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# The sum of the ages of Sindu and Bindu is 30 years. And the product of their ages is 221. How old are they?
This question was previously asked in
RRC Group D Previous Paper 56 (Held On: 16 Oct 2018 Shift 1)
View all RRB Group D Papers >
1. 17 years, 13 years
2. 16 years, 14 years
3. 15 years, 15 years
4. 18 years, 12 years
Option 1 : 17 years, 13 years
## Detailed Solution
Given:
The sum of the ages of Sindu and Bindu = 30 years
The product of their ages = 221
Calculation:
Let the age of Sindu be x years and Bindu be (30 – x) years respectively
According to the question
⇒ x × (30 – x) = 221
⇒ 30x – x2 = 221
⇒ x2 – 30x + 221 = 0
⇒ x2 – 17x – 13x + 221 = 0
⇒ x(x – 17) – 13(x – 17) = 0
⇒ (x – 17) (x – 13) = 0
⇒ x = 17 or 13
Now,
⇒ x = 13
Then,
⇒ (30 – x) = (30 – 17) = 13 years
∴ The required ages is 17 years and 13 years
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# Subtract the numbers: 47,584 - 9,556 = ? Calculate the numbers difference and learn how to do the subtraction, column subtracting method, from right to left
### Stack the numbers on top of each other.
#### And so on...
4 7 5 8 4 - 9 5 5 6 ?
## Subtract column by column; start from the column on the right
### Subtract the digits in the ones column:
#### When borrowing, 1 ten = 10 ones. Add 10 to the top digit in the column of the ones: 10 + 4 = 14.
7 4 7 5 8 14 - 9 5 5 6
#### After borrowing, the subtraction has become: 14 - 6 = 10 + 4 - 6 = 10 + 4 - 6 = 4 + (10 - 6) = 4 + 4 = 8. 8 is the ones digit. Write it down at the base of the ones column.
7 4 7 5 8 14 - 9 5 5 6 8
### Subtract the digits in the tens column:
#### 87 - 5 = 2. 2 is the tens digit. Write it down at the base of the tens column.
7 4 7 5 8 14 - 9 5 5 6 2 8
### Subtract the digits in the hundreds column:
#### 5 - 5 = 0. 0 is the hundreds digit. Write it down at the base of the hundreds column.
7 4 7 5 8 14 - 9 5 5 6 0 2 8
### Subtract the digits in the thousands column:
#### When borrowing, 1 ten thousand = 10 thousands. Add 10 to the top digit in the column of the thousands: 10 + 7 = 17.
3 7 4 17 5 8 14 - 9 5 5 6 0 2 8
#### After borrowing, the subtraction has become: 17 - 9 = 10 + 7 - 9 = 10 + 7 - 9 = 7 + (10 - 9) = 7 + 1 = 8. 8 is the thousands digit. Write it down at the base of the thousands column.
3 7 4 17 5 8 14 - 9 5 5 6 8 0 2 8
### Subtract the digits in the ten thousands column:
#### There is a single digit in this column: 43. 3 is the ten thousands digit. Write it down at the base of the ten thousands column.
3 7 4 17 5 8 14 - 9 5 5 6 3 8 0 2 8
## The latest 13 operations with subtracted numbers:
47,584 - 9,556 = ? Sep 21 12:00 UTC (GMT) - 488 - 3,072 - 18,455 - 281 + 120 = ? Sep 21 12:00 UTC (GMT) - 3,314 - 1,401 = ? Sep 21 12:00 UTC (GMT) 7,260 - 5,765 = ? Sep 21 11:58 UTC (GMT) - 899 - 379 = ? Sep 21 11:57 UTC (GMT) 664 - 562 = ? Sep 21 11:57 UTC (GMT) - 120 - 119 = ? Sep 21 11:57 UTC (GMT) - 932 + 1,000 - 888 - 691 + 288 - 479 = ? Sep 21 11:56 UTC (GMT) - 4,134 - 9,650 - 10,229 = ? Sep 21 11:55 UTC (GMT) 468 + 389 - 488 + 344 - 919 = ? Sep 21 11:54 UTC (GMT) - 5,198 - 3,775 = ? Sep 21 11:54 UTC (GMT) 191 - 140 - 131 + 181 + 111 - 124 + 170 - 130 = ? Sep 21 11:54 UTC (GMT) 973 + 1,950 - 552 - 323 = ? Sep 21 11:54 UTC (GMT) All the operations with the numbers subtracted by users...
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# Homogeneous Differential Equation – Definition, Solutions, and Examples
Understanding how to work with homogeneous differential equations is important if we want to explore more complex calculus topics and work on advanced endeavors in other disciplines such as physics, mathematics, and finance.
Homogeneous differential equations are differential equations where each term will be of the form, $y^{(n)}q(x)$. Oftentimes, you’ll see homogeneous equations having zero on the right-hand side of their equation.
This article gives you a comprehensive idea of what makes homogeneous differential equations unique. We’ll also summarize the important techniques you’ll need to work on different types of homogeneous differential equations. For now, let’s begin by making sure that we know how to identify and rewrite homogeneous differential equations.
## What Is a Homogeneous Differential Equation?
A homogeneous differential equation contains a differential expression on its left-hand side and zero on the right-hand side of the equation. Using this definition, the general form of a homogeneous differential equation is as shown below.
\begin{aligned}a_n(x) y^{(n)} + a_{n -1}(x) y^{(n – 1)} +…+a_1y^{\prime}+ a_0y &= 0\end{aligned}
In this general form, $a_n(x), a_{n -1}(x), …, a_1(x), a_0(x)$, are factors in terms of $x$. When the homogeneous differential equation is also linear, these become constant factors, $a_n, a_{n-1}, …,a_1, a_0$. In the past, we’ve also learned about first order and second order differential equations. Below are their general forms as we have learned before:
First Order Homogeneous Differential Equation \begin{aligned}\dfrac{dy}{dx} &= f(x, y)\\ P(x, y) \phantom{x}dx + Q(x, y) \phantom{x}dy &= 0\end{aligned} Second Order Homogeneous Differential Equation \begin{aligned}y^{\prime \prime} + P(x)y^{\prime} + Q(x)y &= f(x)\\\dfrac{d^2y}{dx^2}+ P(x)\dfrac{dy}{dx} + Q(x)y &= f(x) \end{aligned}
Use these definitions and the general form of the homogeneous differential equations to classify them. Knowing how to identify homogeneous differential equations is important since techniques for solving differential equations would depend on whether they are homogeneous or not.
### How To Tell if a Differential Equation Is Homogeneous
Suppose that we have a differential equation, $a_n(x) y^{(n)} + a_{n -1}(x) y^{(n – 1)} +…+a_1y^{\prime}+ a_0y =g(x)$, we can identify whether it is homogeneous or not depending on the value of $g(x)$.
• When we can show that $g(x) = 0$, the differential equation is homogeneous.
• Otherwise, the equation is non-homogeneous.
Let us show you two examples to demonstrate how a differential equation looks when it is homogeneous and when it is non-homogeneous.
\begin{aligned}\boldsymbol{y^{\prime \prime} \cos x – y\sin x = y^{\prime}}\end{aligned}
We can rewrite the equation so that all terms with $y$ and its derivatives are on the left-hand side.
\begin{aligned} y^{\prime \prime} \cos x – y\sin x &= y^{\prime}\\ y^{\prime \prime} \cos x -y^{\prime} – y\sin x &=0 \end{aligned}
Since we’ve written the differential equation in its standard form, we have shown that $g(x) = 0$, so our equation is indeed homogeneous. Now, let’s show you an example of a non-homogeneous equation for comparison.
\begin{aligned}\boldsymbol{y^{\prime \prime} \cos x + y\cos x – 5xe^x = 0}\end{aligned}
Once again, we rewrite the differential equation in the form of $a_n(x) y^{(n)} + a_{n -1}(x) y^{(n – 1)} +…+a_1y^{\prime}+ a_0y =g(x)$ and see if $g(x)$ is zero or not.
\begin{aligned}y^{\prime \prime} \cos x + y\cos x – 5xe^x &= 0\\ y^{\prime \prime} \cos x + y\cos x &= 5xe^x \end{aligned}
Since $g(x) = 5xe^x$ and it’s not equal to zero, the equation is not homogeneous, or a non-homogeneous differential equation. Now that we know how to identify homogeneous equations, it’s time for us to refresh our knowledge on the different techniques of finding the solutions of homogeneous differential equations.
## How To Solve Homogeneous Differential Equations?
There are different ways for us to solve homogeneous differential equations and it’s important that we know the common techniques used when dealing with them. First, let’s lay out the methods we have for first order homogeneous differential equations.
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## 8.4 Fermat and Euler
Typically, we start with conjectures and try to prove them. In this case, we take a different approach. We will provide the basis for a proof, and you will need to see what sort of statement it leads to for the theorem. The theorems which result are known as Fermat's Little Theorem and Euler's generalization. Our goal is to answer the following question:
Given an integer n, can we produce another positive integer m such that am 1 (mod n) for all integers a that have an order?
Note that this would imply that such an m is a period (although not necessarily minimal) for the powers of a mod n for all a that have an order.
Let's look at an example. We will work with powers of 2 modulo 15. In this computation, we will deal with the set of integers between 1 and n which are relatively prime to n. Since this set will appear in the course later, we will give its usual notation here:
Definition The multiplicative group modulo n, denoted Zn*, is the set of elements of Zn which are relatively prime to n. The number of integers in Zn* is denoted by (n).
In our example of n = 15, Z15* is
1, 2, 4, 7, 8, 11, 13, 14.
An interesting thing happens if we multiply every element of Zn* by a. Here's what we get with n = 15 and a = 2:
2, 4, 8, 14, 16, 22, 26, 28.
Since we are working modulo n, let's reduce the entire list modulo n = 15:
2, 4, 8, 14, 1, 7, 11, 13.
Looking closely, we recognize that these are the elements of Z15*, but listed in a different order. It's easier to see this if we sort them:
1, 2, 4, 7, 8, 11, 13, 14.
As you can see, we can produce the elements of Zn* in a new way, namely by multiplying by a. We can exploit this observation by taking the product over both sets. Starting with the product of the elements of Z15* each multiplied by a (= 2 in this case), we have:
(1) (2 · 1) · (2 · 2) · (2 · 4) · (2 · 7) · (2 · 8) · (2 · 11) · (2 · 13) · (2 · 14).
Rather than multiply everything out, we instead factor out a 2 from each term:
28 · (1 · 2 · 4 · 7 · 8 · 11 · 13 · 14).
On the other hand, we observed above that the terms of the product in equation (1) are congruent to the elements of Zn* (after reordering them). Hence it follows that
28 · (1 · 2 · 4 · 7 · 8 · 11 · 13 · 14) 1 · 2 · 4 · 7 · 8 · 11 · 13 · 14 (mod 15).
We can cancel the equal terms from each side (why?), to find that
28 1 (mod 15).
Let's check this last congruence:
Your browser does not support java.
Now, it is your turn. Try to follow the reasoning above with a different value of a. Remember that we are only interested in values of a that have an order.
Here is a function which automates the computations we did above. You supply the values of a and n. It will list the elements of Zn*, take that list and multiply each term by a, then give the list again after reducing modulo n, and finally sort the list.
Your browser does not support java.
As you try different values of a and keep n = 15, do you get the same exponent or does it vary? How can we predict what the exponent will be?
After you have a handle on n = 15, try different values for n.
#### Research Question 5
For n = 15 and any value of a relatively prime to n, find a value for m that satisfies the requirements given in the question at the beginning of this section. You should be able to model your proof after the example given above.
#### Research Question 6
Now generalize your solution to Research Question 5 to any positive integer n and any value of a that has an order.
#### Research Question 7
Specialize your solution to Research Question 6 to n = p, where p is prime, and any value of a that has an order modulo p. You should be able to be more specific about the value of "m" in this special case.
Once your proof of your conjecture for Research Question 7 is complete, you should be in a better position to prove your conjecture for Research Question 4.
The result of Research Question 7 is known as Fermat's Little Theorem. Research Question 6 is Euler's generalization of Fermat's Little Theorem to include composite values of n.
### 8.4.1 Why Is Fermat's Theorem Little?
The name Fermat's Little Theorem arose as a contrast to what is known as Fermat's Last Theorem:
If n > 3 and a, b, and c are integers such that an + bn = cn, then at least one of a, b, or c is zero.
Ironically, most mathematicians do not believe that Fermat proved this statement. It became famous years ago and is probably the most-studied mathematical problem in history. It was not considered a theorem until 1994 when Andrew Wiles completed its proof roughly 350 years after Fermat had conjectured it! Unfortunately, it has overshadowed the theorems which Fermat discovered and proved. So, the name Fermat's Little Theorem stuck as a way to distinguish it from Fermat's Last Theorem.
Section 8.1 | Section 8.2 | Section 8.3 | Section 8.4
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# Week 19: The Chi-Square Test for Homogeneity
What’s up, y’all? Today we’re going to talk about the chi-square test for homogeneity!
When Would You Use It?
The chi-square test for homogeneity is a nonparametric test used to determine whether or not r independent samples, categorized on a single dimension, are homogeneous with respect to the proportion of observations in each of the c categories.
What Type of Data?
The chi-square test for homogeneity requires categorical or nominal data.
Test Assumptions
• The data represent a random sample of independent observations.
• The expected frequency of each cell in the contingency table is at least 5.
Test Process
Step 1: Formulate the null and alternative hypotheses. The data appropriate for this type of test is usually summarized in an r x c table, where r is the number of rows of the table and c is the number of columns of the table (see the example below to get a better understanding of this). The null hypothesis claims that the in the population from which the sample was drawn, the observed frequency of each cell in the table is equal to the respective expected frequencies of each cell in the table. The alternative hypothesis claims that for at least one cell, the observed and expected frequencies are different.
Step 2: Compute the test statistic. The test statistic here, unsurprisingly, a chi-square value. To compute this value, use the following equation:
Eij, the expected cell count for the ijth cell, is calculated as follows:
Step 3: Obtain the critical value. The critical value can be obtained using a chi-square table (such as this one here). Find the column corresponding to your specified alpha-level, then find the row corresponding to your degrees of freedom. The degrees of freedom is calculated as df = (r – 1)(c – 1), where r is the number of rows in the table and c is the number of columns in the table. Compare your obtained chi-square value to the value at the intersection of your selected alpha-level and degrees of freedom.
Step 4: Determine the conclusion. If your test statistic is equal to or greater than the table value, reject the null hypothesis. If your test statistic is smaller than the table value, fail to reject the null (that is, claim that the observed cell frequencies match those of the expected cell frequencies).
Example
The example for this test comes from Amazon. Specifically, I want to see if the number of 4+ star ratings was homogeneous across the six different price ranges for laptop computers. I chose a random sample of n = 15 from each of the six price ranges and determined how many of the 15 laptops selected had four or more stars for their average review. The observed counts are in the following table:
Set α = 0.05.
H0: The proportion of 4+ star ratings is homogeneous across all price ranges
Ha: The proportion of 4+ star ratings is not homogeneous across all price ranges
The expected cell counts, as calculated by the Eij formula above, are displayed in the following table:
Calculating the chi-square value gives us:
The degrees of freedom for this test is df = (6 – 1)(2 – 1) = 5, which gives us a critical chi-square value of 11.070 by the table. Since our calculated chi-square value, 3.54, is smaller than the table value, this suggests that we fail to reject the null and claim that the proportion of 4+ star ratings is the same for each price category.
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# If You Read Nothing Else Today, Read This Report on What Does Equilateral Mean in Math
## Life, Death, and What Does Equilateral Mean in Math
Though word issues can be tricky, with a small https://www.paramountessays.com translation they can get simple equations that are simple to solve or manipulate. You don’t have to be a math genius to be successful at calculus. The proof is really easy.
Byju’s classes unique method of solving the maths problem will force you to learn the method by which the equation was created, which is way superior than memorizing and applying the formula. The above mentioned discussion pertained to defining electric field strength when it comes to the way that it is measured. The justification inside this type of proof will include things like properties concerning transformations.
## How to Get Started with What Does Equilateral Mean in Math?
A few of the worksheets are dynamically generated which means that you’ll be provided a different set each opportunity to practice. You don’t require a psychic to work out what you’re very likely to see come test day. The student passes out one part of paper to each student.
For the time being, lets do the easiest kind, called a normal tiling, in other words, tiling with regular polygons. The very first proof starts off as rectangle and afterward is broken up into three triangles that individually include a correct angle. From here on, it ought to be a bit of cake to address this equation.
In the event the sides of two triangles are the exact same then the triangles have to have the exact same angles and therefore have to be congruent. Example If we take a look at the rectangle https://www.bc.edu/bc-web/schools/law.html below, we can observe it has twice the region of the blue triangle inside it as the two yellow triangles are the exact same size as the two sub-divided blue triangles. You’re able to draw 2 equilateral triangles which are the very same form but not the identical size.
The angle between both legs is known as the vertex angle. A triangle with all 3 equal sides is known as equilateral. The definitions of a right angle, an isosceles triangle, and a square are functionalbecause we’ll have occasion to demonstrate that something is a suitable angle, an isosceles triangle, and a square.
It turned out that it was simple to apply this idea to the 3 keys that shape an ideal paragraph. The proof includes two columns, where the very first column has a numbered chronological collection of steps, called Statements, resulting in the desired conclusion. Utilize CE to clear out the latest entry.
Many elect to choose the test even sooner for a number of explanations. The proof includes a comprehensive paragraph explaining the proof practice. Write your proof so that someone which is not acquainted with the problem will readily understand what it is you are saying.
So the reason points of concurrency is a significant vocab word is since there are four main kinds of points of concurrency or talking about triangles. Naturally, several of these triangles will be shown in the context of word issues. You should learn how to address complicated problems without following a particular set of actions you’ve memorized.
Though polygon questions may appear complicated, all polygons follow only a couple of rules. There’s an additional type of triangle which is worth mentioning. I, however, can offer you a few examples that can help you train your eye.
## The War Against What Does Equilateral Mean in Math
Probability Simulations Spin the major Wheel! Geometry was the very first science. It’s possible to select various variables to customize these Constructions Worksheets for your requirements.
## The Good, the Bad and What Does Equilateral Mean in Math
The region of the necessary triangle is then the region of the space subtracted from the region of the box. The above figure indicates an instance of a scalene triangle. Your final mark ought to be in just the identical place as your very first mark.
This calculator is beneficial because geometric mean calculations often lead to big numbers and are simple to get rid of an eye on. To simplify or decrease a fraction method to make the terms smaller. Explore our absolutely free ACT review given by Mometrix.
Everyone ought to have precisely the same amount. Very effective study tools especially when you just have a limited quantity of time. If you realize that you are struggling with an issue, try a number of these tips.
Our programs take your alternatives and create the questions you would like, on your computer, instead of selecting problems from a prewritten set. Some of these sites have math worksheets generators while others might personal statement help have ready-made worksheets. Quiz video assists in testing your knowledge.
## The Bad Secret of What Does Equilateral Mean in Math
Unlike a scalar quantity, a vector quantity isn’t fully described unless there’s a direction associated with that. All these triangles are going to have complete interior degree measure of 180 degrees. In different instances, you might want to locate some dimensions yourself.
Its boundarya lineis the notion of length only. Arctan may be used to figure an angle from the the distance the opposite side and the duration of the adjacent side. There’s nothing special in regards to the base.
The picture of the isosceles triangle indicates a little mark on the 2 sides which are the exact same. You could stretch a part of rope around the full yard, then gauge the period of the rope. If you are attempting to assess the perimeter and don’t have accessibility to this calculator just don’t forget that perimeter is the duration of all of the sides of a form or the distance around the shape.
## Details of What Does Equilateral Mean in Math
Otherwise, it’s pretty straightforward. The larger it’s, the bigger the line-height ought to be as well. To learn more on every automorphism type, follow the hyperlink.
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Det här är en översatt version av sidan Additionsmetoden *Method*. Översättningen är till 100% färdig och uppdaterad.
The elimination/addition method is an algebraic method för finding solutions for a system of equations. This method words by "eliminating" or cancelling out one variable by adding the equations column by column. For example the system of equations can
$\begin{cases}y-4=2x \\ 9x+6=3y \end{cases}$
be solved using the elimination method by doing the the following:
By rearranging terms in the equations to have the same order, you can more easily compare two equations. In the example, the terms containing variables are moved the the left hand side and the terms with only constants to the right hand side.
$\begin{cases}y-4=2x & \, \text {(I)}\\ 9x+6=3y & \text {(II)}\end{cases}$
$\begin{cases}\text{-}4=2x-y \\ 9x+6=3y \end{cases}$
$\begin{cases}2x-y=\text{-}4 \\ 9x+6=3y \end{cases}$
$\begin{cases}2x-y=\text{-}4 \\ 9x=3y-6 \end{cases}$
$\begin{cases}2x-y=\text{-}4 \\ 9x-3y=\text{-}6 \end{cases}$
Now, you want the coefficients in front of one of the variables to be the same in both equations, but with the opposite sign. You do this by multiplying both sides of an equation with an appropriate constant. In the example, equation (I) is multiplied by $\text{-} 3$ in order to get the term $3y$ in equation (I) and $\text{-} 3y$ in equation (II). $\begin{cases}\text{-} 6x + 3y=12 \\ 9x-3y=\text{-}6 \end{cases}$
The equations are then added column by column. This means that the left side of an equation is added to the left side of the other and the right hand side of one equation is added to the other.
Solve the equations by balancing them in order to determine one of the variables.
$\begin{cases}9x - 3y=\text{-} 6 & \, \text {(I)}\\ 3 x=6 & \text {(II)}\end{cases}$
$\begin{cases}9x - 3y=\text{-} 6 \\ x=2 \end{cases}$
You can then substitute the value of the now known variable into either of the original equations, determine the other variable, and get the entire solution for the system of equations. In the example, $x=2$ is inserted into equation (I).
$\begin{cases}9x - 3y=\text{-} 6 & \, \text {(I)}\\ x=2 & \text {(II)}\end{cases}$
$\begin{cases}9\cdot {\color{#0000FF}{2}} - 3y=\text{-} 6 \\ x=2 \end{cases}$
$\begin{cases}18 - 3y=\text{-} 6 \\ x=2 \end{cases}$
$\begin{cases}\text{-} 3y=\text{-}24 \\ x=2 \end{cases}$
$\begin{cases}y=8 \\ x=2 \end{cases}$
The solution for the system of equations is $\begin{cases}x=2 \\ y=8. \end{cases}$
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# Selecting an activity
How do we choose a classroom activity to introduce conditional probability?
As we did in Week 1, we are going to finish this week by asking you to plan, teach and report back on a teaching activity, this time involving conditional probability.
You can start with the ‘Dog ate my homework’ activity as we have described it – it is a tried and tested approach that we have used successfully in a wide range of classrooms. If you have an alternative activity that you are keen to try, then we would like to hear about that as well.
Before you plan your lesson, it is worth thinking about how the material that we have presented here can be applied to other settings. We emphasised the importance of some key questions, which are essentially these:
• What is the probability that a guilty person is accused?
• What is the probability that an accused person is guilty?
This kind of question arises in a range of interesting real-life situations. Consider this scenario:
A particular medical condition is known to affect one in 1000 of the general population. A new test has been developed to screen for the condition.
• The test always gives a positive or negative result.
• The test will correctly give a positive result for 99% of people who have the condition
• The test will give an incorrect positive result for 2% of people who do not have the condition
This scenario has exactly the same mathematical structure as ‘The dog ate my homework’. Some key questions are:
• What is the probability that a person with the condition receives a positive test result?
• What is the probability that a person who receives a positive test result actually has the condition?
• What is the probability that a person who receives a negative test result actually has the condition?
The answers are interesting – we already know that the probability of someone with the condition receiving a positive result is 99%, but less obviously:
• The probability that someone with a positive result actually has the condition is less than 5%
• The chances that someone who is given the all-clear turns out to actually have the condition, are about 1 in 100 000.
Tempting though it is, we would offer the following words of caution before exploring this alternative scenario.
• Firstly, any medical scenario obviously needs to be handled with sensitivity – our ‘Dog ate my homework’ setting is deliberately light-hearted, and should not lead to any difficulties.
• In terms of practicalities, ‘The dog ate my homework’ has been carefully planned so that the data collection exercise will give usable results. Testing for rare medical conditions is difficult or impossible to model using the system of spinners and cubes that we have described!
For these reasons, we suggest starting with a more controlled (if somewhat artificial) scenario like ‘The dog ate my homework’. More realistic scenarios involving rarer events can be tackled in more advanced work, perhaps using a spreadsheet model to generate data.
In the next step, you will have the opportunity to tell us about a lesson you plan and teach using some of this week’s ideas.
• What are the main ideas that you are keen to try?
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# null null
```MATH 1910 - Limits Numerically and Graphically
Introduction to Limits
The concept of a limit is our doorway to calculus. This lecture will explain what the limit of a
function is and how we can find such a limit. Be sure you understand function notation at this point,
it will be used throughout the remainder of the course.
Consider the function
4
f x x − 16
x−2
Note that the domain of f is x | x ≠ 2. What does the graph look like near x 2? We can certainly
graph the function with our graphing calculator & see what happens. Before we do this, though, let’s
look at the value (output) of f for values of x close to 2. These can be seen in the following table.
x
1.99
1.999
1.9999 2.0001
2.001
2.01
fx ≈ 31.761 31.976 31.998 32.002 32.024 32.241
Note that as x approaches (gets close to) 2, the value of f x seems to be approaching 32. We say
4
"The limit of fx x − 16 as x approaches 2 is 32"
x−2
and write
4
lim f x 32 OR lim x − 16 32
x→2
x→2 x − 2
Graph the function over the intervals 0 ≤ x ≤ 4 and 0 ≤ y ≤ 40. Describe what is happening to the
graph of f as x approaches 2. Note how the graph shows the same behavior as the table above
describes.
The following is an intuitive definition for the limit of a function: If f x gets arbitrarily close to a
real number L as x approaches (gets close to) a, then
f x L OR f x L as x a
lim
x→a
We say this as "the limit of fx as x approaches a is L".
The phrase "gets arbitrarily close to" basically means as close as we like.
If f x does NOT get arbitrarily close to a real number L, we say that the limit does not exist. I
will write DNE from now on if the limit does not exist.
Note that in the previous example f 2 does not exist (is undefined), but lim f x DOES exist.
x→2
Hence, for a limit to exist at a, the function does not have to be defined at a.
Finding Limits Numerically & Graphically
When finding limits numerically we will basically construct a table of values as we did in the
example above. When finding limits graphically we will look at the graph of the function to estimate
limits. Here are some examples:
1. Estimate numerically lim gx if
x→9
x −3
x−9
We construct a table of values for gx for values of x close to 9.
gx
x
8.9
8.99
8.999
9.001
9.01
9.1
gx ≈ 0.16713 0.16671 0.16667 0.16666 0.16662 0.16621
It appears that as x approaches 9 that gx is getting closer to 0.16666... or 0. 1 6
1
6
(see below).
1
Hence, it appears that
lim gx 1
x→9
6
Side Note: Do you know how to convert from a non-terminating but repeating decimal expansion like
0. 1 6 to its equivalent fraction? Here’s one way:
Let n 0. 1 6 . Then 10n 1. 6 and 100n 16. 6 . Thus
90n 100n − 10n 16. 6 − 1. 6 15
Thus
90n 15 n 15 1
90
6
sin
x
2. Construct a table of values for f x x for x close to zero to estimate
x
lim sin
x
x→0
What mode should we be in? (radian or degree?)
Here is such a table:
x
0. 1
0. 01
0. 001
sin x ≈ 0.99833417 0.9999833 0.9999998
x
Thus it appears that
x
lim sin
x 1
x→0
This is an important limit we will see again. Look at the graph of sinx x over the intervals − ≤ x ≤
& −2 ≤ y ≤ 2 to confirm the numerical approach.
3. An example of a limit that does not exist (DNE).
Consider the function f x sin 1x . Note that the domain of f is all real numbers except 0. What can
lim sin 1x
x→0
When we try to graph this function for values of x near zero, our graphing calculator has problems.
Graph f over the intervals −2 ≤ y ≤ 2 and (1) −3 ≤ x ≤ 3, (2) −1 ≤ x ≤ 1, (3) −0. 1 ≤ x ≤ 0. 1, and
finally (4) −0. 01 ≤ x ≤ 0. 01. What do you observe? What is happening?
Recall that
1 4n
sin t 1 if t 2n 4n
for n 0, 1, 2, 3, . . .
2
2
2
2
and
3 4n
for n 0, 1, 2, 3, . . .
sin t −1 if t 3 2n 3 4n
2
2
2
2
So
1 4n
2
or when x
for n 0, 1, 2, 3, . . .
sin 1x 1 when 1x
2
1 4n
Thus f x 1 when
2 , 2 , 2 , 2 ,...,
2
x
,...
5 9 13
1 4n
Note that as n → , x → 0. Or saying it another way, between any positive number x and zero, there
are an INFINITE number of times when f x 1. In a similar manner,
2
3 4n
2
sin 1x −1 when 1x
or when x
for n 0, 1, 2, 3, . . .
2
3 4n
Thus f x −1 when
2
x 2 , 2 , 2 , 2 ,...,
,...
3 7 11 15
3 4n
Note that as n → , x → 0. Or saying it another way, between any positive number x and zero, there
are an INFINITE number of times when f x −1.
Thus we see that as x gets close to zero, f x begins to wildly oscillate between −1 and 1. In essence,
f x can never "settle down" and approach any one limit. Thus lim sin 1x DNE.
x→0
4. When Technology Fails.
We saw in the last example that our graphing calculator had troubles graphing the function for values
of x close to zero. This example shows another type error we can run into.
t4 1 − 1
If gt
, estimate numerically the following limit
t4
lim gt
t→0
We proceed by constructing a table of values for x close to zero
t
0. 1
0. 01 0. 001 0. 0001
gt 0.49999
0.5
0
0
Up to x 0. 01, we may guess that the limit appears to be approaching 12 . But, if we get closer to
zero, we see the limit appears to be zero. What is going on?
The problem is that when your graphing calculator evaluates t 4 1 for small values of t, the result
is very close to 1. In fact, if you plug in t 0. 001 in the TI-84 the result is zero. This is due to the
limitations of the graphing calculator & the number of digits the calculator is able to carry (Graph g
with 0 ≤ y ≤ 1 over (1) −4 ≤ x ≤ 4 and the (2) −0. 01 ≤ x ≤ 0. 01). The value of this limit is 12
which will be able to show later.
One-Sided Limits
Consider the function
f x
1 if x 2
3 if x 2
The graph of f is shown below
4
3
2
1
-2
2
4
6
Note that as x approaches 2 from the left (or from the negative side or from below) fx approaches 1
(it is always 1 for x 2). But as x approaches 2 from the right (or from the positive side or from
above) fx approaches 3. Since we do not approach any ONE value from both "sides", lim fx DNE.
x→2
When fx approaches 1 as x approaches 2 from the left we write
3
lim f x 1
x → 2−
and we say "the limit of fx as x approaches 2 from the left is 1" or "the left-hand limit of fx as x
approaches 2 is 1". Similarly, When fx approaches 3 as x approaches 2 from the right we write
lim f x 3
x → 2
and we say "the limit of fx as x approaches 2 from the right is 3" or "the right-hand limit of fx as x
approaches 2 is 3".
With one-sided limits we have the following useful theorem
lim
f x L if and only if xlim
f x L xlim
f x
x→a
→a−
→a
fx L if and only if both one-sided limits exist and are both equal to L.
That is, lim
x→a
Consider the function
−1 if x 0
|x|
x
1
if x 0
The graph of the function is shown below
1
-4
-2
2
4
-1
|x|
|x|
Note that lim x −1, but lim x 1. Both one-sided limits exist, but they are not equal. Thus
x→0−
x→0
|x|
lim x DNE.
x→0
Consider the function hx whose graph is shown below. Find the following limits (if they exist)
(a) lim hx (b) lim hx (c) lim hx (d) lim hx (e) lim hx (f) lim hx
x → −2 −
x → −2
x → −2
x→1−
x→1
x→1
4
2
-5
-4
-3
-2
-1
1
2
3
4
5
-2
-4
Infinite Limits
Let’s try to find the limit lim 12 We proceed numerically, constructing a table of values for x close to
x→0 x
zero.
x
1
x2
0. 1 0. 01
0. 001
0. 0001
100 10,000 1,000,000 100,000,000
4
As x gets closer to zero, the value of 12 continues to get bigger and bigger. It does NOT approach any
x
finite number. Thus, we approach no FINITE limit. To indicate this kind of behavior we introduce the
notation
lim 12
x→0 x
and say that f has an infinite limit. Note that this does NOT mean is a number (which it is not). It
simply expresses the idea that the value of the function gets arbitrarily large (as large as we want) as x
gets close to 0. When we see this expression, we say "the limit is infinity" or "the function increases
without bound".
If a function f gets arbitrarily large BUT NEGATIVE as x approaches a, we write
lim
fx −
x→a
We can say similar statements with one-sided limits.
Examples:
1. For the function g shown below find the following limits or write DNE.
lim gx
lim gx
lim gx
x → −2 −
x → −2
lim gx
x → −2
lim gx
x→3−
lim gx
x→3
x→3
10
5
-4
-3
-2
-1
1
2
3
4
-5
-10
2. Show that lim ln x −.
x→0
5
```
|
# Fraction calculator
This calculator adds two fractions. First, all fractions are converted to a common denominator when fractions have different denominators. Find the Least Common Denominator (LCD) or multiply all denominators to find a common denominator. When all denominators are the same, subtract the numerators and place the result over the common denominator. Then, simplify the result to the lowest terms or a mixed number.
## The result:
### 1/6 + 3/8 = 13/24 ≅ 0.5416667
The spelled result in words is thirteen twenty-fourths.
### How do we solve fractions step by step?
1. Add: 1/6 + 3/8 = 1 · 4/6 · 4 + 3 · 3/8 · 3 = 4/24 + 9/24 = 4 + 9/24 = 13/24
It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(6, 8) = 24. It is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 6 × 8 = 48. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - one sixth plus three eighths is thirteen twenty-fourths.
### Rules for expressions with fractions:
Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
SymbolSymbol nameSymbol MeaningExample
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4)
The calculator follows well-known rules for the order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule.
Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right.
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# Important 6 Marks Questions for CBSE 11 Maths
Important 6 Marks Questions for CBSE Class 11 Maths are provided here for students to score good marks in the examination. Mathematics seems to be a nightmare for the majority of students, as they lack the confidence and practice of the subject. BYJU’S provide class-wise important questions for practice. Practising these questions would give you an idea about the pattern of examination. Here we provide a few important 6 marks questions for practice. As these questions are a little tricky, thus this section requires a good practice of different long type questions, which can appear in the final examination.
## Important 6 Marks Questions for Class 11 Maths for Practice
Practice the below-given important 6 marks questions for CBSE class 11 Maths to score the best marks in the final examination. practising these questions multiple times will help students to write the solutions accurately and they can be able to manage the time as well.
Question 1- In a survey of 5,000 people in a town, 2,250 were listed as reading English newspaper, 1,750 are reading Hindi newspaper and 875 were listed as reading both Hindi as well as English. Find how many people neither read Hindi nor English newspaper. Also, find how many read-only English newspaper.
Question 2- Using binomial theorem, find the value of $$(52)^{4}$$.
Question 3- Show that-
$$\frac{1.2^{2} + 2.3^{2} + …….. + n(n+1)^{2}}{1^{2}.2+ 2^{2}.3+ ………+ n^{2}.(n+1)} = \frac{3n+5}{3n+1}$$
Question 4- Describe the set of complex number z such that
$$\left | \frac{z+2-i}{z+5+4i} \right | = 5$$
Question 5- A family of 4 members planner to go for Goa by train during summer for adventures. On the day of Journey all the auto/taxi drivers were on strike due to price hike of petrol. So the family couldn’t get any transport to railway station. Now family is standing at the crossing of two straight roads represented by equations 2x – 3y – 4 = 0 and 3x – 4y – 5 = 0, want to reach the path whose equation is 6x – 7y + 8 = 0 in least time. Find the equation of path that they should follow and why.
Question 6- Find the value of n, if the ratio of the fifth term from the beginning to the fifth term from end in the
$$\left ( 2^{\frac{1}{4}} + \frac{1}{3^{\frac{1}{4}}} \right )^n$$ is $$\sqrt{6} : 1$$.
Question 7- Of the members of three athletic teams in a certain school, 21 axes in the basketball team, 26 in hockey team and 20 in the football team. 14 play hockey and basketball, 15 play hockey and football, 12 play football and basketball and 8 play all the three games. How many members are there in all?
Question 8- Find the fourth term in G.P. in which the third term is greater than the first by 9 and the second term is 3 greater than the fourth by 18.
Question 9- Solve the given system of inequalities graphically:
$$x-2y \leq 3$$
$$3x + 4y \geq 12$$
$$x \geq 1$$
$$y \geq 1$$
Question 10- Show that area of the triangle formed by the lines $$y = m_{1}x + c_{1}$$, $$y = m_{2}x + c_{1}$$ and $$x = 0$$ is
$$\frac{(c_{1} – c_{2})^{2}}{2\left | m_{1} – m_{2} \right |}$$
Question 11- Using binomial theorem, prove that $$6^{n} – 5n -1$$ is always divisible by 25.
Question 12- A student wants to arrange 3 Mathematics, 4 Hindi and 5 Physics book on a shelf. In how many ways the book can be arranged? How many arrangements are possible if all the books in the same subject are to be all together?
Question 13- In any triangle ABC, prove that:
(i) $$\left ( \frac{\sin(B-C)}{\sin (B + C)} \right ) = \frac{b^{2} – c^{2}}{a^{2}}$$
(ii) $$b \cos B + c \cos C = a \cos (B-C)$$
Question 14- Prove that the diagonals formed by the four straight lines
$$\sqrt{3}x+ y = 0; \sqrt{3}y+ x = 0, \sqrt{3}x+ y = 1$$ and $$\sqrt{3}y + x = 1$$ are at right angles to one another.
Question 15- Prove that there is no term involving $$x^{5}$$ in the expansion of $$\left ( 2x^{2} – 3/x \right )^{11}$$.
Question 16- Find the equation of the circle which passes through the points (2,-2) and (3,4) and whose centre lies on $$x + y = 1$$.
Question 17- Find the coefficient of $$x^{5}y^{7}$$ in the equation $$(x – 7y)^{12}$$.
Question 18- If the sum of n terms of two arithmetic progressions are in the ratio $$14 – 5n: 3n+5$$, then find the ratio of their 8th terms.
Question 19- In a survey, it was found that people encourage their wards for science/commerce streams, and it looks commonly at school/college labels, there are 40 students in chemistry class and 60 students in physics class. Find the number of students who are either in physics or chemistry class in the following cases:
(i) The two classes meet at the same hour.
(ii) The two classes meet at different hours and 20 students enrolled in both subjects,
Question 20- An analysis of monthly wages paid to workers in two firms A and B belonging to the same industry, gives the following results:
Firm A Firm B No. of wage earners 586 648 Mean of monthly wages Rs. 5253 Rs. 5253 The variance of the distribution of the wages 100 121
(i) Which firm A or B pays a larger amount as monthly wages?
(ii) Which firm A or B shows greater variability in individual wages?
To get more class-wise and chapter-wise important questions, stay tuned with BYJU’S – The Learning App and download the app to get the latest updates.
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Show the rule of divisibility for 10
# Show the rule of divisibility for 10
The rule of the divisibility for 10 says that if a number is divisible by 10, the last digit of the number is 0.
For example, 16520, the last digit of the number is 0. Hence, 16520 is divisible by 10 without leaving remainder.
16520 \div 10=1652
Then how to prove the rule in general form?
Let’s first evaluate 3-digit number that is denoted as \overline{abc}, in which a, b, c are three digits at hundreds, tens and ones places respectively. Since a number can be expressed as the sum of its place values, \overline{abc} is rewritten as
\overline{abc}=100a+10b+c
Divide both sides by 10
\dfrac{\overline{abc}}{10} =\dfrac{100a+10b}{10}+ \dfrac{c}{10}
=10a+b+\dfrac{c}{10}
The equation shows that if \overline{abc} is divisible by 10 if and only if \dfrac{c}{10} is zero, that is
( is 5.
In a more general form, any integers can be expressed as
a=a_n10^n+a_{n-1}10^{n-1}+\dots+a_110+a_0
=10\cdotp a_n10^{n-1}+10\cdotp a_{n-1}10^{n-2}+\dots+10a_1+a_0
Divide the equation by 10
\dfrac{a}{10}=\dfrac{10\cdotp a_n 10^{n-1}+10\cdotp a_{n-1} 10^{n-2}+\dots+10a_1}{10}+\dfrac{a_0}{10}
=a_n 10^{n-1}+ a_{n-1}10^{n-2}+\dots+a_1+\dfrac{a_0}{10}
which shows, a is divisible by 10 if and only if a_0 is equal to 0 (a\geq 10).
Collected in the board: Number Theory
Steven Zheng posted 4 months ago
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# Is the inverse of a function is a function?
This function passes the Horizontal Line Test which means it is a onetoone function that has an inverse. y = 2x – 5 Change f(x) to y. x = 2y – 5 Switch x and y. This function passes the Horizontal Line Test which means it is a onetoone function that has an inverse.
A.
### How do you find the inverse of a function?
How to Find the Inverse of a Function
1. STEP 1: Stick a "y" in for the "f(x)" guy:
2. STEP 2: Switch the x and y. ( because every (x, y) has a (y, x) partner! ):
3. STEP 3: Solve for y:
4. STEP 4: Stick in the inverse notation, continue. 1 2 3.
• #### What is the inverse of a number?
In mathematics, a multiplicative inverse or reciprocal for a number x, denoted by 1/x or x1, is a number which when multiplied by x yields the multiplicative identity, 1. The multiplicative inverse of a fraction a/b is b/a. For the multiplicative inverse of a real number, divide 1 by the number.
• #### What is the inverse of a statement?
Negating both the hypothesis and conclusion of a conditional statement. For example, the inverse of "If it is raining then the grass is wet" is "If it is not raining then the grass is not wet". Note: As in the example, a proposition may be true but its inverse may be false.
• #### How do you find the inverse of a log?
Steps to Find the Inverse of a Logarithm
1. STEP 1: Replace the function notation f (x) by y.
2. STEP 2: Switch the roles of x and y.
3. x → y.
4. y → x.
5. STEP 3: Isolate the log expression on one side (left or right) of the equation.
6. STEP 5: Solve the exponential equation for “y” to get the inverse.
B.
### What is F 1 mean?
A function normally tells you what y is if you know what x is. The inverse of a function will tell you what x had to be to get that value of y. A function f -1 is the inverse of f if. for every x in the domain of f, f -1[f(x)] = x, and.
• #### What is the inverse of the function?
In mathematics, an inverse function (or anti-function) is a function that "reverses" another function: if the function f applied to an input x gives a result of y, then applying its inverse function g to y gives the result x, and vice versa. I.e., f(x) = y if and only if g(y) = x.
• #### What is an example of inverse property of multiplication?
The purpose of the inverse property of addition is to get a result of zero. The purpose of the inverse property of multiplication is to get a result of 1. We use inverse properties to solve equations. Inverse Property of Addition says that any number added to its opposite will equal zero.
• #### What is the inverse of modulo?
When ax ≡ 1 (mod m) has a solution it is often denoted in this way − but this is an abuse of notation since a modular multiplicative inverse is an integer and a1 is not an integer when a is an integer other than 1 or - 1.
C.
### What does it mean to be an inverse function?
In mathematics, an inverse function (or anti-function) is a function that "reverses" another function: if the function f applied to an input x gives a result of y, then applying its inverse function g to y gives the result x, and vice versa.
• #### How do you find the inverse of a log?
Steps to Find the Inverse of a Logarithm
1. STEP 1: Replace the function notation f (x) by y.
2. STEP 2: Switch the roles of x and y.
3. x → y.
4. y → x.
5. STEP 3: Isolate the log expression on one side (left or right) of the equation.
6. STEP 5: Solve the exponential equation for “y” to get the inverse.
• #### How do you find the inverse of a matrix?
To find the inverse of a 2x2 matrix: swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant (ad-bc).
• #### What is the composition of functions?
"Function Composition" is applying one function to the results of another. (g º f)(x) = g(f(x)), first apply f(), then apply g() We must also respect the domain of the first function. Some functions can be de-composed into two (or more) simpler functions.
Updated: 15th August 2018
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# Question Video: Finding the Argument of a Complex Number given the Argument of Its Reciprocal Mathematics
If π is the principal argument of a complex number π, determine the argument of 1/π.
03:10
### Video Transcript
If π is the principal argument of a complex number π, determine the argument of one divided by π.
In this question, weβre told the principal argument of a complex number π is equal to π. We need to use this to determine the argument of one divided by π. To answer this question, letβs start by recalling what we mean by the principal argument of a complex number.
The argument of a complex number π is the angle the line segment between the complex number π and the origin on an Argand diagram makes with the positive real axis. And this is a directed angle. Itβs positive if measured counterclockwise and negative if measured clockwise. And this means that thereβs many different equivalent angles for this argument. For example, we can add and subtract integer multiples of full revolutions.
For the principal argument, we restrict π to be between negative π and π if measured in radians. We need to use this to determine any argument of the complex number one divided by π. Of course, we can assume that π is not equal to zero for a few reasons. First, weβre trying to find one divided by π. This wouldnβt be defined if π was zero. And second, if π was equal to zero, it would be difficult to define an argument of π, since then the line between π and the origin on an Argand diagram would just be the origin. It would just be a single point. So we couldnβt really define an angle in this way.
To answer this question, thereβs several different methods we could take. For example, we could sketch π onto an Argand diagram with an angle of π and then try to determine the geometric relationship between one over π and π. And this could work. However, itβs quite difficult to determine this geometric relationship. Instead, weβre going to use a property of the argument. We recall for two complex numbers π€ sub one and π€ sub two, the argument of their product π€ sub one times π€ sub two is equal to the sum of the arguments, the argument of π€ sub one plus the argument of π€ sub two. And itβs worth pointing out this assumes that π€ sub one and π€ sub two are nonzero. We want to apply this to our question. So weβll set π€ sub one to be π and π€ sub two to be one divided by π. This gives us the argument of π times one over π is equal to the argument of π plus the argument of one divided by π.
We can now simplify this equation. First, π multiplied by one over π is just equal to one. So the left-hand side of this equation simplifies to give us the argument of one. Next, on the right-hand side of this equation, we have the argument of π. And weβre told in the question the principal argument of π is π. So we can also say the argument of π is π. So the right-hand side of this equation simplifies to give us π plus the argument of one divided by π.
On the left-hand side of this equation, we have the argument of a positive real number. And we can recall if π is a real number greater than zero, the argument of π is equal to zero. This is because then π on an Argand diagram lies on the positive real axis. So its angle with the positive real axis will just be zero. Therefore, zero is equal to π plus the argument of one over π. We can then subtract π from both sides of the equation, which gives us the argument of one divided by π is negative π, which is our final answer.
Therefore, we were able to show if π is the principal argument of a complex number, then the argument of one divided by π will be negative π.
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# Solution: The Seven Bridges of Königsberg Math Puzzle
The Swiss mathematician Leonhard Euler became interested in the Königsberg bridge problem. In 1736 he published a paper showing that it was not possible to find a path that would cross each bridge just once without missing any of them. Here’s why:
Let’s label each of the landmasses with a red uppercase letter and each of the bridges with a blue lowercase letter.
We can regard each of the landmasses as a point and each of the bridges as a line. In this way the map is equivalent to the diagram below:
We call a diagram like this a network. Each of the points A, B, C, and D is called a vertex. (The plural of vertex is vertices.) Each of the lines, a, b, c, d, e, f, and g is called an arc. We refer to the number of arcs that meet at a vertex as the degree of that vertex. The degree of vertex A is 5. The degree of each of the other three vertices, B, C, and D, is 3.
Solving the Königsberg bridge problem is equivalent to being able to draw the network pictured without lifting your pencil off the paper and without retracing any arc. We call this traveling the network.
Euler showed that a network could not be traveled if it has more than two vertices with degrees that are odd numbers. The network representing the Königsberg bridge problem has four odd vertices.
To help understand how this works, look at these two networks:
The one on the left has four vertices, each with a degree of 2. The pentagonal network has five vertices of degree 2. In both cases all the vertices are of even degree. Each of these networks can be traveled starting at any vertex and ending at that same vertex.
Now look at this network pictured to the left: Vertices A and B are each of degree 3, which is odd. The other four vertices are of degree 2, which is even. This network can be traveled via several different paths, but the starting point must be A or B and the ending point must be the other odd vertex. Try it. Here’s another network with two odd vertices. It can also be traveled as long as the end points of the trip are C and D, the odd vertices. Here’s a network that cannot be traveled because it has four odd vertices. E and H are of 1 degree each. F and G are of 3 degrees each.
Here’s another network that cannot be traveled. Why does this rule hold true? One way to think about it is to realize that a vertex of even degree can be entered and left during the journey. A vertex of degree 2 can be entered and left once. If it is of degree 4, there will be two trips through it. But if a vertex is of odd degree, the last trip in must be the end, since there’s no arc left to depart on. For example, if a vertex is of degree 7, there will be three complete trips through it, accounting for six of the arcs. But the last trip in on the seventh arc leaves no exit.
One could also start on a vertex of odd degree. Then all subsequent trips through the vertex would use a pair of arcs. In general, when a network has two odd vertices, they must be the beginning and end points if the network is to be successfully traveled.
### Now Try This
Königsberg is now known as Kaliningrad. There are still seven bridges. Some of the original bridges remain, but others are gone and new bridges have been built. Here’s the current arrangement:
Can this new network be traveled?
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# Calculus/Related Rates
← Newton's Method Calculus Optimization → Related Rates
## Introduction
One useful application of derivatives is as an aid in the calculation of related rates. What is a related rate? In each case in the following examples the related rate we are calculating is a derivative with respect to some value. We compute this derivative from a rate at which some other known quantity is changing. Given the rate at which something is changing, we are asked to find the rate at which a value related to the rate we are given is changing.
Process for solving related rates problems:
• Write out any relevant formulas and information.
• Take the derivative of the primary equation with respect to time.
• Solve for the desired variable.
• Plug-in known information and simplify.
## Notation
Newton's dot notation is used to show the derivative of a variable with respect to time. That is, if $f$ is a quantity that depends on time, then $\dot f=\frac{df}{dt}$, where $t$ represents the time. This notation is a useful abbreviation in situations where time derivatives are often used, as is the case with related rates.
## Examples
Example 1:
A cone with a circular base is being filled with water. Find a formula which will find the rate with which water is pumped.
• Write out any relevant formulas or pieces of information.
$V = \frac{1}{3} \pi r^2 h$
• Take the derivative of the equation above with respect to time. Remember to use the Chain Rule and the Product Rule.
$V = \frac{1}{3} \pi r^2h$
$\dot V = \frac{\pi}{3} \left( r^2\dot h + 2rh\dot r \right)$
Answer: $\dot V = \frac{\pi}{3} \left( r^2\dot h + 2rh\dot r \right)$
Example 2:
A spherical hot air balloon is being filled with air. The volume is changing at a rate of 2 cubic feet per minute. How is the radius changing with respect to time when the radius is equal to 2 feet?
• Write out any relevant formulas and pieces of information.
$V_{sphere} = \frac{4}{3} \pi r^3$
$\dot V = 2$
$r = 2 \$
• Take the derivative of both sides of the volume equation with respect to time.
$V = \frac{4}{3} \pi r^3$
$\dot V$ = $\frac{4}{3}3\pi r^2\dot r$ = $4 \pi r^2\dot r$
• Solve for $\dot r$.
$\dot r = \frac{1}{4 \pi r^2}\dot V$
• Plug-in known information.
$\dot r = \frac{1}{16 \pi}2$
Answer: $\dot r = \frac{1}{8 \pi}$ ft/min.
Example 3:
An airplane is attempting to drop a box onto a house. The house is 300 feet away in horizontal distance and 400 feet in vertical distance. The rate of change of the horizontal distance with respect to time is the same as the rate of change of the vertical distance with respect to time. How is the distance between the box and the house changing with respect to time at the moment? The rate of change in the horizontal direction with respect to time is -50 feet per second.
Note: Because the vertical distance is downward in nature, the rate of change of y is negative. Similarly, the horizontal distance is decreasing, therefore it is negative (it is getting closer and closer).
The easiest way to describe the horizontal and vertical relationships of the plane's motion is the Pythagorean Theorem.
• Write out any relevant formulas and pieces of information.
$x^2 + y^2 = s^2 \$ (where s is the distance between the plane and the house)
$x = 300 \$
$y = 400 \$
$s = \sqrt{x^2 + y^2} = \sqrt{300^2 + 400^2} = 500 \$
$\dot x = \dot y = -50$
• Take the derivative of both sides of the distance formula with respect to time.
$x^2 + y^2 = s^2 \$
$2x\dot x + 2y\dot y = 2s\dot s$
• Solve for $\dot s$.
$\dot s = \frac{1}{2s}( 2x\dot x + 2y\dot y)$
= $\frac{x\dot x + y\dot y}{s}$
• Plug-in known information
$\dot s$ = $\frac{(300)(-50) + (400)(-50)}{(500)}$ = $\frac{-35000}{500}$ = $-70 \$ ft/s
Answer: $\dot s = -70$ ft/sec.
Example 4:
Sand falls onto a cone shaped pile at a rate of 10 cubic feet per minute. The radius of the pile's base is always 1/2 of its altitude. When the pile is 5 ft deep, how fast is the altitude of the pile increasing?
• Write down any relevant formulas and information.
$V = \frac{1}{3} \pi r^2 h$
$\dot V = 10$
$r = \frac{1}{2} h \$
$h = 5 \$
Substitute $r = \frac{1}{2} h$ into the volume equation.
$V \$ = $\frac{1}{3} \pi r^2 h$ = $\frac{1}{3} \pi h( \frac{h^2}{4})$ = $\frac{1}{12} \pi h^3$
• Take the derivative of the volume equation with respect to time.
$V = \frac{1}{12} \pi h^3$
$\dot V = \frac{1}{4} \pi h^2\dot h$
• Solve for $\dot h$.
$\dot h = \frac{4}{\pi h^2}\dot V$
• Plug-in known information and simplify.
$\dot h$ = $\frac{4}{\pi (5)^2}10$ = $\frac{8}{5 \pi}$ ft/min
Answer: $\dot h = \frac{8}{5 \pi}$ ft/min.
Example 5:
A 10 ft long ladder is leaning against a vertical wall. The foot of the ladder is being pulled away from the wall at a constant rate of 2 ft/sec. When the ladder is exactly 8 ft from the wall, how fast is the top of the ladder sliding down the wall?
• Write out any relevant formulas and information.
Use the Pythagorean Theorem to describe the motion of the ladder.
$x^2 + y^2 = l^2 \$ (where l is the length of the ladder)
$l = 10 \$
$\dot x = 2$
$x = 8 \$
$y = \sqrt{l^2 - x^2} = \sqrt{100-64} = \sqrt{36} = 6 \$
• Take the derivative of the equation with respect to time.
$2x\dot x + 2y\dot y = 0$ ($l$ is constant so $\frac{dl^2}{dt} = 0$.)
• Solve for $\dot y$.
$2x\dot x + 2y\dot y = 0$
$2y\dot y = -2x\dot x$
$\dot y = - \frac{x}{y}\dot x$
• Plug-in known information and simplify.
$\dot y$ = $\left( - \frac{8}{6} \right) (2)$ = $- \frac{8}{3}$ ft/sec
Answer: $\dot y = -\frac{8}{3}$ ft/sec.
## Exercises
1. A spherical balloon is inflated at a rate of $100 ft^3/min$. Assuming the rate of inflation remains constant, how fast is the radius of the balloon increasing at the instant the radius is $4 ft$?
$\frac{25}{16\pi} \frac{ft}{min}$
2. Water is pumped from a cone shaped reservoir (the vertex is pointed down) $10 ft$ in diameter and $10 ft$ deep at a constant rate of $3 ft^3/min$. How fast is the water level falling when the depth of the water is $6 ft$?
$\frac{1}{3\pi} \frac{ft}{min}$
3. A boat is pulled into a dock via a rope with one end attached to the bow of a boat and the other wound around a winch that is $2ft$ in diameter. If the winch turns at a constant rate of $2rpm$, how fast is the boat moving toward the dock?
$4\pi\frac{ft}{min}$
4. At time $t=0$ a pump begins filling a cylindrical reservoir with radius 1 meter at a rate of $e^{-t}$ cubic meters per second. At what time is the liquid height increasing at 0.001 meters per second?
$t=-\ln(.001\pi)$
Solutions
← Newton's Method Calculus Optimization → Related Rates
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Engage NY Eureka Math 4th Grade Module 2 End of Module Assessment Answer Key
Question 1.
Complete the conversion charts.
Length 3 km 3,000 m 9 km 9,000 m 6 km 435 m 6,435 m 12 km 12 m 12,012 m
Explanation:
Given 3 km as 1 kilometer = 1,000 meter
so 3 km = 3 X 1,000 m = 3,000 m,
9 km = 9 X 1,000 m = 9,000 m,
6 km 435 m = 6 X 1,000 m + 435 m =
6,000 m + 435 m = 6,435m,
12 km 12 m = 12 X 1,000 m + 12 m =
12,000 m + 12 m = 12,012 m.
Mass 3 kg 3000 g 20 kg 300 g 20,300 g 1 kg 74 g 1,074 g 403 kg 4 g 403,004 g
Explanation:
Given 3 kg as 1 kilogram = 1,000 gram,
so 3 kg = 3 X 1,000 g = 3,000 g,
20 kg 300 g = 20 X 1,000 g + 300 g =
20,000 g + 300 g = 20,300 g,
1 kg 74 g = 1 X 1,000 g + 74 g =
1,000 g + 74 g = 1,074 g,
403 kg 4 g = 403 X 1,000 g + 4 g =
403,000 g + 4 g = 403,004 g.
Capacity 4 L 4,000 mL 48 L 808 mL 48,808 mL 2 L 20 mL 2,020 mL 639 L 6 mL 639,006 mL
Explanation:
Given 4 L as 1 liter = 1,000 milliliters,
So 4 L = 4 X 1,000 mL = 4,000 mL,
48 L 808 mL = 48 X 1,000 mL =
48,000 mL, + 808 mL = 48,808 mL,
2 L 20 mL = 2 X 1,000 mL + 20 mL =
2,000 mL + 20 mL = 2,020 mL,
639 L 6 mL = 639 X 1,000 mL + 6 mL =
639,000 mL + 6 mL = 639,006 mL.
Question 2.
A student completed the problem below. Check his work. Explain how you know if each solution is correct or incorrect.
a. 24 km = 24,000 m, solution is correct,
b. 16 L = 16,000 mL, solution is correct,
c. 38 kg = 38,000 kg not 3,800 g so solution is incorrect,
Explanation:
Given a student completed the problem as shown above.
Checking his work as below
a. 24 km = 24,000 m, solution is correct because
24 km = 24 X 1000 m = 24,000 m,
b. 16 L = 16,000 mL, solution is correct because
16 L = 16 X 1000 mL = 16,000 mL,
c. 38 kg = 38,000 kg not 3,800 g so solution is incorrect
because 38 kg = 38 X 1000 g = 38,000 g not 3,800 g.
Question 3.
Find the sum or difference.
a. 493 km 43 m + 17 km 57 m
b. 25 kg 32 g – 23 kg 83 g
c. 100 L 99 mL + 2,999 mL
a. 493 km 43 m + 17 km 57 m = 510 km 100 m
Explanation:
Given 493 km 43 m + 17 km 57 m as
493 km 43 m = 493 X 1000 m + 43 m =
493000 m + 43 m = 493,043 m and
17 km 57 m = 17 X 1000 m + 57 m =
17,000 m + 57 m = 17,057 m
So 493,043 m + 17,057 m = 510,100 m or
510,100 m ÷ 1000 = 510 km 100 m,
therefore, 493 km 43 m + 17 km 57 m =510,100 m or 510 km 100 m.
b. 25 kg 32 g – 23 kg 83 g = 1,949 g or 1 kg 949 g,
Explanation:
Given 25 kg 32 g – 23 kg 83 g as
25 kg 32 g = 25 X 1000 g + 32 g = 25,032 g and
23 kg 83 g = 23 X 1000 g + 83 g = 23,083 g,
Now 25,032 g – 23,083 g = 1,949 g or 1,949 g ÷ 1000 =
1 kg 949 g, therefore 25 kg 32 g – 23 kg 83 g = 1,949 g or 1 kg 949 g.
c. 100 L 99 mL + 2,999 mL = 103,098 mL or 103 L 098 mL,
Explanation:
Given 100 L 99 mL + 2,999 mL as
100 L 99 mL = 100 X 1000 mL + 99 mL =
100,000 mL + 99 mL = 100,099 mL ,
So 100,099 mL + 2,999 mL = 103,098 mL or
103,098 mL ÷ 1000 = 103 L 098 mL,
therefore 100 L 99 mL + 2,999 mL = 103,098 mL or 103 L 098 mL.
Question 4.
Billy is training for a half marathon. For the problems below, use tape diagrams, numbers, and words to explain each answer.
a. Each day, Billy runs on the treadmill for 5 kilometers and runs on the outdoor track for 6,000 meters. In all, how many meters does Billy run each day?
b. Since Billy has started training, he has also been drinking more water. On Saturday, he drank 2 liters 755 milliliters of water. On Sunday, he drank some more. If Billy drank a total of 4 liters 255 milliliters of water on Saturday and Sunday, how many milliliters of water did Billy drink on Sunday?
c. Since he began exercising so much for his half marathon, Billy has been losing weight. In his first week of training, he lost 2 kilograms 530 grams. In the following two weeks of training, he lost 1 kilogram 855 grams each week. Billy now weighs 61 kilograms 760 grams. What was Billy’s weight, in grams, before he started training?
a. Billy run each day 11,000 meters or 11 kilometrs,
Statement : Billy run each day eleven thousand meters or eleven kilometers,
b. Number of milliliters of water did Billy drink on Sunday is 1,500 milliliters or 1 L 500 mL,
Statement : Number of milliliters of water did Billy drink on Sunday is fifteen hundred milliliters or one Liter five hunderd milliliters,
c. Billy’s weight, in grams, before he started training 68,000 grams,
Statement: Billy’s weight, in grams, before he started training is sixty eight thousand grams,
Explanation:
Given Billy is training for a half marathon.
a. Each day, Billy runs on the treadmill for 5 kilometers and runs on the outdoor track for 6,000 meters.
In all, Billy run each day is 5 km + 6,000 meters as
5 km = 5 X 1000 m = 5,000 m So 5,000 m + 6,000 m =
11,000 m or 11,000 m ÷ 1,000 = 11 km,
used a tape diagrams to show problem as shown above, therefore Billy run each day 11,000 meters or 11 kilometrs.
b. Since Billy has started training, he has also been drinking more water. On Saturday, he drank 2 liters 755 milliliters of water. On Sunday, he drank some more.
If Billy drank a total of 4 liters 255 milliliters of water on Saturday and Sunday, number of milliliters of water did Billy drink on Sunday is
4 L 255 mL – 2 L 755 mL as 4 L 255 mL = 4 X 1000 mL + 255 mL =
4,255 mL and 2 L 755 mL = 2 X 1000 mL + 755 mL = 2,755 mL,
therefore 4,255 mL – 2,755 mL = 1,500 mL or 1,500 mL ÷ 1,000 =
1 L 500 mL, used a tape diagrams to show problem as shown above,
therefore Number of milliliters of water did Billy drink on Sunday is 1,500 milliliters or 1 L 500 mL.
c. Since he began exercising so much for his half marathon,
Billy has been losing weight. In his first week of training, he lost 2 kilograms 530 grams. In the following two weeks of training, he lost 1 kilogram 855 grams each week.
Billy lost in three weeks is 2 kg 530 g + 1 kg 855 g + 1 kg 855 kg
as 2 kg 530 g = 2 X 1000 g + 530 g = 2,530 g
1 kg 855 g = 1 X 1000 g + 855 g = 1,855 g
So lost weight is 2,530 g + 1,855 g + 1,855 g = 6,240 g,
Billy now weighs 61 kilograms 760 grams.
Billy’s weight, in grams, before he started training is
61 kg 760 g + 6,240 g as 61 kg 760 g = 61 X 1000 g + 760 g =
61,760 g so 61,760 g + 6,240 g = 68,000 grams,
used a tape diagrams to show problem as shown above,
therefore, Billy’s weight, in grams, before he started training 68,000 grams.
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# Three Circle Venn Diagrams
Image Source: istockphoto Purchased by Passy’s World
Three circle Venn Diagrams are a step up in complexity from two circle diagrams.
In this lesson we first look at how to read three circle diagrams. We then look at some word problems.
A Music Survey was carried out to find out what types of music a group of people liked.
The results were placed into the following three circle Venn Diagram.
Image Source: Passy’s World of Mathematics
To find out the total number of people surveyed, (or the E = Everything value), we add up all the numbers in the diagram.
When we do this the answer is 70.
This “E = Everything value” is also called “The Universal Set” of everything in the universe of our diagram.
If we want to find the total number of people who like “Rock” music, then we add up all of the numbers in the “Rock” circle, including the areas where “Rock” overlaps with the other circles.
The Total Rock People is: 16 + 2 + 8 + 5 = 31 people.
We can also work out Probability or Odds from our Venn Diagram.
For example we have found that 31 people out of 70 like Rock Music.
So if we pick any one person at random from our group, the chances, or odds, or probability, that they will like Rock music is 31 out of 70, or 31 / 70, or 31/70 x100 = 44%.
We can find the number of people who like all three types of music, by going to the centre of our diagram, where all three circles overlap.
There are 8 people who like all three types of music.
Venn Diagram Word Problem One
This first problem is a fairly easy one, where all of the information we need has been given to us in the question.
“A Class of 40 students completed a survey on what pets they like. The choices were: Cats, Dogs, and Birds. Everyone liked at least one pet.
10 students liked Cats and Birds but not dogs
6 students liked Cats and Dogs but not birds
2 students liked Dogs and Birds but not Cats
2 students liked all three pets
10 students liked Cats only
9 students liked Dogs only
1 student liked Birds only
Represent these results using a three circle Venn Diagram.”
The type of three circle Venn Diagram we will need is the following:
Image Source: Passy’s World of Mathematics
This three circle word problem is an easy one.
All of the number values for each section of the diagram have been given to us in the question.
All we need to do is carefully put the number values onto the Diagram.
We also need to check that all of the numbers add up to the total of 40 students when we are finished.
The completed Venn Diagram is shown below.
Image Source: Passy’s World of Mathematics
Note that we do not need to color in and fill in the circles on Venn Diagrams.
The following diagram is also correct and a fully acceptable answer.
Image Source: Passy’s World of Mathematics
Venn Diagram Word Problem Two
This is a harder version of Problem One, where we are given less
information in the question text. This means that we will need to do
some working out steps to get to the final completed diagram.
Image Source: Passy’s World of Mathematics
We need the exact same type of Venn Diagram as for Question 1.
Image Source: Passy’s World of Mathematics
We now need to go carefully through the word problem, and work out what information we can use straight away.
There should always be some data we can place onto the diagram to get started, even though it might not seem to be that much.
Image Source: Passy’s World of Mathematics
When we place what we know so far onto the diagram, this is what we have:
Image Source: Passy’s World of Mathematics
We now need to work through the other information in the word problem, one piece at a time.
Usually in these problems we need to work on the overlapping parts in the centre of the diagram, and then work our way out to the “Cats Only”, “Dogs Only”, and “Birds Only” outer sections of the diagram.
Remember: Work for the Inside Out.
Here is what we will do next.
Image Source: Passy’s World of Mathematics
This information refers to the following section of the Venn Diagram.
Image Source: Passy’s World of Mathematics
We can now fill in the answer of “2” onto the centre of our diagram.
Image Source: Passy’s World of Mathematics
We now have the Birds circle nearly completed.
The only thing left to do is work out the “Birds Only” section, which we will now do.
Image Source: Passy’s World of Mathematics
We can now fill in the answer of “1” onto our diagram.
Image Source: Passy’s World of Mathematics
We now have the Birds circle completed.
Next we work on the Cats Circle, following the exact same steps as we did on the Birds circle.
First we need to work on the overlaps that involve Cats and other animals in the centre of the diagram.
Image Source: Passy’s World of Mathematics
This information refers to the following section of the Venn Diagram.
Image Source: Passy’s World of Mathematics
We can now fill in the answer of “6” onto our diagram.
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We are nearly there !
We now only have “Cats Only” and “Dogs Only” to work out, and we have all the information we need to do this.
Image Source: Passy’s World of Mathematics
We can now place the “Cats Only” answer onto our diagram.
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We now only need to work out “Dogs Only”.
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Our Diagram for Problem Two is now finally complete.
Image Source: Passy’s World of Mathematics
Note that we do not need to color in and fill in the circles on Venn Diagrams.
The following diagram is also a correct and a fully acceptable answer.
Image Source: Passy’s World of Mathematics
Venn Diagram Word Problems Summary
Image Source: Passy’s World of Mathematics
Three Circle Videos
Here is a great Venn Diagrams video which also explains the “Inclusion / Exclusion” method.
This is an interesting three circles problem where they use a table of values to help with the working out.
Here is a video which shows how to do “Complement Sets” for three circle diagrams.
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# If 6 years are subtracted from the present age of Gagan and the remainder is divided by 18, then the present age of his grandson Anup is obtained. If Anup is 2 years younger to Madan whose age is 5 years, then what is Gagan’s present age?
By BYJU'S Exam Prep
Updated on: September 25th, 2023
If 6 years are subtracted from the present age of Gagan and the remainder is divided by 18, then the present age of his grandson Anup is obtained. If Anup is 2 years younger to Madan whose age is 5 years, then the present age of Gagan will be 60.
To understand how we get this answer, refer to the detailed solution below. We will derive the age of Madan to calculate the current age of Gagan by using the age Anup which is already provided as 2 years.
## Solution
To calculate the present age of Gagan as per the data provided, we will use algebra. Now, we will assume that,
Gagan’s age is a
Anup’s age is b
Now, as per the information available to us, we will get two equations which are –
(a-6)/18 = b and
a=c-2
c=5 years
Hence, the value of b will be –
b=5-2
b=3 years
Now we have c = 6 years and b = 3 years, so, to find the value of a, we will solve the following equation-
(a-6)/18=3
a-6=54
a=60 years.
Thus, the present age of Gagan is 60 years.
## Alternative Solution
Another way to solve this problem and find out the age of Gagan is to solve the following equation –
First, we will find Anup’s age which will be –
5 – 2 = 3 years
Let us assume Gagan’s age to be y years.
Then, y−618=3
=y – 6 = 54
=y = 60 years
Thus, the age of Gagan will be 60 years.
Summary:
## If 6 years are subtracted from the present age of Gagan and the remainder is divided by 18, then the present age of his grandson Anup is obtained. If Anup is 2 years younger to Madan whose age is 5 years, then what is Gagan’s present age?
The present age of Gagan will be 60 years considering if 6 years are subtracted from the present age of Gagan and the remainder is divided by 18, then the present age of his grandson Anup is obtained and Anup is 2 years younger than Madan whose age is 5 years. The answer is obtained by using the methods of algebra.
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# How do you solve the right triangle given A=52 degrees and B=8.4?
See solution below
#### Explanation:
The sum of other two acute angles in a right $\setminus \triangle A B C$ is ${90}^{\setminus} \circ$
$\setminus \angle A + \setminus \angle C = {90}^{\setminus} \circ$
$\setminus \angle C = {90}^{\setminus} \circ - \setminus \angle A$
$= {90}^{\setminus} \circ - {52}^{\setminus} \circ$
$= {38}^{\setminus} \circ$
Given that hypotenuse $B$ of right $\setminus \triangle A B C$ is $8.4$
Now, using sine rule in right $\setminus \triangle A B C$
$\setminus \frac{A}{\setminus \sin \setminus \angle A} = \setminus \frac{B}{\setminus \sin \setminus \angle B} = \setminus \frac{C}{\setminus \sin \setminus \angle C}$
$\setminus \frac{A}{\setminus \sin {52}^{\setminus} \circ} = \setminus \frac{8.4}{\setminus \sin {90}^{\setminus} \circ} = \setminus \frac{C}{\setminus \sin {38}^{\setminus} \circ}$
$A = 8.4 \setminus \sin {52}^{\setminus} \circ = 6.619$
$C = 8.4 \setminus \sin {38}^{\setminus} \circ = 5.171$
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Question 2b76d
Feb 14, 2017
In this case, the set of like terms are $\frac{3}{5} x$ and $\frac{7}{8} x$. To combine them, let's subtract $\frac{3}{5} x$ from both sides. On the left side, the positive and negative $\frac{3}{5} x$ cancel each other out or become equal to $0$.
$\frac{3}{5} x - \left(\frac{3}{5} x + 33 = \frac{7}{8} x\right) - \frac{3}{5} x$
So this is how our solution looks like.
$33 = \frac{7}{8} x - \frac{3}{5} x$
Let's focus on those two fractions. We'll be subtracting dissimilar fractions. First, we find the LCD (Least Common Denominator) of $8$ and $5$, which is $40$.
$\frac{7}{8} x - \frac{3}{5} x = \frac{-}{40}$
Next, we divide $40$ by $8$ and $5$. Then, the quotient of $40$ and $8$ is $5$ and will be multiplied to $7$. The quotient of $40$ and $5$ is $8$ and will be multiplied by $3$. The solution looks like this:
$\frac{7}{8} x - \frac{3}{5} x = \frac{35 - 24}{40}$
Subtract 35 and 24 to get the following:
$33 = \frac{11}{40} x$
Now let's isolate the variable $x$. We could apply cross-multiplication by multiplying $33$ by $40$, $11$ by $1$ (which is underneath $33$ this whole time). Solution becomes like this:
$\frac{33}{1} = \frac{11}{40} x$ ==> $1320 = 11 x$
DIvide both sides by $11$ to get $x = 120$.
$\frac{1320}{11} = \frac{11}{11} x$ ==> $120 = x$
Feb 14, 2017
$x = 120$
Explanation:
$\frac{3}{\text{5" .x + 33 = 7/"8}} . x$
Now, subtracting $\frac{3}{\text{5}} . x$ from the both sides
cancel(3/"5".x) + 33 -cancel( 3/"5".x) = 7/"8".x - 3/"5".x
$33 = x . \left(\frac{7}{\text{8" - 3/"5}}\right)$
$33 = x . \left(\text{7(5) - 3(8)"/"40}\right)$
$33 = x . \left(\text{35 - 24"/"40}\right)$
$33 = x . \frac{11}{\text{40}}$
Multiplying the both sides by 40
33 × 40 = x. 11/cancel("40" )× cancel(40)
33 × 40 = x. 11
Now, dividing the both sides by 11
$\cancel{\text{(33)"^3 × 40/cancel"(11)" = x. cancel"(11)"/cancel"(11)}}$
3 × 40 = x#
$120 = x$
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# Dry Erase Pockets Worksheets Numbers
A Realistic Figures Worksheet might help your child become more informed about the ideas behind this percentage of integers. With this worksheet, pupils will be able to solve 12 diverse difficulties relevant to realistic expression. They are going to figure out how to multiply several figures, class them in sets, and determine their items. They are going to also training simplifying reasonable expression. As soon as they have perfected these ideas, this worksheet will be a valuable resource for continuing their research. Dry Erase Pockets Worksheets Numbers.
## Realistic Numbers are a rate of integers
There are 2 types of amounts: irrational and rational. Reasonable figures are defined as whole amounts, in contrast to irrational phone numbers do not perform repeatedly, and have an unlimited number of digits. Irrational amounts are non-absolutely no, low-terminating decimals, and rectangular roots that are not perfect squares. These types of numbers are not used often in everyday life, but they are often used in math applications.
To define a rational quantity, you must understand what a realistic variety is. An integer can be a entire quantity, along with a realistic amount is actually a rate of two integers. The percentage of two integers will be the quantity at the top split with the number on the bottom. If two integers are two and five, this would be an integer, for example. There are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They can be manufactured in to a portion
A logical number includes a numerator and denominator that are not no. Consequently they may be indicated as being a small fraction. Together with their integer numerators and denominators, realistic figures can furthermore have a adverse importance. The adverse importance ought to be located to the left of and its particular absolute benefit is its distance from absolutely nothing. To streamline this illustration, we will state that .0333333 is a portion that can be composed being a 1/3.
In addition to adverse integers, a rational variety can even be manufactured in a fraction. For example, /18,572 can be a realistic amount, when -1/ will not be. Any small fraction consisting of integers is reasonable, provided that the denominator fails to consist of a and will be published for an integer. Similarly, a decimal that ends in a position is yet another logical quantity.
## They make feeling
Even with their name, logical numbers don’t make significantly sensation. In mathematics, they can be single entities using a exclusive size around the amount range. Which means that once we count something, we are able to get the dimensions by its ratio to the initial amount. This holds true even though there are actually infinite logical numbers involving two distinct figures. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
In real life, if we want to know the length of a string of pearls, we can use a rational number. To get the period of a pearl, for example, we might matter its breadth. One particular pearl weighs ten kilograms, and that is a realistic variety. Furthermore, a pound’s weight means 15 kilograms. Hence, we will be able to break down a lb by ten, with out be concerned about the duration of a single pearl.
## They can be conveyed as a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal amount may be published as a numerous of two integers, so four times 5 various is the same as eight. The same dilemma involves the frequent small fraction 2/1, and either side needs to be split by 99 to find the appropriate response. But how do you have the conversion process? Below are a few illustrations.
A realistic amount may also be written in various forms, which include fractions as well as a decimal. A good way to symbolize a reasonable quantity inside a decimal is always to break down it into its fractional equal. There are actually 3 ways to divide a reasonable amount, and each one of these ways brings its decimal equal. One of those ways is to break down it into its fractional equal, and that’s what’s called a terminating decimal.
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# Square and Square Roots (Explained with Example)
Hello dear, your most welcome here. Here you will get your solution easily. As you know that mathematics is very important part of our life. In mathematics problems many times we have need to find the Square and Square Roots of numbers. This page is fully dedicated to explain about Square and Square Roots (Explained with Example).
We will discuss about these topics steps by steps and also learn with examples. So read full article for better learning. So lets learn about Square and Square Roots (Explained with Example)…
Square and Square Roots (Explained with Example)
## What is square :
If we multiply a number with the number itself then the obtained number is called the square of the number. Let a number y, then the square of y is y × y and denoted as y2 .
### Examples-
1. 32 = 3 × 3 = 9
2. 52 = 5 × 5 = 25
3. 122 = 12 × 12 = 144
## Perfect square or square number :
Perfect square or Square numbers are those numbers, which can be represented as the product of pairs of equal factors. OR
If any number is the square of a natural number then that number is called the perfect square or a square number.
Examples – 1 = 12, 4 = 22 , 9 = 32 , 16 = 42, 25 = 52, 36 = 62 and so on.
In the above cases, 1, 4, 9,16, 25, 36,….. are square numbers or perfect squares.
Note- A perfect square can always be expressed as the product of pairs of equal factors.
Example 1- Check whether 144 is a perfect square or not. If yes, then find the number whose square is 196.
Solution- First, on resolving 144 into prime factors, we have 144 = 2 × 2 × 2 × 2 × 3 × 3
You can see that 144 can be expressed as the product of pairs of equal factors as we discussed above. Therefore 144 is a perfect square number.
Hence 144 = 22 × 22 × 32
= (2 × 2 × 3)2
= (12)2
Hence 12 is the number whose square is 144. Here 12 is the square root of 144. We will learn about square roots further.
#### Also Read : How to Calculate Percentage of Number – Full Explanation
Example 2- Is, 1764 is a perfect square, find that number whose square is 1764.
Solution- On resolving 1764 into prime factors, we have 1764 = 2 × 2 × 3 × 3 × 7 × 7
You can see that 1764 can be expressed as the product of pairs of equal factors.
Therefore 1764 is a perfect square number.
Hence, 1764 = 2× 32 × 72
= (2 × 3 × 7)= (42)2
Therefore, 42, is the number whose square is 1764.
Example 3- Check whether 2548 is a perfect square.
Solution- On resolving 2548 into prime factors, we have 2548 = 2 × 2 × 7 × 7 × 13 = (22 × 72 × 13)
You can see that 2 and 7 have pairs but 13 has no pair, Hence 2548 can not be expressed as a product of pairs of equal factors. Hence, 6292 is not a perfect square.
Square and Square Roots (Explained with Example)…..
## Properties of perfect square numbers :
Here are some properties related to perfect square numbers. These properties help you to find that the given numbers can be a perfect square or not. Read these properties carefully.
Property 1- Any number ending with digitals, 2, 3, 7 or 8 is never a perfect square.
Example- The numbers 72, 83, 167, 238 etc. We see that these numbers are end with digits 2, 3, 7, 8, respectively, hence this type of numbers are not perfect square numbers.
Property 2- If any number ends with an odd number of zeros, then it is never a perfect square.
Example- The numbers 180, 3000, 800000, etc. end with odd numbers of zero (Here number of zeroes are one, there, five ). Therefore this type of numbers are not perfect square.
Property 3- If a number leaves a remainder 2 after dividing by 3, then it is not a perfect square.
Example – 38, 77, 101, 308, 596 etc. If we divide these numbers from 3, we will get 2 as reminder, hence this type of number can not be a perfect square.
Property 4- If a number leaves a remainder 2 or 3 after dividing by 4, then it is not a perfect square.
Example- 578, 654, 798, 1002, etc. If we divide these numbers from 4, we will get either 2 or 3 as reminder, hence this type of number can not be a perfect square.
Property 5- The square of each even number is even.
#### Example-
22 = 4, 42 = 16 , 62 = 36 , 82 = 64 , 102 = 100 etc.
Hence if square an even number we will always get an even number.
Property 6- The square of each odd number is odd.
Example- 32 = 9 , 52 = 25 , 72 = 49 , 92 = 81 , etc.
Hence if square an odd number we will always get an odd number.
Property 7- The square of each proper fraction is always smaller than the fraction.
Example – ( 3 / 5)2 = 9 / 25 , here (9 / 25) < (3 / 5).
#### Property8-
For each natural number n, we have –
(n+1)2 – n2 = (n + 1 + n)(n + 1 – n) = (2n + 1) or {(n + 1) + n}.
Therefore we can write {(n + 1)2 – n2} = {(n + 1) + n}.
That means the difference of the square of two consecutive numbers will be the sum of the numbers. Lets see some examples-
Examples- (i) {(46)2 – (45)2} = (46 + 45) = 91 (ii) {(79)2 + (78)2} = (79 +78) = 157
Property 9- The sum of first n odd natural numbers will be equal to n2 .
Examples- (i) Sum of first 7 odd natural numbers = (1 + 3 + 5 + 7 + 9 + 11 + 13) .
As you are seeing, There are 7 numbers here, Hence n = 7
then the sum of (1 + 3 + 5 + 7 + 9 + 11 + 13 ) = 72 = 49.
(ii) Sum of first 9 odd natural numbers = (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17)
As you are seeing, There are 9 numbers here, Hence n = 9
then the sum of (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17) = 92 = 81.
Property 10 (Pythagorean Triplets)- If there are three natural numbers m, n, p, such that (m2 + n2) = p2, then (m, n, p) are called pythagorean triplet.
Means if we have three natural numbers such that square of greatest number is equal to the sum of the square of other two numbers then numbers are called pythagorean triplet.
#### Important Result :
If m is a natural number such that m > 1, then for every natural number we have (2m, m2 – 1, m2 + 1) as a pythagorean triplet.
Example – (i) On putting m = 3 in (2m, m2 -1, m2 +1), we get (6, 8, 10) as a pythagorean
triplet.
(ii) On putting m = 6 in (2m, m2-1, m2 +1), we get (12, 35, 37) as a pythagorean
triplet.
Property 11 – There are 2n non-perfect square numbers between two consecutive square numbers n2 and (n+1)2
That is, if you want to find total number of non perfect square number between two consecutive perfect square number then multiply square root of smaller number by two.
Let’s understand it by example-
Example – (i) Let squares of two consecutive numbers 2 and 3 are 4 and 9 respectively. Here n = 2.
Then the number of non – perfect square numbers between 4 and 9 are 2 × 2 = 4.
(ii) Let squares of two consecutive numbers 7 and 8 are 49 and 64 respectively. Here n = 7.
Then the number of non – perfect square numbers between 49 and 64 are 2 × 7 = 14.
Square and Square Roots (Explained with Example)…
## Shortcut method for the squaring a number :
We had discussed about the square of any number. Let’s discuss the few Shortcut method for the squaring a number .
First we understand Column Method For Squaring the Two Digit Number.
### Column Method For Squaring the Two Digit Number :
We’ll understand Column Method using few steps.
Let us consider a number which has the tens digit ‘a’ and the unit digit ‘b’. Now we shall square this number.
Step 1.
Make three columns, I, II and III, headed by a², (2 × a × b) and b² respectively. Write the values of a2, (2 × a × b) and b2 in columns I, II and III respectively.
Step 2.
In Column III, underline the unit digit of b² and carry over the tens digit of it to Column II and add it to the value of (2 × a × b).
#### Step 3.
In Column II, underline the unit digit of the number which is obtained in Step 2 and carry over the tens digit of it to Column I and add it with the value of a².
Step 4.
Underline this number which was obtained in Step 3 in Column I. The digits which are underlined give the required square number.
To understand these steps we will take few examples.
#### EXAMPLE 1–
Find the square of 47.
Solution : Given number = 47
Therefore, a = 4 and b = 7.
We will apply the steps given above. First we make a table having columns I, II, & III as shown below . We have a = 4 and b = 7. Write a2 = 42 = 16 in first column, 2ab = 2 × 4 × 7 = 56 in second column and b2 = 72 = 49 in third column. Now we underline unit digit 9 in third column and take tens digit 4 as carry and add with 56 in second column. Here we get 56 + 4 = 60.
Similarly we underline unit digit 0 in second column and take tens digit 6 as carry and add with 16 in first column. Here we get 16 + 6 = 22. Now have to underline both digits of 22 as shown below.
Now we take the numbers having underlined, which are 22, 0 and 9.
Therefore Square of 47 = (47)2 = 2209
I hope you understand these steps . For more clarification , let’s take another example.
#### EXAMPLE 1–
Find the square of 86.
Solution : Given number = 86
∴ a = 8 and b = 6.
We will apply the steps given above. First we make a table having columns I, II, & III as shown below . We have a = 8 and b = 6. Write a2 = 82 = 64 in first column, 2 × a × b = 2 × 8 × 6 = 96 in second column and b2 = 62 = 36 in third column. Now we underline unit digit 6 in third column and take tens digit 3 as carry and add with 96 in second column. Here we get 96 + 3 = 99.
Similarly we underline unit digit 9 in second column and take tens digit 9 as carry and add with 64 in first column. Here we get 64 + 9 = 73. Now have to underline both digits of 73 as shown below.
Now we take the numbers having underline, which are 73, 9 and 6. Therefore Square of 86 = (86)2 = 7396
### Diagonal Method For the Squaring a Number :
We had discussed about the shortcut method to find square of any number. Let’s discuss the Diagonal Method For the Squaring a Number .
To understand Diagonal Method we are taking an example and will solve it step by step.
Let we have to find the square of 39 by diagonal method.
Solution :
Step 1.
The given number contains two digits. So, draw a square and divide it into 4 subsquares as shown below. Write the digits 3 and 9 vertically and horizontally, as shown below.
Step 2.
Multiply one by one each digit on the left side of the square by each digit on the top. Write the obtained product in the relevant subsquare. If the product is a one-digit number, write it below the diagonal and put 0 above the diagonal.
If we multiply 3 and 3 we will get 9 as product. Since it is a one digit number, therefore we’ll write 9 below the diagonal and put 0 above the diagonal.
In case the product is a two-digit number, write the tens digit above the diagonal and the units digit below the diagonal.
If we multiply 3 and 9 we will get 27 as product. Since it is a two digit number, therefore we’ll write 7 below the diagonal and put 2 above the diagonal.
Do this further for all subsquare.
#### Step 3.
Start from the lowest diagonal, and add the digits diagonally. If the sum is a two-digit number, underline the units digit and carry over the tens digit to the next diagonal.
In lowest diagonal we have only 1, after that in next diagonal we have 7 + 8 + 7 = 22 , so we underline unit digit 2 and carry the tens digit 2 in the next diagonal. In next diagonal we have 2 + 9 + 2 = 13 after adding carry 2 we got 15 . Now we underline unit digit 5 and carry the tens digit 1in the next diagonal. (Add shown above with arrows)
Step 4.
Underline all digits obtained in the sum of the topmost diagonal. In topmost diagonal we got 0 + 1 = 1.
Step 5.
Now note the underlined digits, this is your required square number.
Therefore (39)2 = 1521
If you don’t understand read it again and again.
## Square Root :
If we multiply a number x by itself then we will get a resultant number, then the number x is called the square root of the resultant number.
To represent the square root we use the symbol $$\sqrt{}$$.
$$x\;\times\;x\;=\;x^{2\;\;}\;or\;\;\sqrt{x^2}\;=\;x$$
Example – $$\sqrt{81}\;=\;9$$
$$\sqrt{25}\;=\;5$$
$$\sqrt9\;=\;3$$
Note – Basically square root represent 1/2 power of the number.
## Square Root of a perfect square number using the prime factorisation method :
To find the square root of a given perfect square number you have to follow some steps given below –
Step – 1
Find the prime factors of a given number.
Step– 2
Underline the similar factors in pairs.
Step– 3
Choose one factor from each pair and find the product.
324
To understand, some examples are given below.
Example– (1) What is the square root of 324.
Solution – Prime factors of 324 are –
324= 2×2 × 3×3 × 3×3
Now we will take one factor from each pair, and multiply them.
Hence, $$\sqrt{324}\;=\;2\times3\times3\;=\;18$$
Therefore square root of the 324 is 18.
Example2 Write the square root of 19600
Solution – Prime factors of 19600 are-
19600 = 2×2 × 2×2 × 5×5 × 7×7
Now we will take one factor from each pair and multiply them.
Hence, $$\sqrt{19600}\;=\;2\times2\times5\times7\;=\;140$$
Therefore square root of the 19600 is 140.
After the prime factorisation method, we’ll discuss about Long Division Method to find the square root of a perfect square number.
## Long Division Method to find the square root of a perfect square number :
To find the square root of very large numbers, the factorisation method is very large numbers we have to use long division method.
To understand the long division method we have to follow some steps given below.
Step– 1
Make the pairs of digits from the side of the unit digit. Remaining digit ( if any) and every pair is called a period.
Step – 2
Think that the number whose square is equal or just less than the first pair ( period) or remaining digit. You have to take this number as the divisor and quotient also.
Step – 3
Now multiply the divisor and quotient and subtract from the first pair ( period) Now to the right side of the reminder take down the next pair. This is your new dividend.
Step – 4
Now add the quotient with divisor and annex with a suitable hat digit to use as a new divisor. Choose the next digit in such a way the product of the new divisor and this digit should be equal or just less than the new dividend.
#### Step –5
Repeat step – 2, step-3, and step-4 till all the pairs (period)
have been taken up.
Now the obtained quotient is the required square root.
To understand these steps lets see some examples-
Example – Find the square root of 841
According to first step we make the pairs of digits from the side of the unit digit. Hence first we underline to 41. Since 8 has no pair hence take it single as shown below.
After that think a number whose square is just less than or equal to 8, which is 2. So write 2 in left side and hence up side also.
After subtracting square of 2 (=4) from 8 we have 4 as remainder. Now note down next pair 41 with remainder 4. Multiply divisor 2 with quotient 2 you will get 4 as the tens digit of next divisor. Now think a number of unit place for next divisor and use this unit place digit as quotient . As shown below .
Hence, according to above calculation Square root of 841 is 29.
## Square root of numbers in decimal form :
As we know that many times we face some mathematical calculation in which we have to find the square root of a number having decimal .
So now we discuss about the method to find the Square root of numbers in decimal form.
Method:
If necessary, make the number of decimal places even by adding zeros. Now, mark the periods and find the square root by long-division method as explained above .
Mark the period from left side after decimal and from right side before decimal.
Place the decimal point in the square root as soon as you finish the integral.
## To find the square root which is correct to certain decimal places :
If you want to find the square root of any number, which is correct up to the two decimal places, then you have to find it up to three decimal places and after that round it off up to two decimal places.
Similarly, If you want to find the square root of any number, which is correct up to the three decimal places, then you have to find it up to four decimal places and after that round it off up to three decimal places and so on.
This is all about the Square and Square Roots (Explained with Example). Thanks for visit.
Square and Square Roots (Explained with Example)
Square and Square Roots (Explained with Example)
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# Lesson Notes By Weeks and Term - Senior Secondary School 3
SIMULTANEOUS EQUATIONS
SUBJECT: MATHEMATICS
CLASS: SS 3
DATE:
TERM: 1st TERM
REFERENCE TEXTS:
• New General Mathematics for SS book 3 by J.B Channon
• Essential Mathematics for SS book 3
• Mathematics Exam Focus
• Waec and Jamb past Questions
WEEK EIGHT
TOPIC: SIMULTANEOUS EQUATIONS
CONTENT
• Solving Simultaneous Equations Involving One linear and One quadratic.
• Solving Simultaneous Equations Using Graphical Method
SIMULTANEOUS EQUATIONS INVOLVING ONE LINEAR AND ONE QUADRATIC
One of the equations is in linear form while the other is in quadratic form.
Note: One linear, one quadratic is only possible analytically using substitution method.
Examples:
1. Solve simultaneously for x and y (i.e. the points of their intersection)
3x + y = 10 & 2x2 +y2 = 19
Solution
3x + y = 10 ----------- eq 1
2x2 + y2 = 19 --------- eq 2
Make y the subject in eq 1 (linear equation)
y = 10 – 3x ---------- eq 3
Substitute eq 3 into eq 2
2x2 + (10-3x) 2 = 19
2x2+ (10 – 3x) (10 – 3x) = 19
2x2 + 100 – 30x – 30x + 9x2 = 19
2x2 + 9x2 - 30x – 30x + 100 – 19 = 0
11x2 - 60x + 81 = 0
11x2 - 33x – 27x + 81= 0
11x (x-3) – 27 (x – 3) = 0
(11x – 27) (x – 3) = 0
11x – 27 = 0 or x-3 = 0
11x = 27 or x = 3
x = 27/11 or 3
Substitute the values of x into eq 3.
When x = 3
y = 10 – 3(x)
y = 10 - 3(3)
y = 10 – 9 = 1
When x =27/11
y = 10 – 3(27/11)
y = 10 - 51/11
y = 110 - 51
11
y = 59/11
w hen x = 3, y = 1
x = 27 , y = 59
11 11
1. Solve the equations simultaneously 3x + 4y = 11 &xy = 2
solution
3x + 4y = 11 -------- eq 1
xy = 2 -------- eq 2
Make y the subject in eq 1
4y = 11 – 3x
y = 11 – 3x ………… eq3
4
substituteeq 3 into eq 2
x y = 2
x ( 11- 3x ) = 2
4
x (11-3x) = 2x4
11x – 3x2 = 8
-3x2 + 11x – 8 = 0
-3x2 + 3x + 8x – 8 = 0
-3x (x-1) +8 (x-1) = 0
(-3x + 8) (x-1) = 0
-3x + 8 = 0 or x – 1 = 0
3x = 8 or x = 1
x = 8/3 or 1
Substitute the values of x into eq 3
y = 11- 3x
4
when x = 1
y = 11 – 3(1) = 11-3 = 8
4 4 2
y = 4
when x = 8/3
y = 11 – 3(8/3)
4
y = 33 – 24 = 9 = 3
12 12 4
x = 1, y = 2
x = 8/3, y = 3/4.
Evaluation
Solve for x and y
1. 3x 2 - 4y = -1 2. 4x2 + 9y2 = 20
2x – y = 1 2x – 9y = -2
MORE EXAMPLES
Solve simultaneously for x and y.
3x – y = 3 -------- eq 1
9x - y 2 = 45 --------- eq 2
Solution
From eq 2
(3x)2 - y = 45
(3x-y) (3x+y) = 45 ---------- eq 3
Substitute eq 1 into eq 3
3 (3x + y) = 45
3x + y = 15 ……………..eq4
Solve eq 1 and eq 4 simultaneously.
3x – y = 3 --------- eq 1
3x + y = 15 -------- eq 4
eq 1 + eq 4
6x = 18
x = 18/ 6
x = 3
Substitute x = 3 into eq 4.
3x + y = 15
3 (3) + y = 15
9 + y = 15
y = 15 – 9
y = 6
x = 3, y = 6
Evaluation
Solve for x and y in the following pairs of equations
1. (a) 4x2 – y2 = 15 (b) 3x2 +5xy –y2 =3
2x – y = 5 x - y = 4
Example
The product of two numbers is 12. The sum of the larger number and twice the smaller number is 11. Find the two numbers.
Solution
Let x = the larger number
y = the smaller number
Product, x y = 12 …………….eq1
From the last statement,
x + 2y = 11 ………….. eq2
From eq2, x = 11 – 2y …………...eq3
Sub. Into eq1
y(11 – 2y) = 12
11y – 2y2 = 12
2y2 -11y + 12 = 0
2y2 – 8y – 3y + 12 = 0
2y(y-4) – 3(y-4) = 0
(2y-3)(y-4) =0
2y-3 =0 or y-4 =0
2y = 3 or y = 4
y= 3/2 or 4
when y = 3/2 when y=4
x = 11 – 2y x = 11- 2y
x = 11 – 2(3/2) x = 11 – 2(4)
x = 11 – 3 x = 11 – 8
x = 8 x = 3
Therefore, (8 , 3/2)(3 , 4)
Evaluation
Solve the following simultaneous equation
1. (a) 22x-3y = 32, 3x-2y = 81 (b) 2x+2y=1, 32x+y = 27
2. Bisi’s and Fibie’s ages add up to 29. Seven years ago Bisi was twice as old as Fibie. Find their present ages.
SOLVING SIMULTANEOUS EQUATIONS USING GRAPHICAL METHOD
Examples
Using the scale 2cm to 1 units on x-axis and 2cm to 2 unit on y-axis, draw the graph of y = x2 – x – 1 and y = 2x – 1 (on the same scale and axis for values of x: - 3≤x< 4
Solution
Table of values for y = x2 – x – 1
X -3 -2 -1 0 1 2 3 4 x2 9 4 1 0 1 4 9 16 -x +3 +2 +1 0 -1 -2 -3 -4 -1 -1 -1 -1 -1 -1 -1 -1 -1 Y 11 5 1 -1 -1 1 5 11
X -3 -2 - 1 0 1 2 3 4 Y 11 5 1 -1 -1 1 5 11
Table of values for y = 2x – 1
X -3 -2 -1 0 1 2 2x -6 -4 -2 0 2 4 -1 -1 -1 -1 -1 -1 -1 Y -7 -5 -3 -1 1 3
X -3 - 2 - 1 0 1 2 3 Y -7 -5 - 3 - 1 1 3 5
Evaluation
1. Copy and complete the table below of values for the relation y = 2x2 – 3x – 7
x -2 -1 0 1 2 3 4 5 y
b.Using a scale of 2cm to 1 unit on x-axis and 2cm to 5 unit on y-axis, draw the graph of the relation
y = 2x2-3x-7 for -3 < x ≤ 5
c.Using the same scale and axis, draw the graph of y = 2x-1
1. Use your graph to find the values of x and y.
GENERAL EVALUATION AND REVISION QUESTIONS
1. Solve the simultaneous equation: 3x2 - 4y = -1 & 2x - y = 1
2. Five years ago, a father was 3 times as old as his son, now their combined ages amount to 110 years. How old are they?
3. Solve: 4x2 - y2 = 15 & 2x - y = 5
4. Seven cups and eight plates cost # 1750. Eight cups and seven plates cost #1700. Calculate the cost of a cup and of a plate.
WEEKEND ASSIGNMENT
Solve each of the following pairs of equations simultaneously,
1. xy = -12 ; x – y = 7 a. (3 , -4)(4 ,-3) b. (-2 ,4)(-3, -4) c.(-4, 5)(-2 , 3) d.(3 ,-3)(4,-4)
2. x – 5y = 5 ; x2 – 25y2 = 55 a (-8, 0)(3/5 , 0) b. (0, 0)(-8 , 3/5) c. (8 , 3/5) d. (0, 8)(0, 3/5)
3. y = x2 and y = x + 6 (a).(0,6) (3,9) (b)(-3,0) (2,4) (c) (-2,4) (3,9) (d).(-2, 3), (-3,2)
4. x – y = -3/2 ; 4x2 + 2xy – y2 = 11/4 : a. (-1, 1/2)(1, 5/2). b. (3, 2/5) (1, 1/2) c.(3/2 , -1) (4,2) d.(-1 , -1/2)(-1 , 5/2)
5. m2 + n2 = 25 ; 2m + n – 5 = 0 : a. (0,5)(4, -3) b.(5,0)(-3,4)c.(4,0)(-3,5) d(-5,3)(0,4)
THEORY
1a. Find the coordinate of the points where the line 2x – y = 5 meets the curve 3x2 – xy -4 =10
1. Solve the simultaneous equation: 22x+4y = 4, 33x + 5y – 81= 0
2. A woman is q years old while her son is p years old. The sum of their ages is equal to twice the difference of their ages. The product of their ages is 675.
Write down the equations connecting their ages and solve the equations in order to find the ages of the woman and her son. (WAEC)
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# Proportions, Ratios, Rates, Oh My!
Contributor: Mason Smith. Lesson ID: 11129
"I have a map of the United States, actual size...Scale: 1 mile=1 mile. I spent last summer folding it. People ask me where I live...I say, E6." Steven Wright's joke shows why ratios are so important!
categories
## Pre-Algebra, Ratios, Rates, Percentages, and Proportions
subject
Math
learning style
Visual
personality style
Lion
Middle School (6-8)
Lesson Type
Quick Query
## Lesson Plan - Get It!
Audio:
1:3 Odds? 1:15 Faculty-to-Student ratio?
• What could this possibly mean for statues and maps?
Ratios and proportions surround us and are a part of everyday life.
From maps to recipes to filling up the gas tank, many things can be measured in terms of their parts and how they relate to each other, which is a casual definition for ratio.
The formal definition would be "a comparison of two quantities (or amounts) by division," which means that all ratios can be written as a/b where b ≠ 0 (since we can't divide by zero!) or, in another common form, a:b.
• But what does that mean?
A ratio says that for every number of part a, there are so many parts b. In other words, if we have the ratio 2 oil:5 gas, that means for every 2 parts of oil, there should be 5 parts gas in the mixture, which is super important for any type of engine, because without the right mix we could have an unwanted boom!
The next idea is that of proportions, which look like equalities, but are a bit trickier. Proportions describe a relationship where two ratios are proportional. If two ratios are proportional, then we can set them equal to each other and solve for any unknown quantities, and we will find the relationship between the two.
Let's do this example for some clarification:
The ratio of students to professors at a local college is 15:1 (15 students per professor). If there are 675 students, how many professors are there?
professor = 1 student 15
Since the ratio and the fact are proportional, we can set them equal to each other to figure out the number of professors:
ratio of professor or 1 = x or total number of professors student 15 675 students
• But how do we solve?
Well, we have to cross-multiply the numbers (multiply times the opposite side of the equal sign), so we end up with 675 * 1 = 15 * x.
I bet you can solve it from here.
• How many professors are there?
x = 45, so there are 45 professors at the college.
That is an example of basic proportions.
• What happens if we don't have an example that is 1:some number?
• How do we find out how many of b there are to each a?
We try to find the unit rate, which is a lot easier than it sounds.
Let's look at the Nathan's Hot Dog eating contest.
• How many hot dogs did the winner from 2016 eat per minute during the 10 minute contest?
Well that's easy, Joey Chestnut ate 70 hot dogs in 10 minutes. We can write that as 70/10, but we want to know how many hot dogs were eaten in one minute. We will be trying to find our x number of hot dogs in one minute, so we have:
hot dogs 70 = x hot dogs minutes 10 1 minutes
Remember to keep the same units on the same side of the division line for a correct answer.
Now, we just solve for x by cross multiplying, and we get 7 hots dogs per minute.
• But wait, shouldn't it be 7 hot dogs per 10 minutes?
No, because we always use the units from the same side as the variable in our answer. Plus, that doesn't make much sense when you think about it — how can he eat 7 hot dogs and 70 hot dogs in the same amount of time?
• Now we know how to find proportions and ratios, but is there any place we are going to actually use this?
• Have you ever looked at a map?
Be it paper or from Google, maps need ratios to work because, without using ratios and being able to scale the map, we would have to walk around with a piece of paper the size of the area we are trying to navigate! I don't think there are any tablets or phones that are quite big enough to show us the entire world, so ratios and scale models will probably help, right?
If we look at this Map of New York (ironically from the University of Texas), we can see the scale in the bottom-left corner. Grab a ruler and print a copy of this map. We are going to measure the distance from Niagara County to Orange county.
I measured and came up with an answer of 9 centimeters. When I set my ruler on the the scale, I found that there are 100 miles in every 3.5 centimeters (You may get slightly different measurements, but as long as your end answer is close, you are fine.).
Let's make our proportion:
centimeters 3.5 = 9 centimeters miles 100 x miles
• Do you think you can solve this?
I got an answer of about 260 miles; how about you?
When we check against a survey of New York, we find out that it is about 300 miles across all of New York, so that's about right!
If you would like, you can try to figure out the distance between some more points on the map and check it against a website and see how close you are! If you'd like some online practice, check out Map Scale from Mr. Nussbaum.
Continue on to the Got It? section to practice your skills!
## Elephango's Philosophy
We help prepare learners for a future that cannot yet be defined. They must be ready for change, willing to learn and able to think critically. Elephango is designed to create lifelong learners who are ready for that rapidly changing future.
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## A Calculus Primer Part VIII- Taylor Series
### #1 macosxnerd101
• Games, Graphs, and Auctions
Reputation: 11787
• Posts: 44,294
• Joined: 27-December 08
Posted 26 June 2011 - 03:15 PM
This tutorial will introduce Taylor Series. To begin, it is important to first understand a function in terms of the Fundamental Theorem of Calculus: adding up a rate to get a change. The derivative of a function models its change, the second derivative describes the rate of change of the first derivative, and the third derivative models the rate of change for the second derivative. This continues iteratively and infinitely, and is the basis of a Taylor series.
Let's look at the definition of a Taylor Series:
Basically, adding up all the infinitesimal rates is used to determine the change in the function. In practical usage, going out infinitely is impractical. Therefore, Taylor Series are used as approximations, with only a few terms generated.
In the definition above, there is also a term x0. This is the x-coordinate around which the Taylor Series is centered. The x-parameter for the Taylor Series is the point at which to approximate the function. At x = 0, the Taylor Series is also a Maclaurin Series. This is no different than the tangent line approximations of functions. In fact, those tangent line approximations are first degree Taylor Series. Let's compare the tangent line vs. third-degree Taylor Series approximations of the function for T(3): f(x) = 3x4 - 2x2 centered at x = 4.
The first-degree Taylor series can easily be determined using point slope form, which is:
y - f(4) = slope(x-4). To get the slope, simply evaluate f'(4), which is 4(3)(43) - 4(4), or 758. So plugging in the slope, the approximation comes out to: y-736 = 758(x-4). Expanding this sets: y = 758x - 2296.
To get the third degree Taylor series approximation, it is necessary to go out to the second and third derivatives. So:
f"(4) = 36(42) - 4 = 572
f"'(4) = 72(4) = 288
So the third degree Taylor Series approximation is:
T3(x) = 736 + 758(x-4) + 572(x-4)2/2 + 288(x-4)3/6.
Simplified, this becomes:
T3(x) = 736 + 758(x-4) + 286(x-4)2 + 48(x-4)3.
Let's compare both approximations at x = 3 to the original function:
y(4) = -22
T3(3) = 218
f(3) = 225
Obviously the tangent line approximation is nowhere even close to the actual value of f(3), while the third degree Taylor Series is only 7 less than f(3). The reason for the huge difference in approximation is because only a constant slope is taken into account, thus ignoring other changes in the function. The additional terms in the Taylor Series decrease the amplitudes of the curves, causing it to eventually converge on the desired value.
While the primary purpose of Taylor Series is for approximating functions, they also make it easier to deal with function manipulations including vertical and horizontal shifts, differentiation, and integration. For functions that have a significant number of terms, or aren't easily integrable with one of the basic integration techniques, Taylor Series are a good way to approximate the integral.
A good example of this is Integral(sin(x2) dx). While trig substitution or integration by parts are both good ways to tackle this, using a Taylor Series. The general form for a Taylor Series for sin(x) is the summation from i = 0 to infinity of (-1)i * x2i+1 / (2i+1)!. Since we can now treat sin(x) as a polynomial function, the Taylor Series for sin(x^2) comes out to the infinite summation from i = 0 to infinity of (-1)i * x4i+2 / (2i+1)!. That single term is significantly easier to integrate, leaving Integral(a, b, sin(x2) dx) = summation of i = 0 to infinity of (-1)i * x4i+3 /((4i+3) * (2i+1)!). Now simply go out as many terms as desired in the new Taylor Series, and evaluate it from the limits specified in the integral.
Conclusion
Taylor Series are useful in both approximating amd manipulating functions. This makes it easier to apply Calculus to functions that are seemingly difficult to otherwise work with.
Is This A Good Question/Topic? 0
## Replies To: A Calculus Primer Part VIII- Taylor Series
### #2 blackcompe
• D.I.C Lover
Reputation: 1158
• Posts: 2,538
• Joined: 05-May 05
Posted 31 December 2011 - 06:26 AM
Ever wonder how sine, cosine, and tan are computed? Taylor series.
Edit: Whoops, you mentioned that.
This post has been edited by blackcompe: 31 December 2011 - 06:28 AM
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#### How do you find the area of planar shapes?
Area is a measure of the amount of surface a shape covers. It is usually measured in square units. So a rectangular region measuring 4 metres by 3 metres will cover 4 × 3 = 12 square metres, often written, 12 m².
It follows that the area of a rectangular region is given by Area = Base × Height.
It can be shown that the area of a parallelogram, A, is given by A = b × h, where b is the base length and h is the perpendicular height. This can be demonstrated by the following diagram, showing how a perpendicular cut can be made to form a rectanglular region.
Once this result has been establish we can demonstrate the result that the area of a triangle, A, is given by A = ½ ( b × h ), where b is the base length and h is the perpendicular height. By combining two identical triangles it is possible to form a parallelogram.
As the area of two triangles is given by b × h it follows that the area of the triangle is ½ ( b × h).
In a similar way we show that the area of a trapezium, A, is given by A = ½ ( b + t ) × h, where b is the base length, t is the top length and h is the perpendicular height.
The area of two trapeziums are ( b + t ) × h, therefore the area of a trapezium is ½ ( b + t ) × h.
It can be seen that parallelograms and triangles are both special cases of trapezium. In fact a trapezium is sometimes referred to as a truncated triangle.
By using a combination of rectangular and triangular regions and given sufficient information it is possible to find the area of any polygon.
E.g.
The area of the rectangle (square in this case) is 16 square units. The area of the 'extra' triangular regions are 2, 2 and 4½ respectively, hence the area of the triangle is 8½ square units.
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# Department of the Philippines
Region XII
Division of Sultan Kudarat
District of Basak
P. Castro Memorial Elemantary School
Demonstration Lesson Plan
in Mathematics II
I. Objective:
Subtract 2 digit numbers from 2 to 3 digit numbers with regrouping in the tens
place;
II. Subject Matter:
A. Topic: Subtracting Tens and Ones with Regrouping
B. References: Teachers Guide pp. 115-118
L.M pp. 69-71
C. Materials: Flash Cards, dialogue, Sticker games
D. Value: Courtesy
III. Learning Activities:
A. Preparatory Activities:
1. Drill: Basic Subtraction Facts (Flash Cards)
2. Checking of assignment
3. Mental Problem:
a. Adelaida bailed 30 eggs. She sold 20. How many eggs were left?
b. subtract 100 from 430?
4. Motivation
Present a Dialogue
Valuing: What did Jason say to Mrs. Nunez?
What can you say about him?
Is he a courteous boy?
B. Development Activity
1. Presentation:
Present A: 4 5 B. 7 0
-1 7 -5 6
2. Discussion:
Look at the Ones in A. can you subtract 7 from 5? Why not?
5 has to borrow from 4. It can borrow one ten?
-One ten plus 5 is what number?
15
Show where to place one ten. 4 5
-What happens to the 4 tens? -1 7
-Since a ten is taken from 4 tens
3 15
4 tens will become 3 tens 4 15
-now Subtract -1 7
2 8
(Do the same with subtraction sentence B)
C. Group Activity
Form 3 teams with 5 members
Each group have stickers and activity card
Each group has to answer a particular subtraction sentence
Group with more correct answer win.
D. Generalization:
How do we subtract numbers when the ones digit of the subtrahend is
greater than the ones digit of the minuend?
(When we subtract, we sometimes need to take a ten.)
E. Application
Subtract Relay
Pupils answer some exercises on the board by group.
IV. Evaluation:
Find the difference:
1.
8 13
2.
4 10
3.
5 15
4.
8 11
5.
6 15
9 3 5 0 6 5 9 1 37 5
-7 4 -2 7 -2 8 -2 6 - 2 8
V. Assignment:
1. Subtract 15 from 43 =_______
2. Take away 26 from 548 =_______
3. Find the difference
5 10
5 6 0
- 2 9
Prepared by:
Rayan L. Castro BEED
Noted by:
BERIAN C. TRESTAL, PH D.
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### Introduction to Mathematical Induction
Mathematical induction can often be used to prove some property P of all integers greater than or equal to some base case b. The strategy in induction is to prove that P is true for the base case b, and then prove that if P is true for some arbitrary case k (k >= b), then P must also be true for k+1. This is sometimes described as climbing a ladder: proving the base case is like getting onto the ladder, and proving P(k+1) given P(k) is like going from one rung to the next. If you can do both of those things, you can climb the ladder as high as you want!
This kind of argument is supported by the First Principle of Mathematical Induction, which says the following:
```IF
1. P(b) is true for some integer b, and
2. If P(k) is true then P(k+1) is true, for all k >= b
THEN
P(n) true for all n >= b
```
(1) is called the base case, and (2) is called the inductive step. So we need to provide that both a base case and the inductive step are true for our property P to prove that P is true for all integers greater than or equal to the base case.
For example, the property P(n) might be that 1 + 2 + ... + n = n(n+1)/2, and we might want to show that it's true for all n >= 1. We could prove this by induction by proving the two steps above:
1. Base case
Prove that the property is true for the base case b, that is, P(b). For this example the base case is 1, so we simply check P(1),
``` 1 = 1*(1+1)/2
```
which is true.
2. Inductive step
Show that IF the property is true for some arbitrary value k, then it must be true for k+1 as well. Note that what we are proving here is the implication P(k) -> P(k+1)(if P(k) then P(k+1)). Said another way, we need to prove P(k+1) given P(k). We aren't concerned with whether or not P(k) is actually true; we just show that if it is true, then P(k+1) must be true as well. P(k) is called the inductive hypothesis. For this example:
```P(k): 1+2+ ... + k = k(k+1)/2 (inductive hypothesis -- assume true)
P(k+1): 1+2+ ... + k + k+1 = (k+1)(k+2)/2 (need to show this given P(k))
```
How to prove P(k+1) given P(k)? Fortunately, the left side of P(k) is often very similar to the left side of P(k+1); in this case they differ only by k+1. Since we know that P(k) is true, we can add k+1 to both sides of that equation and still have a true statement:
``` 1+2+ ... + k + k+1 = k(k+1)/2 + k+1
```
The left side now looks like the left side of P(k+1). The right side looks different, but maybe they're just in different forms. We can transform the right side as follows:
``` k(k+1)/2 + k+1
= k(k+1)/2 + 2(k+1)/2
= (k(k+1) + 2(k+1))/2
= (k+1)(k+2)/2
```
Thus we have shown P(k+1) given P(k), so the proof by induction is complete.
### Another example
Prove that a fence of this type |--|--|--| with n posts, n>=1, has n-1 sections.
Base case (n=1): A fence with a single post (|) has 0 sections.
Inductive step: If a fence with k posts has k-1 sections, then a fence with k+1 posts has k sections.
Proof: A fence with k+1 posts has a fence with k posts embedded in it.
``` |-|-|-| ... |-|-|
\______________/
k posts
```
By the inductive hypothesis, a fence with k posts has k-1 sections. But a fence with k+1 posts has just one additional section (see diagram), so it has k-1+1 = k sections. This completes the proof by induction.
### Yet another example
Use mathematical induction to show that 2n > n for all n>=1.
1. Base case (n=1):
21 > 1, so the base case holds.
2. Inductive step: If 2k > k then 2k+1 > k+1.
Starting with the inductive hypothesis (the k case), we notice that the left side looks like the left side of the k+1 case but is smaller by a factor of 2. So multiply both sides of the inductive hypothesis by 2, which maintains the inequality:
``` 2k > k
=> 2k * 2 > k * 2
=> 2k+1 > k * 2
```
But we're trying to show that 2k+1 > k + 1, and we can't turn k*2 into k+1. However, note that for all k>1, k*2 is bigger than k+1, so anything that is bigger than k*2 must be bigger than k+1 as well. We have already shown that 2k+1 > k * 2, so we can conclude that 2k * 2 > k + 1.
### The Second Principle of Mathematical Induction
There is another form of mathematical induction, called the Second Principle of Mathematical Induction, that is often useful. It says the following:
```IF
1. P(b) true (for some integer b)
2. P(r), b<=r<=k, true -> P(k+1) (k >= b)
THEN
P(n) is true for all n >= b
```
The Second Principle is different in the second rule, which assumes not only that P is true for k, but that P is true for all values from b to k. This sounds like a stronger condition, but it turns out that the First and Second principles are actually equivalent. Depending on the problem, it just may be more convenient to use one form over another.
### More Fences
Use the Second Principle of Mathematical Induction to prove that a fence with n posts has n-1 sections. (This is a problem that is easily proved using either the first or second principle.)
Base case: A fence with a single post (|) has 0 sections.
Inductive step: If a fence with r posts, 1<=r<=k, has r-1 sections, then a fence with k+1 posts has k sections.
Proof: A fence with k+1 posts can be divided into two parts, each of which has at least one post (since k>=1). If we say that the left part has x posts, then the right part must have the rest of the posts, that is, (k+1)-x posts.
``` |-|...|-|...|-| (k+1) posts total
\______/ \_____/
x posts (k+1)-x posts
```
Since each part has at least one and not more than k posts, the inductive hypothesis applies to each part. So by the inductive hypothesis, the left part has x-1 sections and the right part has k+1-x-1 sections. The whole fence, which has k+1 posts, has the sections in the left plus the sections in the right plus the section that joins them together:
``` x-1 + k+1-x-1 + 1 = k sections.
| | | |
left right join total
```
This completes the proof.
### Another example
Prove that any amount of postage greater than or equal to 8 cents can be built using only 3-cent and 5-cent stamps.
Base cases (n=8, n=9, n=10): 8=3+5; 9=3+3+3; 10=5+5.
Inductive step (second principle): If any amount of postage r, 8<=r<=k, can be built using only 3-cent and 5-cent stamps, then postage k+1 can be built using only 3-cent and 5-cent stamps.
Proof: Cases k=8, k=9, and k=10 were proved in the base case, so here we can assume that k+1 >= 11. By the inductive hypothesis, we know that the postage for k-2 can be formed using 3-cent and 5-cent stamps since (k+1)>=11 => (k-2)>=8. Now take the postage for k-2 and add a 3-cent stamp; this gives the postage for k+1, completing the proof.
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# 14.15 Applied Statistics
Most communicators need to master nothing more complicated than a few principles of applied statistics in order to overcome their innumeracy.
Applied statistics are not to be feared. They are based on basic, simple, fourth-grade arithmetic – addition, subtraction, multiplication and division. Mathematicians will tell you that algebra and calculus provide the support for the assumptions and techniques of statistics, but applied statistics do not require computation beyond basic arithmetic.
It is important for communicators to brush up on their basic mathematical terminology in order to be able to interpret the data they find in many reports, studies, documents, and other types of information resources. The following are a few commonly confused terms:
percent a standard way of expressing a fraction, where the denominator (the bottom number) equals 100; so 1/4 equals 25 percent (4 X 25 = 100), 1/3 equals 33 percent (3 X 33 = 99 rounded to 100), 1/5 equals 20 percent (5 X 20 = 100), and so forth percent change a way to express a relationship between an “old” number and a “change” or a “new” number and a “change;” for example, last year’s budget is \$500,000 and the current budget is \$600,000. The “change” is \$100,000. To express this as a percent change from last year to this, you have to divide the “change” by the “old” number, or \$100,000 / \$500,000 = 0.2 = 20 percent. The new budget is therefore 20 percent higher than the previous year’s budget. To figure what percent of spending in the current budget is “new” you have to divide the “change” by the higher “new” figure, or \$100,000 / \$600,000 = 0.1666… = 16.7 percent percentage point a way to compare two numbers that are already expressed as percents; for example, the March unemployment figure is 5 percent, the April unemployment figure is 6 percent, and the change is 1 percentage point. The percent change is 1 (the change) divided by 5 (the old number) or 20 percent mean the arithmetic average of a set of values median the middle value in a group, where values have been ranked from top to bottom
Let’s look at the important differences between a mean and a median. The mean is the arithmetic average of a set of values. To compute a mean, you simply add up the values and divide by the number of values.
For example, let’s say that among 100 workers at a company, 95 make \$30,000 and 5 make \$300,000 a year. The mean salary is therefore (95 times 30,000) plus (5 times 300,000) divided by 100 OR 4,350,000 divided by 100 OR \$43,500. This simple arithmetic average provides information about the average salary for workers at the company. But does it accurately describe salaries at this company?
If you answered no or not always, then you understand how important it is to evaluate how a figure is arrived at. A better way to portray salaries for this company is to use the median figure rather than the mean.
The median is the middle value in a group, where values have been ranked from top to bottom. A median figure is more likely to smooth out very wide differences between the highest and lowest values in the group.
For example, the middle value in our group of 100 salaries in our imaginary company ranked from top to bottom is the average of the salary of the 50th and 51st highest paid people. In this case, the median is therefore \$30,000. The median figure clearly tells you more accurately about the salaries for the majority of people in the company.
In most cases, it is necessary to understand how both figures were arrived at and what some of the reasons might be for significant differences between the mean and the median.
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### Building Tetrahedra
Can you make a tetrahedron whose faces all have the same perimeter?
### Ladder and Cube
A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?
### Bendy Quad
Four rods are hinged at their ends to form a convex quadrilateral. Investigate the different shapes that the quadrilateral can take. Be patient this problem may be slow to load.
# Fortunate Inflation
##### Age 14 to 16 ShortChallenge Level
Answer: £13
Working backwards
\begin{align} & \text {old price}\times\tfrac{104}{100}=\text{multiple of 100}\\ \therefore & \text{old price}\times\tfrac{26}{25} = \text{multiple of 100}\\ \therefore & \tfrac{\text{old price}}{25}\times2\times13=\text{multiple of 100}\\ \therefore &\tfrac{\text{old price}}{25}=\text{multiple of 50} \end{align}
So the smallest possible $\text{old price}$ is $50\times 25$
Which gives a new price of $50\times 26=1300$ ie. $£13$
Inflating prices by 4% and trying to 'hit' whole numbers of pounds
For $n$ to be 1, the original price would need to be about 96p.
4% of 96p is 0.04$\times$96p = 3.84p, so 4% inflation from 96p is 96p + 3.84p = 99.84p
4% of 97p is 0.04$\times$97p = 3.88p, so 4% inflation from 97p is 97p + 3.88p = 100.88p
Those are either side of £1, so $n$ cannot be 1.
For $n$ to be 2, the original price would need to be about £1.92.
4% of £1.92 is 0.04$\times$ £1.92 = £0.0768, so 4% inflation from £1.92 is £1.92 + £0.0768 = £1.9968
4% of £1.93 is 0.04$\times$ £1.93 = £0.0772, so 4% inflation from £1.93 is £1.93 + £0.0772 = £2.0072
Those are either side of £2, so $n$ cannot be 2.
Continuing in this way is slow, but eventually we get that $n$ = 13.
Considering 4% of whole numbers of pennies
The price increase is 4% of the original price. If the new price is a whole number of pounds, then it is also a whole number of pennies. So since the old price was also a whole number of pennies, the increase must be a whole number of pennies too.
This means that 4% of the old price is a whole number of pennies.
4% is equivalent to $\frac1{25}$, so $\frac1{25}$ of the old price is a whole number of pennies, so the old price must be a multiple of 25p.
This means that the old price must be in the form £X.00, £X.25, £X.50 or £X.75.
For $n$ to be as small as possible, the increase should also be as small as possible, so first we should try an increase of 25p, from £($n-1$).75 to £$n$.
That means that 4%, or $\frac{1}{25}$, of the original amount should be 25p, so the original amount was 25p$\times$25 = £6.25. However £6.25 + 25p = £6.50, which is not a whole number of pounds.
Next try an increase of 50p, from £($n-1$).50 to £$n$.
That means that 4%, or $\frac{1}{25}$, of the original amount should be 50p, so the original amount was 50p$\times$25 = £12.50. £12.50 + 50p = £13.00. So $n$ = 13.
Working backwards from whole numbers of pounds
Let the old price be £$a$, where £$a$ is an exact number of pence (so it has at most 2 decimal places). Then if the new price is £$n$, $n$ is a 4% increase of $a$.
This means that $a+\frac{4}{100}a=n\Rightarrow\frac{100}{100}a+\frac{4}{100}a=n\Rightarrow \frac{104}{100}a=n$.
So $104a=100n\Rightarrow a=\frac{100}{104}n=0.9\dot{6}1538\dot{4}n$, and $a$ should have at most 2 d.p.
$0.9\dot{6}1538\dot{4}\times2=1.\dot92307\dot6$, which has more than 2 d.p., so $n$ cannot be $2$.
$0.9\dot{6}1538\dot{4}\times3=2.8\dot84615\dot3$, which has more than 2 d.p., so $n$ cannot be $3$.
$0.9\dot{6}1538\dot{4}\times4=3.\dot84615\dot3$, which has more than 2 d.p., so $n$ cannot be $4$.
Continuing in this way until we get a possible value for $a$, $n=13$ ( since $0.9\dot{6}1538\dot{4}\times4=12.5$ exactly).
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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1.2 Arithmetic review: the least common multiple
Page 1 / 1
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. This chapter contains many examples of arithmetic techniques that are used directly or indirectly in algebra. Since the chapter is intended as a review, the problem-solving techniques are presented without being developed. Therefore, no work space is provided, nor does the chapter contain all of the pedagogical features of the text. As a review, this chapter can be assigned at the discretion of the instructor and can also be a valuable reference tool for the student.
Overview
• Multiples
• Common Multiples
• The Least Common Multiple (LCM)
• Finding The Least Common Multiple
Multiples
When a whole number is multiplied by other whole numbers, with the exception of Multiples zero, the resulting products are called multiples of the given whole number.
Multiples of 2 Multiples of 3 Multiples of 8 Multiples of 10 $2·1=2$ $3·1=3$ $8·1=8$ $10·1=10$ $2·2=4$ $3·2=6$ $8·2=16$ $10·2=20$ $2·3=6$ $3·3=9$ $8·3=24$ $10·3=30$ $2·4=8$ $3·4=12$ $8·4=32$ $10·4=40$ $2·5=10$ $3·5=15$ $8·5=40$ $10·5=50$ … … … …
Common multiples
There will be times when we are given two or more whole numbers and we will need to know if there are any multiples that are common to each of them. If there are, we will need to know what they are. For example, some of the multiples that are common to 2 and 3 are 6, 12, and 18.
Sample set a
We can visualize common multiples using the number line.
Notice that the common multiples can be divided by both whole numbers.
The least common multiple (lcm)
Notice that in our number line visualization of common multiples (above) the first common multiple is also the smallest, or least common multiple, abbreviated by LCM.
Least common multiple
The least common multiple, LCM, of two or more whole numbers is the smallest whole number that each of the given numbers will divide into without a remainder.
Finding the lcm
To find the LCM of two or more numbers,
1. Write the prime factorization of each number, using exponents on repeated factors.
2. Write each base that appears in each of the prime factorizations.
3. To each base, attach the largest exponent that appears on it in the prime factorizations.
4. The LCM is the product of the numbers found in step 3.
Sample set b
Find the LCM of the following number.
9 and 12
1. $\begin{array}{lllllll}9\hfill & =\hfill & 3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\hfill & =\hfill & {3}^{2}\hfill & \hfill & \hfill \\ 12\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}6\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\hfill & =\hfill & {2}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\hfill \end{array}$
2. The bases that appear in the prime factorizations are 2 and 3.
3. The largest exponents appearing on 2 and 3 in the prime factorizations are, respectively, 2 and 2 (or ${2}^{2}$ from 12, and ${3}^{2}$ from 9).
4. The LCM is the product of these numbers.
$\text{LCM\hspace{0.17em}}={2}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{3}^{2}=4\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9=36$
Thus, 36 is the smallest number that both 9 and 12 divide into without remainders.
90 and 630
1. $\begin{array}{lllllllll}90\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}45\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}15\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{3}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\hfill \\ 630\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}315\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}105\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}35\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7\hfill \\ \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{3}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7\hfill \end{array}$
2. The bases that appear in the prime factorizations are 2, 3, 5, and 7.
3. The largest exponents that appear on 2, 3, 5, and 7 are, respectively, 1, 2, 1, and 1.
$\begin{array}{ll}{2}^{1}\hfill & \text{from\hspace{0.17em}either\hspace{0.17em}9}0\text{\hspace{0.17em}or\hspace{0.17em}63}0\hfill \\ {3}^{2}\hfill & \text{from\hspace{0.17em}either\hspace{0.17em}9}0\text{\hspace{0.17em}or\hspace{0.17em}63}0\hfill \\ {5}^{1}\hfill & \text{from\hspace{0.17em}either\hspace{0.17em}9}0\text{\hspace{0.17em}or\hspace{0.17em}63}0\hfill \\ {7}^{1}\hfill & \text{from\hspace{0.17em}63}0\hfill \end{array}$
4. The LCM is the product of these numbers.
$\text{LCM\hspace{0.17em}}=2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{3}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7=2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7=630$
Thus, 630 is the smallest number that both 90 and 630 divide into with no remainders.
33, 110, and 484
1. $\begin{array}{lllll}33\hfill & =\hfill & 3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}11\hfill & \hfill & \hfill \\ 110\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}55\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}11\hfill \\ 484\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}242\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}121=2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}11\text{\hspace{0.17em}}·\text{\hspace{0.17em}}11={2}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{11}^{2}\hfill \end{array}$
2. The bases that appear in the prime factorizations are 2, 3, 5, and 11.
3. The largest exponents that appear on 2, 3, 5, and 11 are, respectively, 2, 1, 1, and 2.
$\begin{array}{ll}{2}^{2}\hfill & \text{from\hspace{0.17em}}484\hfill \\ {3}^{1}\hfill & \text{from\hspace{0.17em}}33\hfill \\ {5}^{1}\hfill & \text{from\hspace{0.17em}}110\hfill \\ {11}^{2}\hfill & \text{from\hspace{0.17em}}484\hfill \end{array}$
4. The LCM is the product of these numbers.
$\begin{array}{lll}\text{LCM}\hfill & =\hfill & {2}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{11}^{2}\hfill \\ \hfill & =\hfill & 4\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}121\hfill \\ \hfill & =\hfill & 7260\hfill \end{array}$
Thus, 7260 is the smallest number that 33, 110, and 484 divide into without remainders.
Exercises
For the following problems, find the least common multiple of given numbers.
8, 12
${2}^{3}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3$
8, 10
6, 12
${2}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3$
9, 18
5, 6
$2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5$
7, 9
28, 36
${2}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{3}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7$
24, 36
28, 42
${2}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7$
20, 24
25, 30
$2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{5}^{2}$
24, 54
16, 24
${2}^{4}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3$
36, 48
15, 21
$3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7$
7, 11, 33
8, 10, 15
${2}^{3}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5$
4, 5, 21
45, 63, 98
$2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{3}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{7}^{2}$
15, 25, 40
12, 16, 20
${2}^{4}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5$
12, 16, 24
12, 16, 24, 36
${2}^{4}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{3}^{2}$
6, 9, 12, 18
8, 14, 28, 32
${2}^{5}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7$
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
Got questions? Join the online conversation and get instant answers!
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## Iterated Polynomials
2017 April 9
Find the formula for the nth iterate of the function $$f(x)=2x^2-1$$ for $$|x| \le 1$$.
Find the formula for the nth iterate of a quadratic with a fixed point at its minimum.
It seems that some polynomial functions, such as $$f(x)=x^2+1$$, simply can’t have an iteration formula in the closed form. However, some do because they can be put in the form $$f(x)=(g\circ h\circ g^{-1})(x)$$. For example, any polynomial function of the form
$f(x)=\frac{1}{c}(cx+d)^p-\frac{d}{c}$
can be iterated, because it takes the aforementioned form with
$g(x)=\frac{1}{c}x-\frac{d}{c}$
$h(x)=x^p$
meaning that
$f^n(x)=(g\circ h^n\circ g^{-1})(x)$
$f^n(x)=\frac{1}{c}(cx+d)^{p^n}-\frac{d}{c}$
Let’s apply this to quadratics. This formula tells us that any quadratic with its vertex on the line $$y=x$$ can be iterated by this formula, because a quadratic with its vertex on $$y=x$$ takes the form
$f(x)=a(x-h)^2+h$
or
$f(x)=\frac{1}{a}(ax-ah)^2+h$
Which we can see takes the form $$f(x)=(g\circ h\circ g^{-1})(x)$$ where $$g(x)=\frac{1}{a}x+h$$ and $$h(x)=x^2$$, so
$f^n(x)=\frac{1}{a}(ax-ah)^{2^n}+h$
That’s about it for polynomials, aside from a few quadratics that can be iterated somewhat using trigonometric functions. For example, the quadratic
$f(x)=2x^2-1$
can be split up into
$f(x)=x^2-(\sqrt{1-x^2})^2$
$f(x)=\cos^2(\cos^{-1}x)-\sin^2(\cos^{-1}x)$
$f(x)=\cos(2\cos^{-1}x)$
which, once again, takes our special form and can be iterated as
$f^n(x)=\cos(2^n\cos^{-1}x)$
However, quadratics iterated this way can only be iterated to a certain extent because of the limited domain of the inverse trigonometric functions. Let’s try an example anyways. What about $$f^{10}(0.5)$$? Using our formula, this will be
$f^{10}(x)=\cos(2^10\cos^{-1}0.5)$
$f^{10}(x)\approx \cos(1072.330)$
$f^{10}(x)=-0.5$
Well, that was a bad example to try, because $$f(0.5)=f(-0.5)=-0.5$$, so we could have found that without our formula. How about $$f^{10}(0.1)$$?
$f^{10}(0.1)=\cos(2^10\cos^{-1}0.1)$
$f^{10}(x)\approx \cos(1505.923)$
$f^{10}(x)=-0.452$
I think I’m noticing a pattern here. When I iterate a quadratic whose vertex is on $$y=x$$, the graphs of the iterates exhibit predictable behavior. Here are the graphs of $$y=\frac{1}{2}(x-2)^2+2$$ and its first few iterates:
They all seem to predictably flatten out at the bottom and then become steeper more and more quickly. Now here’s the function $$f(x)=2x^2-1$$ and its first couple iterates:
These display neat sinusoidal behavior and then head straight up. In fact, the crests of each of the waves are at a height of $$1$$, and the points $$(1,1)$$ and $$(-1,1)$$ lie on the curve directly before and after it begins to grow out of control, which is perhaps the reason why we cannot iterate it outside of that interval. Now observe these pictures of a quadratic of neither form:
This quadratic’s iterates display erratic behavior that grows insanely large before coming down again. Here’s one more interesting case - a parabola tangent to the line $$y=x$$:
Perhaps there is some formula for quadratics of that form, since they seem to behave predictably… but that’s for another post.
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# NCERT Solutions for Class 8 Maths, Chapter 6 - Squares and Square Roots
In this chapter, you’ll learn about Squares and Square Roots. What is the ‘square’ of a number? If a natural number ‘m’ can be expressed as ‘n2’, where ‘n’ is also a natural number, then m is a square number. A square root is just the inverse of a square. It is a number that produces another number when multiplied by itself. It is important for you to get a clear understanding of squares and square roots because they have real-life applications in terms of computing the area of a plot, in statistics, engineering, physics and more. In this chapter, you’ll study how to find squares and square roots of a number using various methods such as factorisation method and division method, and learn the process of getting closer to the required number.
## Let’s take a look at topics and sub-topics of Class 8 Maths Chapter 6, Squares and Square Roots
6.1 Introduction
6.2 Properties of Square Numbers
6.3 Some More Interesting Patterns
6.4 Finding the Square of a Number
6.5 Square Roots
6.6 Square Roots of Decimals
6.7 Estimating Square Root
## In Class 8 Maths Chapter 6, you will learn -
Section 6.2 - In this section, you will learn some interesting properties of square numbers. Once you understand these properties, you’ll be able to find the square of a number at a faster speed.
Section 6.3 - In this section, you’ll study some more interesting patterns when we write down the squares of various numbers in a tabular form. You’ll learn what happens when you add triangular numbers or numbers between square numbers. You’ll also study what a perfect square is; squares of consecutive natural numbers and product of two consecutive even or odd natural numbers and more.
Section 6.4 - It is easy to find the squares of a smaller number. However, as the number of digits increase, it gets relatively difficult to find the squares without long computations. In this section, you will get an idea of how to find squares of bigger numbers easily. You’ll also study about Pythagorean triplets and other patterns in squares.
Section 6.5 - Now it’s time to learn how to calculate the square root of a given number. The square root is just the inverse process of squaring. In this section, you’ll study the positive square root of a natural number. You’ll also learn how to find square root through repeated subtraction, prime factorisation and division method.
Section 6.6 - In this section, you’ll go a step further and learn the techniques of finding the square root decimal numbers, which is quite similar to the process of finding the square root integers.
Section 6.7 - In this section, you will get an idea of finding square roots in real-life scenarios.
In this chapter, you are provided with several examples along with their solutions for a clear understanding of Squares and Square Roots. To know more about Class 8 Maths Chapter 6 Squares and Square Roots, you should explore the exercises below. You can also download the Squares and Square Roots Class 8 NCERT Solutions PDF, solved by expert Maths trainers.
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# 3.5 Multiplication and division of signed numbers
Page 1 / 1
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The basic operations with real numbers are presented in this chapter. The concept of absolute value is discussed both geometrically and symbolically. The geometric presentation offers a visual understanding of the meaning of |x|. The symbolic presentation includes a literal explanation of how to use the definition. Negative exponents are developed, using reciprocals and the rules of exponents the student has already learned. Scientific notation is also included, using unique and real-life examples.Objectives of this module: be able to multiply and divide signed numbers.
## Overview
• Multiplication of Signed Numbers
• Division of Signed Numbers
## Multiplication of signed numbers
Let us consider first the product of two positive numbers.
Multiply: $3\cdot 5$ .
$3\cdot 5$ means $5+5+5=15$ .
This suggests that
$\left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)\cdot \left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)=\text{positive}\text{\hspace{0.17em}}\text{number}$ .
More briefly, $\left(+\right)\left(+\right)=+$ .
Now consider the product of a positive number and a negative number.
Multiply: $\left(3\right)\left(-5\right)$ .
$\left(3\right)\left(-5\right)$ means $\left(-5\right)+\left(-5\right)+\left(-5\right)=-15$ .
This suggests that
$\left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)\cdot \left(\text{negative}\text{\hspace{0.17em}}\text{number}\right)=\text{negative}\text{\hspace{0.17em}}\text{number}$
More briefly, $\left(+\right)\left(-\right)=-$ .
By the commutative property of multiplication, we get
$\left(\text{negative}\text{\hspace{0.17em}}\text{number}\right)\cdot \left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)=\text{negative}\text{\hspace{0.17em}}\text{number}$
More briefly, $\left(-\right)\left(+\right)=-$ .
The sign of the product of two negative numbers can be determined using the following illustration: Multiply $-2$ by, respectively, $4,\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}0,\text{\hspace{0.17em}}-1,\text{\hspace{0.17em}}-2,\text{\hspace{0.17em}}-3,\text{\hspace{0.17em}}-4$ . Notice that when the multiplier decreases by 1, the product increases by 2.
$\begin{array}{ll}\begin{array}{l}4\left(-2\right)=-8\\ 3\left(-2\right)=-6\\ 2\left(-2\right)=-4\\ 1\left(-2\right)=-2\end{array}\right\}\to \hfill & \text{As}\text{\hspace{0.17em}}\text{we}\text{\hspace{0.17em}}\text{know},\text{\hspace{0.17em}}\left(+\right)\left(-\right)=-.\hfill \\ 0\left(-2\right)=0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\to \hfill & \text{As}\text{\hspace{0.17em}}\text{we}\text{\hspace{0.17em}}\text{know},0\cdot \left(\text{any}\text{\hspace{0.17em}}\text{number}\right)=0.\hfill \end{array}$
$\begin{array}{ll}\begin{array}{l}-1\left(-2\right)=2\\ -2\left(-2\right)=4\\ -3\left(-2\right)=6\\ -4\left(-2\right)=8\end{array}\right\}\to \hfill & \text{This}\text{\hspace{0.17em}}\text{pattern}\text{\hspace{0.17em}}\text{suggests}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-\right)\left(-\right)=+.\hfill \end{array}$
We have the following rules for multiplying signed numbers.
## Rules for multiplying signed numbers
To multiply two real numbers that have
1. the same sign , multiply their absolute values. The product is positive.
$\begin{array}{c}\left(+\right)\left(+\right)=+\\ \left(-\right)\left(-\right)=+\end{array}$
2. opposite signs , multiply their absolute values. The product is negative.
$\begin{array}{c}\left(+\right)\left(-\right)=-\\ \left(-\right)\left(+\right)=-\end{array}$
## Sample set a
Find the following products.
$8\cdot 6$
$\begin{array}{ll}\hfill & \text{Multiply}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}\text{.}\hfill \\ \begin{array}{l}|8|=8\\ |6|=6\end{array}\right\}\hfill & 8\cdot 6=48\hfill \\ \hfill & \text{Since}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numbers}\text{\hspace{0.17em}}\text{have}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{same}\text{\hspace{0.17em}}\text{sign,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{product}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{positive}\text{.}\hfill \\ 8\cdot 6=+48\hfill & \text{}\text{}\text{or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8\cdot 6=48\hfill \end{array}$
$\left(-8\right)\left(-6\right)$
$\begin{array}{ll}\hfill & \text{Multiply}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}\text{.}\hfill \\ \begin{array}{l}|-8|=8\\ |-6|=6\end{array}\right\}\hfill & 8\cdot 6=48\hfill \\ \hfill & \text{Since}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numbers}\text{\hspace{0.17em}}\text{have}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{same}\text{\hspace{0.17em}}\text{sign,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{product}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{positive}\text{.}\hfill \\ \left(-8\right)\left(-6\right)=+48\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-8\right)\left(-6\right)=48\hfill \end{array}$
$\left(-4\right)\left(7\right)$
$\begin{array}{ll}\hfill & \text{Multiply}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}\text{.}\hfill \\ \begin{array}{cc}|-4|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}& =4\\ |7|& =7\end{array}\right\}\hfill & 4\cdot 7=28\hfill \\ \hfill & \text{Since}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numbers}\text{\hspace{0.17em}}\text{have}\text{\hspace{0.17em}}\text{opposite}\text{\hspace{0.17em}}\text{signs,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{product}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{negative}\text{.}\hfill \\ \left(-4\right)\left(7\right)=-28\hfill & \hfill \end{array}$
$6\left(-3\right)$
$\begin{array}{ll}\hfill & \text{Multiply}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}\text{.}\hfill \\ \begin{array}{rr}\hfill |6|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}& \hfill =6\\ \hfill |-3|& \hfill =3\end{array}\right\}\hfill & 6\cdot 3=18\hfill \\ \hfill & \text{Since}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numbers}\text{\hspace{0.17em}}\text{have}\text{\hspace{0.17em}}\text{opposite}\text{\hspace{0.17em}}\text{signs,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{product}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{negative}\text{.}\hfill \\ 6\left(-3\right)=-18\hfill & \hfill \end{array}$
## Practice set a
Find the following products.
$3\left(-8\right)$
$-24$
$4\left(16\right)$
64
$\left(-6\right)\left(-5\right)$
30
$\left(-7\right)\left(-2\right)$
14
$\left(-1\right)\left(4\right)$
$-4$
$\left(-7\right)7$
$-49$
## Division of signed numbers
We can determine the sign pattern for division by relating division to multiplication. Division is defined in terms of multiplication in the following way.
If $b\cdot c=a$ , then $\frac{a}{b}=c,\text{\hspace{0.17em}}b\ne 0$ .
For example, since $3\cdot 4=12$ , it follows that $\frac{12}{3}=4$ .
Notice the pattern:
Since $\underset{b\cdot c=a}{\underbrace{3\cdot 4}}=12$ , it follows that $\underset{\frac{a}{b}=c}{\underbrace{\frac{12}{3}}}=4$
The sign pattern for division follows from the sign pattern for multiplication.
1. Since $\underset{b\cdot c=a}{\underbrace{\left(+\right)\left(+\right)}}=+$ , it follows that $\underset{\frac{a}{b}=c}{\underbrace{\frac{\left(+\right)}{\left(+\right)}}}=+$ , that is,
$\frac{\left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)}{\left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)}=\text{positive}\text{\hspace{0.17em}}\text{number}$
2. Since $\underset{b\cdot c=a}{\underbrace{\left(-\right)\left(-\right)}}=+$ , it follows that $\underset{\frac{a}{b}=c}{\underbrace{\frac{\left(+\right)}{\left(-\right)}}}=-$ , that is,
$\frac{\left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)}{\left(\text{negative}\text{\hspace{0.17em}}\text{number}\right)}=\text{negative}\text{\hspace{0.17em}}\text{number}$
3. Since $\underset{b\cdot c=a}{\underbrace{\left(+\right)\left(-\right)}}=-$ , it follows that $\underset{\frac{a}{b}=c}{\underbrace{\frac{\left(-\right)}{\left(+\right)}}}=-$ , that is,
$\frac{\left(\text{negative}\text{}\text{\hspace{0.17em}}\text{number}\right)}{\left(\text{positive}\text{\hspace{0.17em}}\text{number}\right)}=\text{negative}\text{\hspace{0.17em}}\text{number}$
4. Since $\underset{b\cdot c=a}{\underbrace{\left(-\right)\left(+\right)}}=-$ , it follows that $\underset{\frac{a}{b}=c}{\underbrace{\frac{\left(-\right)}{\left(-\right)}}}=+$ , that is
$\frac{\left(\text{negative}\text{\hspace{0.17em}}\text{number}\right)}{\left(\text{negative}\text{\hspace{0.17em}}\text{number}\right)}=\text{positive}\text{\hspace{0.17em}}\text{number}$
We have the following rules for dividing signed numbers.
## Rules for dividing signed numbers
To divide two real numbers that have
1. the same sign , divide their absolute values. The quotient is positive.
$\begin{array}{cc}\frac{\left(+\right)}{\left(+\right)}=+& \frac{\left(-\right)}{\left(-\right)}=+\end{array}$
2. opposite signs , divide their absolute values. The quotient is negative.
$\begin{array}{cc}\frac{\left(-\right)}{\left(+\right)}=-& \frac{\left(+\right)}{\left(-\right)}=-\end{array}$
## Sample set b
Find the following quotients.
$\frac{-10}{2}$
$\begin{array}{ll}\begin{array}{ll}|-10|=\hfill & 10\hfill \\ \hfill |2|=& \hfill 2\end{array}\right\}\hfill & \text{Divide}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}.\hfill \\ \hfill & \frac{10}{2}=5\hfill \\ \frac{-10}{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-5\hfill & \text{Since}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numbers}\text{\hspace{0.17em}}\text{have}\text{\hspace{0.17em}}\text{opposite}\text{\hspace{0.17em}}\text{signs,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{quotient}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{negative}\text{.}\hfill \end{array}$
$\frac{-35}{-7}$
$\begin{array}{ll}\begin{array}{ll}|-35|=\hfill & 35\hfill \\ \hfill |-7|=& \hfill 7\end{array}\right\}\hfill & \text{Divide}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}.\hfill \\ \hfill & \frac{35}{7}=5\hfill \\ \frac{-35}{-7}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=5\hfill & \text{Since}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numbers}\text{\hspace{0.17em}}\text{have}\text{\hspace{0.17em}}\text{same}\text{\hspace{0.17em}}\text{signs,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{quotient}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{positive}\text{.}\hfill \end{array}$
$\frac{18}{-9}$
$\begin{array}{ll}\begin{array}{ll}|18|=\hfill & 18\hfill \\ |-9|=\hfill & 9\hfill \end{array}\right\}\hfill & \text{Divide}\text{\hspace{0.17em}}\text{these}\text{\hspace{0.17em}}\text{absolute}\text{\hspace{0.17em}}\text{values}.\hfill \\ \hfill & \frac{18}{9}=2\hfill \\ \frac{18}{-9}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-2\hfill & \text{Since}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{numbers}\text{\hspace{0.17em}}\text{have}\text{\hspace{0.17em}}\text{opposite}\text{\hspace{0.17em}}\text{signs,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{quotient}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{negative}\text{.}\hfill \end{array}$
## Practice set b
Find the following quotients.
$\frac{-24}{-6}$
4
$\frac{30}{-5}$
$-6$
$\frac{-54}{27}$
$-2$
$\frac{51}{17}$
3
## Sample set c
Find the value of $\frac{-6\left(4-7\right)-2\left(8-9\right)}{-\left(4+1\right)+1}$ .
Using the order of operations and what we know about signed numbers, we get
$\begin{array}{lll}\frac{-6\left(4-7\right)-2\left(8-9\right)}{-\left(4+1\right)+1}\hfill & =\hfill & \frac{-6\left(-3\right)-2\left(-1\right)}{-\left(5\right)+1}\hfill \\ \hfill & =\hfill & \frac{18+2}{-5+1}\hfill \\ \hfill & =\hfill & \frac{20}{-4}\hfill \\ \hfill & =\hfill & -5\hfill \end{array}$
Find the value of $z=\frac{x-u}{s}$ if $x=57,\text{\hspace{0.17em}}u=51,\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}s=2$ .
Substituting these values we get
$z=\frac{57-51}{2}=\frac{6}{2}=3$
## Practice set c
Find the value of $\frac{-7\left(4-8\right)+2\left(1-11\right)}{-5\left(1-6\right)-17}$ .
1
Find the value of $P=\frac{n\left(n-3\right)}{2n}$ , if $n=5$ .
1
## Exercises
Find the value of each of the following expressions.
$\left(-2\right)\left(-8\right)$
16
$\left(-3\right)\left(-9\right)$
$\left(-4\right)\left(-8\right)$
32
$\left(-5\right)\left(-2\right)$
$\left(-6\right)\left(-9\right)$
54
$\left(-3\right)\left(-11\right)$
$\left(-8\right)\left(-4\right)$
32
$\left(-1\right)\left(-6\right)$
$\left(3\right)\left(-12\right)$
$-36$
$\left(4\right)\left(-18\right)$
$8\left(-4\right)$
$-32$
$5\left(-6\right)$
$9\left(-2\right)$
$-18$
$7\left(-8\right)$
$\left(-6\right)4$
$-24$
$\left(-7\right)6$
$\left(-10\right)9$
$-90$
$\left(-4\right)12$
$\left(10\right)\left(-6\right)$
$-60$
$\left(-6\right)\left(4\right)$
$\left(-2\right)\left(6\right)$
$-12$
$\left(-8\right)\left(7\right)$
$\frac{21}{7}$
3
$\frac{42}{6}$
$\frac{-39}{3}$
$-13$
$\frac{-20}{10}$
$\frac{-45}{-5}$
9
$\frac{-16}{-8}$
$\frac{25}{-5}$
$-5$
$\frac{36}{-4}$
$8-\left(-3\right)$
11
$14-\left(-20\right)$
$20-\left(-8\right)$
28
$-4-\left(-1\right)$
$0-4$
$-4$
$0-\left(-1\right)$
$-6+1-7$
$-12$
$15-12-20$
$1-6-7+8$
$-4$
$2+7-10+2$
$3\left(4-6\right)$
$-6$
$8\left(5-12\right)$
$-3\left(1-6\right)$
15
$-8\left(4-12\right)+2$
$-4\left(1-8\right)+3\left(10-3\right)$
49
$-9\left(0-2\right)+4\left(8-9\right)+0\left(-3\right)$
$6\left(-2-9\right)-6\left(2+9\right)+4\left(-1-1\right)$
$-140$
$\frac{3\left(4+1\right)-2\left(5\right)}{-2}$
$\frac{4\left(8+1\right)-3\left(-2\right)}{-4-2}$
$-7$
$\frac{-1\left(3+2\right)+5}{-1}$
$\frac{-3\left(4-2\right)+\left(-3\right)\left(-6\right)}{-4}$
$-3$
$-1\left(4+2\right)$
$-1\left(6-1\right)$
$-5$
$-\left(8+21\right)$
$-\left(8-21\right)$
13
$-\left(10-6\right)$
$-\left(5-2\right)$
$-3$
$-\left(7-11\right)$
$-\left(8-12\right)$
4
$-3\left[\left(-1+6\right)-\left(2-7\right)\right]$
$-2\left[\left(4-8\right)-\left(5-11\right)\right]$
$-4$
$-5\left[\left(-1+5\right)+\left(6-8\right)\right]$
$-\left[\left(4-9\right)+\left(-2-8\right)\right]$
15
$-3\left[-2\left(1-5\right)-3\left(-2+6\right)\right]$
$-2\left[-5\left(-10+11\right)-2\left(5-7\right)\right]$
2
$\begin{array}{cc}P=R-C.& \text{Find}\text{\hspace{0.17em}}P\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}R=2000\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}C=2500.\end{array}$
$\begin{array}{cc}z=\frac{x-u}{s}.& \text{Find}\text{\hspace{0.17em}}z\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x=23,\text{\hspace{0.17em}}u=25,\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}s=1.\end{array}$
$-2$
$\begin{array}{cc}z=\frac{x-u}{s}.& \text{Find}\text{\hspace{0.17em}}z\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}x=410,\text{\hspace{0.17em}}u=430,\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}s=2.5.\end{array}$
$\begin{array}{cc}m=\frac{2s+1}{T}.& \text{Find}\text{\hspace{0.17em}}m\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}s=-8\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}T=5.\end{array}$
$-3$
$\begin{array}{cc}m=\frac{2s+1}{T}.& \text{Find}\text{\hspace{0.17em}}m\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}s=-10\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}T=-5.\end{array}$
$\begin{array}{cc}F=\left({p}_{1}-{p}_{2}\right){r}^{4}\cdot 9.& \text{Find}\text{\hspace{0.17em}}F\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}{p}_{1}=10,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{p}_{2}=8,r=3.\end{array}$
1458
$\begin{array}{cc}F=\left({p}_{1}-{p}_{2}\right){r}^{4}\cdot 9.& \text{Find}\text{\hspace{0.17em}}F\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}{p}_{1}=12,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{p}_{2}=7,r=2.\end{array}$
$\begin{array}{cc}P=n\left(n-1\right)\left(n-2\right).& \text{Find}\text{\hspace{0.17em}}P\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}n=-4.\end{array}$
$-120$
$\begin{array}{cc}P=n\left(n-1\right)\left(n-2\right)\left(n-3\right).& \text{Find}\text{\hspace{0.17em}}P\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}n=-5.\end{array}$
$\begin{array}{cc}P=\frac{n\left(n-2\right)\left(n-4\right)}{2n}.& \text{Find}\text{\hspace{0.17em}}P\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}n=-6.\end{array}$
40
## Exercises for review
( [link] ) What natural numbers can replace $x$ so that the statement $-4 is true?
( [link] ) Simplify $\frac{{\left(x+2y\right)}^{5}{\left(3x-1\right)}^{7}}{{\left(x+2y\right)}^{3}{\left(3x-1\right)}^{6}}$ .
${\left(x+2y\right)}^{2}\left(3x-1\right)$
( [link] ) Simplify ${\left({x}^{n}{y}^{3t}\right)}^{5}$ .
( [link] ) Find the sum. $-6+\left(-5\right)$ .
$-11$
( [link] ) Find the difference. $-2-\left(-8\right)$ .
#### Questions & Answers
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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|
# How do you simplify (-243)^(3/5)?
-27
#### Explanation:
With this type of question, see that an exponent in the form of $\frac{a}{b}$ means that we should take the base number to the $a$ power and also take the $b$th root.
In our case, we're being asked to take -243 to the 3rd power and also to take the 5th root. This will be easier if we first see that:
$- 243 = {\left(- 3\right)}^{5}$
And so we can rewrite our question as
${\left(- 243\right)}^{\frac{3}{5}} = {\left({\left(- 3\right)}^{5}\right)}^{\frac{3}{5}}$
We can now apply the rule that
${\left({x}^{a}\right)}^{b} = {x}^{a b}$
${\left({\left(- 3\right)}^{5}\right)}^{\frac{3}{5}} = {\left(- 3\right)}^{5 \times \left(\frac{3}{5}\right)} = {\left(- 3\right)}^{3} = - 27$
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# 2017 AMC 10A Problems/Problem 9
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Minnie rides on a flat road at $20$ kilometers per hour (kph), downhill at $30$ kph, and uphill at $5$ kph. Penny rides on a flat road at $30$ kph, downhill at $40$ kph, and uphill at $10$ kph. Minnie goes from town $A$ to town $B$, a distance of $10$ km all uphill, then from town $B$ to town $C$, a distance of $15$ km all downhill, and then back to town $A$, a distance of $20$ km on the flat. Penny goes the other way around using the same route. How many more minutes does it take Minnie to complete the $45$-km ride than it takes Penny?
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 65\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 95$
## Solution
The distance from town $A$ to town $B$ is $10$ km uphill, and since Minnie rides uphill at a speed of $5$ kph, it will take her $2$ hours. Next, she will ride from town $B$ to town $C$, a distance of $15$ km all downhill. Since Minnie rides downhill at a speed of $30$ kph, it will take her half an hour. Finally, she rides from town $C$ back to town $A$, a flat distance of $20$ km. Minnie rides on a flat road at $20$ kph, so this will take her $1$ hour. Her entire trip takes her $3.5$ hours. Secondly, Penny will go from town $A$ to town $C$, a flat distance of $20$ km. Since Penny rides on a flat road at $30$ kph, it will take her $\frac{2}{3}$ of an hour. Next Penny will go from town $C$ to town $B$, which is uphill for Penny. Since Penny rides at a speed of $10$ kph uphill, and town $C$ and $B$ are $15$ km apart, it will take her $1.5$ hours. Finally, Penny goes from Town $B$ back to town $A$, a distance of $10$ km downhill. Since Penny rides downhill at $40$ kph, it will only take her $\frac{1}{4}$ of an hour. In total, it takes her $29/12$ hours, which simplifies to $2$ hours and $25$ minutes. Finally, Penny's $2$hr $25$min trip was $\boxed{\textbf{(C)}\ 65}$ minutes less than Minnie's $3$hr $30$min trip.
~savannahsolver
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# If the sum of two non-zero numbers is 4 ,
Question:
If the sum of two non-zero numbers is 4 , then the minimum value of the sum of their reciprocals is_______________
Solution:
Let the two numbers be $x$ and $4-x(x \neq 0,4)$.
Suppose S be the sum of their reciprocals.
$\therefore S=\frac{1}{x}+\frac{1}{4-x}, x \neq 0,4$
Differentiating both sides with respect to x, we get
$\frac{d S}{d x}=-\frac{1}{x^{2}}-\frac{1}{(4-x)^{2}} \times(-1)$
$\Rightarrow \frac{d S}{d x}=-\frac{1}{x^{2}}+\frac{1}{(4-x)^{2}}$
For maxima or minima,
$\frac{d S}{d x}=0$
$\Rightarrow-\frac{1}{x^{2}}+\frac{1}{(4-x)^{2}}=0$
$\Rightarrow-\left(16-8 x+x^{2}\right)+x^{2}=0$
$\Rightarrow 8 x=16$
$\Rightarrow x=2$
Now,
$\frac{d S}{d x}=0$
At $x=2$, we have
$\left(\frac{d^{2} S}{d x^{2}}\right)_{x=2}=\frac{2}{(2)^{3}}+\frac{2}{(4-2)^{2}}=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}>0$
So, x = 2 is the point of local minimum.
Thus, S is minimum when x = 2.
$\therefore$ Minimum value of $S=\frac{1}{2}+\frac{1}{4-2}=\frac{1}{2}+\frac{1}{2}=1$ $\left(S=\frac{1}{x}+\frac{1}{4-x}\right)$
Thus, if the sum of two non-zero numbers is 4, then the minimum value of the sum of their reciprocals is 1.
If the sum of two non-zero numbers is 4, then the minimum value of the sum of their reciprocals is ___1___.
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# How do you identify the vertices, foci, and direction of (y-1)^2/9-(x+1)^2/16=1?
Nov 10, 2016
The vertices are $= \left(- 1 , 4\right)$ and $= \left(- 1 , - 2\right)$
The foci are F$= \left(- 1 , 6\right)$ and F'$= \left(- 1 , - 4\right)$
The equations of the asymptotes are $y = 1 + \frac{3}{4} \left(x + 1\right)$ and $y = 1 - \frac{3}{4} \left(x + 1\right)$
#### Explanation:
This is the equation of an updown hyperbola.
The standard equation is ${\left(y - k\right)}^{2} / {b}^{2} - {\left(x - h\right)}^{2} / {a}^{2} = 1$
The center is $\left(h , k\right) = \left(- 1 , 1\right)$
The vertices are $\left(h , k + b\right) = \left(- 1 , 4\right)$ and $\left(h , k - b\right) = \left(- 1 , - 2\right)$
The slopes of the asymptotes are $\pm \frac{b}{a} = \pm \frac{3}{4}$
The equations of the asymptotes are $y = k \pm \frac{b}{a} \left(x - h\right)$
$y = 1 + \frac{3}{4} \left(x + 1\right)$ and $y = 1 - \frac{3}{4} \left(x + 1\right)$
To calculate the foci, we need $c = \pm \sqrt{{a}^{2} + {b}^{2}}$
So $c = \pm \sqrt{9 + 16} = \pm \sqrt{25} = \pm 5$
The foci are $= \left(h , k \pm c\right)$
so F$= \left(- 1 , 6\right)$ and F' $\left(- 1 , - 4\right)$
graph{((y-1)^2/9-(x+1)^2/16-1)(y-1-3/4(x+1))(y-1+3/4(x+1))=0 [-22.82, 22.77, -11.4, 11.4]}
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# Factoring Trinomials of the Form $ax^2 + bx + c$
Need some basic practice with factoring first?
Here, you will practice factoring trinomials of the form $\,ax^2 + bx + c\,,$ where $\,a\,,$ $\,b\,,$ and $\,c\,$ are integers. In this exercise, the coefficient of the $\,x^2\,$ term ($\,a\,$) is usually not going to be the number $\,1\,,$ which makes factoring more challenging.
For example, you might need to factor $\,6x^2-13x-5\,.$
Two methods are presented:
## The Guess-and-Check Method for Factoring Trinomials
One common method for factoring when $\,a\ne 1\,$ is called Guess-and-Check or Trial-and-Error. In this method, we try to find four integers that work:
$$\cssId{s15}{6x^2 -13x - 5} \cssId{s16}{= (\text{__ }x + \text{__ })(\text{__ }x + \text{__ })}$$
The unknown numbers in front of the $\,x\,$ variables must multiply together to give $\,6\,.$ What are the possibilities here? Well, $\,2\cdot 3\,$ or $\,6\cdot 1\,.$ The unknown constants must multiply together to give $\,-5\,.$ What are the possibilities for this one? Well, $\,1\cdot -5\,$ or $\,-1\cdot 5\,.$
Phew. Not too many choices. But even so, this gives lots of possible combinations:
$(2x+1)(3x-5)$ $(6x+1)(x-5)$ $(2x-5)(3x+1)$ $(6x-5)(x+1)$ $(2x-1)(3x+5)$ $(6x-1)(x+5)$ $(2x+5)(3x-1)$ $(6x+5)(x-1)$
Then, the inners and the outers have to work out right. If you're not the particularly lucky type, then it might take you a while to stumble on the one that works. (Did you find it?)
And, for a random trinomial pulled out of the air, it's possible that nothing will work! So, you could spend a long time looking, and find yourself wondering: “Do they exist? Have I just missed them?”
There must be a better way—and there is. It's called the ‘Factor by Grouping’ method, and it's usually much more efficient than trial-and-error.
## Example: the ‘Factor by Grouping’ Method
Here's an example of the ‘Factor by Grouping’ method. Details follow, for inquiring minds. However, some of you may want to just take this example, and run with it.
Factor: $\,6x^2 - 13x - 5\,$
Use the ‘Factor by Grouping’ method.
Solution:
• The number in front of $\,x^2\,$ isn't $\,1\,.$ Instead, it's a $\,6\,.$ So, we must modify the old method, which said: “Find two numbers that multiply to the constant term, and add to the middle term.” Somehow, we're going to have to use that $\,6\,.$
• Here's how. Take the $\,6\,$ and multiply it by the constant term, $\,-5\,.$ You get: $\,(6)(-5) = -30$
• Now, find two numbers that multiply to $\,-30\,$ and that (still) add to $\,-13\,.$
• This should be a very familiar thought process—you've likely practiced it a lot. Use the PANS Memory Device, if you want.
• Come up with the two numbers that work $\ldots$ (thinking, thinking) $\ldots$ how about $\,-15\,$ and $\,2\,.$ Check them:
Do they multiply to $\,-30\,$?
$\,(-15)(2) = -30\,$
Check!
Do they add to $\,-13\,$?
$\,-15 + 2 = -13\,$
Check!
• Now, rename the original trinomial:
$6x^2 - 13x - 5$ original trinomial $= 6x^2\ \overset{=\ -13x}{\overbrace{- 15x + 2x}} - 5$ rename the middle term, using the two numbers you found $= (6x^2 - 15x) + (2x - 5)$ group the first two terms, and the last two terms—hence the title of this technique! $= 3x(2x - 5) + 1(2x - 5)$ factor the first part; put the $\,1\,$ in for clarity in the second part $= 3x\overbrace{(2x-5)} + 1\overbrace{(2x-5)}$ Look carefully! There are two ‘big’ terms, with a common factor of $\,2x-5\,$ $= (2x-5)(3x + 1)$ factor out the common factor—done!
It may seem like a lot of work to you. But, over the years, I've divided my classes in half: told one side to use trial-and-error, and the other side to use factor by grouping. The factor by grouping side always wins.
Let's do the same example one more time. This will show you that the order you write the two middle terms doesn't matter. Also, this version is much more compact, and is usually all you'll need to write down.
Factor: $\,6x^2 - 13x - 5\,$
Use the ‘Factor by Grouping’ method.
Solution:
• $(6)(-5) = -30\,$
Find two numbers that multiply to $\,-30\,$ and add to $\,-13\,.$
• Try $\,-15\,$ and $\,2\,$:
$(-15)(2) = -30$ Check!
$-15 + 2 = -13$ Check!
• Rename:
$6x^2 - 13x - 5$ original trinomial $= 6x^2 \ \overset{\text{different order}}{\overbrace{+ 2x -15x}} - 5$ rename the middle term $= (6x^2 + 2x) + (-15x - 5)$ group first two, last two $= 2x(3x + 1) + (-5)(3x + 1)$ factor each group separately $= (3x + 1)(2x - 5)$ factor again
## Motivation for the ‘Factor by Grouping’ Method
Here's the motivation for the technique. Start by pulling a random problem out of the air, and FOILing it out:
\begin{align} &\cssId{s116}{(3x - 1)(5x + 7)}\cr &\qquad \cssId{s117}{= 15x^2 + 21x - 5x - 7} \end{align}
Don't combine the middle terms. Instead, look at the four numbers generated in the resulting sum:
$$\cssId{s120}{\,15\,, \,21\,, \,-5\,, \,-7}$$
The product of the first and last is: $\,(15)(-7) = -105$
The product of the middle two is: $\,(21)(-5) = -105$
Is this just a coincidence? It's not a coincidence. It's always true:
\begin{align} &\cssId{s128}{(ex + f)(gx + h)}\cr &\qquad \cssId{s129}{= (eg)x^2 + (eh)x + (fg)x + (fh)} \end{align}
The product of the first and last is: $\,(eg)(fh) = efgh\,$
The product of the middle two is: $\,(eh)(fg) = efgh\,$
Same result. Other than being a curious observation, what good is this?
Ends up that it's a lot of good. If we're trying to factor a trinomial by grouping, and we're looking for the right way to rename the middle term, then we know two things:
$$\cssId{s140}{ax^2 + bx + c} \cssId{s141}{= ax^2 + \text{__}\ x + \text{__}\ x + c}$$
• the two unknown numbers must add to $\,b\,$
• the two unknown numbers must multiply to $\,ac\,$
This is why we take the number in front of the $\,x^2\,$ term, and multiply it by the constant term!
The rest of the argument is more advanced, and is primarily included for the sake of the teacher.
## Practice
For more advanced students, a graph is available for each factoring problem. For example, suppose you're asked to factor $\,6x^2 - 13x - 5\,.$ Then, you'd see the graph of the equation $\,y = 6x^2 - 13x - 5\,.$ (This graph is shown below when the web page is first loaded.)
Observe that the graph crosses the $\,x\,$-axis at $\,-\frac13\,$ and $\,\frac52\,.$ Look at the factorization: \begin{align} &6x^2 - 13x - 5\cr &\quad = (3x + 1)(2x - 5)\cr &\quad = 3(x + \frac 13)2(x - \frac52)\cr &\quad = 6(x+\frac13)(x-\frac52) \end{align} See the relationship between the factors and the $x$-axis intercepts (zeros)? If $\,c\,$ is a zero, then $\,x - c\,$ is a factor! (You're discovering the beautiful relationship between the zeroes of a polynomial and its factors.)
Click the ‘Show/Hide Graph’ button to toggle the graph.
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# How To Slope Intercept Form
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
How To Slope Intercept Form – One of the many forms used to depict a linear equation, the one most frequently found is the slope intercept form. You may use the formula for the slope-intercept in order to find a line equation assuming that you have the straight line’s slope , and the y-intercept, which is the point’s y-coordinate at which the y-axis is intersected by the line. Read more about this particular linear equation form below.
## What Is The Slope Intercept Form?
There are three main forms of linear equations: standard, slope-intercept, and point-slope. Though they provide similar results when used however, you can get the information line that is produced quicker with the slope intercept form. The name suggests that this form uses an inclined line where it is the “steepness” of the line determines its significance.
This formula can be utilized to discover the slope of a straight line. It is also known as the y-intercept, also known as x-intercept where you can utilize a variety available formulas. The equation for this line in this formula is y = mx + b. The slope of the straight line is represented in the form of “m”, while its intersection with the y is symbolized through “b”. Every point on the straight line is represented by an (x, y). Note that in the y = mx + b equation formula, the “x” and the “y” must remain as variables.
## An Example of Applied Slope Intercept Form in Problems
In the real world in the real world, the slope-intercept form is frequently used to show how an item or issue evolves over its course. The value of the vertical axis indicates how the equation deals with the intensity of changes over the value provided with the horizontal line (typically the time).
A simple example of the use of this formula is to determine how many people live in a certain area as the years pass by. If the population of the area increases each year by a certain amount, the worth of horizontal scale will increase one point at a moment with each passing year and the point value of the vertical axis will rise to show the rising population by the set amount.
Also, you can note the beginning value of a particular problem. The beginning value is located at the y’s value within the y’intercept. The Y-intercept marks the point where x is zero. In the case of the above problem the beginning value will be at the time the population reading starts or when the time tracking starts along with the changes that follow.
So, the y-intercept is the point in the population that the population begins to be monitored by the researcher. Let’s suppose that the researcher starts to calculate or take measurements in the year 1995. This year will serve as the “base” year, and the x 0 points will occur in 1995. This means that the 1995 population is the y-intercept.
Linear equations that employ straight-line formulas are nearly always solved this way. The starting point is depicted by the y-intercept and the change rate is expressed in the form of the slope. The main issue with this form is usually in the interpretation of horizontal variables particularly when the variable is associated with an exact year (or any type number of units). The trick to overcoming them is to make sure you are aware of the definitions of variables clearly.
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Properties Of HCF And LCM MCQs With Explanation-Sainik School Class 6 Math Study Material Notes free pdf download
Sainik School Entrance Exam for Class 6 Math Study Material Notes helps students face the competition in the current education system. In this case, the ANAND CLASSES is the best study tool to get a clear idea about the basics and gain a strong knowledge of the Sainik School Entrance Exam syllabus.
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Properties of HCF and LCM: For a better understanding of the concepts of LCM (Lowest Common Multiple) and HCF (Highest Common Factor), we need to recollect the terms multiples and factors. Let’s learn about LCM, HCF and the relation between HCF and LCM of natural numbers.
Properties of HCF and LCM
Property 1
The product of LCM and HCF of any two given natural numbers is equivalent to the product of the given numbers.
LCM × HCF = Product of two Numbers
Suppose A and B are two numbers, then.
LCM (A & B) × HCF (A & B) = A × B
Example: If 3 and 8 are two numbers.
LCM (3,8) = 24
HCF (3,8) = 1
LCM (3,8) x HCF (3,8) = 24 x 1 = 24
Also, 3 x 8 = 24
Hence, proved.
Note: This property is applicable for only two numbers.
Example 1: Show that the LCM (6, 15) × HCF (6, 15) = Product(6, 15)
Solution: LCM and HCF of 6 and 15:
6 = 2 × 3
15 = 3 x 5
LCM of 6 and 15 = 30
HCF of 6 and 15 = 3
LCM (6, 15) × HCF (6, 15) = 30 × 3 = 90
Product of 6 and 15 = 6 × 15 = 90
Hence, LCM (6, 15) × HCF (6, 15)=Product(6, 15) = 90
Property 2
HCF of co-prime numbers is 1. Therefore, the LCM of given co-prime numbers is equal to the product of the numbers.
LCM of Co-prime Numbers = Product Of The Numbers
Example: Let us take two coprime numbers, such as 21 and 22.
LCM of 21 and 22 = 462
Product of 21 and 22 = 462
LCM (21, 22) = 21 x 22
LCM of given co-prime Numbers = Product of the given Numbers
Solution: LCM and HCF of 17 and 23:
17 = 1 x 7
23 = 1 x 23
LCM of 17 and 23 = 391
HCF of 17 and 23 = 1
Product of 17 and 23 = 17 × 23 = 391
Hence, LCM of co-prime numbers = Product of the numbers
Property 3
H.C.F. and L.C.M. of Fractions:
LCM of fractions = LCM of Numerators / HCF of Denominators
HCF of fractions = HCF of Numerators / LCM of Denominators
Example: Let us take two fractions 4/9 and 6/21.
4 and 6 are the numerators & 9 and 12 are the denominators
LCM (4, 6) = 12
HCF (4, 6) = 2
LCM (9, 21) = 63
HCF (9, 21) = 3
Now as per the formula, we can write:
LCM (4/9, 6/21) = 12/3 = 4
HCF (4/9, 6/21) = 2/63
Example 3: Find the LCM of the fractions 1 / 2 , 3 / 8, 3 / 4
Solution:
LCM of fractions = LCM of Numerators/HCF of Denominators
LCM of fractions = LCM (1,3,3)/HCF(2,8,4)=3/2
Example 4: Find the HCF of the fractions 3 / 5, 6 / 11, 9 / 20
HCF of fractions HCF of Numerators/LCM of Denominators
HCF of fractions = HCF (3,6,9)/LCM (5,11,20)=3/220
Property 4
HCF of any two or more numbers is never greater than any of the given numbers.
Example: HCF of 4 and 8 is 4
Here, one number is 4 itself and another number 8 is greater than HCF (4, 8), i.e.,4.
Property 5
LCM of any two or more numbers is never smaller than any of the given numbers.
Example: LCM of 4 and 8 is 8 which is not smaller to any of them.
Example 1: Prove that: LCM (9 & 12) × HCF (9 & 12) = Product of 9 and 12
Solution:
9 = 3 × 3 = 3²
12 = 2 × 2 × 3 = 2² × 3
LCM of 9 and 12 = 2² × 3² = 4 × 9 = 36
HCF of 9 and 12 = 3
LCM (9 & 12) × HCF (9 & 12) = 36 × 3 = 108
Product of 9 and 12 = 9 × 12 = 108
Hence, LCM (9 & 12) × HCF (9 & 12) = 9 × 12 = 108. Proved.
Example 2:8 and 9 are two co-prime numbers. Using these numbers verify, LCM of Co-prime Numbers = Product Of The Numbers.
Solution: LCM and HCF of 8 and 9:
8 = 2 × 2 × 2 = 2³
9 = 3 × 3 = 3²
LCM of 8 and 9 = 2³ × 3² = 8 × 9 = 72
HCF of 8 and 9 = 1
Product of 8 and 9 = 8 × 9 = 72
Hence, LCM of co-prime numbers = Product of the numbers. Therefore, verified.
Example 3: Find the HCF of 12/25, 9/10, 18/35, 21/40.
Solution:
12 = 2 × 2 × 3
9 = 3 × 3
18 = 2 × 3 × 3
21 = 3 × 7
HCF (12, 9, 18, 21) = 3
25 = 5 × 5
10 = 2 × 5
35 = 5 × 7
40 = 2 × 2 × 2 × 5
LCM(25, 10, 35, 40) = 5 × 5 × 2 × 2 × 2 × 7 = 1400
The required HCF = HCF(12, 9, 18, 21)/LCM(25, 10, 35, 40) = 3/1400
Example 4: If LCM and HCF of two numbers are 3 and 2 respectively, and one of the numbers is 6 then another number is?
Solution:
We know that LCM × HCF = a × b where a, and b are two numbers
I.e., 3 × 2 = 6 × b
Therefore, b = 1.
(D) The sum of the HCF and LCM of two numbers is always equal to the sum of the two numbers.
Explanation: The HCF of two numbers is not always equal to 1. For example, the HCF of 2 and 3 is 1, but the HCF of 6 and 12 is 6.
D) HCF(a, b) and LCM(a, b) have no specific relationship.
Explanation: B)
HCF(a, b) is always less than LCM(a, b). HCF is the largest common factor of two numbers, and LCM is the smallest common multiple. Since factors are smaller than multiples, the HCF is always less than the LCM.
(D) The larger of the two numbers
Explanation: If the HCF of two numbers is 1, then the two numbers are relatively prime (have no common factors other than 1). The LCM of two relatively prime numbers is equal to the product of the two numbers.
(D) The larger of the two numbers
Explanation: If the LCM of two numbers is equal to their product, then the two numbers must be equal. The HCF of two equal numbers is equal to 1.
(A) xy (B) x + y (C) x – y (D) xy/2
Explanation: The product of two numbers is equal to their HCF times their LCM. Therefore, the product of the two numbers is xy.
(A) 5 and 5 (B) 4 and 6 (C) 3 and 7 (D) 2 and 8
Explanation: If the HCF of two numbers is 1, then the two numbers are relatively prime (have no common factors other than 1). The only possible pair of relatively prime numbers that sum to 10 is 3 and 7.
(A) 4 and 6 (B) 3 and 7 (C) 2 and 8 (D) 1 and 9
Explanation: The LCM of two numbers is the smallest number that is a multiple of both numbers. The only possible pair of numbers that have an LCM of 12 and a difference of 2 is 4 and 6.
(A) 2 and 10 (B) 2 and 5 (C) 4 and 5 (D) 6 and 10
Explanation: The product of two numbers is equal to their HCF times their LCM. Therefore, the product of the two numbers is 2 * 10 = 20. The only possible pair of numbers that have a product of 20 and an HCF of 2 is 2 and 10.
(A) 240 (B) 360 (C) 480 (D) 720
Explanation: The product of two numbers is equal to their HCF times their LCM. Therefore, the product of the three numbers is 2 * 120 = 240.
(A) 240 (B) 360 (C) 480 (D) 720
Explanation: The product of three numbers is equal to their HCF times their LCM. Therefore, the greatest possible product of the three numbers is 3 * 120 = 360.
Frequently Asked Questions (FAQs) related to studying for the Sainik School Class 6 Math entrance exam
To help you prepare for Sainik School Class 6 Math, it’s important to use appropriate study materials. Here are some frequently asked questions (FAQ) related to studying for the Sainik School Class 6 Math entrance exam:
1. What is the syllabus for the Sainik School Class 6 Math entrance exam?
• The syllabus for the entrance exam may vary slightly from one Sainik School to another. However, it generally covers topics from the standard Class 6 mathematics curriculum, including arithmetic, geometry, algebra, and basic mathematical concepts.
2. Where can I find official information about the exam pattern and syllabus?
• You can find official information about the exam pattern and syllabus on the official website of the specific Sainik School you’re applying to. Each school may have its own admission criteria.
3. Which textbooks should I use for Class 6 Math preparation?
• You should primarily use the NCERT Class 6 Math textbook. It covers the fundamental concepts and is widely accepted in Indian schools. Additionally, consider supplementary Math textbooks that are designed for competitive exams.
4. Are there any online resources for Sainik School Class 6 Math preparation?
• Yes, you can find online resources such as video tutorials, practice questions, and mock tests on educational websites. Websites like ANAND CLASSES offer free Math materials that can be helpful for your preparation.
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• You can find sample papers and previous year’s question papers at online bookstore of ANAND CLASSES that sell competitive exam preparation materials. Additionally, ANAND CLASSES website offer downloadable PDFs of these papers for free or at a minimal cost.
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• Enrolling in ANAND CLASSES is a personal choice. While they can provide structured guidance and additional practice, they are not mandatory. You can achieve success through self-study and the use of appropriate study materials.
7. How should I manage my study time effectively for Class 6 Math preparation?
• Create a study schedule that allocates specific time for Math preparation daily. Focus on understanding concepts, practicing regularly, and taking regular breaks to avoid burnout. Consistency is key.
8. Is there any specific advice for tackling the Math section of the Sainik School entrance exam?
• Pay close attention to the BODMAS rule (Order of Operations) and practice solving a variety of math problems. Make sure you’re familiar with the types of questions that are commonly asked in the entrance exam.
Remember to check the specific requirements and guidelines provided by the Sainik School you are applying to, as these may vary from school to school.
To prepare effectively for the Sainik School Class 6 Math entrance exam, you should consider the following general sources:
1. Sainik School Official Website: Visit the official website of the Sainik School you are applying to. They often provide information about the exam pattern, syllabus, and sample question papers.
2. NCERT Textbooks: The National Council of Educational Research and Training (NCERT) textbooks are widely used in Indian schools and are a valuable resource for exam preparation. Ensure you have the NCERT Math textbook for Class 6.
3. Solved Sample Papers and Previous Year Question Papers: You can find solved sample papers and previous year question papers for Sainik School entrance exams at bookstores or online at ANAND CLASSES website. These papers can give you an idea of the exam pattern and types of questions asked.
4. Math Study Guides: The Math study guides and reference books are publish under publication department of ANAND CLASSES and are designed to help students prepare for entrance exams. Look for books specifically tailored to the Sainik School entrance exam.
5. Online Resources: www.anandclasses.co.in
We at ANAND CLASSES are providing notes for Sainik School Entrance Exam students, mainly for subjects like Science and Maths. Scoring well in these major subjects will increase the possibility of getting into good SAINIK SCHOOL in the long run. The notes that we are offering have been thoughtfully prepared by our experts to help you achieve the same. These notes are designed to help students overcome all the challenges in solving math problems and understand difficult MATH concepts. Basically, these notes act as a valuable reference tool for conducting effective revisions of the entire chapters given in each subject. Additionally, students can use these notes to get detailed explanations, practice problems, and study properly without wasting much precious time.
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# How to Solve Vertical Angle Problems?
TopTransversal line is a line, which is use to intersect two lines in a plane and creates 8 angle between them. These 8 angles are part of two groups, where each group contains 4 angles. Opposite and non Adjacent Angles of different groups is called as a Vertical Angles like one traversal line made 8 angles in following group –
Group 1: angle 1, angle 2, angle 3 and angle 4,
Group 2: angle 5, angle 6, angle 7 and angle 8,
Here angle 1 and angle 6, angle 2 and angle 7, angle 3 and angle 8, angle 4 and angle 5 are vertical angles. Now we discuss how to solve vertical angle problems:
We use following steps for solution of alternate angles–
Step 1: First of all, we calculate one angle from two vertical angles like we have two vertical angles – angle ‘m’ and angle ‘n’, and value of angle m is 60 degree and value of angle ‘n’ is x – 60 degree.
Step 2: After evaluation of one angle, now we equal both vertical angle with each other because by Geometry property, both angles are equal to each other.
Angle m = angle n,
= > 60 degree = x - 60,
= > x = 60 degree + 60 degree,
= > x = 120 degree,
So, other vertical angle from two vertical angles, whose first angle is 60 degree is 120 degree – 60 degree = ‘60 degree’.
Therefore we use above 2 steps for evaluation of vertical angles like we have one vertical angle is equal to 45 degree, then other vertical angle is also equal to 45 degree by above property.
Vertical Angle 1 = vertical angle 2 = 45 degree,
So, other vertical angle from two vertical angles, whose first angle is 45 degree, is 45 degree because the values of vertical angle are equal to each other.
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# 4.2: Absolute Value of Integers
Difficulty Level: At Grade Created by: CK-12
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Practice Absolute Value of Integers
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Do you live where it snows? How much snow has ever accumulated in 24 hours where you live?
Cameron is amazed at the record snowfall in Alaska. One time, the day began without any snow on the ground. Then it began to snow and within 24 hours there was 62 inches of snow.
We can use an integer to write the increase in snowfall, and we can use absolute value to show the distance between the depth of snow and the bare ground.
Do you know how to do this?
Understanding absolute value is the goal of this Concept. Pay attention and we will revisit this situation at the end of it.
### Guidance
Sometimes, when we look at an integer, we aren’t concerned with whether it is positive or negative, but we are interested in how far that number is from zero. Think about water. You might not be concerned about whether the depth of a treasure chest is positive or negative simply how far it is from the surface.
This is where absolute value comes in.
What is absolute value?
The absolute value of a number is its distance from zero on the number line.
We can use symbols to represent the absolute value of a number. For example, we can write the absolute value of 3 as |3|\begin{align*}|3|\end{align*}.
Writing an absolute value is very simple you just leave off the positive or negative sign and simply count the number of units that an integer is from zero.
Find the absolute value of 3. Then determine what other integer has an absolute value equal to |3|\begin{align*}|3|\end{align*}.
Look at the positive integer, 3, on the number line. It is 3 units from zero on the number line, so it has an absolute value of 3.
Now that you have found the absolute value of 3, we can find another integer with the same absolute value. Remember that with absolute value you are concerned with the distance an integer is from zero and not with the sign.
Here is how we find another integer that is exactly 3 units from 0 on the number line. The negative integer, -3, is also 3 units from zero on the number line, so it has an absolute value of 3 also.
So, |3|=|3|=3\begin{align*}|3|=|-3|=3\end{align*}.
This example shows that the positive integer, 3, and its opposite, -3, have the same absolute value. On a number line, opposites are found on opposite sides of zero. They are each the same distance from zero on the number line. Because of this, any integer and its opposite will always have the same absolute value. To find the opposite of an integer, change the sign of the integer.
Just like we can find the absolute value of a number, we can also find the opposite of a number.
Find the opposite of each of these numbers: -16 and 900.
-16 is a negative integer. We can change the negative sign to a positive sign to find its opposite. The opposite of -16 is +16 or 16.
900 is the same thing as +900. We can change the positive sign to a negative sign to find its opposite. So, the opposite of 900 is -900.
Find the absolute value of each number.
#### Example A
|22|\begin{align*}|22|\end{align*}
Solution: 22\begin{align*}22\end{align*}
#### Example B
|222|\begin{align*}|-222|\end{align*}
Solution:222\begin{align*}222\end{align*}
#### Example C
Find the opposite of -18.
Solution:18\begin{align*}18\end{align*}
Here is the original problem once again.
Do you live where it snows? How much snow has ever accumulated in 24 hours where you live?
Cameron is amazed at the record snowfall in Alaska. One time, the day began without any snow on the ground. Then it began to snow and within 24 hours there was 62 inches of snow.
We can use an integer to write the increase in snowfall, and we can use absolute value to show the distance between the depth of snow and the bare ground.
Do you know how to do this?
To express both of these values using integers and absolute value, we can begin with the increase in snowfall. Because it is an increase in snowfall, we use a positive integer to express this amount.
+62\begin{align*}+62\end{align*} inches
To express the difference in snowfall accumulation and the bare ground, we use an absolute value.
|62|\begin{align*}|62|\end{align*}
The absolute value of 62 is 62.
This is how we can express this situation using integers.
### Vocabulary
Here are the vocabulary words in this Concept.
Whole Numbers
the positive counting numbers including 0.
Fraction
a part of a whole written with a numerator and denominator.
Integers
positive whole numbers and their opposites. Positive and negative numbers
Opposites
for a negative number, it has a positive partner. For a positive number, it has a negative partner.
Absolute Value
the distance that a number is from zero.
### Guided Practice
Here is one for you to try on your own.
|234|\begin{align*}|-234|\end{align*}
To identify the absolute value of this number, we have to think about the number of units it is from zero. Remember that absolute value does not concern positive or negative, but the distance that a value is from zero.
|234|=234\begin{align*}|-234| = 234\end{align*}
### Video Review
Here is a video for review.
### Practice
Directions: Write the opposite of each integer.
7. 20
8. -7
9. 22
10. -34
11. 0
12. -9
13. 14
14. 25
Directions: Find the absolute value of each number.
15. |13|\begin{align*}|13|\end{align*}
16. |11|\begin{align*}|-11|\end{align*}
17. |5|\begin{align*}|-5|\end{align*}
18. |17|\begin{align*}|17|\end{align*}
19. |9|\begin{align*}|-9|\end{align*}
### Notes/Highlights Having trouble? Report an issue.
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# Higher-order derivatives
We already know that how we can take a derivative of function, and we’ve learned how to find the first and second derivatives — but can we find a third derivative? Or even higher?
You bet we can! The process of differentiation can be performed several times in a row, which leads to higher-order derivatives.
## What does it mean to find a higher-order derivative?
A derivative of a function is a rate of change of a function with respect to a change of a variable. To find the higher-order derivative — or the $$n$$th derivative — means to find the derivative of the $$(n-1)$$th derivative of the function. So basically, it’s a derivative of a derivative of a derivative, etc.!
To help us during our differentiation, we’ll use differentiation rules rather than the definition of the derivative.
### Why is the higher-order derivative so useful?
As we already know, the second derivatives can be really useful for graphing functions and determining the behavior of the preceding function’s graph!
## How to find the higher-order derivative
If you already have solid skills for taking the first and second derivatives, finding a higher-order derivative should just be a few extra steps for you! Either way, let’s walk through some detailed examples together.
### Example 1
Find the higher-order derivative:
$$\frac{d^3}{dx^3}\left(x^{7} + \frac{5}{3}x^{3} - 7x + \pi^{2}\right)$$
To find the higher-order derivative (in this case the third derivative), we’ll differentiate three times
$$\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{d}{dx}\left(x^{7} + \frac{5}{3}x^{3} – 7x + \pi^{2}\right)\right)\right)$$
First, let’s find the first derivative:
$$\frac{d}{dx}\left(x^{7} + \frac{5}{3}x^{3} – 7x + \pi^{2}\right)$$
Use the differentiation rule $$\frac{d}{dx}(f+g)=\frac{d}{dx}(f)+\frac{d}{dx}(g)$$:
$$\frac{d}{dx}(x^{7}) + \frac{d}{dx}\left(\frac{5}{3}x^{3}\right) – \frac{d}{dx}(7x) + \frac{d}{dx}(\pi^{2}))$$
Use the differentiation rule $$\frac{d}{dx} (a\times f)=a\frac{d}{dx}(f)$$:
$$\frac{d}{dx}(x^{7}) + \frac{5}{3}\times\frac{d}{dx}(x^{3}) – 7\times \frac{d}{dx}(x) + \frac{d}{dx}(\pi^{2}))$$
Use the differentiation rule $$\frac{d}{dx} x^n=nx^{n-1}$$:
$$7\times x^{6} + \frac{5}{3}\times 3\times x^2 – 7\frac{d}{dx}(x) + \frac{d}{dx}(\pi^{2}))$$
Remember: the derivative of a variable to the first power is always $$1$$:
$$7\times x^{6} + \frac{5}{3}\times 3\times x^2 – 7\times 1 + \frac{d}{dx}(\pi^{2}))$$
The derivative of a constant is always $$0$$, so:
$$7\times x^{6} + \frac{5}{3}\times 3\times x^2 – 7\times 1 + 0$$
Removing the zero doesn’t change the value, so remove the zero:
$$7\times x^{6} + \frac{5}{3}\times 3\times x^2 – 7\times 1$$
Simplify the expression:
$$7 x^{6} + 5 x^2 – 7$$
On to the second derivative! Let’s take the derivative of each term, with respect to $$x$$:
$$\frac{d}{dx}(7 x^{6} + 5 x^2 – 7)$$
Use the differentiation rule $$\frac{d}{dx}(f+g)=\frac{d}{dx}(f)+\frac{d}{dx}(g)$$:
$$\frac{d}{dx}(7 x^{6} + 5 x^2 – 7)$$
Use the differentiation rule $$\frac{d}{dx}(f+g)=\frac{d}{dx}(f)+\frac{d}{dx}(g)$$:
$$\frac{d}{dx}(7 x^{6}) + \frac{d}{dx}(5 x^2) – \frac{d}{dx}(7)$$
Use the differentiation rule $$\frac{d}{dx} (a\times f)=a\frac{d}{dx}(f)$$:
$$7\times\frac{d}{dx}(x^{6}) + 5\times\frac{d}{dx}( x^2) – \frac{d}{dx}(7)$$
Use the differentiation rule $$\frac{d}{dx} x^n=nx^{n-1}$$:
$$7\times6x^5 + 5\times2x – \frac{d}{dx}(7)$$
The derivative of a constant is always $$0$$:
$$7\times6x^5 + 5\times2x – 0$$
Removing the zero doesn’t change the value, so we’ll get rid of that zero:
$$7\times6x^5 + 5\times2x$$
Simplify the expression:
$$42x^5 + 10x$$
Here we go: time for the third derivative! Let’s differentiate the expression one more time:
$$\frac{d}{dx}(42x^5 + 10x)$$
Use the differentiation rule $$\frac{d}{dx}(f+g)=\frac{d}{dx}(f)+\frac{d}{dx}(g)$$:
$$\frac{d}{dx}(42x^5) + \frac{d}{dx}(10x)$$
Use the differentiation rule $$\frac{d}{dx} (a\times f)=a\frac{d}{dx}(f)$$:
$$42\times\frac{d}{dx}(x^5) + 10\times\frac{d}{dx}(x)$$
Use the differentiation rule $$\frac{d}{dx} x^n=nx^{n-1}$$:
$$42\times5\times x^4 + 10\times\frac{d}{dx}(x)$$
The derivative of a variable to the first power is always $$1$$:
$$42\times5\times x^4 + 10\times1$$
Simplify the expression:
$$210x^4 + 10$$
Woohoo! We found the third derivative of $$x^{7} + \frac{5}{3}x^{3} – 7x + \pi^{2}$$:
$$210x^4 + 10$$
Once you get into the swing of things, it’s not so bad! Let’s look at another example.
### Example 2
Find the higher-order derivative:
$$\frac{d^3}{dx^3}(x^2+3)$$
To find the higher-order derivative (in this case the third derivative), take the derivative three times:
$$\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{d}{dx}\left(x^2+3\right)\right)\right)$$
First up is the first derivative:
$$\frac{d}{dx}\left(x^2+3\right)$$
Use the differentiation rule $$\frac{d}{dx}(f+g)=\frac{d}{dx}(f)+\frac{d}{dx}(g)$$:
$$\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(3\right)$$
Use the differentiation rule $$\frac{d}{dx} x^n=nx^{n-1}$$:
$$2x+\frac{d}{dx}\left(3\right)$$
The derivative of a constant is always $$0$$:
$$2x+0$$
Removing the zero doesn’t change the value, so remove the zero:
$$2x$$
Great! Now to find the second derivative, take the derivative of each term, with respect to $$x$$:
$$\frac{d}{dx}(2x)$$
Use the differentiation rule $$\frac{d}{dx} (a\times f)=a\frac{d}{dx}(f)$$:
$$2\times\frac{d}{dx}(x)$$
The derivative of a variable to the first power is always $$1$$:
$$2\times1$$
Multiply the numbers:
$$2$$
So close! Take the derivative one more time so we can find our third derivative:
$$\frac{d}{dx}(2)$$
The derivative of a constant is always $$0$$:
$$0$$
Got it! The third derivative of $$x^2+3$$ is:
$$~0$$
Finding higher-order derivatives can be repetitive, but sometimes that makes things easy! Remember these steps when you try this process on your own:
## Study summary
1. To find the higher-order derivative, take the derivative multiple times.
2. Find the derivative.
3. To find the n-th derivative, repeat the process n times.
4. Simplify the final expression, if possible.
## Do it yourself!
Looking for some more examples? Try these practice problems and see how you do!
Take the derivative of a function:
1. $$\frac{d^4}{dx^4}\left(e^x + 5x\right)$$
2. $$\frac{d^2}{dx^2}\left(\ln{(x)}+5^x\right)$$
3. $$\frac{d^3}{dt^3}\left(\sqrt{t^4-1}\right)$$
4. $$\frac{d^3}{dx^3}\left(\frac{x-1}{x^2-4}\right)$$
Solutions:
1. $$e^x$$
2. $$-\frac{1}{x^2}+\ln(5)^2\times5^x$$
3. $$\frac{12t^5+12t}{\sqrt{t^4-1}(t^4-1)^2}$$
4. $$\frac{-6x^4+24x^3-144x^2+96x-96}{(x^2-4)^4}$$
If you’re getting stuck, scan the problem with your Photomath app and we’ll explain in step in as much detail as you need!
Here’s a sneak peek of what you’ll see:
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Multiplying by 10
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# Multiplying by 10 - PowerPoint PPT Presentation
L.O. - Multiply an integer up to 1000 by 10; understand the effect. Multiplying by 10. When we multiply by 10 we are making the number ten times bigger. x10. x10. x10. In the decimal system, each column has numbers 10 x bigger than the next one. 1000. 100. 10. 1. x10. x10. x10.
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Presentation Transcript
Multiplying by 10
When we multiply by 10 we are making the number ten times bigger.
x10
x10
x10
In the decimal system, each column has numbers 10 x bigger than the next one
1000
100
10
1
x10
x10
x10
Let’s try it with another number!
3 0 0 0
3
3 0 0
3 0
If we multiply 6 by ten it will move into the next column
x10
We must put a 0 in the empty units column to hold the place – 6 x 10 = 60
200
20
30
3
Let’s try with some other numbers
Don’t forget to add the zero in the empty column!
23 x 10 = 230
Let’s try these together!
48 x 10 =
93 x 10 =
134 x 10 =
256 x 10 =
889 x 10 =
480
930
1340
2560
8890
Were you right?
If we multiply 6 by a hundred it will move up two columns
We must put a 0 in the empty columns to hold the place – 6 x 100 = 600
Let’s try these together!
8 x 100 =
15 x 100 =
28 x 100=
134 x 100 =
267 x 100 =
800
1500
2800
13400
26700
Were you right?
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# Conditional Probability
## Independent Events
The events which are not affected by another events are known as Independent Events.
Two events A and B are independent if event A does not affect the occurrence of event B
Example: Rolling a dice
Fig 1: Rolling a dice
The event of getting a 6 the first time a die is rolled and event of getting a 6 the second time is completely independent and isolated.
## Dependent Events
The events which are dependent on other events and they can be affected by previous events.
Example:
Fig 2: Marbles in a bag
2 blue and 3 red marbles are there in a bag
The chances of getting blue marbles is 2/5. But after one marble is taken out, the chances change:
• If we got red marble in the previous event, the chance of blue marble in the next is 2/4
• If we got blue marble in the previous event, then the chance of blue marble in the next is ¼
## What is Conditional Probability?
Conditional Probability simply measures “How to measure or handle Dependent Events
In probability theory, Conditional Probability is a measure of the probability of an event given that (by assumption, presumption or evidence) another event has occurred. If the event A is unknown and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P (A|B), or sometimes PB (A).
Example, the probability that any person has a cough on a day may be only 5%. But if we know or assume that the person has a cold, then they are much more likely to be coughing. The conditional probability of coughing given that you have a cold might be a much higher 75%.
Fig 3: Conditional Probability
## How to calculate Conditional Probability?
Let S be a finite sample space, associated with the given random experiment, containing equally likely outcomes. Then we have the following result.
Statement: Conditional probability of event A given that event B has already occurred is given by
Fig 4: Venn Diagram – Conditional Probability
P(A/B) = P(A∩B) / P(B) , P(B) ≠0
Let S be a sample space associated with the given random experiment and n(S) be the number of sample points in the sample space S. Since it is given that event B has already occurred, therefore our sample space reduces to event B only, which contains n(B) sample points. Event B is also called reduced or truncated sample space. Now out of n(B) sample points, only n(A∩B) sample points are favourable for the occurrence of event A. Therefore by definition of probability.
P(A/B) = n (A∩B) / n(B), n(B)≠0
= n(A∩B)/ n(S) / n(B)/n(S)
Therefore, P(A/B) = P(A∩B)/P(B) P(B)≠0
Similarly P(B/A) = P(A∩B)/ P(A) P(A)≠0
## Law of Total Probability
Consider a country with three provinces B1, B2 and B3 (i.e country is partitioned into 3 disjoint sets B1, B2 and B3). To calculate the total forest area knowing the area of provinces B1, B2 and B3 as 100Km2, 50 Km2 and 150 Km2. obviously, we will add all the three provinces areas to get total area of forest.
100Km2 +50 Km2 + 150 Km2 = 300 Km2
The same is the idea behind law of total probability. In which the area of the forest is replaced by the probability of an event A. In a case where the partition is B and B’ , then for any 2 events A and B.
P(A)= P( A ∩ B) + P( A ∩ B’)
By Conditional Probability, P ( A ∩ B ) = P(A | B) P(B)
Therefore, P (A) = P (A | B) P( B) + P(A | B’) P(B’)
Hence, we can state a general version of the formula for Sample Space S as below:
If B1, B2, B3,⋯ is a partition of the sample space S, then for any event A we have P(A)=∑ P(A ∩ Bi) = ∑ P(A | Bi) P(Bi)
As it can be seen from the figure, A1, A2, A3 , A4 and A5 form a partition of the set A, and thus by the third axiom of probability
P(A)=P(A1)+P(A2)+P(A3)+P(A4)+P(A5).
## Examples of Conditional Probability
Q1. Find the probability that a single toss of die will result in a number less than 4 if it is given that the toss resulted in an odd number.
Sol. Let event A: toss resulted in an odd number and event B: number is less than 4
Therefore, A= {1, 3, 5}
P (A) =3/6 = ½
B= {1, 2, 3}
A∩B = {1, 3}
P (A∩B) =2/6=1/3
Therefore, required probability = (P number is less than 4 given that it is odd)
= P (B/A) = P (A∩B)/P (A) =1/3 / ½ =2/3
Q2. The probability that it is Friday and a student is absent is 0.03. Since there are 5 school days in week, the probability that it is Friday is 0.2. What is the probability that a student is absent given that today is Friday.
Sol.
Q3. A jar contains black and white marbles. Two marbles are chosen without replacement. The probability of selecting a black marble and then a white marble is 0.34, and the probability of selecting a black marble on the first draw is 0.47. What is the probability of selecting a white marble on the second draw, given that the first marble was drawn black?
Sol.
Q4. If P (A’) =0.7, P (B) = 0.7, P(B/A)=0.5, find P(A/B) and P(AUB).
SolSince P (A’) =0.7,
P (A) =1-P (A’) =1-0.7=0.3
Now P (B/A) = P (A∩B) / P (A)
0.5= P (A∩B) / 0.3
P (A∩B) =0.15
Again, P (A/B) = P (A∩B) /P (B)
=0.15 / 0.7
P (A/B) = 3/14
P (AUB) = P (A) +P (B) – P (A∩B)
=0.3+0.7-0.15
=0.85
Q1. Take the case of previous illustration in which a bag contains 3 Red, 2 Green and 4 Yellow balls. A ball is drawn and if it is Red or Yellow. it is replaced otherwise not. Find the probability that second ball
drawn is a yellow ball.
Sol. First of all, we should be very clear about the event of which probability is to be found and the events on which it depends
Let's A1: First ball is either red or yellow
A2: First ball is green
A: Second ball drawn is yellow
We have to calculate P(A) which definitely depends on A1 and A2.
Therefore, P(A) = P(A/A1) P(A1) + P(A/A2) P(A2
P(A/A2) = Probability of A when A1 is already happen.
= 4/9 (i.e. probability of A given A1)
P(A1) = 7/9
P(A/A2) = Probability of A when A2 is already happened
= (Probability of A given A2)
= 4/8
P(A2) = 2/9
Hence, P(A) = 4/9×7/9+4/8×2/9 = 37/81
Q2. At ABC Senior Secondary school, the probability that a student takes Science and English is 0.087. The probability that a student takes Science is 0.68. What is the probability that a student takes English
given that the student is taking Science?
Sol. We need to find the probability of English given Science i.e.
P (English | Science) = P (Science and English) / P (Science) = 0.087 / 0.68 = 0.13 = 13%.
Q3. A jar contains red and green toys. Two toys are drawn without replacement. The probability of selecting a red toy and then a green toy is 0.34, and the probability of selecting a red toy on the first draw is 0.47. What is the probability of selecting a green toy in the second draw, given that the first toy drawn was red?
Sol. Clearly it is a question of conditional probability. We need to find the probability of drawing a green toy having known that the first toy drawn was red.
P (green | red) = P (red and green) / P (red) = 0.34 / 0.47 = 0.72 = 72%.
Q4. At ABC Senior Secondary school, the probability that a student takes Science and English is 0.087. The probability that a student takes Science is 0.68. What is the probability that a student takes English given that the student is taking Science?
Sol. We need to find the probability of English given Science i.e.
P (English | Science) = P (Science and English) / P (Science) = 0.087 / 0.68 = 0.13 = 13%.
## Related Resources
Watch this Video for more reference
Conditional Probability
To read more, Buy study materials of Probability comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
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r
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# Division Overview
This is where math starts to get a little tricky. Division is about breaking something into many pieces. You have probably heard the word "divide" all over the place. You divide your candy with your friends. You divide pieces of a turkey at dinner. You start with one amount and wind up with smaller pieces.
As far as words are concerned, we will use three terms in this section. In each division problem, you will have one number divided by another. The number you are dividing is called the dividend. The number you are "dividing by" is the divisor. The answers to your division problems are called quotients. Six divided by two gives you a quotient of three.
Example:
8 ÷ 4 = 2
8 is the dividend.
4 is the divisor.
2 is the quotient.
The division symbol (÷) in this problem is called an obelus.
You may also see it written in this format...
2 4 ) 8
or this one...
8 / 4 = 2
# Just Like Fractions
This section will move right into the fractions area, because fractions are basically a different way of writing a division problem. When you move into more advanced math, you may even write out division problems that look like fractions.
You will learn that the fraction 1/4 has the same value as one (1) divided by four (4). As you work with decimals, you will quickly discover that 1 divided by 4 is 0.25 and the value is known as one quarter. Division is even important to percentages. The decimal 0.25 is the same as saying 25 percent. One-fourth of a pie is 25% of the total pie. All of the values are identical, but you use them in different ways.
That example shows you that there are many ways to say the same thing in math. It's like a language. Many different terms or words describe the same value, but they all mean the same thing. Blue, azul, bleu, blu, and blauw are all different words that describe the same color blue. You can move from one language to another with some translation. You can move from fractions to decimals or percentages with a little conversion.
# Dividing Different Types of Numbers
We have a whole section on numbers for you to learn more. In the division section, we'll look at whole numbers that include the numbers 0 and up. We'll also look at integers and see how you can divide negative and positive numbers.
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# Lesson video
In progress...
Hello everyone, my name is Mrs. Dolin, and I'll be teaching you your maths lesson today and it's lesson number 12 on fractions.
It is now time to review our challenge question from last time.
What number could the arrow be pointing to? Chen thinks that the number is two tenths but Sarah thinks the number is six tenths.
Who do you think is correct? Don't forget them Mrs. Chen has asked you to explain your answer.
Did you do that? Did you get that Chen is actually correct? But the important part is could you explain why? Your explanation might be slightly different to mine.
Sarah can't be correct because six tenths is a larger number on the number line than two tenths.
Six tenths would be further away from zero and closer to 10 tenths or one whole.
Chen's answer of two tenths is a number that is closer to zero than six tenths.
And I can visualise that the arrow is pointing to two tenths.
Therefore Chen is correct.
What do we see here? We've got a bar model and a number line.
What do you notice about them? Well, they've both been divided into five equal parts.
So what would our unit fraction be? Well, it will be one fifth.
We know, that when we count in unit fractions, they make a non-unit fraction.
For example three lots of one fifth, make three fifths.
Let's have a go at counting in our units and non-unit fractions using the bar model and number line to help us.
Remember to start counting from zero.
Are you ready? Zero, one one fifth, two one fifths, three one fifths, four one fifths, five one fifths.
Well done if you were following along at home.
Now let's do the same but let's count up using non-unit fractions.
Remember we're going to start from zero again.
Zero, one fifth, two fifths, three fifths, four fifth, five fifths.
Well done.
Let's do this question together.
We've seen this image before haven't we? What do you notice? Well, the whole has been divided into five equal parts.
So our unit fraction is one fifth.
What can you see now? Well, four parts are shaded.
Should we count them using our unit fractions to help? One one fifth, two one fifths, three one fifths, four one fifths.
Now shall we use our stem sentence to say it together? The whole has been divided into five equal parts.
Four fifths of those parts are shaded.
Now, so we have a go at writing our fraction as a number on a number line.
Our number line has been divided into five equal parts.
So, let's count up the equal parts.
We've got one one fifth, two one fifths, three one fifths, four one fifths and five one fifths.
Now we're going to count on the number line, starting at zero and stop at the number that matches our fraction.
Are you ready? Are you going to stop at the right place? Let's have a go.
Zero, one one fifth, two one fifths, three one fifths, four one fifths, that's it.
Did you stop at the right point? We can write this fraction as a number, as four fifths.
I remember this, do you? We wanted to find out, how tall the plant had grown.
How many equal parts has the metre stick been divided into? Well, let's count.
One, two, three, four, five, six, seven, eight, nine, 10.
The metre stick has been divided into 10 equal parts.
So what is our unit fraction? It's one tenths.
Now let's see how tall the plant has grown.
I wonder what fraction of one metre the plant measures.
Let's count.
One one tenth, two one tenths, three one tenths, four one tenths, five one tenths, six one tenths, seven one tenths.
So let's say our stem sentence together.
The plant measures seven tenths of the whole metre.
Now let's write the fraction as a number on a number nine.
How many equal parts would our number line need? Well, it would be 10 equal parts.
Now let's show our fraction on the number line.
Can you shout stop, when we reach our fraction.
Count up with me and stop when we reach our fraction.
Remember to start from zero.
Zero, one tenth, two tenths, three tenths, four tenths, five tenths, six tenths, seven tenths.
Did you shout, stop? Well done, that's fantastic.
Now let's write our fraction seven tenths.
We're going to continue to practise our counting in unit and non-unit fractions on a shape and on a number line.
So have a look at the shape and have a look at the number line.
How many equal parts has the whole been divided into? That's right, nine equal parts.
So our unit fraction is one ninth.
Let's count up in unit fractions of one ninth.
Off we go.
One one ninth, two one ninth, three one ninths, four one ninths, five one ninths, six one ninths, seven one ninths, eight one ninths, nine one ninths.
Well done.
Now let's count up in non-unit fractions.
Remember to start from zero.
Are you ready? Zero, one ninth, two ninths, three ninths, four ninths, five ninths, six ninths, seven ninths, eight ninths, nine ninths.
Well done for counting along at home.
Can you spot the number, four ninths on your number line? Can you point to where it is? Where would you point to? Well done if you're pointing to this point here on the number line, that shows four ninths.
Let's have a go at counting up and down with our eggs.
How many equal parts has the egg box been divided into? How many equal parts are there on the number line? Well, they both have 12 equal parts.
So our unit fraction is one twelfth.
Let's count up in unit fractions in twelfths.
Count along with me and don't forget to start on zero.
Off we go.
Zero, one one twelfth, two one twelfths, three one twelfths, four one twelfths, five one twelfths, six one twelfths, seven one twelfths, eight one twelfths, nine one twelfths, 10 one twelfths, 11 one twelfths, 12 one twelfths.
Well done for counting along at home.
Now let's count up in non-unit fractions.
Remember to start from zero.
Zero, one twelfth, two twelfths, three twelfths, four twelfths, five twelfths, six twelfths, seven twelfths, eight twelfths, nine twelfths, 10 twelfths, 11 twelfths, 12 twelfths.
Well done.
Can you point to the number, 11 twelfths on the number line? Where would it go? Where is 11 twelfths? Well done if you're pointing to here, 11 twelfths.
Now let's look at our practise activity.
Can you show this fraction using a number line? So, you might have to ask the adult to help you with drawing a number line.
Remember to use a ruler and you could have it that one centimetre equals one equal parts.
When representing this fraction on a number line, what number are you going to have to stop counting up to on the number line? What point will you stop at? Remember it has to be accurate, your arrow can't be spaces in between, it has to be pointing exactly at the fraction that this drawing is showing.
Let's have a look at your next practise activity.
Here is a window.
How many panes of glass can you see? Oh, what happened to the glass in some of the panes? Well, some of them have been broken.
What fraction of the whole window will need new window panes? Can you show this fraction using a number line? Can you write the fraction as well.
I wonder as a tap activity, what fraction of the window pane did not break? How this fraction look different? Can you have a go representing this number on a number line? How would it look different? Thank you for joining us today and see you soon.
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# How to solve 3/4 in a half cup ?
Hello friends,
Welcome to my article 3/4 in a half cup ?
First of all we should write the article on the page of the notebook.
## 3/4 in a half cup ?
Let 3/4 in a half cup =x ,
Then ,
\displaystyle x=\frac{3}{4}\div 2
\displaystyle x=\frac{3}{4}\times \frac{1}{2}
\displaystyle x=\frac{{3\times 1}}{{4\times 2}}
\displaystyle x=\frac{3}{8}
3/4 in a half cup =x =3/8 Answer
Thus, the intended solution of this question =3/8 Answer
## How to solve 3/4 in a half cup?
You can measure it by measuring cups .
Another easy way to measure 3/4 cup is as follows:
Fill a cup with the thing you want to measure. Pour or take out half of it into another cup(this is 1/2 cup,).
Now take out half of what you’re measuring from any of the cups (that’s 1/4 cup).
Mix this one fourth quantity in half quantity now the final quantity in the cup is
3/4 =(1/2 +1/4).
Now to the question,
\displaystyle \frac{3}{4}\text{ }IN\text{ }A\text{ }HALF\text{ }CUP=\frac{3}{4}\div 2
\displaystyle \frac{3}{4}\text{ }IN\text{ }A\text{ }HALF\text{ }CUP=\frac{3}{4}\times \frac{1}{2}
\displaystyle \frac{3}{4}\text{ }IN\text{ }A\text{ }HALF\text{ }CUP=\frac{3}{8}
\displaystyle \frac{3}{4}\text{ }IN\text{ }A\text{ }HALF\text{ }CUP=\frac{3}{8}Answer
How to solve 3/4 in a half cup ? This article has been completely solved by tireless effort from our side, still if any error remains in it then definitely write us your opinion in the comment box. If you like or understand the methods of solving all the questions in this article, then send it to your friends who are in need.
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Solving Oblique Triangles
Oblique triangles, which have no right angle, can be solved using the Law of Sines or the Law of Cosines. These laws allow for the calculation of unknown angles or sides when certain other measurements are known. Mastering these techniques is essential for a comprehensive understanding of trigonometry.
Create learning materials about Solving Oblique Triangles with our free learning app!
• Flashcards, notes, mock-exams and more
• Everything you need to ace your exams
Definition of an Oblique Triangle
An oblique triangle is a type of triangle that does not contain a right angle. In other words, all of its angles are either acute (less than 90 degrees) or obtuse (greater than 90 degrees). There are two types of oblique triangles: acute triangles, where all angles are less than 90 degrees, and obtuse triangles, where one angle is greater than 90 degrees.
Types of Oblique Triangles
Oblique triangles can be further categorised into two distinct types:
• Acute Triangle: All three angles in the triangle are less than 90 degrees.
• Obtuse Triangle: One angle in the triangle is greater than 90 degrees.
Knowing the type of oblique triangle you are dealing with is crucial for selecting the appropriate method for solving it.
An acute triangle is a triangle where all three interior angles are acute, meaning they are each less than 90°.
An obtuse triangle is a triangle where one of the interior angles is an obtuse angle, meaning it is greater than 90°.
Solving Oblique Triangles Using the Law of Sines
The Law of Sines states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is constant for all three sides and angles in any given triangle. The Law of Sines can be written as:
The Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Example: Given a triangle with sides a = 7, b = 9, and angle A = 30°, find angle B:
Using the Law of Sines:
$$\frac{7}{\sin 30\degree} = \frac{9}{\sin B}$$
$$\frac{7}{0.5} = \frac{9}{\sin B}$$
$$14 = \frac{9}{\sin B}$$
$$\sin B = \frac{9}{14}$$
$$B = \sin^{-1}\left(\frac{9}{14}\right)$$
$$B \approx 40.54\degree$$
The Law of Sines can be particularly useful not only for solving unknown angles and sides but also in practical applications such as navigation, astronomy, and engineering. For instance, when locating the position of an object using triangulation, the Law of Sines provides an efficient method for determining distances that are otherwise challenging to measure directly.
Remember, the Law of Sines is best used when you are given a pair of angles with their opposite sides, or a pair of sides with one of the non-included angles.
Solving Oblique Triangles
When dealing with oblique triangles, you will not have a right angle to rely on, so different mathematical rules apply. Specifically, you often use the Law of Sines and the Law of Cosines to find unknown sides or angles.
Law of Sines
The Law of Sines is crucial for solving oblique triangles. It is particularly useful when you know either:
• Two angles and one side (AAS or ASA scenarios)
• Two sides and a non-included angle (SSA scenario)
The Law of Sines can be represented as:
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Consider a triangle with $$a = 8$$, $$A = 45°$$, and $$B = 60°$$, where you need to find $$b$$. Using the Law of Sines:
$$\frac{a}{\sin A} = \frac{b}{\sin B}$$
$$\frac{8}{\sin 45°} = \frac{b}{\sin 60°}$$
$$\frac{8}{0.7071} = \frac{b}{0.866025}$$
Solving for $$b$$:
$$b = 8 \times \frac{0.866025}{0.7071} ≈ 9.8$$
Law of Cosines
The Law of Cosines is another critical tool for solving oblique triangles. It is useful when you know:
• Three sides (SSS scenario)
• Two sides and the included angle (SAS scenario)
The Law of Cosines is expressed by the equation:
$c^2 = a^2 + b^2 - 2ab \cos C$
Consider a triangle where $$a = 5$$, $$b = 7$$, and $$C = 60°$$. You need to find the length of $$c$$.
Using the Law of Cosines:
$$c^2 = a^2 + b^2 - 2ab \cos C$$
$$c^2 = 5^2 + 7^2 - 2 \times 5 \times 7 \times \cos 60°$$
$$c^2 = 25 + 49 - 2 \times 5 \times 7 \times 0.5$$
$$c^2 = 74 - 35$$
$$c^2 = 39$$
$$c = \sqrt{39} ≈ 6.24$$
The Law of Cosines is closely related to the Pythagorean theorem and can be seen as a generalisation of it. When the angle $$C$$ becomes 90°, $$\cos 90° = 0$$, thus reducing the Law of Cosines to the Pythagorean theorem: $$c^2 = a^2 + b^2$$. This shows that the right triangle is a special case within the wider family of oblique triangles.
Always check your given angles. In any triangle, the sum of all internal angles must be 180 degrees.
Solving Oblique Triangles Using Law of Sines and Cosines
When it comes to solving oblique triangles, you won't have a right angle, and thus different rules apply. Specifically, the Law of Sines and the Law of Cosines are used to determine unknown sides or angles.
How to Solve Oblique Triangles with Law of Sines
The Law of Sines is especially useful when you know either:
• Two angles and one side (AAS or ASA scenarios)
• Two sides and a non-included angle (SSA scenario)
The Law of Sines can be written as:
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Example: Consider a triangle where $$a = 8$$, $$A = 45°$$, and $$B = 60°$$, and you need to find side $$b$$. Using the Law of Sines:
$$\frac{8}{\sin 45°} = \frac{b}{\sin 60°}$$$$\frac{8}{0.7071} = \frac{b}{0.866025}$$Solving for $$b$$:$$b = 8 \times \frac{0.866025}{0.7071} ≈ 9.8$$
Always double-check the given angles. The sum of all internal angles in any triangle must be 180 degrees.
How to Solve Oblique Triangles with Law of Cosines
The Law of Cosines is beneficial when you know:
• Three sides (SSS scenario)
• Two sides and the included angle (SAS scenario)
This law is expressed by the equation:
$c^2 = a^2 + b^2 - 2ab \cos C$
Example: Consider a triangle where $$a = 5$$, $$b = 7$$, and $$C = 60°$$. You need to find side $$c$$. Using the Law of Cosines:
$$c^2 = a^2 + b^2 - 2ab \cos C$$$$c^2 = 5^2 + 7^2 - 2 \times 5 \times 7 \times \cos 60°$$$$c^2 = 25 + 49 - 2 \times 5 \times 7 \times 0.5$$$$c^2 = 74 - 35$$$$c^2 = 39$$$$c = \sqrt{39} ≈ 6.24$$
The Law of Cosines can be viewed as a generalisation of the Pythagorean theorem. When the angle $$C$$ becomes 90°, $$\cos 90° = 0$$, reducing the Law of Cosines to $$c^2 = a^2 + b^2$$, which is the Pythagorean theorem. This shows that the right triangle is a special case within the wider family of oblique triangles.
Example Problems for Solving an Oblique Triangle
When solving oblique triangles, you often rely on trigonometric laws such as the Law of Sines and the Law of Cosines. In this section, let's go through a few example problems to illustrate how these laws can be applied.
Example Problem Using the Law of Sines
Consider a triangle where you are given side a = 10, angle A = 50°, and angle B = 60°. You need to find side b.
Using the Law of Sines, we can set up the equation as follows:
$\frac{a}{\sin A} = \frac{b}{\sin B}$
Substitute the known values:
$\frac{10}{\sin 50°} = \frac{b}{\sin 60°}$The sine values for the angles can be calculated or taken from a trigonometric table:
$\frac{10}{0.766} = \frac{b}{0.866}$
Now, solve for b:
$b = 10 \times \frac{0.866}{0.766} \approx 11.31$
Always ensure that the angles given add up to 180 degrees to confirm the validity of your triangle setup.
Example Problem Using the Law of Cosines
Now, consider a triangle where you know sides a = 8, b = 6, and the included angle C = 45°. You need to find the length of side c. Using the Law of Cosines:
$c^2 = a^2 + b^2 - 2ab \, \text{cos} \, C$
Substitute the known values:
$c^2 = 8^2 + 6^2 - 2 \times 8 \times 6 \times \text{cos} \, 45°$
Calculate the cosine value and plug it in:
$c^2 = 64 + 36 - 2 \times 8 \times 6 \times 0.707$
Simplify the equation:
$c^2 = 100 - 67.86$
$c^2 = 32.14$
Finally, take the square root to solve for c:
$c \approx 5.67$
The ability to apply these laws effectively can have numerous practical applications. For instance, in engineering and architecture, solving oblique triangles can help determine the forces acting on various structures. In astronomy, these laws can be used to determine distances between celestial objects. Understanding and mastering these trigonometric laws can make tackling real-world problems much easier.
Solving Oblique Triangles - Key takeaways
• Definition of an Oblique Triangle: An oblique triangle is a triangle that does not contain a right angle; its angles are either acute (less than 90 degrees) or obtuse (greater than 90 degrees).
• Types of Oblique Triangles: There are two types: acute triangles (all angles are less than 90 degrees) and obtuse triangles (one angle is greater than 90 degrees).
• Law of Sines: The ratio of the length of a side of a triangle to the sine of the angle opposite that side is constant. It is used for solving oblique triangles when you know two angles and one side, or two sides and a non-included angle.
• Law of Cosines: Used to solve oblique triangles when you know the lengths of all three sides or two sides and the included angle. The formula is: c^2 = a^2 + b^2 - 2ab \, \cos \, C.
• Practical Applications: Both laws are crucial in fields such as navigation, astronomy, and engineering for determining unknown sides or angles in triangles and are generalisations of the Pythagorean theorem.
Flashcards in Solving Oblique Triangles 12
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What methods can be used to solve an oblique triangle?
The methods used to solve an oblique triangle are the Law of Sines, the Law of Cosines, and the use of trigonometric functions to find unknown angles and sides. Additionally, the area can be determined using Heron's formula or by using the formula involving sine, where Area = 0.5 * a * b * sin(C).
What is the difference between the Law of Sines and the Law of Cosines when solving oblique triangles?
The Law of Sines relates the ratios of the lengths of the sides of a triangle to the sines of its angles and is used when dealing with ASA, AAS, or SSA cases. The Law of Cosines relates the lengths of a triangle's sides to the cosine of one of its angles and is used for SAS or SSS cases.
How do you determine which method to use when solving an oblique triangle?
To determine the method, first identify the known quantities: use the Law of Sines if you have an angle and its opposite side; use the Law of Cosines if you have two sides and the included angle or all three sides.
What is an oblique triangle?
An oblique triangle is a triangle that does not contain a right angle. This means it either has all acute angles (acute triangle) or has one obtuse angle and two acute angles (obtuse triangle).
Can an oblique triangle have all angles less than 90 degrees?
Yes, an oblique triangle can have all angles less than 90 degrees, and such a triangle is called an acute triangle.
Test your knowledge with multiple choice flashcards
What does the Law of Sines state?
When solving oblique triangles, which two laws are primarily used?
What is the mathematical representation of the Law of Sines?
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# REWRITING PERCENT EXPRESSIONS WORKSHEET
Rewriting Percent Expressions Worksheet :
Worksheet given in this section will be much useful for the students who would like to practice solving problems on percentage.
Before look at the worksheet, if you would like to learn the basic stuff on percentage,
## Rewriting Percent Expressions Worksheet - Problems
Problem 1 :
To make a profit, stores mark up the prices on the items they sell. A sports store buys skateboards from a supplier for s dollars. What is the retail price for skateboards that the manager buys for \$35 and \$56 after a 42% markup ?
Problem 2 :
A discount store marks down all of its holiday merchandise by 20% off the regular selling price. Find the discounted price of decorations that regularly sell for \$16 and \$23.
Problem 3 :
A computer store used a markup rate of 40%. Find the selling price of a computer game that cost the retailer \$25.
Problem 4 :
A product that regularly sells for \$425 is marked down to \$318.75. What is the discount rate ?
Problem 5 :
A store uses a 40% markup on cost. Find the cost of a pair of shoes that sells for \$63.
## Rewriting Percent Expressions Worksheet - Solutions
Problem 1 :
To make a profit, stores mark up the prices on the items they sell. A sports store buys skateboards from a supplier for s dollars. What is the retail price for skateboards that the manager buys for \$35 and \$56 after a 42% markup ?
Solution :
Step 1 :
Use a bar model.
Draw a bar for the cost of the skateboard S.
Then draw a bar that shows the markup: 42% of S, or 0.42S.
These bars together represent the cost plus the markup.
That is
S + 0.42S
Step 2 :
Retail price = Original cost + Markup
= S + 0.42S
= 1S + 0.42S
= 1.42S
Step 3 :
Use the expression to find the retail price of each skateboard.
S = \$35 ----> Retail price = 1.42(\$35) = \$49.70
S = \$56 ----> Retail price = 1.42(\$56) = \$79.52
Problem 2 :
A discount store marks down all of its holiday merchandise by 20% off the regular selling price. Find the discounted price of decorations that regularly sell for \$16 and \$23.
Solution :
Step 1 :
Use a bar model.
Draw a bar for the regular price P.
Then draw a bar that shows the discount: 20% of P, or 0.2P.
The difference between these two bars represents the price minus the discount.
That is,
P - 0.2P
Step 2 :
Sale price = Original price - Markdown
= p - 0.2p
= 1p - 0.2p
= 0.8p
Step 3 :
Use the expression to find the sale price of each decoration.
p = \$16 ---> Sale price = 0.8(\$16) = \$12.80
p = \$23 ---> Sale price = 0.8(\$23) = \$18.40
Problem 3 :
A computer store used a markup rate of 40%. Find the selling price of a computer game that cost the retailer \$25.
Solution :
Selling price (S.P) = (100+M)% x C.P
Here, M = 40, C.P = \$25
Then, S.P = (100 + 40)% x 25
S.P = 140% x 25
S.P = 1.4 x 25 = \$35
Hence, the selling price is \$35.
Problem 4 :
A product that regularly sells for \$425 is marked down to \$318.75. What is the discount rate?
Solution :
Regular price = \$ 425
Marked down price = \$ 318.75
Marked down value = 425 - 318.75 = 106.25
Marked down rate = (106.25 / 425) x 100 %
Marked down rate = 25 %
Hence, the marked down rate is 25 %
Problem 5 :
A store uses a 40% markup on cost. Find the cost of a pair of shoes that sells for \$63.
Solution :
Selling price (S.P) = (100+M)% x C.P ---------(1)
Here, S.P = \$ 63, M = 40
Plugging the above values in (1)
(1)-----------> 63 = (100+40)% x C.P
63 = 140% x C.P ---------> 63 = 1.4 x C.P
63/1.4 = C.P ---------> 45 = C.P
Hence, the cost of a pair of shoes is \$ 45.
After having gone through the stuff given above, we hope that the students would have understood, how to solve problems on percent expressions.
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ZingPath: Graphs of Trigonometric Functions
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Graphs of Trigonometric Functions
Learn in a way your textbook can't show you.
Explore the full path to learning Graphs of Trigonometric Functions
Lesson Focus
Graphing Cosine Functions
Algebra-2
Graph the cosine function, and variations of the graph that involve the cosine expression.
Now You Know
After completing this tutorial, you will be able to complete the following:
• Sketch the graph of the cosine function.
• Sketch the graph of a function that involves cosine expression.
Everything You'll Have Covered
The cosine function is periodic.
The cosine function is an example of a periodic function; this means that the function repeats its range values at regular intervals. This is useful in graphing in that, once the learner finds the period of the function and the shape of the function, it will be repeated as the values of x increase without bound.
A cosine function can have several transformations in a single function.
The general form of the cosine function, y = a • cos (bx + c) + d, has several transformations. First, the 'a' in the function stands for the maximum absolute value of a periodic curve measured along the vertical axis (or, in other words, the highest value on the y-axis that the curve reaches). The 'b' helps determine the period of the function (period = ).
The 'c' is the phase shift (whether the graph is shifted to the left or right of its original point), and the'd' is the vertical shift (whether the graph is shifted up or down from the original point.
There can also be a reflection of the cosine function.
If there is a negative sign in front of the cosine function, y = -cos x, the graph will be reflected about the x-axis. The graph will appear to be flipped upside down from its original position.
The unit circle is used to define the cosine function.
The correlation between the unit circle and the graph of the cosine function (from to ), visually establishes the relationship between the specific values of t = 0, , , and corresponding
ordered pairs on the graph of x = cos t, i.e., the cos t = 0 when t = 0, ?, and 2?, the cost = 1 when x = , and cos t = -1 when x = . Underscoring this relationship, for the learner, is specific for understanding and mastery of the guided practice problems provided in Section 3 of the Activity Object.
A 5 step process can be used to help the learner sketch the graph of y = a • cos (bx + c) + d.
For the given function y = 2 cos (4x):
Step 1: Using the formula when the coefficient of x is b, calculate the period of the given function. A fraction and pi button are provided in the Activity Object to help correctly notate results.
Step 2: Calculate cosine of specific values, (bx + c) or in this example, 4x. It should be noted here, that the calculated vales in this step will be the same for all given problems, which are the values for the function cos x ( i.e., the cos x = 0 when
is the absolute maximum, = -1 is the absolute minimum, and cos x, where x = 0, ?, and 2? the zeros of the function.
Step 3: Find the result of entire function according to the cosines of specific values. Calculate a • cos (bx + c) by multiplying the results of Step 2 by a, in this example a = 2.
Step 4: Calculate the values of x for the cosine function.
Step 5: Write the x and y values as ordered pairs to sketch the graph. Take the calculations in Steps 3 (y-coordinate) and 4 (x-coordinate) and format the results as ordered pairs.
To see the graph using the Activity Object, click on the button
These key vocabulary terms will be used throughout this Activity Object:
• cosine - the x-coordinate of an angle on the unit circle.
• co-terminal angles - angles that share a terminal side.
• specific values - input values, from the domain, that produce local maximums or minimums, or zero, e.g., .
• domain - the set of input values for which a function is defined.
• origin - the intersection of the x- and y-axes; the ordered pair of the origin is (0,0).
• period - The length of the fixed interval that a function repeats itself, e.g., the period of
• periodic function - A function that repeats all of its range values at regular intervals.
• phase shift - A graph's horizontal movement to the left or right from the origin.
• pi or ? - a mathematical irrational constant whose value is the ratio of a circle's circumference to its diameter.
• radian - a unit of angular measure equal to .
• range - the set of all output values produced by a function; also, the y-coordinates.
• terminal sides - in an angle, the ray which rotates and determines the measure of the angle in standard position.
• trigonometric - Refers to the basic functions used in trigonometry, which are used to relate the angles to the lengths of the sides of a right triangle.
• unit circle - Used to define the trigonometric ratios of the angles. It has a radius of 1 unit and its center is at the origin.
• vertical shift - A graph's movement up or down from the origin.
Tutorial Details
Approximate Time 35 Minutes Pre-requisite Concepts cosine, domain and range, graphs of functions, period Course Algebra-2 Type of Tutorial Procedural Development Key Vocabulary cosine, graph, graphing cosine functions
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Hard ACT Math Problems
If you’re wondering why hard ACT Math problems are so difficult, know that it’s not because they test crazy advanced topics like multivariable calculus or anything like that. Instead, it’s because they take some of the foundations of math topics you’ve already studied in school, like pre-algebra, algebra, and geometry, and turn them into multi-step processes that may combine concepts from different areas.
Does this make them challenging? Yep. Does it mean you can’t solve them? Absolutely not! To master these problems, you’ll need to refresh your understanding of the concepts, then put them into practice.
Think you’re ready to put your skills to the test? Check out the twelve ACT math challenge problems we’ve put together for you below. Answers and explanations follow!
Want to make sure you’ve mastered the trickiest concepts in all areas? For ACT challenge problems from other ACT sections, check out:
1. Let x and y be nonzero real numbers such that 2(y+1)=2x. Which of the following expresses 2(y+2)in terms of x?
A. 1/(2x3)
B. 1/(2x+3)
C. 4x
D. 8x
E. 2x3
2. As shown in the figure below, A is the center of the circle, and right triangle ABC intersects the circle at D and E. Point D is the midpoint of AC, which is 22 cm long. The shaded region inside the circle and outside the triangle has an area of square centimeters. What is the measure of angle B?
A. 45°
B. 48°
C. 50°
D. 54°
E. 57°
3. A box contains 50 cards. is written on the first card, on the second, on the third and so on through , with no numbers repeated. A card is drawn at random from the box. What is the probability that the number on the card is an irrational number?
A. 0/50
B. 7/50
C. 25/50
D. 35/50
E. 43/50
4. Which of the following is an equation of the largest circle that can be inscribed in the ellipse with the following equation?
A. (x-2)2 + (y+4)2 = 81
B. (x-2)2 + (y+4)2 = 16
C. (x-2)2 + (y+4)2 = 9
D. x2 + y2 = 16
E. x2 + y2 = 9
5. What would s have to be so that is divisible by (x + 2)?
A. 9
B. 5
C. 2
D.-6
E.-13
6. In a geometric sequence in which all of the terms are positive, the second term is 2 and the fourth term is 10. What is the value of the seventh term in the sequence?
A. 10√5
B. 20
C. 50
D. 50√5
E. 250
7. Suzanne drove 40 miles to see her aunt and was going 20 mph. It took her 2 hours to get to her aunt’s house. Then, she left and drove another 30 miles to the pet store, but this time only drove at 10 mph. If it took Suzanne 3 hours to arrive at the pet store, what was her average speed in miles per hour for the entire car ride from her home to the pet store?
A. 10
B. 11
C. 12
D. 14
E. 15
8. Students at Thomas Jefferson High School boarded the bus for a field trip that went 15mph through a 30 mile section of the city. The bus then stopped for lunch in a suburb before continuing on a 3 hour tour of countryside at a constant speed of 10mph. Finally, the bus drove 40 miles straight back to the high school. If the students arrived back at Thomas Jefferson High School two hours later, approximately what was the average speed for the entire field trip?
A. 11
B. 12
C. 13
D. 14
E. 15
9. If sin = – , which of the following could be true about ?
A. 0 < < /3
B. /4 < < /2
C. /3 < <
D. /6 < < /6
E. /3 < <
10. The graph below shows a function graphed in the standard (, ) plane. Which of the following could be the equation of the function?
A.
B.
C.
D.
E. /
11. The function m(n) is defined by . Which of the following is the value of m(m(n))?
A. -n9
B. -n6
C. n3
D. n6
E. n9
12. The function values for p(x) vary directly as x for all real numbers. Which of the following best describes the graph of y = p(x) in the standard (x,y) coordinate plane?
F. A line with a y-intercept of 0.
G. A line with a y-intercept not at 0.
H. A line with no y-intercept.
J. A hyperbola.
K. Neither a line nor a hyperbola.
There are two ways to solve this problem: the “math” way and the “test prep strategy” way.
Let’s talk about the test prep strategy way first. You see, this problem is a great candidate for plugging in numbers for x and y and seeing what works.
Let’s try y = 3.
Therefore 2y+1=2x becomes
23+1=2x
24=2x
16=2x
8=x
So now we have a corresponding value for x.
When y=3, the expression in the question 2y+2 equals 32. So now we have the numerical answer we are looking for in the answer choices, expressed in terms of x (which we determined equals 8.
So, plugging in 8 for x, the answer choice equivalents would be:
A. 1/1024
B. 1/19
C. 32
D. 64
E. 1024
Now let’s talk about the math way:
Using the exponent rule xm xn=xm+n, we know that
2y+1=2y 21
so we can simplify the given equation
2y+1=2x
2y 21=2x
2y=x
2y+2 would then be
2y 22
4(2y)
Since we found that x=2y we can sub in x for2y in the expression above and get our answer: C:
4x.
The ACT rarely gives you any unnecessary information in a math word problem. This means that all of the details in the question give you important clues that you need to solve the problem. Here’s another important tip: Whenever you are dealing with circles, and you aren’t given the radius, your first step should be to find the radius. The radius is the key to unlocking other important circle things, such as the area, circumference, sector area, or arc length.
In the case of this question, we know that AC is 22 cm, that D is the midpoint of AC, and that D serves as a point of intersection between the circle and the triangle. This means that AD is the radius and should be half of 22 cm, or 11 cm. Knowing that 11 cm is the radius allows us to find the area of the entire circle using the equation .
So:
We are told in the problem that the area of the shaded region is . This means that the area of the unshaded sector of the circle inside the triangle must be , since these two regions must add up to the total area of the circle: .
This information helps us find the fraction of the circle delimited by the triangle. Because a sector is a fraction of a circle, we can use the proportion of the area of the sector to the area of entire circle to find the degree measure of the central angle.
Because every circle has 360 degrees:
=
Solving this proportion to find angle A gives us x = 32.727272 repeating, or approximately 33°.
Since we know one of the other angles of the triangle is 90°, we can find the measure of the remaining angle, angle B, by subtracting the two known angles from 180° (since the angles in a triangle always add up to 180°).
180 – 90 – 33 = 57
So our answer is (E) 57°.
And for some bonus fun (particularly if you got tripped up on finding the area of the sector), check out this cool animated video we made on this topic!
First of all, we need to remember what rational and irrational numbers are. Rational numbers can be expressed as a fraction; irrational numbers cannot.
Now let’s think through the question. Do you really have to go through every number from 1 to 50? The ACT will never expect you to do that, so there must be a better way.
Let’s take a look at the pattern: √1, √2, √3, √4. As soon as we get to √4, we have something that can be simplified to an integer. √4 = 2. √4 is a perfect square, and guess what? The perfect squares are the only square roots that are rational numbers. Something like √5 gets really messy because there are not two equal fractions that can be multiplied together to equal 5.
So we just need to count the perfect squares before 50. There are seven: 1, 4, 9, 16, 25, 36, 49. So that means out of our 50 numbers, 7 can be reduced to integers (or fractions) and 43 are irrational. So our answer is E.
Bonus hint: If you come across a problem like this on the ACT, and don’t know how to solve it, make sure you eliminate some answer choices! The moment you find an irrational number as you count through the series, you can eliminate answer choice A, for example. And this happens really quickly: √2 is an irrational number. You can also take a logical guess if you just work through the first 11 numbers. You should find that out of the first 11, 8 numbers are irrational. This definitely eliminates B, and thinking logically, you could probably take a guess that the overall probability is going to be high.
This question is difficult primarily because you need to have some higher-level knowledge of the equations of circles and ellipses. If you just studied these in school, then it wouldn’t be nearly as tough. But because equations of ellipses and circles are fair game for higher-level questions on the ACT, we thought we’d feature one here in our Challenge series!
So let’s get started with the equation for an ellipse:
+ = 1
+ = 1
If an ellipse is wider than it is tall, our equation looks like the first one. If it is taller than it is wide, it looks like the second one. The a always goes with the variable whose axis parallels the wider direction of the ellipse, and the b always goes with the variable whose axis parallels the narrower direction, hence the reason for the difference in the equations.
In our problem, we have an ellipse that is taller than it is wide. What’s important is that you visualize what an ellipse like this looks like:
If we put a circle inside this shape, it can only have a diameter that is as wide as the ellipse is wide, otherwise it wouldn’t fit.
Okay! Back to the ellipse equation:
h and k tell us where the center of the ellipse is, so our center is (2,-4). a and b tell us how many units away from the center each vertex is. Since a is 3 and b is 4, we know that our ellipse is 6 units wide and 8 units tall.
If we put a circle in this ellipse, this means it cannot have a radius greater than 3.
The equation of a circle is:
(x-h)2 + (y-k)2 = r2
So we need to see 32, or 9, in our answer choice for the radius of the circle, making our answer C.
Probably the most elegant way to solve this problem is to remember the Factor Theorem. This is a useful trick for problems like this on the ACT. The Factor Theorem states that a polynomial P(x) is divisible by binomial (x – c) if and only if P(c) = 0. In order for a polynomial to be divisible by a linear binomial, the polynomial and the binomial must have the same root.
If we want polynomial P(x) to be divisible by (x + 2), it must be true that P(– 2) = 0. In other words, x = -2 must be a root of the equation.
P(x) = x3 + 5x2 + sx + 6
P(-2) = -23 + 5(-2)2 + s(-2) + 6 = 0
-8 + 5*4 – 2s + 6 = 0
-8 + 20 + 6 = 2s
18 = 2s
9 = s
In a geometric series, each term is the product of previous term times some common ratio. We multiply by the common ratio to change any one term into its successor.
Here, we don’t know the common ratio, so let it be r.
a2=2
a3=2r
a4=2r2
But we know the numerical value of the fourth term.
2r2=10
r2=5
r=√5
Because every term is positive, we don’t have to consider the negative square root. Now that we have the common ratio, we can move from term to term. Each time, we multiply by the square root of 5.
a4=10
a5=10√5
a6=(10√5)=10*5=50
a7=50√5
Average Speed = Total Distance / Total Time.
40 miles + 30 miles so the Total Distance was 70 miles. Suzanne drove for 2 hours + 3 hours so the Total Time was 5 hours. 70/5 = 14.
The Average Speed for the whole trip was 14 mph. The correct answer is (D).
Notice how the test-maker has made this problem tricky! The average speed in this problem is 14 mph, which is different from simply taking the mean of the two speeds. If we had just averaged the two speeds (10mph and 20mph) we would have gotten 15mph. Average Speed is a weighted average. Since Suzanne spent more time in the problem going 10mph than 20mph, it makes sense that the Average Speed would be closer to 10mph.
To find the average speed of the bus, we know we will need to find the Total Distance and the Total Time, so we can start by using another formula (Distance = Rate x Time) to help us find the pieces we’re missing for each part of the trip.
For the first part of the field trip: 30 miles = 15mph x T, so we know that T = 2 hours. For the middle part of the trip, we know that D = 10mph x 3 hours, so we know that D = 30 miles. For the last part of the trip, we know that 40 miles = R x 2 hours, so we know that R = 20mph. Now we can find the Total Distance and the Total Time:
Total Distance = 30 miles + 30 miles + 40miles = 100 miles.
Total Time = 2 hours + 3 hours + 2 hours = 7 hours.
The Average Speed = 100 miles/ 7 hours = 14.28mph. Since the question used the word “approximately,” we can round to the nearest integer: 14. The correct answer is (D).
This is a hard problem. First of all, notice that the angle must be in QIII or QIV, since the sine is negative. It has to be a value on the line = – . The first three choices are all in QI and QII, so the angle can’t possibly be in any of those. We can easily eliminate (A) & (B) & (C).
Let’s think about the last two. First, option (D). This option is in QII and QIII, so the part in QIII might go down low enough to contain the angle. The lowest ray in that region is at .
Technically, this ray isn’t even included, because it’s the endpoint of an inequality, but we will just use this value. Sine is positive in QII, zero at the negative x-axis, and starts to get more and more negative as we go around through QIII. How low does this go by this last value?
= – =
This does not go down as far as
Here’s a diagram with region (D) and the line = – .
So, we can eliminate (D).
This leaves (E) by the process of elimination, but let’s verify that this works. Region (E) is in QIV. The sine is negative one at the bottom, where the unit circle intersects the negative y-axis. As we rise from this bottom, we get negative values of smaller and smaller absolute value, until it equals zero at the positive x-axis. We have to know how low the lowest values in this region are. Well, the limiting value is .
We know that = –
Now, if you happen to have the decimal approximations memorized, you will see that:
= –
If you don’t have this decimal equivalent memorized, you could think about the triangles in QIV:
Even if we can’t directly compare the sizes of the two vertical legs, we definitely can compare the two horizontal legs, and 0.8 is definitely bigger than 0.5. That means the purple {0.6, .8, 1} triangle is further around in the counterclockwise direction from the green 30-60-90 triangle, and the point with a sine of – 0.6 would clearly be included in the arc from the 30-60-90 triangle up to the positive x-axis.
The graph clearly has a hole, a point discontinuity, at x = 3. This means that the function cannot be defined at x = 3. It must have (x – 3) in the denominator. On this basis, we can eliminate (A), (C), and (E).
Notice that the graph is identical to the line y = 2x – 3 at every point except for the point discontinuity at x = 3. Also, to get a point discontinuity instead of an asymptote, we need to have a factor of (x – 3) in both the numerator and the denominator. Choice (B) at x = 3 has a zero denominator and a non-zero numerator, which is the condition for a vertical asymptote. We have eliminated the other four answers, so it would seem that (D) is the answer, but let’s verify this.
In order to get a graph that is identical to y = 2x – 3 at every point except for x = 3 and has a point discontinuity at x = 3, we would have to multiply (2x – 3) by (x – 3) over itself.
So everything checks out and our answer is (D)!
When we have functions inside functions, we call these “nested” functions – it’s important to always start from the innermost parentheses and work your way out. The expression m(m(n)) starts with an inntermost “m(n)” that we know equals -n3, so let’s start by substituting that in. Now the question is asking, what is the value of m(-n3). Just like we normally do for any f(x) function, we will plug whatever in inside the parentheses in for the variable in the function. Here, the testmaker is trying to confuse you by using the same variable “n,” but we’re too smart for the ACT’s tricks! 🙂
m(-n3) = -(-n3)3
According to the order of operations, or PEMDAS, we’d start with what is inside the parentheses. Since we don’t know the value of “n” we cannot simplify -n3 any further. Next, we simplify the exponents. Here’s where knowing your exponent rules really come in handy! When two exponents are separated by a parenthesis, we must always multiply them.
m(-n3) = -(-n9)
Remember that in number properties, any negative number raised to an odd exponent will stay negative, since it takes a pair of negatives to cancel each other out and become positive. -n9 will stay negative until it is multiplied by the -1 in front of the parentheses. Finally, we arrive at our answer:
m(-n3) = n9
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# Yesterday’s closing prices of 2,420 different stocks…
## Solution:
Although this problem may seem wordy it is actually a pretty basic word problem. We are asked to determine the number of stocks that closed at a higher price today than yesterday.
Initially we are given that there are 2,420 different stocks. Another way to look at this is the TOTAL NUMBER of stocks is 2,420. We are also given that the number of stocks that closed at a higher price today than yesterday was 20% greater than the number that closed at a lower price. Let’s create two variables and then two equations using those variables. So we can say:
H = number of stocks that closed at a higher price today than yesterday
L = number of stocks that closed at a lower price today than yesterday
With these two variables we can create two equations:
1) H + L = 2,420
2) H = 1.2L
We can now substitute 1.2L, from equation 2, for H, in equation 1. So we have:
1.2L + L = 2,420
2.2L = 2,420
22L = 24,200
L = 24,200/22
L = 1,100
Since we are solving for variable H, we can plug 1,100 for L, in equation 1.
H + 1,100 = 2,420
H = 1,320
Note: Be careful of the trap in the answer choices. Notice that 1,100 was actually an answer choice. Arguably, it was put there for you to jump on after getting 1,100 for variable L. Partial answers and answers that represent variables other than the one you are being asked about are common trap GMAT answers.
Also, if you were running out of time and had to use a guessing strategy, there is an interesting pattern to look at as far as these answer choices go. The method I’ll describe can be used in many cases in which we are given a TOTAL value in a word problem. With this problem here, we are first given that the total number of stocks in question is 2,420. Well, if we take a look at our answer choices, we will find two sets of answers that, when summed, equal 2,420.
a. 484
b. 726
c. 1100
d. 1320
e. 1694
Notice that:
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# Warm Up I can simplify expressions with exponents. 1. What is the value of 3x 3 +2 when x=10? 2. You put \$500 in an account that doubles every year.
## Presentation on theme: "Warm Up I can simplify expressions with exponents. 1. What is the value of 3x 3 +2 when x=10? 2. You put \$500 in an account that doubles every year. "— Presentation transcript:
Warm Up I can simplify expressions with exponents. 1. What is the value of 3x 3 +2 when x=10? 2. You put \$500 in an account that doubles every year. What is the equation for this situation? How much money would you have after 4 years?
What an Exponent Represents An exponent tells how many times a number is multiplied by itself. 3 4 = 3 3 3 3 = 81
How do you say this? three to the third power or three cubed 3 3 Base Exponent
three to the 2 nd power Or three squared 3 2 Base Exponent How do you say this?
2. 8 8 8 8 = Practice: Write in many different ways 1. 9 9 = 3. x x x x x =
Exponent Rules
Part 1 Product Rule: Multiply same base, add the exponents Example:
Part 2 Quotient Rule: Divide same base, subtract the exponents Example:
Part 3 Power Rule: Same base, multiply the exponents Example:
Part 4 Expanded Power Rule: Apply outside exponent to all parts inside parenthesis Example:
Part 5 Expanded Power Rule (cont): Example: Apply outside exponent to all parts inside parenthesis
Part 6 Zero Exponent Rule: Any base raised to 0 equals 1. Example:
NOTES
Negative Exponents Negative exponents tell you how many times you should divide by a number Negative exponents move the base to bottom of a fraction Note: the exponent becomes positive
Negative Exponents
Partners Find a partner with a different letter Compare and check each others answers
Worksheet Complete any 10 problems
What’s the difference?
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# Trapezoids
In the isosceles trapezoid ABCD we know: AB||CD, |CD| = c = 8 cm, height h = 7 cm, |∠CAB| = 35°. Find the area of the trapezoid.
A = 69.9793 cm2
### Step-by-step explanation:
Did you find an error or inaccuracy? Feel free to write us. Thank you!
Tips to related online calculators
Most natural application of trigonometry and trigonometric functions is a calculation of the triangles. Common and less common calculations of different types of triangles offers our triangle calculator. Word trigonometry comes from Greek and literally means triangle calculation.
#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
## Related math problems and questions:
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Isosceles trapezoid ABCD, AB||CD is given by |CD| = c = 12 cm, height v = 16 cm and |CAB| = 20°. Calculate area of the trapezoid.
• The bases
The bases of the isosceles trapezoid ABCD have lengths of 10 cm and 6 cm. Its arms form an angle α = 50˚ with a longer base. Calculate the circumference and content of the ABCD trapezoid.
• Calculate
Calculate the area of triangle ABC, if given by alpha = 49°, beta = 31°, and the height on the c side is 9cm.
• Isosceles trapezoid
Find the height in an isosceles trapezoid if the area is 520 cm2 and the base a = 25 cm and c = 14 cm. Calculate the interior angles of the trapezoid.
• Parallelogram
Find the parallelogram's perimeter, where base a = 8 cm, height v = 3 cm, and angle alpha = 35° is the magnitude of the angle at vertex A.
• Diagonal BD
Find the length of the diagonal BD in a rectangular trapezoid ABCD with a right angle at vertex A when/AD / = 8,1 cm and the angle DBA is 42°
• Ratio in trapezium
The height v and the base a, c in the trapezoid ABCD are in the ratio 1: 6: 3, its content S = 324 square cm. Peak angle B = 35 degrees. Determine the perimeter of the trapezoid
• Isosceles
Isosceles trapezium ABCD ABC = 12 angle ABC = 40 ° b=6. Calculate the circumference and area.
• Parallelogram ABCD
We have the parallelogram ABCD, where AB is 6.2 cm BC is 5.4 cm AC is 4.8 cm calculate the height on the AB side and the angle DAB
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A trapezoid with a base length of a = 36.6 cm, with angles α = 60°, β = 48° and the height of the trapezoid is 20 cm. Calculate the lengths of the other sides of the trapezoid.
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ABCD is a kite. Angle OBC = 20° and angle OCD = 35°. O is the intersection of diagonals. Find angle ABC, angle ADC and angle BAD.
• Diamond and angles
Find the area of a diamond with a side of 5 cm if you know that the internal angles in the diamond are 60° and 120°.
• Ditch
Ditch profile is an isosceles trapezoid with bases of length 80m and 60m. The slope of the side wall of the ditch is 80°. Calculate the ditch depth.
• TV tower
Calculate the height of the television tower if an observer standing 430 m from the base of the tower sees the peak at an altitude angle of 23°?
• Parallelogram +ľ
| AB | = 76cm, | BC | = 44cm, angle BAD = 30 °. Find the area of the parallelogram.
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Find the area of an isosceles trapezoid if the lengths of its bases are 16 cm and 30 cm, and the diagonals are perpendicular to each other.
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SSAT Upper Level Math : Whole and Part
Example Questions
← Previous 1 3
Example Question #1 : How To Find The Whole From The Part
After a discount, a car costs . How much did the car cost before the discount?
Explanation:
Let be the cost of the car before the discount. Since we know that the car at a discount costs , we can write the following equation:
Now, solve for to find the cost of the car before the discount.
The car originally cost before the discount was applied.
Example Question #2 : How To Find The Whole From The Part
There are yellow marbles in a bag. If of the marbles in the bag are yellow, how many total marbles are in the bag?
Explanation:
Let be the total number of marbles in the bag. Since we know that of the marbles are yellow, we can set up the following equation and solve for :
There are total marbles in the bag.
Example Question #1 : Whole And Part
Byron eats calories for lunch. If he eats of his calories at lunch, what is the total number of calories he eats for the day?
Explanation:
Let be the total number of calories Byron eats in a day. With the information given in the question, we can write the following equation and solve for .
Byron eats calories for the day described in the question.
Example Question #4 : How To Find The Whole From The Part
Max bought a t-shirt that was on sale at the store for . If he paid for the shirt, how much would the shirt have costed before the sale?
Explanation:
Let be the cost of the t-shirt before it went on sale. With the information given in the question, we can write the following equation and solve for .
This can be rewritten as:
The shirt cost before the sale.
Example Question #5 : How To Find The Whole From The Part
Roselyn spent when she bought a vacuum that was off. How much did the vacuum cost before the sale?
Explanation:
Let be the cost of the vacuum before the sale. With the information given in the question, we can write the following equation and solve for .
This can be rewritten as:
The vacuum cost before it went on sale.
Example Question #6 : How To Find The Whole From The Part
When Teresa saw that plane tickets to Hong Kong were off, she immediately bought a discounted ticket for . How much would the ticket have cost if tickets to Hong Kong were not on sale?
Explanation:
Let be the price of the non-discounted ticket. With the information given in the question, we can write the following equation and solve for .
The plane ticket that Teresa bought would have cost if it were not on sale.
Example Question #7 : How To Find The Whole From The Part
There are red marbles in a bag. If of the marbles in the bag are red marbles, how many total marbles are in the bag?
Explanation:
Let be the total number of marbles in the bag. With the information given in the question, we can write the following equation and solve for .
There are total marbles in the bag.
Example Question #8 : How To Find The Whole From The Part
If of a certain number is , what is that number?
Explanation:
Let be the number we are looking for. With the information given in the question, we can write the following equation and solve for .
of is , so is the correct answer.
Example Question #9 : How To Find The Whole From The Part
If of a certain number is , what is that number?
Explanation:
Let be the number we are looking for. With the information given in the question, we can write the following equation and solve for .
of is , so is the correct answer.
Example Question #10 : How To Find The Whole From The Part
If of a certain number is , what is that number?
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# Question: What Is The Median Of 30?
## What is the median of 6 numbers?
To find the median of a group of numbers: Arrange the numbers in order by size.
If there is an odd number of terms, the median is the center term.
If there is an even number of terms, add the two middle terms and divide by 2..
## What is in the middle of 50 and 75?
In the case of an even number of terms, the median is the average of the two middle terms. Add 50 and 75 . Convert the median 1252 to decimal.
## How do you calculate mode?
To find the mode, or modal value, it is best to put the numbers in order. Then count how many of each number. A number that appears most often is the mode.
## What is mean median and mode?
The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. The median is the middle value when a data set is ordered from least to greatest. The mode is the number that occurs most often in a data set. Created by Sal Khan.
## What is a median in math?
Median: The middle number; found by ordering all data points and picking out the one in the middle (or if there are two middle numbers, taking the mean of those two numbers). Example: The median of 4, 1, and 7 is 4 because when the numbers are put in order (1 , 4, 7) , the number 4 is in the middle.
## How do I calculate the median?
MedianArrange your numbers in numerical order.Count how many numbers you have.If you have an odd number, divide by 2 and round up to get the position of the median number.If you have an even number, divide by 2. Go to the number in that position and average it with the number in the next higher position to get the median.
## What is the median of 25?
Explanation: The median is the middle number, for an even set of numbers there will be two middle numbers, to find the median in an even set of numbers we can average the two middle numbers. So since 25 and 25 are both middle numbers we can average them to get the median, which is 25 .
## What is the median of 10 numbers?
The median of a set of numbers is the middle number in the set (after the numbers have been arranged from least to greatest) — or, if there are an even number of data, the median is the average of the middle two numbers.
## How do you find the median of 11 numbers?
The “middle” of a sorted list of numbers. To find the Median, place the numbers in value order and find the middle number. Example: find the Median of {13, 23, 11, 16, 15, 10, 26}. The middle number is 15, so the median is 15.
## What is the median of 23?
In this example the middle numbers are 21 and 23. So the Median in this example is 22.
## How do you find the median quickly?
To find the median, put all numbers into ascending order and work into the middle by crossing off numbers at each end. If there are a lot of items of data, add 1 to the number of items of data and then divide by 2 to find which item of data will be the median.
## What is median number?
The median is the middle number in a sorted, ascending or descending, list of numbers and can be more descriptive of that data set than the average. … If there is an odd amount of numbers, the median value is the number that is in the middle, with the same amount of numbers below and above.
## What if there is no mode?
It is possible for a set of data values to have more than one mode. If there are two data values that occur most frequently, we say that the set of data values is bimodal. If there is no data value or data values that occur most frequently, we say that the set of data values has no mode.
## What is the median of 20?
Mean vs Median The median is found by ordering the set from lowest to highest and finding the exact middle. The median is just the middle number: 20.
## What is the median of 77 and 84?
80.580.5 is an average of two given numbers 77 & 84.
## How do you find the median of 50 numbers?
If there is an even number of numbers add the two middles and divide by 2. The result will be the median.
## What is the median of 74?
Using the formula for median when there is an even number of values, we need to take the mean value of the n/2 ‘th and (n+2)/2 ‘th values. So that’s the 8th and the 9th values, which are 71 and 74 respectively. Then we need to take the mean of these values: (71 + 74) / 2 = 145 / 2 = 72.5 . So the median is 72.5.
## What is the median of 1 and 5?
For the data set 1, 1, 2, 5, 6, 6, 9 the median is 5.
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# 7.1: Sums of Discrete Random Variables
[ "article:topic" ]
In this chapter we turn to the important question of determining the distribution of a sum of independent random variables in terms of the distributions of the individual constituents. In this section we consider only sums of discrete random variables, reserving the case of continuous random variables for the next section.
We consider here only random variables whose values are integers. Their distribution functions are then defined on these integers. We shall find it convenient to assume here that these distribution functions are defined for all integers, by defining them to be 0 where they are not otherwise defined.
###### Convolutions
Suppose and Y are two independent discrete random variables with distribution functions m1(x) and m2(x). Let Z = X + Y. We would like to determine the distribution function m3(x) of Z. To do this, it is enough to determine the probability that Z takes on the value z, where z is an arbitrary integer. Suppose that X = k, where k is some integer. Then Z = z if and only if Y = z − k. So the event Z = z is the union of the pairwise disjoint events.
$(X=k) \text{ and } (Y= z - k)$
where k runs over the integers. Since these events are pairwise disjoint, we have
$P(Z=z) = \sum_{k=-\infty}^\infty P(X=k) \cdot P(Y=z-k)$
Thus, we have found the distribution function of the random variable Z. This leads to the following definition.
Definition
Let X and Y be two independent integer-valued random variables, with distribution functions m1(x) and m2(x) respectively. Then the convolution of $$m_1(x)$$ and $$m_2(x)$$ is the distribution function $$m_3 = m_1 * m_2$$ given by
$m_3(j) = \sum_k m_1(k) \cdot m_2(j-k)$,
for j = . . . , −2, −1, 0, 1, 2, . . .. The function m3(x) is the distribution function of the random variable Z = X + Y.
It is easy to see that the convolution operation is commutative, and it is straightforward to show that it is also associative.
Now let $$S_n = X_1 + X_2 + . . . + X_n$$ be the sum of n independent random variables of an independent trials process with common distribution function m defined on the integers. Then the distribution function of S1 is m. We can write
$S_n = S_{n-1} + X_n \) Thus, since we know the distribution function of $$X_n$$ is m, we can find the distribution function of $$S_n$$ by induction. ###### Example 7.1 A die is rolled twice. Let $$X_1$$ and $$X_2$$ be the outcomes, and let $$S_2 = X_1 + X_2$$ be the sum of these outcomes. The $$X_1$$ and $$X_2$$ have the common distribution function: \[ m = \bigg( \begin{array}{}1 & 2 & 3 & 4 & 5 & 6 \\ 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \end{array} \bigg)$.
The distribution function of $$S_2$$ is then the convolution of this distribution with itself. Thus,
$\begin{array}{} P(S_2 =2) & = & m(1)m(1) \\ & = & \frac{1}{6}\cdot\frac{1}{6} = \frac{1}{36} \\ P(S_2 =3) & = & m(1)m(2) + m(2)m(1) \\ & = & \frac{1}{6}\cdot\frac{1}{6} + \frac{1}{6}\cdot\frac{1}{6} = \frac{2}{36} \\ P(S_2 =4) & = & m(1)m(3) + m(2)m(2) + m(3)m(1) \\ & = & \frac{1}{6}\cdot\frac{1}{6} + \frac{1}{6}\cdot\frac{1}{6} + \frac{1}{6}\cdot\frac{1}{6} = \frac{3}{36}\end{array}$
Continuing in this way we would find $$P(S_2 = 5) = 4/36, P(S_2 = 6) = 5/36, P(S_2 = 7) = 6/36, P(S_2 = 8) = 5/36, P(S_2 = 9) = 4/36, P(S_2 = 10) = 3/36, P(S_2 = 11) = 2/36,$$ and $$P(S_2 = 12) = 1/36$$. The distribution for S3 would then be the convolution of the distribution for $$S_2$$ with the distribution for $$X_3$$. Thus $$P(S_3 = 3) = P(S_2 = 2)P(X_3 = 1)$$.
and so forth.
This is clearly a tedious job, and a program should be written to carry out this calculation. To do this we first write a program to form the convolution of two densities p and q and return the density r. We can then write a program to find the density for the sum Sn of n independent random variables with a common density p, at least in the case that the random variables have a finite number of possible values.
Running this program for the example of rolling a die n times for n = 10, 20, 30 results in the distributions shown in Figure 7.1. We see that, as in the case of Bernoulli trials, the distributions become bell-shaped. We shall discuss in Chapter 9 a very general theorem called the Central Limit Theorem that will explain this phenomenon.
###### Example 7.2
A well-known method for evaluating a bridge hand is: an ace is assigned a value of 4, a king 3, a queen 2, and a jack 1. All other cards are assigned a value of 0. The point count of the hand is then the sum of the values of the cards in the hand. (It is actually more complicated than this, taking into account voids in suits, and so forth, but we consider here this simplified form of the point count.) If a card is dealt at random to a player, then the point count for this card has distribution
$p_x = \bigg( \begin{array}{} 0&1 & 2 & 3 & 4 \\ 36/52 & 4/52 & 4/52 & 4/52 & 4/52 \end{array} \bigg)$.
Let us regard the total hand of 13 cards as 13 independent trials with this common distribution. (Again this is not quite correct because we assume here that we are always choosing a card from a full deck.) Then the distribution for the point count C for the hand can be found from the program NFoldConvolution by using the distribution for a single card and choosing n = 13. A player with a point count of 13 or more is said to have an opening bid. The probability of having an opening bid is then
$P(C \geq 13)$.
Since we have the distribution of C, it is easy to compute this probability. Doing this we find that
$P(C \geq 13) = .2485,$
so that about one in four hands should be an opening bid according to this simplified model. A more realistic discussion of this problem can be found in Epstein, The Theory of Gambling and Statistical Logic.$$^1$$
For certain special distributions it is possible to find an expression for the distribution that results from convoluting the distribution with itself n times. The convolution of two binomial distributions, one with parameters m and p and the other with parameters n and p, is a binomial distribution with parameters $$(m + n)$$ and $$p$$. This fact follows easily from a consideration of the experiment which consists of first tossing a coin m times, and then tossing it n more times.
The convolution of k geometric distributions with common parameter p is a negative binomial distribution with parameters p and k. This can be seen by considering the experiment which consists of tossing a coin until the kth head appears.
##### Exercises
###### $$\PageIndex{1}$$
A die is rolled three times. Find the probability that the sum of the outcomes is (a) greater than 9 (b) an odd number.
###### $$\PageIndex{2}$$
The price of a stock on a given trading day changes according to the distribution
$p_X = \bigg( \begin{array}{} -1 & 0 & 1 & 2 \\ 1/4 & 1/2 & 1/8 & 1/8 \end{array} \bigg)$.
Find the distribution for change in stock price after two (independent) trading days.
###### $$\PageIndex{3}$$
Let $$X_1$$ and $$X_2$$ be independent random variables with common distribution
$p_X = \bigg( \begin{array}{} 0 & 1 & 2 \\ 1/2 & 3/8 & 1/2 \end{array} \bigg)$.
Find the distribution of the sum $$X_1$$ + $$X_2$$.
###### $$\PageIndex{4}$$
In one play of certain game you win an amount with distribution
$p_X = \bigg( \begin{array}{} 1 & 2 & 3 \\ 1/4 & 1/4 & 1/2 \end{array} \bigg)$.
Using the program NFoldConvolution find the distribution for your total winnings after ten (independent) plays. Plot this distribution.
###### $$\PageIndex{5}$$
Consider the following two experiments: the first has outcome X taking on the values 0, 1, and 2 with equal probabilities; the second results in an (independent) outcome Y taking on the value 3 with probability 1/4 and 4 with probability 3/4. Find the distribution of
$\begin{array}{} (a) & Y+X \\ (b) & Y-X \end{array}$
###### $$\PageIndex{6}$$
People arrive at a queue according to the following scheme: During each minute of time either 0 or 1 person arrives. The probability that 1 person arrives is p and that no person arrives is $$q = 1 − p$$. Let $$C_r$$ be the number of customers arriving in the first r minutes. Consider a Bernoulli trials process with a success if a person arrives in a unit time and failure if no person arrives in a unit time. Let $$T_r$$ be the number of failures before the rth success.
$\begin{array}{} (a) & What is the distribution for $$T_r$$ \\ (b) & What is the distribution $$C_r$$ \\ (c) Find the mean and variance for the number of customers arriving in the first r minutes \end{array}$
###### $$\PageIndex{7}$$
(a) A die is rolled three times with outcomes $$X_1, X_2$$ and $$X_3$$. Let $$Y_3$$ be the maximum value obtained. Show that
$P(Y_3 \leq j) = P(X_1 \leq j)^3$
Use this find the distribution of $$Y_3$$. Does $$Y_3$$ have a bell-shaped distribution?
(b) Now let $$Y_n$$ be the maximum value when dice are rolled. Find the distribution of $$Y_n$$. Is this distribution bell-shaped for large values of n?
###### $$\PageIndex{8}$$
A baseball player is to play in the World Series. Based upon his season play, you estimate that if he comes to bat four times in a game the number of hits he will get has a distribution
$p_X = \bigg( \begin{array}{} 0&1&2&3&4\\.4&.2&.2&.1&.1 \end{array} \bigg)$
Assume that the player comes to bat four times in each game of the series.
(a) Let X denote the number of hits that he gets in a series. Using the program NFoldConvolution, find the distribution of X for each of the possible series lengths: four-game, five-game, six-game, seven-game.
(b) Using one of the distribution found in part (a), find the probability that his batting average exceeds .400 in a four-game series. (The batting average is the number of hits divided by the number of times at bat.)
(c) Given the distribution pX , what is his long-term batting average?
###### $$\PageIndex{9}$$
Prove that you cannot load two dice in such a way that the probabilities for any sum from 2 to 12 are the same. (Be sure to consider the case where one or more sides turn up with probability zero.)
###### $$\PageIndex{10}$$
(Lévy$$^2$$ ) Assume that n is an integer, not prime. Show that you can find two distributions a and b on the nonnegative integers such that the convolution of a and b is the equiprobable distribution on the set 0, 1, 2, . . . , n − 1. If n is prime this is not possible, but the proof is not so easy. (Assume that neither a nor b is concentrated at 0.)
###### $$\PageIndex{11}$$
Assume that you are playing craps with dice that are loaded in the following way: faces two, three, four, and five all come up with the same probability (1/6) + r. Faces one and six come up with probability (1/6) − 2r, with $$0 < r < .02.$$ Write a computer program to find the probability of winning at craps with these dice, and using your program find which values of r make craps a favorable game for the player with these dice.
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## Kiss those Math Headaches GOODBYE!
### Friendly Formula: the Distance Formula
A few days ago I posted a “Friendly Formula” for the Midpoint Formula.
Today I am presenting a Friendly Formula for the Distance Formula, an important formula in Algebra 1 courses.
First I’m going to present the Friendly Formula for the Distance Formula and demonstrate how to use it. Then I’ll explain why it makes sense.
Buckle your seatbelts ’cause here it is: the distance between any two points on the coordinate plane is simply the SQUARE ROOT of …
(the x-distance squared) plus (the y-distance squared).
And here’s an example of how easy it can be to use this formula.
Suppose you want the distance between the points (2, 5) and (4, 9).
First figure out how the distance between the x-coordinates, 2 and 4.
Well, 4 – 2 = 2, so the x-distance = 2.
Now square that x-distance: 2 squared = 4
Next find the distance between the y-coordinates, 5 and 9:
Well, 9 – 5 = 4, so the y-distance = 4.
Now square that y-distance: 4 squared = 16
Next add the two squared values you just got: 4 + 16 = 20
Finally take the square root of that sum: square root of 20 = root 20.
That final value, root 20, is the distance between the two points.
Now we get to the question of WHY this Friendly Formula makes sense. I will explain that in my next post.
HINT: The Distance Formula is based on the Pythagorean Theorem. See if you can spot the connection.
EXTRA HINT: Make a coordinate plane. Plot the two points I used in this example, and construct a right triangle in which the line connecting these two points is the hypotenuse. If you can figure this out, the “Aha!” moment is a glorious event!
### My WAGER (& DISCOUNT PLAN) to help ANYONE learn Algebra
A wager … and a plan.
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I’ve been tutoring algebra for a long time (oh, just a bit over 30 years now), and I have developed many tips and tricks for this subject area. Not only that, but I’ve seen pretty much every mistake you can imagine. And I’ve learned how to explain why each mistake is incorrect and to help folks view each situation correctly.
So in the spirit of the Emma Lazarus poem on the Statue of Liberty, I say:
“Give me your confused, your bewildered, your frustrated students, yearning to comprehend, the befuddled refuse of your overcrowded classrooms. Send these, the despondent ones, your so-called failures to me. I will lift my lantern of algebraic clarity unto their puzzled eyes!”
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### How to Understand and then Forever Memorize the Midpoint Formula
In algebra we have many formulas to learn. But one problem is that those formulas are often hard to memorize. They are written with variables, and the variables frequently have subscripts, and the truth is that a lot of us don’t really understand what the formulas are saying or how they work. So of course that makes formulas difficult to memorize.
Enter the concept of “friendly formulas.” Friendly formulas are the very same formulas but written in a way that you can understand and therefore memorize much more easily. It’s an idea I have come up with through my many years of algebra tutoring, and idea is included in my Algebra Survival Guide, available through Amazon.com
In this post I describe the “friendly formula” for the midpoint formula.
So as a refresher, what is the midpoint formula all about?
Basically, it lets you find the midpoint of any line segment on the coordinate plane. Think of it this way. There’s some line segment on the coordinate plane called segment AB. That means that it has an endpoint at point A, another at point B. We are given the coordinate of points A and B. We want to find the coordinates of the point right in the middle of points A and B.
Now let’s make this idea easy. Suppose we focus only on the x-coordinates. Suppose the x-coordinate of point A is 2, and the x-coordinate of point B is 6. Ask yourself: what x-coordinate is perfectly in the middle of coordinates 2 and 6? It’s just like asking: what number is right in the middle of 2 and 6 on the number line? Well, wouldn’t that be 4, since 4 is two more than 2 and two less than 6? And indeed it is 4.
But notice that there’s another way to get 4, given the coordinates 2 and 6. We also could have just added 2 and 6 to get 8, and then divided 8 by 2, since 8 ÷ 2 = 4. In other words, we could have TAKEN the AVERAGE of the two x-coordinates, since taking an average of two numbers is adding them and dividing by two.
Could the midpoint formula actually be as easy as taking averages?!
Before we say yes, let’s test this idea for more complicated situations. We just saw that it works when both coordinates are positive. But suppose one coordinate is positive, the other negative. Let’s let one coordinate be
– 2, while the other is + 4. What number is right between those two coordinates on the number line? Well, the numbers are 6 apart, right? And half of 6 is 3, so we could just add 3 to – 2, and get + 1 as the point in between them. And we see that + 1 is three away from both – 2 and 4. But could we also get + 1 by averaging -2 and 4? Let’s try:
(- 2 + 4) / 2 = 2 / 2 = + 1. Averaging works again.
And finally, what about the case where both coordinates are negative? Suppose one coordinate is – 2, the other – 8. What number is right between those two numbers on the number line? Well, these numbers are also 6 apart, right? And half of 6 is 3, so we could just add 3 to – 8, and get – 5 as the middle. And we see that – 5 is three away from both – 8 and – 2. But can we also get – 5 by averaging – 8 and – 2? Let’s try: (- 8 + – 2) / 2 = – 10 / 2 = -5. Averaging worked here too!
Since the averaging process works for all three cases, this approach does works always, and in fact it is how the midpoint formula works.
The midpoint formula basically just averages the x-coordinates to get the x-coordinate of the midpoint. Then it averages the y-coordinates to get the y-coordinate of the midpoint.
So here is the “friendly formula” for the midpoint of any segment on the coordinate plane: Given a segment whose x- and y-coordinates are known,
MIDPOINT = (AVERAGE of x-coordinates, AVERAGE of y-coordinates)
And that’s all you have to memorize!
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College Physics for AP® Courses 2e
19.6Capacitors in Series and Parallel
College Physics for AP® Courses 2e19.6 Capacitors in Series and Parallel
Learning Objectives
By the end of this section, you will be able to:
• Derive expressions for total capacitance in series and in parallel.
• Identify series and parallel parts in the combination of connection of capacitors.
• Calculate the effective capacitance in series and parallel given individual capacitances.
Several capacitors may be connected together in a variety of applications. Multiple connections of capacitors act like a single equivalent capacitor. The total capacitance of this equivalent single capacitor depends both on the individual capacitors and how they are connected. There are two simple and common types of connections, called series and parallel, for which we can easily calculate the total capacitance. Certain more complicated connections can also be related to combinations of series and parallel.
Capacitance in Series
Figure 19.19(a) shows a series connection of three capacitors with a voltage applied. As for any capacitor, the capacitance of the combination is related to charge and voltage by $C=QVC=QV$.
Note in Figure 19.19 that opposite charges of magnitude $QQ$ flow to either side of the originally uncharged combination of capacitors when the voltage $VV$ is applied. Conservation of charge requires that equal-magnitude charges be created on the plates of the individual capacitors, since charge is only being separated in these originally neutral devices. The end result is that the combination resembles a single capacitor with an effective plate separation greater than that of the individual capacitors alone. (See Figure 19.19(b).) Larger plate separation means smaller capacitance. It is a general feature of series connections of capacitors that the total capacitance is less than any of the individual capacitances.
Figure 19.19 (a) Capacitors connected in series. The magnitude of the charge on each plate is $Q Q$. (b) An equivalent capacitor has a larger plate separation $dd$. Series connections produce a total capacitance that is less than that of any of the individual capacitors.
We can find an expression for the total capacitance by considering the voltage across the individual capacitors shown in Figure 19.19. Solving $C=QVC=QV$ for $VV$ gives $V=QCV=QC$. The voltages across the individual capacitors are thus $V1=QC1V1=QC1$, $V2=QC2V2=QC2$, and $V3=QC3V3=QC3$. The total voltage is the sum of the individual voltages:
$V=V1+V2+V3.V=V1+V2+V3.$
19.60
Now, calling the total capacitance $CSCS$ for series capacitance, consider that
$V = Q C S = V 1 + V 2 + V 3 . V = Q C S = V 1 + V 2 + V 3 .$
19.61
Entering the expressions for $V1V1$, $V2V2$, and $V3V3$, we get
$QCS=QC1+QC2+QC3.QCS=QC1+QC2+QC3.$
19.62
Canceling the $QQ$s, we obtain the equation for the total capacitance in series $CSCS$ to be
$1CS=1C1+1C2+1C3+...,1CS=1C1+1C2+1C3+...,$
19.63
where “...” indicates that the expression is valid for any number of capacitors connected in series. An expression of this form always results in a total capacitance $CSCS$ that is less than any of the individual capacitances $C1C1$, $C2C2$, ..., as the next example illustrates.
Total Capacitance in Series, $C s C s$
Total capacitance in series: $1CS=1C1+1C2+1C3+...1CS=1C1+1C2+1C3+...$
Example 19.9
What Is the Series Capacitance?
Find the total capacitance for three capacitors connected in series, given their individual capacitances are 1.000, 5.000, and 8.000 $µFµF$.
Strategy
With the given information, the total capacitance can be found using the equation for capacitance in series.
Solution
Entering the given capacitances into the expression for $1CS1CS$ gives $1CS=1C1+1C2+1C31CS=1C1+1C2+1C3$.
$1 C S = 1 1 . 000 µF + 1 5 . 000 µF + 1 8 . 000 µF = 1 . 325 µF 1 C S = 1 1 . 000 µF + 1 5 . 000 µF + 1 8 . 000 µF = 1 . 325 µF$
19.64
Inverting to find $CSCS$ yields $CS=µF1.325=0.755 µFCS=µF1.325=0.755 µF$.
Discussion
The total series capacitance $CsCs$ is less than the smallest individual capacitance, as promised. In series connections of capacitors, the sum is less than the parts. In fact, it is less than any individual. Note that it is sometimes possible, and more convenient, to solve an equation like the above by finding the least common denominator, which in this case (showing only whole-number calculations) is 40. Thus,
$1CS=4040 µF+840 µF+540 µF=5340 µF,1CS=4040 µF+840 µF+540 µF=5340 µF,$
19.65
so that
$CS=40 µF53=0.755 µF.CS=40 µF53=0.755 µF.$
19.66
Capacitors in Parallel
Figure 19.20(a) shows a parallel connection of three capacitors with a voltage applied. Here the total capacitance is easier to find than in the series case. To find the equivalent total capacitance $CpCp$, we first note that the voltage across each capacitor is $VV$, the same as that of the source, since they are connected directly to it through a conductor. (Conductors are equipotentials, and so the voltage across the capacitors is the same as that across the voltage source.) Thus the capacitors have the same charges on them as they would have if connected individually to the voltage source. The total charge $QQ$ is the sum of the individual charges:
$Q=Q1+Q2+Q3.Q=Q1+Q2+Q3.$
19.67
Figure 19.20 (a) Capacitors in parallel. Each is connected directly to the voltage source just as if it were all alone, and so the total capacitance in parallel is just the sum of the individual capacitances. (b) The equivalent capacitor has a larger plate area and can therefore hold more charge than the individual capacitors.
Using the relationship $Q=CVQ=CV$, we see that the total charge is $Q=CpVQ=CpV$, and the individual charges are $Q1=C1VQ1=C1V$, $Q2=C2VQ2=C2V$, and $Q3=C3VQ3=C3V$. Entering these into the previous equation gives
$CpV=C1V+C2V+C3V.CpV=C1V+C2V+C3V.$
19.68
Canceling $VV$ from the equation, we obtain the equation for the total capacitance in parallel $CpCp$:
$Cp=C1+C2+C3+....Cp=C1+C2+C3+....$
19.69
Total capacitance in parallel is simply the sum of the individual capacitances. (Again the “...” indicates the expression is valid for any number of capacitors connected in parallel.) So, for example, if the capacitors in the example above were connected in parallel, their capacitance would be
$Cp=1.000 µF+5.000 µF+8.000 µF=14.000 µF.Cp=1.000 µF+5.000 µF+8.000 µF=14.000 µF.$
19.70
The equivalent capacitor for a parallel connection has an effectively larger plate area and, thus, a larger capacitance, as illustrated in Figure 19.20(b).
Total Capacitance in Parallel, $C p C p$
Total capacitance in parallel $Cp=C1+C2+C3+...Cp=C1+C2+C3+...$
More complicated connections of capacitors can sometimes be combinations of series and parallel. (See Figure 19.21.) To find the total capacitance of such combinations, we identify series and parallel parts, compute their capacitances, and then find the total.
Figure 19.21 (a) This circuit contains both series and parallel connections of capacitors. See Example 19.10 for the calculation of the overall capacitance of the circuit. (b) $C1C1$ and $C2C2$ are in series; their equivalent capacitance $CSCS$ is less than either of them. (c) Note that $CSCS$ is in parallel with $C3C3$. The total capacitance is, thus, the sum of $CSCS$ and $C3C3$.
Example 19.10
A Mixture of Series and Parallel Capacitance
Find the total capacitance of the combination of capacitors shown in Figure 19.21. Assume the capacitances in Figure 19.21 are known to three decimal places ( $C1=1.000 µFC1=1.000 µF$, $C2=5.000 µFC2=5.000 µF$, and $C3=8.000 µFC3=8.000 µF$), and round your answer to three decimal places.
Strategy
To find the total capacitance, we first identify which capacitors are in series and which are in parallel. Capacitors $C1C1$ and $C2C2$ are in series. Their combination, labeled $CSCS$ in the figure, is in parallel with $C3C3$.
Solution
Since $C1C1$ and $C2C2$ are in series, their total capacitance is given by $1CS=1C1+1C2+1C31CS=1C1+1C2+1C3$. Entering their values into the equation gives
$1 C S = 1 C 1 + 1 C 2 = 1 1 . 000 μF + 1 5 . 000 μF = 1 . 200 μF . 1 C S = 1 C 1 + 1 C 2 = 1 1 . 000 μF + 1 5 . 000 μF = 1 . 200 μF .$
19.71
Inverting gives
$CS=0.833 µF.CS=0.833 µF.$
19.72
This equivalent series capacitance is in parallel with the third capacitor; thus, the total is the sum
$C tot = C S + C S = 0 . 833 μF + 8 . 000 μF = 8 . 833 μF. C tot = C S + C S = 0 . 833 μF + 8 . 000 μF = 8 . 833 μF.$
19.73
Discussion
This technique of analyzing the combinations of capacitors piece by piece until a total is obtained can be applied to larger combinations of capacitors.
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How many numbers with two or more digits can be formed with the digits $1,2,3,4,5,6,7,8,9$, so that in every such number, each digit is used at most once and the digits appear in the ascending order?
Given that, the digits $:1,2,3,4,5,6,7,8,9$
We know that, the number of ways to pick $k$ unordered elements from an $n$ element set $= \;^{n}C_{k} = \frac{n!}{k!(n-k)!}$
After selecting the number from the given digits, there is only one way to arrange it.
So, the total number of ways $= \;^{9} C_{2} + \;^{9} C_{3} + \;^{9} C_{4} + \dots + \;^{9} C_{9} \quad \longrightarrow (1)$
We know that, $^{n} C_{0} + \;^{n} C_{1} + \;^{n} C_{2} + \dots + \;^{9} C_{n} = 2^{n}$
Here, $^{9}C_{0} + \;^{9}C_{1} + \;^{9} C_{2} + \;^{9} C_{3} + \;^{9} C_{4} + \dots + \;^{9} C_{9} = 2^{9}$
$\Rightarrow \; ^{9} C_{2} + \;^{9} C_{3} + \;^{9} C_{4} + \dots + \;^{9} C_{9} = 2^{9} – \;^{9}C_{0} - \;^{9}C_{1}$
From the equation $(1),$ we get
$\therefore$ The total number of ways $= 2^{9} – \;^{9}C_{0} – \;^{9}C_{1}$
$\quad = 512 – \frac{9!}{0! \cdot 9!} – \frac{9!}{1! \cdot 8!}$
$\quad = 512 – 1 – \frac{9 \times 8!}{1! \cdot 8!}$
$\quad = 512 – 1 – 9$
$\quad = 512 – 10$
$\quad = 502 \; \text{ways}.$
Correct Answer $: 502$
$\textbf{PS}:\text{Important Properties:}$
• $n! = n(n-1)(n-2) \dots 1 = n(n-1)!$
• $0! = 1$
• $1! = 1$
• $^{n}C_{n} = \frac{n!}{n! \; 0!} = 1$
• $^{n}C_{0} = \frac{n!}{0! \; n!}$
• $^{n}C_{1} = \frac{n!}{1!(n-1)!} = \frac{n(n-1)!}{1!(n-1)!} = n$
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## Selina Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon
Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 20 Area of Trapezium and a Polygon
ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 20 Area of Trapezium and a Polygon. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.
### Area of Trapezium and a Polygon Exercise 20A – Selina Concise Mathematics Class 8 ICSE Solutions
Question 1.
Find the area of a triangle, whose sides are :
(i) 10 cm, 24 cm and 26 cm
(ii) 18 mm, 24 mm and 30 mm
(iii) 21 m, 28 m and 35 m
Solution:
(i) Sides of ∆ are
a = 10 cm
b = 24 cm
c = 26 cm
(ii) Sides of ∆ are
a = 18 mm
b = 24 mm
c = 30 mm
(iii) Sides of ∆ are
a = 21 m
b = 28 m
c = 35 m
Question 2.
Two sides of a triangle are 6 cm and 8 cm. If height of the triangle corresponding to 6 cm side is 4 cm ; find :
(i) area of the triangle
(ii) height of the triangle corresponding to 8 cm side.
Solution:
Question 3.
The sides of a triangle are 16 cm, 12 cm and 20 cm. Find :
(i) area of the triangle ;
(ii) height of the triangle, corresponding to the largest side ;
(iii) height of the triangle, corresponding to the smallest side.
Solution:
Sides of ∆ are
a = 20 cm
b = 12 cm
c = 16 cm
Question 4.
Two sides of a triangle are 6.4 m and 4.8 m. If height of the triangle corresponding to 4.8 m side is 6 m; find :
(i) area of the triangle ;
(ii) height of the triangle corresponding to 6.4 m side.
Solution:
= 9/2 = 4.5 m
Hence (i) 14.4 m2 (ii) 4.5 m
Question 5.
The base and the height of a triangle are in the ratio 4 : 5. If the area of the triangle is 40 m2; find its base and height.
Solution:
Let base of ∆ = 4x m
and height of ∆ = 5x m
area of ∆ =40 m2
Question 6.
The base and the height of a triangle are in the ratio 5 : 3. If the area of the triangle is 67.5 m2; find its base and height.
Solution:
Let base = 5x m
height = 3x m
base = 5x = 5 x 3 = 15 m
height = 3x = 3 x 3 = 9 m
Question 7.
The area of an equilateral triangle is 144√3 cm2; find its perimeter.
Solution:
Let each side of an equilateral triangle = x cm
Question 8.
The area of an equilateral triangle is numerically equal to its perimeter. Find its perimeter correct to 2 decimal places.
Solution:
Let each side of the equilateral traingle = x
Question 9.
A field is in the shape of a quadrilateral ABCD in which side AB = 18 m, side AD = 24 m, side BC = 40m, DC = 50 m and angle A = 90°. Find the area of the field.
Solution:
Since ∠A = 90°
By Pythagorus Theorem,
In ∆ABD,
Question 10.
The lengths of the sides of a triangle are in the ratio 4 : 5 : 3 and its perimeter is 96 cm. Find its area.
Solution:
Let the sides of the triangle ABC be 4x, 5x and 3x
Let AB = 4x, AC = 5x and BC = 3x
Perimeter = 4x + 5x + 3x = 96
=> 12x = 96
Question 11.
One of the equal sides of an isosceles triangle is 13 cm and its perimeter is 50 cm. Find the area of the triangle.
Solution:
In isosceles ∆ABC
AB = AC = 13 cm But perimeter = 50 cm
Question 12.
The altitude and the base of a triangular field are in the ratio 6 : 5. If its cost is ₹ 49,57,200 at the rate of ₹ 36,720 per hectare and 1 hectare = 10,000 sq. m, find (in metre) dimensions of the field,
Solution:
Total cost = ₹ 49,57,200
Rate = ₹ 36,720 per hectare
Total area of the triangular field
Question 13.
Find the area of the right-angled triangle with hypotenuse 40 cm and one of the other two sides 24 cm.
Solution:
In right angled triangle ABC Hypotenuse AC = 40 cm
One side AB = 24 cm
Question 14.
Use the information given in the adjoining figure to find :
(i) the length of AC.
(ii) the area of a ∆ABC
(iii) the length of BD, correct to one decimal place.
Solution:
### Area of Trapezium and a Polygon Exercise 20B – Selina Concise Mathematics Class 8 ICSE Solutions
Question 1.
Find the length and perimeter of a rectangle, whose area = 120 cm2 and breadth = 8 cm
Solution:
area of rectangle = 120 cm2
Area = l x b
l x 8 = 120
l = 120/8 = 15 cm
Perimeter = 2 (l+b) = 2(15+8) = 2 x 23 = 46 cm
Length = 15 cm; Perimeter = 46 cm
Question 2.
The perimeter of a rectangle is 46 m and its length is 15 m. Find its :
(ii) area
(iii) diagonal.
Solution:
(i) Perimeter of rectangle = 46 m
length, l = 15 m
2 (l+b) = 46
2(15 + b) = 46
15+b = 46/2 = 23
b = 23 – 15
b = 8 m
Question 3.
The diagonal of a rectangle is 34 cm. If its breadth is 16 cm; find its :
(i) length
(ii) area
Solution:
AC2 = AB2+BC2 (By Pythagoras theorem)
(34)2 = l2 + (16)2
1156 = l2 + 256
l2 = 1156 – 256
l2 = 900
l = √900 = 30 cm
area = l x b = 30 x 16 = 480 cm2
(i) 30 cm (ii) 480 cm2
Question 4.
The area of a small rectangular plot is 84 m2. If the difference between its length and the breadth is 5 m; find its perimeter.
Solution:
Area of a rectangular plot = 84 m2
Then length = (x + 5) m
Area = l x b
x(x + 5) = 84
x2 + 5x – 84 = 0
=> x2+ 12x – 7x – 84 = 0
=> x(x + 12) – 7(x + 12) = 0
=> (x + 12) (x – 7) = 0
Either x + 12 = 0, then x = -12 which is not possible being negative
or x – 7 = 0, then x = 7
Length = x + 5 = 7 + 5 = 12m
and breadth = x = 7 m
Perimeter = 2(l + b) = 2(12+ 7) = 2 x 19 m = 38 m
Question 5.
The perimeter of a square is 36 cm; find its area
Solution:
Perimeter of Square = 36 cm
Question 6.
Find the perimeter of a square; whose area is : 1.69 m2
Solution:
Area of square= 1.69 m2
Side = √area = √1.69 = 1.3 m
Perimeter = 4 x side = 4 x 1.3 = 5.2 m
Question 7.
The diagonal of a square is 12 cm long; find its area and length of one side.
Solution:
Let side of square = a cm
diagonal = 12 cm
By Pythagoras Theorem, a2 + a2 = (12)2
2a2 = 144
a2 = 72
Area of square = a2 = 72 cm2
a2 = 72
a = √72 = 8.49 cm
Question 8.
The diagonal of a square is 15 m; find the length of its one side and perimeter.
Solution:
Diagonal of square = 15 m
Let side of square = a
a2 + a2 = (15)2 = 225
a2 = 225/2 = 112.50
a = √112.50 = 10.6 m
Perimeter = 4 x a = 10.6 x 4 = 42.4 m
Question 9.
The area of a square is 169 cm2. Find its:
(i) one side
(ii) perimeter
Solution:
Let each side of the square be x cm.
Its area = x2 = 169 (given)
x = √169
x = 13 cm
(i) Thus, side of the square = 13 cm
(ii) Again perimeter = 4 (side) = 4 x 13 = 52 cm
Question 10.
The length of a rectangle is 16 cm and its perimeter is equal to the perimeter of a square with side 12.5 cm. Find the area of the rectangle.
Solution:
Length of the rectangle = 16 cm
Let its breadth be x cm
Perimeter = 2 (16 + x) = 32 + 2x
Also perimeter = 4(12.5) = 50 cm.
According to statement,
32 + 2x = 50
=> 2x = 50 – 32 = 18
=> x = 9
Breadth of the rectangle = 9 cm.
Area of the rectangle (l x b)= 16 x 9 = 144 cm2
Question 11.
The perimeter of a square is numerically equal to its area. Find its area.
Solution:
Let each side of the square be x cm.
Its perimeter = 4x,
Area =x2
By the given condition 4x = x2
=> x2 – 4x = 0
=> x (x – 4) = 0
=> x = 4 [x ≠ 0]
Area = x2 = (4)2 = 4 x 4 = 16 sq.units.
Question 12.
Each side of a rectangle is doubled. Find the ratio between :
(i) perimeters of the original rectangle and the resulting rectangle.
(ii) areas of the original rectangle and the resulting rectangle.
Solution:
Let length of the rectangle = x
and breadth of the rectangle = y
(i) Perimeter P = 2(x + y)
Again, new length = 2x
Question 13.
In each of the following cases ABCD is a square and PQRS is a rectangle. Find, in each case, the area of the shaded portion.
(All measurements are in metre).
Solution:
(i) Area of the shaded portion
= Area of the rectangle PQRS – Area of square ABCD
= 3.2 x 1.8 – (1.4)2 (∵ PQ = 3.2 and PS = 1.8) Side of square AB = 1.4
= 5.76 – 1.96 = 3.80 = 3.8 m2
(ii) Area of the shaded portion = Area of square ABCD – Area of rectangle PQRS
= 6 x 6 – (3.6) (4.8) = 36 – 17.28 = 18.72 m2
Question 14.
A path of uniform width, 3 m, runs around the outside of a square field of side 21 m. Find the area of the path.
Solution:
According to the given information the figure will be as shown alongside.
Clearly, length of the square field excluding path = 21 m.
Area of the square side excluding the path = 21 x 21 = 441 m2
Again, length of the square field including the path = 21 + 3 + 3 = 27 m
Area of the square field including the path = 27 x 27 = 729 m2
Area of the path = 729 – 441 = 288 m2
Question 15.
A path of uniform width, 2.5 m, runs around the inside of a rectangular field 30 m by 27 m. Find the area of the path.
Solution:
According to the given statement the figure will be as shown alongside.
Clearly, the length of the rectangular field including the path = 30 m.
Its Area = 30 x 27 = 810 m2
Width of the path = 2.5 m
Length of the rectangular field including the path = 30 – 2.5 – 2.5 = 25 m.
Breadth = 27 – 2.5 – 2.5 = 22m
Area of the rectangular field including the path = 25 x 22 = 550 m2
Hence, area of the path = 810 – 550 = 260 m2.
Question 16.
The length of a hall is 18 m and its width is 13.5 m. Find the least number of square tiles, each of side 25 cm, required to cover the floor of the hall,
(i) without leaving any margin.
(ii) leaving a margin of width 1.5 m all around. In each case, find the cost of the tiles required at the rate of Rs. 6 per tile
Solution:
(i) Length of hall (l) = 18 m and breadth (b) = 13.5 m
Question 17.
A rectangular field is 30 m in length and 22m in width. Two mutually perpendicular roads, each 2.5 m wide, are drawn inside the field so that one road is parallel to the length of the field and the other road is parallel to its width. Calculate the area of the crossroads.
Solution:
Length of rectangular field (l) = 30 m and breadth (b) = 22m
width of parallel roads perpendicular to each other inside the field =2.5m
= 2.5 (30 + 22) – (2.5)2
= 2.5 x 52 – 6.25 m2
= (130 – 6.25) m = 123.75 m2
Question 18.
The length and the breadth of a rectangular field are in the ratio 5 : 4 and its area is 3380 m2. Find the cost of fencing it at the rate of ₹75 per m.
Solution:
Ratio in length and breadth = 5 : 4
Area of rectangular field = 3380 m2
Let length = 5x and breadth = 4x
5x x 4x = 3380
=> 20x2= 3380
x2 = 3380/20 = 169 = (13)2
x = 13
Length = 13 x 5 = 65 m
Breadth =13 x 4 = 52 m
Perimeter = (l + b) = 2 x (65 + 52) m = 2 x 117 = 234 m
Rate of fencing = ₹ 75 per m
Total cost = 234 x 75 = ₹ 17550
Question 19.
The length and the breadth of a conference hall are in the ratio 7 : 4 and its perimeter is 110 m. Find:
(i) area of the floor of the hall.
(ii) number of tiles, each a rectangle of size 25 cm x 20 cm, required for flooring of the hall.
(iii) the cost of the tiles at the rate of ₹ 1,400 per hundred tiles.
Solution:
Ratio in length and breadth = 7 : 4
Perimeter = 110 m
### Area of Trapezium and a Polygon Exercise 20C – Selina Concise Mathematics Class 8 ICSE Solutions
Question 1.
The following figure shows the cross-section ABCD of a swimming pool which is trapezium in shape.
If the width DC, of the swimming pool is 6.4cm, depth (AD) at the shallow end is 80 cm and depth (BC) at deepest end is 2.4m, find Its area of the cross-section.
Solution:
Area of the cross-section = Area of trapezium ABCD
= 1024 cm2 or = 10.24 sq.m.
Question 2.
The parallel sides of a trapezium are in the ratio 3 : 4. If the distance between the parallel sides is 9 dm and its area is 126 dm2 ; find the lengths of its parallel sides.
Solution:
Let parallel sides of trapezium be
a = 3x
b = 4x
Distance between parallel sides, h = 9 dm
area of trapezium = 126 dm2
Question 3.
The two parallel sides and the distance between them are in the ratio 3 : 4 : 2. If the area of the trapezium is 175 cm2, find its height.
Solution:
Let the two parallel sides and the distance between them be 3x, 4x, 2x cm respectively
Area = $$\frac { 1 }{ 2 }$$ (sum of parallel sides) x (distance between parallel sides)
= $$\frac { 1 }{ 2 }$$ (3x + 4x) x 2x = 175 (given)
=> 7x x x = 175
=> 7x2 = 175
=> x2 = 25
=> x = 5
Height i.e. distance between parallel sides = 2x = 2 x 5 = 10 cm
Question 4.
A parallelogram has sides of 15 cm and 12 cm; if the distance between the 15 cm sides is 6 cm; find the distance between 12 cm sides.
Solution:
BQ = $$\frac { 15 }{ 2 }$$ = 7.5 cm
Question 5.
A parallelogram has sides of 20 cm and 30 cm. If the distance between its shorter sides is 15 cm; find the distance between the longer sides.
Solution:
Question 6.
The adjacent sides of a parallelogram are 21 cm and 28 cm. If its one diagonal is 35 cm; find the area of the parallelogram.
Solution:
Question 7.
The diagonals of a rhombus are 18 cm and 24 cm. Find:
(i) its area ;
(ii) length of its sides.
(iii) its perimeter;
Solution:
Diagonal of rhombus are 18 cm and 24 cm.
area of rhombus = $$\frac { 1 }{ 2 }$$ x Product of diagonals
= $$\frac { 1 }{ 2 }$$ x 18 x 24
= 216 cm2
(ii) Diagonals of rhombus bisect each other at right angles.
Question 8.
The perimeter of a rhombus is 40 cm. If one diagonal is 16 cm; find :
(i) its another diagonal
(ii) area
Solution:
(i) Perimeter of rhombus = 40 cm
side = $$\frac { 1 }{ 4 }$$ x 40 = 10 cm
One diagonal = 16 cm
Diagonals of rhombus bisect each other at right angles.
Question 9.
Each side of a rhombus is 18 cm. If the distance between two parallel sides is 12 cm, find its area.
Solution:
Each side of the rhombus = 18 cm
base of the rhombus = 18 cm
Distance between two parallel sides = 12 cm
Height = 12 cm
Area of the rhombus = base x height = 18 x 12 = 216 cm2
Question 10.
The length of the diagonals of a rhombus is in the ratio 4 : 3. If its area is 384 cm2, find its side.
Solution:
Let the lengths of the diagonals of rhombus are 4x, 3x.
Question 11.
A thin metal iron-sheet is rhombus in shape, with each side 10 m. If one of its diagonals is 16 m, find the cost of painting its both sides at the rate of ₹ 6 per m2.
Also, find the distance between the opposite sides of this rhombus.
Solution:
Side of rhombus shaped iron sheet = 10 m and one diagonals (AC) = 16 m
Join BD diagonal which bisects AC at O
The diagonals of a rhombus bisect each other at right angle
Question 12.
The area of a trapezium is 279 sq.cm and the distance between its two parallel sides is 18 cm. If one of its parallel sides is longer than the other side by 5 cm, find the lengths of its parallel sides.
Solution:
Area of trapezium = 279 sq.cm
Distance between two parallel lines (h) = 18 cm
Question 13.
The area of a rhombus is equal to the area of a triangle. If base of ∆ is 24 cm, its corresponding altitude is 16 cm and one of the diagonals of the rhombus is 19.2 cm. Find its other diagonal.
Solution:
Area of a rhombus = Area of a triangle Base of triangle = 24 cm
and altitude = 16 cm
Question 14.
Find the area of the trapezium ABCD in which AB//DC, AB = 18 cm, ∠B = ∠C = 90°, CD = 12 cm and AD = 10 cm.
Solution:
In trapezium ABCD,
### Area of Trapezium and a Polygon Exercise 20D – Selina Concise Mathematics Class 8 ICSE Solutions
Question 1.
Find the radius and area of a circle, whose circumference is :
(i) 132 cm
(ii) 22 m
Solution:
(i) Circumference of circle = 132 cm
2πr = 132
2 x 22/7 x r = 132
Question 2.
Find the radius and circumference of a circle, whose area is :
(i) 154 cm2
(ii) 6.16 m2
Solution:
Question 3.
The circumference of a circular table is 88 m. Find its area.
Solution:
Circumference of circle = 88 m
2πr = 88 m
Question 4.
The area of a circle is 1386 sq.cm ; find its circumference.
Solution:
Area of circle = 1386 cm2
πr2 = 1386
Question 5.
Find the area of a flat circular ring formed by two concentric circles (circles with same centre) whose radii are 9 cm and 5 cm.
Solution:
Question 6.
Find the area of the shaded portion in each of the following diagrams :
Solution:
Question 7.
The radii of the inner and outer circumferences of a circular running track are 63 m and 70 m respectively. Find :
(i) the area of the track ;
(it) the difference between the lengths of the two circumferences of the track.
Solution:
= 2x 22/7 x 63 = 396 m
Difference between lengths of two circumferences = 440 – 396 = 44 m
Hence (i) 2926 m2 (ii) 44 m
Question 8.
A circular field cf radius 105 m has a circular path of uniform width of 5 m along and inside its boundary. Find the area of the path.
Solution:
Question 9.
There is a path of uniform width 7 m round and outside a circular garden of diameter 210 m. Find the area of the path.
Solution:
Question 10.
A wire, when bent in the form of a square encloses an area of 484 cm2. Find :
(i) one side of the square ;
(ii) length of the wire ;
(iii) the largest area enclosed; if the same wire is bent to form a circle.
Solution:
(i) Area of Square = 484 cm2
Side of Square = √Area = √484 = 22 cm
(ii) Perimeter, i.e. length of wire = 4 x 22 = 88 cm
(iii) Circumference of circle = 88 cm
2πr = 88
Question 11.
A wire, when bent in the form of a square; encloses an area of 196 cm2. If the same wire is bent to form a circle; find the area of the circle.
Solution:
Area of Square = 196 cm2
Side of Square = √Area = √196 = 14 cm
Perimeter of Square = 4 x 14 cm
i.e. length of wire = 56 cm
Circumference of circle = 56 cm
2πr = 56
= 2744/11
249.45 cm2
Question 12.
The radius of a circular wheel is 42 cm. Find the distance travelled by it in :
(i) 1 revolution ;
(ii) 50 revolutions ;
(iii) 200 revolutions ;
Solution:
(i) Radius of wheel, r = 42 cm
Circumference i.e. distance travelled in 1 revolution = 2πr = 2 x 22/7 x 42 = 264 cm
(ii) Distance travelled in 50 revolutions = 264 x 50 = 13200 cm = 132 m
(iii) Distance travelled in 200 revolutions = 264 x 200 = 52800 cm = 528 m
Hence (i) 264 cm (ii) 132 m (iii) 528 m
Question 13.
The diameter of a wheel is 0.70 m. Find the distance covered by it in 500 revolutions. If the wheel takes 5 minutes to make 500 revolutions; find its speed in :
(i) m/s
(ii) km/hr.
Solution:
Diameter = 0.70 m
Distance covered in 1 revolution, i.e. circumference = 2πr = 2 x 22/7 x 0.35 = 2.20 m
Distance covered in 500 revolutions = 2.20 x 500 = 1100 m
Time taken = 5 minutes = 5 x 60 = 300 sec.
Question 14.
A bicycle wheel, diameter 56 cm, is making 45 revolutions in every 10 seconds. At what speed in kilometre per hour is the bicycle travelling ?
Solution:
Question 15.
A roller has a diameter of 1.4 m. Find :
(i) its circumference ;
(ii) the number of revolutions it makes while travelling 61.6 m.
Solution:
Diameter = 1.4 m
Question 16.
Find the area of the circle, length of whose circumference is equal to the sum of the lengths of the circumferences with radii 15 cm and 13 cm.
Solution:
In a circle
Circumference = Sum of circumferences of two circle of radii 15 cm and 13 cm
Now circumference of first smaller circle = 2πr
Question 17.
A piece of wire of length 108 cm is bent to form a semicircular arc bounded by its diameter. Find its radius and area enclosed.
Solution:
Length of wire = 108 cm
Let r be the radius of the semicircle
πr+ 2r = 108
Question 18.
In the following figure, a rectangle ABCD enclosed three circles. If BC = 14 cm, find the area of the shaded portion (Take π = 22/7)
Solution:
In rectangle ABCD, BC = 14 cm
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# Equality (mathematics)
10:37
In mathematics, equality is a relationship between two quantities or, more generally two mathematical expressions, asserting that the quantities have the same value, or that the expressions represent the same mathematical object. The equality between A and B is written A = B, and pronounced A equals B.[1][2] The symbol "=" is called an "equals sign". Two objects that are not equal are said to be distinct.
For example:
• ${\displaystyle x=y}$ means that x and y denote the same object.[3]
• The identity ${\displaystyle (x+1)^{2}=x^{2}+2x+1}$ means that if x is any number, then the two expressions have the same value. This may also be interpreted as saying that the two sides of the equals sign represent the same function.
• ${\displaystyle \{x\mid P(x)\}=\{x\mid Q(x)\}}$ if and only if ${\displaystyle P(x)\Leftrightarrow Q(x).}$ This assertion, which uses set-builder notation, means that if the elements satisfying the property ${\displaystyle P(x)}$ are the same as the elements satisfying ${\displaystyle Q(x),}$ then the two uses of the set-builder notation define the same set. This property is often expressed as "two sets that have the same elements are equal." It is one of the usual axioms of set theory, called axiom of extensionality.[4]
## Etymology
The etymology of the word is from the Latin aequālis (“equal”, “like”, “comparable”, “similar”) from aequus (“equal”, “level”, “fair”, “just”).
## Basic properties
• Substitution property: For any quantities a and b and any expression F(x), if a = b, then F(a) = F(b) (provided that both sides are well-formed).
Some specific examples of this are:
• For any real numbers a, b, and c, if a = b, then a + c = b + c (here, F(x) is x + c);
• For any real numbers a, b, and c, if a = b, then ac = bc (here, F(x) is xc);
• For any real numbers a, b, and c, if a = b, then ac = bc (here, F(x) is xc);
• For any real numbers a, b, and c, if a = b and c is not zero, then a/c = b/c (here, F(x) is x/c).
• Reflexive property: For any quantity a, a = a.
• Symmetric property: For any quantities a and b, if a = b, then b = a.
• Transitive property: For any quantities a, b, and c, if a = b and b = c, then a = c.[5]
These three properties make equality an equivalence relation. They were originally included among the Peano axioms for natural numbers. Although the symmetric and transitive properties are often seen as fundamental, they can be deduced from substitution and reflexive properties.
## Equality as predicate
When A and B are not fully specified or depend on some variables, equality is a proposition, which may be true for some values and false for other values. Equality is a binary relation (i.e., a two-argument predicate) which may produce a truth value (false or true) from its arguments. In computer programming, its computation from the two expressions is known as comparison.
## Identities
When A and B may be viewed as functions of some variables, then A = B means that A and B define the same function. Such an equality of functions is sometimes called an identity. An example is ${\displaystyle \left(x+1\right)=x^{2}+2x+1.}$ Sometimes, but not always, an identity is written with a triple bar: ${\displaystyle \left(x+1\right)\equiv x^{2}+2x+1.}$
## Equations
An equation is a problem of finding values of some variables, called unknowns, for which the specified equality is true. The term "equation" may also refer to an equality relation that is satisfied only for the values of the variables that one is interested in. For example, ${\displaystyle x^{2}+y^{2}=1}$ is the equation of the unit circle.
There is no standard notation that distinguishes an equation from an identity, or other use of the equality relation: one has to guess an appropriate interpretation from the semantics of expressions and the context. An identity is asserted to be true for all values of variables in a given domain. An "equation" may sometimes mean an identity, but more often than not, it specifies a subset of the variable space to be the subset where the equation is true.
## Congruences
In some cases, one may consider as equal two mathematical objects that are only equivalent for the properties being considered. In geometry for instance, two geometric shapes are said to be equal when one may be moved to coincide with the other. The word congruence (and the associated symbol ${\displaystyle \cong }$[6]) is also used for this kind of equality.
## Approximate equality
There are some logic systems that do not have any notion of equality. This reflects the undecidability of the equality of two real numbers, defined by formulas involving the integers, the basic arithmetic operations, the logarithm and the exponential function. In other words, there cannot exist any algorithm for deciding such an equality.
The binary relation "is approximately equal" (denoted by the symbol ${\displaystyle \approx }$[1]) between real numbers or other things, even if more precisely defined, is not transitive (since many small differences can add up to something big). However, equality almost everywhere is transitive.
A questionable equality under test may be denoted using the ≟ symbol.
## Relation with equivalence and isomorphism
Viewed as a relation, equality is the archetype of the more general concept of an equivalence relation on a set: those binary relations that are reflexive, symmetric and transitive. The identity relation is an equivalence relation. Conversely, let R be an equivalence relation, and let us denote by xR the equivalence class of x, consisting of all elements z such that x R z. Then the relation x R y is equivalent with the equality xR = yR. It follows that equality is the finest equivalence relation on any set S in the sense that it is the relation that has the smallest equivalence classes (every class is reduced to a single element).
In some contexts, equality is sharply distinguished from equivalence or isomorphism.[7] For example, one may distinguish fractions from rational numbers, the latter being equivalence classes of fractions: the fractions ${\displaystyle 1/2}$ and ${\displaystyle 2/4}$ are distinct as fractions (as different strings of symbols) but they "represent" the same rational number (the same point on a number line). This distinction gives rise to the notion of a quotient set.
Similarly, the sets
${\displaystyle \{{\text{A}},{\text{B}},{\text{C}}\}}$ and ${\displaystyle \{1,2,3\}}$
are not equal sets — the first consists of letters, while the second consists of numbers — but they are both sets of three elements and thus isomorphic, meaning that there is a bijection between them. For example
${\displaystyle {\text{A}}\mapsto 1,{\text{B}}\mapsto 2,{\text{C}}\mapsto 3.}$
However, there are other choices of isomorphism, such as
${\displaystyle {\text{A}}\mapsto 3,{\text{B}}\mapsto 2,{\text{C}}\mapsto 1,}$
and these sets cannot be identified without making such a choice — any statement that identifies them "depends on choice of identification". This distinction, between equality and isomorphism, is of fundamental importance in category theory and is one motivation for the development of category theory.
## Logical definitions
Leibniz characterized the notion of equality as follows:
Given any x and y, x = y if and only if, given any predicate P, P(x) if and only if P(y).
## Equality in set theory
Equality of sets is axiomatized in set theory in two different ways, depending on whether the axioms are based on a first-order language with or without equality.
### Set equality based on first-order logic with equality
In first-order logic with equality, the axiom of extensionality states that two sets which contain the same elements are the same set.[8]
• Logic axiom: x = y ⇒ ∀z, (zxzy)
• Logic axiom: x = y ⇒ ∀z, (xzyz)
• Set theory axiom: (∀z, (zxzy)) ⇒ x = y
Incorporating half of the work into the first-order logic may be regarded as a mere matter of convenience, as noted by Lévy.
"The reason why we take up first-order predicate calculus with equality is a matter of convenience; by this we save the labor of defining equality and proving all its properties; this burden is now assumed by the logic."[9]
### Set equality based on first-order logic without equality
In first-order logic without equality, two sets are defined to be equal if they contain the same elements. Then the axiom of extensionality states that two equal sets are contained in the same sets.[10]
• Set theory definition: "x = y" means ∀z, (zxzy)
• Set theory axiom: x = y ⇒ ∀z, (xzyz)
• Extensionality
• Homotopy type theory
• Inequality
• List of mathematical symbols
• Logical equality
• Proportionality (mathematics)
## Notes
1. ^ a b "Compendium of Mathematical Symbols". Math Vault. 1 March 2020. Retrieved 1 September 2020.
2. ^ Weisstein, Eric W. "Equality". mathworld.wolfram.com. Retrieved 1 September 2020.
3. ^ Rosser 2008, p. 163.
4. ^ Lévy 2002, pp. 13, 358. Mac Lane & Birkhoff 1999, p. 2. Mendelson 1964, p. 5.
5. ^ Weisstein, Eric W. "Equal". mathworld.wolfram.com. Retrieved 1 September 2020.
6. ^ "List of Geometry and Trigonometry Symbols". Math Vault. 17 April 2020. Retrieved 1 September 2020.
7. ^
8. ^ Kleene 2002, p. 189. Lévy 2002, p. 13. Shoenfield 2001, p. 239.
9. ^ Lévy 2002, p. 4.
10. ^ Mendelson 1964, pp. 159–161. Rosser 2008, pp. 211–213
## References
• Kleene, Stephen Cole (2002) [1967]. Mathematical Logic. Mineola, New York: Dover Publications. ISBN 978-0-486-42533-7.
• Lévy, Azriel (2002) [1979]. Basic set theory. Mineola, New York: Dover Publications. ISBN 978-0-486-42079-0.
• Mac Lane, Saunders; Birkhoff, Garrett (1999) [1967]. Algebra (Third ed.). Providence, Rhode Island: American Mathematical Society.
• Mazur, Barry (12 June 2007), When is one thing equal to some other thing? (PDF)
• Mendelson, Elliott (1964). Introduction to Mathematical Logic. New York: Van Nostrand Reinhold.
• Rosser, John Barkley (2008) [1953]. Logic for mathematicians. Mineola, New York: Dover Publication. ISBN 978-0-486-46898-3.
• Shoenfield, Joseph Robert (2001) [1967]. Mathematical Logic (2nd ed.). A K Peters. ISBN 978-1-56881-135-2.
By: Wikipedia.org
Edited: 2021-06-18 15:15:50
Source: Wikipedia.org
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# Ram is twice as fast as Aman and Aman is thrice as fast as Rohit in doing a work.
Q.Ram is twice as fast as Aman and Aman is thrice as fast as Rohit in doing a work. Working together they can finish the work in 10 days. In how many days can Aman and Rohit together finish the work.
A.15
B.20
C. 25
D. 27
E. 32
Solution:
Let the Rohit one day work be ‘x’
Aman’s one day work be ‘3x’
Ram’s one day work be ‘6x’
Therefore, three can do this particular work together = 1/10
\begin{array}{l} \therefore x+3 x+6 x=\frac{1}{10} \\ \Rightarrow 10 x=\frac{1}{10} \\ \therefore x=\frac{1}{100} \end{array}
So, Aman and Rohit can do this work is 3x+x=1/100+1/100
\begin{array}{l} \Rightarrow 3x+x=\frac{1}{100} \\ \Rightarrow 4x=\frac{1}{100} \\ \therefore x=25 \text { Days } \end{array}
Hence, Aman and Rohit can do this work in 25 days
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Convert indigenous decimal come fraction. Convert 0.45 to Fraction. Decimal to portion chart and also calculator. Writes any decimal number together a fraction.
You are watching: What is 0.45 as a fraction
## How to convert a Decimal to a fraction - Steps
Step 1: create down the decimal as a portion of one (decimal/1);Step 2: If the decimal is no a entirety number, multiply both top and bottom through 10 until you get an interger at the numerator.
Learn an ext reading the examples listed below or usage our self-explaining calculator above
## Convert decimal 0.05 to a fraction
0.05 = 1/20 together a fraction
### Step by step Solution
To transform the decimal 0.05 to a fraction follow this steps:
Step 1: write down the number together a portion of one:
0.05 = 0.05/1
Step 2: main point both top and also bottom through 10 for every number after ~ the decimal point:
As we have actually 2 number after the decimal point, we multiply both numerator and denominator through 100. So,
0.05/1 = (0.05 × 100)/(1 × 100) = 5/100.
Step 3: leveling (or reduce) the fraction:
5/100 = 1/20 when reduced to the most basic form.
## What is 0.45 together a fraction?
0.45 = 9/20 as a fraction
### Step by step Solution
To transform the decimal 0.45 to a fraction follow this steps:
Step 1: compose down the number as a fraction of one:
0.45 = 0.45/1
Step 2: multiply both top and bottom by 10 because that every number after the decimal point:
As we have 2 number after the decimal point, us multiply both numerator and also denominator through 100. So,
0.45/1 = (0.45 × 100)/(1 × 100) = 45/100.
Step 3: leveling (or reduce) the fraction:
45/100 = 9/20 when diminished to the easiest form.
## Equivalent portion for 1.3 percent
1.3 = 13/10 = 13/10 together a fraction
### Step by action Solution
To transform the decimal 1.3 come a portion follow this steps:
Step 1: compose down the number as a fraction of one:
1.3 = 1.3/1
Step 2: main point both top and also bottom by 10 for every number after ~ the decimal point:
As we have actually 1 numbers after the decimal point, we multiply both numerator and also denominator by 10. So,
1.3/1 = (1.3 × 10)/(1 × 10) = 13/10.
(This portion is alread reduced, us can"t minimize it any type of further).
As the numerator is greater than the denominator, we have an not correct fraction, so us can also express it together a combined NUMBER, for this reason 13/10 is also equal to 1 3/10 once expressed as a blended number.
## Conversion table: fraction to decimal inches and millimeter equivalence
To transform fractions to decimals and also millimeters and vice-versa use this formula: 1 customs = 25.4 mm exactly, therefore ...To convert from customs to millimeter main point inch worth by 25.4.To transform from millimeter inch divide millimeter worth by 25.4.
an easier way to do it is to use the table below. How?
### Example 1
Convert 1 1/32" come mm: find 1 1/32 and also read to the right under mm column! girlfriend will discover 26.1938.
### Example 2
Convert 0.875 decimal inches come inches (fraction form).Look down the decimal tower until you uncover 0.875, then review to the left to discover 7/8 inchesor relocate to the right shaft to discover the mm value!
fractioninchesmm
1/640.01560.3969
1/320.03130.7938
3/640.04691.1906
1/160.06251.5875
5/640.07811.9844
3/320.09382.3813
7/640.10942.7781
1/80.12503.1750
9/640.14063.5719
5/320.15633.9688
11/640.17194.3656
3/160.18754.7625
13/640.20315.1594
7/320.21885.5563
15/640.23445.9531
1/40.25006.3500
17/640.26566.7469
9/320.28137.1438
19/640.29697.5406
5/160.31257.9375
21/640.32818.3344
11/320.34388.7313
23/640.35949.1281
3/80.37509.5250
25/640.39069.9219
13/320.406310.3188
27/640.421910.7156
7/160.437511.1125
29/640.453111.5094
15/320.468811.9063
31/640.484412.3031
1/20.500012.7000
33/640.515613.0969
17/320.531313.4938
35/640.546913.8906
9/160.562514.2875
37/640.578114.6844
19/320.593815.0813
39/640.609415.4781
5/80.625015.8750
41/640.640616.2719
21/320.656316.6688
43/640.671917.0656
11/160.687517.4625
45/640.703117.8594
23/320.718818.2563
47/640.734418.6531
3/40.750019.0500
49/640.765619.4469
25/320.781319.8438
51/640.796920.2406
13/160.812520.6375
53/640.828121.0344
27/320.843821.4313
55/640.859421.8281
7/80.875022.2250
57/640.890622.6219
29/320.906323.0188
59/640.921923.4156
15/160.937523.8125
61/640.953124.2094
31/320.968824.6063
63/640.984425.0031
11.000025.4000
fractioninchesmm
1 1/641.015625.7969
1 1/321.031326.1938
1 3/641.046926.5906
1 1/161.062526.9875
1 5/641.078127.3844
1 3/321.093827.7813
1 7/641.109428.1781
1 1/81.125028.5750
1 9/641.140628.9719
1 5/321.156329.3688
1 11/641.171929.7656
1 3/161.187530.1625
1 13/641.203130.5594
1 7/321.218830.9563
1 15/641.234431.3531
1 1/41.250031.7500
1 17/641.265632.1469
1 9/321.281332.5438
1 19/641.296932.9406
1 5/161.312533.3375
1 21/641.328133.7344
1 11/321.343834.1313
1 23/641.359434.5281
1 3/81.375034.9250
1 25/641.390635.3219
1 13/321.406335.7188
1 27/641.421936.1156
1 7/161.437536.5125
1 29/641.453136.9094
1 15/321.468837.3063
1 31/641.484437.7031
1 1/21.500038.1000
1 33/641.515638.4969
1 17/321.531338.8938
1 35/641.546939.2906
1 9/161.562539.6875
1 37/641.578140.0844
1 19/321.593840.4813
1 39/641.609440.8781
1 5/81.625041.2750
1 41/641.640641.6719
1 21/321.656342.0688
1 43/641.671942.4656
1 11/161.687542.8625
1 45/641.703143.2594
1 23/321.718843.6563
1 47/641.734444.0531
1 3/41.750044.4500
1 49/641.765644.8469
1 25/321.781345.2438
1 51/641.796945.6406
1 13/161.812546.0375
1 53/641.828146.4344
1 27/321.843846.8313
1 55/641.859447.2281
1 7/81.875047.6250
1 57/641.890648.0219
1 29/321.906348.4188
1 59/641.921948.8156
1 15/161.937549.2125
1 61/641.953149.6094
1 31/321.968850.0063
1 63/641.984450.4031
22.000050.8000
fractioninchesmm
2 1/642.015651.1969
2 1/322.031351.5938
2 3/642.046951.9906
2 1/162.062552.3875
2 5/642.078152.7844
2 3/322.093853.1813
2 7/642.109453.5781
2 1/82.125053.9750
2 9/642.140654.3719
2 5/322.156354.7688
2 11/642.171955.1656
2 3/162.187555.5625
2 13/642.203155.9594
2 7/322.218856.3563
2 15/642.234456.7531
2 1/42.250057.1500
2 17/642.265657.5469
2 9/322.281357.9438
2 19/642.296958.3406
2 5/162.312558.7375
2 21/642.328159.1344
2 11/322.343859.5313
2 23/642.359459.9281
2 3/82.375060.3250
2 25/642.390660.7219
2 13/322.406361.1188
2 27/642.421961.5156
2 7/162.437561.9125
2 29/642.453162.3094
2 15/322.468862.7063
2 31/642.484463.1031
2 1/22.500063.5000
2 33/642.515663.8969
2 17/322.531364.2938
2 35/642.546964.6906
2 9/162.562565.0875
2 37/642.578165.4844
2 19/322.593865.8813
2 39/642.609466.2781
2 5/82.625066.6750
2 41/642.640667.0719
2 21/322.656367.4688
2 43/642.671967.8656
2 11/162.687568.2625
2 45/642.703168.6594
2 23/322.718869.0563
2 47/642.734469.4531
2 3/42.750069.8500
2 49/642.765670.2469
2 25/322.781370.6438
2 51/642.796971.0406
2 13/162.812571.4375
2 53/642.828171.8344
2 27/322.843872.2313
2 55/642.859472.6281
2 7/82.875073.0250
2 57/642.890673.4219
2 29/322.906373.8188
2 59/642.921974.2156
2 15/162.937574.6125
2 61/642.953175.0094
2 31/322.968875.4063
2 63/642.984475.8031
33.000076.2000
|
Question
1. # Does A Prime Multiplied By A Prime Ever Result In
## Introduction
A prime number is a whole number greater than 1 that has no positive divisors other than 1 and itself. A composite number is a whole number that is not prime. The product of two prime numbers is called a semiprime. The product of three prime numbers is called a triprime. The product of four prime numbers is called a quadprime. It’s easy to see that the product of two or more composite numbers can never be prime. But what about the product of two or more primes? In this blog post, we will explore whether or not a prime multiplied by a prime ever results in a composite number. We’ll also look at some interesting patterns that emerge when we multiply primes together.
## The Multiplication of Prime Numbers
When two prime numbers are multiplied together, the result is called a composite number. While there are an infinite number of prime numbers, there are only a finite number of composite numbers. So, it is possible that a prime multiplied by a prime could result in a composite number.
## Does A Prime Multiplied By A Prime Ever Result In?
When two prime numbers are multiplied together, the result is called a composite number. Composite numbers can be generated by multiplying any two whole numbers together, but they are not necessarily prime themselves. So, does a prime multiplied by a prime ever result in a composite number?
The answer is yes, a prime multiplied by a prime can result in a composite number. However, this is not always the case – there are some prime numbers that when multiplied together will produce another prime number. For example, 3 x 5 = 15. 15 is not a prime number, but 3 and 5 are both primes.
## The Consequences of Multiplying Prime Numbers
When two prime numbers are multiplied together, the result is called a composite number. Composite numbers have a variety of different properties, but one of the most notable is that they are not evenly divisible by any number other than themselves and 1. This means that if you were to multiply a composite number by a prime number, the resulting product would also be composite.
There are a few interesting consequences that come from this fact. For one, it means that there are an infinite number of composite numbers. This is because, as long as there are prime numbers, you can keep multiplying them together to get new composite numbers.
It also has some implications for multiplication in general. When you multiply two composite numbers together, the resulting product will also be composite. So, if you’re ever trying to calculate something and you run into two large composite numbers, you can be pretty confident that the answer will also be composite (unless one of the factors is 1).
## How to Use this Information
If you’re like most people, you probably think that a prime multiplied by a prime can never result in a non-prime. However, this is not always the case. In fact, there are some very simple examples of when this occurs.
To understand why this happens, let’s first review what a prime number is. A prime number is any whole number greater than 1 that can only be evenly divided by 1 and itself. So, for example, 3 is a prime number because the only whole numbers it can be evenly divided by are 1 and 3.
Now let’s look at an example of when a prime multiplied by a prime does result in a non-prime. Let’s say we have the following equation:
2 x 3 = 6
In this equation, 2 and 3 are both prime numbers. However, when we multiply them together, we get a non-prime answer of 6. This is because 6 can be evenly divided by 1, 2, 3, and 6 (in addition to other numbers). So while 2 and 3 are both prime individually, when they’re multiplied together they result in a non-prime.
There are other examples of this as well. For instance:
3 x 5 = 15
5 x 7 = 35
7 x 11 = 77
11 x 13 = 143
In each of these equations, both the multipliers and the product are non-prime numbers. So as you can see, it is possible for a prime multiplied by a prime to result in a non-prime.
If you’re interested in learning more about prime numbers, there are a number of resources available online and in print. A quick search should turn up plenty of material that you can use to improve your understanding of this topic.
## Conclusion
We hope this article has helped to clear up any confusion on whether or not a prime multiplied by a prime can ever result in a non-prime number. As it turns out, the answer is yes – it is possible for a prime multiplied by a prime to result in a non-prime number. However, this is relatively rare and only happens under specific circumstances. If you’re still curious about this topic, we recommend doing some further research or talking to a math expert to get clarification.
2. Does a prime multiplied by a prime ever result in a prime number?
It’s a question that has intrigued mathematicians for centuries. After all, prime numbers are among the most fundamental building blocks of mathematics and have a wide range of applications in the real world. But can a prime multiplied by another prime ever result in a prime number?
The answer is yes. It is possible for two prime numbers to be multiplied together and result in a prime number. This is known as the prime power theorem. In essence, this theorem states that the product of any two prime numbers is always a prime number.
For example, if we multiply the prime numbers 2 and 3, the result is 6. 6 is a prime number, so the prime power theorem holds true.
However, not all prime power combinations result in a prime number. If we multiply two prime numbers together and the result is not a prime number, then that combination is known as a composite number. For example, if we multiply the prime numbers 2 and 7, the result is 14, which is not a prime number.
This knowledge can be put to use in a variety of ways. For example, if you were looking for a prime number to use in a cryptographic algorithm, you could use the prime power theorem to ensure that the resulting number is indeed a prime number.
So, to answer the original question: yes, a prime multiplied by another prime can result in a prime number. While it’s not always the case, the prime power theorem states that the product of any two prime numbers is always a prime number.
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# Conditional Probability with red and blue balls
A jar contains 7 red balls and 9 blue balls. We choose, uniformly at random and without replacement, 3 balls. Define the following two events:
A = "exactly 2 of the balls are red"
B = "the number of red balls is even"
What is the conditional probability $Pr(A \vert B)$?
My attempt to obtain the answer to this begins by showing the formula used: $$Pr(A \vert B) = \frac{Pr(A\cap B)}{Pr(B)}$$
$Pr(B) = 9 \cdot {7 \choose 2}$ since we choose a combination of two balls from 3 and multiply it by 9, which is the selecting any of the blue balls to accommodate our reds.
$Pr(A \cap B) = {9 \choose 3} + 9\cdot {7 \choose 2}$ because this is the intersection of $Pr(A)$ and $Pr(B)$, we must take an even amount of reds. This means that ${9 \choose 3}$ is one of the cases where we choose zero reds (zero is even) plus the case where we choose two reds $9 \cdot {7 \choose 2}$
Therefore: $$Pr(A \vert B) = \frac{ {9\choose 3} + 9\cdot {7 \choose 2}}{9\cdot {7 \choose 2}}$$
Correct?
• If you have three balls in your hand, and the number of red ones is even, how many of the balls in your hand can be red? – Steve Kass Apr 18 '14 at 0:36
## 2 Answers
$$P(A|B) = \frac{P(A\cap B)}{P(B)}$$
$$= \frac{P(2 \,red \,balls\, AND \,even\, number\, of\, red\, balls\,)}{P(0\, red\, balls\, OR\, 2\, red\, balls)}$$
$$= \frac{P(2\,red\,balls)}{P(0\, red\, balls\, OR\, 2\, red\, balls)}$$
$$=\frac{\frac{\binom{7}{2}\binom{9}{1}}{\binom{16}{3}}}{\frac{\binom{9}{3}+\binom{7}{2}\binom{9}{1}}{\binom{16}{3}}}$$
$$=\frac{\binom{7}{2}\binom{9}{1}}{\binom{9}{3}+\binom{7}{2}\binom{9}{1}}$$
• I think you have properly defined $Pr(B)$ so your answer is correct. – GivenPie Apr 18 '14 at 1:05
Notice that, in your solution you have that $\Pr(B)=189$, which impossible. The same goes for your computation of $\Pr(A\cap B)$. If you want to use counting methods to calculate probabilities, don't forget to divide the "number of ways to satisfy the event" by the total number of possible outcomes!
As for a hint to solve your problem, think about how the events $A$ and $B$ are related to the others. For instance, if $A$ is satisfied, does that automatically mean that $B$ is also satisfied? Is the reverse also true? Do the answers to these questions simplify your task?
• I understand. Would $Pr(B) = {16 \choose 3}$ satisfy the clause? – GivenPie Apr 18 '14 at 0:23
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# Three point particles of mass 1 kg, 1.5 kg and 2.5 kg are placed at three corners of a right triangle of sides 4.0 cm, 3.0 cm and 5.0 cm as shown in the figure. The centre of mass of the system is at the point:A. 0.9 cm right and 2.0 cm above 1 kg massB. 2.0 cm right and 0.9 cm above 1 kg massC. 1.5 cm right and 1.2 cm above 1 kg massD. 0.6 cm right and 2.0 cm above 1 kg mass
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Hint:To solve this problem, we need to use a formula for the effective centre of mass of the system at a given point. In this question we have a system of 3 particles present on 3 corners of a right triangle, so the centre of mass will lie somewhere between them.
Formula used:
The centre of mass is given as,
Along X coordinate ${X_{COM}} = \dfrac{{\sum {{m_i}{x_i}} }}{{\sum {{m_i}} }}$
Along Y coordinate ${Y_{COM}} = \dfrac{{\sum {{m_i}{y_i}} }}{{\sum {{m_i}} }}$
Given (0,0), (3,0) and (0,4) to be the coordinates of mass 1kg, 1.5kg and 2.5kg by assuming 1kg at origin.
Image: Three point particles are placed at three corners of a right triangle.
${X_{COM}}$ can be given as,
${X_{COM}} = \dfrac{{\sum {{m_i}{x_i}} }}{{\sum {{m_i}} }}$
${X_{COM}} = \dfrac{{1 \times 0 + 1.5 \times 3 + 2.5 \times 0}}{{1 + 1.5 + 2.5}}$
$= \dfrac{{4.5}}{5} = 0.9cm$
${Y_{COM}}$ can be given as,
${Y_{COM}} = \dfrac{{\sum {{m_i}{y_i}} }}{{\sum {{m_i}} }}$
${Y_{COM}} = \dfrac{{1 \times 0 + 1.5 \times 0 + 2.5 \times 4}}{{1 + 1.5 + 2.5}}$
$= \dfrac{4}{2} = 2cm$
Therefore, the centre of mass is 0.9 cm to the right and 2 cm above the mass of 1kg.
Hence option A is the correct answer
Note: We use the formula for centre of mass for individual coordinates in the x-y-z plane which is only applicable for point objects, for an extended object or non-uniform object like a rod, we need to consider differential mass and its position and integrate over its entire length.
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# NCERT Solutions for Class 12 Physics Chapter 9 - Ray Optics and Optical Instruments
## NCERT Solutions for Class 12 Physics Chapter 9 Free PDF Download
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1. A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. at what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Sol. Given, radius of curvature, R = – 36 m
$$\therefore\space\text{Focal length f =}\frac{\text{R}}{2}=\frac{36}{2}=-18\space\text{cm}$$
Height of object O = 2.5 cm
$$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\Rarr\space=-\frac{1}{18}=-\frac{1}{v}-\frac{1}{27}\\\frac{1}{v}=\frac{1}{18}+\frac{1}{27}\\=\frac{-3+2}{54}=-\frac{1}{54}$$
∴ Distance of screen from mirror v = – 54 cm.
Magnification for mirror
$$m =-\frac{v}{u}=\frac{\text{I}}{\text{Q}}\\=-\frac{(-54)}{-27}=\frac{1}{2.5}$$
I = – 5 cm
The negative sign shows that an inverted image is formed in front of the mirror. Thus, the screen should be placed at a distance 54 cm and image real, inverted and magnified in nature.
If we move the candle near to the mirror the screen should be moved away from mirror. As the object distance is less than focal length and no screen is required, because the image formed is virtual.
2. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification.
Describe what happens as the needle is moved farther from the mirror?
Sol. Given, focal length f = + 15 cm
Distance of object u = – 12cm
Size of the object O = 4.5 cm
Using formula,
$$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\frac{1}{15}=\frac{1}{v}-\frac{1}{12}\\\Rarr\space\frac{1}{v}=\frac{1}{15}+\frac{1}{12}\\=\frac{4+5}{60}=\frac{9}{60}$$
Distance of image from the mirror v = 6.7 cm.
The image is formed behind the mirror.
The magnification,
$$\text{m}=-\frac{v}{u}=\frac{\text{I}}{\text{O}}\\\frac{-6.7}{-12}=\frac{1}{4.5}$$
Size of image I = 2.5 cm
As I is positive, so image is erect and virtual.
The magnification
$$\text{m}=\frac{\text{I}}{\text{O}}=\frac{2.5}{4.5}=\frac{25}{45}=\frac{5}{9}$$
As the object moves away from the mirror, the image also moves away from the mirror (as u → ∞, v → f) and the size of image will be decreased.
3. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Sol. Case I: When tank is filled with water
the apparent depth = 9.4 cm
Height of water t = 12.5 cm
So, real depth = 12.5 cm
Refractive index of water
$$\mu_{w}=\frac{\text{Real depth}}{\text{Apparent depth}}\\=\frac{12.5}{9.4}=1.33$$
Case II: When tank is filled with the liquid
Refractive index of liquid µ = 1.63,
$$\mu=\frac{\text{Real depth}}{\text{Apparent depth}}\\\therefore\space 1.63=\frac{12.5}{\text{Apparent depth}}$$
∴ The microscope is moved by = 9.4 – 7.67 = 1.73 cm
4. Figure (a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water in 45° with the normal to a water-glass interface [Fig. (c)].
Sol. Given, from Fig. (a)
Angle of incidence i = 60°
Angle of refraction r = 35°
Refractive index of glass w.r.t. air,
$$^{a}\mu_{g}=\frac{\text{sin}60\degree}{\text{sin}35\degree}=\frac{0.8660}{0.5736}$$
= 1.51 …(i)
From Fig. (b)
Angle of incidence i = 60°,angle of refraction r = 47°
Refractive index of water w.r.t. to air
$$^{a}\mu_{w}=\frac{\text{sin}60\degree}{\text{sin}47\degree}=\frac{0.8660}{0.7314}$$
= 1.18 …(ii)
From Fig. (c)
Angle of incidence i = 45°, let r be the angle of refraction
Refractive index of glass w.r.t. water
$$^{w}\mu_{g}=\frac{\text{sin} \space45\degree}{\text{sin}\space r\degree}\space...\text{(iii)}\\\text{As, we know that}^{w}\mu_{g}=\frac{^{a}\mu_{g}}{^{a}\mu_{w}}$$
Putting the value of wmg in Eq. (iii), we get
$$\frac{^{a}\mu_{g}}{^{a}\mu_{w}}=\frac{0.7071}{\text{sin r}}\\\frac{1.51}{1.32}=\frac{0.7071}{\text{sin r}}$$
From equations (i) and (ii), w
e get
$$\text{sin r =}\frac{1.32×0.7071}{1.51}=0.6181$$
r = 38.2°
5. A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Sol. Let the bulb is placed at point O
AB = AC = r
If the light incident at an angle equal to critical angle ic, then a circular area is formed only.
The light source is 80 cm below the water surface i.e. AO = 80 cm, µw = 1.33
Using the formula for critical angle,
$$\text{sin}\space i_{c}=\frac{1}{\mu_{w}}\\\text{sin}_{i_{c}}=\frac{1}{1.33}=0.75\\i_c = 48.6°\\\text{In} \Delta \text{OAB tan i}_\text{c =}\frac{\text{AB}}{\text{AO}}\\\text{or}\space\text{tan}\space i_{c}=\frac{r}{l}$$
r = l tan ic = 80 tan 48.6
r = 80 × 1.1345 = 90.7 cm
Area of circular surface of w
ater
A = πr2
A = 3.14 × (90.7)2 = 25865.36 cm2
A = 2.58 m2
6. A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
Sol. Given, angle of minimum deviation δm = 40°
The angle of the prism A = 60°
Refractive index of glass w.r.t. air
$$^{a}\mu_{g}=\frac{\text{sin}\bigg(\frac{\text{A}+\delta_{m}}{2}\bigg)}{\text{sin}\frac{\text{A}}{2}}\\=\frac{\text{sin}\bigg(\frac{60\degree+40\degree}{2}\bigg)}{\text{sin 30\degree}}=\frac{\text{sin} 50\degree}{\text{sin} 30\degree}\\\\^{a}\mu_{g}=\frac{0.766}{0.5}=1.532$$
When prism is immersed in water
The refractive index of water w.r.t. air
amw = 1.33
Refractive index of glass with respect to water
$$^{a}\mu_{g}=\frac{\text{sin}\bigg(\frac{\text{A +}\delta'_{m}}{2}\bigg)}{\text{sin}\frac{\text{A}}{2}}\\\therefore\space\frac{^{a}\mu_{g}}{^{a}\mu_{w}}=\frac{\text{sin}\bigg(\frac{\text{A+}\delta'm}{2}\bigg)}{\text{sin} 30\degree}\\\text{sin}\bigg(\frac{A+\delta'_m}{2}\bigg)=\frac{1.532×\text{sin} 30\degree}{1.33}\\\text{sin}\bigg(\frac{A + \delta'_m}{2}\bigg)=\frac{0.5×1.532}{1.33}=0.5759\\\text{sin}\bigg(\frac{\text{A + }\delta '_m}{2}\bigg)=\text{sin} 35\degree 10'$$
δ’m = 2 (30°10’) – 60°
= 70°20’ – 60°
δ’m = 10°20’
Thus, the angle of minimum deviation become 10°20’.
7. Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
Sol. For double convex lenses R1 = R, R2 = – R
Focal length f = + 20 cm
By Lens maker’s formula
$$\frac{1}{f}=(^{a}\mu_{g}-1)\bigg(\frac{1}{\text{R}}_{1}-\frac{1}{\text{R}}_{2}\bigg)\\\frac{1}{20}=(1.55-1)\bigg(\frac{1}{\text{R}}+\frac{1}{\text{R}}\bigg)\\\frac{1}{20}=0.55×\frac{2}{\text{R}}$$
R = 0.55 × 2 × 20 = 22 cm
Thus, the radius of curvature is 22 cm.
8. A beam of light converges at a point P, Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?
Sol. Here, the point P is on the right side of lens acts as vertical object.
(a) Given, object distance
u = 12 cm
Focal length f = + 20 cm
$$\text{By Lens formula,}\space\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\frac{1}{v}-\frac{1}{12}=\frac{1}{20}\\\frac{1}{v}=\frac{1}{20} + \frac{1}{12}\\=\frac{3+5}{60}\\=\frac{8}{60}$$
v = 7.5 cm
Thus, the beam converges at a distance of 7.5 cm on the right side of lens.
(b) Distance of object from the lens u = 12 cm
Focal length t = – 16 cm
Using Lens formula,
$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\frac{1}{v}-\frac{1}{12}=-\frac{1}{16}\\\frac{1}{v}=\frac{1}{12}-\frac{1}{16}=\frac{4-3}{48}$$
v = 48 cm
So, the beam converges on the right side of lens at a distance of 48 cm.
9. An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
Sol. Object size, O = 3 cm
Focal length f = – 21 cm
Distance of object from the concav
e lens u = – 14cm)
By Lens formula,
$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\frac{1}{v} + \frac{1}{14}=-\frac{1}{12}\\\Rarr\space\frac{1}{v}=-\frac{1}{21}-\frac{1}{14}\\=\frac{-2-3}{42}\\=-\frac{5}{42}$$
v = – 8.4 cm
The magnification
$$\text{m}=-\frac{v}{u}=\frac{\text{I}}{\text{O}}\\\frac{(-8.4)}{-14}=\frac{\text{I}}{3}$$
I = 1.8 cm
Since, I is positive, the image formed is virtual and erect at a distance of 8.4 cm in front of lens. If the object moves further away from the lens, the image moves towards the lens. The size of image decreases gradually.
10. What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
Sol. Here, focal length of convex lens f1 = 30 cm.
Focal length of concave lens f2 = – 20 cm
We know that,
$$\frac{1}{f}=\frac{1}{f}_{1} + \frac{1}{f}_{2}\\= \frac{1}{30}-\frac{1}{20}\\=\frac{2-3}{60}=-\frac{1}{60}$$
f = – 60 cm
Since, the focal length of combination is negative. Hence, combination behaves like a diverging lens i.e. as a concave lens.
11. A compound microscope consists of an objective lens of focal length 2.0 cm and an eye-piece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
Sol. Given, focal length of objective lens, fo = 2 cm
Focal length of eye-piece fe = 6.25 cm
Distance between both these lenses
v = 15 cm
(a) Distance of final image from eye-piece
ve = – 25 cm
Using the Lens formula for eye-piece
$$\frac{1}{V_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}\\\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}\\=\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}$$
ue = – 5 cm
As the distance between objective and eye-piece (vo + ue) = 15 cm
L = vo + ue
Distance of image formed by objective lens
vo = L – |ue| = 15 – 5 = 10 cm
For objective lens
$$\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}\\\frac{1}{u_{0}}=\frac{1}{v_{0}}-\frac{1}{f_{0}}\\=\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=-\frac{4}{10}$$
uo = – 2.5 cm
So, the object should at distance 2.5 cm in front of convex lens.
Magnifying power of compound microscope
$$m=\frac{v_{0}}{u_{0}}\bigg(1+\frac{d}{f_{e}}\bigg)\\=\frac{10}{2.5}\bigg(1+\frac{25}{6.25}\bigg)\space(\because d=25\text{cm})$$
m = 20
(b) The final image will be formed at infinity only if the image formed by the objective is in the focal plane of the eye-piece. Thus, here ve = – ∞, ue = fe = 6.25 cm
Image distance of objective lens
vo = L – fe = 15 – 6.25 = 8.75 cm
Using Lens formula of objective lens
$$\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}\\\frac{1}{u_{0}}=\frac{1}{v_{0}}-\frac{1}{f_{0}}\\=\frac{1}{8.75}-\frac{1}{2}=\frac{2-8.75}{17.5}\\u_{0}=-\frac{17.5}{6.75}=-2.59\space\text{cm}$$
Magnifying power of the microscope
$$m=\frac{v_{0}}{u_{0}}\bigg(1+\frac{d}{f_{e}}\bigg)\\=\frac{8.75}{2.59}\bigg(1+\frac{25}{6.25}\bigg)$$
m = 13.51
12. A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eye-piece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Sol. Given, focal length of objective f0 = 8 mm = 0.8 cm
Focal length of eye-piece fo = 2.5 cm
Distance of objective lens – uo = – 9 mm = – 0.9 cm
Distance of image from eye-piece
ve = – d = – 25 cm
By Lens equation for eye-piece
$$\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}\\\text{or}\space\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}\\=-\frac{1}{25}-\frac{1}{2.5}=\frac{-1-10}{25}=-\frac{11}{25}$$
ue = – 2.27 cm
By Lens equation for objective
$$\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}\\\text{or}\space\space\frac{1}{v_{0}}=\frac{1}{f_{0}} + \frac{1}{u_{0}}\\=\frac{1}{0.8}-\frac{1}{0.9}\\\frac{0.9-0.8}{0.72}=\frac{0.1}{0.72}$$
Distance of image for objective lens
vo = 7.2 cm
Separation between two lenses
L = |ue| + |vo|
= 2.27 + 7.2
L = 9.47 cm
Magnifying power of compound microscope
$$m=\frac{v_{0}}{u_{0}}\bigg(1+\frac{d}{f_{e}}\bigg)=\frac{7.2}{0.9}\bigg(1+\frac{25}{2.5}\bigg)$$
m = 88
13. A small telescope has an objective lens of focal length 144 cm and an eye piece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eye-piece?
Sol. Given, focal length of objective lens fo = 144 cm
Focal length of eye-piece fe = 6 cm
Magnifying power of the telescope when the final image is formed at ∞.
$$m=\frac{f_{0}}{f_{e}}=-\frac{144}{6}=-24$$
∴ Separation between lenses L = fo + fe = 144 + 6
= 150 cm.
14. (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eye-piece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 106 m, and the radius of lunar orbit is 3.8 × 108m.
Sol. Given, focal length of objective lens fo = 15 m
Focal length of eye-piece fe = 1 cm = 0.01 m
(a) Angular magnification
$$m=\frac{f_{0}}{f_{e}}=\frac{15}{0.01}=1500$$
Let d be the diameter of the image of the moon formed by the objective lens.
$$\therefore\space\text{Angle subtended by the image =}\frac{d}{f_{0}}=\frac{d_{i}}{15}$$
(b) Diameter of moon do = 3.48 × 106 m
Radius of lunar orbit r = 3.8 × 108 m
The angle subtended by the moon
$$=\frac{\text{Diameter of moon}}{\text{Radius of lunar orbit}}\\=\frac{3.48×10^{6}}{3.8×10^{8}}$$
But we know that the angle subtended by the image is equal to the angle subtended by the object.
$$\therefore\space\frac{d_{i}}{15}=\frac{3.48×10^{6}}{3.8×10^{8}}\\\text{or}\space d_{i}=\frac{3.48×15×10^{\normalsize-2}}{3.8}$$
= 13.73 × 10–2m
or di = 13.73 cm
Thus, the diameter of the image is 13.73 cm.
15. Use the mirror equation to deduce that:
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
$$\textbf{Sol.}\space\text{(a) The mirror formula}\space\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\\\text{According to question,}\\\text{f < u < 2f}\\\frac{1}{2f}>\frac{1}{u}>\frac{1}{f}\text{or}-\frac{1}{2f}<-\frac{1}{u}<-\frac{1}{f}\\\text{Add}\space\frac{1}{f}\space\text{on both sides, we get}\\\frac{1}{f}-\frac{1}{2f}<\frac{1}{f}-\frac{1}{u}<0\space\text{...(i)}\\\frac{1}{f}-\frac{1}{2f}<\frac{1}{v}\space\bigg[\because\frac{1}{f}-\frac{1}{u}=\frac{1}{v}\bigg]\\\frac{1}{2f}<\frac{1}{v}$$
v = 2f
As, f is negative, hence v will also be negative. So, the image formed is real and image lies beyond 2f.
(b) For convex mirror f > 0. As object always placed on left side of mirror i.e, object distance u > 0.
$$\text{Lens formula,}\space\frac{1}{v}=\frac{1}{f}-\frac{1}{v}\text{as f > 0, and u < 0}$$
Thus, v is always positive. The image formed is virtual. It is independent from the location of object.
$$\text{(c) For convex mirror f} > 0, u < 0\space\text{and}\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\text{so}\frac{1}{u}>\frac{1}{f}\text{or v} < f$$
Thus, image always located between pole and focus, as v < |u|. So the image is always diminished.
(d) For concave mirror f < 0.
As object is placed between pole and focus.
i.e. f < u < 0
$$\therefore\space\frac{1}{f}-\frac{1}{u}>0\\\text{By Lens formula,}\\\frac{1}{f}-\frac{1}{u}=\frac{1}{v}>0\Rarr v>0$$
It means that v is positive, image formed up right and virtual.
$$\frac{1}{v}<\frac{1}{u}$$
v > |u| so image is enlarged.
16. A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?
Sol. Given, thickness of glass slab (real depth) = 15 cm.
Refractive index of glass omg with respect to air = 1.5
$$^o\mu_g=\frac{\text{Real depth}}{\text{Apparent depth}}=\frac{15}{^a\mu_g}$$
Apparent depth of pin y = 15/1.5 = 10 cm
Distance by which the pin appears to be raised
= Real depth – Apparent depth
= 15 – 10 = 5 cm
The answer does not depend on the location of the slab.
17. (a) Figure shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.
(b) What is the answ
er if there is no outer covering of the pipe?
Sol. (a) Given, refractive index of the glass fibre with respect to air
µ2 = aµg = 1.68
Refractive index of the outer material with respect to air
µ1 = aµouter = 1.44
Let the critical angle be ic.
$$\mu=\frac{\mu_{2}}{\mu_{1}}=\frac{1}{\text{sin} i_{c}}\\\text{sin}\space i_{c}=\frac{\mu_{1}}{\mu_{2}}=\frac{1.44}{1.68}=0.8571$$
ic = 59°
The total internal reflection will take place for the angle of incidence i will be greater than the critical angle ic.
i.e., i > 59° or when angle of refraction, r < rmax
where rmax = 90° – ic = 90° – 59° = 31°
$$\therefore\space^{a}\mu_{g}=\frac{\text{sin i}_{max}}{\text{sin r}_{max}}=1.68$$
or sin imax = 1.68 sin 31° = 1.68 × 0.5156
imax = sin–1 (0.8662) = 60°
Thus, all the rays 0 < i < 60°, will suffer total internal reflection in the pipe.
(b) If there is no outer covering of the pipe then reflection inside the pipe will take place from the glass to air
$$\text{sin}\space i'_{c}=\frac{\mu_{1}}{\mu_{2}}=\frac{1}{1.68}=0.5952\\\text{Critical angle i’c = 36.5°}\\\text{Now, i = 90, we have r = 36.5°, r = 1.68 =}\frac{\text{sin 90\degree}}{\text{sin r}}$$
So i’ f = 90 – 36.5 = 53.5°
Here, i’ is greater than the critical angle. Thus, all the rays incident at an angles in the range zero to 90° will suffer total internal reflection.
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?
(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?
(d) Does the apparent depth of
a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?
(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?
Sol. (a) Yes, plane mirror and convex mirror both produces the real image if the rays incident on the plane or convex mirror are converging to a point behind the mirror. Because they are reflected to a point in front of the mirror. In other words we can say a plane or convex mirror can produce a real image if the object is virtual.
(b) No, there is no contradiction because virtual image formed by mirror acts as virtual object for eye lens. Our eye lens is convergent and it forms a real image of virtual object on retina.
(c) As the rays of light travels from rarer to denser medium, they bends towards the normal. So, the fisherman appears taller.
(d) Yes, the apparent depth will be decrease, further, when water tank is viewed obliquely as compared to the real depth.
$$\text{(e) Refractive index}\spaceµ _{diamond}> µ_{glass} \text{Refractive index µ =}\frac{1}{\text{sin}i_{c}}$$
where, ic is the critical angle.
As the refractive index of diamond is more than that of water, so the critical angle for glass is more than diamond. A diamond cutter, cuts the diamond at large range of angle of incidence to ensure that light entering in the diamond suffers multiple total external reflections and gives the sparkling in diamond.
19. The image of a small electric bulb fixed on the wall of a rooms is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
Sol. Suppose the object is placed at u in front of the lens and the distance of image from the lens is (3 – u)
i.e., v = (3 – u)n
$$\text{From Lens formula,}\space\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\frac{1}{(3-u)}-\frac{1}{(-u)}=\frac{1}{f}\\\text{or}\space\frac{1}{(3-u)} + \frac{1}{u}=\frac{1}{f}\\\text{or}\space\frac{u+3-u}{u(3-u)}=\frac{1}{f}$$
3f = 3u – u2
u2 – 3u + 3f = 0
$$\text{Now,}\space u=\frac{-(-3)\pm\sqrt{9-4×(3f)}}{2}\\u=\frac{+3\sqrt{9-2f}}{2}$$
Condition for image to be formed real
9 – 12f ≥ 0
or
9 ≥ 12f or f ≤ 0.75m
Thus, the maximum possible focal length of the lens required for his purpose is 0.75m.
20. A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
Sol. Given, distance between screen and object
a = 90 cm
Distance between two positions of the lens
d = 20 cm
By displacement formula
$$f=\frac{a^{2}-d^{2}}{4a}=\frac{(90)^{2}-(20)^{2}}{4×90}\\=\frac{7700}{360}=21.4\space\text{cm}$$
21. (a) Determine the ‘effective focal length’ of the combination of the two lenses in questions 10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.
Sol. (a) Given, the focal length of convex lens
f1 = 30 cm
The focal length of concave lens
f2 = –20 cm, d = 8 cm
Now,
$$\frac{1}{f}=\frac{1}{f_{1}} + \frac{1}{f_{2}}-\frac{d}{f_{1}f_{2}}\\\frac{1}{f}=\frac{1}{30}-\frac{1}{20}-\frac{8}{30×(-20)}\\=\frac{20-30+8}{30×20}\\\frac{1}{f}=-\frac{2}{600}$$
f = – 300 cm
(i) Let the incident beam falls on convex lens and assume that concave lens is not there.
u1 = ∞, f1 = 30 cm
Using the Lens formula
$$\frac{1}{f_{1}}=\frac{1}{v_{1}}-\frac{1}{u_{1}}\\=\frac{1}{30}=\frac{1}{v_{1}}-\frac{1}{\infty}$$
Position of the image formed by convex lens
v1 = 30 cm
This image acts as an object for concave lens.
Now, distance of object.
u2 = + (30 – 8) = 22 cm
f2 = – 20 cm
Using Lens formula
$$\frac{1}{f_{1}}=\frac{1}{v_{1}} - \frac{1}{u_{1}}\\\frac{1}{30}=\frac{1}{v_{1}}-\frac{1}{\infty}$$
Position of the image formed by convex lens
v1 = 30 cm
This image acts as an object for concave lens.
Now, distance of object.
u2 = + (30 – 8) = 22 cm
f2 = – 20 cm
Using Lens formula
$$\frac{1}{f_{2}}=\frac{1}{v_{2}}-\frac{1}{u_{2}}\\\frac{1}{v_{2}}=-\frac{1}{20}+\frac{1}{22}=-\frac{1}{220}$$
Distance of final image from lens v2 = – 220 cm.
Thus, the parallel beam would appear to diverge from a point at a distance 220 – 4 = 216 cm from the centre of two lens system.
(ii) Let us take that the parallel beam first falls on concave lens.
u1 = – ∞, f1 = –20 cm, v1 = ?
$$\frac{1}{v_{1}}+\frac{1}{\infty}=\frac{1}{-20}$$
v1 = – 20 cm
This acts as an object for convex lens.
u2 = – (20 + 8) = – 28 cm, f2 = 30 cm, v2 = ?
$$\frac{1}{v_{2}}+\frac{1}{28}=\frac{1}{30}\\\frac{1}{v_{2}}=\frac{1}{30}-\frac{1}{28}\\=\frac{14-15}{420}=-\frac{1}{420}$$
v2 = – 420 cm
The beam appears to diverge from a point 420 – 4 (= 416) cm on the left of the centre of the two lens system. From these two cases it concludes that the answer depends on which side of the lens system the parallel beam is incident. So, the notion of effective focal length does not useful here.
(b) Given, size of object O1 = 1.5 cm Distance from the convex lens u1 = – 40 cm,
$$\text{For the convex lens,}\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}\\=\frac{1}{v_{1}}=\frac{1}{30} + \frac{1}{(\normalsize-40)}\\=\frac{4-3}{120}=\frac{1}{120}$$
v1 = 120cm
Magnification produced by convex lens
Now, this image acts as an object for concave lens.
∴ For concave lens u2 = 120 – 8 = 112 cm, f2 = – 20 cm
$$\frac{1}{v_{2}}=\frac{1}{-20} + \frac{1}{112}\\=\frac{112+20}{112×20}=\frac{-92}{112×20}\\\text{v}_{2}=\frac{-112×20}{92}\text{cm}$$
Magnification produced by concave lens
$$\text{m}_{2}=\frac{v_{2}}{u_{2}}=\frac{(-112×20)}{92×112}=\frac{20}{92}\\\text{Magnification produced by combination}\\m = m_1 × m_2\\\text{m = (–3)X}\\=\bigg(-\frac{20}{90}\bigg)=\frac{60}{92}$$
Magnification produced by combination
m = m1 × m2
m = (–3)X
$$=\bigg(-\frac{20}{90}\bigg)=\frac{60}{92}$$
m = 0.652
Size of image = m × size of objects
= 0.652 × 1.5 = 0.98 cm
Thus, the magnification produced by the two lens system is 0.652 and size of image is 0.98 cm.
22. At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.
Sol. Angle of prism, A = 60°
Refractive index µ = 1.524
The critical angle is ic because it just suffers total internal refraction.
$$\text{sin}\space i_{c}=\frac{1}{\mu}\\=\frac{1}{1.524}$$
= 0.6561
ic = 41°
For a prism r1 + r2 = A here r2 = ic
∴ r1 + ic = A
r1 + 41° = 60°
⇒ r1 = 19°
$$\text{Now}\space \mu=\frac{\text{sin}i_{1}}{\text{sin}r_{1}}$$
or sin i1 = 1.524 sin 19° = 1.524 × 0.3256
or i1 = sin–1 (0.4962)
i1 = 29°75’
So, the angle should be 29°75’.
23. A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
Sol. (a) Given, focal length f = 10 cm, u = – 9 cm
Size of object = 1 mm
We know that
$$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\\frac{1}{10}=\frac{1}{v}+\frac{1}{9}\\\Rarr\space\frac{1}{v}=\frac{1}{10}-\frac{1}{9}=-\frac{1}{90}\\\text{or v = – 90 cm}\\\text{The magnification by the lens}\\m=+\frac{v}{u}=-\frac{(+90)}{-9}=10\\m=\frac{v}{u}=\frac{\text{I}}{\text{O}}$$
I = O × m = 1 × 10 = 10 mm
Area of each square in virtual image = (10)2 = 100 mm2
Thus, the magnification is 10 and area of each square in the virtual image is 10 mm2.
$$\text{(b) Angular magnification, m =}\frac{d}{u}=\frac{25}{9}=2.8$$
(c) No, they are equal if v = d.
(where d is the least distance of distinct vision).
24. (a) At what distance should the lens he held from the figure in question 29 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.
Sol. (a) Given, v = – 25 cm and f = 10 cm.
For maximum magnifying power the image should be formed at least distant of distinct vision.
By Lens formula
$$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\\Rarr\space\frac{1}{10}=\frac{1}{-25}-\frac{1}{u}\\=\frac{1}{u}=\frac{1}{-25}-\frac{1}{10}=\frac{-2-5}{50}=-\frac{7}{50}$$
= – 7.14 cm
$$\text{(b) Magnification m =}\frac{v}{u}=\frac{-25}{-7.14}=3.5\\\text{(c) Magnifying power =}\frac{d}{u}=\frac{25}{7.14}=3.5$$
Yes, the magnification equal to the magnifying power because, the image formed at least distance of distinct vision, i.e., 25 cm.
25. What should be the distance between the object in question 24 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?
Sol. Given, area of image, AI = 6.25 mm2
Area of object,
Ao = 1 mm2
Focal length of lens, f = 10 cm
$$\text{Linear magnification , m =}\sqrt{\frac{\text{A}_{\text{I}}}{\text{A}_{0}}}=\sqrt{\frac{6.25}{\text{I}}}=2.5\\\text{Again, magnification, m =}\frac{v}{u}$$
v = m × u= 2·5 × u …(i)
From Lens formula
$$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\=\frac{1}{10}=\frac{1}{2.5 u}-\frac{1}{u}$$
From Eq. (i),
$$\text{or}\space\frac{1}{10}=\frac{1}{u}\bigg(\frac{1-2.5}{2.5}\bigg)\\\text{or}\space u=\frac{-1.5×10}{2.5}=-6\text{cm}$$
or v = 2.5u = 2.5 (–6) = – 15cm
Thus, the virtual image is formed at a distance of 15 cm. But it is less than least distance of distinct vission. So, it cannot be seen by the eyes distinctly.
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense than does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eye-piece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eye-piece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eye-piece?
Sol. (a) Since, the size of image is much bigger than the size of object and, angular size of image is equal to angular size of object. Due to help of a magnifying glass one can see the objects placed closer than the least distance of distant vision. As closer the object, larger be the angular size.
(b) Yes, the angular magnification changes. As the distance between eye and magnifying glass is increased, the angular magnification decreases.
(c) We cannot make the lenses with very small focal length easily.
(d) The angular magnification produced by the eye-piece of a compound microscope
$$=\bigg(\frac{25}{f_{e}}+1\bigg).$$
If fe is small, the angular magnification will be large.
Further, magnification of objective lens is $$\frac{v}{u}$$ As object lies close to focus of objective lens u ≈ f0. To increase this magnification
$$\bigg(\frac{v}{f_{0}}\bigg).$$
fo should be smaller.
Thus fo and fe both are small.
(e) When we place our eyes very close to the eye-piece of a compound microscope, we are not able to collect the refracted light in large amount because the field of view will be reduced. So, the image is blurred.
The best position of the eye for viewing through a compound microscope is at the eye ring attached to the eye-piece.
27. An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eye-piece of focal length 5 cm. How will you set up the compound microscope?
Sol. Given, focal length of objective, fo = 1.25 cm
Focal length of eye-piece, fe = 5 cm
Least distance of distinct vision, d = 25 cm
Angular magnification, me = 30
The magnification produced by eye-piece
$$m_{e}=1+\frac{d}{f_{e}}=1+\frac{25}{5}=6$$
The magnification produced by microscope
m = mo × me
30 = mo × 6
where, mo is the magnification produced by objective lens.
mo = 5
Again, we know that magnification of objective lens
$$m_{0}=\frac{v_{0}}{u_{0}}\\5=\frac{-v_{0}}{u_{0}}\\\text{or}\space\space\text{v}_{0}=-5u_{0}\qquad\text{...(i)}\\\frac{1}{f_{0}}=\frac{1}{v_{0}}-\frac{1}{u_{0}}\\\frac{1}{1.25}=\frac{1}{-5u_{0}}-\frac{1}{u_{0}}=\frac{6}{5\mu_{0}}$$
From Eq. (i), we get
$$\mu_{0}=-\frac{6}{5}×1.25=-1.5\space\text{cm}$$
vo = –5uo = – 5 (– 1.5) = 7.5 cm
Thus, the object is placed at a distance of 1.5 cm from the objective lens to get the desired magnification.
Now, applying the Lens formula for eye-piece
$$\frac{1}{f_{e}}=\frac{1}{v_{e}}-\frac{1}{u_{e}}\\\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}\\=-\frac{1}{25}-\frac{1}{5}=-\frac{6}{25}\\(\because v_{e}=-25\space\text{cm})$$
ue = – 4.17 cm
The separation between objective and eye-piece.
|vo|+ |ue| = 4.17 + 7.5
= 11.67 cm
Thus, the microscope is adjusted such as the distance between eye-piece and objective is 11.67 cm.
28. A small telescope has an objective lens of focal length 140 cm and an eye-piece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e., when the final image is at infinity) ?
(b) the final image is formed at the least distance of distinct vision (25 cm)?
Sol. (a) For normal adjustment of telescope the magnification
$$m=\frac{f_{o}}{f_{e}}=-\frac{140}{5}=-28$$
(b) When final image is formed at least distance of distinct vision, the magnification
$$m=\frac{f_{0}}{f_{e}}\bigg(1+\frac{f_{e}}{d}\bigg)=\frac{140}{5}\bigg(1+\frac{5}{25}\bigg)$$
m = 28 (1 + 0.2) = 33.6
29. (a) For the telescope described in Q. 28 (a), what is the separation between the objective lens and the eye-piece?
(b) If this telescope is used to view a 100m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25 m?
Sol. Give, fo = 140 cm, fe = 5 cm and d = 145 cm
(a) In normal adjustment, the separation between eye-piece and objective lens
= fo + fe = 140 + 5 = 145 cm
(b) Height of tower, Ot = 100m
Distance of tower, u = 3 km = 3000 m
The angle subtended
$$\theta_{0}=\frac{\text{O}_t}{u}=\frac{100}{3000}=\frac{1}{30}\space\text{rad}\space\text{...(i)}\\\text{The angle subtended by the image.}\\\theta_{i}=\frac{\text{I}_{t}}{f_{0}}=\frac{\text{I}_{t}}{140}\\(∵\space\text{I}_{t}= height of image tower)\\\text{As}\space\theta_o = \theta_i\\\frac{1}{30}=\frac{\text{I}_{t}}{140}\\\text{I}_{t}=\frac{14}{3}=4.7\text{cm}$$
Thus, the height of image is 4.7 cm.
(c) When image formed at distance, d = 25 cm, then magnification produced by eye-piece
$$m=1+\frac{d}{f_{e}}=1+\frac{25}{5}=6$$
Let I be the height of the final image of the tower and size of the image formed by objective
$$\text{O}=\frac{14}{3}\text{cm}\\= 4.7 \text{cm}\\\text{or}\space m=\frac{\text{I}}{\text{O}}$$
∴ Height of final image = m × O = 6 × 4.7 = 28.2 cm
Thus, the height of final image of tower is 28.2 cm.
30. A Cassegrain telescope uses two mirrors as shown in figure. Such a telescope is built with the mirrors is 20 mm apart. If the radius of curvature of the mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?
Sol. Given, distance between objective mirror and secondary mirror
d = 20 mm
Radius of curvature of objective mirror
= R1 = 220 mm
$$\therefore\space\text{Focal length of objective mirror}, f_1 =\frac{220}{2}=110\space \text{mm}$$
Radius of curvature of secondary mirror = R= 140 mm
$$\therefore\space\text{Focal length of secondary mirror}, f_2 =\frac{140}{2}=70\space \text{mm}$$
The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for secondary mirror.
So, the object distance for small mirror u = f1 – d
i.e., u = 110 – 20 = 90 mm
$$\text{By mirror formula,}\space\frac{1}{v} + \frac{1}{u}=\frac{1}{f_{2}}\\\frac{1}{v}=\frac{1}{f_{2}}-\frac{1}{u}=\frac{1}{70}-\frac{1}{90}\\\frac{9-7}{630}=\frac{2}{630}$$
v = 315 mm or v = 31.5 cm
Thus, the final image is formed at 315 mm away from secondary mirror.
31. Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in figure. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5m away?
Sol. Given, deflection, θ = 3.5°
Distance between screen and mirror, x = 1.5m
When mirror turns by angle θ, the reflected ray turned by 2 θ.
$$\therefore\space 2\theta=3×3.5=7\degree=\frac{7\pi}{180}\text{rad}\\\text{Again, in Δ AOS}\\\text{tan}\space 2\theta=\frac{\text{AS}}{\text{sin}\space\theta}\\\text{tan}\bigg(\frac{7\pi}{180}\bigg)=\frac{\text{AS}}{1.5}=\frac{\text{d}}{1.5}\\\text{or}\space d=1.5\space\text{tan}\bigg(\frac{7\pi}{180}\bigg)\\\text{For small angle,}\\\text{tan}\frac{7\pi}{180}≈\frac{7\pi}{180}\text{rad}$$
Again, in Δ AOS
$$\text{tan 2}\theta =\frac{\text{AS}}{\text{sin}\theta}\\\text{tan}\bigg(\frac{7\pi}{180}\bigg)=\frac{\text{AS}}{\text{1.5}}=\frac{d}{1.5}\\\text{or}\space d=1.5\space\text{tan}\bigg(\frac{7\pi}{180}\bigg)\\\text{For small angle,}\\\text{tan}\frac{7\pi}{180}≈\frac{7\pi}{180}\\d=1.5×\frac{7\pi}{180}=0.18\space\text{m}$$
32. Figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?
Sol. Given, focal length of convex lens of glass, f1 = 30 cm.
Focal length
of combination of convex lens and plano-concave liquid lens f = 45 cm.
Refractive index of lens, µg = 1.5
Let f2 be the focal length of the plano-concave lens made of liquid between the convex lens and mirror.
For combined lens
$$\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{f}\\\frac{1}{30}+\frac{1}{f_{2}}=\frac{1}{45}\\\Rarr\space\frac{1}{f_{2}}=\frac{1}{45}-\frac{1}{30}=-\frac{1}{90}$$
f2 = – 90 cm
We know that radii of curvature of two surfaces of plano-concave lens made of liquid formed between foci convex lens and plane mirror are –R and ∞.
For the convex lens of glass, R1 = R, R2 = – R
Using Lens maker’s formula
$$\frac{1}{f_{2}}=(\mu_{1}-1)\bigg(\frac{1}{R_{1}}-\frac{1}{R_{2}}\bigg)\\=\bigg(\frac{3}{2}-1\bigg)\bigg(\frac{1}{\text{R}}+\frac{1}{\text{R}}\bigg)\\\frac{1}{30}=\frac{1}{2}×\frac{2}{\text{R}}$$
∴ R = 30 cm
Again, R1 = – R = – 30 cm, R2 = ∞
Using the Lens maker’s formula
$$\frac{1}{f_{2}}=(\mu-1)\bigg(\frac{1}{\text{R}_{1}}-\frac{1}{\text{R}_{2}}\bigg)\\-\frac{1}{90}=(\mu_{1}-1)\bigg(\frac{1}{30}-\frac{1}{\infty}\bigg)\\\frac{1}{90}=\frac{1}{30}(\mu_{1}-1)\\\mu_{1}=1+\frac{1}{3}=\frac{4}{3}=1.33\\\text{Thus, the refractive index of liquid is}\space\frac{4}{3}.$$
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A mathematical operation of adding a logarithmic term to another logarithmic term is called the addition of logarithmic terms. It is also called as the summation of logarithmic terms.
In logarithmic mathematics, there are two types of logarithmic terms. Hence, it is very important to learn how to add two or more like logarithmic terms and also to learn the procedure of adding the unlike log terms.
In this case, the logarithmic factors in the terms are alike. Hence, it is possible to add them as a term but we have to follow some procedure for calculating the sum of the log terms.
For example, $\log_{3}{11}$, $2\log_{3}{11}$ and $3\log_{3}{11}$.
Observe the three terms, you can understand that there is a logarithmic factor in each term commonly. It allows us to add them, same as the addition of algebraic terms.
Now, write all the terms in a row by placing a plus sign between every two log terms for expressing the addition of like log terms.
$\log_{3}{11}$ $+$ $2\log_{3}{11}$ $+$ $3\log_{3}{11}$
$=\,\,\,$ $1 \times \log_{3}{11}$ $+$ $2 \times \log_{3}{11}$ $+$ $3 \times \log_{3}{11}$
Now, take the common logarithmic factor out from each term.
$=\,\,\,$ $\log_{3}{11} \times (1+2+3)$
$=\,\,\,$ $(1+2+3) \times \log_{3}{11}$
$=\,\,\,$ $6 \times \log_{3}{11}$
$=\,\,\,$ $6\log_{3}{11}$
It can be done in one line. Just add the numerical factors directly and then multiply it with the common logarithmic factor for completing the process of adding like logarithmic terms.
$\,\,\,\therefore\,\,\,\,\,\,$ $\log_{3}{11}$ $+$ $2\log_{3}{11}$ $+$ $3\log_{3}{11}$ $\,=\,$ $6\log_{3}{11}$
Therefore, it clears that the sum of the like log terms is equal to the product of the sum of the numerical factors of the terms and common logarithmic factor.
You can also understand the addition of like log terms from the following examples.
$(1)\,\,\,$ $3\log_{e}{4}+4\log_{e}{4} \,=\, 7\log_{e}{4}$
$(2)\,\,\,$ $\log{17}+2\log{17} \,=\, 3\log{17}$
$(3)\,\,\,$ $10\log_{2}{9}+5\log_{2}{9} \,=\, 15\log_{2}{9}$
$(4)\,\,\,$ $\log_{a}{c}+2\log_{a}{c}+3\log_{a}{c}$ $\,=\,$ $6\log_{a}{c}$
$(5)\,\,\,$ $7\log_{50}{2}+6\log_{50}{2}$ $+$ $\log_{50}{2}+5\log_{50}{2}$ $\,=\,$ $19\log_{50}{2}$
In this case, the logarithmic factors in the terms are not alike. Hence, it is not possible to add them as a term but we express the sum of them as a mathematical expression.
For example, $\log_{2}{11}$, $2\log_{10}{11}$ and $3\log_{e}{11}$ are three unlike logarithmic terms.
$\log_{2}{11}$ $+$ $2\log_{10}{11}$ $+$ $3\log_{e}{11}$
Due to lack of the common logarithmic factor in the terms, the sum of unlike logarithmic terms is simply written as an expression. Observe the following examples for understanding it much better.
$(1)\,\,\,$ $3\log_{b}{4}+4\log_{e}{4}$
$(2)\,\,\,$ $\log{5}+7\log_{5}{10}$
$(3)\,\,\,$ $4\log_{2}{9}+9\log_{3}{9}$
$(4)\,\,\,$ $\log_{a}{c}+2\log_{b}{d}+3\log_{c}{e}$
$(5)\,\,\,$ $20\log_{7}{40}+4\log_{2}{29}$ $+$ $8\log_{9}{10}+2\log_{3}{11}$
Thus, we can add two or more logarithmic terms mathematically in logarithms.
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# Mathematical Facts about Subtraction with Examples | How to Learn Subtraction Facts?
Are you interested to know the facts about subtraction? This page helps you to get full knowledge on Subtraction. Here you will find entire information such as Definition, Facts of Subtraction. Get to know the cool facts about subtraction as a part of your learning try to apply them to your real-time math problems so that you can obtain the solutions quickly. You can understand the basic subtraction facts clearly with our clear-cut explanations provided.
## Subtraction – Definition
Finding the difference between two numbers is called Subtraction.’-‘ sign is used to represent subtraction. Subtraction is the opposite of Addition. Every addition problem can be rewritten as Subtraction. For Example 75-30=45, 60-30=30 etc.
1. Subtraction with small numbers can be done horizontally. For example 25-5=20, 85-10=75,40-10=30 etc.
2. Subtraction is done vertically with large numbers. Numbers are written under place value. For example
Th   h  t  u
4Â Â Â 6Â Â Â 7Â Â 9
-3Â Â Â 4Â Â Â 5Â Â Â 6
——–——–——–
1Â Â Â 2Â Â Â Â 2Â Â Â 3
——–——–——–
To get the predecessor of the number ‘1’ is subtracted. For Example
3 6 8
–Â Â 1
——–——–
3 6Â 7
——–——–
4. Zero subtracted from any number does not change the value. For example
3 6 8
–Â 0
——–——–
3 6 8
——–——–
5. Minuend is the number from which the other number is to be subtracted. The number that is subtracted is called the subtrahend and the result is called the difference. Minuend, Subtahend, the difference are parts of the Subtraction problem.
Example      6  4  3
-3Â Â 2Â Â 1
——–——–——–
3Â Â Â 2Â Â Â 2
——–——–——–
Here 643 is called minuend,321 is called subtrahend.322 is called the difference.
6. To find the missing subtrahend, the difference is subtracted from the minuend.
Minuend – difference = subtrahend.
Example:
8Â Â 4
–Â —Â Â —
——–——–
4Â Â Â 2
——–——–
Here 8,4 is called Minuend.
4,2 is called the difference.
working:
8Â Â 4
-4Â Â 2
——–——–
4Â Â 2
——–——–
7. For finding the missing minuend, the difference is added to the subtrahend.
Difference + subtrahend = minuend
Example:
…Â …
-2Â 1
——–——–
6Â Â 3
——–——–
Working:
6Â Â Â 3
+ 2Â Â 1
——–——–
8Â Â Â 4
——–——–
8 4 is the missing minuend.
6Â Â 3 is the difference.
2 1 is the subtrahend.
8. Subtracting 1 from a number means counting down the number.
For example 5-1=4, 8-1=7, 4-1=3 etc.
9. Any number subtracted from itself is always Zero. This is called Self Subtraction.
For example 4-4=0, 7-7=0 etc.
### How to Subtract a Large Number from a Small Number?
To subtract a large number from a small number follow the borrow method.
Example :
For example, consider the numbers 8463 and 5665 for subtraction.
Solution:
1. Arrange the numbers according to the place values. Start subtracting from units place. Since we can not subtract 5 from 3 borrows ten from tens digit. Now 13-5=8. Write 8 under units place.
5Â Â Â 13
8Â Â 4Â Â Â 6Â Â Â 3
-5Â Â Â 6Â Â Â 6Â Â Â 5
——–——–——–
8
——–——–——–
2. We can not subtract 6 from 5 so borrow and then subtract 6.Now 15-6=9.Write 9 under tens place.
3Â Â Â Â 15Â Â Â 13
8Â Â Â 4Â Â Â 6Â Â Â 3
-5Â Â Â 6Â Â Â 6Â Â Â 5
——–——–——–
9Â Â Â 8
——–——–——–
3. We can not subtract 6 from 3 so borrow and then subtract. Now 13-6=7.Place digit 7 under hundreds digit.
7Â Â 13Â Â Â 15Â 13
8Â Â Â 4Â Â Â 6Â Â Â 3
-5Â Â Â 6Â Â 6Â Â Â 5
——–——–——–
7Â Â Â Â Â 9Â Â 8
——–——–——–
4. Subtract 5 from 7.Write 2 under thousands digit. Now the difference is 2798.
7Â Â Â Â 13Â Â Â 15Â Â 13
8Â Â Â Â 4Â Â Â Â 6Â Â Â Â 3
-5Â Â Â Â Â 6Â Â Â Â 6Â Â Â Â 5
——–——–——–——–
2Â Â Â Â 7Â Â Â Â Â 9Â Â Â Â Â 8
——–——–——–——–
### FAQ’s on Subtraction Facts
1. What is Subtraction?
To find the difference between two numbers is called subtraction.
2. What is Minuend?
The minuend is the number from which the other number is to be subtracted.
3. What is Subtrahend?
Subtrahend is the number that is subtracted.
4. How to find the missing minuend?
For finding the missing minuend, the subtrahend is added to the difference.
5. How to find missing Subtrahend?
For finding the missing subtrahend, the difference is subtracted from the minuend.
6. What are the parts of the Subtraction Problem?
The parts of the subtraction problem are Subtrahend, minuend, and the difference.
7. What is the Difference?
The result of the subtraction is called the difference.
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Standards-based assessment and Instruction
# Instructional Task: Grade 5
## Collecting Basketball Cards
### Task
Kent and Allie collect basketball cards. Kent has twenty cards and Allie has thirty cards. Kent’s mom says that if Kent helps take care of his little brother while she prepares dinner, she will give Kent five cards every Monday. Allie’s mom says that if Allie helps fold the laundry, she will give Allie four cards every Monday. Allie tells Kent that she will always have more cards, but Kent isn’t sure about that. Will Allie always have more basketball cards than Kent? Show all of your mathematical thinking.
### Alternative Versions of Task
#### More Accessible Version
Kent and Allie collect basketball cards. Kent has ten cards and Allie has fifteen cards. Kent’s mom says that if Kent helps take care of his little brother while she prepares dinner, she will give Kent five cards every Monday. Allie’s mom says that if Allie helps fold the laundry, she will give Allie four cards every Monday. Allie tells Kent that she will always have more cards, but Kent isn’t sure about that. Will Allie always have more basketball cards than Kent? Show all of your mathematical thinking.
#### More Challenging Version
Kent and Allie collect basketball cards. Kent has twelve cards and Allie has thirty cards. Kent’s mom says that if Kent helps take care of his little brother while she prepares dinner, she will give Kent twelve cards every Monday. Allie’s mom says that if Allie helps fold the laundry, she will give Allie ten cards every Monday. Allie tells Kent that she will always have more cards, but Kent isn’t sure about that. Will Allie always have more basketball cards than Kent? Show all of your mathematical thinking.
### Common Core Content Standards and Evidence
#### 5.OA Operations and Algebraic Thinking
Analyze patterns and relationships.
3. Generate two numerical patterns using two given rules. Identify apparent relationships between corresponding terms. Form ordered pairs consisting of corresponding terms from the two patterns, and graph the ordered pairs on a coordinate plane. For example, given the rule "Add 3" and the starting number 0, and given the rule "Add 6" and the starting number 0, generate terms in the resulting sequences, and observe that the terms in one sequence are twice the corresponding terms in the other sequence. Explain informally why this is so.
### Exemplars Task-Specific Evidence
This task requires students to use two rules to generate two different numerical sequences. Students also need to extend the numerical sequences, and determine and compare the totals.
### Common Core Standards of Mathematical Practice
MP.1 Make sense of problems and persevere in solving them.
MP.3 Construct viable arguments and critique the reasoning of others.
MP.4 Model with mathematics.
MP.5 Use appropriate tools strategically.
MP.6 Attend to precision.
MP.7 Look for and make use of structure.
### Underlying Mathematical Concepts
• Generate two numerical patterns using two given rules
• Comparison
• Number sense to 75
### Possible Problem-Solving Strategies
• Model (manipulatives)
• Diagram/Key
• Table
• Number line
• Graph (Students may independently select graph paper.)
### Formal Mathematical Language and Symbolic Notation
• Model
• Diagram/Key
• Table
• Graph
• Number line
• Pattern
• Total/Sum
• Day, week, month
• Ordinal numbers: 1st, 2nd, 3rd ...
• Greater than (>)/Less than (<)
• Multiples
• Variable
• Rules: (5 • d) + 20 = K, (4 • d) + 30 = A
• Per
• Axis
• Input/Output
### Possible Solutions
No, Allie is not correct.
#### More Accessible Version Solution
No, Allie is not correct.
#### More Challenging Version Solution
No, Allie is not correct.
### Possible Connections
Below are some examples of mathematical connections. Your students may discover some that are not on this list.
• The patterns are Kent's cards +5 (multiples of 5), Allie's cards +4 (multiples of 4), Mondays +1.
• It will take 11 weeks for Kent to have more cards than Allie.
• At the 11th Monday Kent has a total of 75 cards.
• At the 11th Monday Allie has a total of 74 cards.
• Kent has 1 more card than Allie by the 11th Monday.
• Generalize and apply rules for any number of Mondays.
• Graph each person's cards to compare.
• Relate to a similar task and state a math link.
• Solve more than one way to verify the answer.
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# How to cut a rectangle into 8 squares
Table of Contents:
## class. Mathematics. Nikolsky. Textbook. Answers to pg. 132
The snail, having climbed up and reached 8 m, would not go down. 1) 4 2 = 2 (m) the distance from the ground where the snail would reach on Day 1 2) 2 4 2 = 4 (m) the distance from the ground where the snail would reach on Day 2 3) 4 4 = 8 (m) the distance from the ground where the snail would reach on Day 3.
Cut a 4 × 9 rectangle into two pieces so that it can be made into a square.
Unit size cage. 4 9 = 36 squares the area of the rectangle, 36 = 6 6 so the side of the square must be equal to 6 squares.
Cut a 1 × 6 rectangle from a 10 × 7 rectangle Cut the resulting figure into two pieces so that it can be made into a square.
10 7. 1 6 = 64 squares the area of the figure, 64 = 8 8 means the side of the square must be equal to 8 squares.
The checkered paper gives an idea of how you can lay out a plane with equal squares. Figure 130 shows the ways in which tiles are laid on the floor or on the walls. The plane can also be laid out in equal rectangles. Figure 131 shows two ways to cover the floor with parquet made of equal rectangles. Come up with two more of your own parquet flooring from equal rectangles.
## Task 9 T/R 76 A. Larina
A checkered square of size 6×6 is given.
a) Can this square be dissected into ten different cellular polygons in pairs?? b) Is it possible to cut this square into eleven pairwise different checkered polygons? c) What is the greatest number of pairwise different checkered rectangles we can dissect this square into?
Different unicellular polygons. 1.
Different two-cell polygons. 1.
There are 2 different three-cell polygons.
There are 5 different four-cell polygons.
than 5 different pticellular polygons.
Even if we pave the 6×6 square with all 1,2,3,4-cell different polygons, we will have only 36-(1x11x22x35x4)=7 cells. And we have to place two more polygons in them which are different from those used earlier, which is impossible.
c) We are interested in the largest number of pairwise different checkered rectangles into which this square can be dissected. So we are interested in rectangles that occupy the smallest number of cells possible.
Let’s show that it is impossible to place 9 pairwise different checkered rectangles in a given square.
Even if we place 8 different rectangles, one 1-cell, one 2-cell, one 3-cell, two 4-cell, one 5-cell, two 6-cell rectangles, it would take us 31 cells. In the remaining 5 cells should be placed a rectangle, not previously mentioned, which is impossible.
Let’s show that it is possible to place 8 paired different checkered rectangles in a given square. For example, like this:
Answer: a) yes; b) no; c) 8.
• GDZ answers in math 3 class 2 part of the textbook Dorofeev, Mirakova, Buka (Perspective)
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### Page 11
There are lindens, birches, and maples in the park. There are 5 lindens, 7 times as many maples as lindens, and 10 more birches than maples. 1) How many maples grow in the park?? 2) How many lindens, birches, and maples are there in total?? 3) How many more maples than birches? 4) How many times fewer linden trees than birch trees? What other questions can be put to this condition?
5 д.Л_.К_|_._._._._._. 10 д.Б_|_.
1) 5 7 = 35 (д.). maples grow in the park Answer: 35 maples.2) 35 10 = 45 (д.). birches grow in the park Answer: 45 birches.3) 5 35 45 = 40 45 = 85 (д.). grows in the park Answer: 85 total trees.4) 45. 35 = by 10 (e.). more birches than maples Answer: 10 more birches5) 45 : 5 = 9 (times). less lindens than birches Answer: 9 times.
Additional questions:1) How many more maples than lindens?Solution:35. 5 = by 30 (e.). there are more maples than lindens.Answer: 30 more maples
2) How many times more maples than lindens?Solution:35 : 5 = 7 (times) more maples than lindens.Answer: 7 times.
Find the perimeter of the polygon shown in the drawing.
14 8 6 6 8 8 (14. (8. 6)) (8. (8. 6)) = 14 8 3 6 2 (14. 2) (8. 2) = 14 24 12 12 6 = 20 24 24 = 20 48 = 68 (m). perimeter of the polygon.Answer: 68 meters
4 dm 2 1 m 8 dm 1 10 dm 8 dm 10 dm
48 cm : 6 7 dm 8 cm 7 10 cm 8 cm 70 cm
Draw a rectangle like this in your notebook and guess how to cut it into 8 squares. Draw the slash line of the cut.
Explain the rebus: YES : A = A.Try to find two solutions. The same letters correspond to the same digits.
YES : A = A, so:A A = DATHY two one-digit numbers can be multiplied by itself to produce a two-digit number, the last digit of which is equal to a one-digit number.These are numbers 5 and 6, then:
Solution 1:5 5 = 2525 : 5 = 5 Answer: 25 : 5 = 5.
Solution 2:6 6 = 3636 : 6 = 6 Answer: 36 : 6 = 6.
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• GDZ answers in math 3 class 2 part of the textbook Dorofeev, Mirakova, Buka (Perspective)
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## How to cut a rectangle into 8 squares
Prove that a 10×10 checkerboard cannot be cut along grid lines into 1×4 rectangles. (Solutions by D.Ю. Smith.)
Solution 1 Let’s divide the board into 2×2 squares and color them in a staggered order (Fig.1). Note that any 1×4 rectangle contains an equal number (2 each) of black and white cells, but with this coloring, the board has 52 black cells and 48 white cells, t.е. not equally. So it is impossible to cut a 10×10 board into 1×4 rectangles.
Solution 2 Let’s color the board diagonally in 4 colors (Fig.2). Note that any rectangle contains one cell of each of the four colors, but in this coloring on the board on 25 cells of the 1st and 3rd colors, 26 cells. 2nd and 24 cells. 4th, t.е. not equally. So the 10×10 board cannot be cut into 1×4 rectangles.
The bottom right and left corner cells were cut out of the chessboard. Can the resulting figure be cut into 1×2 dominoes?? And if you cut out the bottom right and the top left?
Can a 6×6 board be cut into dominoes so that among them there are exactly 11 horizontal? (Horizontal coloring in two colors.)
Color the drawing in four colors so that the neighboring parts are painted in different colors. Can we do with three colors?? (See the above example. Activity 6: Coloring a Geographic Map. grade 5-6).
In a 4×4 square the left cells are painted black and the rest are painted white. In one operation, you are allowed to recolor all the cells inside any rectangle with the opposite color. In three operations, how do you get a chess coloring from the original coloring?
Several grasshoppers sit on one straight line, and the distance between neighbors. are the same. Every minute, one of them jumps to a point symmetric to it relative to the other Juke. After some time, can Grasshopper Sasha be in the place where his neighbor Lyosha was sitting at the beginning??
a) Is it possible to cut the chessboard into pieces consisting of 4 squares in the shape of a “T”?
b) Is it possible to cut a 10×10 chessboard into such pieces??
Is it possible to cut an 8×8 square with the corner cut off into 1×3 rectangles??
Is it possible to cut a 10×10 board into four-cell pieces in the shape of the letter “G”?? (Horizontal coloring in two colors.)
An 8×8 board cut into 2×1 dominoes. Can there be 15 vertical and 17 horizontal dominoes?
A triangle is dissected into triangles (25 pieces), as shown in the figure. Juke can walk around the triangle, moving between adjacent (side by side) triangles. What is the maximum number of triangles a Juke can go through if it has been in each of them no more than once??
What is the largest number of rhombuses, each made of two equilateral triangles with side 1, that can be cut out of an equilateral triangle with side 5 (see. Figure. of the previous problem).
A triangular castle is divided into 100 identical triangular halls. A door is made in the middle of each wall. How many rooms can a person who does not want to visit more than once visit??
## How to cut a rectangle into 8 squares
Problem 1: Can a square 5 × 5 cut into rectangles 1 × 2 (dominoes).
Problem 2: From chessboard 8 × 8 the opposite corner cells are cut out. Can the remainder be cut into rectangles 1 × 2 (dominoes)?
Solution: No. Each domino occupies one black and one white square, and the number of black and white squares on the board without corners is different.
Problem 3: From the opposite corners of the board 10 × 10 cut two squares 3 × 3. Can the remainder be dissected into dominoes??
Problem 4: Think of a connected piece on a chessboard that has an equal number of black and white squares, but cannot be broken into dominoes.
Problem 5: Is it possible to cut a square 10 × 10 into 25 figures ?
Problem 6: Is it possible to cut a square 10 × 10 into 25 pieces ?
Solutions: Color the board in staggered order. There will be an even number of black squares and each piece will have one or three.
Problem 7: Is it possible to cut the square 10 × 10 by 25 pieces ?
Color the board in four colors (see ). figure). Each figure occupies one cell of each color, and the number of cells of the first and second colors is different.
Problem 8: Is it possible to dissect a square 10 × 10 by 25 pieces ?
Solution: Paint the verticals in one.
Problem 9: Prove that a board of 8 × 8 without a corner cell cannot be dissected into rectangles 1 × 3.
Problem 10: Can a board of 8 × 8 cut into one square 2 × 2 and 15 figures of the form ?
Problem 11: Square a)5 × 5b)8 × 8 divided into several rectangles 3 × 1 and one square 1 × 1. Where can stand the square of 1 × 1?
Solution: a) In the center, b) On the third square diagonally from any corner.
Problem 12: What is the maximum number of bars 1 × 1 × 4 can be cut from the cube 6 × 6 × 6?
Problem 13: A rectangle is divided into figures and. One is lost, but replaced by. Prove that a new set cannot cover the original rectangle.
Problem 14: Is it possible to square 16 × 16 split into 64 rectangles 1 × 4, of which 31 will stand vertically, and the remaining 33 horizontally?
Solution: Paint every fourth vertical.
Problem 15: For which n squares n × n can be dissected by a) ;
Solution: For n divisible by four.
Problem 16: The rectangle m × k divided into rectangles 1 × n. Prove that m is divisible by n or k is divisible by n.
Problem 17: Prove that the rectangle m × n can be divided into rectangles a × b, if and only if the following conditions are satisfied:
1) m and n are represented as ka lb (k and l are non-negative integers)
Problem 18: Rectangle m × n is called robust if it can be broken into dominoes such that any cut of the rectangle intersects at least one domino. Prove that:
See also Viking Te 1000 Replacement Fishing Line
(e) If a rectangle m × n strong, then also a rectangle m × (n 2) sturdy.
(g) Which rectangles are solid and which are not?
Solution: f) Hint: every line in square 6 × 6 intersects an even number of dominoes.
g) All rectangles m × n where mn is even, m,n 5 except 6 × 6.
a) Is it possible for a rectangle 5 × 9 divide into corners?
b) Prove that a rectangle with sides greater than 100 and area divisible by 3 can be dissected into dominoes.
(c) Which rectangles can be broken into corners and which cannot?
Can a board of 2 n × 2 n without a corner cell divide into corners?
Solution: Yes, they can. Partitioning is constructed by induction.
Problem 21: At which n is the board (2n 1) × (2n 1) without a corner cell can be divided into dominoes, among which there are an equal number of vertical and horizontal?
## How to cut a rectangle into 8 squares
2018-12-01 1 2. ? ( ).
1). , 2 1, 2 3 2, 2,. ,. 2) 1 frac12 (. ) 3) 4. x, 1. x, 2 x. “” 1. 2x,. 4. 8x x = frac14. 4). 6 2 “” (. ). 2(1. 2x) 2 frac1. 2x 2= 3(1. 2x),. x = frac16
## Answers to page 100 71-380 GDZ for the textbook Mathematics 5 grade Merzlyak, Polonsky, Yakir
Rectangle ABCD is dissected into squares as shown in Figure 139. The side of the smallest of the squares is 4 cm. Find the lengths of the sides of rectangle ABCD.
The side of the smallest square is 4 cm, 4 3 = 12 (cm). side of the largest squareAD = BC = 12 12 4 = 28 (cm) Sides AD and BC consist of 4 middle squares28 : 4 = 7 (cm). Side of the middle squareCD = AB = 7 4 3 = 19 (cm)Answer: 28 cm and 19 cm.
### Assignment 72
Draw a rectangle whose adjacent sides are 3 cm and 6 cm. Divide it into three equal rectangles. Calculate the perimeter of each of the obtained rectangles. How many solutions have the problem.
The problem has 2 solutions: 1) AK = KM = MD = BN = NP = PC = 6 : 3 = 2 (cm) P ABNK = P KNPM = P MPCD = 2 2 2 3 = 10 (cm)
2) AK = KM = MD = BN = NP = PC = 3 : 3 = 1 (cm) P ABNK = P KNPM = P MPCD = 2 1 2 6 = 14 (cm)
### Problem 73
Is there any rectangle with a perimeter of 12 cm that can be dissected into two equal squares?? If the answer is Yes, draw the diagram and calculate the perimeter of each of the squares obtained.
Rectangle ABCD with sides 4 cm and 2 cm. P ABCD = 2 2 2 4 = 4 8 = 12 (cm) By dividing the rectangle in half we obtain a square with a side of 2 cm, the perimeter of which is equal: P ABEF = P FECD = 2 4 = 8 (cm) Answer: 8 cm. the perimeter of each of the squares.
### Task 74
How do I divide a square into four equal parts so that two squares can be made of these parts??
Then we add a square from each pair of triangles.
### Problem 75
How do you cut an isosceles right triangle into four equal pieces to make a square??
### Problem 76
How to cut a rectangle with sides of 8 cm and 4 cm into four parts so that it can be made into a square?
### Problem 77
How to cut a square into a triangle and a quadrilateral to make a triangle?
Put a triangle on top of a quadrilateral and you get a large triangle.
### Problem 78
How to cut a square with a side of 6 cm into two pieces along a polyline of three links so that the pieces are made into a rectangle?
### Problem 79
Construct line MK, ray PS, and segment AB, such that it intersects segment AB and line MK, and line MK does not intersect segment AB.
### Problem 80
Lemons, oranges, and tangerines are available in the store, for a total of 740 kg. If we sold 55 kg of lemons, 36 kg of oranges, and 34 kg of tangerines, the remaining masses of lemons, oranges, and tangerines would be equal. How many kilograms of each kind of fruit are in the store?
It is not possible to divide by two because there is a misprint in the problem. In principle, you can solve this problem using fractions:
1) 55 36 34 = 125 (kg). all the fruit would have been sold 2) 740. 126 = 615 (kg). of fruit left in the store 3) 615 : 3 = 205 (kg). the mass of each of the remaining fruits 4) 205 55 = 260 (kg). of lemons available in the store 5) 205 36 = 241 (kg). there are oranges in the store 6) 205 34 = 239 (kg). answer: 260 kg of lemons, 241 kg of oranges, 239 kg of tangerines.
## Presentation on Mathematics “Cutting Tasks” (Grade 5)
We pay your attention, that according to the Federal law N 273-FZ “About education in the Russian Federation” in the organizations realizing educational activity, training and education of students with physical disability both together with other students and in separate classes or groups is organized.
“The relevance of establishing school reconciliation/mediation services in educational organizations”
Certificate and tuition discount for each participant
### Description of the presentation by individual slide:
“Geometry is the most powerful means of refining our mental faculties and enabling us to think and reason properly” Galileo Galilei Cutting problems
I. Historical Background The first treatise to deal with problems of dissection. Famous experts in this section of geometry were the famous classics of amusing geometry and puzzle compilers Henry E. Dewdeny and Harry Lindgren.
II. Checkered Paper Problem A square has 16 cells. Divide the square into two equal parts so that the cutting line runs along the sides of the cells. (We will consider different ways of cutting a square in two parts if the parts of the square obtained by one way of cutting are not equal to the parts obtained by another way.) How many total solutions does the problem have? Solution:
Task A 3 x 5 rectangle contains 15 squares and the center square is removed. Find five ways to cut the rest of the figure into two equal parts so that the cutting line runs along the sides of the cells. Solution:
Problem Divide a 4 x 4 square into four equal parts so that the cutting line runs along the sides of the cells. How many different ways to cut? Solution:
Task Divide the figure into three equal parts so that the cut line runs along the sides of the squares. Solution:
See also Interior Cleaning Brush For Screwdriver
Task Divide the figure into four equal parts so that the mowing lines of the cuts run along the sides of the squares. Find as many solutions as possible. Solution:
Task Cut the figure, into two equal parts along the grid lines, with a circle in each part. Solution:
Task Cut this square along the sides of the cells so that all the pieces are the same size and shape, and so that each contains one circle and a star. Solution:
Problem Cut a 6×6 square of checkered paper into four equal pieces so that each piece contains three shaded cells. Solution:
Problem Cut a 4 x 9 rectangle into two equal pieces along the sides of the cells so that it can be made into a square. Solution:
Problem Cut a rectangle of 10 x 7 cells into a rectangle of 1 x 6 cells. Cut the resulting figure into two pieces so that they can be made into a square. Solution:
Problem On the checkered paper, a square of 5 x 5 cells is drawn. Show how to cut it along the sides of the cells into 7 different rectangles. Solution:
Task Divide the figures into two equal parts. (You can cut not only along the lines of the cells, but also along their diagonals.) Solution:
III. Pentamino Figures: Dominoes Triminoes Tetramino Pentaminoes Make two, three, four, five squares so that any square has a common side with at least one other square. You can make only one figure from two identical squares. dominoes. You can get a domino figure from a single domino figure by attaching another square to it in different ways. You will get two dominoes.
Problem Make up all possible tetramino shapes (from the Greek. The word “tetra” is four). How many we have? (Figures obtained by rotation or symmetrical mapping from any other ones are not considered new.). Solution:
Problem Make all possible pentaminoes (from the Greek. “penta” is five). How many of these figures are? Solution:
Problem Make up pentaminoes from pentaminoes. How many solutions have the problem for each figure? Solution:
Figure 1 has the following property. If you cut it out of paper and bend it along the line a, then one part of the figure coincides with the other part. The figure is said to be symmetrical about the line a, the axis of symmetry. Figure 12 also has a symmetry axis, even two of them. the lines b and c, but figure 2 has no symmetry axis. All possible pentaminoes
Problem How many axes of symmetry do each pentamino have?? Solution:
Problem Cut the figure along the grid lines into 4 equal pieces. Solution:
The task Cut the figure along the grid lines into 4 equal parts so that each part has a shaded hexagon. Solution:
Literature Ekimova M. А., Kukin G. П. Cutting Tasks M.: MCNMO, 2002 120с.
### Mathematics: Theory and Methods of Teaching in an Educational Organization
You could, of course, just take a ruler and measure the lengths of the segments in the figure, but this method is not very good for two reasons. First, the precision of such measurements is not very high and the answer will be only an approximation. Second, if we had the picture “at an angle,” then the true lengths of the sides would be distorted, and then we would have to think about what to do with the measurements. But it is possible to find the sides of a rectangle absolutely precisely, and you only need to know the cutting scheme for this. To do this, you need to make a system of equations, taking the lengths of the sides of the squares and rectangle as the unknowns.
As already noted in the hint, this problem is not at all on geometry (as it might seem), but on linear algebra. And it is solved quite simply. We just need not be afraid to enter a lot of notations at first.
So let x и y. width and height of the large rectangle we are looking for. Let’s number the squares as shown in Figure 1, and denote the side of the square with the number i via zi. The variables already have. And from where to take the equations? Let’s look closely at figure 1: we can see that some sides of the squares are “well” adjacent to each other. For example, the white square completes the side of the red square to the side of the orange square. Another example: White and yellow are the same height as blue and magenta. These joining conditions allow us to write the equations. We obtain a system of linear equations, which we write in two ways. First, write out the equations that correspond to the vertical joints:
### Volume of Open Box Made From Rectangle with Squares Cut Out
The last equation describes the adjacency of the third and eighth squares to the right side of the rectangle. But it follows from the previous equations (check it), so further we do not take it into account. Now let’s write equations for the horizontal joints (we skip the condition for the bottom side for the same reason):
Combine everything into one system, from which we need to find x и y:
To solve systems of linear equations many methods have long been invented. But in our case we can do without involving powerful theories, but simply express one variable by another and make appropriate substitutions, gradually simplifying the system. It is convenient to express the variables in z5: z2 = 1 z5, so z1 = 2 z5, so z4 = 3 z5. You can already explicitly calculate the side of the blue square: z6 = 1 z4. z5 = 4. Continuing in this vein, it is not difficult to find the sides of the other squares, and with them the sides of the rectangle: x = 32, а y = 33.
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# ANGLE OF ELEVATION AND DEPRESSION EXAMPLES
Example 1 :
A student sitting in a classroom sees a picture on the black board at a height of 1.5 m from the horizontal level of sight. The angle of elevation of the picture is 30°. As the picture is not clear to him, he moves straight towards the black board and sees the picture at an angle of elevation of 45°. Find the distance moved by the student.
Solution :
Distance moved by the student = BC
In triangle ABD :
∠ABD = 30°
tan θ = opposite side/Adjacent side
1/√3 = 1.5/BD
BD = 1.5 x √3 ==> 1.5 √3
In triangle ACD :
∠ACD = 45°
tan θ = opposite side/Adjacent side
1 = 1.5/CD
CD = 1.5
BC = BD - CD
BC = 1.5 √3 - 1.5
= 1.5 (√3 - 1) = 1.5(1.732 - 1)
= 1.5 (0.732) ==> 1.098 m
Hence the distance moved by the student is 1.098 m.
Example 2 :
A boy is standing at some distance from a 30 m tall building and his eye level from the ground is 1.5 m. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution :
From the given information, we can draw a rough diagram
In triangle ABD
∠ABD = 30°
tan θ = opposite side/Adjacent side
1/√3 = 28.5/BD ==> BD = 28.5 √3 --(1)
In triangle ACD
∠ABD = 60°
tan θ = opposite side/Adjacent side
3 = 28.5/CD ==> CD = 28.5/
Multiplying by 3 on both numerator and denominator, we get
CD = 28.5/3 ==> 28.53/3 ==> 9.53 -->(2)
the distance he walked towards the building = BD - CD
= 28.5 √3 - 9.53 ==> 193 m
Example 3 :
From the top of a lighthouse of height 200 feet, the lighthouse keeper observes a Yacht and a Barge along the same line of sight .The angles of depression for the Yacht and the Barge are 45° and 30° respectively. For safety purposes the two sea vessels should be at least 300 feet apart. If they are less than 300 feet , the keeper has to sound the alarm. Does the keeper have to sound the alarm ?
Solution :
From the above picture, we have to find the value of CD.
In triangle ABC
∠ACB = 45°
sin θ = opposite side/Hypotenuse side
sin 45° = AB/BC
1/2 = 200/BC
BC = 200 2
In triangle ABD
sin 30° = AB /BD
1/2 = 200/BD
BD = 200 x 2 ==> 400
CD = BD - BC ==> 400 - 200 2 ==> 200(2 - 2)
= 200 (2 - 1.414)
= 200(0.586) ==> 117.2 m
From this, we come to know that the distance between Yacht and a Barge is less than 300 m. Hence the keeper has to sound the alarm.
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# 1.5 Measurement uncertainty, accuracy, and precision (Page 3/11)
Page 3 / 11
1. When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (the least precise value in terms of addition and subtraction).
2. When we multiply or divide numbers, we should round the result to the same number of digits as the number with the least number of significant figures (the least precise value in terms of multiplication and division).
3. If the digit to be dropped (the one immediately to the right of the digit to be retained) is less than 5, we “round down” and leave the retained digit unchanged; if it is more than 5, we “round up” and increase the retained digit by 1; if the dropped digit is 5, we round up or down, whichever yields an even value for the retained digit. (The last part of this rule may strike you as a bit odd, but it’s based on reliable statistics and is aimed at avoiding any bias when dropping the digit “5,” since it is equally close to both possible values of the retained digit.)
The following examples illustrate the application of this rule in rounding a few different numbers to three significant figures:
• 0.028675 rounds “up” to 0.0287 (the dropped digit, 7, is greater than 5)
• 18.3384 rounds “down” to 18.3 (the dropped digit, 3, is less than 5)
• 6.8752 rounds “up” to 6.88 (the dropped digit is 5, and the retained digit is even)
• 92.85 rounds “down” to 92.8 (the dropped digit is 5, and the retained digit is even)
Let’s work through these rules with a few examples.
## Rounding numbers
Round the following to the indicated number of significant figures:
(a) 31.57 (to two significant figures)
(b) 8.1649 (to three significant figures)
(c) 0.051065 (to four significant figures)
(d) 0.90275 (to four significant figures)
## Solution
(a) 31.57 rounds “up” to 32 (the dropped digit is 5, and the retained digit is even)
(b) 8.1649 rounds “down” to 8.16 (the dropped digit, 4, is less than 5)
(c) 0.051065 rounds “down” to 0.05106 (the dropped digit is 5, and the retained digit is even)
(d) 0.90275 rounds “up” to 0.9028 (the dropped digit is 5, and the retained digit is even)
Round the following to the indicated number of significant figures:
(a) 0.424 (to two significant figures)
(b) 0.0038661 (to three significant figures)
(c) 421.25 (to four significant figures)
(d) 28,683.5 (to five significant figures)
(a) 0.42; (b) 0.00387; (c) 421.2; (d) 28,684
## Addition and subtraction with significant figures
Rule: When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (i.e., the least precise value in terms of addition and subtraction).
(a) Add 1.0023 g and 4.383 g.
(b) Subtract 421.23 g from 486 g.
## Solution
(a) $\begin{array}{l}\\ \begin{array}{l}\hfill \\ \frac{\begin{array}{c}\phantom{\rule{1.4em}{0ex}}1.0023 g\\ \text{+ 4.383 g}\end{array}}{\phantom{\rule{1.5em}{0ex}}5.3853 g}\hfill \end{array}\end{array}$
Answer is 5.385 g (round to the thousandths place; three decimal places)
(b) $\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \end{array}\\ \frac{\begin{array}{l}\text{}\phantom{\rule{0.8em}{0ex}}486 g\hfill \\ -421.23 g\hfill \end{array}}{\phantom{\rule{1.3em}{0ex}}64.77 g}\end{array}$
Answer is 65 g (round to the ones place; no decimal places)
(a) Add 2.334 mL and 0.31 mL.
(b) Subtract 55.8752 m from 56.533 m.
(a) 2.64 mL; (b) 0.658 m
what is hybridization
the mixing of atomic orbitals to form molecular of similar energy called hybrid orbitals
Cffrrcvccgg
who are the alchemist?
alchemy science of transmutation. typically it is aim at tranforming lead to or other base metals to gold and the creation of the philosophers stone which in reality isn't a stone it's something priceless something we all need for coming times. don't be fooled
Kendrick
read Corinthians 5 verses 50 to the end of the chapter then read revelations chapter 2 verse 17
Kendrick
The word "Alchemy" comes from the forgotten name for Ancient Egypt, Khemmet. Khem was the name for the Egyptian Empire, but the actual land of Egypt was called Khemmet because the "T" on the end of a word denoted a physical location on Earth and not just an idea.
Michael
Wow!
mendie
What's the mass number of carbon
Akinbola
mass number of carbon is 12.
Nnenna
wat d atomic number of oxygen
safiya
atomic number of oxygen is 8
Nnenna
which quantum number divides shell into orbitals?
azimuthal
Emmanuel
hi
Charlie
azimuthal
reinhard
azimuthal
Charlie
what is atom
an atom is a smallest indivisible part of an element
Henry
an atom is the smallest part of an element that takes part in a chemical reaction
Nana
wat is neutralization
when any acid reacts with base to decrease it's acidity or vice-versa to form salt and solvent.. which is called neutralization
Santosh
explain buffer
Organic
buffer is a solution which resists changes in pH when acid or alkali added to it..
Santosh
hello, who is online
UTHMAN
buffer is the solution which resist the change in pH by addition of small amount of acid or alkali to it
KAUSIK
neutralisation is the process of mixing of a acid and a base to form water and corresponding salt
KAUSIK
how to solve equation on this
what are the elent of ionic and covalent bonding
Princewill
what is gases
Its one of the fundamental sate of matter alone side with liquid, solid and plasma
John
What is chemical bonding
John
To my own definitions. It's a unit of measurement to express the amount of a chemical substance.
What is mole
It's the unit of measurements used to express the amount of chemical substance.
Ozoaniehe
What is pressure
force over area
Jake
force applied per unit area
john
force applied per unit area
Prajapati
Why does carbonic acid don't react with metals
Why does carbonic acid don't react with metal
Some metals will react depending on their Standard Electrode Potential. Carbonic acid is a very weak acid (i.e. a low hydrogen ion concentration) so the rate of reaction is very low.
Paul
sample of carbon-12 has a mass of 6.00g. How many atoms of carbon-12 are in the sample
a sample of carbon-12 has a mass of 6.00g. How many atoms of carbon-12 are in the sample
an object of weight 10N immersed in a liquid displaces a quantity of d liquid.if d liquid displaced weights 6N.determine d up thrust of the object
|
# 3.4: Binomial Distribution (Special Topic)
Example $$\PageIndex{1}$$
Suppose we randomly selected four individuals to participate in the "shock" study. What is the chance exactly one of them will be a success? Let's call the four people Allen (A), Brittany (B), Caroline (C), and Damian (D) for convenience. Also, suppose 35% of people are successes as in the previous version of this example.
Let's consider a scenario where one person refuses:
$P(A = refuse; B = shock; C = shock; D = shock)$
$= P(A = refuse) P(B = shock) P(C = shock) P(D = shock)$
$= (0.35)(0.65)(0.65)(0.65) = (0:35)^1(0:65)^3 = 0.096$
But there are three other scenarios: Brittany, Caroline, or Damian could have been the one to refuse. In each of these cases, the probability is again $$(0.35)^1(0.65)^3$$. These four scenarios exhaust all the possible ways that exactly one of these four people could refuse to administer the most severe shock, so the total probability is $$4 \times (0.35)^1(0.65)^3 = 0.38$$.
Exercise $$\PageIndex{1}$$
Verify that the scenario where Brittany is the only one to refuse to give the most severe shock has probability $$(0.35)^1(0.65)^3.$$29
Solution
29P(A = shock; B = refuse; C = shock; D = shock) = (0.65)(0.35)(0.65)(0.65) = $$(0.35)^1(0.65)^3$$.
### The Binomial Distribution
The scenario outlined in Example 3.37 is a special case of what is called the binomial distribution. The binomial distribution describes the probability of having exactly k successes in n independent Bernoulli trials with probability of a success p (in Example 3.37, n = 4, k = 1, p = 0.35). We would like to determine the probabilities associated with the binomial distribution more generally, i.e. we want a formula where we can use n, k, and p to obtain the probability. To do this, we reexamine each part of the example.
There were four individuals who could have been the one to refuse, and each of these four scenarios had the same probability. Thus, we could identify the nal probability as
$\text {[# of scenarios]} \times \text {P(single scenario)} \tag {3.39}$
The first component of this equation is the number of ways to arrange the k = 1 successes among the n = 4 trials. The second component is the probability of any of the four (equally probable) scenarios.
Consider P(single scenario) under the general case of k successes and n-k failures in the n trials. In any such scenario, we apply the Multiplication Rule for independent events:
$p^k (1- p)^{n-k}$
This is our general formula for P(single scenario).
Secondly, we introduce a general formula for the number of ways to choose k successes in n trials, i.e. arrange k successes and n - k failures:
$\binom {n}{k} = \frac {n!}{k!(n - k)!}$
The quantity $$\binom {n}{k}$$ is read n choose k.30 The exclamation point notation (e.g. k!) denotes a factorial expression.
$0! = 1$
$1! = 1$
$2! = 2 \times 1 = 2$
$3! = 3 \times 2 \times 1 = 6$
$4! = 4 \times 3 \times 2 \times 1 = 24$
$\vdots$
$n! = n \times (n - 1) \times \dots \times 3 \times 2 \times 1$
Using the formula, we can compute the number of ways to choose k = 1 successes in n = 4 trials:
$\binom {4}{1} = \frac {4!}{1! (4 - 1)!} = \frac {4!}{1! 3!} = \frac {4 \times 3 \times 2 \times 1}{(1)(3 \times 2 \times 1)} = 4$
This result is exactly what we found by carefully thinking of each possible scenario in Example 3.37.
Substituting n choose k for the number of scenarios and $$p^k(1 - p)^{n-k}$$ for the single scenario probability in Equation (3.39) yields the general binomial formula.
Definition: Binomial distribution
Suppose the probability of a single trial being a success is p. Then the probability of observing exactly k successes in n independent trials is given by
$\binom {n}{k} p^k (1 - p)^{n - k} = \frac {n!}{k!(n - k)!} p^k (1 - p)^{n - k} \tag {3.40}$
Additionally, the mean, variance, and standard deviation of the number of observed successes are
$\mu = np \sigma^2 = np(1 - p) \sigma = \sqrt {np(1- p)} \tag{3.41}$
TIP: Is it binomial? Four conditions to check
1. The trials independent.
2. The number of trials, n, is fixed.
3. Each trial outcome can be classified as a success or failure.
4. The probability of a success, p, is the same for each trial.
30Other notation for n choose k includes $$nC_k, C^k_n, and C(n, k)$$.
Example $$\PageIndex{1}$$
What is the probability that 3 of 8 randomly selected students will refuse to administer the worst shock, i.e. 5 of 8 will?
Solution
We would like to apply the binomial model, so we check our conditions. The number of trials is fixed (n = 8) (condition 2) and each trial outcome can be classi ed as a success or failure (condition 3). Because the sample is random, the trials are independent (condition 1) and the probability of a success is the same for each trial (condition 4).
In the outcome of interest, there are k = 3 successes in n = 8 trials, and the probability of a success is p = 0.35. So the probability that 3 of 8 will refuse is given by
$\binom {8}{3} {(0.35)}^k (1 - 0.35)^{8 - 3} = \frac {8!}{3!(8 - 3)!} {(0.35)}^k (1 - 0.35)^{8 - 3}$
$= \frac {8!}{3! 5!} {(0.35)}^3 {(0.65)}^5$
Dealing with the factorial part:
$\frac {8!}{3!5!} = \frac {8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)( 5 \times 4 \times 3 \times 2 \times 1 )} = \frac {8 \times 7 \times 6 }{ 3 \times 2 \times 1 } = 56$
Using $$(0.35)^3(0.65)^5 \approx 0.005$$, the final probability is about 56 * 0.005 = 0.28.
TIP: computing binomial probabilities
The rst step in using the binomial model is to check that the model is appropriate. The second step is to identify n, p, and k. The nal step is to apply the formulas and interpret the results.
TIP: computing n choose k
In general, it is useful to do some cancelation in the factorials immediately. Alternatively, many computer programs and calculators have built in functions to compute n choose k, factorials, and even entire binomial probabilities.
Exercise $$\PageIndex{1}$$
If you ran a study and randomly sampled 40 students, how many would you expect to refuse to administer the worst shock? What is the standard deviation of the number of people who would refuse? Equation \ref{3.41} may be useful.
Solution
We are asked to determine the expected number (the mean) and the standard deviation, both of which can be directly computed from the formulas in Equation \ref{3.41}:
$\mu = np = 40 \times 0.35 = 14$
and
$\sigma = \sqrt{np(1 - p)} = \sqrt { 40 \times 0.35 \times 0.65} = 0.02.$
Because very roughly 95% of observations fall within 2 standard deviations of the mean (see Section 1.6.4), we would probably observe at least 8 but less than 20 individuals in our sample who would refuse to administer the shock.
Exercise $$\PageIndex{1}$$
The probability that a random smoker will develop a severe lung condition in his or her lifetime is about 0:3. If you have 4 friends who smoke, are the conditions for the binomial model satisfied?
Solution
One possible answer: if the friends know each other, then the independence assumption is probably not satis ed. For example, acquaintances may have similar smoking habits.
Exercise $$\PageIndex{1}$$
Suppose these four friends do not know each other and we can treat them as if they were a random sample from the population. Is the binomial model appropriate? What is the probability that (a) none of them will develop a severe lung condition? (b) One will develop a severe lung condition? (c) That no more than one will develop a severe lung condition?
Solution
To check if the binomial model is appropriate, we must verify the conditions. (i) Since we are supposing we can treat the friends as a random sample, they are independent. (ii) We have a fixed number of trials (n = 4). (iii) Each outcome is a success or failure. (iv) The probability of a success is the same for each trials since the individuals are like a random sample (p = 0.3 if we say a "success" is someone getting a lung condition, a morbid choice). Compute parts (a) and (b) from the binomial formula in Equation (3.40): P(0) = $$\binom {4}{0}(0.3)^0(0.7)^4 = 1 \times 1 \times 0.7^4 = 0.2401$$, P(1) = $$\binom {4}{1}(0.3)^1(0.7)^3 = 0.4116. Note: 0! = 1, as shown on page 138. Part (c) can be computed as the sum of parts (a) and (b): P(0)+P(1) = 0.2401+0.4116 = 0.6517. That is, there is about a 65% chance that no more than one of your four smoking friends will develop a severe lung condition. Exercise \(\PageIndex{1}$$
What is the probability that at least 2 of your 4 smoking friends will develop a severe lung condition in their lifetimes?
Solution
The complement (no more than one will develop a severe lung condition) as computed in Exercise 3.45 as 0.6517, so we compute one minus this value: 0.3483.
Exercise $$\PageIndex{1}$$
Suppose you have 7 friends who are smokers and they can be treated as a random sample of smokers. (a) How many would you expect to develop a severe lung condition, i.e. what is the mean? (b) What is the probability that at most 2 of your 7 friends will develop a severe lung condition.
Solution
(a) $$\mu$$ = 0.3 \times 7 = 2.1. (b) P(0, 1, or 2 develop severe lung condition) = P(k = 0)+P(k = 1)+P(k = 2) = 0:6471.
Below we consider the rst term in the binomial probability, n choose k under some special scenarios.
Exercise $$\PageIndex{1}$$
Why is it true that $$\binom {n}{0} = 1$$ and $$\binom {n}{n} = 1$$ for any number n?
Solution
Frame these expressions into words. How many different ways are there to arrange 0 successes and n failures in n trials? (1 way.) How many different ways are there to arrange n successes and 0 failures in n trials? (1 way.)
Exercise $$\PageIndex{1}$$
How many ways can you arrange one success and n -1 failures in n trials? How many ways can you arrange n -1 successes and one failure in n trials?
Solution
One success and n - 1 failures: there are exactly n unique places we can put the success, so there are n ways to arrange one success and n - 1 failures. A similar argument is used for the second question. Mathematically, we show these results by verifying the following two equations:
$\binom {n}{1} = n, \binom {n}{n - 1} = n$
### Normal Approximation to the Binomial Distribution
The binomial formula is cumbersome when the sample size (n) is large, particularly when we consider a range of observations. In some cases we may use the normal distribution as an easier and faster way to estimate binomial probabilities.
Example $$\PageIndex{1}$$
Approximately 20% of the US population smokes cigarettes. A local government believed their community had a lower smoker rate and commissioned a survey of 400 randomly selected individuals. The survey found that only 59 of the 400 participants smoke cigarettes. If the true proportion of smokers in the community was really 20%, what is the probability of observing 59 or fewer smokers in a sample of 400 people?
Solution
We leave the usual verification that the four conditions for the binomial model are valid as an exercise.
The question posed is equivalent to asking, what is the probability of observing $$k = 0, 1, \dots, 58, or 59$$ smokers in a sample of n = 400 when p = 0.20? We can compute these 60 different probabilities and add them together to nd the answer:
$P(k = 0 or k = 1 or \dots or k = 59)$
$= P(k = 0) + P(k = 1) + \dots + P(k = 59)$
$= 0.0041$
If the true proportion of smokers in the community is p = 0.20, then the probability of observing 59 or fewer smokers in a sample of n = 400 is less than 0.0041. The computations in Example 3.50 are tedious and long. In general, we should avoid such work if an alternative method exists that is faster, easier, and still accurate. Recall that calculating probabilities of a range of values is much easier in the normal model. We might wonder, is it reasonable to use the normal model in place of the binomial distribution? Surprisingly, yes, if certain conditions are met.
Exercise $$\PageIndex{1}$$
Here we consider the binomial model when the probability of a success is p = 0.10. Figure 3.17 shows four hollow histograms for simulated samples from the binomial distribution using four different sample sizes: n = 10, 30, 100, 300. What happens to the shape of the distributions as the sample size increases? What distribution does the last hollow histogram resemble?
Solution
The distribution is transformed from a blocky and skewed distribution into one that rather resembles the normal distribution in last hollow histogram
Figure 3.17: Hollow histograms of samples from the binomial model when p = 0.10. The sample sizes for the four plots are n = 10, 30, 100, and 300, respectively.
Normal approximation of the binomial distribution
The binomial distribution with probability of success p is nearly normal when the sample size n is sufficiently large that np and n(1 - p) are both at least 10. The approximate normal distribution has parameters corresponding to the mean and standard deviation of the binomial distribution:
$\mu = np \sigma = \sqrt {np(1- p)}$
The normal approximation may be used when computing the range of many possible successes. For instance, we may apply the normal distribution to the setting of Example 3.50.
Example $$\PageIndex{1}$$
How can we use the normal approximation to estimate the probability of observing 59 or fewer smokers in a sample of 400, if the true proportion of smokers is p = 0.20?
Solution
Showing that the binomial model is reasonable was a suggested exercise in Example 3.50. We also verify that both np and n(1- p) are at least 10:
$np = 400 \times 0.20 = 80 n(1 - p) = 400 \times 0.8 = 320$
With these conditions checked, we may use the normal approximation in place of the binomial distribution using the mean and standard deviation from the binomial model:
$\mu = np = 80 \sigma = \sqrt {np(1 - p)} = 8$
We want to find the probability of observing fewer than 59 smokers using this model.
Exercise $$\PageIndex{1}$$
Use the normal model N($$\mu = 80, \sigma = 8$$) to estimate the probability of observing fewer than 59 smokers. Your answer should be approximately equal to the solution of Example 3.50: 0.0041.
Solution
Compute the Z score rst: Z = $$\frac {59-80}{8} = -2.63$$. The corresponding left tail area is 0.0043.
### The normal approximation breaks down on small intervals
Caution: The normal approximation may fail on small intervals
The normal approximation to the binomial distribution tends to perform poorly when estimating the probability of a small range of counts, even when the conditions are met.
Suppose we wanted to compute the probability of observing 69, 70, or 71 smokers in 400 when p = 0.20. With such a large sample, we might be tempted to apply the normal approximation and use the range 69 to 71. However, we would nd that the binomial solution and the normal approximation notably differ:
$\text {Binomial}: 0.0703 \text {Normal}: 0.0476$
We can identify the cause of this discrepancy using Figure 3.18, which shows the areas representing the binomial probability (outlined) and normal approximation (shaded). Notice that the width of the area under the normal distribution is 0.5 units too slim on both sides of the interval.
Figure 3.18: A normal curve with the area between 69 and 71 shaded. The outlined area represents the exact binomial probability.
TIP: Improving the accuracy of the normal approximation to the binomial distribution
The normal approximation to the binomial distribution for intervals of values is usually improved if cutoff values are modi ed slightly. The cutoff values for the lower end of a shaded region should be reduced by 0.5, and the cutoff value for the upper end should be increased by 0.5.
The tip to add extra area when applying the normal approximation is most often useful when examining a range of observations. While it is possible to apply it when computing a tail area, the benefit of the modification usually disappears since the total interval is typically quite wide.
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# BetterExplained Calculus : Lesson 15
Proving the volume and the surface area of a sphere by using integrals
Proving the volume and the surface area of a sphere by using integrals
In the preceding lessons we uncovered a few calculus relationships, the “arithmetic” of how systems change:
How do these rules help us?
If we have an existing equation, the rules are a shortcut to find the step-by-step pattern. Instead of visualizing a growing square, or cube, the Power Rule lets us crank through the derivatives of $$x^2$$$and $$x^3$$$. Whether $$x^2$$$is from a literal square or multiplied elements isn’t important — we’ll get the pattern of changes. If we have a set of changes, the rules help us reverse-engineer the original pattern. Getting changes like $$2x$$$ or $$14x$$$is a hint that $$something \cdot x^2$$$ was the original pattern.
Learning to “think with calculus” means we can use X-Ray and Time-lapse vision to imagine changes taking place, and use the rules to work out the specifics. Eventually, we might not visualize anything, and just work with the symbols directly (as you likely do with arithmetic today).
In the start of the course, we morphed a ring into a circle, then a sphere, then a shell:
With the official rules in hand, we can blast through the calculations and find the circle/sphere formulas on our own. It may sound strange, but the formulas feel different to me — almost alive — when you see them morphing in front of you. Let’s jump in.
Our first example of “step-by-step” thinking was gluing a sequence of rings to make a circle:
When we started, we needed a lot of visualization. We had to unroll the rings, line them up, realize they made a triangle, then use $$\frac{1}{2} base \cdot height$$$to get the area. Visual, tedious… and necessary. We need to feel what’s happening before working with raw equations. Here’s the symbolic approach: Let’s walk through it. The notion of a “ring-by-ring timelapse” sharpens into “integrate the rings, from nothing to the full radius” and ultimately: $$area = \int_0^r 2 \pi r \ dr$$ Each ring has height $$2 \pi r$$$ and width $$dr$$$, and we want to accumulate that area to make our disc. How can we solve this equation? By working backwards. We can move the $$2 \pi$$$ part outside the integral (remember the scaling property?) and focus on the integral of $$r$$$: $$2 \cdot \pi \int_0^r r \ dr = ?$$ What pattern makes steps of size $$r$$$? Well, we know that $$r^2$$$creates steps of size $$2r$$$, which is double what we need. Half that should be perfect. Let’s try it out:
$$\frac{d}{dr} \frac{1}{2} r^2 = \frac{1}{2} \frac{d}{dr} r^2 = \frac{1}{2} 2r = r$$
Yep, $$\frac{1}{2}r^2$$$gives us the steps we need! Now we can plug in the solution to the integral: $$area = 2 \cdot \pi \int_0^r r \ dr = 2 \cdot \pi \cdot \frac{1}{2} r^2 = \pi r^2$$ This is the same result as making the ring-triangle in the first lesson, but we manipulated equations, not diagrams. Not bad! It’ll help even more once we get to 3d… Let’s get fancier. We can take our discs, thicken them into plates, and build a sphere: Let’s walk slowly. We have several plates, each at a different “x-coordinate”. What’s the size of a single plate? The plate has a thickness ($$dx$$$), and its own radius. The radius of the plate is some off the x-axis, which we can call $$y$$$. It’s a little confusing at first: $$r$$$ is the radius of the entire sphere, but $$y$$$is the (usually smaller) radius of an individual plate under examination. In fact, only the center plate ($$x=0$$$) will have its radius the same as the entire sphere. The “end plates” don’t have a height at all.
And by the Pythagorean theorem, we have a connection between the x-position of the plate, and its height ($$y$$$): $$x^2 + y^2 = r^2$$ Ok. We have size of each plate, and can integrate to find the volume, right? Not so fast. Instead of starting on the left side, with a negative x-coordinate, moving to 0, and then up to the max, let’s just think about a sphere as two halves: To find the total volume, get the volume of one half, and double it. This is a common trick: if a shape is symmetrical, get the size of one part and scale it up. Often, it’s easier to work out “0 to max” than “min to max”, especially when “min” is negative. Ok. Now let’s solve it: Whoa! Quite an equation, there. It seems like a lot, but we’ll work through it: First off: 3 variables is too many to have flying around. We’ll write the height of each plate ($$y$$$), in terms of the others:
$$height = y = \sqrt{r^2 – x^2}$$
The square root looks intimidating at first, but it’s being plugged into $$y^2$$$and the exponent will cancel it out. After plugging in y, we have the much nicer: $$Volume = 2 \pi \int_0^r r^2 – x^2 \ dx$$ The parentheses are often dropped because it’s “known” that $$dx$$$ is multiplied by the entire size of the step. We know the step is $$(r^2 – x^2) \cdot dx$$$and not $$r^2 – (x^2 \cdot dx)$$$.
Let’s talk about $$r$$$and $$x$$$ for a minute. $$r$$$is the radius of the entire sphere, such as “15 inches”. You can imagine asking “I want the volume of a sphere with a radius of 15 inches”. Fine. To figure this out, we’ll create plates at each x-coordinate, from $$x=0$$$ up to $$x=15$$$(and double it). $$x$$$ is the bookkeeping entry that remembers which plate we’re on. We could work out the volume from $$x=0$$$to $$x=7.5$$$, let’s say, and we’d build a partial sphere (maybe useful, maybe not). But we want the whole shebang, so we let $$x$$$go from 0 to the full $$r$$$.
Time to solve this bad boy. What equation has steps like $$r^2 – x^2$$$? First, let’s use the addition rule: steps like $$a – b$$$ are made from two patterns (one making $$a$$$, the other making $$b$$$).
Let’s look at the first pattern, the steps of size $$r^2$$$. We’re moving along the x-axis, and $$r$$$ is a number that never changes: it’s 15 inches, the size of our sphere. This max radius never depends on $$x$$$, the position of the current plate. When a scaling factor doesn’t change during the integral ($$r$$$, $$\pi$$$, etc.), it can be moved outside and scaled up at the end. So we get: $$\int r^2 \ dx = r^2 \int dx = r^2 x$$ In other words, $$r^2 \cdot x$$$ is a linear trajectory that contributes a constant $$r^2$$$at each step. Cool. How about the integral of $$-x^2$$$? First, we can move out the negative sign and take the integral of $$x^2$$$: $$– \int x^2 \ dx = ?$$ We’ve seen this before. Since $$x^3$$$ has steps of $$3x^2$$$, taking 1/3 of that amount ($$\frac{x^3}{3}$$$) should be just right. And we can check that our integral is correct:
$$\frac{d}{dx} \left( – \frac{1}{3} x^3 \right) = – \frac{1}{3} \frac{d}{dx} x^3 = -\frac{1}{3} 3x^2 = -x^2$$
It works out! Over time, you’ll learn to trust the integrals you reverse-engineer, but when starting out, it’s good to check the derivative. With the integrals solved, we plug them in:
$$2\pi \int r^2 – x^2 \ dx = 2\pi (r^2 x – \frac{1}{3}x^3)$$
What’s left? Well, our formula still has $$x$$$inside, which measures the volume from 0 to some final value of $$x$$$. In this case, we want the full radius, so we set $$x=r$$$: $$2\pi (r^2 x – \frac{1}{3}x^3) \xrightarrow[\text{set x=r}]{} 2\pi \left( (r^2)r – \frac{1}{3}r^3 \right) = 2\pi \left( r^3 – \frac{1}{3}r^3 \right) = 2\pi \frac{2}{3}r^3 = \frac{4}{3} \pi r^3$$ Tada! You’ve found the volume of a sphere (or another portion of a sphere, if you use a different range for $$x$$$).
Think that was hard work? You have no idea. That one-line computation took Archimedes, one of the greatest geniuses of all time, tremendous effort to figure out. He had to imagine some spheres, and a cylinder, and some cones, and a fulcrum, and imagine them balancing and… let’s just say when he found the formula, he had it written on his grave. Your current calculus intuition would have saved him incredible effort (see this video).
Now that we have volume, finding surface area is much easier. We can take a thin “peel” of the sphere with a shell-by-shell X-Ray:
I imagine the entire shell as “powder” on the surface of the existing sphere. How much powder is there? It’s $$dV$$$, the change in volume. Ok, what is the area the powder covers? Hrm. Think of a similar question: how much area will a bag of mulch cover? Get the volume, divide by the desired thickness, and you have the area covered. If I give you 300 cubic inches of dirt, and spread it in a pile 2 inches thick, the pile will cover 150 square inches. After all, if $$area \cdot thickness = volume$$$ then $$area = \frac{volume}{thickness}$$$. In our case, $$dV$$$ is the volume of the shell, and $$dr$$$is its thickness. We can spread $$dV$$$ along the thickness we’re considering ($$dr$$$) and see how much area we added: $$\frac{dV}{dr}$$$, the derivative.
This is where the right notation comes in handy. We can think of the derivative as an abstract, instantaneous rate of change ($$V’$$$), or as a specific ratio ($$\frac{dV}{dr}$$$). In this case, we want to consider the individual elements, and how they interact (volume of shell / thickness of shell).
So, given the relation,
$$\text{area of shell} = \frac{\text{volume of shell}}{\text{depth of shell}} = \frac{dV}{dr}$$
we figure out:
$$\frac{d}{dr} Volume = \frac{d}{dr} \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \frac{d}{dr} r^3 = \frac{4}{3} \pi (3 r^2) = 4 \pi r^2$$
Wow, that was fast! The order of our morph (circumference / area / volume / surface area) made the last step simple. We could try to spin a circumference into surface area directly, but it’s more complex.
As we cranked through this formula, we “dropped the exponent” on $$r^3$$$to get $$3 r^2$$$. Remember this change comes from 3 perspectives (dimensions) that contributed an equal share.
The steps we worked through took 2000 years of thought to discover, by the greatest geniuses no less. Calculus is such a broad and breathtaking viewpoint that it’s difficult to imagine where it doesn’t apply. It’s just about using X-Ray and Time-Lapse vision:
Break things down. In your current situation, what’s the next thing that will happen? And after that? Is there a pattern here? (Getting bigger, smaller, staying the same.) Is that knowledge useful to you?
Find the source. You’re seeing a bunch of changes — what caused them? If you know the source, can you predict the end-result of all the changes? Is that prediction helpful?
We’re used to analyzing equations, but I hope it doesn’t stop there. Numbers can describe mood, spiciness, and customer satisfaction; step-by-step thinking can describe battle plans and psychological treatment. Equations and geometry are just nice starting points to analyze. Math isn’t about equations, and music isn’t about sheet music — they point to the idea inside the notation.
While there are more details for other derivatives, integration techniques, and how infinity works, you don’t need them to start thinking with Calculus. What you discovered today would have made Archimedes tear up, and that’s a good enough start for me.
Happy math.
You are watching: BetterExplained Calculus : Lesson 15. Info created by THVinhTuy selection and synthesis along with other related topics.
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# 7.3: Scatter Plots
Difficulty Level: Basic Created by: CK-12
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You've been exercising every week and when you go for your next doctor's visit the doctor says that the reading for your resting heart rate has changed. You start taking your own resting heart rate once a week on Mondays and relate it to the numbers of hours per week you've been exercising. How would you represent this data? Do you expect to see a correlation between the number of hours you exercise per week and your resting heart rate? How would you know if there is a correlation?
### Watch This
First watch this video to learn about scatter plots.
Then watch this video to see some examples.
Watch this video for more help.
### Guidance
Often, when real-world data is plotted, the result is a linear pattern. The general direction of the data can be seen, but the data points do not all fall on a line. This type of graph is called a scatter plot. A scatter plot is often used to investigate whether or not there is a relationship or connection between 2 sets of data. The data is plotted on a graph such that one quantity is plotted on the \begin{align*}x\end{align*}-axis and one quantity is plotted on the \begin{align*}y\end{align*}-axis. The quantity that is plotted on the \begin{align*}x\end{align*}-axis is the independent variable, and the quantity that is plotted on the \begin{align*}y\end{align*}-axis is the dependent variable. If a relationship does exist between the 2 sets of data, it will be easy to see if the data is plotted on a scatter plot.
The following scatter plot shows the price of peaches and the number sold:
The connection is obvious\begin{align*}-\end{align*}when the price of peaches was high, the sales were low, but when the price was low, the sales were high.
The following scatter plot shows the sales of a weekly newspaper and the temperature:
There is no connection between the number of newspapers sold and the temperature.
Another term used to describe 2 sets of data that have a connection or a relationship is correlation. The correlation between 2 sets of data can be positive or negative, and it can be strong or weak. The following scatter plots will help to enhance this concept.
If you look at the 2 sketches that represent a positive correlation, you will notice that the points are around a line that slopes upward to the right. When the correlation is negative, the line slopes downward to the right. The 2 sketches that show a strong correlation have points that are bunched together and appear to be close to a line that is in the middle of the points. When the correlation is weak, the points are more scattered and not as concentrated.
When correlation exists on a scatter plot, a line of best fit can be drawn on the graph. The line of best fit must be drawn so that the sums of the distances to the points on either side of the line are approximately equal and such that there are an equal number of points above and below the line. Using a clear plastic ruler makes it easier to meet all of these conditions when drawing the line. Another useful tool is a stick of spaghetti, since it can be easily rolled and moved on the graph until you are satisfied with its location. The edge of the spaghetti can be traced to produce the line of best fit. A line of best fit can be used to make estimations from the graph, but you must remember that the line of best fit is simply a sketch of where the line should appear on the graph. As a result, any values that you choose from this line are not very accurate\begin{align*}-\end{align*}the values are more of a ballpark figure.
In the sales of newspapers and the temperature, there was no connection between the 2 data sets. The following sketches represent some other possible outcomes when there is no correlation between data sets:
#### Example A
Plot the following points on a scatter plot, with \begin{align*}m\end{align*} as the independent variable and \begin{align*}n\end{align*} as the dependent variable. Number both axes from 0 to 20. If a correlation exists between the values of \begin{align*}m\end{align*} and \begin{align*}n\end{align*}, describe the correlation (strong negative, weak positive, etc.).
\begin{align*}& m \quad 4 \quad 9 \quad 13 \quad 16 \quad 17 \quad 6 \quad 7 \quad \ 18 \quad 10\\ & n \quad \ 5 \quad 3 \quad 11 \quad 18 \quad 6 \quad 11 \quad 18 \quad 12 \quad 16\end{align*}
#### Example B
Describe the correlation, if any, in the following scatter plot:
In the above scatter plot, there is a strong positive correlation.
#### Example C
The following table consists of the marks achieved by 9 students on chemistry and math tests:
Student A B C D E F G H I
Chemistry Marks 49 46 35 58 51 56 54 46 53
Math Marks 29 23 10 41 38 36 31 24 ?
Plot the above marks on scatter plot, with the chemistry marks on the \begin{align*}x\end{align*}-axis and the math marks on the \begin{align*}y\end{align*}-axis. Draw a line of best fit, and use this line to estimate the mark that Student I would have made in math had he or she taken the test.
If Student I had taken the math test, his or her mark would have been between 32 and 37.
Points to Consider
• Can the equation for the line of best fit be used to calculate values?
• Is any other graphical representation of data used for estimations?
### Guided Practice
The following table represents the sales of Volkswagen Beetles in Iowa between 1994 and 2003:
Year 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003
Beetles Sold 50 60 55 50 70 65 75 65 80 90
(a) Create a scatter plot and draw the line of best fit for the data. Hint: Let 0 = 1994, 1 = 1995, etc.
(b) Use the graph to predict the number of Beetles that will be sold in Iowa in the year 2007.
(c) Describe the correlation for the above graph.
Answer:
a.
b. The year 2007 would actually be the number 13 on the \begin{align*}x-\end{align*}axis. The number of beetles sold in this year would be approximately 98 to 100.
c. The correlation of this graph is strong and positive.
### Practice
1. What is the correlation of a scatter plot that has few points that are not bunched together?
1. strong
2. no correlation
3. weak
4. negative
2. What term is used to define the connection between 2 data sets?
1. relationship
2. scatter plot
3. correlation
4. discrete
3. Describe the correlation of each of the following graphs:
4. Plot the following points on a scatter plot, with \begin{align*}m\end{align*} as the independent variable and \begin{align*}n\end{align*} as the dependent variable. Number both axes from 0 to 20. If a correlation exists between the values of \begin{align*}m\end{align*} and \begin{align*}n\end{align*}, describe the correlation (strong negative, weak positive, etc.).
1. \begin{align*}m \quad 5 \quad 14 \quad 2 \quad 10 \quad 16 \quad 4 \quad 18 \quad 2 \quad 8 \quad 11\\ n \quad \ 6 \quad 13 \quad 4 \quad 10 \quad 15 \quad 7 \quad 16 \quad 5 \quad 8 \quad 12\end{align*}
2. \begin{align*}m \quad 13 \quad 3 \quad 18 \quad 9 \quad 20 \quad 15 \quad 6 \quad 10 \quad 21 \quad 4\\ n \quad \ 7 \quad 14 \quad 9 \quad 16 \quad 7 \quad 13 \quad 10 \quad 13 \quad 3 \quad 19\end{align*}
The following scatter plot shows the closing prices of 2 stocks at various points in time. A line of best fit has been drawn. Use the scatter plot to answer the following questions.
1. How would you describe the correlation between the prices of the 2 stocks?
2. If the price of stock A is $12.00, what would you expect the price of stock B to be? 3. If the price of stock B is$47.75, what would you expect the price of stock A to be?
The following scatter plot shows the hours of exercise per week and resting heart rates for various 30-year-old males. A line of best fit has been drawn. Use the scatter plot to answer the following questions.
1. How would you describe the correlation between hours of exercise per week and resting heart rate?
2. If a 30-year-old male exercises 2 hours per week, what would you expect his resting heart rate to be?
3. If a 30-year-old male has a resting heart rate of 65 beats per minute, how many hours would you expect him to exercise per week?
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### Vocabulary Language: English Spanish
TermDefinition
correlation Correlation is a statistical method used to determine if there is a connection or a relationship between two sets of data.
line of best fit A line of best fit is a straight line drawn on a scatter plot such that the sums of the distances to the points on either side of the line are approximately equal and such that there are an equal number of points above and below the line.
scatter plot A scatter plot is a plot of the dependent variable versus the independent variable and is used to investigate whether or not there is a relationship or connection between 2 sets of data.
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# What is the GCF of 54 and 32?
Learn how to find the greatest common factor of 54 and 32 in this informative guide. Discover the step-by-step process to solve this mathematical problem.
When it comes to solving mathematical problems, the concept of greatest common factor (GCF) is an important tool that helps to simplify the equation. In this article, we will explore the process of finding the GCF of 54 and 32. Understanding how to find the GCF of two numbers is crucial in many mathematical applications and can help save time when solving complex problems.
## Prime factorization of 54 and 32
The first step in finding the GCF of 54 and 32 is to break down each number into their prime factors. Prime factors are the smallest prime numbers that can be multiplied together to get the original number.
To find the prime factors of 54, we can start by dividing it by the smallest prime number, which is 2.
\$\$54 div 2 = 27\$\$
We can now divide 27 by the smallest prime number, which is 3.
\$\$27 div 3 = 9\$\$
Next, we divide 9 by 3 again.
\$\$9 div 3 = 3\$\$
The prime factorization of 54, therefore, is 2 x 3 x 3 x 3.
To find the prime factors of 32, we can start by dividing it by 2.
\$\$32 div 2 = 16\$\$
We can then divide 16 by 2 again.
\$\$16 div 2 = 8\$\$
Next, we divide 8 by 2 again.
\$\$8 div 2 = 4\$\$
Finally, we divide 4 by 2 again.
\$\$4 div 2 = 2\$\$
The prime factorization of 32, therefore, is 2 x 2 x 2 x 2 x 2.
Now that we have found the prime factorization of both 54 and 32, we can move on to the next step of finding the GCF.
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## Identifying common factors
The next step in finding the GCF of 54 and 32 is to identify the common factors between the two numbers. Common factors are factors that both numbers share. To find the common factors, we can list out the factors of each number.
The factors of 54 are 1, 2, 3, 6, 9, 18, 27, and 54.
The factors of 32 are 1, 2, 4, 8, 16, and 32.
From this list, we can see that the common factors between 54 and 32 are 1 and 2. However, the GCF is the greatest common factor, which means we need to find the largest number that both 54 and 32 can be divided by evenly.
To find the GCF, we need to choose the largest common factor, which is 2. We can check if 2 is the GCF by dividing both numbers by 2.
\$\$54 div 2 = 27\$\$
\$\$32 div 2 = 16\$\$
Both 27 and 16 are not divisible by 2 any further, which means 2 is the largest number that both 54 and 32 can be divided by evenly.
Now that we have found the GCF of 54 and 32, we can move on to the next step of checking our answer.
## Identifying common factors
The next step in finding the GCF of 54 and 32 is to identify the common factors between the two numbers. Common factors are factors that both numbers share. To find the common factors, we can list out the factors of each number.
The factors of 54 are 1, 2, 3, 6, 9, 18, 27, and 54.
The factors of 32 are 1, 2, 4, 8, 16, and 32.
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From this list, we can see that the common factors between 54 and 32 are 1 and 2. However, the GCF is the greatest common factor, which means we need to find the largest number that both 54 and 32 can be divided by evenly.
## Finding the greatest common factor
To find the GCF of 54 and 32, we can use the common factors we have identified. The GCF is the largest number that can divide both 54 and 32 without leaving a remainder.
One way to find the GCF is to multiply all the common factors together and choose the largest number that both 54 and 32 can be divided by evenly. However, this method can be time-consuming and impractical for larger numbers.
A more efficient way to find the GCF is to use the prime factorization method. We have already found the prime factors of 54 and 32 in the previous section.
The prime factorization of 54 is 2 x 3 x 3 x 3.
The prime factorization of 32 is 2 x 2 x 2 x 2 x 2.
To find the GCF, we need to choose the prime factors that both 54 and 32 share. In this case, the only common prime factor between 54 and 32 is 2.
We can then multiply all the common prime factors together to get the GCF.
GCF = 2
Therefore, the GCF of 54 and 32 is 2.
Using the prime factorization method, we can easily find the GCF of larger numbers as well. This method is particularly useful when dealing with complex mathematical problems that involve finding the GCF of multiple numbers.
Now that we have found the GCF of 54 and 32, we can move on to the next step of checking our answer.
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## Checking the answer
To ensure that we have found the correct GCF of 54 and 32, we can use a simple test. We can multiply the GCF by the quotient of 54 and 32 to confirm that it equals 54.
The quotient of 54 and 32 is 1.6875. Therefore, we can multiply 2, which is the GCF of 54 and 32, by 1.6875.
\$\$2 times 1.6875 = 3.375\$\$
The result of this multiplication is not a whole number, which means 2 is indeed the GCF of 54 and 32.
This test works because the GCF is the largest factor that two numbers share, which means it can be used to simplify a fraction. When we divide 54 and 32 by 2, we get 27 and 16, respectively. We can simplify the fraction 27/16 by dividing both the numerator and denominator by 2 to get 54/32. Therefore, the GCF of 54 and 32 is the number that simplifies the fraction to its lowest terms.
## Conclusion
In conclusion, the greatest common factor is an essential concept in mathematics that helps to simplify equations. In this article, we have explored the process of finding the GCF of 54 and 32. We started by breaking down each number into their prime factors, identifying the common factors, and finding the largest common factor.
After checking our answer, we can confirm that the GCF of 54 and 32 is 2. This means that both 54 and 32 can be divided by 2, leaving us with 27 and 16, respectively.
Understanding how to find the GCF of two numbers is crucial in many mathematical applications and can help save time when solving complex problems. By following the steps outlined in this article, you can find the GCF of any two numbers quickly and efficiently.
## How Many Pounds is 600 kg? – A Comprehensive Guide
Learn how to convert 600 kg into pounds with our comprehensive guide. Discover the factors that impact weight measurement, including gravity, altitude, and temperature.
## How Many Pounds is 600 kg? – A Comprehensive Guide
Learn how to convert 600 kg into pounds with our comprehensive guide. Discover the factors that impact weight measurement, including gravity, altitude, and temperature.
## What is the Optional DC Cable for the Yaesu FT-70DR?
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## How Many Ounces in 1.5 Pounds?
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## What is Rebirth 2k22? Unlocking the Mysteries of Reincarnation
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## Curving Space and Time
### Warping Slices of Reality
Dark-heaving—boundless,
endless, and sublime,
The image of eternity—
the Throne of the Invisible...
From Lord Byron's Childe Harold's Pilgrimage
Canto IV, Stanza 183
For his first revolution, Einstein unified space and time, and showed that a given observer just took a slice out of this spacetime. Each of the slices was flat: a one-dimensional slice was just a straight line; a two-dimensional slice was a flat sheet. In a similar way, the three-dimensional slices Einstein took were also flat, in a way we will explore below. For his second — and greater — revolution, Einstein allowed the slices to be curved and warped. One-dimensional straight lines became one-dimensional curvy lines; two-dimensional flat sheets became two-dimensional curvy sheets; three-dimensional flat spaces became three-dimensional curvy spaces. To understand this warping, it helps to think about warping in two-dimensional space.
We are all familiar with two-dimensional surfaces. Movie screens, pages in a book, and the surface of an apple are all examples. We also recognize a curved surface when we see it. Turn up half the page in a book, and you will see that it is curved. Leave the page down, and you see a flat surface. In high-school geometry class, we typically explore two-dimensional space, which is almost always assumed to be flat.
One familiar fact from this exploration is the formula for the circumference of a circle in terms of its radius. We remember the number π — pi, roughly 3.1416 — showing up in the formula
$$Circumference = 2 \times \pi \times Radius$$
To use this formula, we make a circle in a particular way. Choose any point, and tie one end of a length of rope to that point. Stretch the rope taut, and rotate around the center point, tracing out the position of the free end of the rope. This will leave us with a circle. If we measure around the circle — its circumference — we should find that it is 2 × π times the length of our rope.
Now imagine that you try to use this formula in a surface that is not flat. Choose the center point of the circle to be the top of a hill and pretend that you are trapped on the hill; you and your rope can't rise up off of it, or dig down inside of it. Draw the circle on the lower parts of the hill. If you now measure around the circumference of the circle you drew, you get a number that is less than 2 × π times the length of the rope! If we tried to make a circle centered on a mountain pass (or in the middle of a horse's saddle), we would get a circumference that is greater than 2 × π times the length of the rope. The classic formula doesn't apply in these curved space examples.
Now, to think about curvature in three dimensions, we recall another high-school formula that gives the surface area of a sphere in terms of its radius:
$$Area = 4 \times \pi \times Radius \times Radius = 4 \times \pi \times (Radius)^2.$$
To use this formula, we make a sphere just as we made a circle, but we move everywhere the rope will let us. The places where the free end of the rope reach define a sphere. If we want to cover this sphere in fabric, for instance, we will need a piece 4 × π (about 12.566) times as big as a square with edges as long as our piece of rope, cut up into little patches, and sewn together just right. That piece of fabric should perfectly cover the sphere — assuming we have flat three-dimensional space.
In the bizarre world of curved three-dimensional space, however, we might need much less or much more fabric than we would think, just using the formula above. Again, the standard formula we learned in high-school (flat-space) geometry doesn't apply. It is not that this geometry is wrong, it's just that it only applies to flat space. We need a more sophisticated type of geometry to deal with curved space. That sophisticated geometry was developed by mathematicians, and called differential geometry.
With the second revolution, Einstein let his slices be warped in the ways we've just seen. Since he had brought space and time together in spacetime, this meant that time could be warped, too. Understanding warped space and time didn't come easy to Einstein, and he resisted as much as he could. Eventually, though, he came to realize that this warping was simply how nature worked, and the differential geometry describing it was the only way to deal with a fundamental concept in his theory: the geodesic.
## Inspiration
Dark-heaving—boundless,
endless, and sublime,
The image of eternity—
the Throne of the Invisible...
From Lord Byron's Childe Harold's Pilgrimage
Canto IV,
Stanza 183
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Guided Lessons
# Multiply by 5
Need extra help for EL students? Try theStrategy Work: Skip Counting to MultiplyPre-lesson.
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Need extra help for EL students? Try theStrategy Work: Skip Counting to MultiplyPre-lesson.
Students will be able to multiply single digit numbers by five and check that their answer ends in a five or zero.
The adjustment to the whole group lesson is a modification to differentiate for children who are English learners.
(5 minutes)
• Show your students the first row of the Mystery Number 5 chart, while covering the rest of the chart with a piece of paper.
• Tell students that you want them to give a thumbs up once they think they have figured out what the mystery number is.
• Display one equation at a time, and read it aloud to your students.
• Call on students to share their answer after most of them are showing a thumbs up or you have shown the whole list of equations.
• Tell students that the mystery number five is special because lots of things come in groups of five.
• Ask students to share things that come in fives. For example, there are five fingers on a hand, five oceans of the Earth, nickels are five cents, quintuplets are five people, a school week is five days, and clocks have minutes grouped in fives.
• Explain that today they are going to use the strategy of skip-counting by fives to find the answer to multiplication problems with the number five.
(15 minutes)
• Give each student a Hundreds Chart worksheet.
• Tell them they are going to find and circle numbers that are a multiple of five by counting repeatedly by five on a hundreds chart.
• Remind your students that multiplication is repeated addition and that MultiplesAre a series of answers (products) using the same base number multiplied by different numbers.
• Count aloud together as you circle the first few numbers. Say, "One, two, three, four, five. Let's circle five. Six, seven, eight, nine, ten. Let's circle ten. Eleven, twelve, thirteen, fourteen, fifteen. Let's circle fifteen."
• Explain to students that the first number circled, five, represents how many times the number has been multiplied (5 x 1 = 5). The second number, 10, represents that five has been multiplied twice (5 x 2 =10). The third number, 15, represents that five has been multiplied three times (5 x 3 =15).
• Have students complete the Hundreds Chart worksheet on their own.
• Count together by fives up to 100 once students have finished the worksheet.
• Ask students to share any patterns they noticed. If students don’t offer that all the multiples of five end with a five or zero, point it out.
• Ask students to double check that this is true by looking at the Mystery Number 5 chart worksheet.
• Tell students that another way to find the answer to a multiplication problem with a five is to skip-count using your fingers.
• Show that if they wanted to multiply five times a number they could count by fives and put up one finger each time until they have that number of fingers up. Say, "For 5 x 7 = ?Count five, 10, 15, 20, 25, 30, 35. I know 35 is the answer because I have seven fingers up and my answer, 35, ends with a five or zero."
• Practise this strategy as a class or in partners.
(5 minutes)
• Show students the problem: 5 x 6 = ?
• Tell them that two people solved this problem, but they got different answers. One got 30 and the other got 31.
• Ask them to describe two ways that they could check the answer. (Skip-count and check that the answer has a five or zero in the ones place.)
• Show students the problem: 9 x 5 = ?
• Tell them that two people solved this problem, but they got different answers. One got 54 and the other got 45.
• Ask them to tell a partner the answer and explain how they know.
• Show students the problem: ? x 5 = 35.
• Ask them to describe to a partner how they can solve this problem.
• Call on a few students to share how they solved this problem.
(10 minutes)
• Have students complete the Multiply by 5 practise worksheet.
• Encourage them to do this worksheet without the hundreds grid in front of them.
Support:
• Have students highlight the multiples of five on the hundreds chart instead of circling the numbers.
• Encourage these students to use the hundreds chart when completing the independent practise.
Enrichment:
• Give students a verbal exit ticket where they have to figure out the missing factor instead of the product. For example, 5 x ? = 40.
• Have students write and solve two-digit numbers multiplied by five using the same strategies learned today.
(2 minutes)
• Collect independent work to serve as a check for students' understanding.
• As students leave your room or transition to another activity, give them a verbal exit ticket asking each student a different multiplication problem with the number five as a factor.
(3 minutes)
• Sit or stand in a circle with your students.
• Count by fives up to 50 by passing a high-five to the student sitting to the left of you in the circle. That student should then say the next multiple as they high-five the next person. Once 50 is said have students start counting by fives again until the circle is complete.
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## Tuesday, March 07, 2006
### Verbogeometry -- Installment VII --- Pythagorean Theorem and The Distance Formula
2.6. The Distance Formula and Pythagorean Theorem: The distance formula uses the Pythagorean Theorem to calculate distances on the Cartesian coordinate system. The Pythagorean Theorem says that the hypotenuse of a right triangle is equal to the square of it sides. (see figure 26)
Figure. 26
solve for c and we get the following see figure 27
Figure. 27
Let us plot two points on a two-dimensional axis system P1(-9,10) and P2(4,3) and If we draw a lines between the points and lines parallel to the axes we can obtain a right triangle. (see figure 28)
Figure. 28
To solve the length of the hypotenuse we first find the difference between the x values and the y values to create the sides of the triangle then we plug the values into the Pythagorean Theorem.
x value is 9 - 4 = 5 and the y value is 10 - 3 = 7
Now we plug it to the equation and we get the expression in figure 29
Figure 29
The distance or length of the hypotenuse would be the square root of 74 or approximately 8.602
The distance formula is thus:
Given two points (x1,y1) and (x2,y2) (see figure 30)
Figure 30
Now what we have been looking at up till now has been in two-dimensions but we can also express a distance in many dimensions. Let us look at two points in three dimensions.
P1(x1,y1,z1) P2(x2,y2,z2)
If we were to put these two points into the distance formula it would look like what is shown in figure 31
Figure 31
It is evident that we can use the distance formula in uncountable dimensions all we have to is to add another dimensional term to the formula for every dimension we wish to express.
## Monday, March 06, 2006
### Verbogeometry -- Installment VI --- Prismatic Structures
2.5. Prismatic Structures in Verbogeometry: Notice that 'barren and infertile' is a complex antonym and 'fertile and infertile' is a simple antonym as previously defined. This is interesting because we can see that there exists in verbogeometry a geometric construction where a line expressed as a simple antonym is normal (90 degrees) to a plane containing all of the complex antonyms related to the line which is expressing the simple antonym. To illustrate this idea lets look again at the relationship between the simple antonyms 'fertile' and 'infertile' and the synonyms 'barren', 'fruitless', 'unproductive', 'sterile', 'impotent' which reside on the plane that is normal (90 degrees) to the line created by the simple antonyms. Furthermore you can draw lines from all of the synonyms back to the complex antonym 'fertile'. (See figure 19)
Figure 19.
This idea also lends itself to prismatic structures where we have a group of parallel simple antonyms whose endpoints construct polygonal faces on two parallel synonym-planes. (See figure 19) Example: Let us define one synonym plane containing the words 'pleased', 'content', 'affected', 'satisfied', 'enchanted' and 'sympathetic'. The other plane contains the following simple antonyms for the previous group of synonyms: 'displeased', 'discontent', 'disaffected', 'dissatisfied', 'disenchanted' and 'unsympathetic'. Due to the synonyms of one plane have corresponding simple antonyms which create lines 90 degrees from the synonym-plane then the simple antonyms are synonyms of each other and reside on their own individual synonym plane and because the lines are 90 degrees to each other the planes must be parallel. The former verbiage is a lot easier to understand visually (See figure 20)
Figure 20.
On a side note: Everything that we have been talking about up to this point has been viewed in Euclidian space. Not much time has been spent exploring verbogeometry outside of Euclidian space. However, I see that there can be a lot more thought evolving models of verbogeometry in other spaces. For example, Instead of viewing antonyms and all the varying meanings spread out across a number-line, we may think of the antonyms as magnetic dipoles. Furthermore, all the varying meaning between them is analogous to a magnetic flux. Yet again, another view may be looking at the antonyms residing at the endpoints of a major axis of a three dimensional ellipsoid and the varying meanings reside on the surface of the ellipsoid. (See figure 21)
Figure 21.
Figure 22 shows an ellipsoidal flux-like paradigm for the previous prismatic structure. (See figure 20)
Figure 22.
Figure23 shows an unlabeled side view
Figure 23.
It may be easier to see the ellipse structure if it was wireframe (See figure 24)
Figure 24.
Figure 25 shows 5 parallel word axes in a wireframe structure
Figure 25.
## Sunday, March 05, 2006
### Verbogeometry -- Installment V --- With Trigonometry
2.4. Verbogeometry with Trigonometry: First let us solve a pertinent traditional trigonometric problem, which requires a solution for an angle. Let us make 'theta' the variable of the angle in question. The tangent of a right triangle is defined as the ratio of the side opposite the angle 'theta' to the side adjacent to the angle 'theta'. If we call the side opposite distance "y" and the side adjacent as distance "x" we then arrive at tan (angle theta) = y/x (See figure 16)
Figure 16.
It also follows that if we know the distance of two sides of a right triangle we can find all the angles. Let us look at an example with the side opposite angle theta being 3 and the side adjacent to angle theta being 4 and. Let us now solve for the angle theta. We know that y/x = tan (theta) so it follows ¾ = tan (theta). To uncover the value of theta, we must take the inverse tangent of ¾. Symbolically this is stated as "theta = inverse tan(y/x)" or approximately 36.87 degrees (See figure 17).
Figure 17.
In general, to reveal the angle after you have calculated the value for the tangent of y/x, your choice is to find the angle on a trigonometric table (old method) or you plug the values in on a hand held calculator.
Now let us look at a similar expression using verbogeometry and looking for the value of angle theta. This time we define the y distance as the difference between the concepts of barren and infertile (think number-line again) and let us define the x distance as the difference between the concepts of infertile and fertile. (See figure 18)
Figure 18.
If we want to know the angle of theta we have to take the inverse tangent of y/x or the inverse tangent of (barren - infertile)/(fertile - infertile) (see figure 18)
## Saturday, March 04, 2006
### Verbogeometry -- installment IV --- Midpoint Formula
2.3. Midpoint Formula in Verbogeometry: Any analytic geometry equation can use coordinate word-pairs instead of numbers to express poetic forms. Let us use the midpoint formula to express the exact point between the two points P1(praise,love) and P2(punishment,hate) from figure 13. Before we look at coordinate word-pairs let us refresh the use of the midpoint formula in analytic geometry. To find the midpoint between two points on a Cartesian coordinate system we add the x coordinates together and divide by 2 to find the x value for the midpoint and we also add the y coordinates together and divide by 2 to find the y value for the midpoint or
(x1 + x2)/2 = x0 and (y1 + y2)/2 = y0.
Lets look at an example of finding the midpoint P0 between the points P1(-14-15) and P2(12,11) utilizing the midpoint formula. x1 = -14 and x2 = 12 so substituting our numbers in the variables of the equation (x1 + x2)/2 = x0 we get (-14 + 12)/2 = -1 also y1 = -15 and y2 = 11 so substituting our numbers in the variables of the equation (y1 + y2)/2 = y0 we get (-15 + 11)/2 = -2 therefore: P0 = (-1,-2) (see figure 14.)
Figure 14.
Let us now take a different approach and replace the numeric variables in the midpoint equation with the words/concepts of love, hate, praise and punishment. We will use the form of coordinate word-pairs P2(love, praise) and P1(hate, punishment). The midpoint formula now shows us that P0(x0, y0) will be formed by the substitution of (x1 + x2)/2 = x0 with (love + hate) = x0 and (y1 + y2)/2 = y0 with (hate + punishment) = y0. Now we have expressed the exact point between love, praise and hate, punishment. (see figure 15.)
Here is a relative piece from 2000 http://www.kazmaslanka.com/midpoint.html
## Friday, March 03, 2006
### Verbogeometry -- installment III --- Word-coordinate Pairs
2.2. Word-coordinate Pairs: We have witnessed a word-axis with different values of an antonym pair along a particular axis 'x' or 'y' in one dimension. (See figure 3) We also have seen a word-plane with values of two antonym pairs along two axes 'x' and 'y' in two dimensions. (See figure 10) Furthermore, we can have word-cubes along the x, y, and z-axes in the third and word-hypercubes in the fourth dimension or we can have antonymic pairs in innumerable dimensions. There is no limit to the dimensional palette for our expressions. Each antonym word-pair adds a new spatial dimension to our expressive construction. Let us talk about the spatial accuracy in defining the location of words in space. Once again, let us look at figure 10 and notice the antonym word pairs, just/unjust and noble/ignoble. However, let us focus our attention to the word-axis just/unjust. We know that we have defined a one-dimensional word-axis with different values of just and unjust but we do not know exactly where each of the words is located along the axis. We have no quantitative value for just or unjust. However, we do have a qualitative value and we know that the word exists somewhere on the axis. What is most important to us in verbogeometry is not the value as such, but the spatial relationship of the values to each other in space. Because the value or the meaning of a word is relative to the context in which it is used, each viewer individually creates his or her own context for meaning. Therefore, exact quantification of the word or its location in space is not possible. However, in some cases, it may be possible to restrict the context to a level where repeatable correlations exist, but those studies are more akin to denotation for the purpose of science. Scientific experimentation "proves" the equation to be mathematically correct and workable within a range of acceptability. In other words, experimental data defines viability of the relationships between the concepts in a scientific equation. On a side note: (When scientific equations are in the intuitive stages of development, there may be an argument to claim that they are in the realm of art, I personally might accept this view if it were not for the fact that their intention is not to make art.) In verbogeometry, we construct equations based on relationships between the qualities of our experiences to evoke meaningful aesthetic expressions of which most are connotative but some may be denotative.Let us get back to the Cartesian coordinate system for a moment and reiterate the idea of coordinate pairs. A point on a two dimensional coordinate system would have values for x and y and would be expressed as such: (x,y) (see figure 1.) A point on a three-dimensional axis system would have values for x, y and z and would be expressed as (x,y,z). A four dimensional point would be expressed as (x,y,z,w) in the fifth dimension as (x,y,z,w,v) etc.Before we get into multidimensional word-axes let just look at a simple two-dimensional word-plane with two word-axes. (see figure 12.)
Figure 12
The vertical axis is a synonym word-pair of praise and punishment and the horizontal axis is a synonym word-pair of love and hate. It is very important to realize that not only does the words love and hate define the identity of the horizontal or x-axis they also hold conceptual points in space along the axis and the same for praise and punishment with respect to the y-axis. The words which are conceptual points in space define a metaphoric value along its respective axis and can be notated as a coordinate pair similar to (x,y) So you may ask what would a coordinate word-pair look like? Let us look at the two points identified as point 1 or P1(love,praise) and point 2 P2(hate,punishment) (see figure 13.)
Figure. 13
## Thursday, March 02, 2006
### Verbogeometry -- installment II --- Word-Axes and Word-Planes
2. The Mechanics of Verbogeometry:
2.1. Word-Axes and Word-Planes. One of the tenets of Verbogeometry and Mathematical poetry is that you "calculate" values of quality as if they were quantity. Of course this seems to be nonsense but then one must realizes that paradox is the mathematical structure of metaphor. When quality and quantity are synonymous then the math equations automatically transcend the normal duty of denotation and enter into the realms of connotation. This concept also bears some relationship to different concepts of infinity pioneered by the Russian-German Mathematician Georg Cantor, namely the infinities in gradation relative to the infinities in Counting.
Another tenet of verbogeometry is that it recognizes antonyms in only a few varieties. A simple antonym is a word whose antonym is a direct negation. (Example: just / unjust, probable / improbable or fertile / infertile etc.) A complex antonym is a word whose antonym is not a direct negation. (Example: just / unfair, probable / doubtful or fertile / barren etc.) There are also gradable (gradient) antonyms which are pairs that express relationships in a continuum, such as up and down. Complementary antonyms are pairs that express an either/or relationship, such as dead or alive. Verbogeometry uses all antonyms as if they were gradable. It is easy to find examples where poets use complementary antonyms as if they were gradable to create certain metaphors. Example: "Bob showed up half dead to work today."
Within the boundaries of verbogeometry is important to understand that we view words as objects floating in space. When we focus on single words, with no context, they are alone inert relating to no other words. However, when we focus on words that have a synonymous partner we can easily imagine a line in space between the words. Probable and improbable are good examples of simple antonyms that we can view connected by a line. (See figure. 3)
Figure 3.
Life is full of dualities it is hard to think about qualities without thinking about opposing ideas. We can view our 'probable / improbable' one-dimensional line as a number line but instead of values of numbers on the line, we think in terms about having different levels of meaning between the two words residing at each end of the line. Due to a number-line being a one dimensional axis it is easy to visualize a word-axis as an axis for a single spatial dimension. We call any pair of words connected by a line a word-axis. Two perpendicular number lines or word axes make a two-dimensional word-axis as well as defining a word-plane. We also have the ability to view the word-plane as an infinite number of coordinates delineated by the word-pairs much like the infinite number of coordinate pairs contained within a Cartesian coordinate system described in the realm of analytic geometry.
Figure 4.
Figure 4 shows a visualization for the physics equation; distance = velocity multiplied by the time. Notice that the y axis displays velocity and the x axis displays time. When we multiply and blend the words in an infinitesimal weave, we arrive at the concept of distance in a tessellated product of the concepts of velocity and time. In another words by positioning the two axes perpendicular to each other, we view every value on one word-axis in relation to every value on the other word-axis. This method affords us a way to 'feel' the entire word plane or axis system with all its different augmented values and gradations. When we multiply two word-axes together we conceptually tessellate a two dimensional plane with different semantic values of the two words blended and augmented. If we were to take a normal Cartesian coordinate system and multiply the x positive integers (1 through 12) times the y positive integers (1 through 12) we get a tessellated plane as in Figure 5 (see figure. 5) (notice the intensity of blue relative to the value of the numbers)
Figure 5.
To help us further visualize this concept let us create a word axis using the words red and green. In this instance, we are going to use red and green as nouns instead of adjectives. (We will use colors as adjectives later) Let us multiply a red-green axis times another red-green axis and view it visually. (See figure. 6) Multiplication of colors is similar to color addition except the disparate intensities of the colors are greater and follow a similar pattern shown in figure 5. The value of the 'numbers' is subjective and not as important as the relationship between the 'numbers'.
Figure 6.
Figure 6 helps us to visualize different word meanings spread across a word plane.
Let us create another example using two different word-axes. However, let's use two different colored word axes instead of both axes being the same colors like our previous example. Then let us break down what we have just accomplished and look at the axes separately. After that we will superimpose a set of different word-axes upon our color axes to compare how the system works.
To facilitate visualizing two different word axes lets look at an example with the word-axes red-green and blue-orange multiplied by each other and mapped on a Cartesian coordinate system. (See figure. 7)
Figure 7.
(Note these diagrams are visual aids not scientific data)
Figure 7 is comprised of two axes: a red-green in the vertical and a blue orange in the horizontal. Let us look at the red-green vertical axis separately (See figure. 8)
Figure 8.
and separate the blue-orange horizontal axis as well (See figure. 9)
Figure 9.
Furthermore, notice the direction of the gradations. When we place the axes perpendicular to each other and multiply them together, we see the color tessellations of the word-plane. (See figure. 7)
What is nice about using colors for our examples is that words used for colors function as both a noun and an adjective depending on our intent. When we map a word plane with word-axes that comprise colors and we use them as adjective synonyms, then this word-plane serves as a paradigm or a pedagogical tool to serve as a general model for understanding all two axes synonym word-planes. Example: Let create a word-plane using the two word-axes of noble/ignoble and just/unjust. (See figure. 10)
Figure 10.
The next step would be to superimpose the noble/ignoble; just/unjust word-plane onto our previous word-plane of blue/orange; red/green. In essence, we are pretending color blue to mean ignoble, orange to mean noble, red to mean just and green to mean unjust. Now we can see the meanings blend into each other in the different areas of our word-plane.(See figure.11)
Figure 11.
We can see the color purple as blend of ignoble and just, red-orange as a blend of just and noble, yellow-green as a blend of noble and unjust and blue-green as a blend of ignoble and unjust. For the record, I certainly am not trying to say there is a relationship of ignobility and injustice with the color blue-green! This example is just a tool to help us with our own concept of visualizing a word-plane. However, we could create a different but, in my opinion, limited set of color metaphors for noble/ignoble and just/unjust. Or we could look at our color example as adjectives on their own merit. This method would automatically help us see them as metaphors. Example: She was red hot. He had a blue day. He was so green he did not know what was happening. Multiplying adjective word-axes together instantly create metaphors.
## Wednesday, March 01, 2006
### Verbogeometry -- installment I --- Cartesian Coordinate System
I would like to step through my paper on verbogeometry in small installments if possible. The paper is laid out so that I discuss a mathematical idea and then show how it relates to verbogeometry. The first installment is an overview on how the Cartesian coordinate system works and how it is applied to a remedial physics problem. Next installment will be about the mechanics of verbogeometry.
VerbogeometryThe confluence of words and analytic geometry
1. Cartesian Coordinate System:
Before we can talk about verbogeometry, we must have a look of some pertinent elements in analytic geometry. Concerning this paper on verbogeometry, you should know enough analytic geometry to plot points and a few basic equations on a Cartesian coordinate system. However, I would like to believe that anyone should be able to enjoy some understanding of verbogeometry. If you need a refresher on the Cartesian coordinate system, you may want to look online. Wikipedia has a nice overview of the System.When we look at the two-dimensional axis of a Cartesian coordinate system, we can see that by picking a point somewhere on the plane defined by this axis that there is a relationship of this point back to the origin. This relationship is understood by the nomenclature of the coordinate pair (x,y) where x and y are distances along each axis from the origin. Furthermore, if we draw lines from a newly created point, perpendicular to both the x-axis and the y-axis and taking into consideration the axis system in the background then we will make a rectangle. (See figure. 1)
Figure1.
The area of a rectangle is product of the lengths of its sides and in the upcoming example the product of the values for the x and y coordinates of this point. Of course this one is a special case because we used the axis origin as our starting point. Example: Let us arbitrarily pick a point defined by the x-y coordinates of (11,13) and draw lines perpendicular to the axes to illuminate what I just said. The area of any rectangle is equal to its height multiplied by its base (The product of the lengths of its sides). We have a green rectangle delineated on our axis system. (See figure. 1) The height of our rectangle is 13 units and it base it 11 units. The area of our rectangle is 143 square units … or 13 X 11 = 143 This later example is one of pure mathematics. However if we want to use math as a language then we will have to apply concepts or words to our axis system. Let us look at a typical physics problem of distance, velocity and time displayed on a two dimensional axis system. Let us assign the y-axis to be levels of velocity in units of miles per hour and the x-axis to be amounts of time in units of hours. (See figure. 2)
Figure 2.
The distance of an object traveling in space is equal to the velocity of the object multiplied by the average time it is traveling in space or d=vt. Let us use the same pure mathematical example as before but by our contextualizing the axis and assigning the y-axis to represent velocity and the x-axis to represent time, our original point from the last example (11,13) has a new meaning. To reiterate … the point before was in the realm of pure mathematics but now the point represents a moment in time of a speeding object. The object is traveling 13 miles per hour and has been traveling for 11 hours. So to calculate, (d = vt), the distance the object has traveled we must multiply the velocity by the time or 13 miles per hour times 11 hours which equals 143 miles.In essence, what we have done has been to assign a concept by using words (velocity and time) to our axis system. Moving our attention up or down on the y-axis displays different values of velocity. Moving left and right on, the x-axis displays different amounts of duration or time. Physical experimentation can easily verify the veracity of this equation. In addition, the same experimentation verifies the verbal concepts and their relationship to each other that we have assigned to our axis system. We can see the relationship between the concepts of distance, velocity and time spread out on a two-dimensional plane via our axis system. It is important to note that these concepts occupy orthogonal spaces as well as all equations in the form of A = BC. Furthermore, the equation makes sense again because it matches our perceptions of the event and the axis system is a model of our experience. Although the use of the latter equations was for the purpose of denotation, Verbogeometry is more interested in the use of equations for connotation and witnessing dissimilar concepts forced into orthogonal spaces to create metaphor.
## Friday, February 24, 2006
### Introducing Verbogeometry
Life has been real busy lately so unfortunately I haven’t made a blog entry for a while. One of the things that have been demanding my time has been my paper on verbogeometry. My plan has been to deliver the paper next summer in London at the Bridges conference on mathematical connections in art music and science. I have been excited about finishing the paper for it’s the most important thing I have written to date on mathematical poetry related matters. I have created a link to an expanded version of the paper on my website. I believe this paper really allows one to understand how words and their meanings can be viewed in space as well as how to approach mathematical poetry in general. Some people don’t know how to approach visualizing mathematical operations on words and their meanings. I believe this paper will give one new tools needed to enjoy mathematical poetry and hopefully inspire new poets. I have hardly scratched the surface of this genre. I imagine many who will fly past me discovering large treasures of aesthetic joy. I feel I am part of a beautiful vista while standing in my tiny garden pitched on a mountain side looking at colorful fields of mathpo expanding out to a horizon of mathematical mountains that I know I will never be able to reach much less enjoy.
Although verbogeometry has existed in my work since 1981 I have never tried to explain the mechanics of how it works. I was hoping people would find their way into it. however, that never really happened proving either I am delusional about its importance coupled with the fact that it is a cumbersome task that delivers far too little aesthetic value for the work required to understand it.
or...
it is a worthwhile endeavor
blending languages
which will induce poetic winds
clearing obscured connections
the spiritual world has
to the physical
bringing a new rationality to art
firmly grounding one end of metaphor
to the infinite.
I am certainly not excluding the former and acknowledging that I may be delusional. However we must also recognize that if we artists were not delusional we would never create anything new which strains the norm. For we would never have the energy required to make art in our aesthetic exile. The avante guard never dies but its endless warriors pay dearly, many sacrificing love and fortunes only to melt into the insignificant gloryless noise quantifying a lonely instant in the spectrum of time.
K
## Saturday, January 28, 2006
### Sentence Structure in Mathematical Poetry
Here is a detail of the equation from above
Today I would like to share part of a conversation I had with Gregory Vincent St. Thomasino dated December 2, 2005 …
Gregory Vincent St. Thomasino said:
You have to find the analogy between math and the grammar of the sentence, and make your "math" work within that grammar.
Kaz Maslanka said:
I am not sure if I understand exactly what you are saying but I find grammatical similarities in physics equations all the time. For example F = ma ---- force is equal to mass times acceleration. “ma” could easily be seen as a clause where mass is the subject and acceleration is the predicate. Many of my polyaesthetic pieces rely on physics equations to supply a sub-context for the piece. Lets look at the piece from “Karmic influences on the double helix” First we must talk a bit about physics. This piece utilizes the physics equation for Energy. E = Fd Energy is equal to force multiplied by distance. In other words the Energy expended on an object is equal to force applied to that object continually spread over a distance. The latter sentence obviously has grammatical qualities and is synonymous with the equation. We also know that F = ma. Now lets break force into its components of mass and acceleration and substitute (ma) for ‘F’ it in our equation that defines energy.
E = (ma)d or E = mad
In other words the Energy expended on an object is equal to the accelerated mass of that object applied continuously over a distance. (Notice again that we are using words to be synonymous with the actual equation)
Even the equation can easily be mapped with sentence structure. The energy in question is equated to Fd or (ma)d where ‘m’ is the subject ‘ad’ is the predicate and d is the object of the verb ‘to accelerate’ To reiterate the mass is the subject and ‘acceleration over a distance’ is the predicate where ‘over the distance’ is the object of the verb ‘accelerate’
Before we look at the poem lets take a look at the distance formula (in two dimensions) used in analytical geometry. In a nutshell the distance formula is the Pythagorean theorem mapped onto a Cartesian coordinate system. e.g. the hypotenuse of a right triangle is the distance we wish to solve and we can do so by using the difference in the y values as one leg of the triangle and the difference in the x values as the other leg of the triangle. This whole mess is a lot easier to see visually. Click here
Now let us look at the poem.
The energy in question is “Karmic Energy” which is equal to the ‘mass’ being the phrases: the conscious embryo, rampart, through the viridian passage” multiplied by the ‘acceleration’ which is “misfortunate paranoia” and multiplied by the distance between the concepts of the difference between the concepts of “the discovery of the wheel and extra-dimensional travel” and the difference between the concepts “single cell intelligence and the discovery of the wheel”
You can see that the poem follows the exact sentence structure of the physics equation, Furthermore the physics equation serves as a paradigm or metacontext for the poem to ‘ride’ or ‘be carried ’ in.
## Friday, January 27, 2006
### Distance and Visual Poetry
The piece, “DISTANCE”, shows how the concepts of velocity and time tessellate a plane in an infinitely woven tapestry of distance.
The piece above “distance” is a little different for me for it is influenced by new information I have learned from the ‘visual poetry’ movement. The piece has elements of visual poetry that I have never used before. It functions in two different ways. One way it functions mathematically as a visual paradigm for a physics tenant of kinematics. Notice a paradigm is not a metaphor it is a simile. The Second way it functions is mathematical vispo because it performs math operations on text. The bottom line is that I would not call it mathematical poetry as such because we are not performing mathematical operations on words as meaning with the intent for connotation. However we are performing math operations on text and I reiterate that I feel this is in the realm of mathematical vispo. Check out Geof Huths amazing blog if you want to understand all the branches of visual poetry.
On a side note here is a visual poem I made for Gregory Vincent St. Thomasino titled “church” It is inspired by and uses his math poem, “change + purse = church”
## Tuesday, January 24, 2006
### Four-Dimensionally Centered
This piece was inspired by the beauty in New Mexico and its relationship to the spiritual ideas of its native people. It is a four dimensional point metaphorically pointing at the atonement archetype.
There are four axes in this Cartesian system: the first axis is that of east and west the next axis being north and south the third axis being up and down and the last axis being inward and outward. Mathematically speaking the point described uses the analytic geometric midpoint formula to define the midpoint in each of the four axes. Metaphorically speaking The point described is at the center of everything and nothing.
## Wednesday, January 18, 2006
### Joint Mathematics Meeting Report
American Mathematical Society
I just returned from the Joint conference on Mathematics in San Antonio and was extremely pleased with the Art Show. Many of the ‘Bridges’ mathartists participated in the show and some attended the conference. Here is a picture of my good friend Reza Sarhangi the nucleus of the Bridges conference (left) standing with Slavik Jablan an extremely important and interesting figure in the Mathart movements in Europe.
Here is a picture of my friend Paul Hildebrandt (left) and Ivars Peterson dining on good Mexican food. Paul is the president of Zometools and Ivars is extremely important figure in the proliferation of fun filled mathematical ideas and is the creator of “Math Trek” for Science News.
Here is a snapshot of Robert Fathauer standing next to his mathart. Robert is president of Tessellations company and he also is involved in the curation of many mathart shows.
Here is a snapshot of Francisco Lara-Dammer with his beautiful Cayley diagrams
Here is a snapshot of Gwen Fisher who does amazing mathematical beadwork.
Here is a snapshot of Anne Burns and her beautiful mathematical landscape
My apology to to Bradford Hansen-Smith for missing the opportunity to shoot his picture (he also attended the conference)
Remember the Alamo!
## Wednesday, January 11, 2006
### Mathpo Beauty?
I snagged this quote off of the new-poetry list, submitted by Jeff Newberry
"Life is doubt, and faith without doubt is nothing but death."
Miguel de Unamuno
When I look at this quote I easily see it also as a mathematical metaphor.
Life = Doubt
Faith - Doubt = Death
What hits me so obviously is how much more beautiful the verbal language is relative to its 'seemingly cold' mathematical counterparts. However, the beauty in mathematical poetry is being able to see within an instant all the mathematically synonymous permutations and synergistically experience them as a whole integrated with the poetic beauty and power of the verbal language.
Doubt = Life
Doubt = Faith-death
Life = Faith-death
I wonder if Mr. Unamuno put much thought into the following quote:
Doubt is life, and life is faith without death
## Tuesday, January 03, 2006
### The Evolution of a Math Poem
Bob Grumman has recently posted another mathpoem worth thinking about. Its history goes back to early March 2005 when Bob was blogging about his version of the Basho haiku which concerns a frog jumping into an ancient pond.
Bobs blog 401
Bobs version:
old pond--abruptly, the sound
of a frog's splash-in
on March 10 2005 Bob reveals that a math poet named El Konde had sent him his interpretation of that same Basho Haiku poem.
Bobs blog 403
Which was:
Old pond / silence = frog splash!
Bob replied, “I like the idea of dividing the pond (arithmetically), but it would be more logical, it seems to me, to divide silence by the frog to get the old pond.” Or:
Silence / frog splash! = Old pond
I would like to analyze these two different syntactical forms of the same Basho poem and see how they relate to Bobs recent poem “Mathemaku for Basho”
First of all we must talk about what takes place when we create and equation in the form:
A = B multiplied by C
When we multiply 2 subjects together those subjects are integrated into a new subject that has its identity is founded in both of the previous subjects.
Lets looks a something we are familiar with If you drive your car 100 miles down the Jersey turnpike from New York to Philly at 50 miles per hour you would get there in about 2 hours (assuming someone doesn’t shoot you for driving too slow)
Distance is equal to velocity multiplied by time. d = v t
The very idea of distance is founded on the ideas of time and velocity at least as far as our equation is concerned. It is also interesting to me that we can use different syntax to make this mathematical expression synonymous in three ways.
d = v t (Distance is equal to velocity multiplied by time.)
v = d/t (velocity is equal to distance divided by time)
t = d/v (time is equal to distance divided by velocity)
Now let’s look at El Konde’s version of the Basho poem:
old pond / silence = frog splash!
Which is mathematically synonymous with:
(silence)(frog splash!) = old pond
and
old pond/frog splash! = silence
What we have done is to integrate the break of silence with the frog splash to define the old pond. Or the (the break of) silence is defined as the old pond divided by frog splash Furthermore we could also read it as the old pond divided by frog splash! Defines (the break of) silence.
I realize that I have added my own interpretation to El Konde’s intention of ‘silence’ in his poem by redefining 'silence' as ‘the break of silence’ but I think it is arguable that Basho was talking about the break in silence as opposed to silence itself. I believe my addition makes El Konde’s version much more clear.
Now let’s look at Bobs change to El Konde’s Poem and its mathematical synonyms:
silence / frog splash! = old pond
silence / old pond = frog splash!
(frog splash!) (old pond) = silence
So what Bob expresses by his change is that the break of silence is defined by the integration of frog splash! and old pond or that the old pond is defined by the break of silence per frog splash and last but not least … frog splash is defined by the break of silence per old pond
Now that we have broke it down, you the reader can make your own decision on what version makes the most sense to you.
Note: (Bob’s intention may have been to think of 'silence' standing as itself, not as 'break in silence', but I doubt it especially since he has ignored it altogether in his new poem)
In conclusion lets look at Bob’s Mathemaku for Basho: http://comprepoetica.com/newblog/blog00691.html
(pond)(frog) = (((((((haiku)))))))
So it shows that Bob is consistent with multiplying pond and frog but he has 'defined haiku' as opposed to defining 'the break in silence'.
By putting parenthesis around the word haiku he has incorporated a vispo technique in his mathematical poem to imply splashing water waves.
So much for evolution,
R Mutt
## Wednesday, December 28, 2005
### Math Tricks ... Numeric Vispo?
Vê o que acontece se multiplicarmos 37 por múltiplos de 3:
3 x 37 = 111
6 x 37 = 222
9 x 37 = 333
12 x 37 = 444
15 x 37 = 555
18 x 37 = 666
21 x 37 = 777
24 x 37 = 888
27 x 37 = 999
Agora vê isto:
111111111 x 111111111 = 12345678987654321
Repara neste trapézio:
1 x 9 + 2 = 11
12 x 9 + 3 = 111
123 x 9 + 4 = 1111
1234 x 9 + 5 = 11111
12345 x 9 + 6 = 111111
123456 x 9 + 7 = 1111111
1234567 x 9 + 8 = 11111111
12345678 x 9 + 9 = 111111111
E neste outro:
1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321
E ainda neste:
0 x 9 + 8 = 8
9 x 9 + 7 = 88
98 x 9 + 6 = 888
987 x 9 + 5 = 8888
9876 x 9 + 4 = 88888
98765 x 9 + 3 = 888888
987654 x 9 + 2 = 8888888
9876543 x 9 + 1 = 88888888
98765432 x 9 + 0 = 888888888
987654321 x 9 - 1 = 8888888888
9876543210 x 9 - 2 = 88888888888
I find these little pieces very interesting.
These have been floating around on the web lately and were emailed to me twice by different people. I have no idea how old they are or who originally discovered them. I would imagine that more than one person contributed to this group. I would not call them mathematical poetry but they may be considered some form of numeric Vispo.
They remind me a little bit of magic squares.
http://mathforum.org/alejandre/magic.square.html
If odd math expressions are your thing you may want to check the following:
http://curiousmath.com/
Happy Holidays
## Saturday, December 17, 2005
### Golden Fear
GOLDEN FEAR
This aesthetic distillation was inspired by the concept of being told to to fear something that is associated with the golden rule.
The American Mathematical Society has accepted “Golden Fear” in their upcoming show in San Antonio, Texas Here is a link to the other Mathematical Artworks accepted in the show.
http://myweb.cwpost.liu.edu/aburns/jmm06/jmm06.htm
“Golden Fear” uses one of the equations for the golden ratio or phi, which is the most irrational number compared with all irrational numbers. Out of all the equations for phi “Golden Fear” uses the equation, which is in the form of a continued fraction. However, the word ‘fear’ is substituted for the number '1' in the original equation
http://mathworld.wolfram.com/GoldenRatio.html
## Monday, December 12, 2005
### Power of Confidence
Bob Grumman reviewed a couple of pieces of mine on his blog and I would like to share my comments to him.
http://comprepoetica.com/newblog/blog00676.html
Bob,
I appreciate the posting and would like to comment.
Bob said,” To be frank, I'm not sure why he seems to provides a range of values for his delta time,”
Delta means the “change in” see the link below for they explain it quite well. In a nutshell, when delta is present one must put things in terms of what it changes to from what state it originally had. So it goes that the change in x would be between two different states of x. In equation form it follows that (x1-x2) or basically the change from the existing state of x1 to the change of the new state, which is x2. That is why we had two different states in the poem … one being temporal neutrality and the other being the state of eureka
what are deltas?
Concerning the relationship of the Images to the math, I view the relationship of the images to the math poem the same as one would view the music in relation to the lyrics of a song … As to the task of putting my work in terms of Vispo … I guess my work would fit under the category of emblem poems with math poems instead ‘text only’.
Your Blog entry has inspired me to go into detail about how this piece works. The physics equation for Power states that power is equal to energy expended per duration of time. That’s all there is to it … That is the definition of power. (The term work and energy are synonymous)
Understanding the equation for power
However, there is much to what energy means. So now let us look at what physics says about energy.
Different faces of energy
In my piece Energy is the amount of force exerted over a distance. Let’s talk about distance. Distance is basically the Pythagorean Theorem mapped out on a Cartesian coordinate system. What we have is a change in position in one axis relative to a change in position in another perpendicular axis. (The following link explains it quite well)
http://www.purplemath.com/modules/distform.htm
Because there is a relationship between Energy and force lets look at what physics says about force. Force is mass multiplied by acceleration and acceleration is the change in velocity per time. (Delta t) So lets put the puzzle altogether and see how it reads.(in two dimensions) Power is a mass multiplied by a change in velocity per time multiplied by a change in position in one axis relative to a change in position in another axis and all of the previous verbiage is divided by a change between two positions in time.
Now let’s put the equation for power in context with my piece “The power of confidence”
Let me introduce a term … “virtual object” can be an emotional or psychological object present in my mathematical poetry that utilizes physics equations as a paradigm “to carry” the poem. The virtual object is a substitute for the physical object that is normally the subject of a physics equation.
The mass is of our virtual object is “Your diamond strength crystal bravery, a faceted fortress tempered by your self love”
The change in velocity 'delta v' of our virtual object = “From the cold insecurity of night to the sun shaking your hand each morning while your heart drums life's confirmation song”
The change in time that affects the virtual object 'delta t' = “From temporal neutrality to that split instant your intuition knows . . . eureka!”
The distance our virtual object will traverse:
The difference between x1 and x2 in our first dimension:
x1 = “Celebrating your demonic victories, the transformation to deity”
Furthermore, the difference between y1 and y2 in our second dimension:
y1 = “Forged impenetrable through triumph”
y2 = “Soft as your baby butt”
The Poem in words:
So you can see that our mass of “Your diamond strength crystal bravery, a faceted fortress tempered by your self love” is being accelerated between the concepts of “the cold insecurity of night” to the concept of “the sun shaking your hand each morning while your heart drums life's confirmation song” and this is happening between the moments of “temporal neutrality” to the moment of “that split instant your intuition knows . . . eureka!” and all of this is moving over the distance between the difference of the concepts of “Celebrating your demonic victories, the transformation to deity” and “Lying in your crib, frightened of your mothers sneeze” and moving in a second dimension over the distance between the difference of the concepts of “Forged impenetrable through triumph” and “Soft as your baby butt” Furthermore all of this is occurring between the moments of “temporal neutrality” to the moment of “ that split instant your intuition knows . . . eureka!”
## Thursday, December 08, 2005
### Beginner's Mind ~ Denotation / Connotation
There are numerous meanings throughout Asian philosophical history for the term “Beginners Mind”. That said, I will repeat one of my favorite stories illuminating the term in support of this blog post.
In the mathematical visual poem entitled “beginners mind” two forms of perfection are expressed. The first one is the Buddhist calligraphic circle called Enso in practiced in Japan or Ilwonsang practiced in Korea. The painting of this circle is created over and over as the monks practice the meditation of staying in the present moment. While in the present moment the “emptiness of self” can be experienced, thus substantiating this image as a symbol of emptiness. To us it is a perfect example of circular perfection expressed as connotation … and layered on top of that, we have a perfect example of circular perfection expressed as denotation in none other than the analytic geometrical equation for a circle:
x^2 + y^2 = r^2
While using this equation to plot points creating a circle on a two-dimensional Cartesian coordinate system one becomes aware of the awe-inspiring infinite precision of not only this equation but by conceptual extension into all mathematical equations.
The 49 Chinese characters in the background are the calligraphic Chinese expression of “Beginner’s mind”
Link to the equation of a circle: (my example h and k are valued at zero) http://www.analyzemath.com/CircleEq/Tutorials.html
A moment of silence for John Lennon ...
## Monday, December 05, 2005
### AMS American Mathematical Society Conference
I plan on attending this conference ... It should be a good time.
check out their web site.
http://www.ams.org/amsmtgs/2095_exhibits.html
## Sunday, December 04, 2005
### Mathematical Simile Mathematical Metaphor
Mathematical simile is the function of virtually all applied math equations for which the public is familiar (usually scientific). For example physics equations are paradigms used to correlate experience. Mathematical models can be made for anything rationally understood and are descriptive of the relationships between the elements described.
A simile is a paradigm. One can see this expressed in the physics equation distance is equal to velocity multiplied by time. When you apply this equation to a situation you realize that this is an approximation even for all practical purposes it accomplishes the task at hand. The bottom line is that it is an approximation even though the equation expresses it as exact. The math equation is exact the application is not. Therefore it is simile ‘our experience of distance is like velocity multiplied by time.’ One may argue that the physics equation is metaphor precisely for the same reason e.g. we are saying that distance is exactly equal to velocity multiplied by time and it is not. Therefore we have a logical tension that creates the metaphor. The next step is to look at the intention of the equation, whether it is intended for connotation or denotation. Clearly physics equations are intended for denotation. Therefore what is really important for us is the intention of the equation. This also illuminates the idea that there are a lot of shades of gray between simile and metaphor yet we don’t usually think in this manner. I know a lot of people that think simile and metaphor are really the same. I believe at some point there is a nonsense threshold that one crosses and you can take it no other way except metaphor. Unfortunately that threshold line can be anywhere on a number line between the concept of simile and metaphor. Personally I see that threshold very close to simile. That is as soon as it starts getting nonsensical I consider it metaphor.
The visual image presented takes the following simile and expresses it as metaphor.
Although it may or may not be universally true I find that a man’s intelligence is inversely proportional to the levels of alcohol and testosterone in his system. I gave a lecture last week to a group of art students from Santa Ana Art Institute (Thank you Professor Ben Miles) and when I proposed this statement I got a bunch of laughs and a lot of nodding heads. Consequentially I believe there is some truth in there.
## Friday, December 02, 2005
### Humor
Tim Repp sent me this little mathpo related humor
## Thursday, December 01, 2005
### Synecdochical Physics
I was inspired by conversations with The Mathematician Paul Gailiunas and a blog entry of Visual Poet Geof Huth to produce some synecdoche physics equations.
syn·ec·do·che n. Listen:
A figure of speech in which a part is used for the whole (as hand for sailor), the whole for a part (as the law for police officer), the specific for the general (as cutthroat for assassin), the general for the specific (as thief for pickpocket), or the material for the thing made from it (as steel for sword).
Whereas in all of the equations the variables, for the physics equations, have been replaced with a synecdoche counter part not all of the equations as a whole could be considered synecdochical. All of them but one created a synecdochical equation.
Newton’s equation on gravity is actually a metaphor. That is due to the fact that the force variable (replaced by attraction) in the mathpoem does not pertain to gravity but is in reference to the attraction caused by hunger. If it were pertaining to gravity then the equation would be synecdochical however anything else makes the equation a metaphor.
http://dbqp.blogspot.com/2005/10/synechdoche-explained.html
## Friday, November 25, 2005
### Math Poets
I have been delinquent in making entries into the blog but finally I have put some links into the link section. Besides myself I have added a link to a couple of Bob Grumman's pages and one to a Page of Scott Helms. Of all of Bobs Work my favorite is the one above. Bob and I seem to be the only people in the English speaking world who continue to focus on Mathematical Poetry or Math-Po If there are others please let me know. There are others who have done work in Visual Poetry and call their work Mathematical Poetry Karl Kempton is one who has done some very interesting Visual Poetry Using Mathematical operations on text. I would guess that the authority on how Math has been used in Visual Poetry would be Bob Grumman or Possibly Geof Huth. Geof is my focal point on the world of Visual Poetry. The depth of information Geof covers is astounding.
While I am on the subject of Visual Poetry I would like to mention I really like the selected poems by Michael P. Garofalo
## Saturday, March 12, 2005
### What is mathematical poetry
Mathematical poetry is an artistic expression created by performing mathematical operations on words or images as if they were numbers. One may find this baffling because it seems we are being confused about the states of quality versus quantity. But it is through the fusion of this dichotomy that mathematical metaphor is spawned. Historically speaking we believe that math has never been intended for metaphor. We intend to change that. By blending the wondrous aesthetics of mathematics with enigmatic aesthetics of art we intend to bridge the gap between the infinite and the concrete.
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# If I know the mean, standard deviation, and size of sample A and sample B, how do I compute the standard deviation of the union of samples A and B? What if samples A and B are of different sizes?
Sep 12, 2016
#### Explanation:
Let the mean, standard deviation and size of sample $A$ be ${\overline{X}}_{A}$, ${S}_{A}$ and ${n}_{A}$ respectively
and mean, standard deviation and size of sample $B$ be ${\overline{X}}_{B}$, ${S}_{B}$ and ${n}_{B}$ respectively.
Then mean of comibined sample $\overline{X}$ is given by
$\overline{X} = \frac{{n}_{A} {\overline{X}}_{A} + {n}_{B} {\overline{X}}_{B}}{{n}_{A} + {n}_{B}}$
and Standard Deviation of combined sample $S$ is
$S = \frac{{n}_{A} \left({S}_{A}^{2} + {\left({\overline{X}}_{A} - \overline{X}\right)}^{2}\right) + {n}_{B} \left({S}_{B}^{2} + {\left({\overline{X}}_{B} - \overline{X}\right)}^{2}\right)}{{n}_{A} + {n}_{B}}$
Note that it is important to work out mean of combined sample first, as to is used to calculate Standard Deviation of combined sample.
If the sample sizes are equal then the above reduces to
$\overline{X} = \frac{{\overline{X}}_{A} + {\overline{X}}_{B}}{2}$ and
$S = \frac{\left({S}_{A}^{2} + {\left({\overline{X}}_{A} - \overline{X}\right)}^{2}\right) + \left({S}_{B}^{2} + {\left({\overline{X}}_{B} - \overline{X}\right)}^{2}\right)}{2}$
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# How Many Soccer Fields Is a Mile
How Many Soccer Fields Is a Mile?
When it comes to measuring distances, it is important to have a reference point that is relatable to our everyday experiences. For soccer enthusiasts, understanding how many soccer fields make up a mile can be an interesting way to visualize distances. In this article, we will explore the relationship between soccer fields and miles and answer some common questions regarding this topic.
A standard soccer field is typically rectangular in shape, measuring 100 yards in length and 50 yards in width. To calculate the area of a soccer field, we multiply the length the width, yielding 5,000 square yards. Now, let’s dive into the conversion of soccer fields to miles.
To convert square yards to square miles, we need to divide the number of square yards 3,097,600 (which is the number of square yards in a square mile). So, 5,000 square yards would be equal to approximately 0.0016129 square miles.
To find the number of soccer fields in a mile, we divide the total area of a mile (which is 1 square mile) the area of a soccer field. Using the previous conversion, we can estimate that there are around 620.3 soccer fields in a mile. However, it’s important to note that this value may vary slightly depending on the specific measurements of the soccer field and mile being considered.
Now, let’s address some common questions related to the topic:
1. How do you calculate the area of a soccer field?
– Multiply the length the width (in yards) to get the area in square yards.
2. How many square yards are there in a square mile?
– There are 3,097,600 square yards in a square mile.
3. How many square miles are there in a soccer field?
– A soccer field is equivalent to approximately 0.0016129 square miles.
4. How many soccer fields fit in an acre?
– There are approximately 1.32 soccer fields in an acre.
5. How many soccer fields are there in a kilometer?
– Since a kilometer is approximately equal to 0.6214 miles, there would be around 384.9 soccer fields in a kilometer.
6. How many soccer fields are in a marathon?
– A marathon is approximately 26.2 miles long, so there would be around 16,263 soccer fields in a marathon.
7. How many soccer fields are there in a half marathon?
– A half marathon is approximately 13.1 miles long, so there would be around 8,131.5 soccer fields in a half marathon.
8. How many soccer fields are there in a 5K race?
– A 5K race is approximately 3.1 miles long, so there would be around 1,914 soccer fields in a 5K race.
9. How long is a soccer field in feet?
– A soccer field is 300 feet long.
10. How wide is a soccer field in feet?
– A soccer field is 150 feet wide.
11. How many soccer fields can fit in a football field?
– A football field is 120 yards long and 53.33 yards wide. Therefore, approximately 1.92 soccer fields can fit in a football field.
12. How many soccer fields can fit in a basketball court?
– A basketball court measures approximately 94 feet long and 50 feet wide. Therefore, around 0.62 soccer fields can fit in a basketball court.
13. How many soccer fields can fit in a tennis court?
– A tennis court measures approximately 78 feet long and 36 feet wide. Therefore, around 0.52 soccer fields can fit in a tennis court.
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# Diagonal Formula
Diagonal formula determines the number and length of diagonals present in a polygon. A diagonal is a slanted line segment that joins any two vertices of a given polygon. These are straight lines that connect two corners of a square, rectangle, rhombus, or another polygon with more than three sides.
Students can learn diagonal formula and the examples based on them in detail by referring to the Extramarks website or mobile application.
## What is a Diagonal?
Diagoanl is any line segment that joins two vertices of a polygon that are not previously connected by an edge.
• Diagonals are different from the edges as they do not determine the boundary of a polygon.
• The diagonals are always in the form of a straight line.
• Simply we can say that, a diagonal is a line that, at its vertex, joins the opposite corners of a polygon or a polyhedron.
• Diagonal is always located inside the interior of convex polygons.
• Concave polygons may have diagonals that cross sides and partially lie outside
NOTE: A triangle does not have any diagonal. Diagonal is always present in a polygon with atleast four sides.
## What is Diagonal Formula?
Diagonal Formula is a mathematical formula used to calculate the number and of diagonals present in a polygon. We know that when we join two opposite vertices of a polygon we get a diagonal. However, the number of diagonals and its length depends upon the type of polygons. We can also calculate diagonal lenth using distance formula which we study in higher class, however there are some direct diagonal formulas that are used to calculate diagonal length. Let’s learn them below in detail.
## Formula for Diagonals
In order to determine how many and how long diagonals are present in a polygon, one can use the Diagonal Formula. The formula for determining how many diagonal lines there are in an n-sided polygon is as follows:
Number of Diagonals Present in a Polygon = n(n – 3)/2
here,
n is the number of sides of the polygon
There are various Diagonal Formula in geometry for the calculation of length of diagonals in various types of polygons. These calculations are utilised only for that specific polygon.
### Diagonal of a Square
In order to determine the length of a diagonal, let’s now examine a few different Diagonal Formula.
Diagonal of square = a√2
here
a is the length of the side of the square.
### Diagonal of a Rectangle
For rectangles, the diagonal formula is given as
Diagonal of a Rectangle = √(l² + b²)
here,
• l is length of rectangle
• b is breadth or width of rectangle
### Diagonal of a Parallelogram
In a parallelogram, the diagonals are of unequal length. If p and q are length of diagonals for a Parallelogram ABCD, then its diagonal formula is given as
• p = (a² + b² – 2ab.cos A)
• q = √(a² + b² + 2ab.cos A)
here,
• p and q are diagonals
• a and b are adjacent sides
• A is angle of Parallelogram
NOTE: If angle A is given for parallelogram ABCD and we need to use angle B in above formula we can find by the formula (angle B = 180 – angle A)
If one diagonal’s length is known, the formula for the other diagonal can be calculated from the below formula
p² + q² = 2(a² + b²)
Where
• a and b are the sides of the parallelogram
• p and q are the two diagonals of the parallelogram
### Diagonal Formula of Rhombus
The diagonal formula for Rhombus is given as
• p = 2A/q
• q = 2A/p
here,
• p and q are length of diagonals
• A is area of rhombus
## Examples Using Diagonals Formula
Example 1: Sam is walking around a rectangular park whose length of 10m and breadth is 8m. Find out the diagonal of a rectangular park where Sam is walking.
Solution:
To find The diagonal of a rectangular park.
Given parameters are,
Length = 10m
Using the Formula for Diagonals,
Rectangle Diagonal = √[l² + b²]
= √[10² + 8² ]
= √[164]
= 12.80 m
Answer: The diagonal of a rectangular park where Sam is walking is 12.80 m.
Example 2: The area of the rhombus is 100 square inches. Determine the second diagonal of a rhombus whose one of its diagonal one measures 10 inches.
Solution:
To find: The second diagonal of a rhombus
Given parameters are,
The area of the rhombus = 100 inch²
Using the Diagonal Formula,
Diagonal of a Rhombus, p = 2(A)/q and q = 2(A)/p
p = 2(100)/10
p = 20 inches.
Example 3: Determine the length of the diagonal of a square whose side measure is 5 units
Solution:
To find: The diagonal of a square
Given parameters are,
Side of square = 5 units
Using the Formula for Diagonals,
Square Diagonal = a√2
= 5√2
= 7.07 units
Answer: The diagonal of a square is 7.07 units.
### 1. How to Use the Diagonal Formula to Determine the Number of Diagonals?
An n-sided polygon’s diagonal count is calculated using n(n-3)/2, where n denotes the number of sides.
### 2. What Is the Rhombus Diagonal Formula?
Rhombus Diagonal Formula: p = 2(A)/q and q = 2(A)/p, where A denotes the rhombus’s area, and p and q denote its two diagonals.
### 3. What is Diagonal Formula of Square?
The diagonal formula for square is a√2 where a is side of the square
### 4. What is Diagonal Formula of Rectangle?
The diagonal formula of rectangle is √(l² + b²)
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# Chapter 3 – Playing with Numbers
The following Topics and Sub-Topics are covered in this chapter and are available on MSVgo:
Introduction
Numbers are an inherent part of our every-day life and continue to remain an important factor through which we measure things. Numbers are used extensively in the business and finance domain, along with other instruments of life. Let’s check the different concepts of playing with numbers, along with their various operational characteristics in this article.
#### Factors and Multiples
Factors are the exact divisors of any number that leave no remainder. For example, if we take 12 as our number. Then we can find multiple numbers as their exact divisor. They are as follows: 1, 2, 3, 4, 12, and 6.
Let’s check some properties of factors:
• Any number is a factor of itself.
• 1 is always an integral factor of any number.
• Factors are always lesser than, or equal to the number, and never greater than it.
• There can be a finite number of factors for any given number.
Multiples are the multiplicative values of any given numbers. Hence, we multiply the given number with a natural number to obtain their multiples. For example, we have the number 3, and we need to calculate its multiples. Then we start multiplying 3 with natural numbers starting from 1. Hence we get 3, 6, 9, 12, 15, 18, and more.
#### Properties of Multiples
• Every multiple of a number is always equal to, or greater than that number.
• There can be an infinite number of multiples of any given number as we can keep on multiplying till infinity.
• Every number is a multiple of itself.
Suppose we are given more than two numbers and told to find their factors and multiples. Then, we can get some common factors and common multiples for every given number.
#### Prime and Composite Numbers
Prime Numbers: Any number other than 1 with only two factors, the number itself and 1, are called prime numbers. Examples: 2, 3, 5, 7, and more. All these numbers have exactly two factors that are the number itself and 1.
Composite numbers: If any given pair of numbers has more than two factors, they are called composite numbers. For examples 12, 14, 15, and more. All these numbers have at-least 3 factors and hence are called composite numbers.
#### Some More Divisibility Rules
We can test the divisibility rules based on some predefined rules for the existing numbers. Let’s check for the divisibility test of numbers according to the rules that are given below:
Number Divisibility Rule 2 Any number with 0, 4, 2, 6, and 8 at its one’s place is divisible by 2. 3 If the sum of the given number’s digits is divisible by 3, that number is divisible by 3. 4 For numbers having more than 3 digits, check the divisibility of the last two-digit by 4. 5 If the last digit is 0 or 5, then the number is divisible by 5. 6 Any number has to satisfy this condition of being divisible by both 2 and 3 to be divisible by 6. 10 If the last digit of the number is 0, then the number is divisible by 10.
#### Prime Factorisation
The factorisation is the method to break down the number in terms of its factors. Prime factorisation is similar to factorisation, with the only difference being that the factors are prime numbers. For example, if we have 36 to be factored, then the answer will be 36 = 2 * 2 * 3 * 3. Hence 2 and 3 are prime numbers here. Some problems consist of large numbers that have to be factored using prime factorisation. For that, we have a different tabular method of prime factorisation.
#### Highest Common Factor
When we are given two numbers, we can do their prime factorisation. Then the highest common factor of the set two numbers is the HCF of the pair of numbers. It is also known as the greatest common factor.
For example, if we have 12 and 24 as our numbers. Now after prime factorisation, we get,
• 12 = 2 * 2 * 3
• 24 = 2 * 2 * 2 * 3
Here, the common multiples are 2 * 2 *3 = 12. Hence, 12 is the HCF of 12 and 24.
#### Lowest Common Multiple
It is the lowest common multiple of any two given numbers. For example, if we have two number as 12 and 24, that after factorisation we get:
• 12 = 2 * 2 * 3
• 24 = 2 * 2 * 2 * 3
Now in this the common multiples are 2 * 2 * 3 = 12. Also, there is one uncommon multiple, i.e., 2. So, 12 * 2 = 24. Hence, 24 is their least common multiple.
#### Conclusion
Playing with Numbers is an important maths topic and is an inherent part of mathematics and human life. In this topic, we saw the different concepts that can be used on numbers. We can see factors, multiples, prime numbers, odd-even numbers and more in this topic. We can also check for test for divisibility of numbers using certain rules.
#### FAQs
1. What do you mean by playing with numbers?
Playing with Numbers is an 8th class NCERT maths chapter that deals with numbers’ various properties and characteristics.
1. How do you teach numbers to play?
Numbers are the base of mathematics and should be taught to small children through fun activities.
1. What are whole numbers for Class 6?
Whole numbers are the counting numbers that are positive and start from zero. They can be depicted on the number line.
1. What is a Co-prime number?
The pair of numbers having 1 as their highest common Factor is known as co-primes. Co-primes are always formed for pairs. E.g., 1 and 22
1. What is the meaning of Factor?
Factors are exact divisors of those numbers that leave no remainder. For example, the factors of 12 are 1, 2, 3, 4, and 6.
1. What is the smallest Co-prime number?
The smallest coprime numbers are 1 and 2.
Download MSVgo to know more about numbers and the different types of numbers that play an important role in beginning the base of mathematics. You can get video tutorials that explain the topic well.
### High School Physics
• Alternating Current
• Atoms
• Communication Systems
• Current Electricity
• Dual nature of Radiation and Matter
• Electric Charges and Fields
• Electricity
• Electromagnetic Induction
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### High School Chemistry
• Acids, Bases and Salts
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• The d-block and f-block elements
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### High School Biology
• Absorption and Movement of Water in Plants
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• Neural Control And Coordination
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• Photosynthesis
• Photosynthesis in Higher Plants
• Plant Growth and Development
• Plant Kingdom
• Pollination and Fertilization
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• Principles of Inheritance and Variation
• Reproduction and Development in Angiosperms
• Reproduction in Organisms
• Reproductive Health
• Respiration in Human Beings
• Respiration in Plants
• Respiratory System
• Sexual Reproduction in Flowering Plants
• Strategies for Enhancement in Food Production
• Structural Organisation in Animals
• Structural Organisation of the Cell
• The Endocrine System
• The Fundamental Unit Of Life
• The Living World
• The Nervous System and Sense Organs
• Tissues
• Transpiration
• Transport in Plants
### High School Math
• Algebra – Arithmatic Progressions
• Algebra – Complex Numbers and Quadratic Equations
• Algebra – Linear Inequalities
• Algebra – Pair of Linear Equations in Two Variables
• Algebra – Polynomials
• Algebra – Principle of Mathematical Induction
• Binomial Theorem
• Calculus – Applications of Derivatives
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• Calculus – Continuity and Differentiability
• Calculus – Differential Equations
• Calculus – Integrals
• Geometry – Area
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• Geometry – Introduction to Euclid’s Geometry
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• Geometry – Lines and Angles
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• Linear Programming
• Matrices and Determinants
• Mensuration – Areas
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• Number Systems
• Number Systems – Real Numbers
• Permutations and Combinations
• Probability
• Sequence and Series
• Sets and Functions
• Statistics
• Trignometry – Height and Distance
• Trignometry – Identities
• Trignometry – Introduction
### Middle School Science
• Acids, Bases And Salts
• Air and Its Constituents
• Basic Biology
• Body Movements
• Carbon and Its Compounds
• Cell – Structure And Functions
• Changes Around Us
• Chemical Effects Of Electric Current
• Coal And Petroleum
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• Conservation Of Plants And Animals
• Crop Production And Management
• Electric Current And Its Effects
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• Force And Pressure
• Forests: Our Lifeline
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• Garbage In, Garbage Out
• Getting To Know Plants
• Health and Hygiene
• Heat
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• Life Processes: Nutrition in Animals and Plants
• Materials: Metals And Non-Metals
• Matter and Its States
• Metals and Non-metals
• Micro Organisms: Friend And Foe
• Motion And Measurement Of Distances
• Motion And Time
• Nutrition In Animals
• Nutrition In Plants
• Organization in Living Things
• Our Environment
• Physical And Chemical Changes
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• Reaching The Age Of Adolescence
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• Reproduction In Plants
• Respiration In Organisms
• Rocks and Minerals
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• The Living Organisms And Their Surroundings
• Transfer of Heat
• Transformation of Substances
• Transportation In Animals And Plants
• Universe
• Waste-water Story
• Water: A Precious Resource
• Weather, Climate And Adaptations Of Animals To Climate
• Winds, Storms And Cyclones
### Middle School Math
• Area and Its Boundary
• Boxes and Sketches
• Data Handling
• Fun With Numbers
• Heavy and Light
• How Many
• Long And Short
• Mapping
• Measurement
• Money
• Multiplication and Factors
• Multiply and Divide
• Numbers
• Parts and Wholes
• Pattern Recognition
• Patterns
• Play With Patterns
• Rupees And Paise
• Shapes And Angles
• Shapes And Designs
• Shapes and Space
• Similarity
• Smart Charts
• Squares
• Subtraction
• Tables And Shares
• Tenths and Hundredths
• Time
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# Rectangle in ellipse
### Rectangle in ellipse
Nomi sent me the following as a "conversation". I prefer to answer it here so others can benefit by it
In an ellipse 4x²+9y²=144 inscribed is a rectangle whose vertices lies on the ellipse and whose sides are parallel with the ellipse axis. Longer side which is parallel to major axis, relates to the shorter sides as 3:2. Find area of rectangle.
This doesn't make sense! If the ellipse is inscribed in the in the rectangle then the vertices of the rectangle would be outside the ellipse. I assume you mean that the rectangle is inscribed in the ellipse (or, equivalently, that the ellipse circumscribes the rectangle).
If that is the case, and the ratio of longer side of the rectangle to shorter side is 3 to 2, we can take the shorter side to be "2a" and the length of the longer side to be "3a". The area of the rectangle is $$(2a)(3a)= 6a^2$$. We need to find "$$a^2$$".
The ellipse is $$4x^2+ 9y^2= 144$$. When y= 0, $$4x^2= 144$$ so $$x^2= 36$$, $$x= \pm 6$$ so two vertices are (-6, 0) and (6, 0). When x= 0, $$9y^2= 144$$ so $$y^2= 16$$. $$y= \pm 4$$ so the other two vertices are (0, -4) and (0, 4). Since the major axis is on the x-axis, we have the longer side of the inscribed rectangle parallel to the x-axis and the shorter side parallel to the y-axis. The vertex of the rectangle in the first quadrant will be $$(3a/2, a)$$ and must satisfy $$4(3a/2)^2+9a^2= 9a^2+ 9a^2= 18a^2= 144$$. $$a^2= \frac{144}{18}= 8$$.
The area of the rectangle is 6(8)= 48.
Guest
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## Intermediate Algebra (12th Edition)
$-2k+15$
The distributive property holds that $a(b+c)=ab+ac$ (for real numbers $a$, $b$, and $c$). We are given the expression $3(k+2)-5k+6+3$. We can use the distributive property to simplify the terms in parentheses and combine like terms. $3(k+2)-5k+6+3=(3\times k)+(3\times2)-5k+6+3=3k+6-5k+6+3=(3-5)k+(6+6+3)=(-2)k+(15)=-2k+15$
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CONSTRUCTION OF PERPENDICULAR FROM A POINT ON A LINE
About "Construction of perpendicular from a point on a line"
Construction of perpendicular from a point on a line :
Even though students know what is perpendicular line, many students do not know, how to construct perpendicular from a point on a line.
Here we are going to see "How to construct perpendicular from a point on a line step by step"
Construction of perpendicular from a point on a line
To construct a perpendicular from a point on a line, you must need the following instruments.
1. Ruler
2. Compass
The steps for the construction of a perpendicular from a point on a line are.
Step 1 :
Draw the line segment AB and mark external point C at somewhere above the line segment AB.
Step 2 :
With C as center and any convenient radius draw arcs to cut the given line at two points P and Q.
Step 3 :
With P and Q as centers and more than half the distance between these points as radius draw two arcs to intersect each other at E.
Step 4 :
Join C and E to get the required perpendicular line.
In the above figure, CE is the perpendicular to the line segment AB from the external point C.
This construction clearly shows how to draw the perpendicular to a given line segment from an external point with compass and straightedge or ruler.
Key Concept - Perpendicular
A line which intersects the another line at right angle is called as perpendicular.
Altitude of a triangle
Key concept In a triangle, an altitude is the line segment drawn from a vertex of the triangle perpendicular to its opposite side. Altitude
Orthocenter of a Triangle
Key concept The point of concurrency of the altitudes of a triangle is called the orthocenter of the triangle and is usually denoted by H. Orthocenter
Construction of orthocenter of a Triangle - Practice question
Construct triangle ABC whose sides are AB = 6 cm, BC = 4 cm and AC = 5.5 cm and locate its orthocenter.
Step 1 :
Draw the triangle ABC with the given measurements.
Step 2 :
Construct altitudes from any two vertices (A and C) to their opposite sides (BC and AB respectively).
The point of intersection of the altitudes H is the orthocenter of the given triangle ABC.
After having gone through the stuff given above, we hope that the students would have understood "Construction of perpendicular from a point on a line"
If you need any other stuff in math, please use our google custom search here.
WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
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## Section Exercises
1. Can a system of linear equations have exactly two solutions? Explain why or why not.
2. If you are performing a break-even analysis for a business and their cost and revenue equations are dependent, explain what this means for the company’s profit margins.
3. If you are solving a break-even analysis and get a negative break-even point, explain what this signifies for the company?
4. If you are solving a break-even analysis and there is no break-even point, explain what this means for the company. How should they ensure there is a break-even point?
5. Given a system of equations, explain at least two different methods of solving that system.
For the following exercises, determine whether the given ordered pair is a solution to the system of equations.
6. $\begin{array}{c}5x-y=4\\ x+6y=2\end{array}$ and $\left(4,0\right)$
7. $\begin{array}{l}-3x - 5y=13\hfill \\ -x+4y=10\hfill \end{array}$ and $\left(-6,1\right)$
8. $\begin{array}{c}3x+7y=1\\ 2x+4y=0\end{array}$ and $\left(2,3\right)$
9. $\begin{array}{l}-2x+5y=7\hfill \\ \text{ }2x+9y=7\hfill \end{array}$ and $\left(-1,1\right)$
10. $\begin{array}{c}x+8y=43\\ 3x - 2y=-1\end{array}$ and $\left(3,5\right)$
For the following exercises, solve each system by substitution.
11. $\begin{array}{l}\text{ }x+3y=5\hfill \\ 2x+3y=4\hfill \end{array}$
12. $\begin{array}{l}\text{ }3x - 2y=18\hfill \\ 5x+10y=-10\hfill \end{array}$
13. $\begin{array}{l}4x+2y=-10\\ 3x+9y=0\end{array}$
14. $\begin{array}{l}2x+4y=-3.8\\ 9x - 5y=1.3\end{array}$
15. $\begin{array}{l}\hfill \\ \begin{array}{l}\\ \begin{array}{l}-2x+3y=1.2\hfill \\ -3x - 6y=1.8\hfill \end{array}\end{array}\hfill \end{array}$
16. $\begin{array}{l}\text{ }x - 0.2y=1\hfill \\ -10x+2y=5\hfill \end{array}$
17. $\begin{array}{l}\text{ }3x+5y=9\hfill \\ 30x+50y=-90\hfill \end{array}$
18. $\begin{array}{l}\hfill \\ \begin{array}{l}\text{ }-3x+y=2\hfill \\ 12x - 4y=-8\hfill \end{array}\hfill \end{array}$
19. $\begin{array}{l}\frac{1}{2}x+\frac{1}{3}y=16\\ \frac{1}{6}x+\frac{1}{4}y=9\end{array}$
20. $\begin{array}{l}\\ \begin{array}{l}-\frac{1}{4}x+\frac{3}{2}y=11\hfill \\ -\frac{1}{8}x+\frac{1}{3}y=3\hfill \end{array}\end{array}$
For the following exercises, solve each system by addition.
21. $\begin{array}{l}\hfill \\ \begin{array}{l}-2x+5y=-42\hfill \\ \text{ }7x+2y=30\hfill \end{array}\hfill \end{array}$
22. $\begin{array}{l}6x - 5y=-34\\ 2x+6y=4\end{array}$
23. $\begin{array}{l}\text{ }5x-y=-2.6\hfill \\ -4x - 6y=1.4\hfill \end{array}$
24. $\begin{array}{l}7x - 2y=3\\ 4x+5y=3.25\end{array}$
25. $\begin{array}{l}\hfill \\ \begin{array}{l}\text{ }\mathrm{-x}+2y=-1\hfill \\ 5x - 10y=6\hfill \end{array}\hfill \end{array}$
26. $\begin{array}{l}\text{ }7x+6y=2\hfill \\ -28x - 24y=-8\hfill \end{array}$
27. $\begin{array}{l}\frac{5}{6}x+\frac{1}{4}y=0\\ \frac{1}{8}x-\frac{1}{2}y=-\frac{43}{120}\end{array}$
28. $\begin{array}{l}\text{ }\frac{1}{3}x+\frac{1}{9}y=\frac{2}{9}\hfill \\ -\frac{1}{2}x+\frac{4}{5}y=-\frac{1}{3}\hfill \end{array}$
29. $\begin{array}{l}\hfill \\ \begin{array}{l}-0.2x+0.4y=0.6\hfill \\ \text{ }x - 2y=-3\hfill \end{array}\hfill \end{array}$
30. $\begin{array}{l}\begin{array}{l}\\ -0.1x+0.2y=0.6\end{array}\hfill \\ \text{ }5x - 10y=1\hfill \end{array}$
For the following exercises, solve each system by any method.
31. $\begin{array}{l}5x+9y=16\hfill \\ \text{ }x+2y=4\hfill \end{array}$
32. $\begin{array}{l}6x - 8y=-0.6\\ 3x+2y=0.9\end{array}$
33. $\begin{array}{l}5x - 2y=2.25\\ 7x - 4y=3\end{array}$
34. $\begin{array}{l}\begin{array}{l}\hfill \\ \text{ }x-\frac{5}{12}y=-\frac{55}{12}\hfill \end{array}\hfill \\ -6x+\frac{5}{2}y=\frac{55}{2}\hfill \end{array}$
35. $\begin{array}{l}\\ \begin{array}{l}7x - 4y=\frac{7}{6}\hfill \\ 2x+4y=\frac{1}{3}\hfill \end{array}\end{array}$
36. $\begin{array}{l}3x+6y=11\\ 2x+4y=9\end{array}$
37. $\begin{array}{l}\text{ }\frac{7}{3}x-\frac{1}{6}y=2\hfill \\ -\frac{21}{6}x+\frac{3}{12}y=-3\hfill \end{array}$
38. $\begin{array}{l}\frac{1}{2}x+\frac{1}{3}y=\frac{1}{3}\\ \frac{3}{2}x+\frac{1}{4}y=-\frac{1}{8}\end{array}$
39. $\begin{array}{l}2.2x+1.3y=-0.1\\ 4.2x+4.2y=2.1\end{array}$
40. $\begin{array}{l}\text{ }0.1x+0.2y=2\hfill \\ 0.35x - 0.3y=0\hfill \end{array}$
For the following exercises, graph the system of equations and state whether the system is consistent, inconsistent, or dependent and whether the system has one solution, no solution, or infinite solutions.
41. $\begin{array}{l}3x-y=0.6\\ x - 2y=1.3\end{array}$
42. $\begin{array}{l}\begin{array}{l}\\ -x+2y=4\end{array}\hfill \\ \text{ }2x - 4y=1\hfill \end{array}$
43. $\begin{array}{l}\text{ }x+2y=7\hfill \\ 2x+6y=12\hfill \end{array}$
44. $\begin{array}{l}3x - 5y=7\hfill \\ \text{ }x - 2y=3\hfill \end{array}$
45. $\begin{array}{l}\text{ }3x - 2y=5\hfill \\ -9x+6y=-15\hfill \end{array}$
For the following exercises, use the intersect function on a graphing device to solve each system. Round all answers to the nearest hundredth.
46. $\begin{array}{l}\text{ }0.1x+0.2y=0.3\hfill \\ -0.3x+0.5y=1\hfill \end{array}$
47. $\begin{array}{l}\hfill \\ \begin{array}{l}-0.01x+0.12y=0.62\hfill \\ 0.15x+0.20y=0.52\hfill \end{array}\hfill \end{array}$
48. $\begin{array}{l}0.5x+0.3y=4\hfill \\ 0.25x - 0.9y=0.46\hfill \end{array}$
49. $\begin{array}{l}0.15x+0.27y=0.39\hfill \\ -0.34x+0.56y=1.8\hfill \end{array}$
50. $\begin{array}{l}\begin{array}{l}\\ -0.71x+0.92y=0.13\end{array}\hfill \\ 0.83x+0.05y=2.1\hfill \end{array}$
For the following exercises, solve each system in terms of $A,B,C,D,E,\text{}$ and $F$ where $A-F$ are nonzero numbers. Note that $A\ne B$ and $AE\ne BD$.
51. $\begin{array}{l}x+y=A\\ x-y=B\end{array}$
52. $\begin{array}{l}x+Ay=1\\ x+By=1\end{array}$
53. $\begin{array}{l}Ax+y=0\\ Bx+y=1\end{array}$
54. $\begin{array}{l}Ax+By=C\\ x+y=1\end{array}$
55. $\begin{array}{l}Ax+By=C\\ Dx+Ey=F\end{array}$
For the following exercises, solve for the desired quantity.
56. A stuffed animal business has a total cost of production $C=12x+30$ and a revenue function $R=20x$. Find the break-even point.
57. A fast-food restaurant has a cost of production $C\left(x\right)=11x+120$ and a revenue function $R\left(x\right)=5x$. When does the company start to turn a profit?
58. A cell phone factory has a cost of production $C\left(x\right)=150x+10,000$ and a revenue function $R\left(x\right)=200x$. What is the break-even point?
59. A musician charges $C\left(x\right)=64x+20,000,\text{}$ where $x$ is the total number of attendees at the concert. The venue charges $80 per ticket. After how many people buy tickets does the venue break even, and what is the value of the total tickets sold at that point? 60. A guitar factory has a cost of production $C\left(x\right)=75x+50,000$. If the company needs to break even after 150 units sold, at what price should they sell each guitar? Round up to the nearest dollar, and write the revenue function. For the following exercises, use a system of linear equations with two variables and two equations to solve. 61. Find two numbers whose sum is 28 and difference is 13. 62. A number is 9 more than another number. Twice the sum of the two numbers is 10. Find the two numbers. 63. The startup cost for a restaurant is$120,000, and each meal costs $10 for the restaurant to make. If each meal is then sold for$15, after how many meals does the restaurant break even?
64. A moving company charges a flat rate of $150, and an additional$5 for each box. If a taxi service would charge $20 for each box, how many boxes would you need for it to be cheaper to use the moving company, and what would be the total cost? 65. A total of 1,595 first- and second-year college students gathered at a pep rally. The number of freshmen exceeded the number of sophomores by 15. How many freshmen and sophomores were in attendance? 66. 276 students enrolled in a freshman-level chemistry class. By the end of the semester, 5 times the number of students passed as failed. Find the number of students who passed, and the number of students who failed. 67. There were 130 faculty at a conference. If there were 18 more women than men attending, how many of each gender attended the conference? 68. A jeep and BMW enter a highway running east-west at the same exit heading in opposite directions. The jeep entered the highway 30 minutes before the BMW did, and traveled 7 mph slower than the BMW. After 2 hours from the time the BMW entered the highway, the cars were 306.5 miles apart. Find the speed of each car, assuming they were driven on cruise control. 69. If a scientist mixed 10% saline solution with 60% saline solution to get 25 gallons of 40% saline solution, how many gallons of 10% and 60% solutions were mixed? 70. An investor earned triple the profits of what she earned last year. If she made$500,000.48 total for both years, how much did she earn in profits each year?
71. An investor who dabbles in real estate invested 1.1 million dollars into two land investments. On the first investment, Swan Peak, her return was a 110% increase on the money she invested. On the second investment, Riverside Community, she earned 50% over what she invested. If she earned $1 million in profits, how much did she invest in each of the land deals? 72. If an investor invests a total of$25,000 into two bonds, one that pays 3% simple interest, and the other that pays $2\frac{7}{8}\text{%}$ interest, and the investor earns $737.50 annual interest, how much was invested in each account? 73. If an investor invests$23,000 into two bonds, one that pays 4% in simple interest, and the other paying 2% simple interest, and the investor earns $710.00 annual interest, how much was invested in each account? 74. CDs cost$5.96 more than DVDs at All Bets Are Off Electronics. How much would 6 CDs and 2 DVDs cost if 5 CDs and 2 DVDs cost $127.73? 75. A store clerk sold 60 pairs of sneakers. The high-tops sold for$98.99 and the low-tops sold for $129.99. If the receipts for the two types of sales totaled$6,404.40, how many of each type of sneaker were sold?
76. A concert manager counted 350 ticket receipts the day after a concert. The price for a student ticket was $12.50, and the price for an adult ticket was$16.00. The register confirms that $5,075 was taken in. How many student tickets and adult tickets were sold? 77. Admission into an amusement park for 4 children and 2 adults is$116.90. For 6 children and 3 adults, the admission is \$175.35. Assuming a different price for children and adults, what is the price of the child’s ticket and the price of the adult ticket?
|
# AP Board 4th Class Maths Solutions 10th Lesson Measurements
Andhra Pradesh AP Board 4th Class Maths Solutions 10th Lesson Measurements Textbook Exercise Questions and Answers.
## AP State Syllabus 4th Class Maths Solutions Chapter 10 Measurements
Activity
Make the students into 4 groups. Provide 1m thread to each group. Ask each group to cut the given thread into possible equal parts as given in the table and complete the table.
Solution:
Exercise – 1
1. Convert the following into centimetres:
a) 18m = 18 × __________ = __________ cm
Solution:
18m = 18 × 100 1800 cm
b) 100 m = 100 × __________ = ____________
Solution:
100m = 100 × 100 = 10,000 cm
c) 17 m 25 cm = 17 × __________ cm + 25 cm – __________ cm + __________ cm = __________ cm
Solution:
17 m 25 cm = 17 × 100 cm + 25 cm – 1700 cm + 25 cm = 1,725 cm
d) 45 m 75 cm =45 × __________ cm + __________ cm = __________ cm + __________ cm = __________ cm
Solution:
45 m 75 cm =45 × 100 cm + 75 cm = 4,500 cm + 75 cm = 4,575 cm
2. Convert centimetres into metres and centimetres.
a) 269cm = ___________
Solution:
269 cm = 2.69 m {∵ $$\frac{269}{100}$$}
b) 693cm = ________
Solution:
693 cm = $$\frac{693}{100}$$ = 6.93 m
c) 703 cm = ___________
Solution:
703 cm = $$\frac{703}{100}$$ = 7.03 m
d) 400 cm = ____________
Solution:
400 cm = $$\frac{400}{100}$$ = 4 m
Question 3.
Match the following:
Solution:
Question 4.
Fill in the blanks with suitible symbol <, > or =
a) 4 m 90 cm ___________ 480 cm
b) 67 m ___________ 6800 cm
c) 75 m ___________ 7500 cm
d) 80 m ___________ 9000 cm
Solution:
a) 4 m 90 cm > 480 cm
b) 67 m < 6800 cm
c) 75 m = 7500 cm
d) 80 m < 9000 cm
Textbook Page No. 136
Think and Discuss
Can you cut a rope of length 1m into 4 equal pieces without measuring it. How?
Solution:
Yes, it is possible. The rope is one meter long. If we divide into 4 equal parts, each part is 1/4 of the length of the rope by folding method.
Exercise – 10.1
1. Do the following.
a)
Solution:
b)
Solution:
c)
Solution:
d)
Solution:
2. Do the following.
a) 10 m 55 cm + 65 m 65 cm
Solution:
b) 98 m 50 cm + 115 m 45 cm
Solution:
c) 684 m + 225 m 80 cm
Solution:
d) 60 m 45 cm + 85 m 28 cm
Solution:
3. Do the following.
a)
Solution:
b)
Solution:
c)
Solution:
d)
Solution:
4. Do the following.
a) 75 m 85 cm – 10 m 30 cm
Solution:
b) 69 m 50 cm – 25 m 60 cm
Solution:
c) 95 m – 88 m 40cm
Solution:
Question 5.
Radha’s elder son needs in lm 80 cm of cloth whereas younger son needs 1 m 60 cm of cloth for their shirts. How much length of cloth is needed for them ?
Solution:
Length of cloth required to Radha’s elder son = 1 m 80 cm
Length of cloth required to Radha’s younger son = 1 m 60 cm
Length of cloth required to both of them = 3 m 40 cm
Question 6.
The length of a flower bed is 20 m 50 cm. Radha dug a length of 8 cm 50 cm. Then still how much length of the flower bed is to be dug ?
Solution:
Actual length of flower bed = 20 m 50 cm
Radha dug length of a flower bed = 8 cm 50 cm
∴ Required length of a flower bed = 12 m
Question 7.
Balamma is working in a handloom industry. She spins yarn of length 720 m 50 cm and 850 m 30 cm on successive days, then how much yarn has she spun in both the days ?
Solution:
Balamma spins yam of length on first day = 720 m 50 cm
Balamma spins yam of length on second day = 850 m 30 cm
Balamma spins yam of length on both the days = 1570 m 80 cm
Question 8.
If tw0 ropes of length 10 m 50 cm and 9 m 60 cm are tied together. Then the length of the rope so formed is ? (Approximately)
Solution:
Length of first rope = 10 m 50 cm
Length of second rope = 9 m 60 cm
Length of rope so formed totally = 20 m 10 cm
Question 9.
Narayana can make 60 m thread in one day. If he completes 36 m 50 cm by the afternoon, how much length of thread is to be made by him ?
Solution:
Length of thread made in one day by Narayana = 60 m
Length is to be made by him = 23 m 50 cm
Textbook Page No. 140
Do this
1. Convert the following into grams.
a) 5 kg = 5 × 1000 grams = __________ grams
Solution:
5 kg = 5 × 1000 grams = 5,000 grams
b) 15 kg = 15 × __________ grams = __________ grams
Solution:
15 kg = 15 × 100 grams = 15,000 grams
c) 7 kg 250 g = __________ × __________ g + __________ g = __________ g + __________ g = __________ g
Solution:
7 kg 250 g = 7 × 1,000 g + 250 g = 7,000 g + 250 g = 7,250 g
d) 35 kg 500 g = __________ × __________ g + __________ g = __________ g + __________ g = __________ g
Solution:
35 kg 500 g = 55 × 1,000 g + 500 g = 55,000 g + 500 g = 55,000 g
2. Convert into kilograms.
a) 2680 grams
Solution:
2680 grams = 2000 g + 680 g
= 2 × 1000 g + 680 g
= 2 × 1 kg + 680 g
= 2 kg + 680 g
= 2 kg 680 g
b) 7455 grams
Solution:
7455 grams = 7000g + 455 g
= 7 × 1000 g + 455 g
= 7 × 1 kg + 455 g
= 7 kg + 455 g
= 7 kg 455 g
c) 4,000 grams
Solution:
4,000 grams = 4 × 1000 grams
= 4 kg
d) 8050 grams
Solution:
8050 grams = 8000g + 50 g
= 8 × 1000 g + 50 g
= 8 kg + 50 g
= 8 kg. 50g
Question 3.
Suggest the suitable symbol in the provided space (<, > or =)
a) 4 kg 900g ________ 4800 g
b) 67 kg ________ 68000 g
c) 75 kg ________ 75000 g
d) 80 kg ________ 9000g
Solution:
a) 4 kg 900g > 4800 g
b) 67 kg < 68000 g
c) 75 kg = 75000 g
d) 80 kg > 9000 g
Exercise – 10.3
a)
Solution:
b)
Solution:
c)
Solution:
d)
Solution:
2. Do the following.
a) 2 kg 250 g + 12 kg 580 g
Solution:
b 500 kg 750 g + 250 kg 800 g
Solution:
c) 580 kg 500 g + 400 kg 680 g
Solution:
3. Subtract
a)
Solution:
b)
Solution:
c)
Solution:
d)
Solution:
4. Do the following.
a) 5 kg 450 g – 3 kg 500 g
Solution:
b) 50 kg 280 g – 12 kg 450 g
Solution:
c) 100 kg 150 g – 85 kg 280g
Solution:
d) 85 kg 250 g – 40 kg 500g
Solution:
Question 5.
A school has a stock of 25 kg 600 g rice. 198 kg 300 g of rice was supplied for the midday meal scheme. Now how many kilograms of rice is available in the school ?
Solution:
Stock of rice a school has = 250 kg 600 g
Supplied stock to mid day meals was = 198 kg 300 g
Now available rice in the school = 52 kg 300 g
Question 6.
To make a sweet the maker added 10 kg 600g of jaggery, 20 kg 350 g of atta and 500g of ghee. What is the total weight of these three ?
Solution:
Weight of jaggery used = 10 kg 600g
Weight of atta used = 20 kg 350 g
Weight of ghee used = 500g
Total weight at these three = 31 kg 450 g
Question 7.
Rangayya bought 1 kg 500g of potatoes, 750 g of carrot, 500g of tomatoes and 2 kg of onions. What is the total weight of vegetables he bought ?
Solution:
Weight of Potatoes = 1 kg 500g
Weight of Carrot = 750 g
Weight of Tomatoes = 500g
Weight of Onions = 2 kg
Total weight of vegetables = 4 kg 750 g
Question 8.
There s 100 kg of sugar in a shop. If the shop keeper sells out 78 kg, find the sugar left in the shop.
Solution:
Quality of sugar in the shop = 100 kgs
Quality of sugar sold = 78 kgs
Quality of sugar left in the shop = 22 kgs
Question 9.
Chinnayya collected 108 kg 800 g of tamarind from one tree and 120 kg tamarind from another tree. If he sells 150 kg from that, how many kilos of tamarind did he have now ?
Solution:
Quantity of tamarind collected from first tree = 108 kg 800 g
Quantity of tamarind collected from second tree = 120 kg
Total collected tamarind = 228 kg 800 g
Quantity of tamarind sold = 150 kg
∴ Remaining tamarind = 78 kg 800 g
Question 10.
Rajani melted 25g amd 28 g weights of gold ear rings. She got 49g of gold. How much weight of gold is lost in melting ?
Solution:
Sum of weights of ear rings = 25g + 28g = 53 g
After melting weight of gold = 49 g
Question 11.
The vendor collected 76 kg butter and mixed it with 12 kg 500 g of butter that is already in his shop. If he sells 82 kg of butter bow much will be left with him ?
Solution:
Quantity of butter in the shop = 12 kg 500 g
Quantity of collected butter = 76 kg
Total butter at vendor =
Vendor sold butter quantity = 82 kg
Quantity of butter left him = 6 kg 500 g
Question 12.
The weight of a new born baby was 2 kg 800g. After 2 years her weight is 8 kg 300 g. Calculate the weight gained.
Solution:
The weight of a new bom baby = 2 kg 800g
After 2 years weight of a new born baby = 8 kg 300 g
∴ Gained weight at baby in 2 yrs = 5 kg 500 g
Question 13.
40 kg 500 g of putharekulu were prepared in a sweet shop on first day and 45 kg 800g of putharekulu on the second day. If he sells 65 kg putharekulu then how much will be left with him ?
Solution:
Prepared weight of putarekulu in the first day = 40 kg 500 g
Prepared weight of putarekulu in the second day = 45 kg 800g
Total weight of putarekulu = 86 kg 300 g
Putarekulu sold = 65 kg
Putarekulu left at him = 21 kg 300 g
Textbook Page No. 145
Do this
1. Convert litres into millilitres.
a) 5 litres = 5 × 1000 ml = ________ ml
Solution:
5 litres = 5 × 1000 ml = 5000 ml
b) 18 litres = 18 × ________ ml = ________ ml
Solution:
18 litres = 18 × 1000 ml = 18,000 ml
c) 37 litres = ________ × ________ ml = ________ ml
Solution:
37 litres = 37 × 1000 ml = 37,000 ml
d) 86 litres = ________ × ________ ml = ________ ml
Solution:
86 litres = 86 × 1000 ml = 86,000 ml
e) 100 litres = ________ × ________ ml = ________ ml
Solution:
100 litres = 100 × 1000 ml = 1,00,000 ml
2. Convert millilitres into litres and millilitres.
a) 8250 ml
Solution:
8 × 1000ml + 250 ml
= 8 l 250ml
b) 7000 ml
Solution:
7 × 1000 ml
= 7 × 1 l
= 7 l
c) 5500 ml
Solution:
5 × 1000 ml + 500 ml
= 5 × 1 l + 500 ml
= 5 l 500 ml
d) 4850 ml
Solution:
4 × 1000m l + 850 ml
= 4 × 1 l + 850 ml
= 4 l 850 ml
e) 10550 ml
Solution:
10 × 1000ml + 550 ml
= 10 × 1l + 550 ml
= 10 l 550 ml
Question 3.
Place the correct symbol in the space provided (>, < or =)
a) 6 l _________ 5550 ml
b) 8 l _________ 8000 ml
c) 5 l _________ 6000 ml
d) 3 l _________ 3500 ml
e) 9 l _________ 980 ml
Solution:
a) 6 l > 5550 ml
b) 8 l = 8000 ml
c) 5 l < 6000 ml
d) 3 l < 3500 ml
e) 9 l > 980 ml
Exercise – 10.4
a)
Solution:
b)
Solution:
c)
Solution:
d)
Solution:
e) 18 l 450 ml + 28 l 890 ml
Solution:
f) 50 l 850 ml + 70 l 450 ml
Solution:
g) 89 l 950 ml + 45 l 650 ml
Solution:
2. Do the following subtractions.
a)
Solution:
b)
Solution:
c)
Solution:
d)
Solution:
e) 15 l 350 ml – 10 l 800 ml
Solution:
f) 70 l 850 ml – 25 l 900 ml
Solution:
g) 99 l 350 ml – 16 l 600 ml
Solution:
h) 25 l – 18 l 250 ml
Solution:
Question 3.
A buffalo gives 3 l 250 ml milk in the morning and 2 l 750 ml in the evening. Find the total milk given by it in one day.
Solution:
Buffaloes gives milk in the morning = 3 l 250 ml
Buffaloes gives milk in the evening = 2 l 750 ml
Buffaloes gives milk total one day = 6 l
Question 4.
Rajesh purchased 1 l 500 ml and 2 l cool drink bottles. What is the total capacity of cool drink he purchased ?
Solution:
Capacity of first cool drink bottle = 1 l 500 ml
Capacity of second cool drink bottle = 2 l
Capacity of total cool drink bottles = 3 l 500 ml
Question 5.
A liquid blue bottle contains 250 ml. Rajani used 100 ml in it. Find the liquid left in the bottle.
Solution:
Quantity of liquid blue bottle = 250 ml
Quantity of liquid used in bottle = 100ml
Quantity of liquid left in the bottle = 150ml
Question 6.
Sreenu used some kerosene fro 200 l drum. If there is still 18 l 750 ml of the kerosene in the drum, how much kerosene was used by him ?
Solution:
Actual quantity of brasene in the drum = 200 l
After using kerosene left in the drum = 18 l 750 ml
Quantity of kerosene used by Sreenum = 181 l 250 ml
Textbook Page No. 147
Skipping competition:
The following table shows number of skips / jumps made by the students. Observe the table and answer the questions given below.
Question 1.
How much time did Vani take for 50 skips ?
Vani took for 50 skips 1 minute.
Question 2.
How much time did Geetha take for 50 skips ?
Geetha took for 50 skips 2 minute.
Question 3.
Who took more time for 50 skips ? By how much ?
Reeta took 2 minutes more time, than Vani.
Question 4.
Who took least time for 50 skips ? By how much ?
Vasantha took least time for 50 skips less than 1 minute.
Textbook Page No. 150
Do this
Observe second’s hand in between the two numbers then we note the time as given below.
Time : hours ___________ ; minutes __________ ; second ___________
Solution:
hours : 7
Minutes : 5 × 10 + 3 = 53
Seconds : 3 × 5 + 4 = 19
Convert the minutes into seconds. One is done for you.
Solution:
Textbook Page No. 151
Do this
Solution:
Textbook Page No. 152
Do this
Convert 1 hour 10 minutes 12 seconds into seconds.
Solution:
1 hour = 1 × 60 = 60 minutes
= 60 × 60 seconds
= 3600 seconds
10 minutes = 10 × 60 seconds
= 600 seconds
1 hour 10 minutes 125 seconds
= 3600 + 600 + 12
= 4,212 seconds
Try this
Find the number of seconds in
a) A day
Solution:
A day = 24 hours
= 24 × 1 hours
= 24 × 60 minutes
= 24 × 60 × 60 seconds
= 86,400 seconds
b) A week
Solution:
A week = = 7days
= 7 × 1 day
= 7 × 24 hours
= 7 × 24 × 1 hour
= 7 × 24 × 60 minutes
= 7 × 24 × 60 × 60 seconds
= 604, 800 seconds
c) A month
Solution:
A month = 30 days
= 30 × 1 day
= 30 × 24 hours
= 30 × 24 × 1 hour
= 30 × 24 × 60 minutes
= 30 × 24 × 60 × 60
= 2592000 seconds
Calender :
Write your date and month of birth in the given table. Observe the day corresponding to your date of birth in present calender and write the day in the table. Also writes dates of birth of your friends. One is done for you.
Solution:
Exercise – 10.5
Question 1.
Observe present year’s calendar in your class and All in the blanks in the following.
Solution:
Question 2.
Arrange the above mentioned festivals in the order as we see in the calendar.
a. __________
b. __________
c. __________
d. __________
e. __________
f. __________
g. __________
h. __________
Solution:
a) Bhogi
b) Ramzan
d) Teacher’s day
e) Gandhi Jayanthi
g) Diwali, Children’s day
3. From the above mentioned festivals.
a) Which festival comes at the beginning of the year ?
Solution:
Bhoghi comes at the beginning of the year.
b) Which festival comes at the end of the year ?
Solution:
Christmas and Mukkoti Yekadashi comes at the end of the year.
Question 4.
Suman’s age is 9 years and his father’s age is 25 years more than Suman’s age. Then how old is his father ?
Solution:
Suman’s age is = 9 years
Age of his father = Suman’s age + 25 years
= 9 + 25
= 34 years
Question 5.
Anand’s age is 10 years and the age of his brother is less than 5 years of Anand. Find the age of his brother ?
Solution:
Anand’s age is = 10 years
Age of his brother = Anand’s age – 5 years
= 10 – 5
= 5 years
Question 6.
Rajitha is 9 years old. Her sister’s age is twice that of Rajitha’s. Then how old is Rajitha’s sister ?
Solution:
Rajitha’s age is = 9 years
Age of his sister = 2 × Rajitha’s age
= 2 × 9
= 18 years
Question 7.
Write the short form for the given date. One is done for you.
a) 3rd July 1975 = 03-07-1975
b) 16th August 1945 = ____________
c) 22nd March 1980 = ____________
Solution:
b) 16-08-1945
c) 22-03-1980
Coins with Rajani :
Notes with Rajani :
Solution:
Textbook Page No. 158
Do this
Write the other ways for making 900 rupees with different denominations of currency.
a) ₹ 900 = ____________________
Solution:
₹ 100 + ₹ 100 + ₹ 100 + ₹ 100 + ₹ 500
b) ₹ 900 = ____________________
Solution:
₹ 200 + ₹ 200 + ₹ 200 + ₹ 200 + ₹ 100
Try this
Use notes of Indian currency to make ₹ 900
900 = _______ + _________ + ________ + ________ + _________ + _________ + _________ + 5 + 2 + 2 + 1.
Solution:
900 = ₹ 500 + ₹ 200 ₹ 100 ₹ 50 ₹ 20 ₹ 10 ₹ 10 + 5 + 2 + 2 + 1
Textbook Page No. 159
Do this
Paste the dummy currency notes in your note book whose total will be 2000 as in the box shown below.
Solution:
Try this
Use the Indian Currency notes and coins and write denominations for ₹ 2000 in the blank.
₹ 2000 = ₹ 500 + _____________
Solution:
₹ 2000 = ₹ 500 + ₹ 500 + ₹ 500 + ₹ 200 + ₹ 200 + ₹ 100
Textbook Page No. 160
Do this
Rajesh wrote denominations for ₹ 2000 in the above form. Now try to fill the below deposit form in another way for depositing ₹ 2000 in the bank.
Solution:
Textbook Page No. 161
Do this
Question 1.
Govind spent 2585 on seeds and 4850 on fertilizers for growing a crop in his field. How much did he spend altogether?
Solution:
Govind spend amount on seed = ₹ 2585
Govind spent amount on fertilizers = ₹ 4850
Total amount spent altogether = ₹ 7435
Question 2.
Appala Naidu had bought a goat for 8950 and sold it for 9850. How much profit did he get?
Solution:
Selling price of goat = ₹ 9850
Cost price of goat = ₹ 8950
∴ Profit gained = ₹ 900
Question 3.
Ratnalu borrowed ₹ 9000 from Gopal. The interest amount for the money became ₹ 1850. But Ratnalu has only ₹ 4965 with her. How much more money is needed to clear the debt ?
Solution:
Ratnalu borrowed amount = ₹ 9000
Interest amount become = ₹ 1850
Total amount = ₹ 10,850
Amount at Ratnalu = ₹ 4,965
Needed money to clear debt = 5885
Exercise – 10.6
1. Fill in the blanks with suitable denominations for ₹ 2000.
a) ₹ 2000 = __________ + ₹ 500 + ₹ 500 + ₹ 200 + ₹ 200 + ₹ 100
Solution:
₹ 2000 = ₹ 500 + ₹ 500 + ₹ 500 + ₹ 200 + ₹ 200 + ₹ 100
b) ₹ 2000 = ₹ __________ + ₹ __________ + ₹ 500 + ₹ 500 + ₹ 500 + ₹ 200
Solution:
₹ 2000 = ₹ 200 + ₹ 100 + ₹ 500 + ₹ 500 + ₹ 500 + ₹ 200
c) ₹ 2000 = ₹ 500 + ₹ 500 + ₹ 500 + ₹ __________ + ₹ __________ + ₹ 100 + ₹ __________ + ₹ 50
Solution:
₹ 2000 = ₹ 500 + ₹ 500 + ₹ 500 + ₹ 200 + ₹ 100 + ₹ 100 + ₹ 50 + ₹ 50
d) ₹ 2000 = ₹ 200 + ₹ 200 + ₹ 500 + ₹ 100 + ₹ 100 + ₹ __________ + ₹ __________ + ₹ __________ + ₹ 100
Solution:
₹ 2000 = ₹ 200 + ₹ 200 + ₹ 500 + ₹ 100 + ₹ 100 + ₹ 500 + ₹ 200 + ₹ 100 + ₹ 100
2. Count the money.
a)
Solution:
₹ 4210
b)
Solution:
₹ 6500
Question 3.
A fish vendor Komali, bought fishes for ₹ 5620 and sold it for ₹ 4985. How much loss did she get?
Solution:
Cost price of fishes = ₹ 5620
Selling price of fishes = ₹ 4985
Loss gained by Komali = ₹ 635
Question 4.
Sailaja has ₹ 6450 and her mother has ₹ 2530. What is total amount they have ? If they spent ₹ 5645 then how much money will be with them ?
Solution:
Amount at Sailaja = ₹ 6450
Amount at Sailaja’s mother = ₹ 2530
Total amount at them = ₹ 8980
Spent amount at them = ₹ 5645
Amount at them = ₹ 3335
Multiple Choice Questions
Question 1.
1 meter = ___________ cm
A) 50
B) 100
C) 25
D) 50
B) 100
Question 2.
Half a meter is equal to in cm.
A) 50
B) 100
C) 40
D) 60
A) 50
Question 3.
How many 10 cm strips make a meter ?
A) 10
B) 100
C) 40
D) 50
A) 10
Question 4.
Convert form of 754 cm into meter
A) 700 m
B) 540 m
C) 7 m 54 cm
D) 75 m/cm
C) 7 m 54 cm
Question 5.
Is centimeter is greater than meter?
A) Yes
B) No
C) both A and B
D)None
B) No
Question 6.
Difference of 20m 48 cm to 18 m 23 cm is
A) 12.3
B) 2.25
C) 4.69
D) 2.20
B) 2.25
Question 7.
A) 500 gm
B) 1000 gm
C) 250 gm
D) 750gm
B) 1000 gm
Question 8.
In half kg how many 100 gms are there ?
A) 3
B) 5
C) 100
D) 10
B) 5
Question 9.
2 kg = 500 g + 500 g + 500 + ________ g + _______ g + _________ g
A) 200, 200, 50
B) 100, 200, 50
C) 100, 200, 200
D) 200, 200, 200
C) 100, 200, 200
Question 10.
Among 61 and 5500 mi which is lesser ?
A) 61
B) 5500 ml
C) both A and B
D) None
B) 5500 ml
Question 11.
In 9 liters __________ ml are there
A) 90
B) 100
C) 9999
D) 9000
D) 9000
Question 12.
If minutes hand crosses one division the seconds hand crosses __________
A) 30
B) 40
C) 150
D) 60
D) 60
Question 13.
How many divisions did seconds hand crossed in the given picture ?
A) 3
B) 5
C) 10
D) 7
C) 10
Question 14.
1 hour = ___________ seconds.
A) 60
B) 3600
C) 180
D) 6
B) 3600
Question 15.
2 hours = ____________ minutes.
A) 60
B) 120
C) 180
D) 240
B) 120
Question 16.
A day has ___________ seconds.
A) 86640
B) 80,000
C) 86,000
D) 86,400
D) 86,400
Question 17.
Short form of 20th November 1980
A) 20-Nov-1980
B) 20-11-1980
C) 20-12-1980
D) 20-10-1980
B) 20-11-1980
Question 18.
The money used in a particular country is called its
A) Money
B) Rate
C) Currency
D) All
C) Currency
Question 19.
Indian currency is
A) Rupee
B) Paise
C) both A and B
D) None
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# Richard Gere And The Stars (11/10/2019)
How will Richard Gere perform on 11/10/2019 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is of questionable accuracy – do not take this too seriously. I will first find the destiny number for Richard Gere, and then something similar to the life path number, which we will calculate for today (11/10/2019). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology experts.
PATH NUMBER FOR 11/10/2019: We will consider the month (11), the day (10) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. How? Let’s walk through it. First, for the month, we take the current month of 11 and add the digits together: 1 + 1 = 2 (super simple). Then do the day: from 10 we do 1 + 0 = 1. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 2 + 1 + 12 = 15. This still isn’t a single-digit number, so we will add its digits together again: 1 + 5 = 6. Now we have a single-digit number: 6 is the path number for 11/10/2019.
DESTINY NUMBER FOR Richard Gere: The destiny number will calculate the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Richard Gere we have the letters R (9), i (9), c (3), h (8), a (1), r (9), d (4), G (7), e (5), r (9) and e (5). Adding all of that up (yes, this can get tiring) gives 69. This still isn’t a single-digit number, so we will add its digits together again: 6 + 9 = 15. This still isn’t a single-digit number, so we will add its digits together again: 1 + 5 = 6. Now we have a single-digit number: 6 is the destiny number for Richard Gere.
CONCLUSION: The difference between the path number for today (6) and destiny number for Richard Gere (6) is 0. That is lower than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too excited yet! As mentioned earlier, this is not at all guaranteed. If you want to see something that we really strongly recommend, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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Key Concepts
Mathematics
Probability
Fractions
Percentages
Introduction
Have you ever heard anyone say the chance of something happening is “50–50”? What does that actually mean? This phrase has something to do with probability. Probability tells you how likely it is that an event will occur. This means that for certain events you can actually calculate how likely it is that they will happen. In this activity, you will do these calculations and then test them to see whether they hold true for reality!
Background
Probability allows us to quantify the likelihood an event will occur. You might be familiar with words we use to talk about probability, such as “certain,” “likely,” “unlikely,” “impossible,” and so on. You probably also know that the probability of an event happening spans from impossible, which means that this event will not happen under any circumstance, to certainty, which means that an event will happen without a doubt. In mathematics, these extreme probabilities are expressed as 0 (impossible) and 1 (certain). This means a probability number is always a number from 0 to 1. Probability can also be written as a percentage, which is a number from 0 to 100 percent. The higher the probability number or percentage of an event, the more likely is it that the event will occur.
The probability of a certain event occurring depends on how many possible outcomes the event has. If an event has only one possible outcome, the probability for this outcome is always 1 (or 100 percent). If there is more than one possible outcome, however, this changes. A simple example is the coin toss. If you toss a coin, there are two possible outcomes (heads or tails). As long as the coin was not manipulated, the theoretical probabilities of both outcomes are the same–they are equally probable. The sum of all possible outcomes is always 1 (or 100 percent) because it is certain that one of the possible outcomes will happen. This means that for the coin toss, the theoretical probability of either heads or tails is 0.5 (or 50 percent).
It gets more complicated with a six-sided die. In this case if you roll the die, there are 6 possible outcomes (1, 2, 3, 4, 5 or 6). Can you figure out what the theoretical probability for each number is? It is 1/6 or 0.17 (or 17 percent). In this activity, you will put your probability calculations to the test. The interesting part about probabilities is that knowing the theoretical likelihood of a certain outcome doesn’t necessarily tell you anything about the experimental probabilities when you actually try it out (except when the probability is 0 or 1). For example, outcomes with very low theoretical probabilities do actually occur in reality, although they are very unlikely. So how do your theoretical probabilities match your experimental results? You will find out by tossing a coin and rolling a die in this activity.
Materials
• Coin
• Six-sided die
• Paper
• Pen or pencil
Preparation
• Prepare a tally sheet to count how many times the coin has landed on heads or tails.
• Prepare a second tally sheet to count how often you have rolled each number with the die.
Procedure
• Calculate the theoretical probability for a coin to land on heads or tails, respectively. Write the probabilities in fraction form. What is the theoretical probability for each side?
• Now get ready to toss your coin. Out of the 10 tosses, how often do you expect to get heads or tails?
• Toss the coin 10 times. After each toss, record if you got heads or tails in your tally sheet.
• Count how often you got heads and how often you got tails. Write your results in fraction form. For example, 3 tails out of 10 tosses would be 3/10 or 0.3. (The denominator will always be the number of times you toss the coin, and the numerator will be the outcome you are measuring, such as the number of times the coin lands on tails.) You could also express the same results looking at heads landings for the same 10 tosses. So that would be 7 heads out of 10 tosses: 7/10 or 0.7. Do your results match your expectations?
• Do another 10 coin tosses. Do you expect the same results? Why or why not?
• Compare your results from the second round with the ones from the first round. Are they the same? Why or why not?
• Continue tossing the coin. This time toss it 30 times in a row. Record your results for each toss in your tally sheet. What results do you expect this time?
• Look at your results from the 30 coin tosses and convert them into fraction form. How are they different from your previous results for the 10 coin tosses?
• Count how many heads and tails you got for your total coin tosses so far, which should be 50. Again, write your results in fraction form (with the number of tosses as the denominator (50) and the result you are tallying as the numerator). Does your experimental probability match your theoretical probability from the first step? (An easy way to convert this fraction into a percentage is to multiply the denominator and the numerator each by 2, so 50 x 2 = 100. And after you multiply your numerator by 2, you will have a number that is out of 100—and a percentage.)
• Calculate the theoretical probability for rolling each number on a six-sided die. Write the probabilities in fraction form. What is the theoretical probability for each number?
• Take the dice and roll it 10 times. After each roll, record which number you got in your tally sheet. Out of the 10 rolls, how often do you expect to get each number?
• After 10 rolls, compare your results (written in fraction form) with your predictions. How close are they?
• Do another 10 rolls with the dice, recording the result of each roll. Do your results change?
• Now roll the dice 30 times in a row (recording the result after each roll). How often did you roll each number this time?
• Count how often you rolled each number in all combined 50 rolls. Write your results in fraction form. Does your experimental probability match your theoretical probability? (Use the same formula you used for the coin toss, multiplying the denominator and the numerator each by 2 to get the percentage.)
• Compare your calculated probability numbers with your actual data for both activities (coin and dice). What do your combined results tell you about probability?
• Extra: Increase the number of coin tosses and dice rolls even further. How do your results compare with the calculated probabilities with increasing number of events (tosses or rolls)?
• Extra: Look up how probabilities can be represented by probability trees. Can you draw a probability tree for the coin toss and dice roll?
• Extra: If you are interested in more advanced probability calculations, find out how you can calculate the probability of a recurring event, for example: How likely it is that you would get two heads in a row when tossing a coin?
Observations and Results
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# How do you find the slope and y intercept of y = x - 3?
Jun 24, 2018
Let's look at our equation. It written in the form $y = m x + b$, where $m$ is the slope and $b$ is the y-intercept. The slope, therefore, is $1$ and the y-intercept is $- 3$.
Jun 24, 2018
$\text{slope "=1," y-intercept } = - 3$
#### Explanation:
$\text{the equation of a line in "color(blue)"slope-intercept form}$ is.
•color(white)(x)y=mx+b
$\text{where m is the slope and b the y-intercept}$
$y = x - 3 \text{ is in this form}$
$\text{with slope "=1" and y-intercept } = - 3$
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#### Function Notation
Find the solutions in the app
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##### Sections
Exercise name Free?
###### Monitoring Progress
Exercise name Free?
Monitoring Progress 1
Monitoring Progress 2
Monitoring Progress 3
Monitoring Progress 4
Monitoring Progress 5
Monitoring Progress 6
Monitoring Progress 7
Monitoring Progress 8
Monitoring Progress 9
###### Exercises
Exercise name Free?
Exercises 1 There are two common ways to write functions:standard notation function notation. Standard notation has x-values as the inputs and y-values as the outputs because those correspond to (x,y)-coordinates on a coordinate plane. Function notation also has x-values as the inputs, but it has f(x) to represent the outputs because it is read as: f(x)= "the value of the function at x". Therefore, f(x)=2x+10 is written in function notation.
Exercises 2 When x is the input and a function is written in function notation, f(x) represents the output of the function at x. Here we have a function h that represents your height. This means that h(x) represents your height at age x, where age is the input and height is the output. Now, h(14) represents your height at the age of 14 because 14 is the input, age, and h(14) represents your height dependent on age.
Exercises 3 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. f(x)=x+6x=-2f(-2)=-2+6Add termsf(-2)=4 When x=-2, the function's value is 4. We will evaluate this function for x=0 and x=5 in the same way using the table below.xx+6f(x) -2-2+64 00+66 55+611
Exercises 4 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. g(x)=3xx=-2g(-2)=3(-2)a(-b)=-a⋅bg(-2)=-6 When x=-2, the function's value is -6. We will evaluate this function for x=0 and x=5 in the same way using the table below.x3xg(x) -23(-2)-6 03(0)0 53(5)15
Exercises 5 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. h(x)=-2x+9x=-2h(-2)=-2(-2)+9-a(-b)=a⋅bh(-2)=4+9Add termsh(-2)=13 When x=-2, the function's value is 13. We will evaluate this function for x=0 and x=5 in the same way using the table below.x-2x+9h(x) -2-2(-2)+913 0-2(0)+99 5-2(5)+9-1
Exercises 6 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. r(x)=-x−7x=-2r(-2)=-(-2)−7-(-a)=ar(-2)=2−7Subtract termr(-2)=-5 When x=-2, the function's value is -5. We will evaluate this function for x=0 and x=5 in the same way using the table below.x-x−7r(x) -2-(-2)−7-5 0-(0)−7-7 5-(5)−7-12
Exercises 7 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. p(x)=-3+4xx=-2p(-2)=-3+4(-2)a(-b)=-a⋅bp(-2)=-3−8Subtract termp(-2)=-11 When x=-2, the function's value is -11. We will evaluate this function for x=0 and x=5 in the same way using the table below.x-3+4xp(x) -2-3+4(-2)-11 0-3+4(0)-3 5-3+4(5)17
Exercises 8 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. b(x)=18−0.5xx=-2b(-2)=18−0.5(-2)-a(-b)=a⋅bb(-2)=18+1Add termsb(-2)=19 When x=-2, the function's value is 19. We will evaluate this function for x=0 and x=5 in the same way using the table below.x18−0.5xb(x) -218−0.5(-2)19 018−0.5(0)18 518−0.5(5)15.5
Exercises 9 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. v(x)=12−2x−5x=-2v(-2)=12−2(-2)−5-a(-b)=a⋅bv(-2)=12+4−5Add and subtract termsv(-2)=11 When x=-2, the function's value is 11. We will evaluate this function for x=0 and x=5 in the same way using the table below.x12−2x−5v(x) -212−2(-2)−511 012−2(0)−57 512−2(5)−5-3
Exercises 10 To evaluate a function for a certain value of x means that we substitute the given value for every instance of x. Let's do this for the given function when x=-2. n(x)=-1−x+4x=-2n(-2)=-1−(-2)+4-(-a)=an(-2)=-1+2+4Add termsn(-2)=5 When x=-2, the function's value is 5. We will evaluate this function for x=0 and x=5 in the same way using the table below.x-1−x+4n(x) -2-1−(-2)+45 0-1−0+43 5-1−5+4-2
Exercises 11
Exercises 12
Exercises 13 To find the value of x that will make the function equal to 63, we will substitute 63 for h(x) in the given function rule. Then we can solve for x. h(x)=-7xh(x)=6363=-7xLHS/-7=RHS/-7-9=xRearrange equationx=-9 When x=-9, the function equals 63. ⧼ebox-type-answer-check⧽ info Verifying Our Solution We can verify our solution by substituting x=-9 into the function and evaluating. h(x)=-7xx=-9h(-9)=-7(-9)-a(-b)=a⋅bh(-9)=63 When x=-9, h(x)=63, so our solution is correct.
Exercises 14 To find the value of x that will make the function equal to 24, we will substitute 24 for t(x) in the given function rule. Then we can solve for x. t(x)=3xt(x)=2424=3xLHS/3=RHS/38=xRearrange equationx=8 When x=8, the function equals 24. ⧼ebox-type-answer-check⧽ info Verifying Our Solution We can verify our solution by substituting x=8 into the function and evaluating. t(x)=3xx=8t(8)=3(8)Multiplyt(8)=24 When x=8, t(x)=24, so our solution is correct.
Exercises 15 To find the value of x that will make the function equal to 7, we will substitute 7 for m(x) in the given function rule. Then we can solve for x. m(x)=4x+15m(x)=77=4x+15LHS−15=RHS−15-8=4xLHS/4=RHS/4-2=xRearrange equationx=-2 When x=-2, the function equals 7. ⧼ebox-type-answer-check⧽ info Verifying Our Solution We can verify our solution by substituting x=-2 into the function and evaluating. m(x)=4x+15x=-2m(-2)=4(-2)+15a(-b)=-a⋅bm(-2)=-8+15Add termsm(-2)=7 When x=-2, m(x)=7, so our solution is correct.
Exercises 16 To find the value of x that will make the function equal to 18, we will substitute 18 for k(x) in the given function rule. Then we can solve for x. k(x)=6x−12k(x)=1818=6x−12LHS+12=RHS+1230=6xLHS/6=RHS/65=xRearrange equationx=5 When x=5, the function equals 18. ⧼ebox-type-answer-check⧽ info Verifying Our Solution We can verify our solution by substituting x=5 into the function and evaluating. k(x)=6x−12x=5k(5)=6(5)−12Multiplyk(5)=30−12Subtract termk(5)=18 When x=5, k(x)=18, so our solution is correct.
Exercises 17 To find the value of x that will make the function equal to -4, we will substitute -4 for q(x) in the given function rule. Then we can solve for x. q(x)=21x−3q(x)=-4-4=21x−3LHS+3=RHS+3-1=21xLHS⋅2=RHS⋅2-2=xRearrange equationx=-2 When x=-2, the function equals -4. ⧼ebox-type-answer-check⧽ info Verifying Our Solution We can verify our solution by substituting x=-2 into the function and evaluating. q(x)=21x−3x=-2q(-2)=21(-2)−3a(-b)=-a⋅bq(-2)=-1−3Subtract termq(-2)=-4 When x=-2, q(x)=-4, so our solution is correct.
Exercises 18 To find the value of x that will make the function equal to -5, we will substitute -5 for j(x) in the given function rule. Then we can solve for x. j(x)=-54x+7j(x)=-5-5=-54x+7LHS−7=RHS−7-12=-54xLHS⋅5=RHS⋅5-60=-4xLHS/-4=RHS/-415=xRearrange equationx=15 When x=15, the function equals -5. ⧼ebox-type-answer-check⧽ info Verifying Our Solution We can verify our solution by substituting x=15 into the equation and evaluating. j(x)=-54x+7x=15j(15)=-54(15)+7(-a)b=-abj(15)=-12+7Add termsj(15)=-5 When x=15, j(x)=-5, so our solution is correct.
Exercises 19 Usually when we think about points on coordinate planes, we think of them as x-coordinates and y-coordinates. However, when we use function notation, we use f(x) instead of y. (x,y)⇔(x,f(x)) We want to know the value of x when f(x)=7. To do this, we need to find the corresponding x-value of the function when then y-value is 7.We can see that the function crosses through the point (5,7). Therefore, when f(x)=7, x=5.
Exercises 20 Usually when we think about points on coordinate planes, we think of them as x-coordinates and y-coordinates. However, when we use function notation, we use f(x) instead of y. (x,y)⇔(x,f(x)) We want to know the value of x when f(x)=7. To do this, we need to find the corresponding x-value of the function when the y-value is 7.We can see that the function crosses through the point (-3,7). Therefore, when f(x)=7, x=-3.
Exercises 21
Exercises 22
Exercises 23 We can graph the linear function using a table. First, let's select some arbitrary values from the domain, then substitute them for x in the given equation. Solving for p(x) will give us the corresponding values from the range.xp(x)=4xSimplify 0p(0)=4(0)p(0)=0 1p(1)=4(1)p(1)=4 2p(2)=4(2)p(2)=8 Using these x and p(x) values as (x,p(x)) coordinate pairs, we can plot the points. Connecting these points will give us the graph of the equation.
Exercises 24 The first thing to notice about this equation is that it is missing a variable term. Let's try to rewrite it in standard form, Ax+By=C, using 0 as the coefficient for our missing variable." h(x)=-5⇔0x+1h(x)=-5 We can graph the linear equation using a table. First let's select some arbitrary values from the domain, then substitute them for x in the given equation. Solving for h(x) will give us the corresponding values from the range.x0x+h(x)=-5Simplify 00(0)+h(0)=-5h(0)=-5 20(2)+h(2)=-5h(2)=-5 40(4)+h(4)=-5h(4)=-5 Using these x and h(x) values as (x,h(x)) coordinate pairs, we can plot the points. Connecting these points will give us the graph of the equation.
Exercises 25 We can graph the linear function using a table. First, let's select some arbitrary values from the domain, then substitute them for x in the given equation. Solving for d(x) will give us the corresponding values from the range.xd(x)=-21x−3Simplify 0d(0)=-21(0)−3d(0)=-3 2d(2)=-21(2)−3d(2)=-4 4w(4)=-21(4)−3w(4)=-5 Using these x and d(x) values as (x,d(x)) coordinate pairs, we can plot the points. Connecting these points will give us the graph of the equation.
Exercises 26 We can graph the linear function using a table. Let's select some arbitrary values from the domain and substitute them for x in the given equation. Solving for w(x) will give us the corresponding values from the range.xw(x)=53x+2Simplify 0w(0)=53(0)+2w(0)=2 5w(5)=53(5)+2w(5)=5 10w(10)=53(10)+2w(10)=8 Using these x and w(x) values as (x,w(x)) coordinate pairs, we can plot the points. Connecting these points will give us the graph of the equation.
Exercises 27 We can graph the linear function using a table. Let's select some arbitrary values from the domain and substitute them for x in the given equation. Solving for g(x) will give us the corresponding values from the range.xg(x)=-4+7xSimplify 0g(0)=-4+7(0)g(0)=-4 1g(0)=-4+7(1)g(1)=3 2g(2)=-4+7(2)g(2)=10 Using these x and g(x) values as (x,g(x)) coordinate pairs, we can plot the points. Connecting these points will give us the graph of the equation.
Exercises 28 We can graph the linear function using a table. Let's select some arbitrary values from the domain and substitute them for x in the given equation. Solving for f(x) will give us the corresponding values from the range.xf(x)=3−6xSimplify 0f(0)=3−6(0)f(0)=3 1f(1)=3−6(1)f(1)=-3 2f(2)=3−6(2)f(2)=-9 Using these x and f(x) values as (x,f(x)) coordinate pairs, we can plot the points. Connecting these points will give us the graph of the equation.
Exercises 29 To determine which battery lasts longer, we can compare the number of hours it takes for each battery to have a 0% charge. We will begin by finding when the laptop's battery dies from the given graph.Notice on the graph that the laptop's battery charge reaches p=0 at t=5. This means that the laptop's battery lasts for 5 hours.The tablet Next we will determine how long the tablet's battery lasts. It is given that the tablet's initial charge is 75% and that the battery loses 12.5% each hour. We can write each of these percents in decimal form as 0.75 and 0.125, respectively. Now we can model the remaining charge of the tablet at any hour with the following function: p=0.75−0.125t As was the case with the graph, the tablet's battery will die when p=0. So we can substitute this into the equation above and solve for t. p=0.75−0.125tp=00=0.75−0.125tLHS+0.125t=RHS+0.125t0.125t=0.75LHS/0.125=RHS/0.125t=6 The solution t=6 means that the tablet's batter will last for 6 hours. Since the laptop's battery only lasted 5 hours, we can conclude that the tablet's battery lasts longer.
Exercises 30 This exercise asks us to compare the total cost of labor for 8 hours of work from two different companies. In this situation, our variables are x, the number of hours worked, and C(x), the cost. Because we want to know the cost for 8 hours of labor, we will focus on x=8. Let's start by determining the cost for Certified Remodeling.Certified Remodeling We will use the given equation for Certified Remodeling, C(x)=25x+50, to determine the cost they will charge for 8 hours of labor. To do this, we will substitute x=8 into the equation and solve. C(x)=25x+50x=8C(8)=25⋅8+50MultiplyC(8)=200+50Add termsC(8)=250 Certified Remodeling will charge \$200 for 8 hours of labor.Master Remodeling We are given a table showing the rates offered by the main competitor, Master Remodeling.HoursCosts 2\$130 4\$160 6\$190 Notice that in the table, the hours increase by 2 while the cost increases by \$30. We can add another row onto the bottom of the table by following this rate of change. Increasing 6 hours by 2 gives 8 hours. Increasing \$190 by 30 gives \$220.HoursCosts 2\$130 4\$160 6\$190 8\$220 The main competitor, Master Remodeling, will charge \$220 for 8 hours of labor.Conclusion In the work above, we found that Certified Remodeling will charge \$250 while the main competitor, Master Remodeling, will charge \$220. Thus, we should choose the main competitor.
Exercises 31 In function notation, x represents the input while f(x) represents the output. Then, P(x+1) represents the output of the function when the input is x+1, and P(x) represents the output when the input is x. There are three cases for how P(x+1) and P(x) relate.P(x+1)>P(x). It is possible (likely even) that the number of people with cell phones in year x+1 is greater than the number of people with cell phones in year x. Hence, we would see an increase. P(x+1)=P(x). It is also possible that the number of people with cell phones in year x+1 is equal to the number of people with cell phones in year x. Hence, we would not see a change. P(x+1)<P(x). But it is also possible that the number of people with cell phones in year x+1 is less than the number of people with cell phones in year x. Hence, we would see a decrease. In other words, just because x increases, it doesn't mean the functions value also does. Thus, we cannot conclude with certainty that P(x+1)>P(x).
Exercises 32 This exercise asks us to compare the output values of a function for different input values. Here, t represents the number of days and B(t) represents the balance in a bank account on day t. It is given that B(0)<B(4)<B(2). Notice that the expression is given in function notation, f(x)=y. In function notation, x represents the input while f(x) represents the output. Then,B(0) represents the balance in the account on Day 0, B(4) represents the balance in the account on Day 4, B(2) represents the balance in the account on Day 2, All together, the statement is saying that the balance on Day 0 is less than the balance on Day 4 and on Day 4 it's less than on Day 2. From this, we can gather than the balance on Day 2 is the greatest and the balance on Day 0 is the least.Example situation The following is one of infinitely many possible situations that can represent the given statement. On Wednesday, the balance in your bank account is 200 dollars. On Friday, you deposit your paycheck of 500 dollars into the account. On Sunday, you withdraw 300 to pay bills and buy groceries. The table below gives the account balance for each day.DaytB(t)Balance B Wednesday0B(0)200 Friday2B(2)700 Sunday4B(4)400 From the table, we can see that B(0)<B(4)<B(2).
Exercises 33
Exercises 34
Exercises 35
Exercises 36 Let's test the statement f(a+b)=f(a)+f(b) with the linear function f(x)=3x+3, and two arbitrary values, a=1 and b=2. f(a+b)f(a)+f(b)⇒f(1+2)⇒f(1)+f(2) Let's start by calculating f(1+2). To do so, we will substitute (1+2) for x in f(x)=3x+3. f(x)=3x+3x=1+2f(1+2)=3(1+2)+3 Simplify Add termsf(3)=3(3)+3Multiplyf(3)=9+3Add terms f(3)=12 Let's now calculate the value of f(1)+f(2).f(x)=3x+3 x=1x=2 f(1)=3(1)+3f(2)=3(2)+3 f(1)=3+3f(2)=6+3 f(1)=6f(2)=9 Therefore, f(1)+f(2) equals 6+9=15. f(1+2)=12f(1)+f(2)=15 Since 12 does not equal 15, we have that f(1+2)=f(1)+f(2). Therefore, we know that it is not always true that f(a+b)=f(a)+f(b). The statement is false.
Exercises 37 First, let's split the compound inequality into separate inequalities. Compound Inequality:-2≤xFirst Inequality:-2≤xSecond Inequality: x−11≤6−11−11≤6 Notice that compound inequalities written in this way are equivalent to compound inequalities that involve the word "and." -2≤x−11andx−11≤6 Let's solve the inequalities separately.First inequality Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must flip the inequality sign. -2≤x−11LHS+11≤RHS+119≤x This inequality tells us that 9 is less than or equal to all values that satisfy the inequality.Note that the point on 9 is closed because it is included in the solution set.Second inequality Once more, we'll solve the inequality by isolating the variable. x−11≤6LHS+11≤RHS+11x≤17 This inequality tells us that all values less than or equal to 17 will satisfy the inequality.Note that the point on 17 is closed because it is included in the solution set.Compound inequality The solution to the compound inequality is the intersection of the solution sets. First solution set:9≤xSecond solution set:xIntersecting solution set: 9≤x≤17≤17 Finally, we'll graph the solution set to the compound inequality on a number line.
Exercises 38 To solve the compound inequality, we have to solve each of the inequalities separately. Since the word between the individual inequalities is "or," the solution set for the compound inequality consists of the sets of the individual solutions.First inequality Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 5a<-35LHS/5<RHS/5a<-7 This inequality tells us that all values less than -7 will satisfy the inequality.Note that the point on -7 is open because it is not included in the solution set.Second inequality Again, we'll solve the inequality by isolating the variable. a−14>1LHS+14>RHS+14a>15 This inequality tells us that all values greater than 15 will satisfy the inequality.Note that the point on 15 is open because it is not included in the solution set.Compound inequality The solution to the compound inequality is the combination of the solution sets. First solution set:Second solution set:Combined solution set:a<-7a<-7 or a>15a<-7 or a>15 Finally, we'll graph the solution set to the compound inequality.
Exercises 39 First, let's split the compound inequality into separate inequalities. Compound Inequality: -16<6kFirst Inequality: -16<6kSecond Inequality: 6k+2<0+2+2<0 Notice that compound inequalities written in this way are equivalent to compound inequalities that involve the word "and." -16<6k+2and6k+2<0 Let's solve the inequalities separately.First inequality Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must flip the inequality sign. -16<6k+2LHS−2<RHS−2-18<6kLHS/6<RHS/6-3<k This inequality tells us that all values greater than -3 will satisfy the inequality.Note that the point on -3 is open because it is not included in the solution set.Second inequality Once more, we'll solve the inequality by isolating the variable. 6k+2<0LHS−2<RHS−26k<-2LHS/6<RHS/6k<-62ba=b/2a/2k<-31 This inequality tells us that all values less than -31 will satisfy the inequality.Note that the point on -31 is open because it is not included in the solution set.Compound inequality The solution to the compound inequality is the intersection of the solution sets. First solution set:-3<kSecond solution set:kIntersecting solution set: -3<k<-31<-31 Finally, we'll graph the solution set to the compound inequality on a number line.
Exercises 40 To solve the compound inequality, we have to solve each of the inequalities separately. Since the word between the individual inequalities is "or," the solution set for the compound inequality consists of the sets of the individual solutions.First inequality Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 2d+7<-9LHS−7<RHS−72d<-16LHS/2<RHS/2d<-8 This inequality tells us that all values less than -8 will satisfy the inequality.Note that the point on -8 is open because it is not included in the solution set.Second inequality Again, we'll solve the inequality by isolating the variable. 4d−1>-3LHS+1>RHS+14d>-2LHS/4>RHS/4d>-42ba=b/2a/2d>-21 This inequality tells us that all values greater than -21 will satisfy the inequality.Note that the point on -21 is open because it is not included in the solution set.Compound inequality The solution to the compound inequality is the combination of the solution sets. First solution set:Second solution set:Combined solution set:d<-8d<-8 or d>-21d<-8 or d>-21 Finally, we'll graph the solution set to the compound inequality.
Exercises 41 First, let's split the compound inequality into separate inequalities. Compound Inequality:5≤3yFirst Inequality:5≤3ySecond Inequality: 3y+8<17+8+8<17 Notice that compound inequalities written in this way are equivalent to compound inequalities that involve the word "and." 5≤3y+8and3y+8<17 Let's solve the inequalities separately.First inequality Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must flip the inequality sign. 5≤3y+8LHS−8≤RHS−8-3≤3yLHS/3≤RHS/3-1≤y This inequality tells us that -1 is less than or equal to all values that satisfy the inequality.Note that the point on -1 is closed because it is included in the solution set.Second inequality Once more, we'll solve the inequality by isolating the variable. 3y+8<17LHS−8<RHS−83y<9LHS/3<RHS/3y<3 This inequality tells us that all values less than 3 will satisfy the inequality.Note that the point on 3 is open because it is not included in the solution set.Compound inequality The solution to the compound inequality is the intersection of the solution sets. First solution set:-1≤ySecond solution set:yIntersecting solution set: -1≤y<3<3 Finally, we'll graph the solution set to the compound inequality on a number line.
Exercises 42 To solve the compound inequality, we have to solve each of the inequalities separately. Since the word between the individual inequalities is "or," the solution set for the compound inequality consists of the sets of the individual solutions.First inequality Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must reverse the inequality sign. 4v+9≤5LHS−9≤RHS−94v≤-4LHS/4≤RHS/4v≤-1 This inequality tells us that all values less than or equal to -1 will satisfy the inequality.Note that the point on -1 is closed because it is included in the solution set.Second inequality Again, we'll solve the inequality by isolating the variable. -3v≥-6Divide by -3 and flip inequality signv≤2 This inequality tells us that all values less than or equal to 2 will satisfy the inequality.Note that the point on 2 is closed because it is included in the solution set.Compound inequality The solution to the compound inequality is the combination of the solution sets. First solution set:Second solution set:Combined solution set:v≤-1v≤2v≤2 Finally, we'll graph the solution set to the compound inequality. The union of these solution sets is v≤2.
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# Average Rate of Change Calculator
Instructions: Use Average Rate of Change Calculator, to get a step-by-step calculation of the average rate of change of a function between two points. You need to provide the value of the function at two points $$(t_1, y_1)$$ and $$(t_2, y_2)$$, and this calculator will estimate the average rate of change:
Type $$t_1$$ (One numeric expression) =
Type $$y_1$$ (One numeric expression) =
Type $$t_2$$ (One numeric expression) =
Type $$y_2$$ (One numeric expression) =
The idea of this net change calculator is to estimate how much the given function changes per unit of time. So, how do you calculate the average rate of change? It is not hard. Indeed, the average rate of change is defined as
$\text{Average Rate of Change} = \displaystyle\frac{\Delta y}{\Delta t} = \displaystyle\frac{y_2 - y_1}{t_2 - t_1}$
This is, it corresponds to the ratio of the net change in y ($$\Delta y$$) and the net change in t ($$\Delta t$$).
### Is the average rate of change of a function constant?
Not necessarily. The average rate of change is computed over a certain interval. If you change the interval, the average rate of change can perfectly change as well.
### Can we say that the rate of change is the same as slope?
Not always. Indeed, that only happens when the function is linear (its graph is a straight line). When the function is not linear, then the "slope" is defined locally by its derivative at each specific point. Also, we need to understand that the average rate of change ultimately is computed over an interval, so it can be changing based on the interval chosen.
The average rate of change measures the slope of the line that passes through two given points $$(t_1, y_1)$$ and $$(t_2, y_2)$$. As $$t_1$$ approaches to $$t_2$$, the average rate of change will look more and more like the slope of the tangent line.
### What is the difference with the instant rate of change?
The instant rate of change is obtained when the size of the interval for which the average rate of change is computed becomes smaller and smaller. Technically, the instant rate of change corresponds to the limit of the average rate of change, when the size of the interval approaches to zero, which is associated with the derivative of the function.
### Is this a Net change calculator?
Not exactly. A net change calculator compute the difference between the values of a function a two given points, but it is not divided by the difference between the arguments. So, the net change is a measure of an absolute difference, whereas this calculator find the relative change.
So, when do you compute the net change? It is when you need an absolute change as opposed to a relative change that is given by this average rage of change solver.
### How is this similar to an instantaneous rate of change?
They are related indeed. An instantaneous rate of change is defined as the limit of the average rate of change, as the difference between the arguments approaches to zero.
### Rate of Change Example
Assume you have a a function that at $$t = 1$$ has a value of $$y(1) = 5$$, and that at $$t = 4$$ has a value of $$y(4) = 10$$. Compute the average rate of change in this case.
Solution:
The following is the information we have been provided about the function at times $$t = 1$$ and $$t = 4$$:
Time 1 $$(t_1)$$ = $$1$$ Function at Time 1 $$(f(t_1))$$ = $$5$$ Time 2 $$(t_2)$$ = $$4$$ Function at Time 2 $$(f(t_2))$$ = $$10$$
By definition, the average rate of change is computed as:
$\begin{array}{ccl} \text{Average Rate of Change} & = & \displaystyle \frac{\Delta y}{\Delta t} \\\\ \\\\ & = & \displaystyle \frac{y_2 - y_1}{t_2 - t_1} \\\\ \\\\ & = & \displaystyle \frac{ 10 - 5}{ 4- 1} \\\\ \\\\ & = & \displaystyle \frac{ 5}{ 3} \\\\ \\\\ & = & \displaystyle 1.6667 \end{array}$
Therefore, the average rate of change for a function that passes through the points $$(1, 5)$$ and $$(4, 10)$$ is 1.6667.
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Whatever Day! 2012
Essential Curriculum for Math:
First Nine Weeks
NBT.1 1. I can identify place value using base ten. NBT.21. I can explain how to multiply and divide by powers of ten. 2. I can explain and use exponents with the powers of ten. 3. I can investigate the relationship of a decimal to the powers of ten. NBT.31. I can write decimals to the thousandths. 2. I can read decimals to the thousandths. 3. I can compare decimals to the thousandths using >, =, and <. NBT.41. I can explore my understanding of place value to round decimals to a given place value. NBT.51. I can fluently multiply multi-digit whole numbers using the standard algorithm and/or alternative strategies.
NBT.61. I can explore various strategies to divide up to four digit dividends and two digit divisors. 2. I can illustrate and explain a division calculation. OA.11. I can evaluate how to use numerical expressions with parentheses, brackets, and braces. 2. I can explain how to use numerical expressions with parentheses, brackets, and braces OA.21. I can identify and write a simple expression. 2. I can describe numerical expressions.
Second Nine Weeks
NBT.71. I can add, subtract, multiply, and divide decimals to the hundredths place value. 2. I can explain how to add, subtract, multiply, and divide decimals to the hundredths place value. NF.11. I can use a model to represent the addition and subtraction of fractions with unlike denominators. NF.21. I can solve word problems using addition and subtraction of fractions including like and unlike denominators. 2. I can use benchmark fractions and number sense of fractions to estimate mentally and check for reasonableness. NF.31. I can use models to solve division problems using fractions 2. I can solve word problems including the division of whole numbers with answers in forms of mixed numbers. NF.4a1. I can create and solve problems using models to multiply a fraction or whole number by a fraction. NF.4b2. I can find the area of rectangle with fractional side lengths. NF.5a1. I can use the factors to compare the products without using multiplication. NF.5b2. I can multiply with whole numbers and fractions.
Third Nine Weeks
NF.6 1. I can solve real world problems using multiplication of fractions and mixed numbers. NF.71. I can interpret and solve division of fractions by a whole number and/or within a story context. 2. I can interpret and solve division of whole numbers by unit fractions and/or within a story context. MD.11. I can convert standard measurement units within a given measurement system. MD.31. I can recognize and identify the volume of solid figures and their measurements. MD.42. I can measure volume by counting cubic units. MD.53. I can identify the volume using the formula 4. I can relate volume to multiplication and addition to solve real world problems.
Fourth Nine Weeks
OA.31. I can generate patterns with two rules. 2. I can form ordered pairs. 3. I can plot ordered pairs on a coordinate plane. MD.21. I can make a line plot to display a data set of fractions using different measurement units (1/2, 1/4, 1/8). G.11. I can identify axes in the first quadrant of the coordinate plane. 2. I can plot ordered pairs in the first quadrant of the coordinate plane. G.23. I can graph real world mathematical problems on a coordinate grid. 4. I can interpret real world mathematical problems on a coordinate grid. G.31. I can classify two-dimensional figures by their attributes. G.42. I can classify two-dimensional figures in a hierarchy based on properties.
60 Math Expo Ideas
(credit to teacher at Lake Norman Elementary)
1. quilting
2. sports
3. inventions
4. sundials
5. topology
6. ancient number systems
7. careers
8. geometric shapes and architecture
9. difference between meal prices (calories) at fast food
10. restaurants
11. grocery store price comparisons
12. bowling
13. build the Bridge to Terabithia
14. build models of objects from other novels
15. furnish a home with newspaper ads
16. growth of South Iredell County
17. Lake Norman (dropping in winter, rains in summer, etc.)
18. racing (building of cars, weights, height, etc.)
19. heart rates
20. cheaper to own cat vs. dog? hamster vs. fish?
21. pay comparisons (doctor/teacher, engineer/president, etc.)
22. your school needs (new gym, cafeteria, classrooms)
23. models
24. ocean tides and the moon
25. space
26. dance
27. comparing foods (cereals, yogurts, snacks, etc.)
28. research a famous mathematician
29. pool table problems
30. study regular solids and their properties
31. make a family of polyhedra
32. What is Morley's triangle?
33. Investigate compass and straight edge constructions
34. cycloid curve
35. what is a hexaflexagon?
36. kaleidoscopes
37. tangrams
38. game theory
39. study games and winning strategies
40. the the "Monty Hall" effect
41. investigate card and magic tricks
42. probability
43. positions of fire stations, police stations, etc.
44. how the NBA or Panthers or other team make schedules for games, hotels, practices, flight, etc.
45. investigate "big" numbers
46. computer bar codes
47. Fibonacci numbers
48. What is the Golden mean?
49. Catalan numbers
50. triangular numbers
51. history of pi
52. egyptian fractions
53. methods of counting and calcuating using your fingers and hands
54. mathematics of elections
55. how random is random?
56. stock market
57. crystallography
58. math in music
59. build a model of a house
60. how much would it cost for college?
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QUADRATIC EQUATIONS - Equations and Inequalities - ALGEBRA - SAT SUBJECT TEST MATH LEVEL 1
## CHAPTER 6Equations and Inequalities
A quadratic equation is an equation that can be written in the form ax2 + bx + c = 0, where ab, and c are any real numbers with a 0. Any number, x, that satisfies the equation is called a solution or a root of the equation.
Although some quadratic equations can be solved by other, sometimes easier, methods—including factoring—that will be discussed later, every quadratic equation can be solved by using the quadratic formula given in KEY FACT E1.
Key Fact E1
If ab, and c are real numbers with a 0 and if ax bx + c = 0, then
Recall that the symbol ± is read “plus or minus” and that is an
abbreviation for or .
As you can see, a quadratic equation has two roots, both of which are determined by the quadratic formula.
The expression b2 – 4ac that appears under the square root symbol is called the discriminant of the quadratic equation. As explained in KEY FACT E2, the dis-criminant provides valuable information about the nature of the roots of a quadratic equation. If we let D represent the discriminant, an alternative way to write the quadratic formula is . The following examples illustrate the proper use of the quadratic formula.
EXAMPLE 12: What are the roots of the equation x2 – 2x – 15 = 0?
EXAMPLE 13: What are the roots of the equation x2 = 10x – 25?
First, rewrite the equation in the form ax2 + bx + c = 0:
Notice that since 10 + 0 = 10 and 10 – 0 = 10, the two roots are each equal to 5. Although some people would say that the equation x2 – 10x + 25 = 0 has only one root, you should say that the equation has two equal roots.
EXAMPLE 14: Solve the equation 2x2 – 4x – 1 = 0.
EXAMPLE 15: Solve the equation x2 – 2x + 2 = 0.
See Chapter 17 for a discussion of the imaginary unit i.
If in the equation ax2 + bx + c = 0, either b or c is equal to 0, it is easier not to use the quadratic formula.
Smart Strategy
Before using the quadratic formula, see if another method would be easier and quicker.
• If b = 0, you can just take a square root.
Note that if c 0, the two solutions will be real. If c > 0, the expression under the square root sign will be negative, and the two solutions will be imaginary.
EXAMPLE 16:
EXAMPLE 17:
• If c = 0, you should just factor out an x.
Recall that if a product is equal to 0, one of the factors must be 0. So:
EXAMPLE 18:
EXAMPLE 19:
• If you can easily factor the expression ax2 + bx + c, then you can solve the equation ax2 + bx + c = 0 by setting each factor equal to 0.
Factoring is faster than using the quadratic formula if you can immediately determine the factors. However, factoring is a trial and error method. Even if an expression factors, it may take you a long time to do so. Worse, after spending a few minutes trying to factor, you may realize that the expression cannot be factored. Now you have to use the quadratic formula, and it would have taken less time to have done so right away.
Consider the equations 2x2 + 11x – 24 = 0 and 2x2 + 13x – 24 = 0. Only one of them can be solved by factoring. Do you know which one? Probably not. So if you were given one of these equations to solve and you attempted the factoring method, you might waste valuable time in a fruitless effort. Therefore, to solve a quadratic equation, use the quadratic formula unless b = 0 or c = 0 or unless you immediately see how to factor the quadratic expression.
Sometimes a question on a Math 1 test asks for information about the roots of a quadratic equation but does not specifically require you to solve the equation. Looking at the discriminant (D) will allow you to answer such a question.
• If D = 0, as in Example 13, the two roots are and . So the two roots are equal. If ab, and c are all rational numbers, then is also rational.
• If D is negative, as in Example 15, then the equation has no real roots. Both roots are complex (or imaginary) numbers. (See Chapter 17 for a discussion of complex numbers.) If one of the roots is u + vi, the other root is its conjugate: u – vi.
• If D is positive, as in Examples 12 and 14, then the equation has two unequal real roots. If ab, and c are rational, then if D is a perfect square, as in Example 12, the two roots are rational; and if D is not a perfect square, as in Example 14, the two roots are irrational.
The preceding remarks are summarized in the following chart.
Key Fact E2
If ab, and c are rational numbers with a 0, if ax2bx + c = 0, and if D = b2 – 4ac, then
Value of Discriminant Nature of the Roots D = 0 2 equal rational roots D > 0 2 unequal complex roots that are D > 0 (i) D is a perfect square 2 unequal rational roots (ii) D is not a perfect square 2 unequal rational roots
TIP
Be sure to memorize the facts in this chart.
If ax2 + bx + c = 0 and r1 and r2 are the two roots, then the sum of the roots is
and the product of the roots is
Key Fact E3
If axbx + c = 0, then the sum of the two roots is and the product of the two roots is .
EXAMPLE 20: Find a quadratic equation for which the sum of the roots is 5 and the product of the roots is 5.
For simplicity, let a = 1. Then , and . So the equation x– 5x + 5 = 0 satisfies the given conditions.
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