x tables.
The statutory requirements for Year 2 Multiplication and division :
Pupils should be taught to:
• recall and use multiplication and division facts for the 2, 5 and 10 multiplication tables, including recognising odd and even numbers
• calculate mathematical statements for multiplication and division within the multiplication tables and write them using the multiplication (×), division (÷) and equals (=) signs
• show that multiplication of two numbers can be done in any order (commutative) and division of one number by another cannot
• solve problems involving multiplication and division, using materials, arrays, repeated addition, mental methods, and multiplication and division facts, including problems in contexts.
This is the year that children begin to learn multiplication tables, starting with the two, five and ten times tables. Remember, it is not enough to just count up in fives (5, 10, 15, 20 etc.) the whole sentence needs to be learned (5 times 2 equals 10 etc).
Perhaps the main change is that tables are meant to be learned up to 12 times, rather than just 10.
As with addition, it is important to know that multiplication can be done in any order, but not division. The relationship between multiplication and division is now stressed more, with inverse operations for checking made more explicit in the Guidance.
Year 2 multiplication and division statutory requirements
## New Maths Curriculum: Year 1 Multiplication and Division
Let’s look at the new targets for year 1 multiplication and division. A lot of the work this year should be practically based.
The statutory requirements for Year 1 Multiplication and Division are:
Multiplication and division
Pupils should be taught to:
• solve one-step problems involving multiplication and division, by calculating the answer using concrete objects, pictorial representations and arrays with the support of the teacher.
Main changes
Multiplication and division problems including arrays are now included, whilst previously these were expectations for Year 2 onwards.
Grouping and sharing small quantities forms the major part of the work.
Doubling small numbers is an important part of this year’s work. Plenty of opportunity should also be given to counting, using 2p, 5p and 10p pieces, both counting up and down. Similar activities can be carried out with pairs of socks, gloves etc. Once again, it is better to use real life objects when beginning to combine groups.
Practice should be given at counting up in twos before moving on to fives or tens and can be shown as arrays or number patterns.
Remember, division can be understood in two ways:
1. Sharing equally e.g. 6 bars of chocolate are shared between 2 people.
How many bars does each person get?
This concept is best introduced with practical apparatus – bowls of sweets, counters, buttons etc. are ideal for sharing into equal groups, and then counting the number in the group.
2. Grouping, or repeated subtraction e.g. 8 divided by 2
can be seen as how many lots of two can I take from eight?
Again, this is best done with apparatus at first and it is a slightly different process to sharing equally (where one item is given to each person in turn).
This can be answered by repeatedly taking two from the pile and then counting how many lots of two have been taken altogether.
Multiplication and division in Year 1
## Mental division practice
being able to divide mentally is essential and is made much easier by knowing times tables.
Here is a follow up page to one published earlier, giving more practice with simple division. If children have a good knowledge of the 2x, 4x, 5x and 10x tables they should find these quite straightforward. The only potentially tricky ones are where the missing number is in the middle of the number sentence
eg 24 divided by ? = 6
This requires a good understanding of what the number sentence means, but all that is required to answer is a knowledge of ‘what multiplied by 6 makes 24’.
This worksheet can be found in our Year 3 Knowing Number Facts section.
Y3 division practice 2
## Year 5 Knowing multiplication facts to help with division.
There is no doubt that most children find division harder than multiplication, yet there is little real reason for this to be so. Most division questions can be turned on their head and knowledge of times tables used to work them out. For example:
30 ÷ 5 =?? can be thought as what number times 5 makes 30?
?? ÷ 6 = 7 can be thought of as 6 times 7.
60 ÷ ?? = 10 can be thought of as what number times 10 makes 60?
The key to success with division is to have a really good knowledge of times tables.
This worksheet looks at these types of division question, all with easy numbers and no remainders and is a good assessment sheet to see if division is understood and tables known. if times tables are not known sufficiently well children will spend more time than necessary trying to work these out.
This can be found in our Year 5 Knowing Number Facts category.
Know division facts (1)
## Resource of the Week: short division of decimals
This week I would like to look at a page of practice on using the short method of division of decimals. The numbers being divided are just units and tenths which helps with getting the method correct.
There are arguments for and against putting the decimal point in before you start, or leaving it until you have reached that point in the question. it does not matter as long as it is inserted correctly.
One of the best ways to be fluent with this method is to talk it through out loud. If we look at question 2 which is 4.5 divided by 3, the verbal stages are:
a. How many 3s in 4?
b. 1 times 3 is 3 so there is 1 with a remainder of 1.
c. Place the 1 on the answer line, immediately above the 3.
d. Place the decimal point just above the answer line so it can be clearly seen.
e. The remainder 1 is placed just in front of the 5 (usually written smaller).
f. How many 3s in 15?
g. 3 x 5 is 15 so the answer is 5.
h. Place the 5 on the answer line, immediately above the 5 (tenths).
Division of decimals (1)
## Division: 3-digit numbers divided by 8
Here is another page for practising the short division method, this time just using the eight times table. The key to this method is, of course, a good knowledge of the 8 times table, as without this the solution can take an awfully long time and lots of errors may occur. The table has been written out as a helpful starting point, so that the method can be concentrated on.
For example: 8)659
‘How many eights in 65?’
‘8 times 8 is 64, 9 times 8 is 72 which is too much.’
‘8 goes into 65, 8 times with 1 left over.’
Write the 8 in the answer above the 5.
Write the remainder 1 beside the 9 units, making 19.
‘How many eights in 19?’
‘2 times 8 is 16.’
‘8 goes into 19, 2 times with 3 left over.’
Write the 2 in the answer in the units and write rem 3 next to it.
Division: 3 digits by 8
## Year 5 division problems: calculating mentally
Here is the second page which looks at Year 5 division problems that can be answered by using mental skills. The questions cover a range of concepts involving division, including:
remainders
divisibility rule for 9
and factors.
Watch out for errors that occur in questions such as number 9, where the number of whole lengths of wire is calculated and the remainder is irrelevant.
This page, and other similar worksheets can be found in the Year 5 calculating category.
Division problems (2)
## Resource of the week: division as sharing
Our resource of the week this week looks at early understanding of division. Division is usually the hardest of the four rules for children to learn, but in the early stages it is quite straightforward. Children often come across division for the first time when sharing, usually between two. The key concept,of course, is that the sharing is done equally. So, for example, 6 sweets shared between 2 children implies that the sharing is equal and they both receive the same number.
By far the best way to practice sharing is to use practical apparatus or use real life situations: e.g. share the strawberries equally between two, share or deal the cards equally etc. Usually this is done on a ‘one for you and one for me’ type process until there are none left. This maths worksheet replicates a practical situation, with the ultimate aim that children begin to remember the answers, which, of course are the inverse of multiplication.
Be careful when sharing that it is not always done into two, as some children begin to think that to share something it can always be divided by 2 and no other number.
Division as sharing (pg 1)
## Dividing by 3 table
In recent years there has been a call for children to not only learn the times tables but also the equivalent division tables. Of course this is the inverse of the 3x table and for children who know this there should not be much of a problem learning this. There are advantages to be had as it can lead to quicker mental arithmetic when carrying out written long and short division. So, here have have a quick look at learning facts connected to dividing by three and an exercise putting this knowledge into practice.
Divide by 3 table
## Divide 3-digit numbers by 6
This is a short division page which concentrates just on dividing 3-digit numbers by 6. Short division, as its name suggests is a shortened version of the long division method where subtractions and remainders are done mentally.
The key to answering these is to have a set speech for the process.
For example: 6)457
‘How many sixes in 45?’
‘7 times 6 is 42.’
‘6 goes into 45, 7 times with 3 left over.’
Write the 7 in the answer above the 5.
Write the remainder three beside the 7 units, making 37.
‘How many sixes in 37?’
‘6 times 6 is 36.’
‘6 goes into 37, 6 times with 1 left over.’
Write the 6 in the answer in the units and write rem 1 next to it.
Division: 3-digit numbers by 6
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## How Do You Do Addition and Subtraction?
There are many ways we can do addition and subtraction such as counting on or backwards, using a number line or hundred square or using base ten blocks. But can you subtract by adding? Yes, you can! Let’s learn all about the think addition to subtract strategy.
## How To Subtract By Adding
You may know that there are many strategies to help solve addition equations, but did you know that there are many to solve subtraction equations as well? One of those is the think addition strategy, also known as subtract using addition!
You may be wondering, how does addition help subtraction? Addition and subtraction are related, meaning that we can use what we know about one to help us with the other! When using addition to solve subtraction equations;
• First: create the subtraction equation
• Then: think of an addition fact to help you subtract. You can do this by looking at the subtraction equation!
• Next: take the smaller number from the subtraction equation to create the new addition equation
• Finally: count up from the smaller number to find the answer!
## Think Addition to Subtract - Practice
What is the easiest way to add and subtract numbers? Take a look at the example below and use the think addition or subtract using addition strategy to solve!
Remember to create a new addition equation, and to count up to find the answer. We can create thirty plus what equals forty-six as our new addition equation.
Thirty plus sixteen equals forty-six, which means that forty-six minus thirty also equals sixteen!
## Think Addition to Subtract - Summary
• First: create the subtraction equation
• Then: Use the smaller number to create the new addition equation
• Next: Count up to solve
Looking for extra practice? After this video check out the interactive exercises and a think addition to subtract worksheet.
"Come on, Imani! We've got to practice for the race." Imani is helping Mr. Squeaks train for his upcoming race. He starts at the twenty-three marker and runs to the ten marker, but how far did he actually run? Mr. Squeaks and Imani can "subtract by adding" to find this out! There are many different strategies and tools to help with addition, and there are many for subtraction as well! But, did you know that one of the strategies to help with subtraction is to actually use addition? Some people use this strategy because they prefer adding to subtracting. When using addition to subtract, start with the subtraction equation. What is the subtraction equation here? It's twenty-three minus ten. Then, remake it as an addition equation by starting with the smaller number. Our addition equation is ten plus something equals twenty-three. Next, count up from the first number to the answer, twenty-three. You can use many tools or strategies to count up, let's use a hundred square here.
Ten plus what equals twenty-three?
Finally, the amount you count up is the answer to your subtraction equation. Ten plus thirteen equals twenty-three! That means twenty-three minus ten ALSO equals thirteen. Mr. Squeaks has been practicing A LOT! If he just sprinted from the thirty-eight marker to the twenty marker, how many markers did he run? We can subtract by adding to find out! First, set up the subtraction equation. Here it is thirty-eight minus twenty. Then, remake it as an addition equation by starting with the smaller number. That means our equation becomes twenty plus something equals thirty-eight. Next, count up from the first number to the answer, thirty-eight. Remember, there are many ways to solve addition equations, let's try place value this time. Twenty plus what is thirty-eight?
"Woah! Now that's what I'm talking about, thanks Imani!"
Would you like to apply the knowledge you’ve learnt? You can review and practice it with the tasks for the video Subtract by Adding.
• ### Solve 20 - 12 by adding.
Hints
The first step is to write out the original subtraction equation.
Solution
Step one: Write out the original subtraction: 20 - 12.
Step two: Rewrite the equation as an addition: 12 + ___ = 20.
Step three: Solve the addition equation by counting up from 12 to 20.
Step four: Fill in answer as: 12 + 8 = 20. Then switch this around to 20 - 12 = 8.
Hints
In order to solve with addition, rephrase the subtraction equation as an addition problem, starting with the smaller number.
Re-arranging a subtraction equation as an addition problem can make it easier to count up to the solution and solve the equation.
Solution
1) 10 - 2 = ?
2) 2 + ? = 10
3) 2 + 8 = 10
A good technique to solve with addition can be to count up from 2 to 10. The difference is your answer!
• ### Word problems to equations.
Hints
When making a subtraction equation, write the larger number first and subtract the smaller number next.
If Mr. Squeaks begins at the 6 marker and runs until the 12 marker this problem would be written as 12 - 6 = ? to find how far he ran.
Solution
• Mr. Squeaks started at the 23 marker and ran to the 10 marker becomes 23 - 10 = ___
• Mr. Squeaks started at the 30 marker and ran to the 18 marker becomes 30 - 18 = ___
• Mr. Squeaks started at the 5 marker and ran to the 15 marker becomes 15 - 5 = ___
• Mr. Squeaks started at the 10 marker and ran to the 50 marker becomes 50 - 10 = ___
Hints
When remaking an equation, the larger number in the subtraction equation will be the solution in the addition equation.
In order to solve with addition, rephrase the subtraction equation as an addition problem.
Solution
When remaking an equation, the larger number in the subtraction equation will be the solution in the addition equation.
• 10 - 4 can be re-written as 4 + ___ = 10. 10 - 4 = 6 and 4 + 6 = 10.
• 18 - 6 can be re-written as 6 + ___ = 18. 18 - 6 = 12 and 6 + 12 = 18.
• 10 - 6 can be re-written as 6 + ___ = 10. 10 - 6 = 4 and 6 + 4 = 10.
• 5 - 5 can be re-written as 5 + ___ = 5. 5 - 5 = 0 and 5 + 0 =5.
• 20 - 5 can be re-written as 5 + ___ = 20. 20 - 5 = 15 and 5 + 15 = 20.
• ### Tools for adding when subtracting.
Hints
A hundreds chart can be a good way to see the numbers while you are counting.
A number line allows you to count and add quickly with tens before you start adding ones.
A place value chart will let you easily see how many tens and ones you still need to add to the total.
There are three correct choices.
Solution
• A number line allows you to count and add quickly with tens before you start adding ones.
• A hundreds chart can be a good way to see the numbers while you are counting.
• A place value chart will let you easily see how many tens and ones you still need to add to the total.
• ### Solving subtraction equations by adding.
Hints
In order to solve with addition, rephrase the subtraction equation as an addition problem.
Re-arranging a subtraction equation as an addition problem can make it easier to count up to the solution and solve the equation. Remember to rewrite again as a subtraction equation as the last step.
Solution
When remaking an equation, the larger number in the subtraction equation will be the solution in the addition equation. Then fill in the rest of the numbers step by step.
Problem 1
1) 10 - 6 = ?
2) 6 + ___ = 10
3) 6 + 4 = 10
4) 10 - 6 = 4
Problem 2
1) 16 - 14 = ?
2) 14 + ___ = 16
3) 14 + 2 = 16
4) 16 - 14 = 2
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# Acceleration main formula
Like displacement $(s)$ and velocity $(v)$, the acceleration $(a)$ is a vector quantity.
Often we use the symbols $\Tred{u}$ for initial velocity, $\Tblue{v}$ for final velocity and $\Delta v$ for change in velocity.
If the velocity and acceleration are in the same direction, the magnitude of acceleration $\Torange{a}$ can be calculated using:
\begin{align*}\Torange{\text{acceleration}}&=\frac{\Tblue{\text{final velocity}}-\Tred{\text{initial velocity}}}{\text{time}} \\ \Torange{a} & = \frac{\Tblue{v}-\Tred{u}}{t} = \frac{\Delta v}{t}\end{align*}
The unit of acceleration is metres per second squared $(\text{m/s}^2)$. This is equal to the unit of velocity $(\text{m/s})$ divided by the unit of time $(\text{s})$: $$\frac{\text{m/s}}{\text{s}}=\frac{\text{m}}{\text{s}^2} = \text{m/s}^{2} = \text{m s}^{-2}$$
The average velocity during constant acceleration $a$ along a straight line is:\begin{align*}\text{average velocity}=&\frac{\Tred{\text{initial velocity}} + \Tblue{\text{final velocity}}}{2} \\ v_{avg}=& \frac{\Tred{u} +\Tblue{v}}{2} \end{align*}
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# CAT 2021 Set-2 | Quantitative Aptitude | Question: 6
1 vote
47 views
For a real number $x$ the condition $|3x – 20| + |3x – 40| = 20$ necessarily holds if
1. $9 < x < 14$
2. $6 < x < 11$
3. $7 < x < 12$
4. $10 < x < 15$
retagged
1 vote
Given that, $|3x-20| + |3x-40| = 20 ; x \in \mathbb{R} \quad \longrightarrow (1)$
We know that $,|x| = \left\{\begin{matrix} x\;;&x\geq 0 \\ -x\;; &x<0 \end{matrix}\right.$
We can open mod as positive and negative. There are four such cases.
$\textbf{Case 1:}\;\text{ Positive, Positive}$
$\Rightarrow 3x – 20 + 3x – 40 = 20$
$\Rightarrow 6x – 60 = 20$
$\Rightarrow 6x = 80$
$\Rightarrow \boxed{x = \frac{40}{3} = 13.33}$
$\textbf{Case 2:}\;\text{ Positive, Negative}$
$\Rightarrow 3x – 20 – (3x – 40) = 20$
$\Rightarrow 3x – 20 – 3x + 40 = 20$
$\Rightarrow \boxed{20 = 20\; {\color{Green} {\text{(True)}}}}$
$\textbf{Case 3:}\;\text{ Negative, Positive}$
$\Rightarrow \;– (3x – 20) + 3x – 40 = 20$
$\Rightarrow\; – 3x + 20 + 3x – 40 = 20$
$\Rightarrow \boxed{- 20 = 20\;\color{Red}{\text{(False)}}}$
$\textbf{Case 4:}\;\text{ Negative, Negative}$
$\Rightarrow \;– (3x – 20) – (3x – 40) = 20$
$\Rightarrow \;– 3x + 20 – 3x + 40 = 20$
$\Rightarrow \;– 6x =\; – 40$
$\Rightarrow \boxed{x = \frac{20}{3} = 6.66}$
$\therefore$ $\boxed{7 < x < 12}$
Correct Answer $: \text{C}$
10.1k points 4 8 30
edited
## Related questions
1
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The number of distinct pairs of integers $(m,n)$ satisfying $|1 + mn| < |m + n| < 5$ is
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If $3x + 2|y| + y = 7$ and $x + |x| + 3y = 1,$ then $x + 2y$ is $\frac{8}{3}$ $1$ $– \frac{4}{3}$ $0$
1 vote
Consider the pair of equations: $x^{2} – xy – x = 22$ and $y^{2} – xy + y = 34.$ If $x>y,$ then $x – y$ equals $7$ $8$ $6$ $4$
Anil, Bobby and Chintu jointly invest in a business and agree to share the overall profit in proportion to their investments. Anil's share of investment is $70 \%.$ His share of profit decreases by $₹ \; 420$ if the overall profit goes down from $18 \%$ to $15 \%.$ Chintu's share of ... goes up from $15 \%$ to $17 \%.$ The amount, $\text{in INR},$ invested by Bobby is $2400$ $2200$ $2000$ $1800$
If a rhombus has area $12 \; \text{sq cm}$ and side length $5 \; \text{cm},$ then the length, $\text{in cm},$ of its longer diagonal is $\sqrt{13} + \sqrt{12}$ $\sqrt{37} + \sqrt{13}$ $\frac{\sqrt{37} + \sqrt{13}}{2}$ $\frac{\sqrt{13} + \sqrt{12}}{2}$
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# The median BM of triangle ABC is the diameter of the circle intersecting side BC in its middle. The length of the side AC is 4
The median BM of triangle ABC is the diameter of the circle intersecting side BC in its middle. The length of the side AC is 4. Find the radius of the circumscribed circle of triangle ABC
Draw a segment MP, as shown in the figure. BM is the diameter of the small circle (according to the condition of the problem), therefore the triangle BMP is right-angled with the hypotenuse BM (by the property of the circumscribed circle).
Consider the triangles BMP and CPM:
MP – common side
BP = PC (by condition of the problem)
∠BPM = ∠CPM, because ∠BPM is direct and ∠CPM is adjacent to it.
Therefore, the triangles BMP and CPM are equal (by the first sign). It follows that BM = MC = MA.
Consider the triangle BMC. Because MB = MC, then this triangle is isosceles, therefore ∠MCP = ∠PBM (by the property of isosceles triangles).
The triangle ABM has a similar situation, ∠BAM = ∠ABM. Those. it turns out that ∠BAM + ∠MCP = ∠ABC. From the theorem on the sum of the angles of a triangle it follows that 180 ° = ∠BAM + ∠MCP + ∠ABC
180 ° = ∠ABC + ∠ABC
180 ° = 2 * ∠ABC
90 ° = ∠ABC
It follows that the triangle ABC is rectangular. By the property of the circumscribed circle, it follows that the point M is the center of the circle => R = AC / 2 = 4/2 = 2.
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# How to find scale factor
This calculator help us find the scale factor between two lengths, simply enter two lengths, it will automatically calculate the scale factor, supports different length units (mm, cm, m, km, in, ft, yd, mi), in addition corresponding visual graphic and formula, easy understanding the calculation process and the result.
### How to use the scale factor calcutor
1. Enter the length of A and B
2. Number accept decimal or fraction, eg. 6, 12, 4.7, 1/2, 5 3/8
3. If the length units are different, choose the right unit
4. The result(scale factor) will be calculated automatically.
### How to figure out scale factor ?
In two similar geometric figures, the scale factor is the ratio of their corresponding sides, dividing the two corresponding lengths of the sides will gives the ratio, for example
#### What is the scale factor between 4 cm and 10 cm ?
4 and 10 are divisible by 2
Length A : 4 ÷ 2 = 2
Length B : 10 ÷ 2 = 5
so the scale factor from A to B is 2:5
#### If 12 inches equals 3 inches, what is the scale factor ?
12 and 3 are divisible by 3
12 ÷ 3 = 4
3 ÷ 3 = 1
12:3 ratio simplified is 4:1
so the scale factor of 12 inches to 3 inches is 4:1
#### If 1/4 inches equals 2 feet, what is the scale factor ?
1⁄4 in = 1 ÷ 4 = 0.25 in
2 ft = 12 × 2 = 24 in
1 ÷ 0.25 = 4
24 × 4 = 96
0.25:24 ratio simplified is 1:96
so the scale factor of 1⁄4 inches to 2 feet is 1:96
If you want to know what is the length under conversion at different scales, try this scale length conversion tool, it helps us calculate the length quickly.
When enlarging a shape or image, we use a scale factor to tell us how many times bigger we want each line/side to become. For example, if we enlarged a rectangle by scale factor 2, each side would become twice as long. If we enlarged by a scale factor of 10, each side would become 10 times as long.
The same idea works with fractional scale factors. A scale factor of 1/2 would make every side 1/2 as big (this is still called an enlargement, even though we have ended up with a smaller shape).
## Enlarging with a Scale Factor of 5
In the diagram above, the left-hand triangle has been enlarged by a scale factor of 5 to produce the triangle on the right. As you can see, each of the three side lengths of the original triangle have been multiplied by 5 to produce the side lengths of the new triangle.
## Scale Factors with Area
But what effect does enlarging by a scale factor have on the area of a shape? Is the area also multiplied by the scale factor?
Let's look at an example.
## Enlarging an area by a scale factor
In the diagram above, we have started with a rectangle of 3cm by 5cm and then enlarged this by a scale factor of 2 to get a new rectangle of 6cm by 10cm (each side has been multiplied by 2).
Look at what has happened to the areas:
Original area = 3 x 5 = 15cm 2
New area = 6 x 10 = 60cm 2
The new area is 4 times the size of the old area. By looking at the numbers we can see why this has happened.
The length and the height of the rectangle have both been multiplied by 2, hence when we find the area of the new rectangle we now have two lots of x2 in there, hence the area has been multiplied by 2 twice, the equivalent of multiplying by 4.
## 1683: The Siege of Vienna
More formally, we can think of it like this:
After an enlargement of scale factor n:
New area = n x original length x n x original height
= n x n x original length x original height
= n 2 x original area.
So to find the new area of an enlarged shape, you multiply the old area by the square of the scale factor.
This is true for all 2-d shapes, not just rectangles. The reasoning is the same; area is always two dimensions multiplied together. These dimensions are both being multiplied by the same scale factor, hence the area is multiplied by the scale factor squared.
## Enlarging a Volume by a Scale Factor
What about if we enlarge a volume by a scale factor?
Look at the diagram above. We have enlarged the left hand cuboid by a scale factor of 3 to produce the cuboid on the right. You can see that each side has been multiplied by 3.
The volume of a cuboid is height x width x length, so:
Original volume = 2 x 3 x 6 = 36cm 3
New volume = 9 x 6 x 18 = 972cm 3
By using division we can quickly see that the new volume is actually 27 times larger than the original volume. But why is this?
When enlarging the area we needed to take into account how two multiplied sides were both being multiplied by the scale factor, hence we ended up using the square of the scale factor to find the new area.
For volume it is a very similar idea, however this time we have three dimensions to take into consideration. Again, each of these is being multiplied by the scale factor, so we need to multiply our original volume by the scale factor cubed.
More formally, we can think of it like this:
After an enlargement of scale factor n:
New volume = n x original length x n x original height x n x original width
I’m trying to scale a point from a map (Latitude, Longitude) to an image (x, y). For that i need to find the scale factor between the 2 NON-SIMILAR rectangles (i think).
I’ll clarify, let’s say:
Rectangle 1: A(40.0, 50.0) B(40.0, 56.0) C(43.0, 56.0) D(43.0, 50.0)
(Latitude Delta = 3, Longitude Delta = 6).
Rectangle 2: E(0, 0) F(500, 0) G(500, 300) H(0, 300)
(X Delta = 500, Y Delta = 300).
How can i scale a point P(41.5, 52.5) from rectangle 1 to point (x, y) on rectangle 2?
UPDATE:
I’m trying to display the user current location (Lat, Lon) on a custom image (not a map image, a drawing of my own) therefor i can’t use maps (MKMapKit, Google, Tom-Tom).
I have the user current location (via CoreLocation) and an image (800×460).
The area that i’m mapping is small so i don’t need to worry about the earth’s curve.
I’m trying to find a formula that’ll help me scale my user (Lat, Lon) location into my image (on my iPhone screen)
Well , getting the rectangle that is similar with the first one but fits in the second one is not hard. Here’s how:
I’m not sure this is what you are trying to achieve.
If you’re happy with the distortions it introduces you can just treat x ( latitude ) and y ( longitude ) coordinates separately.
If you think you’re not going to be happy with such distortions, what do you propose to do about the incommensurate scales in latitude and longitude ? 1°lat != 1°long in the latitudes in your question.
If you need help scaling from a line of one length to a line of another, update your question.
EDIT following comment by OP.
OK, so you want to map an interval of 3° in latitude to an interval of 300 pixels (I guess) in Y, and an interval of 6° in longitude to an interval of 500 pixels in X.
I’m still not entirely sure if you are concerned about introducing further distortions; if you’re not it’s very easy. I’ll suppose that the bottom left of your screen area is pixel (1,1) and that that is where you want to map (40°,56°) to. For every minute of longitude east of that point, move (500/360) pixels to the right. For every minute of latitude north of the corner, move (300/180) pixels up.
This rough-and-ready projection will not preserve the appearance of 2D shapes (for example a square will project to a rectangle) but there are many map projections in widespread use which do not preserve shapes.
Similar triangles are objects that have the same shape and angle size, but their side lengths are different. The corresponding sides of the triangles, however, are in the same length ratio, also called the scale factor. Multiplying the smaller triangle’s side lengths by the scale factor will give you the side lengths of the larger triangle. Similarly, dividing the larger triangle’s side lengths by the scale factor will give you the side lengths of the smaller triangle.
Set up ratios of the corresponding sides of the triangles. For example, the ratio of small to big triangle sides in two triangles is 5/10, 10/20 and 20/40.
Divide both numbers in one of the ratios by their highest common factor. This will give you the scale factor of the bigger triangle to the smaller triangle. In the example, 5 is the highest common factor in the 5/10 ratio. Dividing 5 and 10 by 5 gives you a ratio of 1/2.
Multiply the other sides in the larger triangle by the the ratio calculated in Step 2. In the example, when you multiply 20 by 1/2 and 40 by 1/2, you get 10 and 20, respectively. This confirms that the scale factor of the bigger triangle to the smaller triangle is 1/2.
Divide one of the sides in the bigger triangle by its corresponding side in the smaller triangle to determine the scale factor for the smaller triangle to the bigger triangle. In the example, if you divided 40 by 20 you would get a scale factor of 2.
Multiply the other sides in the smaller triangle by the the scale factor calculated in Step 4. In the example, when you multiply 5 by 2 and 10 by 2, you get 10 and 20, respectively. This confirms that the scale factor of the smaller triangle to the bigger triangle is 2.
When we are trying to find a scale factor, we will have two objects that are basically the same, but one is bigger than the other.
(note: these squares are not to scale)
The scale factor is simply what we could multiply one of the objects by to get the other one. For example, if we had the area of the square on the right, we could multiply the scale factor to find the area of of square on the left.
We need to find an corrosponding legnth from our squares, and make it a ratio.
So we can look at our example here, and we can see that the first square last a legnth of 1, whereas the second square has a legnth of 5. Let’s make this a ratio!
This is basically saying that the first square is 1/5 the size of the second square. And it is our scale factor.
The scale factor could also be 5/1, or just 5, and this is just saying that the second square is 5 times the size of the first square.
How we might use this scale factor is to find the height of the second square in the example picture. The height is unknown (x), but we can figure this out.
We know that the first square is 1/5 the size of the second square, or in other words, the second square is 5 times bigger than the first square. Knowing this we could multiply the hieght of the first square(4) by 5 to find the height of the second square, so the height of the second square is 20!
Also, scale factors doesn’t have to just apply to squares, it can apply to triangles, circles, etc!
When we are trying to find a scale factor, we will have two objects that are basically the same, but one is bigger than the other.
(note: these squares are not to scale)
The scale factor is simply what we could multiply one of the objects by to get the other one. For example, if we had the area of the square on the right, we could multiply the scale factor to find the area of of square on the left.
We need to find an corrosponding legnth from our squares, and make it a ratio.
So we can look at our example here, and we can see that the first square last a legnth of 1, whereas the second square has a legnth of 5. Let’s make this a ratio!
This is basically saying that the first square is 1/5 the size of the second square. And it is our scale factor.
The scale factor could also be 5/1, or just 5, and this is just saying that the second square is 5 times the size of the first square.
How we might use this scale factor is to find the height of the second square in the example picture. The height is unknown (x), but we can figure this out.
We know that the first square is 1/5 the size of the second square, or in other words, the second square is 5 times bigger than the first square. Knowing this we could multiply the hieght of the first square(4) by 5 to find the height of the second square, so the height of the second square is 20!
Also, scale factors doesn’t have to just apply to squares, it can apply to triangles, circles, etc!
A scale factor is a number by which a quantity is multiplied, changing the magnitude of the quantity. Scale factors are often used in geometric contexts, as part of figure models, and more.
The larger penguin model above is 3 times larger than the smaller penguin; To change the larger penguin into the smaller one, we would use a scale factor of . The pentagon shown in green is enlarged by a scale factor of 2 to produce the pentagon shown in blue.
## Scaling geometric figures
The scale factor tells us what to multiply each side length of a geometric figure by to produce a scaled, similar figure.
Triangle ABC is similar to triangle DEF (△ABC
△DEF), which means that the corresponding side lengths of the triangles are proportional:
Any of the three ratios can be used to determine the scale factor.
Find the lengths of sides b and d for the triangles below given that △ABC
Since the triangles are similar, . We can find the scale factor using the ratio of a pair of corresponding sides: . This is the scale factor we multiply a side length of triangle ABC by to find its corresponding side length in DEF. We would multiply a side length of DEF by instead to find its corresponding side length in ABC.
We could also set the ratios of the corresponding sides equal to find b and d.
Recommended practice includes specification of scale factors for each plant input and output variable, which is especially important when certain variables have much larger or smaller magnitudes than others.
The scale factor should equal (or approximate) the span of the variable. Span is the difference between its maximum and minimum value in engineering units, that is, the unit of measure specified in the plant model. Internally, MPC divides each plant input and output signal by its scale factor to generate dimensionless signals.
The potential benefits of scaling are as follows:
Default MPC tuning weights work best when all signals are of order unity. Appropriate scale factors make the default weights a good starting point for controller tuning and refinement.
When choosing cost function weights, you can focus on the relative priority of each term rather than a combination of priority and signal scale.
Improved numerical conditioning. When values are scaled, round-off errors have less impact on calculations.
Once you have tuned the controller, changing a scale factor is likely to affect performance and the controller may need retuning. Best practice is to establish scale factors at the beginning of controller design and hold them constant thereafter.
You can define scale factors at the command line and using the MPC Designer app.
### Determine Scale Factors
To identify scale factors, estimate the span of each plant input and output variable in engineering units.
If the signal has known bounds, use the difference between the upper and lower limit.
If you do not know the signal bounds, consider running open-loop plant model simulations. You can vary the inputs over their likely ranges, and record output signal spans.
If you have no idea, use the default scale factor (=1).
### Specify Scale Factors at Command Line
After you create the MPC controller object using the mpc command, set the scale factor property for each plant input and output variable.
For example, the following commands create a random plant, specify the signal types, and define a scale factor for each signal.
### Specify Scale Factors Using MPC Designer
After opening MPC Designer and defining the initial MPC structure, on the MPC Designer tab, click I/O Attributes .
In the Input and Output Channel Specifications dialog box, specify a Scale Factor for each input and output signal.
Scale a measurement to a larger or smaller measurement, which is useful for architecture, modeling, and other projects. You can also add the real size and scaled size to find the scale factor.
• Find Scale Size
• Find the Scale Factor
## Scaled Results:
#### Scaled Size
• Scale Conversion Calculator
• How to Scale Up or Down
• How to Find the Scale Factor
• How to Reduce the Scale Factor
• Commonly Used Architectural Scales
• Commonly Used Model Scales
## How to Scale Up or Down
Making a measurement smaller or larger, known as scale conversion, requires a common scale factor, which you can use to multiply or divide all measurements by.
To scale a measurement down to a smaller measurement, for instance, when making a blueprint, simply divide the real measurement by the scale factor. The scale factor is commonly expressed as 1:n or 1/n, where n is the factor.
For example, if the scale factor is 1:8 and the real measurement is 32, divide 32 ÷ 8 = 4 to convert.
To convert a scaled measurement up to the actual measurement, simply multiply the smaller measurement by the scale factor. For example, if the scale factor is 1:8 and the smaller length is 4, multiply 4 × 8 = 32 to convert it to the larger actual size.
## How to Find the Scale Factor
A scale factor is a ratio of two corresponding measurements or lengths. You can use the factor to increase or decrease the dimensions of a geometric shape, drawing, or model to different sizes. You can find the scale factor in a few easy steps.
### Step One: Use the Scale Factor Formula
Since the scale factor is a ratio, the first step to finding it is to use the following formula:
scale factor = scaled size / real size
So, the scale factor is a ratio of the scaled size to the real size.
### Step Two: Simplify the Fraction
The next step is to reduce or simplify the fraction.
If you’re scaling down, then the ratio should be shown with a numerator of 1. If you’re scaling up, then the ratio should be shown with a denominator of 1.
To find the final scale factor when you’re scaling up, reduce the ratio to a fraction with a denominator 1. To do this, divide both the numerator and the denominator by the denominator.
If you’re scaling down, then reduce the fraction so that the numerator is 1. You can do this by dividing both the numerator and the denominator by the numerator.
Our fraction simplifier can help with this step too, if needed.
### Step Three: Rewrite the Fraction as a Ratio
Finally, rewrite the fraction as a ratio by replacing the fraction bar with a colon. For instance, a scale factor of 1/10 can be rewritten as 1:10.
For example, let’s find the scale factor used on an architectural drawing where ½” on the drawing represents 12″ on the final building.
Replace the values in the formula above.
Since the drawing is scaled down, then the scale factor should be reduced to a fraction with a denominator of 1.
Multiply both the numerator and denominator by 2 to simplify.
And finally, rewrite the fraction as a ratio.
scale factor = 1 / 24 = 1:24
Thus the scale factor for this drawing is 1:24.
## How to Reduce the Scale Factor
If you already know the scale factor, but it is not in the form of 1:n or 1/n, then some additional work is needed to reduce or simplify it. If the ratio is 2:3, for example, then you’ll need to reduce it to so that the numerator is 1.
Use our ratio calculator to reduce a ratio. You can also reduce a ratio by dividing both the numerator and the denominator by the numerator.
For example: reduce 2/3 by dividing both numbers by 2, which would be 1/1.5 or 1:1.5.
1. Click View tab Viewports panel Scale Monitor. Find. The Scale Monitor dialog box is displayed.
2. In the drawing area, move the cursor over the scale area (or a viewport ) and check the Scale Monitor dialog box.
3. Press ENTER to exit this command.
## What is the formula for scale factor?
The basic formula to find the scale factor of a figure is: Scale factor = Dimensions of the new shape ÷ Dimensions of the original shape. This can also be used to calculate the dimensions of the new figure or the original figure by simply substituting the values in the same formula.
## How do you scale proportionally in AutoCAD?
With a calculator, divide the intended length by the measured length. Enter the SCALE (Command). Select a base point, such as 0,0,0. Enter the obtained scale factor to adjust all objects in the drawing model to their correct size.2 mar. 2021
## What is the scale factor for MM to inches?
Conversion TableCurrent base unitsDesired base unitsScale factormillimetersinches.03937centimetersmillimeters10centimetersmeters.01centimetersinches.39378 autres lignes
## How do you calculate scale?
To scale an object to a smaller size, you simply divide each dimension by the required scale factor. For example, if you would like to apply a scale factor of 1:6 and the length of the item is 60 cm, you simply divide 60 / 6 = 10 cm to get the new dimension.
## What is the scale factor for 1 20?
1″ = 20′ Multiply the feet by 12. 20 x 12 = Scale Factor 240.2 fév. 2021
## What is scale factor in AutoCAD?
Scale Factor. Multiplies the dimensions of the selected objects by the specified scale. A scale factor greater than 1 enlarges the objects. A scale factor between 0 and 1 shrinks the objects. You can also drag the cursor to make the object larger or smaller.15 déc. 2015
## How do I reduce scale size in AutoCAD?
This will show you how to change scale in AutoCAD without changing the dimension. How to scale down in AutoCAD – Window select the object(s) in AutoCAD, type SCALE, and then specify a number between 0 and 1. Hit Enter. The size of the object(s) will SCALE DOWN by that factor.
## What is the scale factor of 1 5?
If the scale factor is 1/5, that means the original triangle would have dimensions 5 times larger than the current model. So if the current base is 5 units in size, the original triangle would be 5 times larger.17 nov. 2017
## What does scale factor mean in maths?
A scale factor in math is the ratio between corresponding measurements of an object and a representation of that object.27 sept. 2020
## What is a scale factor in math in 7th grade?
VOCABULARY. ● Scale Factor: The ratio of any two corresponding lengths in two similar. geometric figures. ● Corresponding Angles: Angles in matching locations of two shapes.7 avr. 2020
## How do you scale a survey in AutoCAD?
How do you scale a survey in AutoCAD? Press Ctrl + A on your keyboard to select all elements in the drawing. Type ‘scale’ in to the command bar and press enter. AutoCAD will ask ‘SCALE Specify base point:’, type ‘0,0’ (without the quotes) and press enter.
## How do you match the scale of two drawings in AutoCAD?
1. Draw a line that is at the proper length (Ex: If the dimension shows 25′, draw a line at that length).
2. Type ALIGN into the command line and press Enter. …
3. Select the image to be scaled and press Enter.
## How do you set limits in AutoCAD?
1. At the Command prompt, enter limits.
2. Enter the coordinates for a point at the lower-left corner of the grid limits.
3. Enter the coordinates for a point at the upper-right corner of the grid limits.
|
# Fractions: Multiplying proper fractions
#### Everything You Need in One Place
Homework problems? Exam preparation? Trying to grasp a concept or just brushing up the basics? Our extensive help & practice library have got you covered.
#### Learn and Practice With Ease
Our proven video lessons ease you through problems quickly, and you get tonnes of friendly practice on questions that trip students up on tests and finals.
#### Instant and Unlimited Help
Our personalized learning platform enables you to instantly find the exact walkthrough to your specific type of question. Activate unlimited help now!
0/4
##### Intros
###### Lessons
1. Introduction to Multiplying proper fractions:
2. Simplify fractions: Method A - By using greatest common factors
3. Simplify fractions: Method B - By using common factors
4. How to multiply fractions with cross-cancelling?
5. How to multiply proper fractions?
0/5
##### Examples
###### Lessons
1. Multiplying Single-digit Proper Fractions
Multiply the proper fractions. Give the answers in the lowest form:
1. $\frac{1}{4} \times \frac{3}{4}$
2. $\frac{2}{7} \times \frac{5}{8}$
3. $\frac{5}{7} \times \frac{7}{{10}}$
2. Word Problems: Multiplying Proper Fractions
In April last year, $\frac{2}{5}$ of the month has records of rainfall. $\frac{1}{4}$ of these rainy days fell on weekends.
1. What fraction of rainfall happened on weekends in April last year?
2. How many of these rainy days were on weekends?
0%
##### Practice
###### Topic Notes
Unlike the previous sections, this section will only deal with multiplications with fractions only, specifically proper fractions. To do so, we will need to multiply the numerators together and then the denominators together. We also will learn the trick of simplifying fractions in order to reduce the answers to the lowest form.
In this lesson, we will learn:
• Multiplying Single-digit Proper Fractions
• Word Problems: Multiplying Proper Fractions
• Multiplying Proper Fractions Involving Multiple-digit Numbers and Negative Numbers
• Proper fractions:
• Fractions in which the values of the numerators are less than that of the denominators.
• Fractions that are greater than 0 but less than 1.
• The product of two proper fractions must also be a proper fraction.
|
# Square of a Trinomial Expansion | Perfect Square Trinomial Definition, Formula & Solved Examples
Do you want to expand trinomials easily without any confusion and hassle? You should refer to this page. Here, we have explained how to expand the Square of a Trinomial and perfect square trinomial definition and formulas. Students who need more subject knowledge about square trinomials and solve any kind of trinomial expansions must go with this article completely. In this article, you will also get some worked-out examples on Square of a Trinomial and Perfect square trinomial. So, let’s continue your read and learn the concept of square trinomial.
## Perfect Square Trinomial Definition & Formula
An expression obtained from the square of the binomial equation is a perfect square trinomial. When the trinomial is in the form ax² + bx + c then it is said to be a perfect square, if and only if it meets the condition b² = 4ac.
The Perfect Square Trinomial Formula is as follows,
(ax)²+2abx+b² = (ax+b)²
(ax)²−2abx+b² = (ax−b)²
### How to Expand the Square of a Trinomial?
Here, we are discussing the expansion of the square of a trinomial (a + b + c).
Let (b + c) = x
(i) Then (a + b + c)2 = (a + x)2 = a2 + 2ax + x2
= a2 + 2a (b + c) + (b + c)2
= a2 + 2ab + 2ac + (b2 + c2 + 2bc)
= a2 + b2 + c2 + 2ab + 2bc + 2ca
Therefore, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca.
(ii) (a – b – c)2 = [a + (-b) + (-c)]2
= a2 + (-b2) + (-c2) + 2 (a) (-b) + 2 (-b) (-c) + 2 (-c) (a)
= a2 + b2 + c2 – 2ab + 2bc – 2ca
Therefore, (a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca.
(iii) (a + b – c)2 = [a + b + (-c)]2
= a2 + b2 + (-c)2 + 2ab + 2 (b) (-c) + 2 (-c) (a)
= a2 + b2 + c2 + 2ab – 2bc – 2ca
Therefore, (a + b – c)2 = a2 + b2 + c2 + 2ab – 2bc – 2ca.
(iv) (a – b + c)2 = [a + (- b) + c]2
= a2 + (-b2) + c2 + 2 (a) (-b) + 2 (-b) (-c) + 2 (c) (a)
= a2 + b2 + c2 – 2ab – 2bc + 2ca
Therefore, (a – b + c)2 = a2 + b2 + c2 – 2ab – 2bc + 2ca.
### Solved Examples on Square Trinomial
1. Expand (x+4y+6z)2
Solution:
Given trinomial expression is (x+4y+6z)2
We know that (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca.
Here a=1x, b=4y, c=6z
Now, substitute the values in the expression of (a + b + c)2
Then (x+4y+6z)2 = (1x)2 + (4y)2 + (6z)2 + 2(1x)(4y) + 2(4y)(6z) + 2(6z)(1x)
= x2 + 16y2 + 36z2 + 8xy + 48yz + 12zx
Hence, (x+4y+6z)2 = x2 + 16y2 + 36z2 + 8xy + 48yz + 12zx.
2. Is the trinomial x– 6x + 9 a perfect square?
Solution:
Given trinomial is x2 – 6x + 9, now calculate the expression and find it is a perfect square or not.
x2 – 6x + 9 = x2 – 3x – 3x + 9
= x(x – 3) – 3(x – 3)
= (x – 3)(x – 3)
Otherwise,
x2 – 6x + 9 = x2 – 2(3)(x) + 32 = (x – 3)2
The factors of the given equation are a perfect square.
Therefore, the given trinomial is a perfect square.
3. Simplify a + b + c = 16 and ab + bc + ca = 40. Find the value of a2 + b2 + c2.
Solution:
As per the given question, a + b + c = 16
Now, by squaring both sides, we get
(a+ b + c)2 = (16)2
a2 + b2 + c2 + 2ab + 2bc + 2ca = 256
a2 + b2 + c2 + 2(ab + bc + ca) = 256
a2 + b2 + c2 + 2 × 40 = 256 [Given, ab + bc + ca = 40]
a2 + b2 + c2 + 80 = 256
At this step, we have to subtract 80 from both sides
a2 + b2 + c2 + 80 – 80 = 256 – 80
a2 + b2 + c2 = 176
Hence, the square of a trinomial formula will help us to expand and get the result for a2 + b2 + c2 is 176.
|
# All You Need to Know About Hailstone Sequence in Python
Lets us first understand what a sequence is in the hailstone sequence python. A sequence is an ordered series of numbers that follows a particular pattern. A sequence can be something as simple as a series of odd numbers. A hailstone sequence is a series of numbers that increase and decrease while following a pattern and eventually end when a repeating pattern is found.
Contents
## Introduction to Hailstone Sequence
The Hailstone sequence is also known as The Collatz Conjecture. It was a problem proposed by mathematician L. Collatz. It is known as the hailstone sequence because the sequence resembles the pattern of formation of hailstones. Hailstones get blown up by the winds in the cloud, come down when they form ice pellets, and again get blown by the winds. Similarly, the sequence increases and decreases its value alternatively. Thus, according to the sequence, we may start with any whole number but will always end up with the sequence of 4, 2, 1.
We start with a given number, ‘N..’ If the number N is even, we will divide it by 2. Otherwise, if N is odd, multiply N by 3 and add 1 to it (3*N + 1). Thus, irrespective of the number we start with, the sequence will never be infinite and always come to an end.
Let us understand the hailstone sequence by taking an example.
## Example of Hailstone Sequence
Let us taking N = 5. Since N is odd, We will perform (3*N + 1) = (3*5 + 1) = 16
Now, 16 is even, so we will divide it by 2 = 16/2 = 8.
We will follow the same pattern for the entire series.
8 is even, so 8/2 = 4 4 is even, so 4/2 = 2 2 is even, so 2/2 = 1 1 is odd, so (3*1 + 2) = 4. Now, here as you can see, the sequence is repeating itself. It generated 4 again. So the sequence stops here.
Therefore, the sequence for N = 5 is 5, 16, 8, 4, 2, 1
Let us take another example of N = 6
6 is even, so 6/2 = 3 3 is odd, so (3*3 + 1) = 10 10 is even, so 10/2 = 5 5 is odd, so (3*5 + 1) = 16 16 is even, so 16/2 = 8 8 is even, so 8/2 = 4 4 is even, so 4/2 = 2 2 is even, so 2/2 = 1 1 is odd, so (3*1 + 1) = 4.
The hailstone sequence for N = 6 is 6, 3, 10, 5, 16, 8, 4, 2, 1
As seen, for n = 6 too, the sequence is ending with 4, 2, 1 and is finite.
## Computing the Sequence
Let us write a program in python to calculate the hailstone sequence. We will try to implement it both, recursively and non-recursively. The sequence would be generated by taking the initial number of the sequence from the user as an input. The output would be the hailstone sequence in python generated along with the number of steps taken to find the entire sequence.
## Without Recursion(while loop)
First, we will define a function named hailstone().
```def hailstone(n):
hailstone_list = []
hailstone_list.append(int(n))
while n != 1:
if n % 2 == 0:
n = n/2
hailstone_list.append(int(n))
else:
n = 3*n + 1
hailstone_list.append(int(n))
return hailstone_list
```
It will accept one argument n – which is the beginning number of the sequence and will return a list. There will be a while loop that will stop the execution if n becomes 1. We have a list named hailstone_list which will store the sequence. If n is even, n would be divided by 2 else it will be multiplied by 3 and then 1 will be added to it. The numbers would be appended to the list.
```n = int(input("Enter n:"))
hailstone_list = []
hailstone_list = hailstone(n)
print("Hailstone Sequence for n = {} is : {}".format(n, hailstone_list))
print("Number of steps is : ",len(hailstone_list))
```
Here, n would be taken as input from the user. We call hailstone function by passing n as the argument and store the return value inside an empty list hailstone_list. Then the print statements will print the hailstone sequence and the number of steps.
```Enter n:7
Hailstone Sequence for n = 7 is : [7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1]
Number of steps is : 17```
Here also the sequence has terminated with 4, 2, 1.
Also, Read | Understanding Collatz Sequence in Python
## With Recursion
Now, we will be implementing the hailstone sequence in python with the help of recursion. If n is one, we will return the hailstone_list. If n is an even number, we shall call the hailstone() function by recursion and pass n/2 as the argument. Else if n is an even number, we shall call hailstone() function by recursion but pass (3*n + 1) as the argument to the function.
```hailstone_list = []
def hailstone(n):
hailstone_list.append(int(n))
if n == 1:
return hailstone_list
while n != 1:
if n % 2 == 0:
return hailstone(int(n/2))
else:
return hailstone(int(3*n + 1))
n = int(input("Enter n:"))
hailstone_list = []
hailstone_list = hailstone(n)
print("Hailstone Sequence for n = {} is : {}".format(n, hailstone_list))
print("Number of steps is : ",len(hailstone_list))
```
The output for different values of n is:
N = 7:
```Enter n:7
Hailstone Sequence for n = 7 is : [7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1]
Number of steps is : 17```
N = 6:
```Enter n:6
Hailstone Sequence for n = 6 is : [6, 3, 10, 5, 16, 8, 4, 2, 1]
Number of steps is : 9```
Even for large values of n, the sequence shall always end with 1.
```Enter n:19840
Hailstone Sequence for n = 19840 is : [19840, 9920, 4960, 2480, 1240, 620, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1]
Number of steps is : 93```
## FAQ’s
#### Q. What is the longest hailstone sequence?
A. The longest hailstone sequence is for the number 77031 whose length is 351.
```Enter n:77031
Hailstone Sequence for n = 77031 is : [77031, 231094, 115547, 346642, 173321, 519964, 259982, 129991, 389974, 194987, 584962, 292481, 877444, 438722, 219361, 658084, 329042, 164521, 493564, 246782, 123391, 370174, 185087, 555262, 277631, 832894, 416447, 1249342, 624671, 1874014, 937007, 2811022, 1405511, 4216534, 2108267, 6324802, 3162401, 9487204, 4743602, 2371801, 7115404, 3557702, 1778851, 5336554, 2668277, 8004832, 4002416, 2001208, 1000604, 500302, 250151, 750454, 375227, 1125682, 562841, 1688524, 844262, 422131, 1266394, 633197, 1899592, 949796, 474898, 237449, 712348, 356174, 178087, 534262, 267131, 801394, 400697, 1202092, 601046, 300523, 901570, 450785, 1352356, 676178, 338089, 1014268, 507134, 253567, 760702, 380351, 1141054, 570527, 1711582, 855791, 2567374, 1283687, 3851062, 1925531, 5776594, 2888297, 8664892, 4332446, 2166223, 6498670, 3249335, 9748006, 4874003, 14622010, 7311005, 21933016, 10966508, 5483254, 2741627, 8224882, 4112441, 12337324, 6168662, 3084331, 9252994, 4626497, 13879492, 6939746, 3469873, 10409620, 5204810, 2602405, 7807216, 3903608, 1951804, 975902, 487951, 1463854, 731927, 2195782, 1097891, 3293674, 1646837, 4940512, 2470256, 1235128, 617564, 308782, 154391, 463174, 231587, 694762, 347381, 1042144, 521072, 260536, 130268, 65134, 32567, 97702, 48851, 146554, 73277, 219832, 109916, 54958, 27479, 82438, 41219, 123658, 61829, 185488, 92744, 46372, 23186, 11593, 34780, 17390, 8695, 26086, 13043, 39130, 19565, 58696, 29348, 14674, 7337, 22012, 11006, 5503, 16510, 8255, 24766, 12383, 37150, 18575, 55726, 27863, 83590, 41795, 125386, 62693, 188080, 94040, 47020, 23510, 11755, 35266, 17633, 52900, 26450, 13225, 39676, 19838, 9919, 29758, 14879, 44638, 22319, 66958, 33479, 100438, 50219, 150658, 75329, 225988, 112994, 56497, 169492, 84746, 42373, 127120, 63560, 31780, 15890, 7945, 23836, 11918, 5959, 17878, 8939, 26818, 13409, 40228, 20114, 10057, 30172, 15086, 7543, 22630, 11315, 33946, 16973, 50920, 25460, 12730, 6365, 19096, 9548, 4774, 2387, 7162, 3581, 10744, 5372, 2686, 1343, 4030, 2015, 6046, 3023, 9070, 4535, 13606, 6803, 20410, 10205, 30616, 15308, 7654, 3827, 11482, 5741, 17224, 8612, 4306, 2153, 6460, 3230, 1615, 4846, 2423, 7270, 3635, 10906, 5453, 16360, 8180, 4090, 2045, 6136, 3068, 1534, 767, 2302, 1151, 3454, 1727, 5182, 2591, 7774, 3887, 11662, 5831, 17494, 8747, 26242, 13121, 39364, 19682, 9841, 29524, 14762, 7381, 22144, 11072, 5536, 2768, 1384, 692, 346, 173, 520, 260, 130, 65, 196, 98, 49, 148, 74, 37, 112, 56, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1]
Number of steps is : 351```
#### Q. Is the hailstone sequence solved?
A. Hailstone sequence has been checked for two to the sixtieth power numbers (2^60). All the sequences generated up to that number has turned out to be a finite sequence ending with the same pattern – 4, 2, 1.
That was all about hailstone sequence in python. If you have any questions or thoughts to share, let us know in the comments.
Till then, Happy Learning!
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Contact The Learning Centre
# Probability
### Example 3: using the probability rules
A treasure chest contains $$30$$ diamonds, $$10$$ gold coins, $$20$$ sapphires and $$5$$ silver medallions. If an item is selected at random, find the probability that it is:
a) a gold coin?
b) not a sapphire?
c) either a gold coin or a silver medallion?
d) neither a gold coin nor a silver medallion?
Solution:
a) \begin{eqnarray*}
&&P(\mbox{Gold}) \\
&=& \frac{\mbox{Number of Gold coins}}{\mbox{Total number of items in the chest}} \\
&=& \frac{10}{65} \\
&=& \frac{2}{13} \approx 0.1538
\end{eqnarray*}
b) \begin{eqnarray*}
&&P(\mbox{Not a Sapphire}) \\
&=& \frac{65 - 20}{65} \\
&=& \frac{45}{65} \\
&=& \frac{9}{13} \approx 0.6923
\end{eqnarray*}
c) \begin{eqnarray*}
&&P(\mbox{Gold or Silver}) \\
&=& \frac{10}{65} + \frac{5}{65} \\
&=& \frac{15}{65} \\
&=& \frac{3}{13} \approx 0.2308
\end{eqnarray*}
d) \begin{eqnarray*}
&&P(\mbox{Not gold or silver}) \\
&=& 1 - \frac{3}{13} \\
&=& \frac{10}{13} \approx 0.7692
\end{eqnarray*}
|
# Common Core: High School - Functions : Symmetry and Periodicity of Trigonometric Functions: CCSS.Math.Content.HSF.TF.B.4
## Example Questions
← Previous 1
### Example Question #1 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4
Find the following trigonometric exact value.
Explanation:
This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.
For the purpose of Common Core Standards, "Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions." falls within the Cluster A of "Extend the domain of trigonometric functions using the unit circle" concept (CCSS.MATH.CONTENT.HSF-TF.A.4).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Recall the unit circle.
Step 2: Identify the coordinate pair that represents the extension of .
Step 3: Calculate the exact value of .
Recall that,
therefore,
.
### Example Question #2 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4
Find the following trigonometric exact value.
Explanation:
This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.
For the purpose of Common Core Standards, "Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions." falls within the Cluster A of "Extend the domain of trigonometric functions using the unit circle" concept (CCSS.MATH.CONTENT.HSF-TF.A.4).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Recall the unit circle.
Step 2: Identify the coordinate pair that represents the extension of .
Step 3: Calculate the exact value of .
Recall that,
therefore,
.
### Example Question #3 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4
Find the following trigonometric exact value.
Explanation:
This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.
For the purpose of Common Core Standards, "Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions." falls within the Cluster A of "Extend the domain of trigonometric functions using the unit circle" concept (CCSS.MATH.CONTENT.HSF-TF.A.4).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Recall the unit circle.
Step 2: Identify the coordinate pair that represents the extension of .
Step 3: Calculate the exact value of .
Recall that,
therefore,
.
### Example Question #4 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4
Find the following trigonometric exact value.
Explanation:
This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.
For the purpose of Common Core Standards, "Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions." falls within the Cluster A of "Extend the domain of trigonometric functions using the unit circle" concept (CCSS.MATH.CONTENT.HSF-TF.A.4).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Recall the unit circle.
Step 2: Identify the coordinate pair that represents the extension of .
Step 3: Calculate the exact value of .
Recall that,
therefore,
.
### Example Question #5 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4
Find the following trigonometric exact value.
Explanation:
This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.
For the purpose of Common Core Standards, "Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions." falls within the Cluster A of "Extend the domain of trigonometric functions using the unit circle" concept (CCSS.MATH.CONTENT.HSF-TF.A.4).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Recall the unit circle.
Step 2: Identify the coordinate pair that represents the extension of .
Step 3: Calculate the exact value of .
Recall that,
therefore,
.
### Example Question #6 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4
Find the following trigonometric exact value.
Explanation:
This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.
For the purpose of Common Core Standards, "Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions." falls within the Cluster A of "Extend the domain of trigonometric functions using the unit circle" concept (CCSS.MATH.CONTENT.HSF-TF.A.4).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Recall the unit circle.
Step 2: Identify the coordinate pair that represents the extension of .
Step 3: Calculate the exact value of .
Recall that,
therefore,
.
### Example Question #7 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4
Find the following trigonometric exact value.
Explanation:
This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.
For the purpose of Common Core Standards, "Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions." falls within the Cluster A of "Extend the domain of trigonometric functions using the unit circle" concept (CCSS.MATH.CONTENT.HSF-TF.A.4).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Recall the unit circle.
Step 2: Identify the coordinate pair that represents the extension of .
Step 3: Calculate the exact value of .
Recall that,
therefore,
.
### Example Question #8 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4
Find the following trigonometric exact value.
Explanation:
This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.
For the purpose of Common Core Standards, "Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions." falls within the Cluster A of "Extend the domain of trigonometric functions using the unit circle" concept (CCSS.MATH.CONTENT.HSF-TF.A.4).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Recall the unit circle.
Step 2: Identify the coordinate pair that represents the extension of .
Step 3: Calculate the exact value of .
Recall that,
therefore,
.
### Example Question #9 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4
Find the following trigonometric exact value.
Explanation:
This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.
For the purpose of Common Core Standards, "Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions." falls within the Cluster A of "Extend the domain of trigonometric functions using the unit circle" concept (CCSS.MATH.CONTENT.HSF-TF.A.4).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Recall the unit circle.
Step 2: Identify the coordinate pair that represents the extension of .
Step 3: Calculate the exact value of .
Recall that,
therefore,
.
### Example Question #10 : Symmetry And Periodicity Of Trigonometric Functions: Ccss.Math.Content.Hsf.Tf.B.4
Find the following trigonometric exact value.
Explanation:
This question tests one's ability to recognize and use the unit circle to calculate the exact trigonometric value.
For the purpose of Common Core Standards, "Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions." falls within the Cluster A of "Extend the domain of trigonometric functions using the unit circle" concept (CCSS.MATH.CONTENT.HSF-TF.A.4).
Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.
Step 1: Recall the unit circle.
Step 2: Identify the coordinate pair that represents the extension of .
Step 3: Calculate the exact value of .
Recall that,
therefore,
.
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# Law of Cosines by Ancient Sliding
### What Might This Be About?
20 November 2014, Created with GeoGebra
### The Law of Cosines
As is common, we shall denote the sides of $\Delta ABC$ opposite vertices $A,B,C$ as $a,b,c$ and the enclosed angles as $\alpha,\beta,\gamma.$
$c^{2}=a^{2}+b^{2}-2ab\cos\gamma.$
### Proof
This simple and elegant proof mimics the sliding argument that was used, for example, in Proof 69 of the Pythagorean theorem but applies it to a modification of Proof 9. The foundation of the proof below lies on the equality of blue areas (where in the left part the blue figure is a square with side $c$):
This is the same as the assertion in the Proof 55 of the Pythagorean theorem, though much more intuitive and straightforward.
The equality of the blue areas is sufficient to establish the Pythagorean theorem; for the Cosine Law we'll find explicit formulas for the areas of the parallelograms. The main tool here is an identity already used in another proof of the Law of Cosines:
(1)
$a=c\cos\beta+b\cos\gamma.$
The formula is explained in the following diagram:
Note that the formula works also when either $\beta$ or $\gamma$ is obtuse. The next step is to identify the angles in the configuration:
The next diagram now elucidates the expressions for the areas of small blue parallelograms.
In the diagram, the distance from $F$ to $DH$ equals $c\sin(90^{\circ}-\beta)=c\cos\beta,$ which, along with (1), gives the distance as $h=a-b\cos\gamma$ so that the area of the lower parallelogram equals
$a\cdot h=a(a-b\cos\gamma )=a^{2}-ab\cos\gamma.$
Similarly, the area of the upper parallelogram equals $b^{2}-ab\cos\gamma$ which together add up to the Law of Cosines.
### Acknowledgment
The proof has been posted by Mike Todd as a comment on another page devoted to the Law of Cosines.
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# Measuring the Distance between Two Lines in Geometry
In geometry, it is important to be able to accurately measure the distance between two lines. This skill can come in handy when solving complex equations and making calculations. To measure the distance between two lines, you need to know what a line segment is and how to calculate its length using the Pythagorean theorem. Let's take a closer look at how to measure the distance between two lines.
## What is a Line Segment?
A line segment is a section of a line that runs between two points, with each point being one endpoint of the segment. The distance between the endpoints is referred to as the length of the line segment. For example, if we were to draw a straight line from point A (1,2) to point B (4,6), then any section of this particular line would be considered a line segment.
## Calculating Length Using the Pythagorean Theorem
The most common way of calculating the length of a line segment—or any side of an object—is by using the Pythagorean theorem. This theorem states that in any right triangle (one with an angle of 90 degrees), the square of the hypotenuse (the longest side) will always equal the sum of squares of other two sides. In other words, c2 = a2 + b2 where c represents hypotenuse and a & b are other sides. We can use this formula to calculate length for any triangle or shape by finding distances for all three sides first and then proceeding with calculation accordingly.
## Calculating Distance Between Two Lines
Once you’ve determined what type of shape you’re dealing with and calculated its individual side lengths, it’s time to find out how far apart those lines are from each other. To do this, simply subtract one side length from another—in other words, subtract one endpoint’s coordinates from another endpoint’s coordinates—and then use your new result in conjunction with Pythagorean theorem mentioned above in order to get final result for that particular shape or equation!
## Conclusion
Measurement is an integral part of geometry; it allows us to solve complex problems and calculations quickly and accurately. Measuring the distance between two lines requires knowledge about what constitutes a line segment as well as how calculate its length using the Pythagorean theorem. By understanding these concepts, students can confidently measure distances between lines in geometric equations!
## FAQ
### How do you find the distance between 2 lines?
To find the distance between two lines, you must calculate the length of each line using the Pythagorean theorem. Then subtract one endpoint’s coordinates from another endpoint’s coordinates and use this result in conjunction with the Pythagorean theorem to get your final result.
### What is a line segment?
A line segment is a section of a line that runs between two points, with each point being one endpoint of the segment. The distance between the endpoints is referred to as the length of the line segment.
### What is the distance between two lines called?
The distance between two lines is referred to as the length of the line segment. This is the distance between the endpoints of each line.
### How do you find the distance of a line in geometry?
To find the distance of a line in geometry, you must calculate the length of the line using the Pythagorean theorem. This involves finding the distances for each side of a triangle or shape and then using this information to calculate the overall distance of that particular line.
### How do you find the distance between two points in geometry?
To find the distance between two points in geometry, you must subtract one endpoint’s coordinates from another endpoint’s coordinates and then use this result in conjunction with Pythagorean theorem to get your final result. This will give you the straight-line distance between those two points.
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# How do you simplify 4|3-(-3)| + (-1)?
The absolute value acts as a bracket, so analyze what's inside there first, then it's order of operations as normal, to get to $23$.
#### Explanation:
Starting with the original:
$4 | 3 - \left(- 3\right) | + \left(- 1\right)$
The absolute value acts as a set of brackets, so let's focus there first:
$| 3 - \left(- 3\right) |$
So $3 - \left(- 3\right) = 6$
So let's substitute that in:
$4 | 6 | + \left(- 1\right)$
Next, $| 6 | = 6$
So we have:
$4 \cdot 6 + \left(- 1\right) = 24 - 1 = 23$
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# Vertex of Absolute Value Equations
by FARAWAY
(PA)
CALCULATING THE VERTEX OF AN ABSOLUTE VALUE OF THE EQUATION
WRITE DOWN WHAT IS INSIDE THE ABSOLUTE VALUE SYMBOL AND SET EQUAL TO ZERO
SOLVE FOR "X"
SUBSTITUTE THE VALUE OF X INTO THE ORIGINAL EQUATION AND SOLVE FOR "Y"
WRITE THE ORDERED PAIR
Y=(X)-2
y=-10(X=4)-5
Y= -(X-5)
Y= (X)
THE LAST QUESTION COULD BE ANY NUMBER BUT THE SAME NUMBER THAT IS POSITIVE FOR BOTH X AND Y?
OR IT CAN BE ANY NEGATIVE NUMBER BUT THE SAME NUMBER THAT IS NEGATIVE ON BOTH SIDE BUT THE ANSWER FOR BOTH OF THESE WILL BE POSITIVE?
IS THAT CORRECT FOR Y=(X)?
Karin from Algebra Class Says:
Your explanation for finding the vertex of an absolute value equation is exactly correct!
Let's look at Example 1: y = |x| - 2
x = 0 x is inside the absolute value
symbol, so set it equal to 0.
y = |0| - 2 Substitute the value for x back
into the equation and solve for
y.
y = 0-2
y = -2
The vertex for y = |x| - 2 is (0, -2)
Example 2: y= -10|x+4| - 5
Step 1: Set x + 4 = 0 & Solve.
x + 4 = 0
x + 4 - 4 = 0 - 4
x = -4
Step 2: Substitute -4 for x back into the equation and solve for y.
y =-10|x+4| - 5
y =-10|-4+4| - 5
y =-10(0) -5
y = -5
The vertex for y= -10|x+4| - 5 is (-4, -5)
Example 3: y = -|x-5|
Step 1: set x - 5 = 0
x - 5 = 0
x- 5 + 5 = 0 + 5
x = 5
Step 2: Substitute 5 for x and solve for y.
y = -|x-5|
y = -|5-5|
y = -(0)
y = 0
The vertex for y = -|x-5| is (5,0)
It looks like you are on the right track!
Good luck,
Karin
### Comments for Vertex of Absolute Value Equations
Average Rating
Nov 19, 2019 Rating Clear & to the point! by: Kofi Yankey Very well explained and demonstrated. I also agree to point out that a "-" outside the absolute value sign means it doesn't hold water or is upside down. And a positive absolute value would show that it does "hold water" or V-shaped upwards.Excellent!!!
Apr 17, 2018 Rating LEL by: Billy Thx m8. 1 needed th!s b@d1y
Sep 25, 2017 Rating Good job by: Vobo Caleb Wow! It works like magic. Thanks
Mar 05, 2013 Rating Good by: Anonymous Very nicely elucidated on. One suggestion: In the last problem for instance, Y =-⎮X-5⎮ it might be a good idea to state that the graph points downwards, i.e., that it will not hold water. Otherwise, you did a beautiful Job Sincerely, Harry Dunleavy
Dec 04, 2012 Rating thanks soo much by: An 8th grade elgabra student this REALY helped i was sick for 3days of school and my teacher wouldnt catch me up so i had to do it on my own and this helped me ace 2 of my worksheets!!! thanks alot
Feb 07, 2012 Rating :) by: Maria thanks x 1000! :D
Mar 02, 2011 Rating nice by: Anonymous thumbs up nice work job done
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# Calculating Simple Compound Interest Simple Interest Simple interest
• Slides: 8
Calculating Simple & Compound Interest
Simple Interest ¡ Simple interest (represented as I in the equation) is determined by multiplying the interest rate by the principal by the number of periods. (Same amount every year)
¡ ¡ ¡ Principal sum (P)- The initial amount of money invested or borrowed (ie. \$10, 000) Interest rate (r)- The amount charged or given to the principal sum (ie. 5%) Interest period (t)- The number of years you plan to invest or borrow (ie. 5 years) Simple Interest equation: I = Prt
Let’s try it Tim, a grade 9 student at Preston High school received \$1000 from his grandma for his birthday. After learning about savings in his BBI class, he decides to go to his bank and put the money into a savings account at an interest rate of 3% annually until he goes to university in 4 years. Calculate the simple interesting using the simple interest equation: I=Prt
I = Prt P = \$1000 ¡ r= 3% annually ¡ t= 4 years ¡ I = 1000 x 0. 03 x 4 = 120 Therefore Tim will have made \$120 in interest over the 4 years resulting in a total of \$1120.
Compound Interest ¡ Interest calculated on the amount saved or borrowed plus any interest already accumulated
Calculating Compound Interest, using the Tim example from before Principal (P) 1000 1030 1060. 90 Interest (Pxr) = 1000 x. 03 = 30 = 1030 x. 03 = 30. 90 =1060. 90 x. 03 Total = 1030 = 1060. 90 = 1092. 73 = 31. 83 1092. 73 = 1092. 73 x. 03 = 1125. 51 = 32. 78
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# Class 8 NCERT Solutions – Chapter 9 Algebraic Expressions and Identities – Exercise 9.3
• Last Updated : 20 Nov, 2020
### Question 1. Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r
Solution:
(4p) * (q + r) = 4pq + 4pr
(ii) ab, a – b
Solution:
(ab) * (a – b) = a2 b – ab2
(iii) a + b, 7a2b2
Solution:
(a + b) * (7a2b2) = 7a3b2 + 7a2b3
(iv) a2 – 9, 4a
Solution:
(a2 – 9) * (4a) = 4a3 – 36a
(v) pq + qr + rp, 0
Solution:
(pq + qr + rp ) * 0 = 0
Explanation: Anything multiplied to 0 will give zero.
Solution:
### Question 3. Find the product.
(i) (a2) x (2a22) x (4a26
Solution:
(1 x 2 x 4 ) (a2 x a22 x a26 )
= (8) (a50)
= 8a50
Explanation: when two number numbers of same base are multiplied their power gets added up. [ ax x ay = ax+y ]
(ii) (2/3 xy) x (-9/10 x2y2)
Solution:
(2/3 x -9/10) (xy x x2y2)
= (-3/5) (x3y3)
= -3/5 x3y3
Explanation: when two number numbers of same base are multiplied their power gets added up. [ ax x ay = ax+y ]
(iii) (-10/3pq3) * (6/5p3q)
Solution:
(-10/3 x 6/5) (pq3 x p3q)
= (-4) (p4q4)
= -4p4q4
Explanation: when two number numbers of same base are multiplied their power gets added up. [ ax x ay = ax+y]
(iv) x * x2 * x3 * x4
Solution:
(x) (x2) (x3) (x4)
= x10
Explanation: when two number numbers of same base are multiplied their power gets added up. [ ax x ay = ax+y]
### Question 4.
(a) Simplify 3x (4x – 5) + 3 and find its values for
(i) x = 3
Solution:
First we will simplify the given equation and the put the value of x as required.
3x (4x – 5) + 3
⇒ 12x2 – 15x + 3
⇒ 12 (3)2 – 15 (3) + 3 [ putting the value of x = 3 ]
⇒ 12 (9) – 15 (3) + 3
⇒ 108 – 45 + 3
⇒ 66
(ii) x = 1/2
Solution:
3x (4x – 5) + 3
⇒ 12x2 – 15x + 3
⇒ 12 (1/2)2 – 15 (1/2) + 3
⇒ 12 (1/4) – 15 (1/2) +3
⇒ 3 – 15/2 + 3
⇒ -3/2
(b) Simplify a (a2 + a + 1) + 5 and find its value for
(i) a = 0
Solution:
a (a2 + a + 1) + 5
⇒ a3 + a2 + a + 5
⇒ (0)3 + (0)2 + (0) + 5 [ anything to the power 0 gives 0 only ]
⇒ 5
(ii) a = 1
Solution:
a (a2 + a + 1) + 5
⇒ a3 + a2 + a + 5
⇒ (1)3 + (1)2 + (1) + 5 [ 1 to the power any number gives 1
⇒ 8
(iii) a = -1
Solution:
a (a2 + a + 1) + 5
⇒ a3 + a2 + a + 5
⇒ (-1)3 + (-1)2 + (-1) + 5 [ if -1 has even power then it is 1 or else if it has odd power it is -1]
⇒ -1 + 1 -1 + 5
⇒ 4
### Question 5.
(a) Add: p (p – q), q (q – r) and r (r – p)
Solution:
p (p – q) + q (q – r) + r (r – p)
⇒ p2 – pq + q2 – qr + r2 – rp
⇒ p2 + q2+ r2 – pq – rp – qr
(b) Add: 2x (z – x – y) and 2y (z – y – x)
Solution:
2x (z – x – y) + 2y (z – y – x)
⇒ 2xz – 2x2 – 2xy + 2yz – 2y2 – 2yx
⇒ 2xz – 4xy + 2yz – 2x2 – 2y2
(c) Subtract: 3l (l – 4m + 5n) from 4l (10n – 3m + 2l)
Solution:
4l (10n – 3m + 2l) – 3l (l – 4m + 5n)
⇒ 40ln – 12lm + 8l2 – 3l2 + 12lm – 15ln
⇒ 25ln + 5l2
(d) Subtract: 3a (a + b + c) – 2b (a – b + c) from 4c (–a + b + c)
Solution:
4c (– a + b + c) – [3a (a + b + c) – 2b (a – b + c)]
⇒ -4ac + 4bc + 4c2 – [3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc]
⇒ -4ac + 4bc + 4c2 – 3a2 – ab – 3ac – 2b2 + 2bc
⇒ -3a2 – 2b2 + 4c2 – 7ac + 6bc – ab
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# Reduce the equation
Question:
Reduce the equation 2x – 3y – 5 = 0 to slope-intercept form, and find from it the slope and y-intercept.
Solution:
Given equation is $2 x-3 y-5=0$
We can rewrite it as 2x – 5 = 3y
$\Rightarrow 3 y=2 x-5$
$\Rightarrow y=\frac{2}{3} x-\frac{5}{3}$
This equation is in the slope-intercept form i.e. it is the form of
$\mathrm{y}=\mathrm{m} \times \mathrm{X}+\mathrm{c}$, where $\mathrm{m}$ is the slope of the line and $\mathrm{c}$ is $\mathrm{y}$-intercept of the line
Therefore, $\mathrm{m}=\frac{2}{3}$ and $\mathrm{c}=-\frac{5}{3}$
Conclusion:
Slope is $\frac{2}{3}$ and $y-$ intercept is $-\frac{5}{3}$
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# Coordinate Graphing of Real World Problems
Patterns are all around you. From the six-pack of sodas to the dozen eggs you might buy, there is something to notice everywhere you look.
You may not have realized it before, but your powers of observation are a mathematical skill. By graphing your observations on a coordinate graph, you can make a visual picture of the relationships you observe between numbers.
### Learning Outcomes
By the end of this lesson, your children will be able to use a coordinate graph to show real-world situations involving related numbers. They will also be able to look at information displayed on a coordinate graph and understand the real-world situation that it describes.
### Warm Up
Even though algebra and geometry are two different categories of math that involve some very different skills, they overlap in a unique way when you study coordinate graphing. Coordinate graphing allows you to visually display relationships you observe between numbers. A coordinate graph is like a picture of what you see a pattern doing.
One tool that is very useful for organizing information about patterns is a function table. It is often called a "T-chart" because it is a two-column chart that looks like a "T". You will be learning more about function tables in this lesson.
### Pre-assessment worksheet
Have your children take the Pre-Test below to see if they are ready for this lesson. If they get 11 or less correct, review the introduction with them before continuing on to the lesson.
## Main Lesson: Coordinate Graphing of Real-World Problems
There are many situations in life that involve two sets of numbers that are related to each other. For example, If you know the price of one ticket for a show, you can calculate the cost for any number of people to attend. Similarly, if you know how much gas will cost for one gallon, you can calculate how many gallons you will be able to purchase with the money in your wallet.
Often, the hardest part of figuring out a complicated problem with lots of data is keeping it all organized so you don't lose track of what you are doing. Using a function table, or "T-chart" can help you organize information.
A function table has two columns, because it is used to show the relationship between two different strings of numbers. Each function table has a rule, called a "function" that generates a pattern for one string of numbers (often named by the variable "Y") when another string of number (often named by the variable "X") is used.
The examples below start with one that shows how a function table can be used to figure out how much it would cost for "x" number of people to attend an afternoon movie that costs \$5 per person.
Using "x" as a place holder for an unknown number allows us to generalize the information and write a rule, or function, that works for any number of people. By plugging in a value for x in the function above the table, the corresponding value for y can be determined. In this case, Y is the total cost of going to the movies.
Each ticket costs \$5, so start plugging in the values for X, one at a time.
5x = y X Y 1 5 \$5 x 1 person = \$5 2 10 \$5 x 2 person = \$10 3 15 \$5 x 3 person = \$15 4 20 \$5 x 4 person = \$20
If the evening show costs \$9, we could make a second table, like this:
9x = y X Y 1 9 \$9 x 1 person = \$9 2 18 \$9 x 2 person = \$18 3 27 \$9 x 3 person = \$27 4 36 \$9 x 4 person = \$36
Remind your children that the numbers which are on the same level on the function table are the numbers that are related to each other. Children will sometimes become confused because they try to match numbers which are on different levels on the function table.
While function tables come in handy for small problems like figuring out the cost of a group of people attending a movie together, their real value can be seen when applying patterns to very large numbers. Small problems typically have many ways to be solved. Pictures can be used, or tally marks. Numbers can be added repeatedly. But no one wants to draw 218 pictures of something, or count out 1,053 tally marks, to solve a problem. When you understand a pattern dealing with a small number, the pattern can also be applied to help you figure out a much larger number.
So, how does your information get from a function table to a visual display on a coordinate graph? Let's look another example.
To figure out how many books you can read in x number of days if you read two books per day, first make a function table:
2x = y X Y 0 0 = (0,0) 1 2 = (1,2) 2 4 = (2,4) 3 6 = (3,6) 4 8 = (4,8) 5 10 = (5,10) 6 12 = (6,12)
Each corresponding pair of numbers on the same level of the function table is used to write an ordered pair. The first number tells you how many units to move across the horizontal line (x axis). The second number tells you how many units to move up the vertical line (y axis). Draw a point where the lines cross. Join the points to form a line.
Take time to point out numbers that are related to each other when they come up in the course of your daily life. Even a library fine can be a learning opportunity. "Wow, ten cents a day for seven days added up to seventy cents! How much would it have cost if we kept the book until it was three weeks overdue instead of just one week overdue?"
Click on the link below and print out the worksheet that will allow your children to practice generating and plotting ordered pairs on a coordinate grid.
You will also find several more coordinate geometry worksheets listed here.
### Recap
• Real-life situations can be represented using a function table and a coordinate graph.
• The terms that are on the same level of a function table are related to each other, and can be written as an ordered pair.
• The first number in an ordered pair tells the number to move left or right on the x axis.
• The second number in an ordered pair tells the number to move up or down on the y axis.
### Test Questions
Review these points with your children and then print out the Post Test worksheet below.
At least 14 out of 18 correct will show that your children are ready to move on.
All Geometry 24 Worksheets Terms/ Definitions Formulas/ Equations 2D Shapes 3D Shapes Quadrilaterals Measuring Angles Using a Protractor Adding and Subtracting Angles Angle Properties Finding Angles Symmetry Area Volume Surface Area Perimeter Coordinate System Coordinate Graphing Pythagoras' Theorem Distance Between Two Points Congruent Triangles Similar Triangles Transformations Dilations
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Click the links below for information and help on dealing with bullying.
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# How do you simplify 3sqrt[3] + 2sqrt[27] –sqrt[12]?
Apr 8, 2016
$7 \sqrt{3}$
#### Explanation:
Given an expression to simply
$3 \sqrt{3} + 2 \sqrt{27} - \sqrt{12}$
Factorize second and third terms as
3sqrt3+2sqrt(3xx3xx3)-sqrt(2xx2xx3
We know that if there are 2 same digits/numbers under the square root symbol, these can be taken outside the symbol with the condition that the digit/number is written only once outside the symbol. Following this we obtain
$3 \sqrt{3} + 2 \times 3 \sqrt{3} - 2 \sqrt{3}$,
simplifying and taking out common factor $\sqrt{3}$ gives us
or $3 \sqrt{3} + 6 \sqrt{3} - 2 \sqrt{3}$
or $\sqrt{3} \left(3 + 6 - 2\right)$,
simplifying numbers inside the parenthesis we obtain
$7 \sqrt{3}$
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# Tower of Hanoi
The Tower of Hanoi is a simple mathematical problem or puzzle. You are given three rods and a number of discs of different sizes. The puzzle starts with all discs on a single rod. Your aim is to move all of them to a different rod according to various rules:
• Only one disc can be moved at a time
• No disc can sit atop a smaller disc.
It is not hard to show that with N discs, we can achieve the goal in 2N – 1 moves. The simplest proof is to observe that with N discs we need to perform the following three steps: (i) shift the top N-1 discs to an empty rod (ii) shift the bottom disc to the other empty rod, (iii) shift the top N-1 discs onto the bottom disc. By mathematical induction one easily establishes the formula 2N – 1. Note that we are essentially reducing the problem with N discs to a problem with N-1 discs.
With similar reasoning one can show that any random position of discs can be obtained (as long as no disc covers a smaller disc). The proof is left as an exercise for the reader.
The Tower of Hanoi is an example of shifting a large pile of items with limited resources. If you are not familiar with this puzzle, you will probably be surprised by the fact that only three rods are required no matter how many discs you start with. Avid readers of this blog may have come across terms like “Tower-Of-Hanoi manoeuvres” from previous posts, so if you were unsure what the fuss was all about, then now you know 😊.
In Spider Solitaire we are often confronted with the problem of shifting large sequences of cards with limited resources. A simple example is shown below: A complete suit of Spades is visible but can we actually clear the suit with only one empty column?
The answer is yes. We can shift the Eight of Diamonds onto the Nine of Diamonds in column six, build the J-0-9 of Spades onto the K-Q in column 2, move the 8-7-6-5 of Spades from column five onto the 9 of Spades, swap the 4H and 4S on top of both the Spade Fives and finally add the Ace of Spades from Column three to complete the suit.
Going back to the Hanoi puzzle, with a small number of rods a monkey could probably luck his way into a solution by making random moves, but once you get a decent size pile of discs the random move strategy doesn’t work so well! Also, with random moves it is difficult to prove that e.g. 30 moves or less is impossible given five discs. Similar considerations apply to Spider Solitaire. Since the above example is relatively simple, a monkey could probably complete a suit of Spades by repeated trial and error, assuming he only makes moves that are “reversible”. But with a more complex problem, the monkey won’t do so well.
If you want more practice with “Tower-of-Hanoi manoeuvres” I recommend the following exercise: set up the diagram above, ignoring any face-down cards or cards not in sequence (for instance in column two you keep only the K-Q of spades). Then try to minimise the number of in-suit builds using only reversible moves (you should be able to get pretty close to zero). From this new position pretend you’ve just realised your mistake and try to clear the Spades using only reversible moves. This exercise should give you an idea of why empty columns are so valuable.
Note that all this carries the assumption of no 1-point penalty per move (commonly used in many implementations of Spider Solitaire). If there was such a penalty then we would have to think twice about performing an extra 50 moves just for the sake of one more in-suit build. But for now we’ll keep things simple.
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Evaluating Exponents with Negative Bases | KS3 Maths Resources
## What you need to know
Things to remember:
• If the base number is negative and the power is odd, the answer will be negative.
• If the base number is negative and the power is even, the answer will be positive.
An exponent is a number written to the top right of a number and tells us how many times we multiply a number by itself.
$$6^2 = 6\times 6$$
$$6^3 = 6\times 6\times 6$$
$$6^4= 6\times 6\times 6\times 6$$
So, all we’re doing when we have exponents is lots of multiplications. When we do these multiplications, we refer to it as “evaluating the exponent”.
Evaluate the following exponents:
$$5^2=5\times5=25$$
$$7^3=7\times7\times7=343$$
$$8^4=8\times8\times8\times8=4096$$
Let’s look at some examples with a base number of just -1, to keep things simple.
Hint: Remember, when we multiply two negative numbers, we get a positive answer, and when we multiply a negative number and positive number, we get a negative answer.
$$(-1)^1=-1$$
$$(-1)^2=-1\times-1=1$$
$$(-1)^3=-1\times-1\times-1=-1$$
$$(-1)^4=-1\times-1\times-1\times-1=1$$
$$(-1)^5=-1\times-1\times-1\times-1\times-1=-1$$
$$(-1)^6=-1\times-1\times-1\times-1\times-1\times-1=1$$
There are two key points we should take away from these examples:
• When the base is negative and the power is odd, the answer is negative
• When the base is negative and the power is even, the answer is positive.
If we evaluate $(-4)^3$ will the answer be positive or negative?
The answer will be negative, because the base number is negative, and the power is odd.
If we evaluate $(8)^7$ will the answer be positive or negative?
The answer will be positive, because the base number is positive.
Hint: The base number is positive, so the answer will always be positive.
If we evaluate $(-9)^2$ will the answer be positive or negative?
The answer will be positive, because the base number is negative, and the power is odd.
Hint: Whenever the power is even, the answer will be positive.
If we evaluate $(-7)^7$ will the answer be positive or negative?
The answer will be negative, because the base number is negative, and the power is odd.
If we evaluate $(-11)^{-4}$ will the answer be positive or negative?
The answer will be positive, because the base number is negative, and the power is odd.
Hint: It doesn’t matter that the power is negative, it is still even.
So, we can actually do evaluate exponents with negative bases in three steps:
Evaluate $(-4)^4$
Step 1: Determine whether the answer will be positive or negative.
The base number is negative, and the power is even, so the answer will be positive.
Step 2: Evaluate the exponent without the negative.
$$4^4=256$$
Step 3: Write the answer appropriately, as positive or negative.
The answer will be positive, so $(-4)^4=256$
Evaluate $(-7)^3$
Step 1: Determine whether the answer will be positive or negative.
The base number is negative, and the power is odd, so the answer will be negative.
Step 2: Evaluate the exponent without the negative.
$$7^3=343$$
Step 3: Write the answer appropriately, as positive or negative.
The answer will be positive, so $(-7)^3=-343$
## KS3 Maths Revision Cards
(78 Reviews) £8.99
## Example Questions
Step 1: Determine whether the answer will be positive or negative.
The base number is negative, and the power is odd, so the answer will be negative.
Step 2: Evaluate the exponent without the negative.
$$3^5=243$$
Step 3: Write the answer appropriately, as positive or negative.
The answer will be negative, so $(-3)^5=-243$
Step 1: Determine whether the answer will be positive or negative.
The base number is negative, and the power is even, so the answer will be positive.
Step 2: Evaluate the exponent without the negative.
$$2^8=256$$
Step 3: Write the answer appropriately, as positive or negative.
The answer will be positive, so $(-2)^8=256$
## KS3 Maths Revision Cards
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• All of the major KS2 Maths SATs topics covered
• Practice questions and answers on every topic
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# Roman Numbers: Learn 7 Roman Symbols
Roman numbers are a part of a numeral system that was used in ancient Rome to represent any number using 7 different symbols. The seven different symbols used to represent any number in the roman number system include I, V, X, L, C, D, and M. The roman symbols:
• I indicate 1
• V indicates 5
• X indicates 10
• L indicates 50
• C indicates 100
• D indicates 500
• M indicates 1000
## Roman Numerals Rules
1. If the same roman symbol is repeated, then its value is added as many times as it occurs.
For example:
VVV = 5 + 5 + 5 = 15
XX = 10 + 10 = 20
2. When a small roman symbol appears after a larger or equal symbol, then it is added.
For example:
VII = 5 + 1 + 1 = 7
LXX = 50 + 10 + 10 = 70
3. When a small roman symbol appears before another larger symbol, then it is subtracted.
For example:
IV = V – I = 5 -1 = 4
IX = 10 -1 = 9
4. The same romal symbol cannot be used more than 4 times in a row.
For example:
XXXX = 10 + 10 + 10 + 10 = 40 is incorrect because the same symbol is added 4 times.
## How Big Numbers are Represented in Roman Symbols?
Numbers larger than 1000 in roman symbols are formed by placing a dash over the symbol, meaning 1000 times. For example:
V = 5000
X = 10,000
L = 50,000
C = 100,000
D = 500,000
M = 1,000,000
Note: These numbers are not commonly used.
## How to Convert Number 1724 Into Roman Symbols?
Solution:
Split 1724 as 1000 + 700 + 20 + 4
Now, convert the numbers that are split into roman symbols as shown below:
Here,
1000 = M
500 = DCC
20 = XX
5 = IV
Therefore, 1724 = MDCCXXIV
Some of the most important tips associated with the learning of Roman numerals have been explained as follows:
1. First of all, it is very much advisable for the kids to understand the very basic symbols, for example, V will be five 5, X will be 10, L will be 50, C will be a hundred, D will be 500, and so on. It is advisable for people to be clear about all these kinds of technicalities.
2. Kids should utilize a mnemonic method of memorizing the value order of the symbols so that everybody will be on the right track in dealing with things. In this particular case, learning the things with the help of phrases is a good idea and one of the best possible types of phrases is My Dear Cat Loves Xtra Vitamins Intensely.
3. Learning different kinds of digits in one’s place is very much important so that overall goals are very easily achieved. In this particular case, kids will have a clear idea about the basic technicalities and formulation of the bigger numbers.
4. It is important to be clear about multiple rules and regulations of this particular case for example there might not be a case of more than three symbols of the same symbol in the row which is the main reason kids need to be clear about the valuation process so that overall goals are very easily achieved.
5. Adding the smaller symbol values which are placed after the large symbol values is important for easy understanding.
6. Having a clear idea about the writing of the compound numbers is another very important thing to be taken into consideration and further everyone will have a crystal-clear idea about the formulation of the big numbers in the whole process.
7. Learning to write the larger numbers in this particular case is very much important because of the utilization in the normal life and further checking out the work simultaneously is important so that one can diagnose and correct the mistakes very successfully.
## Angles: Learn the Definition and Types of Angles
Do you know ancient mathematics used the concept of line to represent any straight object? A line is a one-dimensional figure that has length but no width. It is made up of different sets of points that are extended in opposite directions infinitely. When two rays (part of a line that has a fixed starting point but no ending point) intersect each other in the same plane, it forms an angle. The point of intersection where the rays meet is termed a vertex. Here, we will learn the definitions and different types of angles
## Angle Definition
In Geometry, an angle can be defined as a figure formed when two rays, lines, or, line segments are joined together at a common point. The common point is known as the vertex. Angles are generally measured in radians or degrees.
## What are the Different Parts of An Angle?
The two main parts of an angle are:
1. Arms: The two rays of an angle that intersect at a common point are termed arms of an angle.
2. Vertex: A common ending point where two rays meet is known as a vertex.
## What are the Different Types of Angles?
The different types of angles are:
• Acute Angle: It is an angle whose measure is greater than 0 degrees but less than 90 degrees.
• Right Angle: It is an angle whose exact measure is 90 degrees.
• Obtuse Angle: It is an angle whose measure is greater than 90 degrees but less than 180 degrees.
• Straight Angle: It is an angle whose exact measure is 180 degrees.
• Reflex Angle: It is an angle whose measure is greater than 180 degrees but less than 360 degrees.
• Complete Angle: It is an angle whose exact measure is 360 degrees.
## What are Positive and Negative Angles?
Positive angles are the angles that are rotated from the base in the anti-clockwise direction. Negative angles are the angles that are rotated from the base in the clockwise direction.
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# 2007 AIME II Problems/Problem 8
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
A rectangular piece of paper measures 4 units by 5 units. Several lines are drawn parallel to the edges of the paper. A rectangle determined by the intersections of some of these lines is called basic if
(i) all four sides of the rectangle are segments of drawn line segments, and
(ii) no segments of drawn lines lie inside the rectangle.
Given that the total length of all lines drawn is exactly 2007 units, let $N$ be the maximum possible number of basic rectangles determined. Find the remainder when $N$ is divided by 1000.
## Solution
### Solution 1
Denote the number of horizontal lines drawn as $x$, and the number of vertical lines drawn as $y$. The number of basic rectangles is $(x - 1)(y - 1)$. $5x + 4y = 2007 \Longrightarrow y = \frac{2007 - 5x}{4}$. Substituting, we find that $(x - 1)\left(-\frac 54x + \frac{2003}4\right)$.
FOIL this to get a quadratic, $-\frac 54x^2 + 502x - \frac{2003}4$. Use $\frac{-b}{2a}$ to find the maximum possible value of the quadratic: $x = \frac{-502}{-2 \cdot \frac 54} = \frac{1004}5 \approx 201$. However, this gives a non-integral answer for $y$. The closest two values that work are $(199,253)$ and $(203,248)$.
We see that $252 \cdot 198 = 49896 > 202 \cdot 247 = 49894$. The solution is $\boxed{896}$.
### Solution 2
We realize that drawing $x$ vertical lines and $y$ horizontal lines, the number of basic rectangles we have is $(x-1)(y-1)$. The easiest possible case to see is $223$ vertical and $223$ horizontal lines, as $(4+5)223 = 2007$. Now, for every 4 vertical lines you take away, you can add 5 horizontal lines, so you basically have the equation $(222-4x)(222+5x)$ maximize.
Expanded, this gives $-20x^{2}+222x+222^{2}$. From $-\frac{b}{2a}$ you get that the vertex is at $x=\frac{111}{20}$. This is not an integer though, so you see that when $x=5$, you have $-20*25+222*5+222^{2}$ and that when x=6, you have $-20*36+222*6+222^{2}$. $222 > 20*11$, so the maximum integral value for x occurs when $x=6$. Now you just evaluate $-20*36+222*6+222^{2}\mod 1000$ which is ${896}$.
## See also
2007 AIME II (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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## Introduction to Multiplication Strategies
In our previous grades, we have read about the concept of multiplication. Let's revise with an example: you visit a bakery shop and buy 19 cakes for 3 rupees each. It's time to pay now, you will add 3 rupees to 19 times to get the final amount. Instead of addition here we can use the concept of multiplication. We are going to multiply $3 \times 19 =57$. So 57 is the final amount. In this article, we are going to learn about multiplication, how to teach multiplication and multiplication activities for class 3.
## What is Multiplication?
Multiplication is an operation that represents the basic idea of adding the same number over and over again. Multiplied numbers are called factors, and the result obtained after multiplying two or more numbers is known as the product of those numbers. Multiplication is used to simplify the task of repeatedly adding the same number.
For example, what is 36 times 9? We know that 36 times 9 is written in the form of a multiplication sentence as 36 × 9 = 324. Here, 36 and 9 are the factors and 324 is the product. So, 36 times 9 is 324.
## Multiplication Formula
A multiplication expression is expressed as multiplicand × multiplier = product. Where:
• Multiplicand: first number (factor).
• Multiplier: Second number (factor).
• Product: The final result after multiplying the multiplicand and the multiplier.
• Multiplication symbol: "×" (connects the first number and second number).
## Multiplication Strategies for Grade 3
How to teach multiplication efficiently is important to help children understand multiplication. Teaching these four multiplication strategies for class 3 with the help of multiplication activity for class 3 is essential before children can remember facts.
### Make Equal Groups
One of the first multiplication strategies for 3rd grade is to start with equal grouping. This allows children to visually understand why two elements are multiplied to get a product.
It is helpful for children to start with manipulative devices such as yellow and red tokens. Start by writing the multiplication facts on your screen or blackboard.
In this example, we use 3x4.
Equal Group
• Ask the children to form groups of three and place four game pieces in each group. Spend a few minutes creating groups and then discussing what they did. Ask children to share what they create.
• Explain that there are three groups and each has four counters. It's the same as saying 3×4. Ask the children to tell you the total number of game pieces they used.
• The answer is 12, as 3 × 4=12.
• This is a good time to point out that adding numbers gives you a total. Multiplication is the same.
### Make an Array
The next multiplication strategy I will teach you is to create an array: Creating a 3x4 array using a counter, similar to the group.
• So, create 3 rows with 4 counters in each row. Alternatively, you can create 4 rows of 3 counters each.
• When I teach an array, I don't care whether I display the rows or the columns first. Once we arrive at the commutativity of multiplication, we see that we can commute the factors anyway.
• It explains that creating three lines, including 4 in each line is the same as 3x4.
After using counters to create equal groups and arrays, let's move on to using numbers only. However, repeated addition can be combined with equal group and array formation.
Giving equal groups and formations is a useful strategy in the Army as it helps keep track of how many there are when counting totals.
When teaching repeated addition, tell students that it is easier to flip the factors. For example, if you find the product of 5 and 3, you can add 5+5+5 or 3+3+3+3+3. To show that counting by 5 is easier, he adds 5 three times.
### Skip Counting
Similar to repeated addition, skip counting can be used in conjunction with creating equal groups and arrays.
Students can label groups and skip how many people are in each group. This helps the students to keep track of their numbers.
The 4×3 example teaches students that they can skip 4 times 3 times or 3 times 4 times.
At the beginning of the school year, let students choose which strategy they like best.
Skipping numbers, however, is a strategy that children must use before they can memorize the facts of multiplication because this is the fastest strategy.
Sometimes you don't have time to draw a 9x8 array. So skipping the count is more efficient in this case.
## Solved Questions
Q1. Solve 3 x 4 using an array.
Ans: We use counters to create an array for 3×4. So we will create 3 rows with 4 counters in each row. Or we can have 4 rows with 3 counters in each.
Using an Array
Thus, the result of the multiplication is 12.
Q2. Using repeated addition, solve 3 x 5.
Ans. Using repeated addition, we get,
3 + 3 + 3 + 3 + 3 or 15
Thus, the result of the multiplication is 15.
## Q3. Using an equal group, solve 3 x 3.
Ans. Using the concept of equal groups, we can transform the given product into 3 groups of 3 stars each, as shown below:
Using Equal Group
Thus, the result of the multiplication is 9.
## Practice Questions
Q1. Multiply 30 x 2.
Ans. 60
Q2. Multiply 70 x 6.
Ans. 420
Q3. 6 x 9.
Ans. 54
Q4. 9 x 3.
Ans. 27
Summary
Last updated date: 02nd Oct 2023
Total views: 78.6k
Views today: 0.78k
## FAQs on Multiplication Strategies Grade 3
1. Who discovered matrix multiplication?
Matrix multiplication was first described by the French mathematician Jacques Philippe Marie Binet in 1812.
2. What is another name for skip counting?
This technique is also called counting by twos (threes, fours, etc.).
3. What sign can we use in place of the cross?
We can use * or . in the place of x in multiplication.
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## Plotting Points on the Coordinate Plane
### Learning Outcomes
• Define the components of the Cartesian coordinate system.
• Plot points on the Cartesian coordinate plane.
An old story describes how seventeenth-century philosopher/mathematician René Descartes invented the system that has become the foundation of algebra while sick in bed. According to the story, Descartes was staring at a fly crawling on the ceiling when he realized that he could describe the fly’s location in relation to the perpendicular lines formed by the adjacent walls of his room. He viewed the perpendicular lines as horizontal and vertical axes. Further, by dividing each axis into equal unit lengths, Descartes saw that it was possible to locate any object in a two-dimensional plane using just two numbers—the displacement from the horizontal axis and the displacement from the vertical axis.
While there is evidence that ideas similar to Descartes’ grid system existed centuries earlier, it was Descartes who introduced the components that comprise the Cartesian coordinate system, a grid system having perpendicular axes. Descartes named the horizontal axis the x-axis and the vertical axis the y-axis.
The Cartesian coordinate system, also called the rectangular coordinate system, is based on a two-dimensional plane consisting of the x-axis and the y-axis. Perpendicular to each other, the axes divide the plane into four sections. Each section is called a quadrant; the quadrants are numbered counterclockwise as shown in the figure below.
The Cartesian coordinate system with all four quadrants labeled.
### Try It
The center of the plane is the point at which the two axes cross. It is known as the origin or point $\left(0,0\right)$. From the origin, each axis is further divided into equal units: increasing, positive numbers to the right on the x-axis and up the y-axis; decreasing, negative numbers to the left on the x-axis and down the y-axis. The axes extend to positive and negative infinity as shown by the arrowheads in the figure below.
Each point in the plane is identified by its x-coordinate, or horizontal displacement from the origin, and its y-coordinate, or vertical displacement from the origin. Together we write them as an ordered pair indicating the combined distance from the origin in the form $\left(x,y\right)$. An ordered pair is also known as a coordinate pair because it consists of and y-coordinates. For example, we can represent the point $\left(3,-1\right)$ in the plane by moving three units to the right of the origin in the horizontal direction and one unit down in the vertical direction.
An illustration of how to plot the point (3,-1).
When dividing the axes into equally spaced increments, note that the x-axis may be considered separately from the y-axis. In other words, while the x-axis may be divided and labeled according to consecutive integers, the y-axis may be divided and labeled by increments of 2 or 10 or 100. In fact, the axes may represent other units such as years against the balance in a savings account or quantity against cost. Consider the rectangular coordinate system primarily as a method for showing the relationship between two quantities.
### A General Note: Cartesian Coordinate System
A two-dimensional plane where the
• x-axis is the horizontal axis
• y-axis is the vertical axis
A point in the plane is defined as an ordered pair, $\left(x,y\right)$, such that x is determined by its horizontal distance from the origin and y is determined by its vertical distance from the origin.
### Example: Plotting Points in a Rectangular Coordinate System
Plot the points $\left(-2,4\right)$, $\left(3,3\right)$, and $\left(0,-3\right)$ in the coordinate plane.
### Key Takeaways
You can use an online graphing tool to practice plotting points on the Cartesian Coordinate plane. Watch the following short video to learn how!
Now try plotting the following points with an online graphing tool.
$(4,0)$
$(-1,5)$
$(0,-10)$
$(2,7)$
$(3,-5)$
$(-4,-7)$
## Contribute!
Did you have an idea for improving this content? We’d love your input.
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Introductory Statistics
# 9.6Hypothesis Testing of a Single Mean and Single Proportion
Introductory Statistics9.6 Hypothesis Testing of a Single Mean and Single Proportion
## Stats Lab
### Hypothesis Testing of a Single Mean and Single Proportion
Class Time:
Names:
Student Learning Outcomes
• The student will select the appropriate distributions to use in each case.
• The student will conduct hypothesis tests and interpret the results.
Television SurveyIn a recent survey, it was stated that Americans watch television on average four hours per day. Assume that σ = 2. Using your class as the sample, conduct a hypothesis test to determine if the average for students at your school is lower.
1. H0: _____________
2. Ha: _____________
3. In words, define the random variable. __________ = ______________________
4. The distribution to use for the test is _______________________.
5. Determine the test statistic using your data.
6. Draw a graph and label it appropriately.Shade the actual level of significance.
1. Graph:
Figure 9.21
2. Determine the p-value.
7. Do you or do you not reject the null hypothesis? Why?
8. Write a clear conclusion using a complete sentence.
Language SurveyAbout 42.3% of Californians and 19.6% of all Americans over age five speak a language other than English at home. Using your class as the sample, conduct a hypothesis test to determine if the percent of the students at your school who speak a language other than English at home is different from 42.3%.
1. H0: ___________
2. Ha: ___________
3. In words, define the random variable. __________ = _______________
4. The distribution to use for the test is ________________
5. Determine the test statistic using your data.
6. Draw a graph and label it appropriately. Shade the actual level of significance.
1. Graph:
Figure 9.22
2. Determine the p-value.
7. Do you or do you not reject the null hypothesis? Why?
8. Write a clear conclusion using a complete sentence.
Jeans SurveySuppose that young adults own an average of three pairs of jeans. Survey eight people from your class to determine if the average is higher than three. Assume the population is normal.
1. H0: _____________
2. Ha: _____________
3. In words, define the random variable. __________ = ______________________
4. The distribution to use for the test is _______________________.
5. Determine the test statistic using your data.
6. Draw a graph and label it appropriately. Shade the actual level of significance.
1. Graph:
Figure 9.23
2. Determine the p-value.
7. Do you or do you not reject the null hypothesis? Why?
8. Write a clear conclusion using a complete sentence.
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Transcript
```Discrete Probability Distributions
6- 1
Chapter
Six
McGraw-Hill/Irwin
Random variable
A result from an experiment that, by chance, can
take different values.
Egs.
1. No. of heads you would get in 3 tosses of a coin
2. No. of employees absent on a shift
3. No. of students at CSUN in a semester
4. No. of minutes to drive home from CSUN
5. Inches of rainfall in LA during a year
6. Tire pressure in PSI of a car tire
Examples 1-3 are ‘discrete’
4-6 are ‘continuous’ random variables
In this chapter, we focus on the ‘discrete’.
Probability
Distribution
A listing of all possible outcomes of an
experiment and the corresponding probability.
c
P
F
T
Possible Outcomes
N
= 1/8
= 3/8
= 3/8
= 1/8
Probability Distribution
Note the similarity to histogram
Mean of Discrete Probability Distribution
[ xP( x)]
where
represents the mean
o x is each outcome
o P(x) is the probability of the various outcomes x.
Calculating Mean/Expected Value of Heads in 3
coin-toss experiment
[ xP( x )]
μ = 0 * .125 + 1 * .375 + 2 * .375 + 3 * .125
= 1.5
(which by intuition makes sense because for each toss you have 50%
is a weighted average.
It is also referred to as Expected Value, E(X).
Binomial Probability Distribution
•An outcome of an experiment is classified into one of
two mutually exclusive categories, such as a success or
failure (bi means two).
•The data collected are the results of counts (hence, a
discrete probability distribution).
•The probability of success stays the same for each trial
(independence).
Binomial Probability Distribution
P( x)n Cx (1 )
x
n x Can you logically
explain this formula?
n is the number of trials
x is the number of observed successes
π is the probability of success on each trial
Cx
n
n!
x!(n-x)!
Let us re-visit the 3-toss coin experiment
n=3
π = .5
X = 0,1,2,3 (number of heads)
P(0) = 3C0 * (.5)0 * (1 - .5)3-0
P(1) = 3C1 * (.5)1 * (1 - .5)3-1
P(2) = 3C2 * (.5)2 * (1 - .5)3-2
P(3) = 3C3 * (.5)3 * (1 - .5)3-3
= .125
= .375
= .375
= .125
You can also
use
Appendix A
(page 489)
Practice time!
Do Self-Review 6-3, Page 168
Use Appendix A, Page 490
Cumulative Binomial Probability Distribution
Problem 21, Page 173
Use table in Page 491
a. 0.387 ( n=9; x=9; straight from table )
b. 0.001 ( P(x<5) = P(x ≤ 4) )
c. 0.992 ( 1 – P(x ≤ 5) = 1 – 0.008 )
d. 0.946 [ (P(x=7) + (P(x=8) + (P(x=9) ]
Mean of the Binomial Distribution
n
Logic: If you throw a coin say 100 times, how many
times you would expect to get a head?
For each throw, the probability of getting a head is .5.
So, in 100 trials, you would expect 50 heads (which is
100*.5; ie. nπ )
Variance of the Binomial Distribution
n (1 )
2
For the previous example,
= 100*.5*(1-.5)
σ 2 = 25
S.D. σ = 5
```
Related documents
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# 180 Days of Math for First Grade Day 105 Answers Key
By accessing our 180 Days of Math for First Grade Answers Key Day 105 regularly, students can get better problem-solving skills.
## 180 Days of Math for First Grade Answers Key Day 105
Directions: Solve each problem.
Question 1.
Count the dots in the frame. How many more dots are needed to make twenty?
10 more dots,
Explanation:
Given to count the dots in the frame,
Counted there are 10 dots in the frame,
more number of dots are needed to make
twenty are 20 – 10 = 10 more dots.
Question 2.
20,
Explanation:
Given to add 10 to 10 we will get
10
+10
20.
Question 3.
13 – 8 =
5,
Explanation:
Given to subtract 8 from 13 we will get
13
-8
5.
Question 4.
6,
Explanation:
Given to find missing number in __- 5 = 1,
let x be the missing number so x – 5 = 1,
x = 1 + 5 = 6.
Question 5.
Name the solid.
Cube,
Explanation:
Given solid has 6 square faces so it cube.
Question 6.
Write the day that comes after Thursday.
Friday,
Explanation:
The day that comes after Thursday if Friday.
Question 7.
Record the data in the chart with tally marks.
• Ten children like to swim.
• Six children like to run through the sprinklers.
• Five children like to stay inside.
• Three children like to play with water toys.
Favorite Summer Activity
Water Toys Sprinklers Stay Inside Swim
Explanation:
Recorded the data in the chart above with tally marks as
Three children like to play with water toys.
Six children like to run through the sprinklers,
Five children like to stay inside and
Ten children like to swim.
Question 8.
There are 4 children. How many ears are there?
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# How do you simplify ((3x^-2 y)^-2)/( 4xy^-2)^-1 ?
##### 1 Answer
Mar 15, 2018
color(red)(((3x^-2 y)^-2)/( 4xy^-2)^-1 = color(blue)((4x^5)/(9y^4)
#### Explanation:
Given:
color(red)(((3x^-2 y)^-2)/( 4xy^-2)^-1 Expression 1
We can simplify the above exponent problem as follows:
color(green)(Step" 1"
Rule 1
color(blue)((a^m)^n = a^(mn)
Using this rule, we can write Expression 1 as
$\frac{{3}^{- 2} {x}^{4} {y}^{- 2}}{{4}^{- 1} {x}^{- 1} {y}^{2}}$
We can rewrite the above expression, combining the like terms as
$\left(\frac{{3}^{-} 2}{{4}^{-} 1}\right) \left(\frac{{x}^{4}}{{x}^{-} 1}\right) \left(\frac{{y}^{-} 2}{{y}^{2}}\right)$ Expression 2
color(green)(Step" 2"
Rule 2
color(blue)(a^m/a^n = a^(m-n)
Using this rule, we can simplify Expression 2 as
$\left(\frac{{3}^{-} 2}{{4}^{-} 1}\right) \left({x}^{4 - \left(- 1\right)}\right) \left({y}^{- 2 - \left(2\right)}\right)$
$\Rightarrow \left(\frac{{3}^{-} 2}{{4}^{-} 1}\right) \left({x}^{4 + 1}\right) \left({y}^{- 2 - 2}\right)$
$\Rightarrow \left(\frac{{3}^{-} 2}{{4}^{-} 1}\right) \left({x}^{5}\right) \left({y}^{-} 4\right)$ Expression 3
color(green)(Step" 3"
Rule 3
color(blue)(a^-m = 1/a^m
Using this rule, we can simplify Expression 3 as
$\left(\frac{\frac{1}{3} ^ 2}{\frac{1}{4} ^ 1}\right) \left({x}^{5}\right) \left({y}^{-} 4\right)$
Use the rule color(brown)((1/m)/(1/n)=(1/m)(n/1)
$\Rightarrow \left(\frac{1}{3} ^ 2\right) \left({4}^{1} / 1\right) \left({x}^{5}\right) \left({y}^{-} 4\right)$
$\Rightarrow \left(\frac{4}{9}\right) \left({x}^{5}\right) \left(\frac{1}{{y}^{4}}\right)$
$\Rightarrow \frac{4 {x}^{5}}{9 {y}^{4}}$
Hence,
color(red)(((3x^-2 y)^-2)/( 4xy^-2)^-1 = color(blue)((4x^5)/(9y^4)
Hope you find this solution useful.
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# Video: Differentiating a Combination of Trigonometric and Linear Functions
Chris O’Reilly
Given that 𝑦 = 10𝑥 − 2 cos 9𝑥, determine 𝑑𝑦/𝑑𝑥.
02:18
### Video Transcript
Given that 𝑦 equals 10𝑥 minus two cos nine 𝑥, determine 𝑑𝑦 𝑑𝑥.
Well in order to solve this problem, we can actually differentiate each part separately. However, it’s the second part of our function that we’re gonna have to have a look at and see if there’s some general rules that can help us. Well first of all, we can say that if 𝑦 equals cos 𝑥, well we know that 𝑑𝑦 𝑑𝑥 is gonna be equal to negative sin 𝑥.
Okay, so this is useful because actually it helps us differentiate cos 𝑥. However, the term that we’re looking in is in a slightly different form, because our term is actually more in this form which is 𝑦 is equal to 𝑎 cos 𝑓 𝑥.
So what it actually means is a constant, so in our case which would be negative two, and then multiplied by cosine of our function itself. Well if we have it in this form, we can say that 𝑑𝑦 𝑑𝑥 is gonna be equal to negative 𝑎 multiplied by the derivative of our function multiplied by the sine of our function.
Okay, great. So now we’ve got this general term. We can use this to help us actually differentiate and determine 𝑑𝑦 𝑑𝑥. So we’ve got 𝑦 is equal to 10𝑥 minus two cos nine 𝑥. So we’re gonna get 𝑑𝑦 𝑑𝑥 is equal to 10. We get 10 because if we differentiate the first term, it gives us 10.
Well just to remind us how we do that, how to differentiate, what we do is we actually multiply the coefficient by the exponent, so 10 by one which gives us 10, and then we reduce the exponent by one. So we had 𝑥 power of one. Well one minus one is zero. But we know that anything to the power of zero is just one, so we get 10𝑥 to the power of zero which is just equal to 10.
Okay, great. So we can get on to the rest of our differentiation. So then we’re gonna have minus negative two multiplied by the derivative with respect to 𝑥 of nine 𝑥 multiplied by sin nine 𝑥, which is gonna be equal to 10 minus negative two multiplied by nine because actually if we differentiate with respect to 𝑥 nine 𝑥, we just get nine multiplied by sin nine 𝑥, which is gonna be equal to 10 minus negative 18 sin nine 𝑥.
So therefore, we can say that given that 𝑦 equals 10𝑥 minus two cos nine 𝑥, 𝑑𝑦 𝑑𝑥 is gonna be equal to 10 plus 18 sin nine 𝑥.
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# wikiHow to Solve a Linear Diophantine Equation
Solving a linear Diophantine equation means that you need to find solutions for the variables x and y that are integers only. Finding integral solutions is more difficult than a standard solution and requires an ordered pattern of steps. You must first find the greatest common factor of the coefficients in the problem, and then use that result to find a solution. If you can find one integral solution to a linear equation, you can apply a simple pattern to find infinitely many more.
### Part 1 Setting up the Equation
1. 1
Write the equation in standard form. A linear equation is one that has no exponents greater than 1 on any variables. To solve a linear equation in this style, you need to begin by writing it in what is called “standard form.” The standard form of a linear equation looks like ${\displaystyle Ax+By=C}$, where ${\displaystyle A,B}$ and ${\displaystyle C}$ are integers.
• If the equation is not already in standard form, you need to use the basic rules of algebra to rearrange or combine the terms to create the standard form. For example, if you begin with ${\displaystyle 23x+4y-7x=-3y+15}$, you can combine similar terms to reduce the equation to ${\displaystyle 16x+7y=15}$.
2. 2
Reduce the equation if possible. When the equation is in standard form, check all three terms ${\displaystyle A,B}$ and ${\displaystyle C}$. If there is a common factor in all three terms, then reduce the equation by dividing all terms by that factor. If you reduce evenly across all three terms, then any solution you find for the reduced equation will also be a solution for the original equation.
• For example, if all three terms are even, you can at least divide by 2, as follows:
• ${\displaystyle 42x+36y=48}$ (all terms are divisible by 2)
• ${\displaystyle 21x+18y=24}$ (all terms now are divisible by 3)
• ${\displaystyle 7x+6y=8}$ (this equation is as reduced as possible)
3. 3
Check for the impossibility of a solution. In some cases, you may be able to tell immediately if there is no solution to your problem. If you see a common factor on the left side of the equation that is not shared on the right side, then there can be no solution to the problem.
• For example, if both ${\displaystyle A}$ and ${\displaystyle B}$ are even, then the sum of the left side of the equation would have to be even. But if ${\displaystyle C}$ is odd, then there will be no integer solution to the problem.
• ${\displaystyle 2x+4y=21}$ will have no integer solution.
• ${\displaystyle 5x+10y=17}$ can have no integer solution, because the left side of the equation is divisible by 5, but the right side is not.
### Part 2 Using the Euclidean Algorithm
1. 1
Review the Euclidean algorithm. The Euclidean algorithm is a system of repeated divisions, using the remainder each time as the divisor of a new division. The last divisor that divides evenly is the greatest common factor (GCF) of the two numbers.[1]
• For example, the following steps illustrate the Euclidean algorithm being used to find the GCF of 272 and 36:
• ${\displaystyle 272=7*36+20}$....divide the larger number (272) by the smaller (36) and note the remainder (20)
• ${\displaystyle 36=1*20+16}$....divide the previous divisor (36) by the previous remainder (20). Note the new remainder (16).
• ${\displaystyle 20=1*16+4}$....Repeat. Divide the previous divisor (20) by the previous remainder (16). Note the new remainder (4).
• ${\displaystyle 16=4*4+0}$....Repeat. Divide the previous divisor (16) by the previous remainder (4). Since the remainder is now 0, conclude that 4 is the GCF of the original two numbers 272 and 36.
2. 2
Apply the Euclidean algorithm to the coefficients A and B. With your linear equation in standard form, identify the coefficients A and B. Apply the Euclidean algorithm to find their GCF. Suppose you need to find integral solutions for the linear equation ${\displaystyle 87x-64y=3}$.[2]
• The steps of the Euclidean algorithm for the coefficients 87 and 64 are as follows:
• ${\displaystyle 87=1*64+23}$
• ${\displaystyle 64=2*23+18}$
• ${\displaystyle 23=1*18+5}$
• ${\displaystyle 18=3*5+3}$
• ${\displaystyle 5=1*3+2}$
• ${\displaystyle 3=1*2+1}$
• ${\displaystyle 2=2*1+0}$
3. 3
Identify the greatest common factor (GCF). Because the Euclidean algorithm for this pair continues all the way down to dividing by 1, the GCF between 87 and 64 is 1. This is another way of saying that 87 and 64 are relatively prime.[3]
4. 4
Interpret the result. When you complete the Euclidean algorithm to find the GCF of ${\displaystyle A}$ and ${\displaystyle B}$, you need to compare that result with the number ${\displaystyle C}$ of the original equation. If the greatest common factor of ${\displaystyle A}$ and ${\displaystyle B}$ is a number that can divide into ${\displaystyle C}$, then your linear equation will have an integral solution. If not, then there will be no solution.[4]
• For example, the sample problem ${\displaystyle 87x-64y=3}$ will have an integral solution, since the GCF of 1 can be evenly divided into 3.
• Suppose, for example, that the GCF had worked out to be 5. The divisor 5 cannot go evenly into 3. In that case, the equation would have no integral solutions.
• As you will see below, if an equation has one integral solution, then it also has infinitely many integral solutions.
### Part 3 Renaming the GCF to find the Solution
1. 1
Label the steps of the GCF reduction. To find the solution of the linear equation, you will use your work on the Euclidean algorithm as the basis for a repeated process of renaming and simplifying values.[5]
• Begin by numbering the steps of the Euclidean algorithm reduction, as reference points. Thus, you have the following steps:
• ${\displaystyle {\text{Step 1}}:87=(1*64)+23}$
• ${\displaystyle {\text{Step 2}}:64=(2*23)+18}$
• ${\displaystyle {\text{Step 3}}:23=(1*18)+5}$
• ${\displaystyle {\text{Step 4}}:18=(3*5)+3}$
• ${\displaystyle {\text{Step 5}}:5=(1*3)+2}$
• ${\displaystyle {\text{Step 6}}:3=(1*2)+1}$
• ${\displaystyle {\text{Step 7}}:2=(2*1)+0}$
2. 2
Begin with the last step that has a remainder. Rewrite that equation so the remainder stands alone, as equal to the rest of the information in the equation.[6]
• For this problem, Step 6 is the last one that showed a remainder. That remainder was 1. Rewrite the equation in Step 6 as follows:
• ${\displaystyle 1=3-(1*2)}$
3. 3
Isolate the remainder of the previous step. This procedure is a step-by-step process of moving “up” the steps. Each time, you will be revising the right side of the equation in terms of the numbers in the higher step.[7]
• You can revise Step 5 to isolate its remainder as:
• ${\displaystyle 2=5-(1*3)}$ or ${\displaystyle 2=5-3}$
4. 4
Perform a substitution and simplify. You should notice that your revision of Step 6 contains the number 2, and your revision of Step 5 is equal to 2. Substitute the equality in Step 5 into the place of the 2 in your Step 6 revision:[8]
• ${\displaystyle 1=3-(1*2)}$….. (This is the Step 6 revision.)
• ${\displaystyle 1=3-(5-3)}$….. (Substitute in place of the value 2.)
• ${\displaystyle 1=3-5+3}$….. (Distribution of the negative sign)
• ${\displaystyle 1=2(3)-5}$…..(Simplify)
5. 5
Repeat the process of substitution and simplification. Moving through the Euclidean algorithm steps in reverse, repeat the process. Each time, you will revise the previous step, and substitute its value into your latest result.[9]
• The last step was Step 5. Now revise Step 4 to isolate its remainder as:
• ${\displaystyle 3=18-(3*5)}$
• Substitute that value in place of the 3 in your latest simplification step and then simplify:
• ${\displaystyle 1=2(18-3*5)-5}$
• ${\displaystyle 1=2(18)-6(5)-5}$
• ${\displaystyle 1=2(18)-7(5)}$
6. 6
Continue repeating substitution and simplification. This process will repeat, step by step, until you reach the original step of the Euclidean algorithm. The purpose of this procedure is to wind up with an equation that will be written in terms of 87 and 64, which are the original coefficients of the problem you are trying to solve. Continuing in this manner, the remaining steps are as follows:[10]
• ${\displaystyle 1=2(18)-7(5)}$
• ${\displaystyle 1=2(18)-7(23-18)}$…..(Substitution from Step 3)
• ${\displaystyle 1=2(18)-7(23)+7(18)}$
• ${\displaystyle 1=9(18)-7(23)}$
• ${\displaystyle 1=9(64-2*23)-7(23)}$…..(Substitution from Step 2)
• ${\displaystyle 1=9(64)-18(23)-7(23)}$
• ${\displaystyle 1=9(64)-25(23)}$
• ${\displaystyle 1=9(64)-25(87-64)}$…..(Substitution from Step 1)
• ${\displaystyle 1=9(64)-25(87)+25(64)}$
• ${\displaystyle 1=34(64)-25(87)}$
7. 7
Rewrite the result in terms of the original coefficients. When you return to the first step of the Euclidean algorithm, you should notice that the resulting equation contains the two coefficients of the original problem. Rearrange the numbers so they align with the original equation.[11]
• In this case, the original problem you are trying to solve is ${\displaystyle 87x-64y=3}$. Thus, you can rearrange your last step to put the terms in that standard order. Pay particular attention to the 64 term. In the original problem, that term is subtracted, but the Euclidean algorithm treats it as a positive term. To account for the subtraction, you need to change the multiplier 34 to a negative. The final equation looks like this:
• ${\displaystyle 87(-25)-64(-34)=1}$
8. 8
Multiply by the necessary factor to find your solutions. Notice that the greatest common divisor for this problem was 1, so the solution that you reached is equal to 1. However, that is not the solution to the problem, since the original problem sets 87x-64y equal to 3. You need to multiply the terms of your last equation by 3 to get a solution:[12]
• ${\displaystyle 87(-25*3)-64(-34*3)=1*3}$
• ${\displaystyle 87(-75)-64(-102)=3}$
9. 9
Identify the integral solution to the equation. The values that must be multiplied by the coefficients are the x and y solutions to the equation.
• In this case, you can identify the solution as the coordinate pair ${\displaystyle (x,y)=(-75,-102)}$.
### Part 4 Finding Infinitely Many More Solutions
1. 1
Recognize that infinitely many solutions exist. If a linear equation has one integral solution, then it must have infinitely many integral solutions. Here is a brief algebraic statement of the proof:[13]
• ${\displaystyle Ax+By=C}$
• ${\displaystyle A(x+B)+B(y-A)=C}$ ….. (Adding a B to x while subtracting A from y results in the same solution.)
2. 2
Identify your original solution values for x and y. The pattern of infinite solutions begins with the single solution that you identified.[14]
• In this case, your solution is the coordinate pair ${\displaystyle (x,y)=(-75,-102)}$.
3. 3
Add the y-coefficient B to the x solution. To find a new solution for x, add the value of the coefficient of y.[15]
• In this problem, beginning with the solution x=-75, add the y coefficient of -64, as follows:
• ${\displaystyle x=-75+(-64)=-139}$
• Thus, a new solution for the original equation will have the x value of -139.
4. 4
Subtract the x-coefficient A from the y solution. To make the equation remain balanced, when you add to the x term, you must then subtract from the y term.
• For this problem, beginning with the solution y=-102, subtract the x coefficient of 87, as follows:
• ${\displaystyle y=-102-87=-189}$
• Thus, a new solution for the original equation will have the y coordinate of -189.
• The new ordered pair should be ${\displaystyle (x,y)=(-139,-189)}$.
5. 5
Check the solution. To verify that your new ordered pair is a solution to the equation, insert the values into the equation and see if it works.[16]
• ${\displaystyle 87x-64y=3}$
• ${\displaystyle 87(-139)-64(-189)=3}$
• ${\displaystyle 3=3}$
• Because the statement is true, the solution works.
6. 6
Write a general solution. The values for x will fit a pattern of the original solution, plus any multiple of the B coefficient. You can write this algebraically as follows:[17]
• x(k)=x+k(B), where x(k) represents the series of all x solutions, and x is the original x value that you solved.
• For this problem, you can say:
• ${\displaystyle x(k)=-75-64k}$
• y(k)=y-k(A), where y(k) represents the series of all y solutions, and y is the original y value that you solved.
• For this problem, you can say:
• ${\displaystyle y(k)=-102-87k}$
## Community Q&A
Search
• How do you solve the linear congruence of 28x=38(mod42)?
wikiHow Contributor
Introduce a second variable to convert the modular equation to an equivalent diophantine equarion. So 28x = 38 + 42y for some integers x and y. Simplify to 14 (2x - 3y) = 38. But 2x - 3y is an integer. The left side is always a multiple of 14, but 38 is not. So that equation has no solutions mod 42.
• Are all equations with integral coefficients diophantine equations?
wikiHow Contributor
No. Both ordinary and diophantine equations can have any type of integer or non-integer coefficients. Diophantine-ness refers to the domain of the variable(s) - it's those that have to be integers.
• How do I find solutions to word problems involving linear Diophantine equations?
wikiHow Contributor
Figure out what the question is asking. Cross out any irrelevant information, then put all the values into your equation.
200 characters left
If this question (or a similar one) is answered twice in this section, please click here to let us know.
## Article Info
Featured Article
Categories: Featured Articles | Algebra
In other languages:
Español: resolver una ecuación diofántica lineal, Русский: решить линейное диофантово уравнение, Italiano: Risolvere un’Equazione Diofantea Lineare, Português: Resolver Equações Diofantinas Lineares
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# Algebra 1 : How to find the slope of parallel lines
## Example Questions
← Previous 1 3
### Example Question #1 : How To Find The Slope Of Parallel Lines
What is the slope of a line parallel to the line described by 3x + 8y =16?
Explanation:
First, you should put the equation in slope intercept form (y = mx + b), where m is the slope.
Isolate the y term
3x + 8y – 3x = 16 – 3x
8y = 16 – 3x
Rearrange terms
8y = –3x +16
Divide both sides by 8
The slope of the line is -3/8. A parallel line will have the same slope, thus -3/8 is the correct answer.
### Example Question #2 : How To Find The Slope Of Parallel Lines
What is the slope of a line parallel to ?
Explanation:
When two lines are parallel, they have the same slope. With this in mind we take the slope of the first line which is and make it the slope of our parallel line.
If , then .
### Example Question #3 : How To Find The Slope Of Parallel Lines
What is the slope of a line that is parallel to ?
Explanation:
Parallel lines have identical slopes. To determine the slope of the given line, transform into the format, or . The slope of the given line is , so its parallel line must also be .
### Example Question #4 : How To Find The Slope Of Parallel Lines
Which equation described a line parallel to the line that connects points (–8,9) and (3,–4)?
Explanation:
In order for two lines to be parellel, their slopes have to be the same.
Find the slope of the line connecting those two points using the general slope formula, , where the points are and . In our case, the points are (–8,9) and (3,–4). The slope is calculated below.
Match this slope value with one of the possible choice of equations. The correct equation is because its slope is the same.
### Example Question #5 : How To Find The Slope Of Parallel Lines
Which of the following are NOT parallel to each other?
Explanation:
Four of the answers are not in slope-intercept form.
For the lines to be parallel, all must share the same slopes.
To identify the slopes, this is the term of:
The only equation that does not have a slope of is
.
### Example Question #6 : How To Find The Slope Of Parallel Lines
Lines A and B are parallel. Line A can be represented by the equation .
Find the slope of line B.
Explanation:
If two lines are parallel, then they have the same slope.
Line A is:
Rewrite this in slope-intercept form where is the slope:
The slope of line A is .
If the slope of line A is , then the slope of line B must be
### Example Question #7 : How To Find The Slope Of Parallel Lines
Find the slope of a line parallel to the line with the equation:
Explanation:
Lines can be written in the slope-intercept format:
In this format, equals the line's slope and represents where the line intercepts the y-axis.
In the given equation:
And it has a slope of:
Parallel lines share the same slope.
The parallel line has a slope of .
### Example Question #8 : How To Find The Slope Of Parallel Lines
Find the slope of a line parallel to the line with the equation:
Explanation:
Lines can be written in the slope-intercept format:
In this format, equals the line's slope and represents where the line intercepts the y-axis.
In the given equation:
And it has a slope of:
Parallel lines share the same slope.
The parallel line has a slope of .
### Example Question #701 : Functions And Lines
Find the slope of a line parallel to the line with the equation:
Explanation:
Lines can be written in the slope-intercept format:
In this format, equals the line's slope and represents where the line intercepts the y-axis.
In the given equation:
And it has a slope of:
Parallel lines share the same slope.
The parallel line has a slope of .
### Example Question #702 : Functions And Lines
Find the slope of a line parallel to the line with the equation:
Explanation:
Lines can be written in the slope-intercept format:
In this format, equals the line's slope and represents where the line intercepts the y-axis.
In the given equation:
And it has a slope of:
Parallel lines share the same slope.
The parallel line has a slope of .
← Previous 1 3
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# Weighted MeanWeighted Average Formula
The mean page showed how to work out the arithmetic mean/average of a set of values.
But there is also what is known as a "weighted mean" or "weighted average".
Which differs slightly from the arithmetic mean in regards to how we obtain the value.
There is a weighted average formula that can be used when necessary.
But first, it helps to have a recap of how to work out the arithmetic mean/average.
Arithmetic Mean Example
For a small group of 5 numbers:
3 , 5 , 7 , 9 , 6
The arithmetic mean/average is established by the numbers summed together, divided by the amount of numbers there is.
Here:
Mean/average = \bf{\frac{3\space+\space5\space+\space7\space+\space9\space+\space6}{5}} = \bf{\frac{30}{5}} = 6
## Weighted Mean
Consider the same list of 5 numbers again.
3 , 5 , 7 , 9 , 6
Each of these 5 numbers, has its own part in the whole group.
If we consider them all to have equal weight in the group,
then they each represent a {\frac{1}{5}} of the whole group on their own.
The sum we did for the arithmetic mean before, \bf{\frac{3\space+\space5\space+\space7\space+\space9\space+\space6}{5}}.
Is actually the same as \bf{\frac{3}{5}} + \bf{\frac{5}{5}} + \bf{\frac{7}{5}} + \bf{\frac{9}{5}} + \bf{\frac{6}{5}}.
=> \bf{\frac{3\space+\space5\space+\space7\space+\space9\space+\space6}{5}} = \bf{\frac{3}{5}} + \bf{\frac{5}{5}} + \bf{\frac{7}{5}} + \bf{\frac{9}{5}} + \bf{\frac{6}{5}} = 6
Dividing by 5, is the same as multiplying by 0.2. ( \bf{\frac{1}{5}} = 0.2 )
So instead of dividing each number by 5 and adding them up, as was done above.
We can multiply each number by 0.2,
and add them up to get the average.
( 3 × 0.2 ) + ( 5 × 0.2 ) + ( 7 × 0.2 ) + ( 9 × 0.2 ) + ( 6 × 0.2 )
= 0.6 + 1 + 1.4 + 1.8 + 1.2 = 6
This multiplication approach, is one way to work out the weighted mean/average.
With the weighted mean, the numbers in a group don't have to have the same weighting as each other.
Above, we had each number with an equal weight of 0.2 in the group.
But we can change that. As long as each number weighting still adds up to 1 all together.
As 1 represents the whole 100% of the complete group of numbers.
So with the same group of 5 numbers as before,
let 3 , 5 , 6 still have a weighting of 0.2 in the group.
But let 7 have a weighting of 0.3,
and 9 have a weighting of 0.1.
Now using the multiplication approach with these new weightings.
( 3 × 0.2 ) + ( 5 × 0.2 ) + ( 7 × 0.3 ) + ( 9 × 0.1 ) + ( 6 × 0.2 )
= 0.6 + 1 + 0.9 + 2.1 + 1.2 = 5.8
Here the weighted mean/average of 5.8 is slightly different from the arithmetic mean of 6.
This is often the case with a set of values, different averages can be larger than or smaller than each other.
Example
In a Science class, 3 different tests over a term, determines the overall mark a student will achieve for that term.
With each test being marked out of 100, one student has the following results:
TEST 1 => \bf{\frac{77}{100}} TEST 2 => \bf{\frac{62}{100}}
TEST 3 => \bf{\frac{81}{100}}
The Science teacher gives the tests different weighting when the overall term grade is awarded.
TEST 1 ( 20% ) TEST 2 ( 30% ) TEST 3 ( 50% )
20% = 0.2 30% = 0.3 50% = 0.5
Weighted Mean = ( 77 × 0.2 ) + ( 62 × 0.3 ) + ( 81 × 0.5 ) = 74.5
The students overall grade in the Science class for the 3 tests was 74.5%.
## Weighted Average Formula
Often the weighting of a different value is not given as a decimal percentage, like in the Science class example above with 0.20.3 and 0.5.
Instead a whole number weighting can be used, and this is where a specific weighted average formula can be handy to use.
We can look at the Science class example again with the 3 tests, same scores as before.
TEST 1 => \bf{\frac{77}{100}} TEST 2 => \bf{\frac{62}{100}} TEST 3 => \bf{\frac{81}{100}}
The teacher could give each separate test a whole number weighting, as opposed to the already seen decimal percentage, such as:
TEST 1 ( 2 ) TEST 2 ( 4 ) TEST 3 ( 5 )
All of this information can be used in the weighted average formula, which is:
x = a test result value.
w = weight of the test result value.
n = sample size of values.
Here we have 3 different test results, so n = 3.
{\frac{\sum_{i=1}^3\space(x_i\space\times\space w_i)}{\sum_{i=1}^3\space w_i}}
What the weighted average formula tells us for the students Science grade, is that we want to work out:
With the 3 Science Tests:
x1 = 77 , w1 = 2
x2 = 62 , w2 = 4
x3 = 81 , w3 = 5
Weighted Mean/Average = \bf{\frac{(77 \space \times \space 2) \space\space + \space\space (62\space \times \space 4) \space\space + \space\space (81 \space \times \space 5)}{2 \space\space + \space\space 4 \space\space + \space\space 5}}
= \bf{\frac{154 \space\space + \space\space 248 \space\space + \space\space 405}{11}} = 73.36
With this new weighting used by the teacher, the student actually achieved a higher average score over the 3 tests.
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# Conditional probability & distribution
Contents:
## What is conditional probability?
Conditional probability is the probability of an event occurring, given that another event has already occurred. It is used to determine the probability of one event happening when we have information about the occurrence of a related event. For example, what is the probability the temperature will dip below freezing today, given that there is a cold front on the way?
## Conditional probability notation and formula
Conditional probability is denoted as P(A|B), which is read as “the probability of event A happening given that event B has occurred.”
It is calculated by dividing the probability of the two events occurring together by the probability of the first event occurring. In mathematical terms, conditional probability can be calculated using the following formula:
P(A|B) = P(A ∩ B) / P(B)
where
• P(A|B) represents the conditional probability of event A occurring, given that event B has occurred.
• P(A ∩ B) denotes the probability of both events A and B happening together (also known as joint probability).
• P(B) is the probability of event B occurring.
Note that conditional probability is only meaningful when event B has a non-zero probability, i.e., P(B) > 0.
## Conditional probability examples
Example 1: the Coin Toss
Consider the probability of obtaining heads when flipping a coin, which is 0.5. The likelihood of getting heads twice consecutively is 0.25. If you already got heads on the first flip, the conditional probability of getting heads again on the second flip is calculated as 0.25 / 0.5 = 0.5.
This means that, given you obtained heads on the first flip, there is a 50% chance of getting heads on the subsequent flip.
In a group of 100 car buyers, 40 purchased alarm systems, 30 opted for vegan leather seats, and 20 bought both an alarm system and vegan leather seats. If we randomly select a car buyer who acquired an alarm system, what is the likelihood they also purchased vegan leather seats?
1. Determine P(A). This is given in the problem as 40%, or 0.4.
2. Identify P(A∩B). This represents the intersection of A and B, where both events occur together. In this case, 20 out of 100 buyers, or 0.2.
3. Plug your values into the formula: P(B|A) = P(A∩B) / P(A) = 0.2 / 0.4 = 0.5.
Thus, given that a buyer bought an alarm system, there is a 0.5 (50%) probability that they also purchased vegan leather seats.
## Real life examples of conditional probability
Conditional probability is used in various real-life situations to make informed decisions and predictions, especially when events are interrelated. Here are a few examples:
1. Medical Diagnosis: Doctors use conditional probability to determine the likelihood of a patient having a particular disease, given specific symptoms or test results. This helps them make more accurate diagnoses and treatment recommendations. For example, a woman aged between 40 and 50 years has roughly a 1% probability of breast cancer. However, if a woman receives a positive mammogram result, this probability alters: the chances of a woman having cancer with a positive mammogram result increase to roughly 8.3%. [1].
2. Weather Forecasting: Meteorologists use conditional probability to predict the chance of rain or other weather events, given certain atmospheric conditions, such as temperature and humidity.
3. Insurance: Insurance companies use conditional probability to calculate the risk of accidents or damage based on factors like age, driving history, and location. This information helps them set appropriate insurance premiums for their clients.
4. Finance: In the stock market, investors use conditional probability to estimate the likelihood of stock price fluctuations given specific economic indicators, such as interest rates or unemployment rates.
5. Marketing: Businesses use conditional probability to analyze customer behavior and preferences. For example, they might evaluate the probability of a customer purchasing a particular product, given that they have already purchased a related item. This information helps businesses develop targeted marketing strategies and product recommendations.
6. Quality Control: In manufacturing, conditional probability can be used to determine the likelihood of a product defect, given specific production parameters, such as machine settings or raw material quality. This information helps companies improve their production processes and maintain high-quality standards.
These examples demonstrate how conditional probability plays a critical role in various industries, helping professionals make better decisions based on the relationships between different events.
## Where does conditional probability come from?
The origins of conditional probability can be traced back to the 17th century when English mathematician Thomas Bayes devised a theorem that established the groundwork for probability theory. Bayes’ theorem asserts that the probability of an event happening, given that another event has already occurred, can be determined by dividing the joint probability of both events by the probability of the initial event.
In the 18th century, French mathematician Pierre-Simon Laplace formulated a method for computing conditional probabilities, which is still in use today. Laplace’s approach is centered around the concept of conditional expectation – the anticipated value of a random variable when another random variable has already been observed.
During the 19th century, German mathematician Carl Friedrich Gauss created a method for calculating conditional probabilities based on the least squares concept. Gauss’s technique remains widely used in various applications such as data analysis and machine learning.
In the 20th century, the advent of computers enabled the easier and faster computation of conditional probabilities. This development led to new applications for conditional probability in fields like weather forecasting, medical diagnosis, and risk assessment.
Currently, conditional probability is an integral element of probability theory and statistics, with applications spanning across weather forecasting, medical diagnosis, risk assessment, marketing, and game theory.
Key milestones in the history of conditional probability include:
• 1763: Thomas Bayes publishes his theorem on conditional probability.
• 1774: Pierre-Simon Laplace introduces a method for calculating conditional probabilities.
• 1809: Carl Friedrich Gauss develops a method for calculating conditional probabilities based on least squares.
• 20th century: The emergence of computers facilitates easier and quicker calculations of conditional probabilities.
• Present day: Conditional probability is a fundamental concept in probability theory and statistics, employed in a wide array of applications.
The conditional probability formula we use today stems from the multiplication rule: P(A and B) = P(A) * P(B|A).
Here’s a step-by-step breakdown of how to derive the conditional probability equation from the multiplication rule:
1. Write down the multiplication rule: P(A and B) = P(A) * P(B|A)
2. Divide both sides of the equation by P(A): P(A and B) / P(A) = [P(A) * P(B|A)] / P(A)
3. Cancel out P(A) on the right side of the equation: P(A and B) / P(A) = P(B|A)
4. Rearrange the equation: P(B|A) = P(A and B) / P(A)
## Conditional probability distribution
A conditional probability distribution represents the probability distribution of a random variable, determined based on the principles of conditional probability after observing the outcome of another related random variable.
Put simply, it refers to the likelihood of a specific event happening, provided that another event has already taken place. It can also be thought of as a probability distribution for a specific subgroup. Essentially, it outlines the likelihood of a randomly chosen individual from that subgroup possessing a particular attribute of interest.
For instance, imagine a biased coin with a 60% chance of landing on heads. Flipping the coin once yields a 0.6 probability of it landing on heads. If the coin is flipped twice and lands on heads both times, the probability of it landing on heads the third time remains 0.6 since the first two flips do not influence the third flip’s probability.
Now, consider a deck of cards where the probability of drawing a certain card is 1/52. If we draw a card and it is not the desired one, the probability of drawing the sought-after card in the next draw is still 1/52, as the first draw does not affect the second draw’s probability.
However, if we know that the first draw resulted in a red card, the probability of the second draw yielding a black card rises to 26/51 because there are now 26 black cards and 51 cards in total left in the deck.
In this case, the probability of drawing a black card on the second draw is conditional upon the fact that a red card was drawn on the first draw, forming a conditional probability distribution.
## Conditional probability distribution definition
More formally, a conditional probability distribution is defined as follows [3]: Let and Y be discrete random variables with joint probability mass function (PMF) given by p(x, y). Then the conditional PMF of X, given that Y = y, is denoted by pX|Y(x | y) and given as
## Conditional probability distribution applications
Conditional probability distributions have various applications, such as:
• Machine learning: They are used in machine learning algorithms for predicting future events.
• Finance: They help in determining the risk of investments.
• Healthcare: They aid in estimating the likelihood of a patient developing a specific disease.
Overall, conditional probability distributions serve as a useful tool for predicting future events and assessing the risk of particular outcomes.
## The Array Distribution
The array distribution refers to the conditional distribution of independent random variables X1 (given X2, … , Xn). It is often used as a synonym for a conditional distribution (e.g., Kotz et al. [4], Giri & Banerjee [5], Wadworth & Bryan [6]). You’re more likely to see the term “array distribution” in applied statistics; in theoretical statistics, the usual term is conditional distribution in theoretical statistics [7].
The Encyclopedia of Statistical Sciences [8] gives an example of an array distribution of X:
The variance of this distribution is called the array variance; if the variance doesn’t depend on X2, … ,Xn , the variation is homoscedastic [9]. Its variance-covariance matrix is called the array-variance-covariance matrix.
The array distribution is denoted by:
You can think of this as the density of y for a fixed value of x.
## References
[1] ICMA Photos, CC BY-SA 2.0 https://creativecommons.org/licenses/by-sa/2.0, via Wikimedia Commons
[3] LibreTexts Statistics. 5.3: Conditional Probability Distributions. Retrieved June 10, 2023 from: https://stats.libretexts.org/Courses/Saint_Mary’s_College_Notre_Dame/MATH_345__-Probability(Kuter)/5%3A_Probability_Distributions_for_Combinations_of_Random_Variables/5.3%3A_Conditional_Probability_Distributions
[4] Kotz, S. et al. (2000). Continuous Multivariate Distributions, Volume 1 Models and Applications. John Wiley & Sons.
[5] Giri, P. & Banerjee, J. (2021). Statistical Tools and Techniques. Academic Publishers.
[6] Wadsworth, G. & Bryan, J. (1960). Introduction to Probability and Random Variables. McGraw-Hill.
[7] Vidakovic , B. et al. (Eds.) (2005). Encyclopedia of Statistical Sciences, Volume 3. Wiley.
[8] Haight (ed.) (1963). Mathematical Theories of Traffic Flow. Elsevier Science
[9] Johnson, Kotz, and Balakrishnan, (1994), Continuous Univariate Distributions, Volumes I and II, 2nd. Ed., John Wiley and Sons.
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# Event planners are planning a fireworks display for the 4th of July. The fireworks will launch from a boat in the bay, and must land in a designated "fire-safe" zone 100 meters away. To be visible above the buildings, the fireworks must reach at least 30 meters high. Write an equation that could model the path of the fireworks.
The path of the fireworks will be in the form of a parabola, whose maxima is 30. We know that the vertex of a downward parabola gives maxima. So our parabola will be a downward opening parabola.
Are you familiar with the vertex form of a parabola?
The vertex form of a parabola whose vertex is at (h, k) is given by `y=a(x-h)^2 +k` . For a downward parabola, the value of a must be negative.
Let us assume that the boat is at origin (0,0).
The zeros of the parabola will be at (0,0) and (100,0) because the span of the parabola should be 100 meters.
We know that a parabola is symmetric along the vertical line that passes through the vertex. We can use this information to find the x-coordinate of the vertex, which will be the mid-point of 0 and 100.
We know that 50 is in the middle of 0 and 100, so the x-coordinate of the vertex is 50. We already know that the y-coordinate of the vertex is 30, because that is the height that the fireworks must reach. Therefore, the vertex is at point (50,30). I have attached an image of the path of the fireworks.
Using those values of h and k, we get:
`y=a(x-50)^2+30`
Our next step is to find the value of a. We have two points on the x-axis, and we can use either point to solve for a. Please go ahead and use the coordinates of the point (0,0) for x and y and solve for a before looking at the next step.
`0=a(0-50)^2+30`
`0=2500a+30`
`0-30=2500a+30-30`
`-30=2500a`
`(-30)/2500=a`
`(-3)/(250)=a`
Upon substituting the value of a in our equation, we get:
`y=(-3)/(250)(x-50)^2+30`
Therefore, the above equation will model the path of the fireworks.
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# Video: Pack 2 • Paper 1 • Question 7
Pack 2 • Paper 1 • Question 7
02:46
### Video Transcript
A student drew this for the graph of 𝑦 equals 𝑥 squared minus two. Identify one error that the student made.
Well, the first thing we’re gonna do to actually help solve this problem is work out some 𝑥 and 𝑦 values that should be there in the graph of 𝑦 equals 𝑥 squared minus two. So for the first value, I’ve got 𝑥 is equal to negative one.
Well, what we’re gonna do is actually substitute negative one for 𝑥 in our equation 𝑦 equals 𝑥 squared minus two. If we do that, we’re gonna get 𝑦 is equal to negative one all squared minus two, which is gonna give us 𝑦 is equal to one minus two. And that’s because if you have negative one squared, it’s negative one multiplied by negative one, which gives us a positive. So we’re gonna get 𝑦 is equal to one minus two. So therefore, 𝑦 is equal to negative one.
So now we actually move on to calculate the next pair of coordinates. We’ve got 𝑥 is equal to zero. Well, now for this one, what we actually do is we substitute 𝑥 is equal to zero into our equation. So we now have 𝑦 is equal to zero squared minus two. Well, zero squared is just zero. So therefore, we’re left with a 𝑦-coordinate of negative two.
So it’s great. We’ve got two points. I’m just gonna do one more cause then we can compare them fully. It’s gonna be the point where 𝑥 is equal to one. So therefore, when we do this, we’re gonna get 𝑦 is equal to one squared because we substitute in 𝑥 for one. So we’re gonna have 𝑦 is equal to one squared minus two. So therefore, 𝑦 is gonna be equal to negative one because if we have one squared, we get one. And one minus two is negative one.
Okay, great! So we’ve now found the three pairs of coordinates. So therefore, if we actually mark on our points, we’ve got one here at negative one, negative one; one at zero, negative two; and one at one, negative one. So we’re gonna take a look at the middle point. So we’ve got a point zero, negative two. And what we want to do in this question is actually work out one error the student has made. And we can actually see that the corresponding point will be this point here.
So therefore, we can see that on the drawing that should be, so if we were gonna draw the graph, the point should be at zero, negative two. However, we can see that, on the student drawing, it’s actually at negative two, zero. So therefore, we can say that one error that the student has made is the fact that the 𝑥- and the 𝑦-coordinates have actually been swapped.
And therefore just to double-check this and make sure that it’s correct, what we can actually do is look at some of the other points we’ve plotted. So we’ve got the point here at one, negative one. Well, if we look at the corresponding point on the student’s graph, we see one at negative one, one. And therefore, yes, that agrees that an error that the student has made is that the 𝑥- and 𝑦-coordinates have been swapped.
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# How to Calculate 4/10 Plus 6/10
Are you looking to work out and calculate how to add 4/10 plus 6/10? In this really simple guide, we'll teach you exactly what 4/10 + 6/10 is and walk you through the step-by-process of how to add two fractions together.
To start with, the number above the line in a fraction is called a numerator and the number below the line is called the denominator.
Why do you need to know this? Well, because the easiest way to add fractions together is to make sure both fractions have the same denominator.
Let's set up 4/10 and 6/10 side by side so they are easier to see:
4 / 10 + 6 / 10
In this calculation, some of the hard work has already been done for us because the denominator is the same in both fractions. All we need to do is add the numerators together and keep the denominator as it is:
4 + 6 / 10 = 10 / 10
You're done! You now know exactly how to calculate 4/10 + 6/10. Hopefully you understood the process and can use the same techniques to add other fractions together. The complete answer is below (simplified to the lowest form):
1
## Convert 4/10 plus 6/10 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:
10 / 10 = 1
### Cite, Link, or Reference This Page
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "How to Calculate 4/10 plus 6/10". VisualFractions.com. Accessed on December 9, 2022. http://visualfractions.com/calculator/adding-fractions/what-is-4-10-plus-6-10/.
• "How to Calculate 4/10 plus 6/10". VisualFractions.com, http://visualfractions.com/calculator/adding-fractions/what-is-4-10-plus-6-10/. Accessed 9 December, 2022.
• How to Calculate 4/10 plus 6/10. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/adding-fractions/what-is-4-10-plus-6-10/.
### Preset List of Fraction Addition Examples
Below are links to some preset calculations that are commonly searched for:
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# How Will Alex O’Loughlin Fare Today? (01/16/2020)
How will Alex O’Loughlin get by on 01/16/2020 and the days ahead? Let’s use astrology to undertake a simple analysis. Note this is for entertainment purposes only – don’t get too worked up about the result. I will first calculate the destiny number for Alex O’Loughlin, and then something similar to the life path number, which we will calculate for today (01/16/2020). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology people.
PATH NUMBER FOR 01/16/2020: We will analyze the month (01), the day (16) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. What does this entail? We will show you. First, for the month, we take the current month of 01 and add the digits together: 0 + 1 = 1 (super simple). Then do the day: from 16 we do 1 + 6 = 7. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 1 + 7 + 4 = 12. This still isn’t a single-digit number, so we will add its digits together again: 1 + 2 = 3. Now we have a single-digit number: 3 is the path number for 01/16/2020.
DESTINY NUMBER FOR Alex O’Loughlin: The destiny number will take the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Alex O’Loughlin we have the letters A (1), l (3), e (5), x (6), O (6), L (3), o (6), u (3), g (7), h (8), l (3), i (9) and n (5). Adding all of that up (yes, this can get tiring) gives 65. This still isn’t a single-digit number, so we will add its digits together again: 6 + 5 = 11. This still isn’t a single-digit number, so we will add its digits together again: 1 + 1 = 2. Now we have a single-digit number: 2 is the destiny number for Alex O’Loughlin.
CONCLUSION: The difference between the path number for today (3) and destiny number for Alex O’Loughlin (2) is 1. That is less than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t go jumping for joy yet! As mentioned earlier, this is not scientifically verified. If you want really means something, check out your cosmic energy profile here. Go see what it says for you now – you may be absolutely amazed. It only takes 1 minute.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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Sequences
Sequences
Definition: A Sequence $\{ a_n \}_{n=1}^{\infty}$ is an ordered list of numbers $a_1, a_2, ...$ where $a_j$ denotes the $j^{\mathrm{th}}$ term of the sequence.
One of the most famous sequences is the Fibonacci Sequence that is recursively defined by $f_{1} = 1$, $f_{2} = 1$, and $f_{n} = f_{n-1} + f_{n-2}$ for $n \in \mathrm{N}$ and $n ≥ 3$. In other words, each number of the Fibonacci sequence is obtained by taking the preceding two terms of the Fibonacci sequence and summing them. The first few terms of the Fibonacci sequence are $\{ 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ... \}$.
Another example of a sequence is when $a_n = \frac{n}{n+1}$. We can calculate the $n^{\mathrm{th}}$ term of this sequence by plugging in some $n \in \mathbb{N}$, for example, the 4th term of this sequence is $a_4 = \frac{4}{4 + 1} = \frac{4}{5}$, and thus $\left \{ a_n \right \} = \left \{ \frac{n}{n + 1} \right \}_{n=1}^{\infty} = \left \{ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, ... \right \}$.
One other great example of a sequence is when $a_n = (-1)^n$ and $\{ (-1)^n \}_{n=1}^{\infty} = \{ -1, 1, -1, 1, -1, ... \}$. This might not be the most interesting sequence as there are only two distinct terms in it, but alas, it definitely is a sequence.
We will now look at some important definitions of sequences, many of which are relatively straightforward.
Definition: The sequence $\{ a_n \}$ is said to be Bounded between $m$ and $M$ where $m ≤ M$ if $\forall n \in \mathbb{N}$, $m ≤ a_n ≤ M$. The sequence is said to be Bounded Above if $\forall n$, $a_n ≤ M$, and Bounded Below if $\forall n$, $m ≤ a_n$.
Definition: The sequence $\{ a_n \}$ is said to be Strictly Increasing denoted $\{ a_n \} \uparrow$ if $a_{n+1} > a_{n}$ and for every term in the sequence and is said to be Strictly Decreasing denoted $\{ a_n \} \downarrow$ if $a_{n+1} < a_{n}$ for every term in the sequence. If a sequence is increasing or decreasing, we classify the sequence as Monotonic.
Definition: The sequence $\{ a_n \}$ is said to be Ultimately Increasing if $a_{n+1} > a_{n}$ whenever $n ≥ N$, and is said to be **Ultimately Decreasing if $a_{n+1} < a_{n}$ whenever $n ≥ N$.
We note that the definitions of a sequence being strictly increasing/decreasing different from a sequence being ultimately increasing/decreasing. A sequence that is increasing/decreasing is increasing/decreasing for ALL terms in the sequence. A sequence that is ultimately increasing/decreasing need not increase/decrease for a finite number of terms, but after some $N^{\mathrm{th}}$ term it must indefinitely increase/decrease.
Definition: The sequence $\{ a_n \}$ is said to be Positive if $a_n > 0$ for all $n \in N$ and the sequence is said to be Negative if $a_n < 0$.
Definition: The sequence $\{ a_n \}$ is said to be Ultimately Positive if $a_n > 0$ whenever $n ≥ N$ and the sequence is said to be Ultimately Negative if $a_n < 0$ whenever $n ≥ N$.
Like the distinction we made above regarding increasing/decreasing sequences and ultimately increasing/decreasing sequences, the same applies here. A sequence that is positive/negative is positive/negative for ALL of its terms. A sequence that is ultimately positive/negative need not always be one or the other for a finite number of terms, but after the $N^{\mathrm{th}}$ term, all the terms in the sequence must be positive/negative.
Example 1
Consider the sequence $\{ \frac{2}{2}, \frac{4}{8}, \frac{8}{26}, \frac{16}{80}, ... \}$. Find a formula for the $n^{\mathrm{th}}$ term of this sequence. Determine whether this set is bounded.
We note that the numerators of this sequence are $2^1, 2^2, 2^3, 2^4, ...$, while the denominators seem to be $3^1 - 1, 3^2 - 1, 3^3 - 1, 3^4 - 1, ...$. Therefore a general formula for the $n^{\mathrm{th}}$ term is:
(1)
\begin{align} a_n = \frac{2^n}{3^n - 1} \end{align}
Since $\forall n \in \mathbb{N}$, $2^n ≤ 3^n - 1$ so the fraction $\frac{2^n}{3^n - 1} ≤ 1$ and thus our sequence is bounded above by $M = 1$, that is $a_n ≤ 1$.
Example 2
Consider the sequence $\{ \frac{1}{1}, \frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{3}}, \frac{-1}{2}, ... \}$. Find a formula for the $n^{\mathrm{th}}$ term of this sequence.
We note that the numerators of this sequence are $1, -1, 1, -1, ...$, while the denominators seem to be the squareroot of successive natural numbers. Therefore a general formula for the $n^{\mathrm{th}}$ term is:
(2)
\begin{align} a_n = \frac{-(-1)^n}{\sqrt{n}} \end{align}
Sequences as Functions
We can represent a sequence in a variety of ways such as on a number line. For example, consider the sequence in example 1 plotted on a number line:
Alternatively, we could represent the sequence as a function whose domain $A := \{ n : n \in \mathbb{N} \}$. For example, the following graph illustrates example 1:
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# Vectors
This is a vector:
A vector has magnitude (size) and direction:
The length of the line shows its magnitude and the arrowhead points in the direction.
We can add two vectors by joining them head-to-tail:
And it doesn't matter which order we add them, we get the same result:
### Example: A plane is flying along, pointing North, but there is a wind coming from the North-West.
The two vectors (the velocity caused by the propeller, and the velocity of the wind) result in a slightly slower ground speed heading a little East of North.
If you watched the plane from the ground it would seem to be slipping sideways a little.
Have you ever seen that happen? Maybe you have seen birds struggling against a strong wind that seem to fly sideways. Vectors help explain that.
Velocity, acceleration, force and many other things are vectors.
## Subtracting
We can also subtract one vector from another:
• first we reverse the direction of the vector we want to subtract,
• then add them as usual:
ab
## Notation
A vector is often written in bold, like a or b.
A vector can also be written as the letters of its head and tail with an arrow above it, like this:
## Calculations
Now ... how do we do the calculations?
The most common way is to first break up vectors into x and y parts, like this:
The vector a is broken up into
the two vectors ax and ay
(We see later how to do this.)
## Adding Vectors
We can then add vectors by adding the x parts and adding the y parts:
The vector (8,13) and the vector (26,7) add up to the vector (34,20)
### Example: add the vectors a = (8,13) and b = (26,7)
c = a + b
c = (8,13) + (26,7) = (8+26,13+7) = (34,20)
When we break up a vector like that, each part is called a component.
## Subtracting Vectors
To subtract, first reverse the vector we want to subtract, then add.
### Example: subtract k = (4,5) from v = (12,2)
a = v + −k
a = (12,2) + −(4,5) = (12,2) + (−4,−5) = (12−4,2−5) = (8,−3)
## Magnitude of a Vector
The magnitude of a vector is shown by two vertical bars on either side of the vector:
|a|
OR it can be written with double vertical bars (so as not to confuse it with absolute value):
||a||
We use Pythagoras' theorem to calculate it:
|a| = √( x2 + y2 )
### Example: what is the magnitude of the vector b = (6,8) ?
|b| = √( 62 + 82 ) = √( 36+64 ) = √100 = 10
A vector with magnitude 1 is called a Unit Vector.
## Vector vs Scalar
A scalar has magnitude (size) only.
Scalar: just a number (like 7 or −0.32) ... definitely not a vector.
A vector has magnitude and direction, and is often written in bold, so we know it is not a scalar:
• so c is a vector, it has magnitude and direction
• but c is just a value, like 3 or 12.4
## Multiplying a Vector by a Scalar
When we multiply a vector by a scalar it is called "scaling" a vector, because we change how big or small the vector is.
### Example: multiply the vector m = (7,3) by the scalar 3
a = 3m = (3×7,3×3) = (21,9)
It still points in the same direction, but is 3 times longer
(And now you know why numbers are called "scalars", because they "scale" the vector up or down.)
## Multiplying a Vector by a Vector (Dot Product and Cross Product)
How do we multiply two vectors together? There is more than one way! The scalar or Dot Product (the result is a scalar). The vector or Cross Product (the result is a vector). (Read those pages for more details.)
## More Than 2 Dimensions
Vectors also work perfectly well in 3 or more dimensions:
The vector (1,4,5)
### Example: add the vectors a = (3,7,4) and b = (2,9,11)
c = a + b
c = (3,7,4) + (2,9,11) = (3+2,7+9,4+11) = (5,16,15)
### Example: what is the magnitude of the vector w = (1,−2,3) ?
|w| = √( 12 + (−2)2 + 32 ) = √( 1+4+9 ) = √14
Here is an example with 4 dimensions (but it is hard to draw!):
### Example: subtract (1,2,3,4) from (3,3,3,3)
(3,3,3,3) + −(1,2,3,4)
= (3,3,3,3) + (−1,−2,−3,−4)
= (3−1,3−2,3−3,3−4)
= (2,1,0,−1)
## Magnitude and Direction
We may know a vector's magnitude and direction, but want its x and y lengths (or vice versa):
<=> Vector a in Polar Coordinates Vector a in Cartesian Coordinates
You can read how to convert them at Polar and Cartesian Coordinates, but here is a quick summary:
From Polar Coordinates (r,θ) From Cartesian Coordinates (x,y) to Cartesian Coordinates (x,y) to Polar Coordinates (r,θ) x = r × cos( θ ) y = r × sin( θ ) r = √ ( x2 + y2 ) θ = tan-1 ( y / x )
## An Example
Sam and Alex are pulling a box.
• Sam pulls with 200 Newtons of force at 60°
• Alex pulls with 120 Newtons of force at 45° as shown
What is the combined force, and its direction?
Let us add the two vectors head to tail:
First convert from polar to Cartesian (to 2 decimals):
Sam's Vector:
• x = r × cos( θ ) = 200 × cos(60°) = 200 × 0.5 = 100
• y = r × sin( θ ) = 200 × sin(60°) = 200 × 0.8660 = 173.21
Alex's Vector:
• x = r × cos( θ ) = 120 × cos(-45°) = 120 × 0.7071 = 84.85
• y = r × sin( θ ) = 120 × sin(-45°) = 120 × -0.7071 = −84.85
Now we have:
Add them:
(100, 173.21) + (84.85, −84.85) = (184.85, 88.36)
That answer is valid, but let's convert back to polar as the question was in polar:
• r = √ ( x2 + y2 ) = √ ( 184.852 + 88.362 ) = 204.88
• θ = tan-1 ( y / x ) = tan-1 ( 88.36 / 184.85 ) = 25.5°
And we have this (rounded) result:
And it looks like this for Sam and Alex:
They might get a better result if they were shoulder-to-shoulder!
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# Divisibility Rule by 3
If the sum of all digits of a given umber is multiple of 3 or is exactly divisible by 3 then we can say that the given number is divisible by 3.
Lets take an example to understand this point more clearly.
Example - Is 543 divisible by 3?
Solution -
Let's sum the all digits i.e. 5 + 4 + 3 = 12.
Now we all know that 12 is divisible by 3 right? Hence 543 is divisible 3.
Example - Is 2861 divisible by 3?
Let's sum the all digits i.e. 2 + 8 + 6 + 1 = 17.
Now we all know that 17 is not divisible by 3 right? Hence 2861 is not divisible 3.
Point to be noted: It can be further added too.
Example - Is 845217 divisible by 3?
Let's sum the all digits i.e.
8 + 4 + 5 + 2 + 1 + 7 = 27 = 2 + 7 = 9
Now we all know that 9 is divisible by 3 right? Hence 845217 is divisible 3.
So
to add further is more easier to know whether the number is divisible by 3 or not.
Do it by your self
Which of the following numbers are divisible by 3?
78451, 65844, 5694, 698754, 95356
Solution
78451 -7 + 8 + 4 + 5 + 1 = 25 = 2 + 5 = 7 hence not divisible by 3
65844 - 6 + 5 + 8 + 4 + 4 = 27 = 2 + 7 = 9 hence divisible by 3
5694 - 5 + 6 + 9 + 4 = 24 = 2 + 4 = 6 hence divisible by 3
698754 - 6 + 9 + 8 + 7 + 5 + 4 = 39 = 3 + 9 = 12 hence divisible by 3
95356 - 9 + 5 + 3 + 5 + 6 = 28 = 2 + 8 = 10 hence not divisible by 3
Related Article
Divisibility Rule by 2
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# How do you solve 2( 9t - 1) + 3= 13?
Jun 10, 2017
See a solution process below:
#### Explanation:
First, subtract $\textcolor{red}{3}$ from each side of the equation to isolate the term with parenthesis while keeping the equation balanced:
$2 \left(9 t - 1\right) + 3 - \textcolor{red}{3} = 13 - \textcolor{red}{3}$
$2 \left(9 t - 1\right) + 0 = 10$
$2 \left(9 t - 1\right) = 10$
Next, divide each side of the equation by $\textcolor{red}{2}$ to isolate the parenthesis while keeping the equation balanced:
$\frac{2 \left(9 t - 1\right)}{\textcolor{red}{2}} = \frac{10}{\textcolor{red}{2}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \left(9 t - 1\right)}{\cancel{\textcolor{red}{2}}} = 5$
$9 t - 1 = 5$
Then, add $\textcolor{red}{1}$ to each side of the equation to isolate the $t$ term while keeping the equation balanced:
$9 t - 1 + \textcolor{red}{1} = 5 + \textcolor{red}{1}$
$9 t - 0 = 6$
$9 t = 6$
Now, divide each side of the equation by $\textcolor{red}{9}$ to solve for $t$ while keeping the equation balanced:
$\frac{9 t}{\textcolor{red}{9}} = \frac{6}{\textcolor{red}{9}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{9}}} t}{\cancel{\textcolor{red}{9}}} = \frac{3 \times 2}{\textcolor{red}{3 \times 3}}$
$t = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \times 2}{\textcolor{red}{\textcolor{b l a c k}{\cancel{\textcolor{red}{3}}} \times 3}}$
$t = \frac{2}{3}$
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# Domain and Range of a Function
A step by step tutorial, with detailed solutions, on how to find the domain and range of real valued functions is presented. First the definitions of these two concepts are presented. A table of domain and range of basic functions might be useful to answer the questions below.
## Definition of the Domain of a Function
For a function f defined by an expression with variable x, the implied domain of f is the set of all real numbers variable x can take such that the expression defining the function is real. The domain can also be given explicitly.
also
Step by Step Calculator to Find Domain of a Function
## Definition of the Range of a Function
The range of f is the set of all values that the function takes when x takes values in the domain.
Also a
Step by Step Calculator to Find Range of a Function is included in this website.
## Examples with Detailed Solutions
### Example 1
Find the domain of function f defined by
Solution to Example 1
x can take any real number except 1 since x = 1 would make the denominator equal to zero and the division by zero is not allowed in mathematics. Hence the domain in interval notation is given by the set
(- ∞ , 1) U (1 , + ∞)
### Matched Problem 1
Find the domain of function f defined by
Answers to matched problems 1,2,3 and 4
### Example 2
Find the domain of function f defined by
Solution to Example 2
The expression defining function f contains a square root. The expression under the radical has to satisfy the condition
2x - 8 >= 0
for the function to take real values.
Solve the above linear inequality
x >= 4
The domain, in interval notation, is given by
[4 , +∞)
### Matched Problem 2
Find the domain of function f defined by:
### Example 3
Find the domain of function f defined by:
Solution to Example 3
The expression defining function f contains a square root. The expression under the radical has to satisfy the condition
-x ≥ 0
Which is equivalent to
x ≤ 0
The denominator must not be zero, hence x not equal to 3 and x not equal to -5.
The domain of f is given by
(-∞ , - 5) ∪ ( - 5 , 0]
### Matched Problem 3
Find the domain of function f defined by:
### Example 4
Find the range of function f defined by:
Solution to Example 4
The domain of this function is the set of all real numbers. The range is the set of values that f(x) takes as x varies. If x is a real number, x2 is either positive or zero. Hence we can write the following:
x
2 ≥ 0
Subtract - 2 to both sides to obtain
x
2 - 2 ≥ - 2
The last inequality indicates that x2 - 2 takes all values greater that or equal to - 2. The range of f is given by
[ -2 , +∞)
A graph of f also helps in interpreting the range of a function. Below is shown the graph of function f given above. Note the lowest point in the graph has a y (= f (x) ) value of - 2.
### Matched Problem 4
Find the range of function f defined by:
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# How do you simplify x^2-5x+3(x^2+7x)?
Apr 18, 2018
Distribute and combine like terms.
#### Explanation:
Distribute the 3 among the terms inside parentheses:
${x}^{2} - 5 x + 3 \left({x}^{2} + 7 x\right) = {x}^{2} - 5 x + 3 {x}^{2} + 21 x$
Combine like terms.
${x}^{2} - 5 x + 3 {x}^{2} + 21 x = 4 {x}^{2} + 16 x$
$= 4 {x}^{2} + 16 x$
Apr 18, 2018
$x = 0 , - 4$
#### Explanation:
Simplify, then factor
${x}^{2} - 5 x + 3 {x}^{2} + 21 x$
$4 {x}^{2} + 16 x$
${x}^{2} + 4 x$
$x \left(x + 4\right)$
$x = 0 , x + 4 = 0$
$\therefore$
$x = 0 , - 4$
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## Series Solutions: Hermite's Equation
Hermite's Equation of order k has the form
y''-2ty'+2ky=0,
where k is usually a non-negative integer.
We know from the previous section that this equation will have series solutions which both converge and solve the differential equation everywhere.
Hermite's Equation is our first example of a differential equation, which has a polynomial solution.
As usual, the generic form of a power series is
We have to determine the right choice for the coefficients (an).
As in other techniques for solving differential equations, once we have a "guess" for the solutions, we plug it into the differential equation. Recall that
and
Plugging this information into the differential equation we obtain:
or after rewriting slightly:
Next we shift the first summation up by two units:
Before we can combine the terms into one sum, we have to overcome another slight obstacle: the second summation starts at n=1, while the other two start at n=0.
Evaluate the 0th term for the second sum: . Consequently, we do not change the value of the second summation, if we start at n=0 instead of n=1:
Thus we can combine all three sums as follows:
Therefore our recurrence relations become:
After simplification, this becomes
Let us look at the special case, where k=5, and the initial conditions are given as . In this case, all even coefficients will be equal to zero, since a0=0 and each coefficient is a multiple of its second predecessor.
What about the odd coefficients? a1=1, consequently
and
Since a7=0, all odd coefficients from now on will be equal to zero, since each coefficient is a multiple of its second predecessor.
Consequently, the solution has only 3 non-zero coefficients, and hence is a polynomial. This polynomial
(or a multiple of this polynomial) is called the Hermite Polynomial of order 5.
It turns out that the Hermite Equation of positive integer order k always has a polynomial solution of order k. We can even be more precise: If k is odd, the initial value problem will have a polynomial solution, while for k even, the initial value problem will have a polynomial solution.
#### Exercise 1:
Find the Hermite Polynomials of order 1 and 3.
#### Exercise 2:
Find the Hermite Polynomials of order 2, 4 and 6.
#### Exercise 3:
Consider the Hermite Equation of order 5:
y''-2ty'+10y=0.
Find the solution satisfying the initial conditions a0=1, a1=0.
[Back] [Next]
[Algebra] [Trigonometry] [Complex Variables]
[Calculus] [Differential Equations] [Matrix Algebra]
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Imaginary and Complex Numbers
Complex numbers
Real and imaginary parts
The complex plane
The set of all complex numbers C
Absolute value, modulus of a complex number
Complex conjugates
Addition and subtraction of complex numbers
Multiplication and division of complex numbers
Complex numbers
A complex number is the sum of a real number and an imaginary number.
A complex number z is written in the form of z = x + yi, where x and y are real numbers, and
The real number x is called the real part of the complex number, and the real number y is the imaginary part.
The real part of z is denoted Re(z) = and the imaginary part is denoted Im(z) = y.
Hence, an imaginary number is a complex number whose real part is zero, while real numbers may be considered to be complex numbers with an imaginary part of zero.
That is, the real number x is equivalent to the complex number x + 0i.
Equality of complex numbers
Two complex numbers are equal if their real parts are equal and their imaginary parts are equal.
The complex plane
Complex numbers are represented by points or position vectors in the coordinate plane called the complex plane (or the Gauss plane). Where, the x-axis is called the real axis and the y-axis is called the imaginary axis.
The representation of a complex number by Cartesian coordinates is called the rectangular form or algebraic form of the complex number.
The standard symbol for the set of all complex numbers is C
Absolute value, modulus of a complex number
The absolute value of a complex number z is defined as the distance from z to the origin in the complex
plane, i.e.,
Complex conjugates
The complex conjugate of the complex number z = x + yi is x - yi that has the same real part x, but differ in the sign of the imaginary part.
That is, the conjugate is the reflection of z about the real axis, as is shown in the above figure.
Addition and subtraction of complex numbers
To add or subtract two complex numbers z1 = a + bi and z2 = c + di, we add or subtract the real parts and the imaginary parts.
Addition: z1 + z2 = (a + bi) + (c + di) = (a + c) + (b + d)i
Subtraction: z1 - z2 = (a + bi) - (c + di) = (a - c) + (b - d)i
Examples: Given are complex numbers, z1 = -3 + 2i and z2 = 4 + 3i, find z1 + z2 and z1 - z2.
Solutions: z1 + z2 = (-3 + 2i) + (4 + 3i) = (-3 + 4) + (2 + 3)i = 1 + 5i
and z1 - z2 = (4 + 3i) - (1 + 5i) = (4 - 1) + (3 - 5)i = 3 - 2i
Given addition and subtraction are shown in the complex plane in the figures below.
z1 + z2 = (-3 + 2i) + (4 + 3i) = 1 + 5i z1 - z2 = (4 + 3i) - (1 + 5i) = 3 - 2i
Multiplication and division of complex numbers
Multiplication: z1· z2 = (a + bi) · (c + di) = ac + bci + adi + bdi2 = (ac - bd) + (ad + bc)i
Division:
Examples: Given are complex numbers, z1 = -3 + 2i and z2 = 4 + 3i, find z1 · z2 and z1 / z2.
Solutions: z1 · z2 = (-3 + 2i) · (4 + 3i) = -3 · 4 + 2 · 4i + (-3) · 3i + 2 · 3 i2 = -18 - i
and
Example: For what real number a the real part of the complex number equals 1.
Solution:
Example: Evaluate the expression where z = 1 - i.
Solution:
Pre-calculus contents A
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# A function f(x) is defined as,
Question:
A function f(x) is defined as,
$f(x)=\left\{\begin{array}{rr}\frac{x^{2}-x-6}{x-3} ; & \text { if } \quad x \neq 3 \\ 5 & ; \text { if } x=3\end{array}\right.$
Show that f(x) is continuous that x = 3.
Solution:
Given:
$f(x)=\left\{\begin{array}{l}\frac{x^{2}-x-6}{x-3}, x \neq 3 \\ 5, \quad x=3\end{array}\right.$
We observe
$(\mathrm{LHL}$ at $x=3)=\lim _{x \rightarrow 3} f(x)=\lim _{h \rightarrow 0} f(3-h)$
$=\lim _{h \rightarrow 0} \frac{(3-h)^{2}-(3-h)-6}{(3-h)-3}=\lim _{h \rightarrow 0} \frac{9+h^{2}-6 h-3+h-6}{-h}=\lim _{h \rightarrow 0} \frac{h^{2}-5 h}{-h}=\lim _{h \rightarrow 0}(5-h)=5$
And, $(\mathrm{RHL}$ at $x=3)=\lim _{x \rightarrow 3^{+}} f(x)=\lim _{h \rightarrow 0} f(3+h)$
$=\lim _{h \rightarrow 0} \frac{(3+h)^{2}-(3+h)-6}{(3+h)-3}=\lim _{h \rightarrow 0} \frac{9+h^{2}+6 h-3-h-6}{h}=\lim _{h \rightarrow 0} \frac{h^{2}+5 h}{h}=\lim _{h \rightarrow 0}(5+h)=5$
Also, $f(3)=5$
$\therefore \lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{-}} f(x)=f(3)$
Hence, $f(x)$ is continuous at $x=3$.
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Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 18 Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are provided here with simple step-by-step explanations. These solutions for Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are extremely popular among Class 10 students for Maths Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.
#### Question 1:
If the mean of 5 observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x. [CBSE 2014]
Mean of given observations =
Hence, the value of x is 7.
#### Question 2:
If the mean of 25 observations is 27 and each observation is decreased by 7, what will be the new mean?
Mean of given observations =
Mean of 25 observations = 27
∴ Sum of 25 observations = 27 × 25 = 675
If 7 is subtracted from every number, then the sum = 675 − (25 × 7)
= 675 − 175
= 500
Then, new mean = $\frac{500}{25}=20$
Thus, the new mean will be 20.
#### Question 3:
Compute the mean of the following frequency distribution:
Class 10 − 30 30 − 50 50 − 70 70 − 90 90 − 110 Frequency 15 18 25 10 2
Here, h = 20
Let the assumed mean, A be 60.
Class Mid-Values(xi) Frequency (fi) ${\mathbit{d}}_{\mathit{i}}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{60}$ ${\mathbit{u}}_{\mathit{i}}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{60}}{\mathbf{20}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 10−30 20 15 −40 −2 −30 30−50 40 18 −20 −1 −18 50−70 60 25 0 0 0 70−90 80 10 20 1 10 90−110 100 2 40 2 4 $\underset{}{\sum {f}_{i}=70}$ $\underset{}{\sum {f}_{i}{u}_{i}=-34}$
We know
Mean $=A+h×\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)$
∴ Mean of the given frequency distribution
Thus, the mean of the given frequency distribution is approximately 50.29.
#### Question 4:
Compute the mean of the following frequency distribution:
Class 0 − 20 20 − 40 40 − 60 60 − 80 80 − 100 100 – 120 120 – 140 Frequency 6 8 10 12 6 5 3
Here, h = 20
Let the assumed mean, A be 70.
Class Mid-Values(xi) Frequency (fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{70}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{70}}{\mathbf{20}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 0−20 10 6 −60 −3 −18 20−40 30 8 −40 −2 −16 40−60 50 10 −20 −1 −10 60−80 70 12 0 0 0 80−100 90 6 20 1 6 100−120 110 5 40 2 10 120−140 130 3 60 3 9 $\underset{}{\sum {f}_{i}=50}$ $\underset{}{\sum {f}_{i}{u}_{i}=-19}$
We know
Mean $=A+h×\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)$
∴ Mean of the given frequency distribution
$=70+20×\left(\frac{-19}{50}\right)\phantom{\rule{0ex}{0ex}}=70-7.6\phantom{\rule{0ex}{0ex}}=62.4$
Thus, the mean of the given frequency distribution is 62.4.
#### Question 5:
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Number of days 0 − 6 6 − 12 12 − 18 18 − 24 24 − 30 30 – 36 36 – 42 Number of students 10 11 7 4 4 3 1
Here, h = 6
Let the assumed mean, A be 21.
Number of days Mid-Values(xi) Number of Students (fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{21}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{21}}{\mathbf{6}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 0−6 3 10 −18 −3 −30 6−12 9 11 −12 −2 −22 12−18 15 7 −6 −1 −7 18−24 21 4 0 0 0 24−30 27 4 6 1 4 30−36 33 3 12 2 6 36−42 39 1 18 3 3 $\underset{}{\sum {f}_{i}=40}$ $\underset{}{\sum {f}_{i}{u}_{i}=-46}$
We know
Mean $=A+h×\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)$
∴ Mean number of days a student was absent
$=21+6×\left(\frac{-46}{40}\right)\phantom{\rule{0ex}{0ex}}=21-6.9\phantom{\rule{0ex}{0ex}}=14.1$
Thus, the mean number of days a student was absent is 14.1.
#### Question 6:
The daily expenses of 40 families has been shown by the following frequency distribution:
Expenses (in ₹) 500 − 700 700 − 900 900 − 1100 1100 − 1300 1300 − 1500 Number of families 6 8 10 9 7
Find the mean daily expenses.
Here, h = 200
Let the assumed mean, A be ₹1000.
Expenses (in ₹) Mid-Values(xi) Number of Families(fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{1000}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{1000}}{\mathbf{200}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 500−700 600 6 −400 −2 −12 700−900 800 8 −200 −1 −8 900−1100 1000 10 0 0 0 1100−1300 1200 9 200 1 9 1300−1500 1400 7 400 2 14 $\underset{}{\sum {f}_{i}=40}$ $\underset{}{\sum {f}_{i}{u}_{i}=3}$
We know
Mean $=A+h×\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)$
∴ Mean daily expenses
$=1000+200×\left(\frac{3}{40}\right)\phantom{\rule{0ex}{0ex}}=1000+15\phantom{\rule{0ex}{0ex}}=₹1015$
Thus, the mean daily expenses is ₹1015.
#### Question 7:
During a medical checkup, the number of heartbeats per minute of 40 patients were recorded and summarised as follows:
Number of heartbeats per minute 64 − 68 68 − 72 72 − 76 76 − 80 80 − 84 84 – 88 Number of patients 6 8 10 12 3 1
Find the mean heartbeats per minute for these patients.
Here, h = 4
Let the assumed mean, A be 78.
Number of Heartbeats per Minute Mid-Values(xi) Number of Patients (fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{78}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{78}}{\mathbf{4}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 64−68 66 6 −12 −3 −18 68−72 70 8 −8 −2 −16 72−76 74 10 −4 −1 −10 76−80 78 12 0 0 0 80−84 82 3 4 1 3 84−88 86 1 8 2 2 $\underset{}{\sum {f}_{i}=40}$ $\underset{}{\sum {f}_{i}{u}_{i}=-39}$
We know
Mean $=A+h×\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)$
∴ Mean heartbeats per minute for these patients
$=78+4×\left(\frac{-39}{40}\right)\phantom{\rule{0ex}{0ex}}=78-3.9\phantom{\rule{0ex}{0ex}}=74.1$
Thus, the mean heartbeats per minute for these patients is 74.1.
#### Question 8:
The table given below shows the age distribution of 1000 persons who visited a marketing centre on a Sunday.
Age (in years) 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 – 60 60 – 70 Number of persons 105 222 220 138 102 113 100
Find the mean age of the persons visiting the marketing centre on that day.
Here, h = 10
Let the assumed mean, A be 35.
Age (in years) Mid-Values(xi) Number of persons (fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{35}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{35}}{\mathbf{10}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 0−10 5 105 −30 −3 −315 10−20 15 222 −20 −2 −444 20−30 25 220 −10 −1 −220 30−40 35 138 0 0 0 40−50 45 102 10 1 102 50−60 55 113 20 2 226 60−70 65 100 30 3 300 $\underset{}{\sum {f}_{i}=1000}$ $\underset{}{\sum {f}_{i}{u}_{i}=-351}$
We know
Mean $=A+h×\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)$
∴ Mean age of the persons visiting the marketing centre on that day
$=35+10×\left(\frac{-351}{1000}\right)\phantom{\rule{0ex}{0ex}}=35-3.51\phantom{\rule{0ex}{0ex}}=31.49$
Thus, the mean age of the persons visiting the marketing centre on that day 31.49 years.
#### Question 9:
The arithmetic mean of the following frequency distribution is 53. Find the value of x.
Class 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 Frequency 12 15 32 x 13
Class Mid-Values(xi) Frequency (fi) fi xi 0−20 10 12 120 20−40 30 15 450 40−60 50 32 1600 60−80 70 x 70x 80−100 90 13 1170 $\underset{}{\sum {f}_{i}=72+x}$ $\underset{}{\sum {f}_{i}{x}_{i}=3340+70x}$
Mean of the given frequency distribution = 53
We know
Mean $=\frac{\sum _{}{f}_{i}{x}_{i}}{\sum _{}{f}_{i}}$
$\therefore \frac{3340+70x}{72+x}=53\phantom{\rule{0ex}{0ex}}⇒3340+70x=3816+53x\phantom{\rule{0ex}{0ex}}⇒70x-53x=3816-3340$
$⇒17x=476\phantom{\rule{0ex}{0ex}}⇒x=\frac{476}{17}=28$
Thus, the value of x is 28.
#### Question 10:
The following distribution shows the daily pocket allowance of children of a locality. If the mean pocket allowance is ₹18, find the missing frequency f.
Daily pocket allowance (in ₹) 11−13 13−15 15−17 17−19 19−21 21−23 23−25 Number of children 7 6 9 13 f 5 4
The given data is shown as follows:
Daily pocket allowance (in ₹) Number of children (fi) Class mark (xi) fixi 11−13 7 12 84 13−15 6 14 84 15−17 9 16 144 17−19 13 18 234 19−21 f 20 20f 21−23 5 22 110 23−25 4 24 96 Total ∑ fi = 44 + f ∑ fixi = 752 + 20f
The mean of given data is given by
$\overline{x}=\frac{\sum _{i}{f}_{i}{x}_{i}}{\sum _{i}{f}_{i}}\phantom{\rule{0ex}{0ex}}⇒18=\frac{752+20f}{44+f}\phantom{\rule{0ex}{0ex}}⇒18\left(44+f\right)=752+20f\phantom{\rule{0ex}{0ex}}⇒792+18f=752+20f\phantom{\rule{0ex}{0ex}}⇒20f-18f=792-752\phantom{\rule{0ex}{0ex}}⇒2f=40\phantom{\rule{0ex}{0ex}}⇒f=20$
Hence, the value of f is 20.
#### Question 11:
If the mean of the following frequency distribution is 54, find the value of p. [CBSE 2006C]
Class 0−20 20−40 40−60 60−80 80−100 Frequency 7 p 10 9 13
The given data is shown as follows:
Class Frequency (fi) Class mark (xi) fixi 0−20 7 10 70 20−40 p 30 30p 40−60 10 50 500 60−80 9 70 630 80−100 13 90 1170 Total ∑ fi = 39 + p ∑ fixi = 2370 + 30p
The mean of given data is given by
$\overline{x}=\frac{\sum _{i}{f}_{i}{x}_{i}}{\sum _{i}{f}_{i}}\phantom{\rule{0ex}{0ex}}⇒54=\frac{2370+30p}{39+p}\phantom{\rule{0ex}{0ex}}⇒54\left(39+p\right)=2370+30p\phantom{\rule{0ex}{0ex}}⇒2106+54p=2370+30p\phantom{\rule{0ex}{0ex}}⇒54p-30p=2370-2106\phantom{\rule{0ex}{0ex}}⇒24p=264\phantom{\rule{0ex}{0ex}}⇒p=11$
Thus, the value of is 11.
#### Question 12:
The mean of the following data is 42, find the missing frequencies and y if the sum of the frequencies is 100.
Class interval 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Frequency 7 10 x 13 y 10 14 9
The given data is shown as follows:
Class interval Frequency (fi) Class mark (xi) fixi 0−10 7 5 35 10−20 10 15 150 20−30 x 25 25x 30−40 13 35 455 40−50 y 45 45y 50−60 10 55 550 60−70 14 65 910 70−80 9 75 675 Total ∑ fi = 63 + x + y ∑ fixi = 2775 + 25x + 45y
Sum of the frequencies = 100
Now, The mean of given data is given by
If x = 12, then y = 37 − 12 = 25
Thus, the value of is 12 and y is 25.
#### Question 13:
The daily expenditure of 100 families are given below. Calculate f1 and f2 if the mean daily expenditure is ₹188.
Expenditure (in ₹) 140−160 160−180 180−200 200−220 220−240 Number of families 5 25 f1 f2 5
The given data is shown as follows:
Expenditure (in ₹) Number of families (fi) Class mark (xi) fixi 140−160 5 150 750 160−180 25 170 4250 180−200 f1 190 190f1 200−220 f2 210 210f2 220−240 5 230 1150 Total ∑ fi = 35 + f1 + f2 ∑ fixi = 6150 + 190f1 + 210f2
Sum of the frequencies = 100
Now, The mean of given data is given by
If f1 = 50, then f2 = 65 − 50 = 15
Thus, the value of f1 is 50 and f2 is 15.
#### Question 14:
The mean of the following frequency distribution is 57.6 and the sum of the observations is 50
Class 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 7 f1 12 f2 8 5
Class Frequency $\left({f}_{i}\right)$ Mid values $\left({x}_{i}\right)$ $\left({f}_{i}×{x}_{i}\right)$ 0-20 7 10 70 20-40 f1 30 30 f1 40-60 12 50 600 60-80 18- f1 70 1260-70 f1 80-100 8 90 720 100-120 5 110 550 $\sum {f}_{i}=50$ $\sum \left({f}_{i}×{x}_{i}\right)=3200-40{f}_{1}$
#### Question 15:
If the mean of the following frequency distribution is 188, find the missing frequencies x and y, if the sum of all frequencies is 100.
Class 0 – 80 80 – 160 160 – 240 240 – 320 320 – 400 Frequency 20 25 x y 10
Here, h = 80
Let the assumed mean, A be 200.
Class Mid-Values(xi) Frequency(fi) ${\mathbit{d}}_{i}\mathbf{=}{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{200}$ ${\mathbit{u}}_{i}\mathbf{=}\frac{{\mathbit{x}}_{i}\mathbf{-}\mathbf{200}}{\mathbf{80}}$ ${\mathbit{f}}_{\mathit{i}}{\mathbit{u}}_{\mathit{i}}$ 0−80 40 20 −160 −2 −40 80−160 120 25 −80 −1 −25 160−240 200 x 0 0 0 240−320 280 y 80 1 y 320−400 360 10 160 2 20 $\underset{}{\sum {f}_{i}=x+y+55}$ $\underset{}{\sum {f}_{i}{u}_{i}=-45+y}$
$\underset{}{\sum {f}_{i}=x+y+55}=100$ (Given)
We know
Mean $=A+h×\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)$
$\therefore 200+80×\left(\frac{-45+y}{100}\right)=188$ (Given)
$⇒\frac{-45+y}{100}=\frac{188-200}{80}=-0.15\phantom{\rule{0ex}{0ex}}⇒-45+y=-15\phantom{\rule{0ex}{0ex}}⇒y=45-15=30$
Substituting the value of y in (1), we get
x + 30 = 45
x = 45 − 30 = 15
Thus, the missing frequencies x and y are 15 and 30, respectively.
#### Question 16:
The following table gives the literacy rate (in percentage) in 40 cities. Find the mean literacy rate, choosing a suitable method.
Literacy rate (%) 45−55 55−65 65−75 75−85 85−95 Number of cities 4 11 12 9 4
Using Direct method, the given data is shown as follows:
Literacy rate (%) Number of cities (fi) Class mark (xi) fixi 45−55 4 50 200 55−65 11 60 660 65−75 12 70 840 75−85 9 80 720 85−95 4 90 360 Total ∑ fi = 40 ∑ fixi = 2780
The mean of given data is given by
Thus, the mean literacy rate is 69.5%.
#### Question 17:
Find the mean, using assumed-mean method:
Marks 0-10 10-20 20-30 30-40 40-50 50-60 Number of students 12 18 27 20 17 6
Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ Deviation $\left({d}_{i}\right)$ ${d}_{i}=\left({x}_{i}-25\right)$ $\left({f}_{i}×{d}_{i}\right)$ 0-10 12 5 -20 -240 10-20 18 15 -10 -180 20-30 27 25=A 0 0 30-40 20 35 10 200 40-50 17 45 20 340 50-60 6 55 30 180 $\sum {f}_{i}=100$ $\sum \left({f}_{i}×{d}_{i}\right)=300$
#### Question 18:
Find the mean, using assumed-mean method:
Class 100-120 120-140 140-160 160-180 180-200 Frequency 10 20 30 15 5
Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ Deviation $\left({d}_{i}\right)$ ${d}_{i}=\left({x}_{i}-150\right)$ $\left({f}_{i}×{d}_{i}\right)$ 100-120 10 110 -40 -400 120-140 20 130 -20 -400 140-160 30 150=A 0 0 160-180 15 170 20 300 180-200 5 190 40 200 $\sum {f}_{i}=80$ $\sum \left({f}_{i}×{d}_{i}\right)=-\text{3}00$
#### Question 19:
Find the mean, using assumed-mean method:
Class 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 20 35 52 44 38 31
Class Frequency $\left({f}_{i}\right)$ Mid Values$\left({x}_{i}\right)$ Deviation $\left({d}_{i}\right)$ ${d}_{i}=\left({x}_{i}-50\right)$ $\left({f}_{i}×{d}_{i}\right)$ 0-20 20 10 -40 -800 20-40 35 30 -20 -700 40-60 52 50=A 0 0 60-80 44 70 20 880 80-100 38 90 40 1520 100-120 31 110 60 1860 $\sum {f}_{i}=220$ $\sum \left({f}_{i}×{d}_{i}\right)=2760$
#### Question 20:
Find the mean of the following frequency distribution using step-deviation method.
Class 0−10 10−20 20−30 30−40 40−50 Frequency 7 10 15 8 10
Let us choose a = 25, h = 10, then di = xi − 25 and ui$\frac{{x}_{i}-25}{10}$.
Using Step-deviation method, the given data is shown as follows:
Class Frequency (fi) Class mark (xi) di = xi − 25 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{25}}{\mathbf{10}}$ fiui 0−10 7 5 −20 −2 −14 10−20 10 15 −10 −1 −10 20−30 15 25 0 0 0 30−40 8 35 10 1 8 40−50 10 45 20 2 20 Total ∑ fi = 50 ∑ fiui = 4
The mean of given data is given by
Thus, the mean is 25.8.
#### Question 21:
Find the mean of the following data, using step-deviation method.
Class 5−15 15−25 25−35 35−45 45−55 55−65 65−75 Frequency 6 10 16 15 24 8 7
Let us choose a = 40, h = 10, then di = xi − 40 and ui = $\frac{{x}_{i}-40}{10}$.
Using Step-deviation method, the given data is shown as follows:
Class Frequency (fi) Class mark (xi) di = xi − 40 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{40}}{\mathbf{10}}$ fiui 5−15 6 10 −30 −3 −18 15−25 10 20 −20 −2 −20 25−35 16 30 −10 −1 −16 35−45 15 40 0 0 0 45−55 24 50 10 1 24 55−65 8 60 20 2 16 65−75 7 70 30 3 21 Total ∑ fi = 86 ∑ fiui = 7
The mean of given data is given by
Thus, the mean is 40.81.
#### Question 22:
The weights of tea in 70 packets are shown in the following table:
Weight (in grams) 200−201 201−202 202−203 203−204 204−205 205−206 Number of packets 13 27 18 10 1 1
Find the mean weight of packets using step-deviation method. [CBSE 2013]
Let us choose a = 202.5, h = 1, then di = xi − 202.5 and ui = $\frac{{x}_{i}-202.5}{1}$.
Using Step-deviation method, the given data is shown as follows:
Weight (in grams) Number of packets (fi) Class mark (xi) di = xi − 202.5 ui = $\frac{{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{202}\mathbf{.}\mathbf{5}}{\mathbf{1}}$ fiui 200−201 13 200.5 −2 −2 −26 201−202 27 201.5 −1 −1 −27 202−203 18 202.5 0 0 0 203−204 10 203.5 1 1 10 204−205 1 204.5 2 2 2 205−206 1 205.5 3 3 3 Total ∑ fi = 70 ∑ fiui = −38
The mean of given data is given by
Hence, the mean is 201.96 g.
#### Question 23:
Find the mean of the following frequency distribution using a suitable method.
Class 20−30 30−40 40−50 50−60 60−70 Frequency 25 40 42 33 10
Let us choose a = 45, h = 10, then di = xi − 45 and ui = $\frac{{x}_{i}-45}{10}$.
Using Step-deviation method, the given data is shown as follows:
Class Frequency (fi) Class mark (xi) di = xi − 45 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{45}}{\mathbf{10}}$ fiui 20−30 25 25 −20 −2 −50 30−40 40 35 −10 −1 −40 40−50 42 45 0 0 0 50−60 33 55 10 1 33 60−70 10 65 20 2 20 Total ∑ fi = 150 ∑ fiui = −37
The mean of given data is given by
Thus, the mean is 42.534.
#### Question 24:
Find the arithmetic mean of each of the following frequency distributions using step-deviation method:
Age (in years) 18-24 24-30 30-36 36-42 42-48 48-54 Number of workers 6 8 12 8 4 2
Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-33\right)}{6}$ $\left({f}_{i}×{u}_{i}\right)$ 18-24 6 21 −2 −12 24-30 8 27 −1 −8 30-36 12 33 = A 0 0 36-42 8 39 1 8 42-48 4 45 2 8 48-54 2 51 3 6 $\sum {f}_{i}=40$ $\sum \left({f}_{i}×{u}_{i}\right)=2$
#### Question 25:
Find the arithmetic mean of each of the following frequency distributions using step-deviation method:
Class 500-520 520-540 540-560 560-580 580-600 600-620 Frequency 14 9 5 4 3 5
Class Frequency$\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-550\right)}{20}$ $\left({f}_{i}×{u}_{i}\right)$ 500-520 14 510 −2 −28 520-540 9 530 −1 −9 540-560 5 550 = A 0 0 560-580 4 570 1 4 580-600 3 590 2 6 600-620 5 610 3 15 $\sum {f}_{i}=40$ $\sum \left({f}_{i}×{u}_{i}\right)=-12$
#### Question 26:
Find the mean age from the following frequency distribution:
Age (in years) 25-29 30-34 35-39 40-44 45-49 50-54 55-59 No. of person 4 14 22 16 6 5 3
Converting the series into exclusive form, we get:
Class Frequency$\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-42\right)}{5}$ $\left({f}_{i}×{u}_{i}\right)$ 24.5-29.5 4 27 −3 −12 29.5-34.5 14 32 −2 −28 34.5-39.5 22 37 −1 −22 39.5-44.5 16 42 = A 0 0 44.5-49.5 6 47 1 6 49.5-54.5 5 52 2 10 54.5-59.5 3 57 3 9 $\sum {f}_{i}=70$ $\sum \left({f}_{i}×{u}_{i}\right)=-37$
#### Question 27:
The following table shows the age distribution of patients of malaria in a village during a particular month.
Age (in years) 5-14 15-24 25-34 35-44 45-54 55-64 No. of cases 6 11 21 23 14 5
Find the average age of the patients.
Converting the series into exclusive form, we get:
Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-29.5\right)}{10}$ $\left({f}_{i}×{u}_{i}\right)$ 4.5-14.5 6 9.5 −2 −12 14.5-24.5 11 19.5 −1 −11 24.5-34.5 21 29.5 = A 0 0 34.5-44.5 23 39.5 1 23 44.5-54.5 14 49.5 2 28 54.5-64.5 5 59.5 3 15 $\sum {f}_{i}=80$ $\sum \left({f}_{i}×{u}_{i}\right)=43$
#### Question 28:
Weight of 60 eggs were recorded as given below:
Weight (in grams) 75−79 80−84 85−89 90−94 95−99 100−104 105−109 Number of eggs 4 9 13 17 12 3 2
Calculate their mean weight to the nearest gram.
Let us choose a = 92, h = 5, then di = xi − 92 and ui = $\frac{{x}_{i}-92}{5}$.
Using Step-deviation method, the given data is shown as follows:
Weight (in grams) Number of eggs (fi) Class mark (xi) di = xi − 92 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{92}}{\mathbf{5}}$ fiui 74.5−79.5 4 77 −15 −3 −12 79.5−84.5 9 82 −10 −2 −18 84.5−89.5 13 87 −5 −1 −13 89.5−94.5 17 92 0 0 0 94.5−99.5 12 97 5 1 12 99.5−104.5 3 102 10 2 6 104.5−109.5 2 107 15 3 6 Total ∑ fi = 60 ∑ fiui = −19
The mean of given data is given by
Thus, the mean weight to the nearest gram is 90 g.
#### Question 29:
The following table shows the marks scored by 80 students in an examination:
Marks Less than 5 Less than 10 Less than 15 Less than 20 Less than 25 Less than 30 Less than 35 Less than 40 Number of students 3 10 25 49 65 73 78 80
Calculate the mean marks correct to 2 decimal places.
Let us choose a = 17.5, h = 5, then di = xi − 17.5 and ui = $\frac{{x}_{i}-17.5}{5}$.
Using Step-deviation method, the given data is shown as follows:
Marks Number of students (cf) Frequency (fi) Class mark (xi) di = xi − 17.5 ui = $\frac{{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{17}\mathbf{.}\mathbf{5}}{\mathbf{5}}$ fiui 0−5 3 3 2.5 −15 −3 −9 5−10 10 7 7.5 −10 −2 −14 10−15 25 15 12.5 −5 −1 −15 15−20 49 24 17.5 0 0 0 20−25 65 16 22.5 5 1 16 25−30 73 8 27.5 10 2 16 30−35 78 5 32.5 15 3 15 35−40 80 2 37.5 20 4 8 Total ∑ fi = 80 ∑ fiui = 17
The mean of given data is given by
Thus, the mean marks correct to 2 decimal places is 18.56.
#### Question 1:
In a hospital, the ages of diabetic patients were recorded as follows. Find the median age. [CBSE 2014]
Age (in years) 0−15 15−30 30−45 45−60 60−75 Number of patients 5 20 40 50 25
We prepare the cumulative frequency table, as shown below:
Age (in years) Number of patients (fi) Cumulative Frequency (cf) 0−15 5 5 15−30 20 25 30−45 40 65 45−60 50 115 60−75 25 140 Total N = ∑ fi = 140
Now, N = 140 $⇒\frac{N}{2}=70$.
The cumulative frequency just greater than 70 is 115 and the corresponding class is 45−60.
Thus, the median class is 45−60.
∴ l = 45, h = 15, f = 50, N = 140 and cf = 65.
Now,
Hence, the median age is 46.5 years.
#### Question 2:
Compute the median from the following data:
Marks 0-7 7-14 14-21 21-28 28-35 35-42 42-49 Number of students 3 4 7 11 0 16 9
Class Frequency (f) Cumulative frequency 0-7 3 3 7-14 4 7 14-21 7 14 21-28 11 25 28-35 0 25 35-42 16 41 42-49 9 50 N=∑f=50
#### Question 3:
The following table shows the daily wages of workers in a factory:
Daily wages (in Rs) 0-100 100-200 200-300 300-400 400-500 Number of workers 40 32 48 22 8
Find the median daily wage income of the workers.
Class Frequency(f) Cumulative frequency 0-100 40 40 100-200 32 72 200-300 48 120 300-400 22 142 400-500 8 150 N=∑f=150
#### Question 4:
Calculate the median from the following frequency distribution:
Class 5-10 10-15 15-20 20-25 20-30 30-35 35-40 40-45 Frequency 5 6 15 10 5 4 2 2
Class Frequency(f) Cumulative frequency 5-10 5 5 10-15 6 11 15-20 15 26 20-25 10 36 25-30 5 41 30-35 4 45 35-40 2 47 40-45 2 49 N=∑f=49
#### Question 5:
Given below is the number of units of electricity consumed in a week in a certain locality:
Consumption (in units) 65-85 85-105 105-125 125-145 145-165 165-185 195-205 Number of consumers 4 5 13 20 14 7 4
Calculate the median
Class Frequency(f) Cumulative frequency 65-85 4 4 85-105 5 9 105-125 13 22 125-145 20 42 145-165 14 56 165-185 7 63 185-205 4 67 N=∑f=67
#### Question 6:
Calculate the median from the following data:
Height (in cm) 135-140 140-145 145-150 150-155 155-160 160-165 165-170 170-175 No. of boys 6 10 18 22 20 15 6 3
Class Frequency(f) Cumulative frequency 135-140 6 6 140=145 10 16 145-150 18 34 150-155 22 56 155-160 20 76 160-165 15 91 165-170 6 97 170-175 3 100 N=∑f=100
#### Question 7:
Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.
Class 0-10 10-20 20-30 30-40 40-50 Frequency 5 25 ? 18 7
Class Frequency (fi) c.f 0-10 5 5 10-20 25 30 20-30 x x+30 30-40 18 x+48 40-50 7 x+55
#### Question 8:
The median of the following data is 16. Find the missing frequencies a and b if the total of frequencies is 70.
Class 0−5 5−10 10−15 15−20 20−25 25−30 30−35 35−40 Frequency 12 a 12 15 b 6 6 4
We prepare the cumulative frequency table, as shown below:
Class Frequency (fi) Cumulative frequency (cf) 0−5 12 12 5−10 a 12 + a 10−15 12 24 + a 15−20 15 39 + a 20−25 b 39 + a + b 25−30 6 45 + a + b 30−35 6 51 + a + b 35−40 4 55 + a + b Total N = ∑fi = 70
Let a and b be the missing frequencies of class intervals 5−10 and 20−25 respectively. Then,
55 a + = 70 ⇒ a + = 15 ...(1)
Median is 16, which lies in 15−20. So, the median class is 15−20.
l = 15, h = 5, N = 70, f = 15 and cf = 24 + a
Now,
$\mathrm{Median}=l+\left(\frac{\frac{N}{2}-cf}{f}\right)×h\phantom{\rule{0ex}{0ex}}⇒16=15+\left(\frac{\frac{70}{2}-\left(24+a\right)}{15}\right)×5\phantom{\rule{0ex}{0ex}}⇒16=15+\left(\frac{35-24-a}{3}\right)\phantom{\rule{0ex}{0ex}}⇒16=15+\left(\frac{11-a}{3}\right)\phantom{\rule{0ex}{0ex}}⇒16-15=\frac{11-a}{3}\phantom{\rule{0ex}{0ex}}⇒1×3=11-a\phantom{\rule{0ex}{0ex}}⇒a=11-3\phantom{\rule{0ex}{0ex}}⇒a=8$
b = 15 a [From (1)]
b = 15 − 8
b = 7
Hence, a = 8 and b = 7.
#### Question 9:
In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000. Find the missing frequencies x and y.
Runs scored 2500−3500 3500−4500 4500−5500 5500−6500 6500−7500 7500−8500 Number of batsmen 5 x y 12 6 2
We prepare the cumulative frequency table, as shown below:
Runs scored Number of batsmen (fi) Cumulative frequency (cf) 2500−3500 5 5 3500−4500 x 5 + x 4500−5500 y 5 + x + y 5500−6500 12 17 + x + y 6500−7500 6 23 + x + y 7500−8500 2 25 + x + y Total N = ∑fi = 60
Let x and y be the missing frequencies of class intervals 3500−4500 and 4500−5500 respectively. Then,
25 x + = 60 ⇒ x + = 35 ...(1)
Median is 5000, which lies in 4500−5500. So, the median class is 4500−5500.
∴ l = 4500, h = 1000, N = 60, f = y and cf = 5 + x
Now,
∴ y = 35 − x [From (1)]
⇒ y = 35 − 15
⇒ y = 20
Hence, x = 15 and y = 20.
#### Question 10:
If the median of the following frequency distribution is 32.5, find the values of f1 and f2.
Class interval 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Total Frequency f1 5 9 12 f2 3 2 40
Class Frequency(f) Cumulative frequency 0-10 ${f}_{1}$ ${f}_{1}$ 10-20 5 ${f}_{1}$+5 20-30 9 ${f}_{1}$+14 30-40 12 ${f}_{1}$+26 40-50 ${f}_{2}$ ${f}_{1}+{f}_{2}$+26 50-60 3 ${f}_{1}+{f}_{2}$+29 60-70 2 ${f}_{1}+{f}_{2}$+31 N=∑f=40
#### Question 11:
Calculate the median for the following data:
Age (in years) 19-25 26-32 33-39 40-46 47-53 54-60 Frequency 35 96 68 102 35 4
First, we will convert the data into exclusive form.
Class Frequency(f) Cumulative frequency 18.5-25.5 35 35 25.5-32.5 96 131 32.5-39.5 68 199 39.5-46.5 102 301 46.5-53.5 35 336 53.5-60.5 4 340 N=∑f=340
#### Question 12:
Find the median wages for the following frequencies distribution:
Wages per day (in Rs) 61-70 71-80 81-90 91-100 101-110 111-120 No. of women workers 5 15 20 30 20 8
Converting the given data into exclusive form, we get:
Class Frequency(f) Cumulative frequency 60.5-70.5 5 5 70.5-80.5 15 20 80.5-90.5 20 40 90.5-100.5 30 70 100.5-110.5 20 90 110.5-120.5 8 98 N=∑f=98
#### Question 13:
Find the median from the following data:
Class 1-5 6-10 11-15 16-20 21-25 26-30 31-35 35-40 41-45 Frequency 7 10 16 32 24 16 11 5 2
Converting into exclusive form, we get:
Class Frequency(f) Cumulative frequency 0.5-5.5 7 7 5.5-10.5 10 17 10.5-15.5 16 33 15.5-20.5 32 65 20.5-25.5 24 89 25.5-30.5 16 105 30.5-35.5 11 116 35.5-40.5 5 121 40.5-45.5 2 123 N=∑f=123
#### Question 14:
Find the median from the following data:
Marks No. of students Below 10 12 Below 20 32 Below 30 57 Below 40 80 Below 50 92 Below 60 116 Below 70 164 Below 80 200
Class Cumulative frequency Frequency (f) 0-10 12 12 10-20 32 20 20-30 57 25 30-40 80 23 40-50 92 12 50-60 116 24 60-70 164 48 70-80 200 36 N=∑f=200
#### Question 1:
Find the mode of the following frequency distribution: [CBSE 2014]
Marks 10−20 20−30 30−40 40−50 50−60 Frequency 12 35 45 25 13
Here the maximum class frequency is 45, and the class corresponding to this frequency is 30−40. So, the modal class is 30−40.
Now,
modal class = 30−40, lower limit (l) of modal class = 30, class size (h) = 10,
frequency (f1) of the modal class = 45,
frequency (f0) of class preceding the modal class = 35,
frequency (
f2) of class succeeding the modal class = 25
Now, let us substitute these values in the formula:
Hence, the mode is 33.33.
#### Question 2:
Compute the mode of the following data: [CBSE 2013]
Class 0−20 20−40 40−60 60−80 80−100 Frequency 25 16 28 20 5
Here the maximum class frequency is 28, and the class corresponding to this frequency is 40−60. So, the modal class is 40−60.
Now,
Modal class = 40−60, lower limit (l) of modal class = 40, class size (h) = 20,
frequency (f1) of the modal class = 28,
frequency (f0) of class preceding the modal class = 16,
frequency (f2
) of class succeeding the modal class = 20.
Now, let us substitute these values in the formula:
Hence, the mode is 52.
#### Question 3:
Heights of students of Class X are given in the foloowing frequency distribution:
Height (in cm) 150−155 155−160 160−165 165−170 170−175 Number of students 15 8 20 12 5
Find the modal height.
Also, find the mean height. Compare and interpret the two measures of central tendency. [CBSE 2014]
Here the maximum class frequency is 20, and the class corresponding to this frequency is 160−165. So, the modal class is 160−165.
Now,
Modal class =
160−165, lower limit (l) of modal class = 160, class size (h) = 5,
frequency (f1) of the modal class = 20,
frequency (f0) of class preceding the modal class = 8,
frequency (f2) of class succeeding the modal class = 12.
Now, let us substitute these values in the formula:
Hence, the mode is 163.
It represents that the height of maximum number of students is 163 cm.
Now, to find the mean let us put the data in the table given below:
Height (in cm) Number of students (fi) Class mark (xi) fixi 150−155 15 152.5 2287.5 155−160 8 157.5 1260 160−165 20 162.5 3250 165−170 12 167.5 2010 170−175 5 172.5 862.5 Total ∑fi = 60 ∑fixi = 9670
Thus, mean of the given data is 161.17.
It represents that on an average, the height of a student is 161.17 cm.
#### Question 4:
Find the mode of the following distribution:
Class interval 10-14 14-18 18-22 22-26 26-30 30-34 34-38 38-42 Frequency 8 6 11 20 25 22 10 4
As the class 26-30 has the maximum frequency, it is the modal class.
#### Question 5:
Given below is the distribution of total household expenditure of 200 manual workers in a city:
Expenditure (in Rs) No. of manual workers 1000-1500 24 1500-2000 40 2000-2500 31 2500-3000 28 3000-3500 32 3500-4000 23 4000-4500 17 4500-5000 5
Find the expenditure done by maximum number of manual workers.
As the class 1500-2000 has the maximum frequency, it is the modal class.
Hence, mode = Rs 1820
#### Question 6:
Calculate the mode from the following data:
Monthly salary (in Rs) No. of employees 0-5000 90 5000-1000 150 10000-15000 100 15000-20000 80 20000-25000 70 25000-30000 10
As the class 5000-10000 has the maximum frequency, it is the modal class.
Hence, mode = Rs 7727.27
#### Question 7:
Compute the mode from the following data:
Age (in years) 0-5 5-10 10-15 15-20 20-25 25-30 30-35 Number of patients 6 11 18 24 17 13 5
As the class 15-20 has the maximum frequency, it is the modal class.
Hence, mode=17.3 years
#### Question 8:
Compute the mode from the following series:
Size 45-55 55-65 65-75 75-85 85-95 95-105 105-115 Frequency 7 12 17 30 32 6 10
As the class 85-95 has the maximum frequency, it is the modal class.
Hence, mode=85.71
#### Question 9:
Compute the mode from the following data:
Class interval 1-5 6-10 11-15 16-20 21-25 26-30 31-35 36-40 41-45 46-50 Frequency 3 8 13 18 28 20 13 8 6 4
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Princess Erika Vegafria
2022-10-04
determine the following functions (v/g)x
v= x² + 5x+4 g= x² + 2x-8
alenahelenash
To determine the value of the function $\left(v/g\right)x$, we need to substitute the given expressions for v and g into the function and simplify the result.
Given:
$v={x}^{2}+5x+4$
$g={x}^{2}+2x-8$
Substituting these expressions into $\left(v/g\right)x$, we get:
$\left(v/g\right)x=\frac{v}{g}=\frac{{x}^{2}+5x+4}{{x}^{2}+2x-8}$
To simplify this expression, we can perform polynomial division.
Dividing ${x}^{2}+5x+4$ by ${x}^{2}+2x-8$:
$\begin{array}{ccc}& x& 3\\ {x}^{2}+2x-8& {x}^{2}& 5x& 4\\ & -\left({x}^{2}& 3{x}^{2}& -8x\right)\\ & & 8x& 12\\ & & -\left(8x& -16\right)\\ & & & 28\end{array}$
The result of the division is $x+3$, with a remainder of $28$. Therefore:
$\left(v/g\right)x=x+3+\frac{28}{{x}^{2}+2x-8}$
And that is the simplified form of the function (v/g)x, where $v={x}^{2}+5x+4$ and $g={x}^{2}+2x-8$.
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Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
Solutions for Session 8, Part C
See solutions for Problems: C1 | C2 | C3 | C4 | C5 | C6 | C7
Problem C1 The triangle with hypotenuse 4: By similarity with the original triangle, we can say: 5/4 = 3/a, so a = 12/5. The triangle with hypotenuse 10: By similarity with the original triangle, we can say: 5/10 = 3/a, so a = 30/5 = 6. The triangle with hypotenuse 1: By similarity with the original triangle, we can say: 5/1 = 3/a, so a = 3/5.
Problem C2 The triangle with hypotenuse 4: By similarity with the original triangle, we can say: 5/4 = 4/b, so b = 16/5. The triangle with hypotenuse 10: By similarity with the original triangle, we can say: 5/10 = 4/b, so b = 40/5 = 8. The triangle with hypotenuse 1: By similarity with the original triangle, we can say: 5/1 = 4/b, so b = 4/5.
Problem C3
a. sin A = 3/5 and cos A = 4/5 b. sin B = 4/5 and cos B = 3/5
Problem C4
We calculate the hypotenuse with the Pythagorean theorem and find that it is 13.
a. sin A = 12/13, cos A = 5/13 b. sin B = 5/13, cos B = 12/13 c. tan A = 12/5, tan B = 5/12
Problem C5
a. sin 30° = cos 60° = 1/2 sin 60° = cos 30° = /2 tan 30° = 1/ tan 60° = These values are constant for any triangle with angles 30°-60°-90°. b. sin 45° = cos 45° = 1/ tan 45° = 1 These values are constant for any triangle with angles 45°-45°-90°.
Problem C6 Angles A and B are adjacent, so sin A = cos B. The hypotenuse is the same for both angles, but the roles of "adjacent side" and "opposite side" switch. The side opposite angle B is adjacent to angle A, and vice versa.
Problem C7 Let x be the length of the ramp. Then we have a right triangle with hypotenuse x, shorter leg 2, and the angle opposite to the shorter leg of 10°. Since sin 10° = 2/x, we have x = 2/sin 10° = 2/0.17 11.765 feet.
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Lesson Objectives
• Demonstrate an understanding of rational expressions
• Demonstrate an understanding of how to find the LCM
• Demonstrate the ability to factor a polynomial
• Learn how to find the LCD for a group of rational expressions
## How to Find the LCD for a Group of Rational Expressions
In our pre-algebra course, we learned how to find the Least Common Multiple (LCM) for a group of numbers. To find the LCM, we begin by factoring each number. We then build a list that contains each prime factor from all numbers involved. When a prime factor is repeated between two or more factorizations, we only include the largest number of repeats from any of the factorizations. The LCM is the product of the numbers in the list. Let's look at an example:
LCM(4, 18, 48)
Step 1) Let's factor each number:
4 » 2 • 2
18 » 2 • 3 • 3
48 » 2 • 2 • 2 • 2 • 3
Step 2) Build our list:
LCM List » 2,2,2,2,3,3
Step 3) Multiply the numbers on the list
2 • 2 • 2 • 2 • 3 • 3 = 144
LCM(4, 18, 48) = 144
When we add or subtract fractions, we need to have a common denominator. The least common denominator (LCD) is the LCM of the denominators. Let's look at an example: $$\frac{1}{12}, \frac{7}{20}$$ What is the LCD for these two fractions (1/12 and 7/20)? We want to find the LCM for the two denominators (12 and 20).
LCM(12,20) = 2 • 2 • 3 • 5 = 60
The LCD for the two fractions is 60.
### LCD of Rational Expressions
When we add or subtract rational expressions, we will need to have a common denominator. To find the LCD for a group of rational expressions, we want to find the LCM of the denominators. The process is similar to finding the LCM with numbers. The only difference is the involvement of variables. Since we already know how to find the number part, let's focus on the variable part. Let's suppose we had the following:
x2, x5, x6
How would we find the LCM? Since the variable (x) is the same in each case, we only need to know the largest number of repeats. In each case, our exponent tells us how many factors of x that we have. Therefore, our LCM will be x raised to the largest power in the group. In this particular case, our largest power in the group is 6.
LCM(x2, x5, x6) = x6
Let's look at some examples.
Example 1: Find the LCD for each group of rational expressions. $$\frac{x - 3}{x^2 + 7x - 18}, \frac{x^2 + 1}{x^2-3x + 2}$$ Step 1) Factor each denominator:
x2 + 7x - 18 = (x - 2)(x + 9)
x2 - 3x + 2 = (x - 2)(x - 1)
Step 2) Build our list:
Notice how the factor (x - 2) appears once in each factorization. This means our list will include one and only one factor of (x - 2):
LCM List » (x - 2)(x + 9)(x - 1)
Step 3) Multiply the factors:
(x - 2)(x + 9)(x - 1) =
x3 + 6x2 - 25x + 18
In most cases, we will leave our LCD in factored form. If asked for the LCD, either form is correct.
Example 2: Find the LCD for each group of rational expressions. $$\frac{9x^5 - 7}{2x^2 - 2}, \frac{15x^9}{4x^2 + 8x + 4}$$ Step 1) Factor each denominator:
2x2 - 2 = 2(x + 1)(x - 1)
4x2 + 8x + 4 = 4(x + 1)2
Step 2) Build our list:
LCM List » 4(x - 1)(x + 1)2
Step 3) Multiply the factors:
4(x - 1)(x + 1)2 =
4x3 + 4x2 - 4x - 4
Example 3: Find the LCD for each group of rational expressions. $$\frac{4x - 11}{3x - 21}, \frac{2x^5 + 9}{x^2 + 5x - 84}$$ Step 1) Factor each denominator:
3x - 21 = 3(x - 7)
x2 + 5x - 84 = (x - 7)(x + 12)
Step 2) Build our list:
LCM List » 3(x - 7)(x + 12)
Step 3) Multiply the factors:
3(x - 7)(x + 12) =
3x2 + 15x - 252
#### Skills Check:
Example #1
Find the LCD. $$\frac{3x}{5x - 20}, \frac{2}{x^2 - 4x}$$
A
$$5(x - 4)$$
B
$$5x(x - 4)$$
C
$$x - 4$$
D
$$30x^2(x - 4)$$
E
$$5x(x - 4)^2$$
Example #2
Find the LCD. $$\frac{5x - 1}{2x^2 - 11x + 5}, \frac{3x^2 - 5}{2x^3 - x^2}$$
A
$$x^2(2x - 1)(x - 5)$$
B
$$(x + 1)(x - 3)$$
C
$$(2x - 1)^2$$
D
$$5x^2(x + 5)(x - 1)$$
E
$$5x(x - 1)^2$$
Example #3
Find the LCD. $$\frac{7x}{3x^3+12x^2-15x}, \frac{x-12}{x^4+4x^3-5x^2}$$
A
$$(x-1)(x+5)$$
B
$$3x^2(x+5)^2$$
C
$$2(x-1)^2$$
D
$$3x^2(x-1)(x+5)$$
E
$$x^2(x - 3)(x + 2)$$
Better Luck Next Time, Your Score is %
Try again?
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# OU on the BBC: The Code - Prima Pizzeria hints and tips
Updated Tuesday 5th July 2011
Improve your chances with the Prima Pizzeria game with these hints and tips from our maths academics on prime numbers as well as further links to improve your learning.
1. Prima Pizzeria is about prime numbers. A prime number is a whole number greater than 1 that is divisible only by itself and 1. For example, 7 is a prime number because it is divisible only by 7 and 1, but 9 is not a prime number because it is divisible by 3. The first ten prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.
2. The only even prime number is 2. That’s because every even number other than 2 is divisible by 2!
3. Prime numbers are the building blocks of all whole numbers. That’s because every whole number greater than 1 can be obtained by multiplying prime numbers together. For instance, 60 = 2 × 2 × 3 × 5.
4. Whole numbers greater than 1 that are not prime numbers are called composite numbers. In Prima Pizzeria you must spread a composite number of toppings evenly over a pizza with a prime number of slices. So given a batch of 60 toppings you must choose a pizza with either 2, 3, or 5 slices, because 2, 3, and 5 are the only prime numbers that divide 60
5. Prima Pizzeria tests your dividing skills! For instance, given a batch of 8 toppings, a pizza with 3 slices is no use, because 3 doesn’t divide 8. Instead you need a pizza with only 2 slices, because 2 is the only prime number that divides 8.
6. Suppose we have a number (which we call A) that is divisible by another number (which we call B). Then B is said to be a factor of A. The numbers 2, 3, and 5 are all factors of 60. There are other factors of 60 too; such as 4, 15, and 20. Prima Pizzeria is about searching for prime number factors of composite numbers.
7. To test whether a number is divisible by 3, add the digits of that number together, and determine whether the resulting number is divisible by 3. For example, consider the number 513. Add 5, 1, and 3 together to obtain 9, which is divisible by 3. So 513 is also divisible by 3, and in fact 513 = 171 × 3. On the other hand, consider the number 514. Add 5, 1, and 4 together to obtain 10, which is not divisible by 3. So 514 is not divisible by 3.
8. You can work out all the prime numbers less than 100 as follows. Create a ten by ten grid containing the numbers 1 to 100, in order. Cross off the number 1. Now cross off all multiples of 2 other than 2 itself, namely 4, 6, 8, …, 100. Next cross off all multiples of 3 other than 3 itself, namely 6, 9, 12, …, 99. Repeat this for the prime numbers 5 and 7. The remaining numbers that haven’t been crossed off are the prime numbers. This method for identifying prime numbers is known as the Sieve of Eratosthenes.
9. Every composite number has a prime factor that is no greater than the square root of the composite number. For example, consider the composite number 100. The square root of 100 is 10, and 5 is a prime factor of 100 which is less than 10.
10. The ancient Greek mathematician Euclid proved that there are infinitely many prime numbers. This means if you tried to list the prime numbers then your list would never end. Euclid established this fact using a proof by contradiction: from the assumption that there are only finitely many prime numbers, he deduced an impossible conclusion. This meant his assumption was false, and on the contrary there are indeed infinitely many prime numbers.
For further information, take a look at our frequently asked questions which may give you the support you need.
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# How do you simplify sqrt (27 / 64)?
Feb 8, 2016
$\frac{3 \sqrt{3}}{8}$
#### Explanation:
We can begin be re writing the square root as:
$\sqrt{\frac{27}{64}} = \frac{\sqrt{27}}{\sqrt{64}}$
Luckily, 64 is a square number so that will allow us to get rid of the square root on the bottom:
$= \frac{\sqrt{27}}{8}$
Now we can transform $\sqrt{27}$ into a surd like so:
$\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3 \sqrt{3}$
So the fraction can be re written as:
$\frac{3 \sqrt{3}}{8}$
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# 2012 AIME II Problems/Problem 14
## Problem 14
In a group of nine people each person shakes hands with exactly two of the other people from the group. Let $N$ be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when $N$ is divided by $1000$.
## Solution 1
Given that each person shakes hands with two people, we can view all of these through graph theory as 'rings'. This will split it into four cases: Three rings of three, one ring of three and one ring of six, one ring of four and one ring of five, and one ring of nine. (All other cases that sum to nine won't work, since they have at least one 'ring' of two or fewer points, which doesn't satisfy the handshaking conditions of the problem.)
Case 1: To create our groups of three, there are $\dfrac{\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{3!}$. In general, the number of ways we can arrange people within the rings to count properly is $\dfrac{(n-1)!}{2}$, since there are $(n-1)!$ ways to arrange items in the circle, and then we don't want to want to consider reflections as separate entities. Thus, each of the three cases has $\dfrac{(3-1)!}{2}=1$ arrangements. Therefore, for this case, there are $\left(\dfrac{\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{3!}\right)(1)^3=280$
Case 2: For three and six, there are $\dbinom{9}{6}=84$ sets for the rings. For organization within the ring, as before, there is only one way to arrange the ring of three. For six, there is $\dfrac{(6-1)!}{2}=60$. This means there are $(84)(1)(60)=5040$ arrangements.
Case 3: For four and five, there are $\dbinom{9}{5}=126$ sets for the rings. Within the five, there are $\dfrac{4!}{2}=12$, and within the four there are $\dfrac{3!}{2}=3$ arrangements. This means the total is $(126)(12)(3)=4536$.
Case 4: For the nine case, there is $\dbinom{9}{9}=1$ arrangement for the ring. Within it, there are $\dfrac{8!}{2}=20160$ arrangements.
Summing the cases, we have $280+5040+4536+20160=30016 \to \boxed{016}$.
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Lesson 1ª
Operations with Algebraic Fractions
10.1 Calculate the result of the sum:
1st. Calculate the least common multiple (l.c.m.) of the denominators: We find that the l.c.m.(2, 3, 4) = 12
2nd. Divide the l.c.m. by each denominator (the value below the fraction), then, multiply the quotient times the numerator (the value above the fraction):
10.2 Calculate the value of:
1st. Calculate the l.c.m. (3, 4 and 5) = 60
2nd. Divide 60 by each denominator, then, multiply the quotient times the numerator. Write its corresponding sign in front.
10.3 Calculate the sum:
Solution:
When you find a whole number (no denominator), you can use 1 as a denominator to make things easier. Dividing or multiplying a number by 1 results in the same number, but it sometimes clears any doubts:
The l.c.m. of the denominators is 5. Each denominator is divided by this number, then, multiply the quotient times the numerator.
10.4 Calculate:
10.5 Calculate:
Solution:
Working with letter variables is simple. The l.c.m. of ‘a’ and ‘b’ is ‘ab’. These two letter variables have nothing in common. Imagine that ‘a’ is equal to 7 and ‘b’ is equal to 5. Since 7 and 5 are prime numbers, they have nothing in common, their l.c.m. is the same as the l.c.m.
Each denominator is divided by this number, then, we multiply the quotient times the numerator:
Dividing is like dividingimagining that 'a' = 7 and 'b' = 5.
We simplify the equal factors in the numerator and denominator and we will get:
10.6 Calculate:
Solution:
This exercise can be written like this:
The l.c.m. of the denominators is 'xy'
We divide this value by each denominator, then, multiply the quotient times the numerator:
We mustn't simplify the xy in the numerator with the xy in the denominator because the numerator is adding. To be able to simplify terms, they need to be multiplying.
10.7 Calculate the sum:
Solution:
The l.c.m. of these denominators is ‘ab’. We divide this value by each denominator, then, multiply the quotient times the numerator:
We find that the numerator is the square of the difference between both numbers:
10.8 Calculate:
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# Equations with Variables on Both Sides
4 teachers like this lesson
Print Lesson
## Objective
To solve equations with variables and integers on both sides.
#### Big Idea
Use of a pan balance to explain solving equations with integers on both sides
## Do Now
15 minutes
Students enter the class silently according to the Daily Entrance Routine. Do Now assignments are at their desk and include a series of pan balance problems. Everyone is coming back from Thanksgiving break and it is best to start with a concrete representation of equations so that students can refresh their knowledge of solving equations. There are two problems on the front of the sheet, representing two different equations. The first equation 3x = 9 is represented in the picture of 3 boxes with an unknown amount of coins inside each and an equivalent expression on the opposite side of the = sign. I made 12 of these cube boxes using attached nets (see resources). I will have these boxes available in class, with the given amount of coins inside each. These will be set up at booths on the side of the room so that I can send student who struggle with the pictured representations. These concrete representations are helpful for review as well as for students still struggling to understand complex relationships within equations. I will be using pictorial and abstract representations in today’s class, thus using the Concrete-Pictorial-Abstract approach. For more information on this approach click here for a great summary provided by an online resource.
Students can earn up to two achievement points for completing both sides of the worksheet correctly within 6 minutes. The set of extra problems on the back of the sheet serves multiple purposes: it keeps students who finish early busy, it allows me to travel to several students to check in about the basic concepts of solving equations before diving into today’s topic and it introduces the topic using a pictorial representation before diving into abstract representations in equations.
I reserve 3-4 minutes to review the last problem. The illustration depicts the equation 4A + 5 = A + 14. In this example, I model with the boxes I made that we can begin by taking away 5 marbles, or chips from each side of the equation. Then I take away 1 A from each side of the equation. I select one other student to show the same steps on the opposite side of the room for more students to see.
## Class Notes
15 minutes
I ask students to organize their Do Now into their binder and get ready for class notes which are distributed. At the beginning I provide the definition for a “pan balance” or “balance scale” as reference and to use this concept to transition into the next topic. I have students write the steps for solving equations which we reviewed in the last Do Now problem:
Step 1: Isolate the variable terms on each side of the equation by using opposite operations (+/–). Remember to keep the equation balanced!
Step 2: Isolate the variables on the opposite side of remaining constants by using opposite operations (+/–). Remember to keep the equation balanced!
We use these steps to review the solution to the equation in the notes which yields a negative answer. I ask students to share their opinion with their neighbor about the order of these steps. Do we have to isolate the variable terms on one side of the equation first, or can we start by isolating the constants on one side of the equation first? The goal is to get students to identify the more basic goal of isolating variables and constants on opposite sides of the equation so that we can figure out how many units the variable is worth. This is also a good opportunity to practice MP4 as we create real world examples for the equation we are solving:
2x + 6 = 2 + x
____–6 –6_____
2x + 4 = x
__–x –x__
x + 4 = 0
_______–4 –4_____
x = –4
Ex 1: Ms. Chavira owes the same amount of money at Bloomingdales and Nike.com. She has 6 dollars in the bank. Kaya owes the same amount as Ms. Chavira to Nike.com only and she has 2 dollars in the bank. If they both pay off their debts they will have the same amount of money. How much do they each owe Nike.com?
I can check to see if students can apply this real-world situation to the equation by identifying expressions and symbols aligning to the context of the situation. Which expression summarizes the first sentence in this problem? What constant represents the amount of money Ms. Chavira/Kaya have in the bank? What part of the equation represents the quantity obtained when Ms. Chavira pays off her debt?
20 minutes
I set up sticky notes labeled 1 – 24 on the blackboard before class. By now students are getting curious about those numbers. At the beginning of the task section I ask students to organize their notes, close their binders, clear their desks, and be ready to move. Then, I ask them to stand up on row at a time, select a sticky note off the board and return to their seat. Students with ODD numbered sticky notes will remain seated on the left side of the room at a table to themselves. Students have the option of moving to a different table even if they were already seated on the left side. Students with even numbers must choose a partner with an odd number or have the option of working independently on the right side of the room. As we near the winter break and as students prepare to take interim assessments for the next two days, I know that I need to provide as many chances to move and to work with different peers as possible so that students don’t lag and get bored.
Students work in pairs or independently to solve the equations on their worksheet. They are split into two levels of difficulty for which students can receive achievement points if scored correctly and within a given amount of time. I will be walking around with counter chips and “x boxes” to concretely represent any equations I can and help students who are struggling. The most common error to look out for with these types of equations is when students incorrectly combine terms on opposite sides of the number line. For example, #3 reads:
n + 8 = 4 + 4n
Some students may incorrectly combine the n's on opposite sides of the equation like this:
5n + 8 =
In these cases, having the algebra balance to concretely show how this action unbalances the equation can be very useful for delivering this concept.
Students will also be notified that today’s classwork is being graded for effort. A rubric which details how they will be graded will be distributed. This rubric includes three categories: completion (student must complete 10 problems), work shown (student must show the steps we discussed during the notes), and partner evaluations (student must listen to their partner and remain on task). Students who choose to work independently will have a “partner” score from me based on how well they remained on task and how well they listened to my feedback.
## Closing
10 minutes
Students are asked to complete the same rubric and grade themselves. This will inform both student and teacher about perceptions of understanding the directions of a task; both can then problem solve to improve learning of the lesson.
After completing the rubric, students are to complete their exit ticket and turn it in at the end of class.
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# Is 7 8 9 a triangle?
Can 7 8 and 9 be sides of a triangle
lengths can form a triangle. Therefore, it is possible to form a triangle with side lengths of 8, 7 and 9 cm.
What type of triangle is 7 8 9
scalene triangle
Hence the triangle with sides 7 cm, 8 cm and 9 cm is a scalene triangle.
What is a 7cm 8cm 9cm triangle
scalene trianle
Triangle with length of sides 7 cm,8 cm and 9 cm has all distinct sides that is all sides are unequal. A triangle with all distinct sides is called a scalene trianle, hence the given triangle is scalene. Was this answer helpful
Does 6 8 9 form a right triangle
Therefore, the given triangle is an acute angled triangle.
Can 7 8 15 make a triangle
In the usual sense of things (Euclidean Geometry), no. You could have a 'degenerate triangle', but that's not exactly a triangle in the way we usually think about a triangle.
How do you check if it is a triangle
All you have to do is use the Triangle Inequality Theorem, which states that the sum of two side lengths of a triangle is always greater than the third side. If this is true for all three combinations of added side lengths, then you will have a triangle.
Is 6cm 8cm 10cm a right triangle
Solution: Given, the length of sides of a triangle are 6, 8 and 10. We have to find whether the given triangle is the right triangle. Therefore, the given triangle is a right triangle.
Is 7cm 24cm 25cm a right triangle
7cm,24cm,25cm will form the sides of a right angled triangle.
Is a 6 8 10 triangle possible
Hence, the correct answer is yes and the given triangle is right triangle.
Is 6 10 8 10 and 1 a right triangle
Summary: A triangle has sides of lengths 6, 8, and 10 is a right triangle.
Can 7 8 14 make a triangle
Yes 7 plus 14 yeah 21. Right we do a plus c 7 plus 16 is that greater than 14. Yes and c plus b is 16 plus 14 greater than 7.. Yes so is this a triangle. Yes now if i asked you to classify.
Is 6 8 10 a right triangle
Hence, the correct answer is yes and the given triangle is right triangle. Sides of a triangle are given below.
Can 6 8 10 make a triangle
Hence, the correct answer is yes and the given triangle is right triangle.
How do you tell if 3 sides make a triangle
Approach: A triangle is valid if sum of its two sides is greater than the third side.
Can 6 8 10 be a triangle
Hence, the correct answer is yes and the given triangle is right triangle. Sides of a triangle are given below.
Is 5cm 12cm 13cm a right triangle
5 cm, 12 cm and 13 cm are the sides of a right-angled triangle.
Will 12cm 7cm 5cm make a triangle
Hence, the construction of a triangle is not possible with sides 7 c m , 5 c m and 12 c m .
Can 5cm 12cm and 13cm make a right triangle
We can also see that the 5, 12, 13 is a Pythagorean triplet, the other Pythagorean triplets are 3, 4, 5. This means that any triangle with sides of Pythagorean triplets is always a right angle triangle.
Can 4 8 12 be a triangle
The sum of 4 and 8 is 12 and 12 is less than 15 . This set of side lengths does not satisfy Triangle Inequality Theorem. These lengths do not form a triangle.
Is a 5 12 13 triangle right
Yes, 5 12 and 13 make a right triangle. They are referred to as Pythagorean triplets, where 5 squared and 12 squared equal 13 squared, which is the application of the Pythagorean theorem.
What 3 numbers make a triangle
Let's say that's. Two that's 2 right it says these two the lengths of these two have to be greater than my third side right what if i have a third side. That's.
Is 17 8 15 a right triangle
If all three sides of a right triangle have lengths that are integers, it is known as a Pythagorean triangle. In a triangle of this type, the lengths of the three sides are collectively known as a Pythagorean triple. Examples include: 3, 4, 5; 5, 12, 13; 8, 15, 17, etc.
Can 6 10 14 be a triangle
SOLUTION: The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Since, 6 + 14 > 10, 6 + 10 > 14, and 10 + 14 > 6, you can form a triangle with side lengths 6 m, 14 m, 10 m.
What is an example of an invalid triangle
Know how to spot an invalid triangle.
5 + 8 > 3 = 13 > 3, so one side passes. 5 + 3 > 8 = 8 > 8. Since this is invalid, you can stop right here. This triangle is not valid.
Is 8cm 15cm 17cm a right triangle
Then this triangle is a right triangle.
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# How to Solve a Math Problem
There are three steps to solving a math problem.
• Figure out what the problem is asking.
• Solve the problem.
• If you get stuck, figure out why you're stuck.
• Check the answer.
### Sample Problem
What does the Mean Value Theorem say about the function f(x) = 4x + x2 on the interval (4, 6)?
Figure out what the problem is asking.
We need to understand what all the words in the problem mean, and what any theorems mentioned say. The MVT says if
f is continuous on [a, b], and
f is differentiable on (a, b),
then there is some c in (a, b) with
.
The problem is actually two problems. Are we allowed to apply the MVT to the given function on the given interval? If so, what does the conclusion of the MVT say for this particular function on this particular interval?
Solve the problem.
To solve the problem we need to do two smaller problems.
Are we allowed to apply the MVT to the given function on the given interval?
First, we need to understand what the problem is asking. This problem is asking "do the hypotheses/assumptions of the MVT hold in this case?"
Next, we solve the problem. The answer is yes. f is a polynomial, since it's continuous and differentiable everywhere. In particular, f is continuous on [4, 6] and differentiable on (4, 6). There's not anything to check for this answer.
Now what does the conclusion of the MVT say for this particular function on this particular interval? First we need to understand what the problem is asking. This problem is asking "if we plug in the right values of a and b and use the conclusion of the MVT, what does it say?" To solve the problem, we plug in a = 4 and b = 6. The MVT says there is some c in (4, 6) with
Again, there's not anything to check for this answer.
Check the answer.
Problems like this that don't involve much calculation, don't have answers to check. A good thing to do, though, would be to read through the answer again and make sure it's believable.
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# Triangle inequality
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{{#invoke:Hatnote|hatnote}}
Three examples of the triangle inequality for triangles with sides of lengths x, y, z. The top example shows the case when there is a clear inequality and the bottom example shows the case when the third side, z, is nearly equal to the sum of the other two sides x + y.
In mathematics, the triangle inequality states that for any triangle, the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side.[1][2] If x, y, and z are the lengths of the sides of the triangle, then the triangle inequality states that
${\displaystyle z\leq x+y,}$
with equality only in the degenerate case of a triangle with zero area. In Euclidean geometry and some other geometries, the triangle inequality is a theorem about distances, and it is written using vectors and vector lengths (norms):
${\displaystyle \|\mathbf {x} +\mathbf {y} \|\leq \|\mathbf {x} \|+\|\mathbf {y} \|,}$
where the length z of the third side has been replaced by the vector sum x + y. When x and y are real numbers, they can be viewed as vectors in 1, and the triangle inequality expresses a relationship between absolute values.
In Euclidean geometry, for right triangles the triangle inequality is a consequence of the Pythagorean theorem, and for general triangles a consequence of the law of cosines, although it may be proven without these theorems. The inequality can be viewed intuitively in either 2 or 3. The figure at the right shows three examples beginning with clear inequality (top) and approaching equality (bottom). In the Euclidean case, equality occurs only if the triangle has a 180° angle and two angles, making the three vertices collinear, as shown in the bottom example. Thus, in Euclidean geometry, the shortest distance between two points is a straight line.
In spherical geometry, the shortest distance between two points is an arc of a great circle, but the triangle inequality holds provided the restriction is made that the distance between two points on a sphere is the length of a minor spherical line segment (that is, one with central angle in [0, π]) with those endpoints.[3][4]
The triangle inequality is a defining property of norms and measures of distance. This property must be established as a theorem for any function proposed for such purposes for each particular space: for example, spaces such as the real numbers, Euclidean spaces, the Lp spaces (p ≥ 1), and inner product spaces.
## Euclidean geometry
Euclid's construction for proof of the triangle inequality for plane geometry.
Euclid proved the triangle inequality for distances in plane geometry using the construction in the figure.[5] Beginning with triangle ABC, an isosceles triangle is constructed with one side taken as BC and the other equal leg BD along the extension of side AB. It then is argued that angle β > α, so side . But so the sum of sides . This proof appears in Euclid's Elements, Book 1, Proposition 20.[6]
### Mathematical expression of the constraint on the sides of a triangle
The triangle inequality, as stated in words, literally translates into three inequalities (given that the side lengths a, b, c are all positive):
${\displaystyle a+b>c,b+c>a,c+a>b.}$
A more succinct form of such a statement can easily be shown to be
${\displaystyle |a-b|
Another way to state it is
${\displaystyle \max(a,b,c)
implying
${\displaystyle 2\max(a,b,c)
and thus that the longest side length is less than the semiperimeter.
### Right triangle
Isosceles triangle with equal sides divided into two right triangles by an altitude drawn from one of the two base angles.
A specialization of this argument to right triangles is:[7]
In a right triangle, the hypotenuse is greater than either of the two sides, and less than their sum.
The second part of this theorem already is established above for any side of any triangle. The first part is established using the lower figure. In the figure, consider the right triangle ADC. An isosceles triangle ABC is constructed with equal sides . From the triangle postulate, the angles in the right triangle ADC satisfy:
${\displaystyle \alpha +\gamma =\pi /2\ .}$
Likewise, in the isosceles triangle ABC, the angles satisfy:
${\displaystyle 2\beta +\gamma =\pi \ .}$
Therefore,
${\displaystyle \alpha =\pi /2-\gamma ,\ \mathrm {while} \ \beta =\pi /2-\gamma /2\ ,}$
and so, in particular,
${\displaystyle \alpha <\beta \ .}$
That means side AD opposite angle α is shorter than side AB opposite the larger angle β. But . Hence:
${\displaystyle {\overline {\mathrm {AC} }}>{\overline {\mathrm {AD} }}\ .}$
A similar construction shows , establishing the theorem.
An alternative proof (also based upon the triangle postulate) proceeds by considering three positions for point B:[8] (i) as depicted (which is to be proven), or (ii) B coincident with D (which would mean the isosceles triangle had two right angles as base angles plus the vertex angle γ, which would violate the triangle postulate), or lastly, (iii) B interior to the right triangle between points A and D (in which case angle ABC is an exterior angle of a right triangle BDC and therefore larger than π/2, meaning the other base angle of the isosceles triangle also is greater than π/2 and their sum exceeds π in violation of the triangle postulate).
This theorem establishing inequalities is sharpened by Pythagoras' theorem to the equality that the square of the length of the hypotenuse equals the sum of the squares of the other two sides.
### Some practical examples of the use of the triangle inequality
Consider a triangle whose sides are in an arithmetic progression and let the sides be a, a + d, a + 2d. Then the triangle inequality requires that
${\displaystyle 0
${\displaystyle 0
${\displaystyle 0
To satisfy all these inequalities requires
${\displaystyle a>0{\text{ and }}-{\frac {a}{3}}[9]
When d is chosen such that d = a/3, it generates a right triangle that is always similar to the Pythagorean triple with sides 3, 4, 5.
Now consider a triangle whose sides are in a geometric progression and let the sides be a, ar, ar2. Then the triangle inequality requires that
${\displaystyle 0
${\displaystyle 0
${\displaystyle 0
The first inequality requires a > 0, consequently it can be divided through and eliminated. With a > 0, the middle inequality only requires r > 0. This now leaves the first and third inequalities needing to satisfy
{\displaystyle {\begin{aligned}r^{2}+r-1&{}>0\\r^{2}-r-1&{}<0.\end{aligned}}\,}
The first of these quadratic inequalities requires r to range in the region beyond the value of the positive root of the quadratic equation r2 + r − 1 = 0, i.e. r > φ − 1 where φ is the golden ratio. The second quadratic inequality requires r to range between 0 and the positive root of the quadratic equation r2r − 1 = 0, i.e. 0 < r < φ. The combined requirements result in r being confined to the range
${\displaystyle \varphi -10.\,}$[10]
When r the common ratio is chosen such that it generates a right triangle that is always similar to the Kepler triangle.
### Generalization of the inequality to any polygon
The triangle inequality can be extended by mathematical induction to arbitrary polygonal paths, showing that the total length of such a path is no less than the length of the straight line between its endpoints. Consequently the length of any polygon side is always less than the sum of the other polygon side lengths.
#### Example of the generalized polygon inequality for a quadrilateral
Consider a quadrilateral whose sides are in a geometric progression and let the sides be a, ar, ar2, ar3. Then the generalized polygon inequality requires that
${\displaystyle 0
${\displaystyle 0
${\displaystyle 0
${\displaystyle 0
These inequalities for a > 0 reduce to the following
${\displaystyle r^{3}+r^{2}+r-1>0\,}$
${\displaystyle r^{3}-r^{2}-r-1<0.\,}$[11]
The LHS polynomials of these two inequalities have roots that are the tribonacci constant and its reciprocal. Consequently r is limited to the range 1/t < r < t where t is the tribonacci constant.
#### Relationship with shortest paths
The arc length of a curve is defined as the least upper bound of the lengths of polygonal approximations.
This generalization can be used to prove that the shortest curve between two points in Euclidean geometry is a straight line.
No polygonal path between two points is shorter than the line between them. This implies that no curve can have an arc length less than the distance between its endpoints. By definition, the arc length of a curve is the least upper bound of the lengths of all polygonal approximations of the curve. The result for polygonal paths shows that the straight line between the endpoints is shortest of all the polygonal approximations. Because the arc length of the curve is greater than or equal to the length of every polygonal approximation, the curve itself cannot be shorter than the straight line path.[12]
### Generalization of the inequality to higher dimensions
In Euclidean space, the hypervolume of an (n − 1)-facet of an n-simplex is less than or equal to the sum of the hypervolumes of the other n facets. In particular, the area of a triangular face of a tetrahedron is less than or equal to the sum of the areas of the other three sides.
## Normed vector space
Triangle inequality for norms of vectors.
In a normed vector space V, one of the defining properties of the norm is the triangle inequality:
${\displaystyle \displaystyle \|x+y\|\leq \|x\|+\|y\|\quad \forall \,x,y\in V}$
that is, the norm of the sum of two vectors is at most as large as the sum of the norms of the two vectors. This is also referred to as subadditivity. For any proposed function to behave as a norm, it must satisfy this requirement.[13]
If the normed space is euclidean, or, more generally, strictly convex, then ${\displaystyle \|x+y\|=\|x\|+\|y\|}$ if and only if the triangle formed by x, y, and x + y, is degenerate, that is, x and y are on the same ray, i.e., x = 0 or y = 0, or x = α y for some α > 0. This property characterizes strictly convex normed spaces such as the p spaces with 1 < p < ∞. However, there are normed spaces in which this is not true. For instance, consider the plane with the 1 norm (the Manhattan distance) and denote x = (1, 0) and y = (0, 1). Then the triangle formed by x, y, and x' + y, is non-degenerate but
${\displaystyle \|x+y\|=\|(1,1)\|=|1|+|1|=2=\|x\|+\|y\|.}$
### Example norms
• Absolute value as norm for the real line. To be a norm, the triangle inequality requires that the absolute value satisfy for any real numbers x and y:
${\displaystyle |x+y|\leq |x|+|y|,}$
which it does.
The triangle inequality is useful in mathematical analysis for determining the best upper estimate on the size of the sum of two numbers, in terms of the sizes of the individual numbers.
There is also a lower estimate, which can be found using the reverse triangle inequality which states that for any real numbers x and y:
${\displaystyle |x-y|\geq {\bigg |}|x|-|y|{\bigg |}.}$
• Inner product as norm in an inner product space. If the norm arises from an inner product (as is the case for Euclidean spaces), then the triangle inequality follows from the Cauchy–Schwarz inequality as follows: Given vectors x and y, and denoting the inner product as x, y:[14]
where the last form is a consequence of:
${\displaystyle \|x\|^{2}+2\|x\|\|y\|+\|y\|^{2}=\left(\|x\|+\|y\|\right)^{2}\ .}$
The Cauchy-Schwarz Inequality turns into an equality if and only if x and y are linearly dependent. The inequality ${\displaystyle \langle x,y\rangle +\langle y,x\rangle \leq 2|\langle x,y\rangle |}$ turns into an equality for linearly dependent ${\displaystyle x}$ and ${\displaystyle y}$ if and only if one of the vectors x or y is a nonnegative scalar of the other.
Taking the square root of the final result gives the triangle inequality.
• p-norm: a commonly used norm is the p-norm:
${\displaystyle \|x\|_{p}=\left(\sum _{i=1}^{n}|x_{i}|^{p}\right)^{1/p}\ ,}$
where the xi are the components of vector x. For p = 2 the p-norm becomes the Euclidean norm:
${\displaystyle \|x\|_{2}=\left(\sum _{i=1}^{n}|x_{i}|^{2}\right)^{1/2}=\left(\sum _{i=1}^{n}x_{i}^{2}\right)^{1/2}\ ,}$
which is Pythagoras' theorem in n-dimensions, a very special case corresponding to an inner product norm. Except for the case p = 2, the p-norm is not an inner product norm, because it does not satisfy the parallelogram law. The triangle inequality for general values of p is called Minkowski's inequality.[15] It takes the form:
${\displaystyle \|x+y\|_{p}\leq \|x\|_{p}+\|y\|_{p}\ .}$
## Metric space
In a metric space M with metric d, the triangle inequality is a requirement upon distance:
${\displaystyle d(x,\ z)\leq d(x,\ y)+d(y,\ z)\ ,}$
for all x, y, z in M. That is, the distance from x to z is at most as large as the sum of the distance from x to y and the distance from y to z.
The triangle inequality is responsible for most of the interesting structure on a metric space, namely, convergence. This is because the remaining requirements for a metric are rather simplistic in comparison. For example, the fact that any convergent sequence in a metric space is a Cauchy sequence is a direct consequence of the triangle inequality, because if we choose any xn and xm such that d(xn, x) < ε/2 and d(xm, x) < ε/2, where ε > 0 is given and arbitrary (as in the definition of a limit in a metric space), then by the triangle inequality, d(xn, xm) ≤ d(xn, x) + d(xm, x) < ε/2 + ε/2 = ε, so that the sequence Template:Mset is a Cauchy sequence, by definition.
This version of the triangle inequality reduces to the one stated above in case of normed vector spaces where a metric is induced via d(x, y) ≔ ‖xy, with xy being the vector pointing from point y to x.
## Reverse triangle inequality
The reverse triangle inequality is an elementary consequence of the triangle inequality that gives lower bounds instead of upper bounds. For plane geometry the statement is:[16]
Any side of a triangle is greater than the difference between the other two sides.
In the case of a normed vector space, the statement is:
${\displaystyle {\bigg |}\|x\|-\|y\|{\bigg |}\leq \|x-y\|,}$
or for metric spaces, |d(y, x) − d(x, z)| ≤ d(y, z). This implies that the norm ‖–‖ as well as the distance function d(x, –) are Lipschitz continuous with Lipschitz constant 1, and therefore are in particular uniformly continuous.
The proof for the reverse triangle uses the regular triangle inequality, and ${\displaystyle \|y-x\|=\|-1(x-y)\|=|-1|\|x-y\|=\|x-y\|}$:
${\displaystyle \|x\|=\|(x-y)+y\|\leq \|x-y\|+\|y\|\Rightarrow \|x\|-\|y\|\leq \|x-y\|,}$
${\displaystyle \|y\|=\|(y-x)+x\|\leq \|y-x\|+\|x\|\Rightarrow \|x\|-\|y\|\geq -\|x-y\|,}$
Combining these two statements gives:
${\displaystyle -\|x-y\|\leq \|x\|-\|y\|\leq \|x-y\|\Rightarrow {\bigg |}\|x\|-\|y\|{\bigg |}\leq \|x-y\|.}$
## Reversal in Minkowski space
In the usual Minkowski space and in Minkowski space extended to an arbitrary number of spatial dimensions, assuming null or timelike vectors in the same time direction, the triangle inequality is reversed:
${\displaystyle \|x+y\|\geq \|x\|+\|y\|\;\forall x,y\in V{\text{ such that }}\|x\|,\|y\|\geq 0{\text{ and }}t_{x},t_{y}\geq 0.}$
A physical example of this inequality is the twin paradox in special relativity.
## Notes
1. Wolfram MathWorld - http://mathworld.wolfram.com/TriangleInequality.html
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Word Problems For Linear Inequalities
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When \ (233\) Is Added To A Number, The Result Is \ (65\) More Than \ (5\) Times The Number.
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Home | | Maths 12th Std | Continuous Distributions
# Continuous Distributions
In this section we learn (i) Continuous random variable (ii) Probability density function (iii) Distribution function (Cumulative distribution function). (iv) To determine distribution function from probability density function. (v) To determine probability density function from distribution function.
Continuous Distributions
In this section we learn
(i) Continuous random variable
(ii) Probability density function
(iii) Distribution function (Cumulative distribution function).
(iv) To determine distribution function from probability density function.
(v) To determine probability density function from distribution function.
Sometimes a measurement such as current in a copper wire or length of lifetime of an electric bulb, can assume any value in an interval of real numbers. Then any precision in the measurement is possible. The random variable that represents this measurement is said to be a continuous random variable. The range of the random variable includes all values in an interval of real numbers; that is, the range can be thought of as a continuum of real numbers
## 1. The definition of continuous random variable
### Definition 11.5 (Continuous Random Variable)
Let S be a sample space and let a random variable X : S that takes on any value in a set I of . Then X is called a continuous random variable if P (X = x) = 0 for every x in I
## 2. Probability density function
### Definition 11.6: (Probability density function)
A non-negative real valued function f ( x) is said to be a probability density function if, for each possible outcome x, x [a,b] of a continuous random variable X having the property
### Theorem 11.2 (Without proof)
A function f (.) is a probability density function for some continuous random variable X if and only if it satisfies the following properties.
(i) f (x) 0 , for every x and (ii) -∞f (x)dx = 1 .
### Note
It follows from the above definition, if X is a continuous random variable,
P(a ≤ X ≤ b) = ba f( x) dx, which means that P( X = a) = aa f (x) dx = 0
That is probability when X takes on any one particular value is zero.
## 3. Distribution function (Cumulative distribution function)
### Definition 11.7 : (Cumulative Distribution Function)
The distribution function or cumulative distribution function F ( x) of a continuous random variable X with probability density f(x) is
Remark
(1) In the discrete case, f (a) = P (X = a) is the probability that X takes the value a.
In the continuous case, f (x) at x = a is not the probability that X takes the value a, that is f (a) ≠ P (X = a) . If X is continuous type, P (X = a) = 0 for a .
(2) When the random variable is continuous, the summation used in discrete is replaced by integration.
(3) For continuous random variable
P(a < X < b) = P(aX < b) = P(a < Xb) = P(aXb)
(4) The distribution function of a continuous random variable is known as Continuous Distribution Function.
## Properties of distribution function
For a continuous random variable X, the cumulative distribution function satisfies the following properties.
(i) 0 F ( x) 1 .
(ii) F ( x) is a real valued non-decreasing. That is, if x < y , then F ( x) F ( y) .
(iii) F ( x) is continuous everywhere.
(iv) lim x → −∞ F (x) = F( − ∞) = 0 and lim x → ∞ F (x) = F (+∞) = 1.
(v) P (X > x) = 1 − P (X≤ x) = 1 − F (x) .
(vi) P(a < X < b) = F(b) −F(a) .
Example 11.11
Find the constant C such that the function is a density function, and compute (i) P(1.5 < X < 3.5) (ii) P ( X 2) (iii) P(3 < X ) .
Solution
Since the given function is a probability density function,
From the given information,
Since f ( x) is continuous, the probability that X is equal to any particular value is zero.
Therefore when the random variable is continuous, either or both of the signs < by and > by can be interchanged. Thus
## 4. Distribution function from Probability density function
Both the probability density function and the cumulative distribution function (or distribution function) of a continuous random variable X contain all the probabilistic information of X . The probability distribution of X is determined by either of them. Let us learn the method to determine the distribution function F of a continuous random variable X from the probability density function f (x) of X and vice versa.
Example 11.12
If X is the random variable with probability density function f(x) given by,
Check:
(i) Whether F ( x) is continuous everywhere.
(ii) From the Fig. 11.16, triangle area = 1/2 bh = 1.
## 5. Probability density function from Probability distribution function.
Let us learn the method to determine the probability density function f ( x) from the distribution function F ( x) of a continuous random variable X .
Suppose F ( x) is the distribution function of a continuous random variable X . Then the probability density function f(x) is given by
f( x) = dF(x) / dx = F ′ ( x) , wherever derivative exists.
### Example 11.13
If X is the random variable with distribution function F (x) given by,
then find (i) the probability density function f (x) (ii) P ( 0.2 ≤ X 0.7).
Solution
### Example 11.15
Let X be a random variable denoting the life time of an electrical equipment having probability density function
Find (i) the value of k (ii) Distribution function (iii) P (X< 2)
(iv) calculate the probability that X is at least for four unit of time (v) P (X = 3) .
### Solution
Tags : Definition, Properties | Probability Distributions | Mathematics , 12th Maths : UNIT 11 : Probability Distributions
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
12th Maths : UNIT 11 : Probability Distributions : Continuous Distributions | Definition, Properties | Probability Distributions | Mathematics
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# How to “make up a formula” of a sum: $S=2+7+12+\cdots+(5n-3)$?
Determine a formula for the following sum with $n\in\mathbb N$:
$$S=2+7+12+\cdots+(5n-3)$$
I had no clue about what to do about it. I just wrote, for the sake of doing something, the following:
$$\sum_{i=0}^n(5i-3) = \left(\sum_{i=0}^n(5i) - 3n\right)$$
Curiously, I got a few points for that. But eventually, there is an annotation by the professor:
And?
So apparently I was on the right path. What was I supposed to do afterward? How do you "make up a formula" of a given sum?
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The sum should be taken from $i = 1$ to $n$, by the way =) – amWhy Dec 1 '12 at 20:44
May I suggest a little reading on this topic? It is the following "tutorial" by D. Gleich: cs.purdue.edu/homes/dgleich/publications/… – Giuseppe Negro Dec 1 '12 at 21:05
So you have $$S = 2 + 7 + 12 + \dots + (5n - 3) = \sum_{i=1}^n (5i-3) = \left(5\sum_{i=1}^{n} i\right) - 3n.$$ So all you need to know is that $$\sum_{i=1}^{n} i = \frac{n(n-1)}{2}.$$
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Hint: you are almost done, you need to take the $5$ out of the sum (its something like $5x+5y=5(x+y)$) and you should recall how to sum $1+\cdots+n$
-
I suspect your professor wanted a closed form for this sum (that is, a formula in terms of $n$ only). You can further simplify your formula to $$5\left(\sum_{i=1}^n i \right) - 3n = 5 \frac{n(n-1)}{2} - 3n = \frac{5}{2}n^2 - \frac{11}{2}n.$$ (Also note your summation should start at $i = 1$, not $i = 0$.)
-
$$S_n=2 + 7 + 12 + \dots + (5n - 3)$$ is sum of first n terms of arithmetic progression with first term $a_1=2$ and difference $d=5$. From formula $$S_n=\frac{n}{2}(2a_1+(n-1)d)=\frac{n}{2}(4+5n-5)=\frac{n}{2}(5n-1)$$
-
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# Mathematics
### Solving Trigonometric Equations
E.g.1
Solve sin x = 0.5 for 0 ≤ x ≤ 360
sin x = 0.5
x = 300
Since y = 0.5 line crosses the sine curve at two points, there are two solutions.
Now, look at the symmetry of the graph; the two values of x are 300 and 1500.
x = 30 and 150.
E.g.2
Solve cos x = -0.5 for 0 ≤ x ≤ 360
cos x = -0.5
x = 1200
Since y = -0.5 line crosses the sine curve at two points, there are two solutions.
Now, look at the symmetry of the graph; the two values of x are 1200 and 2400.
x = 120 and 240.
E.g.3
Solve sin (x +10) = 0.5 for 0 ≤ x ≤ 360
From example 1,
(x + 10) = 300
There are two values for (x +10) that satisfy the equation; they are 300 and 1500
x + 10 = 30 or x + 10 = 150
x = 20 or x = 140
E.g.4
Solve 1 + 2 sin x = 2 for 0 ≤ x ≤ 360
2 sin x = 1
sin x = 0.5
From example 1,
x = 300
There are two values for x that satisfy the equation; they are 300 and 1500
x = 30 or x = 150.
#### Solving Equations - interactive
You can animate the y=a line by clicking the play button or move the slider to a point you want it to be at.
Now, in order to complement what you have just learnt, work out the following questions:
1. Solve 2 sin x = 1.5 for 0 ≤ x ≤ 360
2. Solve 1 + 3 cos x = 2 for 0 ≤ x ≤ 360
3. Solve sin (2x - 10) = 0.7 for 0 ≤ x ≤ 360
4. Solve tan (x -30) = 0.7 for 0 ≤ x ≤ 360
5. Solve cos (2x -20) = 0.5 for 0 ≤ x ≤ 360
6. Solve sin2 x = 0.25 for 0 ≤ x ≤ 360
7. Solve sin 2x = 0.5 for 0 ≤ x ≤ 360
8. Solve 1 + 3 sin 2x = 2 for 0 ≤ x ≤ 360
Maths is challenging; so is finding the right book. K A Stroud, in this book, cleverly managed to make all the major topics crystal clear with plenty of examples; popularity of the book speak for itself - 7th edition in print.
### Recommended - GCSE & iGCSE
This is the best book available for the new GCSE(9-1) specification and iGCSE: there are plenty of worked examples; a really good collection of problems for practising; every single topic is adequately covered; the topics are organized in a logical order.
### Recommended for A Level
This is the best book that can be recommended for the new A Level - Edexcel board: it covers every single topic in detail;lots of worked examples; ample problems for practising; beautifully and clearly presented.
|
Probability
```Probability
Example and Definitions:
Question: A spinner has 4 equal sectors colored yellow, blue,
green and red. What are the chances of landing on blue after
spinning the spinner? What are the chances of landing on red?
Solution: The chances of landing on red are 1 in 4, or one fourth.
This problem asked us to find some probabilities involving a spinner.
Let's look at some definitions and examples from the problem above.
Definition
Example
An experiment is a situation involving chance or probability
that leads to results called outcomes.
In the problem above, the experiment is
spinning the spinner.
An outcome is the result of a single trial of an experiment.
The possible outcomes are landing on
yellow, blue, green or red.
An event is one or more outcomes of an experiment.
One event of this experiment is landing
on blue.
The probability of landing on blue is one
fourth.
Probability is the measure of how likely an event is.
Probability Formula
In order to measure probabilities, mathematicians have devised the following
formula for finding the probability of an event.
Probability Of An Event
The Number Of Ways Event A Can
Occur
P(A) =
The Total Number Of Possible
Outcomes
Example 1: Spinner
Experiment 1: A spinner has 4 equal sectors: blue, red, green and
yellow. After spinning the spinner, what is the probability of
landing on each color?
Outcomes: The possible outcomes for this experiment are blue,
red, green, and
yellow.
Probabilities:
P(yellow)= The number of ways to land on yellow =
25%
The total number of colors
P (blue)=
P (green)=
P (red)=
4
1 or
Example 2: A Die
Experiment 2: A single 6 sided die is rolled. What is the probability of
each outcome? What is the probability of rolling an even number? An
odd number?
Outcomes: The possible outcomes are _1_, _2_, _3_, _4_, _5_, & _6_
Probabilities:
P (1) =
# of ways to roll a 1 =
1
P (2)=
# of sides on a die 6
P(3)=
P (4)=
P (5)=
P(6)=
P(even) =
# of even sides = 3 or 1
# of sides on a die 6
P (odd)=
2
Difference between Events and
Outcomes:
Experiment 2 illustrates the difference between an
outcome and an event.
A single outcome of this experiment is rolling a 1, or
rolling a 2, or rolling a 3, etc. Rolling an even
number (2, 4 or 6) is an event, and rolling an odd
number (1, 3 or 5) is also an event.
Example 3: For all the Marbles
Experiment: A glass jar contains 6 red, 5 green, 8 blue and 3 yellow
marbles. If a single marble is chosen at random, what is the probability it
will be a red marble? A green marble? A red marble? A yellow marble?
Outcomes: The possible outcomes of this experiment are red, green, blue,
and yellow.
Probabilities:
P (red) =
P (green)=
P (blue) =
P (yellow) =
Independent Events
Independent events:
are events whose outcomes do
not influence each other.
The result of A will not affect the probability of B occurring.
Experiment 4: A jar contains 3 red, 5 green, 2 blue and 6 yellow
marbles. A marble is chosen at random from the jar. After replacing it,
a second marble is chosen. What is the probability of choosing a green
and then a yellow marble?
Outcomes: red, green, blue, and yellow
Probabilities:
P(green) =
P(yellow) =
P (green and yellow) = P(green) * P(yellow)
```
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# 3.2 - Identifying Outliers: IQR Method
3.2 - Identifying Outliers: IQR Method
Some observations within a set of data may fall outside the general scope of the other observations. Such observations are called outliers. In Lesson 2.2.2 you identified outliers by looking at a histogram or dotplot. Here, you will learn a more objective method for identifying outliers.
We can use the IQR method of identifying outliers to set up a “fence” outside of Q1 and Q3. Any values that fall outside of this fence are considered outliers. To build this fence we take 1.5 times the IQR and then subtract this value from Q1 and add this value to Q3. This gives us the minimum and maximum fence posts that we compare each observation to. Any observations that are more than 1.5 IQR below Q1 or more than 1.5 IQR above Q3 are considered outliers. This is the method that Minitab Express uses to identify outliers by default.
## Example: Test Scores
A teacher wants to examine students’ test scores. Their scores are: 74, 88, 78, 90, 94, 90, 84, 90, 98, and 80.
Five number summary: 74, 80, 89, 90, 98.
$IQR = 90 - 80 = 10$
The interquartile range is 10.
$1.5 IQR = 1.5 (10) = 15$
1.5 times the interquartile range is 15. Our fences will be 15 points below Q1 and 15 points above Q3.
Lower fence: $80 - 15 = 65$
Upper fence: $90 + 15 = 105$
Any scores that are less than 65 or greater than 105 are outliers. In this case, there are no outliers.
## Example: Books
A survey was given to a random sample of 20 sophomore college students. They were asked, “how many textbooks do you own?” Their responses, were: 0, 0, 2, 5, 8, 8, 8, 9, 9, 10, 10, 10, 11, 12, 12, 12, 14, 15, 20, and 25.
The observations are in order from smallest to largest, we can now compute the IQR by finding the median followed by Q1 and Q3.
$Median = 10$
$Q1 = 8$
$Q3 = 12$
$IQR = 12 - 8 = 4$
The interquartile range is 4.
$1.5 IQR = 1.5 (4) = 6$
1.5 times the interquartile range is 6. Our fences will be 6 points below Q1 and 6 points above Q3.
Lower fence: $8 - 6 = 2$
Upper fence: $12 + 6 = 18$
Any observations less than 2 books or greater than 18 books are outliers. There are 4 outliers: 0, 0, 20, and 25.
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility
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# Parasitic number
(Redirected from Parasitic numbers)
An n-parasitic number (in base 10) is a positive natural number which can be multiplied by n by moving the rightmost digit of its decimal representation to the front. Here n is itself a single-digit positive natural number. In other words, the decimal representation undergoes a right circular shift by one place. For example, 4•128205=512820, so 128205 is 4-parasitic. Most authors do not allow leading zeros to be used, and this article follows that convention. So even though 4•025641=102564, the number 025641 is not 4-parasitic.
## Derivation
An n-parasitic number can be derived by starting with a digit k (which should be equal to n or greater) in the rightmost (units) place, and working up one digit at a time. For example, for n = 4 and k = 7
4•7=28
4•87=348
4•487=1948
4•9487=37948
4•79487=317948
4•179487=717948.
So 179487 is a 4-parasitic number with units digit 7. Others are 179487179487, 179487179487179487, etc.
Notice that the repeating decimal
${\displaystyle x=0.179487179487179487\ldots =0.{\overline {179487}}{\mbox{ has }}4x=0.{\overline {717948}}={\frac {7.{\overline {179487}}}{10}}.}$
Thus
${\displaystyle 4x={\frac {7+x}{10}}{\mbox{ so }}x={\frac {7}{39}}.}$
In general, an n-parasitic number can be found as follows. Pick a one digit integer k such that kn, and take the period of the repeating decimal k/(10n−1). This will be ${\displaystyle {\frac {k}{10n-1}}(10^{m}-1)}$ where m is the length of the period; i.e. the multiplicative order of 10 modulo (10n − 1).
For another example, if n = 2, then 10n − 1 = 19 and the repeating decimal for 1/19 is
${\displaystyle {\frac {1}{19}}=0.{\overline {052631578947368421}}.}$
So that for 2/19 is double that:
${\displaystyle {\frac {2}{19}}=0.{\overline {105263157894736842}}.}$
The length m of this period is 18, the same as the order of 10 modulo 19, so 2 × (1018 − 1)/19 = 105263157894736842.
105263157894736842 × 2 = 210526315789473684, which is the result of moving the last digit of 105263157894736842 to the front.
The step-by-step derivation algorithm depicted above is a great core technique but will not find all n-parasitic numbers. It literally gets stuck in an infinite loop with the derived number equaling the derivation source. An example of this where n = 5 and k = 5. The 42-digit n-parasitic number to be derived is 102040816326530612244897959183673469387755. Check the steps in Table One below. The algorithm begins building from right to left until it reaches step 15—then the infinite loop occurs. Lines 16 and 17 are pictured to show that nothing changes. There is a fix for this problem, and when applied, the algorithm will not only find all n-parasitic numbers in base ten, it will find them in base 8 and base 16 as well. Look at line 15 in Table Two. The fix, when this condition is identified and the n-parasitic number has not been found, is simply to not shift the product from the multiplication, but use it as is, and append n (in this case 5) to the end. After 42 steps, the proper parasitic number will be found.
#### Table One
1. 5 * 5 = 25 - Shift = 55 2. 5 * 55 = 275 - Shift = 755 3. 5 * 755 = 3775 - Shift = 7755 4. 5 * 7755 = 38775 - Shift = 87755 5. 5 * 87755 = 438775 - Shift = 387755 6. 5 * 387755 = 1938775 - Shift = 9387755 7. 5 * 9387755 = 46938775 - Shift = 69387755 8. 5 * 69387755 = 346938775 - Shift = 469387755 9. 5 * 469387755 = 2346938775 - Shift = 3469387755 10. 5 * 3469387755 = 17346938775 - Shift = 73469387755 11. 5 * 73469387755 = 367346938775 - Shift = 673469387755 12. 5 * 673469387755 = 3367346938775 - Shift = 3673469387755 13. 5 * 3673469387755 = 18367346938775 - Shift = 83673469387755 14. 5 * 83673469387755 = 418367346938775 - Shift = 183673469387755 15. 5 * 183673469387755 = 918367346938775 - Shift = 183673469387755 16. 5 * 183673469387755 = 918367346938775 - Shift = 183673469387755 17. 5 * 183673469387755 = 918367346938775 - Shift = 183673469387755
#### Table Two
1. 5 * 5 = 25 - Shift = 55 2. 5 * 55 = 275 - Shift = 755 3. 5 * 755 = 3775 - Shift = 7755 4. 5 * 7755 = 38775 - Shift = 87755 5. 5 * 87755 = 438775 - Shift = 387755 6. 5 * 387755 = 1938775 - Shift = 9387755 7. 5 * 9387755 = 46938775 - Shift = 69387755 8. 5 * 69387755 = 346938775 - Shift = 469387755 9. 5 * 469387755 = 2346938775 - Shift = 3469387755 10. 5 * 3469387755 = 17346938775 - Shift = 73469387755 11. 5 * 73469387755 = 367346938775 - Shift = 673469387755 12. 5 * 673469387755 = 3367346938775 - Shift = 3673469387755 13. 5 * 3673469387755 = 18367346938775 - Shift = 83673469387755 14. 5 * 83673469387755 = 418367346938775 - Shift = 183673469387755 15. 5 * 183673469387755 = 918367346938775 - Shift = 9183673469387755 16. 5 * 9183673469387755 = 45918367346938775 - Shift = 59183673469387755 17. 5 * 59183673469387755 = 295918367346938775 - Shift = 959183673469387755
There is one more condition to be aware of when working with this algorithm, leading zeros must not be lost. When the shift number is created it may contain a leading zero which is positionally important and must be carried into and through the next step. Calculators and computer math methods will remove leading zeros. Look at Table Three below displaying the derivation steps for n=4 and k=4. The Shift number created in step 4, 02564, has a leading zero which is fed into step 5 creating a leading zero product. The resulting Shift is fed into Step 6 which displays a product proving the 4-parasitic number ending in 4 is 102564.
#### Table Three
1. 4 * 4 = 16 - Shift = 64 2. 4 * 64 = 256 - Shift = 564 3. 4 * 564 = 2256 - Shift = 2564 4. 4 * 2564 = 10256 - Shift = 02564 5. 4 * 02564 = 010256 - Shift = 102564 6. 4 * 102564 = 410256 - Shift = 102564
## Smallest n-parasitic numbers
The smallest n-parasitic numbers are also known as Dyson numbers, after a puzzle concerning these numbers posed by Freeman Dyson.[1][2][3] They are: (leading zeros are not allowed) (sequence A092697 in the OEIS)
n Smallest n-parasitic number Digits Period of
1 1 1 1/9
2 105263157894736842 18 2/19
3 1034482758620689655172413793 28 3/29
4 102564 6 4/39
5 102040816326530612244897959183673469387755 42 5/49
6 1016949152542372881355932203389830508474576271186440677966 58 6/59
7 1014492753623188405797 22 7/69
8 1012658227848 13 8/79
9 10112359550561797752808988764044943820224719 44 9/89
## General note
In general, if we relax the rules to allow a leading zero, then there are 9 n-parasitic numbers for each n. Otherwise only if kn then the numbers do not start with zero and hence fit the actual definition.
Other n-parasitic integers can be built by concatenation. For example, since 179487 is a 4-parasitic number, so are 179487179487, 179487179487179487 etc.
## Other bases
In duodecimal system, the smallest n-parasitic numbers are: (using inverted two and three for ten and eleven, respectively) (leading zeros are not allowed)
n Smallest n-parasitic number Digits Period of
1 1 1 1/Ɛ
2 10631694842 Ɛ 2/1Ɛ
3 2497 4 7/2Ɛ = 1/5
4 10309236ᘔ88206164719544 4/3Ɛ
5 1025355ᘔ9433073ᘔ458409919Ɛ715 25 5/4Ɛ
6 1020408142854ᘔ997732650ᘔ18346916306 6/5Ɛ
7 101899Ɛ864406Ɛ33ᘔᘔ15423913745949305255Ɛ17 35 7/6Ɛ
8 131ᘔ8ᘔ 6 /7Ɛ = 2/17
9 101419648634459Ɛ9384Ɛ26Ɛ533040547216ᘔ1155Ɛ3Ɛ12978ᘔ399 45 9/8Ɛ
14Ɛ36429ᘔ7085792 14 12/9Ɛ = 2/15
Ɛ 1011235930336ᘔ53909ᘔ873Ɛ325819Ɛ9975055Ɛ54ᘔ3145ᘔ42694157078404491Ɛ 55 Ɛ/ᘔƐ
## Strict definition
In strict definition, least number m beginning with 1 such that the quotient m/n is obtained merely by shifting the leftmost digit 1 of m to the right end are
1, 105263157894736842, 1034482758620689655172413793, 102564, 102040816326530612244897959183673469387755, 1016949152542372881355932203389830508474576271186440677966, 1014492753623188405797, 1012658227848, 10112359550561797752808988764044943820224719, 10, 100917431192660550458715596330275229357798165137614678899082568807339449541284403669724770642201834862385321, 100840336134453781512605042016806722689075630252, ... (sequence A128857 in the OEIS)
They are the period of n/(10n - 1), also the period of the decadic integer -n/(10n - 1).
Number of digits of them are
1, 18, 28, 6, 42, 58, 22, 13, 44, 2, 108, 48, 21, 46, 148, 13, 78, 178, 6, 99, 18, 8, 228, 7, 41, 6, 268, 15, 272, 66, 34, 28, 138, 112, 116, 179, 5, 378, 388, 18, 204, 418, 6, 219, 32, 48, 66, 239, 81, 498, ... (sequence A128858 in the OEIS)
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# Using the Property a B = a K B K , Fill in the Blanks Substituting Proper Numbers in the Following. - Algebra
Sum
Using the property a/b = (ak)/(bk) , fill in the blanks substituting proper numbers in the following.
9/14 = 4.5/(.......) = (........)/ 42 = (......) /3.5
#### Solution
4.5 = 9 × 0.5, 42 = 14 × 3, 3.5 = 14 × 0.25
Now,
9/14 = (9 xx 0.5)/(14 xx 0.5) = (9 xx 3)/(14 xx 3) = (9 xx 0.25) /(14 xx 0.25)
⇒ 9/14 = 4.5/7 = 27/42 = 2.25/3.5
Concept: Concept of Ratio
Is there an error in this question or solution?
#### APPEARS IN
Balbharati Mathematics 1 Algebra 9th Standard Maharashtra State Board
Chapter 4 Ratio and Proportion
Practice Set 4.2 | Q 1.2 | Page 63
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# Commutative property
In mathematics, a binary operation is commutative if changing the order of the operands does not change the result. It is a fundamental property of many binary operations, and many mathematical proofs depend on it. Most familiar as the name of the property that says "3 + 4 = 4 + 3" or "2 × 5 = 5 × 2", the property can also be used in more advanced settings. The name is needed because there are operations, such as division and subtraction, that do not have it (for example, "3 − 5 ≠ 5 − 3"); such operations are not commutative, and so are referred to as noncommutative operations. The idea that simple operations, such as the multiplication and addition of numbers, are commutative was for many years implicitly assumed. Thus, this property was not named until the 19th century, when mathematics started to become formalized.[1][2] A corresponding property exists for binary relations; a binary relation is said to be symmetric if the relation applies regardless of the order of its operands; for example, equality is symmetric as two equal mathematical objects are equal regardless of their order.[3]
## Common uses
The commutative property (or commutative law) is a property generally associated with binary operations and functions. If the commutative property holds for a pair of elements under a certain binary operation then the two elements are said to commute under that operation.
## Mathematical definitions
The term "commutative" is used in several related senses.[4][5]
1. A binary operation ${\displaystyle *}$ on a set S is called commutative if:
${\displaystyle x*y=y*x\qquad {\mbox{for all }}x,y\in S}$
An operation that does not satisfy the above property is called non-commutative.
2. One says that x commutes with y under ${\displaystyle *}$ if:
${\displaystyle x*y=y*x}$
3. A binary function ${\displaystyle f\colon A\times A\to B}$ is called commutative if:
${\displaystyle f(x,y)=f(y,x)\qquad {\mbox{for all }}x,y\in A}$
## Examples
### Commutative operations in everyday life
• Putting on socks resembles a commutative operation since which sock is put on first is unimportant. Either way, the result (having both socks on), is the same. In contrast, putting on underwear and trousers is not commutative.
• The commutativity of addition is observed when paying for an item with cash. Regardless of the order the bills are handed over in, they always give the same total.
### Commutative operations in mathematics
Two well-known examples of commutative binary operations:[4]
${\displaystyle y+z=z+y\qquad {\mbox{for all }}y,z\in \mathbb {R} }$
For example 4 + 5 = 5 + 4, since both expressions equal 9.
${\displaystyle yz=zy\qquad {\mbox{for all }}y,z\in \mathbb {R} }$
For example, 3 × 5 = 5 × 3, since both expressions equal 15.
• Some binary truth functions are also commutative, since the truth tables for the functions are the same when one changes the order of the operands.
For example, the logical biconditional function p ↔ q is equivalent to q ↔ p. This function is also written as p IFF q, or as p ≡ q, or as Epq.
The last form is an example of the most concise notation in the article on truth functions, which lists the sixteen possible binary truth functions of which eight are commutative: Vpq = Vqp; Apq (OR) = Aqp; Dpq (NAND) = Dqp; Epq (IFF) = Eqp; Jpq = Jqp; Kpq (AND) = Kqp; Xpq (NOR) = Xqp; Opq = Oqp.
### Noncommutative operations in daily life
• Concatenation, the act of joining character strings together, is a noncommutative operation. For example,
EA + T = EAT TEA = T + EA
• Washing and drying clothes resembles a noncommutative operation; washing and then drying produces a markedly different result to drying and then washing.
• Rotating a book 90° around a vertical axis then 90° around a horizontal axis produces a different orientation than when the rotations are performed in the opposite order.
• The twists of the Rubik's Cube are noncommutative. This can be studied using group theory.
• Thought processes are noncommutative: A person asked a question (A) and then a question (B) may give different answers to each question than a person asked first (B) and then (A), because asking a question may change the person's state of mind.
• The act of dressing is either commutative or non-commutative, depending on the items. Putting on underwear and normal clothing is noncommutative. Putting on left and right socks is commutative.
• Shuffling a deck of cards is non-commutative. Given two ways, A and B, of shuffling a deck of cards, doing A first and then B is in general not the same as doing B first and then A.
### Noncommutative operations in mathematics
Some noncommutative binary operations:[6]
#### Division and subtraction
Division is noncommutative, since ${\displaystyle 1\div 2\neq 2\div 1}$.
Subtraction is noncommutative, since ${\displaystyle 0-1\neq 1-0}$. However it is classified more precisely as anti-commutative, since ${\displaystyle 0-1=-(1-0)}$.
#### Truth functions
Some truth functions are noncommutative, since the truth tables for the functions are different when one changes the order of the operands. For example, the truth tables for (A ⇒ B) = (¬A ∨ B) and (B ⇒ A) = (A ∨ ¬B) are
A B A ⇒ B B ⇒ A
F F T T
F T T F
T F F T
T T T T
#### Function composition of linear functions
Function composition of linear functions from the real numbers to the real numbers is almost always noncommutative. For example, let ${\displaystyle f(x)=2x+1}$ and ${\displaystyle g(x)=3x+7}$. Then
${\displaystyle (f\circ g)(x)=f(g(x))=2(3x+7)+1=6x+15}$
and
${\displaystyle (g\circ f)(x)=g(f(x))=3(2x+1)+7=6x+10}$
This also applies more generally for linear and affine transformations from a vector space to itself (see below for the Matrix representation).
#### Matrix multiplication
Matrix multiplication of square matrices is almost always noncommutative, for example:
${\displaystyle {\begin{bmatrix}0&2\\0&1\end{bmatrix}}={\begin{bmatrix}1&1\\0&1\end{bmatrix}}\cdot {\begin{bmatrix}0&1\\0&1\end{bmatrix}}\neq {\begin{bmatrix}0&1\\0&1\end{bmatrix}}\cdot {\begin{bmatrix}1&1\\0&1\end{bmatrix}}={\begin{bmatrix}0&1\\0&1\end{bmatrix}}}$
#### Vector product
The vector product (or cross product) of two vectors in three dimensions is anti-commutative; i.e., b × a = −(a × b).
## History and etymology
Records of the implicit use of the commutative property go back to ancient times. The Egyptians used the commutative property of multiplication to simplify computing products.[7][8] Euclid is known to have assumed the commutative property of multiplication in his book Elements.[9] Formal uses of the commutative property arose in the late 18th and early 19th centuries, when mathematicians began to work on a theory of functions. Today the commutative property is a well-known and basic property used in most branches of mathematics.
The first recorded use of the term commutative was in a memoir by François Servois in 1814,[1][10] which used the word commutatives when describing functions that have what is now called the commutative property. The word is a combination of the French word commuter meaning "to substitute or switch" and the suffix -ative meaning "tending to" so the word literally means "tending to substitute or switch." The term then appeared in English in 1838[2] in Duncan Farquharson Gregory's article entitled "On the real nature of symbolical algebra" published in 1840 in the Transactions of the Royal Society of Edinburgh.[11]
## Propositional logic
### Rule of replacement
In truth-functional propositional logic, commutation,[12][13] or commutativity[14] refer to two valid rules of replacement. The rules allow one to transpose propositional variables within logical expressions in logical proofs. The rules are:
${\displaystyle (P\lor Q)\Leftrightarrow (Q\lor P)}$
and
${\displaystyle (P\land Q)\Leftrightarrow (Q\land P)}$
where "${\displaystyle \Leftrightarrow }$" is a metalogical symbol representing "can be replaced in a proof with."
### Truth functional connectives
Commutativity is a property of some logical connectives of truth functional propositional logic. The following logical equivalences demonstrate that commutativity is a property of particular connectives. The following are truth-functional tautologies.
Commutativity of conjunction
${\displaystyle (P\land Q)\leftrightarrow (Q\land P)}$
Commutativity of disjunction
${\displaystyle (P\lor Q)\leftrightarrow (Q\lor P)}$
Commutativity of implication (also called the law of permutation)
${\displaystyle (P\to (Q\to R))\leftrightarrow (Q\to (P\to R))}$
Commutativity of equivalence (also called the complete commutative law of equivalence)
${\displaystyle (P\leftrightarrow Q)\leftrightarrow (Q\leftrightarrow P)}$
## Set theory
In group and set theory, many algebraic structures are called commutative when certain operands satisfy the commutative property. In higher branches of mathematics, such as analysis and linear algebra the commutativity of well-known operations (such as addition and multiplication on real and complex numbers) is often used (or implicitly assumed) in proofs.[15][16][17]
## Mathematical structures and commutativity
### Associativity
The associative property is closely related to the commutative property. The associative property of an expression containing two or more occurrences of the same operator states that the order operations are performed in does not affect the final result, as long as the order of terms doesn't change. In contrast, the commutative property states that the order of the terms does not affect the final result.
Most commutative operations encountered in practice are also associative. However, commutativity does not imply associativity. A counterexample is the function
${\displaystyle f(x,y)={\frac {x+y}{2}},}$
which is clearly commutative (interchanging x and y does not affect the result), but it is not associative (since, for example, ${\displaystyle f(-4,f(0,+4))=-1}$ but ${\displaystyle f(f(-4,0),+4)=+1}$). More such examples may be found in commutative non-associative magmas.
### Symmetry
Some forms of symmetry can be directly linked to commutativity. When a commutative operator is written as a binary function then the resulting function is symmetric across the line y = x. As an example, if we let a function f represent addition (a commutative operation) so that f(x,y) = x + y then f is a symmetric function, which can be seen in the adjacent image.
For relations, a symmetric relation is analogous to a commutative operation, in that if a relation R is symmetric, then ${\displaystyle aRb\Leftrightarrow bRa}$.
## Non-commuting operators in quantum mechanics
In quantum mechanics as formulated by Schrödinger, physical variables are represented by linear operators such as x (meaning multiply by x), and ${\displaystyle {\frac {d}{dx}}}$. These two operators do not commute as may be seen by considering the effect of their compositions ${\displaystyle x{\frac {d}{dx}}}$ and ${\displaystyle {\frac {d}{dx}}x}$ (also called products of operators) on a one-dimensional wave function ${\displaystyle \psi (x)}$:
${\displaystyle x\cdot {\mathrm {d} \over \mathrm {d} x}\psi =x\cdot \psi '\ \neq \ \psi +x\cdot \psi '={\mathrm {d} \over \mathrm {d} x}\left(x\cdot \psi \right)}$
According to the uncertainty principle of Heisenberg, if the two operators representing a pair of variables do not commute, then that pair of variables are mutually complementary, which means they cannot be simultaneously measured or known precisely. For example, the position and the linear momentum in the x-direction of a particle are represented by the operators ${\displaystyle x}$ and ${\displaystyle -i\hbar {\frac {\partial }{\partial x}}}$, respectively (where ${\displaystyle \hbar }$ is the reduced Planck constant). This is the same example except for the constant ${\displaystyle -i\hbar }$, so again the operators do not commute and the physical meaning is that the position and linear momentum in a given direction are complementary.
## Notes
1. Cabillón and Miller, Commutative and Distributive
2. Flood, Raymond; Rice, Adrian; Wilson, Robin, eds. (2011). Mathematics in Victorian Britain. Oxford University Press. p. 4.
3. Weisstein, Eric W. "Symmetric Relation". MathWorld.
4. Krowne, p.1
5. Weisstein, Commute, p.1
6. Yark, p.1.
7. Lumpkin, p.11
8. Gay and Shute, p.?
9. O'Conner and Robertson, Real Numbers
10. O'Conner and Robertson, Servois
11. D. F. Gregory (1840). "On the real nature of symbolical algebra". Transactions of the Royal Society of Edinburgh. 14: 208–216.
12. Moore and Parker
13. Copi, Irving M.; Cohen, Carl (2005). Introduction to Logic. Prentice Hall.
14. Hurley, Patrick (1991). A Concise Introduction to Logic 4th edition. Wadsworth Publishing.
15. Axler, p.2
16. Gallian, p.34
17. p. 26,87
18. Gallian p.236
19. Gallian p.250
## References
### Books
• Axler, Sheldon (1997). Linear Algebra Done Right, 2e. Springer. ISBN 0-387-98258-2.
Abstract algebra theory. Covers commutativity in that context. Uses property throughout book.
• Copi, Irving M.; Cohen, Carl (2005). Introduction to Logic. Prentice Hall.
• Gallian, Joseph (2006). Contemporary Abstract Algebra, 6e. Boston, Mass.: Houghton Mifflin. ISBN 0-618-51471-6.
Linear algebra theory. Explains commutativity in chapter 1, uses it throughout.
• Goodman, Frederick (2003). Algebra: Abstract and Concrete, Stressing Symmetry, 2e. Prentice Hall. ISBN 0-13-067342-0.
Abstract algebra theory. Uses commutativity property throughout book.
• Hurley, Patrick (1991). A Concise Introduction to Logic 4th edition. Wadsworth Publishing.
### Articles
Article describing the mathematical ability of ancient civilizations.
• Robins, R. Gay, and Charles C. D. Shute. 1987. The Rhind Mathematical Papyrus: An Ancient Egyptian Text. London: British Museum Publications Limited. ISBN 0-7141-0944-4
Translation and interpretation of the Rhind Mathematical Papyrus.
### Online resources
Definition of commutativity and examples of commutative operations
Explanation of the term commute
Examples proving some noncommutative operations
Article giving the history of the real numbers
Page covering the earliest uses of mathematical terms
Biography of Francois Servois, who first used the term
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# What makes something a linear operator?
## What makes something a linear operator?
A linear transformation is a function from one vector space to another that respects the underlying (linear) structure of each vector space. A linear transformation is also known as a linear operator or map. The two vector spaces must have the same underlying field.
## What is linear operator with examples?
Examples: The simplest linear operator is the identity operator I. I|V> = |V>,
What makes a matrix A linear operator?
A linear operator can be written as a matrix in a given basis. If we use the “standard basis” for R2, (1, 0) and (0, 1), then (x, y)= x(1,0)+ y(0, 1) so [xy] is the representation in the standard basis. The operation, in matrix form is [abcd][xy]=[ax+bycx+dy].
READ ALSO: Does a stiff neck get worse before it gets better?
What is a linear differential operator?
From differential calculus we know that acts linearly on (differentiable) functions, that is, We think of the differential operator as operating on functions (that are sufficiently differentiable). The differential operator is linear, that is, for all sufficiently differentiable functions and and all scalars .
### Are all operators linear?
The most basic operators (in some sense) are linear maps, which act on vector spaces. However, when using “linear operator” instead of “linear map”, mathematicians often mean actions on vector spaces of functions, which also preserve other properties, such as continuity.
### Which is not linear operator?
If Y is the set R of real or C of complex numbers, then a non-linear operator is called a non-linear functional. The simplest example of a non-linear operator (non-linear functional) is a real-valued function of a real argument other than a linear function.
Which operator operators are linear?
Linear Operators
• ˆO is a linear operator,
• c is a constant that can be a complex number (c=a+ib), and.
• f(x) and g(x) are functions of x.
READ ALSO: Does any country own the ocean?
Is any matrix A linear operator?
A matrix is a linear operator acting on the vector space of column vectors. Per linear algebra and its isomorphism theorems, any vector space is isomorphic to any other vector space of the same dimension. As such, matrices can be seen as representations of linear operators subject to some basis of column vectors.
## What is a linear operator?
A linear operator is an operator which satisfies the following two conditions: (43) (44) where is a constant and and are functions. As an example, consider the operators and . We can see that is a linear operator because (45) (46) However, is not a linear operator because (47)
## How can you tell if a function is linear or nonlinear?
How Can You Tell if a Function is Linear or Nonlinear From a Table? How Can You Tell if a Function is Linear or Nonlinear From a Table? To see if a table of values represents a linear function, check to see if there’s a constant rate of change. If there is, you’re looking at a linear function!
READ ALSO: How Long Will electronic components last?
Is the range of a linear operator a subspace of Y?
The range of a linear operator is a subspace of Y. Proposition. A linear operator on a normed space X (to a normed space Y) is continuous at every point X if it is continuous at a single point in X. Proof.Exercise. [3, p. 240]. Luenberger does not mention thatY needs to be a normed space too.
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# Difference between revisions of "2012 AMC 8 Problems/Problem 20"
## Problem
What is the correct ordering of the three numbers $\frac{5}{19}$, $\frac{7}{21}$, and $\frac{9}{23}$, in increasing order?
$\textbf{(A)}\hspace{.05in}\frac{9}{23}<\frac{7}{21}<\frac{5}{19}\quad\textbf{(B)}\hspace{.05in}\frac{5}{19}<\frac{7}{21}<\frac{9}{23}\quad\textbf{(C)}\hspace{.05in}\frac{9}{23}<\frac{5}{19}<\frac{7}{21}$
$\textbf{(D)}\hspace{.05in}\frac{5}{19}<\frac{9}{23}<\frac{7}{21}\quad\textbf{(E)}\hspace{.05in}\frac{7}{21}<\frac{5}{19}<\frac{9}{23}$
## Solution 1
The value of $\frac{7}{21}$ is $\frac{1}{3}$. Now we give all the fractions a common denominator.
$\frac{5}{19} \implies \frac{345}{1311}$
$\frac{1}{3} \implies \frac{437}{1311}$
$\frac{9}{23} \implies \frac{513}{1311}$
Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, our final answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.
## Solution 2
Instead of finding the LCD, we can subtract each fraction from $1$ to get a common numerator. Thus,
$1-\dfrac{5}{19}=\dfrac{14}{19}$
$1-\dfrac{7}{21}=\dfrac{14}{21}$
$1-\dfrac{9}{23}=\dfrac{14}{23}$
All three fractions have common numerator $14$. Now it is obvious the order of the fractions. $\dfrac{14}{19}>\dfrac{14}{21}>\dfrac{14}{23}\implies\dfrac{5}{19}<\dfrac{7}{21}<\dfrac{9}{23}$. Therefore, our answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.
## Solution 3
Change $7/21$ into $1/3$; $$\frac{1}{3}\cdot\frac{5}{5}=\frac{5}{15}$$ $$\frac{5}{15}>\frac{5}{19}$$ $$\frac{7}{21}>\frac{5}{19}$$ And $$\frac{1}{3}\cdot\frac{9}{9}=\frac{9}{27}$$ $$\frac{9}{27}<\frac{9}{23}$$ $$\frac{7}{21}<\frac{9}{23}$$ Therefore, our answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.
## Solution 4
When $\frac{x}{y}<1$ and $z>0$, $\frac{x+z}{y+z}>\frac{x}{y}$. Hence, the answer is ${\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$. ~ ryjs
This is also similar to Problem 3 on the AMC 8 2019, but with the rule switched.
## Solution 5
By dividing, we see that 5/19 ≈ 0.26, 7/21=0.33, and 9/23=0.39. When we put this in order, 0.26<0.33<0.39. So our answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.
## See Also
2012 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 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 AJHSME/AMC 8 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# How do you write an equation in standard form given a line that passes through (4,-3) with m=2?
Jun 4, 2015
The equation in standard form is $2 x - y = 5$.
The general equation for the standard form of a line is $\text{A"x+"B"y="C}$, where $\text{A}$ and $\text{B}$ are integers, and $\text{A} \ne 0$, and $\text{B} \ne 0$.
http://www.mathwarehouse.com/algebra/linear_equation/standard-form-equation-of-a-line.php
Start with the equation to find the point-slope form of a line: $y - {y}_{1} = m \left(x - {x}_{1}\right)$, where $m$ is the slope, and ${x}_{1}$ and ${y}_{1}$ are the known point.
$m = 2$
${x}_{1} = 4$
${y}_{1} = - 3$
Solution:
$y - {y}_{1} = m \left(x - {x}_{1}\right)$
Substitute known values into the equation.
$\left(y - \left(- 3\right)\right) = 2 \left(x - 4\right)$ =
$y + 3 = 2 \left(x - 4\right)$ = Point-slope form.
Continue to the Standard Form.
Distribute the $2$.
$y + 3 = 2 x - 8$
Subtract $y$ from both sides.
$3 = 2 x - y - 8$
Add $8$ to both sides.
$11 = 2 x - y$
Flip the equation.
$2 x - y = 11$ = Standard Form
graph{2x-y=11 [-32.48, 32.43, -16.23, 16.24]}
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# In the equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ represents a circle with X-axis as a diameter and radius a, then which of the following is possible.A. $f=2a,g=0,c=3{{a}^{2}}$ B. $f=0,g=a,c=3{{a}^{2}}$ C. $f=0,g=-2a,c=3{{a}^{2}}$ D. none of these
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Hint: We first try to form the given circle in its general form of ${{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}}$ to find the centre and the radius. from there we place the values of the y coordinate of the centre as 0. Then we put the value of $y=0$ in the equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ to get end points of the diameter. We put the possible given options and check the most appropriate one.
The equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ represents a circle with X-axis as a diameter and radius a.
As the diameter is a part of X-axis, it means that the centre is on the X-axis and the y coordinate of the centre is 0.
It’s given that the equation of the circle is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ . We transform it in its general form of ${{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}}$ and get ${{\left( x-g \right)}^{2}}+{{\left( y-f \right)}^{2}}={{\left( \sqrt{{{g}^{2}}+{{f}^{2}}-c} \right)}^{2}}$ . O is the centre.
Equating with the general equation of circle ${{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}}$ , we get the centre as $O\equiv \left( -g,-f \right)$ and the radius as $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$ units.
We can put the value $-f=0$ which means $f=0$ . Also, the radius is a.
This means the centre is $O\equiv \left( -g,0 \right)$ and the radius as $\sqrt{{{g}^{2}}+{{f}^{2}}-c}=a$ units.
Solving the radius, we get ${{g}^{2}}-c={{a}^{2}}$ .
Now we put the value of $y=0$ in the equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ to get the two end points of the diameter. This gives ${{x}^{2}}+2gx+c=0$ . Solving we get
$x=\dfrac{-2g\pm \sqrt{4{{g}^{2}}-4c}}{2}=-g\pm \sqrt{{{g}^{2}}-c}$ .
Now we place the given two options which are possible and they are $f=0,g=a,c=3{{a}^{2}}$ and $f=0,g=-2a,c=3{{a}^{2}}$ . In both cases $f=0,c=3{{a}^{2}}$ . Now we put $c=3{{a}^{2}}$ in ${{g}^{2}}-c={{a}^{2}}$ .
We get ${{g}^{2}}=4{{a}^{2}}$ which gives $g=\pm 2a$ .
Therefore, the correct option is C.
Note:
We need to remember that the general equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ always makes an intercept $2\sqrt{{{g}^{2}}-c}$ units on the X-axis and $2\sqrt{{{f}^{2}}-c}$ units on the Y-axis. We use that to find the diameter length in the circle of ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ .
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# Optimization Problems in Calculus
Contents:
More optimization problems:
## General Optimization Steps
Optimization problems in calculus often involve the determination of the “optimal” (meaning, the best) value of a quantity. For example, we might want to know:
• The biggest area that a piece of rope could be tied around.
• How high a ball could go before it falls back to the ground.
• At which point of a loop does a roller coaster run the slowest.
Very often, the optimization must be done with certain constraints. In the case of the rope, we’re limited by its length. These constraints are usually very helpful to solve optimization problems (for an advanced example of using constraints, see: Lagrange Multiplier). Optimal values are often either the maximum or the minimum values of a certain function.
## Optimization Problems in Calculus: Steps.
Example problem: Find the maximum area of a rectangle whose perimeter is 100 meters. (Note: This is a typical optimization problem in AP calculus).
Step 1: Determine the function that you need to optimize. In the example problem, we need to optimize the area A of a rectangle, which is the product of its length L and width W. Our function in this example is:
A = LW.
Step 2: Identify the constraints to the optimization problem. In our example problem, the perimeter of the rectangle must be 100 meters. This will be useful in the next step.
Step 3: Express that function in terms of a single variable upon which it depends, using algebra. For this example, we’re going to express the function in a single variable. “L.”
1. A rectangle’s perimeter is the sum of its sides, that is, 100m = 2L + 2W.
2. Subtract 2L from both sides of this equation, 2W = 100m – 2L.
3. Divide each side by 2: W = 50m – L.
4. Substitute 50m – L for “W” in A = LW: A = L (50m – L) = 50m L – L².
Step 4: Calculate the derivative of the function with respect to a variable.
• dA/dL = 50m (1) L(1 – 1) – 2 L(2-1) = 50m – 2L.
Step 5: Set the function to zero and compute the corresponding variable’s value. For our sample problem, we set:
• dA/dL = 0 = 50m – 2L.
• So, L = 25m.
Step 6: Use the value from Step 5 to calculate the corresponding optimal value of the function. In our sample problem, A = 50m L – L2 = 50 m (25m) – (25m)2 = 625 m2.
## The Volume of the Largest Rectangular Box Inscribed in a Pyramid
If you’re asked to find the volume of the largest rectangular box in the first octant, with three faces in the coordinate planes and one vertex in a given plane, you’re being asked to find the volume of the largest rectangular box that fits in a pyramid like the one below.
Imagine that point A is embedded in the rectangular face.
Suppose the plane is x + 2y + 3z = 6. What is the volume of the largest box?
Your first step should be to define the volume. Since your box is rectangular, the formula is:
width x depth x height.
We can write this as:
V = xyz
Since x + 2y + 3z = 6, we know z = (6 – x – 2y) / 3. We can substitute that in our volume equation to create a function that tells us the volume in terms of x and y:
V = f(x, y) = xy(6 – x – 2y) / 3
We’re looking for a maximum, so we want to find the point where the derivative of volume with respect to both x and y is zero.
We can call those two derivatives fx and fy, and we can calculate them as
• fx = ⅓y (6 – 2x – 2y) = 0 → y = 0, y = 3 – x
• fy = ⅓x (6 – x – 4y) = 0
At y = 0, equation B becomes (1/3)x (6 – x) = 0, so x = 0 or x = 6.
At y = 3 – x, equation B becomes x(-2 + x) = 0, so the zeros are x = 0 (in which case y = 3) or x = 2 (in which case y = 1).
We’e just found the four critical points: (6, 0), (0,0) (0, 3), (2, 1).
We can draw each of these points on a 3D graph, and use our knowledge of geometry to decide which box is the maximum: it should be fairly evident that it will be (2,1), and since x = 2 and y = 1 leads to z = 2/3, the coordinate of the largest box will be (2, 1, 2/3) .
## Alternative: Second Derivative Test
If you don’t want to rely on geometry, you can use the second derivative test to find out which critical point gives us that box with maximum volume:
D = D(a, b) = fxx(a,b) fyy(a, b) – [fxy(a, b)]2
at (6, 0), (0, 0), (0, 3)
D = 0 – 4 = -4
at (2, 1)
Which leads to the same conclusion.
So the volume V = xyz = 2 · 1 · 2/3 = 4/3.
## Optimization Problems: References
Miech, Ron. Calculus Problems: Volume of the Largest Rectangular Box. Retrieved from http://www.math.ucla.edu/~ronmiech/Calculus_Problems/32A/chap12/section7/820d45/820_45.html on January 1, 2019.
Formula for the volume of a rectangular box.
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# Solving for Xmas: how to make mathematical Christmas cards
With less than a month to go before Christmas Day, it’s time to start thinking about cards. Here Oxford maths don Vicky Neale explains how to impress your friends with these stunning geometrical designs. It’s Christ-maths time, and there’s no need to be afraid.
Christmas is a fantastic opportunity for me to share some maths with friends and family. One of my favourite ways to do this is by stitching geometrical designs on cards. The magic is how the straight lines produce perfect curves.
Here’s how it’s done. Draw two straight lines that intersect. Draw points along each of those lines at equal distances. When you join the dots from one line to the other, as in the star and tree above and in the four images below, you get a parabola. Strictly speaking, the curve is the envelope of the family of straight lines.
Let’s start simple: here are two parabolas in a square:
The angle between the intersecting lines makes no difference - you always get a parabola. Here are three within a triangle.
Four within a square.
And six in hexagonal form.
This idea of using lines to make the envelope of a curve has been used by numerous artists and architects over the years. Barbara Hepworth used it to great effect, and it appears in the Chords suspension bridge in Jerusalem.
There are lots of other curves that occur as envelopes. For example, here’s my Christmas card from last year.
To make this one, you’ll need a circle of evenly spaced dots. I find that 36 dots works well. Pick a number, I suggest a number between 3 and 15. And then simply take each point in turn, count round your chosen number of spaces to another point, and join the two. Here the envelope turns out to be a circle. You’ll get different sizes of circle by choosing different spacings.
Some spacings lead to a star that can be drawn without taking the pencil off the paper, others are made up of multiple stars. Thinking about which spacings are which, and how we can predict the number of stars, leads to some beautiful mathematical ideas. I feel sure this is what my friends consider over their mince pies when my cards arrive...
Now for the practical advice. When I make these cards, I buy the sort of card that is ready-made with two folds. These are available surprisingly cheaply if you order online. I do the stitching on the middle panel, and then can fold over one end to cover the mess on the back (this is a top tip).
The first step is to draw the design on a piece of paper the same size as the card -- this saves time later on, and saves messing up the cards themselves if you need to redesign the layout. Then place a folded tea towel on a flat table (the tea towel protects the table), put the card face-up over it, then hold the design firmly in place and use a pin to make a hole for each point. Remember to make the holes on the middle panel of the card!
Then you can start stitching. I find that a single strand of stranded cotton works well - stranded cotton is relatively cheap, and comes in lots of colours (including sparkly ones, perfect for Christmas). Use a length around 50cm. If you use longer then you’ll get tangled up very quickly. To start and finish, simply use sticky tape on the back to fasten the thread neatly. This all gets covered at the end, when you can use double-sided tape to glue one panel over the back. I have to concentrate extremely hard at this point to make sure I glue the correct side down, otherwise the card opens backwards!
My final favourite family of designs is also based on a circle of dots, but this time I find that 72 dots works best. Imagine the points as being numbered consecutively from 1 to 72. Join each point to its double (so 1 to 2, 2 to 4, 3 to 6, 4 to 8, and so on). If you do this with a ruler, it becomes quite easy to find a pattern without needing the dots to be numbered, simply move one end of the ruler along one dot and the other end two dots each time. This is what you get:
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# 2.3 graphing linear equations in slope-intercept form · 2009-09-18 · 2.3 62 chapter 2 graphing...
of 6 /6
2.3 62 Chapter 2 Graphing Linear Equations and Linear Systems Graphing Linear Equations in Slope-Intercept Form 2.3 How can you describe the graph of the equation y = mx + b? Work with a partner. Graph the equation. Find the slope of the line. Find the point where the line crosses the y-axis. a. y = 1 2 x + 1 b. y = x + 2 x y 1 2 1 3 2 3 2 1 1 2 3 3 x y 1 2 1 3 2 3 2 1 1 2 3 3 c. y = x 2 d. y = 1 2 x + 1 x y 1 2 1 3 2 3 2 1 1 2 3 3 x y 1 2 1 3 2 3 2 1 1 2 3 3 ACTIVITY: Finding Slopes and y-Intercepts 1 1
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2.3
62 Chapter 2 Graphing Linear Equations and Linear Systems
Graphing Linear Equations in Slope-Intercept Form
2.3
How can you describe the graph of the
equation y = mx + b?
Work with a partner.
● Graph the equation.
● Find the slope of the line.
● Find the point where the line crosses the y-axis.
a. y = − 1
— 2
x + 1 b. y = −x + 2
x
y
1−2 −1−3 2 3
−2
−1
1
2
3
−3
x
y
1−2 −1−3 2 3
−2
−1
1
2
3
−3
c. y = −x − 2 d. y = 1
— 2
x + 1
x
y
1−2 −1−3 2 3
−2
−1
1
2
3
−3
x
y
1−2 −1−3 2 3
−2
−1
1
2
3
−3
ACTIVITY: Finding Slopes and y-Intercepts11
Section 2.3 Graphing Linear Equations in Slope-Intercept Form 63
Equation Description of GraphSlope of Graph
Point of Intersection with y-axis
2. y = − 1
— 2
x + 1 Line − 1
— 2
(0, 1)
3. y = −x + 2
4. y = −x − 2
5. y = 1
— 2
x + 1
6. y = x + 2
7. y = x − 2
8. y = 1
— 2
x − 1
9. y = − 1
— 2
x − 1
10. y = 3x + 2
11. y = 3x − 2
12. y = −2x + 3
1a1a
1b1b
1c1c
1d1d
Use what you learned about graphing linear equations in slope-intercept form to complete Exercises 4 – 6 on page 66.
13. IN YOUR OWN WORDS How can you describe the graph of the equation y = mx + b ?
a. How does the value of m affect the graph of the equation?
b. How does the value of b affect the graph of the equation?
c. Check your answers to parts (a) and (b) with three equations that are not in the table.
14. Why is y = mx + b called the “slope-intercept” form of the equation of a line?
Inductive ReasoningWork with a partner. Graph each equation. Then copy and complete the table.
64 Chapter 2 Graphing Linear Equations and Linear Systems
Lesson2.3Lesson Tutorials
Key Vocabularyx-intercept, p. 64y-intercept, p. 64slope-intercept form, p. 64
Intercepts
The x-intercept of a line is the x-coordinate of the point where the line crosses the x-axis. It occurs when y = 0.
The y-intercept of a line is the y-coordinate of the point where the line crosses the y-axis. It occurs when x = 0.
Slope-Intercept Form
Words An equation written in the form y = mx + b is in slope-intercept form. The slope of the line is m and the y-intercept of the line is b.
Algebra y = mx + b
slope y-intercept
EXAMPLE Identifying Slopes and y -Intercepts11Find the slope and y-intercept of the graph of each linear equation.
a. y = −4x − 2
y = −4x + (−2) Write in slope-intercept form.
The slope is −4 and the y-intercept is −2.
b. y − 5 = 3
— 2
x
y = 3
— 2
x + 5 Add 5 to each side.
The slope is 3
— 2
and the y-intercept is 5.
Find the slope and y-intercept of the graph of the linear equation.
1. y = 3x − 7 2. y − 1 = − 2
— 3
xExercises 7–15
y
x(a, 0)
(0, b)
O
y-intercept = b
x-intercept = a
Section 2.3 Graphing Linear Equations in Slope-Intercept Form 65
EXAMPLE Graphing a Linear Equation in Slope-Intercept Form22
Graph y = −3x + 3. Identify the x-intercept.
Step 1: Find the slope and y-intercept.
y = −3x + 3
Step 2: The y-intercept is 3. So, plot (0, 3).
x
y
−1
−2
−2−3
1
2 3
(0, 3)−3
1
y = −3x + 3
Step 3: Use the slope to fi nd another point and draw the line.
slope = rise
— run
= −3
— 1
Plot the point that is 1 unit right and 3 units down from (0, 3). Draw a line through the two points.
The line crosses the x-axis at (1, 0). So, the x-intercept is 1.
slope y-intercept
Study TipYou can check the x-intercept by substituting y = 0 in the equation and solving for x. y = −3x + 3 0 = −3x + 3−3 = −3x 1 = x
EXAMPLE Real-Life Application33
The cost y (in dollars) of taking a taxi x miles is y = 2.5x + 2. (a) Graph the equation. (b) Interpret the y-intercept and slope.
a. The slope of the line is 2.5 = 5
— 2
. Use the slope and y-intercept to graph the equation.
x
y
10 2 3 4 5 6
4
3
0
5
6
7
(0, 2) 2
5
The y-intercept is 2.So, plot (0, 2). Use the slope to plot
another point, (2, 7). Drawa line through the points.
b. The slope is 2.5. So, the cost per mile is \$2.50. The y-intercept is 2. So, there is an initial fee of \$2 to take the taxi.
Graph the linear equation. Identify the x-intercept.
3. y = x − 4 4. y = − 1
— 2
x + 1
5. In Example 3, the cost y (in dollars) of taking a different taxi x miles is y = 2x + 1.5. Interpret the y-intercept and slope.
Exercises 18–23
Exercises2.3
9+(-6)=3
3+(-3)=
4+(-9)=
9+(-1)=
66 Chapter 2 Graphing Linear Equations and Linear Systems
1. VOCABULARY How can you fi nd the x-intercept of the graph of 2x + 3y = 6?
2. CRITICAL THINKING Is the equation y = 3x in slope-intercept form? Explain.
3. OPEN-ENDED Describe a real-life situation that can be modeled by a linear equation. Write the equation. Interpret the y-intercept and slope.
Match the equation with its graph. Identify the slope and y-intercept.
4. y = 2x + 1 5. y = 1
— 3
x − 2 6. y = − 2
— 3
x + 1
A.
x
y
1−2 −1−3 2 3
1
2
3
−3
B.
x
y
1−2−3 2 3
1
2
3
−3
−2
C.
x
y
1−2 −1−3 3
2
3
−3
−2
Find the slope and y-intercept of the graph of the linear equation.
7. y = 4x − 5 8. y = −7x + 12 9. y = − 4
— 5
x − 2
10. y = 2.25x + 3 11. y + 1 = 4
— 3
x 12. y − 6 = 3
— 8
x
13. y − 3.5 = −2x 14. y + 5 = − 1
— 2
x 15. y = 1.5x + 11
16. ERROR ANALYSIS Describe and correct the error in fi nding the slope and y-intercept of the graph of the linear equation.
17. SKYDIVING A skydiver parachutes to the ground. The height y (in feet) of the skydiver after x seconds is y = −10x + 3000.
a. Graph the equation.
b. Interpret the x-intercept and slope.
Help with Homework
y = 4x − 3
The slope is 4 and the y-intercept is 3.
11
Section 2.3 Graphing Linear Equations in Slope-Intercept Form 67
Solve the equation for y.
29. y − 2x = 3 30. 4x + 5y = 13 31. 2x − 3y = 6 32. 7x + 4y = 8
33. MULTIPLE CHOICE Which point is a solution of the equation 3x − 8y = 11?
○A (1, 1) ○B (1, −1) ○C (−1, 1) ○D (−1, −1)
Graph the linear equation. Identify the x-intercept.
18. y = 1
— 5
x + 3 19. y = 6x − 7 20. y = − 8
— 3
x + 9
21. y = −1.4x − 1 22. y + 9 = −3x 23. y − 4 = − 3
— 5
x
24. PHONES The cost y (in dollars) of making a long distance phone call for x minutes is y = 0.25x + 2.
a. Graph the equation.
b. Interpret the slope and y-intercept.
25. APPLES Write a linear equation that models the cost y of picking x pounds of apples. Graph the equation.
26. ELEVATOR The basement of a building is 40 feet below ground level. The elevator rises at a rate of 5 feet per second. You enter the elevator in the basement. Write an equation that represents the height y (in feet) of the elevator after x seconds. Graph the equation.
27. BONUS You work in an electronics store. You earn a fi xed amount of \$35 per day, plus a 15% bonus on the merchandise you sell. Write an equation that models the amount y (in dollars) you earn for selling x dollars of merchandise in one day. Graph the equation.
28. Six friends create a website. The website earns money by selling banner ads. The site has fi ve banner ads. It costs \$120 a month to operate the website.
a. A banner ad earns \$0.005 per click. Write a linear equation that represents the monthly income y (in dollars) for x clicks.
b. Draw a graph of the equation in part (a). On the graph, label the number of clicks needed for the friends to start making a profi t.
22
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# NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Straight Lines
#### NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Straight Lines
The NCERT Solutions for Class 11 Maths Chapter 9, covering Straight Lines, are a helpful resource for students preparing for their Math board exams. The chapter covers important topics like slope, Horizontal and vertical Lines, Point-slope form, Two-point form, Slope-intercept form, Intercept – form, and Normal form. These solutions are designed to help students practice and improve their understanding using straightforward methods and formulas.
### NCERT Solutions for Class 11 Maths Chapter 09
#### EXERCISE 9.3 PAGE NO: 167
1. Reduce the following equations into slope-intercept form and find their slopes and the y-intercepts.
(i) x + 7y = 0
(ii) 6x + 3y – 5 = 0
(iii) y = 0
Solution:
(i) x + 7y = 0
Given:
The equation is x + 7y = 0
The slope-intercept form is represented in the form ‘y = mx + c’, where m is the slope and c is the y-intercept.
So, the above equation can be expressed as
y = -1/7x + 0
∴ The above equation is of the form y = mx + c, where m = -1/7 and c = 0
(ii) 6x + 3y – 5 = 0
Given:
The equation is 6x + 3y – 5 = 0
The slope-intercept form is represented in the form ‘y = mx + c’, where m is the slope and c is the y-intercept.
So, the above equation can be expressed as
3y = -6x + 5
y = -6/3x + 5/3
= -2x + 5/3
∴ The above equation is of the form y = mx + c, where m = -2 and c = 5/3
(iii) y = 0
Given:
The equation is y = 0
The slope-intercept form is given by ‘y = mx + c’, where m is the slope and c is the y-intercept.
y = 0 × x + 0
∴ The above equation is of the form y = mx + c, where m = 0 and c = 0
2. Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y – 12 = 0
(ii) 4x – 3y = 6
(iii) 3y + 2 = 0
Solution:
(i) 3x + 2y – 12 = 0
Given:
The equation is 3x + 2y – 12 = 0
The equation of the line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepted on the x-axis and the y-axis, respectively.
So, 3x + 2y = 12
Now, let us divide both sides by 12; we get
3x/12 + 2y/12 = 12/12
x/4 + y/6 = 1
∴ The above equation is of the form x/a + y/b = 1, where a = 4, b = 6
The intercept on the x-axis is 4.
The intercept on the y-axis is 6.
(ii) 4x – 3y = 6
Given:
The equation is 4x – 3y = 6
The equation of the line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepted on the x-axis and the y-axis, respectively.
So, 4x – 3y = 6
Now, let us divide both sides by 6; we get
4x/6 – 3y/6 = 6/6
2x/3 – y/2 = 1
x/(3/2) + y/(-2) = 1
∴ The above equation is of the form x/a + y/b = 1, where a = 3/2, b = -2
The intercept on the x-axis is 3/2.
The intercept on the y-axis is -2.
(iii) 3y + 2 = 0
Given:
The equation is 3y + 2 = 0
The equation of the line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepted on the x-axis and the y-axis, respectively.
So, 3y = -2
Now, let us divide both sides by -2; we get
3y/-2 = -2/-2
3y/-2 = 1
y/(-2/3) = 1
∴ The above equation is of the form x/a + y/b = 1, where a = 0, b = -2/3
The intercept on the x-axis is 0.
The intercept on the y-axis is -2/3.
3. Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).
Solution:
Given:
The equation of the line is 12(x + 6) = 5(y – 2).
12x + 72 = 5y – 10
12x – 5y + 82 = 0 … (1)
Now, compare equation (1) with the general equation of line Ax + By + C = 0, where A = 12, B = –5, and C = 82
Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
∴ The distance is 5 units.
4. Find the points on the x-axis whose distances from the line x/3 + y/4 = 1 are 4 units.
Solution:
Given:
The equation of the line is x/3 + y/4 = 1
4x + 3y = 12
4x + 3y – 12 = 0 …. (1)
Now, compare equation (1) with the general equation of line Ax + By + C = 0, where A = 4, B = 3, and C = -12
Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.
So, the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
|4a – 12| = 4 × 5
± (4a – 12) = 20
4a – 12 = 20 or – (4a – 12) = 20
4a = 20 + 12 or 4a = -20 + 12
a = 32/4 or a = -8/4
a = 8 or a = -2
∴ The required points on the x-axis are (-2, 0) and (8, 0)
5. Find the distance between parallel lines.
(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0
(ii) l(x + y) + p = 0 and l (x + y) – r = 0
Solution:
(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0
Given:
The parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.
By using the formula,
The distance (d) between parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by
∴ The distance between parallel lines is 65/17
(ii) l(x + y) + p = 0 and l (x + y) – r = 0
Given:
The parallel lines are l (x + y) + p = 0 and l (x + y) – r = 0
lx + ly + p = 0 and lx + ly – r = 0
By using the formula,
The distance (d) between parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by
∴ The distance between parallel lines is |p+r|/l2
6. Find the equation of the line parallel to the line 3x − 4y + 2 = 0 and passing through the point (–2, 3).
Solution:
Given:
The line is 3x – 4y + 2 = 0
So, y = 3x/4 + 2/4
= 3x/4 + ½
Which is of the form y = mx + c, where m is the slope of the given line.
The slope of the given line is 3/4
We know that parallel lines have the same slope.
∴ Slope of other line = m = 3/4
The equation of line having slope m and passing through (x1, y1) is given by
y – y1 = m (x – x1)
∴ The equation of the line having slope 3/4 and passing through (-2, 3) is
y – 3 = ¾ (x – (-2))
4y – 3 × 4 = 3x + 3 × 2
3x – 4y = 18
∴ The equation is 3x – 4y = 18
7. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.
Solution:
Given:
The equation of line is x – 7y + 5 = 0
So, y = 1/7x + 5/7 [which is of the form y = mx + c, where m is the slope of the given line.]
The slope of the given line is 1/7
The slope of the line perpendicular to the line having slope m is -1/m
The slope of the line perpendicular to the line having a slope of 1/7 is -1/(1/7) = -7
So, the equation of the line with slope -7 and the x-intercept 3 is given by y = m(x – d)
y = -7 (x – 3)
y = -7x + 21
7x + y = 21
∴ The equation is 7x + y = 21
8. Find angles between the lines √3x + y = 1 and x + √3y = 1.
Solution: Given:
The lines are √3x + y = 1 and x + √3y = 1
So, y = -√3x + 1 … (1) and
y = -1/√3x + 1/√3 …. (2)
The slope of the line (1) is m1 = -√3, while the slope of the line (2) is m2 = -1/√3
Let θ be the angle between two lines.
So,
θ = 30°
∴ The angle between the given lines is either 30° or 180°- 30° = 150°
9. The line through the points (h, 3) and (4, 1) intersects the line 7x − 9y −19 = 0. At the right angle. Find the value of h.
Solution:
Let the slope of the line passing through (h, 3) and (4, 1) be m1
Then, m1 = (1-3)/(4-h) = -2/(4-h)
Let the slope of line 7x – 9y – 19 = 0 be m2
7x – 9y – 19 = 0
So, y = 7/9x – 19/9
m2 = 7/9
Since the given lines are perpendicular,
m1 × m2 = -1
-2/(4-h) × 7/9 = -1
-14/(36-9h) = -1
-14 = -1 × (36 – 9h)
36 – 9h = 14
9h = 36 – 14
h = 22/9
∴ The value of h is 22/9
10. Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x – x1) + B (y – y1) = 0.
Solution:
Let the slope of line Ax + By + C = 0 be m
Ax + By + C = 0
So, y = -A/Bx – C/B
m = -A/B
By using the formula,
Equation of the line passing through point (x1, y1) and having slope m = -A/B is
y – y1 = m (x – x1)
y – y1= -A/B (x – x1)
B (y – y1) = -A (x – x1)
∴ A(x – x1) + B(y – y1) = 0
So, the line through point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x – x1) + B (y – y1) = 0
Hence, proved.
11. Two lines passing through point (2, 3) intersects each other at an angle of 60o. If the slope of one line is 2, find the equation of the other line.
Solution:
Given: m1 = 2
Let the slope of the first line be m1
And let the slope of the other line be m2.
The angle between the two lines is 60°.
So,
12. Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).
Solution:
Given:
The right bisector of a line segment bisects the line segment at 90°.
End-points of the line segment AB are given as A (3, 4) and B (–1, 2).
Let the mid-point of AB be (x, y).
x = (3-1)/2= 2/2 = 1
y = (4+2)/2 = 6/2 = 3
(x, y) = (1, 3)
Let the slope of line AB be m1
m1 = (2 – 4)/(-1 – 3)
= -2/(-4)
= 1/2
And let the slope of the line perpendicular to AB be m2
m2 = -1/(1/2)
= -2
The equation of the line passing through (1, 3) and having a slope of –2 is
(y – 3) = -2 (x – 1)
y – 3 = – 2x + 2
2x + y = 5
∴ The required equation of the line is 2x + y = 5
13. Find the coordinates of the foot of the perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.
Solution:
Let us consider the coordinates of the foot of the perpendicular from (-1, 3) to the line 3x – 4y – 16 = 0 be (a, b)
So, let the slope of the line joining (-1, 3) and (a, b) be m1
m= (b-3)/(a+1)
And let the slope of the line 3x – 4y – 16 = 0 be m2
y = 3/4x – 4
m2 = 3/4
Since these two lines are perpendicular, m1 × m2 = -1
(b-3)/(a+1) × (3/4) = -1
(3b-9)/(4a+4) = -1
3b – 9 = -4a – 4
4a + 3b = 5 …….(1)
Point (a, b) lies on the line 3x – 4y = 16
3a – 4b = 16 ……..(2)
Solving equations (1) and (2), we get
a = 68/25 and b = -49/25
∴ The coordinates of the foot of perpendicular are (68/25, -49/25)
14. The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c.
Solution:
Given:
The perpendicular from the origin meets the given line at (–1, 2).
The equation of the line is y = mx + c
The line joining the points (0, 0) and (–1, 2) is perpendicular to the given line.
So, the slope of the line joining (0, 0) and (–1, 2) = 2/(-1) = -2
The slope of the given line is m.
m × (-2) = -1
m = 1/2
Since point (-1, 2) lies on the given line,
y = mx + c
2 = 1/2 × (-1) + c
c = 2 + 1/2 = 5/2
∴ The values of m and c are 1/2 and 5/2, respectively.
15. If p and q are the lengths of perpendiculars from the origin to the lines x cos θ − y sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p2 + 4q2 = k2
Solution:
Given:
The equations of the given lines are
x cos θ – y sin θ = k cos 2θ …………………… (1)
x sec θ + y cosec θ = k ……………….… (2)
Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
q = k cos θ sin θ
Multiply both sides by 2, and we get
2q = 2k cos θ sin θ = k × 2sin θ cos θ
2q = k sin 2θ
Squaring both sides, we get
4q2 = k2 sin22θ …………………(4)
Now add (3) and (4); we get
p2 + 4q2 = k2 cos2 2θ + k2 sin2 2θ
p2 + 4q2 = k2 (cos2 2θ + sin2 2θ) [Since, cos2 2θ + sin2 2θ = 1]
∴ p2 + 4q2 = k2
Hence proved.
16. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from vertex A.
Solution:
Let AD be the altitude of triangle ABC from vertex A.
So, AD is perpendicular to BC.
Given:
Vertices A (2, 3), B (4, –1) and C (1, 2)
Let the slope of the line BC = m1
m1 = (- 1 – 2)/(4 – 1)
m1 = -1
Let the slope of the line AD be m2
m1 × m2 = -1
-1 × m2 = -1
m2 = 1
The equation of the line passing through the point (2, 3) and having a slope of 1 is
y – 3 = 1 × (x – 2)
y – 3 = x – 2
y – x = 1
Equation of the altitude from vertex A = y – x = 1
Length of AD = Length of the perpendicular from A (2, 3) to BC
The equation of BC is
y + 1 = -1 × (x – 4)
y + 1 = -x + 4
x + y – 3 = 0 …………………(1)
Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
Now, compare equation (1) to the general equation of the line, i.e., Ax + By + C = 0; we get
[where, A = 1, B = 1 and C = -3]
∴ The equation and the length of the altitude from vertex A are y – x = 1 and
√2 units, respectively.
17. If p is the length of the perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that 1/p2 = 1/a2 + 1/b2
Solution:
The equation of a line whose intercepts on the axes are a and b is x/a + y/b = 1
bx + ay = ab
bx + ay – ab = 0 ………………..(1)
Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
Now, square on both sides; we get
∴ 1/p2 = 1/a2 + 1/b2
Hence, proved.
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# Harry's Job at a Candy Store
Media Type:
Video
Running Time: 3m 27s
Size: 9.5 MB
or
Source: Cyberchase: “A Fraction of a Chance”
### Collection Developed by:
Collection Credits
### Collection Funded by:
Funding for the VITAL/Ready to Teach collection was secured through the United States Department of Education under the Ready to Teach Program.
In this video segment from Cyberchase, Harry gets a new job as the manager of a candy store. Customers who come into the store use fractions to order boxes of different types of chocolates. Harry teaches his cousin, Harley, how the fractional amounts correspond to the number of pieces to place in the box. Harley takes over during Harry’s lunch break, and when Harry returns he finds a disappointed customer was not given the correct amount of chocolates in his order.
Connections
Everyday Math (2004)
Teacher Lesson Guide: pp. 105, 462, 920
Teacher Reference Manual: pp. 145-148.
Student Reference Book: pp. 69
Teaching Tips
Here are some Frame, Focus and Follow-up suggestions for using this video in a math lesson.
What is Frame, Focus and Follow-up?
Frame: When we have a group or a set of something, like a bowl of berries, we can use fractions to talk about how much of the bowl is made up of one particular fruit, such as strawberries. If there are 12 berries in the bowl and 3 are strawberries, what fraction could we use to describe the amount of berries that are strawberries? What fraction of that same bowl would be other kinds of berries?
Focus: As you watch this segment, notice all the different candy orders that Harry has to fill and how the customers use fractions to describe the mix of chocolates that they want. Write down at least one of the arrangements from one of the boxes.
Follow Up: Describe one candy arrangement from one of the orders that Harry filled. Use fractions to describe the arrangement. Why do you think the customers used fractions when they ordered? All of the boxes Harry used had space for eight candies. What is the smallest fraction of the box that could be used when placing an order? What would the largest fraction be?
Transcript
HARRY: I've got a sweet job in an amazing place! And what's really awesome is I'm the manager! Harley! What are you doing here?
HARLEY: I heard about your job, I wanted to check it out. You know I have a thing for chocolate. So, how about some free samples?
HARRY: No way, Harley. Li-lac chocolates. Margaret? You sound awful. I… I hope you feel better. Have some chicken soup. Bye. My employee called in sick. I had something special planned for my lunch break. If I'm the only one here, I can't leave the store!
HARLEY: Let me help. I have great people skills.
HARRY: I don't know…
HARLEY: Hire me for the day and I'll cover for you while you go on your lunch break…
HARRY: Well…
HARLEY: Great. Now that I'm an employee, can I have free samples?
HARRY: Harley, no. You have to pay, just like the customers.
CUSTOMER #1: I'd like a box of truffles, please. 5/8 hazelnut and 3/8 raspberry.
HARRY: Watch and learn. The bottom number shows how many pieces are in a whole box. The top number shows how many are hazelnut or raspberry. 5/8 hazelnut and 3/8 raspberry.
CUSTOMER #2: Hi, may I have a box of truffles, please? 5/8 hazelnut, 2/8 raspberry, and 1/8 mocha.
HARRY: Harley, pay attention.
CUSTOMER #2: Thank you.
HARRY: You're welcome.
HARLEY: Why don't you take your lunch break now?
HARRY: Are you sure you can handle this by yourself?
HARLEY: Of course! If you can do it, anybody can! What I mean is…since you're such a great teacher, I know what to do.
HARRY: Okay. Thanks, Harley.
HARLEY: My pleasure.
CUSTOMER #3: Hi, how are ya. I'd like to get a box of truffles – 4/8 hazelnut, 3/8 mocha and 1/8 caramel.
HARRY: Thanks for covering for me, Harley.
CUSTOMER #3: Hmmph! I just bought this box of truffles, and 4/8 are missing!!! There are only hazelnut truffles in this box!!! I also ordered 3/8 mocha and 1/8 caramel.
HARRY: I am so sorry. Of course, we'll replace the missing ones.
CUSTOMER #3: Hmmph!
HARRY: Do you know how that happened?
HARLEY: No idea.
HARRY: Hey Harley, I have some more chocolate for you.
HARLEY: You're fired!
Standards
to:
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# When calculating to find the mean and standard deviation of these two sets of numbers: (a)4 0 1 4 3 6 (b)5 3 1 3 4 2 Which data set is more spread out?
Sep 16, 2016
Data-set (a) is more spread out.
#### Explanation:
Both the given data sets have exactly $6$ data points and their sum is too $18$, hence mean of both is $\frac{18}{6} = 3$.
As regards which data is more spread out, there are mainly three measures for the same.
First is Range, which is the difference of larges and smallest data point. As the range of data-set (a) is $6 - 0 = 6$ and that of data-set (b) is $5 - 1 = 4$, hence data-set (a) is more spread out.
Another measure is mean deviation, which is the average of difference (i.e. absolute value) between data point and mean.
In data-set (a) absolute value of deviations (note it makes deviations positive and hence does not cancel out, which will happen if we keep positive or negative sign of differences) is $\left\{| 4 - 3 | , | 0 - 3 | . | 1 - 3 | , | 4 - 3 | , | 3 - 3 | , | 6 - 3 |\right\}$ i.e. $\left\{1 , 3 , 2 , 1 , 0 , 3\right\}$ and their mean is $\frac{1 + 3 + 2 + 1 + 0 + 3}{6} = \frac{10}{6} = \frac{5}{3}$. In data-set (b) absolute value of deviations is $\left\{| 5 - 3 | , | 3 - 3 | . | 1 - 3 | , | 3 - 3 | , | 4 - 3 | , | 2 - 3 |\right\}$ i.e. $\left\{2 , 0 , 2 , 0 , 1 , 1\right\}$ and their mean is $\frac{2 + 0 + 2 + 0 + 1 + 1}{6} = \frac{6}{6} = 1$. As mean deviation in data-set (a) is higher, it is more spread out.
The third measure is standard deviation, for which we square the deviations (again to make them positive), take their average and then take square root.
We have already worked out deviations above. In data-set (a) squares of deviations are$\left\{1 , 9 , 4 , 1 , 0 , 9\right\}$ and their mean is $\frac{1 + 9 + 4 + 1 + 0 + 9}{6} = \frac{24}{6} = 4$ and standard deviation is $\sqrt{4} = 2$. In data-set (b) squares of deviations are $\left\{4 , 0 , 4 , 0 , 1 , 1\right\}$ and their mean is $\frac{4 + 0 + 4 + 0 + 1 + 1}{6} = \frac{10}{6} = \frac{5}{3}$ and standard deviation is $\sqrt{\frac{5}{3}} = 1.291$. Again as standard deviation in data-set (a) is higher, it is more spread out.
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## Computing Limits II: The Squeeze Theorem – Application Proof
Posted: 9th February 2013 by seanmathmodelguy in Lectures
One very important application of the squeeze theorem is the proof that $$\displaystyle \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$$. We present this proof next.
This proof is for the limit as $$\theta\to 0^+$$. The case for $$\theta\to 0^-$$ can be proved in exactly the same manner and we leave it as an exercise to the student.
Consider the figure to the right. Notice that the area of the triangle OAP is less than the area of the sector OAP which is in turn less than the area of the triangle OAT. Let’s estimate each of these areas in turn.
\begin{align*}
\mbox{Area of triangle OAP} &=
\frac{1}{2} \mbox{base} \times \mbox{height}\\
&= \frac{1}{2} (1) \sin \theta = \frac {1}{2} \sin \theta.\\
\mbox{Area of sector OAP} &=
\frac{1}{2} r^2 \theta = \frac{1}{2} (1) \theta =
\frac {1}{2} \theta.\\
\mbox{Area of triangle OAT} &=
\frac{1}{2} (1) \tan\theta = \frac{1}{2} \tan\theta.
\end{align*}
Using our initial observation, one has $$\frac{1}{2} \sin \theta \le \frac{1}{2} \theta \le \frac{1}{2} \tan \theta, \hspace{1cm} 0 \le \theta < \frac{\pi}{2}.\hspace{2cm}(1)$$The inequality for $$-\pi/2 < \theta \le 0$$ is $$\tan\theta\le\theta\le\sin\theta$$. From the left hand side of (1) one has $$\sin\theta\le\theta$$ or $$(\sin\theta)/\theta\le 1$$ if $$\theta \ge 0$$. From the right hand side we have $$\theta \le \tan\theta$$ or $$\cos\theta\le (\sin\theta)/\theta$$ if $$\theta \ge 0$$. In other words,$$\cos \theta \le \frac{\sin \theta}{\theta} \le 1$$provided $$0\le\theta<\pi/2$$. Again, we leave it to the student to verify that this last inequality is true for $$-\pi/2<\theta\le 0$$. At this point, we look at the limit as $$\theta\to 0$$ and apply the squeeze theorem:$$\lim_{\theta\to 0} \cos \theta \le \lim_{\theta\to 0} \frac{\sin \theta}{\theta} \le \lim_{\theta\to 0} 1.$$Since the left and right hand sides has a limit of one, we can conclude that $$\lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1.$$This limit is used so often in calculus it is a good idea to remember it. We leave this material with two examples that use this new result.
Examples:
1. Find $$\displaystyle \lim_{x\to 0}\frac{\sin^2 x}{x}$$.
Again, we use the limit laws, $$\lim_{x\to 0} \frac{\sin^2 x}{x} = \left(\lim_{x\to 0} \frac{\sin x}{x}\right) (\lim_{x\to 0} \sin x) = 1\cdot 0 = 0.$$
2. Find $$\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^2}$$.
For this one we need to multiply by the conjugate of the numerator,
\begin{align*}
\lim_{x\to 0}\frac{1-\cos x}{x^2} &=
\lim_{x\to 0}\frac{1-\cos x}{x^2}
\left(\frac{1+\cos x}{1+\cos x}\right) =
\lim_{x\to 0}\frac{1-\cos^2 x}{x^2(1+\cos x)}\\ &=
\lim_{x\to 0}\frac{\sin^2 x}{x^2(1+\cos x)} =
\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^2
\lim_{x\to 0}\frac{1}{1+\cos x}
= 1^2 \cdot \frac{1}{2} = \frac{1}{2}.
\end{align*}
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## Trigonometry (11th Edition) Clone
Published by Pearson
# Chapter 5 - Trigonometric Identities - Section 5.1 Fundamental Identities - 5.1 Exercises - Page 200: 34
#### Answer
$$\sin\theta=-\frac{2}{5}$$ $$\cos\theta=-\frac{\sqrt{21}}{5}$$ $$\tan\theta=\frac{2\sqrt{21}}{21}$$ $$\cot\theta=\frac{\sqrt{21}}{2}$$ $$\sec\theta=-\frac{5\sqrt{21}}{21}$$
#### Work Step by Step
$$\csc\theta=-\frac{5}{2}\hspace{1.5cm}\theta\hspace{0.1cm}\text{in quadrant III}$$ 1) Reciprocal Identities: $$\csc\theta=\frac{1}{\sin\theta}$$ $$\sin\theta=\frac{1}{\csc\theta}=\frac{1}{-\frac{5}{2}}=-\frac{2}{5}$$ 2) Pythagorean Identities: $$\csc^2\theta=\cot^2\theta+1$$ $$\cot^2\theta=\csc^2\theta-1=(-\frac{5}{2})^2-1=\frac{25}{4}-1=\frac{21}{4}$$ $$\cot\theta=\pm\frac{\sqrt{21}}{2}$$ Also, $$\cos^2\theta=1-\sin^2\theta=1-(-\frac{2}{5})^2=1-\frac{4}{25}=\frac{21}{25}$$ $$\cos\theta=\pm\frac{\sqrt{21}}{5}$$ We know that $\theta$ is in quadrant IV, where both $\sin\theta\lt0$ and $\cos\theta\lt0$. Therefore, $$\cos\theta=-\frac{\sqrt{21}}{5}$$ 3) Quotient Identities: $$\cot\theta=\frac{\cos\theta}{\sin\theta}$$ We already have $\sin\theta\lt0$ and $\cos\theta\lt0$.That means, with the above identity, $\cot\theta\gt0$ $$\cot\theta=\frac{\sqrt{21}}{2}$$ Also, we have $$\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{-\frac{2}{5}}{-\frac{\sqrt{21}}{5}}=\frac{2}{\sqrt{21}}=\frac{2\sqrt{21}}{21}$$ 4) Reciprocal Identities: $$\sec\theta=\frac{1}{\cos\theta}=\frac{1}{-\frac{\sqrt{21}}{5}}=-\frac{5}{\sqrt{21}}=-\frac{5\sqrt{21}}{21}$$
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How to Calculate Fractional Exponents
Page content
The Basics
Fractional exponents can look intimidating, but they’re much simpler than they seem. Remember that ½ is really the reciprocal – or the “opposite” of 2. That’s why multiplying 3 times 2 gives you 6, so if you want to get from 6 back to 3, you need to multiply by the reciprocal of 2: ½. So 6 X ½ = 3.
The same thing applies to fractional exponents. Let’s take a simple example. You know that 3^2 = 9. Why? Because 3 X 3 (3 multiplied two times) is 9. So how can you get from 9 back to 3? You know one simple way – take the square root of 9, and you’ll get 3. Guess what? You could also use a fractional exponent of ½, since it’s the reciprocal of 2: 9^(1/2) is also 3.
In other words, this the basic rule of fractional exponents: To raise a base to a fractional power, following the following steps:
**1.**Find the reciprocal of the power.
**2.**Take the resulting root of the base.
A Few Examples
Because this basic rule can be tough to understand, here are a few examples to make it clearer.
1. 16^(1/2) – To find the answer to 16^(1/2), first take the reciprocal of the power. The reciprocal of ½ is 2. Then, take the 2nd root of the base (or the square root). The square root of 16 is 4, so 16^(1/2) = 4.
2. 27^(1/3) – To find the answer to 27^(1/3), first take the reciprocal of the power. The reciprocal of 1/3 is 3. Then, take the 3rd root of the base (or the cubic root). The cubic root of 27 is 3, so 27^(1/3) = 3.
3. 16^(1/4) – To find the answer to 16^(1/4), first take the reciprocal of the power. The reciprocal of 1/4 is 4. Then, take the 4th root of the base. The 4th root of 16 is 2 (2 X 2 X 2 X 2 = 16), so 16^(1/4) = 2.
More Complex Roots
The rule above works if the numerator of the fractional power is 1, but what do you do if the fractional power is more than 1? Easy – treat it as a power.
For example, take the problem 4^(3/2). That’s the same thing as (4^(1/2))^3. So first you’d figure out that 4^(1/2) is 2. Then you’d raise the answer (2) to the third power. 2^3 = 8, so 4^(3/2) = 8 too.
Next time you’re not sure how to calculate fractional exponents, follow the simple steps you just learned. There you go, that wasn’t so hard, was it?
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'If you believe you can do it, you can and you will!'
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# Fractions
Children begin to learn about fractions in Year One when they understand that a fraction is an equal part of a whole. Initially, children learn about halves and quarters. They use shapes to see that the whole has been split into equal amounts. When this understanding is secure, children move on to finding halves and quarters of groups.
In Year Two, children revisit halves and quarters, and thirds are introduced. The language of fractions is introduced: numerator (the top number/how many parts) and denominator (the bottom number/how many parts the whole has been split into). Different fractions are introduced as children become more familiar with the fact that a fraction is an equal part of a whole and that the names of these fractions relate to how many equal parts the whole has been split into. Children are able to recognise pairs of fractions that make a whole, compare them and place them on a number line. They also begin to find unit and non-unit fractions of sets.
When in Year Three, children add and subtract fractions with the same denominator, knowing that only the numerator changes and the denominator stays the same. They begin to find equivalent fractions and to simplify them, understanding that to do so, you must perform the same operation to the numerator and denominator.
In Year Four, children's knowledge of fractions is extended to mixed numbers, which they are able to place on a number line. They continue simplifying fractions, this time using improper fractions. Children also learn how to add fractions that total more than one (mixed numbers) and to subtract fractions from whole numbers. Children are also introduced to decimal numbers, learning about tenths and hundredths as well as equivalent fractions.
Children in Year Five add and subtract fractions with different denominators, using their previous knowledge of finding equivalents to be able to complete the calculation. They also use bar models to help them multiply fractions by whole numbers and learn two methods for multiplying mixed numbers by whole numbers. Children's knowledge of decimals is extended to thousandths and percentages are introduced.
In Year Six, children consolidate and apply their knowledge of fractions to be able to add, subtract, multiply and divide them as well as compare them. They are able to use a formal method to find fractions of amounts. Children know equivalent fractions, decimals and percentages and are able to find percentages of amounts.
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# 3.4 Composition of functions (Page 6/9)
Page 6 / 9
## Finding the domain of a composite function involving radicals
Find the domain of
Because we cannot take the square root of a negative number, the domain of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}\left(-\infty ,3\right].\text{\hspace{0.17em}}$ Now we check the domain of the composite function
The domain of this function is $\text{\hspace{0.17em}}\left(-\infty ,5\right].\text{\hspace{0.17em}}$ To find the domain of $\text{\hspace{0.17em}}f\circ g,\text{\hspace{0.17em}}$ we ask ourselves if there are any further restrictions offered by the domain of the composite function. The answer is no, since $\text{\hspace{0.17em}}\left(-\infty ,3\right]\text{\hspace{0.17em}}$ is a proper subset of the domain of $\text{\hspace{0.17em}}f\circ g.\text{\hspace{0.17em}}$ This means the domain of $\text{\hspace{0.17em}}f\circ g\text{\hspace{0.17em}}$ is the same as the domain of $\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ namely, $\text{\hspace{0.17em}}\left(-\infty ,3\right].$
Find the domain of
$\left[-4,0\right)\cup \left(0,\infty \right)$
## Decomposing a composite function into its component functions
In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to decompose a composite function , so we may choose the decomposition that appears to be most expedient.
## Decomposing a function
Write $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{5-{x}^{2}}\text{\hspace{0.17em}}$ as the composition of two functions.
We are looking for two functions, $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h,\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}f\left(x\right)=g\left(h\left(x\right)\right).\text{\hspace{0.17em}}$ To do this, we look for a function inside a function in the formula for $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ As one possibility, we might notice that the expression $\text{\hspace{0.17em}}5-{x}^{2}\text{\hspace{0.17em}}$ is the inside of the square root. We could then decompose the function as
We can check our answer by recomposing the functions.
$g\left(h\left(x\right)\right)=g\left(5-{x}^{2}\right)=\sqrt{5-{x}^{2}}$
Write $\text{\hspace{0.17em}}f\left(x\right)=\frac{4}{3-\sqrt{4+{x}^{2}}}\text{\hspace{0.17em}}$ as the composition of two functions.
$\begin{array}{l}g\left(x\right)=\sqrt{4+{x}^{2}}\\ h\left(x\right)=\frac{4}{3-x}\\ f=h\circ g\end{array}$
Access these online resources for additional instruction and practice with composite functions.
## Key equation
Composite function $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$
## Key concepts
• We can perform algebraic operations on functions. See [link] .
• When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function.
• The function produced by combining two functions is a composite function. See [link] and [link] .
• The order of function composition must be considered when interpreting the meaning of composite functions. See [link] .
• A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function.
• A composite function can be evaluated from a table. See [link] .
• A composite function can be evaluated from a graph. See [link] .
• A composite function can be evaluated from a formula. See [link] .
• The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function. See [link] and [link] .
• Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions.
• Functions can often be decomposed in more than one way. See [link] .
## Verbal
How does one find the domain of the quotient of two functions, $\text{\hspace{0.17em}}\frac{f}{g}?\text{\hspace{0.17em}}$
Find the numbers that make the function in the denominator $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ equal to zero, and check for any other domain restrictions on $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g,\text{\hspace{0.17em}}$ such as an even-indexed root or zeros in the denominator.
#### Questions & Answers
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
give me treganamentry question
Solve 2cos x + 3sin x = 0.5
madras university algebra questions papers first year B. SC. maths
Hey
Rightspect
hi
chesky
Give me algebra questions
Rightspect
how to send you
Vandna
What does this mean
cos(x+iy)=cos alpha+isinalpha prove that: sin⁴x=sin²alpha
cos(x+iy)=cos aplha+i sinalpha prove that: sinh⁴y=sin²alpha
rajan
cos(x+iy)=cos aplha+i sinalpha prove that: sinh⁴y=sin²alpha
rajan
is there any case that you can have a polynomials with a degree of four?
victor
***sscc.edu/home/jdavidso/math/catalog/polynomials/fourth/fourth.html
Oliver
can you solve it step b step
give me some important question in tregnamentry
Anshuman
what is linear equation with one unknown 2x+5=3
-4
Joel
x=-4
Joel
x=-1
Joan
I was wrong. I didn't move all constants to the right of the equation.
Joel
x=-1
Cristian
Adityasuman x= - 1
y=x+1
gary
what is the VA Ha D R X int Y int of f(x) =x²+4x+4/x+2 f(x) =x³-1/x-1
can I get help with this?
Wayne
Are they two separate problems or are the two functions a system?
Wilson
Also, is the first x squared in "x+4x+4"
Wilson
x^2+4x+4?
Wilson
thank you
Wilson
Please see ***imgur.com/a/lpTpDZk for solutions
Wilson
f(x)=x square-root 2 +2x+1 how to solve this value
factor or use quadratic formula
Wilson
what is algebra
The product of two is 32. Find a function that represents the sum of their squares.
Paul
if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta
how to compute this 1. g(1-x) 2. f(x-2) 3. g (-x-/5) 4. f (x)- g (x)
hi
John
hi
Grace
what sup friend
John
not much For functions, there are two conditions for a function to be the inverse function: 1--- g(f(x)) = x for all x in the domain of f 2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the element that you started with, namely, x.
Grace
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# Factors of 21
The factors of 21 are the numbers that produce the result as 21 when a pair factor is multiplied together. A factor of a number divides the original number uniformly.
The factors of 21 are 1, 3, 7 and 21.
Factor pairs of the number 21 are the numbers that result in the original number multiplying them together. To find the factors of a number, 21, we can use the factorization method.
## How to Find the Factors of 21?
Go through the following steps to find factors of 21.
Step 1: First, write the number 21
Step 2: Find the two numbers, which results in 21 under the multiplication, say 3 and 7, such that 3 × 7 = 21.
Step 3: We know that 3 and 7 are prime numbers with only two factors, i.e., one and the number itself.
The factors of 3 = 3 × 1
The factors of 7 = 7 x 1
So, we cannot further factorize them.
Step 4: Therefore, the factorization of 21 can be expressed as 21 = 3 × 7 × 1
Step 7: Finally, write down all the unique numbers which we can obtain from the above process.
Factors of 21 1, 3, 7, and 21
## Pair Factors of 21
To find the pair factors of 21, multiply the two numbers to get the original number as 21. We can write both positive and negative integers in pairs as shown below:
Positive pairs Negative pairs 1 × 21 = 21; (1, 21) (-1) × (-21) = 21; (-1, -21) 3 × 7 = 21; (3, 7) (-3) × (-7) = 21; (-3, -7) 7 × 3 = 21; (7, 3) (-7) × (-3) = 21; (-7, -3) 21 × 1 = 21; (21, 1) (-21) × (-1) = 21; (-21, -1)
Therefore, the positive pair factors of 21 are (1, 21), (3, 7), (7, 3) and (21, 1)
The negative pair factors of 21 are (-1, -21), (-3, -7), (-7, -3) and (-21, -1).
### Prime Factors of 21
As we know, 21 is a composite number, and it has prime factors. Now let us find the prime factors of 21.
• The first step is to divide the number 21 with the smallest prime number, i.e. 2.
21 ÷ 2 = 10.5
Factors should be whole numbers, so 2 cannot be the factor of 21. Hence, proceed with the next prime number, i.e. 3.
21 ÷ 3 = 7
Thus, 3 is one of the prime factors of 21.
Now, divide 7 by the numbers which are prime.
7 ÷ 2 = 3.5
7 ÷ 3 = 2.333
7 ÷÷ 5 = 1.4
7 ÷ 7 = 1
• We have received number 1 at the end of the division process; thus, we cannot proceed further.
So, the prime factors of 21 are 3 and 7, where 2 and 7 are the prime numbers.
Prime factorisation of 21 is 3 × 7.
Some of the important facts about the factors of 21 are listed below:
• The number of factors of 21 is 4.
• The sum of all factors of 21 is equal to 32.
• The product of all factors of 21 is equal to square of 21 or 21 times of 21.
### Get Factors of More Numbers Here
Learn more about factors, multiples and prime factors of other numbers, visit www.byjus.com also, download BYJU’S – The Learning App.
## Frequently Asked Questions on Factors of 21 – FAQs
### What are the multiples and factors of 21?
The multiples of 21 include 21, 42, 63, 84, 105, 126, 147, 168, 189, 210, etc.,
The factors of 21 are 1, 3, 7 and 21.
### What are multiples of 21?
The multiples of 21 are 21, 42, 63, 84, 105, 126, 147, 168, 189, 210, ….
### What are the factors of 21 and 24?
The factors of 21 are 1, 3, 7 and 21.
The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
### Does 21 only have 2 factors?
The number 21 has 2 more factors other than 1 and 21 as it’s factors are 1, 3, 7 and 21.
### What 3 numbers make 21?
We have multiple group of 3 numbers that add up to 21 such as:
6, 7, 8 (6 + 7 + 8 = 21)
3, 8, 10 (3 + 8 + 10 = 21)
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Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
Observing Student Reasoning and Proof
Introduction | Inscribed Triangles | Problem Reflection #1 | Inscribed Triangle Continued | Problem Reflection #2 | Classroom Practice | Observe a Classroom | Your Journal
Ms. Saleh's tenth-grade geometry class is working on an inscribed triangle problem. Students had previously worked with angles, triangles, and circles and had been introduced to the idea of a proof.
Ms. Selah's instructions: "Work in your groups to construct an inscribed triangle in the top of a semi-circle. Here's an example: You'll need to use your compass, protractor and straightedge to do so, and you may choose any random point C on the semi-circle, you don't have to make the triangle look like the example. After you have your triangle, measure angles A, B, and C for at least five different cases and record your findings in a table. I'm particularly interested in what you find out about angle C. In a few minutes I'll come around to each group to hear your observations."
One group of students created this chart.
Here is a dialogue between Ms. Saleh and members of one group:
Ms. Saleh: Can you describe how you worked on the problem?
Ben: We made a circle with the compass, and then drew lines.
Ms. Saleh: How did you draw the lines?
Ben: We found a place on the top of the circle and drew lines to either corner.
Sonia: See these are our points. {she points to the top of the circle] we actually tried a bunch of different points for C.
Ms. Saleh: What about that point in the middle, O?
Grant: We thought we'd need that because last time we did circles we used it. But we didn't use it to draw with.
Ms. Saleh: Okay, we may want to come back to it later, so it wasn't a mistake to put it in. Now tell me about your chart. Can you draw any conclusions about angle C?
Grant: Well, we measured it in five positions on the circle. It is always close to 90 degrees, although it's never exact.
Ben: No, it was exact once.
Grant: Oh, I see. Yes, but it was just once.
Ms. Saleh: Do you have any ideas about angle C?
Sonia: Yes, it's always the top of the triangle and C always looks like a right triangle. So I think C is a right triangle, no matter what.
Ms. Saleh: Yes, it certainly looks that way. But does the table support the idea that angle C will be a right angle in every case? We have some data, 94 degrees for instance, that isn't a right angle.
Grant: That's just cuz we measured with a protractor and maybe it was off a little.
Ms. Saleh: If we had a more accurate protractor and many more triangles how could we show that angle C is always a right triangle? How many measurements would it take?
Ben: You'd have to do every single triangle ever. And that's not possible, right?
Ms. Saleh: Sonia what do you think? Could we check every possible angle C?
Sonia: No, I couldn't at least. Maybe I could make an argument about why it needs to be 90 degrees.
Ms. Saleh: That's great. Please work on an argument you think will convince one another that angle C is always a right angle. Then I'll come back and you can try to convince me!
Teaching Math Home | Grades 9-12 | Reasoning and Proof | Site Map | © |
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# Interest Formula: Days
Solving for the unknown factor for the formula when time is stated in whole days using either ordinary interest dating or exact interest dating: All Example 1 will use ordinary interest dating and All Example 2 will use exact interest dating in this handout. PXRXT=I P = Principle (The amount borrowed before interest) R = Rate (stated as a one year rate) T = Time (for this handout we will state time in “days/360” ordinary interest dating or “days/365” exact interest dating) I = Interest (Cost to borrow money from another entity) This box has the basic formula:
Interest
Principle Rate Time
This box shows the basic formula to find principle:
Interest
Principle Rate Time
MJC Revised 1/2012
Page 1
Interest Formula: Days
Example 1: Cindy borrows money with a rate of 5% for 180 days with an interest of \$125, using ordinary interest dating. Cindy’s principle is \$5,000.
Example 2: Cindy borrows money with a rate of 5% for 180 days with an interest of \$123.29, using exact interest dating. Cindy’s principle is \$5,000.
MJC Revised 1/2012
Page 2
Interest Formula: Days
This box shows the basic formula to find rate.
Interest
Principle Rate Time
Example 1: Cindy borrows money with a principle of \$5,000 for 180 days with an interest of \$125. Cindy’s rate is 5%.
Example 2: Cindy borrows money with a principle of \$5,000 for 180 days with an interest of \$123.29. Cindy’s rate is 5%.
MJC Revised 1/2012
Page 3
Interest Formula: Days
This box shows the basic formula to find time.
Interest
Principle Rate Time
Example 1: Cindy borrows money with a principle of \$5,000, a rate of 5% and with an interest of \$125. Cindy’s time is 180 days.
MJC Revised 1/2012
Page 4
Interest Formula: Days
Example 2: Cindy borrows money with a principle of \$5,000, a rate of 5% and with an interest of \$123.29. Cindy’s time is 180 days.
Note: When time is less than 1 multiply “T” by 360 to get the number of days for ordinary interest dating and multiply “T” by 365 to get the number of days for exact interest dating.
MJC Revised 1/2012
Page 5
Interest Formula: Days
This box shows the basic formula to find interest.
Interest
Principle Rate Time
PXRXT=I Example 1: Cindy borrows money with a principle of \$5,000, a rate of 5% and with a time of 180 days, using ordinary interest dating. Cindy interest is \$125. \$5,000 X 5% X 180/360 = I \$5,000 X 5%X .5 = I \$250 X .5 = \$125 Example 2: Cindy borrows money with a principle of \$5,000, a rate of 5% and with a time of 180 days, using exact interest dating. Cindy’s interest is \$123.29. \$5,000 X 5% X 180/365 = I \$5,000 X 5%X .49315 = I \$250 X .49315 = \$123.29
Note: In order to get .5 for the time you must divide 180 by 360 which will then results in .5 as the answer for example 1. For example 2 you must divide 180 by 365 which will result in .49315 as the rounded answer.
MJC Revised 1/2012
Page 6
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# How do you use implicit differentiation to find dy/dx given xe^y-y=5?
Mar 5, 2017
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{y}}{1 - x {e}^{y}}$
#### Explanation:
When we differentiate $y$ wrt $x$ we get $\frac{\mathrm{dy}}{\mathrm{dx}}$.
However, we cannot differentiate a non implicit function of $y$ wrt $x$. But if we apply the chain rule we can differentiate a function of $y$ wrt $y$ but we must also multiply the result by $\frac{\mathrm{dy}}{\mathrm{dx}}$.
When this is done in situ it is known as implicit differentiation.
We have:
$x {e}^{y} - y = 5$
Differentiate wrt $x$ (applying product rule):
$\left(x\right) \left({e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(1\right) \left({e}^{y}\right) - \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{y} - \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore {e}^{y} = \frac{\mathrm{dy}}{\mathrm{dx}} - x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}$
$\therefore \left(1 - x {e}^{y}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{y}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{y} / \left(1 - x {e}^{y}\right)$
There is another (often faster) approach using partial derivatives. Suppose we cannot find $y$ explicitly as a function of $x$, only implicitly through the equation $F \left(x , y\right) = 0$ which defines $y$ as a function of $x , y = y \left(x\right)$. Therefore we can write $F \left(x , y\right) = 0$ as $F \left(x , y \left(x\right)\right) = 0$. Differentiating both sides of this, using the partial chain rule gives us
(partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))
So Let $F \left(x , y\right) = x {e}^{y} - y - 5$; Then;
$\frac{\partial F}{\partial x} = {e}^{y}$
$\frac{\partial F}{\partial y} = x {e}^{y} - 1$
And so:
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{e}^{y}}{x {e}^{y} - 1} = \frac{{e}^{y}}{1 - x {e}^{y}}$, as before.
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Solve for b
## Share
a+b=-4 ab=4
To solve the equation, factor b^{2}-4b+4 using formula b^{2}+\left(a+b\right)b+ab=\left(b+a\right)\left(b+b\right). To find a and b, set up a system to be solved.
-1,-4 -2,-2
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.
-1-4=-5 -2-2=-4
Calculate the sum for each pair.
a=-2 b=-2
The solution is the pair that gives sum -4.
\left(b-2\right)\left(b-2\right)
Rewrite factored expression \left(b+a\right)\left(b+b\right) using the obtained values.
\left(b-2\right)^{2}
Rewrite as a binomial square.
b=2
To find equation solution, solve b-2=0.
a+b=-4 ab=1\times 4=4
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as b^{2}+ab+bb+4. To find a and b, set up a system to be solved.
-1,-4 -2,-2
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.
-1-4=-5 -2-2=-4
Calculate the sum for each pair.
a=-2 b=-2
The solution is the pair that gives sum -4.
\left(b^{2}-2b\right)+\left(-2b+4\right)
Rewrite b^{2}-4b+4 as \left(b^{2}-2b\right)+\left(-2b+4\right).
b\left(b-2\right)-2\left(b-2\right)
Factor out b in the first and -2 in the second group.
\left(b-2\right)\left(b-2\right)
Factor out common term b-2 by using distributive property.
\left(b-2\right)^{2}
Rewrite as a binomial square.
b=2
To find equation solution, solve b-2=0.
b^{2}-4b+4=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
b=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 4}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
b=\frac{-\left(-4\right)±\sqrt{16-4\times 4}}{2}
Square -4.
b=\frac{-\left(-4\right)±\sqrt{16-16}}{2}
Multiply -4 times 4.
b=\frac{-\left(-4\right)±\sqrt{0}}{2}
Add 16 to -16.
b=-\frac{-4}{2}
Take the square root of 0.
b=\frac{4}{2}
The opposite of -4 is 4.
b=2
Divide 4 by 2.
b^{2}-4b+4=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
\left(b-2\right)^{2}=0
Factor b^{2}-4b+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(b-2\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
b-2=0 b-2=0
Simplify.
b=2 b=2
Add 2 to both sides of the equation.
b=2
The equation is now solved. Solutions are the same.
x ^ 2 -4x +4 = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.
r + s = 4 rs = 4
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = 2 - u s = 2 + u
Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 2. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(2 - u) (2 + u) = 4
To solve for unknown quantity u, substitute these in the product equation rs = 4
4 - u^2 = 4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 4-4 = 0
Simplify the expression by subtracting 4 on both sides
u^2 = 0 u = 0
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r = s = 2
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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# Equation Solver
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
We can say equation solver is a kind of tool to solve the equation. We can say an equation is of the form A = B. Here A and B are expressions having unknowns. For example, 3 is the unique solution of the equation x + 4 = 7. In other words we can say an equation is a kind mathematical statement that has two expressions divide by an equal sign.
Example 1: Solve the equation 3x + 5 = 13 – x.
Solution: The given equation is 3x + 5 = 13 – x
Now we can add -5 on both sides of the equality
3x + 5 – 5= 13 – x – 5
3x = 8 – x
Now we can add x on both sides of the equality 3x + x = 8 – x + x
4x = 8
Therefore, x = 2.
Example 2: Solve the equation 2x + 21 = 9x – 14 .
Solution: The given equation is 2x + 21 = 9x – 14
Now we can add 14 on both sides of the equality
2x + 21 +14 = 9x – 14 – 14
2x + 35 = 9x
Now we can add – 2x on both sides of the equality 2x + 35 – 2x = 9x – 2x
35 = 7x
Now we can divide this equation by 7, then we get 35/7 = 7x/7
5 = x
Therefore, solution of the given equation is x = 5.
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# Special Matrices: Triangular, Symmetric, Diagonal
We have seen that a matrix is a block of entries or two dimensional data. The size of the matrix is given by the number of rows and the number of columns. If the two numbers are the same, we called such matrix a square matrix.
To square matrices we associate what we call the main diagonal (in short the diagonal). Indeed, consider the matrix
Its diagonal is given by the numbers a and d. For the matrix
its diagonal consists of a, e, and k. In general, if A is a square matrix of order n and if aij is the number in the ith-row and jth-colum, then the diagonal is given by the numbers aii, for i=1,..,n.
The diagonal of a square matrix helps define two type of matrices: upper-triangular and lower-triangular. Indeed, the diagonal subdivides the matrix into two blocks: one above the diagonal and the other one below it. If the lower-block consists of zeros, we call such a matrix upper-triangular. If the upper-block consists of zeros, we call such a matrix lower-triangular. For example, the matrices
are upper-triangular, while the matrices
are lower-triangular. Now consider the two matrices
The matrices A and B are triangular. But there is something special about these two matrices. Indeed, as you can see if you reflect the matrix A about the diagonal, you get the matrix B. This operation is called the transpose operation. Indeed, let A be a nxm matrix defined by the numbers aij, then the transpose of A, denoted AT is the mxn matrix defined by the numbers bij where bij = aji. For example, for the matrix
we have
Properties of the Transpose operation. If X and Y are mxn matrices and Z is an nxk matrix, then
1.
(X+Y)T = XT + YT
2.
(XZ)T = ZT XT
3.
(XT)T = X
A symmetric matrix is a matrix equal to its transpose. So a symmetric matrix must be a square matrix. For example, the matrices
are symmetric matrices. In particular a symmetric matrix of order n, contains at most different numbers.
A diagonal matrix is a symmetric matrix with all of its entries equal to zero except may be the ones on the diagonal. So a diagonal matrix has at most n different numbers other than 0. For example, the matrices
are diagonal matrices. Identity matrices are examples of diagonal matrices. Diagonal matrices play a crucial role in matrix theory. We will see this later on.
Example. Consider the diagonal matrix
Define the power-matrices of A by
Find the power matrices of A and then evaluate the matrices
for n=1,2,....
and
By induction, one may easily show that
for every natural number n. Then we have
for n=1,2,..
Scalar Product. Consider the 3x1 matrices
The scalar product of X and Y is defined by
In particular, we have
XTX = (a2 + b2 + c2). This is a 1 x 1 matrix .
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Author: M.A. Khamsi
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