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# Factorise the equation ${x^2} - 9x + 18$.
Last updated date: 24th Jun 2024
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Hint: The given expression is ${x^2} - 9x + 18$. We can see that the coefficient of $x$ is $- 9$ and constant term is $18$. We can express these numbers as sum and product respectively using the numbers six and three. Then we can rearrange the terms to get the factors. Since the given equation is second degree, we will get two linear (first degree) factors.
Complete step-by-step solution:
We are given the equation, ${x^2} - 9x + 18$.
We can split the term $9x$.
Since $9 = 6 + 3$, we have,
${x^2} - 9x + 18 = {x^2} - (6 + 3)x + 18$
Opening brackets we get,
${x^2} - 9x + 18 = {x^2} - 6x - 3x + 18$
Now we can see the constant term is $18$.
Also $18 = 6 \times 3$
So we can rewrite the equation as,
${x^2} - 9x + 18 = {x^2} - 6x - 3x + 6 \times 3$
Now we can take $x$ common from the first two terms and $- 3$ common from the last two terms of the right side.
Thus we have,
${x^2} - 9x + 18 = x(x - 6) - 3(x - 6)$
So we get a common factor $x - 6$.
This gives,
${x^2} - 9x + 18 = (x - 6)(x - 3)$
Thus we had factorised the given expression.
For the equation ${x^2} - 9x + 18$, $x - 6$ and $x - 3$ are the factors.
Note: There is another method to solve the equation. It is called completing the square method.
For, the given equation can be rewritten as ${x^2} - 9x = - 18$
Now add the square of half of the coefficient of $x$ on both sides.
Here the coefficient of $x$ is $- 9$. Half of $- 9$ is $- \dfrac{9}{2}$.
Squaring we get $\dfrac{{81}}{4}$.
Adding this on both sides we get,
${x^2} - 9x + \dfrac{{81}}{4} = \dfrac{{81}}{4} - 18$
$\Rightarrow {x^2} - 9x + \dfrac{{81}}{4} = \dfrac{{81 - 72}}{4} = \dfrac{9}{4}$
We can observe that the left side is in the form ${a^2} - 2ab + {b^2}$ form which is equal to ${(a - b)^2}$.
So we get,
${x^2} - 9x + \dfrac{{81}}{4} = {(x - \dfrac{9}{2})^2}$
And also $\dfrac{9}{4} = {( \pm \dfrac{3}{2})^2}$.
So taking roots on both sides we can write,
$x - \dfrac{9}{2} = \pm \dfrac{3}{2}$
This gives $x = \dfrac{9}{2} + \dfrac{3}{2} = \dfrac{{12}}{2} = 6$ or $x = \dfrac{9}{2} - \dfrac{3}{2} = \dfrac{6}{2} = 3$
So the factors are $x - 6$ and $x - 3$. |
# Subset is Compatible with Ordinal Successor
## Theorem
Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x \in y$.
Then:
$x^+ \in y^+$
## Proof 1
$\ds x \in y$ $\implies$ $\ds x \ne y$ No Membership Loops $\ds$ $\implies$ $\ds x^+ \ne y^+$ Equality of Successors $\ds x \in y$ $\implies$ $\ds y \notin x$ No Membership Loops $\ds$ $\implies$ $\ds y \notin x^+$ $x ≠ y$ and Definition of Successor Set $\ds y^+ \in x^+$ $\implies$ $\ds y \in x^+$ Successor Set of Ordinal is Ordinal, Ordinals are Transitive, and Set is Element of Successor
The last part is a contradiction, so $y^+ \notin x^+$.
$x^+ \in y^+$
$\blacksquare$
## Proof 2
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
Let $x \in y$.
We wish to show that $x^+ \in y^+$.
Either $x^+ \in y^+$, $y^+ \in x^+$, or $x^+ = y^+$.
Aiming for a contradiction, suppose $y^+ = x^+$.
Then $y \in x$ or $y = x$ by the definition of successor set.
If $y = x$ then $x \in x$, contradicting the fact that Ordinal is not Element of Itself.
If $y \in x$ then since an ordinal is transitive, $y \in y$, again contradicting Ordinal is not Element of Itself.
Thus $y^+ ≠ x^+$.
Aiming for a contradiction, suppose $y^+ \in x^+$.
By definition of successor set, $y^+ \in x$ or $y^+ = x$.
If $y^+ \in x$, then since $y^+$ and $x$ are both ordinals, $y^+ \subsetneqq x$.
Then $y \in x$.
Since $y$ is transitive, $y \in y$, contradicting Ordinal is not Element of Itself.
If instead $y^+ = x$, then $x \in y \in y^+ = x$, so the same contradiction arises because $x$ is transitive.
Thus $y^+ \notin x^+$.
So the only remaining possibility, that $x^+ \in y^+$, must hold.
$\blacksquare$
## Proof 3
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
By Ordinal Membership is Trichotomy, one of the following must be true:
$\ds x^+$ $=$ $\ds y^+$ $\ds y^+$ $\in$ $\ds x^+$ $\ds x^+$ $\in$ $\ds y^+$
We will show that the first two are both false, so that the third must hold.
Two preliminary facts:
$(1)$ $:$ $\ds x$ $\ds \ne$ $\ds y$ $x \in y$ and Ordinal is not Element of Itself $(2)$ $:$ $\ds y$ $\ds \notin$ $\ds x$ $x \in y$ and Ordinal Membership is Asymmetric
By $(1)$ and Equality of Successors:
$x^+ \ne y^+$
Thus the first of the three possibilities is false.
Aiming for a contradiction, suppose $y^+ \in x^+$.
Then:
$\ds y$ $\in$ $\ds y^+$ Definition of Successor Set $\ds \leadsto \ \$ $\ds y$ $\in$ $\ds x^+$ $y^+ \in x^+$ and $x^+$ is an ordinal and therefore transitive. $\ds \leadsto \ \$ $\ds y$ $\in$ $\ds x$ Definition of Successor Set $\, \ds \lor \,$ $\ds y$ $=$ $\ds x$
But we already know that $y \notin x$ by $(2)$ and $y \ne x$ by $(1)$.
So this is a contradiction, and we conclude that $y^+ \notin x^+$.
Thus we have shown that the second possibility is false.
Thus the third and final one must hold: $x^+ \in y^+$.
$\blacksquare$ |
# 180 Days of Math for Third Grade Day 93 Answers Key
By accessing our 180 Days of Math for Third Grade Answers Key Day 93 regularly, students can get better problem-solving skills.
## 180 Days of Math for Third Grade Answers Key Day 93
Directions: Solve each problem.
Question 1.
9 more than 33 is __________.
9 more than 33 means we have to add 9 to 33.
33
+9
42
Thus the sum of 33 and 9 is 42.
Question 2.
6 times 8 is ___________.
Explanation:
By performing the multiplication operation we can find the product of 6 and 8.
6 × 8 = 48
Question 3.
Explanation:
By performing the multiplication operation we can find the product of 10 and 6.
10 × 6 = 60
Question 4.
8 ÷ 4 =
By dividing 8 and 2 we get 2
8/4 = 2
The quotient is 2.
Question 5.
Subtract 3 tens from 73.
3 tens = 30
73 – 30 = 43
Question 6.
4 × = 28
Explanation:
Let the missing number be x
4 × x = 28
x = 28/4
x = 7
So, 4 × 7 = 28
Question 7.
Show a quarter to 10:00 on the clock.
Question 8.
Circle the object that weighs less than one pound.
Question 9.
Match the solid to its top, front, and side views.
The base of the square pyramid is square
The top view of the front view is the triangle.
So, the top, front, and side views are options C.
Question 10.
You have a quart of chocolate milk. How many cups of chocolate milk can you pour? |
Numbers
Vectors
Probability
Statistics
Algebra
Sequences
# Negative Numbers
We often use positive and negative numbers when telling the temperature. In winter, for example, a temperature ℃ is ℃ below zero.
When doing arithmetic, it is very helpful to regard positive numbers as something we have in our possession and the negative numbers as something we owe (a debt) and need to pay back.
A negative number has a minus sign in front of it, as in ℃. Positive numbers are written with or without the positive sign in front of the number.
In , we regard the 10 as something we possess and the as something we possess. So in total, we possess . As .
In -10 + 3, we regard the as a debt and the as something we have. So, after selling the account, we shall still have as a debt. Therefore, .
In , we regard as a debt and the again as a debt. So in total, we have as a debt. Therefore, we should have .
Another way to calculate is by using the number line. Starting from the first number, we move to the right if we are adding and to the left if we are subtracting.
For , start at on the number line
Since we are adding, we move to the right from by steps. This gives .
For , start at and move steps to the right, since we are adding.
For , start at and move steps to the left since we are subtracting.
This take us to . Hence,
For , we start at and we move steps to the right
We are at , so .
We should also know some rules:
With bigger numbers, using the number line becomes impractical. We should then visualise what is happening on the number line. For example, consider .
On the number line we will start at , because we are adding . We should move steps to the right of . With some efforts we see that this will take us to . Therefore .
To calculate , we start at on the number line and because we are subtracting , we should move to the left of by steps. Again, with some effort, we see that this will take us to on the number line. Now . Therefore .
With practice, we shall become more fluent at these and will not even need to visualise the number line.
When we are multiplying numbers (positive, negative), we should remember these rules:
A positivea positivea positive
A positivea negativea positive
A negativea positivea negative
A negativea negativea positive
### Examples
The same rules apply for division |
# 8.2 A single population mean using the student t distribution (Page 4/21)
Page 4 / 21
## Chapter review
In many cases, the researcher does not know the population standard deviation, σ , of the measure being studied. In these cases, it is common to use the sample standard deviation, s , as an estimate of σ . The normal distribution creates accurate confidence intervals when σ is known, but it is not as accurate when s is used as an estimate. In this case, the Student’s t-distribution is much better. Define a t-score using the following formula:
The t -score follows the Student’s t-distribution with n – 1 degrees of freedom. The confidence interval under this distribution is calculated with EBM = $\left({t}_{\frac{\alpha }{2}}\right)\frac{s}{\sqrt{n}}$ where ${t}_{\frac{\alpha }{2}}$ is the t -score with area to the right equal to $\frac{\alpha }{2}$ , s is the sample standard deviation, and n is the sample size. Use a table, calculator, or computer to find ${t}_{\frac{\alpha }{2}}$ for a given α .
## Formula review
s = the standard deviation of sample values.
is the formula for the t -score which measures how far away a measure is from the population mean in the Student’s t-distribution
df = n - 1; the degrees of freedom for a Student’s t-distribution where n represents the size of the sample
T ~ t df the random variable, T , has a Student’s t-distribution with df degrees of freedom
$EBM={t}_{\frac{\alpha }{2}}\frac{s}{\sqrt{n}}$ = the error bound for the population mean when the population standard deviation is unknown
${t}_{\frac{\alpha }{2}}$ is the t -score in the Student’s t-distribution with area to the right equal to $\frac{\alpha }{2}$
The general form for a confidence interval for a single mean, population standard deviation unknown, Student's t is given by (lower bound, upper bound)
= (point estimate – EBM , point estimate + EBM )
=
Use the following information to answer the next five exercises. A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hours with a sample standard deviation of 0.5 hours.
Identify the following:
1. $\overline{x}$ =_______
2. ${s}_{x}$ =_______
3. n =_______
4. n – 1 =_______
Define the random variables X and $\overline{X}$ in words.
X is the number of hours a patient waits in the emergency room before being called back to be examined. $\overline{X}$ is the mean wait time of 70 patients in the emergency room.
Which distribution should you use for this problem?
Construct a 95% confidence interval for the population mean time spent waiting. State the confidence interval, sketch the graph, and calculate the error bound.
CI: (1.3808, 1.6192)
EBM = 0.12
Explain in complete sentences what the confidence interval means.
Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.
Identify the following:
1. $\overline{x}$ =_______
2. ${s}_{x}$ =_______
3. n =_______
4. n – 1 =_______
1. $\overline{x}$ = 151
2. ${s}_{x}$ = 32
3. n = 108
4. n – 1 = 107
Define the random variable X in words.
Define the random variable $\overline{X}$ in words.
$\overline{X}$ is the mean number of hours spent watching television per month from a sample of 108 Americans.
Which distribution should you use for this problem?
Construct a 99% confidence interval for the population mean hours spent watching television per month. (a) State the confidence interval, (b) sketch the graph, and (c) calculate the error bound.
CI: (142.92, 159.08)
EBM = 8.08
Why would the error bound change if the confidence level were lowered to 95%?
Use the following information to answer the next 13 exercises: The data in [link] are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let X = the number of colors on a national flag.
X Freq.
1 1
2 7
3 18
4 7
5 6
Calculate the following:
1. $\overline{x}$ =______
2. ${s}_{x}$ =______
3. n =______
1. 3.26
2. 1.02
3. 39
Define the random variable $\overline{X}$ in words.
What is $\overline{x}$ estimating?
μ
Is ${\sigma }_{x}$ known?
As a result of your answer to [link] , state the exact distribution to use when calculating the confidence interval.
t 38
Construct a 95% confidence interval for the true mean number of colors on national flags.
How much area is in both tails (combined)?
How much area is in each tail?
0.025
Calculate the following:
1. lower limit
2. upper limit
3. error bound
The 95% confidence interval is_____.
(2.93, 3.59)
Fill in the blanks on the graph with the areas, the upper and lower limits of the Confidence Interval and the sample mean.
In one complete sentence, explain what the interval means.
We are 95% confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors.
Using the same $\overline{x}$ , ${s}_{x}$ , and level of confidence, suppose that n were 69 instead of 39. Would the error bound become larger or smaller? How do you know?
The error bound would become EBM = 0.245. This error bound decreases because as sample sizes increase, variability decreases and we need less interval length to capture the true mean.
Using the same $\overline{x}$ , ${s}_{x}$ , and n = 39, how would the error bound change if the confidence level were reduced to 90%? Why?
mean is number that occurs frequently in a giving data
That places the mode and the mean as the same thing. I'd define the mean as the ratio of the total sum of variables to the variable count, and it assigns the variables a similar value across the board.
Samsicker
what is mean
what is normal distribution
What is the uses of sample in real life
change of origin and scale
3. If the grades of 40000 students in a course at the Hashemite University are distributed according to N(60,400) Then the number of students with grades less than 75 =*
If a constant value is added to every observation of data, then arithmetic mean is obtained by
sum of AM+Constnt
Fazal
data can be defined as numbers in context. suppose you are given the following set of numbers 18,22,22,20,19,21
what are data
what is mode?
what is statistics
Natasha
statistics is a combination of collect data summraize data analyiz data and interprete data
Ali
what is mode
Natasha
what is statistics
It is the science of analysing numerical data in large quantities, especially for the purpose of inferring proportions in a whole from those in a representative sample.
Bernice
history of statistics
statistics was first used by?
Terseer
if a population has a prevalence of Hypertension 5%, what is the probability of 4 people having hypertension from 8 randomly selected individuals?
Carpet land sales persons average 8000 per weekend sales Steve qantas the firm's vice president proposes a compensation plan with new selling incentives Steve hopes that the results of a trial selling period will enable him to conclude that the compensation plan increases the average sales per sales
Supposed we have Standard deviation 1.56, mean 6.36, sample size 25 and Z-score 1.96 at 95% confidence level, what is the confidence interval? |
# Integrals and primitives: senior math course in PDF.
In this lesson we will see the definition and the different properties of the integral as well as the geometrical meaning with the areas. In addition, in this course you will see the different ways to calculate an integral using the primitive and the associativity and linearity properties of the integral. In addition,integrals and primitives must be mastered in order to progress well in math .
## I. Definitions and properties of the integral and primitives.
Vocabulary:
A function f is integrable on an interval I when it has primitives on this interval I.
In particular, functions which are derivable on an interval I, are integrable on I, but this condition, although sufficient, is not necessary. The function f defined on by , although not derivable at 0, is integrable on , because it has as primitive on the function F such that: .
However, except in special cases containing indications, in the baccalaureate problems, functions integrable on an interval I will always be functions derivable on I.
Ownership:
If F and G are two primitives of an integrable function on an interval I,
then whatever a I and b I, we have
F(b) – F(a) = G(b) – G(a)Indeed, if F ‘(x) = G ‘(x), then there exists c such that G(x) = F(x) + c.
Therefore: G(b) – G(a) = F(b) + c – [F(a) + c] = F(b) – F(a).
Note:
The difference in the images of b and a for any primitive of f is the same.
This number depends only on f, a and b.
This will allow us to give the following definition:
Definition:
Let f be an integrable function on an interval I,
Let a I and b I be two real numbers of this interval I,
The integral from a to b of the function f is the number F(b) – F(a) where F is any primitive of f on I.
Notation:
which reads“sum from a to b of f from t dt ” and is also called“integral of f between a and b“.
In the writing , the letter t is called: “dumb variable”.
Indeed, we can also write The letter “dummy variable” indicates the name of the “variable of integration” (any other letter in the expression of the function f to be integrated is then considered as constant. The interest of this appears when there are several variables, but this is not in the Terminale program, however, it will be useful in the presence of parameters).
is also written in the condensed form using F :
Example:
The function is derivable on .
It is therefore integrable on and admits primitives on .
For example is a primitive of f on .
So we have:
## II. consequences of the definitionThe first properties.
Properties:
Let f be a function derivable on an interval [a,b].
Ownership:
If f is derivable on I, if a I and b I, we have
From this property, we deduce an important synthesis that links a derivable function, its derivative function and the notion of integral.
Theorem:
If f is a differentiable function on I and if a I.
Then, for all x I, we have .
## III.primitive function of an integrable function.
Theorem:
If f is an integrable function on an interval I and if a I, then
the function defined on I by is the primitive of f which cancels at x = a.
Demonstration:
If G is any primitive of f on I, then , so
Fa‘(x) = G ‘(x) = f(x) .
Indeed, G(a) is a constant, so its derivative is zero and G ‘(x) = f(x) because G is a primitive of f. Conclusion: is also a primitive of f .
Moreover:Fa (a) = G(a) – G(a )= 0, soFa cancels for x=a .
Exercise:
Compute the following integrals, then state the primitives they define.
## IV. Integral of a positive function.
Ownership:
We assume here that f is integrable and positive on [a;b], that is to say that for all x [a;b], we have: f(x) 0.
Call F a primitive of f on [a;b]Then we have F ‘(x) = f(x) for all x [a;b]The derivative f of F being positive on [a;b]the function F is increasing on [a;b].as a < b, then we have: F(a) F(b).
So: .
Ownership:
If a < b and if for all x [a;b] we have f(x) 0 then .
## V. Integrals and areas
### 1. summary table
Ownership:
In an orthogonal frame of reference , we represent graphically a function f derivable and positive on an interval [a;b]. We call D the set of points of the plane whose coordinates (x;y) verify :
a x b and 0 y f(x)Then Area of D = units of areaThe unit of area being that of the rectangle whose sides are the units of length of the abscissa and ordinate.
The set D consists of the points located between the curve representing the function f, the x-axis and the lines of equation x = a and x = b.
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# Lecture 3.2 - Describing Numerical Data - Mean | Summary and Q&A
92.8K views
October 21, 2021
by
IIT Madras - B.S. Degree Programme
Lecture 3.2 - Describing Numerical Data - Mean
## TL;DR
This content explains how to calculate measures of central tendency, such as mean and median, in data analysis.
## Key Insights
• 🔂 Measures of central tendency, such as mean and median, are commonly used in data analysis to summarize data in a single value.
• 😥 The mean is calculated by summing all the data points and dividing by the total number of data points.
• 🖕 The median is the middle value of the data when arranged in order, or the average of the two middle values.
• 🇨🇫 Measures of central tendency can be used for both discrete and continuous data.
• 💨 They provide a useful way to understand and communicate the overall behavior of a dataset.
• 👨🔬 The choice of measure of central tendency depends on the characteristics of the data and the research questions being addressed.
• 😒 Both mean and median have their own advantages and disadvantages, and their use depends on the specific context and goals of the analysis.
## Transcript
पुढील गोष्ट ज्यावर आपण चर्चा करणार आहोत, ती म्हणजे संख्यात्मक सारांश वापरून डेटाचा सारांश कसा काढावा. जेव्हा आम्ही श्रेणीबद्ध व्हेरिएबलबद्दल बोललो, तेव्हा आम्ही दोन वर्णनात्मक उपाय सादर करतो; म्हणजे, मोड अशा दोन्ही प्रकारच्या डेटासाठी केला जात असे. आम्ही पाहिले की जर तुम्हाला मेडिअनचा वापर करून डेटा सारांशित करायचा असेल तर तो एक ऑर्डिनल डेटा असावा ... Read More
### Q: What are measures of central tendency used for in data analysis?
Measures of central tendency are used to summarize data and provide a single representative value for a dataset.
### Q: How is the mean calculated?
The mean is calculated by summing all the data points and dividing by the total number of data points.
### Q: How is the median calculated?
The median is found by arranging the data in order and selecting the middle value. If there is an even number of data points, the average of the two middle values is taken as the median.
### Q: Can measures of central tendency be used for both discrete and continuous data?
Yes, measures of central tendency can be used for both discrete data, which consists of distinct values, and continuous data, which can take on any value within a range.
### Q: What are measures of central tendency used for in data analysis?
Measures of central tendency are used to summarize data and provide a single representative value for a dataset.
## More Insights
• Measures of central tendency, such as mean and median, are commonly used in data analysis to summarize data in a single value.
• The mean is calculated by summing all the data points and dividing by the total number of data points.
• The median is the middle value of the data when arranged in order, or the average of the two middle values.
• Measures of central tendency can be used for both discrete and continuous data.
• They provide a useful way to understand and communicate the overall behavior of a dataset.
• The choice of measure of central tendency depends on the characteristics of the data and the research questions being addressed.
• Both mean and median have their own advantages and disadvantages, and their use depends on the specific context and goals of the analysis.
• Measures of central tendency are affected by extreme values or outliers in the data.
## Summary & Key Takeaways
• Measures of central tendency, such as mean and median, are used to summarize data in data analysis.
• The mean is calculated by summing all the data points and dividing by the total number of data points.
• The median is the middle value of the data when arranged in order, or the average of the two middle values. |
# Finding the greatest common factor
The greatest common factor is exactly as it sounds: the greatest factors of two or more numbers.
Example
Find the common factors for 60 and 30
Begin by factoring both numbers. Find all factors that the numbers have in common.
The product of all common factors is the greatest common factor (GCF)
$5\cdot 3\cdot 2=30$
The greatest common factor is 30
Example
You can also determine the GCF if you have both numbers and variables.
Find the common factors for 36x2y and 16xy
Factorize the numbers and identify all common factors. To get the GCF multiply all common factors.
You can use the greatest common factor to simplify fractions.
A ratio is an expression that tells us the quotient of two numbers.
There are different ways to write a ratio and all examples below are read as "the ratio of 3 to 4".
$3\: to\: 4$
$3:4$
$\frac{3}{4}$
$3\div 4$
Example
You can simplify a ratio by findng the GCF
$\frac{2000}{1500}$
Fraction all numbers and find all common factors
Multiply all common factors to find the GCF
$2\cdot 5\cdot 5\cdot 10=500$
Since the GCF is a factor of both the numerator and the denominator we can divide both the numerator and the denominator by the GCF to produce a simplified fraction.
$\frac{2000\div 500}{1500\div 500}=\frac{4}{3}$
## Video lesson
Find the GCF of 24x3 and 56x3 |
Factors
# Factors of 75 | Prime Factorization of 75 | Factor Tree of 75
Written by Prerit Jain
Updated on: 24 Aug 2023
## Factors of 75
Calculate Factors of
The Factors are
https://wiingy.com/learn/math/factors-of-75/
## What are the factors of 75
The factors of 75 are the numbers that divide evenly the number 75. They are:1, 3, 5, 15, 25, and 75
To find the factors of a number, you can start by dividing it by the smallest possible factor (which is usually 2) and then continue dividing by the next smallest prime numbers (3, 5, 7, etc.) until you can’t divide anymore. Alternatively, you can use a factorization tree or the prime factorization of the number to find its factors.
## How to Find Factors of 75
To find the factors of 75, you can follow any of 6these methods:
1. Factors of 75 using the Multiplication Method
2. Factors of 75 using the Division Method
3. Prime Factorization of 75
4. Factor tree of 75
## Factors of 75 Using the Multiplication Method
• One way to find the factors of 75 is to list out all the pairs of numbers that multiply to give 75.
• The factors of 75 will be all the numbers that appear in these pairs.
• For example, if we list out the pairs of numbers that multiply to give 75, we get: (1, 75), (3, 25), and (5, 15).
• Therefore, the factors of 75 are 1, 3, 5, 15, 25, and 75.
• This method works because if two numbers multiply to give a third number, then the third number is a multiple of the first two numbers.
## Factors of 75 Using the Division Method
• To find the factors of 75 using the division method, start by dividing 75 by the smallest possible factor (usually 2).
• If the result is an integer, then that number is a factor of 75.
• If the result is not an integer, divide it by the next smallest prime number (3).
• Continue dividing by the next smallest prime numbers (4, 5, 6, etc.) until you reach a result that is an integer.
• The integer results are the factors of 75.
For example:
• 75 / 2 = 37.5 (not an integer)
• 75 / 3 = 25 (an integer)
Therefore, the factors of 75 are 3 and 25. This method works because if a number is a factor of 75, then 75 will divide evenly by that number.
## Prime Factorization of 75
Calculate Prime Factors of
The Prime Factors of 75 =
3 x
5 x
5
https://wiingy.com/learn/math/factors-of-75/
• The prime factorization of 75 can be found by breaking it down into its prime factors.
• To begin, we divide 75 by the smallest prime number, which is 2. However, 75 is not divisible by 2.
• Next, we divide 75 by the next prime number, which is 3. 75 divided by 3 equals 25.
• Now, we divide 25 by the next prime number, which is also 5. 25 divided by 5 equals 5.
• At this point, we have obtained the prime factorization of 75:
• 75 = 3 x 5 x 5
• Therefore, the prime factorization of 75 is 3 x 5 x 5.
## Factor tree of 75
https://wiingy.com/learn/math/factors-of-75/
To create a factor tree, we start by writing the number we want to factor at the top of the tree. In this case, that number is 75. We then find two numbers that multiply together to equal 75 and write them as the children of the top node. In this case, those numbers are 5 and 15.
Next, we repeat this process for each of the children. For the child node with the value of 15, we find two numbers that multiply by 15 and write those as the children of the 15 nodes. In this case, those numbers are 5 and 3.
We continue this process until all of the nodes are prime numbers. When we are finished, our factor tree looks like this:
This factor tree shows that 75 can be written as the product of the prime numbers 3 and 5, or as 3 x 5 x 5, which is the prime factorization of 75.
.
## Factor Pairs of 75
Calculate Pair Factors of
1 x 75=75
3 x 25=75
5 x 15=75
15 x 5=75
25 x 3=75
So Pair Factors of 75 are
(1,75)
(3,25)
(5,15)
(15,5)
(25,3)
https://wiingy.com/learn/math/factors-of-75/
The factor pairs of a number are the pairs of numbers that can be multiplied together to equal that number. To find the factor pairs of 75, we can list out all of the numbers that 75 can be evenly divided by, along with the result of the division. These numbers are 1, 3, and 5.
The corresponding factor pairs for 75 are (1, 75), (3, 25), and (5, 15). These are the pairs of numbers that multiply together to give us 75. For example, the factor pair (1, 75) means that 1 and 75 can be multiplied together to equal 75. The factor pair (3, 25) means that 3 and 25 can be multiplied together to equal 75. And the factor pair (5, 15) means that 5 and 15 can be multiplied together to equal 75.
## Factors of 75 – Quick Recap
• Factors of 75: 1, 3, 5, 15, 25, 75
• Negative Factors of 75: -1, -3, -5, -15, -25, -75
• Prime Factors of 75: 3 × 5 × 5
• Prime Factorization of 75: 3 × 5 × 5
## Factors of 75 – Fun Facts
• 75 is a composite number, which means it has more than two factors. In addition to 1 and itself, 75 has three other positive integer factors: 3, 5, and 15.
• 75 is the smallest number that has three distinct prime factors. Its prime factorization is 3 x 5 x 5.
• The factors of 75 can be used to make a variety of interesting patterns when they are arranged in a grid. For example, you can create a 3×5 grid by multiplying the factors 3 and 5 together, or a 5×5 grid by multiplying the factors 5 and 5 together.
• The factors of 75 can be used to make interesting shapes when they are plotted on a coordinate plane. For example, you can plot the factors of 75 as points on a grid and connect them to create polygons.
Also Check: Multiples, Square Root, and LCM
## Solved Examples of Factor of 75
Q.1: If a number is divisible by five and three, what is the greatest common factor of that number?
Solution:
The greatest common factor (GCF) of a number that is divisible by five and three is 15, as both 5 and 3 are divisible by 15 without a remainder.
Q.2: Is 74 a multiple of 75?
Solution: No, 74 is not a multiple of 75 since 75 does not divide into 74 evenly.
Q.3: What two numbers add up to 75?
Solution:
Two numbers that add up to 75 include 28 and 47; 28 + 47 = 75.
Q.4: What is the product of all positive integer divisors for 75?
Solution:
The product of all positive integer divisors for 75 is 421,875; 1 x 3 x 5 x 15 x 25 x75 = 421,875.
Q.5: What two prime factors added together equal 74?
Solution:
Two prime factors that added together equal 74 include 37 and 37; 37 + 37 = 74.
Q.6: What three numbers multiplied together equal 75?
Solution: Three numbers multiplied together to equal 75 are 3,5 and 5; 3 x5x5=75.
Q.7: What Are Some Examples Of Factors Of 75?
Solution: Some examples of factors of 7 5 include1,3,5,15,25, and 75 .
Q.8: How do you calculate the LCM (Least Common Multiple) of 75?
Solution: To find the least common multiple (LCM) of 75, we need to determine the prime factorization of 75. The prime factorization of 75 is 3 x 5 x 5. To calculate the LCM, we take the highest power of each prime factor that appears in the factorization. In this case, we have 3 and two 5s. LCM of 75 = (Highest power of 3) x (Highest power of 5)Highest power of 3 = 3^1 = 3 Highest power of 5 = 5^2 = 25. LCM of 75 = 3 x 25 = 75
Therefore, the least common multiple (LCM) of 75 is 75.
## Frequently Asked Questions on Factors of 75
### What is the greatest common factor of 75?
The greatest common factor (GCF) of 75 is 25, as both 75 and 25 are divisible by 25 without a remainder.
### How many factors does 75 have?
There are 8 factors of 75; 1,3,5,15,25,75,-1,-3,-5,-15,-25,-75.
### What are all the prime factors of 75?
All the prime factors of 75 are 3 and 5; 3 x 5 = 75.
### What two numbers add up to 75?
Two numbers that add up to 75 include 28 and 47; 28 + 47 = 75.
### What number can divide into 75 evenly?
Any number from 1-75 can divide into 75 with no remainders, but some will produce fractions or decimals instead of a whole number result.
### Is there a difference between the factors and multiples of 75?
While factors and multiples have similar definitions in that they both refer to groups or collections of related numbers generated by multiplying or dividing a given number, there is an important difference between them – factors refer to how many times the original number can be divided evenly while multiples make reference to how many times it has been multiplied by itself.
### What is the sum of all positive integer divisors for 75?
The sum of the positive integerdivisorsfor75 is150; 1+3+5+15+25+75 =124&
### What Are Some Examples Of Factors Of 75?
Some examples of factors of 75 include 1,3,5,15,25 and 75.
Written by by
Prerit Jain
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Counting factors
Counting factors is not a bad idea when you are looking for the factors of a number.
You may need this information if you do not want to miss any factors.
To count factors, first we need to get the prime factorization of the number
Example #1:
How many factors does 8 have?
8 = 2 × 2 × 2 = 23
The factors of 8 are 1, 2, 4, 8. There are 4 factors
Looking at 23, we notice that if we add 1 to the exponent, we get the 4
However, just one case is not enough to conclude that when counting factors the number of factors is whatever the exponent is plus 1
Let's look at more examples
Example #2:
How many factors does 25 have?
25 = 5 × 5 = 52
The factors of 25 are 1, 5, and 25. 25 has 3 factors
Again, to get the 3, just add 1 to the exponent of 2
Example #3:
How many factors does 72 have?
72 = 8 × 9 = 2 × 2 × 2 × 3 × 3 = 23 × 32
The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72. There are 12 factors
How do we get the 12? By adding 1 to each exponent and then multiply
(3 + 1) × (2 + 1) = 4 × 3 = 12
So far it seems like adding 1 is a good strategy when counting factors
Another way to see all the factors of 72 are shown below:
20 × 30 = 1 × 1 = 1
20 × 31 = 1 × 3 = 3
20 × 32 = 1 × 9 = 9
21 × 30 = 2 × 1 = 2
21 × 31 = 2 × 3 = 6
21 × 32 = 2 × 9 = 18
22 × 30 = 4 × 1 = 4
22 × 31 = 4 × 3 = 12
22 × 32 = 4 × 9 = 36
23 × 30 = 8 × 1 = 8
23 × 31 = 8 × 3 = 24
23 × 32 = 8 × 9 = 72
As you can see there are 4 choices for the exponents of 2: 0, 1, 2, 3.
And 4 choices for the exponents of 3: 0, 1, 2
4 choices × 3 choices = 12 choices and this is equal to 12 factors
Example #4:
How many factors 12600 have?
When counting factors for big numbers, it may be useful to make a factor tree
Pull out all the prime numbers from the tree and multiply the numbers. This is your prime factorization.
2 × 2 × 2 × 3 × 3 × 5 × 5 × 7
23 × 32 × 52 × 71
Add 1 to each exponent and multiply:
(3 + 1) × (2 + 1) × (2 + 1) × (1 + 1)
4 × 3 × 3 × 2
12 × 3 × 2
36 × 2 = 72
12600 has 72 factors.
Take the counting factors quiz below to see how well you understand this lesson.
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# How do you prove that sum of infinite series 1+1/4+1/9+............... is less than two?
Feb 23, 2017
See explanation...
#### Explanation:
${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2 = 1 + {\sum}_{n = 2}^{\infty} \frac{1}{n} ^ 2 < 1 + {\sum}_{n = 2}^{\infty} \frac{1}{n \left(n - 1\right)}$
Then:
$\frac{1}{n \left(n - 1\right)} = \frac{n - \left(n - 1\right)}{n \left(n - 1\right)} = \frac{1}{n - 1} - \frac{1}{n}$
So:
${\sum}_{n = 2}^{N} \frac{1}{n \left(n - 1\right)} = {\sum}_{n = 2}^{N} \left(\frac{1}{n - 1} - \frac{1}{n}\right)$
$\textcolor{w h i t e}{{\sum}_{n = 2}^{N} \frac{1}{n \left(n - 1\right)}} = {\sum}_{n = 2}^{N} \frac{1}{n - 1} - {\sum}_{n = 2}^{N} \frac{1}{n}$
$\textcolor{w h i t e}{{\sum}_{n = 2}^{N} \frac{1}{n \left(n - 1\right)}} = 1 + {\sum}_{n = 3}^{N} \frac{1}{n - 1} - {\sum}_{n = 2}^{N - 1} \frac{1}{n} - \frac{1}{N}$
$\textcolor{w h i t e}{{\sum}_{n = 2}^{N} \frac{1}{n \left(n - 1\right)}} = 1 + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N - 1} \frac{1}{n}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N - 1} \frac{1}{n}}}} - \frac{1}{N}$
$\textcolor{w h i t e}{{\sum}_{n = 2}^{N} \frac{1}{n \left(n - 1\right)}} = 1 - \frac{1}{N}$
So:
${\sum}_{n = 2}^{\infty} \frac{1}{n \left(n - 1\right)} = {\lim}_{N \to \infty} \left(1 - \frac{1}{N}\right) = 1$
So:
${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2 < 1 + {\sum}_{n = 2}^{\infty} \frac{1}{n \left(n - 1\right)} = 1 + 1 = 2$ |
# Surface Area Formulas and Volume Formulas of 3-D Solids
The meaning of 3-D solids is 3 dimensional solids. In Geometry, a three dimensional shape is defined as a shape or solid having three dimensions- length, breadth (or width) and height (or depth). They have volume. Students can also download the HD printout of Surface Area Formulas and Volume Formulas of 3-D Solids and can stick on their pinboard. This will help them to revise and learn all the formulas for doing sums on 3 d shapes.
## Surface Area Formulas and Volume Formulas of 3-D Solids
Few example of 3-D shapes are cuboid, cube, cone, cylinder, sphere, hemisphere. Below are the formulas for surface area and volume of 3-d solids.
## Surface Area and Volume Formulas of Cuboid
Cuboid is a solid bounded by 6 rectangular plane surfaces.
If l be the length of the cuboid, b its breadth, h be the height and d be the diagonal, then
• Diagonal = (l2 +b2 + h2)
• Lateral Surface Area/ Area of 4 walls = 2h (l +b)
• Total Surface Area = (lb + bh + hl)
• Volume = l x b x h
## Surface Area and Volume Formulas of Cube
Cube is a solid bounded by 6 square plane surfaces.
If a be the edge of the cube, and d be the diagonal, then
• Diagonal = (3)a
• Lateral Surface Area/ Area of 4 walls = 4a2
• Total Surface Area = 6a2
• Volume = a3
## Surface Area and Volume Formulas of Cylinder
A solid obtained by revolving a rectangular lamina about one of its sides is called a right circular cylinder. The cross sections of a cylinder are two congruent circles. Below are the formulas for surface area and volume of a solid cylinder and a hollow cylinder.
### Surface area and volume of Solid Cylinder
Let r be the radius and h be the height of a solid cylinder, then
• Circumference of base = 2πr
• Area of base = πr2
• Curved Surface area = 2πrh
• Total Surface Area = 2πr2 + 2πrh
• Volume = πr2h
### Surface area and volume of Hollow Cylinder
Let R and r be the external and internal radii of a hollow cylinder and h be the height, then
• Thickness of cylinder = R – r
• Area of circular ring (cross-section) = π (R2 _ r2)
• External Curved Surface area = 2πRh
• Internal Curved Surface area = 2πrh
• Total Surface Area = 2πRh + 2πrh + 2π (R2 _ r2)
• Volume of the material = πh (R2 _ r2)
## Surface Area and Volume Formulas of Cone
A solid obtained by revolving a right angled triangular lamina about any side (other than the hypotenuse) is called a right circular cone. Below are the formulas for surface area and volume of a cone.
Let r be the radius, h be the height and l be the slant height of the cone, then
• Slant height (l) = (h2 + r2)
• Area of Base = πr2
• Curved (lateral) surface area = πrl
• Total Surface Area = πr+ πrl
## Surface Area and Volume Formulas of Sphere
A sphere is a three dimensional figure, which is made up of all points in space that lie at a constant distance from a fixed point. The constant distance is called radius and the fixed point is called the centre of the sphere.
### Surface Area and Volume Formulas of a Solid Sphere
Let r be the radius of the solid sphere, then
• Surface Area = 4πr2
• Volume = 4/3 πr3
### Surface Area and Volume Formulas of a Spherical Shell
Let R and r be the external and internal radii of a spherical shell, then
• Thickness of shell = R – r
• Surface Area = 4π (R2 + r2)
• Volume = 4/3 π (R3 – r3)
## Surface Area and Volume Formulas of a Hemisphere
When a sphere is cut by a plane passing through its centre, then the sphere is divided into two equal parts. Each part is called a hemisphere.
Let r be the radius of the hemisphere, then
• Curved Surface Area = 2πr2
• Total Surface Area = 3πr2
• Volume = 2/3 πr3
### Surface Area and Volume Formulas of a Hemipherical Shell
Let R and r be the external and internal radii of a Hemispherical shell, then
• Thickness of shell = R – r
• Area of base
• External Curved Surface Area 2πR2
• Internal Curved Surface Area 2πr2
• Total Surface Area2πR2πr2(R2 – r2)
• Volume = 2/3 π (R3 – r3)
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# A(7, −3), B(5, 3), C(3, −1) are the vertices of
Question:
$A(7,-3), B(5,3), C(3,-1)$ are the vertices of a $\Delta A B C$ and $A D$ is its median. Prove that the median $A D$ divides $\Delta A B C$ into two triangles of equal areas.
Solution:
The vertices of the triangle are A(7, −3), B(5, 3), C(3, −1).
Coordinates of $D=\left(\frac{5+3}{2}, \frac{3-1}{2}\right)=(4,1)$
For the area of the triangle $A D C$, let $A\left(x_{1}, y_{1}\right)=A(7,-3), D\left(x_{2}, y_{2}\right)=D(4,1)$ and $C\left(x_{3}, y_{3}\right)=C(3,-1)$. Then
Area of $\Delta A D C=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$
$=\frac{1}{2}[7(1+1)+4(-1+3)+3(-3-1)]$
$=\frac{1}{2}[14+8-12]=5$ sq. unit
Now, for the area of triangle $A B D$, let $A\left(x_{1}, y_{1}\right)=A(7,-3), B\left(x_{2}, y_{2}\right)=B(5,3)$ and $D\left(x_{3}, y_{3}\right)=D(4,1)$. Then
Area of $\Delta A B D=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$
$=\frac{1}{2}[7(3-1)+5(1+3)+4(-3-3)]$
$=\frac{1}{2}[14+20-24]=5$ sq. unit
Thus, Area $(\Delta A D C)=$ Area $(\Delta A B D)=5$ sq. units.
Hence, $A D$ divides $\Delta A B C$ into two triangles of equal areas. |
# Solving Inequalities: Finding the Value of X in 9(2x + 1) < 9x – 18
In solving inequalities, the goal is to find the value of X that will satisfy the inequality. For example, given the equation 9(2x + 1) < 9x – 18, we need to find the value of X that will make the inequality true. The first step in solving an inequality is to simplify both sides of the equation as much as possible. In this case, we can start by distributing the 9 on the left side to get: 18x + 9 < 9x – 18 Next, we can simplify the right side by adding 18 to both sides: 18x + 9 < 9x Then, we can subtract 9x from both sides to isolate the variable: 9x + 9 < 0 Finally, we can subtract 9 from both sides to get the value of X: 9x < -9 X < -1 Therefore, the solution to the inequality 9(2x + 1) < 9x – 18 is X < -1. This means that any value of X less than -1 will make the inequality true. In conclusion, solving inequalities requires simplifying both sides of the equation and isolating the variable to find the value that satisfies the inequality. With practice, anyone can become proficient in solving inequalities and finding the value of X. |
# Computing Arc Length and Surfaces of Revolution
Multi-Variable Calculus & Calculus Review Chapter 7 Section 4
Mrs. Shak
7.4: Arc Length & Surfaces
• Using definite integrals to compute arc length of curves • Line segments are given by the familiar Distance Formula
Rectifiable curve = curve with finite arc length Function continuously differentiable on [a,b], i.e. f’ continuous on [a,b] Graph of f on [a,b] is a Smooth Curve
1
7.4 : Approximating Arc Length
Arc Length (s) = Sum of little line segments
7.4 : Definition of Arc Length
• Computing Arc Length as Definite Integral
2
7.4: Homework Problem p.483 #4
Find the arc length of the graph of the function over the 3 interval [0,9] 2
y = 2x
+3
Differentiate: Definite Integral: Integrate:
7.4 Area of a Surface of Revolution
The area of a surface of revolution is derived from the formula for the lateral surface area of the frustum of a right circular cone.
NOTE: frustrum is the conic shape (normally cone or pyramid) with the top “chopped off” by a plane parallel to the base
3
7.4 Area of Surface of Revolution
• How to set up lateral surface area of cylinder as an integral? • Can we extend this to the surface area of an object generated by revolving an arbitrary continuous function revolved about the x-axis? What is our new “dx”, or rather “ds”?
7.4 Area of a Surface of Revolution
Suppose the graph of a function f, having a continuous derivative on the interval [a, b], is revolved about the xaxis to form a surface of revolution, as shown.
∆ L = ∆ x 2 + ∆y 2 ∆y = 1 + ∆x ∆x dy ≈ 1 + dx = 1 + f '( x )2 dx = ds dx
2 2
Arclength Formula!!
4
7.4 Area of a Surface of Revolution
7.4 Area of a Surface of Revolution
In these two formulas for S, you can regard the products 2πf(x) and 2πx as the circumferences of the circles traced by a point (x, y) on the graph of f as it is revolved about the x-axis and the y-axis (Figure 7.45). In one case the radius is r = f(x), and in the other case the radius is r = x.
Figure 7.45
5
Example – The Area of a Surface of Revolution
Find the area of the surface formed by revolving the graph of f(x) = x3 on the interval [0, 1] about the x-axis, as shown.
Figure 7.46
Example – Solution
The distance between the x-axis and the graph of f is r(x) = f(x), and because f'(x) = 3x2, the surface area is
6
Homework
• Read Chapter 7.4 (Single Var Textbook) • Do Homework
– Pg 484 (39-46:all, 51, 53): 10 problems – State the function, its derivative, and the complete integral … either integrate by hand or on the calculator
7 |
## betterworld2016.org
Let"s take into consideration again the 2 equations us did very first on the previous page, and also compare the lines" equations with their slope values.
You are watching: What is slope of vertical line
The very first line"s equation to be y = (2/3) x – 4, and the line"s slope was m = 2/3.
The 2nd line"s equation was y = –2x + 3, and the line"s slope was m = –2. In both cases, the number multiply on the variable x was additionally the worth of the slope for the line. This relationship always holds true: If the line"s equation is in the form "y=", then the number multiplied on x is the worth of the slope m.
This connection will become an extremely important when you begin working v straight-line equations.
Now let"s take into consideration those 2 equations and also their graphs.
For the first equation, y = ( 2/3 )x – 4, the slope was m = 2/3, a optimistic number. The graph looked prefer this:
Notice how the line, together we relocate from left to ideal along the x-axis, is edging upward toward the peak of the drawing; technically, the heat is an "increasing" line. And... The slope was positive.
This relationship constantly holds true: If a heat is increasing, climate its slope will certainly be positive; and also if a line"s slope is positive, then its graph will be increasing.
For the second line, y = –2x + 3, the slope was m = –2, a negative number. The graph looked favor this:
Notice just how the line, together we relocate from left to ideal along the x-axis, is edging downward toward the bottom the the drawing; technically, the heat is a "decreasing" line. And... The slope to be negative.
This partnership is constantly true: If a heat is decreasing, climate its slope will certainly be negative; and if a line"s slope is negative, climate its graph will be decreasing.
This relationship between the sign on the slope and also the direction the the line"s graph can aid you examine your calculations: if you calculation a slope together being negative, however you can see indigenous the graph the the equation the the line is actually increasing (so the slope need to be positive), climate you understand you need to re-do your calculations. Being conscious of this link can conserve you point out on a test since it will permit you to check your work-related before friend hand the in.
So now we know: increasing lines have positive slopes, and also decreasing lines have an unfavorable slopes. V this in mind, let"s consider the following horizontal line:
Is the horizontal line edging upward; the is, is it raising line? No, so its steep can"t be positive. Is the horizontal line edging downward; the is, is the a to decrease line? No, so its slope can"t it is in negative. What number is neither optimistic nor negative?
Zero!
So the slope of this (and any kind of other) horizontal heat should, logically, be zero. Let"s do the calculations to confirm this. Using the (arbitrary) points from the line, (–3, 4) and also (5, 4), the slope computes as:
This relationship constantly holds: a steep of zero means that the line is horizontal, and also a horizontal line way you"ll acquire a slope of zero.
(By the way, all horizontal lines are of the form "y = part number", and the equation "y = some number" always graphs as a horizontal line.)
Is the vertical heat going up on one end? Well, yes, kind of. So maybe the slope will be positive...? Is the vertical line going down on the other end? Well, again, type of. So possibly the slope will be negative...?
But is there any type of number the is both optimistic and negative? Nope.
Verdict: upright lines have actually NO SLOPE. The concept of slope merely does no work because that vertical lines. The slope of a vertical line does not exist!
Let"s carry out the calculations to check the logic. Indigenous the line"s graph, I"ll usage the (arbitrary) clues (4, 5) and also (4, –3). Climate the steep is:
We can"t division by zero, i beg your pardon is of course why this slope value is "undefined".
This partnership is constantly true: a vertical line will have no slope, and "the steep is undefined" or "the line has no slope" way that the heat is vertical.
(By the way, every vertical lines room of the type "x = part number", and also "x = part number" way the line is vertical. Any type of time her line entails an undefined slope, the line is vertical; and any time the heat is vertical, you"ll finish up splitting by zero if you try to compute the slope.)
Warning: the is very common to confuse this two types of lines and their slopes, however they are very different.
Just together "horizontal" is not at all the same as "vertical", so also "zero slope" is no at every the very same as "no slope".
Just as a "Z" (with its 2 horizontal lines) is not the exact same as one "N" (with its two vertical lines), so likewise "Zero" steep (for a horizontal line) is no the same as "No" slope (for a vertical line).
See more: What Is An Example Of Gaining Awareness About A Health Problem
The number "zero" exists, for this reason horizontal lines do indeed have a slope. Yet vertical present don"t have any slope; "slope" simply doesn"t have any definition for vertical lines.
It is very common for tests come contain questions regarding horizontals and also verticals. Don"t mix them up! |
## Calculating with polynomials
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If $$f(x)$$ and $$g(x)$$ are two polynomials and $$c$$ is a real number, then the following expressions also are polynomials:
• $$c\cdot {f(x)}$$
• $$f(x)+g(x)$$
• $$f(x)-g(x)$$
• $$f(x)\cdot g(x)$$
Rules of calculation for polynomials
• Multiplying a polynomial by a constant is equivalent to multiplying each term of the polynomial by that constant
• Addition of two polynomials in $$x$$ is equivalent to adding the coefficients of terms with the same power of $$x$$.
• Subtraction of polynomial $$g(x)$$ by a polynomial $$f(x)$$ is the same as subtracting the coefficients of terms in $$g(x)$$ by the coefficients of the same power of $$x$$ in $$f(x)$$.
• Multiplication of two polynomials is obtained by multiplying each term of one polynomial by each term of the other polynomial and adding all the products.
The rules specify how we can add, subtract and multiply polynomials. The quotient $$\frac{f(x)}{g(x)}$$ of two polynomials is not always a polynomial, but does result in a rational function.
Degree of polynomials that result from arithmetic operations
Let $$f(x)$$ and $$g(x)$$ be polynomials of degree respectively $$m$$ and $$n$$, and let $$c$$ be a real number.
• The degree of $$c\cdot f(x)$$ is the degree of $$f(x)$$ if $$c\ne0$$.
• The degree of $$f(x)\cdot g(x)$$ is the sum of the degrees $$f(x)$$ and $$g(x)$$.
• if $$m\gt n$$, then the degree of $$f(x)+g(x)$$ is equal to the degree of $$f(x)$$.
• If $$m=n$$, then the degree of $$f(x)+g(x)$$ is less than or equal to the degree of $$f(x)$$.
Let’s practice!
What is the product of polynomial $$f(x)=x^3+x^2-4\cdot x-4$$ and the constant $$2$$?
In order to compute the product of $$f(x)$$ and $$2$$ we multiply the coefficient of each power of $$x$$ in $$f(x)$$ with $$2$$:
$$\begin{array}{rcl}2\cdot f(x)&=& 2\cdot \left(x^3+x^2-4\cdot x-4\right)\\ &=&{ -8+{ (2\cdot1)\cdot x^2 }+{ (2\cdot-4)\cdot x }+{ (2\cdot-4) }} \\ &=& 2\cdot x^3+2\cdot x^2-8\cdot x-8\end{array}$$
$$2\cdot x^3+2\cdot x^2-8\cdot x-8$$
## The notion of polynomial
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A polynomial is an expression of the form:
$$a_0+a_1x+a_2x^2+\cdots + a_nx^n\tiny,$$,
where $$a_2,\ldots,a_n$$ are numbers (the coefficients of the polynomial) and $$x$$ is the variable.
If $$a_n\ne0$$, then $$n$$ is called the degree of the polynomial. The number $$a_2$$ is then called the leading coefficient of the polynomial.
By way of convention, we say that the polynomial $$0$$ is of degree $$-1$$.
The above polynomial defines a function $$f$$ with rule
$$f(x) =a_0+a_1x+a_2x^2+\cdots + a_nx^n\tiny.$$.
Such a function is called a polynomial function.
Let’s practice!
What is the degree of the polynomial $$f(x)=4+7 \cdot x$$? And what is its leading coefficient?
The degree of the polynomial$$f(x)=4+7 \cdot x$$ is equal to $$1$$. The leading coefficient equals $$7$$. Polynomials of degree $$1$$ are also known as linear functions.
The corresponding graph is a straight line with slope equal to $$7$$ and the intersection with the $$y$$-axis is at $$\left[0,4\right]$$.
## Solving equations with factorization
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If we can factor the left hand side of the equation $$ax^2+bx+c=0$$, then the solution is easily found:
Consider the quadratic equation with unknown $$x$$:
$$ax^2+bx+c=0$$,
where $$a$$, $$b$$ and $$c$$ are real numbers with $$a\ne0$$. If the left hand side can be factored in linear factors:
$$ax^2+bx+c=(x-p)\cdot(x-q)\tiny,$$,
where $$p$$ and $$q$$ are real numbers, then the solution of the equation is $$x=p\lor x=q$$.
Let’s practice!
Solve the equation for $$x$$ below using factorization.
$$x^2-14 x-16=-8 x\tiny.$$
Give your answer in the form of $$x=a\lor x=b$$.
This can be solved by means of the following deduction:
$$\begin{array}{rcl} x^2-14 x-16&=&-8 x \\ &&\phantom{xxx}\color{blue}{\text{the original equation}}\\ x^2-6 x-16&=&0 \\ &&\phantom{xxx}\color{blue}{\text{all terms to the left}}\\ (x-8)\cdot (x+2)&=&0 \\ &&\phantom{xxx}\color{blue}{\text{left hand side factored}}\\ x-8=0 &\lor& x+2=0\\ &&\phantom{xxx}\color{blue}{A\cdot B=0 \text{ if and only if }A=0\lor B=0}\\ x=8 &\lor& x=-2\\ &&\phantom{xxx}\color{blue}{\text{constant terms to the right}} \end{array}$$
$$x=8\lor x=-2$$
## Quadratic function – Completing the square
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A quadratic function $$f$$ is given by $$f(x)=ax^2+bx+c$$ with $$a$$, $$b$$ and $$c$$ being real numbers and $$a\neq 0$$.
The graph associated with such a function is called a parabola.
When $$a\gt0$$, we speak of a parabola opening upwards, and when $$a\lt0$$, we speak of a parabola opening downwards.
The significance of these names becomes evident from the graph below of the parabola opening upwards of $$x^2+8$$ and the parabola opening downwards of $$-x^2-8$$.
To find the zeros of a quadratic function, we need do find the points of intersection of the parabola with the $$x$$-axis. We can do this in three ways:
• completing the square
• $$abc$$-formula
• factorization
In what follows, we will discuss the completing the square method.
An equation like $$\left(x+2\right)^2=3$$ can easily be solved by taking the root:
Solve $$x$$ in $$\left(x+4\right)^2={{16}\over{25}}$$.
Take the root at both sides of the $$=$$ sign: $$x+4=\pm {{4}\over{5}}$$;
Next, solve the equation: $$x=-4\pm {{4}\over{5}}$$;
$$x=-4\pm {{4}\over{5}}$$
However, this will not work in general for a quadratic equation like $$x^2+8x-1=0$$, because the unknown $$x$$ occurs twice. But there is a method to obtain the first form from the second one:
Completing the square is rewriting a quadratic expression in $$x$$ as an expression in which $$x$$ only occurs once, in the base of a second power. To be precise, if $$a$$, $$b$$ and $$c$$ are real numbers, then
$$ax^2+bx+c=a\cdot\left( \left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{(2a)^2}\right)\tiny.$$
With this method we cannot only solve quadratic equations, but also determine what the extreme point of a parabola is:
The quadratic polynomial $$ax^2+bx+c$$ in which $$a\ne0$$, can be written as
$$a\cdot\left(x+\frac{b}{2a}\right)^2 -\frac{b^2-4ac}{4a}\tiny.$$
In particular, $$\left[-\frac{b}{2a},-\frac{{b}^{2}-4a·c}{4a}\right]$$ is an extreme:
• If $$a\gt0$$, then the extreme is the lowest point of the parabola opening upwards.
• If $$a\lt0$$, then the extreme is the highest point of the parabola opening downwards.
In other words, the quadratic function $$ax^2+bx+c$$ in $$x$$ has a minimum or maximum (depending on $$a\gt0$$ or $$a\lt0$$) for $$x=-\frac{b}{2a}$$, which is $$-\frac{b^2-4ac}{4a}$$.
Let’s practice!
Solve $$x$$ by completing the square in the equation
$$x^2-6 x-9=0$$
Give your answer in the form $$x=a\lor x=b$$.
To find this answer, we complete the square in the following manner:
$$\begin{array}{rcl} (x-3)^2-3^2-9&=&0\\ &&\phantom{xx}\color{blue}{\text{ }x^2-6x\text{ completed to a square}}\\ (x-3)^2&=&3^2+9\\ &&\phantom{xx}\color{blue}{\text{everything outside of the brackets moved to the right hand side}}\\ (x-3)^2&=&18\\ &&\phantom{xx}\color{blue}{\text{simplified the right hand side}}\\ x-3=\sqrt{18} &\lor&x-3=-\sqrt{18}\\ &&\phantom{xx}\color{blue}{\text{the root taken at both sides}}\\ x=3\cdot \sqrt{2}+3&\lor& x=3-3\cdot \sqrt{2}\\ &&\phantom{xx}\color{blue}{-3 \text{ subtracted from both sides}}\\ \end{array}$$
$$x=3\cdot \sqrt{2}+3\lor x=3-3\cdot \sqrt{2}$$
## Solving systems of equations by addition
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A general method of solving two linear equations with two unknowns is by eliminating one unknown. Here we discuss a method that is also suitable for larger systems of linear equations, with more unknowns.
The goal is to give the first equation the form $$x = a$$ and the second equation the form $$y = b$$.
• The strategy is to edit the equations in such a way that the new system will be equivalent to the old;
• the new system looks more like a solution than the old one.
Steps that will occur are multiplying all terms in the same equation by the same non-zero number and subtracting one equation over the other.
The addition method for linear equations
A system of two linear equations with unknowns $$x$$ and $$y$$ can be solved as follows.
1. Make sure $$x$$ occurs in the first equation: if this is not the case, then switch the two equations; this way, $$x$$ will occur in the first equation.
2. Replace the second equation by the difference of this equation with a suitable multiple of the first equation, in such a way that $$x$$ no longer occurs in the second equation.
3. Replace the first equation by the difference of this equation with a suitable multiple of the second equation in such a way that $$y$$ no longer occurs in the first equation.
4. The first equation is now a linear equation with $$x$$ as the only unknown, and the second is a linear equation with $$y$$ as the only unknown. These equations can be solved with the theory of linear equations with one unknown.
Let’s practice!
Solve the following system of equations with unknowns $$x$$ and $$y$$.
$$\left\{\begin{array}{l}-5·\left\{x\right\}+12·\left\{y\right\}+3=0\\ 2·\left\{x\right\}-5·y+1=0\end{array}\right.$$
Give the answer in the form $$x=a\land y=b$$ for suitable values of $$a$$ and $$b$$.
There are many ways to get to this solution. We will describe one.
• To make sure that the unknown $$x$$ is present in the first equation, we switch the two equations if this was not the case in the original system: $$-5\cdot x+12\cdot y+3=0\land 2\cdot x-5\cdot y+1=0$$.
• Next we get rid of the term with $$x$$ from the second equation by multiplying the first equation with $$\frac{2}{-5}$$ and subtracting from the second: $$-5\cdot x+12\cdot y+3=0\land -{{y}\over{5}}=0$$.
• By dividing (left and right hand side) in the second equation by $$-\frac{1}{5}$$ we find $$y = 11$$. We now have the system $$-5\cdot{x}+12\cdot y+3=0\land y=11$$.
• If we enter the solution of $$y$$ (the second equation) in the first equation (or stated different: we subtract $$12$$ times the second equation from the first), then we find the system: $$135-5\cdot x=0\land y=11$$.
• The first equation can be solved with the theory of linear equations with one unknown. The result is $$x= 27\land y = 11$$.
$$x= 27\land y = 11$$
## The equation of a line
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Assume that $$a$$, $$b$$, and $$c$$ are constant real numbers: parameters.
The solution to the equation $$a\cdot x+b\cdot y+c=0$$ can be drawn in the plane. They are the points $$\left[x,y\right]$$ satisfying $$a\cdot x+b\cdot y+c=0$$. If $$a\ne0$$ or $$b\ne0$$, then these points form a straight line, or simply just a line.
• If $$b\ne0$$, then the equation can be written as $$y=-\frac{a}{b}\cdot{x}-\frac{c}{b}$$. For, these are the solutions if we consider $$x$$ as a parameter and $$y$$ as unknown. This indicates that for every value of $$x$$ there is a point $$\left[x,y\right]$$ with $$y$$ equal to $$-\frac{a}{b}\cdot{x}-\frac{c}{b}-\frac{a}{b}\cdot{x}-\frac{c}{b}$$.
• If $$a\ne0$$, the line is oblique (by oblique we mean neither horizontal nor vertical).
• If $$a = 0$$, then the value of $$y$$ is constant, equal to $$-\frac{c}{b}$$. In this case the line is horizontal.
• In the exceptional case $$b = 0$$ the equation looks like $$a \cdot{x}+c = 0$$.
• If $$a\ne0$$, then the line is vertical.
• If $$a = 0$$ and
• $$c\ne0$$, then there are no solutions
• $$c = 0$$, then each pair of values of $$\left[x,y\right]$$ is a solution.
A straight line can be described in different ways.
1. The solutions $$\left[x,y\right]$$ to an equation $$a\cdot x+b\cdot y+c=0$$ with unknowns $$x$$ and $$y$$. Here $$a$$, $$b$$ and $$c$$ are real numbers such that $$a$$ and $$b$$ are not equal to zero.
2. The line through two given points in the plane: if $$P = \left[x,y\right]$$ and $$Q = \left[s,t\right]$$ are points in the plane, then the line through $$P$$ and $$Q$$ has equation $$a\cdot x+b\cdot y+c=0$$ with $$a=q-t$$, $$b = s – p$$ and $$c = t \cdot {p} – q \cdot{s}$$.
3. The line through a given point, the base point, and a direction, indicated by the number $$-\frac{a}{b}$$, where $$a$$ and $$b$$ are as in the equation given above; this number is called the slope of the line.
4. The line with function representation $$y = p\cdot x+q$$ if $$b\ne0$$ and $$x = r$$ otherwise; here we have $$p = -\frac{a}{b}$$ (the slope), $$q = -\frac{c}{b}$$ (the intercept), which is the value of $$y$$ for $$x = 0$$ and $$r = -\frac{c}{a}$$ in terms of the above $$a$$, $$b$$ and $$c$$. This can be seen as a special case of the previous description, with base point $$\left[0,q\right]$$. In the case where $$b\ne0$$, the variable $$y$$ is a function of $$x$$, in the other case, $$x$$ is a constant function of $$y$$.
Let’s practice!
The line segment between the two points $$\left[2,\frac{77}{12}\right]$$ and $$\left[6,\frac{63}{4}\right]$$ is drawn in the figure below.
Give the function rule for this line in the form $$y = a\cdot{x} + b$$.
The line is described by the function rule $$y = a\cdot{x} + b$$ where $$a$$ is the slope, given as the quotient of the difference of the $$y$$-values with the difference of the $$x$$-values of two points on the line. Hence $$a=\frac{{{63}\over{4}}-{{77}\over{12}}}{6-2}={{7}\over{3}}$$. The value of $$b$$ follows from $$b = y – a\cdot{x}$$, where $$\left[x,y\right]$$ is a random point on the line. Hence $$b={{77}\over{12}}-{{7}\over{3}}\cdot{2}={{7}\over{4}}$$.
$$y={{7}\over{3}}\cdot{x}+{{7}\over{4}}$$
## Arithmetic operations for continuity
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Let’s discuss some methods to create a new continuous function from known continuous functions.
Continuity of sums, products, quotients and compositions of continuous functions
Let $$a$$ be a real number.
1. Suppose $$f$$ and $$g$$ are functions continuous in $$a$$. Then, the functions $$f + g$$ and $$f \cdot g$$ are also continuous in $$a$$. If $$g(a)\ne0$$, then we have the same for $$\frac{f}{g}$$.
2. Suppose $$f$$ is a function in $$a$$ and that $$g$$ is a function continuous in $$f(a)$$. Then the composition $${g}\circ{f}$$ is continuous in $$a$$.
Continuity of power functions
Let $$d$$ be a real number. If $$f$$ is a function that is continuous in $$a$$ with $$f(a)>0$$, then $$f(x)^d$$ is also continuous in $$a$$.
Let’s practice!
Consider the function rules $$f(x)=x^2$$ and $$g(x)=x+1$$.
What is the function rule of the composition $$f\circ g$$?
$$\begin{array}{rcl} f \circ g(x) &=& f \left(g(x) \right) \\ &&\phantom{xyzuvw}\color{blue}{\text{composition of functions}} \\ &=& g(x)^2 \\ &&\phantom{xyzuvw}\color{blue}{\text{function rule of }f\text{ entered}} \\ &=&\left(x+1\right)^2 \\ &&\phantom{xyzuvw}\color{blue}{\text{function rule of }g\text{ entered}} \\ \end{array}$$
$$f\circ g(x) = \left(x+1\right)^2$$
Let’s practice once more!
Let $$f$$ be the function with rule $$f(x) = {{1}\over{x^2+4}}$$ and $$g$$ the function with rule $$g(x)=x^2$$.
Determine the function rule for $${f – g}$$.
$$\left({f – g}\right)(x)=f(x)-g(x)=\left({{1}\over{x^2+4}}\right)-\left(x^2\right)={{-x^4-4\cdot x^2+1}\over{x^2+4}}$$
$$\left({f – g}\right)(x)\,=\, {{-x^4-4\cdot x^2+1}\over{x^2+4}}$$
## Continuity
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Let $$f$$ be a real function defined on an open interval around point $$a$$.
If $$\displaystyle \lim_{x\to a}f(x) =f(a)$$, then $$f$$ is called continuous at $$a$$. If not, then $$f$$ is called discontinuous at $$a$$.
If $$f$$ is continuous at every point of an open interval $$(c.d)$$, then $$f$$ is called continuous on $$(c.d)$$.
In other words, a function is continuous if the graph can be drawn without lifting the pen from the paper.
Let’s practice!
Consider the function $$f$$ given by
$$\displaystyle f(x)=\begin{cases} 8 x+9&\text{if }x\lt1\\ 9 x^2+8&\text{if }x\geq1\end{cases}$$
Is this function continuous at $$1$$?
The function $$f$$ is left continuous at $$x = 1$$:
$$\begin{array}{rcl} \displaystyle\lim_{x\uparrow1}f(x)&=&\lim_{x\uparrow1}( 8x+9)=8+9=17=f(1)\end{array}$$
The function $$f$$ is right continuous at $$x = 1$$:
$$\begin{array}{rcl}\displaystyle\lim_{x\downarrow1}f(x)&=&\lim_{x\downarrow1} (9 x^2+8)=9+8=17=f(1)\end{array}$$
Since $$f$$ is both right continuous and left continuous at $$1$$, it is continuous at $$1$$.
The graph of $$f$$ is drawn below. It has no jump at $$1$$.
Yes, this function is continuous at $$1$$
## The notion of limit
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Let $$a$$ and $$b$$ be real numbers and let $$f$$ be a real function that is defined on an open interval containing $$a$$.
We say that $$f$$ has limit $$b$$ at $$a$$ if $$f(x)$$ comes closer to $$b$$ as $$x$$ comes closer to $$a$$.
In this case, we write $$\textstyle\lim_{x\to a} f(x) = b$$ or $$\displaystyle\lim_{x\to a} f(x) = b$$.
Let’s practice!
The rational function $$f(x) = \frac{2\cdot x-16}{x-8}$$ is defined everywhere except at $$8$$.
What is the limit of $$f$$ at $$8$$?
For every value of $$x$$, $$f(x)=2$$ is close to (but distinct from) $$8$$.
$$\lim_{x\to 8}f(x)= 2$$
## The range of a function
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Consider the real function $$f$$.
The set of all values $$f(x)$$ for $$x$$ in the domain of $$f$$ is called the range of $$f$$.
If $$p$$ is a real number in the domain of $$f$$ with $$f(p)=0$$, then $$p$$ is called a zero of $$f$$.
The range of a function depends on the domain of the function: the larger we choose the domain, the larger in general the range will be.
Let’s practice!
Consider the real function $$f(x) = \sqrt{x-9}+4$$.
The largest possible domain of $$f$$ has the form $$x\ge a$$ for a certain number $$a$$; with other words: it is of the form $$\left[a,\infty \right]$$.
The range of $$f$$ on this domain is of the form $$\left[b,\infty \right]$$ for a certain number $$b$$. Hence, it consists of all $$y$$ with $$y\ge b$$.
Determine $$a$$ and $$b$$.
The largest possible domain of $$f$$ is the set of values of $$x$$ at which $$f(x)$$ is defined. Because the argument of the root has to be greater than or equal to zero, the function $$f(x) = \sqrt{x-9}+4$$ is only defined for $$x-9\ge0$$, or $$x\ge 9$$. Therefore, the largest possible domain of $$f$$ is $$x\ge 9$$.
The range is the set of values that $$f$$ can take on this domain. The smallest value $$f$$ can take $$f(9) = 4$$. Hence, the range of $$f$$ is $$y\ge 4$$.
$$a = 9$$ and $$b = 4$$ |
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When you go out to buy anything in the marketplace or a mall, you need to pay a certain amount of money to buy the desired commodity of service. You need to pay for everything in the form of money whether you are going for a dinner at a restaurant, want to pay the rent of your house, pay the fees of your school or have a coffee at the coffee shop. Thus, we all need the basic skill of counting money to be able to carry out transactions without any problem or mistake. If you make a mistake in counting money, it can be very costly. Thus, one of the very first things that we want from our children is ‘Learn to Count Money’. Counting money does not require any extra skills but simply the implementation of basic skills that you learn in primary lessons of mathematics.
How to Count Money?
Counting money requires the basic learning of addition, subtraction, multiplication and division of numbers. Counting money would be extremely simple if you remember mathematical tables at least till 10.
Let us take a few examples to Learn to Count Money
Suppose you have two coins or notes with a value of 10 units and 20 units, the total money that you have can be simply obtained by adding the two values 10 and 20. Thus, you have 30 units of money.
However, when you have a large number of currency notes or coins, it would take a lot of time if you just rely on addition. Here, you would need multiplication tables and your skills to multiply numbers. Suppose you have 20 currency notes with a value of 50 units each, the total money that you have is simply the multiplication of the number of currency notes and value of each note. Thus, in this case, you have total money with a value of 20*50 = 1000 units.
Now, moving to a more general case where you may have two or more different currency notes with different values. Suppose you have two types of currency notes with a value of 10 units and 50 units each. Their numbers are 5 and 4 respectively. Here you would need both addition and multiplication to count the total money.
5 Currency Notes of value 10 units each makes a total money = 5 *10 units = 50 units
4 Currency Notes of value 50 units each makes a total money = 4 * 50 units = 200 units
Total Money = (50 +200) units = 250 units
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2022-08-10
What is the probability that not more than 2 wells are pure?
Eliza Beth13
To find the probability that not more than 2 wells are pure, we need some additional information. Specifically, we need to know the total number of wells and the probability that each individual well is pure. Without this information, it is not possible to provide a specific numerical solution. However, I can explain the general approach to solving such a problem.
Let's assume that we have a total of $n$ wells, and the probability of a well being pure is $p$. We want to find the probability that not more than 2 wells are pure.
To solve this, we can consider the cases where 0, 1, or 2 wells are pure, and then sum up the probabilities of these cases.
1. Zero pure wells: The probability that no well is pure can be calculated as $\left(1-p{\right)}^{n}$. This is because for each individual well, the probability of it not being pure is $\left(1-p\right)$, and since all the wells are independent, we multiply these probabilities together.
2. One pure well: The probability that exactly one well is pure can be calculated as $np\left(1-p{\right)}^{n-1}$. Here, we choose one well out of the $n$ wells to be pure, which can be done in $n$ ways. The chosen well has a probability $p$ of being pure, and the remaining $\left(n-1\right)$ wells have a probability $\left(1-p\right)$ of not being pure.
3. Two pure wells: The probability that exactly two wells are pure can be calculated as $\left(\genfrac{}{}{0}{}{n}{2}\right){p}^{2}\left(1-p{\right)}^{n-2}$. Here, we choose two wells out of the $n$ wells to be pure, which can be done in $\left(\genfrac{}{}{0}{}{n}{2}\right)$ ways (using combinations). The chosen wells have a probability ${p}^{2}$ of being pure, and the remaining $\left(n-2\right)$ wells have a probability $\left(1-p\right)$ of not being pure.
To find the probability that not more than 2 wells are pure, we sum up the probabilities from the above three cases:
Please note that this is a general formula, and the specific values of $n$ and $p$ would be needed to obtain a numerical solution.
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Problem 661
# The 54 th and 4 th terms of an arithmetic progression are 61 and 64 ; find the $$23 \mathrm{rd}$$ term.
Expert verified
The 23rd term of the arithmetic progression is $$\frac{3143}{50}$$.
See the step by step solution
## Step 1: Understand the formula for arithmetic progressions
For an arithmetic progression, the formula to find the nth term is: $$a_n = a_1 + (n-1)d$$, where: - $$a_n$$ is the nth term of the progression - $$a_1$$ is the first term of the progression - n is the position of the term in the progression - d is the common difference between consecutive terms in the progression
## Step 2: Write the given information
We are given the 54th term and the 4th term. So, we have: $$a_{54} = 61$$ $$a_{4} = 64$$
## Step 3: Write the formulas for the given terms
Using the formula from Step 1, we can write the formulas for the given terms: $$a_{54} = a_1 + 53d = 61$$ $$a_{4} = a_1+3d = 64$$
## Step 4: Solve for the common difference d
Subtract the second equation from the first equation to eliminate $$a_1$$: $$(a_1 + 53d) - (a_1+3d) = 61 - 64$$ This simplifies to: $$50d = -3$$ Now, divide by 50: $$d = -\frac{3}{50}$$
## Step 5: Solve for the first term a1
Plug the value of d into one of the equations (here we use the equation for $$a_{4}$$): $$a_1 + 3\left(-\frac{3}{50}\right) = 64$$ Multiply out the term: $$a_1 - \frac{9}{50} = 64$$ Add the fraction to the other side: $$a_1 = 64 + \frac{9}{50}$$ Convert 64 to a fraction with the same denominator: $$a_1 = \frac{3200}{50} + \frac{9}{50}$$ Add the fractions together: $$a_1 = \frac{3209}{50}$$
## Step 6: Find the 23rd term a23
Now that we have the common difference and the first term, we can find the 23rd term using the formula from Step 1: $$a_{23} = a_1 + 22d$$ Plug in the values of $$a_1$$ and d: $$a_{23} = \frac{3209}{50} + 22\left(-\frac{3}{50}\right)$$ Multiply out the term: $$a_{23} = \frac{3209}{50} - \frac{66}{50}$$ Combine the fractions: $$a_{23} = \frac{3209-66}{50}$$ Simplify: $$a_{23} = \frac{3143}{50}$$ The 23rd term of the arithmetic progression is $$\frac{3143}{50}$$.
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# Algebra 4. Solving Linear Equations
## Presentation on theme: "Algebra 4. Solving Linear Equations"— Presentation transcript:
Algebra 4. Solving Linear Equations
Mr F’s Maths Notes Algebra 4. Solving Linear Equations
4. Solving Linear Equations
What on earth does Solving Equations mean? Let’s look at each of these three words in turn… Equations – these are just the same as expressions (what we have looked at in the last 3 sections, but with an equals sign (=) thrown in for good measure Linear – this just means we don’t have to worry about annoying powers… just yet! Solving – this means we must find the value of the unknown which makes the equation balance Now, there are a lot of different ways to solve equations, and if you are happy with the way that you have been taught, then stick to it, but this is the way I do them… How Mr F Solves Equations Golden Rule: Whatever you do to one side of the equation, you must do exactly the same to the other side to keep the equation in balance Aim: To be left with your unknown letter on one side of the equals sign, and a number on the other side Method: By doing the same to both sides of the equation… 1. If they are not already, get all your unknown letters on one side of the equation (NOT on the bottom of fractions and avoiding negatives). 2. Begin unwrapping your unknown letter, by thinking about the order that things were done to the letter 3. Use inverse operations to do this until you are left with just your unknown letter on one side, and the answer on the other 4. Check your answer using substitution and you should never ever get one of these wrong!
What are Inverse Operations?...
Inverse operations are the key to solving equations as they allow you to unwrap all the things surrounding your unknown letter and leave you with a simple answer. Inverse operations are just operations which are the opposite of each other, and as such they cancel each other out. Here are the main ones you need to know…. power of 2 Now, the way I am going to set out these first few examples may seem very long and painful, but if you can do it this way for the simple ones, there is no reason why you can do the same for the nightmare stinker ones at the end…
Example 1 1. Right, here we go… now our unknown letter (p) only appears on the left hand side of the equation, there is no negative sign in front of it, and it is not on the bottom of a fraction, so that’s a good start! 2. Okay, what order were things done to p… 3. And so now we can unwrap, starting with the last operation, and doing the inverse (opposite) to both sides: Add three to both sides Notice how the +3 cancels out the -3! Divide both sides by 7 Notice how dividing by 7 cancels out the 7 multiplying the p 4. We have our answer, but it’s so easy to check if we are right, that we might as well do it. Just substitute p = 5 into the questions, and hope the equation balances… When p = 5…
Example 2 1. Okay, so let’s do our checks… our unknown letter (r) only appears on the left hand side of the equation, there is no negative sign in front of it, and it is not on the bottom of a fraction, so we are good to go… after we expand the brackets, of course… 2. Okay, what order were things done to r… Note: if this bit confused you, have another read of 1. Rules of Algebra 3. And so now we can unwrap, starting with the last operation, and doing the inverse (opposite) to both sides: Subtract twelve from both sides Notice how the -12 cancels out the +12! Divide both sides by 6 Notice how dividing by 6 cancels out the 6 on the top! 4. We have our answer, but it’s so easy to check if we are right, that we might as well do it. Just substitute r = 4 into the questions, and hope the equation balances… When r = 4…
Example 3 1. Okay, so our unknown letter (k) only appears on the left hand side of the equation, there is no negative sign in front of it, and it is not on the bottom of a fraction. Phew! 2. Okay, what order were things done to k… Note: just because k is not written first, doesn’t change the order in which things are done to k! Think: BODMAS! 3. And so now we can unwrap, starting with the last operation, and doing the inverse (opposite) to both sides: Subtract six from both sides Again, look at the cancelling out! Multiply both sides by 5 It all cancels out! We have our answer, but it’s so easy to check if we are right, that we might as well do it. Just substitute k = -35 into the questions, and hope the equation balances… When k = -35…
Example 4 1. Okay, so let’s do our checks… our unknown letter (m) only appears on the left hand side of the equation, it’s not on the bottom of a fraction, but wait… it’s got a negative sign in front of it! This is going to make life difficult, but we can sort it out by using inverse operations to cancel out the -3m… We just need to add 3m to both sides! And now we have an equation just like all the others! 2. Okay, what order were things done to m… 3. And so now we can unwrap, starting with the last operation, and doing the inverse (opposite) to both sides: Subtract six from both sides The 6s on the right hand side will cancel! Divide both sides by 3 Substitution to check our answer: When m = 6…
Example 5 1. Okay, we have trouble right away! All of the unknowns (y) are NOT on the same side. No problem, we just need a bit of inverse operations. Top Tip: Collect your letters on the side which starts off with the most letters… so the right hand side! So, we just need to subtract 7y from both sides! And now we have an equation just like all the others! Note: if this bit confused you, have another read of 1. Rules of Algebra 2. Okay, what order were things done to y… 3. And so now we can unwrap, starting with the last operation, and doing the inverse (opposite) to both sides: Add six to both sides The 6s on the right hand side will cancel! Divide both sides by 3 Substitution to check our answer balances!: When y = 3… Left hand side Right hand side
Example 6 1. Problem! Our unknown letter (g) is on the bottom of a fraction! The only way we are going to get that g off the bottom of the fraction is to realise that the 25 is being divided by g – 1 and use inverse operations… So, we just need to multiply both sides by (g – 1) And expand the brackets on the right hand side And now we have an equation just like all the others! 2. Okay, what order were things done to g… 3. And so now we can unwrap, starting with the last operation, and doing the inverse (opposite) to both sides: Add five to both sides The 5s on the right hand side will cancel! Divide both sides by 5 Substitution to check our answer is correct!: When g = 6… |
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# Multiplication and Division of Radicals
## Rationalize the denominator
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Practice Multiplication and Division of Radicals
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Multiplication and Division of Radicals
What if you knew that the area of a rectangular mirror is $12 \sqrt{6}$ square feet and that the width of the mirror is $2 \sqrt{2}$ feet? Could you find the length of the mirror? What operation would you have to perform? If you knew the width and the length of the mirror, could you find its area? What operation would you perform in this case? In this Concept, you'll learn about multiplying and dividing radicals so that you can answer questions like these.
### Guidance
To multiply radicands, the roots must be the same.
$\sqrt[n]{a} \cdot \sqrt[n]{b}= \sqrt[n]{ab}$
#### Example A
Simplify $\sqrt{3} \cdot \sqrt{12}$ .
Solution:
$\sqrt{3} \cdot \sqrt{12}=\sqrt{36}=6$
Dividing radicals is more complicated. A radical in the denominator of a fraction is not considered simplified by mathematicians. In order to simplify the fraction, you must rationalize the denominator.
To rationalize the denominator means to remove any radical signs from the denominator of the fraction using multiplication.
Remember: $\sqrt{a} \times \sqrt{a}= \sqrt{a^2}=a$
#### Example B
Simplify $\frac{2}{\sqrt{3}}$ .
Solution:
We must clear the denominator of its radical using the property above. Remember , what you do to one piece of a fraction, you must do to all pieces of the fraction.
$\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{\sqrt{3^2}}=\frac{2\sqrt{3}}{3}$
Real-World Radicals
#### Example C
A pool is twice as long as it is wide and is surrounded by a walkway of uniform width of 1 foot. The combined area of the pool and the walkway is 400 square-feet. Find the dimensions of the pool and the area of the pool.
Solution:
1. Make a sketch.
2. Let $x=$ the width of the pool.
3. Write an equation. $Area=length \cdot width$
Combined length of pool and walkway $=2x+2$
Combined width of pool and walkway $=x+2$
$\text{Area}=(2x+2)(x+2)$
Since the combined area of the pool and walkway is $400 \ ft^2$ , we can write the equation.
$(2x+2)(x+2)=400$
4. Solve the equation:
$&& & (2x+2)(x+2)=400\\& \text{Multiply in order to eliminate the parentheses}. && 2x^2+4x+2x+4=400\\& \text{Collect like terms}. && 2x^2+6x+4=400\\& \text{Move all terms to one side of the equation}. && 2x^2+6x-396=0\\& \text{Divide all terms by} \ 2. && x^2+3x-198=0$
$x & = \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\& = \frac{-3 \pm \sqrt{3^2-4(1)(-198)}}{2(1)}\\& = \frac{-3\pm \sqrt{801}}{2} \approx \frac{-3\pm 28.3}{2}$
Use the quadratic formula. $x \approx 12.65$ or –15.65 feet
5. We can disregard the negative solution since it does not make sense in this context. Thus, we can check our answer of 12.65 by substituting the result into the area formula.
$\text{Area} = [2(12.65)+2)](12.65+2)=27.3 \cdot 14.65 \approx 400 \ ft^2.$
The answer checks out.
### Guided Practice
Simplify $\frac{7}{\sqrt[3]{5}}$ .
Solution:
In this case, we need to make the number inside the cube root a perfect cube. We need to multiply the numerator and the denominator by $\sqrt[3]{5^2}$ .
$\frac{7}{\sqrt[3]{5}} \cdot \frac{\sqrt[3]{5^2}}{\sqrt[3]{5^2}}=\frac{7\sqrt[3]{25}}{\sqrt[3]{5^3}}=\frac{7\sqrt[3]{25}}{5}$
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both
Multiply the following expressions.
1. $\sqrt{6}\left ( \sqrt{10} + \sqrt{8} \right )$
2. $\left ( \sqrt{a} - \sqrt{b} \right ) \left ( \sqrt{a} + \sqrt{b} \right )$
3. $\left ( 2\sqrt{x}+ 5 \right ) \left ( 2\sqrt{x}+5 \right )$
Rationalize the denominator.
1. $\frac{7}{\sqrt{15}}$
2. $\frac{9}{\sqrt{10}}$
3. $\frac{2x}{\sqrt{5}x}$
4. $\frac{\sqrt{5}}{\sqrt{3}y}$
5. The volume of a spherical balloon is $950 cm^3$ . Find the radius of the balloon. (Volume of a sphere $=\frac{4}{3} \pi R^3$ )
6. A rectangular picture is 9 inches wide and 12 inches long. The picture has a frame of uniform width. If the combined area of picture and frame is $180 in^2$ , what is the width of the frame?
7. The volume of a soda can is $355 \ cm^3$ . The height of the can is four times the radius of the base. Find the radius of the base of the cylinder.
### Vocabulary Language: English Spanish
rationalize the denominator
rationalize the denominator
To remove any radical signs from the denominator of the fraction using multiplication.
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# 1 C FRACTIONS A smaller part of a
• Slides: 25
1
C. FRACTIONS - A smaller part of a whole number. Written with one number over the other, divided by a line. 3 8 11 16 or 3 8 11 16 Any number smaller than 1, must be a fraction. Try thinking of the fraction as “so many of a specified number of parts”. For example: Think of 3/8 as “three of eight parts” or. . . Think of 11/16 as “eleven of sixteen parts”. 1. Changing whole numbers to fractions. Multiply the whole number times the number of parts being considered. Changing the whole number 4 to “sixths”: 4 = 4 x 6 = 24 or 6 6 24 6 2
CHANGING WHOLE NUMBERS TO FRACTIONS EXERCISES 1. 49 to sevenths = 49 x 7 7 = 343 7 or 343 7 2. 40 to eighths = 40 x 8 8 = 320 8 or 320 8 3. 54 to ninths = 54 x 9 9 = 486 9 or 486 9 4. 27 to thirds = 27 x 3 3 = 81 3 or 81 5. 12 to fourths = 12 x 4 4 = 48 4 or 48 6. 130 to fifths = 130 x 5 = 5 650 5 or 650 5 3 4 3
2. Proper and improper fractions. Proper Fraction - Numerator is smaller number than denominator. 3/4 Improper Fraction - Numerator is greater than or equal to denominator. 15/9 3. Mixed numbers. Combination of a whole number and a proper fraction. 4. Changing mixed numbers to fractions. Change 3 7/8 into an improper fraction. • Change whole number (3) to match fraction (eighths). 3 • = 3 x 8 8 24 8 = or 24 8 Add both fractions together. 24 8 + 7 8 = 31 8 4
CHANGING MIXED NUMBERS TO FRACTIONS EXERCISES 1. 4 1/2 = 4 x 2 2 = 8 2 + 1 2 = 9 2 2. 8 3/4 = 8 x 4 4 = 24 4 + 3 4 = 27 4 3. 19 304 = 19 x 16 = 16 + 7 = 311 16 16 = 7 x 12 = 84 12 12 + 11 = 95 12 12 84 14 + 9 = 93 14 14 7/16 16 4. 7 11/12 5. 6 9/14 = 6 x 14 = 14 1/64 = 5 x 64 = 64 6. 5 320 64 + 1 = 321 64 64 5
5. Changing improper fractions to whole/mixed numbers. Change 19/3 into whole/mixed number. . 19/3 = 19 3 = 6, remainder 1 = 6 1/3 (a mixed number) CHANGING IMPROPER FRACTIONS TO WHOLE/MIXED NUMBERS EXERCISES 1. 37/7 = = 37 7 = 5, remainder 2 = 5 2/7 (a mixed number) 2. 44/4 = = 44 4 = 11, no remainder = 11 (a whole number) 3. 23/5 = = 23 5 = 4, remainder 3 = 4 3/5 (a mixed number) 4. 43/9 = = 43 9 = 4, remainder 7 = 4 7/9 (a mixed number) 5. 240/8 = = 240 8 = 30, no remainder = 30 (a whole number) 6. 191/6 = = 191 6 = 31, remainder 5 = 31 5/6 (a mixed number) 6
6. Reducing Fractions Reducing - Changing to different terms. Terms - The name for numerator and denominator of a fraction. Reducing does not change value of original fraction. 7. Reducing to Lower Terms Divide both numerator and denominator by same number. . 3. 3= 1 Example: 3 3 & 1 3 Have same value. 9 =. . 9 9 3= 3 8. Reducing to Lowest Terms - 1 is only number which evenly divides both numerator and denominator. Example: 16 32 = a. . 16. 2 = 8. 32. 2 = 16 . b. 8. . 2 = 4. 16 2= 8 . c. 4. . 2 = 2. 8 2= 4 . d. 2. . 2 = 1. 4 2= 2 7
REDUCING TO LOWER/LOWEST TERMS EXERCISES 1. Reduce the following fractions to LOWER terms: 15. . 5 = 3 15 a. 20. . 5 = 4 20 to 4 ths = • • Divide the original denominator (20) by the desired denominator (4) = 5. . Then divide both parts of original fraction by that number ( 5). . . 4 = 9. . 4 = 10 b. 36 40 to 10 ths = 36 40 c. 24 36 to 6 ths = . 24. 6 = 36. . 6 = 4 6 d. 12 36 to 9 ths = 12 36 . . 4 = 3 9 e. 30 45 to 15 ths = 30 45 . . 3 = 10 15 16 76 . . 4 = 4 19 f. 16 76 to 19 ths = 8
REDUCING TO LOWER/LOWEST TERMS EXERCISES (con’t) 2. Reduce the following fractions to LOWEST terms: 10 = a. . 6. 2= 3 10. . 2 = 5 9 = a. . 3. 3= 1 9. . 3 = 3 64 = a. . 6. 2= 3 64. . 2 = 32 d. 13 32 = Cannot be reduced. e. 32 48 = a. . 32. 2 = 64. . 2 = 16 32 b. . 16. 2 = 8 32. . 2 = 16 16 = a. . 16. 2 = 76. . 2 = 8 38 b. . 8. 2= 4 38. . 2 = 19 a. 6 b. 3 c. 6 f. 76 c. . 8. 8= 16. . 8 = 1 2 9
9. Common Denominator Two or more fractions with the same denominator. 7 6 2 1 8 8 When denominators are not the same, a common denominator is found by multiplying each denominator together. 1 7 5 5 2 3 1 36 24 12 18 9 8 6 6 x 8 x 9 x 12 x 18 x 24 x 36 = 80, 621, 568 is only one possible common denominator. . . but certainly not the best, or easiest to work with. 10. Least Common Denominator (LCD) Smallest number into which denominators of a group of two or more fractions will divide evenly. 10
10. Least Common Denominator (LCD) con’t. To find the LCD, find the “lowest prime factors” of each denominator. 3 1 6 2 8 2 x 2 x 2 2 x 3 5 9 3 x 3 12 2 x 3 x 2 5 7 18 2 x 3 x 3 1 24 36 2 x 2 x 3 x 3 3 x 2 x 2 x 2 The most number of times any single factors appears in a set is multiplied by the most number of time any other factor appears. (2 x 2) x (3 x 3) = 72 Remember: If a denominator is a “prime number”, it can’t be factored except by itself and 1. LCD Exercises (Find the LCD’s) 1 6 1 8 2 x 3 2 x 2 x 2 1 12 2 x 3 x 2 2 x 2 x 3 = 24 1 12 2 x 3 1 16 2 x 2 x 2 x 2 2 x 2 x 3 = 48 1 24 3 x 2 x 2 x 2 3 10 4 15 2 x 5 3 x 5 7 20 2 x 2 x 5 2 x 3 x 5 = 60 11
11. Reducing to LCD can only be done after the LCD itself is known. 3 1 6 2 x 3 8 2 x 2 x 2 2 9 3 x 3 5 12 2 x 3 x 2 5 18 2 x 3 x 3 7 1 24 36 2 x 2 x 3 x 3 3 x 2 x 2 x 2 LCD = 72 Divide the LCD by each of the other denominators, then multiply both the numerator and denominator of the fraction by that result. 1 6 3 8 2 9 5 12 72. . 6 = 12 72. . 8 = 9 72. . 9 = 8 72. . 12 = 6 1 x 12 = 12 6 x 12 = 72 3 x 9 = 27 8 x 9 = 72 2 x 8 = 16 9 x 8 = 72 5 x 6 = 30 12 x 6 = 72 Remaining fractions are handled in same way. 12
Reducing to LCD Exercises Reduce each set of fractions to their LCD. 1 6 1 8 2 x 3 2 x 2 x 2 1 2 x 3 x 2 2 x 2 x 3 = 24. 6 = 4 1 6 1 x 4= 4 6 x 4 = 24. 8 = 3 1 8 1 x 3= 3 8 x 3 = 24. 12 = 2 1 12 12 1 x 2= 2 12 x 2 = 24 1 16 1 12 2 x 3 2 x 2 x 2 x 2 1 24 3 x 2 x 2 x 2 3 10 4 15 2 x 5 3 x 5 . 48. 12 = 4 12 1 x 4= 4 12 x 4 = 48. 16 = 3 1 16 1 x 3= 3 16 x 3 = 48. 24 = 2 1 24 1 x 2= 2 24 x 2 = 48 20 2 x 2 x 5 2 x 3 x 5 = 60 2 x 2 x 3 = 48 1 7 . 60. 10 = 6 3 10 3 x 6 = 18 10 x 6 = 60. 15 = 4 4 15 4 x 4 = 16 15 x 4 = 60. 20 = 3 7 20 7 x 3 = 21 20 x 3 = 60 13
12. Addition of Fractions All fractions must have same denominator. Determine common denominator according to previous process. Then add fractions. 1 4 + 2 4 + 3 4 = 6 4 = 1 1 2 Always reduce to lowest terms. 13. Addition of Mixed Numbers Mixed number consists of a whole number and a fraction. (3 1/3) • • • Whole numbers are added together first. Then determine LCD for fractions. Reduce fractions to their LCD. Add numerators together and reduce answer to lowest terms. Add sum of fractions to the sum of whole numbers. 14
Adding Fractions and Mixed Numbers Exercises Add the following fractions and mixed numbers, reducing answers to lowest terms. 1. 3 4 6 4 3. 9 3 + = 4 15 16 = + 30 32 =1 5 11 2 32 + 9 2 2. = 7 32 4. 39 32 5 7 + 10 4 10 + =1 1 2 5 7 10 = = 11 10 10 + 13 4 = 5+1=6 = 23 20 20 + 6 =7 8 20 + =1 3 15 20 3 20 15
14. Subtraction of Fractions Similar to adding, in that a common denominator must be found first. Then subtract one numerator from the other. 20 24 - 14 24 = 6 24 To subtract fractions with different denominators: ( 5 16 - 1 4 ) • Find the LCD. . . 1 4 - 5 16 2 x 2 x 2 x 2 2 x 2 x 2 = 16 • Change the fractions to the LCD. . . 4 5 16 16 - • Subtract the numerators. . . 5 16 - 4 16 = 1 16 16
15. Subtraction of Mixed Numbers • Subtract the fractions first. (Determine LCD) 10 2 3 - 412 3 x 2 = 6 (LCD) • Divide the LCD by denominator of each fraction. . 6. 3=2 6. . 2 = 3 • Multiply numerator and denominator by their respective numbers. 2 2 = 4 x 3 2 6 1 3 = 3 2 x 3 6 • Subtract the fractions. 4 3 = 1 6 - 6 6 • Subtract the whole numbers. 10 - 4 = 6 • Add whole number and fraction together to form complete answer. 6 1 + 6 = 6 17
15. Subtraction of Mixed Numbers (con’t) Borrowing • Subtract the fractions first. (Determine LCD) 5 1 3 3 8 16 becomes 6 5 16 3 16 1 (LCD) = 16 • Six-sixteenths cannot be subtracted from one-sixteenth, so 1 unit ( 16 ) is borrowed from the 5 units, leaving 4. 16 • Add 16 16 to 1 16 and problem becomes: 4 17 16 - 3 6 16 • Subtract the fractions. 17 - 6 = 11 16 16 16 • Subtract the whole numbers. 4 -3=1 • Add whole number and fraction together to form complete answer. 1 11 + 16 = 1 11 16 18
Subtracting Fractions and Mixed Numbers Exercises Subtract the following fractions and mixed numbers, reducing answers to lowest terms. 1. 2 5 - 6 15 2. 1 5 8 - - 3 5 15 3 12 15 - 6 24 24 4. = =1 15 1 24 =3 8 33 32 20 15 47 47 2 5 - 28 13 = 6 5 = 19 115 28 15 15 - 15 6 15 = 1714 15 - 57 1516 = 4 15 101 16 - 57 16 = 100 3. - 15 2 5 = - 15 615 = 3 5 15 5. 101 = =9 33 1 4 20 16 - 57 1516 = 43 516 - 10 5 12 = 9 5 4 14 12 - 10 12 = 4 12 6. 14 3 4 = 4 13 19
16. MULTIPLYING FRACTIONS • Common denominator not required for multiplication. 3 4 X 4 16 1. First, multiply the numerators. 3 4 X 4 16 = 12 = 2. Then, multiply the denominators. 3 4 X 4 16 = 12 64 = 3. Reduce answer to its lowest terms. 12 64 . . 4 4 = 3 16 20
17. Multiplying Fractions & Whole/Mixed Numbers • Change to an improper fraction before multiplication. 3 4 X 4 1. First, the whole number (4) is changed to improper fraction. 4 1 2. Then, multiply the numerators and denominators. 3 4 X 4 1 = 12 4 3. Reduce answer to its lowest terms. 12 4 . . 4 4 = 3 1 = 3 21
18. Cancellation • Makes multiplying fractions easier. • If numerator of one of fractions and denominator of other fraction can be evenly divided by the same number, they can be reduced, or cancelled. Example: 8 X 5 = 3 16 18 5 3 X 16 = 2 1 X 5 = 5 3 2 6 Cancellation can be done on both parts of a fraction. 1 1 12 X 3 = 21 24 7 2 1 X 1 = 1 14 7 2 22
Multiplying Fractions and Mixed Numbers Exercises Multiply the following fraction, whole & mixed numbers. Reduce to lowest terms. 1. 3 X 4 = 3 4 16 16 2. 26 X 126 = 1 3. 4 2 2 = X 3 5 5 4. 9 X 2 = 1 1 5 5 3 6. 9 X 3 = 27 50 5 10 5. 35 4 4 X 35 = 1 7. 16 X 7 = 7 12 72 9. 5 = 10 X 33 3 11 8. 2 77 5 X 15 = 25 23 23
19. Division of Fractions • Actually done by multiplication, by inverting divisors. • The sign “ “ means “divided by” and the fraction to the right of the sign is always the divisor. Example: 3 4 1 becomes 5 3 5 = 15 = 3 3 X 4 4 4 1 20. Division of Fractions and Whole/Mixed Numbers • Whole and mixed numbers must be changed to improper fractions. Example: 3 316 2 1 8 51 16 17 8 3 becomes 16 X 3 + 3 16 Inverts to 51 X 16 1 1 3 2 X 1 = 2 = 1 2 8 17 = 51 16 = and 2 3 51 X 16 2 X 8 + 1 8 17 1 = 17 8 = 32 X 1 1 Double Cancellation 24
Dividing Fractions, Whole/Mixed Numbers Exercises Divide the following fraction, whole & mixed numbers. Reduce to lowest terms. 8 3 = 1 1 4 6 2. 51 16 3. 18 1 = 144 8 4. 15 5. 14 3 1. 5 7 = 4 3 = 8 7 12 = 8 12 25 57 2 23 25 |
# If-then statement
When we previously discussed inductive reasoning we based our reasoning on examples and on data from earlier events. If we instead use facts, rules and definitions then it's called deductive reasoning.
We will explain this by using an example.
If you get good grades then you will get into a good college.
The part after the "if": you get good grades - is called a hypotheses and the part after the "then" - you will get into a good college - is called a conclusion.
Hypotheses followed by a conclusion is called an If-then statement or a conditional statement.
This is noted as
$p \to q$
This is read - if p then q.
A conditional statement is false if hypothesis is true and the conclusion is false. The example above would be false if it said "if you get good grades then you will not get into a good college".
If we re-arrange a conditional statement or change parts of it then we have what is called a related conditional.
Example
Our conditional statement is: if a population consists of 50% men then 50% of the population must be women.
$p \to q$
If we exchange the position of the hypothesis and the conclusion we get a converse statement: if a population consists of 50% women then 50% of the population must be men.
$q\rightarrow p$
If both statements are true or if both statements are false then the converse is true. A conditional and its converse do not mean the same thing
If we negate both the hypothesis and the conclusion we get a inverse statement: if a population do not consist of 50% men then the population do not consist of 50% women.
$\sim p\rightarrow \: \sim q$
The inverse is not true juest because the conditional is true. The inverse always has the same truth value as the converse.
We could also negate a converse statement, this is called a contrapositive statement: if a population do not consist of 50% women then the population do not consist of 50% men.
$\sim q\rightarrow \: \sim p$
The contrapositive does always have the same truth value as the conditional. If the conditional is true then the contrapositive is true.
A pattern of reaoning is a true assumption if it always lead to a true conclusion. The most common patterns of reasoning are detachment and syllogism.
Example
If we turn of the water in the shower, then the water will stop pouring.
If we call the first part p and the second part q then we know that p results in q. This means that if p is true then q will also be true. This is called the law of detachment and is noted:
$\left [ (p \to q)\wedge p \right ] \to q$
The law of syllogism tells us that if p → q and q → r then p → r is also true.
This is noted:
$\left [ (p \to q)\wedge (q \to r ) \right ] \to (p \to r)$
Example
If the following statements are true:
If we turn of the water (p), then the water will stop pouring (q). If the water stops pouring (q) then we don't get wet any more (r).
Then the law of syllogism tells us that if we turn of the water (p) then we don't get wet (r) must be true.
## Video lesson
Write a converse, inverse and contrapositive to the conditional
"If you eat a whole pint of ice cream, then you won't be hungry" |
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# NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers: Download Free PDF
NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers: This chapter will teach properties identified with real numbers, integers, whole numbers, rational numbers and natural numbers – independent, associative and closure. The chapter manages the part of zero and one, multiplication over addition just as the portrayal of Rational Numbers on the number line, alongside looking for Rational Numbers between two Rational Numbers. Likewise, Additive identity (0) and Multiplicative identity (1) are canvassed in this chapter. This chapter contains two activities, which are questions from all the topics present in the chapter. Introduction to Rational Numbers The numbers engaged with numerous mathematical applications, such as addition, subtraction, and multiplication, which are intrinsically closed with numerous mathematical operations, are called Rational numbers.
## Class 8 Maths Chapter 1 Rational Numbers- Important Topics
Whole Numbers and Natural Numbers
Natural numbers are sets of numbers starting from 1 counting up to limitlessness. The set of natural numbers mentioned here is denoted as ′N′. Whole numbers are a set of numbers starting from 0 and going up to endlessness. So fundamentally they are natural numbers with the number zero added to the said set. The set of whole numbers is represented as ′W′.
Integers
In straightforward terms, Integers are natural numbers and their negatives. The arrangement of Integers is denoted as ′Z′ or ′I′.
Rational Numbers
A rational number is a certain number that can be represented as a small amount of two integers as p/q, where q should be non-zero. The arrangement of rational numbers is denoted as Q.
Properties of Rational Numbers
• Closure Property of Rational Numbers
• Commutative Property of Rational Numbers
• Associative Property of Rational Numbers
• Distributive Property of Rational Numbers
Negatives and Reciprocals
• Negation of a Number
• Reciprocal of a Number
Representing on a Number Line
You will learn how to represent a given rational number an, where a and n are integers, on the number line.
Rational Numbers between Two Rational Numbers
The quantity of rational numbers between any two given rational numbers isn't positive, not normal for whole numbers and natural numbers.
## FAQs for NCERT Solutions For Class 8 Maths
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A. NCERT solutions provide topic-wise solutions of the chapters, which helps the student to prepare for their exams and understand each topic nicely. The basic concepts are cleared by providing step by step methods. Mathematics of class 8 is fundamental and challenging, so it is essential to have a solution handy. Aakash even provides some new methods and techniques to solve the problems which makes the studies easier. There are a lot of Question papers, teaching materials, and online books written by Aakash’s various teachers across the country. The whole purpose of teaching is to prepare the students for their exams as well as for their future.
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# How do you simplify 2/(x-1) + (x-11)/(x^2+3x-4)?
Mar 18, 2016
$\frac{3}{x + 4}$
#### Explanation:
$1$. Factor the denominator of the second fraction.
$\frac{2}{x - 1} + \frac{x - 11}{{x}^{2} + 3 x - 4}$
$= \frac{2}{x - 1} + \frac{x - 11}{\left(x + 4\right) \left(x - 1\right)}$
$2$. To add the two fractions together, find its L.C.D. (lowest common denominator).
$= \frac{2 \textcolor{b l u e}{\left(x + 4\right)}}{\left(x - 1\right) \textcolor{b l u e}{\left(x + 4\right)}} + \frac{x - 11}{\left(x + 4\right) \left(x - 1\right)}$
$= \frac{2 \textcolor{b l u e}{\left(x + 4\right)} + \left(x - 11\right)}{\left(x + 4\right) \left(x - 1\right)}$
$3$. Simplify the numerator.
$= \frac{2 x + 8 + x - 11}{\left(x + 4\right) \left(x - 1\right)}$
$= \frac{3 x - 3}{\left(x + 4\right) \left(x - 1\right)}$
$4$. Factor out $3$ from the numerator.
$= \frac{3 \left(x - 1\right)}{\left(x + 4\right) \left(x - 1\right)}$
$5$. Since the factor, $\left(x - 1\right)$, appears in both the numerator and denominator, they cancel out.
$= \frac{3 \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}}}{\left(x + 4\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}}}$
$= \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \frac{3}{x + 4} \textcolor{w h i t e}{\frac{a}{a}} |}}}$ |
# 17 Angles
You will need a calculator near the end of this module.
# Angle Measure
Angle measurement is important in construction, surveying, physical therapy, and many other fields. We can visualize an angle as the figure formed when two line segments share a common endpoint.
We can also think about an angle as a measure of rotation. One full rotation or a full circle is , so a half rotation or U-turn is , and a quarter turn is . We often classify angles by their size relative to these and benchmarks.
Acute Angle: between and
Right Angle: exactly
Obtuse Angle: between and
Straight Angle: exactly
Reflexive Angle: between and
Exercises
Identify each angle shown below as acute, right, obtuse, straight, or reflexive.
Lines that form a angle are called perpendicular. As shown below, the needle should be perpendicular to the body surface for an intramuscular injection.
When lines cross, they form angles. No surprises there. If we know the measure of one angle, we may be able to determine the measures of the remaining angles using a little logic.
Exercises
Find the measure of each unknown angle.
# Angles in Triangles
If you need to find the measures of the angles in a triangle, there are a few rules that can help.
The sum of the angles of every triangle is .
If any sides of a triangle have equal lengths, then the angles opposite those sides will have equal measures.
Exercises
Find the measures of the unknown angles in each triangle.
# Angles and Parallel Lines
Two lines that point in the exact same direction and will never cross are called parallel lines. If two parallel lines are crossed by a third line, sets of equally-sized angles will be formed, as shown in the following diagram. All four acute angles will be equal in measure, all four obtuse angles will be equal in measure, and any acute angle and obtuse angle will have a combined measure of .
Exercises
1. Find the measures of angles , , and .
# Degrees, Minutes, Seconds
It is possible to have angle measures that are not a whole number of degrees. It is common to use decimals in these situations, but the older method—which is called the degrees-minutes-seconds or DMS system—divides a degree using fractions out of : a minute is of a degree, and a second is of a minute, which means a second is of a degree. (Fortunately, these units work exactly like time; think of degree as hour.) For example, .
We will look at the procedure for converting between systems, but there are online calculators such as the one at https://www.fcc.gov/media/radio/dms-decimal which will do the conversions for you.
If you have latitude and longitude in DMS, like N W , and need to convert it to decimal degrees, the process is fairly simple with a calculator.
Converting from DMS to Decimal Degrees
Enter in your calculator. Round the result to the fourth decimal place, if necessary.[1]
Exercises
Convert each angle measurement from degrees-minutes-seconds into decimal form. Round to the nearest ten-thousandth, if necessary.
Going from decimal degrees to DMS is a more complicated process.
Converting from Decimal Degrees to DMS
1. The whole-number part of the angle measurement gives the number of degrees.
2. Multiply the decimal part by . The whole number part of this result is the number of minutes.
3. Multiply the decimal part of the minutes by . This gives the number of seconds (including any decimal part of seconds).
For example, let’s convert .
1. The degrees part of our answer will be .
2. The decimal part times is minutes. The minutes part of our answer will be .
3. The decimal part times is seconds. The seconds part of our answer will be .
So .
Exercises
Convert each angle measurement from decimal into degrees-minutes-seconds form. |
# Multiplying Complex Numbers, Day 1 of 4
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## Objective
SWBAT use the unit imaginary number and the field axioms to multiply complex numbers. SWBAT represent and interpret multiplication of complex numbers in the complex number plane.
#### Big Idea
Students explore and explain correspondences between numerical and graphical representations of arithmetic with complex numbers.
## Powers of i
10 minutes
I will begin this lesson with a brief look at the powers of the imaginary unit, i. These have already come up once or twice in class, but before we begin multiplying complex numbers, it's important to activate this prior understanding. Beginning with i and i^2, we'll work quickly through the next few powers of i. Along the way, we will constantly reiterate the fact that i^2 = -1, and we'll use this fact to simplify one power of i after another. Before long, the class should have a pretty long list of powers and should clearly see the repeating pattern of i, -1, -i, 1. (MP 8)
Once this pattern is clear, it's time to dive into multiplication of complex numbers.
## Multiplication of Complex Numbers
30 minutes
Now, let the class know that today they're going to be considering multiplication with complex numbers. Just as with addition, there is an arithmetic aspect to this operation and there is a geometric aspect. (Do they recall both the arithmetic and geometric aspects of addition?)
In terms of arithmetic, multiplying complex numbers is pretty straightforward. Use the distributive law, remember that i^2 = -1, and then combine like terms. But what about the geometric interpretation? Will this create a parallelogram like addition does? The problem set Multiplying Complex Numbers will guide students to the answer.
Before you hand out the problems, however, I'd explain the definitions given on the first page. The relationship between "distance" and "absolute value" should be familiar from the real number line, but the "argument" will be something new. For now, students will just have to trust you that these two quantities will help them to think about the geometric interpretation of multiplication.
Now, pass out the problem set and break the class into small groups. Continually move back and forth between group work and whole-class discussion. As groups begin to finish problem 1, initiate a summary discussion of their solutions. The point is that multiplying by i is equivalent to a rotation of 90 degrees. For the summary discussion of #2, the point is that multiplying by a real number is equivalent to scaling the distance. Finally, for the summary of #3, the point is that multiplying by a purely imaginary number is equivalent to scaling and rotating by 90 degrees. (During this first lesson, I would not expect to get much beyond #1.)
Alternatively, you might consider running taking a Socratic approach to this lesson.
## Wrapping Up
5 minutes
The end of class will depend on how far we have progressed today. A 3-2-1 Exit Ticket is an appropriate formative assessment on a day like this one, since it gives students a chance to reflect on the concepts we've covered, and not just the skills . I expect that many students will still be a little mystified by the geometric interpretation (MP2), and perhaps by the powers of i, but I'll have to wait for the exit tickets to know for sure. |
# What is a Binomial Coefficient?
First, let’s start with a binomial. A binomial is a polynomial with two terms typically in the format $$(x + y)$$.
A binomial coefficient is raising a binomial to the power of n, like so $$(x + y)^n$$.
We all remember from school that $$(a + b)^2 = a^2 + 2ab + b^2$$ but what if I was to ask for $$(a+b)^18$$? This is where the binomial formula comes in handy.
# Binomial Therom
The Binomial Theorem is the expected method to use for finding binomial coefficients because it is how a computer would compute it. The theorem is as follows:
Luckily for us, this formula is the same as another formula we’ve seen, according to here
Yes, the combinations formula!
Let’s try an example.
## Example question
What is the coefficient of $$x^6$$ in $$(1+x)^8$$?
Simply plug this into the formula like so
Something that may confuse people is, how do we work out what n and k are? Well, we have n objects overall and we want to choose k of them.
# Pascal’s Triangle
Pascal’s triangle is a triangle created by starting off with a 1, starting every line and ending every line with a 1 and adding the numbers above to make a new number; as seen in this gif.
No one could ever explain a maths topic as well as Numberphile, so here’s a Numberphile video on it
Here’s the Pascal’s Triangle we want.
Pascal’s triangle always starts counting from 0, so to solve $$8\choose 6$$ we simply count 8 rows down, then 6 across. So the row here is the line of the number 1’s on the left hand side, and we start counting from 0. So the eigth row is the one that starts with 1, 8. Notice how the second inner column defines what row we’re on.
Now we count 6 across which is… 28. We just found the binomial coefficient using a super neat and easy to draw up triangle. Of course, the hardest part is adding together all the numbers and if the coefficient is large it may be easier to just use the Binomial theorem, but this method still exists and is useful if you’ve forgotten the binomial theorem.
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# A solution of $9\%$ acid is to be diluted by adding $3\%$ acid solution to it. The resulting mixture is to be more than $5\%$ but less than $7\%$ acid. If there is $460$ litres of the $9\%$ solution, how many litres of $3\%$solution will have to be added?
Last updated date: 01st Mar 2024
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Hint: We know that by considering the quantity of the $3$ percent solution added to be x. Given a quantity of $9$ percent solution. Then the resulting mixture will be we litres. Proceeding by forming the appropriate inequalities according to the problem statement and then solving these inequalities, we will get the value of x i.e. quantity of solution to be added.
Let the required quantity of $3$ percent boric acid solution be litres. Given, quantity of$9$ percent acid solution is $460litres.$
Let us assume that ‘x’ litres of $3\%$ solution is added to $460\text{ }L$ of $9\%$ solution.
Thus, total solution $=\left( 460+x \right)L$ & total acid content in resulting solution $=460\left( \dfrac{9}{100} \right)+\text{ }x\left( \dfrac{3}{100} \right)$
$=\left( 41.4+0.03x \right)\%$
Now, according to the question, the resulting mixture we get should be less than $7\%$ acidic & more than $5\%$acidic Thus, we get, $5\%\text{ }of\text{ }\left( 460\text{ }+\text{ }x \right)\text{ }<\text{ }41.4+0.03x<7\%\text{ }of\text{ }\left( 460+x \right)$
$=\text{ }\left( 23+0.05x \right)<\left( 41.4+0.03x \right)<\left( 32.2+0.07x \right)$
Now we have; $\left( 23+0.05x \right)<\left( 41.4+0.03x \right)\text{ }\And \text{ }~\left( 41.4+0.03x \right)<\left( 32.2+0.07x \right)=\text{ }0.02x<18.4~~\And ~~0.04x>9.2$
Thus, $2x < 1840\text{ }\And \text{ }4x > 920\text{ }=\text{ }230 < x < 920.$
Therefore, the number of litres of the $3\%$ solution of acid must be more than $230$ and less than $920.$
Note:
Remember that in these types of problems, the question statement is very crucial. According to the problem statement, all the inequalities are formed and hence these inequalities are further reduced to the simplest form and evaluation of the variable is aimed. Here, after solving we are getting a range of the values instead of a particular value. |
## Greatest positive rate of change
Directions of Greatest Increase and Decrease. The directional derivative can also be written: where theta is the angle between the gradient vector and u. The directional derivative takes on its greatest positive value if theta=0. Hence, the direction of greatest increase of f is the same direction as the gradient vector.
In this tutorial, learn about rate of change and see the difference between positive and negative rates of change! Further Exploration. Defining Slope. How Do You Find the Rate of Change Between Two Points in a Table? The rate of change is a rate that describes how one quantity changes in relation to another quantity. This tutorial shows you The calculator will find the average rate of change of the given function on the given interval, with steps shown. Show Instructions In general, you can skip the multiplication sign, so `5x` is equivalent to `5*x`. Directions of Greatest Increase and Decrease. The directional derivative can also be written: where theta is the angle between the gradient vector and u. The directional derivative takes on its greatest positive value if theta=0. Hence, the direction of greatest increase of f is the same direction as the gradient vector. This is called the rate of change per month. By finding the slope of the line, we would be calculating the rate of change. We can't count the rise over the run like we did in the calculating slope lesson because our units on the x and y axis are not the same. In most real life problems, your units will not be the same on the x and y axis. Regents Exam Questions F.IF.B.6: Rate of Change 1 Name: _____ www.jmap.org 2 4 The graph below shows the distance in miles, m, hiked from a camp in h hours. Which hourly interval had the greatest rate of change? 1) hour 0 to hour 1 3) hour 2 to hour 3 2) hour 1 to hour 2 4) hour 3 to hour 4
## Most organizational change efforts take longer and cost more money than leaders and managers anticipate. In fact, research from McKinsey and Company shows that 70% of all transformations fail. Why?
Determine whether the velocity is positive, negative, or zero at times t1, t2, t3, and t4 (b) How long would it take the BMW to go from 60 mi/h to 130 mi/h at this rate ? Slowing down means its velocity is decreasing or the change in velocity is ( b) Estimate the time at which the acceleration has its greatest positive value 13 Aug 2015 These apps are great for both keeping a positive mentality (indispensable for any entrepreneur) and making sure you are focused and on-task. Transformational change is still hard, according to a new survey. report a 79 percent success rate—three times the average for all transformations. half of respondents (and the largest share) wish their organizations had spent Senior leaders are more positive than others about the rigor of their transformation efforts. First, I'll look at the polynomial as it stands, not changing the sign on x. This is the There are four sign changes in the positive-root case. This number "four" is The continent's child demographics are changing rapidly: by 2050, Africa will education and protection systems – is the highest of priorities for the continent to science to promote the adoption and maintenance of positive behaviours. Lapse rate, rate of change in temperature observed while moving upward through the Earth's atmosphere. The lapse rate is considered positive when the
### Most organizational change efforts take longer and cost more money than leaders and managers anticipate. In fact, research from McKinsey and Company shows that 70% of all transformations fail. Why?
What happens to sensitivity and specificity if I change the cutting point? Great! That wasn't so hard. Is there anything else I need to know? Sadly, Yes. sensitivity and specificity: it's sensitivity divided by (1-specificity), or the true positive rate
### Rate Of Change - ROC: The rate of change - ROC - is the speed at which a variable changes over a specific period of time. ROC is often used when speaking about momentum, and it can generally be
13 May 2019 The rate of change - ROC - is the speed at which a variable changes over or one that has a positive ROC, normally outperforms the market in 15 Jul 2019 Percentage change is a simple mathematical concept measuring the degree of Positive values indicate a percentage increase whereas negative values it's best to use the formula for calculating percentage increase. Review average rate of change and how to apply it to solve problems. Find the function that represents the greatest average rate of change from 0 to 5. Reply. 13 Nov 2019 of derivatives from the previous chapter (i.e. rates of change) that we will Due to the nature of the mathematics on this site it is best views in In math, slope is the ratio of the vertical and horizontal changes between two Their slopes may be large or small, but they are always positive or negative 17 Oct 2017 In this lesson, learn about how rates of change are calculated and what it means for. velocity is a great example of a rate of change because your position about what it means for a rate of change to be positive or negative. #Algebra #grade8 #grade9 Calculate the #rate of change of a #linear #function represented algebraically, in a #table or in a #graph in mathematical and real
## The query returns the greatest field value in the water_level field key and in the NON_NEGATIVE_DERIVATIVE() returns only positive rates of change or rates
What happens to sensitivity and specificity if I change the cutting point? Great! That wasn't so hard. Is there anything else I need to know? Sadly, Yes. sensitivity and specificity: it's sensitivity divided by (1-specificity), or the true positive rate 31 Jan 2020 Top Trading Partners - January 2020. Data are goods only, on a Census Basis, in billions of dollars, unrevised. 23 Mar 2016 Absolute, positive and negative impact of each driver of change. Climatic change accounted for the second largest percentage of impact, Ideally, you want to focus on fixing the problems that have the biggest impact. out a Pareto Analysis, and explain how to use your findings to prioritize tasks that will deliver the greatest positive impact. and Net Present Value (NPV) and Internal Rate of Return (IRR) to determine which changes you should implement. People, communities, cities, businesses, schools and other organizations are taking action to help fight climate change. What changes will you make? Positive thinking — Harness the power of optimism to help with stress management. You think the best is going to happen, not the worst. Increased life span; Lower rates of depression; Lower levels of distress; Greater resistance to and you assume that the change in plans is because no one wanted to be around you. 21 Feb 2020 Mortgage rates are holding low and everyone is predicting low rates through 2020. The decreased activity and mobility in one of China's largest cities could spark a wider Will the Fed change rates in March? No one U.S. mortgage bonds will continue to deliver positive returns, so investors will pile in.
Directions of Greatest Increase and Decrease. The directional derivative can also be written: where theta is the angle between the gradient vector and u. The directional derivative takes on its greatest positive value if theta=0. Hence, the direction of greatest increase of f is the same direction as the gradient vector. This is called the rate of change per month. By finding the slope of the line, we would be calculating the rate of change. We can't count the rise over the run like we did in the calculating slope lesson because our units on the x and y axis are not the same. In most real life problems, your units will not be the same on the x and y axis. Regents Exam Questions F.IF.B.6: Rate of Change 1 Name: _____ www.jmap.org 2 4 The graph below shows the distance in miles, m, hiked from a camp in h hours. Which hourly interval had the greatest rate of change? 1) hour 0 to hour 1 3) hour 2 to hour 3 2) hour 1 to hour 2 4) hour 3 to hour 4 A rate of change tells you how quickly a quantity is changing. In this lesson, learn about how rates of change are calculated and what it means for a rate of change to be negative. The rate of change calculator is a free online tool that gives the change in slope for the given input coordinate points. BYJU’S online rate of change calculator tool makes the calculations faster and easier where it displays the result in a fraction of seconds. |
# What is 1/3 minus 2/5 as a fraction?
## What is 1/3 minus 2/5 as a fraction?
1/3 – 2/5 = -115 ≅ -0.06666667.
## What is the LCM of 2 5 and 2 3?
LCM of 2, 3, and 5
1. LCM of 2, 3, and 5
2. List of Methods
3. Solved Examples
4. FAQs
What will you get when you multiply 3/5 and 1 2?
1/2 * 3/5 = 310 = 0.3.
What do 2 5 and 3 have in common?
The numbers 2 , 3 and 5 are distinct prime numbers, so they have no common factor larger than 1 .
### What is the common denominator of 1/3 and 2 5?
Originally Answered: What is the LCM of 1/3 and 2/5? Find the LCM of the numerators with common denominators, ie 5 and 6, that is 30. Answer: 2 is the LCM of 1/3 and 2/5.
READ ALSO: Is Cetaphil good for 18 year old?
### What is the least common multiple of 3 and 5?
The Least Common Multiple of 3 and 5 is 15 (15 is a multiple of both 3 and 5, and is the smallest number like that.) So… what is a “Multiple”? We get a multiple of a number when we multiply it by another number.
How to find LCM of 2 numbers?
The product of HCF of two numbers and the LCM of two numbers is equal to the product of two numbers. Therefore, the formula to find LCM of two numbers is, LCM of two numbers = product of two numbers ÷ HCF of two numbers.
What are the common multiples of 3 and 5?
The multiples of 3 are 3, 6, 9, 12, 15, 18,… etc The multiples of 5 are 5, 10, 15, 20, 25,… etc Find the first Common (same) value: The Least Common Multiple of 3 and 5 is 15
#### What is the formula for LCM?
Problem 1. Illustrate the formula lcm(a, b) = a × b / gcd(a, b) by calculating the following examples by using the formula and by calculating the lcm in some other manner and seeing that you get the same value by both methods. It makes perfect sense to talk about the least common multiple of three or more integers. |
# What is the general solution of the differential equation (x+y)dx-xdy = 0?
Dec 30, 2016
$y = x \ln | x | + A x$
#### Explanation:
We can write the equation $\left(x + y\right) \mathrm{dx} - x \mathrm{dy} = 0$ as:
$\text{ } x \mathrm{dy} = \left(x + y\right) \mathrm{dx}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x + y}{x}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \frac{y}{x}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{1}{x} y = 1$
This is a First Order DE of the form:
$y ' \left(x\right) + P \left(x\right) y = Q \left(x\right)$
Which we know how to solve using an Integrating Factor given by:
$I F = {e}^{\int P \left(x\right) \setminus \mathrm{dx}}$
And so our Integrating Factor is:
$I F = {e}^{\int - \frac{1}{x} \setminus \mathrm{dx}}$
$\setminus \setminus \setminus \setminus = {e}^{- \ln | x |}$
$\setminus \setminus \setminus \setminus = {e}^{\ln | \frac{1}{x} |}$
$\setminus \setminus \setminus \setminus = \frac{1}{x}$
If we multiply by this Integrating Factor we will (by its very design) have the perfect differential of a product:
$\text{ } \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{1}{x} y = 1$
$\therefore \frac{1}{x} \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{1}{x} ^ 2 y = \frac{1}{x}$
$\therefore \setminus \setminus \setminus \setminus \setminus \frac{d}{\mathrm{dx}} \left(\frac{1}{x} y\right) = \frac{1}{x}$
Which is now a separable DE, and we can separate the variables to get:
$\text{ } \frac{y}{x} = \int \setminus \frac{1}{x} \setminus \mathrm{dx}$
$\therefore \frac{y}{x} = \ln | x | + A$ (where $A$ is an arbitrary constant)
$\therefore \setminus \setminus y = x \ln | x | + A x$ |
Learning Objective(s)
· Solve algebraic equations with radical terms.
Introduction
When a radical expression appears in an equation, we call that a radical equation. Solving radical equations requires applying the rules of exponents and following some basic algebraic principles. In some cases, it also requires looking out for error generated by raising unknown quantities to an even power.
Squaring Both Sides
A basic strategy for solving radical equations is to isolate the radical term first, and then use the inverse operation (raising the radical term to a power) to pull out the variable. This is the same type of strategy we use to solve other, non-radical equations: rearrange the expression to isolate the variable we want to know, and then solve the resulting equation.
Example Problem Add 3 to both sides to isolate variable term Collect like terms Square both sides to remove variable from the radical Answer x = 64 Simplify
To check our solution, we can substitute 64 in for x in the original equation. Does ? Yes—the square root of 64 is 8, and 8 3 = 5.
Notice how we combined like terms and then squared both sides of the equation in this problem. This is a standard method for removing a radical from an equation. It is important to isolate a radical on one side of the equation and simplify as much as possible before squaring. The more terms there are before squaring, the more additional terms will be generated by the process of squaring, and that gets messy fast.
Another warning: be careful not to just square individual terms in the equation. This method only works if we square both sides of the equation.
Even if we follow the rules, it’s not all sunshine and roses. Let’s try another problem that demonstrates a potential pitfall of squaring both sides to remove the radical:
Example Problem Square both sides to remove the term a – 5 from the radical. a − 5 = 4 Simplified equation a − 5 + 5 = 4 + 5 Add 5 to both sides to isolate variable Answer a = 9 Collect like terms
We get a value of 9 for a. But watch what happens if we substitute 9 in for a in the original equation:
This is incorrect—the square root of 4 is 2, not -2. Our answer that a = 9 does not check out. What happened?
Look back at the original problem: . Notice that the radical is set equal to -2, and recall that a square root of a number can only be positive. This means that no value for a will result in a radical expression whose square root is -2! We could have noticed that right away and concluded that there were no solutions for a. But why did the squaring method work in the first example and not in the second example?
The answer lies in the process of squaring itself. When we raise a number to an even power—whether second, fourth, or 50th power—we can introduce a false solution because the result of an even power is always a positive number. Think about it: 32 and (-3)2 are both 9, and 24 and (-2)4 are both 16. So when we squared -2 and got 4 in this problem, we artificially turned the quantity positive. This is why we were still able to find a value for a—we solved the problem as if we were solving ! Oops.
Maritza is solving the following radical equation: . Which of the following steps should Maritza do first? A) Multiply both sides by h. B) Square both sides. C) Divide both sides by D) Subtract from both sides Show/Hide Answer
Roots Beyond Squares
The technique of eliminating a radical by raising both sides of an equation to the same power can be used with roots beyond squares as well. Consider the equation .
At first glance, this equation may seem unsolvable since the radical is set equal to -3. We don’t want to make that mistake again! But wait—this time the radical is not an even root—it is an odd root, 3. This means that the radical expression can be set equal to a negative number since radicals with odd roots may have negative values as roots.
Example Problem Cube both sides to remove the radical. -27 = b - 2 Simplified equation. Note that cubing preserves the negative sign in front of 27. -27 + 2 = b – 2 + 2 Add 2 to both sides to isolate variable Answer -25 = b Collect like terms
As with all answers we get when we solve radical equations, we should substitute -25 in for b in our original equation to make sure it is a legitimate solution.
-3 = -3
The substitution results in a true statement, so our answer is right.
Summary
A common method for solving radical equations is to raise both sides of an equation to whatever power will eliminate the radical sign from the equation. But be careful—when both sides of an equation are raised to an even power, the possibility exists that negative terms will be artificially changed into positive terms. When solving these problems, it is important to check the answer by substituting the value back into the original equation. |
# How to Calculate 1/21 Times 3/15
Are you looking to work out and calculate how to multiply 1/21 by 3/15? In this really simple guide, we'll teach you exactly what 1/21 times 3/15 is and walk you through the step-by-process of how to multiply two fractions together.
Just a quick reminder here that the number above the fraction line is called the numerator and the number below the fraction line is called the denominator.
To multiply two fractions, all we need to do is multiply the numerators and the denominators together and then simplify the fraction if we can.
Let's set up 1/21 and 3/15 side by side so they are easier to see:
1 / 21 x 3 / 15
The next step is to multiply the numerators on the top line and the denominators on the bottom line:
1 x 3 / 21 x 15
From there, we can perform the multiplication to get the resulting fraction:
1 x 3 / 21 x 15 = 3 / 315
You're done! You now know exactly how to calculate 1/21 x 3/15. 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):
1/105
## Convert 1/21 times 3/15 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:
3 / 315 = 0.0095
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Here the denominators are 5 and 10 so we can Method 2 for
Here the denominators are 5 and 10 so we can method 2
This preview shows page 28 - 37 out of 37 pages.
Here the denominators are 5 and 10, so we can write 28
6.4: Method 2 for Simplifying Complex Fractions The second approach interprets the complex fraction as division and applies the earlier work in dividing fractions in which you invert and multiply . Recall that by the fundamental principle we can always multiply the numerator and denominator of a fraction by the same nonzero quantity. 29
6.4.1: Use the Fundamental Principle to Simplify Complex Fractions Examples: Simplify the followings: 1. 2. 30
6.4.2: Use division to simplify complex fractions Examples: Simplify the following: 1. 2. 31 2 6 5 4 18 3 2 2 2 x x x x x x
6.4.2: Use division to simplify complex fractions The following algorithm summarizes our work with complex fractions. 32
6.5: Solving Rational Equations Rational Equations Applications of algebra will often result in equations involving rational expressions. The objective in this section is to develop methods to find solutions for such equations. The usual technique for solving such equations is to multiply both sides of the equation by the least common denominator (LCD) of all the rational expressions appearing in the equation. The resulting equation will be cleared of fractions, and we can then proceed to solve the equation as before. 33
6.5: Solving Rational Equations Examples: Solve the following rational equations: 1. 2. 3. 34
6.5: Solving Rational Equations Examples: Solve the following rational equations: 1. 2. 3. 35
6.5: Solving Rational Equations Examples: Solve the following rational equations: 1. 2. 36 1 6 6 1 3 3 3 1 2 x x x x x x 1 6 2 7 3 5 x x x
6.5: Solving Rational Equations The following algorithm summarizes our work in solving equations containing rational expressions. 37
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# Slope
### Slope – ACT Math
Calculating slope is one of the most important skills that you will need for coordinate geometry problems on your ACT math test.
The formula for the slope of a line is as follows:
y = mx + b
Remember that m is the slope and b is the y intercept.
Try these free exercises.
Problem 1:
Consider a line that crosses the y axis at 10 and upon which lies point (6, 4). What is the slope of this line?
A. − 1
B. 1
C. − 6
D. 6
E. 10
y = mx + b
Remember that m is the slope and b is the y intercept.
We have got point (6, 4), so we can substitute those values for a and y.
4 = m6 + 10
4 − 10 = m6 + 10 − 10
− 6 = m6
− 1 = m
Problem 2:
Marta has just bought a new pair of tennis shoes. She wants to wear the new running shoes to run up and down a hill near her house.
The measurements to find the slope of the hill can be placed on a two dimensional linear graph on which x = 5 and y = 165.
If the line crosses the y axis at 15, what is the slope of this hill?
A. 10
B. 20
C. 30
D. 36
E. 75
In order to calculate the slope of a line, you need this formula:
y = mx + b
Again remember that m is the slope and b is the y intercept.
Now substitute the values:
y = mx + b
165 = m5 + 15
165 − 15 = m5 + 15 − 15
150 = m5
150 ÷ 5 = m5 ÷ 5
30 = m
Have a look at our other geometry problems.
Go ahead to our free sample ACT math test. |
# ISEE Lower Level Math : How to find the perimeter of a rectangle
## Example Questions
← Previous 1 3 4 5 6
### Example Question #1 : How To Find The Perimeter Of A Rectangle
Dennis is building a fence around his field to keep his cattle from getting off the property. If Dennis’ field is miles long and miles wide, how much fence will Dennis need to surround all of his property?
Explanation:
In order to determine how much fence Dennis will need, we must find the perimeter of his property, which can be found using the formula . When we plug in the and in for , we find that Dennis needs of fence to surround his property.
### Example Question #2 : How To Find The Perimeter Of A Rectangle
Give the perimeter of the rectangle in the above diagram.
Explanation:
Add the length and the height, then multiply the sum by two:
The rectangle has a perimeter of 58 inches.
### Example Question #3 : How To Find The Perimeter Of A Rectangle
Give the perimeter of the above rectangle in feet.
Explanation:
Use the following formula, substituting :
Now, divide this by 12 to convert inches to feet.
6 inches make half of a foot, so this means the perimeter is feet.
### Example Question #4 : How To Find The Perimeter Of A Rectangle
What is the perimeter of a rectangle that has a length of inches and a width of inches?
Explanation:
The perimeter of a shape is the sum of all its sides. To find the perimeter of a rectangle, one can use the formula listed above, , since a rectangle has opposite sides of equal length.
The length of the rectangle is :
The width of the rectangle is :
.
Therefore, the perimeter is .
### Example Question #4 : How To Find The Perimeter Of A Rectangle
What is the perimeter of a rectangle with length equal to 5 and width equal to 3?
30
15
8
16
16
Explanation:
### Example Question #5 : How To Find The Perimeter Of A Rectangle
What is the perimeter of a rectangle with a width of and a length of ?
Explanation:
The perimeter of a rectangle is equal to the sum of all its sides. The formula for finding the perimeter of a rectangle is
The length of the rectangle is eight, and the width is five; .
Therefore, the perimeter of the rectangle is .
### Example Question #101 : Geometry
Mr. Barker is building a rectangular fence. His yard has an area of feet, and the one side of the fence he's already built is feet long.
What will the perimeter of his fence be when he is finished building it?
Explanation:
The perimeter is calculated by adding up all the sides of the rectangle—in this case,
So the perimeter is feet.
### Example Question #7 : How To Find The Perimeter Of A Rectangle
A rectangle has sides 10 cm and 4 cm. What is its perimeter?
Explanation:
A rectangle has two sides of congruent, or equal, sides. Therefore, there are two 10 cm sides and two 4 cm sides. Perimeters is the sum of all the sides, so you must add up the four sides. The answer is 28 cm.
### Example Question #8 : How To Find The Perimeter Of A Rectangle
If a rectangle has a width of 6.5 inches and a length of 9 inches, what is its perimeter?
Explanation:
The perimeter of a rectangle is found by adding together all four sides. Two sides will be equal to the length, and two sides will be equal to the width.
Since the width is 6.5 inches and the length is 9 inches, the perimeter would be:
### Example Question #102 : Geometry
If Margaret is buying a tablecloth for a table that is 4 feet by 2 feet, what should be the area of the tablecloth?
Explanation:
We will need to find the area of the rectangle by multiplying the length by the width:
Use the given dimensions:
← Previous 1 3 4 5 6 |
# Odds: Refresher
https://arbital.com/p/odds_refresher
by Nate Soares Jul 6 2016 updated Oct 13 2016
A quick review of the notations and mathematical behaviors for odds (e.g. odds of 1 : 2 for drawing a red ball vs. green ball from a barrel).
Let's say that, in a certain forest, there are 2 sick trees for every 3 healthy trees. We can then say that the odds of a tree being sick (as opposed to healthy) are $(2 : 3).$
Odds express relative chances. Saying "There's 2 sick trees for every 3 healthy trees" is the same as saying "There's 10 sick trees for every 15 healthy trees." If the original odds are $(x : y)$ we can multiply by a positive number $\alpha$ and get a set of equivalent odds $(\alpha x : \alpha y).$
If there's 2 sick trees for every 3 healthy trees, and every tree is either sick or healthy, then the probability of randomly picking a sick tree from among all trees is 2/(2+3):
If the set of possibilities $A, B, C$ are mutually exclusive and exhaustive, then the probabilities $\mathbb P(A) + \mathbb P(B) + \mathbb P(C)$ should sum to $1.$ If there's no further possibilities $d,$ we can convert the relative odds $(a : b : c)$ into the probabilities $(\frac{a}{a + b + c} : \frac{b}{a + b + c} : \frac{c}{a + b + c}).$ The process of dividing each term by the sum of terms, to turn a set of proportional odds into probabilities that sum to 1, is called normalization.
When there are only two terms $x$ and $y$ in the odds, they can be expressed as a single ratio $\frac{x}{y}.$ An odds ratio of $\frac{x}{y}$ refers to odds of $(x : y),$ or, equivalently, odds of $\left(\frac{x}{y} : 1\right).$ Odds of $(x : y)$ are sometimes called odds ratios, where it is understood that the actual ratio is $\frac{x}{y}.$ |
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Measurement is the foundation of all mathematical concepts and this is not possible to imagine the world without measurements. The perfect measurements will increase the level of accuracy if they are based on international standards. Still, always measurement is suspected to small errors in mathematics or a level of uncertainty too. In simple words, the 100 percent accurate measurements are not possible in the practical world.
## Error Formulas
Errors are simply defined as the difference between the measured value and the actual value. For example, when two operators use the same device for the measurement then this is not necessary that results would be the same. The difference that occurs between the actual value and the measured value is named as the ERROR.
$\ Percentage\;Error=\frac{Approximate\;Value-Exact\;Value}{Exact\;Value}\times 100$
$\ Standard\;Error =SE_{\overline{x}}=\frac{S}{\sqrt{n}}$
Where,
s is the standard deviation
n is the number of observation
$\ Sampling\;Error=\pm \sqrt{\frac{2500}{Sample\;Size}}\times 1.96$
$\ E=Z\left(\frac{\alpha}{2}\right)\left(\frac{\alpha}{\sqrt{n}}\right)$
Here,
$z$ $(\frac{\alpha }{2})$ = represents the critical value.
$z$ $(\frac{\sigma }{\sqrt{n}})$ = represents the standard deviation.
To learn the mathematics concepts deeply, you should know the different terms that could define the errors like sampling error, standard error, marginal error or percent error etc. Let us discuss on each of terms one by one with respective formulas. If you would understand these definitions and formulas deeply then there are chances that you could calculate the values as accurate as possible.
### Sampling Error Formula
The error that arises due to sampling is named as the sampling error. This is the error usually related to the statistical analysis because of the wrong samples of the observations are taken. For example, the weight of 2000 citizens of a country are noted down and you need to calculate the average of weights now then it could be the same as the average weight of two million people.
To determine the weight of the whole population, the sampling technique is used. The difference between sample values and the population is termed as the sampling error. This is not possible to calculate the exact value of the population of you don’t know the value of sampling error and that could be found with sample modeling only.
So, the sampling error Formula in mathematics could be written as below –
$\ Sampling\;Error=\pm \sqrt{\frac{2500}{Sample\;Size}}\times 1.96$
### Percent Error Formula
There could be a manufacturing error in measuring instruments too. This is not possible to assure them the exact. To know what type of error could be available here, we should know about the percentage error formula too. This is the absolute difference between measured value and the actual value and you should multiply the values by hundred too.
$\ Percentage\;Error=\frac{Approximate\;Value-Exact\;Value}{Exact\;Value}\times 100$
### The Margin of Error Formula
The margin of errors is generally found in random sampling or the result of a survey. It is assumed that result of a sample is highly closers to the one would get from the population has been queried. In easy words, the margin of error is the product of critical value with the standard deviation. This is given by E and it could be written as –
$\ E=Z\left(\frac{\alpha}{2}\right)\left(\frac{\alpha}{\sqrt{n}}\right)$
Here,
$\ Z\left(\frac{\alpha}{2}\right)= represents\;the\;critical\;value.$
$\ Z\left(\frac{\alpha}{\sqrt{n}}\right) = represents\;the\;standard\;deviation.$
### Standard Error Formula
Standard Error is the important statistical measure that is related to the standard deviation. The accuracy of a sample that could be presented by the population is given through the standard error formula and it could be written as below –
$\ Standard\;Error =SE_{\overline{x}}=\frac{S}{\sqrt{n}}$
Where,
s is the standard deviation
n is the number of observation
Where,S is the standard deviation, and n is the number of observations. |
Lesson Objectives
• Learn about the whole numbers
• Learn how to create a place value chart
• Learn how to find the place value of a given digit in a number using a place value chart
• Learn how to find the place value of a given digit in a number without a place value chart
## What is Place Value?
When we begin learning basic arithmetic, our study involves the use of the group of numbers known as the whole numbers.
Whole Numbers: {0,1,2,3,4,5,...}
The whole numbers begin with a 0, and increase in increments of 1 indefinitely. There is no largest whole number. This is shown using an ellipsis or three dots "..." after the 5. This simply tells us that the pattern of increasing by 1 will continue forever. The numbers 0 through 9 have a special name, these numbers are referred to as the "digits".
Digits: {0,1,2,3,4,5,6,7,8,9}
Our number system relies on the digits along with place value to construct each number. Single-digit numbers are very easy to understand. The number 5, simply has one digit and its value is 5. When we look at multi-digit numbers, the situation is a bit more complex. In the number 53, the 5 no longer has a value of simply 5, it now represents a value of 50. To understand why we look to our place value system. Under the place value system, a digit obtains its value based on its position or placement in a number. Let's take a look at an example:
Observe how the value of the digit 8 changes in each number:
8 - The 8 means 8 ones, or just 8
83 - The 8 means 8 tens, or 80
859 - The 8 means 8 hundreds, or 800
8032 - The 8 means 8 thousands, or 8000
We can see from our example, that changing the position or placement of a digit, changes its value. Generally, we learn place value with a visual aid known as a place value chart: If we start at the rightmost position of the place value chart, we see it begins with the ones' place. As we move left, we are simply multiplying by ten to get to the next place. Moving to the left of the ones' place is the tens' place, since 1 x 10 = 10. Then we come across the hundreds' place (10 x 10 = 100). This pattern continues out indefinitely. We simply keep multiplying the previous place by ten to obtain the next place to the left.
• 1 - Ones
• 1 x 10 = 10 » Tens
• 10 x 10 = 100 » Hundreds
• 100 x 10 = 1000 » Thousands
• 1000 x 10 = 10,000 » Ten Thousands
• 10,000 x 10 = 100,000 » Hundred Thousands
• 100,000 x 10 = 1,000,000 » Millions
Once we understand how to build a place value chart, using one is very easy. We simply line our number up starting with the rightmost digit going into the rightmost place of the place value chart. We then keep filling in the chart one digit at a time moving left. Once this is done, we can easily give the place for each digit in a given number. Over time, this can be done without a place value chart, but it helps with memorization in the beginning. Let's try a few examples.
Example 1: Write 759 in the place value chart, and give the place of each digit.
• 7 - Hundreds
• 5 - Tens
• 9 - Ones
Example 2: Write 26,041 in the place value chart, and give the place of each digit.
• 2 - Ten Thousands
• 6 - Thousands
• 0 - Hundreds
• 4 - Tens
• 1 - Ones
It is important to note how zero is used as a placeholder. The 0 in the number 26,041, is used to tell us we have 0 hundreds. Without the 0 involved, the number will collapse down to 2641, which is not the same value.
Example 3: Write 8,352,194 in the place value chart, and give the place of each digit.
• 8 - Millions
• 3 - Hundred Thousands
• 5 - Ten Thousands
• 2 - Thousands
• 1 - Hundreds
• 9 - Tens
• 4 - Ones
We can also achieve the same result without a place value chart. We begin by writing a 1 (1's) above the rightmost digit. We can then multiply by 10 as we move to the left (1 x 10 = 10 » 10's, 10 x 10 = 100 » 100's, 100 x 10 = 1000 » 1000's,...). Let's try an example.
Example 4: Give the place of each digit for the number 2509, without a place value chart.
• 2 - Thousands
• 5 - Hundreds
• 0 - Tens
• 9 - Ones
#### Skills Check:
Example #1
State the place value of the underlined digit.
3,582,353
A
tens
B
hundreds
C
millions
D
thousands
E
ones
Example #2
State the place value of the underlined digit.
81,573
A
thousands
B
ten thousands
C
ones
D
hundreds
E
tens
Example #3
State the place value of the underlined digit.
3,924
A
millions
B
tens
C
hundreds
D
thousands
E
ones |
Sometimes a circle can be both inscribed and circumscribed with respect to a polygon. They are then called inscribed or circumscribed circles. Shown below is an inscribed and a circumscribed circle with respect to a triangle.
Let us discuss them in detail.
‘Inscribe’ means to draw inside of any figure, just touching it but will not cross the figure. It is thus the opposite of the circumscribed circle.
In geometry, an inscribed circle, also known as the incircle of a polygon is the largest possible circle that can be drawn inside a regular, cyclic polygon. The inscribed circle will touch each of the three sides of the triangle at exactly one point. The center of such a circle is called the incenter. It is the point where the angle bisectors of the triangle meet. The radius of such a circle is called the inradius.
When a circle inscribes a triangle, the triangle is outside of the circle and the circle touches the sides of the triangle at one point on each side. The sides of the triangle are tangent to the circle. Thus, for a polygon, a circle is not inscribed unless each side of the polygon is tangent to the circle.
Thus, all triangles and regular polygons have inscribed circles.
Given below is a circle inscribed in a triangle
Note that each side of the triangle is tangent to the circle, so if we draw a radius from the center of the circle to the point where the circle touches the edge of the triangle, the radius will form a right angle with the edge of the triangle.
Given below is a link showing how to inscribe a circle in a triangle.
The circumscribed circle of a polygon is a circle that touches all 3 vertices of the polygon. The center of such a circle is called the circumcenter, the point where the perpendicular bisectors of the sides meet. The radius of such a circle is called the circumradius.
Not every polygon though has a circumscribed circle. A polygon that has a circumscribed circle is called a cyclic polygon. It is also called a concyclic polygon since its vertices are concyclic. All triangles and all regular polygons such as square, rectangle, trapezoid, and kite are concyclic.
Since all triangles are cyclic, they always have a circumscribed circle. When a circle circumscribes a triangle, the triangle is inside the circle, and the triangle touches the circles at each of its vertices.
Given below is a diagram showing a circle circumscribes a triangle.
The center of the circumscribed circle can be obtained by drawing the three perpendicular bisectors of the triangle. The point where the three perpendicular bisectors meet is the center of the circumscribed circle.
Given below is a link showing how to circumscribe a circle in a triangle. |
Math in Focus Grade 1 Chapter 11 Practice 3 Answer Key Tally Charts and Bar Graphs
Practice the problems of Math in Focus Grade 1 Workbook Answer Key Chapter 11 Practice 3 Tally Charts and Bar Graphs to score better marks in the exam.
Math in Focus Grade 1 Chapter 11 Practice 3 Answer Key Tally Charts and Bar Graphs
There are some spoons, forks, and knives on the table.
Complete the tally chart. Then answer the questions.
Question
Question 1.
How many spoons are there? ______
Question 2.
How many knives are there? ______
Question 3.
There are 2 more than spoons. ______
Answer: There are 2 more forks than spoons
Question 4.
How many fewer knives than forks are there? ______
Answer: 3 fewer knives are there than forks
Kelly bought balloons for her party. The tally chart shows the different colors of balloons she bought.
Complete the tally chart. Then fill in the blanks.
Question 5.
Question 6.
Kelly bought ___ red balloons.
Answer: Kelly bought 6 red balloons.
Question 7.
She bought ___ blue balloons.
Answer: She bought 9 blue balloons.
Question 8.
Kelly bought ___ more yellow balloons than blue balloons.
Answer: Kelly bought 6 more yellow balloons than blue balloons.
Question 9.
She bought 9 fewer __________ balloons than __________ balloons.
Answer: She bought 9 fewer Red balloons than yellow balloons.
Complete the tally chart.
Question 10.
Abby bought some seed packages. The tally chart shows the different kinds of seeds she bought.
Make a bar graph.
Question 11.
Question 12.
How many packages of sunflower seeds did she buy? ____.
Answer: She bought 5 packages of sunflower seeds
Question 13.
How many more packages of cucumber seeds than pumpkin seeds did she buy? ______
Answer: She bought 7 more packages of cucumber seeds than pumpkin seeds
Question 14.
How many packages of seeds did she buy in all? ____
Answer: she bought 15 packages of seeds in all
Question 15.
She bought 2 more packages of ___ seeds than sunflower seeds.
Answer: She bought 2 more packages of cucumber seeds than sunflower seeds. |
Question Video: Using the Pythagorean Identities to Evaluate a Trigonometric Expression Mathematics
Find (sin π + cos π)Β² given that sin π cos π = 6/7.
02:46
Video Transcript
Find sin π plus cos π squared given that sin π cos π equals six over seven.
The first thing we can do with this question is to write out sin π plus cos π squared as two sets of brackets. We can treat this like any other binomial expansion and multiply out these brackets to find the following four terms: sin π times sin π plus sin π times cos π plus cos π times sin π and plus cos π times cos π.
Now, we know that the order in which sin π and cos π are multiplied together does not matter. This means that cos π sin π is the same as sin π cos π. Upon realizing this, we see that our two middle terms are in fact the same. And we can, therefore, group them together.
Another thing we can do to tidy up our equation, is to see that sin π times sin π is sin π squared. This can equivalently be written as sin squared π, the same being true for cos. Letβs rewrite these two terms in our expression. Now that weβve tidied things up, letβs rearrange our terms.
Looking at the first two terms of our expression, you may recognize one of the Pythagorean identities. This identity states that sin squared π plus cos squared π is equal to one. The second thing we may notice is that our question gives us the value for sin π cos π.
We can, therefore, perform two substitutions on our expression. First, we can replace sin squared π plus cos squared π with one. Next, we can replace sin π cos π with six over seven.
To make our addition easier, we can rewrite one as seven over seven so that our denominators match. Multiplying six over seven by two, we find 12 over seven. Adding seven over seven to 12 over seven, we find our answer is 19 over seven.
We can therefore, say that sin π plus cos π all squared is 19 over seven when sin π times cos π is six over seven. |
## Derivatives of different orders — Leibniz’ Rule: IITJEE Maths training
Let a function $y=f(x)$ be differentiable on some interval $[a,b]$. Generally speaking, the values of the derivative $f^{'}(x)$ depend on x, which is to say that the derivative $f^{'}(x)$ is also a function of x. Differentiating this function, we obtain the so-called second derivative of the function $f(x)$.
The derivative of a first derivative is called a derivative of the second order or the second derivative of the original function and is denoted by the symbol $y^{''}$ or $f^{''}(x)$: $y^{''}=(y^{'})^{'} = f^{''}(x)$
For example, if $y=x^{5}$, then $y^{'}=5x^{4}$ and $y^{''} = (5x^{4})^{'}=20x^{3}$
The derivative of the second derivative is called a derivative of the third order or the third derivative and is denoted by $y^{'''}$ or $f^{'''}(x)$.
Generally, a derivative of the nth order of a function $f(x)$ is called the derivative (first order) of the derivative of the $(n-1)$th order and is denoted by the symbol $y^{(n)}$ or $f^{(n)}(x)$:
$y^{(n)} = (y^{(n-1)})^{'}=f^{(n)}(x)$
(Note: the order of the derivative is taken in parentheses so as to avoid confusion with the exponent of a power.)
Derivatives of the fourth, fifth and higher orders are also denoted by Roman numerals: $y^{IV}$, $y^{V}$, $y^{VI}$, $\ldots$. Here, the order of the derivative may be written without brackets. For instance, if $y=x^{5}$, $y^{'}=5x^{4}$, $y^{''}=20x^{3}$, $y^{'''}=60x^{2}$, $y^{IV}=y^{(4)}=120x$, $y^{V}=y^{(5)}=100$, $y^{(6)}=y^{(7)}=\ldots = 0$
Example 1:
Given a function $y=e^{kx}$, where k is a constant, find the expression of its derivative of any order n.
Solution 1:
$y^{'}=ke^{kx}$, $y^{''}=k^{2}e^{kx}$, $y^{(n)}=k^{n}e^{kx}$
Example 2:
$y=\sin{x}$. Find $y^{(n)}$.
Solution 2:
$y^{'}=\cos{x}=\sin(x+\frac{\pi}{2})$
$y^{''}=-\sin{x}=\sin(x+2\frac{\pi}{2})$
$y^{'''}=-\cos{x}=\sin(x+3\frac{\pi}{2})$
$y^{IV}=\sin{x}=\sin(x+4\frac{\pi}{2})$
$\vdots$
$y^{(n)}=\sin(x+n\frac{\pi}{2})$
In similar fashion, we can also derive the formulae for the derivatives of any order of certain other elementary functions. You can find yourself the formulae for derivatives of the $n^{th}$ order of the functions $y=x^{k}$, $y=\cos{x}$, $y=\ln (x)$.
Let us derive a formula called the Leibniz rule that will enable us to calculate the $n^{th}$ derivative of the product of two functions $u(x), v(x)$. To obtain this formula, let us first find several derivatives and then establish the general rule for finding the derivative of any order:
$y=uv$
$y^{'}=u^{'}v+uv^{'}$
$y^{''}=u^{''}v+u^{'}v^{'}+u^{'}v^{'}+uv^{''}=u^{''}v+2u^{'}v^{'}+uv^{''}$
$y^{'''}=u^{'''}v+u^{''}v^{'}+2u^{''}v^{'}+2u^{'}v^{''}+u^{'}v^{''}+uv^{'''}$
which in turn equals $u^{'''}v+3u^{''}v^{'}+3u^{'}v^{''}+uv^{'''}$
$y^{IV}=u^{IV}v+4u^{'''}v+6u^{''}v^{''}+4u^{'}v^{'''}+uv^{IV}$
The rule for forming derivatives holds for the derivative of any order and obviously consists in the following:
The expression $(u+v)^{n}$ is expanded by the binomial theorem, and in the expansion obtained the exponents of the powers of and are replaced by indices that are the orders of the derivatives, and the zero powers $(u^{0}=v^{0}=1)$ in the end terms of the expansion are replaced by the function themselves (that is, “derivatives of zero order”):
$y^{(n)}=(uv)^{(n)}=u^{(n)}v+nu^{(n-1)}v^{'}+\frac{n(n-1)}{1.2}u^{(n-2)}v^{''}+\ldots + uv^{(n)}$
This is the Leibniz Rule.
A rigorous proof of this formula may be carried out by the method of complete mathematical induction (in other words, to prove that if this formula holds for the nth order, it will also hold for the order $n+1$).
Example:
$y=e^{ax}x^{2}$. Find the derivative $y^{(n)}$.
Solution:
$u=e^{ax}$, $v=x^{2}$
$u^{'}=ae^{ax}$, $v^{'}=2x$
$u^{''}=a^{2}e^{ax}$, $v^{''}=2$
$\vdots$, $\vdots$
$u^{(n)}=a^{n}e^{ax}$, $v^{'''}=v^{IV}=\ldots=0$
$y^{(n)}=a^{n}e^{ax}x^{2}+na^{n-1}e^{ax}.2x+\frac{n(n-1)}{1.2}a^{n-2}e^{ax}.2$, or
$y^{(n)}=e^{ax}[a^{n}x^{2}+2na^{n-1}x+n(n-1)a^{n-2}]$
More calculus in the pipeline…there is no limit to it 🙂
Nalin Pithwa
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# S1 Measures of Dispersion The mean, variance and standard deviation
Document Sample
``` S1 Measures of Dispersion
The mean, variance and
standard deviation
S1 Measures of Dispersion
Objectives:
To be able to find the variance
and standard deviation for
discrete data
To be able to find the variance
and standard deviation for
continuous data
A factory worker counted the numbers of nuts in a
packet. The results are shown in the table
Number Freq CF Calculate P80 and P40
of nuts
6 8 8 P80 = 99/100x80 = 79.2
7 12 20 P80 = 80th term
8 36 56 P80 = 10 nuts
9 18 74 P40 = 99/100x40 = 39.6
10 15 89 P40 = 40th term
11 10 99 P40 = 8 nuts
Calculate the 40th to 80th interpercentile range
40th to 80th interpercentile range = 10 - 8 = 2 nuts
The lengths of a batch of 2000 rods were measured to the nearest
cm. The measurements are summarised below.
Length Number of Cumulative
(nearest cm) rods frequency
Q1=74.5 + 500-250 x 5
738-250
60-64 11 11 Q1=77.06
65-69 49 60
70-74 190 Q2=79.5+1000-738 x 5
250 1370-738
75-79 488 738 Q2=81.57
80-84 632 1370
85-89 470 1840 Q3=84.5+1500-1370 x 5
90-94 137 1977 1840-1370
Q3=85.88
95-99 23 2000
By altering the formula slightly can you work out how to find the 3rd
decile (D3) and the 67th percentile (P67)?
D3=74.5 + 600-250 x 5
738-250
D3=78.09
P67=79.5 + 1340-738 x 5
1370-738
P67=84.26
Variance of discrete data
The deviation (difference) of the data (x) from
_
the mean (x) is one way of measuring the
dispersion (spread) of a set of data.
_
Variance = Σ(x – x)²
n
_
Variance = Σx² – Σx ²
n n
Standard deviation.
As variance is measured in units ² you usually
take the square root. This gives you the
standard deviation.
Standard deviation = √variance
Standard deviation symbol is σ
The marks scored in a test by seven students
are 3,4,6,2,8,8,5. Calculate the variance and
standard deviation.
x x² _
Variance = Σx² – Σx ²
3 9
n n
4 16
36
6 σ ² = 218 – 36 ² = 4.69
2 4 7 7
8 64
64
σ = √4.69 = 2.17
8
5 25
Σx = 36 Σx² = 218
The variance and standard deviation from a
frequency table
Number of Freq of
rods (x) rods (f) fx fx²
35 3 105 3675
36 17 612 22032
37 29 1073 39701
38 34 1292 49096
39 26 1014 39546
Σf=109 Σfx=4096 Σfx²=154050
Variance = Σfx² – Σfx ² Variance = 154050 – 4096 ²
Σf Σf 109 109
σ ²=1.19805 σ = 1.109
Variance and standard deviation from a grouped frequency distribution
Height of Freq
plant (cm) (f)
0<h≤5 4
5 < h ≤ 10 15
10 < h ≤ 15 5
15 < h ≤ 20 2
20 < h ≤ 25 0
25 < h ≤ 30 1
Total
σ ² = 6487.5 – 285 ² = 128.85802 σ = √128.85802
27 27 σ = 11.35
```
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In this article, we have compiled a list of geometry formulas which are quite helpful in solving the questions related to areas, volumes and perimeters of geometrical figures. So, let us get started.
## Area and Perimeter Formulas
In this section, we have compiled a list of area and perimeter formulas of various geometrical figures.
Area of a geometrical figure refers to the amount of space in square length units occupied by the surface of the geometrical figure. On the other hand, a perimeter refers to the distance around the closed geometrical figure or shape.
### Triangle
Triangle is one of the most fundamental shape in geometry that consists of three sides and three vertices. The sum of the interior angles of a triangle is equal to 180 degrees. The three types of triangle are equilateral, isosceles and scalene.
The formula for area of the triangle is:
Area =
The formulas for perimeter of equilateral, isosceles and scalene triangles are a bit different. Basically, they all involve adding the length of three sides of the triangle.
Perimeter of an equilateral triangle = 3 x length of its side
Perimeter of an isosceles triangle = 2 x length + base
Perimeter of the scalene triangle = P = a + b + c
### Square
Square refers to a regular quadrilateral which has four equal sides and angles. It means that all sides of the square are of the same length and all the angles are equal in measure.
The formula for area of the square is:
Area = l x l
The formula for calculating the perimeter of a square is:
Perimeter = 4 x length of its side
### Rectangle
A rectangle is a type of quadrilateral with four right angles. The opposite sides of the rectangle are of the same length and are parallel to each other.
The formula for area of the rectangle is:
Area = l x w, where l is the length of the rectangle, and w is its width
The formula for perimeter of the rectangle is:
Perimeter = 2 (l + w)
### Rhombus
A rhombus refers to a parallelogram with four equal sides and opposite equal angles. It means that all the four sides of the rhombus are of the same length and opposite angles are congruent.
The formula to calculate the area of the rhombus is:
Area = , here D and d represent the diagonals of the rhombus
The formula to calculate the perimeter of the rhombus is:
Perimeter = 4 x length of the side of rhombus
### Rhomboid
A rhomboid is a type of parallelogram in which the adjacent sides have different lengths and angles are not equal to 90 degrees.
The formula to calculate the area of the rhomboid is:
Area = base x height
The formula to calculate the perimeter of the rhomboid is:
Perimeter = 2 . (a + b), where a and b are the sides of the rhomboid
### Area of Trapezoid
Trapezoid is a quadrilateral with one pair of parallel sides as shown in the figure below:
The formula to calculate the area of the trapezoid is:
Area =
### Area of a Regular Polygon
In geometry, a regular polygon is a polygon that is equilateral and equiangular. It means that all sides of the regular polygon are of the same length and all of its angles are of the same measure.
The formula to calculate the area of the regular polygon is:
Area =
The formula to calculate the perimeter of the regular polygon is:
Perimeter = n x l, where n depicts the number of sides of the polygon
### Polygon
If you have an irregular polygon which seems like the figure given below, then you can calculate the area by triangulating the polygon and adding the area of these triangles.
### Circle
A circle is a geometrical figure in which all points are located at equal distance from its center.
Instead of the perimeter, a circle has the circumference. The formula for calculating the circumference of the circle is:
, where has a fixed value and "r" is the radius of the circle
Since the radius of the circle is half of its diameter, hence we can also write the formula of the circumference of the circle as shown below:
The formula to calculate the area of the circle is:
, where r is the radius of the circle
### Circular Sector
A circular sector, also referred to as a disk center or circle center is the portion of the circle that is enclosed by an arc and two radii of the circle.
The formula for computing the area of the circular sector is given below:
The formula for calculating the arc length of the circular sector is:
### Circular Segment
A circular segment refers to the region of the circle that is "cut off" from the remaining circle by a chord or secant.
The formula for calculating the area of the circular segment is:
The circular area of the segment AB = Area of the circular sector AOB - Area of triangle AOB
### Lune of Hippocrates
Area of the lune = area of the semicircle − area of circular segment.
Area of the lune = Area of the right triangle
### Circular Trapezoid
In two given concentric circle, a circular trapezoid refers to the area that lies between two non-crossing chords of the circle.
The formula for calculating the area of the circular trapezoid is given below:
### Area Enclosed between Two Concentric Circles
The formula for calculating the area between two concentric circle us given below:
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## Surface Area and Volume Formulas
In this section of the article, we have compiled a list of surface area and volume formulas of various geometrical figures.
### Tetrahedron
A tetrahedron refers to a solid that contains four plane triangular faces
The formula for calculating the area of the tetrahedron is:
The formula for calculating the volume of the tetrahedron is:
### Octahedron
An octahedron refers to a three-dimensional shape which has eight plane faces.
The formula for calculating the area of the octahedron is:
The formula for calculating the volume of the octahedron is:
### Icosahedron
Icosahedron refers to a solid figure that contains 20 plane faces.
The formula for calculating the area of an icosahedron is:
The formula for calculating the volume of the tetrahedron is:
### Dodecahedron
A dodecahedron refers to a three-dimensional figure with twelve plane faces.
The formula for calculating the area of the dodecahedron is:
The formula for calculating the volume of the dodecahedron is:
### Cube
A three-dimensional shape that contains six equal square is known as a cube.
The formula for calculating the volume of the cube is:
The formula for calculating the surface area of the cube is:
### Cuboid
A three-dimensional geometrical figure that contains six rectangular faces is known as a cuboid
The formula for calculating the area of the cuboid is:
The formula for calculating the volume of the cuboid is:
### Prism
A prism refers to a solid geometrical figure in which two ends are equal, similar and parallel rectilinear figures, and sides are parallelograms as shown in the figure below:
= Perimeter of the base
### Pyramid
In geometry, a pyramid refers to polyhedron that is formed by joining the polygonal based an point, known as apex
= Perimeter of the base
Ap = apothem of the pyramid
ap = apothem of the base
### Truncated Pyramid
A truncated pyramid results from cutting a pyramid by a plane parallel to the base and segregating the portion that contains the apex.
P = Perimeter of the larger base
P' = Perimeter of the smaller base
A = Area of the larger base
A' = Area of the smaller base
### Cylinder
A cylinder as shown below refers to a surface that contains all the points on all the lines that are parallel to a given line and which pass through a fixed plane curve in a plane that is non-parallel to the given line.
### Cone
A cone in geometry refers to a three-dimensional shape that narrows smoothly from a flat base to a point known as a vertex or apex.
### Truncated Cone
a cone section or pyramid lacking an apex and terminating in a plane usually parallel to the base.
It refers to the cone section or a pyramid that lacks an apex and terminates in a plane usually parallel to the base.
### Sphere
Just like circle, a sphere refers to the set of points that are all located at the equal distance "r" from a given point.
### Spherical Wedge
A spherical wedge, also known as ungula is a part of a ball bounded by two plane semi disks and a spherical lune.
### Spherical Cap
A spherical cap, also known as a spherical done is a part of sphere or a ball that is "cut off" by a plane.
### Spherical Segment
A spherical segment refers to a solid defined by cutting a ball or a sphere with a pair of parallel lines.
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# What is the probability of rolling double threes on two dice?
Contents
Therefore the probability of rolling doubles with all three dice is 1/6 + 1/6 + 1/6 = 3/6 or 1/2 (including triples). The probability of rolling triples is 1/36 so the probability of rolling doubles with no triples is 18/36 – 1/36 = 17/36.
## What is the probability of rolling doubles on two dice?
The number of different outcomes on two dice is the number of different outcomes on the first die (6) TIMES the number of different outcomes on the second (6), or 36. The probability of rolling doubles is 1/6, because there are 6 ways to roll doubles. 6/36 = 36.
## What is the probability of rolling double sixes from two dies?
The probability of a double-six in one throw of two die is 1/36 or 0.028.
IT IS SURPRISING: How long has Vanna White been on the wheel of fortune?
## What is rolling doubles in dice?
Dice Rules
If the dice show the same number on each, it is called Doubles. If the player rolls Doubles, he/she rolls again.
## How do you find the probability of a dice?
Probability = Number of desired outcomes ÷ Number of possible outcomes = 3 ÷ 36 = 0.0833. The percentage comes out to be 8.33 per cent. Also, 7 is the most likely result for two dice. Moreover, there are six ways to achieve it.
## What is the probability of rolling a 6 on a dice twice?
The probability of rolling a specific number twice in a row is indeed 1/36, because you have a 1/6 chance of getting that number on each of two rolls (1/6 x 1/6).
## What is the probability of rolling double sixes and double two at the same time when rolling the dice twice?
The fact they are entirely independent of each other means we simply multiply each roll’s probability together: 1/6 × 1/6 × 1/6 × 1/6 = (1/6)4 = 1/1296 = 0.00077.
## What are the odds of rolling three double sixes in a row?
Since there are six sides to a die, you multiply the (1/46656) by 6 to find the probability of rolling the same doubles 3 times in a row. That would be 6/46656, or about 0.013% of the time.
## What happens when you roll doubles 3 times in Monopoly?
If you roll doubles 3 times on the same turn, you must Go to Jail. 6. When you finish your move and action, pass the dice to the player on your left.
## What is the probability of rolling 10 with two dice?
If the first die shows a 4, 5, or 6, the second can show a 6, 5, or 4 respectively. So that’s three ways of making a 10 out of 36 ways of rolling the two dice (6 for the first times 6 for the second). That means the probability is 3/36 which reduces to 1/12 or 0.083.
IT IS SURPRISING: You asked: Are the names of lottery winners published?
## What is the probability of 3 dice?
Two (6-sided) dice roll probability table
Roll a… Probability
3 3/36 (8.333%)
4 6/36 (16.667%)
5 10/36 (27.778%)
6 15/36 (41.667%)
## How do you find the probability of the sum of two dice?
The combinations for rolling a sum of seven are much greater (1 and 6, 2 and 5, 3 and 4, and so on). To find the probability that the sum of the two dice is three, we can divide the event frequency (2) by the size of the sample space (36), resulting in a probability of 1/18.
## When 3 dice are rolled what is the probability of getting a sum of 4?
Specific Probabilities
We divide the total number of ways to obtain each sum by the total number of outcomes in the sample space, or 216. The results are: Probability of a sum of 3: 1/216 = 0.5% Probability of a sum of 4: 3/216 = 1.4% |
# How to simplify or factor this equation
$$1 = x + 2\cdot x$$
How can I simplify this formula for $x$.
-
Are you actually looking for the roots of $x^2+x-1$? If so, you could start reading this and you will be able to proudly solve it on your own. If not, apologies, I have no idea what you are asking then. – 1015 Apr 29 '13 at 19:53
What is $x(2)$? Is it a $x^2$ or something else? – m0nhawk Apr 29 '13 at 19:54
it's x times 2 or x*2 or x(2) – Ben Apr 29 '13 at 19:54
Wait, if this is $1=x+2x=3x$, this is trivially $x=\frac{1}{3}$ (in the real or complex numbers). – 1015 Apr 29 '13 at 19:59
5 star! period. will get +1 from me. – Kaster Apr 29 '13 at 20:00
## 2 Answers
By equivalent transform of terms/equations: \begin{align} 1&=x+2x \\ 1&=1\cdot x+2\cdot x \\ 1&=(1+2)\cdot x \\ 1&=3\cdot x \\ 1/3&=x \end{align} In the last step we have divided by $3$ both sides of the equations. In the previous steps we used the unit and distributive properties:
$1\cdot x=x$ and $(a+b)\cdot x=a\cdot x+b\cdot x$.
-
$$1 = x + 2 \cdot x$$ This can be simplified to: $$1 = x + 2x$$ Remember that you can combine these terms by adding the coefficients (the numbers attached to the $x$'s). $$1 = 1x + 2x$$ $$1 = 3x$$ Now we have to isolate $x$. Divide both sides by 3. $$\frac{1}{3} = \frac{3x}{3}$$ $$\frac{1}{3} = x$$ I hope this post helped!
- |
# How to Calculate and Solve for Bulk Modulus of Elasticity (E) | Rock Mechanics
The image above represents bulk modulus of elasticity (E).
To compute for bulk modulus of elasticity (E), two essential parameters are needed and these parameters are young’s modulus (E) and Poisson’s ratio (v).
The formula for calculating the bulk modulus of elasticity (E):
k = E / 3(1 – 2v)
Where:
k = Bulk Modulus of Elasticity (E)
E = Young’s Modulus
v = Poisson’s Ratio
Let’s solve an example;
Find the bulk modulus of elasticity (E) when the young’s modulus is 66 and the Poisson’s ratio is 58.
This implies that;
E = Young’s Modulus = 66
v = Poisson’s Ratio = 58
k = E / 3(1 – 2v)
k = 66 / 3(1 – 2(58))
k = 66 / 3(1 – 116)
k = 66 / 3(-115)
k = 66 / -345
k = -0.19
Therefore, the bulk modulus of elasticity (E) is -0.19.
Calculating the Young’s Modulus when the Bulk Modulus of Elasticity (E) and the Poisson’s Ratio is Given.
E = k (3 – 6v)
Where:
E = Young’s Modulus
k = Bulk Modulus of Elasticity (E)
v = Poisson’s Ratio
Let’s solve an example;
Find the young’s modulus when the bulk modulus of elasticity (E) is 22 and the Poisson’s ratio is 18.
This implies that;
k = Bulk Modulus of Elasticity (E) = 22
v = Poisson’s Ratio = 18
E = k (3 – 6v)
E = 22 (3 – 6(18))
E = 22 (3 – 108)
E = 22 (-105)
E = -2376
Therefore, the young’s modulus is -2376.
Calculating the Poisson’s Ratio when the Bulk Modulus of Elasticity (E) and the Young’s Modulus is Given.
v = -(E – 3k / 6k)
Where:
v = Poisson’s Ratio
k = Bulk Modulus of Elasticity (E)
E = Young’s Modulus
Let’s solve an example;
Find the Poisson’s ratio when the bulk modulus of elasticity (E) is 12 and the young’s modulus is 40.
This implies that;
k = Bulk Modulus of Elasticity (E) = 12
E = Young’s Modulus = 40
v = -(E – 3k / 6k)
v = -(40 – 3(12) / 6(12))
v = -(40 – 36 / 72)
v = -(4 / 72)
v = -(0.055)
v = -0.055
Therefore, the Poisson’s ratio is -0.055.
Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the bulk modulus of elasticity (E).
To get the answer and workings of the bulk modulus of elasticity (E) using the Nickzom Calculator – The Calculator Encyclopedia. First, you need to obtain the app.
You can get this app via any of these means:
To get access to the professional version via web, you need to register and subscribe for NGN 1,500 per annum to have utter access to all functionalities.
You can also try the demo version via https://www.nickzom.org/calculator
Android (Paid) – https://play.google.com/store/apps/details?id=org.nickzom.nickzomcalculator
Android (Free) – https://play.google.com/store/apps/details?id=com.nickzom.nickzomcalculator
Apple (Paid) – https://itunes.apple.com/us/app/nickzom-calculator/id1331162702?mt=8
Once, you have obtained the calculator encyclopedia app, proceed to the Calculator Map, then click on Geology under Add-on.
Now, Click on Rock Mechanics under Geology
Now, Click on Bulk Modulus of Elasticity (E) under Rock Mechanics
The screenshot below displays the page or activity to enter your values, to get the answer for the bulk modulus of elasticity (E) according to the respective parameters which are the young’s modulus (E) and Poisson’s ratio (v).
Now, enter the values appropriately and accordingly for the parameters as required by the young’s modulus (E) is 66 and Poisson’s ratio (v) is 58.
Finally, Click on Calculate
As you can see from the screenshot above, Nickzom Calculator– The Calculator Encyclopedia solves for the bulk modulus of elasticity (E) and presents the formula, workings and steps too. |
Cube Root Of 12167
Cube Root Of 12167. 1 and 23 factors of cube root of. Web to find cube root by estimation we divide a number into two groups and find ones and tens from each group. Thus, the cube root of 12167 has to be calculated. Starting with the smallest prime number, prime. The group are 4 and 913. The cube root of 216 is expressed as ∛216 in radical form and as. Web the procedure to use the cube root calculator is as follows:
Cube root of 12167 is 23. Here we will show you how to simplify 3rd root of 12167, or convert 64 radical 3 to. The cube root of 216 is expressed as ∛216 in radical form and as. Write down the root of that number (3). 12167 is a natural number. The cube root of 12167 is another number that when multiplied by itself twice,. Making pairs of three from the end we'll get (12) (167) as unit digit is 7 so the digit at unit place is 3 (because 3 3 = 27, unit digit is 7) as, 2 3 < 12 < 3 3.
Web to find cube root by estimation we divide a number into two groups and find ones and tens from each group.
Find the cube root of 7 which is 343 here the once. Cube root of 12167 is 23. The nearest value of cube which is lesser than 12 is 8. We will make groups of three digits starting. Match the last digit of the number (7) to the last digit of a blue cube above (here, 27). Web find the cube root of given number using prime factorisation. Make group of 3 digit, starting from right. Starting with the smallest prime number, prime.
Unit Digit Of Cube Root Of 857375 = Unit Digit Of.
The group are 4 and 913. Web to find the cube root of.
Unit Digit Of Cube Root Of 857375 = Unit Digit Of.
Making pairs of three from the end we'll get (12) (167) as unit digit is 7 so the digit at unit place is 3 (because 3 3 = 27, unit digit is 7) as, 2 3 < 12 < 3 3. Information about find cube root of 12167 with estimation method covers all.
Conclusion of Cube Root Of 12167.
Web the cube root of 12167 is 23. The group are 4 and 913. 23 12167 23 529 23 23 1 ⇒ 12167 = 23 3 ⇒ 12167 3 = 23. Making pairs of three from the end we'll get (12) (167) as unit digit is 7 so the digit at unit place is 3 (because 3 3 = 27, unit digit is 7) as, 2 3 < 12 < 3 3.. Cube root of 12167 is 23. Web find cube root of 857375.
Source |
# A Proof of the Butterfly Theorem Using Ceva’s Theorem
### Theorem (The Butterfly Theorem)
Through the midpoint $M\;$ of a chord $PQ\;$ of a circle, two other chords $AB\;$ and $CD\;$ are drawn. Chords $AD\;$ and $BC\;$ intersect $PQ\;$ at points $X\;$ and $Y,\;$ respectively.
Then $M\;$ is also the midpoint of $XY.$
We introduce the points $A'\;$ and $D'\,$ that are the symmetric of $A\;$ and $D\;$ about $M,\;$ respectively. Hence, $MA' = MA\;$ and $MD' = MD.\;$ Next we connect the point $D'\;$ to $A'\;$ and $B,\;$ and call $E\;$ the intersection of $DB\;$ with the line through $P\;$ and $Q.\;$ Thus we have constructed triangle $MBD'\;$ with cevians $D'A',\;$ $ME,\;$ and $BC.\;$
We show that the segment $DA\;$ cuts the chord $PQ\;$ at the same point $Y\;$ as $BC,\;$ i.e., that the three cevians are concurrent at $Y.\;$ This property will be proved by applying Ceva’s theorem to triangle $MBD'.\;$
### Lemma
In triangle $MBD,\;$ the cevians $DA,\;$ $ME,\;$ and $BC\;$ are concurrent at $Y.\;$
### Proof of Lemma
We set $MA'=MA=\rho_1,\;$ $MB=r_1,\;$ $MC=r_2\;$ and $MD'=MD=\rho_2.\;$ We observe that $\displaystyle\frac{BE}{ED'}=\frac{r_1\sin\beta}{\rho_2\sin\alpha},\;$ the ratio of respective distances of $B\;$ and $D'\;$ from the line $PQ.\;$ Moreover, $A'B=r_1-\rho_1,\;$ and $D'C=\rho_2-r_2.\;$ Now,
(1)
\displaystyle\begin{align} \frac{BE}{ED'}\cdot\frac{D'C}{CM}\cdot\frac{MA'}{A'B} &= \frac{r_1\sin\beta}{\rho_2\sin\alpha}\cdot\frac{\rho_2-r_2}{r_2}\cdot\frac{\rho_1}{r_1-\rho_1}\\ &=\frac{(\rho_2-r_2)\sin\beta}{(r_1-\rho_1)\sin\alpha}, \end{align}
since $\rho_1r_1=\rho_2r_2\;$ by the Intersecting Chords Theorem.
The differences appearing in (1) can be written in terms of the distance $d = OM\;$ of the circle center $O\;$ to the chord $PQ,\;$ and the angles $\alpha\;$ and $\beta.\;$ The above figure shows that the projection of $OM\;$ onto the chord $PQ\;$ has length $d\sin\alpha,\;$ so that we get $\displaystyle\rho_2=\frac{1}{2}CD+d\sin\alpha,\;$ and $\displaystyle r_2=\frac{1}{2}C-d\sin\alpha.\;$ Hence, $\rho_2-r_2=2d\sin\alpha.\;$ Similarly, we find $r_1-\rho_1=2d\sin\beta.$ Substituting these expressions into (1) we obtain
$\displaystyle\frac{BE}{ED'}\cdot\frac{D'C}{CM}\cdot\frac{MA'}{A'B}=1.$
By Ceva's Theorem, the cevians $D'A',\;$ $ME,\;$ and $BC\;$ are concurrent. The common point is clearly $Y.$
### Proof of Butterfly Theorem
We observe that triangle $MA'D'\;$ is congruent by construction to triangle $MAD,\;$ because two sides of the first $(MA',MD')\;$ are equal to two sides of the second $(MA,MD),\;$ and the included angles are equal. It follows that $\angle MD'Y = \angle MDX.\;$ Consequently, triangles $MD'Y\;$ and $MDX\;$ are also congruent, since they have equal two pairs of angles $(\angle MD'Y = \angle MDX,\;$ and $\angle YMD'= \angle XMD,\;$ vertical angles), as well as the included sides $(MD' = MD).\;$ This congruence implies that the corresponding sides $MY\;$ and $MX\;$ are equal. Therefore, $M\;$ is the midpoint of $XY\;$ and the butterfly theorem is proved.
### Remark
It is rather interesting to compare this proof to William Wallace's 1803 Statement. |
# How to Calculate Percent Difference
Save
The percent difference is a mathematical way of comparing two independent quantities that use the same measurement, the way you may compare the number of candies in two separate bags or the height of two similar objects. The purpose is to find out by what percentage the two measurements differ. To do this, you need to find the difference in measurements, the average of the measurements, and then calculate the percent. If you are using a calculator, you input the data manually on your computer, apply the formula and calculate.
## Find the Difference of the Two Numbers
• In math, difference refers to subtraction. To find the difference, subtract one quantity from the other. For example, if you had two bags of candies, and one had 18 candies and the other bag only had 16 candies, you subtract 16 from 18 to find the difference between the two bags.
18 candies - 16 candies = 2 candies
So, the difference between the two bags of candies is 2.
## Find the Average of the Two Numbers
• The average of two numbers is the quantity that exists halfway between them, such as how the number 3 exists between the numbers 2 and 4, but on a larger scale. Calculating the average of two numbers is simple. You add the two together to find their total sum and then you divide that sum by 2.
Using the previous example, the two bags contained 16 and 18 candies individually. To find the total number of candies in the two bags, you add the two bags together.
16 + 18 = 34
Once you have the total number, you divide that total by 2 to find the average of the two bags.
34 / 2 = 17
This means the average of number of candies is 17.
## Find the Percent Difference
• The percent difference between two numbers is the difference of the two numbers divided by their average. Looking at the previous examples, you know that the difference between the two bags of candies is 2 and the average of the two bags is 17. When you divide the two, you get a decimal. This is the percent difference.
2 / 17 = 0.1176
Once you have the decimal, you convert it to a percentage. To do this, you multiply it by 100. Or, you move the decimal point two spaces to the right.
So,
0.1176 x 100 = 11.76 percent
This means that the percent difference between the bag of 16 candies and the bag of 18 candies is 11.76 percent.
## More Examples
• If one boy stands 63 inches tall and another stands 67 inches tall, what is the percent difference between their heights?
The difference between the two heights is 67 inches - 63 inches = 4 inches
The average of the values is (67 + 63) / 2 = 65 inches
The percent difference between the two heights is 100 x (4 / 65) = 6.15 percent
Another example:
If gas costs \$2.14 at one station and \$2.29 at another, what is the percent difference between the two prices?
The difference between the two gas prices is \$2.29 - \$2.14 = \$.15 cents
The average of the two prices is (2.29 + 2.14) / 2 = 2.215 cents
The percent difference is 100 x (.15 / 2.215) = 6.77 percent
## References
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## Mathematical properties – wikiversity harry mileaf electricity 1 7 pdf
#########
1. Commutative Property of Addition/Multiplication – The word, commutative, from French, commuter (to switch), means move around. Such as in problem, a {\displaystyle a} • b {\displaystyle b} = b {\displaystyle b} • a {\displaystyle a} , the a {\displaystyle a} and the b {\displaystyle b} move around about the equal sign, but nonetheless, still equal to the same sum (in this situation, we can say gas house edwards co c {\displaystyle c} ). Now, let’s implement this on real numbers, such as:
2. Associative Property of Multiplication/Addition – Etymology here: The word, associative, is derived from the word associate, which electricity facts for 4th graders comes from the Latin word, associo, which means to unite together, associate. This property basically reflects the same rules as the Commutative property: No matter the order of the parentheses, you will get the same result. The parenthesis can go wherever you like it to be! So:
3. Identity Property – Specifically the Additive Identity Property and the Multiplicative a gas station near me Identity Property, is when the total (number) does not change throughout the equation. The word, identity, comes from the Latin word for idem, which means the same. So, you can think of it as the number is still the same after the math equation. In 5 {\displaystyle 5} + 0 {\displaystyle 0} = 5 {\displaystyle 5} : The total (number) stayed the same throughout the math e payment electricity bill up problem–It is still 5! For the Additive Identity Property, it is always zero (this works for subtraction as well)–For the Multiplicative Identity Property, it is always one (this works for division c gastronomie vitam as well).
4. Inverse Property, also the Additive Inverse Property and the Multiplicative Inverse Property, state that any number to added to its opposite counterpart ( 5 {\displaystyle 5} and − 5 {\displaystyle -5} ) will equal either zero or one. This zero or one depends on which Inverse Property you are using gas and water company. If you are using the Additive Inverse Property, you should be getting zero. If you are using the Multiplicative Inverse Property, you should be getting one. The Additive Inverse Property works with subtraction as well, and the Multiplicative Inverse Property works with division as well. In the Additive Inverse Property, the was electricity invented during the industrial revolution number must to be added to its negative counterpart, such as 29 {\displaystyle 29} and − 29 {\displaystyle -29} , 6 {\displaystyle 6} and − 6 {\displaystyle -6} , 91 {\displaystyle 91} and − 91 {\displaystyle -91} , 47 {\displaystyle 47} and − 47 {\displaystyle -47} , etc.. In the Multiplicative Inverse Property, the number must be multiplied to its reciprocal, which electricity dance moms choreography is the number found by flipping the numerator and the denomerator of a fraction ( 2 {\displaystyle 2} = 2 1 {\displaystyle {\tfrac {2}{1}}} ), such as 3 4 {\displaystyle {\tfrac {3}{4}}} and 4 3 {\displaystyle {\tfrac {4}{3}}} , 9 2 {\displaystyle {\tfrac {9}{2}}} and 2 9 {\displaystyle {\tfrac {2}{9}}} , 82 93 {\displaystyle {\tfrac {82}{93}}} and 93 82 {\displaystyle {\tfrac {93}{82}}} , 2 {\displaystyle 2} and 1 2 {\displaystyle {\tfrac {1}{2}}} , etc..
5. Property of Zero – Specifically the Multiplicative Property of Zero and the Additive Property of Zero, is when zero plays a role in a math equation. Specifically in the Multiplicative Property of Zero, anything multiplying zero gas works park fireworks is zero. This rule is true no matter what number it is. Big or small, this rule works in every and all cases of problems that follow the Multiplicative Property of Zero. For example:
The Additive Property of Zero is part of the property of zero group and the identity group, which will be explained later in this section. The Additive Property of Zero states that british gas jokes any number adding zero will remain the same (thus, identity): r {\displaystyle r} + 0 {\displaystyle electricity electricity lyrics 0} = r {\displaystyle r} . Here are a few problems to help explain:
6. Substitution property is the act of substituting factors with the answer. An example is − 5 x + 8 + x = − 4 x + 8 {\displaystyle -5x+8+x=-4x+8} . The substitution property simple takes − 5 x {\displaystyle -5x} , and adds it to x {\displaystyle x} , which equals − 4 x {\displaystyle -4x} and the constant: 8 {\displaystyle 8} (since there are no like terms). |
# How do you find the derivative of 6x^(- 7/8)?
Mar 5, 2016
$f ' \left(x\right) = - \frac{21}{4} {x}^{- \frac{15}{8}}$
#### Explanation:
Before we derive this, we shall first remind ourselves of 2 important formulas, $f \left(x\right) = k h \left(x\right) \setminus \implies f ' \left(x\right) = k h ' \left(x\right)$ and $f \left(x\right) = {x}^{n} \setminus \implies f ' \left(x\right) = n {x}^{n - 1}$
Now, we have $f \left(x\right) = 6 {x}^{- \frac{7}{8}}$
From above equations, we take $k = 6$ and $g \left(x\right) = {x}^{- \frac{7}{8}}$
So, $f ' \left(x\right) = 6 \cdot \frac{d \left({x}^{- \frac{7}{8}}\right)}{\mathrm{dx}} \setminus \implies f ' \left(x\right) = - 6 \cdot \frac{7}{8} \cdot {x}^{- \frac{15}{8}}$ |
# 3.02 Equivalent and simplified ratios
Lesson
We previously learned how to write ratios that match the information given to us. Writing a ratio can let us compare things mathematically but has only limited use in solving problems. We can build upon this with the use of equivalent ratios and simplified ratios.
#### Exploration
Consider a cake recipe that uses $1$1 cup of milk and $4$4 cups of flour. What is the ratio of milk to flour used in the cake?
Letting the unit be "the number of cups", we can express the information given as the ratio $1:4$1:4.
What if we want to make two cakes? We would need to double the amount of milk and flour we use. This means we will need $2$2 cups of milk and $8$8 cups of flour. Now the ratio of milk to flour is $2:8$2:8.
But how do we get two different ratios from the same recipe? The secret is that the two ratios actually represent the same proportion of milk to flour. We say that $1:4$1:4 and $2:8$2:8 are equivalent ratios.
Now consider if we wanted to make enough cakes to use up $4$4 cups of milk. How many cakes would this make, and how much flour would we need?
### Equivalent ratios
Equivalent ratios are useful for when we want to change the value of one quantity but also keep it in the same proportion to another quantity. After calculating how much the value of the first quantity has increased, we can increase the value of the second quantity by the same multiple to preserve the ratio.
We saw in the cake example that increasing both the amount of milk and the amount of flour by the same multiple preserved the ratio. That's because this is the same as having multiple sets of the same ratio.
Two cakes require twice the ingredients of one cake, but in the same proportion.
And since this is an equivalence relation, we can also say the same for the reverse:
One cake requires half the ingredients of two cakes, but in the same proportion.
Equivalent ratios
Two ratios are equivalent if one of the ratios can be increased or decreased by some multiple to be equal to the other ratio.
#### Worked examples
##### question 1
The ratio of tables to chairs is $1:2$1:2. If there are $14$14 chairs, how many tables are there?
Think: The ratio $1:2$1:2 says that each table has two chairs. We want to increase both sides of this ratio by some multiple to get the equivalent ratio that looks like $\editable{?}:14$?:14. The missing number in this ratio will be the number of tables needed for $14$14 chairs.
Do: The first thing we can figure out is by what multiple the chairs have been increased. We can find this by dividing $14$14 by $2$2. This tells us how many sets of two chairs there are:
Number of pairs of chairs $=$= $14\div2$14÷2 $=$= $7$7
If there are $7$7 sets of two chairs, and each table has two chairs, then we will need $7\times1=7$7×1=7 tables.
##### question 2
The ratio of players to teams is $60:10$60:10. If there are only $12$12 students present, how many teams can be made?
Think: We only have $12$12 out of $60$60 students. What fraction of the students are present? To preserve the ratio, we want to take the same fraction of the usual $10$10 teams.
Do: The fraction of students present is $12$12 out of $60$60 which we can write as:
Fraction of students present $=$= $12$12 out of $60$60 $=$= $\frac{12}{60}$1260 $=$= $\frac{1}{5}$15
If only one fifth of the students are present, they can only make one fifth of the usual number of teams. So we can make $10\times\frac{1}{5}=2$10×15=2 teams with $12$12 students.
In both of these worked examples we were able to preserve the ratio by either increasing or decreasing the initial ratio by some multiple. It is important to note that whenever we performed an operation we applied it to both sides of the ratio.
Tables to Chairs Students to Teams $1$1 : $2$2 $60$60 : $10$10 $\times7$×7 $\times7$×7 $\div5$÷5 $\div5$÷5 $7$7 : $14$14 $12$12 : $2$2
### Simplified Ratios
A ratio is a simplified ratio if there is no equivalent ratio with smaller integer values. This is the same as saying that the two integers in the ratio have a greatest common factor of $1$1.
Since the simplified ratio is the smallest integer valued ratio, this also means that all the ratios equivalent to it are multiples of it. This makes the simplified ratio very useful for solving equivalent ratio questions that don't have very nice numbers.
Simplified ratio
A ratio is a simplified ratio if there is no equivalent ratio with smaller integer values.
#### Worked example
##### question 3
The ratio of sultanas to nuts in a bag of trail mix is always $32:56$32:56. If there are $12$12 sultanas left, how many nuts are left?
Think: Since $12$12 is not a factor of $32$32, we can first turn $32:56$32:56 into a simplified ratio, then find the equivalent ratio that looks like $12:\editable{?}$12:?.
Do: We can simplify the ratio $32:56$32:56 by decreasing both sides of the ratio until the greatest common factor of the two numbers is $1$1. Use the fact that both $32$32 and $56$56 are divisible by $8$8.
$32:56$32:56 $=$= $\frac{32}{8}:\frac{56}{8}$328:568 $=$= $4:7$4:7
Since $4$4 and $7$7 have no common factors except for $1$1, this is a simplified ratio. So we now know that for every $4$4 sultanas there are $7$7 nuts.
To get from $4$4 sultanas to $12$12 sultanas we multiply by $3$3. But we know the ratio of sultanas and nuts is always $4:7$4:7, which means there must be $7\times3=21$7×3=21 nuts left.
Reflect: We started with the ratio $32:56$32:56, then found the simplified ratio $4:7$4:7, and used this to find the equivalent ratio $12:21$12:21 that was relevant to our problem. Notice that the three ratios are all equivalent, but only $4:7$4:7 is a simplified ratio.
Careful!
• The simplified ratio uses only integers. A ratio that uses fractions or decimals is not yet fully simplified and can be increased or decreased by the appropriate multiple to simplify it.
• If the two quantities are in different units, then we need to convert them to the same unit. For example $1$1 km to $350$350 m, we would need to start as $1000$1000 m to $350$350 m and then be simplified.
The application of equivalent and simplified ratios is useful for when we want to keep things in proper proportion while changing their size, or when we want to measure large objects by considering their ratio with smaller objects.
Did you know?
The ratio of the length of your hand to your height is approximately $1:10$1:10. Try measuring your height using the length of your hand. How accurate is this ratio?
#### Practice questions
##### Question 4
Complete the table of equivalent ratios and use it to answer the following questions.
1. Dogs to Cats
$9$9 : $5$5
$18$18 : $10$10
$27$27 : $\editable{}$
$45$45 : $\editable{}$
$\editable{}$ : $50$50
2. If there are $270$270 dogs, how many cats are there expected to be?
$150$150
A
$30$30
B
$270$270
C
$266$266
D
$150$150
A
$30$30
B
$270$270
C
$266$266
D
3. Which of the following is the fully simplified ratio for $270:150$270:150?
$135:75$135:75
A
dogs$:$:cats
B
$2:1$2:1
C
$9:5$9:5
D
$135:75$135:75
A
dogs$:$:cats
B
$2:1$2:1
C
$9:5$9:5
D
##### Question 5
Write $540$540 cents to $\$3.003.00 as a fully simplified ratio.
##### Question 6
The ratio of students to teachers competing in a charity race is $9:4$9:4. If $54$54 students take part in the race, how many teachers are there?
### Outcomes
#### 6.1
Represent relationships between quantities using ratios, and will use appropriate notations, such as a/b , a to b, and a:b |
Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
Result:
1/5 + 3/4 + 1/2 = 29/20 = 1 9/20 = 1.45
Spelled result in words is twenty-nine twentieths (or one and nine twentieths).
How do you solve fractions step by step?
1. Add: 1/5 + 3/4 = 1 · 4/5 · 4 + 3 · 5/4 · 5 = 4/20 + 15/20 = 4 + 15/20 = 19/20
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(5, 4) = 20. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 5 × 4 = 20. In the next intermediate step, the fraction result cannot be further simplified by canceling.
In words - one fifth plus three quarters = nineteen twentieths.
2. Add: the result of step No. 1 + 1/2 = 19/20 + 1/2 = 19/20 + 1 · 10/2 · 10 = 19/20 + 10/20 = 19 + 10/20 = 29/20
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(20, 2) = 20. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 20 × 2 = 40. In the next intermediate step, the fraction result cannot be further simplified by canceling.
In words - nineteen twentieths plus one half = twenty-nine twentieths.
Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
Fractions in word problems:
• Expressions with variable
This is algebra. Let n represent an unknown number and write the following expressions: 1. 4 times the sum of 7 and the number x 2. 4 times 7 plus the number x 3. 7 less than the product of 4 and the number x 4. 7 times the quantity 4 more than the number
• Lengths of the pool
Miguel swam 6 lengths of the pool. Mat swam 3 times as far as Miguel. Lionel swam 1/3 as far as Miguel. How many lengths did Mat swim?
• Fitness center
Every Wednesday, Monica works out for 3/4 of an hour at the fitness center. Every Saturday, he goes to the fitness center again and exercises for 3 times as long. How much time does Wayne spend at the fitness center in all each week?
Divide number 135 into two additions so that one adds 30 more than 2/5 of the other add. Write the bigger one.
• Recipe ingredients
Monica’s cookie recipe calls for Three-fourths of a cup of flour. Her mother’s recipe calls for Two-thirds as much as Monica’s. How many cups of flour does her mother’s recipe require?
• Metal rod
You have a metal rod that’s 51/64 inches long. The rod needs to be trimmed. You cut 1/64 inches from one end and 1/32 inches from the other end. Next, you cut the rod into 6 equal pieces. What will be the final length of each piece?
• Stones in aquarium
In an aquarium with a length of 2 m, 1.5 m wide, and 2.5 m deep, the water is up to three-quarters of the depth. Can we place 2m cubic meters of stones in the aquarium without spilling water? (0 = no, 1 = yes)
• Walk for exercise
Anya, Jose, Cali, and Stephan walk for exercise. Anya's route is 2 1/4 kilometers long. Jose's route is 1 1/2 fewer km. Cali's route is 1 1/2 times as long as Jose's route, and 2 fewer km than Stephan's route. What distance (S) is Stephan's route?
• Farmer 5
Farmer Joe ordered 3 bags of soil last month. Each bag weighed 4 ⅖ kilograms. He used the first bag in a week. At the end of this month, there were 2 ¾ kilograms of soil left in the second bag and ⅞ kilograms of soil left in the third bag. How much soil w |
# Video: AQA GCSE Mathematics Foundation Tier Pack 4 • Paper 2 • Question 13
Simon draws a hexagon. His hexagon has rotational symmetry of order 2, two lines of symmetry, three pairs of parallel sides, and a perimeter of 32 cm. Draw a sketch that could be the same as Simon’s hexagon. Label the lengths of each side.
05:00
### Video Transcript
Simon draws a hexagon. His hexagon has rotational symmetry of order two, two lines of symmetry, three pairs of parallel sides, and a perimeter of 32 centimeters. Draw a sketch that could be the same as Simon’s hexagon. Label the lengths of each side.
If we consider a regular hexagon, that is a hexagon where all sides and all angles are equal, we know that it has three pairs of parallel sides. It meets one of the requirements for Simon’s hexagon. However, when it comes to rotational symmetry, this regular hexagon has an order of six. To think about rotational symmetry, imagine putting your finger on the yellow dot in the middle of this hexagon and then spinning. Rotational symmetry of order six means that there are six ways you could spin this hexagon. And it would look the same. We’re looking for a hexagon that has rotational symmetry of order two where there are only two places where the shape would look the same if you were spinning it.
If we take a regular hexagon and stretch it in width, we keep the three pairs of parallel sides. But this shape no longer has a rotational symmetry of order six. If you put your finger in the middle of the shape and spin it 180 degrees, it would look the same which means this shape does have rotational symmetry of order two. Now, we need to sketch some lines of symmetry. These are the places we could fold our shape in half. And all the edges would line up evenly. Heres one line of symmetry and another line of symmetry. Now that we’ve shown this shape has a rotational symmetry of order two, two lines of symmetry, and three pairs of parallel sides, we need to label it so that it has a perimeter of 32 centimeters.
The perimeter of a hexagon is equal to adding up all six side lengths. And if the perimeter equals 32, all the sides need to add up to 32. In this hexagon, we know that the top and the bottom must be equal to each other. So let’s say that there are 10 centimeters each. Since the perimeter must measure 32 and we’ve used 20 of those centimeters on the top and bottom, this means that the other four sides must be equal to 12 centimeters when theyre added together. Remember how we started with a regular hexagon and we just stretched its width. This means we know the remaining four sides must be equal to each other. And so we divide 12 by four and we get three. The other four sides need to measure three centimeters. And we know that 32 does equal three plus three plus three plus three plus 10 plus 10. And so our hexagon fits all the requirements.
This is not the only possible option for a correct hexagon. We could take our regular hexagon and compress it. It has three pairs of parallel sides, rotational symmetry of order two, two lines of symmetry. Now, we need to give it a perimeter of 32. If we make the top two centimeters, the bottom equals two centimeters. This means that out of our 32, weve used four. And we have 28 remaining centimeters to divide evenly into our four sides. 28 divided by four is seven. If we label the remaining four sides as seven centimeters, then this hexagon also fits the requirements.
Here is our third option that is a less common hexagon that could still fit all the criteria. It has six sides, three sets of parallel sides, rotational symmetry of order two, and two lines of symmetry. In this case, we could label all the small sides four centimeters and the two longer sides eight centimeters. I know that four times four equals 16. And eight times two equals 16. 16 plus 16 equals 32. And so this third hexagon does have a perimeter of 32 centimeters. You could draw any hexagon as long as it meets these four criteria. |
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# What is the length of a rectangle whose perimeter is 48 cm and breadth is 9 cm?
Everything that surrounds us has a definite shape except for liquids. Starting from our mobile phones, laptops to our cupboards, water bottles, microwave oven, kitchen bowls, water tanks, LPG pipelines, etc. all have a definite shape. Clearly, some of these objects are used to store stuff for us. Such shapes are categorized on certain criteria in Mathematics and studied to either create more shapes like them or to improvise the existing ones by either covering them or decorating them. Furthermore, their storage capacity is also worked upon in mathematics to use them to hold things.
### What is Mensuration?
Mensuration is that branch of mathematics that deals with the computation of dimensions and dimensional quantities pertaining to various geometrical shapes. It is the act of measuring, be its length, breadth, height, area, or any such parameter pertinent to all geometrical shapes we see around us and use everyday.
Basic Terminology
• Two-Dimensional Shape: As the name suggests, such shapes have only two proportions: length and breadth. Rectangle, square, triangle fall under this category.
• Three-Dimensional Shape: Those shapes which have three proportions/ measurements are called three-dimensional shapes. Such proportions are length, breadth, and height. Cube, Cuboid, etc. fall under this category of shapes.
• Area: Such a parameter that determines the region occupied by a shape. It can only be calculated for two- dimensional shapes.
• Perimeter: Adding up the lengths of sides of a 2- D shape yields its perimeter. It is used to depict the length of an object around its sides.
Such a geometrical shape that is formed by joining four points in such a way that at least three of them do not lie in the same straight line is called a quadrilateral. If the points happen to be in the same line, then we would get a ray or a line segment instead of an enclosed shape. Quadrilaterals, as the name suggests, have 4 sides, angles, edges, and vertices since the word ‘quad’ means four. The sum of all the four angles of a quadrilateral must always be 360 degrees. Quadrilaterals are assigned different names on the basis of lengths of their sides as either square, rhombus, rectangle, parallelogram, trapezium, etc.
Rectangle
A quadrilateral whose all angles measure 90 degrees and whose parallel sides are equal is called a rectangle. It is a two- dimensional shape which has only two measurements pertaining to it, i.e., length and breadth. The figure below depicts a rectangle ABCD, whose angles are equal to 90 degrees each and the parallel sides are equal in length.
The sides AB and CD represent the length of the rectangle and AC and BD represent the breadth. Like any other quadrilateral, the sum of interior angles of rectangle ABCD is 360 degrees.
Area of a Rectangle
The area of a rectangle is computed by multiplying both its dimensions, keeping in mind they both are of the same unit. Thus, in the figure above, if the area of rectangle ABCD were to be calculated, it would come out to be (AB × CD) sq. units.
Thus, area of rectangle = l × b sq. units.
Perimeter of a Rectangle
The perimeter of a shape is defined as the sum total of all its sides, like for a triangle, it would be the sum of all three lengths of its sides, for a square it would be the sum of lengths of its four sides. Likewise, in the case of a rectangle, its perimeter would be equal to the sum of lengths of all four of its sides. Since two sides of a rectangle represent its length and the other two its breadth, the perimeter is found using the following formula:
Perimeter of a rectangle = l + l + b + b
= 2l + 2b
### What is the length of a rectangle whose perimeter is 48 cm and breadth is 9 cm?
Solution:
Perimeter of a rectangle = 2l + 2b
⇒ 2l + 2b = 48 cm
⇒ 2l + 2(9) = 48
⇒ 2l = 48 – 18
⇒ 2l = 30
⇒ l = 15
Thus, the length of the rectangle is 15 cm.
### Similar Problems
Question 1. Find the length of a rectangle if its perimeter is 100 cm and breadth is 10 cm.
Solution:
Perimeter of a rectangle = 2l + 2b
⇒ 2l + 2b = 100 cm
⇒ 2l + 2(10) = 100
⇒ 2l = 100 – 20
⇒ 2l = 80
⇒ l = 40
Thus, the length of the rectangle is 40 cm.
Question 2. Find the length of a rectangle if its perimeter is 200 cm and breadth is 40 cm.
Solution:
Perimeter of a rectangle = 2l + 2b
⇒ 2l + 2b = 200 cm
⇒ 2l + 2(40) = 200
⇒ 2l = 200 – 80
⇒ 2l = 120
⇒ l = 60
Thus, the length of the rectangle is 60 cm.
Question 3. Find the length of a rectangle if its perimeter is 30 cm and breadth is 5 cm.
Solution:
Perimeter of a rectangle = 2l + 2b
⇒ 2l + 2b = 30 cm
⇒ 2l + 2(5) = 30
⇒ 2l = 30 – 10
⇒ 2l = 20
⇒ l = 10
Thus, the length of the rectangle is 10 cm.
Question 4. Find the length of a rectangle if its perimeter is 15 cm and breadth is 3 cm.
Solution:
Perimeter of a rectangle = 2l + 2b
⇒ 2l + 2b = 15 cm
⇒ 2l + 2(5) = 15
⇒ 2l = 15 – 10
⇒ 2l = 5
⇒ l = 2.5
Thus, the length of the rectangle is 2.5 cm.
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Related Tutorials |
# How do you solve 4x + y = -9 and 3x + 7y = 37?
Jun 12, 2016
Solve one equation for x in terms of y, then substitute in the result for x in the second equation. Solve second equation for y, then plug in the value for y into either equation to solve for x.
x = -4 y = 7
#### Explanation:
Equations:
1) $4 x + y = - 9$
2) $3 x + 7 y = 37$
If we wished to do so here, we could multiply equation 2 by 4/3, and subtract the result from equation 1 (or subtract equation 1 from the result). However, that will require substantial dealing with fractions. The below method should help us avoid some of that.
a) Solve either equation for one variable.
In this case, the easiest way to do this is to solve (1) for $y$.
$4 x + y = - 9 \implies y = - 4 x - 9$
b) Plug the solution for y into the other equation.
Since from (a) we have $y = - 4 x - 9$...
$3 x + 7 y = 37$
$3 x + 7 \left(- 4 x - 9\right) = 37$
$3 x - 28 x - 63 = 37$
$- 25 x = 100$
$x = - 4$
c) Now that we have a solution for x, we can plug it into either of our initial equations (or into our revised version of equation 1, which put y in terms of x) to obtain a value for y. Below we use the revised equation 1 for simplicity's sake.
$y = - 4 x - 9$
$y = - 4 \left(- 4\right) - 9$
$y = 16 - 9$
$y = 7$
Thus our solution is $\left(x , y\right) = \left(- 4 , 7\right)$ |
The Dot Product
Vector dot product is also called a scalar product because the product of vectors gives a scalar quantity. Sometimes, a dot product is also named as an inner product. In vector algebra, dot product is an operation applied on vectors
The Scalar product or dot product is commutative. When two vectors are operated under a dot product, the answer is only a number. A brief explanation on dot products is given below.
When two vectors are combined under addition or subtraction, the result is a vector. When two vectors are combined using the dot product, the result is a scalar. For this reason, the dot product is often called the scalar product. It may also be called the inner product.
Dot product of two vector
If we have two vectors
then the dot product or scalar product between them is defined as
## Formula for vectors Dot Product
Let be two non zero vectors. Then, the scalar product is denoted byand is defined as the scalar
## Dot Product of Two Parallel Vectors
If two vectors have the same direction or two vectors are parallel to each other, then the dot product of two vectors is the product of their magnitude.
Here, θ = 0 degree
so, cos 0 = 1
Therefore,
## Dot Product of Opposite Vectors
If the two vectors are opposite in direction, then θ=π
Here,
cosπ=−1
Now,
Let us know about some properties of Dot product.
Let
• Commutative Property
• Distributive Property
• Associative Property
• property of magnitude
Let us solve an example using S2 for better understanding.
Q: Determine $$x .y$$ given $$x = \begin{bmatrix}1\\2\\3\\4\end{bmatrix}$$ and $$y = \begin{bmatrix}4\\5\\6\\7\end{bmatrix}$$
%use s2
// define vectors x and y
var x = DenseVector(arrayOf(1.0, 2.0, 3.0, 4.0))
var y = DenseVector(arrayOf(4.0, 5.0, 6.0, 7.0))
// P = x.y
val P = x.multiply(y)
println(P)
[4.000000, 10.000000, 18.000000, 28.000000]
Geometrical Connection:
The dot product $$x . y$$ can also be calculated with the help of the angle $$\theta$$ between these two vectors as follows:
$$x . y =\left|x\right|\left|y\right|cos \theta$$
Let us find the dot product of the vectors in the figure below:
The Transpose
The transpose of a matrix is one of the most commonly used methods in matrix transformation. For a given matrix, the transpose of a matrix is obtained by interchanging rows into columns or columns to rows. In this article, we are going to learn the definition of the transpose of a matrix, steps to find the transpose of a matrix, properties and examples with a complete explanation.
The transpose of a matrix is found by interchanging its rows into columns or columns into rows. The transpose of the matrix is denoted by using the letter “T” in the superscript of the given matrix. For example, if “A” is the given matrix, then the transpose of the matrix is represented by A’ or AT.
Definition: If $$A$$ is an $$m \times n$$ matrix, then its transpose $$A^T$$ is defined to be the matrix with $$n$$ rows whose $$i^{th}$$ row is equal to the $$i^{th}$$ column of $$A$$, for each $$i$$ from $$1$$ to $$n$$.
If $$A = \begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}$$, then $$A^T = \begin{bmatrix}1 & 4\\2 & 5\\3 & 6\end{bmatrix}$$
Let us implement the same using S2.
%use s2
// define a matrix
var A = DenseMatrix(arrayOf(
doubleArrayOf(1.0, 2.0, 3.0),
doubleArrayOf(4.0, 5.0, 6.0)))
println(A)
// B = Tranpose of A
val B = A.t()
println("\n\nTranspose of A:\n")
println(B)
2x3
[,1] [,2] [,3]
[1,] 1.000000, 2.000000, 3.000000,
[2,] 4.000000, 5.000000, 6.000000,
Transpose of A:
3x2
[,1] [,2]
[1,] 1.000000, 4.000000,
[2,] 2.000000, 5.000000,
[3,] 3.000000, 6.000000,
Matrix
When we take only the real constants of a linear equation or a system of linear equations and make an array then it becomes a matrix. There are different types of matrix which are identified and defined based on the structure consisting of elements in different ways and existing in different positions. Writing the real constants of a linear equation or a system of linear equations in a rectangular array is a matrix. All the elements of a matrix are called entries [2,3].
Let the following system of linear equations
Then the matrix of the above system of linear equations is
Matrix Properties
Now that we know what the transpose of a matrix is, let’s learn about various matrix properties and their illustrations in S2.
Symmetric Matrix
A symmetric matrix in linear algebra is a square matrix that remains unaltered when its transpose is calculated. That means, a matrix whose transpose is equal to the matrix itself, is called a symmetric matrix.
A matrix in which its $$(i,j)^{th}$$ entry will be necessarily equal to its $$(j,i)^{th}$$ entry is known as a Symmetric Matrix.
In other words, if $$A$$ is an $$n \times n$$ matrix satisfying the equation $$A = A^T$$, we say that $$A$$ is symmetric.
%use s2
// define a matrix
var A = DenseMatrix(arrayOf(
doubleArrayOf(1.0, 2.0, 3.0),
doubleArrayOf(2.0, 4.0, 5.0),
doubleArrayOf(3.0, 5.0, 6.0)))
println(A)
print("\n")
//prints 'true' if A is symmetric, else prints 'false'
val precision = 1e-15
println(MatrixPropertyUtils.isSymmetric(A,precision))
3x3
[,1] [,2] [,3]
[1,] 1.000000, 2.000000, 3.000000,
[2,] 2.000000, 4.000000, 5.000000,
[3,] 3.000000, 5.000000, 6.000000,
true
Skew-Symmetric Matrix
A skew symmetric matrix is defined as the square matrix that is equal to the negative of its transpose matrix. For any square matrix, A, the transpose matrix is given as AT. A skew-symmetric or antisymmetric matrix A can therefore be represented as, A = -AT
If $$A$$ is an $$n \times n$$ matrix satisfying the equation $$A^T = -A$$, we say $$A$$ is Skew-Symmetric.
%use s2
// define a matrix
var A = DenseMatrix(arrayOf(
doubleArrayOf(0.0, -6.0, 4.0),
doubleArrayOf(6.0, 0.0, -5.0),
doubleArrayOf(-4.0, 5.0, 0.0)))
println(A)
print("\n")
//prints 'true' if A is skew-symmetric, else prints 'false'
val precision = 1e-15
println(MatrixPropertyUtils.isSkewSymmetric(A,precision))
3x3
[,1] [,2] [,3]
[1,] 0.000000, -6.000000, 4.000000,
[2,] 6.000000, 0.000000, -5.000000,
[3,] -4.000000, 5.000000, 0.000000,
true
Idempotent Matrix
An idempotent matrix is a matrix which, when multiplied by itself, yields itself. That is, the matrix is idempotent if and only if . For this product to be defined, it must necessarily be a square matrix. Viewed this way, idempotent matrices are idempotent elements of matrix rings.
If $$A$$ is an $$n \times n$$ matrix satisfying the equation $$A = A*A$$ or $$A = A^2$$, we say $$A$$ is Idempotent.
%use s2
// define a matrix
var A = DenseMatrix(arrayOf(
doubleArrayOf(2.0, -2.0, -4.0),
doubleArrayOf(-1.0, 3.0, 4.0),
doubleArrayOf(1.0, -2.0, -3.0)))
println(A)
//prints 'true' if A is idempotent, else prints 'false'
val precision = 6.0
println("\n")
println(MatrixPropertyUtils.isIdempotent(A,precision))
3x3
[,1] [,2] [,3]
[1,] 2.000000, -2.000000, -4.000000,
[2,] -1.000000, 3.000000, 4.000000,
[3,] 1.000000, -2.000000, -3.000000,
true
Orthogonal Matrix
A square matrix with real numbers or elements is said to be an orthogonal matrix if its transpose is equal to its inverse matrix. Or we can say when the product of a square matrix and its transpose gives an identity matrix, then the square matrix is known as an orthogonal matrix
1If $$A$$ is an $$n \times n$$ matrix satisfying the equation $$A.A^T = A^T.A = I$$ where $$I$$ is an Identity Matrix of order $$n$$, we say $$A$$ is Orthogonal.
%use s2
// define a matrix
var A = DenseMatrix(arrayOf(
doubleArrayOf(1.0, 0.0, 0.0),
doubleArrayOf(0.0, 0.0, -1.0),
doubleArrayOf(0.0, -1.0, 0.0)))
println(A)
//prints 'true' if A is orthogonal, else prints 'false'
val precision = 1e-15
println("\n")
println(MatrixPropertyUtils.isOrthogonal(A,precision))
3x3
[,1] [,2] [,3]
[1,] 1.000000, 0.000000, 0.000000,
[2,] 0.000000, 0.000000, -1.000000,
[3,] 0.000000, -1.000000, 0.000000,
true |
### Lesson 25: I can use basic facts to approximate
```Lesson 29: I can solve division word
problems involving multi-digit
division with group size unknown
and the number of groups
unknown.
5th Grade Module 2 – Lesson 29
Unit Conversions
.013
=g=____
gmL
liter
=___
___
mL
inches
0.052
.397
1foot
lb
=kg
oz
____
0.413
1 1kg
=liters
___
5th Grade Module 2 – Lesson 29
Divide Decimals by Two-Digit
Numbers
8.61
4.9
.32
8.4
24 ÷÷÷16
14
40
21===____
____
____
5th Grade Module 2 – Lesson 29
Application Problem
A one-year (52 week) subscription to a weekly
magazine is \$39.95. Greg calculates that he would
save \$219.53 if he subscribed to the magazine instead
of purchasing it each week at the store. What is the
price of the individual magazine at the store?
5th Grade Module 2 – Lesson 29
Application Problem
A one-year (52 week) subscription to a weekly
magazine is \$39.95. Greg calculates that he would
save \$219.53 if he subscribed to the magazine instead
of purchasing it each week at the store. What is the
price of the individual magazine at the store?
219.53 + 39.95 = 259.48
259.48 ÷ 52 = 4.99
5th Grade Module 2 – Lesson 29
Concept Development
Lamar has 1354.5 kilograms of potatoes to deliver in
equal amounts to 18 stores. 12 of the stores are in the
Bronx. How many kilograms of potatoes will he deliver
to stores in the Bronx?
Will the amount delivered to the
Bronx be more or less than half of the
total amount? How do you know?
5th Grade Module 2 – Lesson 29
1354.5
12 stores in Bronx
1354.5 ÷ 18 = 75.25
75.25 x 12 = 903 kg
The stores in the Bronx receive 903 kg of potatoes.
5th Grade Module 2 – Lesson 29
Concept Development
Valerie uses 12 oz of detergent each week for her
laundry. If there are 75 oz of detergent in the bottle, in
how many weeks will she need to buy a new bottle of
detergent?
Solution on next slide…
5th Grade Module 2 – Lesson 29
75 oz
12
?
12
How many 12’s would fit in 75?
Valerie will have to buy detergent after 6 weeks
because she won’t have enough for a 7th week.
5th Grade Module 2 – Lesson 29
Concept Development
Jim Nasium is building a tree house for his two
daughters. He cuts 12 pieces of wood from a board
that is 128 inches long. He cuts 5 pieces that measure
15.75 inches each, and 7 pieces evenly cut from what
is left. Jim calculates that due to the width of his
cutting blade, he will lose a total of 2 inches of wood
after making all the cuts. What is the length of each of
the seven pieces?
Solution on next slide…
5th Grade Module 2 – Lesson 29
128 inches
15.75
15.75
15.75
15.75
15.75
2
7 pieces from what’s left
Each of the 7 pieces is 6.75 inches long.
5th Grade Module 2 – Lesson 29
Concept Development
The area of a rectangle is 56.96 square meters. If the
length is 16 m, what is its perimeter?
Solution on next slide…
5th Grade Module 2 – Lesson 29
16 x ? = 56.96
16
?
Area = 56.96
16
The perimeter is 39.12 meters.
5th Grade Module 2 – Lesson 29
?
Concept Development
A city block is 3 times as long as it is wide. If the total
distance around the block is 0.48 kilometers, what is
the area of the block in square meters?
Solution on next slide…
5th Grade Module 2 – Lesson 29
One side
60
width
60
60
60
.48 km
480 m
length
One side
One side
One side
480 ÷ 8 = 60 m per unit.
60 x 180 = 10,800.
The area of the block is 10,800 square meters.
5th Grade Module 2 – Lesson 29
Complete Pages 2.H.21 & 2.H.22
You will have 15 minutes to work.
5th Grade Module 2 – Lesson 29
• What did the divisor represent in each
equation?
• What did the unknown represent for each?
• How did that change the model you drew?
• Which is easier to draw?
5th Grade Module 2 – Lesson 29
EXIT
TICKET
Page 2.H.23
5th Grade Module 2 – Lesson 29
``` |
# Question 6, Exercise 10.2
Solutions of Question 6 of Exercise 10.2 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
Use the half angle identities to evaluate exactly $\cos {{15}^{\circ }}$.
Because ${{15}^{\circ }}=\dfrac{{{30}^{\circ }}}{2}$, and $\dfrac{\theta }{2}=\dfrac{{{30}^{\circ }}}{2}$, we can find $\cos {{15}^{\circ }}$by using half angle identity as, \begin{align}\cos {{15}^{\circ }}&=\cos \dfrac{{{30}^{\circ }}}{2}=\sqrt{\dfrac{1+\cos {{30}^{\circ }}}{2}}\\ &=\sqrt{\dfrac{1+\dfrac{\sqrt{3}}{2}}{2}}=\dfrac{\sqrt{2+\sqrt{3}}}{2}\end{align}
Use the half angle identities to evaluate exactly $\tan {{67.5}^{\circ }}$.
Because ${{67.5}^{\circ }}=\dfrac{{{135}^{\circ }}}{2}$, the $\dfrac{\theta }{2}=\dfrac{{{135}^{\circ }}}{2}$, so we can find $\tan {{67.5}^{\circ }}$by using half angle identity as, \begin{align}\tan {{67.5}^{\circ }}&=\tan \dfrac{{{135}^{\circ }}}{2}=\sqrt{\dfrac{1-\cos {{135}^{\circ }}}{1+\cos {{135}^{\circ }}}}\\ &=\sqrt{\dfrac{1-\left( -\dfrac{1}{\sqrt{2}} \right)}{1+\left( -\dfrac{1}{\sqrt{2}} \right)}}=\sqrt{\dfrac{1+\dfrac{1}{\sqrt{2}}}{1-\dfrac{1}{\sqrt{2}}}}\\ &=\sqrt{\dfrac{\sqrt{2}+1}{\sqrt{2}-1}}=\sqrt{\dfrac{\sqrt{2}+1}{\sqrt{2}-1}\times \dfrac{\sqrt{2}+1}{\sqrt{2}+1}}\\ &=\sqrt{\dfrac{2+1+2\sqrt{2}}{2-1}}=\sqrt{3+2\sqrt{2}}\end{align}
Use the half angle identities to evaluate exactly $sin{{112.5}^{\circ }}$.
Because ${{112.5}^{\circ }}=\dfrac{{{225}^{\circ }}}{2}$, the $\dfrac{\theta }{2}=\dfrac{{{225}^{\circ }}}{2}$, so we can find $sin{{112.5}^{\circ }}$by using half angle identity as, \begin{align} sin{{112.5}^{\circ }}&=\sin \dfrac{{{225}^{\circ }}}{2}=\sqrt{\dfrac{1-\cos {{225}^{\circ }}}{2}}\\ &=\sqrt{\dfrac{1-\left( -\dfrac{1}{\sqrt{2}} \right)}{2}}=\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}}\\ &=\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}}=\dfrac{\sqrt{2+\sqrt{2}}}{2}\end{align}
Use the half angle identities to evaluate exactly $\cos \dfrac{\pi }{8}$.
Because $\dfrac{\pi }{8}=\dfrac{\dfrac{\pi }{4}}{2}$, the $\dfrac{\theta }{2}=\dfrac{\dfrac{\pi }{4}}{2}$, so we can find $\cos \dfrac{\pi }{8}$by using half angle identity as, \begin{align} \cos \dfrac{\pi }{8}&=\cos \dfrac{\dfrac{\pi }{4}}{2}=\sqrt{\dfrac{1+\cos \dfrac{\pi }{4}}{2}}\\ &=\sqrt{\dfrac{1+\dfrac{1}{\sqrt{2}}}{2}} =\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}}\\ &=\dfrac{\sqrt{2+\sqrt{2}}}{2} \end{align}
Use the half angle identities to evaluate exactly $\tan {{75}^{\circ }}$.
Because ${{75}^{\circ }}=\dfrac{{{150}^{\circ }}}{2}$, the $\dfrac{\theta }{2}=\dfrac{{{150}^{\circ }}}{2}$ lies in first quadrant, so we can find $\tan {{75}^{\circ }}$by using half angle identity as, \begin{align}\tan {{75}^{\circ }}&=\tan \dfrac{{{150}^{\circ }}}{2}\\ &=\sqrt{\dfrac{1-\cos {{150}^{\circ }}}{1+\cos {{150}^{\circ }}}}\\ &=\sqrt{\dfrac{1-\left( -\dfrac{\sqrt{3}}{2} \right)}{1+\left( -\dfrac{\sqrt{3}}{2} \right)}}\\ &=\sqrt{\dfrac{1+\dfrac{\sqrt{3}}{2}}{1-\dfrac{\sqrt{3}}{2}}}\\ &=\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}=}\sqrt{7+2\sqrt{3}}\end{align}
Use the half angle identities to evaluate exactly $\sin \dfrac{5\pi }{12}$.
Because $\dfrac{5\pi }{12}=\dfrac{\dfrac{5\pi }{6}}{2}$, the $\dfrac{\theta }{2}=\dfrac{\dfrac{5\pi }{6}}{2}$, so we can find $\sin \dfrac{5\pi }{12}$by using half angle identity as, \begin{align} \sin \dfrac{5\pi }{12}&=\sin \dfrac{\dfrac{5\pi }{6}}{2}=\sqrt{\dfrac{1-\cos \dfrac{5\pi }{12}}{2}}\\&=\sqrt{\dfrac{1-\left( -\dfrac{\sqrt{3}}{2} \right)}{2}}\\ &=\sqrt{\dfrac{1-\left( -\dfrac{\sqrt{3}}{2} \right)}{2}}\\ &=\sqrt{\dfrac{1+\dfrac{\sqrt{3}}{2}}{2}}=\dfrac{\sqrt{2+\sqrt{3}}}{2}\end{align} |
# 3 Pythagoras' Theorem
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## Transcription
1 3 Pythagoras' Theorem 3.1 Pythagoras' Theorem Pythagoras' Theorem relates the length of the hypotenuse of a right-angled triangle to the lengths of the other two sides. Hypotenuse The hypotenuse is always the longest side: it is always the side opposite the right angle. The diagram opposite shows a right-angled triangle. The length of the hypotenuse is and the other two sides have lengths 3 cm and 4 cm. 3 cm 4 cm In this diagram, a square, A, has been drawn on the 3 cm side. Area of square A = 3 3 = 9 cm 2 In this diagram, a second square, B, has been drawn on the 4 cm side. Area of square B = 4 4 = 1 2 Squares A and B together have total area: Area A + Area B = = A A 4 cm B Finally, a third square, C, has been drawn on the side. Area of square C = 5 5 = A C We can see that Area A + Area B = Area C. B This formula is always true for right-angled triangles. 45
2 3.1 MEP Y8 Practice Book A We now look at a right-angled triangle with sides a, b and c, as shown opposite. Area A = a a = a 2 Area B = b b = b 2 Area C = c c A a c b C So, = c 2 Area A + Area B = Area C B gives us the formula 2 a + b = c for all right-angled triangles. Pythagoras' Theorem states that, for any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the two shorter sides. Hypotenuse If we use the letters a, b and c for the sides of a right-angled triangle, then Pythagoras' Theorem states that a 2 + b = c c b where c is the length of the hypotenuse. a Eample 1 Verify Pythagoras' Theorem for the right-angled triangle opposite: 9 cm 41 cm 40 cm 46
3 Here a = 9 cm, b = 40 cm, c = 41 cm. a a 2 = 9 2 = 9 9 = 81 b 2 = 40 2 = = b = 1681 c 2 = 41 2 = = So a + b = c for this triangle. Eercises 1. Which side is the hypotenuse in each of the following right angled triangles: (a) P X Y R Q Z (c) J K (d) R L T S 2. For each of the three diagrams at the top of the net page: (i) calculate the area of square A, (ii) calculate the area of square B, (iii) calculate the sum of area A and area B, (iv) calculate the area of square C, (v) check that : area A + area B = area C 47
4 3.1 MEP Y8 Practice Book A (a) C C A 13 cm 12 cm 17 cm A 1 B B (c) C 61 cm 60 cm B 11 cm A 3. Using the method shown in Eample 1, verify Pythagoras' Theorem for the right-angled triangles below: (a) 9 cm 26 m (c) 85 mm 10 m 24 m 1 12 cm 84 mm 13 mm 4. The whole numbers 3, 4, 5 are called a Pythagorean triple because = 5. A triangle with sides of lengths 3 cm, 4 cm and is rightangled. Use Pythagoras' Theorem to determine which of the sets of numbers below are Pythagorean triples: (a) 15, 20, 25 10, 24, 26 (c) 11, 22, 30 (d) 6, 8, 9 48
5 3.2 Calculating the Length of the Hypotenuse Pythagoras' Theorem states that, for a right-angled triangle, 2 c = a + b With this result it is very easy to calculate the length of the hypotenuse of a right-angled triangle. Eample 1 Calculate the length of the hypotenuse of a triangle in which the other two sides are of lengths 7 m and 8 m. h c a b 7 m Let h be the length of the hypotenuse. By Pythagoras' Theorem, h 2 = h 2 = h 2 = 113 h = 113 h = m h = 10.6 m, correct to 1 decimal place 8 m Eample 2 Calculate the length of the diagonals of the rectangle opposite: 1 The diagram shows the right-angled triangle that you need to use to find the length of the diagonal. The hypotenuse is the diagonal of the rectangle and this is labelled d on the diagram. d 1 49
6 3.2 MEP Y8 Practice Book A By Pythagoras' Theorem, d 2 = = = 320 d = 320 d = cm d = 17.9 cm, correct to 1 decimal place Eercises 1. Calculate the length of the hypotenuse of each of these triangles: (a) 15 mm 36 mm (c) 9 cm 40 cm (d) 1 30 cm 2. Calculate the length of the hypotenuse of each of the following triangles, giving your answers correct to 1 decimal place. (a) 10 cm (c) (d) 4 m 3.5 m 50
7 3. A rectangle has sides of lengths and 10 cm. How long is the diagonal of the rectangle? 4. Calculate the length of the diagonal of a square with sides of length. 5. The diagram shows a wooden frame that is to be part of the roof of a house: Q (a) (c) Use Pythagoras' Theorem in triangle PQR to find the length PQ. Calculate the length QS. Calculate the total length of wood needed to make the frame. P 3.5 m 3 m R 4 m S 6. An isosceles triangle has a base of length 4 cm and perpendicular height. Giving your answers correct to 1 decimal place, calculate: (a) the length, cm, of one of the equal sides, the perimeter of the triangle. 4 cm 7. One end of a rope is tied to the top of a vertical flagpole of height 5.2 m. When the rope is pulled tight, the other end is on the ground 3.8 m from the base of the flagpole. Calculate the length of the rope, giving your answer correct to 1 decimal place. 5.2 m 3.8 m 8. A rectangular lawn is 12.5 m long and 8 m wide. Matthew walks diagonally across the lawn from one corner to the other. He returns to the first corner by walking round the edge of the lawn. How much further does he walk on his return journey? 9. Which of the rectangles below has the longer diagonal? 3 cm Rectangle A 3. Rectangle B 4. 51
8 3.2 MEP Y8 Practice Book A cm 10 cm (a) Use Pythagoras' Theorem to show that the length of the hypotenuse of this triangle is 10.0 cm correct to 1 decimal place. Maine says that this triangle is isosceles because there are two sides of the same length. 1 cm Is Maine correct? 10.0 cm 10 cm 3.3 Calculating the Lengths of Other Sides Eample 1 Calculate the length of the side marked in the following triangle: 2 24 cm By Pythagoras' Theorem: = = = = 100 = 100 = 10 The length of the side is 10 cm. 52
9 Eample 2 Calculate the perpendicular height of the isosceles triangle shown opposite: The height can be calculated by using half of the original isosceles triangle, as shown: The height has been labelled h on the diagram. By Pythagoras' Theorem: 4 cm h h = cm h = 36 h 2 = 36 4 h 2 = 32 h = 32 h = The perpendicular height of the triangle is 5.7 cm to 1 decimal place. Eercises 1. Calculate the length of the side marked in each of the following triangles: (a) 20 cm 50 cm 30 cm 12 cm (c) (d) 51 cm
10 3.3 MEP Y8 Practice Book A 2. Calculate the length of the side marked in each of the following triangles, giving your answer correct to 1 decimal place: (a) 12 cm 20 cm 10 cm (c) 11 cm (d) 3. Calculate the perpendicular height of this equilateral triangle, giving your answer correct to 1 decimal place. 4 cm 4 cm h 4 cm 4. Calculate the perpendicular height of an equilateral triangle with sides of length, giving your answer correct to 1 decimal place. 5. Calculate the perpendicular height of the isosceles triangle shown opposite, giving your answer correct to 1 decimal place. h 6. The width of a rectangle is and the length of its diagonal is 13 cm. (a) How long is the other side of the rectangle? What is the area of the rectangle? 7. The isosceles triangle at the top of the net page has 2 sides of length cm. Copy and complete the calculation to find the value of correct to 1 decimal place. 54
11 10 cm By Pythagoras' Theorem, + = = = = = = to 1 decimal place. 8. The length of the diagonal of a square is. How long are the sides of the square? 9. The diagram shows part of the framework of a roof. (a) Calculate the length XZ. 5 m X 4 m Calculate the length of YZ, correct to 1 decimal place. W 4 m Z Y 10. A sheet is stretched over a washing line to make a tent, as shown in the diagram. (a) How high is the washing line above the ground? Give your answer to Washing line 1 decimal place. If the same sheet was used and the washing line was now at a height of 1.25 m above the ground, what would be the width of the base of 2 m 2 m the tent? Give your answer correct 3 m to 1 decimal place. 11. A fishing rod is used to catch plastic ducks in a fairground game. The rod is 1 m long. A string with a ring is tied to the end of the rod. The length of the string is 0.4 m. When the ring is level with the lower end of the rod, as shown in the diagram, how far is the ring from that end of the fishing rod? 1 m Rod String 0.4 m Ring 55
12 3.4 Problems in Contet When we use Pythagoras' Theorem to solve problems in contet the first key step is to draw a right-angled triangle. Eample 1 A ladder is 5 m long. The bottom of the ladder is 2 m from the foot of a wall and the top leans against the wall. How high is the top of the ladder above the ground? The first step is to draw a triangle to represent the situation. The height to the top of the ladder has been labelled h. (We assume that the ground is horizontal and the wall is vertical.) Now use Pythagoras' Theorem: h = 5 2 h = 25 h 2 = 25 4 = 21 h = 21 h = m 2 m h The top of the ladder is 4.58 m above the ground (to the nearest cm). Eample 2 A ship sails 300 km due west and then 100 km due south. At the end of this journey, how far is the ship from its starting position? The first step is to draw a diagram showing the ship's journey. The distance from the starting point has been labelled d. N 300 km Start W E 100 km d Now use Pythagoras Theorem: S End d 2 = d 2 = d 2 =
13 d = d = The distance from the starting point is 316 km to the nearest km. Eample 3 Calculate the area of the triangle shown opposite: 10 cm The length of the unknown side has been marked. Using Pythagoras' Theorem, = = = = 64 = 64 = 10 cm Area of the triangle = = 1 2 base perpendicular height = 24 cm 2 10 cm Eercises 1. A hiker walks 300 m due north and then 400 m due east. How far is the hiker now from her starting position? 2. A ladder of length 4 m leans against a wall so that the top of the ladder is 3 m above ground level. How far is the bottom of the ladder from the wall? 3. Two remote-controlled cars set off from the same position. After a short time one has travelled 20 m due north and the other 15 m due east. How far apart are the two cars? 57
15 3.5 Constructions and Angles The formula for Pythagoras' Theorem can be used to decide if a triangle is right-angled. In any triangle, the longest side faces the largest angle the shortest side faces the smallest angle. In a triangle with longest side c, and other two sides a and b, 2 if c = a + b, then the angle opposite c = 90 ; 2 if c < a + b, then the angle opposite c < 90 (so all three angles are acute); 2 if c > a + b, then the angle opposite c > 90 (i.e. the triangle has an obtuse angle). Eample 1 (a) Use a ruler and a pair of compasses to construct this triangle: 3 cm (c) 4 cm Use a protractor to check that the triangle has a right angle. Confirm that Pythagoras' Theorem is true for this triangle. (a) First draw a line with length 4 cm. Then draw an arc of radius 3 cm with centre on the left-hand end of the line. Net draw an arc of radius with centre on the right-hand end of the line. The point where the two arcs cross is the third corner of the triangle. 4 cm 59
16 3.5 MEP Y8 Practice Book A The angle at the bottom left-hand corner measures 90, so the triangle has a right angle cm (c) Here a = 4 cm, b = 3 cm and c =. 4 cm a + b = = = 25 c 2 = 5 2 = 25 Therefore 2 a + b = c So Pythagoras' Theorem is true in this case, confirming that this is a rightangled triangle. Eample 2 Which of these triangles contains a right angle? (a) (c) 5 m 13 m 7 cm 14 cm 12 m 11 cm We use Pythagoras' Theorem to find out if a triangle is right-angled, using c for the longest side. (a) In this triangle, a = 5, b = 12 and c = 13. a + b = c 2 = 13 2 = = 169 = Here a + b = c, so this triangle does contain a right angle. 60
17 In this triangle, a = 6, b = 7 and c = 8. a + b = c 2 = 8 2 = = 64 = 85 2 Here c a + b, so the triangle does not contain a right angle. As 2 c < a + b, the angle opposite c is less than 90, so all the angles in this triangle are acute. (c) Here a = 6, b = 11and c = 14. a + b = c 2 = 14 2 = = 196 = Here c a + b, so the triangle does not contain a right angle. As 2 c > a + b the angle opposite c is greater than 90, so the triangle contains one obtuse angle. Eercises 1. (a) Using a ruler and a pair of compasses, construct a triangle with sides of lengths, and 10 cm. Use a protractor to measure the angles of your triangle. (c) Is the triangle right-angled? (d) Use Pythagoras' Theorem to decide whether the triangle is right-angled. (e) Was your answer to part (c) correct? 2. Repeat question 2 for a triangle with sides of lengths 7 cm, and 11 cm. 3. Decide which of the triangles described below: (a) is right-angled, contains an obtuse angle, (c) contains all acute angles. In each case, show how you reached your conclusion. (i) a triangle with sides of lengths 10 cm, 11 cm and 14 cm (ii) a triangle with sides of lengths 10 cm, 12 cm and 1 (iii) a triangle with sides of lengths 9 cm, 12 cm and 1 61
18 3.5 MEP Y8 Practice Book A 4. (a) Use an accurate construction to find out if a triangle with sides of lengths, 7 cm and 12 cm contains a right angle. Use Pythagoras' Theorem to check your answer to part (a). 5. Ahmed draws a square with sides of length. He then measures a diagonal as 8.2 cm. Use Pythagoras' Theorem to decide if Ahmed has drawn the square accurately. 8.2 cm 6. An isosceles triangle has 2 sides of length. The length of the base is 9 cm. Decide, by calculation, whether the angle θ is a right angle, an acute angle or an obtuse angle. Show clearly how you reached your conclusion. θ 9 cm 7. Measure the lengths of the sides and diagonal of your tetbook. Use your measurements to decide whether the corners of your book are right-angled. 8. A triangle has sides of lengths 21 cm, 2 and cm. (a) Show that the triangle has a right angle if = 35. For what values of will the triangle contain an obtuse angle? 9. An isosceles triangle is known to have one side of length 1 and one side of length 2. (a) Eplain why the triangle cannot contain a right angle, Show, by calculation, that it is possible for the triangle to contain three acute angles. Draw a sketch of the triangle in this case. 10. A right-angled triangle has two sides of lengths 24 cm and 32 cm. Use Pythagoras' Theorem to calculate the length of the other side. [Note: there are 2 possible answers.] 62
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# What Is 1/48 as a Decimal + Solution With Free Steps
The fraction 1/48 as a decimal is equal to 0.0208.
The division is one of the four basic arithmetic operators. The division between two numbers is often represented by fractions. The solution of such fractions can either be a decimal or an integer.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 1/48.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 1
Divisor = 48
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 1 $\div$ 48
This is when we go through the Long Division solution to our problem. Figure 1 shows the solution for the fraction 1/48.
Figure 1
## 1/48 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 1 and 48, we can see how 1 is Smaller than 48, and to solve this division, we require that 1 be Bigger than 48.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
As dividend 1 is smaller than the divisor 48 we multiply it by 10. After multiplying it becomes 10 and still, it is smaller than the denominator, so we again multiply it by 10. This multiplication requires an extra zero in the quotient.
Now, we begin solving for our dividend which is 100.
We take this 100 and divide it by 48; this can be done as follows:
100 $\div$ 48 $\approx$ 2
Where:
48 x 2 = 96
This will lead to the generation of a Remainder equal to 100 – 96 = 4. Now this means we have to repeat the process by Converting the 4 into 40, but it is still smaller than the divisor which is 48, so it is multiplied by 10 again to make it 400. Also, an extra zero is added to the quotient. Now the new dividend is 400 and solving for that:
400 $\div$ 48 $\approx$ 8
Where:
48 x 8 = 384
Finally, we have a Quotient generated after combining the three pieces of it as 0.0208, with a Remainder equal to 16.
Images/mathematical drawings are created with GeoGebra. |
Statistical Methods - Lecture 8 HcWjnyVHiTd8hN_8STvJ2rWaXvhPz4wXYCNGvD4qDkU
# The t TestsLecture 7
## The t Distribution
1. The t-tests are similar to the z-tests
1. However, you assumed you knew the population variance
2. In reality, the variance has to be estimated too!
3. Switch to the t distribution
1. The t-distribution is shorter and fatter because you estimated two parameters, the mean and the variance
2. The Rule of Thumb
1. If the observations are less than 30, then use the t-distribution
2. If the number of observations are equal to or greater than 31, then use the z-distribution as an approximation
1. Example
1. You survey 30 people in Almaty. The average income, = \$600 per month and variance, = 10,000
2. Find the 95% Confidence Interval
1. Use an a = 0.05 and df = 30 – 1 = 29
2. Using Excel, = tinv(a, df)
3. tc = 2.04523
4. If this was a normal distribution, then zc = 1.96
5. The standard error (SE) is
1. The 95% Confidence Interval is
1. There is a 95% chance that the true population mean lies between [562.5, 637.5]
## Testing the Means between Two Samples
1. Testing the Difference of the means of two samples
1. This is more complicated because you are estimating the variance
2. Two methods
1. If the variances are equal, then pool the variances
2. If the variances are unequal, then use a different method to pool the variance
3. Assume the variances are equal
4. Example
1. You survey 80 people at Mega Center
1. The average income is = \$800 per month
2. The estimated variance is = 10,000
2. You survey 60 people at Thieves’ Market
1. The average income is = \$500 per month
2. The estimated variance is = 2,500
3. Note – you should test data to determine if data is normally distributed
4. Variance is calculated at
Variance in Sample 1 (n1 – 1) (80 -1)(10,000) 790,000 Variance in Sample 2 (n2 – 1) (60 – 1)(2,500) 147,500 Total Variance 937,500
1. Total degrees of freedom = n1 – 1 + n2 – 1 = 80 + 60 – 2 = 138
2. The pooled variance is
1. The standard error is
1. The t-statistic is
1. If a = 0.05, then the tc = 1.977304
2. The p-value is 5.01 X 10-15
3. The hypothesis test is
1. Reject the H0 and conclude the population means are different
1. Can use a Confidence Interval for hypothesis test
• If a = 0.05, df = 138, and tc=1.977304
1. Assume variances are unequal
1. Same example
2. The =10,000, n1 = 80, =2,500, and n2 = 60
1. However, we have to adjust the degrees of freedom
1. Round the degrees of freedom to 122
2. The t-statistic is
1. Reject the H0 and conclude the population means are different
## Difference of Means of Paired Observations
1. We have observations that are paired
2. Two treatments, A and B
3. Example: Patients are given two types of blood pressure medicine
Observations Treatment A Treatment B Difference 1 64 84 -20 2 67 51 16 3 49 61 -12 . . . . 23 72 70 2
1. Calculate the average for the differences,
2. Calculate the standard deviation of the differences
1. The standard error is
1. The hypothesis test is
1. The t-statistic is
1. The a = 0.05, df = 23 – 1 = 22, and tc = tdist(a, df) = 2.034
1. Fail to reject the H0 and conclude both treatments are similar
2. The paired test is a more powerful test than the other two
3. Contains more information, because you took the extra step of pairing the observations |
# Using an Abacus/Modern division
## Introduction and first methods
### Euclidean division
If we consider two natural numbers ${\displaystyle a}$ and ${\displaystyle b}$, the division of ${\displaystyle a}$ by ${\displaystyle b}$ (indicated as ${\displaystyle a/b}$ or ${\displaystyle a\div b}$) answers the question of how many times the number ${\displaystyle b}$ is contained in number ${\displaystyle a}$. Number ${\displaystyle a}$ in ${\displaystyle a/b}$ is called the dividend and ${\displaystyle b}$ the divisor. The answer is called the quotient.
Let's take ${\displaystyle a=1225}$ and ${\displaystyle b=35}$ as an example. There is no simpler way to proceed to answer the question than by repeated subtraction counting the number of times we subtract the divisor from the dividend. We can do it directly on the abacus using a column as counter:
1225÷35 = 35, primitive approach
Abacus Comment
ABCDEFGHIJKL
35 1225
+1 -35 subtract 35 from KL, add 1 to counter F
35 1 1190
+1 -35 subtract 35 from KL, add 1 to counter F
35 2 1155
+1 -35 subtract 35 from KL, add 1 to counter F
35 3 1120
... Continue 33 times more...
35 33 70
+1 -35 subtract 35 from KL, add 1 to counter F
35 34 35
+1 -35 subtract 35 from KL, add 1 to counter F
35 35 00 Done, quotient is 35 in EF!
Thus we discover that the number ${\displaystyle 35}$ is contained exactly ${\displaystyle 35}$ times in ${\displaystyle 1225}$, since we cannot continue to subtract ${\displaystyle 35}$ without starting to deal with negative numbers. Therefore, in this example, the quotient is: ${\displaystyle q=35}$.
As we can see, in this case we can write ${\displaystyle 1225=35\times 35}$ or
${\displaystyle a=q\times b}$
which we cannot expect in the general case. If we repeat the process with ${\displaystyle a=1240}$, we would see that after subtracting ${\displaystyle 35}$ times ${\displaystyle 35}$ we would have ${\displaystyle 15}$ left on the abacus, from which we cannot continue subtracting ${\displaystyle 35}$ without entering negative numbers. Therefore we have that ${\displaystyle 1240=35\times 35+15}$; that is, the result of dividing ${\displaystyle 1240}$ by ${\displaystyle 35}$ is a quotient of ${\displaystyle 35}$ leaving a remainder of ${\displaystyle 15}$. In general we will have:
${\displaystyle a=q\times b+r}$
where:
• ${\displaystyle a}$: dividend
• ${\displaystyle b}$: divisor
• ${\displaystyle q}$: quotient
• ${\displaystyle r}$: remainder
In the case that the remainder is zero, the dividend is a multiple of the divisor.
This is the concept of Euclidian division for natural numbers to which the division of numbers with decimal fractions can be reduced.
### Some improvements: Chunking methods
The procedure followed in the previous section is the simplest possible conceptually, but it is extraordinarily long and inefficient. Instead of starting directly by subtracting the divisor ${\displaystyle b}$ (${\displaystyle 35}$) from the dividend, let's start by asking what power of 10 times the divisor we can subtract from the dividend; in our case: can we subtract 3500, 350, or only 35? Clearly we can subtract 350 and we will start subtracting 350 chunks, and when we cannot continue, we will start subtracting 35 chunks as follows:
1225÷35 = 35, a great improvement
Abacus Comment
ABCDEFGHI
35 1225 Start, counter in D,
35 1 875 subtract 35 from GH, add 1 to counter D,
35 2 525 subtract 35 from GH, add 1 to counter D,
35 3 175 subtract 35 from GH, add 1 to counter D,
35 31140 subtract 35 from HI, add 1 to counter E,
35 32105 subtract 35 from HI, add 1 to counter E,
35 33 70 subtract 35 from HI, add 1 to counter E,
35 34 35 subtract 35 from HI, add 1 to counter E,
35 35 00 subtract 35 from HI, add 1 to counter E.
35 35 No remainder. Done, quotient is 35!
Which has been a lot faster (We have intentionally reduced the distance between the counter and the dividend as much as possible. This obscures the process somewhat but brings us closer to what we will routinely do with the modern division method. Please study the above calculation carefully using your own abacus). Let's continue from here looking for even more efficiency.
If we can easily double the divisor and retain it in memory, we can shorten the operation by subtracting one or two times the divisor chunks.
times chunks
1 35
2 70
1225÷35 = 35, something more sophisticated
Abacus Comment
ABCDEFGHI
35 1225 Start, counter in D,
35 2 525 subtract 70 from GH, add 2 to counter D,
35 3 175 subtract 35 from GH, add 1 to counter D,
35 32105 subtract 70 from HI, add 2 to counter E,
35 34 35 subtract 70 from HI, add 2 to counter E,
35 35 00 subtract 35 from HI, add 1 to counter E.
35 35 No remainder. Done, quotient is 35!
Or even better if we can build a table like the one below by doubling the divisor three times[1]:
times chunks
1 35
2 70
4 140
8 280
1225÷35 = 35, a very effective method
Abacus Comment
ABCDEFGHI
35 1225 Start, counter in D,
35 2 525 subtract 70 from GH, add 2 to counter D,
35 3 175 subtract 35 from GH, add 1 to counter D,
35 34 35 subtract 140 from HI, add 4 to counter E,
35 35 00 subtract 35 from HI, add 1 to counter E.
35 35 No remainder. Done, quotient is 35!
which is somewhat shorter and, clearly, nothing could be faster than having a complete multiplication table of the divisor
### Multiplication table
Multiplication table by 35
times chunks
1 35
2 70
3 105
4 140
5 175
6 210
7 245
8 280
9 315
then
1225÷35 = 35, the shortest method!
Abacus Comment
ABCDEFGHI
35 1225 Start, counter in D,
35 3 175 subtract 105 from GH, add 3 to counter D,
35 35 00 subtract 175 from HI, add 5 to counter E.
35 35 No remainder. Done, quotient is 35!
There is no doubt, this is an optimal division method, nothing can be faster and more comfortable ... once we have a chunk table like the one above. But calculating the chunk table is time consuming and requires paper and pencil to write it and this extra work would only be justified if we have a large number of divisions to do with the same common divisor.
In 1617 John Napier, the father of logarithms, presented his invention to alleviate this problem consisting of a series of rods, known as Napier's Bones, with the one-digit multiplication table written on them and that could be combined to get the multiplication table of any number. For example, in our case
1 35 2 70 3 105 4 140 5 175 6 210 7 245 8 280 9 315
There is no doubt that such an invention spread to the East and was used in conjunction with the abacus, but this use must be considered as exceptional; not everyone had Napier bones close at hand. Another tool is needed and that tool is the traditional 1-digit multiplication table that is learned by heart and that we are going to use as an approximation to the specific multiplication table of the divisor (the one used above), this table will guide us to choose the digit of the quotient that we must try.
It should be noted that the above procedures do not exhaust the possibilities of the chunking methods. If you read The Definitive Higher Math Guide on Integer Long Division[2] article, you will be amazed at the variety of division methods that can be performed.
## Modern Division
The modern method of division is so called because throughout the first half of the 20th century its use has displaced that of the traditional method, but in fact it is much older than this, having been displaced by it in the 13th century. A characteristic of the modern method is the use of the 1-digit multiplication table both as a guide to the choice of the interim quotient figure that we have to try and for the calculation of the chunk that we have to subtract from the dividend.
Decimal multiplication table
× 1 2 3 4 5 6 7 8 9
1 1 2 3 4 5 6 7 8 9
2 2 4 6 8 10 12 14 16 18
3 3 6 9 12 15 18 21 24 27
4 4 8 12 16 20 24 28 32 36
5 5 10 15 20 25 30 35 40 45
6 6 12 18 24 30 36 42 48 54
7 7 14 21 28 35 42 49 56 63
8 8 16 24 32 40 48 56 64 72
9 9 18 27 36 45 54 63 72 81
By comparison, the traditional method uses both a special division table as a guide for the interim quotient figure and the multiplication table for calculating the chunk to subtract.
The main reason why the modern method began to displace the traditional method in Japan after the Meiji Restoration is that it can be learned more easily and quickly by those who already know how to divide with paper and pencil, since it does not require memorization of the complex division table. On the other hand, the traditional method makes the division a completely automated process, without the need to think, one only has to follow the rules to obtain the result, which allows the operation to be carried out without any mental fatigue. If you are interested in this topic you can consult the Wikibook: Traditional Abacus and Bead Arithmetic.
### Key point of division with the abacus
One of the key points of learning abacus is to be aware that this instrument allows us to correct some things very quickly and without leaving traces, which makes the abacus an instrument especially suited to trial and error procedures. This is specially useful in the case of division. So, if we have to divide 634263÷79283, instead of busting our brain trying to find the correct quotient figure, we simply choose an approximate provisional or interim figure by simplifying the original problem to 63÷7 and test it by trying to subtract the chunk (interim quotient digit)✕79283 from the dividend; one of the following will occur:
The interim quotient digit is correct
that is, we can subtract the chunk (interim quotient digit) ✕ (divisor) without entering negative numbers but we cannot subtract the quotient one more time because the remainder is less than the divisor.
It is insufficient and we must revise it up
we can subtract the chunk (interim quotient digit) ✕ (divisor) without entering negative numbers but we can still subtract the quotient one more time because the remainder is greater/equal than the divisor. We add one to the interim quotient and subtract the divisor again from the remainder.
It is excessive and we must revise it down
this is the most complex and error-prone situation. We usually discover too late (in the middle of the chunk subtraction) that the interim figure is excessive and we need to go back, subtract one from the quotient, and restore the dividend/remainder by adding what has been subtracted in excess before we can continue.
Therefore, the process of obtaining a digit of the quotient has two phases:
1. Choose an interim quotient digit.
2. Test if it is correct and modify it if not.
Once we have found the correct figure, we will generally have a non-zero remainder that will act as a dividend if we want to extend the division to the next digit of the quotient.
We will see all of this throughout the examples that follow, but first, we need a few words about how to organize the division on the abacus.
### Modern division arrangement
The dividend is the active term with which we are going to work in the abacus, the divisor is inactive and remains unchanged during the operation, in fact it is not essential to enter it in the abacus but it is recommended, especially for beginners. As in the case of multiplication, there are two styles to place dividend and divisor on the abacus, each with its advantages and disadvantages.
The divisor goes to the far right of the abacus while the dividend is to the left, leaving at least two columns free to its left.
A B C D E F 1 2 2 5 3 5
The divisor goes to the extreme left of the abacus while the dividend is to its right, leaving at least four free columns between the two terms.
A B C D E F 3 5 1 2 2 5
In this book we will use the Japanese style for the examples, but feel free to try both.
#### Placing the quotient figure
The interim quotient figure is placed in one of the two columns directly to the left of the dividend. To decide which one we need to compare the divisor with an equal number of the first digits of the dividend, adding zeros to its right if necessary; call this the working dividend:
Working dividend greater than or equal to the divisor
this means that the divisor goes into the working dividend and the quotient, i.e. the number of times the divisor goes into the working dividend, is arranged in the second column to the left of the first digit of the dividend
Example: 827÷46, 82, the working dividend, is greater than 46, then the interim quotient goes to the second column to the left of dividend. Multiplication table suggest us using 2 as interim quotient (simplify 827÷46 to 8÷4)
827÷46
Abacus Comment
ABCDEFGHIJKLM
46 827
46 2 827 Place interim quotient 2 in E
Working dividend smaller than the divisor
this means that the divisor does not go into the working dividend. In this case, we need to include the next digit of the dividend, or a zero if there are no more left, in our working dividend, and the quotient, the number of times the divisor goes into this expanded working dividend, is arranged in the column directly to the left of the first digit of the dividend
Example: 18÷467, 180 is less than 467, then the working dividend is 1800 and the interim quotient goes to the first column to the left of dividend. Multiplication table suggest us using 4 as interim quotient after simplifying 1800÷467 to 18÷4.
Caption text
Abacus Comment
ABCDEFGHIJKLM
467 18
467 418 Place interim quotient 4 in G
### Examples
You should start by doing exercises with single-digit divisors and later try divisors with two, three, etc. figures. With one-digit divisors you should never have to revise up or down. For example you can divide 123456789 by the digits 2, 3, ..., 9. Let's see the division by 9 here.
#### Example: 123456789÷9 = 13717421
• Please read the "->" symbol as: "the multiplication table suggests using ...".
• As you will see, in all cases except the last one, the working dividend is less than the divisor and we need to expand it to two digits.
123456789÷9 = 13717421
Abacus Comment
ABCDEFGHIJKLMNO
9 123456789 12/9 -> 1 as interim quotient
9 1123456789 place i. quotient in E
-9 subtract 9✕1=9 from FG
9 1 33456789 33/9 -> 3 as interim quotient
9 1333456789 place i. quotient in F
-27 subtract 9✕3=27 from GH
9 13 6456789 64/9 -> 7 as interim quotient
9 1376456789 place i. quotient in G
-63 subtract 9✕7=63 from HI
9 137 156789 15/9 -> 1 as interim quotient
9 1371156789 place i. quotient in H
-9 subtract 9✕1=9 from IJ
9 1371 66789 66/9 -> 7 as interim quotient
9 1371766789 place i. quotient in I
-63 subtract 9✕7=63 from JK
9 13717 3789 37/9 -> 4 as interim quotient
9 1371743789 place i. quotient in J
-36 subtract 9✕4=36 from KL
9 137174 189 18/9 -> 2 as interim quotient
9 1371742189 place i. quotient in K
-18 subtract 9✕2=18 from LM
9 1371742 9 9/9 -> 1 as interim quotient
9 13717421 9 place i. quotient in L
-9 subtract 9✕1=9 from MN
9 13717421 No remainder! Done
123456789÷9 = 13717421
123456789 is a curious number, it is precisely the product of 9 by 13717421, a large prime!
#### Example: 1225÷35 = 35 Two-digit divisor. Revising down and up
1225÷35 = 35
Abacus Comment
ABCDEFGHIJ
35 1225 12÷3↦4 as interim quotient
+4 enter interim quotient in F
35 41225 Now try to subtract the chunk 4✕35 from GHI,
-12 first 4✕3 from GH
35 40025 then 4✕5 from HI
-20 Cannot subtract!
-1 Revise down interim quotient digit
35 30025
+3 return the excess subtracted from GHa
35 30325
-15 continue normally, subtract 3✕5 from HI
35 3 175 17÷3↦5 as interim quotient
+5 enter interim quotient in G
35 35175 Try to subtract chunk 5✕35 from HIJ
-15 first 5✕3 from HI
35 35025
-25 then 5✕5 from IJ
35 35 No remainder, done! 1225÷35 = 35
Note:^a We have subtracted 4 × 3 = 12 from FGH, but if the correct quotient digit is 3, we should have subtracted 3 × 3 = 9, so we subtracted 3 in excess (just the first digit of the divisor). We must return this excess before continuing.
Now, suppose that after our "bad experience" revising down the first figure of the quotient, and in an excess of prudence, we choose 4 as as the second interim quotient instead of 5 as suggested by the multiplication table. Then we continue this way:
1225÷35 = 35, second quotient digit, example of revising up
Abacus Comment
ABCDEFGHI
35 3 175 17÷3 -> 5, but we use 4!
+4 enter interim quotient in G
35 34175 Try to subtract chunk 4✕35 from HIJ
-12 first 4✕3 = 12 from HI
-20 then 4✕5 = 20 from IJ
35 34 35 remainder greater or equal to divisor!
+1 revise up G
-35 subtract divisor from remainder HI
35 35 No remainder, Done!
#### Example 1÷327
So far we have considered divisions between natural numbers with quotients and remainders as well as natural numbers, but we can operate with decimal numbers exactly as we do in written calculation with long division. For example, let's find the inverse of 327; that is, 1/327 in an abacus with 13 columns.
1÷327 = 0.00305810...
Abacus Comment
ABCDEFGHIJKLM
327 1 10/3 -> 3 as interim quotient
327 31 enter interim quotient in G
-09 subtract 3✕3=9 from HI
327 3 1
-06 subtract 3✕2=6 from IJ
327 3 4
-21 subtract 3✕7=21 from JK
327 3 19 19/3 -> 6 as interim quotient
327 30619 enter interim quotient in I
-18 subtract 6✕3=18 from JK
327 306 1
-12 cannot subtract 6✕2=12 from KL!
-1 revise down I
+3 return the excess subtracted from JK
327 305 4
-10 continue normally, subtract 5✕2=10 from KL
327 305 30
-35 subtract 5✕7=35 from LM
327 305 265 36/3 -> 8 as interim quotient
327 3058265 enter interim quotient in J
-24 subtract 8✕3=24 from KL
327 3058 25
-16 subtract 8✕2=16 from LM
327 3058 9 cannot continue! Result: 3058
We have obtained ${\displaystyle 3058}$ as the first digits of ${\displaystyle 1/327}$, but ${\displaystyle 1/327\approx 1/300=1/(3\cdot 100)=(1/3)\cdot 0.01\approx 0.33\cdot 0.01=0.0033}$ so our result is actually ${\displaystyle 0.003058}$. See below the rule to find the unit rod of the division.
#### Example: 634263÷79283 = 7,999987..., a tricky case
Finally let's get the first digit of the quotient of this especially malicious division
634263÷79283 = 7,999987...
Abacus Comment
ABCDEFGHIJKLM
79283 634263 63/7 -> try 9
79283 9634263
-63 subtract 9*7=63 from HI
79283 9004263
-81 cannot subtract 9*9=81 from IJ!
-1 revise D down
+7 restore excess subtracted from remainder
79283 8 74263
-72 continue subtracting 8x9=72 from IJ
79283 8 02263
-16 subtract 8*2=16 from JK
79283 8 00663
-64 subtract 8*8=64 from KL
79283 8 00023 cannot subtract 8*3=24 from LM!
-1 revise D down
+7928 restore excess subtracted from remainder
79283 7 79303
-21 continue subtracting 7x3=21 from LM
79283 7 79282 quotient: 7, remainder: 79282
There is no doubt that in this case rounding the divisor 79283 to 80000 would have given us better results since 63÷/8 suggests using 7 (the correct figure) as the interim quotient digit.
634263÷79283 = 7,999987..., rounding divisor
Abacus Comment
ABCDEFGHIJKLM
79283 634263 63/8 -> try 7
7634263
-49 subtract 7*7=49 from HI
79283 7144263
-63 subtract 7*9=63 from IJ
79283 7 81263
-14 subtract 7*2=14 from JK
79283 7 79863
-56 subtract 7*8=56 from KL
79283 7 79303
-21 subtract 7*3=21 from LM
79283 7 79282 quotient: 7, remainder: 79282
## The unit rod and decimals
The counterpart of the rule to find the unit rod of multiplication is the following rule for division:
The column of the units of the quotients is located ${\displaystyle n+1}$ columns to the left of the column of the units of the dividend; where ${\displaystyle n}$ is the number of digits of the divisor to the left of its decimal point (which could be negative!).
The following table shows the ${\displaystyle n}$ values for some divisors:
Multiplier n
32.7 2
3.27 1
0.327 0
0.00327 -2
Example: 1/327 (we have seen it above)
1/327 unit rod
Abacus Comment
ABCDEFGHIJKLM
327 1 divisor has 3 digits. n=3
. dividend unit rod
...
327 3058 9 End of division. Result: 3058
. dividend unit rod
<--- shift it n+1 = 4 positions to the left
. unit rod of quotient
3058 so that this...
.003058 ... should be read 0.003058
## Multiplication and division as inverse operations
In written calculations we can always review our calculation to make sure that we have not made mistakes and that the result obtained is correct. In calculations with the abacus this is not possible since the abacus does not keep memory of the past and of intermediate results. We can resort to some sanity tests such as casting nines or elevens out, but the traditional way of checking the results with the abacus has been either to repeat the calculations or to undo the calculations.
Undoing additions and subtractions is as simple as starting from the result and subtracting what we have added, adding what we have subtracted; If we do both the calculation and the verification correctly, we should end up with a reset abacus. To verify a multiplication we will use division and, reciprocally, to verify a division we will use multiplication, adding the remainder if there is one. After doing this we will return the abacus to its starting state with the two original operands in their initial positions. Let's see an example:
Testing 2461÷64 by multiplication
Abacus Comment
ABCDEFGHIJ
64 2461 24/6 -> 4 as interim quotient
42461 enter interim quotient in F
-24 subtract 4✕6=24 from GH
64 4 61
-16 cannot subtract 4✕4=16 from HI
-1 revise down interim quotient digit
64 3 61
+6 return the excess subtracted from GH
64 3 661
-12 continue normally, subtract 3✕4=12 from HI
64 3 541 54/6 -> 9, but we will use 8
64 38541
-48 subtract 8✕6=48 from HI
64 38 61
-32 subtract 8✕4=32 from IJ
64 38 29 quotient: 38, remainder 29
revision by multiplication start here!
64 38509
64 38541
64 3 541 clear G
64 32341
64 32461
64 2461 clear F. Initial status!
It has been suggested in this book to use the number 123456789 for your first exercises in both multiplication and division by a single digit. Try combining them with the reverse operation; for example: divide 123456789 by 9 to get 13717421 and multiply this result by 9 to get 123456789 back to the same starting position on the abacus. Or start by multiplying 123456789 by 9 to get 1111111101 and then divide by 9 to get back to where you started. Try all the digits from 2 to 9.
## References
1. Wilson, Jeff. "Long Division Teaching Aid, "Double Division"". Double Division. Archived from the original on March 02, 2021. {{cite web}}: Check date values in: |archivedate= (help); Text "year 2005" ignored (help); Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
2. "The Definitive Higher Math Guide on Integer Long Division (and Its Variants)". Math Vault. Archived from the original on May 14, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help); Unknown parameter |name= ignored (help)
## External resources
Exercise sheets
• "The generator". Practicing the soroban. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
• Kojima, Takashi (1954), "Division", The Japanese Abacus: its Use and Theory, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0278-9
• Heffelfinger, Totton (2004). "Division". 算盤 Abacus: Mystery of the Bead. Archived from the original on June 29, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
• Siqueira, Edvaldo (2004). "Decimals & Division". 算盤 Abacus: Mystery of the Bead. Archived from the original on May 6, 2021. {{cite web}}: Unknown parameter |accesdate= ignored (|access-date= suggested) (help)
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# Comparing Fractions
Comparing fractions means that you want to be able to tell if the one fraction is less than, greater than, or equal to another.
We can use symbols like we would with whole numbers.
Let's Review:
3 is less than 8 would be written as 3 < 8.
14 is greater than 2 would be written as 14 > 2.
17 is equal to 17 would be written as 17 = 17.
We can do the same thing with fractions.
Let's begin with fractions that common denominators.
1.)
When the denominators are the same we can simply compare the numerators.
3 > 2, so
2.)
When the denominators are not the same, there are several different ways that you can compare.
Method 1:Get common denominators.
Now that we have common denominators, we can compare the numerators.
36 > 10, therefore .Furthermore,
3.)
Here is another example where the denominators are not the same. Let's look at another method.
Method 2:Use cross products.
When you cross multiply you need to be sure to multiply up! Then write the product at the top of the fraction.
Now we will compare the products. 30 < 72 and therefore
4.)
Yet another example where the denominators are not the same!
Method 3:Try simplifying the fractions.
Both fraction simplified to the same thing! So we can say that
5.)
Method 4:Change the fractions to decimals.
In this case, you may know the decimal equivalents without much effort.
Now we can see that 0.75 < 0.80. This means that |
Find equation of the line through the point (0, 2) making an angle $$\dfrac{ 2π}{3}$$ with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.
Asked by Abhisek | 1 year ago | 94
##### Solution :-
Point (0, 2) and θ = $$\dfrac{2π}{3}$$
We know that m = tan θ
m = tan ($$\dfrac{2π}{3}$$) = $$- \sqrt{3}$$
We know that the point (x, y) lies on the line with slope m through the fixed point (x0, y0), if and only if, its coordinates satisfy the equation y – y0 = m (x – x0)
y – 2 = $$- \sqrt{3}$$(x – 0)
y – 2 = $$- \sqrt{3}x$$
$$-\sqrt{3}x$$ + y – 2 = 0
Given, equation of line parallel to above obtained equation crosses the y-axis at a distance of 2 units below the origin.
So, the point = (0, -2) and m = $$- \sqrt{3}$$
From point slope form equation,
y – (-2) = $$- \sqrt{3}$$ (x – 0)
y + 2 = $$- \sqrt{3}x$$
$$\sqrt{3}x$$ + y + 2 = 0
The equation of line is $$\sqrt{3}x$$+ y – 2 = 0 and
the line parallel to it is $$\sqrt{3}x$$ + y + 2 = 0.
Answered by Pragya Singh | 1 year ago
### Related Questions
#### Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.
Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.
#### Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices
Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.
#### Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.
Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0. |
## A Multiplication Based Logic Puzzle
### 500 Pick Your Pony! Who’ll Win This Number of Factors Horse Race?
Today I factor the number 500. How many factors does it have? Each number between 401 and 500 has at least 2 factors, but no more than 24 factors.
What if we had a horse race between the number of factors? Click on the graphic below to see a gif of the numbers racing against each other. Before you click, pick your pony. Will 2, 4, 6, 8, 9, 10, 12, 14, 16, 18, 20, or 24 be the number of factors of more integers between 401 and 500 than any other number? Click on the graphic to find out!
Did you see the lead change a couple of times? How did your pony do? Which pony will you choose in the 501 to 600 race?
While we’re on the subject of horse racing, you’ll get a chuckle from reading the mathemagical site’s place-your-bets.
Remarkably, only 37 of these one hundred numbers have reducible square roots. That’s only 37%, which is significantly lower than in the 40% or 39% of previous hundreds as this graphic illustrates:
—————————————————————————————————
• 500 is a composite number.
• Prime factorization: 500 = 2 x 2 x 5 x 5 x 5, which can be written 500 = (2^2) x (5^3)
• The exponents in the prime factorization are 2 and 3. Adding one to each and multiplying we get (2 + 1)(3 + 1) = 3 x 4 = 12. Therefore 500 has exactly 12 factors.
• Factors of 500: 1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250, 500
• Factor pairs: 500 = 1 x 500, 2 x 250, 4 x 125, 5 x 100, 10 x 50, or 20 x 25
• Taking the factor pair with the largest square number factor, we get √500 = (√100)(√5) = 10√5 ≈ 22.36067977
—————————————————————————————————
If you didn’t want to click, you can still see the horse race below, but the numbers from 401 to 500 will be much clearer if you click.
make animated gifs like this at MakeAGif
#### Comments on: "500 Pick Your Pony! Who’ll Win This Number of Factors Horse Race?" (6)
1. Congratulations! A milestone! I bet 2, so my pony lost.
Liked by 1 person
• ivasallay said:
Thank you! Your pony ran very fast and placed 3rd which is not quite the same as losing.
Liked by 1 person
2. Congratulations on reaching 500 posts!
I had figured on six winning out, myself.
Liked by 1 person
• ivasallay said:
Thank you, and also much thanks for participating and commenting!
Like
3. I’m rather out of circulation at the moment, but I couldn’t miss number 500. We’ll all look forward to another special, at 1000.
Liked by 1 person
• ivasallay said:
Life can keep us very busy at times, and that’s a good thing. Thank you for taking the time to read and comment!
Like |
Rajib Singha Mar 24, 2019
In this tutorial on multiplying decimals, some simple instructions and examples are given. So keep a pen and paper handy.
Steps for Multiplying Decimals
Here are some instructions to multiply decimals with the help of a simple example: 0.25 x 0.25.
• Here, we would multiply the numbers first, meaning 25 x 25 (ignoring the decimals).
• The result that we get is 625.
• Now, we would count the number of decimal places in the numbers which we intend to multiply. In this example, the number of decimal places is 4.
• Coming back to 625, here, we would put the decimal point by starting from the right and moving towards the left, by 4 places. So, this gives the answer 0.0625
So, what we get from 0.25 x 0.25 is 0.0625. You may be wondering what is the '0' doing in the answer. Well, as you understand from the above instructions, we had to move left by 4 places. However, there were only three numbers in 625. So to make it 4, the '0' was added to the final answer.
Multiplying Decimals by Decimals
Example #1
4.77 x 1.88
= 477 x 188 [Ignoring the decimals]
= 89676
= 8.9676 [since there are 4 decimal places in total, we had to move by 4 places towards the left].
Example #2
0.35 x 0.4
= 35 x 4
= 140
= 0.140
= 0.14 [0 can be ignored after the decimal point.]
Example #3
5.7 x 4.7
= 57 x 47
= 2679
= 26.79
Example #4
8.559 x 6.569
= 8559 x 6569
= 56224071
= 56.224071
Example #5
2.545 x 3.65
= 2545 x 365
= 928925
= 9.28925
Multiplying Decimals by Whole Numbers
Example #6
1.2 x 5
= 12 x 5
= 60
= 6.0
= 6
Example #7
65.6 x 10
= 656 x 10
= 6560
= 656.0
= 656
Example #8
25 x 9.999
= 25 x 9999
= 249975
= 249.975
Example #9
0.989 x 85
= 989 x 85
= 84065
= 84.065
Example #10
559.5 x 5
= 559 x 5
= 27975
= 2797.5
You see how easy it is to multiply decimals manually, instead of using a calculator. Speaking of a calculator, in my opinion, it should not be used until the need arises for scientific calculations. Always remember, the best tool to master mathematics is practice. Once you know how to use this tool, the magic of numbers will be on your fingertips. |
# Long Division with Remainders
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Long Division with Remainders
CCSS.MATH.CONTENT.4.NBT.B.6
## Information about the videoLong Division with Remainders
### Contents
Mr. Squeaks and Imani are visiting Boulders and Bears National Park. They are practicing division problems with remainders to see if the bears can divide berries, moths, and fish evenly amongst themselves. After they find the quotient and remainders, Mr. Squeaks has an idea about what to do with all the leftovers…
### Long Division with Remainders
In this video, you will learn how to do long division with remainders, how do remainders work, what to do with remainders in division, and how to write remainders.
#### What are Remainders in Division?
Why are there remainders in division? Sometimes, items cannot be divided equally. The remainder is the amount left over after dividing. Why are reminders important? When you have to work with long division and remainders, you must look at what each number in the problem represents in order to solve it. For example, the problem asks if the bears will share the berries evenly, in order to find the answer you have to practice division with remainders and be able to identify if there is a remainder or not.
#### How to do Division with Remainders
You may be wondering, how do I do long division with remainders?In this video you will practice 2 digit by 1 digit division with remainders and 3 digit by 1 digit division with remainders. When teaching division with remainders, it’s easier to find remainders when dividing following a few simple steps. Before we see an example of dividing with remainders, let’s review the parts of division and set up our problem.
#### What Are the Parts of Division?
To set up a long division problem, we write our division symbol and place the dividend, or the number being divided, on the inside. We place the divisor, or the number we are dividing by on the outside.
Now that the problem is set up we can see how to divide with remainders by following these simple steps:
• First, start by seeing how many times the divisor, or three, goes into the first digit. Since three does not go into two, so now we look at twenty-six Three goes into twenty-six eight whole times, so we write it above the one place.
• Second, multiply three times eight to get the product twenty-four, and subtract twenty-four from twenty-six to get two.
• Last, three does not go into two, so two is our remainder. If you’re wondering how to write remainders in division, all you have to do is write an “R” and the remainder at the top next to the answer like in the illustration below.
#### Long Division with Remainders Summary
Remember, sometimes, items cannot be divided equally. The remainder is the amount left over after dividing. When finding remainders in math, including two digit division with remainders and three digit division with remainders, you simply follow the steps for solving a long division problem, writing the remainder at the top next to the answer labeling it with the letter “R”.
Have you practiced yet? On this website you can practice long division with remainders and find a long division with remainders worksheet along with other activities, and exercises.
### TranscriptLong Division with Remainders
Mr. Squeaks and Imani just arrived at Boulders and Bears National Park. They are using some binoculars to watch the bears sort some food for hibernation at a safe distance. They are observing bears to see if they can divide berries, moths, and fish evenly amongst themselves. In order to do this they will use "Long Division with Remainders". Sometimes, items cannot be divided equally(...) the "remainder" is "the amount left over after dividing". For example, Mr. Squeaks and Imani see there are twenty-six berries that need to be divided between three bears. Remember, to set up a long division problem, we write our division symbol (...) and place the DIVIDEND, or the number being divided, HERE. We place the DIVISOR, or the number we are dividing by, HERE. Now, we can start by seeing how many times the DIVISOR, or three, goes into the first digit. Does three go into two? Three DOES NOT go into two, so now we look at both digits. How many times does three go into twenty-six? Three goes into twenty-six eight whole times, so we write it above the ONES place. Three times eight is twenty-four, (...) so we subtract twenty-four from twenty-six. What is twenty-six minus twenty-four? The answer is two (...) three does not go into two and there are no more numbers to bring down... so this is our remainder. We label the remainder with a capital "R" and write it along with the two at the top beside the eight. Out of the twenty-six berries, all three bears will get eight with two leftover. Next, Mr. Squeaks and Imani see the bears will share some moths. There are eighty-one moths to be divided between six bears. We start by seeing how many times six goes into eight. How many times does six go into eight? (...) Six goes into eight one whole time (...) so write one above the TENS place. Six times one is six (...) so subtract six from eight. What is eight minus six? (...) Eight minus six is two. Next, bring down the ONE and see how many times six goes into twenty-one. How many times does six go into twenty-one? (...) Six goes into twenty one three whole times, so write a three above the ONES place. What is our next step? (...) Six times three is eighteen, so we subtract eighteen from twenty-one to get three. Three is our remainder, so we write a capital "R" along with the three next to our answer. Out of eighty-one moths, all six bears will get thirteen with three leftover. Last, Mr. Squeaks and Imani see the bears will share some fish. There are three hundred twenty-eight fish to be divided between nine bears. What is our first step? Calculate how many times nine goes into three. Nine does not go into three, so we calculate how many times nine goes into thirty-two instead. What is our next step? Since nine goes into thirty-two three whole times (...) we write the three above the TENS place... and since nine times three is twenty-seven (...) we subtract twenty-seven from thirty-two. What do we get when we subtract? Thirty-two minus twenty-seven is five. What is our next step? (...) Our next step is to bring down the eight (...) how many times does nine goes into fifty-eight? Nine goes into fifty-eight six whole times (...) so we write a six above the ONES place. What is our next step? (...) Since nine times six is fifty-four, we subtract fifty-four from fifty-eight (...) to get four. Four is our remainder. Last, write the remainder next to our answer using a capital "R" to label it. Out of three hundred twenty-eight fish, all nine bears will get thirty-six with four leftover. Remember (...) sometimes, items cannot be divided equally (...) the "remainder" is "the amount left over after dividing"... and we label our remainder with a capital "R" beside the answer. [shocked/confused] Wait...where did the leftovers go!? I wonder what the bears will do with all their leftovers...
## Long Division with Remainders Exercise
Would you like to practice what you’ve just learned? Practice problems for this video Long Division with Remainders help you practice and recap your knowledge.
• ### What is a remainder?
Hints
If Mr. Squeaks and Imani are going to take the remainder after the bears have shared the food, what could the remainder be?
We find the remainder once we have solved the problem.
Solution
The remainder in division is what is left over after dividing.
• ### Can you figure out the answer?
Hints
How many times did 3 go into 4? How many times did 3 go into 14?
If we subtract 12 from 14, what is the remainder we are left with?
Solution
Here is the complete long division problem.
3 goes into 4 1 whole time so we write 1 at the top.
1 x 3 = 3 so we then subtract 3 from 4 and we are left with 1. We bring the 4 from the ones place down.
3 goes into 14 4 whole times. We write this next to the 1 at the top. 3 x 4 = 12 so we subtract 12 from 14 leaving us with 2.
3 does not go into 2 so 2 is our remainder. We write this after the R at the top.
We can then see that our answer is 14 R 2.
Each bear will get 14 fish and there will be 2 leftover.
• ### Does Mr. Squeaks have the correct remainder?
Hints
Solve 93 $\div$ 4 using a pencil and paper. Do you see any differences between yours and this one?
If 4 goes into 9 2 whole times, what should we subtract from 9?
How many times does 4 go into 13?
There are 4 errors to highlight.
Solution
On the left is the problem solved correclty.
On the right is where the highlighting should be.
_______________________________________________________
First, we need to figure out how many times 4 goes into 9.
4 goes into 9 2 whole times, so we write 2 at the top.
2 x 4 is 8 so we subtract 8 from 9 to get 1.
4 does not go into 1, so we bring the 3 down to make 13.
4 goes into 13 3 whole times.
3 x 4 is 12, so we subtract 12 from 13 to get 1.
The remainder is 1.
• ### Can you solve the problems?
Hints
How many times does the divisor go into the first number of the dividend? Remember, the number we are dividing is the dividend and the number we are dividing by is the divisor.
Remember to subtract from the dividend.
How many are you left with at the end? This is the remainder.
This is how we would start 87 $\div$ 6. Can you complete the rest of the problem?
Solution
Here is how to solve 87 $\div$ 6.
• 6 goes into 8 1 whole time, so we write 1 at the top and subtract 6 from 8.
• 8 - 6 = 2. 6 does not go into 2, so we bring down the 7.
• 6 goes into 27 4 whole times, so we write 4 at the top.
• 4 x 6 is 24, so we subtract 24 from 27.
• 27 - 24 = 3.
• 6 does not go into 3, so this is our remainder.
• 87 $\div$ 6 = 14 R 3
_____________________________________________________
102 $\div$ 7
• 7 does not go into 1, so we see how many times 7 goes into 10.
• 7 goes into 10 1 whole time, so we write 1 at the top.
• 10 - 7 = 3.
• 7 does not go into 3, so we bring the 2 down to get 32.
• 7 goes into 32 4 whole times, so we write 4 at the top.
• 4 x 7 is 28, so we subtract 28 from 32 to get 4.
• 7 does not go into 4, so that is our remainder.
• 102 $\div$ 7 = 14 R 4
249 $\div$ 2
• 2 goes into 2 1 whole time, so we write 1 at the top.
• 2 goes into 4 2 whole times, so we write 2 at the top.
• 2 goes into 9 4 whole times, so we write 4 at the top.
• 4 x 2 is 8, so we subtract 8 from 9 to get 1.
• 2 does not go into 1, so that is our remainder.
• 249 $\div$ 2 = 124 R 1
475 $\div$ 3
• 3 goes into 4 1 whole time, so we write 1 at the top.
• 4 - 3 = 1.
• 3 does not go into 1, so we bring the 7 down to get 17.
• 3 goes into 17 5 whole times, so we write 5 at the top.
• 5 x 3 is 15, so we subtract 15 from 17 to get 2.
• 3 does not go into 2, so we bring the 5 down to get 25.
• 3 goes into 25 8 whole times, so we write 8 at the top.
• 8 x 3 is 24, so we subtract 24 from 25 to get 1.
• 3 does not go into 1, so that is our remainder.
• 475 $\div$ 3 = 158 R 1.
• ### How many berries will Mr. Squeaks get?
Hints
Each of the 5 bears is going to get 14 berries, how many are left?
If we subtract 20 from 21, what are we left with? This is the remainder in this problem.
Solution
Mr. Squeaks will get 1 berry!
We can see that 5 goes into 7 1 whole time. We then subtract 5 from 7 to get 2.
5 doesn't go into 2, so we bring the 1 down and see how many times 5 goes into 21.
5 goes into 21 4 whole times.
4 x 5 is 20, so we then subtract 20 from 21 to find the remainder.
21 - 20 = 1.
We are left with a remainder of 1.
• ### What will Mr. Squeaks and Imani be left with?
Hints
Try solving the division problem using a pencil and paper if you need to.
Look at the divisor, what could the remainder be?
Solution
These are the remainders Mr Squeaks and Imani will be left with.
• 3 fish
• 1 moth
• 2 berries
• 4 insects
_______________________________________________________
To find how many fish there would be for Mr Squeaks and Imani:
• 5 goes into 6 1 whole time, so we write 1 at the top.
• 6 - 5 = 1.
• 5 does not go into 1, so we bring the 8 down to make 18.
• 5 goes into 18 3 whole times, so we write 3 at the top.
• 3 x 5 is 15, so we subtract 15 from 18 to get 3.
• 5 does not go into 3, so that is our remainder.
• 68 $\div$ 5 = 13 R 3
____
To find how many moths there would be for Mr. Squeaks and Imani:
• 4 goes into 9 2 whole times, so we write 2 at the top.
• 2 x 4 is 8, so we subtract 8 from 9.
• 9 - 8 = 1.
• 4 does not go into 1, so we bring the 7 down to make 17.
• 4 goes into 17 4 whole times, so we write 4 at the top.
• 4 x 4 is 16, so we subtract 16 from 17 to get 1.
• 4 does not go into 1, so that is our remainder.
• 97 $\div$ 4 = 24 R 1
__
To find how many berries there would be for Mr. Squeaks and Imani:
• 3 does not go into 1, so we see how many times 3 goes into 12.
• 3 goes into 12 4 whole times, so we write 4 at the top.
• 3 goes into 5 1 whole time, so we write 1 at the top.
• 5 - 3 = 2.
• 3 does not go into 2, so that is our remainder.
• 125 $\div$ 3 = 41 R 2
___
To find how many insects there would be for Mr. Squeaks and Imani:
• 6 does not go into 1, so we see how many times 6 goes into 19.
• 6 goes into 19 3 whole times, so we write 3 at the top.
• 3 x 6 is 18, so we subtract 18 from 19 to get 1.
• 6 does not go into 1, so we bring the 0 down to make 10.
• 6 goes into 10 1 whole time, so we write 1 at the top.
• 10 - 6 = 4.
• 6 does not go into 4, so that is our remainder.
• 190 $\div$ 6 = 31 R 4 |
# Solve distance formula
In this blog post, we will be discussing how to Solve distance formula. Our website can solving math problem.
## Solving distance formula
We can do your math homework for you, and we'll make sure that you understand how to Solve distance formula. Divide the bottom edge into two equal line segments, and the areas of the divided triangles are equal. The angle bisector of a triangle divides the top angle of the triangle into two equal angles, and is equal to half of the original angle. In addition to calculating the area of the triangle, the height of the triangle also constructs two right triangles, which provides favorable conditions for solving the angles of other angles. If there are three special line segments, triangles, in doing the problem, we can carry out simple derivation to match other conditions in the problem. Assuming that the number of cranes is greater than 1, we use the for loop to solve the total number of lifting areas in each configurable case; Then solve the case with the largest lifting area, that is, the case with the largest number of discrete points on the outer contour of the building, which meets the conditions of the optimal case.
this analysis method is called time domain analysis method. If the time variable is transformed into other variables in order to solve the equation by hand, it is called transformation domain analysis method accordingly. This chapter mainly explains the time-domain analysis method, which is also the focus of the postgraduate entrance examination. When we are in contact with ordinary differential equations, we can only solve some special forms of equations, such as first-order linear differential equations, differential equations with separable variables, Bernoulli differential equations, etc.
Pythagorean theorem is learned by students on the basis of having mastered the relevant properties of right triangle. It is one of several important theorems in middle school mathematics. It reveals the quantitative relationship between the three sides of a right triangle. It is one of the main bases for solving a right triangle and is very useful in real life.
When things happen, they will always take their place and directly help children solve their problems. There are two main reasons. The first reason is that parents do not know that there is the latter option. They regard Emotionalization as a solution to problems, thinking that children can learn what to do and what not to do from their parents' outbursts. However, what children learn is the parents' emotional way of solving problems.
And their areas are calculated as follows: If the bottom and height of a parallelogram are 6cm and 4cm respectively, and the area is 6 times 4, 24 square centimeters will be obtained. One of the bottom and high school of a triangle will remain unchanged, and the other will be doubled, and its area will also be 24 square centimeters. 12. When we use data accurate to one decimal place to calculate the areas of these two (triangles), strange things happen again.
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# Differentiate an Inverse Function, Two Methods?
I would like to take the derivative of this inverse function at $\pi$: $f(x) = 2x + \cos{x}$, given that ${f}^{-1}(\pi) = \frac{\pi}{2}$.
I know that there are two methods of doing it. Let me demonstrate the method that I have down pat, using the fact that $\frac{d}{dx}\left[{f}^{-1}(x)\right] = \frac{1}{{f}^{\prime}\left({f}^{-1}(x)\right)}$.
Method 1:
1. $f(x) = 2x + \cos{x}$
2. ${f}^{\prime}(x) = 2 - \sin{x}$
3. Given: ${f}^{-1}(\pi) = \frac{\pi}{2}$
4. $\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{{f}^{\prime}\left({f}^{-1}(\pi)\right)}$
5. $\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{2 - \sin{\left({f}^{-1}(\pi)\right)}}$
6. $\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{2 - \sin{\left(\frac{\pi}{2}\right)}}$
7. $\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = \frac{1}{2 - 1}$
8. $\frac{d}{dx}\left[{f}^{-1}(\pi)\right] = 1$
This method make sense. It is this next method that I am a little sketchy on. For the most part it utilizes some algebra for inverse functions...
Method 2:
1. $f(x) = 2x + \cos{x}$
2. $y = 2x + \cos{x}$
3. $x = 2y + \cos{y}$
The next few steps involve finding the inverse function (can it be done with a function like this?), taking the derivative of that, and plugging in $\pi$ for the answer...
My problem is that I am stuck after this point:
• Can I find the inverse function of this crazy looking function? It is one-to-one, as shown in the graph below.
Thank you for your time.
-
The problem is that method 2 is very inefficient. As you already notice, isolating $y$ from the equation $x=2y+cosy$ is really hard (it might be actually impossible in some cases). – azarel Jan 27 '12 at 4:35
Yes... but for my Calc II class I'm required to know the second method. Hope they won't ask to method 2 on a function like this. Yikes!!! – Oliver Spryn Jan 27 '12 at 4:47
Some of the lines in your first solution are technically wrong, though you arrive at the right answer. In lines $4$ to the end, it looks as if you are differentiating a constant. The derivative of a constant is $0$. The simplest notational trick is to say let $g(x)=f^{-1}(x)$. Then $g'(x)=\dots$. Therefore $g'(\pi)=\dots$. The reason is that it is awkward to put a "prime" on $f^{-1}(x)$. – André Nicolas Jan 27 '12 at 6:33
It is known that an inverse function $exists$ for any one-to-one function, but in many cases it cannot be expressed in terms of elementary functions. So, your first calculation may be the best you can do without using more machinery.
Yes, that would work, IF you can find the inverse, but thats a big IF. I am pretty sure that for your $f$, no "nice" inverse exists. – Ravi Jan 27 '12 at 4:53 |
Informative line
### Conservation Of Momentum And Mechanical Energy
Practice relative velocity problems with examples, learn kinetic energy when work done by external forces and non-conservative internal forces are zero, then total mechanical energy of the system remains constant.
# Relative Velocity
• We know that a quantity has its maximum or minimum value when its derivative is zero.
For ex- Velocity of a body is maximum or minimum when acceleration of that body is zero. $$(a=0)$$
• Similarly, relative displacement (separation) between two objects is maximum or minimum when its derivative i.e., relative velocity between them is zero.
Note : Relative velocity of two objects is said to be zero, if both of them have same speed and same direction.
• Now let us consider a situation, as shown in figure.
• Two blocks attached by an ideal spring, are kept on a smooth horizontal surface, as shown in figure.
• One of them is given a velocity $$'v'$$.
• In both cases, when extension is maximum or minimum, relative velocity of both blocks is zero.
Thus $$\vec v_{AB}=0$$
$$\vec v_{AB}=\vec v_A-\vec v_B$$
$$0=\vec v_A-\vec v_B$$
$$\vec v_A=\vec v_B$$
$$\therefore$$ Velocity of both the blocks is same.
• Assume speed of $$A$$ is $$5\,m/s$$ towards right at the time of maximum compression then, speed of $$B$$ at that time will also be $$5\,m/s$$ towards right.
#### Two blocks attached by an ideal spring, are kept on a smooth horizontal surface, as shown in figure. Now $$A$$ is given a velocity of $$10\,m/s$$ towards right. If velocity of $$A$$ is $$3\,m/s$$ towards right at the time of maximum compression, then calculate the velocity of $$B$$ at that time.
A $$5\,\hat i\;m/sec$$
B $$3\,\hat i\;m/sec$$
C $$2\,\hat i\;m/sec$$
D $$6\,\hat i\;m/sec$$
×
At the time of maximum compression, i.e., minimum separation between blocks, their relative velocity is zero.
Thus,
Relative velocity $$\vec v_{AB}=0$$
$$\Rightarrow\;\vec v_A-\vec v_B=0$$
$$\Rightarrow\;\vec v_A=\vec v_B$$
Since velocity of $$A$$ is $$3\;m/sec$$ towards right, so velocity of $$B$$ is also $$3\;m/sec$$ towards right.
$$\vec v_B=3\,\hat i\,m/sec$$
### Two blocks attached by an ideal spring, are kept on a smooth horizontal surface, as shown in figure. Now $$A$$ is given a velocity of $$10\,m/s$$ towards right. If velocity of $$A$$ is $$3\,m/s$$ towards right at the time of maximum compression, then calculate the velocity of $$B$$ at that time.
A
$$5\,\hat i\;m/sec$$
.
B
$$3\,\hat i\;m/sec$$
C
$$2\,\hat i\;m/sec$$
D
$$6\,\hat i\;m/sec$$
Option B is Correct
# Relative Velocity of Block Placed on a Wedge
• Consider a system consisting of a block and a wedge.
• In the figure shown, a block $$A$$ with $$v$$, is pushed towards a wedge $$B$$.
• All the surfaces are smooth.
• Now, we observe the complete event from the frame of wedge $$B$$.
• From the frame of wedge $$B$$, we see that at maximum height, speed of block $$A$$ with respect to wedge $$B$$ is zero.
• That is block $$A$$ will be at maximum height, when its velocity relative to wedge $$B$$ is zero.
• For a observer on ground, when $$A$$ is at maximum height, the velocity of both $$A$$ and $$B$$ is same.
#### In the figure shown, a block $$A$$ is pushed towards a wedge $$B$$ with a velocity of $$5\,m/s$$, towards right. A wedge is moving with $$5\,m/sec$$ towards left. Finally when block $$A$$ is at maximum height on wedge $$B$$, then wedge $$B$$ will start moving with a speed of $$3\,m/sec$$ towards right. What is the velocity of block $$A$$ at the time when it is at maximum height?
A $$3\,m/s$$ towards left
B $$3\,m/s$$ towards right
C $$5\,m/s$$ towards right
D $$5\,m/s$$ towards left
×
Block $$A$$ will be at maximum height when its relative velocity with respect to wedge is zero.
Since, $$\vec v_{A/B}=0$$,
$$\vec v_B=3\,m/s$$ towards right
So, $$\vec v_{A/B}=\vec v_A-\vec v_B$$
$$\Rightarrow\;\vec v_A=\vec v_{A/B}+\vec v_B$$
$$\Rightarrow\;\vec v_A=0+3\,m/s$$
$$\Rightarrow\;\vec v_A=3\,m/s$$ towards right
### In the figure shown, a block $$A$$ is pushed towards a wedge $$B$$ with a velocity of $$5\,m/s$$, towards right. A wedge is moving with $$5\,m/sec$$ towards left. Finally when block $$A$$ is at maximum height on wedge $$B$$, then wedge $$B$$ will start moving with a speed of $$3\,m/sec$$ towards right. What is the velocity of block $$A$$ at the time when it is at maximum height?
A
$$3\,m/s$$ towards left
.
B
$$3\,m/s$$ towards right
C
$$5\,m/s$$ towards right
D
$$5\,m/s$$ towards left
Option B is Correct
# Calculation of Maximum Height when a Block is Projected towards a Wedge at Rest
### Condition for Conservation of Momentum
• When no external force acts on the system then the momentum of the system is conserved.
### Condition for Conservation of Mechanical Energy
• When there is no loss of energy in the system or none of the condition like jumping or landing, explosion jerk, kinetic friction and non-conservation force are present, then the mechanical energy of the system is conserved.
• Consider a situation in which a block A with velocity v is pushed towards a wedge, as shown in figure.
• All the contact surface are frictionless.
• To find out maximum height and velocity in above mentioned situation or similar situation, following steps to be followed.
1. Draw two separate diagrams to show the system before and after the event.
2. At maximum height, velocities are same.
3. Choose the direction in which momentum can be conserved.
4. Write down momentum conservation equation.
5. Write down energy conservation equation.
#### A block of mass m is projected with speed v towards a wedge B of mass M which is at rest, as shown in figure. If all the surfaces are smooth, then calculate the maximum height attained by the block on the wedge.
A $$\left(\dfrac{M}{M+m}\right)\dfrac{v^2}{2g}$$
B $$(M+m)\dfrac{v^2}{2g}$$
C $$\dfrac{mv^2}{2g}$$
D $$\dfrac{Mv^2}{2g}$$
×
Since at maximum height velocity is same, so
$$v_1^{'} = v_2^{'} = v^{'}$$
Since no external force is acting in horizontal direction, so momentum can be conserved in horizontal direction.
Applying momentum conservation equation in horizontal direction,
$$P_i = P_f$$
$$mv = (M+m)v^{'}$$
$$v^{'} = \dfrac{mv}{M+m}\;\;— (1)$$
There is no loss of energy, so applying energy conservation principle,
$$\dfrac{1}{2}mv^2 = \dfrac{1}{2} (M+m)(v^{'2}) + mgh$$
$$mv^2 = (M+m)(v^{'})^2 + 2 \; mgh$$
$$mv^2 = (M+m) × \left(\dfrac{mv}{M+m}\right)^2 + 2 \; mgh\;\;[from (1)]$$
$$mv^2 = \dfrac{m^2v^2}{M+m} + 2\; mgh$$
$$v^2 = \dfrac{mv^2}{M+m} + 2 \; gh$$
$$\Rightarrow 2 gh = v^2 \left[\dfrac{M+ m - m}{M+m}\right]$$
$$\Rightarrow h_{max} = \left[\dfrac{M}{M+m}\right] \dfrac{v^2}{2g}$$
### A block of mass m is projected with speed v towards a wedge B of mass M which is at rest, as shown in figure. If all the surfaces are smooth, then calculate the maximum height attained by the block on the wedge.
A
$$\left(\dfrac{M}{M+m}\right)\dfrac{v^2}{2g}$$
.
B
$$(M+m)\dfrac{v^2}{2g}$$
C
$$\dfrac{mv^2}{2g}$$
D
$$\dfrac{Mv^2}{2g}$$
Option A is Correct
# Conservation of Momentum and Mechanical Energy
### Condition for Conservation of Momentum
• When no external force acts on the system then the momentum of the system is conserved.
### Condition for Conservation of Mechanical Energy
• When there is no loss of energy in the system or none of the condition like jumping or landing, explosion, jerk, kinetic friction and non-conservation force are present, then the mechanical energy of the system is conserved.
• Consider a situation in which a block A with velocity v is pushed towards a wedge, as shown in figure.
• All the contact surface are frictionless.
• To find out maximum height and velocity in above mentioned situation or similar situation, following steps to be followed.
1. Draw two separate diagrams to show the system before and after the event.
2. At maximum height, velocities are same.
3. Choose the direction in which momentum can be conserved.
4. Write down momentum conservation equation.
5. Write down energy conservation equation.
#### A block A of mass m is kept inside a wedge B of mass M and radius R at position 1. If all the surfaces are smooth, then calculate the velocity of wedge when block is at position 2.
A $$\sqrt {\dfrac{2 \; mgR}{M}}$$
B $$\sqrt{\dfrac{2 \; mgR}{M^2 + m}}$$
C $$\sqrt{\dfrac{mgR}{m^2 + m}}$$
D $$\sqrt{\dfrac{2\,m^2 gR}{M^2 + Mm}}$$
×
Since there is no external force in horizontal direction, so momentum can be conserved in horizontal direction.
Applying conservation of momentum in x-direction,
$$P_i = P_f$$
$$0=mu\hat i - Mv\hat i$$
$$mu = mv$$
$$u = \dfrac{Mv}{m}$$
Since there is no loss of energy, so mechanical energy is conserved.
Applying energy conservation.
$$KE_i + PE_i = KE_f + PE_f$$
$$\Rightarrow 0 + 0 = \dfrac{1}{2} mu^2 + \dfrac{1}{2} Mv^2 - mgR$$
$$\Rightarrow 0 = \dfrac{1}{2} m\left(\dfrac{Mv}{m}\right)^2 + \dfrac{1}{2}\; Mv^2 - mgR$$
$$\Rightarrow mgR = \dfrac{1}{2} \; \dfrac{M^2}{m}v^2 + \dfrac{1}{2} Mv^2$$
$$\Rightarrow 2m^2gR = (M^2 + Mm)v^2$$
$$\Rightarrow v^2 = \dfrac{2m^2gR}{M^2 + Mm}$$
$$\Rightarrow v = \sqrt {\dfrac{2m ^2gR}{M^2 + M m}}$$
### A block A of mass m is kept inside a wedge B of mass M and radius R at position 1. If all the surfaces are smooth, then calculate the velocity of wedge when block is at position 2.
A
$$\sqrt {\dfrac{2 \; mgR}{M}}$$
.
B
$$\sqrt{\dfrac{2 \; mgR}{M^2 + m}}$$
C
$$\sqrt{\dfrac{mgR}{m^2 + m}}$$
D
$$\sqrt{\dfrac{2\,m^2 gR}{M^2 + Mm}}$$
Option D is Correct
# Conservation of Mechanical Energy
• When work is done only by conservative internal force, i.e., work done by external forces and non-conservative internal forces are zero, then total mechanical energy of the system remains constant.
$$\Rightarrow\;\text{Kinetic Energy (K.E.)}+\text{Potential Energy (U) = Constant}$$
Note :
• Kinetic energy $$(KE)$$ is maximum when potential energy $$(U)$$ is minimum.
• Kinetic energy $$(KE)$$ is minimum when potential energy $$(U)$$ is maximum.
#### A block is released from $$A$$, as shown in figure. If all the surfaces are smooth, then kinetic energy of the system is maximum when the block is at
A $$A$$
B $$B$$
C $$C$$
D
×
The system will have maximum kinetic energy when it has minimum potential energy.
Since the system has minimum potential energy at $$B$$, therefore the system will have maximum kinetic energy at $$B$$.
### A block is released from $$A$$, as shown in figure. If all the surfaces are smooth, then kinetic energy of the system is maximum when the block is at
A
$$A$$
.
B
$$B$$
C
$$C$$
D
Option B is Correct
# Common Velocity at the Time of Maximum Compression
• Consider two blocks A and B attached by an ideal spring having spring constant k, are kept on a smooth horizontal surface, as shown in figure.
• Now block A is given a velocity v, as shown.
• Since there is no net external force in horizontal direction therefore, two points has to be noted here,
a) Linear momentum in horizontal direction is conserved.
b) There is no loss of energy, so mechanical energy is conserved.
#### Two blocks A and B, of masses 5 kg and 10 kg respectively, attached by an ideal spring with spring constant $$k= 100\,N/m$$ are kept on a smooth horizontal surface, as shown in figure. If block A is given as velocity of $$10\,m/s$$, then calculate the speed of A and B at the time of maximum compression.
A $$\dfrac{10}{3} \hat i \,m/sec, \,\,\,\dfrac{10}{3} \hat i \,m/sec$$
B $$\dfrac{10}{3} \hat i \,m/sec, \,\,\,\dfrac{5}{3} \hat i \,m/sec$$
C $$\dfrac{8}{3} \hat i \,m/sec, \,\,\,\dfrac{5}{3} \hat i \,m/sec$$
D $$\dfrac{-10}{3} \hat i \,m/sec, \,\,\,\dfrac{10}{3} \hat i \,m/sec$$
×
At the time of maximum compression both the blocks will have same velocity.
Using the principle of conservation of momentum in x- direction,
Initial momentum = Final momentum
$$P_i = P_f$$
$$(5×10)\hat i +(10×0)\hat i = (5×u)\hat i +(10×u)\hat i$$
$$\Rightarrow 50 \,\hat i = 15 \,u \,\hat i$$
$$\Rightarrow u = \dfrac{50}{15} \,m/s$$
$$u = \dfrac{10}{3}\, m/s$$
So, at the maximum compression both of them will move with a speed $$u= \dfrac{10}{3} \,m/s$$ towards right.
### Two blocks A and B, of masses 5 kg and 10 kg respectively, attached by an ideal spring with spring constant $$k= 100\,N/m$$ are kept on a smooth horizontal surface, as shown in figure. If block A is given as velocity of $$10\,m/s$$, then calculate the speed of A and B at the time of maximum compression.
A
$$\dfrac{10}{3} \hat i \,m/sec, \,\,\,\dfrac{10}{3} \hat i \,m/sec$$
.
B
$$\dfrac{10}{3} \hat i \,m/sec, \,\,\,\dfrac{5}{3} \hat i \,m/sec$$
C
$$\dfrac{8}{3} \hat i \,m/sec, \,\,\,\dfrac{5}{3} \hat i \,m/sec$$
D
$$\dfrac{-10}{3} \hat i \,m/sec, \,\,\,\dfrac{10}{3} \hat i \,m/sec$$
Option A is Correct
# Steps to Calculate the Maximum Compression and Elongation in Spring
• Consider two blocks A and B attached by an ideal spring with spring constant k, are kept on a smooth horizontal surface, as shown in figure.
• Now block A is given a velocity v, as shown.
• Since, there is no net external force in horizontal direction therefore, two points need to be noted here,
a) Linear momentum in horizontal direction is conserved.
b) There is no loss of energy, so mechanical energy is conserved.
Consider two blocks A and B attached by an ideal spring are kept on a smooth horizontal surface, as shown in figure.
To calculate maximum compression or elongation in the spring following steps are to be followed:
1. Draw two separate diagram to show system before and after event.
2. Choose the direction in which momentum is conserved.
3. Apply momentum conservation equation.
4. Apply mechanical energy conservation equation.
#### Two blocks A (5 kg) and B (10 kg) attached by an ideal light spring (k = 100 N/m) are kept on a smooth horizontal surface. Calculate maximum elongation in the spring.
A $$\sqrt {30}\,m$$
B $$\sqrt {20}\,m$$
C $$\sqrt {10}\,m$$
D $$\sqrt {40}\,m$$
×
At the time of maximum elongation or compression velocity of both the blocks are same.
Since there is no net external force in horizontal direction, therefore we can conserve linear momentum in horizontal direction.
Applying momentum conservation equation
Initial momentum in horizontal direction = Final momentum in horizontal direction
$$20×10\hat i-10×5\hat i= (5+10)u\,\hat i$$
$$\Rightarrow150=15\,u$$
$$\Rightarrow u=10\,m/sec$$
Energy conservation equation
$$\dfrac {1}{2}×5×10^2+ \dfrac {1}{2}×10×(20)^2= \dfrac {1}{2}×k\,x^2+ \dfrac {1}{2}(5+10)×10^2$$
$$\Rightarrow 500+4000=100×x^2+1500$$
$$\Rightarrow 100\,x^2=4500-1500$$
$$\Rightarrow 10\,x^2=450-150$$
$$\Rightarrow x^2=\dfrac {300}{10}$$
$$\Rightarrow x=\sqrt {30}\,m$$
### Two blocks A (5 kg) and B (10 kg) attached by an ideal light spring (k = 100 N/m) are kept on a smooth horizontal surface. Calculate maximum elongation in the spring.
A
$$\sqrt {30}\,m$$
.
B
$$\sqrt {20}\,m$$
C
$$\sqrt {10}\,m$$
D
$$\sqrt {40}\,m$$
Option A is Correct
# Spring with Two Blocks System
• Consider two blocks A and B attached by an ideal spring with spring constant k, are kept on a smooth horizontal surface, as shown in figure.
• Now block A is given a velocity v, as shown.
• Since, there is no net external force in horizontal direction therefore, two points need to be noted here,
a) Linear momentum in horizontal direction is conserved.
b) There is no loss of energy, so mechanical energy is conserved.
Consider two blocks A and B attached by an ideal spring are kept on a smooth horizontal surface, as shown in figure.
To calculate maximum compression or elongation in the spring following steps are to be followed:
1. Draw two separate diagram to show system before and after event.
2. Choose the direction in which momentum is conserved.
3. Apply momentum conservation equation.
4. Apply energy conservation equation.
#### Two blocks A and B, of masses 5 kg and 10 kg respectively, attached by an ideal spring constant k = 100 N/m, are kept on a smooth horizontal surface. If 5 kg block is given a velocity of 10 m/s in rightward direction, as shown in figure, calculate maximum velocity of 10 kg block.
A $$1 \; m/s \; or \; \dfrac{10}{3} \; m/s$$
B $$20 \; m/s \; or \; 10 \; m/s$$
C $$0 \; m/s \; or \; \dfrac{20}{3} \; m/s$$
D $$\dfrac{10}{3 }\; m/s \; or \; \dfrac{8}{3} \; m/s$$
×
The 10 kg block will have maximum velocity when spring will be relaxed.
Let, at the time of maximum velocity of 10 kg block x=extension in the string.
Since there is no net external force in horizontal direction, therefore we can conserve momentum in horizontal direction.
Momentum conservation equation,
Initial momentum in horizontal direction = Final momentum in horizontal direction
$$10 × 5 \; \hat i = 5 × u \; \hat i + 10 × v \; \hat i$$
$$\Rightarrow 5 0 = 5 \; u + 10 \; v$$
$$\Rightarrow10 = u + 2 v \; — (1)$$
Energy conservation,
$$\dfrac{1}{2} × 5 × (10)^2 = \dfrac{1}{2} × 5 × u^2 + \dfrac{1}{2} × 10 × v^2$$
$$\Rightarrow 500 = 5× (10-2v)^2 + 10 v^2$$
$$\Rightarrow100 = 100 + 4 v^2 - 40 v + 2 v^2$$
$$\Rightarrow6 v^2 = 40 v$$
$$\Rightarrow v = 0 \;or\; v = \dfrac{20}{3}\; m/s$$
### Two blocks A and B, of masses 5 kg and 10 kg respectively, attached by an ideal spring constant k = 100 N/m, are kept on a smooth horizontal surface. If 5 kg block is given a velocity of 10 m/s in rightward direction, as shown in figure, calculate maximum velocity of 10 kg block.
A
$$1 \; m/s \; or \; \dfrac{10}{3} \; m/s$$
.
B
$$20 \; m/s \; or \; 10 \; m/s$$
C
$$0 \; m/s \; or \; \dfrac{20}{3} \; m/s$$
D
$$\dfrac{10}{3 }\; m/s \; or \; \dfrac{8}{3} \; m/s$$
Option C is Correct |
# TDISTS - Editorial
Author: Hazem Issa
Tester: Riley Borgard
Editorialist: Aman Dwivedi
Easy
Tree, Greedy
# PROBLEM
You are given two integers x and y. Your task is to construct a tree with the following properties:
• The number of pairs of vertices with an even distance between them equals x.
• The number of pairs of vertices with an odd distance between them equals y.
By a pair of vertices, we mean an ordered pair of two (possibly, the same or different) vertices.
# EXPLANATION:
The first basic observation that we can draw is that the summation of x and y should be a perfect square.
Why ?
Since every vertex forms a pair with every other vertex of a tree including itself. Hence for any tree of N vertices, the total number of ordered pairs of vertices will be N^2.
As a summation of x and y represent the total number of ordered pairs of vertices which is a perfect square. Hence x+y should be a perfect square.
Now, we are left with finding whether it is possible to have a tree that has x pair of vertices with an even distance between them and y pair of vertices with an odd distance between them.
As its quite clear that any vertex of the tree can either be at an even level or an odd level. Root of the tree being at the even level (0- based indexing). Let’s find out the number of pairs of vertices that have an odd distance between them.
Claim: Any pair of vertices (x, y) has an odd distance between them if the vertices x and y comes from opposite parity levels.
Proof
Let the vertex x is present at an even level and vertex y at an odd level. The distance between two nodes can be obtained in terms of the lowest common ancestor (lca). Below is the formula for the same:
dis(x,y)=dis(root,x)+dis(root,y)-2*dis(root,lca)
Since x is present at even level so dis(root,x) is even while y is at odd level therefore dis(root,y) is odd. Hence we can see:
dis(x,y) = Even + Odd - Even
That means that if the vertices are from opposite parity levels, then the distance between them is odd.
Now let’s calculate the number of pairs of vertices which has an odd distance between them. For odd distances, every vertex that is present at an odd level will form a pair with every vertex that is present at even level and vice versa.
Let us suppose [e_1,e_2,e_3....e_i] and [o_1,o_2,o_3....o_j] are the vertices that are present at even and odd level respectively. Now for every vertex that is present at an even level, it will form a pair with every vertex that is present at the odd level and vice-versa. Hence:
Odd_{pairs}' = e_1*o_1+e_1*o_2+\dots+e_1*o_j+\dots+e_i*o_1+\dots+e_i*o_j
Odd_{pairs}' = (e_1+e_2+e_3+\dots+e_i)*(o_1+o_2+o_3+\dots+o_j)
Odd_{pairs}' = E*O
, where E and O represents the number of nodes that are present at even and odd level respectively.
Since we looking for the ordered pairs of two hence the number of the above pairs will be doubled. As if (x,y) is a pair that has the odd distance between them then (y,x) will also have an odd distance between them. Hence:
Odd_{pairs}= 2*E*O
That means we only care about the number of vertices that are present at even levels and odd levels. The root is the only node that is present at the 0^{th} level, now if we want more nodes at an even level then there must exist at least one vertex at an odd level. Now we can check for every possibility and can find if there exists such an arrangement of vertices that satisfy x and y of the problem. If so, we can output the tree.
# TIME COMPLEXITY:
O(sqrt(X+Y)) per test case
# SOLUTIONS:
Setter
``````#include <bits/stdc++.h>
using namespace std;
int sqr(int n){ return n*n; }
void solve(){
int x ,y; //x even ,y odd
scanf("%d%d",&x,&y);
int n = 1;
while(sqr(n) < x+y)
n++;
if(sqr(n) != x+y){
printf("NO\n");
return;
}
for(int i=1; i<n; i++)
if(sqr(i)+sqr(n-i) == x && 2*i*(n-i) == y)
{
printf("YES\n");
printf("%d\n",n);
for(int j=1; j<=n-i; j++)
printf("1 %d\n",j+1);
for(int j=2; j<=i; j++)
printf("2 %d\n",n-i+j);
return;
}
printf("NO\n");
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
solve();
}
``````
Tester
``````
#include <bits/stdc++.h>
#define ll long long
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define vi vector<int>
#define pii pair<int, int>
#define rep(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;
template<typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;
void solve() {
ll x, y;
cin >> x >> y;
ll n = 1;
while(n * n < x + y) n++;
if(n * n != x + y) {
cout << "NO\n";
return;
}
// (a, b)
// a + b = n
// 2ab = y
// a <= b
for(ll a = 1; 2 * a * a <= y; a++) {
ll b = n - a;
if(1 <= a && a <= n && 1 <= b && b <= n && 2 * a * b == y) {
cout << "YES\n";
cout << n << '\n';
// 1..a red nodes, a+1..n blue nodes
for(ll i = a + 1; i <= n; i++) {
cout << 1 << ' ' << i << '\n';
}
for(ll i = 2; i <= a; i++) {
cout << a + 1 << ' ' << i << '\n';
}
return;
}
}
cout << "NO\n";
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int te;
cin >> te;
while(te--) solve();
}
``````
Editorialist
``````#include<bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
int x,y;
cin>>x>>y;
int n=1;
while(n*n<x+y)
n++;
if(n*n!=x+y)
{
cout<<"NO"<<"\n";
return;
}
if(n==1)
{
if(x==1)
{
cout<<"YES"<<"\n";
cout<<1<<"\n";
}
else
cout<<"NO"<<"\n";
return;
}
int even_lev=1,odd_lev=n-1;
int flag=0;
while(odd_lev>=1)
{
int odd_pairs=even_lev*odd_lev*2;
if(odd_pairs==y)
{
flag=1;
break;
}
even_lev++;
odd_lev--;
}
if(!flag)
{
cout<<"NO"<<"\n";
return;
}
cout<<"YES"<<"\n";
cout<<n<<"\n";
for(int i=1;i<=odd_lev;i++)
cout<<1<<" "<<i+1<<"\n";
for(int i=1;i<even_lev;i++)
cout<<2<<" "<<odd_lev+i+1<<"\n";
}
int32_t main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
int t;
cin>>t;
while(t--)
solve();
return 0;
}
``````
24 Likes
Thanks for such detailed editorial .
4 Likes
I derived the result that for the answer to be tree 2*(x-1)>=y then it is a Yes else NO is my way of thinking correct
Nice Editorial.
2 Likes
I think that result is incorrect. Since it fails for simple cases (5,8) and (5,7). Even I tried but couldn’t find any direct relation between x and y. I think the only correct results were y = 2*E*O and x = E^{2} + O^{2}.
1 Like
It follows from the fact that N = E+O, and (E+O)^2 = (E^2 + O^2) + (2 * E * O), and x and y should be deducible from this.
1 Like
Got TLE during contest just because of using cout<< <<endl; instead of cout<< <<"\n" ; in C++.
Fast I/O is very important.
7 Likes
+1 got accepted due to this.
1 Like
Loved the editorial… and I must say these written editorials feels way better than video editorials.
4 Likes
I found that odd+even must be a perfect square,
now consider no of vertices n = sqrt(odd+even) ,
Then possible values of odd distances are - { 2*(n-1), 2*(n-1)+2*(n-3), 2*(n-1)+2*(n-3)+2*(n-5)…}
My submission
1 Like
Thank You, very well written editorial.
1 Like
Not Sure if worth noting but I also used similar idea with only difference that I directly calculate the value of even or odd nodes instead of looping over it.
Basic idea: Check if valid n by simply summing the odd and even length and see if they are perfect square. If yes n is root of sum
consider the tree to be like a bipartitite graph(Don’t be alrmed with term not using anything complicated) with two sets of vertices even and odd level vertices.
Now one can easily form an equation for even pairs and odd pairs. Lets say I choose even length pairs then
no of even length pairs= even level* even level + odd level*odd level
(Idea is odd level to odd level require even number of edge and even to even require even number of edge)
now we know lhs and we can choose even level=n-odd level then it will reduce to a simple quadratic equation.
solve equation and check for both soution if any solution gets a valid odd even combination and if yes then we know the odd level and evel level nodes in a tree
Now you just need to create a tree with those nodes.
For tree construction I always think that no of levels doesn’t matter you can always choose to create tree of max 3 level with 1 as root with (even level count) childs and then remaining odd level childs in third level. Construct with same
Here is a submitted solution
https://www.codechef.com/viewsolution/45252281
Although taking O(n) for printing it calc the x and y in O(1) (assuming sqrt and other operations constant)
1 Like
Instead of considering odd pairs, is it possible to construct similarly for even pairs alone ?
Yes !! Why not ? Just take care that even pairs will also be formed by itself too i.e. pair (x, x) is considered as even pair
can anyone help me be telling the mistake in my code
def main():
t = int(input())
for k in range(t):
x,y = map(int, input().split())
n = (x+y)**0.5
if x<y or n%1 != 0 or (x-y)**0.5 % 1 != 0 :
print(“NO”)
continue
i = 1
j = 2
print(“YES”)
print(int(n))
while j<=n:
print(i,j)
j += 1
if j<=n :
print(i,j)
j += 1
i += 1
main() |
We shall explore the similarities between the harmonic triangle, which you see below, and Pascal's triangle. You may like to start by trying this Harmonic Triangle problem.
The rule for generating the harmonic triangle is that you add two consecutive entries to give the entry between them in the row above. Equivalently, from any term, to get the next term, subtract that term from the corresponding term on the row above. It is still possible to work downwards row by row because the entry at the left hand end of the $n$th row is ${1\over n}$. For example the third row of the harmonic triangle is: $${1\over 3},\ {1\over 6},\ {1\over 3}$$ and the fourth row is: $${1\over 4},\ {1\over 12},\ {1\over 12}, \ {1\over 4}$$ which is given by $${1\over 4},\ \ \left[{1\over 3}-{1\over 4}\right] ,\ \ \left[{1\over 6}-{1\over 12}\right],\ \ \left[{1\over 3}-{1\over 12}\right]$$ \begin{array}{ccccccccccc} & & & & & \frac{1}{1} & & & & & \\ & & & & \frac{1}{2} & & \frac{1}{2} & & & & \\ & & & \frac{1}{3} & & \frac{1}{6} & & \frac{1}{3} & & & \\ & & \frac{1}{4} & & \frac{1}{12} & & \frac{1}{12} & & \frac{1}{4} & & \\ & \frac{1}{5} & & \frac{1}{20} & & \frac{1}{30} & & \frac{1}{20} & & \frac{1}{5} & \\ \frac{1}{6} & & \frac{1}{30} & & \frac{1}{60} & & \frac{1}{60} & & \frac{1}{30} & & \frac{1}{6} \\ & & & & & ... & & & & & \end{array}
So is it possible to continue generating the harmonic triangle indefinitely using this rule and will all the terms be fractions with unit numerators?
The entries in the harmonic triangle are related by a similar rule to the entries in Pascal's triangle and both sets of entries involve the binomial coefficients. We denote the $r$th entry in the $n$th row of Pascal triangle by the binomial coefficient: $${n-1\choose r-1}= {(n-1)!\over (r-1)!(n-r)!}.$$ For example the sixth row is $${5 \choose 0} = 1 , {5 \choose 1}= 5, {5\choose 2}=10, {5\choose 3}= 10, {5\choose 4}= 5, {5\choose 5} =1$$ \begin{array}{ccccccccccc} & & & & & 1 & & & & & \\ & & & & 1& & 1 & & & & \\ & & & 1 & & 2 & & 1 & & & \\ & & 1 & & 3 & & 3 & & 1 & & \\ & 1& & 4 & & 6 & & 4 & & 1 & \\ 1& & 5& & 10 & & 10 & & 5 & & 1 \\ & & & & & ... & & & & & \end{array}
The rule for generating Pascal's triangle is that you add two consecutive entries to give the entry between them in the row below. This rule is given by $${n-1\choose r-1} + {n-1 \choose r} = {n\choose r}.$$ This is a well known result, it can be proved by simple algebra and the proof is given in many textbooks.
For the harmonic triangle the $r$th entry in the $n$th row is given by: $$H(n, r) = {(r-1)!\over n(n-1)(n-2)...(n-r+1)}= {(r-1)!(n-r)!\over n!}= {1\over n{n-1\choose r-1}}$$ Knowing that the binomial coefficient ${n-1\choose r-1}$ is a whole number it is clear from this formula that all the fractions in the harmonic triangle will have $1$ as numerator.
It remains to prove that the rule for generating the harmonic triangle works in general for all $n$ and $r$ and then we shall have established that the harmonic triangle can be extended to $n$ rows for any $n$ by using the given formula for the $r$th entry in the $n$th row.
The harmonic triangle rule is given by the formula: $$H(n, r) + H(n, r+1) = H(n-1, r)$$ or equivalently, as described above, $$H(n, r+1) = H(n-1, r)-H(n, r).$$ The proof is left to the reader. All you need to do is substitute the formulae in terms of the factorials into the right hand side of this identity and simplify it to get the left hand side. If you want to check that you have managed to do it correctly you can click on Solution above. |
# Square Root of a Perfect Square by Using the Long Division Method
To find the square root of a perfect square by using the long division method is easy when the numbers are very large since, the method of finding their square roots by factorization becomes lengthy and difficult.
### Steps of Long Division Method for Finding Square Roots:
Step I: Group the digits in pairs, starting with the digit in the units place. Each pair and the remaining digit (if any) is called a period.
Step II: Think of the largest number whose square is equal to or just less than the first period. Take this number as the divisor and also as the quotient.
Step III: Subtract the product of the divisor and the quotient from the first period and bring down the next period to the right of the remainder. This becomes the new dividend.
Step IV: Now, the new divisor is obtained by taking two times the quotient and annexing with it a suitable digit which is also taken as the next digit of the quotient, chosen in such a way that the product of the new divisor and this digit is equal to or just less than the new dividend.
Step V: Repeat steps (2), (3) and (4) till all the periods have been taken up. Now, the quotient so obtained is the required square root of the given number.
### Examples on square root of a perfect square by using the long division method
1. Find the square root of 784 by the long-division method.
Solution:
Marking periods and using the long-division method,
Therefore, √784 = 28
2. Evaluate √5329 using long-division method.
Solution:
Marking periods and using the long-division method,
Therefore, √5329 =73
3. Evaluate: √16384.
Solution:
Marking periods and using the long-division method,
Therefore, √16384 = 128.
4. Evaluate: √10609.
Solution:
Marking periods and using the long-division method,
Therefore, √10609 = 103
5. Evaluate: √66049.
Solution:
Marking periods and using the long-division method,
Therefore, √66049 = 257
6. Find the cost of erecting a fence around a square field whose area is 9 hectares if fencing costs $3.50 per metre. Solution: Area of the square field = (9 × 1 0000) m² = 90000 m² Length of each side of the field = √90000 m = 300 m. Perimeter of the field = (4 × 300) m = 1200 m. Cost of fencing =$(1200 × ⁷/₂) = \$4200.
7. Find the least number that must be added to 6412 to make it a perfect square.
Solution:
We try to find out the square root of 6412.
We observe here that (80)² < 6412 < (81)²
The required number to be added = (81)² - 6412
= 6561 – 6412
= 149
Therefore, 149 must be added to 6412 to make it a perfect square.
8. What least number must be subtracted from 7250 to get a perfect square? Also, find the square root of this perfect square.
Solution:
Let us try to find the square root of 7250.
This shows that (85)² is less than 7250 by 25.
So, the least number to be subtracted from 7250 is 25.
Required perfect square number = (7250 - 25) = 7225
And, √7225 = 85.
9. Find the greatest number of four digits which is a perfect square.
Solution
Greatest number of four digits = 9999.
Let us try to find the square root of 9999.
This shows that (99)² is less than 9999 by 198.
So, the least number to be subtracted is 198.
Hence, the required number is (9999 - 198) = 9801.
10. What least number must be added to 5607 to make the sum a perfect square? Find this perfect square and its square root.
Solution:
We try to find out the square root of 5607.
We observe here that (74)² < 5607 < (75)²
The required number to be added = (75)² - 5607
= (5625 – 5607) = 18
11. Find the least number of six digits which is a perfect square. Find the square root of this number.
Solution:
The least number of six digits = 100000, which is not a perfect square.
Now, we must find the least number which when added to 1 00000 gives a perfect square. This perfect square is the required number.
Now, we find out the square root of 100000.
Clearly, (316)² < 1 00000 < (317)²
Therefore, the least number to be added = (317)² - 100000 = 489.
Hence, the required number = (100000 + 489) = 100489.
Also, √100489 = 317.
12. Find the least number that must be subtracted from 1525 to make it a perfect square.
Solution:
Let us take the square root of 1525
We observe that, 39² < 1525
Therefore, to get a perfect square, 4 must be subtracted from 1525.
Therefore the required perfect square = 1525 – 4 = 1521
Square Root
Square Root
Square Root of a Perfect Square by using the Prime Factorization Method
Square Root of a Perfect Square by Using the Long Division Method
Square Root of Numbers in the Decimal Form
Square Root of Number in the Fraction Form
Square Root of Numbers that are Not Perfect Squares
Table of Square Roots
Practice Test on Square and Square Roots
● Square Root- Worksheets
Worksheet on Square Root using Prime Factorization Method
Worksheet on Square Root using Long Division Method
Worksheet on Square Root of Numbers in Decimal and Fraction Form
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# Directly Proportional – Explanation & Examples
## What does Directly Proportional Mean?
Direct proportion is the relationship between two variables whose ratio is equal to a constant value. In other words, direct proportion is a situation where an increase in one quantity causes a corresponding increase in the other quantity, or a decrease in one quantity results in a decrease in the other quantity.
Sometimes, the word proportional is used without the word direct, just know that they have a similar meaning.
## Directly Proportional Formula
Direct proportion is denoted by the proportional symbol (∝). For example, if two variables x and y are directly proportional to each other, then this statement can be represented as x ∝ y. When we replace the proportionality sign (∝) with an equal sign (=), the equation changes to:
x = k * y or x/y = k, where k is called non-zero constant of proportionality
In our day-to-day life, we often encounter situations where a variation in one quantity results in a variation in another quantity. Let’s take a look at some of the real-life examples of directly proportional concept.
• The cost of the food items is directly proportional to the weight.
• Work done is directly proportional to the number of workers. This means that, more workers, more work and les workers, less work accomplished.
• The fuel consumption of a car is proportional to the distance covered.
Example 1
The fuel consumption of a car is 15 liters of diesel per 100 km. What distance can the car cover with 5 liters of diesel?
Solution
• Fuel consumed for every 100 km covered = 15 liters
• Therefore, the car will cover (100/15) km using 1 liter of the fuel
If 1 liter => (100/15) km
• What about 5 liters of diesel
= {(100/15) × 5} km
= 33.3
Therefore, the car can cover 33.3 km using 5 liters of the fuel.
Example 2
The cost of 9 kg of beans is $166.50. How many kgs of beans can be bought for$ 259?
Solution
• $166.50 = > 9 kg of beans • What about$ 1 => 9/166.50 kg
Therefore the amount of beans purchased for $259 = {(9/166.50) × 259} kg • =14 kg Hence, 14 kg of beans can be bought for$259
Example 3
The total wages for 15 men working for 6 days are $9450. What is the total wages for 19 men working for 5 days? Solution Wages of 15 men in 6 days =>$ 9450
The wage in 6 days for 1 worker = >$(9450/15) The wage in 1 day for 1 worker =>$ (9450/15 × 1/6)
Wages of 19 men in a day => $(9450 × 1/6 × 19) The total wages of 19 men in 5 days =$ (9450 × 1/6 × 19 × 5)
= $9975 Therefore, 19 men earn a total of$ 9975 in 5 days.
### Practice Questions
1. If the total daily wages of 7 women or 5 men is $525.What will be the daily wage of 13 and 7 women and men respectively? 2. The fuel consumption of a vehicle is 6.8L/102km. What distance can this vehicle cover in 24 liters of fuel? 3. The cost of ferrying 160 bags of cement for 125 km is Rs. 60. What will be the cost of ferrying 200 bags for 400 km? 4. The wages for 12 men working for 5 days are$7500. Calculate the wages of 17 men working for 6 days.
5. The cost 16 bars of soap each weighing 1.5 kilogram is $672.Calculate the cost of 18 similar bars of soap each weighing 2 kilograms. 6. The drawing scale of a map is represented as 1:20000000. Calculate the actual distance of two regions that are 4 cm apart on the map. 7. A 7-meter flag post casts a shadow of 5 meters. Calculate the height of an electric pole that will cast a shadow of 10 m under the same condition. 8. A train takes 5 hours to cover 200 km. How long will it take to cover 600 km? 9. If the cost of 16 bars of chocolate each weighing 900 g is$84. Find the cost of 27 bars of chocolate each weighing one kilogram.
10. If the daily wages of four women and three men is $480. Calculate the daily wages of seven and eleven men and women respectively. ### Answers Key 1.$ 1710
2. 363 km
3. $240 4.$ 12750
5. $1008 6. 800 km 7. 14 meters 8. 15 hours 9.$ 157.50.
10. \$ 2440. |
# One To One And Onto Functions Problems Pdf
File Name: one to one and onto functions problems .zip
Size: 1539Kb
Published: 23.04.2021
Advanced Functions. In terms of arrow diagrams, a one-to-one function takes distinct points of the domain to distinct points of the co-domain.
The concept of one-to-one functions is necessary to understand the concept of inverse functions. If a function has no two ordered pairs with different first coordinates and the same second coordinate, then the function is called one-to-one. A graph of a function can also be used to determine whether a function is one-to-one using the horizontal line test:. If each horizontal line crosses the graph of a function at no more than one point, then the function is one-to-one. In each plot, the function is in blue and the horizontal line is in red.
## Lecture 18 : One-to-One and Onto Functions.
A function is a way of matching the members of a set "A" to a set "B":. Surjective means that every "B" has at least one matching "A" maybe more than one. Think of it as a "perfect pairing" between the sets: every one has a partner and no one is left out. If every "A" goes to a unique "B", and every "B" has a matching "A" then we can go back and forwards without being led astray. This is not a function because we have an A with many B. It fails the "Vertical Line Test" and so is not a function.
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In other words no element of are mapped to by two or more elements of. In other words, nothing is left out. In this case the map is also called a one-to-one correspondence. Classify the following functions between natural numbers as one-to-one and onto. Prove that the function is one-to-one. Proof: We wish to prove that whenever then. Let us assume that for two numbers.
We know that a function is a set of ordered pairs in which no two ordered pairs that have the same first component have different second components. Given any x , there is only one y that can be paired with that x. The following diagrams depict functions:. With the definition of a function in mind, let's take a look at some special " types " of functions. This cubic function is indeed a "function" as it passes the vertical line test.
## Discrete Mathematics - Functions
Thanks for visiting our website. Our aim is to help students learn subjects like physics, maths and science for students in school , college and those preparing for competitive exams. All right reserved.
### 5.4: Onto Functions and Images/Preimages of Sets
It is usually symbolized as. A single output is associated to each input, as different input can generate the same output. The set X is called domain of the function f dom f , while Y is called codomain cod f. In particular, if x and y are real numbers, G f can be represented on a Cartesian plane to form a curve. A glance at the graphical representation of a function allows us to visualize the behaviour and characteristics of a function.
One-to-one functions focus on the elements in the domain. We do not want any two of them sharing a common image. Onto functions focus on the codomain. We want to know if it contains elements not associated with any element in the domain. The two functions in Example 5. This function maps ordered pairs to a single real numbers.
#### Comparing Cardinalities of Sets
These solutions for Functions are extremely popular among Class 12 Science students for Math Functions Solutions come handy for quickly completing your homework and preparing for exams. Give an example of a function i which is one-one but not onto ii which is not one-one but onto iii which is neither one-one nor onto. Thus, f is not onto. So, f is not one-one. So, for every element in the co-domain, there exists a pre-image in the domain.
Фонтейн вздохнул и обхватил голову руками. Взгляд его черных глаз стал тяжелым и неподвижным. Возвращение домой оказалось долгим и слишком утомительным. Последний месяц был для Лиланда Фонтейна временем больших ожиданий: в агентстве происходило нечто такое, что могло изменить ход истории, и, как это ни странно директор Фонтейн узнал об этом лишь случайно. Три месяца назад до Фонтейна дошли слухи о том, что от Стратмора уходит жена. Он узнал также и о том, что его заместитель просиживает на службе до глубокой ночи и может не выдержать такого напряжения. Несмотря на разногласия со Стратмором по многим вопросам, Фонтейн всегда очень высоко его ценил.
Беккер заглянул в справочник Управления общей бухгалтерской отчетности США, но не нашел в нем ничего похожего. Заинтригованный, он позвонил одному из своих партнеров по теннису, бывшему политологу, перешедшему на службу в Библиотеку конгресса.
Ему не было нужды выискивать Беккера в толпе, выходящей из церкви: жертва в ловушке, все сложилось на редкость удачно. Нужно только выбрать момент, чтобы сделать это тихо. Его глушитель, самый лучший из тех, какие только можно было купить, издавал легкий, похожий на покашливание, звук. Все будет прекрасно. Приближаясь к пиджаку защитного цвета, он не обращал внимания на сердитый шепот людей, которых обгонял.
Не бывает такой диагностики, которая длилась бы восемнадцать часов. Все это вранье, и ты это отлично знаешь. Скажи мне, что происходит. Сьюзан прищурилась.
Все, что я могу, - это проверить статистику, посмотреть, чем загружен ТРАНСТЕКСТ. Слава Богу, разрешено хоть. Стратмор требовал запретить всяческий доступ, но Фонтейн настоял на. - В шифровалке нет камер слежения? - удивился Бринкерхофф.
Черт возьми! - не сдержался Фонтейн, теряя самообладание.
Но этот канадец не знал, что ему надо держаться изо всех сил, поэтому они и трех метров не проехали, как он грохнулся об асфальт, разбил себе голову и сломал запястье. - Что? - Сьюзан не верила своим ушам. - Офицер хотел доставить его в госпиталь, но канадец был вне себя от ярости, сказав, что скорее пойдет в Канаду пешком, чем еще раз сядет на мотоцикл.
И давайте выбираться отсюда. Стратмор поднял руку, давая понять, что ему нужно подумать. Сьюзан опасливо перевела взгляд в сторону люка. Его не было видно за корпусом ТРАНСТЕКСТА, но красноватое сияние отражалось от черного кафеля подобно огню, отражающемуся ото льда.
Я уже говорила, что мы ушли до их прибытия. - Вы хотите сказать - после того как стащили кольцо. - Мы его не украли, - искренне удивилась Росио.
Вообще-то она ничего не имела против этого имени, но Хейл был единственным, кто его использовал, и это было ей неприятно. - Почему бы мне не помочь тебе? - предложил Хейл. Он подошел ближе. - Я опытный диагност. К тому же умираю от любопытства узнать, какая диагностика могла заставить Сьюзан Флетчер выйти на работу в субботний день.
Может быть, хочешь воды. Она не нашлась что ответить. И проклинала. Как я могла не выключить монитор. Сьюзан понимала: как только Хейл заподозрит, что она искала что-то в его компьютере, то сразу же поймет, что подлинное лицо Северной Дакоты раскрыто.
Название показалось ему чересчур земным для такого агрессора. - Червь, - недовольно сказал Джабба. - Никакой усложненной структуры, один лишь инстинкт: жри, опорожняйся и ползи. Вот что это. Простота.
- Он улыбнулся.
В обычных обстоятельствах это насторожило бы Стратмора, но ведь он прочитал электронную почту Танкадо, а там говорилось, что весь трюк и заключался в линейной мутации. Решив, что никакой опасности нет, Стратмор запустил файл, минуя фильтры программы Сквозь строй. Сьюзан едва могла говорить.
ОБЪЕКТ: ДЭВИД БЕККЕР - ЛИКВИДИРОВАН Пора. Халохот проверил оружие, решительно направился вперед и осмотрел площадку. Левый угол пуст. Следуя плану, он бросился в проход и, оказавшись внутри, лицом к правому углу, выстрелил. Пуля отскочила от голой стены и чуть не попала в него .
Ну конечно, - сказала она, все еще не в силах поверить в произошедшее. - Он хотел, чтобы вы восстановили его доброе имя. - Нет, - хмуро сказал Стратмор.
Коммандер, - сказала. - Это еще не конец. Мы еще не проиграли.
2 Response
1. Teresa E. |
## Mode, Median, Mean and Range
8.2 Mod, Median, Mean and Range
We will learn to recognize and determine the mode, median, mean, and range from uncollected data.
MODE
The category with the highest repetition frequency.
EXAMPLE:
Let's say you have the following list of numbers:
• 3, 3, 8, 9, 15, 15, 15, 17, 17, 27, 40, 44, 44
In this case, the mode is 15 because it is the most frequently occurring round number. However, if there is one of the number 15 missing on your list, then you will have four modes which are 3, 15, 17, and 44.
MEDIAN
The data value that is in the middle of a data set arranged in ascending order.
EXAMPLE:
• 3, 9, 15, 17, 44
The middle or median number is 15.
MEAN
Average number.
EXAMPLE:
Let's say you have four test scores 15, 18, 22, and 20.
To get the average, you must add all four scores together, then divide the total by four. The average produced was 18.75.
Written, it looks like this:
$$\frac{15+18+22+20}{4}=\frac{75}{4}=18.75$$
RANGE
The difference between the maximum value and the minimum value.
EXAMPLE:
• 3, 6, 9, 15, 44
The range is simply the smallest number subtracted from the largest number in the set. To calculate the range, you subtract 3 from 44, giving you a range of 41. Having been written, the equation looks like this:
44 - 3 = 41 |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 12.3: Defining Equivalence or Similarity
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
What do we mean when we say that two actors have "similar" patterns of relations, and hence are both members of the same role or social position? Network analysts most broadly defines two nodes (or other more elaborate structures) as similar if they fall in the same "equivalence class". Frankly, that's no immediate help. But it does say that there is something that would cause us to say two actors (or other structures) are members of a "class" that is different from other "classes".
Now it becomes a question of what features of an actor's position place them into a "class" with other actors? In what way are they "equivalent"?
There are many ways in which actors could be defined as "equivalent" based on their relations with others. For example, we could create two "equivalence classes" of actors with out-degree of zero, and actors with out-degree of more than zero. Indeed, a very large number of the algorithms we've examined group sets of actors into categories based on some commonality in their positions in graphs.
Three particular definitions of "equivalence" have been particularly useful in applying graph theory to the understanding of "social roles" and "structural positions". We will look at these in the next three sections on "structural equivalence", "automorphic equivalence", and "regular equivalence". Of these, "automorphic" has rarely been used in substantive work.
The basic ideas of these three kinds of equivalence are easily illustrated with a simple graph (developed by Wasserman and Faust). Consider Figure 12.1, a simple graph of the relations among nine actors "A" to "I".
Figure 12.1: Wasserman-Faust network to illustrate equivalence classes
This graph provides particularly clear examples of how structural, automorphic, and regular equivalence differ. Let's look in more detail at these ideas, starting with the most restrictive notion of what it means for actors to be equivalent.
## Structural Equivalence
Two nodes are said to be exactly structurally equivalent if they have the same relationships to all other nodes. Structural equivalence is easy to grasp (though it can be operationalized in a number of ways) because it is very specific: two actors must be exactly substitutable in order to be structurally equivalent.
In Figure $$\PageIndex{1}$$, there are seven "structural equivalence classes". Can you find them?
• There is no actor who has exactly the same set of ties as actor A (ties to B, C, and D), so actor A is in a class by itself.
• The same is true for actors B, C, and D. Each of these actors has a unique set of ties to others, so they form three classes, each with one member.
• E and F, however, fall in the same structural equivalence class. Each has a single tie; and that tie is to actor B. Since E and F have exactly the same pattern of ties with all other actors, they are structurally equivalent.
• Actor G, again, is in a class by itself. Its profile of ties with the other nodes in the diagram is unique.
• Finally, actors H and I fall in the same structural equivalence class. That is, they have exactly the same pattern of ties to all other actors.
Actors that are structurally equivalent are in identical "positions" in the structure of the diagram. Whatever opportunities and constraints operate on one member of a class are also present for the others. The nodes in a structural equivalence class are, in a sense, in the same position with regard to all other actors.
Because exact structural equivalence is likely to be rare (particularly in large networks), we often are interested in examining the degree of structural equivalence, rather than the simple presence or absence of exact equivalence.
Structural equivalence is the "strongest" form of equivalence that network analysts usually consider. If we soften the requirements just a bit, we can often find some interesting other patterns of equivalence.
## Automorphic Equivalence
The idea of structural equivalence is powerful because it identifies actors that have the same position, or who are completely substitutable. But, even intuitively, you can probably imagine other "less strict" definitions of what it means for two actors to be similar or equivalent.
Suppose that the graph in Figure $$\PageIndex{1}$$ described a franchise group of hamburger restaurants. Actor A is the central headquarters, actors B, C, and D are the managers of three different stores. Actors E and F are workers at one store; G is the lone worker at a second store; H and I are workers at the third store.
Even though actor B and actor D are not structurally equivalent (they doe have the same boss, but not the same workers), they do seem to be "equivalent" in a different sense. Both manager B and manager D report to a boss (in this case, the same boss), and each has exactly two workers. These are different people, but the two managers seem somehow equivalent. If we swapped them, and also swapped the four workers, all of the distances among all the actors in the graph would be exactly identical. In fact, actors B and D form an "automorphic" equivalence class.
In Figure $$\PageIndex{1}$$, there are actually five automorphic equivalence classes: {A}, {B, D}, {C}, {E, F, H, I}, and {G}. These classes are groupings whose members would remain at the same distance from all other actors if they were swapped, and, members of other classes were also swapped.
The idea of automorphic equivalence is that sets of actors can be equivalent by being embedded in local structures that have the same patterns of ties - "parallel" structures. Large scale populations of social actors (perhaps like hamburger restaurant chains) can display a great deal of this sort of "structural replication". The faces are different, but the structures are identical.
Note that the less strict definition of "equivalence" had reduced the number of classes. If we are willing to go one important step further, we can reduce the complexity still further.
## Regular Equivalence
Two nodes are said to be regularly equivalent if they have the same profile of ties with other sets of actors that are also regularly equivalent. This is a complicated way of saying something that we recognize intuitively.
Two mothers, for example, are "equivalent" because each has a certain pattern of ties with a husband, children, and in-laws (for one example - but one that is very culturally relative). The two mothers do not have ties to the same husband (usually) or the same children or in-laws. That is, they are not "structurally equivalent". Because different mothers may have different numbers of husbands, children, and in-laws, they will not be automorphically equivalent. But they are similar because they have the same relationships with some member or members of another set of actors (who are themselves regarded as equivalent because of the similarity of their ties to a member of the set "mother").
This is an obvious notion, but a critical one. Regular equivalence sets describe the "social roles" that are the basic building blocks of all social institutions. Actors that are regularly equivalent to not necessarily fall in the same network positions or locations with respect to other individual actors; rather, they have the same kinds of relationships with some members of other sets of actors.
In Figure $$\PageIndex{1}$$, there are three regular equivalence classes. The first is actor A; the second is composed of the three actors B, C, and D; the third is composed of the remaining five actors E, F, G, H, and I.
The easiest class to see is the give actors across the bottom of the figure (E, F, G, H, and I). These actors are regularly equivalent to one another because a) they have no tie with any actor in the first class (that is, with actor A) and b) each has a tie with an actor in the second class (either B, C, or D). Each of the five actors, then, has an identical pattern of ties with actors in the other classes.
Actors B, C, and D form a class because a) they have a tie with a member of the first class (that is, with actor A) and b) they each have a tie with a member of the third class. B and D actually have ties with two members of the third class, whereas actor C has a tie to only one member of the third class; this doesn't matter, as there is a tie to some member of the third class.
Actor A is in a class by itself, defined by a) a tie to at least one member of class two and b) no tie to any member of class three.
As with structural and automorphic equivalence, exact regular equivalence may be rare in a large population with many equivalence classes. Approximate regular equivalence can be very meaningful though, because it gets at the notion of which actors fall in which social roles, and how social roles (not role occupants) relate to one another. |
## Aug 25, 2015
### IMPORTANT SHORTCUT FORMULAS RELATED TO AVERAGE PROBLEMS
Important Shortcut Formulas related to AVERAGE
Shared By: WWW.GovtJobsPortal.IN
Average: An average or arithmetic mean of given n number of items is a total sum divided by total number of items given.
Therefore, Average= Sum of Total Items Given/Number of Items
Let’s say If we have Given n Items: n1, n2, n3, ….n
Then, Total Average= n1+n2+n3+----+n
n
Important Formulas:
· When a person covers the equal amount of distance at two different intervals of speed, Let’s say one at x km/h and y km/h respectively, Then
Total Average Speed = (2xy)/(x+y)
· When a person covers the equal amount of distance at three different intervals of speed one at x km/h & others two at y km/h and z km/h respectively, Then
Total Average Speed = (3xyz)/(xy+yz+zx)
· If we have given the average of n items equal to x and average of n’ items equal to x’, then,
Total Average (n+n’) will be =(nx+n’x’)
(n+n’)
· Let’s say there are total x members in a family & their total average age is y years. Then a new member joins the family and the new average age of the family becomes n, then the age of new member that joins the family will be =[n+x(n-y)] years
· Let’s say there are total x members in a family & their total average age is y years. Then a member leaves the family and the new average age of the family becomes n, then the age of that member will be =[n-x(y-n)] years
· To Find the Average of first x consecutive natural numbers =(x+1)/2
· To Find the Average of first x natural Even numbers =(x+1)
· To Find the Average of natural even numbers up to x =(x/2)+1
· To Find the Average of first x natural odd numbers= x
· To Find the Average of natural odd numbers up to x = (x+1)/2
Important Points to be Remember:
1. If the value of each item is increased by the same value let’s say x, then their total average will also be increase by that same value x.
2. If the value of each given item is decreased by the same value let’s say x, then their total average will also be decrease by that same value x.
3. If the value of each given item is multiplied by the same value let’s say x, then their total average will also be get multiplied by that same value x.
4. If the value of each given item is divided by the same value let’s say x, then their total average will also be get divided by that same value x.
Let’s Take an Example Now:
Q) If the average age of 20 students in a class given is 16 & When a teacher joins the class then their average age increases by 1. Then what will be teacher’s age?
Normal Method:
[Total No. of students × Average Age] – [Total No. of Students including teacher) × Average Age (Including Teacher)]= Answer.
Therefore, According to Question We get
20×16= 320
21×17= 357,
357-320= 37.
Therefore, 37 years will be the required answer.
Shortcut Method:
[n+x(n-y)] years ( As According to the Formula Given Above )
Where,
n= new Average Age
x=Number of Students
y= Previous Average Age
Therefore, we get
= [17+20(17-16)]
= 17 + 20
=37
Therefore 37 years will be the required answer.
Thank You !! J
RELATED JOBS |
# Question Video: Simplifying Numerical Expressions Involving Square Roots Mathematics • 10th Grade
Simplify √6(−3 − √6) − 3(−6 + √6).
02:02
### Video Transcript
Simplify root six multiplied by negative three minus root six minus three multiplied by negative six plus root six.
In order to simplify this expression, we need to use some of the laws of surds. Firstly, root 𝑎 multiplied by root 𝑎 is equal to 𝑎. Secondly, root 𝑎 multiplied by 𝑏, an integer, is equal to 𝑏 root 𝑎. And root 𝑎 multiplied by root 𝑏 is equal to root 𝑎 times 𝑏.
In order to simplify the expression, we need to expand or multiply out the parentheses. Root six multiplied by negative three is negative three root six. Root six multiplied by negative root six is equal to negative six. Moving on to the second bracket or parenthesis, negative three multiplied by negative six is positive 18. And finally, negative three multiplied by positive root six is negative three root six.
We then need to simplify this expression by grouping the like terms. Negative three root six minus three root six is equal to negative six root six. In the same way, negative six plus 18 is equal to positive 12. This means that our answer is negative six root six plus 12.
Simplifying root six multiplied by negative three minus root six minus three multiplied by negative six plus root six is equal to negative six root six plus 12. |
Print
Maths
Inter-quartile range, cumulative frequency, box and whisker plots - Higher
Page:
Interquartile range
We know that the median divides the data into two halves. We also know that for a set of n ordered numbers the median is the (n + 1) ÷ 2 th value.
Similarly, the lower quartile divides the bottom half of the data into two halves, and the upper quartile also divides the upper half of the data into two halves.
Lower quartile is the (n + 1) ÷ 4 th value.
Upper quartile is the 3 (n + 1) ÷ 4 th value.
Question
Find the median, lower quartile and upper quartile for the following data:
11, 4, 6, 8, 3, 10, 8, 10, 4, 12 and 31.
Ordering the data, we get 3, 4, 4, 6, 8, 8,10, 10, 11, 12 and 31.
The median is the (11 + 1) ÷ 2 = 6th value.
The lower quartile is the (11 + 1) ÷ 4 = 3rd value.
The upper quartile is the 3 (11 + 1) ÷ 4 = 9th value.
Therefore, the median is 8, the lower quartile is 4, and the upper quartile is 11.
3, 4, 4, 6, 8, 8, 10, 10, 11, 12, 31
The interquartile range is the difference between the upper quartile and lower quartile.
In this example, the interquartile range is 11 - 4 = 7.
Question
A survey was carried out to find the number of pets owned by each child in a class.
The results are shown in the table:
Number of petsFrequency
03
15
22
37
410
53
61
>60
Find the interquartile range.
3.
Remember that there is a total of 31 children in the class.
• The lower quartile is the 8th value, which is 1.
• The upper quartile is the 24th value, which is 4.
• Therefore, the interquartile range is 4 - 1 = 3.
Note that the interquartile range ignores extreme values. The range includes extreme values.
Look at this set of data:
• 1, 5, 7, 8, 9, 12, 13, 15, 17, 18, 35.
• The interquartile range is 17 - 7 = 10.
• The range is 35 - 1 = 34.
In cases such as these, it is often preferable to use the interquartile range when comparing the data.
Page:
Back to Statistics and probability index
BBC iD |
# Random Walk III
Alignments to Content Standards: S-CP.B.9
Imagine Scott stood at zero on a life-sized number line. His friend flipped a coin $6$ times. When the coin came up heads, he moved one unit to the right. When the coin came up tails, he moved one unit to the left. After each flip of the coin, Scott's friend recorded his position on the number line. Let $f$ assign to the whole number $n$, when $1 \leq n \leq 6$, Scott's position on the number line after the $n^{\rm th}$ coin flip.
1. How many different outcomes are there for the sequence of $6$ coin tosses?
2. Calculate the probability, before the coin flips have begun, that $f(6) = 0$, $f(6) =1$, and $f(6)= 6$.
3. Make a bar graph showing the frequency of the different outcomes for this random walk.
4. Which number is Scott most likely to land on after the six coin flips? Why?
## IM Commentary
This task is part of a progression, starting with Random Walk and Random Walk II which stressed the function aspect of this situation, transitioning to the probability and statistics side.
The task provides a context to calculate discrete probabilities and represent them on a bar graph. It could also be used to create a class activity where students gather, represent, and analyze data, running simulations of the random walk and recording and then displaying their results.
## Solutions
Solution: 1.
1. If we denote by $H$ an outcome of heads and by $T$ and outcome of tails, then the possible outcomes for the six coin flips can be represented by a sequence of six letters, with each letter being either a $T$ or an $H$. For example $HHHHHH$ would represent the case where all six coin tosses come up heads. There are two choices for each letter and six letters so this gives $2^6 = 64$ possible outcomes.
To see more clearly why the two possibilities for the first toss need to be multiplied by the two possibilities for the second toss observe that each outcome, $H$ or $T$ for the first toss, leads to two possible outcomes when paired with the second toss: $H$ leads to $HH$ and $HT$ while $T$ leads to $TH$ and $TT$. This reasoning for multiplying by two for each additional coin toss leads to $2^6$ possibilities if the coin is tossed six times.
2. For $f(6)$ to be zero, there need to be the same number of steps to the left as to the right. So there must be $3$ heads and $3$ tails. We can make a list of the possibilities. To simplify the list, we will list the possibilities for which tosses came up heads so that $123$ will mean that the first, second, and third tosses were heads while the fourth, fifth, and sixth were tails: $$\begin{array}{c} 123,124,125,126,134,135,136,145,146,156,234,235,236,\\ 245,246,256, 345,346,356,456. \end{array}$$ There are twenty possibilities so the probability that $f(6) = 0$ is $\frac{20}{64}$.
A word is worthwhile regarding the way in which the possible ways three heads could occur was listed above. The list is long and so to avoid duplication and to make sure the list is complete a method is useful. The list above has been formed by choosing the smallest first number (this is why the first ten numbers in the list begin with a $1$), then the smallest possible second number, and finally the smallest third number.
In order for $f(6)$ to be one there would need to be one more occurrence of heads than tails. But this means that the total number of coin tosses would have to be odd: if $n$ is any whole number then $n + (n+1) = 2n+1$ is an odd number. Therefore the probability that $f(6) = 1$ is $0$. The argument for why $f(6)$ can never equal $1$ is essentially the same as the argument in ''Random Walk II'' for why it is not possible for $f(7)$ to be equal to $2$.
Finally for the probability that $f(6) = 6$. This means that Scott must have moved in the positive direction after each of the six coin tosses so all six coin tosses must have been heads. So this can happen in only one way while there are $2^6 = 64$ different possible outcomes for the six coin tosses so the probability that $f(6) = 6$ is $\frac{1}{64}$.
3. The bar graph below shows the probabilities for the different possible outcomes. In part (b) we found that the probability of landing on $0$ is $\frac{20}{64}$. The other possible landing points, with $6$ coin tosses are $\pm 2, \pm 4, \pm 6$. The probabilities for these are, respectively $\frac{15}{64}, \frac{6}{64}$, and $\frac{1}{64}$. This can be found making lists as in part (b) or with combinatorial symbols as in solution 2 below.
4. In part (c) the probability of each possible outcome has been calculated and the most likely spot for Scott to land on after his walk is zero.
Solution: 2. Using Combinatorial symbols to calculate probabilities
Here we give an alternative approach to parts (b) and (c) of the problem. If $a$ and $b$ are positive whole numbers with $a \geq b$ the combinatorial symbol $$\left( \begin{array}{c} a \\b \end{array}\right)$$ denotes the number of collections of $a$ objects which can be chosen from a collection of $b$ objects. Many textbooks use the symbol $_aC_b$ (said ''$a$ choose $b$'') to denote this quantity: $$_aC_b = \left( \begin{array}{c} a \\b \end{array}\right).$$ For example, $$_6C_1 = \left( \begin{array}{c} 6\\1 \end{array}\right) = 6$$ since there are $6$ different objects which could be chosen from a collection of $6$. We can also see that $$_6C_2 = \left( \begin{array}{c} 6 \\2 \end{array}\right) =15$$ because there are $6$ choices of a first object $x_1$, $5$ choices left for the second object $x_2$, but each pair $\{x_1, x_2\}$ gets listed twice, once as $\{x_1,x_2\}$ and once as $\{x_2,x_1\}$. So the thirty choices of two objects gives $15$ different pairs. Finally, we study the case considered in part (b) of the problem $$_6C_3 = \left( \begin{array}{c} 6 \\3 \end{array}\right) = 20$$ as was seen in the first solution. Extending the explanation used for choosing one or two out of six, there are $6$ choices for a first object $x_1$, $5$ choices for a second object $x_2$, and $4$ choice for a third object $x_3$. Then there are $6$ ways to list this triplet: $$\{x_1,x_2,x_3\}, \{x_1,x_3,x_2\}, \{x_2,x_1,x_3\}, \{x_2,x_3,x_1\}, \{x_3,x_1,x_2\}, \{x_3,x_2,x_1\}.$$ It is important to note, for the generalization to follow, that $6 = 3 \cdot 2$: the $6$ different ways of listing $x_1,x_2,x_3$ come from the fact that there are $3$ choices for what goes first and then two choices for what goes second (and only one choice for what goes last).
In general to find $$\left( \begin{array}{c} a \\b \end{array}\right)$$ we can extend the method used above when $a = 6$ and $b = 3$. There are $a$ choices for a first object, $a-1$ for a second, and so on with $a - b+ 1$ for the $b^{\rm th}$ object. Each collection of $b$ objects gets counted $b(b-1)\cdot \ldots \cdot 1$ times, with $b$ choices for the first element of the set, $b-1$ for the second, and so on. Thus we have found $$\left( \begin{array}{c} a \\b \end{array}\right) = \frac{a(a-1) \cdot \ldots \cdot (a-b+1)}{b \cdot (b-1) \cdot \ldots \cdot 1}.$$
Now that we have a general formula, we can apply this to the situation considered here in Scott's random walk. For part (b) we know that $f(6) = 0$ if there are $3$ heads and $3$ tails. So the number of ways this can happen is $$\left( \begin{array}{c} 6 \\3 \end{array}\right) =\frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 20$$ giving a probability of $\frac{20}{64}$ of having $3$ heads and $3$ tails. The probability of landing on $6$ is the same as the probility of having $6$ heads and this is $$\left( \begin{array}{c} 6 \\6 \end{array}\right) = \frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 3 \cdot 1}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 1$$ possibility out of $64$.
All of the probabilities in part (c) can be calculated in the same way using the general formula for calculating combinations. |
# The Significance of Negative Slope
In mathematics, the slope of a line (m) describes how rapidly or slowly change is occurring and in which direction, whether positive or negative. Linear functions-those whose graph is a straight line-have four possible types of slope: positive, negative, zero, and undefined. A function with a positive slope is represented by a line that goes up from left to right, while a function with a negative slope is represented by a line that goes down from left to right. A function with zero slope is represented by a horizontal line, and a function with an undefined slope is represented by a vertical line.
Slope is usually expressed as an absolute value. A positive value indicates a positive slope, while a negative value indicates a negative slope. In the function y = 3x, for example, the slope is positive 3, the coefficient of x.
In statistics, a graph with a negative slope represents a negative correlation between two variables. This means that as one variable increases, the other decreases and vice versa. Negative correlation represents a significant relationship between the variables x and y, which, depending on what they are modeling, can be understood as input and output, or cause and effect.
### How to Find Slope
Negative slope is calculated just like any other type of slope. You can find it by dividing the rise of two points (the difference along the vertical or y-axis) by the run (the difference along the x-axis). Just remember that the "rise" is really a fall, so the resulting number will be negative. The formula for the slope can be expressed as follows:
m = (y2 - y1) / (x2 - x1)
Once you graph the line, you'll see that the slope is negative because the line goes down from left to right. Even without drawing a graph, you will be able to see that the slope is negative simply by calculating m using the values given for the two points. For example, suppose the slope of a line that contains the two points (2,-1) and (1,1) is:
m = 1 - (-1) / (1 - 2)
m = (1 + 1) / -1
m = 2 / -1
m = -2
A slope of -2 means that for every positive change in x, there will be twice as much negative change in y.
### Negative Slope = Negative Correlation
A negative slope demonstrates a negative correlation between the following:
• Variables x and y
• Input and output
• Independent variable and dependent variable
• Cause and effect
Negative correlation occurs when the two variables of a function move in opposite directions. As the value of x increases, the value of y decreases. Likewise, as the value of x decreases, the value of y increases. Negative correlation, then, indicates a clear relationship between the variables, meaning one affects the other in a meaningful way.
In a scientific experiment, a negative correlation would show that an increase in the independent variable (the one manipulated by the researcher) would cause a decrease in the dependent variable (the one measured by the researcher). For example, a scientist might find that as predators are introduced into an environment, the number of prey gets smaller. In other words, there is a negative correlation between number of predators and number of prey.
### Real-World Examples
A simple example of negative slope in the real world is going down a hill. The farther you travel, the farther down you drop. This can be represented as a mathematical function where x equals the distance traveled and y equals the elevation. Other examples of negative slope demonstrate the relationship between two variables might include:
Mr. Nguyen drinks caffeinated coffee two hours before his bedtime. The more cups of coffee he drinks (input), the fewer hours he will sleep (output).
Aisha is purchasing a plane ticket. The fewer days between the purchase date and the departure date (input), the more money Aisha will have to spend on airfare (output).
John is spending some of the money from his last paycheck on presents for his children. The more money John spends (input), the less money he will have in his bank account (output).
Mike has an exam at the end of the week. Unfortunately, he would rather spend his time watching sports on TV than studying for the test. The more time Mike spends watching TV (input), the lower Mike's score will be on the exam (output). (In contrast, the relationship between time spent studying and exam score would be represented by a positive correlation since an increase in studying would lead to a higher score.) |
Question
# Using ruler and compasses only, construct a rhombus whose diagonals are 8 cm and 6 cm. Measure the length of its one side.
Hint: Name the rhombus and its diagonals. First draw the diagonal 8 cm. Bisect 8 cm and draw a perpendicular bisector PQ. Mark 3 cm on top and bottom of the perpendicular to get 3 cm. Join all sides and measure them.
We have to construct a rhombus when only the length of the diagonals of the rhombus is given.
Let us first name the rhombus as ABCD.
Let us consider BD as one diagonal which can be taken equal to 8 cm.
$\therefore BD=8cm$.
Let us consider the other diagonal as AC with the length of 6 cm.
$\therefore AC=6cm$.
Now, we are going to construct rhombus ABCD with diagonal lengths $BD=8cm$and $AC=6cm.$
First draw a line BD of 8 cm.
Take the bisector of BD of 8 cm. Draw the perpendicular bisector PQ of BD. Let’s take the center point of BD as ‘O’ where it is bisected.
Now mark OA = OC = 3 cm, i.e. it will be half of the diagonal of 6 cm.
Now we have to join all the sides to get the rhombus. We have drawn the two diagonals $BD=8cm$and $AC=6cm.$
Let’s join AB, AD, BC and CD.
Now if we measure side AB, we will get it as 5 cm.
Similarly, if we measure AD, BC and CD we will get it as 5cm.
To find the perpendicular bisector, take more than half of the length of side BD. Cut an arc above and below the line segment BC from point B and C. Then join the perpendicular bisector PQ.
Hence, we constructed the required figure of rhombus ABCD.
Note: You can also take base as 6 cm then bisect and draw perpendicular to 6 cm. Then mark 4 cm in top and bottom of the segment. Join them to get 8 cm. Then join all the sides AB, BC, CD and DA. Still you will get the same length as 5 cm for all sides of the rhombus. |
# Intersection of Sets
The intersection of two sets A and B ( denoted by A∩B ) is the set of all elements that is common to both A and B. In mathematical form,
```For two sets A and B,
A∩B = { x: x∈A and x∈B }
```
Similarly for three sets A, B and C,
`A∩B∩C = { x: x∈A and x∈B and x∈C }`
## Intersection of Sets Examples
### Example #1: Intersection of Two Sets With Venn Diagram
If A = {a, b, c, d, e} and B = {d, e, f, g}, find A∩B.
```Here,
A = {a, b, c, d, e}
B = {d, e, f, g}
Now,
A∩B = {a, b, c, d, e} ∩ {d, e, f, g}
∴ A∩B = {d, e}```
### Example #2
Suppose A = { x: x is an integer between 1 and 7} and B = { x: x is an integer between 4 and 9} then, find A∩B.
```Here,
A = {2, 3, 4, 5, 6}
B = {5, 6, 7, 8}
∴ A∩B = {5, 6} = { x: x is an integer between 4 and 7}```
### Example #3: Intersection of Disjoint Sets
If A = {a, b, c} and B = {d, e, f, g}, find A∩B.
```Here,
A∪B = {a, b, c} ∩ {d, e, f, g}
∴ A∪B = {} = ϕ```
### Example #4
If A = { x: x is an integer} and B = { x: x is an even integer} then, find A∩B.
```Here,
A = {..., -3, -2, -1, 0, 1, 2, ...}
B = {..., -2, 0, 2, ...}
∴ A∩B = {..., -4, -2, 0, 2, ... } = {x: x is an even integer}```
### Example #5: Intersection of Three Sets With Venn Diagram
If A = {a, b, c, d, e}, B = {d, e, f, g} and C = {c, e, f, h, i}, find A∩B∩C.
```Here,
A∩B∩C = {a, b, c, d, e} ∩ {d, e, f, g} ∩ {c, e, f, h, i}
The common element among all three sets is 'e'
∴ A∩B = {e}```
## Properties of Intersection of Sets
### 1. Commutative Property
If A and B are two sets then, A∩B = B∩A
```Suppose,
A = {1, 2, 3, 4}
B = {4, 3, 5}
A∩B = {3, 4}
B∩A = {4, 3}
∴ A∩B = B∩A```
### 2. Associative Property
If A, B and C are three sets then, A∩(B∩C)= (A∩B)∩C.
```Suppose,
A = {1, 2, 3, 4}
B = {3, 4, 5, 6}
C = {3, 4, 1, 5, 8, 9}
B∩C = {3, 4, 5}
A∩B = {3, 4}
A∩(B∩C) = {3, 4}
(A∩B)∩C = {3, 4}
∴ A∩(B∩C) = (A∩B)∩C ```
### 3. Identity Property
The intersection of a set and the empty set is always the empty set, i.e, A∩ϕ = ϕ.
```Suppose,
A = {a, b, c}
B = ϕ = { }
A∩B = {1, 2, 3, } ∩ {} = {} = ϕ
∴ A∪ϕ = ϕ``` |
# The sum of three consecutive even integers is 228, how do you find the integers?
Jan 26, 2016
$74$, $76$ and $78$
#### Explanation:
Let the first of your integers be $x$.
As you are only looking at even integers, the next consecutive even integer would be $x + 2$ and the consecutive even integer after that would be $x + 4$.
You know that their sum is $228$, so you have
$x + \left(x + 2\right) + \left(x + 4\right) = 228$
$\iff \textcolor{w h i t e}{\times x} x + x + 2 + x + 4 = 228$
$\iff \textcolor{w h i t e}{\times \times \times \times \times x} 3 x + 6 = 228$
Subtract $6$ from both sides of the equation:
$\iff 3 x = 222$
Divide by $3$ on both sides of the equation:
$\iff x = 74$
Thus, your consecutive even integers are $74$, $76$ and $78$. |
# Diagonal Matrix – Explanation & Examples
A diagonal matrix is a square matrix whose elements, other than the diagonal, are zero. There are certain conditions that must be met for a matrix to be called a diagonal matrix. Firstly, let’s check the formal definition of a diagonal matrix.
A square matrix in which all the elements except the principal diagonal are zero is known as a diagonal matrix.
In this article, we are going to take a close look at what makes a matrix diagonal, how to find diagonal matrices, properties of diagonal matrices, and the determinant of a diagonal matrix. Let’s start!
## What is a Diagonal Matrix?
A matrix to be classified as a diagonal matrix, it has to meet the following conditions:
• square matrix
• all elements (entries) of the matrix, other than the principal diagonal, has to be $0$
A square matrix is said to be:
• lower triangular if its elements above the principal diagonal are all $0$
• upper triangular if its elements below the principal diagonal are all $0$
Lower Triangular Matrix
Upper Triangular Matrix
diagonal matrix is a special square matrix that is BOTH upper and lower triangular since all elements, whether above or below the principal diagonal, are $0$.
## How to find Diagonal Matrix
To find, or identify, a diagonal matrix, we need to see if it is a square matrix and all the elements besides the principal diagonal (diagonal that runs from top left to bottom right) are $0$. Let’s take a look at the matrices shown below:
$A = \begin{pmatrix} 3 & 0 \\ 0 & { – 3 } \end {pmatrix}$
$B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & { – 2 } \end {pmatrix}$
$C = \begin{bmatrix} 10 & 0 \\ 0 & 12 \\ 0 & 12 \end {bmatrix}$
$D = \begin{bmatrix} { – 5 } & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & { 9 } \end {bmatrix}$
Note the following observations about each of the $4$ matrices shown above:
• Matrix $A$ is a square matrix because its has the same number of rows and columns ($2 \times 2$ matrix). The principal diagonal has entries $3$ and $-3$, respectively. All other entries are $0$. Thus, this is a diagonal matrix.
• Matrix $B$ is a square matrix because its has the same number of rows and columns ($3 \times 3$ matrix). The principal diagonal has entries $1$, $5$, and $-2$, respectively. All other entries are $0$. Thus, this is a diagonal matrix.
• A false glance at Matrix $C$ will make us think that it is a diagonal matrix. But, first and foremost, it is not a square matrix. Hence, it cannot be a diagonal matrix.
• This one’s a bit tricky. First of all, it is in fact a square matrix ( $3 \times 3$ ). You will think that it is not a diagonal matrix.
But why? Is it because there’s a $0$ in the middle?
If you take a closer look at the definition of a diagonal matrix, you will see that nowhere does it say that the entries in the diagonal cannot be $0$! The condition is that the elements other than the diagonal has to be $0$. So, even if there are elements that are $0$ in the diagonal, it won’t matter. As long as the elements besides the diagonal are $0$, it will be a diagonal matrix.
Thus, Matrix $D$ is in fact a diagonal matrix!
This brings us to $2$ special types of diagonal matrices:
• Identity Matrix
• Zero Matrix
Identity Matrix
This is a square matrix in which all the entries in the principal diagonal are $1$ and all other elements are $0$. A $2 \times 2$ and a $3 \times 3$ identity matrices are shown below.
$2 \times 2$ identity matrix
$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end {bmatrix}$
$3 \times 3$ identity matrix
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {bmatrix}$
Zero Matrix
A matrix in which all the elements are $0$. A $2 \times 2$ and a $3 \times 3$ zero matrices are shown below.
$2 \times 2$ zero matrix
$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end {bmatrix}$
$3 \times 3$ zero matrix
$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end {bmatrix}$
Now, let’s look at some properties of diagonal matrices.
## Properties of Diagonal Matrices
There are several properties of diagonal matrices but for the purpose of this article, we will look at $3$ properties of diagonal matrices. Below, we take a look at the properties and their examples.
Property 1:
When $2$ diagonal matrices of the same order are added or multiplied together, the resultant matrix is another diagonal matrix with the same order.
Consider the matrices shown below:
$A = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end {bmatrix}$
$B = \begin{bmatrix} 1 & 0 \\ 0 & 5 \end {bmatrix}$
Now, we add both the $2 \times 2$ matrices and show that the resultant matrix is also diagonal.
$A + B = \begin{bmatrix} 3+1 & 0+0 \\ 0+0 & 4+5 \end {bmatrix}$
$A + B = \begin{bmatrix} 4 & 0 \\ 0 & 9 \end {bmatrix}$
Thus, we see that the resultant matrix, $A + B$, is also a diagonal matrix of the order $2 \times 2$.
Let’s check matrix multiplication with the same matrices. We multiply Matrix $A$ and Matrix $B$ and show that the resultant is also a diagonal matrix with the same order. Shown below:
$A \times B = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end {bmatrix} \times \begin{bmatrix} 1 & 0 \\ 0 & 5 \end {bmatrix}$
$A \times B = \begin{bmatrix} {3\times1+0\times0} & 3\times 0 + 0 \times 5 \\ 0 \times 1 + 4 \times 0 & 0\times0 + 4 \times 5 \end {bmatrix}$
$A \times B = \begin{bmatrix} 3 & 0 \\ 0 & 20 \end{bmatrix}$
Thus, we see that the resultant matrix, $A \times B$, is also a diagonal matrix of the order $2 \times 2$.
Property 2:
The transpose of a diagonal matrix is the matrix itself.
If we have a matrix $A$, then we denote its transpose as $A^{T}$. Transposing a matrix means to flip its rows and columns. Let’s show that this property is true by calculating the transpose of Matrix $A$.
$A = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end {bmatrix}$
$A^{T} = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end {bmatrix}$
Interchanging the rows and columns produces the same matrix because of the entries besides the diagonal being $0$.
Property 3:
Under multiplication, diagonal matrices are commutative.
If we have $2$ matrices, $A$ and $B$, this means $AB = BA$. Let’s show this property by using the two matrices from above.
$A \times B = \begin{bmatrix} {3\times1+0\times0} & 3\times 0 + 0 \times 5 \\ 0 \times 1 + 4 \times 0 & 0\times0 + 4 \times 5 \end {bmatrix}$
$A \times B = \begin{bmatrix} 3 & 0 \\ 0 & 20 \end{bmatrix}$
Now,
$B \times A = \begin{bmatrix} {1\times3+0\times0} & 1\times 0 + 0 \times 4 \\ 0 \times 3 + 5 \times 0 & 0\times0 + 5 \times 4 \end {bmatrix}$
$B \times A = \begin{bmatrix} 3 & 0 \\ 0 & 20 \end{bmatrix}$
Thus, we have seen that $AB = BA$.
## Determinant of Diagonal Matrix
First, let’s look at the determinant of a $2\times2$ matrix.
Consider Matrix $M$ shown below:
$M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$
The determinant of this matrix is:
$det(M) = ad – bc$
One property of a diagonal matrix is that the determinant of a diagonal matrix is equal to the product of the elements in its principal diagonal.
Let’s see if it’s true by finding the determinant of the diagonal matrix shown below.
$N = \begin{pmatrix} 2 & 0 \\ 0 & 8 \end{pmatrix}$
$det(N) = (2\times8)-(0\times0) = 16$
This is in fact the product of the elements in its diagonal, $2\times8=16$.
#### Example 1
For the matrices shown below, identify whether they are diagonal matrix or not.
$A = \begin{bmatrix} {-2} & 0 \\ 0 & {-7} \end {bmatrix}$
$B = \begin{bmatrix} a & 0 \\ 0 & b \\ 0 & d \end {bmatrix}$
$C = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 11 \end {bmatrix}$
$D = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end {bmatrix}$
Solution
• Matrix A is a $2\times2$ matrix with the elements being 0 other than the diagonal. So, this is a diagonal matrix.
• Matrix B is a $3\times2$ matrix. It’s not square, so immediately we can say that it is not a diagonal matrix.
• Matrix C is a square matrix ($3\times3$). Also all the elements besides the diagonal are $0$. So, it is a diagonal matrix. Moreover, an entry of the diagonal is also $0$, it doesn’t matter as long as all the entries except the diagonal are zeros.
• Matrix D is a special type of diagonal matrix. It is a zero matrix. Therefore, it is a diagonal matrix.
#### Example 2
Will multiplying Matrix A and Matrix B result in a diagonal matrix?
$A = \begin{bmatrix} {-9} & 0 \\ 0 & 0 \end {bmatrix}$
$B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end {bmatrix}$
Solution
Matrix A is a diagonal matrix but Matrix B is not. So, multiplying Matrix A and B will not result in a diagonal matrix.
#### Example 3
Find the determinant of the matrix shown below:
$B = \begin{bmatrix} 8 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & {-1} \end {bmatrix}$
Solution
Matrix B is a $3\times3$ diagonal matrix. Recall that the product of all the entries of the diagonal of a diagonal matrix is its determinant. Thus, we simply multiply and find the answer:
$det(B) = 8 \times 4 \times {-1} = – 32$
### Practice Questions
1. Identify which of the following matrices are diagonal matrices.
$J = \begin{pmatrix} 0 & 0 \\ 0 & {-5} \end{pmatrix}$
$K = \begin{pmatrix} 0 & 2 \\ 1 & {-1} \end{pmatrix}$
$L = \begin{bmatrix} -3 & 0 & 0 \\ 0 & {-5} & 0 \\ 0 & 0 & 3 \end{bmatrix}$
2. Calculate the determinant of the matrix shown below:
$T = \begin{bmatrix} -1 & 0 & 0 \\ 0 & {-1} & 0 \\ 0 & 0 & 4 \end{bmatrix}$
3. Given
$A = \begin{pmatrix} 2 & 0 \\ 0 & {-1} \end{pmatrix}$
$B = \begin{pmatrix} 1 & 0 \\ 1 & {-2} \end{pmatrix}$
Is $AB = BA$ ?
1. Matrix J is a square matrix. All the elements other than the principal diagonal are zeros. This is a diagonal matrix.
Matrix K is a square matrix but not all the elements, except the diagonal, are zero. Thus, this is not a diagonal matrix
Matrix L is a square matrix (3\times3). The elements other than the diagonal entries are zero. So, this is a diagonal matrix.
2. This is a diagonal matrix. We can find the determinant of this matrix by taking the product of all the 3 entries of the diagonal. Thus, the determinant is:
$det(T) = -1 \times -1 \times 4 = 4$
3. If two matrices are diagonal, the multiplication of those two matrices are commutative. Looking at Matrix A and B, we can see that Matrix A is diagonal but Matrix B is not. Hence, their multiplication will not be commutative.
Thus, $AB \neq BA$. |
# Orbital Velocity Formula
Orbital velocity is the velocity at which a body revolves around the other body. Objects that travel in the uniform circular motion around the Earth are called to be in orbit. The velocity of this orbit depends on the distance between the object and the centre of the earth.
This velocity is usually given to the artificial satellites so that it revolves around any particular planet.
The orbital velocity formula is given by,
$$\begin{array}{l}V \ orbit = \sqrt{\frac{GM}{R}}\end{array}$$
It is given by
Where,
G = gravitational constant,
M = mass of the body at centre,
R = radius of the orbit.
Orbital Velocity Formula is applied to calculate the orbital velocity of any planet if mass M and radius R are known.
Orbital Velocity is expressed in meter per second (m/s).
Question 1:
Calculate the orbital velocity of the earth so that the satellite revolves around the earth if the radius of earth R = 6.5 × 106 m, the mass of earth M = 5.9722×1024 kg and Gravitational constant G = 6.67408 × 10-11 m3 kg-1 s-2
Solution:
Given:
R = 6.5 × 106 m
M = 5.9722×1024 kg
G = 6.67408 × 10-11 m3 kg-1 s-2
The Orbital velocity formula is given by
Vorbit = √GM / R
       = √6.67408 × 10-11 ×5.9722×1024 / 6.5 × 106
    = √36.68 x 1013/ 6.5 x 106
     = 7.5 x 109km/s
Example 2:
A satellite launch is made for the study of Jupiter. Determine its velocity so that its orbit around the Jupiter.
Given: Radius of Jupiter R = 70.5 × 106 m,
     Mass of Jupiter M = 1.5 × 1027 Kg,
     Gravitational constant G = 6.67408 × 10-11 m3 kg-1 s-2
Solution:
When the given parameters are substituted in the orbital velocity formula, we get
Vorbit = √GM / R
   = √6.67408 × 10-11 × 1.5 × 1027   / 70.5×106
   = √10.0095 x 1016  / 70.5 x 106
   = √0.141 x 1010
   = 3.754 x  104m/s. |
Informative line
# Simplification of Expressions involving Addition and Subtraction
• Simplification means making things easier.
• We can simplify the expressions having various operations.
• Here, we will learn to simplify the expressions having two operations i.e., addition and subtraction.
For example:
Simplify: $$5a+6a-7a+8a$$
Here, we know that $$5a,\,6a,\,-7a$$ and $$8a$$ are all like terms.
We can simplify the expression by taking the common variable out.
Thus, $$5a+6a-7a+8a$$
$$= a(5+6-7+8)$$
$$= a (19-7)$$
$$= a (12)$$
$$= 12a$$
#### Simplify the expression: $$7z+2z-3z-2z$$
A $$2z$$
B $$7z$$
C $$4z$$
D $$3z$$
×
Given expression:
$$7z + 2z-3z-2z$$
We can simplify the expression by taking the common variable out.
Thus, $$7z+2z-3z-2z$$
$$=z(7+2-3-2)$$
$$= z (4)$$
$$= 4z$$
Hence, option (C) is correct.
### Simplify the expression: $$7z+2z-3z-2z$$
A
$$2z$$
.
B
$$7z$$
C
$$4z$$
D
$$3z$$
Option C is Correct
# Simplification of Expressions involving Addition and Multiplication
• Here, we will learn to simplify the expressions involving addition and multiplication.
For example:
Simplify: $$5× a×a+6×5×a$$
Here, in both the terms, $$5$$ and $$a$$ are common.
Thus, we can simplify the expression by taking the commons out i.e., $$5$$ and $$a$$.
$$5×a×a + 6× 5 ×a$$
$$= 5a (a+6)$$
#### Simplify the expression: $$2 × b×b+3×b×b×b$$
A $$4b(2+3b)$$
B $$3b(b+2b)$$
C $$b(2+3b)$$
D $$b^2 (2+3b)$$
×
Given expression:
$$2× b×b + 3 × b× b×b$$
Here, in both the terms, two b's are common.
Thus, we can simplify the expression by taking the commons out i.e., $$b×b$$.
Thus, $$2×b×b + 3× b×b×b$$
$$= b×b (2+3×b)$$
$$= b^2 (2+3b)$$
Hence, option (D) is correct.
### Simplify the expression: $$2 × b×b+3×b×b×b$$
A
$$4b(2+3b)$$
.
B
$$3b(b+2b)$$
C
$$b(2+3b)$$
D
$$b^2 (2+3b)$$
Option D is Correct
# Simplification of Expressions involving Subtraction and Multiplication
• Here, we will learn to simplify the expressions involving subtraction and multiplication.
For example:
Simplify: $$6×b×b×b - 6×b×b$$
Here, $$6$$ and two b's are common in both the terms.
Thus, we can simplify the expression by taking the commons out i.e., $$6$$ and two b's.
$$6×b×b×b-6×b×b$$
$$= 6×b×b(b-1)$$
$$= 6b^2(b-1)$$
#### Simplify the expression: $$3×c×c-6×c$$
A $$3c(c-2)$$
B $$c-2$$
C $$3c$$
D $$6c$$
×
Given expression:
$$3×c×c - 6×c$$
Here, $$3$$ and $$c$$ are common in both the terms.
Thus, we can simplify the expression by taking the commons out i.e., $$3$$ and $$c$$.
Thus, $$3×c×c-6×c$$
$$= 3×c(c-2)$$
$$= 3c (c-2)$$
Hence, option (A) is correct.
### Simplify the expression: $$3×c×c-6×c$$
A
$$3c(c-2)$$
.
B
$$c-2$$
C
$$3c$$
D
$$6c$$
Option A is Correct
# Simplification of Expressions involving Addition and Division
• Here, we will learn to simplify the expressions involving addition and division.
For example:
Simplify: $$\dfrac{6× d×d}{3d} + \dfrac{10×d×d×d}{5d×d}$$
In expressions involving division, similar variables in the numerator and the denominator get canceled by each other i.e.,
$$\dfrac{ \not{6}^2× \not{d}×d}{ \not{3}× \not{d}} + \dfrac{\not10^2×d× \not{d}× \not{d}}{\not5× \not{d}× \not{d}}$$
$$= 2d+2d$$
$$= 2d(1+1)$$ [Taking the commons out i.e, 2d]
$$= 2d(2)$$
$$= 4d$$
#### Simplify the expression: $$\dfrac{2×e×e}{2×e} + \dfrac{6×e×e×e}{3×e}$$
A $$2e$$
B $$e (1+2e)$$
C $$e$$
D $$1+2e$$
×
Given expression:
$$\dfrac{2×e×e}{2×e} + \dfrac{6×e×e×e}{3×e}$$
In expressions involving division, similar variables in the numerator and the denominator get canceled by each other i.e.,
$$\dfrac{ \not{2}× \not{e}×e}{ \not{2}× \not{e}} + \dfrac{ \not{6}^2× \not{e}×e×e}{ \not{3}× \not{e}}$$
$$=e + 2e×e$$
$$= e (1+2e)$$ [$$e$$ is common in both the terms]
Hence, option (B) is correct.
### Simplify the expression: $$\dfrac{2×e×e}{2×e} + \dfrac{6×e×e×e}{3×e}$$
A
$$2e$$
.
B
$$e (1+2e)$$
C
$$e$$
D
$$1+2e$$
Option B is Correct
# Simplification of Expressions involving Subtraction and Division
• Here, we will learn to simplify the expressions involving subtraction and division.
For example:
Simplify: $$\dfrac{4×e×e}{2×e} - \dfrac{9×e×e×e}{3×e×e}$$
In expressions involving division, similar variables in the numerator and the denominator get canceled by each other i.e.,
$$\dfrac{ \not{4}^2 × \not{e}×e}{ \not{2}× \not{e}} - \dfrac{ \not{9}^3× \not{e}× \not{e}×e}{ \not{3}× \not{e}× \not{e}}$$
$$= 2e-3e$$
$$=e (2-3)$$ [Taking $$e$$ as common]
$$=e (-1)$$
$$= -e$$
#### Simplify the expression: $$\dfrac{12×z×z×z}{3×z} - \dfrac{6×z×z}{z}$$
A $$z-3$$
B $$2z-3$$
C $$2z$$
D $$2z(2z-3)$$
×
Given expression:
$$\dfrac{12×z×z×z}{3×z} - \dfrac{6×z×z}{z}$$
In expressions involving division, similar variables in the numerator and the denominator get canceled by each other i.e.,
$$\dfrac{\not12^4× \not{z}×z×z}{\not3× \not{z}} - \dfrac{6× \not{z}×z}{ \not{z}}$$
$$= 4 × z×z- 6×z$$
$$= 2×z (2×z-3)$$ [Taking $$2z$$ as common]
$$= 2z (2z-3)$$
Hence, option (D) is correct.
### Simplify the expression: $$\dfrac{12×z×z×z}{3×z} - \dfrac{6×z×z}{z}$$
A
$$z-3$$
.
B
$$2z-3$$
C
$$2z$$
D
$$2z(2z-3)$$
Option D is Correct
# Multiplication and Division
• Here, we will learn to simplify the expressions involving multiplication and division.
For example:
Simplify: $$\dfrac{3×z×5×z×z×z}{z×3×5×z}$$
In expressions involving division, similar variables in the numerator and the denominator get canceled by each other i.e.,
$$\dfrac{\not {3}×\not {z}×\not {5}×\not {z}×z×z}{\not {z}×\not {3}×\not {5}×\not {z}}$$
$$=z×z$$
$$=z^2$$
#### Simplify the expression: $$\dfrac{10×a×a×a×5}{a×5×a}$$
A $$2a$$
B $$5a$$
C $$10a$$
D $$12a$$
×
Given expression:
$$\dfrac{10×a×a×a×5}{a×5×a}$$
In expressions involving division, similar variables in the numerator and the denominator get canceled by each other i.e.,
$$\dfrac{10×a×\not {a}×\not {a}×\not {5}}{\not {a}×\not {5}×\not {a}}$$
$$= 10a$$
Hence, option (C) is correct.
### Simplify the expression: $$\dfrac{10×a×a×a×5}{a×5×a}$$
A
$$2a$$
.
B
$$5a$$
C
$$10a$$
D
$$12a$$
Option C is Correct |
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# Area and Perimeter of Triangles
## Area is half the base times the height while the perimeter is the sum of the sides.
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Practice Area and Perimeter of Triangles
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Area and Perimeter of Triangles
What if you wanted to find the area of a triangle? How does this relate to the area of a parallelogram? After completing this Concept, you'll be able to answer questions like these.
### Guidance
If we take parallelogram and cut it in half, along a diagonal, we would have two congruent triangles. Therefore, the formula for the area of a triangle is the same as the formula for area of a parallelogram, but cut in half.
The area of a triangle is A=12bh\begin{align*}A=\frac{1}{2} bh\end{align*} or A=bh2\begin{align*}A=\frac{bh}{2}\end{align*}. In the case that the triangle is a right triangle, then the height and base would be the legs of the right triangle. If the triangle is an obtuse triangle, the altitude, or height, could be outside of the triangle.
#### Example A
Find the area of the triangle.
This is an obtuse triangle. To find the area, we need to find the height of the triangle. We are given the two sides of the small right triangle, where the hypotenuse is also the short side of the obtuse triangle. From these values, we see that the height is 4 because this is a 3-4-5 right triangle. The area is A=12(4)(7)=14 units2\begin{align*}A=\frac{1}{2} (4)(7)=14 \ units^2\end{align*}.
#### Example B
Find the perimeter of the triangle from Example A.
To find the perimeter, we would need to find the longest side of the obtuse triangle. If we used the dotted lines in the picture, we would see that the longest side is also the hypotenuse of the right triangle with legs 4 and 10. Use the Pythagorean Theorem.
42+10216+100c=c2=c2=11610.77The perimeter is 7+5+10.77=22.77 units
#### Example C
Find the area of a triangle with base of length 28 cm and height of 15 cm.
The area is 12(28)(15)=210 cm2\begin{align*}\frac{1}{2}(28)(15)=210 \ cm^2\end{align*}.
Watch this video for help with the Examples above.
### Vocabulary
Perimeter is the distance around a shape. The perimeter of any figure must have a unit of measurement attached to it. If no specific units are given (feet, inches, centimeters, etc), write “units.” Area is the amount of space inside a figure. Area is measured in square units.
### Guided Practice
Use the triangle to answer the following questions.
1. Find the height of the triangle.
2. Find the perimeter.
3. Find the area.
1. Use the Pythagorean Theorem to find the height.
82+h2h2h=172=225=15 in
2. We need to find the hypotenuse. Use the Pythagorean Theorem again.
(8+24)2+152h2h=h2=124935.3 in
The perimeter is 24+35.3+1776.3 in\begin{align*}24+35.3+17 \approx 76.3 \ in\end{align*}.
3. The area is 12(24)(15)=180 in2\begin{align*}\frac{1}{2}(24)(15)=180 \ in^2\end{align*}.
### Practice
Use the triangle to answer the following questions.
1. Find the height of the triangle by using the geometric mean.
2. Find the perimeter.
3. Find the area.
Find the area of the following shape.
1. What is the height of a triangle with area 144 m2\begin{align*}144 \ m^2\end{align*} and a base of 24 m?
For problems 6 and 7 find the height and area of the equilateral triangle with the given perimeter.
1. Perimeter 18 units.
2. Perimeter 30 units.
3. Generalize your results from problems 6 and 7 into a formula to find the height and area of an equilateral triangle with side length x\begin{align*}x\end{align*}.
Find the area of each triangle.
1. Find the area of a triangle with a base of 10 in and a height of 12 in.
2. Find the area of a triangle with a base of 5 in and a height of 3 in.
3. An equilateral triangle with a height of 63\begin{align*}6\sqrt{3}\end{align*} units.
4. A 45-45-90 triangle with a hypotenuse of 52\begin{align*}5\sqrt{2}\end{align*} units.
5. A 45-45-90 triangle with a leg of 12 units.
6. A 30-60-90 triangle with a hypotenuse of 24 units.
7. A 30-60-90 triangle with a short leg of 5 units.
### Vocabulary Language: English
Area
Area
Area is the space within the perimeter of a two-dimensional figure.
Perimeter
Perimeter
Perimeter is the distance around a two-dimensional figure.
Perpendicular
Perpendicular
Perpendicular lines are lines that intersect at a $90^{\circ}$ angle. The product of the slopes of two perpendicular lines is -1.
Right Angle
Right Angle
A right angle is an angle equal to 90 degrees.
Right Triangle
Right Triangle
A right triangle is a triangle with one 90 degree angle.
Area of a Parallelogram
Area of a Parallelogram
The area of a parallelogram is equal to the base multiplied by the height: A = bh. The height of a parallelogram is always perpendicular to the base (the sides are not the height).
Area of a Triangle
Area of a Triangle
The area of a triangle is half the area of a parallelogram. Hence the formula: $A = \frac{1}{2}bh \text{ or } A = \frac{bh}{2}$. |
Solution to math problem
Solution to math problem can be found online or in mathematical textbooks. Math can be difficult for some students, but with the right tools, it can be conquered.
The Best Solution to math problem
There is Solution to math problem that can make the technique much easier. Quadratic equations are a common type of algebraic equation that can be difficult to solve. However, there are a number of Quadratic equation solvers that can help to make the process easier. These solvers will typically provide a step-by-step solution, making it easy to see how to solve the equation. In addition, some Quadratic equation solvers will also provide a visual representation of the solution, which can be helpful in understanding the concept. There are a number of Quadratic equation solvers available online, and many of them are free to use.
By breaking the problem down into smaller pieces, you can more easily see how to move forward. In addition, taking steps can help you to avoid getting overwhelmed by the problem as a whole. Instead of seeing an insurmountable obstacle, you can focus on each small task and take comfort in knowing that you're slowly but surely making progress. So next time you're stuck, try approaching the problem from a step-by-step perspective and see if it makes it any easier to solve.
To solve inequality equations, you need to first understand what they are. Inequality equations are mathematical equations that involve two variables which are not equal to each other. The inequalities can be either greater than or less than. To solve these equations, you need to find the value of the variable that makes the two sides of the equation equal. This can be done by using the properties of inequality. For example, if the equation is x+5>9, then you can subtract 5 from both sides to get x>4. This means that the solutions to this inequality are all values of x that are greater than 4. Solve inequality equations by using the properties of inequality to find the value of the variable that makes the two sides of the equation equal.
How to solve using substitution is best explained with an example. Let's say you have the equation 4x + 2y = 12. To solve this equation using substitution, you would first need to isolate one of the variables. In this case, let's isolate y by subtracting 4x from both sides of the equation. This gives us: y = (1/2)(12 - 4x). Now that we have isolated y, we can substitute it back into the original equation in place of y. This gives us: 4x + 2((1/2)(12 - 4x)) = 12. We can now solve for x by multiplying both sides of the equation by 2 and then simplifying. This gives us: 8x + 12 - 8x = 24, which simplifies to: 12 = 24, and therefore x = 2. Finally, we can substitute x = 2 back into our original equation to solve for y. This gives us: 4(2) + 2y = 12, which simplifies to 8 + 2y = 12 and therefore y = 2. So the solution to the equation 4x + 2y = 12 is x = 2 and y = 2.
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Welcome to our IEP Goals page for 4th-grade math skills. Our lesson plans, teaching resources, worksheets, and learning centers are designed to help students master the identification of prime and composite numbers, along with understanding factors and division. Aligned with Learning Standard 4.OA.B.4, our objectives focus on recognizing factors, selecting prime and composite numbers up to 20, and dividing two-digit numbers by one-digit numbers using arrays. These structured activities aim to enhance operations and algebraic thinking skills, supporting your student's academic growth and success in mathematics. Explore our materials to support your student's learning journey.
Learning Standard
### 4.OA.B.4
Find all factor pairs for a whole number in the range 1-100. Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range 1-100 is a multiple of a given one-digit number. Determine whether a given whole number in the range 1-100 is prime or composite.
Target IEP Goal
By (date), when given exercises with factors and multiples, the student will select prime and composite numbers (up to 20), improving operations and algebraic thinking skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
Teach Tastic IEP goals written to be SMART: Specific, Measurable, Attainable, Results-Oriented, and Time-Bound.
IEP Goal Objectives
1
Recognize Factors
By (date), when given exercises with factors and multiples, the student will recognize factors, improving operations and algebraic thinking skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
1
2
Divide 2-Digit by 1-Digit Numbers
By (date), when given exercises with divide by one-digit numbers, the student will divide 2-digit by 1-digit numbers, improving number and operations in base ten skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
1
3
Divide 2-Digit by 1-Digit Numbers Using Arrays
By (date), when given exercises with divide by one-digit numbers, the student will divide 2-digit by 1-digit numbers using arrays, improving number and operations in base ten skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
1
4
Write Division Sentences for Arrays
By (date), when given exercises with understand division, the student will write division sentences for arrays, improving operations and algebraic thinking skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
1
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Solve x^3+2x^2+3x+2=0 | Uniteasy.com
# Solve the cubic equation:
## $$x^3+2x^2+3x+2=0$$
Since the discriminant $$\Delta >0$$, the cubic equation has one real root and two conjugate complex roots.
$$\Delta=0.25925925925926$$
$$\begin{cases} x_1=-1 \\ x_2=-\dfrac{1}{2}+\dfrac{\sqrt{7}}{2}i \\ x_3=-\dfrac{1}{2}-\dfrac{\sqrt{7}}{2}i \end{cases}$$
In decimals,
$$\begin{cases} x_1=-1 \\ x_2=-0.5+1.3228756555323i \\ x_3=-0.5-1.3228756555323i \end{cases}$$
Detailed Steps on Solution
A cubic equation has at least one real root. If the coefficient of leading term is 1, one of solutions could be a factor of the constant term.
## 1. Factorization Method
Find all possible factors for constant
$$1, 2$$
$$-1, -2$$
Substitute the factors to the function $$f(x) = x³ + 2x² + 3x + 2$$ and find the one that makes $$f(x) = 0$$.
According to factor theorem, $$f(n) = 0$$, if and only if the polynomial $$x³ + 2x² + 3x + 2$$ has a factor $$x-n$$, that is, $$x=n$$ is a root of the equation.
Fortunetely, one of the numbers is found to make the equation equal.
$$f(-1) = (-1)³ + 2(-1)² + 3(-1) + 2 = 8$$
then we get the first root
$$x_1 = -1$$
And the cubic equation can be factored as
$$(x +1)(ax^2+bx+c) = 0$$
Next we can use either long division or synthetic division to determine the expression of trinomial
### Long division
Divide the polynomial $$x³ + 2x² + 3x + 2$$ by $$x + 1$$
x² +x 2 x + 1 x³ +2x² +3x 2 x³ +x² x² +3x x² +x 2x 2 2x 2 0
Now we get another factor of the cubic equation $$x² + x + 2$$
## Solve the quadratic equation: $$x² + x + 2 = 0$$
Given $$a =1, b=1, c=2$$,
Use the root solution formula for a quadratic equation, the roots of the equation are given as
\begin{aligned} \\t&=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}\\ & =\dfrac{-1\pm\sqrt{1^2-4\cdot 1\cdot 2}}{2 \cdot 1}\\ & =\dfrac{-1\pm\sqrt{-7}}{2}\\ & =-\dfrac{1}{2}\pm\dfrac{\sqrt{7}}{2}i\\ \end{aligned}
Since the discriminat is less than zero, we get another two complex roots:
That is,
$$\begin{cases} t_2 =-\dfrac{1}{2}+\dfrac{\sqrt{7}}{2}i \\ t_3=-\dfrac{1}{2}-\dfrac{\sqrt{7}}{2}i \end{cases}$$
Another method to find the roots of a general cubic equation is simplifing it to a depressed form. This method is applicable to all cases, especially to those difficult to find factors.
## 2. Convert to depressed cubic equation
The idea is to convert general form of cubic equation
$$ax^3+bx^2+cx+d = 0$$
to the form without quadratic term.
$$t^3+pt+q = 0$$
By substituting $$x$$ with $$t - \dfrac{b}{3a}$$, the general cubic equation could be transformed to
$$t^3+\dfrac{3ac-b^2}{3a^2}t+\dfrac{2b^3-9abc+27a^2d}{27a^3} = 0$$
Compare with the depressed cubic equation. Then,
$$p = \dfrac{3ac-b^2}{3a^2}$$
$$q = \dfrac{2b^3-9abc+27a^2d}{27a^3}$$
Substitute the values of coefficients, $$p, q$$ is obtained as
$$p = \dfrac{3\cdot 1\cdot 3-2^2}{3\cdot 1^2}=\dfrac{5}{3}$$
$$q = \dfrac{2\cdot 2^3-9\cdot1\cdot 2\cdot 3+27\cdot 1^2\cdot2}{27\cdot 1^3}=\dfrac{16}{27}$$
### Use the substitution to transform
Let $$p$$ and $$q$$ being the coefficient of the linean and constant terms, the depressed cubic equation is expressed as.
$$t^3 +pt+q=0$$
Let $$x=t-\dfrac{2}{3}$$
The cubic equation $$x³ + 2x² + 3x + 2=0$$ is transformed to
$$t^3 +\dfrac{5}{3}t+\dfrac{16}{27}=0$$
## 3. Cardano's solution
Let $$t=u-v$$
Cube both sides and extract common factor from two middle terms after expanding the bracket.
\begin{aligned} \\t^3&=(u-v)^3\\ & =u^3-3u^2v+3uv^2-v^3\\ & =-3uv(u-v)+u^3-v^3\\ \end{aligned}
Since $$u-v=t$$, substitution gives a linear term for the equation. Rearrange terms.
$$x^3+3uvx-u^3+v^3=0$$
Compare the cubic equation with the original one (1)
$$\begin{cases} 3uv=\dfrac{5}{3}\quad\text{or}\quad v=\dfrac{5}{9u}\\ v^3-u^3=\dfrac{16}{27}\\ \end{cases}$$
$$v=\dfrac{5}{9u}$$ gives relationship between the two variables. Substitute the value of $$v$$ to the second equation
$$\Big(\dfrac{5}{9u}\Big)^3-u^3=\dfrac{16}{27}$$
Simplifying gives,
$$u^3-\dfrac{125}{729}\dfrac{1}{u^3}+\dfrac{16}{27}=0$$2
Let $$m=u^3$$, then the equation is transformed to a quadratic equation in terms of $$m$$. Once the value of $$m$$ is determined, $$v^3$$ could be determined by $$v^3=\dfrac{16}{27}+u^3$$.
$$m^2+\dfrac{16}{27}m-\dfrac{125}{729}=0$$
Sovling the quadratic euqation will give two roots (some may be equal). Here we only cosider one case with positive sign before the square root radical since the negative case will produce the same result.
\begin{aligned} \\u^3=m&=-\dfrac{8}{27}+\dfrac{1}{2}\sqrt{\Big(\dfrac{16}{27}^2\Big)-4\cdot \Big(-\dfrac{125}{729}\Big)}\\ & =-\dfrac{8}{27}+\dfrac{1}{2}\sqrt{\dfrac{256}{729}+\dfrac{500}{729}}\\ & =-\dfrac{8}{27}+\dfrac{1}{2}\sqrt{\dfrac{28}{27}}\\ & =-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}\\ \end{aligned}
$$v^3$$ can be determined by the equation we deduced $$v^3-u^3=\dfrac{16}{27}$$. Then,
\begin{aligned} \\v^3&=\dfrac{16}{27}+u^3\\ & =\dfrac{16}{27}-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}\\ & =\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}\\ \end{aligned}
Now we have,
$$u^3=-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}$$ and $$v^3=\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}$$
Evaluating the simplest cubic equation $$x^3-A=0$$, it has 3 roots, in which the first root is a real number . The second and third are expressed in the product of cubic root of unity and the first one.
If $$ω = \dfrac{-1+i\sqrt{3}}{2}$$, then its reciprocal is equal to its conjugate, $$\dfrac{1}{ω}=\overline{ω}$$.
$$\begin{cases} r_1=\sqrt[3]{A}\\ r_2=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt[3]{A}\\ r_3=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{A}\\ \end{cases}$$
Similary, taking cubic root for $$u^3$$ and $$v^3$$ also gives 3 roots.
$$\begin{cases} u_1=\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\\ u_2=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\\ u_3=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\\ \end{cases}$$
For $$v_2$$ and $$v_3$$, the complex numbers before radicals are the conjugates of those for $$u_2$$ and $$u_3$$, which can be verified by the reciprocal property of the cubic root of unity from the equation $$v=\dfrac{5}{9u}$$. The radicand can be taken as the negative conjugate of that in $$u_1$$, $$u_2$$ and $$u_3$$, which is the same in value.
$$\begin{cases} v_1=\sqrt[3]{\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\\ v_2=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\\ v_3=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt[3]{\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\\ \end{cases}$$
Verification for the redicand in $$v$$.
\begin{aligned} \\v_1&=\dfrac{5}{9u_1}\\ & =\dfrac{5}{9}\cdot \dfrac{1}{\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}}\\ & =\dfrac{5}{9}\cdot \dfrac{1}{\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}}\cdot \dfrac{\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}}{\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}}\\ & =\dfrac{5}{9}\cdot \dfrac{\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}}{\sqrt[3]{\Big(-\dfrac{8}{27}\Big)^2-\Big(\dfrac{1}{3}\sqrt{\dfrac{7}{3}}\Big)^2}}\\ & =\dfrac{5}{9}\cdot \dfrac{\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}}{\sqrt[3]{\dfrac{64}{729}-\dfrac{7}{27}}}\\ & =\dfrac{5}{9}\cdot \dfrac{\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}}{\sqrt[3]{\dfrac{-1\cdot 5^3}{9^3}}}\\ & =-\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\\ \end{aligned}
Since $$x=u-v$$, combining the real and imaginary parts gives 3 results for $$t$$
\begin{aligned} \\t_1&=u_1-v_1\\ & =\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}+\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\\ \end{aligned}
\begin{aligned} \\t_2&=u_2-v_2\\ & =\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\Big(\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\Big)\\ & =\dfrac{1}{2}\Big(-\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\Big)+\dfrac{\sqrt{3}}{2}\Big(\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\Big)i\\ \end{aligned}
\begin{aligned} \\t_3&=u_3-v_3\\ & =\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\Big(\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt[3]{\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\Big)\\ & =\dfrac{1}{2}\Big(-\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\Big)+\dfrac{\sqrt{3}}{2}\Big(-\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}+\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\Big)i\\ \end{aligned}
## 4. Vieta's Substitution
In Cardano' solution, $$t$$ is defined as the difference of $$u$$ and $$v$$. If we substitute the value of $$v$$ (4) into (2), we get the equation. $$t=u-\dfrac{5}{9u}$$. And then substitute the equation to the cubic equation $$t^3+\dfrac{5}{3}t+\dfrac{16}{27}=0$$. This method is called Vieta's Substitution for solving a cubic equation, which simplied the Cardano' solution. The substitution expression can be obtained by the following formula directly.
$$t=u-\dfrac{p}{3u}$$
Substitute the expression $$t=u-\dfrac{5}{9u}$$ to the cubic equation
$$\Big(u-\dfrac{5}{9u}\Big)^3+\dfrac{5}{3}\Big(u-\dfrac{5}{9u}\Big)+\dfrac{16}{27}=0$$
Expand brackets and cancel the like terms
$$u^3-\cancel{\dfrac{5}{3}u^2\dfrac{1}{u}}+\cancel{\dfrac{25}{27}u\dfrac{1}{u^2}}-\dfrac{125}{729}\dfrac{1}{u^3}+\cancel{\dfrac{5}{3}u}-\cancel{\dfrac{25}{27}\dfrac{1}{u}}+\dfrac{16}{27}=0$$
Then we get the same equation as (2)
$$u^3-\dfrac{125}{729}\dfrac{1}{u^3}+\dfrac{16}{27}=0$$
The rest of the steps will be the same as those of Cardano's solution
## $$t^3+\dfrac{5}{3}t+\dfrac{16}{27}=0$$
Move the linear term and constant of (1) to its right hand side. We get the following form of the equation.
$$t^3=-\dfrac{5}{3}t-\dfrac{16}{27}$$3
Let the root of the cubic equation be the sum of two cubic roots
$$t=\sqrt[3]{r_1}+\sqrt[3]{r_2}$$4
in which $$r_1$$ and $$r_2$$ are two roots of a quadratic equation
$$z^2-\alpha z+ β=0$$5
Using Vieta's Formula, the following equations are established.
$$r_1+r_2 = \alpha \quad \text{and} \quad r_1r_2 = β$$
To determine $$\alpha$$, $$β$$, cube both sides of the equation (4)
$$t^3=3\sqrt[3]{r_1r_2}(\sqrt[3]{r_1}+\sqrt[3]{r_2})+r_1+r_2$$
Substituting, the equation is simplified to
$$t^3=3\sqrt[3]{β}t+\alpha$$
Compare the cubic equation with (3), the following equations are established
$$\begin{cases} 3\sqrt[3]{β}=-\dfrac{5}{3}\\ \alpha=-\dfrac{16}{27}\\ \end{cases}$$
Solving for $$β$$ gives
$$β=-\dfrac{125}{729}$$
So the quadratic equation (5) is determined as
$$z^2+\dfrac{16}{27}z-\dfrac{125}{729}=0$$6
$$\begin{cases} r_1=-\dfrac{8}{27}+\dfrac{\sqrt{21}}{9}\approx0.21287878092102\\ r_2=-\dfrac{8}{27}-\dfrac{\sqrt{21}}{9}\approx-0.80547137351361\\ \end{cases}$$
Therefore, one of the roots of the cubic equation could be obtained from (4).
$$t_1=\sqrt[3]{-\dfrac{8}{27}+\dfrac{\sqrt{21}}{9}}+\sqrt[3]{-\dfrac{8}{27}-\dfrac{\sqrt{21}}{9}}$$
in decimals,
$$t_1=-0.33333333333333$$
However, since the cube root of a quantity has triple values,
The other two roots could be determined as,
$$t_2=\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{-\dfrac{8}{27}+\dfrac{\sqrt{21}}{9}}+\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{-\dfrac{8}{27}-\dfrac{\sqrt{21}}{9}}$$
$$t_3=\dfrac{-1-i\sqrt{3}}{2}\sqrt[3]{-\dfrac{8}{27}+\dfrac{\sqrt{21}}{9}}+\dfrac{-1+i\sqrt{3}}{2}\sqrt[3]{-\dfrac{8}{27}-\dfrac{\sqrt{21}}{9}}$$
Combining the real and imaginary parts results in the same result as that obtained by Cardano's solution.
For the equation $$t^3 +\dfrac{5}{3}t+\dfrac{16}{27}$$, we have $$p=\dfrac{5}{3}$$ and $$q = \dfrac{16}{27}$$
### Calculate the discriminant
The nature of the roots are determined by the sign of the discriminant.
\begin{aligned} \\\Delta&=\dfrac{q^2}{4}+\dfrac{p^3}{27}\\ & =\dfrac{\Big(\dfrac{16}{27}\Big)^2}{4}+\dfrac{\Big(\dfrac{5}{3}\Big)^3}{27}\\ & =\dfrac{64}{729}+\dfrac{125}{729}\\ & =\dfrac{64\cdot 1+125\cdot 1}{729}\\ & =0.25925925925926\\ \end{aligned}
### 5.1 Use the root formula directly
If the discriminant is greater than zero, we can use the root formula to determine the roots of the cubic equation.
$$t_{1,2,3} =\begin{cases} \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } +\sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}& \\ ω\cdotp \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + \overline{ω} \sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}&\ \\ \overline{ω}\cdotp \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } + ω\cdotp \sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }} \end{cases}$$
in which, $$ω = \dfrac{-1+i\sqrt{3}}{2}$$ and $$\overline{ω} =\dfrac{-1-i\sqrt{3}}{2}$$
Substitute the values of $$p, q$$ and $$\Delta$$ which we have calculated. Then,
\begin{aligned} \\t_1&=\sqrt[3]{-\dfrac{8}{27}+\sqrt{\dfrac{189}{729}}}+\sqrt[3]{-\dfrac{8}{27}-\sqrt{\dfrac{189}{729}}}\\ & =\sqrt[3]{-\dfrac{8}{27}+\sqrt{\dfrac{7\cdot\cancel{27}}{27\cdot\cancel{27}}}}+\sqrt[3]{-\dfrac{8}{27}-\sqrt{\dfrac{7\cdot\cancel{27}}{27\cdot\cancel{27}}}}\\ & =\sqrt[3]{-\dfrac{8}{27}+\sqrt{\dfrac{7}{3\cdot 3^2}}}+\sqrt[3]{-\dfrac{8}{27}-\sqrt{\dfrac{7}{3\cdot 3^2}}}\\ & =\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}+\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\\ \end{aligned}
If we denote
$$R = -\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }$$
$$\overline{R} = -\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }$$
then,
$$\sqrt[3]{R} = \sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}$$, $$\sqrt[3]{\overline{R}} =\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}$$
\begin{aligned} \\t_2&= ω\cdotp \sqrt[3]{R}+ \overline{ω} \sqrt[3]{\overline{R} }\\ & =\dfrac{-\sqrt[3]{R}-\sqrt[3]{\overline{R} }}{2} +\dfrac{\sqrt{3}( \sqrt[3]{R} - \sqrt[3]{\overline{R} }) }{2} i\\ & =\dfrac{1}{2}\Big(-\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\Big)\\&+\dfrac{\sqrt{3}}{2}\Big(\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\Big)i\\ \end{aligned}
\begin{aligned} \\t_3&= \overline{ω}\cdotp \sqrt[3]{R}+ ω\cdotp \sqrt[3]{\overline{R}}\\ & =\dfrac{-\sqrt[3]{R}-\sqrt[3]{\overline{R} }}{2} +\dfrac{\sqrt{3}(- \sqrt[3]{R} + \sqrt[3]{\overline{R} }) }{2}i \\ & =\dfrac{1}{2}\Big(-\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\Big)\\&-\dfrac{\sqrt{3}}{2}\Big(\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\Big)i\\ \end{aligned}
## Roots of the general cubic equation
Since $$x = t - \dfrac{b}{3a}$$, substituting the values of $$t$$, $$a$$ and $$b$$ gives
$$x_1 = t_1-\dfrac{2}{3}$$
$$x_2 = t_2-\dfrac{2}{3}$$
$$x_3 = t_3-\dfrac{2}{3}$$
## 6. Summary
In summary, we have tried the method of factorization, cubic root formula to explore the solutions of the equation. The cubic equation $$x³ + 2x² + 3x + 2=0$$ is found to have one real root and two complex roots. Exact values and approximations are given below.
$$\begin{cases} x_1=\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}+\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\dfrac{2}{3} \\ x_2=\dfrac{1}{2}\Big(-\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\Big)+\dfrac{\sqrt{3}}{2}\Big(\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\Big)i-\dfrac{2}{3} \\ x_3=\dfrac{1}{2}\Big(-\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\Big)-\dfrac{\sqrt{3}}{2}\Big(\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\Big)i-\dfrac{2}{3} \end{cases}$$
in decimal notation,
$$\begin{cases} x_1=-1 \\ x_2=-0.5+1.3228756555323i \\ x_3=-0.5-1.3228756555323i \end{cases}$$
Using the method of factorization, the roots are derived to the following forms
$$\begin{cases} x_1=-1 \\ x_2=-\dfrac{1}{2}+\dfrac{\sqrt{7}}{2}i \\ x_3=-\dfrac{1}{2}-\dfrac{\sqrt{7}}{2}i \end{cases}$$
The decimal results by using factorization method are
$$\begin{cases} x_1=-1 \\ x_2=-0.5+1.3228756555323i \\ x_3=-0.5-1.3228756555323i \end{cases}$$
## 7. Math problems derived by the cubic equation
It is found that the methods of factorizaiton and using cubic root formula yield the roots for the cubic equation in different forms. However, their deicmal results show that they are equal in values. Comparing the real roots, we get the equation.
$$\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}+\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\dfrac{2}{3}=-1$$
Similarly, comparison of the imaginary parts of complex roots yields the following equation.
$$\dfrac{\sqrt{3}}{2}\Big(\sqrt[3]{-\dfrac{8}{27}+\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}-\sqrt[3]{-\dfrac{8}{27}-\dfrac{1}{3}\sqrt{\dfrac{7}{3}}}\Big)=\dfrac{\sqrt{7}}{2}$$
## 8. Graph for the function $$f(x) = x³ + 2x² + 3x + 2$$
Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = x³ + 2x² + 3x + 2$$ has one intersection point with the x-axis.
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Recall that $${n\choose k}={n!\over k!\,(n-k)!}={n(n-1)(n-2)\cdots(n-k+1)\over k!}.$$ The expression on the right makes sense even if $n$ is not a non-negative integer, so long as $k$ is a non-negative integer, and we therefore define $${r\choose k}={r(r-1)(r-2)\cdots(r-k+1)\over k!}$$ when $r$ is a real number. For example, $${1/2\choose 4}={(1/2)(-1/2)(-3/2)(-5/2)\over 4!}={-5\over128} \quad\hbox{and}\quad {-2\choose 3}={(-2)(-3)(-4)\over 3!}=-4.$$ These generalized binomial coefficients share some important properties of the usual binomial coefficients, most notably that \eqalignno{ {r\choose k}&={r-1\choose k-1}+{r-1\choose k}.& (3.1.1)\cr } Then remarkably:
Theorem 3.1.1 (Newton's Binomial Theorem) For any real number $r$ that is not a non-negative integer, $$(x+1)^r=\sum_{i=0}^\infty {r\choose i}x^i$$ when $-1< x< 1$.
Proof. It is not hard to see that the series is the Maclaurin series for $(x+1)^r$, and that the series converges when $-1< x< 1$. It is rather more difficult to prove that the series is equal to $(x+1)^r$; the proof may be found in many introductory real analysis books. $\qed$
Example 3.1.2 Expand the function $(1-x)^{-n}$ when $n$ is a positive integer.
We first consider $(x+1)^{-n}$; we can simplify the binomial coefficients: \eqalign{ {(-n)(-n-1)(-n-2)\cdots(-n-i+1)\over i!} &=(-1)^i{(n)(n+1)\cdots(n+i-1)\over i!}\cr &=(-1)^i{(n+i-1)!\over i!\,(n-1)!}\cr &=(-1)^i{n+i-1\choose i}=(-1)^i{n+i-1\choose n-1}.\cr } Thus $$(x+1)^{-n}=\sum_{i=0}^\infty (-1)^i{n+i-1\choose n-1}x^i =\sum_{i=0}^\infty {n+i-1\choose n-1}(-x)^i.$$ Now replacing $x$ by $-x$ gives $$(1-x)^{-n}=\sum_{i=0}^\infty {n+i-1\choose n-1}x^i.$$ So $(1-x)^{-n}$ is the generating function for ${n+i-1\choose n-1}$, the number of submultisets of $\{\infty\cdot1,\infty\cdot2,\ldots,\infty\cdot n\}$ of size $i$. $\square$
In many cases it is possible to directly construct the generating function whose coefficients solve a counting problem.
Example 3.1.3 Find the number of solutions to $\ds x_1+x_2+x_3+x_4=17$, where $0\le x_1\le2$, $0\le x_2\le5$, $0\le x_3\le5$, $2\le x_4\le6$.
We can of course solve this problem using the inclusion-exclusion formula, but we use generating functions. Consider the function $$(1+x+x^2)(1+x+x^2+x^3+x^4+x^5)(1+x+x^2+x^3+x^4+x^5)(x^2+x^3+x^4+x^5+x^6).$$ We can multiply this out by choosing one term from each factor in all possible ways. If we then collect like terms, the coefficient of $x^k$ will be the number of ways to choose one term from each factor so that the exponents of the terms add up to $k$. This is precisely the number of solutions to $\ds x_1+x_2+x_3+x_4=k$, where $0\le x_1\le2$, $0\le x_2\le5$, $0\le x_3\le5$, $2\le x_4\le6$. Thus, the answer to the problem is the coefficient of $x^{17}$. With the help of a computer algebra system we get \eqalign{ (1+x+x^2)(1&+x+x^2+x^3+x^4+x^5)^2(x^2+x^3+x^4+x^5+x^6)\cr =\;&x^{18} + 4x^{17} + 10x^{16} + 19x^{15} + 31x^{14} + 45x^{13} + 58x^{12} + 67x^{11} + 70x^{10}\cr &+67x^9 + 58x^8 + 45x^7 + 31x^6 + 19x^5 + 10x^4 + 4x^3 + x^2,\cr } so the answer is 4. $\square$
Example 3.1.4 Find the generating function for the number of solutions to $\ds x_1+x_2+x_3+x_4=k$, where $0\le x_1\le\infty$, $0\le x_2\le5$, $0\le x_3\le5$, $2\le x_4\le6$.
This is just like the previous example except that $x_1$ is not bounded above. The generating function is thus \eqalign{ f(x)&=(1+x+x^2+\cdots)(1+x+x^2+x^3+x^4+x^5)^2(x^2+x^3+x^4+x^5+x^6)\cr &=(1-x)^{-1}(1+x+x^2+x^3+x^4+x^5)^2(x^2+x^3+x^4+x^5+x^6)\cr &={(1+x+x^2+x^3+x^4+x^5)^2(x^2+x^3+x^4+x^5+x^6)\over 1-x}. } Note that $(1-x)^{-1}=(1+x+x^2+\cdots)$ is the familiar geometric series from calculus; alternately, we could use example 3.1.2. Unlike the function in the previous example, this function has an infinite expansion: \eqalign{ f(x)&= x^2+4x^3 + 10x^4 + 20x^5 +35x^6 + 55x^7+ 78x^8 \cr &+ 102x^9 + 125x^{10}+ 145x^{11} + 160x^{12} + 170x^{13}+176x^{14} \cr &+ 179x^{15} +180x^{16} + 180x^{17} + 180x^{18} + 180x^{19} + 180x^{20} +\cdots. } Here is how to do this in Sage. $\square$
Example 3.1.5 Find a generating function for the number of submultisets of $\{\infty\cdot a,\infty\cdot b,\infty\cdot c\}$ in which there are an odd number of $a$s, an even number of $b$s, and any number of $c$s. As we have seen, this is the same as the number of solutions to $x_1+x_2+x_3=n$ in which $x_1$ is odd, $x_2$ is even, and $x_3$ is unrestricted. The generating function is therefore \eqalign{ (x+x^3+x^5&+\cdots)(1+x^2+x^4+\cdots)(1+x+x^2+x^3+\cdots)\cr &=x(1+(x^2)+(x^2)^2+(x^2)^3+\cdots)(1+(x^2)+(x^2)^2+(x^2)^3+\cdots){1\over 1-x}\cr &={x\over (1-x^2)^2(1-x)}.\cr } $\square$
## Exercises 3.1
For some of these exercises, you may want to use the sage applet above, in example 3.1.4, or your favorite computer algebra system.
Ex 3.1.1 Prove that ${r\choose k}={r-1\choose k-1}+{r-1\choose k}$.
Ex 3.1.2 Show that the Maclaurin series for $(x+1)^r$ is $\sum_{i=0}^\infty {r\choose i}x^i$.
Ex 3.1.3 Concerning example 3.1.4, show that all coefficients beginning with $x^{16}$ are 180.
Ex 3.1.4 Use a generating function to find the number of solutions to $\ds x_1+x_2+x_3+x_4=14$, where $0\le x_1\le3$, $2\le x_2\le5$, $0\le x_3\le5$, $4\le x_4\le6$.
Ex 3.1.5 Find the generating function for the number of solutions to $\ds x_1+x_2+x_3+x_4=k$, where $0\le x_1\le\infty$, $3\le x_2\le\infty$, $2\le x_3\le5$, $1\le x_4\le5$.
Ex 3.1.6 Find a generating function for the number of non-negative integer solutions to $3x+2y+7z=n$.
Ex 3.1.7 Suppose we have a large supply of red, white, and blue balloons. How many different bunches of 10 balloons are there, if each bunch must have at least one balloon of each color and the number of white balloons must be even?
Ex 3.1.8 Use generating functions to show that every positive integer can be written in exactly one way as a sum of distinct powers of 2.
Ex 3.1.9 Suppose we have a large supply of blue and green candles, and one gold candle. How many collections of $n$ candles are there in which the number of blue candles is even, the number of green candles is any number, and the number of gold candles is at most one? |
# Question #06bd6
Aug 24, 2017
$\textcolor{g r e e n}{\left(a + b\right) \left(c + d\right)}$
#### Explanation:
When you factor by grouping, you break the expression down into two parts.
$a c + a d + b c + b d$ becomes $\left(a c + a d\right) + \left(b c + b d\right)$
We didn't change anything, we just grouped the left two terms together and the right two terms together.
Now we can factor out a common variable in the broken down expression.
$\textcolor{\mathmr{and} a n \ge}{a} \left(c + d\right) + \textcolor{\mathmr{and} a n \ge}{b} \left(c + d\right)$
Notice how after we factored out a common variable respective to each grouped portion, we had the same expression left inside the parentheses.
Once you recognize this, you can confirm that the expression can be factored by graphing.
The next step is to group the factored out variables by themselves.
$\textcolor{\mathmr{and} a n \ge}{\left(a + b\right)} \left(c + d\right)$
And that is the factored form. Let's review the steps :
1.) Break the expression up into two distinct portions enclosed in parentheses (break up the first two terms into one portion and the last two terms into another)
2.) Look for a common factor/multiple to take out from each portion (they don't have to be the same thing).
3.) Once you factor both portions of the expression, check to make sure the remaining values inside the parentheses match. If they do, the expression can be factored by grouping. If not, they can't.
4.) Assuming that the parentheses match, take the factored value from the first portion and multiply it to the factored value of the second portion.
5.) Now you should have two distinct values each within parentheses.
If these steps don't make sense, look at the way I factored the example above again. |
Integral of cot^2(x) (trigonometric identity)
Integral of cot^2(x) (trigonometric identity)
The derivative of cot(2x) is -2csc2(2x)
## How to calculate the derivative of cot(2x)
Note that in this post we will be looking at differentiating cot(2x) which is not the same as differentiating cot2(x). Here is our post dealing with how to differentiate cot^2(x).
The chain rule is useful for finding the derivative of a function which could have been differentiated had it been in x, but it is in the form of another expression which could also be differentiated if it stood on its own.
In this case:
• We know how to differentiate cot(x) (the answer is cosec2(x))
• We know how to differentiate 2x (the answer is 2)
This means the chain rule will allow us to differentiate the expression cot(2x).
#### Using the chain rule to find the derivative of cot(2x)
To perform the differentiation cot(2x), the chain rule says we must differentiate the expression as if it were just in terms of x as long as we then multiply that result by the derivative of what the expression is actually in terms of (in this case the derivative of 2x).
Let’s call the function in the argument of cot, g(x), which means the function is in the form of cot(x), except it does not have x as the angle, instead it has another function of x (2x) as the angle
If:
g(x) = 2x
It follows that:
cot(2x) = cot(g(x))
So if the function f(x) = cot(x) and the function g(x) = 2x, then the function cot(2x) can be written as a composite function.
f(x) = cot(x)
f(g(x)) = cot(g(x)) (but g(x) = 2x))
f(g(x)) = cot(2x)
Let’s define this composite function as F(x):
F(x) = f(g(x)) = cot(2x)
We can find the derivative of cot(2x) (F'(x)) by making use of the chain rule.
The Chain Rule:
For two differentiable functions f(x) and g(x)
If F(x) = f(g(x))
Then the derivative of F(x) is F'(x) = f’(g(x)).g’(x)
Now we can just plug f(x) and g(x) into the chain rule. But before we do that, just a quick recap on the derivative of the cot function.
The derivative of cot(x) with respect to x is -csc2(x)
The derivative of cot(z) with respect to z is -csc2(z)
In a similar way, the derivative of cot(2x) with respect to 2x is -csc2(2x).
We will use this fact as part of the chain rule to find the derivative of cot(2x) with respect to x.
How to find the derivative of cot(2x) using the Chain Rule:
F'(x) = f'(g(x)).g'(x) Chain Rule Definition = f'(g(x))(2) g(x) = 2x ⇒ g'(x) = 2 = (-csc2(2x)).(2) f(g(x)) = cot(2x) ⇒ f'(g(x)) = -cosec2(2x) = -2csc2(2x)
Using the chain rule, the derivative of cot(2x) is -2csc2(2x)
Finally, just a note on syntax and notation: cot(2x) is sometimes written in the forms below (with the derivative as per the calculation above). Just be aware that not all of the forms below are mathematically correct.
cot2x ► Derivative of cot2x = -2csc2(2x) cot 2 x ► Derivative of cot 2 x = -2csc2(2x) cot 2x ► Derivative of cot 2x = -2csc2(2x) cot (2x) ► Derivative of cot (2x) = -2csc2(2x)
## The Second Derivative Of cot(2x)
To calculate the second derivative of a function, you just differentiate the first derivative.
From above, we found that the first derivative of cot(2x) = -2csc2(2x). So to find the second derivative of cot(2x), we just need to differentiate -2csc2(2x).
We can use the chain rule to find the derivative of -2csc(2x) (bearing in mind that the derivative of csc^2(x) is -2csc2(x)cot(x)) and it gives us a result of 8csc2(2x)cot(2x)
► The second derivative of cot(2x) is 8csc2(2x)cot(2x)
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